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Alkenes are found throughout nature. They form the basis of many natural products, such as terpenes, which play a variety of roles in the lives of plants and insects. The C=C bonds of alkenes are very different from the C=O bonds that are also common in nature. The C=C bonds of alkenes are electron-rich and nucleophilic, in contrast to the electron-poor C=O bonds of carbohydrates, fatty acids and proteins. That difference plays a role in how terpenes form in nature.
Alkenes, or olefins, are also a major product of the petroleum industry. Reactions of alkenes form the basis for a significant porion of our manufacturing economy. Commonly used plastics such as polyethylene, polypropylene and polystyrene are all formed through the reactions of alkenes. These materials continue to find use in our society because of their valuable properties, such as high strength, flexibility and low weight.
Alkenes undergo addition reactions like carbonyls do. Often, they add a proton to one end of the double bond and another group to the other end. These reactions happen in slightly different ways, however.
Alkenes are reactive because they have a high-lying pair of π-bonding electrons. These electrons are loosely held, being high in energy compared to σ-bonds. The fact that they are not located between the carbon nuclei, but are found above and below the plane of the double bond, also makes these electrons more accessible.
Alkenes can donate their electrons to strong electrophiles other than protons, too. Sometimes their reactivity pattern is a little different than the simple addition across the double bond, but that straightforward pattern is what we will focus on in this chapter.
Exercise \(1\)
The reaction of 2-methylpropene (or isobutylene) with HBr, as depicted above, is really a 2-step process. Draw this mechanism again and in each of the two steps label both the nucleophile and the electrophile (so, that's four labels).
Answer
El = electrophile; Nu = nucleophile
Exercise \(2\)
Draw a reaction progress diagram for the reaction of 2-methylpropene with hydrogen bromide.
Answer
Exercise \(3\)
Predict the rate law for the reaction of 2-methylpropene with hydrogen bromide.
Answer
Rate = d[alkyl bromide] / dt = k [alkene] [HBr]
Exercise \(5\)
Acids other than HBr can add to alkenes. Based on the reaction with HBr, draw products for the following reactions.
6.02:
Many of the reactions of alkenes begin with a protonation step. The cation that forms then undergoes a second step in which it combines with the counterion from the acid.
In the first step, the alkene's π bond is the nucleophile and the proton is the electrophile. In the second step, the bromide is the nucleophile and the cation is the electrophile.
If you are familiar with nucleophilic aliphatic substitution, you will already know that the presence of a cationic intermediate signals some potential complications in this reaction.
One issue is the problem of stereochemical control. A carbocation is trigonal planar, because the carbon with the positive charge has only three groups attached to it. Because the cation is trigonal planar, the bromide ion that combines with it can approach from either side. It can come from above or below the trigonal plane.
That fact may have no effect whatsoever. However, if the alkene (and the cation it forms) is prochiral, meaning it has the potential to form a new chiral center during this reaction, then there is a choice of which enantiomer to make.
A prochiral carbocation is easy to recognize because the cationic carbon has three different groups attached to it. The fourth group added, the nucleophile, would result in four different groups attached to that carbon, making it a chiral center. In order to recognize a prochiral alkene, you can picture what the alkene would look like after the reaction has taken place: will there be four different groups?
Exercise \(1\)
Which of the following alkenes are prochiral?
Exercise \(2\)
Addition of the nucleophile to one face of the alkene will result in a stereocentre with R configuration. That face is called the re face. Adding it to the other will lead to formation of S configuration. That face is called the si face.
In the following alkenes, identify whether we are looking at the re face or the si face in terms of the product we would get through addition of HBr.
Exercise \(3\)
Draw the products of the following reactions, paying attention to stereochemistry.
In addition to the problem of stereochemistry, electrophilic additions of alkenes also present potential regiochemical complications. As in aliphatic nucleophilic substitutions, formation of a cation often opens the door to rapid rearrangement via 1,2-hydride shifts. There may be one hydride shift or there may be many of them in a row.
These hydride shifts happen pretty easily. Overlap of a hydrogen atom with the empty p orbital of the the adjacent cation leads to a short hop from one carbon to the next.
A hydride shift from one secondary carbon to the next, as illustrated in the above example, is thermodynamically pretty neutral. Because the barrier is low, it happens quickly, but there isn't a driving force fo the hydride to shift one way or the other. Instead, both cations result. There is a mixture.
However, in a case in which the cation can form in a more stable position, such as a tertary position, there is a driving force for the reaction to go one way. The barrier would be too high for it to get back.
As a result, when the counterion combines with the cation, it may do so in a position away from the original double bond.
Exercise \(4\)
Draw the products of the following reactions, paying attention to regiochemistry. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/IV%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_2/06%3A_Electrophilic_Addition_to_Alkenes/6.01%3A_.txt |
Alkenes can donate their electrons to strong electrophiles and the resulting carbocations combine with the counterion of the electrophile to undergo an overall addition reaction. However, there may be some cases in which the counterion does not combine with the carbocation.
Hydrobrominations of the type we have looked at only occur under certain conditions. Other conditions can lead to other products. For example, a solvent such as water can also participate in the reaction.
The oxygen-based cation (or oxonium ion) that results can easily lose its charge through loss of a proton. As a result, a molecule of water adds to the alkene overall. The alkene becomes an alcohol. This reaction is called an "acid-catalyzed hdration" of an alkene.
Exercise \(1\)
Explain how the hydration of an alkene in the presence of acid is a catalytic reaction.
Answer
If the acid is regenerated at the end of the reaction, it isn't a reagent. It is a catalyst. It makes addition of water to the double bond occur much more quickly than if water acted alone, since water would never manage to protonate the alkene.
Exercise \(2\)
In many cases, equilibrium mixtures of multiple products may result from the addition of acids to alkenes. Show mechanisms, with curved arrows, for the following reactions.
1. The conversion of 2-butene to 2-chlorobutane with aqueous HCl.
2. The conversion of 2-butene to 2-butanol with aqueous HCl.
3. The conversion of 2-chlorobutane to 2-butene with aqueous HCl.
4. The conversion of 2-butanol to 2-butene with aqueous HCl.
Answer
On the other hand, in the absence of any solvent, the bromide ion might still have some competition in the second step. The neat reaction (neat means "without solvent" of an alkene with a small amount of acid can result in polymerization. The alkene, which acted as a nucleophile in the first step, can also act as a nucleophile in the second step.
It is important to remember that in any reaction, millions of molecules are involved. Even if one alkene molecule reacts with acid in the first step of a reaction, there are still plenty of other alkene molecules around to act as nucleophile in the second step.
Exercise \(3\)
Provide a mechanism for the polymerization shown above. Assume there are four 2-methylpropene molecules and one hydrogen bromide molecule to begin.
Answer
Exercise \(4\)
Chain reactions involve an initiation step, in which a reactive species is generated; propagation steps, in which the reactive species reacts to make a new reactive species; and a termination step, in which the reactive species reacts to make a stable molecule. Label each of the steps in your mechanism from the previous question.
6.04:
Electrophilic addition to alkenes generally takes place via donation of the π-bonding electron pair from the alkene to an electrophile. So far, we have only looked at protic electrophiles, but the reaction proceeds with others, as well. For instance, alkenes react quite easily with bromine.
Dripping a solution of bromine into a solution of alkene provides a clear sign of reaction. The red-brown colour of bromine disappears almost instantly.
Although bromine isn't an obvious electrophile, most of the common diatomic elements can behave that way; the exception is dinitrogen. A fleeting asymmetry of electrons can polarize the molecules to one end. That event leaves one atom partially positive and the other end partially negative. Because these elements tend to form somewhat stable anions, the partially negative atom can be displaced failry easily.
As before, a nucleophile connects with the cation in a second step. In this case, a dibromide compound is formed. However, it isn't formed in quite the way that is shown below.
We know that the mechanism shown above does not convey the whole picture because it isn't consistent with the stereochemistry of the reaction. The stereochemical outcome is shown below. The enantiomer is formed as well.
Exercise \(1\)
Assign configurations R or S to the product shown in the above mechanism.
Answer
Two products are formed and they are enantiomers.
However, although two enantiomers are formed in the reaction, the corresponding diastereomer is not. The following step does not occur.
Exercise \(2\)
Assign configurations R or S to the product shown in the above mechanism and explain why it is not an enantiomer to the compound in problem EA4.1.
Answer
They are diastereomers. One chiral center has the same configuration in both compounds but the others are opposite.
Instead, the cation that forms in the reaction appears to be stabilized by lone pair donation from bromine. The intermediate species below is called a cyclic bromnium ion. The bromine prevents approach of the nucleophilic bromide from one side, ensuring formation of product through anti addition only. The trans product forms as a result.
Exercise \(3\)
Additional evidence of the stabilized bromonium comes from the observation of just one product in the bromination of 1-hexene. How many products would be expected in the absence of a stabilized cation? Explain with a mechanism.
Answer
The second bromine could occupy any of the secondary positions if there were a true carbocation. That doesn't happen; the second bromine occupies only the position next to the other bromine.
Exercise \(4\)
Bromonium ions like the one shown below have been isolated and characterized by X-ray crystallography in at least one case. Explain why the intermediate was isolated in this case, rather than a dibromo product.
Exercise \(5\)
In some cases, the reaction of an alkene with bromine does not provide dibromo products. Show products of the reaction of cyclohexene under the following conditions and justify your choices with mechanisms.
a) Br2 in water b) Br2 and NH4+ Cl- in THF
Answer
The nucleophile in the second step changes under different conditions.
Exercise \(6\)
Provide products for the reaction of bromine with each of the following compounds. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/IV%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_2/06%3A_Electrophilic_Addition_to_Alkenes/6.03%3A_.txt |
Alkenes can coordinate to transition metals to form alkene complexes. In some cases, coordination of the alkene to a metal leaves it susceptible to reaction with a nucleophile such as water.
The classic case of nucleophilic donation to a coordinated alkene occurs with mercury (II) salts such as mercuric chloride, HgCl2, or mercuric acetate, Hg(OAc)2. The reaction, or rather the sequence of reactions, is called oxymercuration - demercuration or oxymercuration - reduction.
Exercise \(1\)
Compare the product of the reaction above to that obtained from treatment of 1-pentene with aqueous sulfuric acid.
We will break the two different reactions in this sequence apart and focus only on the first one: oxymercuration. This reaction qualifies as an electrophilic addition because, as in the previous cases, it begins with donation of a π-bonding pair to an electrophile. In this case, we will consider the electrophile to be aqueous Hg2+ ion.
That electrophilic addition (from the alkene's perspective) results in the formation of an alkene complex. In reality, the mercury ion is also coordinated by several water molecules, but we will ignore them for simplicity.
You may know that alkene complexes are not observed with d0 transition metals. Although π-to-metal donation is the key event in the formation of such complexes, the alkene is just a little more sticky if the metal has d electrons. These electrons are able to "back-donate" into the alkene portion of the complex, adding extra stability to the interaction.
This situation is something like formation of a cyclic bromonium ion.
Note that the overall transfer of electrons is still from alkene to metal. That imbalance isn't apparent in a Lewis structure sense, in which case you can draw the structure so that there appears to be an equal trade. In a computational chemistry approach, in which we rely on basic principles of quantum mechanics and let computers churn out high-level calculations, we would still predict a little bit of positive charge on the alkene.
In the structure below, we have over-emphasized that charge, just to see what happens next in the reaction. When we draw it that way, it looks a lot more like simple addition of electrophile , such as H+, to alkene. We know it's more subtle than that. We'll get back to the real mechanism after a small detour.
Of course, the next step is donation of a lone pair from a nucleophile to the almost-cationic carbon.
That looks easy. After that, deprotonation would result in the formation of a hydroxy group.
Exercise \(2\)
Suppose deprotonation is carried out by the acetate ion in solution. Draw a mechanism for this step.
Answer
How do we know the reaction doesn't happen through this simple cation?
Partly we know that because we know about alkene complexes. There are thousands of examples, structurally characterized by NMR spectroscopy and X-ray crystallography. In addition, we know it isn't a simple cation because nothing like the following scenario plays out during oxymercuration.
There are no hydride shifts. The cation stays put. The hydroxy group forms right where the alkene used to be. That means there is not a full carbocation like the one shown above.
If there isn't a real carbocation, though, why does the nucleophile end up at one particular end of the alkene? The hydroxy does end up at the position that would form the more stable cation. ( In other words, this reaction results in what is called "Markovnikov addition".)
There are a couple of reasons that could play a role. Foremost, the alkene isn't bound symmetrically. One end is held a little closer to the mercury than the other. Mostly that's because of sterics. Any other ligands on the mercury (such as those water molecules) push that more crowded end away a little bit. That slight asymmetry allows a little more charge to build on the more substitutted end of the alkene, which is therefore more electrophilic.
The final part of the reaction sequence is displacement of mercury from the hydroxyalkylmercury complex, effected through addition of sodium borohydride. The details of the reaction are usually dismissed in textbooks because they have little to do with electrophilic addition, the topic we are focusing on. However, the result is that the mercury is replaced by a hydrogen atom. The metal is converted to silvery, liquid, elemental mercury.
Exercise \(3\)
Suppose the demercuration reaction takes place via addition of hydride nucleophile to mercury, followed by reductive elimination. Draw this mechanism.
Answer
Exercise \(4\)
When mercuration takes place in a ethanol instead of water, an ether product results rather than an alcohol. Work through the mechanism and show the result of mercuration-demercuration in ethanol.
Answer
Exercise \(5\)
Show the products of the following reactions.
Exercise \(6\)
Predict the products of the following reactions.
6.06:
Metal alkoxides can undergo 1,2-elimination (or beta-elimination) to give organic carbonyl compounds. This reaction is the reverse of a nucleophilic addition of a metal hydride to an organic carbonyl.
Metal alkyls can also undergo 1,2-elimination. In this case, an alkene is formed. Like the carbonyl compounds formed from 1,2-elimination, the alkene usually remains bound to the metal. It can dissociate to form the free alkene, though. For more information on alkene binding, take a look at this page.
These reactions are sometimes called "beta hydride eliminations", emphasizing that the hydrogen attached to the metal usually acts as a nucleophile. Thus, when the hydrogen transfers to the metal, it is forming a hydride.
• Alkyl groups that are coordinated to metals can also undergo elimination.
• In order to undergo elimination, an alkyl group must have a hydrogen.
• When a metal alkyl undergoes elimination, it forms an alkene.
Exercise \(1\)
Draw the elimination products in the following cases.
The reverse of a 1,2-elimination is a 1,2-insertion. Just like aldehydes and ketones, alkenes can undergo 1,2-insertions (also called beta hydride insertions). In terms of electrophiles and nucleophiles, this reaction is a little harder to imagine. However, we can still think of the hydride as a nucleophile. Maybe the alkene is an electrophile. Given that it is donating its pi electrons to the metal, we can think of it as "activated", a little bit like an activated carbonyl.
The formalisms of drawing a beta alkene insertion are tricky. If we use the metallacycle drawing of a bound alkene, it might look like this:
More often, bound alkenes are drawn as shown as in the picture below. In that case, we could try to show the pi bond forming a new carbon-metal bond. The bond between the metal and alkene on the picture to the left does not really stand for a separate pair of electrons in this case; it just stands for the pi bond donating to the metal.
• The reverse of elimination is insertion.
• A coordinated alkene is sometimes considered electrophilic because it is giving electrons to the metal.
• A coordinated alkene is activated, like a coordinated carbonyl compound.
Exercise \(2\)
Draw the insertion products in the following cases.
Exercise \(3\)
Alkenes can be converted into other compounds through the use of organometallic reagents, such as "Schwartz's reagent" (below). In this case, the resulting alkyl compound can easily be converted into a long-chain alkyl halide or alcohol through the addition of appropriate reagents. Provide a mechanism for the reaction shown below.
Exercise \(4\)
1,2-alkyl insertions and -eliminations are also known in a few cases, although they are much slower than 1,2-hydride insertions and -eliminations. Show the 1,2-insertion products for the following cases.
Exercise \(7\)
Fill in the missing insertion / elimination products.
Exercise \(8\)
The following multi-part problem is based on an article in the primary research literature. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/IV%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_2/06%3A_Electrophilic_Addition_to_Alkenes/6.05%3A_.txt |
Alkenes can be treated with aqueous acids or, much more efficiently, with aqueous mercuric salts, followed by sodium borohydride, to produce alcohols. In those cases, the hydroxyl group is found in different places. Treatment with acid often results in a mixture of alcohols in which the OH groups are found in the most substituted positions in the structure, regardless of the position of the original alkene. Oxymercuration - demercuration results in the hydroxyl being fixed at the most substituted end of the former C=C bond.
Those reactions are generally called hydration reactions because they result in the overall addition of H-OH across the double bond.
• Hydration reactions place an OH on one end and a H on the other end of a former double bond.
Hydroboration - oxidation is a two-step sequence of reactions that also results in hydration of a double bond. However, this reaction is complementary to oxymercuration - demercuration.
Instead of leaving an OH group at the most substituted end of the double bond, the hydroxy group is placed at the least substituted end of the double bond.
Let's modify that statement a little bit. In reality, the reaction scheme above just shows the major product. The minor product has the hydroxy group at the more substituted position of the double bond. These two products might be found in different ratios, maybe even as close as 55:45, but the least substituted product always predominates. We will see more efficient hydroboration methods soon, leading to ratios above 95:5, or almost entirely the least substituted product.
The product of non-hydrogen addition (i.e. OH group addition) at the most substituted end of the alkene is called a Markovnikov addition product. The product of non-hydrogen addition at the least substituted end of the alkene is called an anti-Markovnikov addition product.
• Oxymercuration - demercuration results in Markovnikov hydration.
• Hydroboration - oxidation results in anti-Markovnikov hydration.
This selectivity is important in synthetic applications. We use natural products all the time as pharmaceuticals, vitamins and other health and beauty applications, but we can't always obtain these compounds directly from nature, for a number of reasons. It could be that the organism needs to be killed in order to harvest its products, or that there isn't enough of the source in nature ro meet demand. Frequently it is more economical to produce commercially useful compounds from convenient chemical feedstocks. Over the last century and a half, those feedstocks have come from coal tar and, later, petroleum. Currently, there is rapid progress underway to develop chemical feedstocks from sources such as vegetable and algal oil (i.e. oil from seaweed).
These feedstocks are just compounds that can be converted synthetically into pharmaceuticals as well as in plastics, paints, coatings and other materials. Frequently, the starting materials for these processes contain C=C bonds that can be functionalized through electrophilic addition. Thus, electrophilic addition and related reactions are among the most important in the world, economically speaking. It's very valuable to be able to control the outcome of these reactions in order to make processes more efficient, producing fewer wasteful by-products.
• Regioselectivity, or control over where a reaction occurs, is very important.
• The Markovnikov versus anti-Markovnikov additions available in hydration are good examples of regiochemical control.
Exercise \(1\)
List advantages and disadvantages of producing materials based on
1. petroleum
2. vegetable oil
3. algae
4. harvesting desired compounds directly from nature
Again, we are going to focus on the first of the two reactions in this sequence. That part is where the placement of the new substituent is decided. After the addition of the borane, an alkylborane is formed. The major isomer results from anti-Markovnikov addition.
It seems pretty clear at this point that this reaction must proceed like other electrophilic additions to alkenes. The π electrons donate to the electrophile. In this case that's boron, which is strongly Lewis acidic because it lacks an octet. The boron ends up at the least substituted end of the double bond.
That outcome would certainly be favored over this one:
That boron is beginning to look less electrophilic and more nucleophilic. We can easily imagine a hydride nucleophile being delivered to the carbocation.
So far, the picture of how the alkylboration reaction works fits pretty well within our electrophilic addition framework. Unfortunately, there are some problems with this model.
First of all, maybe 55% of the boration takes place in a Markovnikov sense, but the other 45% is added to form an anti-Markovnikov product. Certainly the secondary cation is favored over the primary one, but if the reaction is proceeding through a carbocation, then the primary one shouldn't happen at all.
Something is wrong with our model.
Another hole is torn in the argument when we look at the results of stereochemical studies. We could, for example, take the following deuterium-labelled hexene and treat it with borane. We could look at the products via 1H NMR spectroscopy, and if we could see the coupling constant between the two protons shown in the structure, then we would know their relative arrangement in space. We would know their stereochemistry.
If we did that experiment, then we would see that the hydrogen and the boron from the borane are added to the same face of the alkene. We don't get addition of boron to one face and hydrogen to the other. This type of addition is called syn addition; it is the opposite of anti addition.
• Hydroboration results in syn addition to the alkene.
Exercise \(2\)
Show how a cationic intermediate and conformational changes would allow both syn and anti addition of borane to propene.
Answer
That result means that, although some elements of our mechanism may reflect reality, we at least have a problem with timing. How can the hydride be delivered before the conformation has a chance to change? It has to happen pretty quickly. What if it happens at the same time as π bond donation to the boron?
This reaction would best be described as a concerted addition. Two groups are added to the two ends of the double bond at the same time. During the transition state, two bonds would be breaking and two would be forming at the same time.
Exercise \(3\)
We can further improve our model of how the alkylboration works if we consider that disiamylborane and 9-BBN are much more effecient than borane in terms of regioselectivity. These reagents can produce close to 100% anti-Markovnikov addition. Explain how with the help of drawings.
The subsequent reaction in this series involves removal of the boron and replacement with a hydroxyl group. The mechanism of this reaction may not be worth memorizing becuase it doesn't fit well within categories we have looked at so far.
The important thing to know is that the oxygen ends up in exactly the same place as the boron. There is no change in stereochemistry at that position. Overall, the hydrogen and the hydroxy effectively group undergo syn addition, although they are added in different steps.
Exercise \(4\)
Borane is frequently used in THF because borane alone is not very stable; it is quite pyrophoric, bursting into flame upon contact with air. In THF, borane forms an equilibrium with a Lewis-acid-base complex. Show this equilibrium reaction.
Answer
Exercise \(5\)
Provide reagents for the following reactions
Exercise \(6\)
Provide reagents for the following reactions
Exercise \(7\)
Show products of the following reactions. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/IV%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_2/06%3A_Electrophilic_Addition_to_Alkenes/6.07%3A_.txt |
Epoxidation is the addition of a single oxygen atom across a C=C double bond.
Earlier, we saw that alkenes can donate their pi electrons to electrophiles such as "Br+". In the bromonium ion that results, a lone pair on the bromine can donate back to the incipient carbocation, leading to a more stable intermediate.
We have also seen that addition to alkenes can sometimes be concerted, happening all at once, rather than one step at a time. For example, in hydroboration, the boron and the hydrogen add to the double bond at the same time.
The boron is adding just slightly ahead of the hydrogen. The initial interaction is donation from the pi bond to the Lewis acidic boron. However, as soon as positive charge starts to build up on carbon, and negative charge starts to build up on boron, the hydride is immediately donated. Time is not allowed for the charged intermediate to fully form before proceeding.
Really, that's what is happening to the bromine, too. As the alkene starts to donate its pi electrons to the bromine and begins to build up positive charge, the bromine's lone pair is drawn back to the alkene. As a result, the intermediate that we imagine with a full positive charge on carbon and no charge on bromine exists too fleetingly to be considered an intermediate at all. As soon as it begins to form, it is already turning into something else.
That sort of concerted addition happens with some other electrophiles, too. If an atom is electrophilic, but also has a lone pair to donate, that cyclic transition state can lead to the product in one step.
Alkene epoxidation is another example of this kind of reaction. An epoxidation is the transfer of an oxygen atom from a peroxy compound to an alkene. Peroxides contain O-O bonds, which are relatively weak and reactive.
To simplify a little bit, just look at the reaction from the point of view of the alkene. It's just picking up an oxygen atom, because the peroxide had an extra one.
When the oxygen atom is transferred, it forms an epoxide (sometimes called an oxirane). It is a three-membered ring containing two carbons and an oxygen.
• Epoxidation results in transfer of an oxygen atom from a peroxide to an alkene.
• Peroxides are compounds containing weak O-O bonds.
Like in a bromination, the electrophile is deceptive. It is an oxygen atom, which we more naturally think of as a nucleophile. However, just as Br2 contains an atom attached to a good leaving group (Br -), so do the kinds of oxygen compounds used in epoxidation. Most often, these are "peroxy acids", carboxylic acids containing an extra oxygen.
As in the bromination, as soon as the alkene begins donating to the electrophile, a lone pair can donate back, so that an unstable cation does not have to form.
The entire mechanism is believed to be concerted, based on a number of lines of experimental evidence. A number of things need to be accomplished; in addition to the oxygen donation, the leaving group must leave, and a proton must be transferred.
The reaction mechanism can be cleaned up slightly because it is thought to be an example of a pericyclic reaction. Pericylcic reactions frequently involving three pairs of electrons moving in a circle. Like the three pairs of electrons in a benzene ring, this structure is thought to be unusually stable.
Apart from peroxy acids, many other peroxides can be involved in epoxidations, as well as some metal oxides. In some cases, the reaction is extremely slow, but works better in the presence of a catalyst.
Exercise \(1\)
Predict the products of the following reactions.
Epoxidation reactions display an almost counter-intuitive selectivity. Unlike hydrogenation reactions, which are generally easier with less-substituted alkenes, epoxidations are much faster with more-substituted alkenes. In the case of hydrogenations, the selectivity can be understood as a combination of steric factors (the alkene must bind to a catalyst) as well as thermodynamicic factors (more substituted alkenes are more stable, so they are less likely to react). However, in epoxidations, the more electron-rich the alkene, the more easily it can be induced to react with the peroxide. More substituted alkenes are generally more electron-rich than those that are substituted only with electron-poor hydrogens.
Exercise \(2\)
Predict the products of the following reactions.
The Sharpless epoxidation is one of the most common methods of catalytically adding an oxygen across a double bond. The method generally employs a titanium catalyst; similar approaches use vanadium catalysts or other metallic species. As mentioned before, metal ions can sometimes accelerate epoxidations.
The Sharpless epoxidation is important partly because it selectively epoxidizes allylic alcohols: compounds containing a C=C-C-OH unit. That means that, in addition to being able to selectively epoxidize more-substituted double bonds in the presence of less-substituted double bonds, we can also select double bonds that are close to alcohols.
Exercise \(3\)
Circle the allylic alcohol in each of the following compounds.
How does the metal catalysis selectively identify that position? Remember, one of the important strategies in enzyme catalysis is approximation: the act of bringing two things together. An alcohol is a potential lone pair donor, so it could become a ligand for a metal ion. Ti4+ and V4+ happen to be very oxophilic -- they bind well to oxygen -- and so they are particularly suited for this task.
In this scheme, we're not worrying about exactly how the titanium ion gets to peroxide to give up its extra oxygen to the alkene; that's complicated. However, the fact that both the allylic alcohol and the peroxide can bind to the titanium gets them closer together, and makes them more likely to react with each other.
That's only part of the story of the Sharpless epoxidation. The other reason this method is important is its stereoselectivity. To get stereoselectivity, a chiral ligand is added for the titanium. It's usually diethyl tartrate (DET) or diisopropyl tartrate (DIT). Tartrate is chiral; there is a D-(-)-enantiomer and a L-(+)-enantiomer. The D and L are common symbols used to designate enantiomers in sugars; they relate the structure back to the biochemical grandparents of all sugars, D-glyceraldehyde and L-glyceraldehyde. The (-) and (+) symbols refer to the characteristics of this particular compound in polarimetry; the (+) enantiomer rotates plane polarized light in a clockwise direction, whereas the (-) enantiomer rotates plane polarized light in a counter-clockwise direction.
Exercise \(4\)
Assign stereochemical configurations (R and S) to the tartrates to confirm that they are enantiomers of each other.
Answer
D-(-)-tartrate is the (2S,3S)-isomer. L-(+)-tartrate is the (2R,3R)-isomer.Each chiral center is configured opposite to the corresponding one in the other molecule, so the molecules are enantiomers.
If the D-(-)-isomer is added, one possible enantiomer of the product is obtained. If the L-(+)-isomer is added, the other possible enantiomer is obtained.
In general, we would get one enantiomer if the oxygen were added to one face of the alkene and the other enantiomer if the oxygen were added to the other face. In the drawing below, the face of the alkene towards us is sometimes called the "re face" (pronounced, ray face). The face of the alkene away from us is called the "si face" (see face). These words sound related to R and S configuration, and they sort of are like that, but they are used to describe two different faces of a flat molecule. Adding oxygen to the re face gives on enantiomer; addign oxygen to the si face gives another.
How does this preferred reactivity work? How does the metal manage to add the oxygen to one face but not to the other? Tartrates are oxygen-rich and so they bind very well to titanium. Remember, if we have a reaction site and we make it chiral, one enantiomer of the product is generally preferred. Enzymes are very compicated, chiral molecules, and they are good at producing one enantiomer of a product. By comparison, the titanium DET complex is a relatively simple chiral molecule, but it uses the same idea.
Now, it is really very difficult to look at these conditions and predict exactly which enantiomer would be formed in a reaction. However, we can look at a factor that might illustrate an underlying reason for the preference. In quadrant analysis, we look at the general shape formed by that bidentate tartrate ligand on the metal. In the pictures below, the red ball is the metal atom. The tartrate ligand extends up and to the right as it sits on the metal, and also down and to the left; it is cartooned in blue. As a result, if we think of the metal as sitting in the middle of a square, alternating corners of that square are filled, and the other corners are empty.
Imagine the alkene approaching that metal. The alkene will probably have a preferred orientation in which it will bind. Just for example, maybe it needs to bind with the double bond in the same plane as the ring formed by the titanium and the oxygens (horizontally in the picture). If it does that, it can reduce steric interactions with the ligand by binding one face of the alkene preferentially to the metal, keeping the biggest substituent on the alkane (the black ball) in a relatively open space. The alkene could also bind if rotated upside down compared to the first picture, but the same face would still be towards the titanium.
On the other hand, if the alkene tries to bind through the other face of the pi bond, the largest substituent would be in a more crowded space. That might be less favorable.
Overall, if the alkene has a preferred face that it will bind to the metal, then anything delivered from that metal will land on that face, and not the opposite one.
There are lots of variations on this model. Maybe it isn't steric interactions that influence how the alkene approaches the metal. Maybe it is some other factor, like hydrogen bonding, that pulls in the alkene oriented in one direction and not another. Nevertheless, although the details of a particular case make the outcome very difficult to predict, the general idea is a familiar one: a chiral molecule will fit preferentially one way with another molecule, because of its asymmetric shape.
Exercise \(5\)
Although you may not be able to predict off the top of your head which enantiomer is formed in a Sharpless epoxidation, given one result, you may be able to guess another. Given the reaction on the left, see what you can tell about the reaction on the right.
Exercise \(6\)
Fill in the boxes in the following synthesis.
Exercise \(7\)
Fill in the boxes in the following synthesis.
Exercise \(8\)
Fill in the boxes in the following synthesis. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/IV%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_2/06%3A_Electrophilic_Addition_to_Alkenes/6.08%3A_.txt |
Earlier, we saw that alkenes can donate their pi electrons to oxygen electrophiles in peroxides. The result is transfer of an oxygen atom from the peroxide to the alkene. An epoxide or oxirane ring is formed.
Another, related example is cyclopropanation of an alkene. In alkene cyclopropanation, an alkene is converted into a cyclopropane. A carbon atom is donated to the alkene.
A classic example is the addition of dichlorocarbene to an alkene.
Carbenes are electrophiles because the carbon does not have an octet. The carbon has only two bonds and one lone pair. That's just three electrons, not eight.
On the other hand, there is a lone pair. The carbene can be nucleophilic, too.
The reaction involves addition of the alkene to an electrophilic carbene. At the same time, that lone pair can donate back, so that a carbocation does not actually form.
The carbene is sometimes formed through an unusual "alpha"-elimination in the presence of strong base. Strong bases are often alkyllithium reagents, such as CH3Li, but KOH will work with some compounds.
The reaction is unusual because a proton is abstracted from one carbon atom, and a leaving group departs from the same atom. It is much more common to see the leaving group depart from the next atom over, in a beta-elimination.
Carbenes are frequently formed from diazo compounds. These include diazomethane, CH2N2.
In diazo compounds, an N2 group is hanging on by a thread. It can easily leave, resulting in a carbene. This reaction is often promoted by metal catalysts, such as copper (II) salts.
The carbene, CH2, is even less stable than CCl2. However, in the presence of metal salts, it can be stabilised as a metal carbene complex.
Exercise \(1\)
Why is a carbene more stable with chlorine atoms attached?
Answer
The chlorines can (weakly) share their electrons to fill the octet on carbon.
Exercise \(2\)
Some metal carbene complexes, such as the one shown below, are particularly stable. Explain why.
Exercise \(3\)
Free carbenes, those not attached to a metal ion, are typically so unstable that they can only be generated briefly in solution before they react with a nucleophile, such as an alkene. Arduengo carbenes, such as the one below, are stable enough to be put in a bottle and stored in the refrigerator. Explain why.
Exercise \(4\)
Predict the products of the following reactions.
6.10:
There are a number of other additions to alkenes that occur via concerted mechanisms. Alkene oxidations are among the most synthetically useful of these reactions because they are able to convert simple hydrocarbon starting materials into oxygen-containing compounds. The resulting heteroatomic functional groups may open up new avenues of synthetic utility or they may reflect aspects of a target natural product.
The three most common alkene oxidations are epoxidation, dihydroxylation and oxidative cleavage.
Epoxidation
Epoxidation is a method for converting an alkene into an epoxide. The reagent required is always a peroxo species. A peroxo species looks very much like a normal oxygen-containing compound, but with an extra oxygen in it. Historically, the most common such reagent was m-chloroperbenzoic acid (mCPBA).
However, other reagents can also be used, such as hydrogen peroxide (H2O2) or potassium hydrogen persulfate (KHSO5), marketed under the trade name Oxone. The latter methods are considered "greener" or more environmentally friendly, because the side poducts (water or sulfate, respectively) are pretty innocuous. These methods are generally slower and are often used with a catalyst. Catalysts used with hydrogen peroxide include Lewis acidic species such as sodium tungstate (Na2WO4) needed to activate the peroxide. A similar reaction using titanium (IV) and chiral ligands leads to an enantiomerically pure epoxide; this reaction is called "Sharpless epoxidation". With oxone, ketones are used as oxygen transfer catalysts in a method referred to as "Shi oxidation".
The electrophilicity of peroxy compounds continues a theme seen in halogens such as chlorine and bromine. When two oxygen atoms are connected to each other, one of the can act as an electrophile, just as when two halogens are connected together.
During the epoxidation, the peroxy compound simply delivers its extra oxygen to the double bond. The oxygen atom both accepts a pair of electrons from the double bond and donates an electron pair to the double bond at the same time.
The reaction has something in common with pericyclic reactions. In pericyclic reactions and other reactions that take place under control of orbital symmetry, it is common to see six electrons circulating in a ring as a central feature of the mechanism. This picture is reminiscent of the aromatic structure of benzene. In fact, that aromatic stabilization is thought to play a role in stabilizing the transition states of various reactions. In this case, the three electron pairs involve delivery of the oxygen, proton transfer and π donation to the carbonyl in mCPBA. However, a similar set of arrows might not be found in the reaction of hydrogen peroxide.
Exercise \(1\)
Show how the carbonyl in mCPBA may help activate the donor oxygen toward reaction with the alkene.
Answer
Exercise \(2\)
Show how a titanium (IV) ion may help activate the donor oxygen in hydrogen peroxide toward reaction with the alkene.
Answer
The electrophilic nature of the peroxy compound is seen in the selectivity of the epoxidation reaction. Alkenes that are more electron rich tend to react much more quickly than other ones. For example, more substituted alkenes, often regarded as being electron-rich, can be selectively epoxidized in the presence of other alkenes.
Furthermore, although enones can sometimes be epoxidized, the reaction is generally slower than with regular alkenes. Of course, the carbonyl attached to the alkene in an enone makes the alkene very electron-poor.
Part of the evidence for a concerted mechanism for epoxidation comes from the stereochemistry of the reaction. In general, if a cis-alkene is epoxidized, the two groups that were cis to each other in the alkene remain cis to each other in the epoxide. If the groups start out trans to each other, they remain trans in the epoxide. Just as in hydroboration, there is no opportunity for these stereochemical relationships to change.
Dihydroxylation
Dihydroxylation is the addition of an OH group to both sides of an alkene. Typically, when reagents such as osmium tetroxide are used, the hydroxyl groups are added to the same face of the double bond. This reaction is therefore called a syn-dihydroxylation.
Osmate esters can be isolated from this reaction, resulting from the concerted addition of osmium tetroxide to the alkene. Once again, this step can be compared to a pericyclic reaction. However, the osmate ester is usually decomposed in situ through the addition of a "reducing agent" such as sodium sulfite.
Once again, the concerted nature of the reaction is seen in the stereochemistry of the product. The fact that both oxygens, which come from the osmium, are delivered to the same face of the alkene suggests that they are added at the same time.
It isn't uncommon for oxygen atoms to form additional π bonds to transition metals such as osmium. In this case, we could think of the osmium as forming an 18 electron complex as a result. Whether or not that resonance contributor is an important representation of osmium tetroxide, it is a helpful device to think of how the oxygen might form an initial attraction to the alkene.
Because of osmium tetroxide's high cost, potent toxicity and alarming propensity to rapidly sublime, other reagents are preferred. It is quite common to still use a catalytic amount of osmium tetroxide, though, along with a co-oxidant. Co-oxidants can be things like Fe(III) salts and air, although hydrogen peroxide is often used.
Oxidative Cleavage: Ozonolysis
Ozonolysis results in the complete cleavage of a double bond into two parts. The resulting fragments are each capped by an oxygen atom.
Once again, this reaction starts out with a concerted addition of the ozone to the alkene.
However, the first-formed adduct, termed a molozonide, quickly rearranges to a second product, termed an ozonide. Both of these species can be isolated. However, in practice this is rarely done because of the appalling tendency of molozonides and ozonides to explode unexpectedly. The ozonide is instead decomposed through the addition of a reducing agent, such as dimethylsulfide or zinc, leaving two oxygen-containing fragments behind.
Exercise \(3\)
Fill in the reagents for the following alkene oxidations.
Exercise \(4\)
Fill in the products of the following alkene oxidations.
Exercise \(5\)
Fill in the blanks in the following synthesis. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/IV%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_2/06%3A_Electrophilic_Addition_to_Alkenes/6.09%3A_.txt |
We have already seen the general concept that alkenes can be polymerised through a series of electrophilic additions. In this section, we will look at this topic in more depth.
In polymerisation, a large group of monomers are pulled together into a single, large molecule. In most cases, the monomers are connected in a row, forming a long chain. This process is sometimes called "enchainment".
Looking at the structure of the polymer, we can see where the monomers have ended up. They are linked together along the chain in a repetitive fashion. When the monomers become enchained, they turn into the "repeat units" of the polymer.
The structure of a polymer is often depicted using the repeat unit in parentheses. A subscript n stands for an integer, meaning that n of these repeat units are linked together in a chain.
Of course, this chain has to end somewhere. At either end of the chain there will be something else attached; these parts of the polymer are called the "end groups". The identity of these groups can vary; it depends on how the polymer was made. One of these end groups gets incorporated into the polymer when the polymer starts growing; it was the "initiator". The other end group gets incorporated when the polymer stops growing, in a chain termination step. We'll see more about termination soon.
Let's take a look at how a macromolecule grows via cationic polymerisation. The initiator is a cation that reacts with the alkene. When it does so, it forms a new cation from the old alkene.
This sort of reaction is called a chain reaction because one reactive species reacted to form a new reactive species; this cycle then keeps repeating in a chain.
The process keeps repeating so that more and more monomers become enchained. In the picture below, only three monomers have been enchained so far, but you get the idea.
Most cationic polymerisations are initiated by a proton. That means that one of the end groups can be thought of as a proton. Alternatively, the first monomer can be thought of as having a slightly different structure than the repeat units that follow.
Perhaps the most obvious method of providing a proton is to add a protic acid, such as sulfuric acid (H2SO4) or trifluoromethanesulfonic acid (CF3SO3H or often abbreviated as TfOH). However, although that seems easy on paper, polymerisations typically don't work extremely well under those conditions. Usually, something goes wrong and the polymer stops growing when it is still a fairly short chain.
More commonly, commercial polymerisation involves a two-component mixture that together can initiate the reaction. One of these components is a Lewis acid. Some common Lewis acids used for this purpose include BF3, AlCl3, TiCl4 or SnCl4. The other component is usually a small amount of water or alcohol.
The role of the Lewis acid is to activate the water or alcohol, providing a proton to initiate a growing chain. Because the water or alcohol is the actual source of the proton, it is referred to as the initiator, whereas the Lewis acid is called a co-initiator.
The other end group in the polymer is formed during termination, when something happens that stops the polymer from going. In termination, the cation in the growing chain would be lost. That event can happen through a few different pathways.
The simplest thing that might happen to stop chain growth is for some anion to connect irreversibly with the cation of the growing chain. That might happen accidentally while the polymerisation is supposed to be occurring. Alternatively, it might be forced to happen on purpose. When the polymer has grown for a long enough time to reach the desired weight, we might simply add some aqueous acid (like dilute HCl, for example). In that case, chloride ions might connect with the cations. Also, water molecules might connect with the cations, leaving the chains terminated by hydroxyl groups after loss of a proton.
Other accidental pathways for chain termination include "chain transfer" events. In chain transfer, a proton is lost, either to an anion or to an alkene. This event results in an elimination reaction, forming an alkene at the end of the chain. However, the transferred proton results in initiation of a new growing chain. These accidental terminations can be a problem because some chains just get started growing after others have already been growing for a long time. As a result, the material obtained has a variety of chain lengths. The material has a high polydispersity, which means the properties of the material are difficult to control.
Termination events also raise the possibility that there will be some leftover monomer at the end of polymerisation. We wouldn't want these small molecules in the material we are making, because they would slowly leach out of the material over time. Sometimes, these monomers are removed via a precipitation step. The polymer mixture is simply dumped into a solvent in which the monomers will dissolve but the polymer will not. The solvent is then poured off, taking the monomers with it, leaving behind a purified polymer.
Exercise \(1\)
The following monomers are readily polymerised via cationic polymerisation. Show why, using drawings of the intermediates invlved in the reaction in each case. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/IV%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_2/06%3A_Electrophilic_Addition_to_Alkenes/6.11%3A_.txt |
The trouble with reactive intermediates such as cations is that they are so reactive. Sometimes, cations do unexpected things. For example, we have already seen how they frequently undergo 1,2-hydride shifts. They can undergo nucleophilic addition, but they may undergo elimination instead. Sometimes, there may be competition between different nucleophiles racing to combine with the cation; you may end up with a mixture of products.
In other words, all kinds of things could go wrong during a cationic polymerisation.
What are the consequences of a polymerisation gone bad? The biggest problem is that the polydispersity gets too high. The polymer chains obtained in the reaction are not uniform in size; there are some much, much longer chains as well as some much, much shorter chains.
Remember, the polydispersity is an important component of how we think about the size of polymer chains. Because polymers are a collection of molecules formed through tandem growth of chains -- lots of chains growing at the same time -- we will always have a range of chain lengths. The polydispersity tells us how broad is that distribution. If the polydispersity is too high, it means there is too great a range of chain lengths (and molecular weights) in the material. Because the physical properties of polymers depend in part on the chain length, when the polydispersity gets too high, we have less and less control over the properties of the material. It might not perform the way we want it to perform.
Why do these unexpected reactions of cations lead to a wider range of molecular weights? It's because they often stop a chain from growing. We call these reactions "random termination events".
As a result of random termination events, some chains stop growing when they are still too short. Other chains keep growing and gobble up the rest of the monomer. There is extra monomer left over, because some chains didn't use theirs, so the chains that keep growing get extra long. There is a very wide range of chain lengths.
A secondary symptom of this problem is that the measured molecular weight of the polymer is sometimes higher than expected. That doesn't make any sense at first; if the chains stop growing, how can the molecular weight be higher than we thought it would? Wouldn't it be lower?
Well, sometimes that's true. If all of the chains were to stop growing early, then the molecular weight wouldn't reach our expectations, and there would be a lot of monomer left over. However, we're talking about random termination. Some chains stop growing. Others don't.
Remember that at the end of a polymerization, the polymer is usually precipitated. Any leftover monomers stay dissolved and are decanted off. At the same time, many of the very short chains also stay dissolved, and they are lost, too. We are left with the bigger chains, and the average weight of the chains has shifted higher. In addition, there are physical methods for measuring the size of polymers that are more strongly influenced by the bigger molecules, and that fact may artificially inflate our estimate of molecular weight.
In contrast to this scenario, a "living polymerisation" is a process in which the polymer chains keep growing uniformly and the molecular weight agrees closely with expectations. Furthermore, after a living polymerization has finished, you can add more monomer and the chains will start growing again. The growing chains don't die out and stop growing. When they run out of monomer, they remain "dormant" (it's like they are just sleeping) until more monomer is available.
A key strategy for living polymerisation is to limit the number of reactive species at a given time. The fewer reactive species there are, the fewer random termination events will occur.
That scenario might seem to pose a different problem: if there are fewer reactive species, then there are fewer growing chains. If there are fewer growing chains, won't the polymerisation be much slower? Yes, it will be slower. We are looking for an optimum point at which polymerisation proceeds at a reasonable rate but the termination rate is very low.
However, there is a way to keep the number of reactive species low, but still have lots of growing chains, or at least potentially growing chains. We just exploit an equilibrium in which chains spend part of their time actually growing and part of their time dormant.
The idea is to have an anion that can cap the cation, so that we have an alkyl halide rather than a cation. In a very simple case, we might think about adding some halide salts to the polymerisation reaction. The halide anions might bind reversibly to the cationic intermediate, sending it into a dormant phase. Occasionally, the halide would dissociate again, and the polymer would grow again.
However, the halide must leave once in a while, producing a cation that can undergo polymerisation. In order to help that halide leave, a Lewis acid might be employed. The following example shows tin(IV) chloride added to the mix.
The halide coordinates to the Lewis acid, which polarises the carbon-halogen bond. Tin, iron or titanium compounds are some exampes of Lewis acids that are sometimes used for this purpose, but there are others, as well.
This method may actually change the mechanism of the reaction slightly. Maybe the halide ion never actually leaves completely, and the cation never fully forms. Instead, there may be enough polarization in the presence of the Lewis acid so that the alkene donates to the incipient (almost-formed) cation.
Exercise \(1\)
Provide a mechanism, with arrows, for the following living cationic polymerisation reaction. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/IV%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_2/06%3A_Electrophilic_Addition_to_Alkenes/6.12%3A_.txt |
Polyethylene and polypropylene are two enormously important materials on the market. The fact that their use continues to persist despite legitimate environmental concerns is a testament to how useful these materials have become over the years.
Polyethylene and polypropylene can be thought of as polymers of the very simple alkenes, ethene and propene. In fact, that's exactly where these materials come from.
Exercise \(1\)
Draw the first intermediate formed after the initiation step in the cationic polymerization of:
1. ethene
2. propene
Unlike other polymers of alkenes that we have looked at, polyethylene and polypropylene are not polymerised via cationic methods.
Instead, these monomers are enchained through a process called "Ziegler-Natta polymerisation." This process is named after a German and an Italian chemist who are independently credited with its development in the 1950's.
In Ziegler-Natta polymerisation, monomers are treated with a catalyst, such as a mixture of titanium chloride (or related compounds, like oxovanadium chloride) with triethylaluminum (or trimethylaluminum). Other components are often added, such as magnesium chloride, to modify the catalyst and improve performance. The mixture described here produces a heterogeneous catalyst; it is an insoluble solid.
Now, the catalyst isn't really titanium chloride, because all of these components react together to make something new. Exactly what they make may be hard to determine. It's a complicated gmish.
Despite the complicated catalyst mixture, we do know a little bit about the mechanism of reaction. Partly this information comes from studies of model compounds. Model compounds are simpler than the industrial catalysts, but they still have some structural features in common with their hard-working cousins. They have enough in common to be able to carry out polymerization catalysis, although maybe not as well as the industrial heavyweights.
So, what do we think happens? It seems pretty clear that one of the things that the trialkylaluminum does is provide an alkyl group to titanium. That shouldn't be too surprising. The triethylaluminum, like ethyllithium or ethylmagnesium bromide, ought to be a source of nucleophilic ethyl groups. The titanium tetrachloride ought to be a pretty good electrophile, complete with halide leaving groups. We can imagine at least one of those chlorides getting replaced by an ethyl ligand.
Exercise \(2\)
Provide a mechanism for the ethylation of titanium tetrachloride with triethylaluminum.
The next step, presumably, could be the binding of an alkene ligand to the transition metal.
Now we are looking at an organo transition metal compound. We should be thinking about organometallic reaction mechanisms. For example, a 1,2-insertion of the alkene into the metal-carbon bond would provide a new metal alkyl.
From there, and using these same elementary steps in succession, it is easy to imagine how polymerization of propene might occur. Bnding of an additional alkene, followed by the 1,2-insertion of the propene into the metal-carbon bond, results in formation of the propene dimer (with a methyl end group).
• The mechanism of Ziegler-Natta polymerization involves alkene binding and insertion into metal-carbon bonds
Let's pause for a moment and look a little more closely at the role of the aluminum compound. This topic is peripheral to electrophilic addition although important to the subject of catalysis.
In catalysis, "promoters" and "supports" are sometimes added to improve catalyst function. They might do so in a number of ways. They may take part directly in the reaction, providing additional Lewis acidic or Lewis basic sites, in much the same way that amino acid residues surrounding the active site of an enzyme may help catalyse a reaction. They may play a more subtle role, affecting physical properties of the catalyst (such as its solubility) or even tuning up the chemical properties of the catalyst. For example, maybe the promoter adds a little more electron density to the catalyst, making it a little less electrophilic. That might make the catalyst more stable; maybe it becomes more selective, reacting more carefully instead of with wild abandon. Maybe it makes the catalyst last longer.
Although classic Ziegler-Natta polymerization involves heterogeneous catalysis, lots of variations have been developed, including model systems to study the basics of the reaction as well as other, working homogeneous catalysts. One very successful variation, developed by Walter Kaminsky at University of Hamburg, uses Cp2ZrCl2 as catalyst and methylalumoxane (MAO) as a promoter. This zirconium species, in which the zirconium atom is wedged between two cyclopentadienly ligands, is commonly called a "zirconocene".
MAO is another poorly-defined species. It is obtained by treating trimethylaluminum with a trace of moisture. If you remember anything about Grignard reagents or alkyllithiums, you might think that isn't such a good idea. In truth, it is an even worse idea with trimethylaluminum with either of those other two metal alkyls. The trimethylaluminum is quickly decomposed into something else, a poorly-defined species called "methylalumoxane".
What we know about the structure of MAO may be a little bit fuzzy. Once again, some light can be shed on the subject via model studies. In some beautiful work done in Andrew Barron's lab at Rice University, alkyl aluminum oxide clusters were obtained via the careful treatment of tri(tbutyl)aluminum with water. Aluminum oxide clusters resulted, containing two, three, four or six aluminum atoms, depending on the reaction conditions. A drawing of one example, a hexamer, is shown below.
Some of the alkyl groups have been replaced by oxide ligands. We can imagine something similar would happen with trimethylaluminum. The aluminum forms bonds with oxygen, which bridges between different aluminum atoms. The structure is probably oligomeric itself, forming large clusters of aluminum oxide, although it must retain some methyl groups as well.
Exercise \(3\)
Provide a mechanism for the formation of an oxy-bridged aluminum dimer via treatement of trimethylaluminum with water.
Exercise \(4\)
Why does tri(tbutyl)aluminum with water produce well-defined structures, whereas trimethylaluminum with water leads to a mess?
Just as in the original Ziegler-Natta catalyst, the "extra stuff" plays an important role. The MAO may tune up the qualities of the zirconium catalyst, in addition to providing an alkyl group.
Kaminsky's "zirconocene" catalysts are used commercially to produce polypropylene. They have been particularly important in developing ways to control the stereochemistry of the reaction.
Consider a polypropylene chain. Each methyl group that hangs from the zig-zagging backbone of the polymer could have two possible orientations. It could be coming forward, shown with a wedge. It could be going backward, shown with a dash.
The stereochemical relationship between those wedges and dashes is called "tacticity". Tacticity basically comes in three flavours: random, alternating and same. A random stereochemical arrangment is described as an "atactic" polymer. If all the methyl groups are on the same side in a regular zig-zag projection of the backbone (either all wedges or all dashes) the arrangement is described as "isotactic". If instead the methyl groups alternate (wedge-dash-wedge-dash), the arranement is called "syndiotactic".
In a collaboration with Hans-Herbert Brintzinger at Konstanz University, Kaminsky developed zirconocene catalysts that could control the tacticity of the polymer chain. Use of the original zirconocene, Cp2ZrCl2, resulted in an atactic polymer. However, modified zirconocene catalysts selectively make either the isotactic or the syndiotactic polypropylene. We will look at two examples. The first one is sometimes described as a C2 catalyst and leads to formation of isotactic polypropylene. The second is sometimes described as a Cs catalyst and leads to formation of syndiotactic polypropylene. (The labels, Cs and C2, are symmetry point groups that describe the shape of the catalyst, but we won't go into that idea any further.)
Exercise \(5\)
One of these catalysts (C2 vs. Cs) is chiral; the other one isn't. Which is which?
Let's look at the C2 catalyst. We'll strip it down to just the zirconocene part, leaving off the chlorides. The chlorides are likely replaced by methyl groups or else lost via dissociation (presumably leaving Cp'2ZrCl+ ion).
We will rotate the catalyst fragment to look at it from the "front": the more open part of the zirconocene "wedge", which is the direction from which a newly coordinating propene would approach.
Here is the view from in front of the wedge. This is the surface that the propene will interact with as it approaches.
In our discussion, we will use "quadrant analysis", a standard tool for trying to analyse stereocontrol in transition metal catalysis. In quadrant analysis, we try to imagine differences in steric barriers in each of four quadrants around the metal centre. How will the arrangement of bulky groups influence the approach of a substrate?
In the C2 catalyst we are using, it looks like there will be more room in the upper right and lower left quadrants. The upper left and lower right are blocked by those rings. When the propene is approaching, the polymer chain will present the largest obstacle, because it is extending a significant distance away from the zirconium atom. To minimize steric interactions,the chain may extend into the relatively empty upper right quadrant.
Alternatively, the polymer change could extend into the relatively open lower left quadrant, but that would really give us the same drawing, just rotated by 180 degrees.
Exercise \(6\)
Make a drawing of the complex with the polymer in the lower left quadrant.
As we think about bringing the propene into the empty coordination site next to the polymer chain, there are two questions we need to consider about orientation. The first is about which end of the alkene to bring into the wedge. The propene has two different ends: one end sports two hydrogen atoms, whereas the other end has a hydrogen and a methyl. It seems likely that the propene will fit best if the narrow end, the one with the two hydrogen atoms, extends into that narrow wedge.
Now we need to think about which of the propene's two faces will coordinate to the zirconium. To think about the faces of propene, hold your hand out flat, with the thumb forming a right angle with the rest of your hand. The back of your hand represents one face of the propene; the palm of your hand represents the other. The propene will enter in such a way as to minimise steric interactions.
It looks like the easiest way is as shown below. The methyl group is placed in the lower left quadrant. You can think of it as keeping the methyl group away from the upper ring or keeping it away from the polymer chain, which is also in an upper quadrant.
If your left hand is propene, we have coordinated the back of your hand to the zirconium, with the thumb pointing down. If we had coordinated your palm, the thumb would be pointed up.
Once again, if we started with the polymer in the lower left quadrant, the entire drawing just rotates 180 degrees. It is still the same face that coordinates (the back of your left hand).
Exercise \(7\)
Make a drawing of the complex with the polymer in the lower left quadrant and the propene coordinating to the right.
The next event is 1,2-insertion of the alkene into the metal-alkyl bond. That leaves us with the following structure. Notice that we have formed a new stereocentre. Because the methyl group of coordinated propene was pointing down, and the hydrogen adjacent to it was therefore up, then in the new stereocenter the methyl is still in the lower of two possible positions and the hydrogen is in the upper of two possible positions. Furthermore, the methyl is pushed back and to the left because the alkyl came from the right.
In order to help keep track of that insertion step, here is a drawing with colour labels. The narrow end of the propene is now attached to the zirconium via a sigma bond. The wider end of the proene has formed the new chiral center. The carbon that used to form a sigma bond to zirconium is now just another carbon along the growing polymer chain. The youngest part of the polymer is found at the growing end.
When another propene approaches to occupy the empty position, it will coordinate using the same face as the previous propene.
The 1,2-insertion produces a new chiral centre. The C2 catalyst is producing an isotactic polymer chain. The is an example of "site control" of polymerization. The chiral C2 catalyst has influenced the stereochemistry of the growing chain.
Exercise \(8\)
Assign configuration (R or S) to each of the chiral centres along the polymer chain in the above drawing to confirm isotacticity. Keep in mind that "polymer" stands for a long chain of carbons.
Now let's look at the Cs catalyst. We'll strip it down like we did before.
Where will a growing polymer chain go? Obviously it should go into one of the upper quadrants.
It isn't obvious whether it should be upper right or upper left. This time, it makes a difference, because the picture would not be the same.
Exercise \(9\)
Draw the catalyst site with the polymer chain in the upper left quadrant.
However, once the propene approaches, a clear preference occurs. The chiral centre on the polymer chain is shown with the hydrogen up towards the Cp ring, since it is small and won't cause too much steric interaction. The polymer chain is extending forward, into the big, open space of the wedge. The leaves the methyl pointing to one side of the wedge.
Which way will the propene approach? Will it come in on the same side as the methyl, or the opposite side? Probably the opposite side, as shown below.
Researchers suspect the methyl group points "down" in the above drawing, rather than "up", because the polymer chain is a bigger steric obstacle than the lower aromatic ring, and the polymer chain is in an upper quadrant. Once again, the approach of the propene is sterochemically controlled, although this time the stereochemistry of an existing chiral centre in the growing polymer chain influenced how things proceeded.
Once again, upon 1,2-insertion, a new chiral centre is formed. The Cs catalyst is forming a syndiotactic polymer chain.
Because of the influence of an existing choral centre on the stereochemical outcome of the reaction, this catalyst is considered to work through "chain-end control". The catalyst site simply amplifies the influence of that chiral centre upon the chiral centre that forms next. It does so by bringing the reactants together into a small space where the steric differences of two subtly different pathways become more important.
Exercise \(10\)
Label the configurations (R or S) of the chiral centres in the above drawing to confirm syndiotacticity.
6.14:
Exercise 6.1.1:
El = electrophile; Nu = nucleophile
Exercise 6.1.2:
Exercise 6.1.3:
$Rate = \frac{d[alkyl \: bromide]}{dt} = k[alkene][HBr] \nonumber$
Exercise 6.1.4:
Exercise 6.2.1:
a, e, f, g are prochiral.
Exercise 6.2.2:
a) re b) si c) re d) si e) either; if the Br adds on one end of the double dond it is re, but at the other it is si f) re
Exercise 6.2.3:
Exercise 6.2.4:
Exercise 6.3.1:
If the acid is regenerated at the end of the reaction, it isn't a reagent. It is a catalyst. It makes addition of water to the double bond occur much more quickly than if water acted alone, since water would never manage to protonate the alkene.
Exercise 6.3.2:
Exercise 6.3.3:
Exercise 6.4.1:
Two products are formed and they are enantiomers.
Exercise 6.4.2:
They are diastereomers. One chiral center has the same configuration in both compounds but the others are opposite.
Exercise 6.4.3:
The second bromine could occupy any of the secondary positions if there were a true carbocation. That doesn't happen; the second bromine occupies only the position next to the other bromine.
Exercise 6.4.5:
The nucleophile in the second step changes under different conditions.
Exercise 6.4.6:
Exercise 6.5.2:
Exercise 6.5.3:
Exercise 6.5.4:
Exercise 6.5.5:
Exercise 6.5.6:
Exercise 6.6.1:
Exercise 6.6.2:
Exercise 6.6.3:
Exercise 6.6.4:
Exercise 6.6.5:
Exercise 6.6.6:
Exercise 6.6.7:
Exercise 6.6.8:
Exercise 6.6.9:
Exercise 6.6.10:
Exercise 6.7.2:
Exercise 6.7.3:
Crowding is more severe in the structure on the left than in the structure on the right. The structure on the right, representing an approach to the transition state of the reaction, is more favorable than the other one.
Exercise 6.7.4:
Exercise 6.7.5:
Exercise 6.7.6:
Exercise 6.7.7:
Exercise 6.8.1:
Exercise 6.8.2:
Exercise 6.8.3:
Exercise 6.8.4:
D-(-)-tartrate is the (2S,3S)-isomer. L-(+)-tartrate is the (2R,3R)-isomer. Each chiral center is configured opposite to the corresponding one in the other molecule, so the molecules are enantiomers.
Exercise 6.8.5:
Exercise 6.8.6:
Exercise 6.8.7:
Exercise 6.8.8:
Exercise 6.9.1:
The chlorines can (weakly) share their electrons to fill the octet on carbon.
Exercise 6.9.2:
The oxygen can π-donate to help fill the octet on the carbon.
Exercise 6.9.3:
Not only can the nitrogens π-donate to help fill the octet on carbon, but this is an aromatic system. It is planar, cyclic, fully conjugated, with an odd number of electron pairs in the π-system.
Exercise 6.9.4:
Exercise 6.10.1:
Exercise 6.10.2:
Exercise 6.10.3:
Exercise 6.10.4:
Exercise 6.10.5: | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/IV%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_2/06%3A_Electrophilic_Addition_to_Alkenes/6.13%3A_.txt |
Aromatics, or arenes, are derivatives of benzene or other compounds with aromatic ring systems. That is, they are cyclic, planar, fully conjugated and have an odd number of π-electron pairs. Like alkenes, aromatics have π-electrons that are loosely held and are easily attracted to electrophiles. However, aromatics don't undergo the typical reactions of alkenes.
For example, bromine will not add across the double bond of benzene.
Instead, a bromine atom replaces one of the hydrogen atoms on the benzene. This reaction is greatly accelerated in the presence of Lewis acids, such as ferric chloride.
A similar reaction happens with chlorine. If treated with chlorine gas and a metal catalyst, a chlorine atom from chlorine gas can replace a hydrogen atom on benzene. However, the same thing doesn't work as smoothly with the other halogens, iodine and fluorine.
The reactions of chlorine and bromine with benzene and other aromatics can be catalysed by a variety of Lewis acidic metal catalysts. So can the reactions of alkyl halides and acyl halides, which we don't normally think of as electrophiles for alkene addition.
There are some limitations on what kind of groups can be added in this way. The carbon attached to the halide should be tetrahedral. Typically, it is much easier to add secondary or tertiary alkyls than primary ones. That is, the carbon attached to the halogen had best be attached to two or three other carbons as well. Methyls are very, very difficult to add in this way.
There is an exception. The carbon attached to the halogen need not be tetrahedral, provided it is a carbonyl carbon. That reaction is called an acylation.
In these cases, it is the alkyl or acyl, rather than the halogen, that replaces a hydrogen atom on the benzene. Remember, benzene is most likely acting as a nucleophile in this reaction, even though it is following a different pathway than an alkene would. It is reacting with the most electrophilic part of the alkyl halide or acyl halide.
Aromatics have a limited repertoire of electrophiles with which they commonly undergo reaction. In addition to these Lewis acid-catalysed reactions, there are also reactions strong acidic media, such as a mixture of nitric and sulfuric acid.
Another acidic medium, referred to as "fuming sulfuric acid", is really a mixture of sulfuric acid and sulfur trioxide.
Just as with the acid-catalysed reactions, the nitro group and the sulfonate group just replace a hydrogen atom on the benzene ring. The overall reaction involves bond formation between a benzene carbon and the electrophile, and bond cleavage between the same carbon and a proton.
Exercise \(1\)
Fill in the missing reagents in the following reactions.
7.02%3
Bromine will not add across the double bond of benzene.
Instead, a bromine atom can replace one of the hydrogen atoms on the benzene. This reaction is especially easy in the presence of a catalyst.
How does that outcome happen? Why does that outcome happen?
There has been a good deal of study of these reactions and there is strong evidence of the steps through which they occur. As expected, the reaction involves donation of π electrons from the benzene. For the moment, we'll assume the electrophile is a bromine cation; we will deal with its exact structure later.
The problem is, that initial step results in the loss of aromaticity. The aromatic system confers a little extra stability on the π system, so the molecule is motivated to restore the aromaticity. The easiest way to do that, and get rid of a positive charge at the same time, would be to deprotonate the cation. Some base will pick up the proton; it is likely a bromide ion in this case. We will see later where that bromide comes from.
• Electrophilic aromatic substitution proceeds through a cationic intermediate.
• The intermediate forms via donation of π electrons from the arene to an electrophile.
• Aromaticity is restored through loss of a proton from this cation.
How do we know that the mechanism unfolds this way? There are three basic steps that are clearly accomplished during the course of the reaction: the C-H bond is broken, the C-Br bond is formed, and the Br-Br bond is broken.
When is the C-H bond broken? That question can be answered by looking for what is called an "isotope effect". The most common isotope of hydrogen is 1H, or protium, but 2H is also available; it is called "deuterium". Deuterium is often represented by the symbol D and protium by the symbol H. Deuterium is twice as heavy as the common protium. That mass difference leads to a lower vibrational frequency of a C-D bond than a C-H bond. The C-H bond vibrates more rapidly and energetically than a C-D bond; as a consequence, the C-H bond is more easily broken than the C-D bond.
If we take a sample of ordinary benzene, C6H6, and a sample of deuterated benzene, C6D6, we can measure how quickly they each undergo a bromination reaction. Very often, a reaction that involves C-H bond cleavage will slow down if a C-D bond is involved. This outcome is observed in E2 eliminations, for instance. This slowing of the reaction with the heavier isotope is called the deuterium isotope effect.
However, no deuterium isotope effect is observed during bromination, or other aromatic electrophilic substitution reactions. That absence of an isotope effect usually means the C-H bond cleavage is a sort of an afterthought. The hard part of the reaction is already done. Both the C-H and C-D bonds are broken so quickly and easily, by comparison, that we don't really notice the difference between them.
There is even more evidence. In a few exceptional cases, the cationic intermediate in this reaction is stable enough to be isolated and crystallized. X-ray diffraction shows that there is a tetrahedral carbon in the ring, indicating that the C-H bond has not broken yet.
The C-H bond is broken at the end of the reaction. When is the Br-Br bond broken?
That question is a little harder to answer. We can't use the same isotope strategy that we used with the C-H bond. Although deuterium is twice as heavy as protium, producing a substantial isotope effect, 81Br is only 2.5% more massive than 79Br. Any difference in rates involving these isotopes is undetectable. The exact nature of the bromine species in the reaction is complicated, and may even be different under different conditions.
Exercise \(1\)
In the case of uncatalyzed bromination reactions, the addition of salts such as NaBr has no effect on the reaction rate, indicating that the arene reacts directly with Br2 rather than Br+. Explain this line of reasoning.
Answer
In the case of uncatalyzed bromination reactions, there is clear evidence that the Br-Br bond-breaking step does not start the reaction off. If that were the first step, there would presumably be an equilibrium between Br2 and Br+/Br- ions. That equilibrium would be shifted back toward Br2 if bromide salts were added. In that case, the amount of bromine cation would be suppressed and the reaction would slow down. No such salt effects are observed, however. That evidence suggests that, in the uncatalyzed reaction, the aromatic reacts directly with Br2.
Exercise \(2\)
Given each of the following electrophiles, provide a mechanism for electrophilic aromatic substitution.
a) NO2+ b) CH3CH2+ c) SO3H+ d) CH3CO+
Answer a
In each case, a base must remove the proton from the cationic intermediate. An anion that would be present in solution has been chosen for this role.
Answer b
In each case, a base must remove the proton from the cationic intermediate. An anion that would be present in solution has been chosen for this role.
Answer c
In each case, a base must remove the proton from the cationic intermediate. An anion that would be present in solution has been chosen for this role.
Answer d
In each case, a base must remove the proton from the cationic intermediate. An anion that would be present in solution has been chosen for this role. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/IV%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_2/07%3A_Electrophilic_Aromatic_Substitution/7.01%3.txt |
The mechanism of electrophilic aromatic substitution follows two elementary steps. First, donation of a pair of π electrons to the electrophile results in a loss of aromaticity and formation of a cation. Second, removal of a proton from that cation restores aromaticity.
How does the electrophile form in the first place? The details of that part of the reaction vary from case to case. With the catalysed bromine reaction, the Lewis acid activates the halogen to render it more electrophilic. The activation may even go so far as to form a bromine cation, as suggested earlier. Otherwise, the positive charge on the bromine atom that ligates the Lewis acid can be nullified, indirectly, when the arene donates to the terminal bromine atom.
The appearance of a bromide ion to deprotonate the cation simply results fom the equilibrium of the Lewis acid-base complex.
Exercise \(1\)
Show the mechanism for chlorination of benzene in the presence of ferric chloride.
The reactions of alkyl and acyl halides also involve Lewis acid catalysts; frequently, aluminum chloride (AlCl3) is employed. These two reactions are called Friedel-Crafts reactions after the French and American co-discoverers of the reaction. Typically, Friedel-Crafts reactions are believed to occur through initial formation of cationic electrophiles, which then react with aromatics in the same way as halogen electrophiles.
The alkyl cation is a potent electrophile. It is able to temporarily disrupt the aromaticity of the aromatic ring, forming an arenium ion.
The arenium ion intermediate is probably deprotonated by halide ion; some amount of these ions would be in equilibrium with the Lewis acid-base adduct.
Because Friedel-Crafts alkylations occur via alkyl cations, the reactions of primary alkyl halides are generally pretty slow. Those cations just aren't stable enough to form. Although it sometimes seems like formation of a stable intermediate would slow things down, because it would not be motivasted to react further, usually the reverse is true. Because an intermediate is at high energy, it is inherently difficult to form in the first place. This difficulty acts as a blockade on the reaction, so that it doesn't proceed very easily.
Forming the intermediate more easily, therefore, allows the reaction to proceed more quickly. For example, the tert-butyl cation is relatively stable, so that intermediate is formed relatively easily. Alkylation of aromatics rings with tertiary alkyl halides is especially easy to accomplish.
In some cases, formation of a cation probably does not happen at all. Instead, the activated Lewis acid-base complex acts as the electrophile directly. This pathway seems to occur with methyl electrophiles. However, there are also indications that primary alkyl halides undergo this mechanism in parallel with the cationic mechanism.
It is worth noting that in some cases, multiple products may result via rearrangements. These observations provide additional evidence for the cationic nature of the intermediates as well as competing pathways.
Exercise \(2\)
Show the mechanism for the Friedel-Crafts alkylation of benzene with 2-chloropropane in the presence of aluminum chloride.
Answer
Exercise \(3\)
Why is the Friedel-Crafts reaction of 1-chloropropane so much slower than the reaction of 2-chloropropane? Explain using a mechanism and intermediates.
Answer
The primary cation formed is very unstable. As a result, there is a high barrier to cation formation.
Exercise \(4\)
Show why Friedel-Crafts alkylation of benzene with 2-chloropentane results in the formation of two different products.
Answer
Exercise \(5\)
Show the mechanism for the Friedel-Crafts acylation of benzene with ethanoyl chloride (acetyl chloride) in the presence of aluminum chloride.
Answer
Exercise \(6\)
Show why the Friedel-Crafts acylation of benzene with pentanoyl chloride results in only one product, with no rearrangement.
Answer
The cation that results is stabilized via π-donation from oxygen.
Nitration and sulfonation reactions differ from the other substitutions that we have seen because they do not utilize Lewis acid catalysis. These reactions depend on equilibria that occur in strongly acidic media. When nitric acid is dissolved in sulfuric acid, there is spectroscopic evidence than NO2+ forms, providing an electrophile. That electrophile adds readily to the aromatic ring.
Similarly, when sulfuric acid is concentrated by boiling off residual water, sulfur trioxide results. Under these acidic conditions, the latter easily forms SO3H+, the electrophile in sulfonation.
Exercise \(7\)
Provide a mechanism for the formation of NO2+ from nitric and sulfuric acid.
Answer
Exercise \(8\)
Provide a mechanism for the formation of SO3H+ from sulfuric acid.
Answer | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/IV%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_2/07%3A_Electrophilic_Aromatic_Substitution/7.03%3.txt |
Because the benzene acts as a nucleophile in electrophilic aromatic substitution, substituents that make the benzene more electron-rich can accelerate the reaction. Substituents that make the benzene moor electron-poor can retard the reaction.
In the mid-twentieth century, physical organic chemists including Christopher Ingold conducted a number of kinetic studies on electrophilic aromatic substitution reactions. In the table below, you can see that some substituents confer a rate of reaction that is much higher than that of benzene (R = H). Phenol, C6H5OH, undergoes nitration a thousand times faster than benzene does. Nitrobenzene, C6H5NO2, undergoes the reaction millions of times more slowly.
Rate of nitration in benzene derivatives
R in C6H5R Relative rate
OH 1,000
CH3 25
H 1
CH2Cl 0.71
I 0.18
F 0.15
Cl 0.033
Br 0.030
CO2Et 0.0037
NO2 6 x 10-8
NMe3+
1.2 x 10-8
These observations are consistent with the role of the aromatic as a nucleophile in this reaction. Substituents that draw electron density away from the aromatic ring slow the reaction down. These groups are called deactivating groups in this reaction. Substituents that readily donate electron desnity to the ring, or that effectively stabilize the cationic intermediate, promote the reaction. These groups are called activating groups in this reaction.
The roles of these groups are related to their electronic interactions with the electrons in the ring. Some groups might be π-donors, providing additional electron density to the benzene ring via conjugation.
Other groups may be π-acceptors, drawing electron density away from the ring via conjugation.
Still others may be σ-acceptors, drawing electron density away from the ring via a simple inductive effect which arises from the electronegativity of a substituent.
In some cases, there may be multiple effects, and the overall influence of the substituents is determined by the balance of the effects. One effect may be stronger in one case than the other, so it wins out in one case and loses in another.
Exercise \(1\)
Explain why a fluorine atom would slow down an electrophilic substitution on an adjacent benzene ring.
Exercise \(2\)
Show, with structures, how the OH group in phenol makes the benzene ring more nucleophilic.
Answer
Exercise \(3\)
Show, with structures, how the CO2Et group makes the benzene ring less nucleophilic.
Answer
Exercise \(4\)
Show, with structures, how a methyl group stabilizes the cationic intermediate during a nitration reaction.
Answer
In general, deactivating groups fall into two classes. Π-acceptors, such as carbonyls, if placed directly adjacent to the aromatic ring, slow down the reaction. Highly electronegative atoms, typically halogens, attached directly to the aromatic ring also slow down the reaction.
• deactivating groups make electrophilic aromatic substitutionslower than in benzene
• π-acceptors are deactivating groups
• halogens are deactivating groups
Activating groups also fall into two categories. Π-donors, typically oxygen or nitrogen atoms, accelerate the reaction. This observation is true even though these atoms are also highly electronegative. Alkyl groups attached to the aromatic ring also accelerate the reaction.
• activating groups make electrophilic aromatic substitution faster than in benzene
• oxygen and nitrogen π-donors are activating groups
• alkyls are activating groups
Note that halogens are also π-donors, but they are less effective in this regard than nitrogen or oxygen. That's because nitrogen and oxygen are similar in size to carbon, and they form effective π-overlap with the adjacent carbon on the benzene ring. In the halogens, electronegativity wins by default, because their π-donating effects are not good enough to make them activators.
Conversely, nitrogen and oxygen are both very electronegative, but their exceptional π-donating ability makes them activators rather than deactivators. Thus, in many cases, there is a subtle balance between activating and directing effects. In some cases, the activating effect is more pronounced, and that is what is observed. In other cases, it is the deactivating effect that wins out.
Alkyl groups behave almost as sigma donors, although that may be a misleading way to think about them.
Instead, their mild activating effect arises from hyperconjugation, in which a pair of C-H bonding electrons can weakly interact with a cationic site, providing a little extra stability to the cation.
Exercise \(5\)
One of the groups in the table, CH2Cl, does not quite fit the general rules. It is very slightly deactivating. Explain why this group acts in this way.
Answer
This is a substituted alkyl group. An alkyl group should be moderately activating, but the presence of a halogen exerts an inductive electron-withdrawing effect. The cation-stabilizing effect of the alkyl substituent is completely counteracted by the halogen.
Exercise \(6\)
Predict whether each of the following groups would be activating or deactivating towards electrophilic aromatic substitution.
a) NH2 b) CN c) OCH3 d) SMe2+ e) C(O)CH3
Answer a
a) activating
Answer b
b) deactivating
Answer c
c) activating
Answer d
d) deactivating
Answer e
e) deactivating
Exercise \(7\)
Explain the trend in activating effects among the different halogens. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/IV%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_2/07%3A_Electrophilic_Aromatic_Substitution/7.04%3.txt |
In addition to exerting an effect on the speed of reaction, substituents on the benzene ring also influence the regiochemistry of the reaction. That is, they control where the new substituent appears in the product.
Remember, there are three different position on the bezene ring where a new substituent can attach, relative to the original substituent. Substitution could actually occur on five positions around the ring, but two pairs are related by symmetry. Isomerism in disubstituted benzenes can be described by numbering the substituents (1,2- etc) or by the relationships ortho-, meta- and para-. There are two positions ortho- to the initial substituent and two positions meta- to it.
Ingold and colleagues investigated the question of regiochemistry in nitration. They reported the following observations:
Substitution patterns during nitration of benzene derivatives
R in C6H5R % o- product % m- product % p- product
CH3 56 3 41
F 10 0 90
Cl 30 0 70
Br 38 0 62
OH 10 0 90
CHO 19 72 9
CO2Et 28 68 3
CN 17 81 2
NO2 6 94 0
In looking at the table, you might see that there are two groups of substituents. One group reacts to make mixtures of ortho- and para- products. There may be different ratios of ortho- to para- and there may be small amounts of meta-, but don't get bogged down in the details right now. Focus on the bigger picture. Some groups are "ortho-/para-directors".
The other group reacts to makemostly meta-substituted products. There may be small amounts ofortho- and para- products, but don't worry about that. Focus on the bigger picture. Some groups are "meta-directors".
These regiochemical effects are very closely related to the activating and directing effects we have already seen. If we want to understand this data, we need to think about things like π-donation, π-acceptance, inductive effects and cation stability.
Exercise \(1\)
Show resonance structures for the cationic intermediate that results during nitration of toluene (methylbenzene). Explain why a mixture of ortho- and para- substitution results.
Answer
The tertiary cations that result during ortho- and meta- substitution offer extra stability, leading to preferential formation of these cations.
Exercise \(2\)
Show resonance structures for the cationic intermediate that results during nitration of chlorobenzene. Explain why a mixture of ortho- and para- substitution results.
Answer
The π-donation that occurs in the cations arising from ortho- and meta- substitution results in extra stability, leading to preferential formation of these cations.
Exercise \(3\)
Show resonance structures for the cationic intermediate that results during nitration of acetophenone (C6H5COCH3). Explain why mostly meta- substitution results.
Answer
The cation directly adjacent to the carbonyl is destabilized by the electron withdrawing effect of the ketone. By default, the other intermediate is preferentially formed.
Exercise \(4\)
Show resonance structures for the cationic intermediate that results during nitration of acetanilide (C6H5NH(CO)CH3). Explain why a mixture of ortho- and para- substitution results.
Answer
The π-donation that occurs in the cations arising from ortho- and meta- substitution results in extra stability, leading to preferential formation of these cations.
In general, we can divide these substituents into three groups:
• π-acceptors are meta- directors.
• π-donors are ortho-/para- directors.
• alkyls are ortho-/para- directors.
Note that, once again, we may have two competing effects in one substituent, such as a halogen. In halogens, although the net effect may be to slow the reaction down, that weak π-donation is still enough to tilt the balance of products in favor of ortho- and para- substitution.
For example, fluorine has a lone pair. It can be a π-donor.
We could illustrate the effects with the following cartoon. It shows negative charge buildup, illustrated in red, on three of the carbons on the benzene ring.
Fluorine is also very electronegative. It can be an electron withdrawing group. Because it withdraws electrons through its sigma bond rather than through resonance effects, we think of it as "inductively" electron withdrawing. Nonetheless, a considerable amount of electron density from the benzene is attracted to the fluorine.
Again, we could illustrate the effects with a cartoon. It shows positive charge buildup, illustrated in blue, on the carbons that are closest to the fluorine. Remember, electrostatic interactions decrease rapidly with distance, so the farther the carbons are from the fluorine, the lower the effect that they will experience.
The sum of these effects can be illustrated in a composite cartoon that shows different amounts of positive or negative charge built up on different carbons.
Thus, although fluorine is generally an electronegative and deactivating group, it may still manage to place some extraelectron density on particular carbons.
Whether something is overall activating or deactivating depends on a similar balance of factorsIn general, we can think about competing effects in terms of a see saw.
For an electronegative π-donor like fluorine or another halogen, the stronger those σ-withdrawing effects, the slower the reaction. It pulls more electron density away from the benzene than it puts back. As a result, the benzene is less nucleophilic than it would be if the halogen were not there.
On the other hand, the better the π-donation, the faster the reaction. Good π-donors include oxygen and nitrogen. They are just the right size to lend a lone pair to a neighbouring carbon in the benzene ring. Halogens, on the other hand, are relatively clumsy donors. Either they are just a little too electronegative to be useful, like fluorine, or else they are a little too big to share wtih carbon. That's the case with chlorine, bromine and iodine.
Remember, the same argumants are true in carbonyl chemistry. The general reactivity of carboxyloid derivatives depends on a similar balance between donating and withdrawing effects. As a result, esters and amides are relatively stable and unreactive, whereas chlorides are very reactive.
Of course, a π-acceptor would work in the opposite way. That would be true if the atom attached to the benzene is multiply bonded to another atom, especially if that other atom is something electronegative (an oxygen or a nitrogen).
Retrosynthetic Planning
Sometimes in synthetic work it is useful to look at a compound and imagine what it might be made from. This practice is used in industrial chemistry, when a researcher might be trying to decide on the most economical routed to make a particular compound. There may be readily available materials that a useful pharmaceuticla can be made from. By working backwards one step at a time, the researcher may more easily see different possibilities for starting materials. In a similar way, a biological chemist might be able to identify what biological precursors could lead to the formation of a particular compound in biology.
For example, by looking at the structure of m-bromonitrobenzene, you could imagine that the compound could be made by the bromination of nitrobenzene, but not by the nitration of bromobenzene. That's because nitro groups are meta-directors, but bromo-groups are ortho-, para-directors.
This method of working backward from the target compound is called "retrosynthetic analysis". By making a retrosynthetic plan, we can more efficiently arrive at the possible ways to make a specific compound.
Exercise \(5\)
Fill in the major organic products of the following reactions.
Exercise \(6\)
Fill in the starting materials and reagents needed to obtain the major product shown via electrophilic aromatic substitution.
Exercise \(7\)
Given two different substituents on a benzene, there can sometimes be a conflict in predicting which substitution pattern will result. Generally, the group with the stronger activating effect wins out. Predict the major products of the following reactions.
Problem AR5.8
Exercise \(9\)
Provide a retrosynthetic plan for each of the following compounds, going back to benzene.
Exercise \(10\)
Turn the retrosynthetic plans from the previous problem into syntheses.
Answer
Problem AR5.12.
7.06%3
Exercise 7.1.1:
Exercise 7.2.1:
In the case of uncatalyzed bromination reactions, there is clear evidence that the Br-Br bond-breaking step does not start the reaction off. If that were the first step, there would presumably be an equilibrium between Br2 and Br+/Br- ions. That equilibrium would be shifted back toward Br2 if bromide salts were added. In that case, the amount of bromine cation would be suppressed and the reaction would slow down. No such salt effects are observed, however. That evidence suggests that, in the uncatalyzed reaction, the aromatic reacts directly with Br2.
Exercise 7.2.2:
In each case, a base must remove the proton from the cationic intermediate. An anion that would be present in solution has been chosen for this role.
a)
Exercise 7.3.2:
Exercise 7.3.3:
The primary cation formed is very unstable. As a result, there is a high barrier to cation formation.
Exercise 7.3.4:
Exercise 7.3.5:
Exercise 7.3.6:
The cation that results is stabilized via π-donation from oxygen.
Exercise 7.3.7:
Exercise 7.3.8:
Exercise 7.4.2:
Exercise 7.4.3:
Exercise 7.4.4:
Exercise 7.4.5:
This is a substituted alkyl group. An alkyl group should be moderately activating, but the presence of a halogen exerts an inductive electron-withdrawing effect. The cation-stabilizing effect of the alkyl substituent is completely counteracted by the halogen.
Exercise 7.4.6:
a) activating b) deactivating c) activating d) deactivating e) deactivating
Exercise 7.5.1:
The tertiary cations that result during ortho- and meta- substitution offer extra stability, leading to preferential formation of these cations.
Exercise 7.5.2:
The π-donation that occurs in the cations arising from ortho- and meta- substitution results in extra stability, leading to preferential formation of these cations.
Exercise 7.5.3:
The cation directly adjacent to the carbonyl is destabilized by the electron withdrawing effect of the ketone. By default, the other intermediate is preferentially formed.
Exercise 7.5.4:
The π-donation that occurs in the cations arising from ortho- and meta- substitution results in extra stability, leading to preferential formation of these cations.
Exercise 7.5.5:
Exercise 7.5.6:
Exercise 7.5.7:
In cases leading to mixtures of ortho and para products, only one product was chosen, based on minimal steric interactions.
Exercise 7.5.8:
Exercise 7.5.9:
Exercise 7.5.10:
Exercise 7.5.11:
Exercise 7.5.12: | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/IV%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_2/07%3A_Electrophilic_Aromatic_Substitution/7.05%3.txt |
Topics required for successful completion are listed under each link.
Abyssomicin, Sorensen, 2006
alkene addition
carbonyl addition
carboxylic substitution
nucleophilic substitution
oxidation of alcohols
pericyclics
Amphilectolide, Trauner, 2014
carbonyl addition
carbonyl reduction
"Claisen condensation"-type carboxylic substitution
cuprate addition
ester reduction
nitrile hydrolysis
Annamoxic Acid, Corey, 2004
alkene addition
carbonyl addition
elimination
pericyclics
Anominine, Nicolaou, 2012
aldol reaction
aliphatic nucleophilic substitution
carbonyl addition
carboxyl substitution
elimination
oxidation of alcohols
radicals
Aplykurodinone, Danishefsky, 2010
aliphatic nucleophilic substitution
carbonyl addition
carboxylic substitution
elimination
oxidation of alcohols
Atisine, Fukumoto, 1990
esterification & amidification
Michael addition
reduction of esters
Wittig & Horner-Wadsworth-Emmons reactions
Aurantioclavine, Ellman, 2010
aliphatic nucleophilic substitution
carbonyl addition
carboxylic substitution
Aza-epothilone B, Danishefsky, 2000
with bioassay study
aliphatic nucleophilic substitution
carbonyl addition
carboxylic substitution
Brevenal, Crimmins, 2010
(partial synthesis)
carbonyl addition, anionic & semianionic nucleophiles
carbonyl addition, aldols
carbonyl addition, ylides
Brevetoxin, Crimmins, 2009
aliphatic nucleophilic substitution
electrophilic addition to alkenes
carbonyl addition
Briarellin, Crimmins, 2011
nucleophilic addition to epoxide
aliphatic nucleophilic substitution
silyl protecting groups
carboxylic substitution
carbonyl addition
electrophilic addition to alkenes
ylide addition
oxidation of alcohols
Bryostatin, Burke, 2004
dihydroxylation
Horner-Wadsworth-Emmons or Wittig reaction
hydroboration - oxidation
oxidation of alcohols
reduction of esters and ketones
ring-closing olefin metathesis
Callipeltoside, MacMillan, 2008
aliphatic nucleophilic substitution
electrophilic addition to alkenes
carbonyl addition
carboxylic substitution
oxidation of alcohols
Cassiol, Stoltz, 2008
aliphatic nucleophilic substitution
carbonyl addition
carboxylic substitution
Cavicularin, Beaudry, 2013
Diels Alder
etherification
palladium coupling
Wittig or Tebbe reactions
Clavolonine, Fujioka, 2011
aliphatic nucleophilic substitution and elimination
Deoxytetracycline, Stork, 1996
carbonyl addition (anionic and neutral nucleophiles)
carboxylic substitution
conjugate addition
Dedihydrosemofalin, Overman, 2003
alpha-alkylation
aza-Cope rearrangement
catalytic hydrogenation
Diels Alder
iminium ion formation
oxidation of alcohols
ozonolysis
reduction of esters
silylation of alcohols
silylation of enolates
silyl ether cleavage
Wittig & Horner-Wadsorth-Emmons reactions
Dendrobine, Carreira, 2012
carbonyl addition
carboxylic substitution
Doliculide, Ghosh, 2001
aliphatic nucleophilic substitution
electrophilic addition to alkenes
Dynemicin, Danishefsky, 1996
aliphatic nucleophilic substitution
alkene oxidation
carbonyl addition
carboxylic substitution
elimination
oxidation of alcohols
pericyclics
Echinopine, Nicolaou, 2010
aliphatic nucleophilic substitution
carbonyl addition
carboxylic substitution
Epothilone, Altmann, 2008
aldol reaction
alpha deprotonation
Grignard reactions
Wittig and Horner-Wadsworth-Emmons reactions
Fischerindole, Baran, 2008
aliphatic nucleophilic substitution
carbonyl addition,
carboxylic substitution,
electrophilic addition to alkenes
oxidation of alcohols
radicals
Ginkgolide, Corey, 1988
aliphatic nucleophilic substitution
carbonyl addition
carboxylic substitution
conjugate addition
electrophilic addition to alkenes
enolates
rearrangements
radicals
Glucosylceramide, Overkleeft, 2007
contains a bioassay exercise
aliphatic nucleophilic substitution
carbonyl addition
carboxylic substitution
elimination
oxidation of alcohols
Juvenile Hormone 1, Corey, 1968
aliphatic nucleophilic substitution
carbonyl addition
carboxylic substitution
cuprate addition
electrophilic addition to alkenes
oxidation of alcohols
Wittig reaction
Lepistine, Yokoshima & Fukuyama, 2014
aliphatic nucleophilic substitution
elimination
Longifolene, Corey, 1961
aliphatic nucleophilic substitution
alkene oxidation
carbonyl addition
carboxylic substitution
elimination
oxidation of alcohols
Lycopladine, Weinreb, 2015
acetal/ketal cleavage
acetal/ketal formation
catalytic hydroformylation of alkenes
Grignard reactions
reductive amination
Manzacidin A, Kawabata, 2013
carboxylic substitution: esterification and amidification
conjugate addition
Nahuoic Acid, Smith, 2017
acetal / ketal formation
alkene reduction
carbonyl reduction
cuprate addition
Diels Alder reaction
epoxide opening (organometallic)
ester hydrolysis
keto-enol tautomerism
methoxymethyl ether formation
olefin metathesis
silyl ether formation & cleavage
syn dihydroxylation
thioacetal / thioketal cleavage
Nakadomarin, Kerr, 2007
aliphatic nucleophilic substitution
carbonyl addition
carboxylic substitution
conjugate addition
elimination
oxidation of alcohols
Norzoanthamine, Theodorakis, 2011
Robinson annulation
conjugate addition
anionic addition to carbonyls
carboxylic substitution
Octalactin, Buszek, 1994
alkene reduction
epoxidation
hydroboration / oxidation
Lindlar's catalyst
silyl ether cleavage
aldol condensation
aliphatic nucleophilic substitution
carbonyl addition
carboxylic substitution
elimination
oxidation of alcohols
pericyclics
Perhydrohistrionicotoxin, Corey, 1975
aliphatic nucleophilic substitution
carbonyl addition
carboxylic substitution
conjugate addition
elimination
enolates
oxidation of alcohols
radicals
Periplanone, Still, 1979
aldol condensation
aliphatic nucleophilic substitution
carbonyl addition
carboxylic substitution,
elimination
oxidation of alcohols
pericyclics
Platensimicin, Ghosh, 2009
aliphatic nucleophilic substitution
electrophilic addition to alkenes
Polyanthellin, Johnson, 2009
electrophilic addition to alkenes: halogenation, hydroboration, oxymercuration, epoxidation, cyclopropanation
aliphatic nucleophilic substitution
Polycavernoside, Sasaki, 2017
epoxidation
epoxide ring-opening
esterification
oxidation of alcohols
silyl ether cleavage
Prostaglandin A2, Corey, 1972
aliphatic nucleophilic substitution
carbonyl addition
carboxylic substitution
elimination
oxidation of alcohols
Quinine, Stork, 2001
aliphatic nucleophilic substitution
carbonyl addition
carboxylic substitution
Quinocarcin, Stoltz, 2008
carbonyl addition
carboxylic substitution
Salvileucalin, Reisman, 2011
carbonyl addition
carboxylic substitution
Saxitoxin, Du Bois, 2006
Contains a bioassay exercise.
aliphatic nucleophilic substitution
carbonyl addition
carboxylic substitution
electrophilic addition to alkenes
oxidation of alcohols
Scholarisine, Smith, 2012
aliphatic nucleophilic substitution
carbonyl addition
carboxylic substitution,
Serotobenine, Fukuyama & Kan, 2008
aliphatic nucleophilic substitution
alkene oxidation
carbonyl addition
carboxylic substitution
elimination
oxidation of alcohols
pericyclics
Solamine, Stark, 2006
aliphatic nucleophilic substitution
alkene oxidation
carbonyl addition
carboxylic substitution
elimination
oxidation of alcohols
Spongidepsin, Cossy, 2006
activation of carboxylic acids
esterification & amidification
Grignard reactions
hydrolysis of amides
reduction of esters
Wittig & Horner-Wadsworth-Emmons reactions
Strychnine, Overman, 1993
aliphatic nucleophilic substitution
alkene oxidation
carbonyl addition
carboxylic substitution
conjugate addition
elimination
enolates
Taiwaniaquinol, Fillion, 2005
aldol condensation
alpha alkylation
conjugate addition
electrophilic aromatic substitution
etherification
palladium coupling
Taxol, Nicolaou, 1994
aliphatic nucleophilic substitution
alkene oxidation
carbonyl addition
carboxylic substitution
elimination
oxidation of alcohols
pericyclics
radicals
Tetrodotoxin, Kishi, 1972
aliphatic nucleophilic substitution
alkene oxidation
carbonyl addition
carboxylic substitution
electrophilic addition to alkenes
oxidation of alcohols
pericyclics
rearrangements
Tubingensin, Li, 2012
carbonyl addition
carboxylic substitution, aldol
Tulearin, Cossy, 2009
carbonyl addition, anionic & semianionic nucleophiles
carbonyl addition, aldols
carbonyl addition, ylides
Vermiculine, Corey, 1975
carbonyl addition
carboxylic substitution
Vibsanin, Takeo, 2015
addition to carbonyl (organometallic)
allylic substitution
ester hydrolysis
ester reduction
esterification
ether formation
Mitsunobu reaction
silyl ether formation and cleavage
Wittig / Horner-Wadsworth-Emmons
Vinigrol, Berriault, 2005
acetals & ketals
Claisen rearrangement
Diels Alder
syn-dihydroxylations
esterification
Grignard reagents
hydride additions
oxidation of alcohols
silyl ethers
Vittatalactone, Breit, 2010
Aliphatic nucleophilic substitution
Zaragozic acid, Nicolaou, 1994
aliphatic nucleophilic substitution
alkene oxidation
carbonyl addition
carboxylic substitution
organo-transition metal reactions
oxidation of alcohols
radicals
sulfur ylides
Zincophorin, Hsung, 2007
aliphatic nucleophilic substitution
electrophilic addition to alkenes | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/IV%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_2/08%3A_Organic_Synthesis/8.01%3A_Road_Maps_in_Tot.txt |
Electron transfer is one of the most basic processes that can happen in chemistry. It simply involves the movement of an electron from one atom to another. Many important biological processes rely on electron transfer, as do key industrial transformations used to make valuable products. In biology, for example, electron transfer plays a central role in respiration and the harvesting of energy from glucose, as well as the storage of energy during photosynthesis. In society, electron transfer has been used to obtain metals from ores since the dawn of civilization.
Oxidation state is a useful tool for keeping track of electron transfers. It is most commonly used in dealing with metals and especially with transition metals. Unlike metals from the first two columns of the periodic table, such as sodium or magnesium, transition metals can often transfer different numbers of electrons, leading to different metal ions. Sodium is generally found as Na+ and magnesium is almost always Mg2+, but manganese could be Mn2+, Mn3+, and so on, as far as Mn7+.
At first glance, "oxidation state" is often synonymous with "charge on the metal". However, there is a subtle difference between the two terms. For example, in a coordination complex, a metal atom that is ostensibly an ion with a charge of +2 may have very little charge on it at all. Instead, the positive charge may be delocalized onto the ligands that are donating their own electrons to the metal. Oxidation state is used instead to describe what the charge on the metal ion would be if the coordinated ligands were removed and the metal ion left by itself.
Oxidation state is commonly denoted by Roman numerals after the symbol for the metal atom. This designation can be shown either as a superscript, as in MnII, or in parentheses, as in Mn(II); both of these descriptions refer to a Mn2+ ion, or what might have been a Mn2+ ion before it got into a bonding situation.
Exercise \(1\)
Translate the following oxidation state descriptions into charges on the metal.
a) AgI b) Ni(II) c) MnVII d) Cr(VI) e) Cu(III) f) FeIV g) OsVIII h) Re(V)
Answer a
a) Ag+
Answer b
b) Ni2+
Answer c
c) Mn7+
Answer d
d) Cr6+
Answer e
e) Cu3+
Answer f
f) Fe3+
Answer g
g) Os8+
Answer h
h) Re5+
Exercise \(2\)
1. Provide the valence shell electron configuration for each metal species in the previous question (e.g. oxygen's is 2s22px22py12pz1).
2. Draw an energy level diagram showing the occupation of valence s, p and d levels for each metal species in the previous question.
The oxidation state of a metal within a compound can be determined only after the other components of the compound have been conceptually removed. For example, metals are frequently found in nature as oxides. An oxide anion is O2-, so every oxygen in a compound will correspond to an additional 2- charge. In order to balance charge, the metal must have a corresponding plus charge.
For example, sodium oxide has the formula Na2O. If the oxygen ion is considered to have a 2- charge, and there is no charge overall, there must be a corresponding charge of +2. That means each sodium ion has a charge of +1.
Exercise \(3\)
Determine the charge on the metal in each of the following commercially valuable ores. Note that sulfur, in the same column of the periodic table as oxygen, also has a 2- charge as an anion.
a) galena, PbS b) cassiterite, SnO2 c) cinnabar, HgS d) pyrite, FeS2 e) haematite, Fe2O3 f) magnetite, Fe3O4
Answer a
a) Pb2+ & S2-
Answer b
b) Sn4+ & 2 O2-
Answer c
c) Hg2+ & S2-
Answer d
d) Fe4+ & 2 S2-
Answer e
e) 2 Fe3+ & 3 O2-
Answer f
f) 2 Fe3+ & 1 Fe2+ & 4 O2-
Exercise \(4\)
Sphalerite is a common zinc ore, ZnS. However, sphalerite always has small, variable fractions of iron in place of some of the zinc in its structure. What is the likely oxidation state of the iron?
Answer
Probably Fe2+, to replace Zn2+ ions.
Exercise \(5\)
Sometimes non-metals such as carbon are thought of in different oxidations states, too. For example, the coke used in smelting metal ores is roughly C, in oxidation state 0. Determine the oxidation state of carbon in each case, assuming oxygen is always 2- and hydrogen is always 1+.
a) carbon monoxide, CO b) carbon dioxide, CO2 c) methane, CH4 d) formaldehyde, H2CO e) oxalate, C2O42-
Answer a
a) C2+
Answer b
b) C4+
Answer c
c) C4-
Answer d
d) C0
Answer e
e) C3+
Exercise \(6\)
Sometimes it is useful to know the charges and structures of some of the earth's most common anions. Draw Lewis structures for the following anions:
a) hydroxide, HO- b) carbonate, CO32- c) sulfate, SO42- d) nitrate, NO3-
e) phosphate, PO43- f) silicate, SiO44- g) inosilicate, SiO32-
Exercise \(7\)
Use your knowledge of common anions to determine the oxidation states on the metals in the following ores.
a) dolomite, MgCO3 b) malachite, Cu2CO3(OH)2 c) manganite, MnO(OH)
d) gypsum, CaSO4 e) rhodochrosite, MnCO3 f) rhodonite, MnSiO3
Answer a
a) Mg2+
Answer b
b) Cu2+
Answer c
c) Mn3+
Answer d
d) Ca2+
Answer e
e) Mn2+
Answer f
f) Mn2+
Exercise \(8\)
In mixed-metal species, the presence of two different metals may make it difficult to assign oxidation states to each. For the following ores, propose one solution for the oxidation states of the metals.
a) chalcopyrite, CuFeS2 b) franklinite, ZnFe2O4 c) beryl, Be3Al2(SiO3)6 d) bornite or peacock ore, Cu5FeS4
e) turquoise, CuAl6(PO4)4(OH)8
Answer a
a) Cu(II), Fe(II)
Answer b
b) Zn(II), Fe(III)
Answer c
c) Be(II), Al(III)
Answer d
d) Cu(I), Fe(III)
Answer e
e) Cu(II), Al(III)
Exercise \(9\)
Feldspars are believed to make up about 60% of the earth's crust. The alkali, alkaline earth and aluminum metals in these tectosilicates are typically found in their highest possible oxidation states. What are the charges on the silicates in the following examples?
a) orthoclase, KAlSi3O8 b) anorthite, CaAl2Si2O8 c) celsian, BaAl2Si2O8 d) albite, NaAlSi3O8
Exercise \(10\)
Frequently, minerals are solid solutions in which repeating units of different compositions are mixed together homogeneously. For example, labradorite is a variation of anorthite in which about 50% of the aluminum ions are replaced by silicon ions and about 50% of the calcium ions are replaced by sodium ions. Show that this composition would still be charge neutral overall. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/V%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_3/01%3A_Reduction_and_Oxidation_Reactions/1.01%3A_O.txt |
"Oxidation state" implies a description that can change: a metal can go from one oxidation state to another. For example, a Cu(I) can become a Cu(0). It does so by an electron transfer from one place to another. In the case of Cu(I)/Cu(0), an electron would have to be donated by some other species.
A loss of electrons is called an "oxidation", whereas a gain of electrons is called a "reduction" (as immortalized in the mnemonic: LEO the lion goes GER). Since an electron always goes from somewhere to somewhere else, one thing is always oxidized when something else is reduced. (Note that this is a little like proton transfer reactions: a proton is always transferred from one basic site to another, and is never really by itself.) These paired processes are called "reduction-oxidation" reactions, or "redox" for short.
So the reduction of Cu(I) to Cu(0) is just a "half-reaction"; it needs a corresponding oxidation to make it happen. Li could donate an electron, for example, to become Li+. In a biological setting, a Fe(III) in an important protein may need an extra electron to become Fe(II) in order to do its job (your life may be at stake here). It could get the electron from a nearby Cu(I), which becomes Cu(II).
Exercise $1$
Balance the following half reactions by adding the right number of electrons to one side or the other.
a) Cu → Cu(I) b) Fe(III) → Fe
c) Mn → Mn3+ d) Zn2+ → Zn
e) F- → F2 f) H2 → H+
Answer a
a) Cu → Cu+ + e-
Answer b
b) Fe3+ + 3 e-→ Fe
Answer c
c) Mn → Mn3+ + 3 e-
Answer d
d) Zn2+ + 2 e- → Zn
Answer e
e) 2 F- → F2 + 2 e-
Answer f
f) H2 → 2 H+ + 2 e-
Sometimes, redox reactions work out very neatly: one participant needs an electron or two, and the other participant has one or two electrons to give. For example, a Cu2+ ion in need of two electrons to become a Cu atom might get them from a Zn atom, which would become a Zn2+ ion.
In other words (or other symbols):
$\ce{Cu^{2+} + 2e^{-} -> Cu} \nonumber$
$\ce{Zn^{2+} -> Zn + 2 e^{-}} \nonumber$
Adding those together:
$\ce{Cu^{2+} + Zn -> Cu + Zn^{2+}} \nonumber$
Note that the electrons on each side just cancel each other, much like adding the same thing to both sides of an equals sign.
So, overall, a copper ion could get a couple of electrons from a zinc atom, leaving a copper atom and a zinc ion. Wouldn't the opposite reaction also be possible?
$\ce{Cu + Zn^{2+} -> Cu^{2+} + Zn} \nonumber$
On paper, yes. In reality, it doesn't work as well in this direction. In order to help keep track of what redox reactions are actually able to occur, chemists have compiled something called the activity series of metals. The activity series just lists metals in the order in which they are most likely to give up an electron. Metals appearing at the top of the series give up an electron most easily. Metals at the bottom don't give up electrons so readily.
Metal Ion Formed Reactivity with Acids
Cs Cs+ reacts with water
Rb Rb+
K K+
Li Li+
Ba Ba+
Sr Sr+
Ca Ca2+
Mg Mg2+ reacts with HCl
Al Al3+
Mn Mn2+
Zn Zn2+
Cr Cr2+
Fe Fe2+
Cd Cd2+
Co Co2+
Ni Ni2+
Sn Sn2+
Pb Pb2+
H2 H+
Sb Sb2+ reacts with oxidizing acids
Bi Bi3+ (HNO3, etc)
Cu Cu3+
Hg Hg2+
Ag Ag+
Au Au3+
Pt Pt2+
The activity series was put together using a variety of information about redox reactions. Sometimes, maybe one metal was simply placed with another metal ion to see whether the reaction occurred. Often other reactions were studied in order to infer how easily a particular metal would give up its electrons. Some of the trends in the activity series have simple explanations and others do not. We will see later that there are many different factors that govern redox reactions.
We have seen a few reactions in which one metal atom simply transfers an electron or two to another metal ion. Other times, things may be slightly more complicated. There may be an issue of conserving matter, for instance. For example, hydrogen gas, H2, can be oxidised to give proton, H+. We can't have more hydrogen atoms before the reaction than afterwards; matter can't just be created or destroyed. To solve that problem, the oxidation of hydrogen gas, H2, produces two protons, not just one, and so two electrons are involved as well.
Alternatively, two half-reactions may actually involve different numbers of electrons, and so proportions of each half reaction need to be adjusted in order to match the number of electrons properly.
Exercise $2$
Put the following pairs of half reactions together to make a full reaction in each case. Make sure you have balanced the half reactions first.
a) Cu(I) → Cu(II) and Fe(III) → Fe(II) b) Cu(I) → Cu(0) and Ag(0) → Ag(I)
c) F2 → F- and Fe → Fe3+ d) Mo3+ → Mo and Mn → Mn2+
Answer a
a) Cu(I) + Fe(III) → Cu(II) + Fe(II)
Answer b
b) Cu(I) + Ag(0) → Cu(0) + Ag(I)
Answer c
c) 3 F2 + 2 Fe → 6 F- + 2 Fe(III)
Answer d
d) 2 Mo3+ + 3 Mn → 2 Mo + 3 Mn2+
In many cases, redox reactions do not just involve simple metal ions or atoms. Often, the metal atom is found within a compound or a complex ion. For example, one of the most common oxidizing agents in common use is permanganate ion, MnO4-, which is usually converted to manganese dioxide, MnO2 during a reaction.
That means the half reaction here is:
$\ce{MnO4^{-} -> MnO2} \nonumber$
In order to sort out how many electrons are being traded, we need to know the oxidation state of manganese before and after the reaction. That turns out to be Mn(VII) before and Mn(IV) afterwards. That means 3 electrons are added to permanganate to produce manganese dioxide.
Now we have:
$\ce{MnO4^{-} + 3e^{-} -> MnO2} \nonumber$
But now we just have a mass balance problem again. There are oxygen atoms before the reaction that have just disappeared after the reaction. Where could those oxygen atoms have gotten to? On this planet, the simplest answer to that question is always water. So maybe 2 waters were produced as part of the reaction.
That gives us:
$\ce{MnO4^{-} + 3e^{-} -> MnO2 + 2H2O} \nonumber$
Only now we have more problems. First of all, now we have some hydrogen atoms on the right that we didn't have on the left. Where did these things come from? Also, there is this niggling problem of negative charges that we had before the reaction that we don't have after the reaction. Charge, like matter, doesn't just appear or disappear. It has to go someplace, and we have to explain where.
We'll kill two birds with one stone. Maybe some protons were added to the reaction at the beginning, giving us those hydrogen atoms for the water and balancing out the charge.
We are left with:
$\ce{MnO4^{-} + 3e^{-} + 4H^{+} -> MnO2 + 2H2O} \nonumber$
It all works out. There are four negative charges on the left, and four plus charges, so no charge overall. There are no charges on the right. There is one manganese on the left and on the right. There are four oxygens on the left and on the right. There are four hydrogens on the left and on the right.
An Alternative Situation is Possible
Now let's take a little detour. We're going to go back in time, to the point where we realized we had a problem with our oxygen atoms.
Now we have:
$\ce{MnO4^{-} + 3e^{-} -> MnO2} \nonumber$
But now we just have a mass balance problem again. There are oxygen atoms before the reaction that have just disappeared after the reaction. Where could those oxygen atoms have gotten to? And while we're at it, there are four negative charges on the left and none on the right. Charge doesn't just appear or disappear during a reaction. It has to be balanced.
One solution for this problem involves the production of hydroxide ions, HO-, during the reaction. Hydroxide ions are pretty common; there are a few in every glass of water. The charge in the reaction would be balanced if four hydroxide ions were produced, and it would explain where those oxygen atoms went.
That gives us:
$\ce{MnO4^{-} + 3e^{-} -> MnO2 + 4OH^{-}} \nonumber$
Now the charge is balanced! But the oxygen atoms aren't. And where did those hydrogen atoms come from?
Well, on this planet, a good source of hydrogen and oxygen atoms is water. Maybe the reaction needs water.
That means:
$\ce{MnO4^{-} + 3e^{-} + 2H2O -> MnO2 + 4OH^{-}} \nonumber$
One manganese on each side. Four hydrogens on each side. Six oxygens on each side. Four negative charges on each side. Nothing has appeared or disappeared during the reaction. We know where everything went.
There is actually a shortcut to get to this solution. If we already know how to balance the reaction in acididc media, we just add enough hydroxides to neutralize the acid (H+ + -OH = H2O). But we have to add those hydroxides to both sides. Some of the waters will then cancel out to leave the balanced reaction.
Start with acid
$\ce{MnO4^{-} + 3e^{-} + 4H^{+} -> MnO2 + 2H2O} \nonumber$
Add base to both sides
$\ce{MnO4^{-} + 3e^{-} + 4H^{+} + 4 OH^{-} -> MnO2 + 2H2O + 4 OH^{-}} \nonumber$
Neutralize
$\ce{MnO4^{-} + 3e^{-} + 4H2O -> MnO2 + 2H2O + 4OH^{-}} \nonumber$
Cancel the extra waters
$\ce{MnO4 + 3 e^{-} + 2H2O -> MnO2 + 4OH^{-}} \nonumber$
Now we're back. We have seen two different outcomes to this problem. The moral of the story is that sometimes there is more than one right answer. In the case of redox reactions, sometimes things happen a little differently depending on whether things are occurring under acidic conditions (meaning, in this context, that there are lots of protons around) or in basic conditions (meaning there aren't many protons around but there is hydroxide ion).
Apart from helping us to keep track of things, the presence of acids (protons) and bases (hydroxide ions) in redox reactions are common in reality. For example, batteries rely on redox reactions to produce electricity; the electricity is just a current of electrons trying to get from one site to another to carry out a redox reaction. Many batteries, such as car batteries, contain acid. Other batteries, like "alkaline" batteries, for example, contain hydroxide ion.
Exercise $3$
Balance the following half reactions. Assume the reactions are in acidic conditions.
a)MnO2 → Mn(OH)2 b) NO → N2O
c)HPO32- → H2PO2- d) Sn(OH)62- → HSnO2-
Answer a
a) MnO2 + 2 H+ + 2 e- → Mn(OH)2
Answer b
b) 2 NO + 2 e- + 2 H+ → N2O + H2O
Answer c
c) HPO32- + 2 e- + 3 H+ → H2PO2- + H2O
Answer d
d) Sn(OH)62- + 2 e- + 3 H+ → HSnO2- + 4 H2O
Exercise $4$
State whether or not a reaction would occur in the following situations.
1. Copper wire is covered in nitric acid.
2. Copper wire is covered in zinc chloride.
3. An aluminum sheet is covered in lead chloride.
4. A silver coin is submerged in hydrochloric acid.
5. calcium metal is dropped in water.
6. an iron bar is dipped in hydrochloric acid.
7. a manganese(II) complex is treated with sodium amalgam
8. chromium metal is covered in water
Answer a
a) yes
Answer b
b) no
Answer c
c) yes
Answer d
d) no
Answer e
e) yes
Answer f
f) yes
Answer g
g) yes
Answer h
h) no
Exercise $5$
a) Given the standard reduction potentials below (vs. NHE), determine the cell potential for the reduction of O2 to O2- by Cu(I) ion.
$\ce{Cu^{2+} + e^{-} -> Cu^{+}}\) $E^{\circ} = + 0.153V \nonumber$ \(\ce{Cu^{+} + e^{-} -> Cu}$ $E^{\circ} = + 0.521V$
$\ce{O2 + e^{-} -> O2^{-}}$ $E^{\circ} = -0.33 V$
b) Is the reaction favored under these conditions?
c) Reactions in the presence of metals often involve substrate binding. Show, with arrows and structures, the substrate binding step that would be involved in this case.
d) Explain how this factor may influence the reduction potential of the oxygen.
e) Explain how the following observation supports your answer.
$\ce{O2 + H^{+} e^{-} -> HO2}\) $E^{\circ} = -0.13V \nonumber$ f) Metal ions in biology are usually coordinated by amino acid residues. Show a Cu(I) ion coordinated to two aspartic acid residues and two histidine residues. g) Explain how this factor may influence the reduction potential of the copper. Answer a a) Erxn = - 0.153 V - 0.33 V = - 0.483 V Answer b b) No. Answer c c) Answer d d) The O2 is activated as an electrophile. Addition of an electron may become easier. Answer e e) The reduction potential is 0.2 V more positive when the resulting superoxide ion is stabilised by binding a proton. A similar shift could occur when coordinated to copper. Answer f f) Answer g The two aspartate ions would make the copper complex less cationic. That may make it easier to remove an electron from the copper complex. Exercise \(6$
The use of Cl2 in water treatment has been linked to production of harmful chlorinated species in waterways. ClO2 has been proposed as an alternative, so researchers at Purdue decided to investigate its potential reactivity with common aqueous ions such as iron. (Margerum, Inorg. Chem. 2004, 43, 7545-7551.)
$\ce{ClO2 + e^{-} -> ClO2^{-}}\) $E^{\circ} = 0.95V \nonumber$ $\ce{Fe^{3+} e^{-} -> Fe^{2+}}$ \(E^{\circ}= 0.77V \nonumber$
1. Provide an overall reaction from these half-reactions.
2. Calculate the potential for the overall reaction.
3. Is the reaction favored?
4. Provide a Lewis structure for ClO2 (connectivity O-Cl-O).
5. Propose a reason for the overall thermodynamics (i.e. favored or disfavored) of the reaction.
6. Does the mechanism seem more likely to proceed by an inner sphere mechanism or by an outer sphere mechanism? Explain why.
7. The addition of ClO2- to Fe3+ (aq) is known to produce the complex (H2O)5Fe(ClO2)2+. This complex has a broad absorption band between 480 and 540 nm in the UV-Visible spectrum. Upon mixing ClO2 and Fe2+ (aq), a broad absorbance appears at 500 nm within 20 ms. Explain this observation.
Answer a
a) Fe2+ + ClO2 → Fe3+ + ClO2-
Answer b
b) Erxn = 0.95 V - 0.77 V = 0.18 V
Answer c
c) yes.
Answer d
d)
Answer e
e) ClO2 has an unpaired electron. ClO2- has electrons paired.
Answer f
f) It could be inner sphere: the oxygen in the ClO2 could coordinate to the iron.
Answer g
g) The mixture quickly forms (H2O)5Fe(ClO2)2+ via inner sphere electron transfer. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/V%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_3/01%3A_Reduction_and_Oxidation_Reactions/1.02%3A_R.txt |
Iron and copper are two common metals in biology, and they are both involved in electron relays in which electrons are passed from one metal to another to carry out transformations on substrates in cells. But which one passes the electron to which? Or can it go either way?
How easily one metal can pass an electron to another, or how easily one metal can reduce another, is a pretty well-studied question. There are some preferences, some of which can be easily understood.
In order to look at this question, electrochemists typically measure the voltage produced when a circuit is set up that includes an electron made of the metal in question and an electrode made of a "standard hydrogen electrode". In a standard hydrogen electrode, hydrogen gas (H2) reacts on a platinum surface to produce two electrons and two protons. The electrons travel along a wire to the other electrode. The other electrode sits in an aqueous solution containing a salt of the metal in question. How easily is the metal ion reduced by accepting the electrons from the standard hydrogen electrode? The more downhill energetically this process is, the more positive is the voltage measured in the circuit.
For example, we might put some copper(II) salts, such as CuSO4, into the solution together with a copper electrode. Then we would see whether the copper in solution is spontaneously reduced to copper metal. That would happen, essentially, if the copper ion is more easily reduced than a proton. Thus, an electron released by hydrogen flows from the platinum electrode to the copper electrode and is picked up by copper ions waiting in solution.
Otherwise, what would happen if the reaction were not spontaneous? If no reaction occurred at all, maybe there would be no voltage. However, if the opposite reaction were to occur -- if copper were able to provide an electron to convert protons into molecular hydrogen -- then current would flow through the circuit in the opposite direction. A voltage would register, but it would be negative.
As it happens, the reaction is spontaneous, the hydrogen does send electrons through the wire, turning into protons in the process, and at the other end of the wire, the copper ion is converted into copper metal, putting another layer on the surface of the electrode. The voltage is about 0.34V.
This is a comparative, rather than absolute, measurement. We are measuring the intrinsic potential of an electron to be transferred from one species, a hydrogen molecule, to another, a Cu(II) ion.
It is also an intensive, rather than extensive, property. It does not matter how much copper we have, or how much hydrogen; the electron still has the same natural tendency to flow from the hydrogen to the copper.
The results of many such studies, carefully measured under specific conditions for maximum reproducibility, are gathered in a table of reduction potentials. The reactions are referred to as "half-reactions" because they each provide only half the picture of what is going on. The electron in each reaction doesn't come from nowhere; every reaction in the table would involve transfer of an electron from elemental hydrogen to form a proton.
Half Reaction (species are aqueous unless noted otherwise) Potential, Volts
$\ce{Li^{+} + e^{-} -> Li (s)} \nonumber$
- 3.04
$\ce{Al(OH)3 (s) + 3e^{-} -> Al (s) + 3OH^{-}} \nonumber$
- 2.31
$\ce{Al^{3+} + 3e^{-} -> Al (s)}$ - 1.662
$\ce{Zn^{2+} + 2e^{-} -> Zn (s)}$ - 0.762
$\ce{Fe^{2+} + 2e^{-} -> Fe (s)} \nonumber$
- 0.44
$\ce{CO2(g) + 2H^{+} 2e^{-} -> CO(g) H2O} \nonumber$
- 0.11
$\ce{SnO (s) 2H^{+} + 2e^{-} -> Sn(s) + H2O} \nonumber$
- 0.10
$\ce{2H^{+} + 2e^{-} -> H2 (g)} \nonumber$
0.00
$\ce{Fe3O4(s) + 8H^{+} + 8e^{-} -> 3Fe(s) + 4H2O} \nonumber$
+ 0.085
$\ce{Cu^{2+} + 2e^{-} -> Cu^{0}} \nonumber$
+ 0.34
$\ce{Ag2O (S) + H2O + 2e^{-} -> 2Ag (s) + 2OH^{-} aq} \nonumber$
+ 0.342
$\ce{O2 (g) + 2H2O + 4e^{-} -> 4OH^{-}} \nonumber$
+ 0.40
$\ce{CO(g) + 2H^{+} + 2e^{-} -> C(s) + H2O} \nonumber$
+ 0.52
$\ce{Cu^{+} + e^{-} -> Cu(s)} \nonumber$
+ 0.52
$\ce{MnO4^{-} + 2H2O + 3e^{-} -> MnO2(s) + 4OH^{-}} \nonumber$
+ 0.59
$\ce{Fe^{3+} + e^{-} -> Fe^{2+} (aq)} \nonumber$
+ 0.77
$\ce{Ag^{+} + e^{-} -> Ag(s)} \nonumber$
+ 0.796
$\ce{MnO2 + 4H^{+} + 2e^{-} -> Mn^{2+} + 2H2O} \nonumber$
+ 1.23
$\ce{MnO4^{-} + 4H^{+} + 3e^{-} -> MnO2(s) + 2H2O} \nonumber$
+ 1.70
$\ce{Au^{+} + e^{-} -> Au(s)} \nonumber$
+ 1.83
$\ce{F2(g) + e^{-} -> 2F^{-}} \nonumber$
+ 2.87
Much more extensive tables of reduction potentials can be found; for example, see the following Wikipedia data page.
A positive reduction potential indicates a spontaneous reaction. That makes sense, for instance, in the reaction of fluorine to give fluoride ion. For that reaction, E0 = 2.87 V. Of course, fluorine is a very electronegative element, and it will spontaneously accept an electron to obtain a noble gas configuration.
A negative reduction potential, on the other hand, indicates a reaction that would not occur spontaneously. For example, we would not expect lithium cation to accept an electron. We are used to thinking about alkali metals easily giving up their electrons to become cations. The reduction of lithium ion has a reduction potential E0 = -3.04 V. This reaction would only occur if it were driven by an expenditure of energy.
The opposite reaction, on the other hand, would be the oxidation of lithium metal to give a lithium cation. That reaction would occur spontaneously, and would have a spontaneous "oxidation potential". In fact, that value is + 3.04 V. The oxidation potential is always the same magnitude of the reduction potential for the reverse reaction, but with the opposite sign.
These signs may seem counter-intuitive if you are used to thinking of free energy changes. A negative free energy change means energy is lost in a reaction. A positive free energy change means energy must be put into a reaction to drive it forward. In fact, reduction potential and free energy are closely linked by the following expression:
$\Delta G = -nFE^{0} \nonumber$
in which n = number of electrons transferred in the reaction; F = Faraday's constant, 96 500 Coulombs/mol.
So, a positive reduction potential translates into a negative free energy change.
Note that reduction potentials are pretty sensitive to changes in the environment. Factors that may stabilize one particular metal ion may not have the exact same effect on another, and so the preference for one state versus another will be altered slightly under different conditions.
For example, permanganate ion (MnO4-) has a more positive reduction potential under "acidic conditions" (with excess protons in solution) compared to "basic conditions" (with a paucity of protons in solution and instead an excess of hydroxide ion). The reduction potential under acidic conditions is +1.23V, compared to +0.59 V under basic conditions.
Tables of reduction potentials are also useful in assessing the opposite reaction. For example, lithium metal spontaneously reduces protons to produce hydrogen gas, becoming lithium ion in turn. The potential for that reaction is simply the opposite of the reduction potential of lithium ion; this is called the oxidation potential of lithium metal. The more positive a metal's oxidation potential, the more easily it is oxidized. However, we don't need a separate table of those values; they are just the opposite of the reduction potentials. Reactions with negative reduction potentials easily go backwards, reducing the proton to hydrogen gas by taking an electron from the reducing agent.
However, the most important use of standard reduction potentials is combining them to find out the potential of new reactions.
For example, when it says in the table that
$\ce{Cu^{+} + e^{-} -> Cu(s)}\) $E^{0}= 0.53V \nonumber$ It really means that is the potential produced for a specific reaction involving electron transfer between hydrogen and copper ion: $\ce{H2(g) + 2Cu^{+} -> 2Cu(s) + 2H^{+}}$ $E^{0} = 0.53V \nonumber$ And when it says that $\ce{Fe^{2+} + 2e^{-} -> Fe(s)}$ $E^{0}=-0.44V \nonumber$ It really means that $\ce{H2(g) + Fe^{2+} -> Fe(s) + 2H^{+}}$ $E^{0} = -0.44V \nonumber$ But now we know that reaction is endergonic, with a negative reduction potential and a positive free energy change. However, the reverse reaction $\ce{Fe(s) + 2H^{+} -> H2(g) + Fe^{2+}}$ $E^{0}=0.44V \nonumber$ has a positive reduction potential and would proceed easily. Now, if we combine those previous reactions, simply by adding them together $\ce{H2(g) + 2Cu^{+} + Fe(s) + 2H^{+} -> H2(g) + Fe^{2+} + 2Cu(s) + 2H^{+}}$ $E^{0} = 0.44 + 0.53V \nonumber$ and simplifying $\ce{2Cu^{+} + Fe(s) -> Fe^{2+} + 2Cu(s)}$ $E^{0} = 0.97V \nonumber$ That means the hydrogen reaction doesn't need to be involved at all. It's just a common reference point for all the other reactions. If we know how far uphill (or downhill) any two reactions are compared to that one, then we know how they compare to each other, too. We should note that electrochemistry is a business that demands great care. There are a number of factors that can cause variations in the potential that is measured, and so we need to be very careful to control for those factors. For example, if there is a buildup of charge in one solution or another (because we are taking cations out of solution in one case and putting them into solution in the other), the ability to remove more electrons at one electrode and deliver them at another may be hindered. For that reason, a "salt bridge" is incorporated into the design of the system; this bridge allows ions to diffuse from one cell to the other in order to keep charge balanced. Also, the solutions are maintained at a standard concentration to make sure measurements are always made in comparable circumstances. Finally, non-reactive electrolytes (salts) are added to solution to aid in conductivity and maintain a constant ionic strength. Exercise \(1$
Calculate oxidation states to confirm that the manganese ion is being reduced in the following reaction:
$\ce{MnO4^{-} + 2H2O + 3e^{-} -> MnO2(s) + 4OH^{-}} \nonumber$
Answer
MnO4- : 4 x O2- (= 8-) + Mn7+ = 1- overall
MnO2 : 2 x O2- (= 4-) + Mn4+ = neutral overall
difference = 3 e-
Exercise $2$
Balance the following half reactions by adding the right number of electrons to one side or the other, based on oxidation state. Then add water molecules and protons to help balance oxygens and overall charge.
1. S2O82- → 2 SO42-
2. HPO32- → P
3. Ti2O3 → 2 TiO
4. N2 → 2 NH2OH
Answer a
a) SO42- : 4 x O2- (= 8-) + S6+ = 2- overall
S2O82- : 8 x O2- (= 16-) + 2 x S7+ (= 14+) = 2- overall
difference = 1 e- per S, or 2 e- overall
S2O82- + 2 e- → 2 SO42-
Answer b
b) HPO32- : 3 x O2- (= 6-) + H+ + P3+ = 2- overall
P : P(0)
difference = 3 e-
HPO32- + 3 e- + 5 H+ → P + 3 H2O
Answer c
c) Ti2O3 : 3 x O2- (= 6-) + 2 x Ti3+ (= 6+) = neutral overall
TiO : O2- + Ti2+ = neutral overall
difference = 1 e- per Ti, or 2 e- overall
Ti2O3 + 2 e- + 2 H+ → 2 TiO + H2O
Answer d
d) N2 : N(0)
NH2OH : O2- + 3 x H+ (= 3+) + N- = neutral overall
difference = 1 e- per N, or 2 e- overall
N2 + 2 e- + 2 H+ + 2 H2O → 2 NH2OH
Exercise $3$
Why is the reduction potential of Li+ so negative?
Answer
Lithium is an alkali metal, in the first column of the periodic table. It has a relatively low ionization energy because it has a noble gas configuration as a cation. That noble gas configuration is stable because of the relatively large number of nuclear protons and a relatively short distance between the nucleus and the outermost shell of electrons. In lithium metal, the outermost electron is relatively far from the nucleus and so it is at a relatively high energy, and easily lost.
Exercise $4$
Why is the reduction potential of F2 so positive?
Answer
Fluorine is a halogen, with a relatively high electron affinity. It easily gains an electron to get to a noble gas configuration as a fluoride anion. That noble gas configuration is stable because of the relatively large number of nuclear protons and a relatively short distance between the nucleus and the outermost shell of electrons.
Exercise $5$
Rank the following metals from most easily oxidized to least easily oxidized: gold, iron, aluminum, copper, lithium.
Answer
From most easily oxidized to least easily oxidized: Li > Al > Fe > Cu > Au
Exercise $6$
Calculate reduction potentials for the following reactions:
1. $\ce{Au + Ag^{+} -> Au^{+} + Ag}$
2. $\ce{Zn + Fe^{2+} -> Zn^{2+} + Fe}$
3. $\ce{Li + Cu^{+} -> Li^{+} + Cu}$
4. $\ce{Ag + Fe^{3+} -> Ag^{+} + Fe^{2+}}$
Answer a
a) E0 = + 0.796 (Ag+/Ag) - 1.83 (Au/Au+) = -1.034 V (no forward reaction)
Answer b
b) E0 = - 0.44 (Fe2+/Fe) + 0.762 (Zn/Zn2+) = + 0.0322 V (forward reaction)
Answer c
c) E0 = + 0.52 (Cu+/Cu) + 3.04 (Li/Li+) = + 3.56 V (forward reaction)
Answer d
d) E0 = + 0.77 (Fe3+/Fe2+) - 0.796 (Ag/Ag+) = - 0.026 V (no forward reaction)
Exercise $7$
In general, if one reaction is combined with the reverse of a reaction above it in the table, will the overall reaction be spontaneous? What about if a reaction is combined with the reverse of a reaction below it?
Answer
When the table of standard reduction potentials is displayed with the most negative value at the top and the most positive value at the bottom, any given half-reaction will go forward if it is coupled with the reverse of a half-reaction that lies above it in the table. The opposite is not the case; no half reaction will go forward if it is coupled with the reverse of a half-reaction below it in the table. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/V%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_3/01%3A_Reduction_and_Oxidation_Reactions/1.03%3A_R.txt |
Reduction potentials are relative. Standard reduction potentials are reported relative to the reduction of protons in a standard hydrogen electrode (SHE). But what would we see if we used some other sort of electrode for comparison?
For example, instead of a hydrogen electrode, we might use a fluorine electrode, in which we have fluoride salts and fluorine gas in solution with a platinum electrode. (Maybe we just don't think working with hydrogen is dangerous enough.)
The "reduction potentials" we measure would all be relative to this reaction now. They would tell us: how much more motivated is this ion to gain an electron than fluorine?
The resulting table would look something like this:
Half Reaction (species are aqueous unless noted otherwise) Potential, Volts, Relative to Fluorine Reduction
Li+ + e- → Li (s) - 5.91
Al(OH)3(s) + 3e- → Al(s) + 3OH- - 5.18
Sc3+ + 3e- → Sc (s) - 4.95
Al3+ + 3e- → Al (s) - 4.53
V2+ + 2e- → V (s) - 4.00
Zn2+ + 2e- → Zn (s) - 3.632
Fe2+ + 2e- → Fe (s) - 3.31
CO2(g) + 2H+ + 2 e- → CO(g) + H2O - 2.98
Cu+ + e- → Cu (s) - 2.98
2 H+ + 2 e- → H2(g) - 2.87
Fe3O4(s) + 8 H+ + 8 e- → 3Fe(s) + 4 H2O - 2.79
Cu2+ + 2e- → Cu0 - 2.53
CO(g) + 2H+ + 2 e- → C(s) + H2O - 2.35
MnO4- + 2 H2O + 3e- → MnO2(s) + 4 OH- - 2.28
Fe3+ + e- → Fe2+ - 2.10
Ag+ + e- → Ag (s) - 2.074
MnO2 + 4 H+ + 2e- → Mn2+ + 2 H2O - 1.64
MnO4- + 4 H+ + 3e- → MnO2(s) + 2 H2O - 1.17
Au+ + e- → Au (s) - 1.04
F2 (g) + 2e- → 2F- 0.00
The trouble is, there isn't much out there that would be more motivated than fluorine to gain an electron. All of the potentials in the table are negative because none of the species on the left could take an electron away from fluoride ion. However, if we turned each of these reactions around, the potentials would all become positive. That means gold, for instance, could give an electron to fluorine to become Au(I). This electron transfer would be spontaneous, and a voltmeter in the circuit between the gold electrode and the fluorine electrode would measure a voltage of 1.04 V.
That potential is generated by the reaction:
$\ce{2Au (s) + F2(g) -> 2Au^{+} + 2F^{-}} \nonumber$
Notice that, because 2 electrons are needed to reduce the fluorine to fluoride, and because gold only supplies one electron, two atoms of gold would be needed to supply enough electrons.
The reduction potentials in the table are, indirectly, an index of differences in electronic energy levels. The electron on gold is at a higher energy level than if it were on fluoride. It is thus motivated to spontaneously transfer to the fluorine atom, generating a potential in the circuit of 1.04V.
Silver metal is even more motivated to donate an electron to fluorine. An electron from silver can "fall" even further than an electron from gold, to a lower energy level on fluoride. The potential in that case would be 2.074 V.
There are a couple of things to note here. The first is that, if potential is an index of the relative energy level of an electron, it doesn't matter whether one electron or two is transferred. They are transferred from the same, first energy level to the same, second energy level. The distance that the electron falls is the same regardless of the number of electrons that fall. A reduction potential reflects an inherent property of the material and does not depend on how many electrons are being transferred.
• Reduction potentials reflect the energy difference between two states
• The difference between those two states involves an exchange of one or more electrons
• Frequently, the reduction potential simply reflects the energy level of the electron in one state versus the other (although we will see other factors later)
• The number of electrons transferred does not affect the energy difference between the two states, and so does not affect the reduction potential
• The number of electrons would affect the overall energy gained / released during the electron transfer reaction (because ΔG = - nFE)
Another important note is that, if reduction potentials provide a glimpse of electronic energy levels, we may be able to deduce new relationships from previous information. For example, if an electron on gold is 1.04 V above an electron on fluoride, and an electron on silver is 2.074 V above an electron on fluoride, what can we deduce about the relative energy levels of an electron on silver vs. gold?
The answer is that the electron on silver is 1.034 V above the electron on gold. We know this because reduction potentials are "state functions", reflecting an intrinsic property of a material. It doesn't matter how we get from one place to another; the answer will always be the same. That means that if we transfer an electron from silver to gold indirectly, via fluorine, the overall potential will be the same as if we transfer the electron directly from silver to gold. So, the electron drops from silver to fluorine (a drop of 2.074 V). The electron then hops (under duress) up to gold (a climb of 1.04 V). The net drop is only 1.034 V.
That's the same value we would expect to measure if we took a standard solution of gold salts and a gold electrode and connected it, via a circuit, to a standard solution of silver salts and a silver electrode.
In fact, the table above doesn't reflect any experimental measurements; it's simply the table of standard reduction potentials from the previous page, with the reduction potential of fluorine subtracted from all the other values.
In other words, mapping out the distance from iron to SHE (in terms of the reduction potential for Fe2+ + 2 e- → Fe(s)), together with the distance from SHE to fluorine gives the potential relative to IFE (imaginary fluorine electrode).
Exercise $1$
1. Show an energy diagram showing the relative energy levels when an electron is transferred from silver to gold(I) ion. What is the potential for this reaction?
2. Show an energy diagram showing the relative energy levels when an electron is transferred from copper to silver(I) ion. What is the potential for this reaction?
3. Show how the results of (a) and (b) can be used to determine the potential for the transfer of an electron from copper to gold(I) ion.
Answer a
Answer b
Answer c
Exercise $2$
Frequently in biology, electron transfers are made more efficient through a series of smaller drops rather than one big jump. To think about this, consider the transfer of an electron from lithium to fluorine. (Neither of these species is likely to be found in an organism, but this transfer is a good illustration of a big energy difference.)
1. Write an equation for this reaction.
2. What would be the potential for this reaction?
3. This is a highly exothermic reaction; however, our imaginary organism that attempts to harness this reaction as an energy source is likely to burst into flames every time the reaction occurs. Comment on the organism's Darwinian fate.
4. Propose, instead, a series of reactions that the organism could use to slow down the release of energy. Suppose the organism, apart from fluorine and lithium, also harbors reserves of scandium(III), silver(I), copper(I), vanadium(II) and gold(I).
5. Show an energy diagram for this series of reactions and label the potential for each step. Explain why this approach would more efficiently harness the energy of lithium's electron.
Answer a
a) 2 Li + F2 → 2 Li+ + 2 F-
Answer b
b) Eo = +5.91 V
Answer c
c) Things look pretty grim.
Answer d
d)
Answer e
e) This scheme would result in the release of a small amount of energy at each stage. Each step could be harnessed to perform a task more efficiently, with less heat loss.
Exercise $3$
An "activity series" is a ranking of elements in terms of their "activity" or their ability to provide electrons. The series is normally written in a column, with the strongest reducing metals at the top. Beside these elements, we write the ion produced when the metal loses its electron(s). Looking at the table of reduction potentials relative to fluorine on this page, construct an activity series for the available elements.
Answer
Exercise $4$
Construct an activity series for the alkali metals using the following standard reduction potentials (relative to SHE): Fr, -2.9 V; Cs, -3.026 V; Rb, -2.98 V; K, -2.931 V; Na, -2.71 V; Li, -3.04 V.
Answer
Exercise $5$
In reality, the energy gap that leads to a reduction potential is sometimes more complicated than following an electron as it moves from one level to another. Use the activity series you have constructed for the alkali metals to compare and contrast the redox potential with your expectations of energy level / ease of electron donation based on standard periodic trends.
Answer
There are really two significant departures from expectation here. Lithium is much more active than expected based on electronegativity. The larger alkali metals, cesium, rubidium and francium, are all less active than expected on that basis.
We will see that another factor the influences activity in redox is the stability of ions in aqueous solution. Lithium cation is a small ion; water molecules bind very strongly to the ion because the electrons get relatively close to lithium's nucleus. That strong binding stabilizes this ion especially, tipping the malance of the reaction more strongly towards oxidation of lithium. The larger alkali metal ions are not nearly as stabilized by water ligands in aqueous solution, so the balance of their reactions does not tilt as strongly towards aqueous ions. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/V%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_3/01%3A_Reduction_and_Oxidation_Reactions/1.04%3A_R.txt |
In general, the ions of very late transition metals -- those towards the right-hand end of the transition metal block, such as copper, silver and gold -- have high reduction potentials. In other words, their ions are easily reduced. Alkali metal ions -- on the very left edge of the periodic table, such as potassium or cesium -- have very negative reduction potentials. These ions are very difficult to reduce. These trends are not surprising, because alkali metals are generally at the lower end of the electronegativity scale and are typically found as cations, not as neutral atoms. Late transition metals are comparatively electronegative in this case, and so we would expect their ions to attract electrons more easily than alkali metal ions.
The nice thing about redox is you can always look at it from either direction. Oxidation is simply the opposite of reduction. How easily does an alkali metal lose an electron? If the standard reduction potential of lithium is very negative, then the oxidation potential of lithium ion is very positive. If it is uphill to transfer an electron from hydrogen to lithium cation, it must be downhill to transfer an electron from a lithium atom to a proton. After all, hydrogen is more electronegative than any of the alkalis. Of course, since a late transition metal is generally more electronegative than an alkali metal, copper or silver or gold ought to be more difficult to oxidize than sodium or potassium.
The large trends in redox chemistry are not surprising, then. It's simply a matter of the electron moving to a lower energy level on another atom.
If we look a little more closely, though, there are plenty of exceptions to the general trend. For example, in the coinage triad, gold has the most positive reduction potential, followed by silver, then copper. That's exactly the opposite of expectations; copper, at the top of the column, should be the most electronegative and have the most positive reduction potential, not the least. What's going on in those cases?
Well, there's more going on than just moving an electron. Remember, in the measurement of a reduction potential, we are generally working with a metal electrode in an aqueous solution of ions.
What else is going on in this reaction? Well, the atom that gets reduced starts out as an ion in water, but an ion in water doesn't sit around on its own. It's a Lewis acid, an electrophile. Water is a nucleophile, a potential ligand. So the ion in solution is actually a coordination complex. It swims around for a while, then bumps into the cathode, where it picks up the electron. But the resulting ion doesn't stay in solution; it gets deposited at the electrode, along with others of its kind. It becomes part of a metal solid.
So there are three different things happening here: ligand dissociation, electron transfer and solid formation. If we could get some physical data on each of those events, we might be able to explain why these reduction potentials are contrary to expectations.
The kinds of data we have available for these individual steps may actually fit the opposite reaction better. We can estimate the energy involved in the removal of a metal atom from the solid, the loss of an electron from the metal, and the binding of water to the resulting ion. These data come from measurement of the heat of vaporization of the metal, the ionization energy of the metal, and the enthalpy of hydration of the metal.
We can use these data to construct a thermodynamic cycle. The cycle describes an alternate pathway from copper metal to aqueous Cu(I) ion. The alternative pathway, if we've chosen it correctly, ought to give us a pretty good idea of the enthalpy change involved in the reduction of copper.
The trouble is, these data all involve the gas phase. If they really applied to this situation, it would be as if metal atoms sprayed out into the air above the electrode, shot their electrons back, grabbed some water molecules that drifted by, and then dropped down into the solution. Of course that doesn't happen; we don't see a little, sparkly, metallic mist appear when we connect the circuit, or little lightning bolts from the cloud of metal atoms to the electrode, and we don't see a splash or a fizz or little tendrils of steam as the resulting ions drop into the water.
That doesn't matter. The data we have here are still very useful. That's because what we are looking at -- the energy difference between two states -- is a state function. That means it doesn't matter how we get from one state to the other; the overall difference will always be the same. So if the reaction did happen via the gas phase, the energy change would be exactly as it is when it happens directly at the electrode - solution interface. We can do a sort of thought experiment using the data we know, and even though those steps don't really happen the way they do in the experiments that gave rise to the data, they will eventually lead to the right place. This sort of imaginary path to mimic a reaction we want to know more about employs an idea called "Hess's Law". It is frequently used to gain insight into reactions throughout chemistry.
• Hess' Law states that the enthalpy of a chemical process is the same, whether the process takes place in one step or several steps.
• In other words, enthalpy is a state function.
Here are the data for copper, silver, and gold
element copper silver gold
reduction potential, Eo, V +0.520 +0.7996 +1.83
1st ionization energy, kJ/mol 745.5 731 890
heat of sublimation, kJ/mol 313 265 355
enthalpy of hydration, kJ/mol -593 -473 -615
covalent radius, Angstroms 1.32 1.45 1.36
Taking all of these data together, we can get a better picture of the overall energy changes that would occur during a reduction or, more directly, an oxidation.
The first thing to note is that copper has a higher ionization energy than silver. As expected, Cu+ really is harder to form than Ag+, because copper is more electronegative than silver. But wait a minute -- Au+ appears to be the hardest to form of all three. It's as if gold were the most electronegative of these three elements -- but it's at the bottom of this column.
Gold really is more electronegative than copper or silver. Take a look at the electronegativity chart below.
There are a few deviations from expectation in periodic trends, but this one is probably attributable to a phenomenon called "the lanthanide contraction". Notice the covalent radii of gold and silver in the table above. Normally, we expect atoms to get bigger row by row, as additional layers of electrons are filled in. Not so for the third row of transition metals. To see the probable reason for that, you have to look at the whole periodic table, and remember for the first time ever that the lanthanides and actinides -- the two orphaned rows at the bottom -- actually fit in the middle of the periodic table. The lanthanides, in particular (lanthanum, La, to ytterbium, Yb), go in between lutetium (Lu) and Hafnium (Hf).
As a result, the third row of transition metals contain many more protons in their nuclei, compared to the second row transition metals of the same column. Silver has ten more protons in its nucleus than rubidium, the first atom in the same row as silver, but gold has twenty four more than cesium. The third row "contracts" because of these additional protons.
So the exceptionally positive reduction potential of Au+ (and, by relation, the exceptionally negative oxidation potential of gold metal)may be a result of the lanthanide contraction.
What about copper versus silver? Copper still has a higher electronegativity than silver, but copper metal is more easily oxidized. It's not that copper is more easily pulled away from the metallic bonds holding it in the solid state; copper's heat of vaporization is a little higher than silver's. That leaves hydration. In fact, copper ion does have a higher enthalpy of hydration than silver; more energy is released when water binds to copper than when water binds to silver. The difference between these two appears to be all about the solvation of the copper ion, which is more stable with respect to the metal than is silver ion.
Why would that be? Well, copper is smaller than silver. A simple look at Coulomb's law reminds us that the closer the electrons of the donor ligand are to the cation, the more tightly bound they will be. Looking at it a slightly different way, copper is smaller and "harder" than silver, and forms a stronger bond with water, which is a "hard" ligand.
Taking a look at a Hess's Law cycle for a redox reaction is a useful approach to get some additional insight into the reaction. It lets us use data to assess the influence of various aspects of the reaction that we can't evaluate directly from the reduction potential, because in the redox reaction all of these factors are conflated into one number.
Exercise \(1\)
Compare the reduction potentials of lithium, sodium and potassium ions. Can you use data on heat of vaporization, ionization energy and enthalpy of solvation to determine what factors are responsible for the order of ease of oxidation of these metals?
Answer
Li+/Li: E0 = - 3.04 V; ΔHvap = 147 kJ/mol; IE = 520 kJ/mol; ΔHh = -520 kJ/mol
Na+/Na: E0 = - 2.71 V; ΔHvap = 97 kJ/mol; IE = 495 kJ/mol; ΔHh = -406 kJ/mol
K+/K: E0 = - 2.931 V; ΔHvap = 77 kJ/mol; IE = 419 kJ/mol; ΔHh = -320 kJ/mol
Potassium should be the easiest of the three to oxidize. It is easier to oxidize than sodium. However, lithium's high heat of hydration reverses the trend and tips the balance of reaction in favour of ion formation.
Exercise \(2\)
Compare the reduction potentials of copper, nickel and zinc ions. Can you use data on heat of vaporization, ionization energy and enthalpy of solvation to determine what factors are responsible for the order of ease of oxidation of these metals?
Answer
Cu2+/Cu: E0 = + 0.340 V; ΔHvap = 300 kJ/mol; IE = 745 kJ/mol & 1958 kJ/mol; ΔHh = - 2099 kJ/mol
Ni2+/Ni: E0 = - 0.25 V; ΔHvap = 377 kJ/mol; IE = 737 kJ/mol & 1753 kJ/mol; ΔHh = - 2096 kJ/mol
Zn2+/Zn: E0 = - 0.7618 V; ΔHvap = 123 kJ/mol; IE = 906 kJ/mol & 1733 kJ/mol; ΔHh = - 2047 kJ/mol
In this case, zinc may be considered the outlier. Copper should be easier to reduce than nickel based solely on electronegativity. However, zinc's very low heat of vaporization suggests that formation of the solid metal is less favoured in that case, helping to tilt the balance toward zinc ion instead.
Exercise \(3\)
Born Haber cycles are another example of thermodynamic cycles based on Hess' Law. These particular constructions are used to calculate the lattice energy of an ionic solid: the amount of energy released when ions in the gas phase condense to form an ionic lattice. This quantity is not easily measured directly.
The alternative pathway taken in the Born Haber cycle imagines that the separate ions in the lattice are first formed from the individual elements. The heat of formation of the ionic solid from the elements is usually known (or easily found on the internet, Herr Born and Herr Haber's favorite tool for evening relaxation), as are other physical parameters such as enthalpy of sublimation, ionization energies, electron affinities and so on.
Construct diagrams for the Born Haber cycle and estimate the lattice energy in each of the following cases.
a) LiCl b) CaF2 c) HgO d) ZnS
Some useful data can be found below. Note that enthalpy is a state function, so enthalpy of sublimation is roughly equal to enthalpy of fusion plus enthalpy of vaporization.
mp (°C) bp (°C) ΔHfus (kJ/mol) ΔHvap (kJ/mol) IE1 (kJ/mol) IE2 (kJ/mol) Eea1 (kJ/mol) Eea2 (kJ/mol)
lithium 180 1330 3 136 520 7298 -50 -
calcium 842 1484 9 155 590 1155 - -
zinc 419 907 7 115 906 1733 - -
mercury -39 356 2 59 1007 1810 85 -
oxygen -218 -183 0.4 7 1314 3388 -226 879
fluorine -219 -188 - 7 1681 3374 -347 -
sulfur 115 444 2 45 1000 2252 -100 435
chlorine -101 -34 6 20 1250 2298 -368 -
Answer a
Answer b
Answer c
Answer d | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/V%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_3/01%3A_Reduction_and_Oxidation_Reactions/1.05%3A_F.txt |
Metal ores are typically salts, such as oxides, carbonates or sulfides Conversion of these ores into metals requires oxidation/reduction reactions.
That's not always the case. Some early forays into metallurgy involved native gold (native meaning the metal is found in its elemental state in nature). Gold is relatively soft. It could be easily worked and shaped by heating it. Occasionally, native silver and copper can also be found.
Exercise \(1\)
Explain, with the help of a table of standard reduction potentials, why silver and gold can sometimes be found as elements rather than salts.
Answer
Both reduction potentials are very positive.
Ag+ + e- → Ag (s) E0 = + 0.796 V
Au+ + e- → Au (s) E0 = + 1.83 V
That means both metals are likely to be found in the reduced state.
However, a major leap forward came when people learned to make alloys, mixing in small amounts of other metals to make harder, sturdier materials. For example, the addition of tin to copper ushered in "the bronze age". Tin itself had to be made from its ore via smelting; the earliest evidence for this process comes from what is now Turkey, where it was performed over eight thousand years ago. However, alloys were apparently not discovered until several thousand years later.
In smelting, ore is heated to a high temperature in the presence of carbon sources, such as charcoal or coke. The partial combustion of the carbon source produces carbon monoxide which acts as a reducing agent.
Exercise \(2\)
Show the half-reactions involved in the reduction of tin oxide with carbon monoxide. Use them to come up with a balanced reaction for the process, and calculate the standard potential for the reaction.
Answer
SnO (s) + 2H+ + 2 e- → Sn (s) + H2O E0 = -0.10 V
CO(g) + H2O → CO2(g) + 2H+ + 2 e- E0 = - (-0.11 V)
SnO (s) + CO (g) → Sn (s) + CO2 (g) ΔE0 = + 0.01 V
Another major advance in metallurgy involved the conversion of iron ores into iron and steel. There is evidence that iron smelting in sub-Saharan Africa and Sri Lanka about three thousand years ago. Archaeological evidence in Sri Lanka shows that smelters were located on mountainsides facing the ocean, where constant winds provided ample oxygen to produce fires hot enough for smelting.
In the United States, the discovery of iron ores in the states along the Great Lakes, the use of the Great Lakes as a transportation network, and the availability of anthracite coal in Pennsylvania fueled the development of an American steel industry and the rise of a major industrial power. The fact that the great lakes states are still referred to as the "rust belt" is a testament to the manufacturing prowess of the region throughout the twentieth century, which proceeded from having all the necessary features for an iron-based economy in close geographic proximity.
Exercise \(3\)
Show the half-reactions involved in the reduction of iron oxide with carbon monoxide. Assume the iron oxide is prsent as magnetite, Fe3O4. Use them to come up with a balanced reaction for the process, and calculate the standard potential for the reaction.
Answer
Fe3O4(s) + 8 H+ + 8 e- → 3Fe(s) + 4 H2O E0 = +0.085 V
CO(g) + H2O → CO2(g) + 2H+ + 2 e- E0 = - (-0.11 V)
Fe3O4(s) + 4CO (g) → 3Fe(s) + 4CO2 (g) ΔE0 = + 0.195 V
Exercise \(4\)
Show the half-reactions involved in the reduction of aluminum oxide with carbon monoxide. Assume the aluminum is present as an ion in Al2O3. Use the half-reactions to come up with a balanced reaction for the process, and calculate the standard potential for the reaction.
Answer
Al3+(aq) + 3e- → Al (s) E0 = -1.662 V
CO(g) + H2O → CO2(g) + 2H+ + 2 e- E0 = - (-0.11 V)
2 Al3+(aq) + 3CO (g) + 3H2O → 2Al(s) + 3CO2 (g) ΔE0 = - 1.772 V
Aluminum is a very important material in our economy. It is lightweight, strong, and forms a very hard oxide coating when exposed to the elements, rather than the rust that results from weathering steel. In contrast to the steel industry, the aluminum industry is a far-flung operation in which ore mined on one continent may be shipped to another for processing. However, aluminum metal isn't accessible via smelting. So how is it done?
Just as a thermodynamically favored redox reaction can produce a voltage in a circuit, if we already have a voltage produced by another source, we can drive an unfavorable redox reaction to completion. We can drive the reaction backwards.
Quebec is a major producer of aluminum, despite being endowed with virtually no aluminum ore. Bauxite, the major aluminum-containing ore, is a mixture of minerals of formulae Al(OH)3 or AlO(OH) found amalgamated with other clays and minerals. It is found near the earth's surface in tropical and sub-tropical areas, left behind after of millenia of erosion and drainage of more soluble materials from underlying bedrock. The major producers of bauxite are Vietnam, Australia and Guinea, as well as a number of countries in South America.
Why ship bauxite all the way to the taiga to make aluminum? Aluminum production requires a lot of electrons, and those electrons can't be provided by coal or coke. Instead, they usually come from massive hydroelectric generating stations, such as the 16,000 megawatt James Bay Project in northern Quebec. To make aluminum, you go where electricity is cheap and plentiful.
The bauxite is first processed to help remove all those other materials that come mixed with the aluminum ore. It is dissolved in base, filtered and re-precipitated with acid. The residue is heated to drive off water, leaving pure alumina (Al2O3).
Instead of performing this redox reaction in aqueous solution, it is done in the molten state. Alumina has a melting point around 2,000 oC, but that temperature drops to a much more manageable 1,000 oC if a "flux" is added. To avoid contaminating the aluminum ions, cryolite has often been used as the flux, because it is also an aluminum salt (Na3AlF6).
The alumina is melted in an iron vat, which conveniently functions as one of the electrodes in the redox reaction. It is the cathode, supplying electrons. Graphite anodes draw electrons out of the bath to complete a circuit. Two reactions occur: aluminum ions are reduced to aluminum at the cathode, which drops to the bottom of the vat and is drained away periodically. Oxide ions are oxidized to molecular oxygen at the anodes. However, at these temperatures, the oxygen quickly reacts with the carbon anodes to produce carbon dioxide -- that is, the anodes actually disappear as the reaction proceeds.
Exercise \(5\)
Take a look at the redox reaction happening in the vat.
1. Provide a half reaction for reduction of aluminum ion.
2. Provide a half reaction for oxidation of the oxide anion.
3. Provide an overall, balanced reaction.
4. Calculate the standard potential for this reaction. (Don't worry about the lack of an aqueous solution; we'll just get an estimate of the real potential. Also, you can use the value of the reduction potential of oxygen to yield hydroxide as an approximation)
Answer
Al3+ + 3e- → Al (s) E0 = -1.662 V
2 O2- → O2(g) + 4e- E0 = - (.40 V, estimated)
4 Al3+(aq) + 6 O2- (aq) → 4 Al (s) + 3O2 (g) ΔE0 = - 2.062 V
Exercise \(6\)
Cryolite (Na3AlF6) is added to get alumina to melt at a lower temperature. Unlike bauxite, it's a somewhat rare mineral found in Greenland and Quebec. Presumably, the aluminum ions in the cryolite also get reduced. Wouldn't the rare cryolite quickly get used up? Explain why this isn't a problem.
Answer
Even if the aluminum ions from the cryolite are reduced to Al0, they will be replenished by new aluminum ions from the bauxite ore.
1.07: R
A measurement of a reduction potential involves connection of a circuit between two half-cells. Electrons produced by an oxidation reaction in one half-cell must flow to the other half-cell, where a reduction reaction takes place. If that flow of electrons is used to do work, we are using a chemical reaction to produce and harness electricity. We have a battery. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/V%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_3/01%3A_Reduction_and_Oxidation_Reactions/1.06%3A_R.txt |
In any reaction, it is useful to quantify things. How much product will there be? How much of each reactant do we need to add? What ratios do we need? There are an awful lot of reactions for which this process is straightforward, but sometimes it can be tricky. Redox reactions are sometimes on the tricky side (although certainly not always). For that reason, it's good to have a reliable method for balancing redox reactions: determining the ratios of reactants needed to give the products in the proper amounts.
Suppose, for example, we have a reaction in which silver oxide (Ag2O) reacts with manganese ion (Mn3+) to produce manganese dioxide and silver. That's:
$\ce{Mn^{3+} + Ag2O -> MnO2 + Ag} \nonumber$
What would the balanced reaction look like?
The first thing to do is make sure you are working with one half-reaction at a time. So that's:
$\ce{Mn^{3+} -> MnO2} \nonumber$
and
$\ce{Ag2O -> Ag} \nonumber$
We start out by balancing the atoms involved, one at a time. First we look at the atoms other than hydrogen or oxygen, and we just balance those by adding the right coefficient.
$\ce{Mn^{3+} -> MnO2} \nonumber$
and
$\ce{Ag2O -> Ag} \nonumber$
Second, we balance the oxygen atoms by adding water to one side or the other.
$\ce{Mn^{3+} + 2H2O -> MnO2} \nonumber$
and
$\ce{Ag2O -> 2Ag + H2O} \nonumber$
Third, we balance any hydrogens by adding protons.
$\ce{Mn^{3+} + 2H2O -> 4H^{+} + MnO2} \nonumber$
and
$\ce{Ag2O + 2H^{+} -> 2Ag + H2O} \nonumber$
Fourth, we balance the charge by adding electrons.
$\ce{Mn^{3+} + 2H2O -> 4H^{+} + MnO2 + e^{-}} \nonumber$
and
$\ce{Ag2O + 2H^{+} + 2e^{-} -> 2Ag+ H2O} \nonumber$
Fifth, we multiply so that the number of electrons is the same in both reactions.
$2 \times (\ce{Mn^{3+} + 2H2O -> 4H^{+} + MnO2 + e^{-}}) \nonumber$
or
$\ce{2Mn^{3+} + 4H2O -> 8H^{+} + 2MnO2 + 2e^{-}} \nonumber$
and
$\ce{Ag2O + 2H^{+} + 2e^{-} -> 2Ag + H2O} \nonumber$
Sixth, we simply add these two reactions together. The reaction arrow functions like an equals sign. The left side adds to the left side, and the right side adds to the right.
$\ce{2Mn^{3+} + 4H2O + Ag2O + 2H^{+} + 2e^{-} -> 8H^{+} + 2MnO2 + 2e^{-} + 2Ag + H2O} \nonumber$
At that point, gratifyingly, the equation simplifies. Notice that we have added the same number of electrons to each side; they cancel out. That's perfect, because it means we have supplied just the right number of electrons from one half reaction to satisfy the other half reaction.
$\ce{2Mn^{3+} + 4H2O + Ag2O + 2H^{+} -> 8H^{+} + 2MnO2 + 2Ag + H2O} \nonumber$
Also if we subtract one water from each side, things get slightly simpler.
$\ce{2Mn^{3+} + 3H2O + Ag2O + 2H^{+} -> 8H^{+} + 2MnO2 + 2Ag} \nonumber$
Subtracting two protons from each side makes it simpler still.
$\ce{2Mn^{3+} + 3H2O + Ag2O -> 6H^{+} + 2MnO2 + 2Ag} \nonumber$
This method works for any redox reaction, no matter how complicated.
Exercise $1$
Balance the following reactions.
1. $\ce{Cu + MoO2 -> Cu2O + Mo}$
2. $\ce{NH2OH + Ag2O -> N2 + Ag}$
3. $\ce{Fe3O4 + CO -> Fe + CO2}$
4. $\ce{I2 + MnO4^{-} -> IO3^{-} + MnO2}$
5. $\ce{H3Mo7O24 + S2O3^{2-} -> Mo + So3^{2-}}$
Answer a
a) Cu --> Cu2O MoO2 --> Mo
2 Cu --> Cu2O MoO2 --> Mo
2 Cu + H2O --> Cu2O MoO2 --> Mo + 2 H2O
2 Cu + H2O --> Cu2O + 2 H+ 4H+ + MoO2 --> Mo + 2 H2O
2 Cu + H2O --> Cu2O + 2 H+ + 2e- 4e- + 4H+ + MoO2 --> Mo + 2 H2O
2x (2 Cu + H2O --> Cu2O + 2 H+ + 2e- ) 4e- + 4H+ + MoO2 --> Mo + 2 H2O
adding:
4 Cu + 2 H2O --> 2 Cu2O + 4 H+ + 4e-
4e- + 4H+ + MoO2 --> Mo + 2 H2O
equals
4 Cu + MoO2 --> 2 Cu2O + Mo
Answer b
b) NH2OH --> N2 Ag2O --> Ag
2 NH2OH --> N2 Ag2O --> 2 Ag
2 NH2OH --> N2 + 2 H2O Ag2O --> 2 Ag + H2O
2 NH2OH --> N2 + 2 H2O + 2 H+ 2 H+ + Ag2O --> 2 Ag + H2O
2 NH2OH --> N2 + 2 H2O + 2 H+ + 2 e- 2 H+ + Ag2O + 2 e- --> 2 Ag + H2O
adding:
2 NH2OH --> N2 + 2 H2O + 2 H+ + 2 e-
2 H+ + Ag2O + 2 e- --> 2 Ag + H2O
equals
2 NH2OH + Ag2O --> N2 + 2 Ag + 3 H2O
Answer c
c) Fe3O4 --> Fe CO --> CO2
Fe3O4 --> 3 Fe CO --> CO2
Fe3O4 --> 3 Fe + 4 H2O H2O + CO --> CO2
8 H+ + Fe3O4 --> 3 Fe + 4 H2O H2O + CO --> CO2 + 2 H+
8 e- + 8 H+ + Fe3O4 --> 3 Fe + 4 H2O H2O + CO --> CO2 + 2 H+ + 2 e-
8 e- + 8 H+ + Fe3O4 --> 3 Fe + 4 H2O 4x (H2O + CO --> CO2 + 2 H+ + 2 e- )
adding:
8 e- + 8 H+ + Fe3O4 --> 3 Fe + 4 H2O
4 H2O + 4 CO --> 4 CO2 + 8 H+ + 8 e-
equals
Fe3O4 + 4 CO --> 3 Fe + 4 CO2
Answer d
d) I2 --> IO3- MnO4- --> MnO2
I2 --> 2 IO3- MnO4- --> MnO2
6 H2O + I2 --> 2 IO3- MnO4- --> MnO2 + 2 H2O
6 H2O + I2 --> 2 IO3- + 12 H+ 4H+ + MnO4- --> MnO2 + 2 H2O
6 H2O + I2 --> 2 IO3- + 12 H+ + 10 e- 3 e- + 4H+ + MnO4- --> MnO2 + 2 H2O
3x (6 H2O + I2 --> 2 IO3- + 12 H+ + 10 e-) 10x ( 3 e- + 4H+ + MnO4- --> MnO2 + 2 H2O)
adding
18 H2O + 3 I2 --> 6 IO3- + 36 H+ + 30 e-
30 e- + 40 H+ + 10 MnO4- --> 10 MnO2 + 20 H2O
equals
3 I2 + 4 H+ + 10 MnO4- --> 6 IO3- + 10 MnO2 + 2 H2O
Answer e
e) H3Mo7O24 --> Mo S2O32- --> SO32-
H3Mo7O24 --> 7 Mo S2O32- --> 2 SO32-
H3Mo7O24 --> 7 Mo + 24 H2O 3H2O + S2O32- --> 2 SO32-
45 H+ + H3Mo7O24 --> 7 Mo + 24 H2O 3H2O + S2O32- --> 2 SO32- + 6 H+
45 e- + 45 H+ + H3Mo7O24 --> 7 Mo + 24 H2O 3H2O + S2O32- --> 2 SO32- + 6 H+ + 4 e-
4 x (45 e- + 21 H+ + H3Mo7O24 --> 7 Mo + 24 H2O) 45 x (3H2O + S2O32- --> 2 SO32- + 6 H+ + 4 e- )
adding:
180 e- + 180 H+ + 4 H3Mo7O24 --> 28 Mo + 96 H2O
135 H2O + 45 S2O32- --> 90 SO32- + 270 H+ + 180 e-
equals
4 H3Mo7O24 + 45 S2O32- + 39 H2O --> 28 Mo + 90 SO32- + 90 H+
checking:
28 Mo --> 28 Mo; 90 S --> 90 S; 90 H --> 90 H; 270 O --> 270 O
In the event that the reaction is described as occuring under basic conditions, we can simply "neutralize" our protons at the end, by adding hydroxide to both sides.
$\ce{2Mn^{3+} + 3H2O + Ag2O + 6 OH^{-} -> 6H^{+} + 6OH^{-} + 2MnO2 + 2Ag} \nonumber$
Which of course means
$\ce{2Mn^{3+} + 3H2O + Ag2O + 6OH^{-} -> 6H2O + 2MnO2 + 2Ag} \nonumber$
Simplifying to
$\ce{2Mn^{3+} + Ag2O + 6OH^{-} -> 3H2O + 2MnO2 + 2Ag} \nonumber$
Exercise $2$
Balance the following reactions under basic conditions.
1. $\ce{Fe(OH)2 + N2H4 -> Fe2O3 + NH4^{+}}$
2. $\ce{MnO4^{-} + V^{3+} -> HMnO4^{-} + VO2^{+}}$
Answer a
a) Fe(OH)2 --> Fe2O3 N2H4 --> NH4+
2 Fe(OH)2 --> Fe2O3 N2H4 --> 2 NH4+
2 Fe(OH)2 --> Fe2O3 + H2O N2H4 --> 2 NH4+
2 Fe(OH)2 --> Fe2O3 + H2O + 2 H+ 4 H+ + N2H4 --> 2 NH4+
2 Fe(OH)2 --> Fe2O3 + H2O + 2 H+ + 2 e- 2 e- + 4 H+ + N2H4 --> 2 NH4+
adding:
2 Fe(OH)2 --> Fe2O3 + H2O + 2 H+ + 2 e-
2 e- + 4 H+ + N2H4 --> 2 NH4+
equals:
2 Fe(OH)2 + 2 H+ + N2H4 --> Fe2O3 + H2O + 2 NH4+
in basic conditions:
2 Fe(OH)2 + 2 H+ + 2 -OH + N2H4 --> Fe2O3 + H2O + 2 NH4+ + 2 -OH
2 Fe(OH)2 + 2 H2O + N2H4 --> Fe2O3 + H2O + 2 NH4+ + 2 -OH
2 Fe(OH)2 + H2O + N2H4 --> Fe2O3 + 2 NH4+ + 2 -OH
Answer b
b) MnO4- --> HMnO4- V3+ --> VO2+
MnO4- --> HMnO4- V3+ + H2O --> VO2+
H+ + MnO4- --> HMnO4- V3+ + H2O --> VO2+ + 2 H+
e- + H+ + MnO4- --> HMnO4- V3+ + H2O --> VO2+ + 2 H+ + e-
adding:
e- + H+ + MnO4- --> HMnO4-
V3+ + H2O --> VO2+ + 2 H+ + e-
equals:
MnO4- + V3+ + H2O --> HMnO4- + VO2+ + H+
under basic conditions:
MnO4- + V3+ + H2O + -OH --> HMnO4- + VO2+ + H+ + -OH
MnO4- + V3+ + H2O + -OH --> HMnO4- + VO2+ + H2O
MnO4- + V3+ + -OH --> HMnO4- + VO2+ | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/V%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_3/01%3A_Reduction_and_Oxidation_Reactions/1.08%3A_B.txt |
How does an electron get from one metal to another? This might be a more difficult task than it seems. In biochemistry, an electron may need to be transfered a considerable distance. Often, when the transfer occurs between two metals, the metal ions may be constrained in particular binding sites within a protein, or even in two different proteins.
That means the electron must travel through space to reach its destination. Its ability to do so is generally limited to just a few Angstroms (remember, an Angstrom is roughly the distance of a bond). Still, it can react with something a few bond lengths away. Most things need to actually bump into a partner before they can react with it.
This long distance hop is called an outer sphere electron transfer. The two metals react without ever contacting each other, without getting into each others' coordination spheres. Of course, there are limitations to the distance involved, and the further away the metals, the less likely the reaction. But an outer sphere electron transfer seems a little magical.
Barrier to Reaction: A Qualitative Picture of Marcus Theory
So, what holds the electron back? What is the barrier to the reaction? Rudy Marcus at Caltech has developed a mathematical approach to understanding the kinetics of electron transfer, in work he did beginning in the late 1950's. We will take a very qualitative look at some of the ideas in what is referred to as "Marcus Theory". An electron is small and very fast. All those big, heavy atoms involved in the picture are lumbering and slow. The barrier to the reaction has little to do with the electron's ability to whiz around, although even that is limited by distance. Instead, it has everything to do with all of those things that are barely moving compared to the electron.
Imagine an iron(II) ion is passing an electron to an iron(III) ion. After the electron transfer, they have switched identities; the first has become an iron(III) and the second has become an iron(II) ion.
Nothing could be simpler. The trouble is, there are big differences between an iron(II) ion and an iron(III) ion. For example, in a coordination complex, they have very different bond distances. Why is that a problem? Because when the electron hops, the two iron atoms find themselves in sub-optimal coordination environments.
Exercise \(1\)
Suppose an electron is transferred from an Fe(II) to a Cu(II) ion. Describe how the bond lengths might change in each case, and why. Don't worry about what the specific ligands are.
Answer
The bonds to iron would contract because the increased charge on the iron would attract the ligand donor electrons more strongly. The bonds to copper would lengthen because of the lower charge on the copper.
Exercise \(2\)
In reality, a bond length is not static. If there is a little energy around, the bond can lengthen and shorten a little bit, or vibrate. A typical graph of molecular energy vs. bond length is shown below.
1. Why do you think energy increases when the bond gets shorter than optimal?
2. Why do you think energy increases when the bond gets longer than optimal?
3. In the following drawings, energy is being added as we go from left to right. Describe what is happening to the bond length as available energy increases.
Answer a
a) Most likely there are repulsive forces between ligands if the bonds get too short.
Answer b
b) Insufficient overlap between metal and ligand orbitals would weaken the bond and raise the energy.
Answer c
c) The range of possible bond lengths gets broader as energy is increased. The bond has more latitude, with both longer and shorter bonds allowed at higher energy.
Exercise \(3\)
The optimum C-O bond length in a carbon dioxide molecule is 1.116 Å. Draw a graph of what happens to internal energy when this bond length varies between 1.10 Å and 1.20 Å. Don't worry about quantitative labels on the energy axis.
Answer
Exercise \(4\)
The optimum O-C-O bond angle in a carbon dioxide molecule is 180 °. Draw a graph of what happens to internal energy when this bond angle varies between 170 ° and 190 °. Don't worry about quantitative labels on the energy axis.
Answer
The barrier to electron transfer has to do with reorganizations of all those big atoms before the electron makes the jump. In terms of the coordination sphere, those reorganizations involve bond vibrations, and bond vibrations cost energy. Outside the coordination sphere, solvent molecules have to reorganize, too. Remember, ion stability is highly influenced by the surrounding medium.
Exercise \(5\)
Draw a Fe(II) ion and a Cu(II) ion with three water molecules located somewhere in between them. Don't worry about the ligands on the iron or copper. Show how the water molecules might change position or orientation if an electron is transferred from iron to copper.
Answer
The water molecules may pivot toward the more highly charged Fe(III), or they may shift closer to it because of the attraction between the ion and the dipole of the water molecule.
Keep in mind that such adjustments would happen in non-polar solvents, too, although they would involve weaker IMFs such as ion - induced dipole interactions.
Thus, the energetic changes needed before electron transfer can occur involve a variety of changes, including bond lengths of several ligands, bond angles, solvent molecules, and so on. The whole system, involving both metals, has some optimum set of positions of minimum energy. Any deviations from those positions requires added energy. In the following energy diagram, the x axis no longer defines one particular parameter. Now it lumps all changes in the system onto one axis. This picture is a little more abstract than when we are just looking at one bond length or one bond angle, but the concept is similar: there is an optimum set of positions for the atoms in this system, and it would require an input of energy in order to move any of them move away from their optimum position.
It is thought that these kinds of reorganizations -- involving solvent molecules, bond lengths, coordination geometry and so on -- actually occur prior to electron transfer. They happen via random motions of the molecules involved. However, once they have happened, there is nothing to hold the electron back. Its motion is so rapid that it can immediately find itself on the other atom before anything has a chance to move again.
Consequently, the barrier to electron transfer is just the amount of energy needed for all of those heavy atoms to get to some set of coordinates that would be accessible in the first state, before the electron is transferred, but that would also be accessible in the second state, after the electron is transfered.
Exercise \(6\)
Describe some of the changes that contribute to the barrier to electron transfer in the following case.
Answer
The reactants and products are very similar in this case. However, the Fe(III) complex has shorter bonds than the Fe(II) complex because of greater electrostatic interaction between the metal ion and the ligands. These changes in bond length needed in order to get ready to change from Fe(III) to Fe(II) (or the reverse) pose a major barrier to the reaction.
In the drawing below, an electron is transferred from one metal to another metal of the same kind, so the two are just switching oxidation states. For example, it could be an iron(II) and an iron(III), as pictured in the problem above. In the blue state, one iron has the extra electron, and in the red state it is the other iron that has the extra electron. The energy of the two states are the same, and the reduction potential involved in this transfer is zero. However, there would be some atomic reorganizations needed to get the coordination and solvation environments adjusted to the electron transfer. The ligand atoms and solvent molecules have shifted in the change from one state to another, and so our energy surfaces have shifted along the x axis to reflect that reorganization.
That example isn't very interesting, because we don't form anything new on the product side. Instead, let's picture an electron transfer from one metal to a very different one. For example, maybe the electron is transferred from cytochrome c to the "copper A" center in cytochrome c oxidase, an important protein involved in respiratory electron transfer.
Exercise \(7\)
In the drawing above, some water molecules are included between the two metal centres.
1. Explain what happens to the water molecules in order to allow electron transfer to occur, and why.
2. Suppose there were a different solvent, other than water, between the complexes. How might that affect the barrier to the reaction?
Answer a
a) The drawing is an oversimplification, but in general the water molecules are shown reorienting after the electron transfer because of ion-dipole interactions. In this case, the waters are shown orienting to present their negative ends to the more positive iron atom after the electron transfer. In reality, in a protein there are lots of other charges (including charges on the ligand) that may take part in additional ion-dipole interactions.
Answer b
b) Because electron transfer is so fast, atomic and molecular reorganisations are actually thought to happen before the electron transfer. The water molecules would happen to shift into a position that would provide the greatest possible stabilisation for the ions and then the electron would be transferred. A less polar solvent than water would be less able to stabilize ions and the electron would be slower to transfer as a result. In addition, a less polar solvent than water would be a poor medium to transmit an electron, which is charged and therefore stabilized by interactions with polar solvents.
The energy diagram for the case involving two different metals is very similar, except that now there is a difference in energy between the two states. The reduction potential is no longer zero. We'll assume the reduction potential is positive, and so the free energy change is negative. Energy goes down upon electron transfer.
Compare this picture to the one for the degenerate case, when the electron is just transferred to a new metal of the same type. A positive reduction potential (or a negative free energy change) has the effect of sliding the energy surface for the red state downwards. As a result, the intersection point between the two surfaces also slides downwards. Since that is the point at which the electron can slide from one state to the other, the barrier to the reaction decreases.
What would happen if the reduction potential were even more positive? Let's see in the picture below.
The trend continues. According to this interpretation of the kinetics of electron transfer, the more exothermic the reaction, the lower its barrier will be. It isn't always the case that kinetics tracks along with thermodynamics, but this might be one of them.
But is all of this really true? We should take a look at some experimental data and see whether it truly works this way.
Oxidant k (M-1s-1) (margin of error shown in parentheses)
Co(diene)(NH3)23+ 0.12 3.0(4)
Co(diene)H2O)NCS2+ 0.38 11(1)
Co(diene)(H2O)23+ 0.53 800(100)
Co(EDTA) 0.60 6000(1000)
As the reduction potential becomes more positive, free energy gets more negative, and the rate of the reaction dramatically increases. So far, Marcus theory seems to get things right.
Exercise \(8\)
1. Plot the data in the above table.
2. How would you describe the relationship? Is it linear? Is it exponential? Is it direct? Is it inverse?
3. Plot rate constant versus free energy change. How does this graph compare to the first one?
Answer a
a) Here is a plot of the data.
Answer b
b) It doesn't look linear. If we plot the y axis on a log scale, things become a little more linear.
It looks closer to a logarithmic relationship than a linear one.
Answer c
c) Assuming one electron transfer:
The graph takes the same form but in the opposite direction along the x axis.
Marcus Inverted Region
When you look a little closer at Marcus theory, though, things get a little strange. Suppose we make one more change and see what happens when the reduction potential becomes even more positive.
So, if Marcus is correct, at some point as the reduction potential continues to get more positive, reactions start to slow down again. They don't just reach a maximum rate and hold steady at that plateau; the barrier gets higher and higher and the reactions get slower and slower. If you feel a little skeptical about that, you're in good company.
Marcus always maintained that this phenomenon was a valid aspect of the theory, and not just some aberration that should be ignored. The fact that nobody had ever actually observed such a trend didn't bother him. The reason we didn't see this kind of thing, he said, was that we just hadn't developed technology that was good enough to measure these kind of rates accurately.
But technology did catch up. Just take a look at the following data (from Miller, J. Am.Chem. Soc. 1984, 3047).
Don't worry that there are no metals involved anymore. An electron transfer is an electron transfer. Here, an electron is sent from the aromatic substructure on the right to the substructure on the left. By varying the part on the left, we can adjust the reduction potential (or the free energy change, as reported here.
Exercise \(9\)
1. Plot the data in the above table.
2. How would you describe the relationship?
Answer a
a)
Answer b
b) We can see two sides of an inverted curve. The reaction gets much faster as the free energy becomes more negative, but at some point the rate begins to decrease again.
As the reaction becomes more exergonic, the rate increases, but then it hits a maximum and decreases again. Data like this means that the "Marcus Inverted Region" is a real phenomenon. Are you convinced? So were other people. In 1992, Marcus was awarded the Nobel Prize in Chemistry for this work.
Exercise \(10\)
Take a look at the donor/acceptor molecule used in Williams' study, above. a) Why do you suppose the free energy change is pretty small for the first three compounds in the table? b) Why does the free energy change continue to get bigger over the last three compounds in the table?
Answer
The acceptor compound becomes an anion when it accepts an electron. The first three compounds do not appear to be strongly electrophilic; they can accept electrons simply because of resonance stability of the resulting anion. The last three have electron withdrawing groups (chlorines and oxygens) that would stabilize the anion even further.
Exercise \(11\)
The rates of electron transfer between cobalt complexes of the bidentate bipyridyl ligand, Co(bipy)3n+, are strongly dependent upon oxidation state in the redox pair. Electron transfer between Co(I)/Co(II) occurs with a rate constant of about 109 M-1s-1, whereas the reaction between Co(II)/Co(III) species proceeds with k = 18 M-1s-1.
1. What geometry is adopted by these complexes?
2. Are these species high spin or low spin?
3. Draw d orbital splitting diagrams for each complex.
4. Explain why electron transfer is so much more facile for the Co(I)/Co(II) pair than for the Co(II)/Co(III) pair.
Answer a
a) octahedral; bpy is a bidentate ligand.
Answer b
b) Co is first row; Co(I) and Co(II) have relatively low charge. Usually we would expect them to be high spin. Co(III) is at a cut-off point in the first row; it is just electronegative enough that it is usually low spin.
Answer c
c)
Answer d
d) In a transfer from Co(II) to Co(III), there is additional reorganization needed because the metal changes between high and low spin. Not only does one electron have to move from one metal to another metal, but additional electrons have to shuffle from one orbital to another on the same metal to accommodate the change. These reorganizations have a barrier, slowing the reaction. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/V%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_3/01%3A_Reduction_and_Oxidation_Reactions/1.09%3A_O.txt |
In some cases, electron transfers occur much more quickly in the presence of certain ligands. For example, compare the rate constants for the following two electron transfer reactions, involving almost exactly the same complexes:
$\ce{Co(NH3)6^{3+} + Cr^{2+} -> Co^{2+} + Cr^{3+} + 6 NH3} \: \: k = 10^{-4}M^{-1}s^{-1} \nonumber$
$\ce{Co(NH3)5Cl^{2+} + Cr^{2+} -> Co^{2+} + CrCl^{2+} + 6NH3} \: \: k=6 \times 10^{5} M^{-1}s^{-1} \nonumber$
(Note: aqua ligands are omitted for simplicity. Ions, unless noted otherwise, are aqua complexes.)
Notice two things: first, when there is a chloride ligand involved, the reaction is much faster. Second, after the reaction, the chloride ligand has been transferred to the chromium ion. Possibly, those two events are part of the same phenomenon.
Similar rate enhancements have been reported for reactions in which other halide ligands are involved in the coordination sphere of one of the metals.
In the 1960’s, Henry Taube of Stanford University proposed that halides (and other ligands) may promote electron transfer via bridging effects. What he meant was that the chloride ion could use one of its additional lone pairs to bind to the chromium ion. It would then be bound to both metals at the same time, forming a bridge between them. Perhaps the chloride could act as a conduit for electron transfer. The chloride might then remain attached to the chromium, to which it had already formed a bond, leaving the cobalt behind.
Electron transfers that occur via ligands shared by the two metals undergoing oxidation and reduction are termed "inner sphere" electron transfers. Taube was awarded the Nobel Prize in chemistry in 1983; the award was based on his work on the mechanism of electron transfer reactions.
Exercise $1$
Take another look at the two electron transfer reactions involving the cobalt and chromium ion, above.
1. What geometry is adopted by these complexes?
2. Are these species high spin or low spin?
3. Draw d orbital splitting diagrams for each complex.
4. Explain why electron transfer is accompanied by loss of the ammonia ligands from the cobalt complex.
5. The chloride is lost from the cobalt complex after electron transfer. Why does it remain on the chromium?
Answer a
a) octahedral
Answer b
b) In the first row, 2+ complexes are almost always high spin. However, 3+ complexes are sometimes low spin.
Answer c
c)
Answer d
d) The Co(II) complex is high spin and labile. The ligands are easily replaced by water.
Answer e
e) The Cr(III) complex is only d3; it is inert.
Other ligands can be involved in inner sphere electron transfers. These ligands include carboxylates, oxalate, azide, thiocyanate, and pyrazine ligands. All of these ligands have additional lone pairs with which to bind a second metal ion.
Exercise $2$
Draw an example of each of the ligands listed above bridging between a cobalt(III) and chromium(II) aqua complex.
Answer
Exercise $3$
Explain, with structures and d orbital splitting diagrams, how the products are formed in the following reaction, in aqueous solution.
$\ce{Fe(OH2)6^{2+} + (SCN)Co(NH3)5^{2+} -> (NCS)Fe(OH2)5^{2+} + Co(OH2)6^{2+} + 5NH3} \nonumber$
Answer
How does the electron travel over the bridge?
Once the bridge is in place, the electron transfer may take place via either of two mechanisms. Suppose the bridging ligand is a chloride. The first step might actually involve an electron transfer from chlorine to the metal; that is, the chloride could donate one electron from one of its idle lone pairs. This electron could subsequently be replaced by an electron transfer from metal to chlorine.
Sometimes, we talk about the place where an electron used to be, describing it as a "hole". In this mechanism, the electron donated from the bridging chloride ligand leaves behind a hole. The hole is then filled with an electron donated from the other metal.
Alternatively, an electron might first be transferred from metal to chlorine, which subsequently passes an electron along to the other metal. In the case of chlorine, this idea may be unsatisfactory, because chlorine already has a full octet. Nevertheless, some of the other bridging ligands may have low-lying unoccupied molecular orbitals that could be populated by this extra electron, temporarily.
Exercise $4$
For the iron / cobalt electron transfer in problem Exercise $3$ (RO9.3.), show
1. an electron transfer mechanism via a hole migration along the bridge
2. an electron transfer mechanism via an electron migration along the bridge
Answer a
Answer b
Exercise $5$
One of the many contributions to the barrier for electron transfer between metal ions is internal electronic reorganization.
a) Draw d orbital splitting diagrams for each of the following metal ions in an octahedral environment.
Ru(II) or Os(II)
Ru(III) or Os(III)
Co(II)
Co(III)
Flash photolysis is a method in which an electron can be moved instantly “uphill” from one metal to another (e.g. from M2II to M1III, below); the electron transfer rate can then be measured as the electron “drops” back from M1II to M2III.
b) Explain the relative rates of electron transfer reaction in this system, as measured by flash photolysis in the table below.
M1II M2III kobs s-1
Os Ru > 5 x 109
Os Co 1.9 x 105
c) Does the reaction above probably occur via an inner sphere or by an outer sphere pathway? Why?
Answer a
a)
Answer b
b) The electron transfer between Os(II) and Ru(III) will not involve any electron reorganization because both are low spin to begin with. However, the electron transfer between Os(II) and Co(III) will result in cobalt changing from low spin to high spin. The need to move electrons between different d orbitals on the cobalt will add to the barrier, slowing down the reaction.
Answer c
c) The pathway is probably inner sphere because of the bridging ligand. Furthermore, the conjugation in the bridging ligand would help in conducting an electron from one end of the ligand to the other, either through an electron mechanism or a hole mechanism.
Exercise $6$
Outer sphere electron transfer rates depend on the free energy change of the reaction (ΔG°) and the distance between oxidant and reductant (d) according to the relation
Rate constant = $k = Ae^{(- \Delta G)} e^{-d}$
a) What happens to the rate of the reaction as distance increases between reactants?
One potential problem in measuring rates of intramolecular electron transfer (i.e. within a molecule) is competition from intermolecular electron transfer (between molecules).
b) What would you do in the flash photolysis experiment above to discourage intermolecular electron transfer?
c) How could you confirm whether you were successful in discouraging intermolecular reaction?
Answer a
a) The rate decreases exponentially as distance increases.
Answer b
b) You might keep the concentration low in order to increase the distance between molecules, reducing the likely hood of an outer-sphere electron transfer.
Answer c
c) If you ran the experiment at a series of dilutions, intramolecular electron transfer would be unaffected but outer sphere electron transfer would not. If the rates were the same across a number of different concentrations, the reaction would probably be intramolecular.
Exercise $7$
Stephan Isied and coworkers at Rutgers measured the following electron transfer rates between metal centers separated by a peptide. (Chem Rev 1992, 92, 381-394)
1. The proline repeating unit is crucial in ensuring a steady increase in distance between metal centers with increased repeat units, n. Why?
2. An inner sphere pathway in this case is expected to be somewhat slow because of the lack of conjugation in the polyproline bridge. Explain why.
3. Plot the data below, with logk on the y axis (range from 4-9) and d on the x axis (12-24 Angstroms).
n d (Å) kobs (s-1 )
1 12.2 5 x 108
2 14.8 1.6 x 107
3 18.1 2.3 x 105
4 21.3 5.1 x 104
5 24.1 1.8 x 104
d) A linear relationship is in agreement with Marcus theory; logk = - c x d. Is your plot linear?
Isied offers a number of possible explanations for the data, all of which involve two competing reaction pathways.
e) Suggest one explanation for the data.
Answer a
a) Rings are frequently used to introduce conformational rigidity (or decrease conformational flexibility), limiting the range of potential shapes a molecule could adopt. If the molecule can't wiggle around as much, then the distance between the ends of the molecule should be more constant.
Answer b
b) Although the ligand is bridging, it would be difficult to picture either an electron or hole mechanism of inner sphere electron transfer. There are few pi bonds or lone pairs to use as places to put electrons or temporarily remove electrons from, shuttling the electrons from place to place along the ligand. A conjugated system would be much more likely to carry out inner sphere electron transfer.
Answer c
c)
Answer d
d) The data is not linear.
Answer e
e) The data appear to show two lines that cross. That's a classic symptom of two competing mechanisms. The faster mechanism, to the left, is probably an intramolecular electron transfer. The slower mechanism, to the right, may be an intermolecular electron transfer. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/V%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_3/01%3A_Reduction_and_Oxidation_Reactions/1.10%3A_I.txt |
Cyclic voltammetry is a commonly used method of measuring the reduction potential of a species in solution. The species may be a coordination complex or a redox-active organic compound, for example. Cyclic voltammetry provides additional data that can be interpreted to make conclusions about the reduction / oxidation reaction and the stability of the species resulting from the electron transfer.
In cyclic voltammetry, rather than measuring the voltage produced by a reaction as we discussed before, a voltage is instead applied to the solution. The voltage is changed over time and current through a circuit is monitored. When the voltage reaches a point at which a reduction/oxidation is induced, current begins to flow. A cyclic voltammogram is a plot of current versus applied voltage.
In the experiment, the species of interest is dissolved along with some electrolyte, which promotes conductivity in the solution. Three electrodes are inserted into the solution. The working electrode, where the reduction / oxidation reaction takes place, is mostly covered with an insulator, but has a small disc of electrode exposed so the reaction can take place in a carefully controlled area. The counter electrode completes the circuit. A reference electrode with a known potential is also used in order to measure the potential applied to the cell. Unusually, the solution must not be stirred.
To begin the experiment, a potential is applied that is much more positive than the potential of the reference electrode. This step ensures that the species of interest is completely oxidized to begin with. The voltage is then swept in the negative direction at a constant rate. That is, the potential at the working electrode gets lower and lower, possibly until it becomes negative compared to the reference electrode. At some point, the voltage sweep is reversed, and it becomes more and more positive until it returns to the initial setting.
The (simulated) results of such an experiment are shown below.
At point A, the potential is very positive but is then swept to lower and lower values. At point B, current begins to flow as the voltage reaches a point that allows reduction to occur. Current keeps increasing until, at point C, all of the species in the vicinity of the working electrode has already been reduced. This point is called "cathodic peak potential". Current then begins to decrease, although some still keeps flowing as more of the species slowly diffuses over to the working electrode (point D).
The reasons for two of the features of the experimental design are now apparent. The reason for the very small exposed surface of the working electrode (usuallyan exposed disc about 1 mm wide) is to limit the area in which reaction takes place, so that we can observe when a controlled population of species has been reduced. The reason for not stirring the solution is similar; if we stirred the solution, more species would be continually and quickly fed to the working electrode and we would never observe a point at which the reaction was "finished".
At some point, the potential is increased again (point E). Current keeps decreasing; that trend is reversed as the previously reduced species is again oxidized. This time, current flows in the opposite direction, and a negative peak is observed. At point F, all of the species in the vicinity of the working electrrode has been oxidized, and current begins to "drop" again. This point is called "anodic peak potential". The "formal potential" is the mean of anodic and cathodic peak potential.
Exercise \(1\)
What is different about the following cyclic voltammogram compared to the previous one? Explain what is happening in this sample.
Answer
The compound undergoes a two-electron oxidation. The oxidized species then undergoes a two-electron reduction.
Exercise \(2\)
What is different about the following cyclic voltammogram? Explain what is happening in this sample.
Exercise \(3\)
Estimate the reduction potentials in the following cyclic voltamograms. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/V%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_3/01%3A_Reduction_and_Oxidation_Reactions/1.11%3A_C.txt |
The idea of oxidation states is not normally applied to organic compounds, but it can be useful to do so. When we do, we can gain some insight into certain reactions of organic molecules.
For example, carbon dioxide, CO2, can be thought of as having carbon in an oxidized state. If we apply the usual oxidation state rule, carbon dioxide is overall neutral and contains two oxygens, each with 2- charge. To counter that charge, carbon must be in oxidation state 4+.
On the other hand, methane, CH4, can be thought of as having carbon in a reduced state. If we apply the usual oxidation state rule here, methane is overall neutral but contains four protons. That means the carbon must be in a 4- oxidation state. Of course, the carbon does not behave as if it has a minus four charge. But we will see that this sort of exercise can be useful for book-keeping purposes.
Exercise \(1\)
Assign the formal oxidation state to carbon in the following molecules.
a) methanol, CH3OH b) formaldehyde or methanal, CH2O c) carbonate, CO32- d) hydrogen cyanide, HCN
e) ethane, CH3CH3 f) ethene, CH2CH2 g) ethyne, CHCH
The general trend here is that the more bonds there are to oxygen, the more oxidized is carbon. The more bonds there are to hydrogen, the more reduced is carbon.
Oxidation Levels in Organic Compounds
Before we look at redox reactions in organic compounds, we should take a look at a slightly different convention for assessing oxidation states in organic molecules. This convention is the use of oxidation levels rather than oxidation states. An oxidation state refers to an individual atom. An oxidation level refers to a molecule; it reflects the overall sum of the oxidation states of the atoms in that molecule.
Why would we look at things that way? Clearly, we lose some of the detail. Given an oxidation level, you don't immediately know what atom would have what charge. But because these compounds are covalent, rather than ionic, the oxidation state isn't a completely reliable measure of charge, anyway. It is useful for relative comparisons (which carbon would have more buildup of positive charge?) but not absolute measurement (exactly how much charge is on this carbon?). Oxidation level drops that detail but instead gives us a quick comparison between two molecules, such as the reactant and the product of a redox reaction, and that allows us to quickly assess what has happened over the course of the reaction: is this an oxidation, or a reduction?
How does it work? Instead of a focus on atoms, there is a focus on bonds. Look at the number of C-H bonds in a molecule and assign a value of -1 for each bond. Add those numbers into a total. Next, look at the number of C-X bonds in the molecule, in which X is a heteroatom (something other than carbon or hydrogen); that is most often oxygen, but sometimes nitrogen, a halogen, or some other electronegative element. Each of these bonds counts as +1. Add those into the running total. The result is the oxidation level.
• Each C-H bond counts as -1
• Each C-X bond (such as C-O, C-N etc) counts as +1
For example, look at methanol, CH3OH.
C-H bonds 3-
C-X bonds 1+
Oxidation Level 2-
Note that we don't worry about O-H bonds. We are just looking at carbon, which is the redox-active atom in organic molecules.
Compare that result to methanal, CH2O.
C-H bonds 2-
C-X bonds 2+
Oxidation Level 0
Also compare it to formate ion, HCO2-.
C-H bonds 1-
C-X bonds 3+
Oxidation Level 2+
In these cases, the oxidation level is the same as the oxidation state on the carbon atom. That's because there is only one carbon atom in the molecule. If we have additional carbons, oxidation level becomes a much more efficient way of comparing compounds.
Exercise \(2\)
Compare oxidation levels in the following pairs of compounds.
1. CH2CHCH3 and CH3CH2CH3
2. CH3CH2CHO and CH3CH2CH2OH
3. CH3CH(OH)CH2CH2OH and CH3COCH2CHO
Answer a
a) CH2CHCH3
C-H bonds 6-
C-X bonds 0
Oxidation Level 6-
CH3CH2CH3
C-H bonds 8-
C-X bonds 0
Oxidation Level 8-
Propane results from a two-electron reduction of propene.
Answer b
b) CH3CH2CHO
C-H bonds 6-
C-X bonds 2+
Oxidation Level 4-
CH3CH2CH2OH
C-H bonds 7-
C-X bonds 1_
Oxidation Level 6-
Propanol results from a two-electron reduction of propanal.
Answer c
c) CH3CH(OH)CH2CH2OH
C-H bonds 8-
C-X bonds 2+
Oxidation Level 6-
CH3COCH2CHO
C-H bonds 6-
C-X bonds 4+
Oxidation Level 2-
3-oxobutanal results from a four-electron oxidation of 1,3-butanediol.
Mechanism of Organic Redox
Adding a hydrogen nucleophile to a carbonyl electrophile is routinely referred to as a reduction. For example, adding sodium borohydride to methanal would result in reduction to form methanol. Of course, a hydride is really a proton plus two electrons. We could write an equation for the reduction of methanal that looks a lot like the redox reactions we see in a table of standard reduction potentials.
\[\ce{CH2O + H^{-} + H^{+} -> CH3OH}\]
or
\[\ce{CH2O + 2e^{-} + 2H^{+} -> CH3OH}\]
It stands to reason that the opposite reaction, the conversion of methanol to methanal, is a two electron oxidation.
\[\ce{CH3OH -> CH2O + 2e^{-} + 2H^{+}}\]
We know how to accomplish the reduction of methanal, at least on paper. We just add a complex metal hydride, such as lithium aluminum hydride or sodium borohydride, to the carbonyl compound. After an acidic aqueous workup to remove all the lithium and aluminum compounds, we get methanol. For practical reasons, methanol may be difficult to isolate this way, but that's the general idea of the reaction.
How do we accomplish the reverse reaction: the conversion of an alcohol into a carbonyl?
One way would be to provide a hydride acceptor in the reaction, so that we could catch hydride as it comes off the methanol. The most well-known such entity is NAD+, of course. There are biological oxidations that employ NAD+ for this reason.
More generally, the reaction can be accomplished in a number of ways, on paper, by separating out the two tasks involved. We need something to accept the two protons: that's a base. We need something to accept the two electrons: that's an oxidizing agent.
For the latter task, there are a number of high oxidation state transition metal compounds that are quite willing to accept two electrons. One of the most widely employed is Cr(VI), which accepts two electrons to become Cr(IV). A number of other methods are available, having been developed partly to avoid the toxicity of chromium salts, but let's look at the chromium case as an example.
A simple chromium(VI) compound is chromium trioxide. A simple base is pyridine. If we took these two reagents together with benzyl alcohol in a solvent such as dichloromethane, what would happen? OK, you might not want to try this, because chromium trioxide has an alarming capacity to cause spontaneous combustion in organic compounds, but we can do it on paper.
Is chromium trioxide a nucleophile or an electrophile? That Cr(VI) is pretty electrophilic, surely. So what part of the benzyl alcohol is nucleophilic? The oxygen atom. When we mix these things, the oxygen atom would likely coordinate to the chromium.
When the oxygen atom coordinates to the chromium, the oxygen gets a positive formal charge. It is now motivated to lose a proton. That's what the pyridine is for.
Now we have accomplished one of the goals of the reaction. We have removed a proton from benzyl alcohol. We have one more proton and two electrons left. The second proton will have to come from the carbon attached to the oxygen; that's the place where we need to form a carbonyl.
But wait a minute. You can't take two protons off the same molecule, can you? And certainly not from two atoms that are right next to each other. Doesn't that generate an unstable dianion?
Not this time. The chromium is there to accept two electrons. We won't generate an anion at all, as far as the benzyl alcohol is concerned. It is oxidized to benzaldehyde.
Exercise \(3\)
A completely different outcome to this reaction would be obtained in aqueous solution because of the equilibrium that exists between a carbonyl and the geminal diol (or hydrate) in water. Instead of obtaining an aldehyde, a carboxylic acid would be obtained via a second reduction. Provide a mechanism for this reaction.
Oxidation of alcohols is strongly dependent on conditions. In general, there needs to be a hydrogen on the alcoholic carbon (H-C-O-H). If there is no hydrogen on that carbon, the alcohol is pretty difficult to oxidise to a ketone. If there is one hydrogen on that carbon -- that is, if the alcohol is secondary -- then the alcohol becomes a ketone.
If there are two hydrogens on that alcoholic carbon (H2C-OH), i.e. if the alcohol is a primary one, then two different products may result. Removal of just one hydrogen from the alcoholic carbon, and replacement with an additional bond to oxygen, results in formation of an aldehyde. On the other hand, replacement of the second hydrogen from the alcoholic carbon, and replacement with another oxygen, would lead to formation of a carboxylic acid.
That second case -- replacement of the second hydrogen with an oxygen -- only happens in aqueous media. The aldehyde that forms after the first oxidation (H-C=O) must become hydrated (H-C(OH)2) in order for the second oxidation to occur, making the carboxylic acid (HO-C=O). As a result, if a primary alcohol is oxidised via a reagent that requires water as a solvent, the carboxylic acid results. If an organic-soluble reagent is used, the reaction stops at the aldehyde.
• Oxidation of secondary alcohols results in formation of ketones.
• Oxidation of primary acohols results in aldehydes if mild oxidants are used in organic solvents.
• Oxidation of primary alcohols results in carboxylic acids under aqueous conditions.
The most common methods for mildly converting primary alcohols to aldehydes are Swern oxidations and Dess-Martin oxidations. Dess-Martin oxidations employ a high-oxidation state iodine compound -- that's I(V), compared to the more commonly encountered I- ion. The reduction product is an I(III) compound. Swern oxidations employ sulfur in a moderately high oxidation state of zero; the sulfur is reduced to S2- in the Me2S side product.
In both cases, the oxidation mechanism is similar to the one illustrated with chromium oxide and pyridine. The oxygen donates to the oxidizing atom (the chromium, the sulfur, or the iodine). Deprotonation of the carbon leads to formation of the C=O bond and reduction of the oxidising agent by two electrons.
With Swern oxidations, the mechanism has an added "priming" step, because the thionyl oxidant (S=O) must first be activated; the thionyl oxygen donates to the carbonyl of the oxalyl chloride, (COCl)2. The sulfur is then ready to accept an alcohol donor. Once the alcohol undergoes oxidation, the oxygen from the thionyl group is completley transferred to the oxalyl group, forming both carbon dioxide and carbon monoxide in a subsequent disproportionation reaction.
Oddly enough, although PCC looks like a good candidate for aqueous oxidation (it is ionic, after all), it is frequently absorbed onto a solid surface (usually alumina, Al2O3) and used as a heterogeneous catalyst in organic solvent. Like most heterogeneous catalysts, it works rather slowly, but it is pretty selective for aldehyde formation.
Most "ionic" oxidants really are used in the presence of water, and they do convert primary alcohols into carboxylic acids. Examples include sodium chromate, Na2CrO4, dichromate, Na2Cr2O7, and potassium permanganate, KMnO4. These reagents are much harsher than Swern conditions or DMP, however, and they can lead to extensive decomposition of the reactant. Chromates are sometimes prepared as a solution called Jones reagent, in which the oxidant is pre-mixed with sulfuric and and water; the reagent also contains some acetone, to help solubilise the organic compound to be oxidised.
Exercise \(4\)
Fill in the missing products.
Answer
Exercise \(5\)
Fill in the missing reagents.
Answer
Multiple answers may be possible. However, (a) and (e) require aqueous oxidizing conditions whereas (b) excludes aqueous oxidizing conditions.
Exercise \(6\)
Fill in the blanks in the following synthesis.
Answer | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/V%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_3/01%3A_Reduction_and_Oxidation_Reactions/1.12%3A_O.txt |
Exercise 1.1.1:
a) Ag+ b) Ni2+ c) Mn7+ d) Cr6+ e) Cu3+ f) Fe3+ g)Os8+ h) Re5+
Exercise 1.1.3:
a) Pb2+ & S2- b) Sn4+ & 2 O2- c) Hg2+ & S2- d) Fe4+ & 2 S2- e) 2 Fe3+ & 3 O2- f) 2 Fe3+ & 1 Fe2+ & 4 O2-
Exercise 1.1.4:
Probably Fe2+, to replace Zn2+ ions.
Exercise 1.1.5:
a) C 2+ b) C4+ c) 4- d) C0 e) C3+
Exercise 1.1.7:
a) Mg2+ b) Cu2+ c) Mn3+ d) Ca2+ e) Mn2+ f) Mn2+
Exercise 1.1.8:
a) Cu(II), Fe(II) b) Zn(II), Fe(III) c) Be(II), Al(III) d) Cu(I), Fe(III) e) Cu(II), Al(III)
Exercise 1.2.1:
1. $\ce{Cu -> Cu^{+} + e^{-}}$
2. $\ce{Fe^{3+} + 3e^{-} -> Fe}$
3. $\ce{Mn -> Mn^{3+} + 3e^{-}}$
4. $\ce{Zn^{2+} + 2e^{-} -> Zn}$
5. $\ce{2F^{-} -> F2 + 2e^{-}}$
6. $\ce{H2 + 2H^{+} + 2e^{-}}$
Exercise 1.2.2:
1. $\ce{ Cu(I) + Fe(III) -> Cu(II) + Fe(II)}$
2. $\ce{Cu(I) + Ag(0) -> Cu(0) + Ag(I)}$
3. $\ce{3F2 + 2Fe -> 6F^{-} + 2Fe(III)}$
4. $\ce{2Mo^{3+} + 3Mn -> 2Mo + 3Mn^{2+}}$
Exercise 1.2.3:
1. $\ce{MnO2 + 2H^{+} + 2e^{-} -> Mn(OH)2}$
2. $\ce{2NO + 2e^{-} + 2H^{+} -> N2O + H2O}$
3. $\ce{HPO3^{2-} + 2e^{-} + 3H^{+} -> H2PO2^{-} + H2O}$
4. $\ce{Sn(OH)6^{2-} + 2e^{-} + 3H^{+} -> HSnO2^{-} + 4H2O}$
Exercise 1.2.4:
a) yes b) no c) yes d) no e) yes f) yes g) yes h) no
Exercise 1.2.5:
1. $E_{rxn} = -0.153V-0.33V=-0.483V$
2. No.
d) The O2 is activated as an electrophile. Addition of an electron may become easier.
e) The reduction potential is 0.2 V more positive when the resulting superoxide ion is stabilised by binding a proton. A similar shift could occur when coordinated to copper.
f)
g) The two aspartate ions would make the copper complex less cationic. That may make it easier to remove an electron from the copper complex.
Exercise 1.2.6:
1. $\ce{Fe^{2+} + ClO2 -> Fe^{3+} + ClO2^{-}}$
2. $E_{rxn} = 0.95V - 0.77V = 0.18V$
3. yes.
e) ClO2 has an unpaired electron. ClO2- has electrons paired
f) It could be inner sphere: the oxygen in the ClO2 could coordinate to the iron.
g) The mixture quickly forms (H2O)5Fe(ClO2)2+ via inner sphere electron transfer.
Exercise 1.3.1:
MnO4- : $4 \times O^{2-} \: (=8^{-}) + Mn^{7+} = 1^{-}$ overall
MnO2 : $2 \times O^{2-} \: (=4^{-}) + Mn^{4+} =$ neutral overall
difference = 3 e-
Exercise 1.3.2:
a) SO42- : $4 \times O^{2-} \: (=8^{-}) + S^{6+} = 2^{-}$ overall
S2O82- : $8 \times O^{2-} \: (=16^{-}) + 2 \times S^{7+} \: (=14^{+}) = 2^{-}$ overall
difference = 1 e- per S, or 2 e- overall
$\ce{S2O8^{2-} + 2e^{-} -> 2SO4^{2-}} \nonumber$
b) HPO32- : $3 \times O^{2-} \: (=6^{-}) + H^{+} + P^{3+} = 2^{-}$ overall
P : P(0)
difference = 3 e-
$\ce{HPO3^{2-} + 3e^{-} + 5H^{+} -> P + 3H2O} \nonumber$
c) Ti2O3 : $3 \times O^{2-} \: (=6^{-}) + 2 \times Ti^{3+} \: (= 6^{+})$ = neutral overall
TiO : O2- + Ti2+ = neutral overall
difference = 1 e- per Ti, or 2 e- overall
$\ce{Ti2O3 + 2e^{-} + 2H^{+} -> 2TiO + H2O} \nonumber$
d) N2 : N(0)
NH2OH : $O^{2-} + 3 \times H^{+} \: (=3^{+}) + N^{-}$ = neutral overall
difference = 1 e- per N, or 2 e- overall
$\ce{N2 + 2e^{-} + 2H^{+} + 2H2O -> 2NH2OH} \nonumber$
Exercise 1.3.3:
Lithium is an alkali metal, in the first column of the periodic table. It has a relatively low ionization energy because it has a noble gas configuration as a cation. That noble gas configuration is stable because of the relatively large number of nuclear protons and a relatively short distance between the nucleus and the outermost shell of electrons. In lithium metal, the outermost electron is relatively far from the nucleus and so it is at a relatively high energy, and easily lost.
Exercise 1.3.4
Fluorine is a halogen, with a relatively high electron affinity. It easily gains an electron to get to a noble gas configuration as a fluoride anion. That noble gas configuration is stable because of the relatively large number of nuclear protons and a relatively short distance between the nucleus and the outermost shell of electrons.
Exercise 1.3.5:
From most easily oxidized to least easily oxidized: Li > Al > Fe > Cu > Au
Exercise 1.3.6:
1. $E^{0} = + 0.796 (\frac{Ag^{+}}{Ag}) - 1.83 (\frac{Au}{Au^{+}}) = -1.034V$ (no forward reaction)
2. $E^{0} = -0.44 (\frac{Fe^{2+}}{Fe}) + 0.762 (\frac{Zn}{Zn^{2+}}) = +0.0322V$ (forward reaction)
3. $E^{0} = +0.52 (\frac{Cu^{+}}{Cu}) + 3.04(\frac{Li}{Li^{+}}) = +3.56V$ (forward reaction)
4. $E^{0} = +0.77 (\frac{Fe^{3+}}{Fe^{3+}}) - 0.796(\frac{Ag}{Ag^{+}}) = -0.026V$ (no forward reaction)
Exercise 1.3.7:
When the table of standard reduction potentials is displayed with the most negative value at the top and the most positive value at the bottom, any given half-reaction will go forward if it is coupled with the reverse of a half-reaction that lies above it in the table. The opposite is not the case; no half reaction will go forward if it is coupled with the reverse of a half-reaction below it in the table.
Exercise 1.4.1:
a)
b)
c)
Exercise 1.4.2:
1. $\ce{2Li + F2 -> 2Li^{+} + 2F^{-}}$
2. Eo = +5.91 V
3. Things look pretty grim.
e) This scheme would result in the release of a small amount of energy at each stage. Each step could be harnessed to perform a task more efficiently, with less heat loss.
Exercise 1.4.3:
Exercise 1.4.4:
Exercise 1.4.5:
There are really two significant departures from expectation here. Lithium is much more active than expected based on electronegativity. The larger alkali metals, cesium, rubidium and francium, are all less active than expected on that basis.
We will see that another factor the influences activity in redox is the stability of ions in aqueous solution. Lithium cation is a small ion; water molecules bind very strongly to the ion because the electrons get relatively close to lithium's nucleus. That strong binding stabilizes this ion especially, tipping the malance of the reaction more strongly towards oxidation of lithium. The larger alkali metal ions are not nearly as stabilized by water ligands in aqueous solution, so the balance of their reactions does not tilt as strongly towards aqueous ions.
Exercise 1.5.1:
Li+/Li: $E^{0} = -3.04V$; $\Delta H_{vap} = 147 \frac{kJ}{mol}$; $IE = 520 \frac{kJ}{mol}$; $\Delta H_{h} = -520 \frac{kJ}{mol}$
Na+/Na: $E^{0} = -2.71V$; $\Delta H_{vap} = 97 \frac{kJ}{mol}$; $IE = 495 \frac{kJ}{mol}$; $\Delta H_{h} = -406 \frac{kJ}{mol}$
K+/K: $E^{0} = -2.931V$; $\Delta H_{vap} = 77 \frac{kJ}{mol}$; $IE= 419 \frac{kJ}{mol}$; $\Delta H_{h} = -320 \frac{kJ}{mol}$
Potassium should be the easiest of the three to oxidize. It is easier to oxidize than sodium. However, lithium's high heat of hydration reverses the trend and tips the balance of reaction in favor of ion formation.
Exercise 1.5.2:
Cu2+/Cu: $E^{0} = + 0.340V$; $\Delta H_{vap} = 300 \frac{kJ}{mol}$; $IE = 745 \frac{kJ}{mol} \& 1958 \frac{kJ}{mol}$; $\Delta H_{h} = -2099 \frac{kJ}{mol}$
Ni2+/Ni: $E^{0} = -0.25V$; $\Delta H_{vap} = 377 \frac{kJ}{mol}$; $IE = 737 \frac{kJ}{mol} \& 1753 \frac{kJ}{mol}$; $\Delta H_{h} = -2096 \frac{kJ}{mol}$
Zn2+/Zn: $E^{0} = -0.7618V$; $\Delta H_{vap} = 123 \frac{kJ}{mol}$; $IE = 906 \frac{kJ}{mol} \& 1733 \frac{kJ}{mol}$; $\Delta H_{h} = -2047 \frac{kJ}{mol}$
In this case, zinc may be considered the outlier. Copper should be easier to reduce than nickel based solely on electronegativity. However, zinc's very low heat of vaporization suggests that formation of the solid metal is less favored in that case, helping to tilt the balance toward zinc ion instead.
Exercise 1.5.3:
a)
b)
c)
d)
Exercise 1.6.1:
Both reduction potentials are very positive.
$\ce{Ag^{+} + e^{-} -> Ag (s)} \: E^{0} = + 0.796V \nonumber$
$\ce{Au^{+} + e^{-} -> Au (s)} \: E^{0} = +1.83V \nonumber$
That means both metals are likely to be found in the reduced state.
Exercise 1.6.2:
$\ce{SnO (s) + 2H^{+} + 2e^{-} -> Sn (s) + H2O} \: E^{0} = -0.10V \nonumber$
$\ce{CO (g) + H2O -> CO2 (g) + 2H^{+} + 2e^{-}} \: E^{0} = -(-0.11V) \nonumber$
$\ce{SnO (s) + CO (g) -> Sn (s) + CO2 (g)} \: \Delta E^{0} = + 0.01V \nonumber$
Exercise 1.6.3:
$\ce{Fe3O4 (s) + 8H^{+} + 8e^{-} -> 3Fe(s) + 4H2O} \: E^{0} = + 0.085V \nonumber$
$\ce{CO(g) + H2O -> CO2(g) + 2H^{+} + 2e^{-}} \: E^{0} = -(-0.11V) \nonumber$
$\ce{Fe3O4(s) + 4CO (g) -> 3Fe (s) + 4CO2 (g)} \: \Delta E^{0} = 0.195V \nonumber$
Exercise 1.6.4:
$\ce{Al^{3+}_{(aq)} + 3e^{-} -> Al(s)} \: E^{0}= -1.662V \nonumber$
$\ce{CO(g) + H2O -> CO2(g) + 2H^{+} + 2e^{-}} \: E^{0} = -(-0.11V) \nonumber$
$\ce{2Al^{3+} (aq) + 3CO (g) + 3H2O -> 2Al(s) + 3CO2 (g)} \: \Delta E^{0} = -1.772V \nonumber$
Exercise 1.6.5:
$\ce{Al^{3+} + 3e^{-} -> Al(s)} \: E^{0} = -1.662V \nonumber$
$\ce{2O^{2-} -> O2(g) + 4e^{-}} \: E^{0} = -(.40V, \: estimated) \nonumber$
$\ce{4Al^{3+}(aq) + 6O^{2-} (aq) -> 4Al (s) + 3O2 (g)} \: \Delta E^{0} = -2.062V \nonumber$
Even if the aluminum ions from the cryolite are reduced to Al0, they will be replenished by new aluminum ions from the bauxite ore.
Exercise 1.7.1:
$\Delta G = -nFE^{0}_{cell} \nonumber$
$\Delta G = -2 \times 96485 \frac{J}{V \: mol} \times 1.43V \nonumber$
$\Delta G = -275947 \frac{J}{mol} = -276 \frac{kJ}{mol} \nonumber$
Exercise 1.7.2:
$\Delta G = -nFE^{0}_{cell} \nonumber$
$\Delta G = -2 \times 96485 \frac {J}{V \: mol} \times 1.3V \nonumber$
$\Delta G = -250861 \frac {J}{mol} = -251 \frac {kJ}{mol} \nonumber$
Exercise 1.7.3:
$\ce{Ag2O (s) + H2O + 2e^{-} -> 2 Ag (s) + 2OH^{-} (aq)} E^{0} = +0.342V \nonumber$
$\ce{Zn (s) -> Zn^{2+}(aq) + 2e^{-}} \: E^{0}= -(-0.762V) \nonumber$
$\ce{Ag2O (s) + H2O + Zn(s) -> 2Ag (s) + 2OH^{-} (aq) + Zn^{2+} (aq)} \: \Delta E^{0} = +1.104V \nonumber$
Exercise 1.7.4:
a) Charge on complex calculated as follows:
charge on metal: 3+
charge on ligand donors (4 neutral N, 2 anionic N donors): 2-
other charges on ligand (carboxylate arm): 1-
overall: 0
b) $\Delta GG = RT log K = RT (-pK_{a}) = -8.314 \frac{J} {K \: mol} \times 300K \times (-20.2) = 50382 \frac{J}{mol} = 50.4 \frac{kJ}{mol}$
c) It is more difficult to add the electron to the more negatively charged anion, so the reduction potential is more negative in that case.
d) $\Delta G = -nFE^{o} = -1 \times 96485 \frac{J}{V \: mol} \times (-0.545V) = 52583 \frac{J}{mol} = 52.6 \frac{kJ}{mol}$
e) $\Delta G = -nFE^{o} = -1 \times 96485 \frac{J}{V \: mol} \times (-0.575V) = 55479 \frac{J}{mol} = 55.5 \frac{kJ}{mol}$
f)
$\Delta G_{Fe(II)CO_{2}H/CO_{2}^{-}} = 50.4 + 55.5 -52.6 \frac{kJ}{mol} = 53.3 \frac{kJ}{mol} \nonumber$
$\Delta G = -RT log K_{a} \nonumber$
$K_{a} = e^{(\frac{-\Delta G}{RT})} = e^{(\frac{-53300}{8.314 J/K mol \times 300K})} = e^{-21.4} \nonumber$
pKa = 21.4
g) The Fe(III) complex is easier to deprotonate because it yields an anionic species, whereas the Fe(II) complex yields a dianionic species. The greater charge buildup in the latter would cost more energy.
Exercise 1.8.1:
a) $\ce{Cu -> Cu2O}$ $\ce{MoO2 -> Mo}$
$\ce{2Cu -> Cu2O}\) $\ce{MoO2 -> Mo} \nonumber$ $\ce{2Cu + H2O -> Cu2O}$ $\ce{MoO2 -> Mo + 2H2O} \nonumber$ $\ce{2Cu + H2O -> Cu2O + 2H^{+}}$ $\ce{4H^{+} + MoO2 -> Mo + 2H2O} \nonumber$ $\ce{2Cu + H2O -> Cu2O + 2H^{+} + 2e^{-}}$ $\ce{4e^{-} + 4H^{+} + MoO2 -> Mo + 2H2O} \nonumber$ $2 \times (\ce{2 Cu + H2O -> Cu2O + 2H^{+} + 2e^{-}})$ $\ce{4e^{-} + 4H^{+} + MoO2 -> Mo + 2H2O} \nonumber$ adding: $\ce{4Cu + 2H2O -> 2Cu2O + 4H^{+} + 4e^{-}} \nonumber$ $\ce{4e^{-} + 4H^{+} + MoO2 -> Mo + 2H2O} \nonumber$ equals $\ce{4Cu + MoO2 -> 2Cu2O + Mo} \nonumber$ b) \(\ce{NH2OH -> N2}$ $\ce{Ag2O -> Ag}$
$\ce{2NH2OH -> N2}\) $\ce{Ag2O -> 2Ag} \nonumber$ $\ce{2NH2OH -> N2 + 2H2O}$ $\ce{Ag2O -> 2Ag + H2O} \nonumber$ $\ce{2NH2OH -> N2 + 2H2O + 2H^{+}}$ $\ce{2H^{+} + Ag2O -> 2Ag + H2O} \nonumber$ $\ce{2NH2OH -> N2 + 2H2O + 2H^{+} + 2e^{-}}$ $\ce{2H^{+} + Ag2O + 2e^{-} -> 2Ag + H2O} \nonumber$ adding: $\ce{2NH2OH -> N2 + 2H2O + 2H^{+} + 2e^{-}} \nonumber$ $\ce{2H^{+} + Ag2O + 2e^{-} -> 2Ag + H2O} \nonumber$ equals $\ce{2NH2OH + Ag2O -> N2 + 2Ag + 3H2O} \nonumber$ c) \(\ce{Fe3O4 -> Fe}$ $\ce{CO -> CO2}$
$\ce{Fe3O4 -> 3Fe}\) $\ce{CO -> CO2} \nonumber$ $\ce{Fe3O4 -> 3Fe + 4H2O}$ $\ce{H2O + CO -> CO2} \nonumber$ $\ce{8H^{+} + Fe3O4 -> 3Fe + 4H2O}$ $\ce{H2O + CO -> CO2 + 2H^{+}} \nonumber$ $\ce{8e^{-} + 8H^{+} + Fe3O4 -> 3Fe + 4H2O}$ $\ce{H2O + CO -> CO2 + 2H^{+} + 2e^{-}} \nonumber$ $\ce{8e^{-} + 8H^{+} + Fe3O4 -> 3Fe + 4H2O}$ $4 \times \ce{(H2O + CO -> CO2 + 2H^{+} + 2e^{-})} \nonumber$ adding: $\ce{8e^{-} + 8H^{+} + Fe3O4 -> 3Fe + 4H2O} \nonumber$ $\ce{4H2O + 4CO -> 4CO2 + 8H^{+} + 8e^{-}} \nonumber$ equals $\ce{Fe3O4 + 4CO -> 3Fe + 4CO2} \nonumber$ d) \(\ce{I2 -> IO3^{-}}$ $\ce{MnO4^{-} -> MnO2}$
$\ce{I2 -> 2IO3^{-}}\) $\ce{MnO4^{-} -> MnO2} \nonumber$ $\ce{6H2O + I2 -> 2IO3^{-}}$ $\ce{MnO4^{-} -> MnO2 + 2H2O} \nonumber$ $\ce{6H2O + I2 -> 2IO3^{-} + 12H^{+}}$ $\ce{4H^{+} + MnO4^{-} -> MnO2 + 2H2O} \nonumber$ $\ce{6H2O + I2 -> 2IO3^{-} + 12H^{+} + 10e^{-}}$ $\ce{3e^{-} + 4H^{+} + MnO4^{-} -> MnO2 + 2H2O} \nonumber$ $3 \times (\ce{6H2O + I2 -> 2IO3^{-} + 12H^{+} + 10e^{-}})$ $10 \times (\ce{3e^{-} + 4H^{+} + MnO4^{-} -> MnO2 + 2H2O}) \nonumber$ adding $\ce{18H2O + 3I2 -> 6IO3^{-} + 36H^{+} + 30e^{-}} \nonumber$ $\ce{30e^{-} + 40H^{+} + 10 MnO4^{-} -> 10MnO2 + 20H2O} \nonumber$ equals $\ce{3I2 + 4H^{+} + 10MnO4^{-} -> 6IO3^{-} + 10 MnO2 + 2H2O} \nonumber$ e) \(\ce{H3Mo7O24 -> Mo}$ $\ce{S2O3^{2-} -> SO3^{2-}}$
$\ce{H3Mo7O24 -> 7Mo}\) $\ce{S2Oe^{2-} -> 2SO3^{2-}} \nonumber$ $\ce{H3Mo7O24 -> 7Mo + 24H2O}$ $\ce{3H2O + S2O3^{2-} -> 2So3^{2-}} \nonumber$ $\ce{45H^{+} + H3Mo7O24 -> 7MO + 24 H2O}$ $\ce{3H2O + S2O3^{2-} -> 2SO3^{2-} + 6H^{+}} \nonumber$ $\ce{45e^{-} + 45 H^{+} + H3Mo7O24 -> 7Mo + 24 H2O}$ $\ce{3H2O + S2O3^{2-} -> 2SO3^{2-} + 6H^{+} + 4e^{-}} \nonumber$ $4 \times (\ce{45 e^{-} + 21H^{+} + H3Mo7O24 -> 7Mo + 24 H2O})$ $45 \times (\ce{3H2O + S2O3^{2-} -> 2SO3^{2-} + 6H^{+} + 4e^{-}}) \nonumber$ adding: $\ce{180e^{-} + 180H^{+} + 4H3Mo7O24 -> 28 Mo + 96 H2O} \nonumber$ $\ce{135 H2O + 45 S2O3^{2-} -> 90 SO3^{2-} + 270 H^{+} + 180 e^{-}} \nonumber$ equals $\ce{4H3Mo7O24 + 45 S2O3^{2-} + 39 H2O-> 28 Mo + 90 SO3^{2-} + 90H^{+}} \nonumber$ checking: $\ce{28Mo -> 28Mo}; \: \ce{90S -> 90S}; \: \ce{90H -> 90H}; \: \ce{270O -> 270O} \nonumber$ Exercise 1.8.2: a) \(\ce{Fe(OH)2 -> Fe2O3}$ $\ce{N2H4 -> NH4^{+}}$
$\ce{2Fe(OH)2 -> Fe2O3}\) $\ce{N2H4 -> 2NH4^{+}} \nonumber$ $\ce{2Fe(OH)2 -> Fe2O3 + H2O}$ $\ce{N2H4 -> 2NH4^{+}} \nonumber$ $\ce{2Fe(OH)2 -> Fe2O3 + H2O + 2H^{+}}$ $\ce{4H^{+} + N2H4 -> 2NH4^{+}} \nonumber$ $\ce{2Fe(OH)2 -> Fe2O3 + H2O + 2H^{+} + 2e^{-}}$ $\ce{2e^{-} + 4H^{+} + N2H4 -> 2NH4^{+}} \nonumber$ adding: $\ce{2Fe(OH)2 -> Fe2O3 + H2O + 2H^{+} + 2e^{-}} \nonumber$ $\ce{2e^{-} + 4H^{+} + N2H4 -> 2NH4^{+}} \nonumber$ equals: $\ce{2Fe(OH)2 + 2H^{+} + N2H4 -> Fe2O3 + H2O + 2NH4^{+}} \nonumber$ in basic conditions: $\ce{2Fe(OH)2 + 2H^{+} + 2 OH^{-} + N2H4 -> Fe2O3 + H2O + 2NH4^{+} + 2OH^{-}} \nonumber$ $\ce{2Fe(OH)2 + 2H2O + N2H4 -> Fe2O3 + H2O + 2NH4^{+} + 2OH^{-}} \nonumber$ $\ce{2Fe(OH)2 + H2O + N2H4 -> Fe2O3 + 2NH4^{+} + 2OH^{-}} \nonumber$ b) \(\ce{MnO4^{-} -> HMnO4^{-}}$ $\ce{V^{3+} -> VO^{2+}}$
$\ce{MnO4^{-} -> HMnO4^{-}}\) $\ce{V^{3+} + H2O -> VO^{2+}} \nonumber$ $\ce{H^{+} + MnO4^{-} -> HMnO4^{-}}$ $\ce{V^{3+} + H2O -> VO^{2+} + 2H^{+}} \nonumber$ $\ce{e^{-} + H^{+} + MnO4^{-} -> HMnO4^{-}}$ \(\ce{V^{3+} + H2O -> VO^{2+} + 2H^{+} + e^{-}} \nonumber$
adding:
$\ce{e^{-} + H^{+} + MnO4^{-} -> HMnO4^{-}} \nonumber$
$\ce{V^{3+} + H2O -> VO^{2+} + 2H^{+} + e^{-}} \nonumber$
equals:
$\ce{MnO4^{-} + V^{3+} + H2O -> HmnO4^{-} + VO^{2+} + H^{+}} \nonumber$
under basic conditions:
$\ce{MnO4^{-} + V^{3+} + H2O + OH^{-} -> HMnO4^{-} + VO^{2+} + H^{+} + OH^{-}} \nonumber$
$\ce{MnO4^{-} + V^{3+} + H2O _ OH^{-} -> HMNO4^{-} + VO^{2+} + H2O} \nonumber$
$\ce{MnO4^{-} + V^{3+} + OH^{-} -> HMnO4^{-} + VO^{2+}} \nonumber$
Exercise 1.9.1:
The bonds to iron would contract because the increased charge on the iron would attract the ligand donor electrons more strongly. The bonds to copper would lengthen because of the lower charge on the copper.
Exercise 1.9.2:
1. Most likely there are repulsive forces between ligands if the bonds get too short.
2. Insufficient overlap between metal and ligand orbitals would weaken the bond and raise the energy.
3. The range of possible bond lengths gets broader as energy is increased. The bond has more latitude, with both longer and shorter bonds allowed at higher energy.
Exercise 1.9.6:
The reactants and products are very similar in this case. However, the Fe(III) complex has shorter bonds than the Fe(II) complex because of greater electrostatic interaction between the metal ion and the ligands. These changes in bond length needed in order to get ready to change from Fe(III) to Fe(II) (or the reverse) pose a major barrier to the reaction.
Exercise 1.9.7:
1. The drawing is an oversimplification, but in general the water molecules are shown reorienting after the electron transfer because of ion-dipole interactions. In this case, the waters are shown orienting to present their negative ends to the more positive iron atom after the electron transfer. In reality, in a protein there are lots of other charges (including charges on the ligand) that may take part in additional ion-dipole interactions.
2. Because electron transfer is so fast, atomic and molecular reorganisations are actually thought to happen before the electron transfer. The water molecules would happen to shift into a position that would provide the greatest possible stabilisation for the ions and then the electron would be transferred. A less polar solvent than water would be less able to stabilize ions and the electron would be slower to transfer as a result.
Exercise 1.10.1:
1. octahedral
2. In the first row, 2+ complexes are almost always high spin. However, 3+ complexes are sometimes low spin.
c)
d) The Co(II) complex is high spin and labile. The ligands are easily replaced by water.
e) The Cr(III) complex is only d3; it is inert.
Exercise 1.10.3
Exercise 1.10.4:
a)
b)
Exercise 1.11.1:
The compound undergoes a two-electron oxidation. The oxidized species then undergoes a two-electron reduction.
Exercise 1.12.2:
a) CH2CHCH3
C-H bonds 6-
C-X bonds 0
Oxidation Level 6-
CH3CH2CH3
C-H bonds 8-
C-X bonds 0
Oxidation Level 8-
Propane results from a two-electron reduction of propene.
b) CH3CH2CHO
C-H bonds 6-
C-X bonds 2+
Oxidation Level 4-
CH3CH2CH2OH
C-H bonds 7-
C-X bonds 1_
Oxidation Level 6-
Propanol results from a two-electron reduction of propanal.
c) CH3CH(OH)CH2CH2OH
C-H bonds 8-
C-X bonds 2+
Oxidation Level 6-
CH3COCH2CHO
C-H bonds 6-
C-X bonds 4+
Oxidation Level 2-
3-oxobutanal results from a four-electron oxidation of 1,3-butanediol.
Exercise 1.12.3:
Exercise 1.12.4: | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/V%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_3/01%3A_Reduction_and_Oxidation_Reactions/1.14%3A_S.txt |
• 3.1: Intermediates
First and foremeost, a mechanism is a sequence of intermediates. What happens to the structure of the compound as it undergoes chemical change? Does that change happen all at once, or does it happen in stages? If it happens in stages, what kinds of intermediates are involved? Let's review some different kinds of reactive intermediates that may occur along a reaction pathway. These intermediates are not particularly stable, and so they go on to react further until they form more stable products.
• 3.2: Energetics
• 3.3: Arrow Conventions
• 3.4: Solutions to Selected Problems
03: Understanding Mechanism
By now you are familiar with a range of reaction types in organic, inorganic, and biochemistry. The purpose of this chapter is to help you review some of the tools that we use in communicating how reactions happen.
First and foremeost, a mechanism is a sequence of intermediates. What happens to the structure of the compound as it undergoes chemical change? Does that change happen all at once, or does it happen in stages? If it happens in stages, what kinds of intermediates are involved?
Let's review some different kinds of reactive intermediates that may occur along a reaction pathway. These intermediates are not particularly stable, and so they go on to react further until they form more stable products.
Cations
Cations and anions can be unstable for the simple reason that charge separation costs energy. There are a few cases in which these ions are really quite stable -- alkali cations such as Na+ and halide anions such as Cl- come to mind -- but here we are interested in exploring the less stable, more temporary examples of ions.
Unlike sodium ions, cations of carbon, nitrogen, or oxygen are reactive. These relatively electronegative atoms are not very stable with a positive charge.
Carbocations, or carbenium ions, in which the positive charge is on a carbon atom, are generally unstable. Carbon is in the upper right part of the periodic table, so it is not particularly electropositive like sodium. A positive charge on carbon frequently makes a molecule reactive. Nevertheless, this intermediate is frequently encountered during organic reactions.
Are all carbocations equally unstable? No. In general, there are two main factors that stabilize carbocations. The first, and most important, is the degree of substitution. A tertiary carbocation, in which the carbon with the positive charge is attached to three other carbon atoms, is fairly stable. A primary carbocation, in which the carbon bearing the positive charge is attached to only one other carbon and two hydrogen atoms, is not so stable. A secondary carbocation, with the positive carbon attached to two other carbons and a hydrogen atom, is intermediate in stability. Of course, a methyl cation, in which a positive carbon is attached to three hydrogen atoms, is not very stable at all.
The reasons for these differences are sometimes explained in terms of hyperconjugation. According to this idea, weak interactions between the unoccupied p orbital on the positive carbon and the occupied sigma bonds on the neighbouring carbons can stabilize the cation somewhat. Very loosely, imagine these bonds, which are made of pairs of electrons, can allow a little bit of negative charge to overlap with the cation, lowering its overall positive charge just a tad. More correctly, the empty p orbital can interact with the sigma bonds to produce two molecular orbital combinations; one of these is an in-phase combination and is lower in energy than either of the original orbitals, whereas the other, out-of-phase combination is a little higher in energy. Because only two electrons are involved, from the sigma bond, both can get to a lower energy level this way. They both drop into the lower energy combination. In that sense, the cation is stable not just because the positive charge is any less but because the neighbouring bonds can drop lower in energy.
The second factor that stabilizes positive charge is resonance delocalization. If a double bond is adjacent to a cation, conjugation between filled and empty p orbitals allows the positive charge to be distributed across multiple carbon atoms. This effect lowers the amount of positive charge borne by an one carbon atom. this kind of delocalizing effect is very common in stabilizing reactive intermediates. For this reason, allylic (CH2=CH-CH2+) and benzylic cations (C6H5CH2+) are particularly stable. They are about as stable as a secondary cation along a regular carbon chain, even if they would otherwise be only primary cations.
Of course, other atoms can be cations, too. As seen above, oxygens and nitrogens are very commonly encountered as cations. That is partly because they are very good at donating electrons to neighbouring atoms in need.
Exercise \(1\)
Identify the positive atom in each of the following molecules.
Exercise \(2\)
Within each group, rank the cations from most stable to least stable.
Exercise \(3\)
Sometimes, remote groups provide additional stabilization for a cation. Indicate whether each of the following cations would be more stable or less stable than a benzyl cation, and explain why.
There are other, more subtle factors that can influence the stability of cations. Charge stability is affected by the structure further away from the atom bearing the charge. For example, a triethylammonium cation and a trimethylammonium cation look pretty similar. However, a triethlammonium cation is a little less stable than a trimethylammonium cation.
The difference in these cations is related to the size of the overall molecule. Reactions usually take place in a solvent. The solvent plays an important role; it allows the reactants to move around, moderates heat flow, and may even provide lone pairs or protons to aid in acid/base reactions. A cation or anion most commonly occurs in solution. Because charge stability is a big issue, the solvent will also help to stabilize the charge. To do so, the solvent molecules will arrange themselves in a favorable way around the cation. The bigger the cation, the more solvent molecules will be needed to arrange themelves around it.
Anions
Negatively charged ions are also common intermediates in reactions. Like cations, anions are frequently unstable species. These species are stabilized by a number of different factors, not unlike cation stability.
Carbanions, amide ions and alkoxide ions are examples of anionic intermediates.
Remember, there are just a few key factors that explain a great deal of questions about anion stability.
Within a column of the periodic table, when comparing two atoms with negative charge, the stability of the anions principally depends on polarizability of the atom. Polarizability refers to how easily distorted the electrons are around the atom. The larger the atom, and the further the electrons from the nucleus, the more polarizable it is. The more polarizable the atom, the more stable the anion.
Within a row of the periodic table, the more electronegative an atom, the more stable the anion.
Exercise \(4\)
Within each group, rank the anions from most stable to least stable.
Exercise \(5\)
Sometimes, remote groups provide additional stabilization for a cation. Indicate whether each of the following anions would be more stable or less stable than a phenoxide anion, and explain why.
Radicals
Radicals are species with an unpaired electron. They are reactive because they are short an octet, but the presence of an unpaired electron means they react in a different way from typical electrophiles. Carbon, nitrogen, and oxygen compounds show some typical examples of radical structures.
Note that these radicals do not necessarily have charges. That is because they are bonding to one atom fewer than normal, but they are retaining just one of the electrons from the missing bond. In fact, radicals are often formed by breaking a bond within a normal, "closed-shell" compound, such that each atom involved in the bond takes one of the electrons with it. This is called "bond homolysis" and implies the bond is split evenly between the atoms. In contrast, "bond heterolysis" means the bond is broken unevenly, with one atom taing both of the electrons.
Exercise \(6\)
Confirm that there is no formal charge in each of the species shown above.
Answer a
a) formal charge = # e- in periodic table - # e- nonbonding - (# e- in bonds)/2
formal charge = 4 - 1 - 6/2 = 0
Answer b
b) formal charge = # e- in periodic table - # e- nonbonding - (# e- in bonds)/2
formal charge = 5 - 3 - 4/2 = 0
Answer c
c) formal charge = # e- in periodic table - # e- nonbonding - (# e- in bonds)/2
formal charge = 6 - 5 - 2/2 = 0
Radical ions are also possible. Radical anions can result from the addition of an extra electron to a normal, closed-shell compound. Radical cations can result through the removal of an electron from a normal, closed-shell compound.
Because radicals are electron-deficient species, in the sense that they lack an octet, they are often stabilized by the same factors that would stabilize a cation. In particular, they are stabilized by resonance delocalization, and carbon radicals are more stable on more-substituted carbons than on less-substituted carbons, just like cations. However, they are generally less sensitive that cations to these factors, because they do not actually have a positive charge.
Exercise \(7\)
Within each group, rank the radicals from most stable to least stable.
Carbenes and Nitrenes
We don't often see carbenes and the related nitrenes, but they are important intermediates in synthetic processes involving electrophilic addition to alkenes. Carbenes and nitrenes are two electrons short of an octet, but do not have a formal charge.
Carbenes are unusual because they can be thought of as both electrophiles or nucleophiles. The have lone pairs -- the usual requirement for a nucleophile. They also have an empty orbital, which would typically make them electrophiles.
Because they lack an octet, carbenes and nitrenes can be stabilized through pi-donatin.
Exercise \(8\)
Arrange the following carbenes in order from most stable to least stable.
Coordination Complexes
Exercise \(8\)
In the following pictures, decide whether the ligand is an anionic or neutral donor. Use the correct symbol (a line or an arrow) to stand for the ligand-metal bond. Assign the oxidation state to the metal to satisfy the overall charge. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/V%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_3/03%3A_Understanding_Mechanism/3.01%3A_Intermediat.txt |
Symbolic Notation
Sometimes in mathematics and science we need to convey an idea based on concepts we have learned previously. In mathematics, we use a plus sign (+) to indicate that we are putting these things together; we use a minus sign (-) to indicate we are taking some of these things away. Alternatively, we could use that plus sign to indicate that we have a surplus (a positive number) or that minus sign to show that we have a deficit or a debt (a negative number).
We don't even need to think about what these signs mean because we learned about them in elementary school. We have used them so often that we instantly get the idea of what is going on. In fact, symbolic notation is really how all of mathematics works; when we see the number three, we immediately associate it with a trio of objects (***).
In chemistry, we start using this symbolic notation early on, as well. The most important symbols may be those used for the elements themselves, such as C for carbon or Fe for iron. We also use arrows, lots of different kinds of them, to convey a number of different ideas. You probably already know a little about some of these symbols, but now that you have encountered them a few times, we should review.
Conveying Relationships Between Structures
Chances are, the first arrow you encountered in a chemistry class was a reaction arrow. A reaction arrow just tells you that a change has taken place, and one thing has turned into another. The arrow points from the old thing (the thing that reacted) to the new thing (the thing that forme
The reaction arrow is used in an "equation of reaction". The thing that reacted is called the reactant. The thing that formed is called the product. Normally the reaction is written from left to right, with the reactant on the left and the product on the right, but that isn't always necessary. The important thing is that the arrow points from reactant to product.
The reactants and products themselves are usually drawn using symbolic structures. These structures might be complete Lewis or Kekule structures, or they might be abbreviated line structures in organic compounds. Sometimes we leave out the structure and just use the formulae of the compounds, although we are losing some information that way. Sometimes in biochemistry a simple acronym is used to stand for the compound, such as NADH or AcSCoA, because the structures are relatively complicated. Nevertheless, the idea implied by the reaction arrow is always the same.
Sometimes, you may see a whole series of compounds linked from one to the next by reaction arrows. That means we are dealing with a series of reactions. We might be looking at a synthesis, in which an initial reactant forms a first product, which becomes the reactant for the next reaction, in a slow, building-up process that eventually produces a desired target compound.
The reaction arrow seems pretty simple to draw, although you do have to be careful not to draw the wrong kind of arrow. This one is just a straight line with an arrowhead at one end. Sometimes, the line can have a slight curve to it. That's usually because a series of reactions are connected in a cycle, and the series of reactions must lead back to the beginning. That's how we might show a catalytic cycle, for example.
Just to see how we can sometimes convey the wrong idea by drawing the wrong kind of arrow, consider something called a "retrosynthetic arrow". A retrosynthetic arrow differs from a reaction arrow because a retrosynthetic arrow shows what a compound is made from. In other words, it points from the product of a reaction (or series of reactions) to the reactant. Retrosynthetic arrows are used to illustrate possible sources of a compound of interest, whether the compounds are made in nature, industrially, or in a lab.
That sounds like it could be confusing, but fortunately a retrosynthetic arrow looks very different from a reaction arrow. It is an outline of an arrow, rather than a simple line and arrowhead. By choosing the right kind of arrow, you can clearly convey whether you are considering a reaction that you could do with a compound or wondering where the compound came from originally.
Biosynthetically, cholesterol is made from squalene, a polyunsaturated hydrocarbon. Squalene is an example of a class of compounds called terpenes, recognizable by their five-carbon monomer units.
Reaction arrows can be more subtle in the case of reversible reactions, or equilibrium processes. An equilibrium between two different compounds or groups of compounds can be illustrated using a combination of two arrows: one pointing from left to right and one pointing from right to left. Drawing the arrow that way implies that the reaction could proceed in either direction.
Usually, the individual arrows within that double, equilibrium arrow are drawn with only half an arrowhead.
Sometimes, one of the individual arrows in the equilibrium arrow is drawn a little longer than the other one. This symbol is meant to imply that, although the reaction is reversible, it proceeds more readily in one direction than the other. The left-to-right arrow may be the longer one, indicating that the reaction favors products; presumably it is exothermic (or exergonic, anyway). The right-to-left-arrow might be the longer one, indicating that the reaction mostly stays on the reactant side; it must be endothermic (or at least endergonic).
OR
There is another symbol that people sometimes confuse with an equilibrium arrow. It is a resonance arrow. In a resonance arrow, one straight line has an arrowhead on each end, so that it points in two different directions. It is similar in appearance to an equilibrium arrow, but it has only one stem, whereas an equilibrium arrow has two separate ones.
The resonance arrow indicates that the molecule behaves like both structures at the same time. This situation is termed a "superposition of states" in quantum mechanics.
For example, in a diazomethane molecule, the resonance structures convey that there is a buildup of negative charge on two separate atoms: both the carbon and the terminal nitrogen. In a given situation, the diazomethane may behave one way or the other.
A resonance arrow does not imply that two compounds can change back and forth. It doesn't even mean that they can switch back and forth extremely rapidly. It means that there are two different structures that we can draw for the compound, but that neither one describes the compound satisfactorily. In general, that's because of delocalisation. At least one electron in the compound is not restricted to one position as indicated in the structural drawing; instead, it spreads out to be in two positions at once. Remember, electrons are very small and they enjoy particle-wave duality. They don't have to behave like little objects that have to be moved. They can behave like waves that extend through space.
Curved Arrows: Bond-Making and -Breaking Events
Among the unlearned, curved arrow conventions are just a curly fluorish on a drawing of a reaction. Used properly, these symbols convey meaning to the reader and enhance our understanding of a mechanism.
A curved arrow illustrates the path taken by a pair of electrons during a reaction. The stem of the arrow starts at the electron pair in question; it's usually a lone pair, but it could also be a pi bond in some cases. The arrowhead points to the position that attracts the electrons; that might be a cation or some other electron-deficient site.
For example, in a Lewis acid-base reaction, the arrow would curve from the lone pair on the Lewis base to the electron-deficient atom in the Lewis acid. It might curve from a lone pair on an oxygen atom to a boron atom.
If we follow through to show the product of the reaction, we would find that a new bond is formed. The new bond used to be a lone pair, but now it is shared between the Lewis base and the Lewis acid. The curved arrow shows the direction of electron flow. The curved arrow also highlights the transformation of that lone pair into a bond.
Something similar happens when the Lewis base is really an alkene nucleophile. The pi bond in the alkene is transformed into a new sigma bond.
Curved arrows don't just illustrate bond formation. They can also illustrate bond breaking. In a proton transfer, a sigma bond might be converted into a lone pair. That's the opposite of what we saw in the Lewis acid-base reaction.
Because curved arrows illustrate transformations between non-bonding and bonding electrons, it's important to show the electrons in a mechanism. That means showing lone pairs on heteroatoms such as oxygen and nitrogen. When we do so, it underscores the changes that are occurring in that step. It makes the reaction pathway more clear on paper, and it also reinforces the physical changes in the structure of the molecule over the course of the reaction.
Working with Dative Bonds
A Lewis acid-base reaction results in a Lewis adduct. In the Lewis adduct, one partner has donated a lone pair to the other partner, forming a bond. If the Lewis acid and the Lewis base are both neutral (uncharged), then there will be formal charges on the Lewis acid-base adduct. The Lewis base has shared an entire pair of electrons to form a bond, rather than just contributing one electron to form a regular covalent bond. It will have a formal positive charge. The Lewis acid will contribute no electrons of its own, but will still gain a bond. It will have a formal negative charge. The adduct is a zwitterion: neutral overall, but containing both a positive and a negative formal charge on different atoms.
Formal charges tell us something real about the structure, because there really has been a net transfer of electron density from the Lewis base towards the Lewis acid. However, there is an alternative way of showing this arrangement, using a dative bond formalism. That requires another arrow. This one is short and straight, and points from the lone pair on the neutral donor attom toward the acceptor atom.
The dative bond arrow has something in common with the curved arrow. It illustrates the transfer of electron density from a lone pair toward an electrophile. However, it is meant to be read as static, rather than dynamic. It is structural, rather than reactive.
The advantage of the dative bond formalism is simply that it obviates the use of formal charge. That can be useful in more complicated molecules that would otherwise have lots of formal charges on different atoms. In addition, it reminds us of the transient nature of a dative bond, which usually form reversibly.
There's one more fine point about dative bond formalisms. Usually, the short arrow is only used to illustrate electron donation from a neutral donor. The bond from an anionic donor to an electrophile is generally drawn as a simple straight line, like any other bond. That distinction is sometimes helpful in underscoring the different types of ligands in a more complicated coordination complex.
As a result, if we were to illustrate a proton transfer involving an activated, neutral molecule, the short arrow in the neutral donor would shift to a straight line in the anionic donor.
Radicals and Single-Electron Transfers
Some reactions do not involve the movement of an electron pair, and so we need a symbol that conveys the idea of a single electron participating in a chemical change. Instead of the refular curved arrow that suggests the movement of two electrons, we use a similar arrow having only half an arrowhead. It's more like a fish hook.
In a single electron transfer, one electron is exchanged from a position on one atom or molecule to a position on another. We can show that using the single-electron arrow.
Note that there is a different consequence here than in a two-electron process. A two electron process typically results in the transformation of a lone pair into a bond, or vice versa. A single electron process does not. A bond requires two electrons, so when a single electron moves from one place to another, we are not forming or breaking bonds.
However, there are some one-electron processes that do entail bond formation or bond cleavage. Radical reactions involve unpaired electrons that pair up to make a bond. They might also involve bonds that break to produce unpaired electrons.
Radical reactions are often very different from the typical polar reactions such as acid-base reactions. Rather than being driven by electrostatic attractions, they are motivated by electron pairing and relative bond strengths. So, we might do something that would seem counter-intuitive if we are used to dealing with polar processes. In forming a bond, we would show two of those fish hook arrows coming together; the electrons are not repelling each other because they are becoming spin-paired.
In breaking a bond, we would show two arrows diverging in opposite directions. The electrons that were spin-paired in the bond are retreating, each to a different side of the original bond.
In polar reactions, we never really have arrows moving in opposite directions like that. They are drawn towards a positive position, so they move in the same direction. In radical reactions, to fish hooks meeting together signals bond formation, whereas two fish hooks leading away from each other signals bond breaking.
3.04: Solutions t
a) $\textrm{formal charge} = \# e^{-} \textrm{ in periodic table} - \# e^{-} \textrm{nonbonding} - \frac{\# e^{-} \textrm{in bonds}}{2}$
$\textrm{formal charge} = 4-1- \frac{6}{2} =0 \nonumber$
b) $\textrm{formal charge} = \# e^{-} \textrm{ in periodic table} - \# e^{-} \textrm{nonbonding} - \frac{\# e^{-} \textrm{in bonds}}{2}$
$\textrm{formal charge} = 5-3- \frac{4}{2} =0 \nonumber$
c) $\textrm{formal charge} = \# e^{-} \textrm{ in periodic table} - \# e^{-} \textrm{nonbonding} - \frac{\# e^{-} \textrm{in bonds}}{2}$
$\textrm{formal charge} = 6-5- \frac{2}{2} =0 \nonumber$ | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/V%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_3/03%3A_Understanding_Mechanism/3.03%3A_Arrow_Conve.txt |
The earth is an oxygen-rich place. Elemental oxygen, O2, makes up about 20% of the earth's atmosphere by weight. Oxygen atoms, bound in water, make up about 85% of the hydrosphere (the earth's oceans, rivers and lakes). Oxygen atoms also comprise about 45% of the lithosphere (the earth's crust), mostly bound in the form of silicates, aluminosilicates and carbonates.
In this chapter, we will focus on elemental oxygen and the ways nature has evolved to make use of O2 as a reactant. Dioxygen has a double nature, like Dr. Jeckyll and Mr. Hyde. It is necessary for life, providing a driving force for basic metabolic processes. That driving force ultimately comes from the exergonic conversion of O2 into water. However, nature has also had to evolve mechanisms to protect itself against reactive oxygen species, which would otherwise cause permanent damage to biomolecules. Those reactive oxygen species arise from the reduction of atmospheric oxygen. They also have dual roles, acting as important cell signalling molecules as well as dangers to the cell, and so their regulation is crucial in biochemistry.
Although da Vinci noted in the 1400's that air contained several components, one of which could support combustion, in a sense oxygen was not "discovered" until the turbulent 1770's. It was first isolated from air in experiments performed by Carl Wilhelm Scheele, a German pharmaceutical chemist (although his homeland of Pomerania was at that time part of Sweden, which extended to the south shores of the Baltic sea). This discovery was soon followed by an independent one by the English scientist and fire-and-brimstone preacher, Joseph Priestley. However, early research on oxygen is often associated with Antoine Lavoisier, whose experiments built upon those of Scheele and Priestley and are generally considered to mark the beginnings of modern chemistry.
These men did not lead long, peaceful lives. Priestley was forced to emigrate to Pennsylvania because of his troublemaking ways, Lavoisier was unfortunately executed during the reign of terror following the French Revolution, and Scheele's habit of tasting his experiments led to an early death from kidney failure. In contrast, a fourth scientist who is closely associated with oxygen, Henry Cavendish, died a wealthy (but remarkably reclusive) man of nearly eighty, having achieved some fame in England for his discoveries. Cavendish established the exact proportion of oxygen and nitrogen in the air, demonstrated in an impressively explosive way how oxygen and his other discovery, hydrogen, could be combined to make water, and even found time to accurately weigh the planet Earth.
Oxygen, present at the birth of modern chemistry, continued to make appearances in the development of the field. In the 1840's, Michael Faraday, the great English chemist and physicist, first demonstrated the O2 is paramagnetic, meaning it is attracted to a magnetic field. Sixty years later, American chemist and physicist Robert Mulliken explained this property through "molecular orbital theory" (sometimes called Hund-Mulliken theory after its developers). This explanation should be familiar to college-level chemistry students.
The molecular orbital interaction diagram for O2 shows how two oxygen atoms would combine under symmetry constraints to produce new orbitals in an O2 molecule. The net four electrons in bonding levels (i.e. at lower energy than they were in the atoms, as opposed to antibonding levels, which are at higher energy than they were in the separate atoms) suggests two bonds, since a bond between two atoms generally has two electrons. This prediction is also predicted by simple Lewis theory (i.e. that's what you would draw in a Lewis structure).
However, in contrast to the Lewis structure, in which all of the electrons in O2 are paired, Mulliken's MO picture suggests that there are two unpaired electrons. Because species with unpaired electrons are attracted to magnetic fields, this picture provides a compelling explanation for the paramagnetic behaviour of elemental oxygen.
This species is called a triplet state. A triplet is a state in which there are two net spins in the molecule, which has consequences in spectroscopy. If there were no net spins (i.e. they are all paired), the state would be described as a singlet. If there were one net spin (all other spins paired), it would be called a doublet.
Exercise \(1\)
If O2 accepts one electron, it forms a species called "superoxide" ion.
1. Draw the MO diagram for the species formed when O2 accepts one electron.
2. Is this new species a singlet, doublet or triplet?
3. How would you describe the net number of bonds (the bond order) in this species?
4. Draw a Lewis structure for this species.
Answer
Exercise \(2\)
If O2 accepts two electrons, it forms a species called "peroxide" ion.
1. Draw the MO diagram for the species formed when O2 accepts two electrons.
2. Is this new species a singlet, doublet or triplet?
3. How would you describe the net number of bonds (the bond order) in this species?
4. Draw a Lewis structure for this species.
Answer
In the 1930's, other forms of O2 were discovered by spectroscopists. By monitoring the interaction of light with oxygen, they detected evidence for different electronic states, meaning different arrangements of electrons. W. H. J. Childs and R. Mecke observed a singlet state in which two electrons appeared to be unpaired but with opposite spins. According to Hund's rule, this a higher energy state, because the electrons in singly-occupied orbitals should have the same spin. This state is described by spectroscopists as the 1Σg+ state. The usual form of O2 is described as the 3Σg- state. The 1Σg+ excited state is about 25 kcal/mol higher in energy than the 3Σg- ground state.
Gerhard Herzberg, the German-Canadian Nobel Laureate, observed a second singlet state. In this case, the electrons are spin-paired within the same orbital, also in violation of Hund's rule, since an empty orbital exists at the same energy level. This form of O2 is described as the 1Δg state. It is about 35 kcal/mol higher in energy than the ground state.
Although they are clearly higher in energy than the usual ground state, these two singlet states can arise when an oxygen molecule absorbs extra energy. Singlet oxygen undergoes reactions that are different from the usual triplet oxygen and it is considered a reactive oxygen species; it is linked to LDL cholesterol oxidation and cardiovascular damage, for example.
There is one last historical development in oxygen chemistry that we'll look at here. In the 1960's, Estonian-American chemist Lauri Vaska demonstrated the reversible binding of O2 to a coordination complex, trans-[Ir(CO)Cl(PPh3)2]. This observation was made when Vaska was a research scientist at the Mellon Insitute, now part of Carnegie-Mellon University in Pittsburgh.
Exercise \(3\)
Show a mechanism, with arrows, for the binding of oxygen to Vaska's complex.
This last example is particularly important in this chapter. The binding of oxygen by a transition metal is, of course, an essential step for the survival of every vertebrate. However, haemoglobin is a very complicated molecule. At the time that Vaska made his discovery, there was a need to demonstrate things like how an oxygen molecule could bind to a metal. How did it attach? Did it bind though a lone pair or a pi bond? In what geometry did it bind? Was the Fe-O-O bond linear? Was it angular or bent?
Frequently in the field of bioinorganic chemistry, simple coordination complexes serve as models for the complicated tasks carried out by transition metals in biology. "Model compounds", as they are called, provide the kind of precedent upon which researchers start to build a picture of what is happenning within the complex environment of the cell. That kind of precedent may take the form of structural studies, in which particular coordination environments and geometries may indicate what may be possible in a metalloenzyme. It may also involve reactivity studies, in which a small complex believed to resemble the active site of a metalloprotein is shown to undergo reactions that may be relevant to biological processes.
Visit an overview of oxygen from a biochemical perspective at Henry Jakubowski's Biochemistry Online. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/V%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_3/04%3A_Oxygen_Binding_and_Reduction/4.01%3A_Introd.txt |
Oxygen is vital to life. Very small organisms can get enough oxygen passively from their surroundings, but larger, more complicated organisms need to have better mechanisms for getting oxygen to the cells. Medium-sized organisms such as insects can manage to pump air to their tissues via a system of tubes leading in from pores along their bodies. Organisms bigger than that need a more complicated circulation system involving arteries and veins. Oxygen dissolves pretty well in water, but we can get even more oxygen into our system by binding it to carrier molecules.
The most common carrier molecule for oxygen, used by vertebrates like us, is hemoglobin. Hemoglobin contains a five-coordinate Fe(II) centre in a heme or porphyrin ligand. In the picture, only the coordination complex is shown, stripped of the surrounding protein. Also, there are other groups attached to the porphyrin (the nitrogen-containing ring) but they are left out of the picture for simplicity.
Oxygen binds to the iron in the heme, forming an octahedral iron complex. This form is called oxyhemoglobin; the form without the bound oxygen is called deoxyhemoglobin. Lots of interesting things happen as a result of oxygen binding, structurally speaking. First of all, the heme changes shape. In order to accommodate the change from a pseudo-square planar geometry to an octahedral one, the shape of the heme changes from a distorted bowl to a plane.
Sometimes, keeping track of oxidation states in coordination complexes is easier if using dative bond formalisms. In particular, if a donor arom is neutral, the bond to the metal is shown using a dative bond symbol. That's a short, straight arrow from the donor electron pair to the metal. Bonds between anionic donor atoms and the metal are shown as regular short lines, as we typically draw other bonds.
Looking at the complex that way, it is easier to see that the iron atom is depicted as Fe(II); it has two anionic nitrogen donors from the heme ring. We'll look into the situation more closely later.
Exercise $1$
Draw d orbital splitting diagrams for the iron porphyrin centre in deoxyhemoglobin and in oxyhemoglobin.
Hemoglobin is exceptionally good at transporting oxygen to the tissues not only because it can bind iron tightly under the right conditions, but because it can also let go under the right conditions, releasing oxygen to the tissues. Because hemoglobin is such a complex protein, it has been very difficult to study, although researchers have made impressive strides in undertsanding proteins in recent years.
Instead, bioinorganic chemists have developed model compounds to gain insight into hemoglobin and other important biological compounds. Model compounds are relatively simple compounds that possess a number of characteristics of their more complicated cousins. For example, simple porphyrins are relatively easy to make; if you wanted to study oxygen binding with a simple example, an iron-porphyrin complex would be a great model complex.
The trouble is, that doesn't work very well. Such a complex binds oxygen irreversibly; it never lets go. Part of the problem is that the oxygen "bridges" to other iron-porphyrin complexes, which wouldn't happen in hemoglobin. In hemoglobin, the heme is buried and protected within the protein.
Exercise $2$
Show the complex that would result if an oxygen molecule bridged between two iron porphyrin complexes.
The laboratory of James Collman at Stanford University has been involved in modelling hemoglobin for decades. They succeeded in demonstrating reversible oxygen binding to the "picket fence" porphyrin complex shown below. The bulky tert-butyl groups serve to keep the bound oxygen from bridging to another complex.
That alone is an interesting result. It demonstrates that one of the many roles for the protein in this system is to sterically protect the iron heme complex, modifying its reactivity.
There is just one small problem. It's called the M value. The M value is an index of the discrimination between oxygen binding and carbon monoxide binding. It's actually the ratio of the partial pressures of the two gases needed to half-saturate the hemoglobin (i.e. so that 50% of iron atoms have bound O2 or CO).
$M - \frac{P_{1/2}^{CO}}{P_{1/2}^{O_{2}}} \nonumber$
The lower the M value, the greater the favorability for O2 binding compared to CO binding. In hemoglobin, this value is about 100, although it will vary from one organism to another. That means hemoglobin binds CO about 100 times better than it binds O2. However, the picket fence heme has an M value over 25,000. That means it is much poorer at binding O2, relatively, than hemoglobin.
Why worry about CO binding in these studies? Of course, carbon monoxide poisoning is a serious and potentially fatal condition. There is a deeply problematic consequence of CO poisoning, however, that could be much worse if the M value in an organism was as high as in the picket fence porphyrin. CO is actually a product of the normal breakdown of heme molecules over time in the cell. If our hemoglobin had an M value like that of the picket fence porphyrin, we would all be dead, poisoned by our own metabolic processes.
Not content to rest on their laurels, the Collman lab went back to the drawing board and developed lots of other model complexes. For example, the one shown below has an M value closer to 0.005.
What does that tell us about hemoglobin? It may be nothing, but it could be indicating another role for the surrounding protein in the hemoglobin molecule. The domed or vaulted model compound suggests a protective covering for the oxygen binding site. Is it possible that O2 can fit inside but CO cannot?
It is pretty well-established via other model studies, as well as direct study of oxyhemoglobin, that when oxygen binds to a metal such as iron, the Fe-O-O forms an angle of somewhere around 120 °. The complex has a bent geometry. However, when CO binds to a metal, it does so in a linear fashion. It may be that in a vaulted model complex, the CO simply can't stand up straight, so the complex is destabilized. Presumably, the protein could contribute to a similar destabilization of CO-bound hemoglobin.
Exercise $3$
Why might O2 bind in a bent fashion whereas CO bounds in a linear mode?
This idea has been somewhat controversial. Results from quantum mechanical calculations, for instance, suggest that the Fe-C-O bond actually has a fair amount of leeway. These results suggest that the CO can be "tipped over" and still remain strongly bound. Nevertheless, Collman's results provided a useful starting point for further investigations.
The protein may play other roles in enhancing oxygen selectivity. X-ray crystallographic studies suggest a role for hydrogen bonding between the "distal histidine" site and bound oxygen. The distal site simply refers to a second nearby histidine, other than the one that is bound to the iron (the "proximal histidine"), and a little further away. That histidine could be ideally situated to hydrogen bond with bound oxygen, but out of place for optimal interaction with a bound CO.
There is one more important event to look at in oxygen binding. Evidence indicates that oxyhemoglobin is actually an Fe(III) species, rather than Fe(II). The iron is oxidized by the bound oxygen.
As always, it is useful to map out the movement of electrons, schematically, in this event. Because this event would be a one electron oxidation, we need a different kind of arrow to show where the electrons go. Most of our previous mechanisms have involved electron pairs rather than single electrons. For single electrons, we show a single-headed arrow, rather than a double-headed one.
Here is one way we could show the oxidation of the iron (and the reduction of the oxygen):
Or alternatively, we could show it like this:
Remember, in the structure on the right, the oxygen bound to iron is considered an anionic donor, shown with a regular line bond instead of a dative arrow.
There is something really amazing about that last event. Hemoglobin doesn't form an oxygen complex at all. It forms a complex with superoxide ion, O2-, which it has manufactured itself. When it is ready for delivery to the cells, the superoxide gives back the electron to the iron, and turns back into an everyday oxygen molecule.
Exercise $4$
Describe in words what the curved arrows are showing in the above two schemes.
Exercise $5$
Explain why the oxidation of iron would lead to tighter binding of the oxygen.
Exercise $6$
Modeling studies of Cu(I) complexes like the one below reveal that exposure to O2 results in a square planar Cu(III) peroxide complex (Tolman et al, J. Am. Chem. Soc. 2006, 128, 3445-3458 and references therein).
a) Draw the product of the reaction.
b) A similar copper complex prepared with the following ligand also binds O2, but with a much lower equilibrium binding constant. Draw the copper complex and the O2 adduct and explain the difference in O2 binding constants.
c) DBM (dopamine β-monooxygenase) and a similar copper monooxygenase, PHM (peptidylglycine α-hydroxylating monooxygenase), both contain copper atoms bound to two histidines, a cysteine and a water. Explain the researchers' choice of model compounds above in this context.
d) Explain how the cysteine in these active sites might help to control OH radical levels in the cell. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/V%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_3/04%3A_Oxygen_Binding_and_Reduction/4.02%3A_Oxygen.txt |
Molecular oxygen is exploited in a variety of ways in biology. Apart from its central role in metabolism, it also plays a crucial part in the modification of many compounds through the incorporation of oxygen atoms into the structure.
For example, the cytochromes P450 form an important class of enzymes involved in the oxidation of substrates, chiefly in the liver. One atom of oxygen from an O2 molecule is incorporated into the substrate molecule; the remaining oxygen atom is converted to water. This type of enzyme is called a "monooxygenase" because of the addition of one oxygen atom from O2 into the substrate.
The addition of oxygen atoms to molecules, typically in the form of hydroxyl groups, is vitally important. The reaction may have evolved for a number of reasons. A key reason is to increase the water-solubility of small organic molecules, which are then more easily excreted via the urine or broken down via subsequent reactions. Hydroxylation lets us get rid of foreign substances. Otherwise, these hydrophobic molecules would build up in the tissues.
The cytochrome P450 pathway is a major avenue for the breakdown and excretion of pharmaceuticals, for instance. We can get rid of these substances after they have done their job. In other cases, pharmaceuticals are not active until they are hydroxylated; the reaction acts as an "on" switch. In still other cases, hydroxylation is a dangerous complication, converting a helpful pharmaceutical into a toxin.
Exactly how does the O2 get broken into pieces suitable for incorporation into other molecules? It's a very complicated problem. There is an entire field of chemists who study "small molecule activation", which refers to the breakdown of things like O2, N2, CO or methane for their subsequent conversion into other compounds. In fact, there is an entire field of chemists and biochemists who just focus on cytochrome P450, trying to learn more about how it works. They do so both to understand more about a biological system that is related to human health and to gain insight into how to improve industrial processes involving oxidation of substrates.
Some things are pretty well understood. We'll take a general look, leaving out some important details so things aren't too overwhelming.
The first step is just the binding of dioxygen to a metal. In cytochrome P450, that iron atom looks very much like the one in hemoglobin. In its resting state, it is formally an Fe(III) ion in a porphyrin ring, but with an axial cysteine donor instead of a histidine. In addition, a water molecule is coordinated to form an octahedral complex.
That leaves no place for the dioxygen to bind. However, once the substrate enters the enzyme, a conformational change results in loss of the water molecule. After that, an electron is delivered from a cofactor, leading to an Fe(II) complex.
At this point, things are looking a little more like hemoglobin. Just as in hemoglobin, the O2 binds to the iron, which immediately transfers an electron to the bound dioxygen, forming an Fe(III) superoxide complex.
The difference is that things do not stop there. The addition of a second electron from a cofactor tips things a little further. One more electron turns the superoxide ion into a peroxide ion.
At this point, we have completely severed the first of the two O=O bonds. We are almost there. The subsequent addition of two protons leads to the formation of a water molecule. Now we have one oxygen atom bound to iron.
How do we think about that species? We can think about several resonance structures. The electron-deficient oxygen is attached to an iron atom. The iron possesses a reservoir of electrons. It can donate one to the oxygen atom. It might even donate two electrons, making an iron (V) oxide complex, but that might be going too far.
In the resonance structure with iron (IV), we see an oxygen atom with a single, unpaired electron. This species is called a radical. Radicals are notorious for ripping hydrogen atoms from other molecules. That just makes a new radical. This is the trouble with radicals; they are hard to get rid of, because they are always making more. It also makes an iron hydroxy complex, which can combine with the radical in a "rebound" step to form the hydroxylated compound.
We will see more of this kind of event in the chapter on radicals.
Exercise \(1\)
Fill in the missing intermediates in the following scheme.
Exercise \(2\)
The Karlin lab (Johns Hopkins) has developed a series of model compounds for copper-containing monooxygenases (J. Am. Chem. Soc. 2014, 136, 8063-71).
a) Fill in a reasonable starting material that you might find in the chemistry stockroom.
b) Reaction with O2 gives dimeric copper complexes (LnCu2) of two types: with one bridging peroxide anion or with two bridging oxide anions. Draw both dimers.
c) O2 is reduced / oxidized (select one) by _____ (number of ) electrons to make peroxide anion.
d) O2 is reduced / oxidized (select one) by _____ (number of ) electrons to make oxide anion.
e) Identify the oxidation states of the copper ions in the two dimers.
f) In the ligand, if the group ZR = SEt, the reaction forms the peroxide bridge. If ZR = OEt, it forms the oxide bridge. Suggest a reason why.
g) With Z = S, the group R both influences which dimer forms and affects the Cu-S bond. Predict the combination of features found in each product and provide an explanation.
R = Et peroxide / oxide Cu-S bond / no Cu-S bond
R = Ph peroxide / oxide Cu-S bond / no Cu-S bond | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/V%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_3/04%3A_Oxygen_Binding_and_Reduction/4.03%3A_Oxygen.txt |
In the previous section, we saw the Fe(IV) oxo group of Cytochrome P450 do something unusual. It grabbed a hydrogen atom from a substrate molecule. It did not take a proton. It took the hydrogen atom complete with its electron.
This type of reactivity is characteristic of radicals. Radicals are species that have unpaired electrons. Generally, but not always, they have odd numbers of electrons in their valence shells, leaving one electron without a spin partner. Having an odd number of electrons also leaves these compounds short of a full octet. From that point of view, it is easy to see the driving force for a reaction, from the point of view of the radical species.
However, this reaction is unusual. There are lots of metal oxo compounds on earth, but only some of them engage in this kind of radical behaviour. What makes some oxos different, in particular some biologically important iron oxos?
Let's take another look at that active form of the iron oxo complex in Cytochrome P450.
Earlier, we decided to think about this complex using the Fe(IV) structure. That would be a d4 iron complex. The coordination geometry appears to be octahedral. Chances are that in such a high oxidation state we would have a low-spin iron. Overall, that should give the frontier orbital diagram shown below.
The oxo donor would have lone pairs. It would be a π donor. The same is true of the cysteine ligand trans to the oxo. We should probably modify our d orbital splitting diagram accordingly.
Remember, a π donor brings a pair of electrons from a relatively low-lying ligand orbital. The donor orbital interacts with one of the metal d orbitals that were previously non bonding. In-phase and out-of-phase combinations result. The in-phase combination is lower in energy and the out-of-phase combination is higher in energy than the levels we started with. Because the π donor is donating a lone pair, that lone pair drops in energy. It becomes a π bond. The electrons in the metal orbital involved in the interaction become anti bonding. Overall, the gap between the lower and upper d levels gets smaller.
At least, that's how it would work in a normal case. In a simple example of a coordination complex, all six ligands might be the same. For example, if we had an octahedral Fe(IV) chloride complex such as FeCl62-, all six chloride ligands would be potential π donors. Different p orbitals on the chlorines would be able to interact with each of the three non bonding d orbitals, raising them all in energy via a π bonding interaction.
Note the choice of different orbitals to illustrate pi dontaion along different axes. In two cases, the orientation of the oxygen p orbital may have changes, or the metal d orbital may have changes, or both. There is a need for symmetry matching in these interactions.
Look at a chlorine attached to the right of the metal (along the x axis, at least as defined by the orbital labels above). It would not be able to form π overlap with the dyz orbital. The symmetry isn't right.
In our example of Cytochrome P450, things are different from the example in which all of the ligands were the same. In Cytochrome P450, only the oxygen and the sulfur can donate. As a result, one of the non bonding d orbitals is not participating in π bonding. That's because the π donors are unable to interact with one of the d orbitals. It is too far away and it has the wrong symmetry to overlap with either of the oxygen p orbitals that can participate in the π bond.
The same would be true for the sulfur atom, since it is located along the same axis as the oxo ligand.
The net result is different from the π donor picture we have seen before. Because the π donation is restricted along one axis, not all of the non bonding d orbitals are elevated to π* levels. Only two of them form this interaction. The third remains unchanged.
The overall frontier picture of an Fe(IV) oxo reveals two unpaired electrons in the π antibonding level. As a result, the pi bond may be considered relatively weak compared to some other metal oxo complexes. That may result in increased reactivity. In contrast, a d0 or d2 oxo is usually much more inert than a d4 oxo.
4.05: Soluti
Exercise 4.1.1:
Exercise 4.1.2:
Exercise 4.3.1: | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/V%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_3/04%3A_Oxygen_Binding_and_Reduction/4.04%3A_Metal_.txt |
Nitrogen is the most abundant element in earth's atmosphere. It makes up about 80% of the air around us. It is also a key component of biomolecules. Nitrogen contributes a crucial part of amino acids, which in turn make up proteins, which are the cell's machinery. Nitrogen also provides a crucial part of DNA, which transmits our genetic code and governs the expression of those proteins. Although less well known, many nitrogen-containing natural products such as alkaloids play important roles in biology.
Getting that nitrogen out of the air and into the cell is a Herculean task. Nitrogen in the air is present in its elemental form, which is diatomic nitrogen or dinitrogen, N2. The nitrogen in biomolecules is always found individually; it is always bound to other atoms, especially carbon and hydrogen, but never to other nitrogens. That means the two nitrogen atoms in dinitrogen atom have to be cleaved apart so that they can be combined with other atoms in these useful molecules. That's a problem. N2 is exceptionally stable. Breaking the bond between the two nitrogen atoms costs about 225 kcal/mol. Most of the other bonds in the universe are not nearly so strong.
So how do we break that incredibly strong bond and combine it with other atoms to make molecules? For most of our history, we didn't (we being people). Neither could other animals. Neither could plants. Microbes could do it all along, though.
Certain bacteria, called diazotrophs, contain an enzyme called nitrogenase that can catalytically convert dinitrogen to ammonia. Some diazotrophs, called rhizobacteria, have a symbiotic relationship with specific kinds of plant roots. They provide ammonia or amino acids to the plant and the plant provides them with organic compounds such as malate that can be metabolised to obtain energy.
$\ce{N2 + 6H^{+} + 6 e^{-} -> 2NH3} \nonumber$
All plants need nitrogen to grow, but by evolving to have this symbiotic relationship with rhizobacteria, legume plants have found a distinct advantage over others.
Other bacteria actually live on ammonia. They use it for metabolism the way other organisms use carbohydrates. In doing so, they oxidise the ammonia to nitrites the way other organisms oxidise carbohydrates to carbon dioxide.
$\ce{NH3 + O2 -> NO2^{-} + 3H^{+} + 2e^{-}} \nonumber$
There are even bacteria that oxidise the nitrites.
$\ce{2No2^{-} + O2 -> 2NO3^{-}} \nonumber$
Exercise $1$
Calculate the oxidation state of nitrogen in each of the following compounds.
a) N2 b) NH3 c) NH4+ d) NO2- e) NO3-
Answer a
a) 0
Answer b
b) 3-
Answer c
c) 3-
Answer d
d) 3+
Answer e
e) 5+
Exercise $2$
The oxidation of nitrite must involve electron transfer. Identify:
1. the oxidant
2. the reductant
3. the number of electrons transferred from reductant to oxidant
Answer a
a) O2
Answer b
b) NO2-
Answer c
c) 2 e-
All of these compounds -- ammonia, nitrites, and nitrates -- can be used by plants as sources of nitrogen. People, who have been farming for thousands of years, have developed a number of methods of making sure their crops had plenty of these nutrients available. There are familiar stories of different Native American groups planting corn and squash along with rhizobacter-nurturing beans. In South America, Andean farmers collected guano for their fields.
At the turn of the twentieth century, a great leap forward came with the discovery that ammonia could be manufactured directly from nitrogen on a massive scale. This technology, called the Haber-Bosch process, is responsible for a significant portion of the nitrogen found in food.
The focus of this chapter is on the processes that have allowed the cleavage of the nitrogen bond, by both biological and industiral means.
See the section on nitrogenase at Henry Jakubowski's Biochemistry Online. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/V%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_3/05%3A_Nitrogen_Reduction/5.01%3A_Introduction_to_.txt |
The Haber-Bosch Process is one of the world's most important industrial reactions. It provides for the synthesis of ammonia directly from elemental nitrogen, N2, and hydrogen, H2. Since its development in the early twentieth century, it has led to the production of an enormous quantity of fertilizer, vastly increasing global food production. As a result, it is estimated that a significant fraction of the nitrogen content in the typical human body is ultimately derived from this process.
At the time of its development, the Haber-Bosch Process supplanted a growing dependence on guano, seabird droppings, harvested in order to enrich farmland. Although this method had been practiced by the Inca for centuries, European demand in the late 1800's placed growing pressure on resources in Peru and the Caribbean. Because nitrogen is a limiting factor for plant growth, treatment of soil with nitrogen- and phosphorus-rich guano led to remarkable improvement in crop yields.
In the late 1700's, Henry Cavendish had been able to produce nitrates with an electric arc in air. A Norwegian development, the Birkland-Eyde Process, harnessed the hydroelectric power available in that country in order to scale Cavendish's feat to industrial production, but the reaction was still terribly inefficient. Fritz Haber was able to develop a much more successful procedure for ammonia production (the reduced state, rather than the oxidized nitrates), which was adopted by BASF. Originally the local Baden Aniline and Soda Factory, this German company is now the world's biggest chemical producer.
Haber himself was a fascinating and controversial figure who hoped to improve the human condition through industrial contributions to agriculture; he also developed pesticides. However, he was roundly condemned for developing and implementing chlorine gas against Allied troops during World War I. Although Haber, who was Jewish, died before the Holocaust, many members of his extended family were murdered in concentration camps; in some of these camps, internees were executed by poison derived from Haber's pesticides.
The Haber-Bosch Process is an example of heterogeneous catalysis. Many catalysts operate via homogeneous catalysis, in which the reaction occurs in one phase (in solution). In heterogeneous catalysis, the reaction occurs at the interface between two phases. In this case, the two phases are the gas phase and the solid phase. The reactants, hydrogen and nitrogen, are both gases. They are both introduced into a vessel under enromous pressure and high temperature. The catalyst is a carefully prepared iron oxide supported on a mixture of other metal oxides, although ruthenium and osmium variations have also been used and offer some advantages.
In this case, the reaction involves gas phase reactants and a solid catalyst. The reaction takes place on the surface of the catalyst. We might depict such a surface as a row of atoms. The reaction takes place only on the surface of the atoms (for example, along the top of the picture below). Rows of atoms beneath the surface would not necessarily contribute to the reaction. In the picture below, the second row of atoms might play no role at all in the reaction, other than to provide a place for the top row of atoms to sit.
It might be more convenient to just think of the reaction happening on a flat surface of metal, ignoring its makeup of atoms.
So, molecules come along and they react on that surface. The mechanisms by which molecules react at the surface may be familiar to us. We might find parallels in organometallic chemistry. The surface is, after all, made of metal atoms.
Hydrogen molecules probably become activated through oxidatve addition.
Nitrogen molecules must bind at the surface, too.
The nitrogen is depicted as "end-bound" above, using its lone pair to attach to the metal. It is also possible that the molecule leans over and becomes "side-bound".
How do the nitrogen and hydrogens react further to make ammonia? There are different possibilities. For instance, a 1,2-insertion reaction seems possible.
After that, a reductive elimination would result in formation of a second N-H bond.
That process, so far, results in the conversion of dinitrogen to diazene, N2H2. The diazene could react further to make hydrazine (N2H4) and, ultimately, ammonia.
Of course, the reaction could happen in other ways, too. There are surface studies that suggest the presence of nitrides, imides and amides (M=N, M=NH, M-NH2). The presence of nitrides suggests a series of oxidative additions starting with dinitrogen, proceeding all the way to nitride, N3-.
Once the nitride is in place, N-H bonds could form via reductive eliminations. The first reductive amination would result in an imide, the second would produce an amide, and the third would produce ammonia. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/V%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_3/05%3A_Nitrogen_Reduction/5.02%3A_The_Haber-Bosch_.txt |
A number of organisms take part in the nitrogen cycle, but from the point of view of chemistry, the diazotrophs have received the most attention. That's partly because their conversion of atmospheric nitrogen to ammonia mirrors the twentieth century marvel of the Haber-Bosch Process, which has been responsible for a tremendous increase in global food production over the last century.
The ammonia-manufacturing center of the diazotroph is nitrogenase. The picture below is taken from an X-ray crystal structure of the enzyme (reference cited at the bottom of the page). Rather than showing the individual atoms, as X-ray structures normally do (except hydrogens, which are too small to detect), the structure is depicted in cartoon form. The cartoon image of a protein is meant to convey key substructures: the pink parts are α-helices, the yellow ribbons are β-sheets, and the white threads are loops.
If we delete all of the protein from the image -- that is, the chain of amino acids that make up nitrigenase -- then we are left with a few "ligands". Ligands in this context are simply other molecules or ions found within the protein; often the ligands are bound to amino acid side chains. Looking at the ligands below, we can see that nitrogenase is actually a dimer. The group of molecules in the left half of the picture are the same as the group of molecules in the right half of the picture, although they are rotated upside down with respect to the others.
A couple of the structures in the picture above are metal ions; they look like little balls. To get a better view of the other structures, let's zoom in a little closer.
That's an iron sulfur cluster photobombing the picture. The red spheres are colour-coded for iron, whereas the yellow sphered are colour coded for sulfur. The bars between the atoms may or may not represent bonds; X-ray diffraction doesn't really detect bonds, just the highly electron-dense atoms, but the software inserts bonds when it detects atoms that are close together. The iron-sulfur bonds are certainly real. The red bonds shown between the iron atoms probably are not real, although there is a possibility of magnetic coupling between the iron atoms within the cluster. A drawing of the cluster is provided below. The drawing also includes additional sulfur donors on the iron atoms, but those are from cysteine residues in the protein, which has been rendered invisible in the picture above.
This particular cluster contains 8 iron atoms and 7 sulfide ions (S2-); it's referred to as an [8Fe7S] cluster for short. Iron sulfur clusters are actually pretty common in biology. Probably the most common is a [4Fe4S] cluster, although [3Fe4S] and [2Fe2S] clusters are often seen, too. Iron sulfur clusters are usually charged. The amount of charge on the overall clusters depends on the oxidation states of the iron atoms. The oxidation states can change because these clusters function as electron relays. They can accept an electron from elsewhere in the protein and send it on to where it is needed.
Exercise \(1\)
Indicate the charges on the following clusters.
1. [4Fe4S], with two Fe2+ and two Fe3+
2. [3Fe4S], with two Fe3+ and an Fe2+
3. [2Fe2S], with one Fe2+ and one Fe3+
Answer a
a) S: 4 x 2- = 8- ; CysS: 4 x 1- = 4- ; Fe: 2 x 2+ + 2 x 3+ = 10+ ; total = 2-
Answer b
b) S: 4 x 2- = 8- ; CysS: 3 x 1- = 3- ; Fe: 1 x 2+ + 2 x 3+ = 8+ ; total = 3-
Answer c
c) S: 2 x 2- = 4- ; CysS: 4 x 1- = 4- ; Fe: 2+ + 3+ = 5+ ; total = 3-
There are also a couple of histidines visible in the picture above. They were probably introduced when the researchers were growing the crystals. A far more interesting structure is barely visible on the left. That's actually the iron-molbdenum or Fe-Mo cofactor, also called the M cluster. It's the site of nitrogen reduction in the enzyme. A much clearer view of this structure is provided below. Because we are looking at a dimer, we can see a second M cluster behind.
A drawing of the M cluster is shown below. It bears some resemblance to [8Fe7S]. This time, the atom in the center is believed to be a carbon. A trio of additional sulfides bridge between the two cubes, and in one corner an iron atom has been replaced by a molybdenum.
Here is a picture of the M cluster from a different angle. The molybdenum with its attached ligand is in the upper left, coordinated by yellow sulfides. The gray atom in the middle is probably a carbon atom.
The mechanism by which nitrogenase reduces nitorgen to ammonia is likely somewhat different from the mechanism in the Born-Haber Process. Rather than treatment with hydrogen gas, under biological conditions the mechanism is likely to involve reduction via individual electrons and protons. Exactly how the electrons and protons arrive at their destination is an interesting question.
One possibility is that the dinitrogen is reduced one electron and one proton at a time. The FeS clusters would be able to deliver electrons via an electron relay running through the cell. Amino acid residues in the protein would be able to shuttle protons in as needed.
Exercise \(2\)
Provide a mechanism for the reduction of dinitrogen, N2, to diazene, HN=NH, via alternating additions of electrons and protons. Assume the following model is the site of the reaction, with appropriate electron and proton donors nearby.
Alternatively, the process could more closely resemble the Haber-Bosch Process, using classic "organometallic" reaction mechanism such as oxidative addition, insertion and reductive elimination. In fact, recent evidence suggests H2 may be produced by some nitrogenases under biological conditions. That observation is consistent with the operation of this type of mechanism.
Exercise \(3\)
Show how protonation of an iron atom is equivalent to an oxidative addition.
Answer
Exercise \(4\)
Provide a mechanism for the reduction of dinitrogen, N2, to diazene, HN=NH, via a series of organometallic steps (association/dissociation, oxidative addition/reductive elimination, insertion/elimination). | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/V%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_3/05%3A_Nitrogen_Reduction/5.03%3A_Nitrogenase.txt |
If nitrogen is to be reduced, it first has to be bound. However, nitrogen is remarkably inert. Chemists routinely run reactions under an atmosphere of pure nitrogen because of its lack of reactivity. Fine chemical companies bottle compounds under nitrogen to ensure that the contents remain in pristine condition while sitting on the shelf. If nitrogen doesn't react with anything, how does it react with an iron atom in nitrogenase?
What is the weakness of the dinitrogen molecule? That turns out to be related to its strength. The enormously strong nitrogen-nitrogen triple bond is composed of a sigma bond and two pi bonds. The corresponding antibonding orbital allows nitrogen to act as a pi acceptor.
That means there is a good chance of getting the very unreactive nitrogen molecule to bind to a transition metal. An occupied d orbital on an iron atom could back-donate into that nitrogen orbital, holding the dinitrogen more securely on the iron. The nitrogen can be bound.
Just to make that scenario more likely, the metal could be tuned in order to maximise its ability to backbond with the nitrogen. That means it needs a lot of electron density. An electron-rich metal atom would readily donate electron density into nitrogen's pi acceptor orbital. Then the nitrogen would bind more tightly.
One factor that would help is a low oxidation state. A low oxidation state on the metal would leave it with more electron density to donate to the nitroge pi acceptor orbital.
A secondary factor is strongly donating ancillary ligands. These other ligands would play a supporting role by lending additional electron density to the transition metal so that it could bind to the nitrogen even more tightly.
As with other metalloproteins, researchers have spent a great deal of effort studying nitrogenase. They have also expended a tremendous amount of effort studying model compounds. Model compounds are simpler molecules that incorporate selected aspects of the metalloprotein. By intentionally designing a model compound to include certain features of the metal centre in the protein, researchers can evaluate what role those features play in the reactivity of the metalloprotein.
It's pretty obvious that we might want a model compound for nitrogenase to contain iron atoms. After all, nitrogenase contains a number of iron atoms at its active site. Of course, it also contains molybdenum, or in some cases vanadium. A model compound might contain those atoms, instead. Alternatively, it could contain atoms other than the ones found in the native enzyme. That would be a sort of pushing-the-envelope approach. If an electron-rich metal is important, how electron rich can we go? Or how electron-poor can we get and still be able to bind nitrogen? By exploring things that aren't part of the natural system, we might better see the importance of those things that are.
The same is true with the ancillary ligands, those that support the metal but that may not be directly involved in catalysis. The electron-rich sulfides in nitrogenase may be an important part of a model compound. So could phosphines, whose phosphorus donor atoms are of a similar size to sulfur. Phosphines are commonly used industrially in organometallic catalysis and might make good mimics of the sulfur ligands in nitrogenase.
Exercise \(1\)
Consider the ligand type presented by a phosphine compared to a thioether. What might be the disadvantage of using a phosphine as a stand-in for a sulfur donor?
Answer
Having said all of that, it is worth emphasizing that binding nitrogen is still not easy. Sometimes, researchers who want to study the potential for nitrogen binding in a given complex start with binding carbon monoxide instead. Why carbon monoxide? First of all, it is much easier to bind than dinitrogen. It is a much stronger pi acceptor, because the pi antibonding orbital is much more heavily located on the carbon that sigma donates to the metal. Of course, you may already know that there are some important transition metal reactions that involve binding and reducing carbon monoxide.
Furthermore, carbon monoxide studies can be useful because carbon monoxide acts as a "reporter ligand". It is easily monitored by IR spectroscopy, for instance. Dinitrogen is a poor candidate for IR study because of the non-polar N-N bond. (It can be observed via Raman spectroscopy, which gives similar information but is slightly more complicated to run.) The CO bond is easily detected in the IR spectrum, it is in a region that isn't usually cluttered with other peaks from other bonds, and it is quite sensitive to the oxidation state of the metal. That's because of the strong back-bonding from a filled metal d orbital into the pi antibonding orbital of carbon monoxide. The more back-donation from the metal, the weaker the CO bond, resulting in a drop in the frequency in the IR spectrum.
Exercise \(2\)
Rank the following species in terms of their CO stretching frequency in the IR spectrum.
Exercise \(3\)
Nitrogen can bind to metals in a number of ways. Draw structures that illustrate the following binding modes:
a) an end-bound, terminal nitrogen ligand bound via a lone pair
b) a side-bound, terminal nitrogen ligand bound via donation from a pi bond
c) an end-bound, bridging nitrogen ligand bound via a lone pair
d) a side-bound, bridging nitrogen ligand bound via donation from a pi bond
e) a bridging nitrogen ligand bound via donation from a pi bond and a lone pair
Exercise \(4\)
In naming coordination compounds, the prefixes eta (η) and mu (μ) are sometimes used to indicate ligand binding modes such as the ones described above.
Eta describes the number of ligand atoms bound to a single metal atom, and is generally used when there are pi bonds that could donate, bringing two or more donor atoms close to the metal.
Mu is used to indicate a bridging ligand, and if followed by a number it can describe the number of metal atoms bridged by one ligand.
Use these notations to describe the nitrogen binding modes in the previous question.
Exercise \(5\)
Jonas Peters' lab (Caltech) has developed a new system in an attempt to model the effect of a reported carbon atom in the structure of Fe/Mo cofactor of nitrogenase (J. Am. Chem. Soc. 2014, 136, 1105-1115). The system catalytically produces ammonia in the presence of N2, acid and Na.
a) Fill in the oxidation states.
b) Fill in missing reagents.
c) Fill in d orbital splitting diagrams.
d) Explain the differences in the N-N stretching frequencies.
5.05: Model Studies fo
Model studies for nitrogen reduction seek to develop coordination compounds that can mimic the activity of nitrogenase, which converts atmospheric dinitrogen into ammonia or ammonium ion. Over the past few decades, there have been more and more reports of dinitrogen complexes, so researchers have clearly figured out some of the factors to accomplish that part of the reaction.
It's one thing to be able to bind a dinitrogen ligand to a metal center, but it's quite another thing to be able to conert that dinitrogen into ammonia. That strong N-N triple bond costs about 200 kcal/mol to break; that's a very high cost to pay and it isn't easy to do. The other part of nitrogen fixation involves formation of the N-H bond; this part of the process could actually be exothermic, and so it might be an easier problem to solve.Most studies aiming to make progress towards ammonia production start with reduction of a metal center so that it in turn will have sufficient reducing power to bind dinitrogen. Acids are then added to supply the protons needed to form N-H bonds. Proposed mechanisms of nitrogen fixation in nature involve a series of electron transfer and proton transfer steps, so the addition of acid seems like a reasonable way to model the process.
Exercise \(1\)
Assign the oxidation state on nitrogen in the following molecules:
a) dinitrogen b) diazene, N2H2 c) hydrazine, N2H4 d) ammonia, NH3
Exercise \(2\)
Draw a mechanism with curved arrows and intermediates showing:
i) sodium metal reducing Fe(II) to Fe(0)
ii) binding molecular nitrogen
iii) protonation to form a diazene complex, containing an H2N2 ligand.
Exercise \(3\)
Draw the following possible intermediates of nitrogen reduction:
a) a diazene complex b) a hydrazine complex
c) a diazenyl anion complex d) a hydrazinyl anion complex
e) a bridging hydrazine complex f) a bridging diazene complex (two structures)
g) a bridging hydrazinyl anion complex h) a bridging diazenyl anion complex (three structures)
i) a bridging hydrazinyl dianion complex (two structures) j) a bridging diazenyl dianion complex
5.06: Solutions for Se
Exercise 5.1.1:
a) 0 b) 3- c) 3- d) 3+ e) 5+
Exercise 5.1.2:
a) O2 b) NO2- c) 2 e-
Exercise 5.3.1:
1. S: $4 \times 2^{-} = 8^{-}$; Fe: $2 \times 2^{+} + 2 \times 3^{+} = 10^{+}$; total $= 2^{+}$
2. S: $4 \times 2^{-} = 8^{-}$; Fe: $1 \times 2^{+} + 2 \times 3^{+} = 8^{+}$; total $= 0$
3. S: $2 \times 2^{-} = 4^{-}$; Fe: $2^{+} + 2^{+} = 5^{+}$; total $=1^{+}$
Exercise 5.3.2:
Exercise 5.3.3:
Exercise 5.3.4:
Exercise 5.4.1:
d) As the Fe becomes more reduced, the N-N stretching frequency decreases. That's because the more electron density there is on the Fe, the more it is able to backbond to the N2 (N2 is a π acceptor). | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/V%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_3/05%3A_Nitrogen_Reduction/5.04%3A_Model_Studies_fo.txt |
Radicals are species that have unpaired electrons. They can be atoms or molecules and they can be neutral species or ions. Frequently, radicals are very reactive. However, their mode of reactivity does not fall neatly within the normal patterns of Lewis acids and bases, or nucleophiles and electrophiles.
Radicals play common roles in atmospheric chemistry, including equilibration of the ozone layer. They are also found in a variety of biochemical pathways. In addition, radicals are employed in a number of useful processes, such as the polymerization of methyl methacrylate or vinyl chloride, commonly used to make shatter-resistant "glass" and pipes for plumbing, respectively.
Compounds of p-block elements form radicals if one of the atoms has seven electrons in its valence shell rather than the usual eight.
Exercise \(1\)
Draw structures for the following neutral radicals, making sure to fill in the correct number of electrons.
a) Br b) OH c) CH3CH2 d) CH3CH2S e) CH2CHCH2 f) NO g) (CH3)2NO h) NO2
Answer
A molecule could become a radical in a number of different ways. A bond may break in half via the addition of energy, in the form of either heat or light. Otherwise, it may simply transfer one of its electrons elsewhere. Again, this event may be precipitated by the addition of heat or light energy. Of course, a molecule that receives an additional electron from elsewhere may also become a radical.
Exercise \(2\)
Draw structures for the following cationic radicals, making sure to fill in the correct number of electrons and the formal charge.
a) H2C=O b) CH3NH2 c) CH3OCH3 d) CH3CH2CH2Br e) CH3CH2CHCH2
Answer
Exercise \(3\)
Draw structures for the following anionic radicals, making sure to fill in the correct number of electrons.
a) O2 b) H2CO c) CH3CCCH3 d) cyclo-C6H6
The compounds above are all simple radicals, containing one unpaired electron. Compounds may also have more than one unpaired electron.
Elemental oxygen, O2, is a diradical. Although its Lewis structure does not suggest anything unusual, its molecular orbital diagram reveals that oxygen actually has two unpaired electrons.
Exercise \(4\)
Show two Lewis structures for O2: one illustrating its double bond, and the other illustrating its diradical character.
Answer
Exercise \(5\)
Show, with a molecular orbital interaction diagram, the diradical character of dioxygen.
Answer
A diradical could take two different forms. For example, molecular oxygen has two singly-occupied molecular orbitals. The single electron in each of those orbitals could adopt one of two different spin states. Both could adopt the same spin state (designated with arrow "up", for example), or they could be "spin-paired" (one "up", one "down"). The former situation is called a "triplet state", whereas the latter case is termed a "singlet state". These two situations result in some physical differences, such as different interactions with a magnetic field.
Subsequent pages will focus on the reactivity of radicals, with an emphasis on the stages of radical chain reactions.
Exercise \(6\)
Some transition metal compounds have radical characteristics. Show why, using a low spin Co(II) complex in an octahedral environment as an example.
Answer
6.02: Radical Initiatio
Sometimes, radicals form because a covalent bond simply splits in half. Two atoms that used to be bonded to each other go their separate ways. Each atom takes with it one electron from the former bond. This process is called homolysis, meaning the bond is breaking evenly. In contrast, heterolysis is the term for a bond that breaks via ionization, with one atom getting both electrons from the bond.
• Homolysis describes breaking a bond in half, with one electron going to each side of the former bond.
In pictures, we show this process using curved arrows, but the arrows we use are slightly different from the ones you may be used to seeing in polar reaction chemistry. Instead of a regular arrowhead, we use a half arrowhead. This kind of arrow looks a little more like a fish hook. It is easy to remember the roles of the two kinds of arrows, because a full arrowhead describes the movement of an electron pair, whereas a half arrowhead describes the movement of only one electron.
Why would a covalent bond simply break apart? There are really a number of factors and a number of events that may result in this situation. The simple part of the story is that the bond must have been weak in the first place. There was enough energy available in the form of heat transferred from the surroundings (or sometimes in the form of light) to overcome the stabilization energy of the bond.
What makes a bond weak or strong? That is a complicated question. Many factors influence bond strength. However, two of the main factors responsible for covalent bond strength are the degree of electron sharing because of "overlap" and the degree of bond polarity resulting from "exchange". Most strong covalent bonds rely on a mixture of these two factors.
One fairly common feature in homolysis is a bond between two atoms of the same kind. For example, elemental halogens often undergo homolysis pretty easily. The ease with which these bonds can be split in half is illustrated by their low bond dissociation energies. Not much energy needs to be added in order to overcome the bonds between these atoms.
Bond Bond
Dissociation
Energy
(kcal/mol)
H-H 105
C-C 85
N-N 65
O-O 47
F-F 37
S-S 45
Cl-Cl 57
Br-Br 45
Sn-Sn 45
I-I 35
This propensity for radical formation can be understood in terms of the lack of a polar component in these bonds. These atoms rely solely on atomic overlap to share electrons with each other.
There is a notable exception to the rule that homoatomic bonds are inherently weak, and that is a carbon-carbon bond. Its bond dissociation energy is listed in the table for comparison with the halogens. The relative strength of carbon-carbon bonds gives rise to a multitude of carbon-based "organic" compounds in nature. The formation of bonds between like atoms is called "catenation"; carbon is the world champion.
Exercise \(1\)
Draw structures for the following reagents and show curved arrows to illustrate the initiation of radicals in each case.
a) Br2 b) H2O2 c) (CH3)3CO2H d) (CH3)2SbSb(CH3)2 e) CH3CH2CH2CH2SSCH2CH2CH2CH3
Silicon (BDESi-Si = 53 kcal/mol) and sulfur (BDES-S = 54 kcal/mol) are also capable of catenation, but the bonds that these atoms form between themselves are much weaker than C-C bonds. It seems to be generally true that larger atoms form weaker bonds, at least in the main group of the periodic table. After all, I-I bonds are weaker than Br-Br bonds, which are weaker than Cl-Cl bonds.
It is sometimes argued that this trend is a result of poor spatial overlap between the more diffuse p orbitals nearer the bottom of the periodic table. However, the other side of the equation must not be ignored. Once these bonds break, two new radicals form. Just as ions are more stable on larger, more polarizable atoms, so are radicals.
• Radicals are more stable on larger, more polarizable atoms.
• For example, sulfur radicals are more stable than oxygen radicals.
These trends show up in a comparison of carbon-halogen bond strengths. The average carbon-iodine bond is much weaker than the average carbon-fluorine bond.
Bond BDE (kcal/mol)
C-F 116
C-Cl 78
C-Br 68
C-I 51
Other factors that stabilize radicals can also tilt events in favor of bond homolysis. For example, during catalytic hydrogenations, ether linkages at benzylic positions are often cleaved. A C-O bond is not inherently weak, but a benzylic radical is quite stable. It is the stability of the resulting radical that weakens this particular C-O bond and allows it to be broken so easily.
• Particularly stable radicals form relatively easily.
• For example, benzylic radicals form very easily.
Some compounds are commonly used as radical initiators. For example, peroxides contain weak O-O bonds that can cleave to form radicals. That's the initial event in the formation of a radical from benzoyl peroxide, but the resulting carboxyl radical quickly decomposes in favor for carbon dioxide formation.
AIBN, on the other hand, can cleave to produce a very strong dinitrogen triple bond, leaving behind two radicals.
DMPA is a photoinitiator; it is cleaved by the addition of light.
Exercise \(2\)
Provide mechanisms for radical formation from
a) benzoyl peroxide b) AIBN c) DMPA
Answer a
a)
Answer b
b)
Answer c
c)
Exercise \(3\)
The following initiators form radicals relatively easily. Provide mechanisms for radical formation in each case.
Exercise \(4\)
Chemists have a wide range of initiators available. The following two examples are similar in some ways, but may be useful under different conditions. For what conditions might each initiator best be suited? | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/V%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_3/06%3A_Radical_Reactions/6.01%3A_Introduction_to_R.txt |
Bond strength isn't just about the interaction of the two fragments bonded together. It is also influenced by the stability of those two species on their own. When the bond is broken, what pieces are left over?
The formation of radicals may be driven by the weakness of a particular bond. In terms of radical formation via bond homolysis, the reaction is more product-favored if the bond being broken is weak. In other words, the bond is not very low in energy, so the overall reaction may become more downhill (or at least less uphill). In that case, forward reaction is favored because of reactant destabilization.
However, a downhill reaction could also occur through product stabilization. For example, we have already seen that larger, more polarizable atoms form more stable radicals. Iodine radicals are more stable than bromine radicals, and sulfur radicals are more stable than oxygen radicals.
There are other factors, too. One of the most important factors is resonance. We have seen that the stability of anions and cations is strongly influenced by delocalization. Factors that spread the excess charge onto multiple atoms, rather than allowing charge to concentrate on one atom, make charged species much more stable.
For example, carbon-based anions are relatively unstable, but a delocalized carbanion is within the realm of possibility. Enolate ions are particularly easy to obtain because negative charge is partially delocalized onto a more electronegative oxygen atom. Delocalization also strongly stabilizes radicals. It is one of the most important factors in the stability of carbon-based radicals.
Exercise \(1\)
Illustrate the resonance stabilization in the following radicals
a) allyl, CH2CHCH2 b) benzyl, CH2C6H5 c) cyclopentadienyl, C5H5
Answer
Radicals on carbon atoms are also stabilized when they are in more substituted positions. just as carbocations are more stable if they are on more substituted positions, carbon radicals are also more stable in these positions. A tertiary radical is more stable than a secondary one. A secondary radical is more stable than a primary one.
Exercise \(2\)
Rate constants for the dissociation of the following initiators to form an iodine atom and a radical were measured under a specific set of conditions. For each pair, explain why one compound undergoes homolysis more quickly.
Exercise \(3\)
Predict the relative order of bond dissociation rates for the following initiators, which would each form a tellerium radical and a carbon radical.
6.04: Radical Initiatio
We saw in the section on redox reactions that single electrons can be transferred from one species to another. Because one electron is transferred at a time, radicals can be initiated this way.
Metals that are high in the activity series, such as lithium or sodium, can easily donate their valence electrons to organic compounds. As a result, those organic compounds become radicals.
$\ce{Li -> Li^{+} + e^{-}} \nonumber$
$\ce{Na + Na^{+} + e^{-}} \nonumber$
The possibility for this reaction is most easily illustrated with an example from carbonyl chemistry. Carbonyls are electrophiles. The electrophilic carbon normally accepts a pair of electrons from a nucleophilic donor. However, a single electron could be thought of as a nucleophile, too.
For example, if benzophenone is dissolved in an unreactive solvent, such as ether, over a few pieces of sodium, the sodium can transfer an electron to the carbonyl. The interesting thing about this reaction is that, although benzophenone is a white (or colourless) compound and ether is a colourless liquid, the ether solution turns deep blue after a couple of hours. Benzophenone radical anion is a deep blue colour.
Benzophenone radical has long played an important role in research labs. For many years, sodium has been used as a drying agent for organic solvents. Because of the well-known propensity of sodium to react with water, any traces of water in a flask of ether are destroyed. They are converted to sodium hydroxide and hydrogen gas. However, in the absence of water -- that is, if the sodium has already done its job -- the sodium can transfer electrons to benzophenone in solution, producing a blue colour. Benzophenone thus works as an indicator to let researchers know that the solvent is dry.
After a few more hours, the colour changes once more to a deep purple. That's because a second electron gets transferred to the benzophenone radical, forming benzophenone dianion. That's when you know the solvent is really, really dry. However, all of this has to be done under a nitrogen atmosphere, or else the benzophenone radical anion undergoes additional reaction with oxygen, producing yellow schmutz all over the flask instead of the beautiful purple colour. This drying method also works with benzene or toluene, or with a little modification, saturated hydrocarbons such as pentane. It doesn't work with many other solvents, which might instead react directly with the sodium.
But what would happen if, at this point, we carefully introduced some protons? Maybe it is in the form of an acid, either strong (HCl) or very weak (NH4Cl). The benzophenone dianion would surely get protonated, and since it is a dianion it would get protonated twice.
The overall result is a reduction of the benzophenone to the corresponding alcohol, 1,1-diphenylmethanol. It would be just like we had added sodium borohydride, a source of nucleophilic hydride (that's a proton plus two electrons) and then did an acid workup (adding a second proton). In this case, we have just added the components (two electrons, two protons) in a different order.
Furthermore, this method of reducing things isn't really limited to carbonyls. Alkynes and aromatics are also susceptible to reduction to the radical anion or dianion, although the reaction is more commonly performed using lithium.
We can imagine a similar process of protonation occurring to get an alkene. It looks a little like an alkyne reduction, but instead of using H2 as the source of hydrogen atoms, we have used an alkali metal as the source of the electrons and an alcohol as the source of protons.
Interestingly, these latter reactions are stereospecific. The sites of the anions (and subsequent protonations) are as far apart as possible. That means that the anions are found trans- to each other in the alkenyl anion. As a result, the trans alkene always results from lithium reduction. This reaction is complementary to hydrogenation with Lindlar's catalyst, which always results is cis alkenes.
It may be surprising that the dianion is not required in order to get this "keep-the-negative-charges-far-apart" selectivity. Normally, these reductions are conducted in liquid ammonia, often with a little bit of alcohol added. Under those conditions, the initial radical anion is protonated before the second electron donation. The dianion never actually forms, yet the selectivity is still the same. Repulsion between the lone pair and the radical are enough to account for the stereoselectivity.
These reactions are sometimes called "dissolving metal reductions" because the lithium metal dissolves in the liquid ammonia. In ammonia and some amines, the metal actually undergoes ionization to produce Li+ and e-. This "salt" is called "lithium electride" and it produces a bright blue colour. It is really a coordination complex, with a hexaammine lithium cation and an ammonia-solvated electron. Furthermore, the solution is paramegnetic and highly conducting, because of all of those unpaired electrons floating around on their own.
If you keep adding more and more lithium to the ammonia, at some point the solution becomes diamagnetic and turns gold in colour. At this point, the evidence suggests formation of "lithium lithide", or Li+ Li-. The same thing happens with other alkali metals, such as sodium or potassium, producing bronze-coloured sodium sodide or potassium potasside.
Similar reactions occur with aromatic systems. These reactions are called "Birch reductions". Because of the same electron-electron repulsion problem encountered in alkyne reduction, Birch reductions always result in the radical / anions positioning themselves at the 1,4-positions in the benzene ring.
The result is a cyclohexa-1,4-diene. Just like with the alkyne reduction, the Birch reduction is usually performed with a small amount of alcohol in solution.
Furthermore, the remaining double bonds from a Birch reduction are in the more substituted positions; that's understandable in terms of the trend of alkene stability, in which more substituted double bonds are more stable.
Exercise $1$
Illustrate the products of single electron transfer from lithium to the following compounds
a) 2-hexyne, CH3CC(CH2)2CH3 b) 2-butanone, CH3COC2H5 c) allyl bromide, CH2CHCH2Br
Answer
Exercise $2$
Single electron transfer is much more difficult to a carboxylate anion than to an aldehyde or ketone. Explain why.
Answer
The reduction potential of the negatively charged carboxylate anion would be much less positive than the reduction potential for the neutral aldehyde or ketone. We think of a carboxylate anion as much less electrophilic than aldehydes and ketones for the same reason.
Exercise $3$
Show the products of dissolving metal or Birch reductions in the following cases.
Exercise $4$
Show the mechanism for the Birch reduction of m-xylene (m-dimethylbenzene) with lithium and methanol in liquid ammonia.
Answer | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/V%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_3/06%3A_Radical_Reactions/6.03%3A_Radical_Initiatio.txt |
Radicals are known for engaging in "chain reactions". In a chain reaction, a reactive intermediate is generated. When it reacts, it leaves another reactive intermediate, much like the first. This event is called "propagation".
There are a couple of common ways that propagation occurs. The radical might achieve its stable electron count by snatching another atom, especially a hydrogen atom. That event is called hydrogen atom abstraction. Alternatively, a radical may bond with one of the electrons in a pi bond.
In the abstraction of an atom, the radical forms a bond with that atom. That bond gives the radical an even number of electrons again.
Exercise \(1\)
Show, with arrows, the mechanism for abstraction of a hydrogen atom from ethylbenzene by a hydroxyl radical.
Answer
It's important to note that in a hydrogen atom abstraction, the radical is reacting not just with the proton, but with the entire hydrogen atom. It is taking the electron, too.
Exactly which atom gets abstracted has a lot to do with bond strengths. For example, O-H bonds are quite strong (up to 120 kcal/mol, in water, for example). Thus, an OH radical will frequently abstract hydrogen atoms, because there is an energetic payoff when that happens.
Of course, a bond also has to get broken during an abstraction. That costs some energy. C-H bonds are also pretty strong, so they may be hard to break. However, some C-H bonds in particular are weaker than others. For example, in order to break a benzylic C-H bond, the cost is only about 88 kcal/mol. In the case of hydrogen atom abstraction from ethylbenzene by hydroxyl radical, the trade-off is worth it.
Exercise \(2\)
Draw a reaction progress diagram for the abstraction of a hydrogen atom from ethylbenzene by a hydroxyl radical.
It is not always the case that a reaction is purely determined by the thermochemistry of the bonds involved. Sometimes, there are kinetic factors that block the path to the more stable product, or that lower the path to the less stable product. However, in many atom abstractions, because the old bond is being broken at the same time that the new bond is being formed, both factors matter in the rate determining step. By the time the transition state is reached, the stability of the complex is influenced both by the bond that is being broken and the bond that is being made. As a result, the thermodynamics of the reaction can have a strong influence on the pathway to products.
Bond Dissociation Energy (kcal/mol) Bond Dissociation Energy (kcal/mol)
F-H 136 Br-H 88
Cl-H 103 I-H 71
EtO-H 105 O2N-OMe 42
CH3S-H 87 Cl-OMe 48
PhO-H 87 H3C-OMe 85
Me2N-H 91 H3C-NH2 85
Et3Si-H 96 H3C-F 115
Bu3Ge-H 88 CH3-H2C-Cl 85
Bu3Sn-H 78 CH3-H2C-Br 72
Me3Sn-Cl 100 CH3-H2C-I 57
Exercise \(3\)
Indicate whether a dimethylamine radical is likely to carry out hydrogen atom abstraction from each of the following molecules.
a) Et3SiH b) PhOH c) EtOH d) Bu3SnH e) HF f) HI
Answer a
a) no; a stronger bond would have to be broken and replaced with a weaker bond.
Answer b
b) yes; a weaker bond would be broken and replaced with a stronger one.
Answer c
c) no
Answer d
d) yes
Answer e
e) no
Answer f
f) yes
Exercise \(4\)
Indicate whether a chlorine atom abstraction would be likely to occur in each of the following cases.
1. Chloroethane is exposed to methoxy radical.
2. Chloroethane is exposed to trimethyltin radical.
3. Trimethyltin chloride is exposed to methoxy radical.
Answer a
a) no
Answer b
b) yes
Answer c
c) no
Sometimes, the identity of the two atoms that form a bond does not tell the entire story about bond strengths. In the case of C-H bonds of hydrocarbons, for example, a range of bond strengths have been experimentally determined. There is a 40 kcal/mol difference between the weakest C-H bond in a simple hydrocarbon and the strongest (that's about 175 kJ/mol, for the metric-oriented). These bond strengths are remarkably senstive to subtle structural differences, largely because of relative stabilities of the resulting radicals when the bond is broken.
Bond Bond dissociation energy (approximate; kcal/mol)
HCC-H (aryl) 130
Ph-H (aryl) 110
CH2=CH-H (vinyl) 106
H3C-H 105
CH3CH2-H 98
(CH3)2CH-H 95
(CH3)3C-H 92
PhCH2-H (benzyl) 88
CH2=CH-CH2-H (allyl) 88
Exercise \(5\)
Propose reasons for the trends in bond strengths among the following groups.
1. H3C-H , CH3CH2-H , (CH3)2CH-H, (CH3)3C-H
2. (CH3)2CH-H, PhCH2-H, CH2=CH-CH2-H
3. HCC-H, CH2=CH-H, CH3CH2-H
Answer a
a) The effect is similar to the stability of carbocations. The more substituted radical is more stable. Thus, the trend from most to least stable is tertiary > secondary > primary > methyl radical. The trend likely originates from a hyperconjugation effect, as in carbocations.
Answer b
b) The trend here is that if the radical is delocalized by resonance, it is more stable. The allyl and benzyl radicals are more stable than the isopropyl radical. This trend is also seen in cations.
Answer c
c) The trend here has to do with "hybridization effects" or the atomic orbitals that contribute to the formation of molecular orbitals involved in the relevant bond. In a linear alkyne, the C-H bond can be formed only from some combination involving a hydrogen 1s orbital, carbon 2s orbital and one of the carbon 2p orbitals. This combination is called a "sp" hybrid and the orbital that combines with the hydrogen can be considered 50% 2s, 50% 2p in character.
In a planar alkene, the C-H bond can be formed only from some combination involving a hydrogen 1s orbital, carbon 2s orbital and two of the carbon 2p orbitals (since two of them could lie in this plane). This combination is called a "sp2" hybrid and the orbital that combines with the hydrogen can be considered 33% 2s, 66% 2p in character.
In a tetrahedral alkane, the C-H bond can be formed from some combination involving a hydrogen 1s orbital, carbon 2s orbital and all three of the carbon 2p orbitals. This combination is called a "sp3" hybrid and the orbital that combines with the hydrogen can be considered 25% 2s, 75% 2p in character.
Because a 2s orbital is lower in energy than a 2p orbital, a bond that has greater 2s character is lower in energy than a bond with less 2s character. That means that a bond with greater 2s character is harder to break than a bond with less 2s character. Hence, the alkane C-H bond is weaker than the alkene C-H bond, which is weaker than the alkyne C-H bond.
Exercise \(6\)
Explain why a trialkyltin radical (R3Sn) would not be able to remove a hydrogen atom from propane, but could abstract a chlorine atom from chloroethane.
Answer
The Sn-H bond has a dissociation energy of about 78 kcal/mol, compared to about 98 kcal/mol for the C-H bond in ethane. The formation of the Sn-H bond would not compensate for the energy needed to break the C-H bond. On the other hand, the 100 kcal/mol released upon formation of a Sn-Cl bond would more than make up for the 85 kcal/mol required to breal a C-Cl bond.
We could try to rationalise those differences, although bond strengths are always very complicated issues and we will not be able to explain things satisfactorily without quantum mechanical calculations. Let's start with two basic factors, though: the amount of covalency and the amount of polarity.
The difference between the covalent radii of tin and hydrogen (1.39 vs. 0.31 Å) is much greater than the difference between tin and chlorine (1.39 vs 1.02 Å), so there may be less overlap and less covalency between tin and hydrogen than between tin and chlorine. By comparison, the covalent radius of carbon is about 0.76, which puts it somewhere in between hydrogen and chlorine.
In addition, as measured on the Pauling scale, the electronegativity values of these atoms are: chlorine, 3.16; carbon, 2.55; hydrogen, 2.2; tin, 1.96. The tin-chlorine bond would have a large ionic component; this additional component of bonding would strengthen the Sn-Cl bond.
Addition to an alkene is another common propagation pathway in radical reactions. In this case, a π (pi) bond is broken in the alkene to form a new bond to the radical species. That leaves the second electron from the π bond to form a new radical species.
Π bonds are often weaker than σ bonds, making this pathway energetically accessible in many cases. Perhaps more importantly, the electrons in π bonds are found above and below a flat part of the molecule, leaving them open and accessible for reaction with radicals.
Bond Bond dissociation energy (kcal/mol) Bond Bond dissociation energy (kcal/mol)
H3C-CH3 90 H2C=CH2 174
H3C-NH2 85 H2C=NH ?
H3C-OH 92 H2C=O 179
Exercise \(7\)
Show a mechanism, with curved arrows, for the reaction of pentene with bromine atom.
Answer
Exercise \(8\)
Calculate the strengths of the following pi bonds.
1. In ethene.
2. In methanal.
Answer a
a) 174 - 90 kcal/mol = 84 kcal/mol for the π contribution only
Answer b
b) 179 - 92 kcal/mol = 87 kcal/mol for the π contribution only
Exercise \(9\)
The pi bond in PhCH=NPh has been calculated to have a dissociation energy of 77 kcal/mol.
1. Estimate the missing imine C=N bond dissociation energy in the above table.
2. Explain why your estimate may be unreliable.
Answer a
a) 85 + 77 kcal/mol = 162 kcal/mol for the combined σ + π contribution
Answer b
b) There may be significant differences between the π bond in methanal imine (CH2=NH) and the imine for which we have bond strength data. For example, breaking the bond would result in radicals next to phenyl groups, which may be significantly stabilized. On the other hand, the π bond itself may be significantly stabilized by conjugation. Thus, our estimate is probably not correct, but it is difficult to say whether it is too high or too low. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/V%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_3/06%3A_Radical_Reactions/6.05%3A_Radical_Propagati.txt |
If propagation steps lead to chain reactions, continually leading to the formation of new radicals that engage in further propagation reactions, then where does it end? That self-perpetuating cycle can be part of the beauty of radical reactions, allowing a very small amount of initiator to efficiently start a process that then sustains itself. However, it is important that the reaction has an end point, so that it does not continue to run out of control.
There are a couple of different ways that radical reactions commonly come to an end. These elementary reactions are called terminations steps. The simplest event that could occur is a radical recombination step. In that case, two radicals approach each other and bond together, sharing their previously unpaired electrons.
• In termination steps, two radicals come together to make no radicals.
Energetically, termination steps like this one should be pretty favorable, because they involve formation of a new bond. These steps are mostly limited by the low concentration of radicals in solution. If two radicals don't happen to bump into each other, then they won't react together.
A second event leading to termination also involves the collision of two radicals. However, the trajectory of the two radicals is slightly different, so that the unpaired electrons do not connect with each other. Instead, the unpaired electron on one radical molecule encounters a hydrogen on the other molecule. If the hydrogen is alpha to the radical, it is easily abstracted, forming a double bond.
Researchers are actually able to measure the bond strength of such a hydrogen alpha to a radical. They find that the bond is considerably weakened compared to a similar hydrogen atom in a molecule that does not contain a radical in the same position.
This site is written and maintained by Chris P. Schaller, Ph.D., College of Saint Benedict / Saint John's University (with contributions from other authors as noted). It is freely available for educational use.
6.07: Radical Substitut
There are a couple of classic reactions based on radical chemistry that are often used to illustrate different consequences of the mechanism. One of these reactions is the radical substitution of a halogen atom for a hydrogen atom.
This reaction is commonly observed with chlorine and bromine. Multiple products are possible; only one of these products is shown above.
As a radical reaction, the first step would have to be an initiation step. If the reaction proceeds as written, via the addition of molecular chlorine and light, then initiation would involve direct homolysis of the chlorine-chlorine bond.
Once the chlorine radicals are present, propagation steps would occur. Given the ultimate replacement of a hydrogen by a chlorine atom, radical abstraction of a hydrogen atom seems likely. That event would lead to the formation of an alkyl radical.
The newly-formed alkyl radical would be able to engage in propagation reactions as well. For example, it could react with another chlorine molecule. It would produce a new chlorine radical and the reaction product, chloroproane.
Alternatively, if we wanted to imagine the full range of elementary steps in radical reactions, we could picture a termination step, with another chlorine combining with that alkyl radical to form a chloroalkane product.
Exercise \(1\)
Propose a mechanism, with curved arrows, for the formation of 2-chloropropane via radical chlorination.
Answer
Exercise \(2\)
Propose a mechanism, with curved arrows, for the following variation on radical chlorination.
Answer
Exercise \(3\)
Radical chlorination of pentane also results in multiple products. One of them is shown below. What other isomers are formed?
Exercise \(4\)
Which product would be the major one in radical chlorination of propane? Why?
Answer
2-bromopropane.
Clearly, radical halogenation could result in a mixture of products. That's because there are different hydrogen atoms that could be extracted in the first propagation step. Abstracting a hydrogen atom from the middle carbon of propane would lead ultimately to 2-chloropropane. Abstracting a hydrogen atom from either of the end carbons of propane would lead to formation of 1-chloropropane.
6.08: Radical Addition
Radical addition to alkenes is another classic example of a radical reaction. Like radical substitution, it illustrates some important elements of radical reactivity.
The most common example of radical addition to alkenes seen in college chemistry textbooks is radical addition of hydrogen bromide, HBr. That's because it complements the usual addition of HBr to an alkene.
In the usual addition, HBr adds in a Markovnikov fashion to place the bromine at the more-substituted end of the alkene and the hydrogen on the less substituted end (remember the adage, "the rich get richer and the poor get poorer").
In contrast, addition of HBr under radical conditions leads to the bromine attaching at the least-substituted position, whereas the hydrogen bonds to the most-substituted position. This is a Robin Hood reaction.
How does that reversal in regiochemistry come about? Consider the mechanism of the reaction.
Radical addition of HBr is almost always done in the presence of peroxides. The peroxide acts as an initiator for the reaction. The O-O bond may break through the simple action of thermal energy (maybe even at room temperature).
The alkoxy or hydroxy radicals that result from this initiation step are left to induce radical propagation. Certainly one of the easiest available targets is the relatively weak H-Br bond. Abstraction of a hydrogen atom from HBr produces a bromine radical.
One of the things that the bromine radical could do is add to the double bond of an alkene. When it does so, it will bind to one end or the other of the former double bond. To which end will it go?
Like cations, radicals are considered to be somewhat electron-deficient. They are stabilized by electron-donating factors. That means that, like a cation, the radical will be more favorable on the most-substituted carbon of the formal double bond.
Notice that this step is actually governed by almost the exact same factor that governs the polar addition of HBr. The regiochemistry is governed by the stability of the intermediate. Once that event has transpired, the regiochemistry of the product is fixed. It only remains for the alkyl radical to pluck a hydrogen atom from another HBr molecule, forming the final product and generating another bromine radical. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/V%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_3/06%3A_Radical_Reactions/6.06%3A_Radical_Terminati.txt |
Radical addition to alkenes can be applied to the production of macromolecules. Like other polymerisation reactions involving alkenes, it involves the formation of a reactive intermediate by the action of an initiator on an alkene. A chain reaction results in which other alkenes are enchained into a polymer.
The term "initiator" here is used in a slightly different way than we have used it with other radical reactions. The radical initiator has already undergone its reaction to form a radical. That radical then initiates chain growth. This step is really a propagation step in terms of types of radical elementary reactions, because one radical leads to a new radical.
Polystyrene is one example of a material that is frequently prepared via radical conditions. During the reaction, a radical adds to the double bond of the alkene.
The newly-formed radical, in regular alkene addition, would then react with something to abstract an atom and achieve stable, closed-shell configuration. However, in a polymerisation reaction, alkene molecules are intentionally packed closely together. Either they are very concentrated in solution or else they are neat (with no solvent at all). As a result, the newly-formed radical just gobbles up another alkene.
There are lots of ways to carry out this reaction. One way would be to take some styrene, heat it up until it melts, and add some benzoyl peroxide.
Radical polymerisation is a critical method of preparing polymers. The figure below shows a number of monomers that are commonly polymerised under radical conditions.
Exercise $1$
Look at the monomers listed above.
1. For each of the three groups (fast, medium and slow), identify what structural features the members of the group have in common that distinguishes them from the other groups.
2. Propose a reason why the fast group undergoes more rapid polymerisation than the other groups.
3. Propose a reason why the medium group undergoes more rapid polymerisation than the slow group.
Answer a
a) Fast: disubstituted, conjugated alkenes
Medium: monosubstituted, conjugated alkenes
Slow: alkenes monosubstituted with a heteroatom
Answer b
b) Fast: the radical formed would be both tertiary and conjugated. This stable radical forms very quickly.
Medium: the radical formed would be conjugated, but not tertiary. The radical does not form quite as quickly as in the above group.
Answer c
c) Medium: the radical formed would be both conjugated. This stable radical forms quickly.
Slow: the radical formed would not be conjugated in the usual sense (although heteroatoms next to radicals do provide some stability). The radical does not form very quickly compared to radicals next to double bonds.
Exercise $2$
Provide a mechanism for the polymerisation of styrene in the presence of benzoyl peroxide, up through the first couple of propagation steps.
Answer
The mechanism would begin with initiation of the benzoyl peroxide.
The resulting phenyl radical acts as the initiator for polymerisation.
Additional propagation steps follow.
A radical polymerisation, after the initiator gets going, is just a series of propagation steps in a row. How does it all stop? There must be a termination reaction. That can happen in a couple of different ways.
For example, two growing chains might encounter each other, head-to-head. A collision could bring the unpaired electrons together to form a new bond. Two growing polymer chains would come to an end at once. At the same time, the chain length and molecular weight doubles as two chains combine into one.
Of course, most of the other chains in the polymerisation continue growing normally. As a result, the distribution of molecular weights broadens dramatically. Some chains are twice as long as the others.
Alternatively, when two growing chains collide, one might carry out a hydrogen atom abstraction on the other. The hydrogen abstracted might come from the head of the chain, alpha to the active radical. That C-H bond is a little bit weaker, as bonds alpha to radicals typically are. The chain that picks up the hydrogen is no longer a radical. The chain that loses a hydrogen is no longer a radical, either. Both chains stop growing. Because other chains around them continue to grow, these one lag behind, resulting in a wider distribution of chain lengths.
Of course, a hydrogen atom abstraction could happen elsewhere along the chain. In that case, the chain that abstracted the hydrogen stops growing, as before. However, the chain that lost the hydrogen now has two radicals. That makes two sites of chain growth. This chain starts growing twice as fast as the others. Eventually, that's going to lead to a big difference in molecular weights.
These termination events, all of which are possible and which occur more or less randomly, have profound consequences on the material produced.
There is an effect on chain length and molecular weight. Chains that have abstracted a hydrogen atoms come to a complete halt. They stop growing altogether. On the other hand, chains that have had a hydrogen stolen from them may grow again. Either they have an additional radical introduced, or else they form a π bond, which can react with another radical and start growing again. When they do, it will again be a case of two chains coming together -- an active one and a macromonomer -- leading to a trmendous jump in molecular weight.
The morphology of the polymer is clearly altered by these events. Most growing chains are simply linear: they consist of a series of enchained monomers, all in one row. However, in either of the cases in which a growing chain has lost a hydrogen atom, then continued to grow, the shape will be branched.
These two architectures have very different properties. For example, we would expect a very different Tg in a branched polymer than a linear one, with chains flowing more freely at lower temperatures in the linear one.
Exercise $3$
It has been found that radical polymerisation of styrene with a benzoyl peroxide initiator follows the rate law:
$Rate = k[styrene][peroxide]^{1/2} \nonumber$
Explain why this rate law occurs, based on the mechanism.
Answer
The monomer dependence is straightforward: the more monomer there is present, the faster it can be enchained by the growing radical chain.
The initiator dependence is slightly more complicated, but not much. The more growing radical chains there are, the faster the monomers can be enchained. The more phenyl radical is formed, each initiating a new radical chain, then the more growing radical chains there will be.
Rate = k [monomer][Ph.]
The more benzoyl peroxide there is, the more phenyl radical there will be, but the relationship is not linear, as explained below.
Benzoyl peroxide cleaves in two, ultimately producing two phenyl radicals:
Keq = [Ph.]2/[BP]
so [Ph.]2 = Keq [BP]
and [Ph.] = Keq1/2 [BP]1/2
therefore Rate = k [monomer][Ph.]1/2
Exercise $4$
Provide mechanisms for radical formation from the following initiators:
Exercise $5$
There is a great deal of interest in aqueous polymerisations for environmental reasons. Hydrogen peroxide is an obvious water-soluble initiator; however, it is slow to initiate polymerisation. Show how initiation can be accelerated in the presence of Fe(II) ions.
Answer
Exercise $6$
A number of compounds can be used as inhibitors of radical polymerisation. For example, benzoquinone will stop radical chains from growing. Show a mechanism for its reaction with a growing polymer chain and explain why the product of the reaction is a relatively stable radical.
Exercise $7$
It may be surprising that one of the most effective inhibitors of radical polymerisation is molecular oxygen; it can slow the polymerisation rate by a factor of tens of thousands. The radical that results from the addition of O2 to a growing chain is considered to be resonance-stabilised. Show the mechanism of O2 addition and demonstrate resonance stability in the product.
Answer
Peroxy radicals are relatively stable and react slowly. However, when they do react, they also undergo reactions other than radical polymerisation, so that they are even more effective as inhibitors of polymerisation. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/V%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_3/06%3A_Radical_Reactions/6.09%3A_Radical_Polymeriz.txt |
Chain polymerisation reactions result in the efficient conversion of monomers into high molecular weight polymers. However, chain termination events result in a broadening of the polydispersity index of the material. In other words, instead of producing a material composed of molecules that are all about the same molecular weight, a wide range of sizes of molecules result. Some of the chains are very short and others are very long. That's a problem, because the chain length (and the associated property, molecular weight) strongly influence the properties of the material. If the size of the molecules is not controlled, neither are the properties. If the properties are not controlled, the material won't perform reliably in its intended application.
Exercise \(1\)
The degree of polymerisation of a polymer is simply the average number of monomers incorporated in each polymer chain. Given the following "feed ratios" (ratios of monomer to initiator), what is the expected degree of polymerisation in each case?
1. methyl methacrylate : AIBN 500 : 1
2. styrene : benzoyl peroxide 1000 : 1
3. acrylonitrile : tert-butyl peroxide 600 : 3
Answer a
Because each initiator breaks in half, forming two radicals, one initiator starts two growing chains. Every initiator will, on average, consume half the monomers, assuming no unexpected chain termination events.
a) DP = 500 monomers / 2 growing chains = 250
Answer b
Because each initiator breaks in half, forming two radicals, one initiator starts two growing chains. Every initiator will, on average, consume half the monomers, assuming no unexpected chain termination events.
b) DP = 1000 monomers / 2 growing chains = 500
Answer c
Because each initiator breaks in half, forming two radicals, one initiator starts two growing chains. Every initiator will, on average, consume half the monomers, assuming no unexpected chain termination events.
c) DP = 600 monomers / 3 growing chains = 200
Exercise \(2\)
The measured degree of polymerisation indicates the number average molecular weight of the polymer (Mn). Calculate Mn in each of the following cases.
1. polystyrene, with DP = 1250
2. polymethacrylamide, with DP = 725
3. poly(methyl acrylate), with DP = 1420
Answer a
M0 = molecular weight of monomer in each case. 1 D (Dalton) = 1 amu
a) Mn = DP x M0 = 1250 x 104 D = 130,000 D = 130 kD
Note that end groups were neglected in these calculations; they may contribute a significant percentage of total molecular weight at lower DP. A more accurate calculation would account for the identity of both end groups.
Answer b
M0 = molecular weight of monomer in each case. 1 D (Dalton) = 1 amu
b) Mn = DP x M0 = 725 x 85 D = 61,625 D = approximately 62 kD
Note that end groups were neglected in these calculations; they may contribute a significant percentage of total molecular weight at lower DP. A more accurate calculation would account for the identity of both end groups.
Answer c
M0 = molecular weight of monomer in each case. 1 D (Dalton) = 1 amu
c) Mn = DP x M0 = 1420 x 86 D = 122,120 D = approximately 122 kD
Note that end groups were neglected in these calculations; they may contribute a significant percentage of total molecular weight at lower DP. A more accurate calculation would account for the identity of both end groups.
Exercise \(3\)
Alternatively, if the number average molecular weight of the polymer is measured, that result can be used to establish the degree of polymerisation. What is DP in each of the following cases?
1. polyacrylonitrile, with Mn = 11,450 D
2. poly(vinyl acetate), with Mn = 24,760 D
3. polystyrene, with Mn = 927,000 D
Answer a
a) DP = Mn / M0 = 11,450 D / 53 D = 216
Answer b
b) DP = Mn / M0 = 24,760 D / 86 D = 288
Answer c
c) DP = Mn / M0 = 927,000 D / 104 D = 8,914
Living polymerisation refers to processes in which unexpected chain termination does not occur. The chain keeps growing and growing as long as more monomer is supplied. In extremely hardy cases, the term "immortal" polymerisation is sometimes used.
Typically, strategies for living polymerisation involve controlling the reactivity of the intermediates. Frequently, the concentration of the growing chains is kept low. If the concentration of growing chains is kept low, then unexpected side reactions involving the reactive growing chain will be kept to a minimum.
In radical polymerisation, growing chains with radicals at their growing ends will be surrounded by monomers. The radicals devour and enchain the monomers as they move through the reaction medium.
Typical chain-terminating events in radical polymerisation involve the collision of two growing chains. That event could result in head-to-head radical recombination, formation of a double bond via hydrogen abstraction at the head of a chain ("head biting") or formation of a new radical along the backbone of the polymer ("backbiting").
Exercise \(4\)
Suppose you have three growing polymer chains. The monomers have molecular weight = 100 D. Each chain is currently 8 repeat units long.
1. What is the current molecular weight of each chain?
2. If 15 monomers remain in solution, what is the expected degree of polymerisation of each chain, assuming they all grow at the same rate?
3. What is the expected average molecular weight?
4. What is the expected PDI?
5. Suppose two of the chains join together in a termination step. What will be the molecular weight of the new chain?
6. If the third chain keeps growing, what molecular weight will it reach?
7. What will be the average molecular weight?
8. What will be the PDI (assume it's just the ratio of largest to smallest molecular weight)?
Answer a
a) MW = 8 x 100 D = 800 D
Answer b
b) 15 monomers / 3 chains = 5 new monomers / chain
DP = 8 + 5 = 13
Answer c
c) MW = 13 x 100 D = 1,300 D
Answer d
d) All chains are the same length; PDI = 1.0
Answer e
e) MW = 16 x 100 D = 1,600 D
Answer f
f) MW = (8 + 15) x 100 D = 2,300 D
Answer g
g) Mn = (2,300 + 1,600 D) / 2 = 1,950 D
Answer h
h) PDI = 2,300 D / 1,600 D = 1.44
Exercise \(5\)
Suppose once again you have three growing polymer chains. The monomers have molecular weight = 100 D. Each chain is currently 8 repeat units long.
This time, one chain abstracts a hydrogen atom from the backbone of another. The first chain is terminated; the second now has two sites of growth. All three sites continue to grow at the same rate.
1. What will be the molecular weight of each chain?
2. What will be the average molecular weight?
3. What will be the PDI (assume it's just the ratio of largest to smallest molecular weight)?
Answer a
a) The chain that abstracted the hydrogen: MW = 8 x 100 D = 800 D
The branched chain: MW = 18 x 100 D = 1,800 D
The normal chain: MW = 13 x 100 D = 1,300 D
Answer b
b) Mn = (1,800 + 1,300 + 800 D) / 3 = 1,300 D
Answer c
c) PDI = 1,800 / 800 D = 2.25
Exercise \(6\)
Practically, polymers are purified by precipitation and washing after they are prepared. That means very short oligomers are washed away.
1. What is the average molecular weight of the impure mixture shown above, assuming each monomer has molecular weight = 100 D?
2. After precipitation and washing, if small oligomers are washed away, what is the avergae molecular weight of the isolated polymer?
Answer a
a) Mn = (1,700 + 1,800 + 300 D) / 3 = 3,800 D / 3 = 1,267 D
Answer b
b) Mn = (1,700 + 1,800 D) / 2 = 3,500 D / 2 = 1,750 D
If the concentration of growing chains is limited, then the probability that any of these events will occur is also limited.
Certainly, the rate of chain growth also slows when the concentration of growing chains is lowered. That is the price to pay for a smooth operation.
The key to living radical polymerisation is to reversibly stop chain growth, sending a polymer chain from an active state into a dormant state. While in the dormant state, the polymer chain is less likely to undergo random chain-termination events. It can't grow, either, until it is brought back into an active state.
Exercise \(7\)
An early attempt at living polymerisation employed an alkyl iodide initiator. Thermal initiation (initiation by heating) resulted in a reactive alkyl radical and an iodine atom.
Provide a mechanism for styrene polymerisation, making sure to show the following points:
1. initiation
2. propagation
3. recombination with iodine atom
4. a growing chain
5. a "dormant" chain that could re-initiate and start growing again, but which is currently safe from random termination
Answer
Researchers at IBM found that polystyrene polymerisation proceeded much more smoothly when TEMPO, a relatively stable radical, was added to the reaction. The reaction also proceeded more slowly, resulting in overall lower molecular weight of the polymer. However, the distribution of molecular weight was much more uniform. All of the chains were of similar sizes, relatively speaking. As a result, the properties of the material were much more controlled.
TEMPO helps to control the polymerisation by forming a reversible bond with the growing end of the polymer chain. The radical on TEMPO combines with the radical on the head of the polymer to form a C-O bond. The bond can break again (unusually, for a C-O bond), allowing the polymer chain to resume growing periodically.
Two of the most common methods of inducing living radical polymerisation are RAFT and ATRP. RAFT stands for radical atom fragmentation polymerisation. Like the TEMPO reaction, it involves a reversible radical recombination to form a covalent bond. RAFT was developed by a group of Australian chemists in the late 1990's, including Enzio Rizzardo, Graeme Moad and San Thang of Australia's national science agency, CSIRO. ATRP stands for atom-transfer radical polymerisation. It was developed in the mid 1990's by Krysztof Matyjaszewski at Carnegie Mellon in Pittsburgh and his post-doctoral associate, Jin-Shan Wang, now at Shanghai Jiao Tong University. An independent discovery of the method was made by Mitsuo Sawamoto at Kyoto University in Japan.
RAFT most commonly employs a thioester (or similar compound) as a chain transfer agent. The chain transfer agent intercepts a growing polymer chain, but does so reversibly.
The thioester can react with a radical, placing the growing chain in a dormant state. It can also release a new radical, which can then initiate polymerisation. Eventually, it will regulate the growth of two polymer chains.
In addition, a third species present in equilibrium holds both chains dormant. The chain transfer agent can reversibly release one of these dormant chains at a time.
Exercise \(8\)
Provide a mechanism for conversion of a growing chain into a dormant chain via RAFT.
Answer
In ATRP, a similar process works to keep a fraction of the chains in a dormant state. A key component of this method is a copper(I) complex.
The role of the copper(I) complex is to transfer an electron to an inactive species, producing a copper(II) species and a radical. For example, in initiation of the reaction above, the Cu(I) transfers an electron to the alkyl halide to become Cu(II). In turn, the chlorine atom on the alkyl halide becomes chloride anion, Cl-, and the alkyl portion is left as an initiating radical.
One of the reasons ATRP is so important is that it provides a very reliable and inexpensive way to control polymerisation. In addition, it can be adapted to a wide range of useful processes. For example, Matyjaszeski has developed methods of electrochemically controlling the reaction; polymerisation can literally be turned on and off with a switch. Yusuf Yagci and coworkers at Istanbul Technical University developed a photoinduced ATRP process, in which polymerisation begins when the lights come on and stops when it gets dark. Other researchers, including the Hawker lab at UCSB, have also promoted the utility of this approach.
Exercise \(9\)
Provide a mechanism for initiation using ATRP.
Answer
Exercise \(10\)
Provide a mechanism for conversion of a growing chain to a dormant chain using ATRP. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/V%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_3/06%3A_Radical_Reactions/6.10%3A_Living_Radical_Po.txt |
There are experimental methods for the detection of unpaired electrons. One of the hallmarks of unpaired electrons in materials is interaction with a magnetic field. That interaction can be used to provide information about compounds containing unpaired electrons.
Electron Paramagnetic Resonance
Electron paramagnetic resonance (EPR) or electron spin resonance (ESR) is a spectroscopic method. It depends upon the energetic separation of two spin states that exists only in the presence of a magnetic field.
EPR is very closely related to another common type of spin resonance spectroscopy, NMR.
Spin is a quantum property that has no real analogy that would make sense to us on a macroscopic level. However, we do know that spin has associated with it magnetic properties. An electron can have either of two values for these magnetic properties. There are various labels given to these values: sometimes "up" and "down", sometimes, +1/2 and -1/2. However, whichever value of spin an electron possesses makes no difference energetically.
The two possible spin states are the same energetically -- unless a magnetic field is present. Once that situation arises, there is an energetic separation between the two states.
In the presence of a magnetic field, the two spin states separate into two different energy levels. The amount of separation between the energy levels depends on the magnitude of the magnetic field. The stronger the magnetic field, the greater the separation.
This case is only true for unpaired spins. Remember, unpaired spins interact with a magnetic field. If everything is spin-paired, nothing happens. That's very useful, because it gives us a way to detect those unpaired electrons.
An electron in one spin state can still be excited to the other by the absorbance of a photon. Alternatively, an electron in the higher state can drop down to the lower one if it releases a photon.
In order for any of this to happen, the energy of the photon must exactly match the energy difference between states. This Goldilocks rule is called the resonance condition. The amount of energy supplied to jump from one energy level to the next has to be just right. Too little energy and the electron won't make it. Too much energy and the electron won't make it either. It's waiting for just the right photon.
Just how much energy does a photon have? Remember your Planck-Einstein relationship. It described the energy of a photon:
$E=h \nu \nonumber$
E is the energy of the photon, h is Planck's constant (6.625 x 10-34 Js) and ν is the frequency of the photon (it looks like a Roman vee, but it's the Greek letter, nu).
Alternatively, because of the relationship between wavelength and frequency:
$E= \frac{hc}{\lambda} \nonumber$
The new quantities are c, the speed of light (3.0 x 108 m s-1) and λ, the wavelength of the photon.
So, in general, different wavelengths of light deliver different amounts of energy. Blue light, with a wavelength close to 475 nm, has more energy than red light, with a wavelength close to 700 nm.
In EPR, the general range of electromagnetic radiation, or the general kind of photon, is microwave radiation. The frequency of these photons is about 9 or 10 GHz. (Another type of spectroscopy, rotational spectroscopy, also measures the absorbance of microwaves. It typically uses somewhat higher frequencies of microwaves. Rotational spectroscopy gives structural or bonding information about molecules in the gas phase.)
Depending on the environment of the unpaired electron, it may be more susceptible or less susceptible to the influence of the external magnetic field. That means the energy splitting between the two spin states will vary from one molecule to another. As a result, different molecules in the same magnetic field would absorb different wavelengths of microwave radiation.
Usually, an EPR spectrometer is designed so that it supplies a fixed wavelength of microwave radiation to the sample. The magnetic field is adjusted and the instrument measures what field strength was required for absorption of the photons. An EPR spectrum shows absorbance as a function of magnetic field strength.
There is something a little different about how EPR spectra are usually displayed. It is displayed as a derivative of the plot shown above. That's because of the way that the instrument measures the change in absorbance as it changes the magnetic field; i.e. it measures d(absorbance)/d(magnetic field). That's the slope of the previous plot.
As a result, an EPR spectrum really looks more like this. The part above the baseline reflects the positive slope in the previous plot. The part below the baseline reflects the negative slope in the previous plot.
The magnetic field strength is not typically the value that is reported for the peak position. Instead, something called the g-value is reported. The g-value arises from the equation for the Zeeman effect (the effect of the magnetic field on the splitting between spin energy levels). That relationship is:
$\Delta E = g \beta B \nonumber$
in which ΔE is the energy difference between spin states, g is the g-value, a proportionality constant that depends on how susceptible the electron is to the influence of the magnetic field, β is the Bohr magneton (9.274 x 10-24 J T-1) and B is the applied magnetic field.
That means that, for photon absorption,
$h \nu = g \beta B \nonumber$
and so
$g = \frac{ h \nu}{\beta B} \nonumber$
Remember, h and β are just constants. That means g is a measure of the ratio of the photon absorbed to the magnetic field used. It's a standardisation step. If people have instruments that use slightly different wavelengths of microwave radiation, then the magnetic fields they measure for the same samples would not agree. If everyone just measures the ratio of wavelength to field strength, it should all even out. The g-value is a reproducible measure of the environment of an electron that should be the same from one laboratory to another.
A similar practice is used in NMR spectroscopy, for similar reasons. When we report a chemical shift in ppm instead of Hz, we are correcting for the strength of the magnetic field in the instrument we are using. Otherwise the same sample would give two different shifts on two different instruments.
Coupling in EPR
Coupling is a phenomenon in which magnetic fields interact with each other. In EPR, coupling comes about because of the influence of nearby nuclei on the electron that is being observed.
For example, you may already know that a hydrogen atom's nucleus has an unpaired spin. That's the basis of 1H NMR spectroscopy. If that nucleus has an unpaired spin, it has an associated magnetic field. Because the hydrogen nucleus could have either spin value, +1/2 or -1/2, then it has two possible magnetic fields associated with it.
A nearby electron, placed in an external magnetic field, could now be in either of two different situations. Either the neighbouring proton adds a little bit to the magnetic field, or it subtracts a little bit from the external field.
As a result, the electron can experience two different fields. Remember, we are not dealing with a single molecule in spectroscopy. We are dealing with huge numbers of molecules. Some of the molecules will be in one situation. Some of the molecules will be in the other situation. We will see both situations. There will be absorbance at two different magnetic field strengths.
As a result, the EPR spectrum shows two peaks, like this:
This type of peak in the spectrum is called a doublet, because of the double absorbance. This characteristic of an EPR peak is called its multiplicity. How many lines is the peak split into? Two. It is a doublet.
Things are even more interesting if there are two nearby protons. In that case, both neighbouring protons have spin. Either spin could have value +1/2 or -1/2. Maybe they are both +1/2. Maybe they are both -1/2. Maybe there is one of each. These three possible combinations will have three different effects on the magnetic field experienced by the electron.
As a result, there are three peaks in the spectrum. The spectrum is called a triplet. This triplet is shown below, underneath the diagram that illustrates the spin combinations of the neighbouring hydrogens.
Notice that, because either hydrogen could be up or down in the mixed combination, there are two ways of arriving at that middle state. That combination is twice as likely as the other two, because there is only one way of getting those combination: both hydrogens' spins are up, in one case. Both hydrogens are down in the other. As a result, the middle peak in a triplet is twice as large as the peaks on the edges.
Exercise $1$
Show that, with three neighbouring hydrogens, a quartet would result, in which the ratios of the peaks are 1:3:3:1.
Answer
The combinations are:
a) all spins down (and there is only one way to do that)
b) two of the spins are down, but one is up (and each of the three protons could be up, so there are three ways of doing that)
c) two of the spins are up, but one is down (and each of the three protons could be down, so there are three ways of doing that)
d) all spins up (and there is just one way to do that).
The result is a 1:3:3:1 quartet.
Exercise $2$
Predict the multiplicity in the EPR spectrum for each of the following alkoxy radicals (note that oxygen and carbon have no unpaired spins; assume the same is true for X):
a) X3C-O. b) X2CH-O. c) XCH2-O. d) CH3-O.
Answer a
a) singlet
Answer b
b) doublet
Answer c
c) triplet
Answer d
d) quartet
Exercise $3$
Suppose benzene were reduced by one electron to obtain the benzene radical anion. What would be the multiplicity in the EPR spectrum?
Answer
A septet (in a 1:3:5:7:5:3:1 ratio).
Coupling to Metal Ions
Lots of nuclei other than hydrogen have a net spin. If the unpaired electron happens to be found on a metal, the EPR spectrum may provide confirmation of that structural information. This confirmation may come from both the magnetic field information (similar to chemical shift in NMR) and from the multiplicity.
Nuclear spins of selected metals are shown below.
Metal Spin
V 7/2
Mn 5/2
Fe 0
Co 7/2
Cu 3/2
In each of these metals, the nucleus has different possible magnetic fields. Note that their effects are slightly more complicated than that of a hydrogen atom. For example, copper, with spin 3/2, acts a little like three different hydrogen nuclei (each with spin +/- 1/2) in terms of its effect on a nearby electrom's EPR spectrum. The multiplicity of an unpaired electron on a copper ion should be pretty distinctive.
Exercise $4$
Predict the multiplicity of a peak in the EPR spectrum for an unpaired electron on each of the following metals:
a) vanadium b) manganese c) iron d) cobalt e) copper
Answer a
a) an octet (in a 1:3:5:7:7:5:3:1 ratio)
Answer b
b) a sextet (in a 1:3:5:5:3:1 ratio)
Answer c
c) a singlet
Answer d
d) an octet
Answer e
e) a quartet (in a 1:3:3:1 ratio)
Sometimes things are more complicated, because different isotopes of the same element may have different possible spin values. In fact, that is true with hydrogen and carbon, but the great majority of hydrogen is found as 1H, so in general we can think of it as having spin = 1/2; the great majority of carbon is 12C, with spin = 0.
The natural isotopes of iron, and their nuclear spins, are shown in the table below.
Isotope Spin % Abundance
54Fe 0 5.9
56Fe 0 91.8
57Fe 1/2 2.1
58Fe 0 0.2
The EPR spectrum of an unpaired electron on iron may be slightly more complicated than we first thought. Within the sample, some of the electrons would be situated on iron ions with spin = 1/2, although most would not. This complication may be enough to introduce a slight variation in the appearance of the spectrum, but overall it would still look pretty much like a singlet.
However, in many cases things do get much more complex. Molybdenum provides a good example.
Isotope Spin % Abundance
92Mo 0 14.8
94Mo 0 9.3
95Mo 5/2 15.9
96Mo 0 16.7
97Mo 5/2 9.6
98Mo 0 24.1
100Mo 0
9.6
There are seven naturally occurring isotopes of molybdenum. Five of them have spin = 0, so an unpaired electron on those isotopes would give rise to a simple singlet in the EPR spectrum. The other two isotopes, comprising 25% of the total, have spin = 5/2. Most unpaired electrons on molybdenum would show up as a singlet. However, a significant fraction would show up as a sextet. That means that, in an ideal case, an unpaired electron on molybdenum would give rise to a singlet with a sextet superimposed on it (about a quarter as strong as the singlet).
This situation might look something like the drawing below.
In reality, EPR spectra are enormously complicated in many cases. They often look like fuzzy blobs. There are so many things coupling to so many other things that it becomes almost impossible to decipher by eye. In most cases, computer simulations are run and the experimental data is compared to the computer simulations to obtain structural insight.
Exercise $5$
Vanadium is present in some nitrogenases and so there has been interest in model complexes (e.g. Sandro Gambarotta et al, J. Am. Chem. Soc. 1994, 116, 6927-6928). Gambarotta used the following synthesis, in THF solvent:
$\ce{VCl3 + [(Me3Si)2N]K -> [(Me3Si)2N]3V (THF)} \nonumber$
1. Draw the structure of the product of the reaction.
2. Draw a d orbital splitting diagram for this complex.
3. An EPR spectrum was recorded for this compound. Sketch the spectrum, given that vanadium has nuclear spin I = 7/2.
4. Estimate μeff for this compound.
The compound reacts with N2, forming a N2-bridged dimer.
e) Draw the structure of this product.
f) This compound produces no EPR spectrum. Provide a resonance structure of (e) that explains this observation.
Answer
EPR spectra provided courtesy of Virtual Imagination / Slapdash Chemistry Creations. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/V%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_3/06%3A_Radical_Reactions/6.11%3A_Detection_of_Unpa.txt |
Exercise 6.1.1:
Exercise 6.1.2:
Exercise 6.1.4:
Exercise 6.1.5:
Exercise 6.1.6:
Exercise 6.2.2:
a)
b)
c)
Exercise 6.2.3:
Exercise 6.3.1:
Exercise 6.3.2:
Exercise 6.3.3:
Exercise 6.4.1:
Exercise 6.4.2:
The reduction potential of the negatively charged carboxylate anion would be much less positive than the reduction potential for the neutral aldehyde or ketone. We think of a carboxylate anion as much less electrophilic than aldehydes and ketones for the same reason.
Exercise 6.4.3:
Exercise 6.4.4:
Exercise 6.5.1:
Exercise 6.5.3:
a) no; a stronger bond would have to be broken and replaced with a weaker bond.
b) yes; a weaker bond would be broken and replaced with a stronger one.
c) no
d) yes
e) no
f) yes
Exercise 6.5.4:
a) no b) yes c) no
Exercise 6.5.5:
1. The effect is similar to the stability of carbocations. The more substituted radical is more stable. Thus, the trend from most to least stable is tertiary > secondary > primary > methyl radical. The trend likely originates from a hyperconjugation effect, as in carbocations.
2. The trend here is that if the radical is delocalized by resonance, it is more stable. The allyl and benzyl radicals are more stable than the isopropyl radical. This trend is also seen in cations.
3. The trend here has to do with "hybridization effects" or the atomic orbitals that contribute to the formation of molecular orbitals involved in the relevant bond. In a linear alkyne, the C-H bond can be formed only from some combination involving a hydrogen 1s orbital, carbon 2s orbital and one of the carbon 2p orbitals. This combination is called a "sp" hybrid and the orbital that combines with the hydrogen can be considered 50% 2s, 50% 2p in character.
In a planar alkene, the C-H bond can be formed only from some combination involving a hydrogen 1s orbital, carbon 2s orbital and two of the carbon 2p orbitals (since two of them could lie in this plane). This combination is called a "sp2" hybrid and the orbital that combines with the hydrogen can be considered 33% 2s, 66% 2p in character.
In a tetrahedral alkane, the C-H bond can be formed from some combination involving a hydrogen 1s orbital, carbon 2s orbital and all three of the carbon 2p orbitals. This combination is called a "sp3" hybrid and the orbital that combines with the hydrogen can be considered 25% 2s, 75% 2p in character.
Because a 2s orbital is lower in energy than a 2p orbital, a bond that has greater 2s character is lower in energy than a bond with less 2s character. That means that a bond with greater 2s character is harder to break than a bond with less 2s character. Hence, the alkane C-H bond is weaker than the alkene C-H bond, which is weaker than the alkyne C-H bond.
Exercise 6.5.6:
The Sn-H bond has a dissociation energy of about 78 kcal/mol, compared to about 98 kcal/mol for the C-H bond in ethane. The formation of the Sn-H bond would not compensate for the energy needed to break the C-H bond. On the other hand, the 100 kcal/mol released upon formation of a Sn-Cl bond would more than make up for the 85 kcal/mol required to breal a C-Cl bond.
We could try to rationalise those differences, although bond strengths are always very complicated issues and we will not be able to explain things satisfactorily without quantum mechanical calculations. Let's start with two basic factors, though: the amount of covalency and the amount of polarity.
The difference between the covalent radii of tin and hydrogen (1.39 vs. 0.31 Å) is much greater than the difference between tin and chlorine (1.39 vs 1.02 Å), so there may be less overlap and less covalency between tin and hydrogen than between tin and chlorine. By comparison, the covalent radius of carbon is about 0.76, which puts it somewhere in between hydrogen and chlorine.
In addition, as measured on the Pauling scale, the electronegativity values of these atoms are: chlorine, 3.16; carbon, 2.55; hydrogen, 2.2; tin, 1.96. The tin-chlorine bond would have a large ionic component; this additional component of bonding would strengthen the Sn-Cl bond.
Exercise 6.5.7:
Exercise 6.5.8:
1. 174 - 90 kcal/mol = 84 kcal/mol for the π contribution only
2. 179 - 92 kcal/mol = 87 kcal/mol for the π contribution only
Exercise 6.5.9:
1. 85 + 77 kcal/mol = 162 kcal/mol for the combined σ + π contribution
2. There may be significant differences between the π bond in methanal imine (CH2=NH) and the imine for which we have bond strength data. For example, breaking the bond would result in radicals next to phenyl groups, which may be significantly stabilized. On the other hand, the π bond itself may be significantly stabilized by conjugation. Thus, our estimate is probably not correct, but it is difficult to say whether it is too high or too low.
Exercise 6.7.1:
Exercise 6.9.2:
The mechanism would begin with initiation of the benzoyl peroxide.
The resulting phenyl radical acts as the initiator for polymerisation.
Additional propagation steps follow.
Exercise 6.9.3:
The monomer dependence is straightforward: the more monomer there is present, the faster it can be enchained by the growing radical chain.
The initiator dependence is slightly more complicated, but not much. The more growing radical chains there are, the faster the monomers can be enchained. The more phenyl radical is formed, each initiating a new radical chain, then the more growing radical chains there will be.
$Rate = k[monomer][Ph \cdot] \nonumber$
The more benzoyl peroxide there is, the more phenyl radical there will be, but the relationship is not linear, as explained below.
Benzoyl peroxide cleaves in two, ultimately producing two phenyl radicals:
$K_{eq} = \frac{[Ph \cdot]^{2}}{[BP]} \nonumber$
so $[Ph \cdot]^{2} = K_{eq}[BP]$
and $[Ph \cdot] = K_{eq}^{1/2}[BP]^{1/2}$
therefore $Rate= k[monomer][Ph \cdot]^{1/2}$
Exercise 6.9.4:
Exercise 6.9.5:
Exercise 6.9.6:
Exercise 6.9.7:
Peroxy radicals are relatively stable and react slowly. However, when they do react, they also undergo reactions other than radical polymerisation, so that they are even more effective as inhibitors of polymerisation.
Exercise 6.10.1:
Because each initiator breaks in half, forming two radicals, one initiator starts two growing chains. Every initiator will, on average, consume half the monomers, assuming no unexpected chain termination events.
1. DP = 500 monomers / 2 growing chains = 250
2. DP = 1000 monomers / 2 growing chains = 500
3. DP = 600 monomers / 3 growing chains = 200
Exercise 6.10.2:
M0 = molecular weight of monomer in each case. 1 D (Dalton) = 1 amu
a) $M_{n} = DP \times M_{0} = 1250 \times 104 D = 130000D = 130 kD}$
b) $M_{n} = DP \times M_{0} = 725 \times 85 D = 61625D = approximately \: 62 kD}$
c) $M_{n} = DP \times M_{0} = 1420 \times 86 D = 122120 D = approximately \: 122 kD}$
Note that end groups were neglected in these calculations; they may contribute a significant percentage of total molecular weight at lower DP. A more accurate calculation would account for the identity of both end groups.
Exercise 6.10.3:
1. $DP = \frac{M_{n}}{M_{0}} = \frac{11450D}{53D} = 216}$
2. (DP = \frac{M_{n}}{M_{0}} = \frac{24760D}{86D} = 288}\)
3. (DP = \frac{M_{n}}{M_{0}} = \frac{927000D}{104D} = 8914}\)
Exercise 6.10.4:
1. $MW = 8 \times 100D = 800 D$
2. 15 monomers / 3 chains = 5 new monomers / chain
$DP = 8 + 5 = 13 \nonumber$
c) $MW = 13 \times 100 D = 1300 D$
d) All chains are the same length; PDI = 1.0
e) $MW = 16 \times 100 D = 1600D$
f) $MW = ( 8 + 15) \times 100D = 2300D$
g) $M_{n} = \frac{2300 + 1600D}{2} = 1950D$
h) $PDI = \frac{2300D}{1600D} = 1.44$
Exercise 6.10.5:
a) The chain that abstracted the hydrogen: $MW = 8 \times 100D = 800D$
The branched chain: $MW = 18 \times 100D = 1800 D$
The normal chain: $MW = 13 \times 100D = 1300D$
b) $M_{n} = \frac{1800 + 1300 + 800D}{3} = 1300D$
c) $PDI = \frac{1800}{800D} = 2.25$
Exercise 6.10.6:
1. $M_{n} = \frac{17000 + 1800 + 300D}{3} = \frac{3800D}{3} = 1267$
2. $M_{n} = \frac{1700 + 1800}{2} = \frac{3500D}{2} = 1750D$
Exercise 6.10.7:
Exercise 6.10.8:
Exercise 6.10.9:
Exercise 6.11.1:
The combinations are:
1. all spins down (and there is only one way to do that)
2. two of the spins are down, but one is up (and each of the three protons could be up, so there are three ways of doing that)
3. two of the spins are up, but one is down (and each of the three protons could be down, so there are three ways of doing that)
4. all spins up (and there is just one way to do that).
The result is a 1:3:3:1 quartet.
Exercise 6.11.2:
a) singlet b) doublet c) triplet d) quartet
Exercise 6.11.3:
A septet (in a 1:3:5:7:5:3:1 ratio).
Exercise 6.11.4:
1. an octet (in a 1:3:5:7:7:5:3:1 ratio)
2. a sextet (in a 1:3:5:5:3:1 ratio)
3. a singlet
4. an octet
5. a quartet (in a 1:3:3:1 ratio)
Exercise 6.11.5: | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/V%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_3/06%3A_Radical_Reactions/6.12%3A_Solutions_for_Sel.txt |
Oxidative phosphorylation is a metabolic process in which energy is harnessed for the production of ATP. The process occurs in the mitochondria. Electrons released through the oxidation of glucose are shuttled into the oxidative phosphorylation supercomplex via FMNH2. The electrons are passed through a remarkable electron transport chain along and across the mitochondrial membrane. The electron transport chain releases minute amounts of energy with each electron transfer, and the transport is coupled to the pumping of protons across the mitochondrial membrane. Eventually, the electrons are delivered to molecular oxygen, which is reduced to water. Finally, the protons that have gathered on the edge of the mitochondrial membrane cascade back across, turning a molecular millwheel that drives the manufacture of ATP. The ATP is used to power processes throughout the cell.
The mitochondria are the site of metabolic activity in the eukaryotic cell. The citric acid cycle occurs within the mitochondrial matrix, catalysed by a range of metabolic enzymes. For this reason, mitochondria are sometimes called "the power plants of the cell". Oxidative phosphorylation plays a central role in this production, harvesting electrons from NADH and succinate to manufacture ATP.
The complexes involved in oxidative phosphorylation are embedded in the inner mitochondrial membrane. In the picture below, the lower portion of the outer mitochondrial membrane is visible at the top. The inner mitochondiral membrane stretches across the middle of the picture. Both membranes are formed by lipid bilayers. In contrast, both the intermembrane space and the mitochondrial matrix are aqueous environments. The complexes that take part in oxidative phosphorylation are labelled I-V in the picture.
Each complex is actually a collection of different proteins; Complex I alone is composed of over 40 protein subunits, but in the picture above each complex has been simplified to one monolithic block. Each complex has its own specialised role. Both Complex I and Complex II serve as entry points for electrons into the respiratory electron transport chain. Complex I accepts electrons from NADH, produced in glycolysis and the citric acid cycle. Complex II accepts electrons from succinate, which is one of the intermediates in the citric acid cycle. In fact, Complex II is an integral part of the citric acid cycle, since it carries out a key step in that process.
• Complex I and II are the entry points for electrons into the electron transport chain.
Both Complex I and Complex II release small amounts of energy as electrons roll energetically downhill to sites of higher and higher reduction potential. The electrons are then shuttled to the same acceptor, Complex III, via a lipid-soluble electron carrier molecule. The electron transport chain continues, releasing some more energy before the electrons are passed to the final destination in Complex IV. This time, the trip from Complex III to Complex IV is conducted via a hydrophilic protein, cyctochrome c. Electrons travel through Complex IV, back towards the matrix, and are accepted by molecular oxygen, resulting in its reduction to water.
• Small amounts of energy are released as electrons move to sites of higher reduction potential.
• Complex IV is the final destination of the electron transport chain.
• In Complex IV, electrons are used to convert O2 to H2O.
Complex I, III and IV all use the energy released from the electron transport chain to pump protons from the matrix into the intermembrane space. The proton gradient that results across the inner mitochondrial membrane is used to power ATP production in complex V. In addition, a couple of protons are consumed by Complex I and Complex II as they package electrons into the lipid-soluble carrier, ubiquinone/ubiquinol. However, Complex II does not transport any electrons all the way across the inner mitochondrial membrane.
• Complex I, III and IV all use the energy released from the electron transport chain to pump protons across the inner mitochondrial membrane.
• The proton gradient across the membrane drive a nano-millwheel used to manufacture ATP from ADP and inorganic phosphate.
In the following pages we will take a closer look at each of the complexes that take part in this process.
Visit an overview of oxidative phosphorylation at Henry Jakubowski's Biochemistry Online. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/V%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_3/07%3A_Oxidative_Phosphorylation/7.01%3A_Introduct.txt |
Complex I is a collection of proteins that serves as one of two entry points into oxidative phosphorylation; the other is complex II. Both of these complexes accept electrons from molecules produced in the catabolism of glucose. By shuttling these electrons from one electron acceptor to another, generally moving to higher potential (remember, in redox terminology, that means lower energy) the assembly of proteins that take part in oxidative phosphorylation are able to produce ATP. ATP, in turn, is used to power other metabolic processes.
The main events in Complex I are summarised in the cartoon below. You can see the electrons entering from the matrix at the bottom of the picture (the pathway is shown by the blue arrows). They are delivered by NADH and handed off to FMN; this step will be discussed below. The electrons are transferred via outer sphere electron transfer through a series of iron sulfur clusters and are eventually delivered to the lipid-soluble ubiquinone (Q).
• Complex I marks the beginning of the electron transport chain.
• Electrons are delivered from NADH, pass through Complex I, and finally reach a ubiquinone.
• The energy released during electron transport helps pump protons across the complex.
The picture below is from an X-ray crystal structure of Complex I (the source of the data is cited at the end of this page). Rather than showing every atom, which is the usual result of a crystal structure, the data are displayed in a "cartoon" form, so that you can get a better sense of the overall structure. The picture is also colour-coded to help you see structures more clearly. The pink helices (spirals) along the top are the portion of the complex that is bound in the inner mitochondrial membrane. The membrane, too, would extend acorss the top of the picture from left to right. It is easy to imagine the parallel bundle of α-helices fitting nicely in the midst of the parallel array of phospholipids that form the membrane. The egg-shaped, yellow and pink form to the lower right is the part of the complex that extends out into the mitochondrial matrix. The yellow portions indicate β-sheets, whereas the white threads indicate loops.
Exercise \(1\)
Specific amino acids are likely to be found along the α-helices that bind in the membrane. Indicate different possibilities below.
In the case of Complex I, the electrons are introduced in the form of NADH. Electrons travel through Complex I and eventually are delivered to ubiquinone; ubiquinone carries the electrons to the next stage of the oxidative phosphorylation supercomplex, which is Complex III. Because Complex I takes electrons from NADH and delivers them to ubiquinone, Complex I is also referred to as "NADH:ubiquinone oxidoreductase".
NADH is produced during glycolysis and the TCA cycle. Remember, NADH is a two-electron donor: it donates a hydride ion to a substrate, becoming NAD+. A hydride ion, of course, is just a proton and two electrons.
• NADH produced in glycolsis and the TCA cycle delivers a pair of electrons to Complex I.
• The NADH is delivered to complex I from the mitochondrial matrix (in the interior of the mitochondrion).
By far the most common electron acceptor in the oxidative phosphorylation supercomplex is an iron atom. Of course, the most common oxidation states for iron ions are Fe2+ and Fe3+. An iron in the 3+ oxidation state is able to accept an electron, becoming Fe2+. In contrast, an iron in the 2+ oxidation state could pass an electron on, becoming Fe3+ in the process.
Picture a bucket brigade, in which people passing buckets of water from one to another all act together to put out a fire. The iron atoms act in much the same way, each passing an electron to the next in order to complete the electron transport chain.
• Electron transport is accomplished via many small steps rather than a few large ones.
If we strip away the proteins from Complex I, we can get a picture of some of the other pieces inside. Looking at the X-ray crystal structure data, we can simply ignore every atom in the protein, until we are left with the "ligands". In biochemistry, ligands means the stuff attached to the proteins (as opposed to in inorganic chemistry, where it means the stuff attached to the metals). That's what we see below. The red and yellow shapes that you see are iron sulfide complexes, strung along so that they can pass electrons along through Complex I. The pieces that we see here are found within the hydrophilic part of Complex I; that is, they are found within the yellow and pink egg-shaped part seen in the structure above. This picture is oriented in the same direction as the one above; the iron complexes extend from the lower part of the hydrophilic domain all the way up to the edge of the hydrophobic, membrane-bound portion of the complex.
So, we have a "wire" to carry the electrons through the complex after they are delivered by NADH. We have a mismatch problem, however. NADH is a two electron donor. An Fe3+ ion is a one-electron acceptor. We need an adapter for this electrical connection. The adapter comes in the form of FMN. FMN is the structure with some atoms coloured in blue and red near the bottom right corner of the picture.
• NADH only donates two electrons at a time.
• The iron ions in the electron transport chain can toggle between Fe(III) and Fe(II); they can accept only one electron at a time.
• An adapter is needed to convert two-electron transfer into one-electron transfer.
FMNH2 is a little bit like NADH. Its oxidized form, FMN, can accept two electrons and a proton in the form of a hydride ion, as well as an additional proton. In other words, FMN accepts H- and H+ to become FMNH2. However, a slightly different route is available to FMN. It can also undergo reduction one electron at a time. In reality, the addition of an electron would be shortly preceded by or shortly followed by addition of a proton, in order to keep the overall charge the same. This state, FMNH, is called the semiquinone form.
What's the difference between NAD and FMN? Why is one able to accept only a pair of electrons, whereas the other can accept one at a time? When FMN accepts one electron, it becomes a radical. Radicals are unstable, reactive species. They can be stabilised chiefly by delocalisation. The additional conjugation in FMN compared to NAD allows the odd electron to be delocalised more extensively in FMN. That radical stability is the key difference.
• The presence of extended conjugation stabilises a radical on FMNH.
• The stability of this radical allows FMN to accept one electron at a time.
Exercise \(2\)
Provide a mechanism for the conversion of FMN to FMNH2 in the presence of NADH and a lysine side chain, at pH 7.
Answer
Once FMNH2 has formed, the reverse is true, of course. It can give up one electron at a time. As a result, FMN can take a pair of electrons coming in from NADH and send them out one at a time into the electron transport chain.
The rest of the electron transport chain through Complex I is a series of iron-sulfur clusters. As the name suggests, these clusters consist of iron and sulfur atoms. The most common variety contains four iron atoms and four sulfur atoms arranged at the corners of a cube. These clusters are often referred to as 4Fe4S clusters for obvious reasons: there are 4 iron atoms and 4 sulfur atoms. The sulfur atoms at the corners are really sulfide ions, S2-. In addition to the three sulfur ions, each of the iron atoms is also bound to an anionic cysteine residue, so that the iron has tetrahedral coordination geometry. The iron atoms are present as a combination of Fe2+ and Fe3+ ions.
• Iron sulfur clusters are very common in biological electron transport.
• The iron ions can be Fe(II) or Fe(III).
• The ligands for the iron ions include sulfide ions, S2-.
• The iron sulfide clusters are usually held in place in the protein by cysteine ligands (CysS-).
Here is another view of the ligands, seen from a different point of view this time. See whether you can find a group of four iron atoms (coloured red) bound with four sulfur atoms (coloured yellow).
There are other variations of FeS clusters. A very common one is a 2Fe2S cluster, which of course consists of two iron atoms and two sulfide ions at alternating corners of a diamond. There are a couple of these clusters visible in the picture above. Again, the iron atoms could be Fe2+ or Fe3+ ions, or one of each. Also, each iron is usually bound to two additional cysteine anions to complete a tetrahedral geometry. Those groups aren't shown here, though, because we have left the protein out of the picture. (We will see eventually that other amino acids occasionally bind iron sulfur clusters in place of cysteine.)
Another possibility is a 3Fe4S cluster, a lop-sided affair in which one of the iron atoms is essentially left out of the FeS cube.
Exercise \(3\)
Assuming one iron is present as Fe(III) and the rest as Fe(II), calculate the overall charges on:
a) 2Fe2S b) 3Fe4S c) 4Fe4S
Answer a
a) Iron charges: Fe(II) + Fe(III) = 5+
Ligand charges: 2 sulfides = 2 x 2- = 4- ; 4 cysteines = 4 x 1- = 4- ; total = 8-
Overall: 3-
Answer b
b) Iron charges: 2 x Fe(II) + Fe(III) = 4+ + 3+ = 7+
Ligand charges: 4 sulfides = 4 x 2- = 8- ; 3 cysteines = 3 x 1- = 3- ; total = 11-
Overall: 4-
Answer c
c) Iron charges: 3 x Fe(II) + Fe(III) = 6+ + 3+ = 9+
Ligand charges: 4 sulfides = 4 x 2- = 8- ; 4 cysteines = 4 x 1- = 4- ; total = 12-
Overall: 3-
Exercise \(4\)
Environment plays a role in modulating reduction potentials in proteins. Suppose a 2Fe2S cluster was in a mixed Fe(II)/(III) oxidation state. How would its reduction potential when surrounded by nonpolar amino acid residues compare to its reduction potential when surrounded by polar amino acid residues?
Answer
Upon reduction, the charge on a 2Fe2S cluster will increase from 3- to 4-, assuming it starts in a mixed Fe(II)/(III) state (whereas if it starts in a Fe(III)/(III) state, the overall charge will increase from 2- to 3-). These anions would be stabilised by strong intermolecular interactions such as ion-dipole forces. Both states (oxidised and reduced) will be stabilised by a polar environment, but the more highly charged reduced state will depend even more strongly on stabilisation by the environment. As a result, we might expect the reduction potential to be lower when surrounded by nonpolar amino acid residues, and higher if surrounded by polar residues.
There is a whole series of these clusters in Complex I. The electrons delivered from NADH are sent from one to the next and then to the next. There are a couple of reasons for this arrangement.
Because these clusters are all bound in place by the protein -- completely immobilised -- the electron transfer has to occur via an outer sphere mechanism. Why are there so many clusters? Remember, outer sphere electron transfers have a limited range. The electron, under most circumstances, can only hop so far. By providing a series of conducting FeS clusters, the electron can hop from one to the next, arranged like stepping stones across a river.
Not only that, but the electron transport chain takes electrons from NADH and delivers them, ultimately, to molecular oxygen in Complex IV. The molecular oxygen is converted to water. That transfer from NADH to water is very exothermic. The reaction is very much downhill in energy. In order to make this transfer practically feasible, and in order to harness the enormous amount of energy involved, the electrons are allowed to step downhill just a little at a time.
Occasionally, electrons may even hop back uphill slightly, recapturing energy that has been lost to the surroundings in previous transfers. This damping effect may make the whole process more efficient. Still, the electrons are overall rolling downhill energetically. An electron may jump uphill a few times, but eventually those uphill jumps will be followed by a downhill drop, so that overall the electron has moved to lower energy.
• An occasional uphill electron transfer absorbs energy.
• Re-absorption boosts efficiency by preventing heat loss.
In Complex I, the final destination for the electron transport chain is a ubiquinone, sometimes abbreviated as Q or UQ. Like FMN, UQ is a two electron, two proton acceptor, to become UQH2. Also like FMN, UQ can accept one proton and one electron at a time, to form the semiquinone form, UQH. Once again, this is a radical species.
Exercise \(5\)
Provide a mechanism for the conversion of UQ to UQH2 in the presence of Fe2+ ions and lysine side chains, at pH 7.
Answer
UQ is very different from the FeS clusters or the FMN because it isn't attached to a protein. It isn't tied down. It can move around. That makes it a mobile electron carrier. In addition to being a relatively high-potential electron acceptor (at least compared to other things in Complex I), the role of UQH2 is to deliver electrons to Complex III so that the electron transport chain can continue.
The trouble is, if UQH2 is mobile, what's to keep it from wandering away? How is its pathway limited so that it is more likely to reach its destination? Remember that the oxidative phosphorylation supercomplex is a group of membrane-bound proteins. They are held in a lipid-rich environment. The structure of UQH2, with its long tail, is very lipophilic. If it stays in the membrane, its movements are limited to two dimensions, rather than three, and it is much more likely to reach its destination of Complex III.
• Ubiquinone is a lipid-soluble, mobile electron carrier.
• Its movement is restricted to the mitochondrial membrane.
There is one more important feature of Complex I. Like some of the other complexes involved in oxidative phosphorylation (Complexes III and IV), Complex I pumps protons across the inner mitochondrial membrane. Ultimately, the protons that have gathered on the edge of the inner mitochondrial membrane cascade back across, turning a molecular millwheel that drives the manufacture of ATP. The ATP is used to power processes throughout the cell.
• Protons are actively pumped across the mitochondrial membrane.
• As a result, a charge develops across the membrane.
• The mitochondrial matrix becomes "n-doped" or negatively charged.
• The intermembrane space becomes "p-doped" or positively charged.
This proton pump is an example of active transport. Energy is expended to transport protons across the membrane, despite a buildup of positive charge in the intermembrane space (and a corresponding buildup of negative charge in the matrix). The energy released by the electron transport chain may be responsible for conformational changes in the protein that help this transport occur.
There still seems to be some discussion going on about how, exactly, the proton pump works in this system. However, some things are clear. The transport of proteins across a hydrophobic membrane is likely facilitated by the presence of hydrophilic portions of the protein. There are believed to be channels that open up in the protein, allowing water molecules to move through the protein. Because the protein is embedded in the membrane, these protons are also crossing the membrane at the same time.
Exercise \(6\)
Frequently, specific amino acids can play a role in assisting the transport of protons (or other ions). Fill in some different possibilities for these amino acids.
Exercise \(7\)
In general, the proton that enters the complex on one side of the membrane is probably not the same proton that emerges on the other side. Provide a mechanism with arrows to illustrate this process.
The release of energy over the electron transport chain drives the transport of protons across the membrane. There is another factor that helps, too. Like the electrons, the protons travel from the matrix toward the intermembrane space. Both positive and negative chages are traveling in the same direction. That raises the possibility of coupled transport, in which the flow of electrons through the protein makes it easier for protons to follow along (or vice versa).
In coupled transport, the movement of an proton is quickly followed by the transfer of an electron (or vice versa).
• Electrons and protons are travelling in the same direction through complex I: from the matrix toward the intermembrane space.
• Their opposite charges may lead to a coupled mechanism in which the movement of one makes it easier for the other to follow.
Complex I is not the only entry point for electrons into the electron transport chain. Complex II plays a similar role. Together, they harvest energy from the electron transport chain; that energy is ultimately used to make ATP, which can move through the cell to release energy elsewhere.
Exercise \(8\)
Provide a mechanism for the oxidation of FMNH2 by iron(III).
Exercise \(9\)
It's difficult to measure the reduction potential of an individual site within a protein. However, researchers have been able to estimate these values by measuring EPR spectra under various conditions. For example, here is an approximate picture of potentials in Complex I.
1. Two of the N clusters are probably not directly involved in the electron transport chain. Which ones?
2. Use the data in the diagram to construct a potential energy diagram for the transfer of the electron along the pathway.
Exercise \(10\)
Using the values in the figure above, calculate the energy change when an electron is transferred from the N5 cluster to the N6a cluster.
Answer
Assuming the reduction potentials are:
N5(ox) + e- → N5(red) Eored = -0.40 V
N6a(ox) + e- → N6a(red) Eored = -0.30 V
Then the potential difference for the reaction, ΔEo = -0.30 - (-0.40) V = 0.10 V.
The Faraday relation ΔG = - n F ΔEo gives
ΔG = - 1 x 96,485 J V-1 mol-1 x 0.10 V = 9,649 J mol-1 = 9.7 kJ mol-1 | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/V%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_3/07%3A_Oxidative_Phosphorylation/7.02%3A_Complex_I.txt |
Complex II is another group of proteins that serves as a second entry point into the electron transport chain, which is involved in the additional production of ATP to power cellular processes. The electron transport chain releases minute amounts of energy with each electron transfer, and the transport is coupled to the pumping of protons across the mitochondrial membrane. Eventually, the electrons are delivered to molecular oxygen, which is reduced to water, in Complex IV.
A schematic of Complex II is provided below. It shows electrons provided by succinate traveling first to FAD and then along a series of iron sulfur clusters to a ubiquinone acceptor.
• Complex II, like Complex I, is an entry point into the electron transport chain.
• In Complex II, two electrons are delivered from succinate.
The X-ray crystal structure of Complex II is shown below. It is depicted in cartoon form to highlight the structure, with pink helices, yellow sheets and white loops. Just as in the structure of Complex I, the structure here is shown oriented so that the inner mitochondrial membrane would stretch across the top of the picture; if it were shown, we would see it in cross section, from side to side. Once again, there is a hydrophobic domain in the upper part of the complex that would be embedded in that membrane. Below that lies a much larger hydrophilic domain, which would extend into the mitochondrial matirx.
Complex II actually plays a part in another biochemical process, the citric acid cycle. One of the steps of the citric acid cycle is the oxidation of succinate to fumarate. It is during this oxidation that additional electrons are introduced into the electron transport chain via Complex II. At the same time, two protons are delivered from succinate as well. The loss of two protons and two electrons from succinate corresponds to a formal loss of H2 (formal, meaning a molecule of H2 is not actually formed, but the pieces are equivalent to a full H2 molecule). A double bond forms in the new fumarate molecule, which is said to be "dehydrogenated" with respect to the original succinate. Complex II is the enzyme that carries out this transformation in the citric acid cycle, and so the complex is also called "succinate dehydrogenase".
• Complex II is actually an integral part of the TCA cycle, which takes place in the mitochondrial matrix.
• Conversion of succinate to fumarate is a two-electron, two-proton process.
Shown below, the proteins are stripped out of the picture to reveal some of the working parts inside. Having already seen Complex I, you will probably recognize a few iron sulfur clusters, shown again in red and yellow. At the bottom of the picture, close to the matrix, you can see a flavenoid structure, a little like the FMN you saw in Complex I. At the top, actually buried within the hydrophobic portion (you can just see the edge of it if you look between two of the pink helices in the picture above), is an iron heme complex; it's a little like the active site of hemoglobin. You can see the iron atom, coded in red, and the four nitrogen atoms around it, coloured blue. However, it is not clear what role this heme actually plays in electron transport through Complex II.
In between the highest iron sulfur cluster and the iron heme is a molecule of pentafluorophenol. That's not supposed to be there in the cell. It's an inhibitor; the researchers added it to hold the place of a ubiquinone while they were growing the crystal that they would analyse to get this structure. There is actually a second inhibitor in the structure, an oxaloacetate, which holds the place of the succinate molecule that would deliver electrons to the complex. However, the oxaloacetate is not clearly visible because it's partly obscured by the flavenoid at the bottom of the picture.
• Like in Complex I, electrons eventually arrive at ubiquinone in Complex II.
• Ubiquinone is a mobile, lipid-soluble, two-electron carrier.
Complex II is unique in a another important way. Unlike Complexes I, III, and IV, it takes part in the electron transport chain but does not actively pump protons across the inner mitochondrial membrane. Instead, the electrons that it harvests contribute to building the proton gradient once they arrive at Complex III (and later at Complex IV).
• Complex II does not pump protons directly.
• Complex II does send two protons on to Complex III in the form of the reduced ubiquinol.
In terms of how those electrons make their arrival, the picture is somewhat similar to Complex I. Succinate is a hydride donor in this reaction, so it is donating two electrons. As in Complex I, a series of FeS clusters will eventually relay the electrons, so we have a matching problem. To step down from a two-electron process to a one-electron process, another flavenoid, FAD, acts as an intermediary. Like the related FMN, FAD can accept either one or two electrons at a time, and its reduced form, FADH2, can donate either one or two electrons at a time.
• Just like in Complex I, an adapter is needed to convert the two-electron process of succinate oxidation to the one-electron process of iron reduction.
• The adapter is another flavenoid structure, FAD, which is very similar to the FMN used in Complex I.
After the electrons have arrived and are stored in the form of FADH2, they are passed along, one at a time, through a series of three FeS clusters. Just as in Complex I, these carriers are all bound to the proteins, so they are held in one place and do not move. The outer sphere mechanism by which the electrons are transferred can only operate over a short distance, from one electron carrier to the next.
Complex II also contains a structure we did not see in Complex I: a cytochrome. A cytochromes is a protein that contains a heme complex of iron. They look a little bit like hemoglobin, with which you are undoubtedly familiar by now. As in the FeS clusters, the cytochrome can access either the Fe(II) or the Fe(III) state. Like the FeS clusters, they just accept one electron at a time. We will find out that some cytochromes are mobile, but this one is not. It stays where it is and waits for a ubiquinone to bind nearby.
There are few different variations in the heme family that are commonly encountered in biology. Heme b is a common one and has the simplest structure. Heme a contains a hydrocarbon tail as well as a formyl group (-HC=O) attached to its central porphyrin ring. Heme c is covalently bound to the surrounding protein via cysteine residues.
Exercise \(1\)
Look again at the X-ray crystal structure of the ligand in Complex II. Which type of heme is present?
Answer
Heme b.
Exercise \(2\)
The basic porphyrin structure at the centre of all the hemes is a macrocycle with the formula (C4H2NCH)4. Draw this structure.
Answer
A porphyrin contains four pyrrole rings (five-membered, aromatic ring containing a nitrogen) arranged to form a 16-membered macrocycle.
The exact role of the cytochrome in Complex II is unclear. Although cytochromes form part of the electron transport chain in Complexes III and IV, it might not do so in Complex II. Another possible role is storage of an electron (which does arrive through the electron transport chain). When the ubiquinone binds nearby, the cytochrome may release its electron to the ubiquinone. A second electron would arrive directly from the nearby FeS cluster. That mechanism may allow for more rapid reduction of ubiquinone to ubiquinol, so that it spends less time in the semiquinone stage. As a result, the radical species would be less likely to move away before being fully reduced. Radical species can cause damage, so limiting the movement of radical species through the cell is important.
• The role of the heme group in Complex I is not clear.
• It has been suggested that the heme group may act like a capacitor. It may store one of the two electrons needed to reduce ubiquinone to ubiquinol.
• Storage of an extra electron could result in more rapid reduction of ubiquinone to ubiquinol.
• Rapid, two-electron reduction limits the lifetime of potentially damaging radical intermediates in the cell.
Once two protons and two electrons are delivered to ubiquinone, the resulting ubiquinol travels on to Complex III for the next step in oxidative phosphorylation.
Exercise \(3\)
Provide a mechanism for the reduction of FAD by succinate.
Exercise \(4\)
It's difficult to measure the reduction potential of an individual site within a protein. However, researchers have been able to estimate these values by measuring EPR spectra under various conditions. Assuming the reduction potentials below, draw a reaction progress diagram for transport of an electron all the way from succinate to ubiquinone.
Exercise \(5\)
Using the values in the figure above, calculate the energy change when an electron is transferred from the 4Fe4S cluster to the 3Fe4S cluster.
Answer
Assuming the reduction potentials are:
4Fe4S(ox) + e- → 4Fe4S(red) Eored = -0.15 V
3Fe4S(ox) + e- → 3Fe4S(red) Eored = 0.06 V
Then the potential difference for the reaction, ΔEo = 0.06 - (-0.15) V = 0.21 V.
The Faraday relation ΔG = - n F ΔEo gives
ΔG = - 1 x 96,485 J V-1 mol-1 x 0.21 V = 20,262 J mol-1 = 20 kJ mol-1
Exercise \(6\)
Assuming all iron is present as Fe(III), calculate the overall charges on:
a) 2Fe2S b) 3Fe4S c) 4Fe4S
Answer a
a) Iron charges: 2 x Fe(III) = 6+
Ligand charges: 2 sulfides = 2 x 2- = 4- ; 4 cysteines = 4 x 1- = 4- ; total = 8-
Overall: 2-
Answer b
b) Iron charges: 3 x Fe(III) = 9+
Ligand charges: 4 sulfides = 4 x 2- = 8- ; 3 cysteines = 3 x 1- = 3- ; total = 11-
Overall: 2-
Answer c
c) Iron charges: 4 x Fe(III) = 12+
Ligand charges: 4 sulfides = 4 x 2- = 8- ; 4 cysteines = 4 x 1- = 4- ; total = 12-
Overall: 0
X-ray crystal structures: Zhou, Q., Zhai, Y., Lou, J., Liu, M., Pang, X., Sun, F. Thiabendazole inhibits ubiquinone reduction activity of mitochondrial respiratory complex II via a water molecule mediated binding feature. Protein Cell 2011, 2, 531-542. Images obtained via RCSB Protein Data Bank (3SFD). | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/V%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_3/07%3A_Oxidative_Phosphorylation/7.03%3A_Complex_I.txt |
Complex III accepts electrons from both Complex I and Complex II. The electrons arrive in the form of ubiquinol, UQH2, which delivers two electrons and two protons and is converted back to ubiquinone, UQ. The ubiquinone then shuttles back to Complex I or Complex II to collect more electrons.
• Complex III is the destination for electrons arriving from both Complex I and Complex II.
• Complex III uses energy released in downhill electron transfers to pump more protons across the inner mitochondrial membrane.
• The proton gradient across the membrane is used to drive ATP formation at Complex V.
• Complex III then sends its electrons on to Complex IV.
The path of electrons through Complex III is shown below. Note that the complex is a dimeric structure, with two equivalent paths: one shown on the left and the other on the right. Electrons are delivered from UQH2, ubiquinol, in the middle of the picture, traveling upward to the mobile cytochrome c at the top. A second electron pathway allows travel downward in the picture, toward another molecule of ubiquinone, which recycles the electrons in a loop.
• Complex III features an unusual, diverging electron pathway.
• One electron travels on toward cytochrome c.
• The second electron is eventually recycled into another ubiquinol.
The X-ray structure of Complex III is shown in cartoon form below. The inner mitochondrial membrane would intersect the upper middle portion of the complex. The majority of the complex projects below, into the matrix (the bottom of the picture, in this orientation) although a significant amount also protrudes into the intermembrane space.
Complex III continues the electron transport chain, sending electrons to higher reduction potential (and lower free energy) and towards a meeting with molecular oxygen in Complex IV. Protons are also pumped from the mitochondrial matrix, across the inner mitochondrial membrane and into the intermembrane space. At the same time, a second electron route results in an effective recycling of half the electrons that come into the complex, increasing the number of protons pumped per electron arriving at Complex III.
• It has been suggested that the electron-recycling loop provides a mechanism for drawing more protons up from the matrix.
We can see this unusual arrangement if we ignore the proteins to reveal the ligands, below. The complex can more clearly be seen as a dimer; the left half of the picture is exactly the same as the right, although reflected the opposite way. Electrons enter the complex via the mobile carrier, ubiquinol, visible on the right and left side of the picture about halfway down; you can see the ringand a chain hanging from it.
When UQH2 binds to Complex III, one electron is sent on to an FeS cluster, whereas the second is sent to an iron heme center, called heme BL. The FeS cluster is visible just above the ubiquinol and to the edge of the picture. The heme BL is toward the center of the picture.
Although it isn't apparent in the X-ray picture shown, this FeS cluster is different than the ones that you saw earlier. This one is held in place with different amino acid side chains. You can't see that in the picture because the entire protein has been left out, along with those amino acids that are binding to the cluster. Whereas most FeS clusters are held in place exclusively by cysteines, this cluster is held in place by two cysteiens on one side but by two histidines on the other. This peculiar FeS cluster is called a Rieske cluster.
Rieske clusters are important because they are "high potential FeS clusters". That means that they have exceptionally positive reduction potentials compared to other FeS clusters. The reduction potential of any ligand in a protein is highly dependent on its environment, and so there is a wide range of values in most cases. That's true for iron sulfur clusters; normal ones have reduction potentials that range from -1.0 V to about +0.05 V. Rieske clusters have potentials that range from about 0V to about 0.4 V.
From the FeS cluster, the electron is sent on to another iron heme center in a membrane-bound unit called cytochrome c1. Cytochrome c1 can be seen at the top of the picture. Because this picture is oriented in the same way as the other X-ray structures of the respiratory complex, we can see that this electron is getting transported up toward the intermembrane space. Finally, this electron is transferred to another heme, but this heme is bound in a small, mobile protein, called cytochrome c. Cytochrome c is another mobile electron carrier. It carries the electron on to complex IV. We'll get back to the cytochrome c a little later.
Before we get to that, where does the other electron go? From heme BL, the second electron is passed to another iron porphyrin complex, heme BH. That's toward the bottom of the picture, so this electron is actually getting sent back toward the matrix. Waiting by the heme BL, however, there is another ubiquinone, UQ. It's waiting for the electron. It will actually wait for two, then leave its dock and go back around to the ubiquinol (UQH2) binding site to deliver the electrons again.
Why bother? Why not just send the electrons on their way properly the first time around? It seems they are getting recycled for some reason. It's believed that this diversion allows for additional protons to be pumped across the mitochondrial membrane. When the second ubiquinone gets reduced to ubiquinol via the acceptance of two electrons, it also picks up two more protons from the matrix below. Those protons will eventually get passed along toward the intermembrane space when the ubiquinol gets oxidized again. This recycling, and squeezing out a couple of extra protons to increase the proton gradient, is called the Q loop.
The advantage of the Q-loop is explained in the following diagram. In the diagram, inputs to complex III are shown in red, outputs are shown in blue, and recycled elements are in green. If one ubiquinol simply delivered its electrons and protons and was done, there would be two protons delivered per ubiquinol. That's one proton output per electron that was input.
A second ubiquinol would do exactly the same thing. There would be four protons output for four electrons initially input. That's still one proton output per electron that was input.
If, instead, one electron is recycled each time, then every second ubiquinol leads to the delivery of an extra pair of protons. That's because in picking up the recycled electrons, a ubiquinone has had to travel back to the matrix side of the membrane and pick up two more protons. Overall, that means six protons are delivered for four electrons input, or 1.5 protons output per electron input. Since the proton gradient is what is generating the ATP, then by increasing the number of protons pumped per electron coming in, efficiency is increased.
• The Q-loop increases the number of protons pumped per electron input into the system.
Now let's get back to the outgoing electron carrier, cytochrome c, which will ferry the electrons on to Complex IV. In the picture below, from a different X-ray crystal structure, the Complex has been found with a cyctochrome c docking at the binding site. Cytochrome c is a pink globular protein attached at the top of the picture, on the left half of the dimer as we look at it here. It's sitting at the edge of the intermembrane space, which is aqueous media. Unlike the ubiquinone, which slips along through the lipid bilayer of the membrane, cytochrome c rolls or slides or swims over the top to its destination.
The same view is shown below without the protein. The bottom part of the picture is pretty cluttered with lipids, and there is a sugar hanging around to the upper left, but right at the very top you can see the iron porphyrin that is nestled inside the cytochrome c.
A closer look at an X-ray structure of a cyctochrome c, below, shows a relatively simple picture compared to the complexes we have been looking at. There are a couple of helices, a couple of sheets, some loops, and that iron porphyrin or heme again. A few sulfate ions are floating around nearby, as well as a couple of glycerol molecules, probably introduced during purification or crystallisation of the protein. Note that the heme is found at the edge of the protein. That location may make it easier for electron transfer from Complex III, or to Complex IV.
• Cytochrome c is the final destination of electrons moving through Complex III.
• Cytochrome c is a mobile, one-electron carrier.
• Cytochrome c is actually a small, hydrophilic protein.
• Cytochrome c is found above the mitochondrial membrane, at the edge of the intermembrane space.
What keeps cyctochrome c from wandering off into the intermembrane space? If it did so, it would interrupt the efficient flow of electrons. It is possible that cytochrome c moves back and forth between Complex III and Complex IV via a mechanism called "steering". In steering, the carrier is guided along a pathway via complementary charges. For example, if the membrane is negatively charged because of polar groups on the membrane surface, then positive charges on the surface of the cytochrome c may restrict its movement to the membrane surface. That alone would reduce its mobility from three dimensions (anywhere in the intermembrane space) to two dimensions (anywhere on the membrane surface). It is possible that additional interactions reduce its mobility even further to a one-dimensional track between the two complexes.
• Steering, based on complementary charges, keeps the cytochrome c from being lost in the intermembrane space.
Exercise \(1\)
Suggest possible amino acid residues at the surface of cytochrome c that may help it stay on a negatively charged membrane surface.
Answer
Arginine and lysine are positively charged at neutral pH.
Exercise \(2\)
It's difficult to measure the reduction potential of an individual site within a protein. However, researchers have been able to estimate these values by measuring EPR spectra under various conditions. Assuming the reduction potentials below, draw a reaction progress diagram for transport of an electron all the way from the initial ubiquinone donor all the way to cytochrome c.
Exercise \(3\)
Using the values in the figure above, calculate the energy change when an electron is transferred from the 2Fe2S cluster to the cytochrome c1.
Answer
Assuming the reduction potentials are:
2Fe2S(ox) + e- → 2Fe2S(red) Eored = 0.10 V
cyt c1(ox) + e- → cyt c1(red) Eored = 0.230 V
Then the potential difference for the reaction, ΔEo = 0.23 - (0.10) V = 0.13 V.
The Faraday relation ΔG = - n F ΔEo gives
ΔG = - 1 x 96,485 J V-1 mol-1 x 0.13 V = 12,543 J mol-1 = 12.5 kJ mol-1
Exercise \(4\)
One of the perplexing things about this complex is that, in the electron recycling loop, electron transfer appears to go from an initial ubiquinone to a final ubiquinone, with no overall driving force. However, reduction potentials are very sensitive to environment. Show how nearby arginine residues might make the reduction potential of a ubiquinone more positive.
Answer
The positive arginine residues would confer partial positive charge on the ubiquinone via hydrogen bonding; the ubiquinone would have a more positive reduction potential as a result.
See the focus page on the structure of Complex III at Biochemistry Online.
X-ray crystal structure of Complex III from: Gao, X., Wen, X., Esser, L., Quinn, B., Yu, L., Yu, C.-A., Xia, D. Structural basis for the quinone reduction in the bc1 complex: a comparative analysis of crystal structures of mitochondrial cytochrome bc1 with bound substrate and inhibitors at the Qi site. Biochemistry 2003, 4, 9067-9080. (1NTZ)
X-ray crystal structure of Complex III with bound cytochrome c from: Solmaz, S.R., Hunte, C. Structure of complex III with bound cytochrome c in reduced state and definition of a minimal core interface for electrontransfer. J. Biol. Chem. 2008, 283, 17542-17549. (3CXH)
X-ray crystal structure of cytochrome c from: Enguita, F.J., Pohl, E., Turner, D.L., Santos, H., Carrondo, M.A. Structural evidence for a proton transfer pathway coupled with haem reduction of cytochrome c" from Methylophilus methylotrophus. J. Biol. Inorg. Chem. 2006 11, 189. (1OAE) | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/V%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_3/07%3A_Oxidative_Phosphorylation/7.04%3A_Complex_I.txt |
Complex IV is the final destination in the electron transport chain. Here, the electrons that have been travelling through the other members of the respiratory supercomplex are finally delivered to O2, reducing it to water. That's an impressive feat, because a number of reactive oxygen species must be formed between the initial addition of an electron and the final release of water, but the reaction is controlled in such a way that the possibility of cell damage is minimised. At the same time, more protons are pumped across the inner mitochondrial membrane.
• In Complex IV, electrons are delivered to their final destination, a molecule of O2.
• The O2 is reduced to water.
Exercise \(1\)
Write a balanced redox half-reaction to show how many electrons are needed to reduce an oxygen molecule to water.
Answer
O2 → H2O
O2 → 2 H2O (O balanced)
O2 + 4H+ → 2 H2O (H balanced)
O2 + 4e- + 4H+ → 2 H2O (charge balanced)
The X-ray structure of Complex IV is shown below. Again, the matrix is at the lower end of the picture and the intermembrane space is at the top. That's where the cytochrome c docks, at the top.
The mobile electron carrier, cytochrome c, binds at Complex IV and delivers an electron to a binuclear copper site called CuA. We can see that binuclear copper site when we look inside the protein, below. It is bound just to the protein and nothing else, so we just see the two copper ions on their own at the top of the picture. This pair of copper atoms sends the electron on to a heme, cytochrome a, which you can see below and to the left. From there, the electron proceeds to another binuclear cluster, this time consisting of a heme-bound iron (cytochrome a3) and a nearby copper (CuB). This binuclear site carries out the reduction of dioxygen to water. In the structure, there is a carbon monoxide molecule bound in the active site between cytochrome a3 and CuB. The carbon atoms is attached to the iron and the oxygen is attached to the copper. That's where the oxygen molecule would bind, waiting to be reduced to water.
If you think about that, it means that electrons are traveling in the opposite direction from what we saw in the first three complexes. In Complexes I and II, electrons were delivered from the mitochondrial matrix and traveled up toward the intermembrane space, stopping at a ubiquinone in the mitochondrial membrane. In Complex III, electrons continued in that "upward" direction, from the mitochondrial membrane to the cytochrome c in the intermembrane space. In Complex IV, electrons are reversing course, traveling back toward the mitochondrial matrix. Remember, the mitochondrial matrix is n-doped because of proton pumping, so these electrons are traveling from the positive side of the membrane to the negative side. That must be difficult.
• In Complex IV, electron transport is in the opposite direction from the other complexes.
• Electrons travel from the intermembrane space side to the mitochondrial matrix side, against the charge gradient.
Let's take another look at the important ligands for the complex. A cartoon is shown below. In the cartoon, the O2 molecule is shown binding in that position between the heme a3 and the CuB. That dinuclear metal site is where the oxygen molecule is reduced to water.
Because four electrons are needed to reduce O2 to H2O, four cytochrome c molecules must bind at Complex IV before that reduction can proceed. It is likely that the coordination environment of the oxygen molecule -- between two metals, rather than just bound to one -- allows it be be more rapidly reduced all the way to water rather than forming reactive oxygen species that persist in the cell, such as peroxides.
Exercise \(2\)
Assume the iron in heme a3 starts in a reduced Fe(II) state and the CuB starts in a reduced Cu(I) state. Provide a mechanism for the reduction of oxygen to water, with the addition of four electrons and four protons. Use Fe(II) as the electron donor and lysine as the proton donor.
Answer
In addition to those two metals, there is also a modified histidine-tyrosine conjugate bound to CuB. It has been suggested that this tyrosine provides another source of immediate electrons that may be used in reduction.
Exercise \(3\)
Show why a tyrosine can be a source of both a proton and an electron in biochemical processes.
Answer
In addition to the need to reduce oxygen to water, Complex IV also contributes to the proton gradient, pumping additional protons across the mitochondrial membrane. That task presents additional challenges. A simple coupling mechanism is not possible, because the electrons are moving in the opposite direction. It is though that the mechanism involves conformational changes in the protein that occur as the metals change oxidation states. Subtle changes in coordination environment may result in displacement of amino acid residues nearby. It is easy to imagine that if a particular amino acid shifts upward toward the intermembrane space, it may pull a proton with it.
• Proton pumping and electron transport run in opposite directions in Complex IV and must be uncoupled.
• Proton pumping in Complex IV must rely on conformational changes.
Exercise \(4\)
The CuA site contains two coppers, boound by two bridging cysteines. Both are bound by terminal histidines. In addition, one copper is bound by an additional methionine, whereas the other is bound by a carbonyl from the protein backbone.
1. Draw the coppers in their binding sites.
2. Describe the geometry of each copper.
3. If each copper is Cu(I), what is the coordinated electron count on each copper in the complex?
4. If each copper is Cu(I), what is the overall charge on the complex?
Answer a
a)
Answer b
b) tetrahedral
Answer c
c) Cu(I) = d10
4 donors = 8 e-
total = 18e-
Answer d
d) 2 x Cu(I) = 2+
2 x Cys-S- = 2-
All others neutral
Total = 0
Exercise \(5\)
The copper in the CuB site is bound by two histidines and the histidine-tyrosine conjugate.
1. Draw the copper in the binding site.
2. Describe the geometry of the copper.
3. If copper is Cu(I), what is the coordinated electron count in the complex?
4. If copper is Cu(I), what is the overall charge on the complex?
Answer a
a)
Answer b
b) trigonal planar
Answer c
c) Cu(I) = d10
3 donors = 6 e-
total = 16 e-
Answer d
d) Cu(I) = 1+
histidines neutral
Total = 1+
Exercise \(6\)
It's difficult to measure the reduction potential of an individual site within a protein. However, researchers have been able to estimate these values by measuring EPR spectra under various conditions. Assuming the reduction potentials below, draw a reaction progress diagram for transport of an electron all the way from cytochrome c to molecular oxygen.
Exercise \(7\)
Using the values in the figure above, calculate the energy change when an electron is transferred from heme a to heme a3.
Answer
Assuming the reduction potentials are:
heme a(ox) + e- → heme a(red) Eored = 0.20 V
heme a3(ox) + e- → heme a3(red) Eored = 0.38 V
Then the potential difference for the reaction, ΔEo = 0.38 - (0.20) V = 0.18 V.
The Faraday relation ΔG = - n F ΔEo gives
ΔG = - 1 x 96,485 J V-1 mol-1 x 0.13 V = 17,367 J mol-1 = 17.4 kJ mol-1 | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/V%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_3/07%3A_Oxidative_Phosphorylation/7.05%3A_Complex_I.txt |
Oxidative phosphorylation is a metabolic process in which energy is harnessed for the production of ATP. So far we have looked at the electrons transport chain,which is reponsible for the establishment of a proton gradient across the inner mitochondrial membrane. The actual synthesis of ATP occurs in complex V.
Complex V is very large and contains many different proteins. However, we will simplify the picture here. Schematically, Complex V consists of a proton channel leading from the intermembrane space into the matrix. As protons rush back through the membrane after having been pumped out along the electron transport chain, they cause rotation of a nanoscale millwheel. This millwheel helps bind ADP to phosphate, forming ATP.
• Rotation of a molecular rotor in Complex V helps bring an ADP and a phosphate together to form ATP.
• Rotation of the rotor is powered by protons dropping back through Complex V and into the mitochondiral matrix below.
• The intermembrane space is p-doped (excess positive charge) whereas the mitochondrial matrix is n-doped (excess negative charge), driving the flow of protons toward the mitochondrial matrix.
• For every three protons that drop through the complex, one ATP molecule is formed.
An X-ray crystal structure of Complex V is shown below. At the top of the picture, you can see the membrane-bound portion. Those α-helices run parallel to the fatty acid chains in the lipid layer, so that the protein fits snugly into the membrane. The lower part of the structure is the portion that extends into the matrix.
Exercise \(1\)
Recall that in an α-helix, the amino acid residues project outward around the helix. Suggest some residues that are likely to be found along the membrane-bound portion of the complex.
Looking at the structure from underneath -- that is, from the vantage point of the matrix -- we can see the rotational symmetry of Complex V. Complex V contains a molecular rotor that is used in the manufacture of ATP. This rotor is made from a group of proteins that are able to move past another group of proteins as protons pass through the complex.
The mechanism of the rotor works something like an electric motor. The molecular engine contains two parts: a stationary part, or stator, and a rotating part, or rotor. The two parts have complementary charges. The stationary part has a positively charged arginine residue. The rotating part has a negatively charged aspartate residue. The aspartate residue is attracted to the arginine residue, causing the rotor to begin to rotate.
• The stator is the stationary part of a motor.
• The rotor is the revolving part of a motor.
• A temporary attraction draws a site on the rotor toward the stator.
• When that attraction is turned off, momentum carries that site on the rotor past the stator.
Of course, if that were the entire story, the aspartate (and the rotor) would stop moving when it got as close as possible to the arginine. That would be the end of the story. Instead, a proton arrives as it moves from the intermembrane space toward the matrix. It binds to the aspartate, rendering it neutral. No longer charged, the aspartate continues to move past the arginine.
Meanwhile, another aspartate residue encounters the continuation of the proton channel that leads onward to the matrix. It releases its proton, becoming charged. It is then attracted to the arginine in the stator, continuing the rotation of the rotor.
Exercise \(2\)
You could also imagine a lysine residue at the stator rather than the arginine shown above. Why is an arginine a more reliable positive stator than lysine?
Answer
There is always an equilibrium between the protonated state and the deprotonated state in a charged amino acid residue. For this position, an amino acid is needed that is more reliably in the protonated state; that is, the equilibrium lies more heavily to the protonated side of the equation. Because of the resonance-stabilised cation that results from protonation, arginine is much more likely to remain in a protonated state than lysine. That will make for a more efficient millwheel.
ADP and phosphate bind at the interface of two proteins along this assembly. The rotation apparently is involved in an approximation mechanism. During the conformational changes that result from the molecular rotation, the ADP and phosphate are brought close enough together that they are more likely to bind to each other, forming ATP.
Exercise \(3\)
Why might it be difficult to bring ADP and phosphate together without this mechanism?
Answer
ADP and phosphate are both anions; they would repel normally each other. When bound in the active site, their charges are likely neutralized by complementary charges in the active site.
Exercise \(4\)
Formation of an ATP from ADP costs about 30 kJ/mol. Assuming you could capture all of the energy released by transferring two electrons from NADH to O2, how many units of ATP could be made?
Answer
Assuming the reduction potentials are:
NAD+ + 2e- + 2 H+ → NADH Eored = -0.32 V
0.5 O2 + 2e- + 2 H+ → H2O Eored = 0.816 V
Then the potential difference for the reaction, ΔEo = 0.816 - (-0.32) V = 1.136 V.
The Faraday relation ΔG = - n F ΔEo gives
ΔG = - 2 x 96,485 J V-1 mol-1 x 1.136 V = 219,213 J mol-1 = 219 kJ mol-1
so 219 kJ mol-1 / 30 kJ mol-1 = 7.3
With 100% efficiency, 7 moles of ATP could be produced per mole of NADH. In reality, about half that amount is produced (closer to 3 moles ATP per mole NADH).
See the focus page on the structure of Complex V at Biochemistry Online.
X-ray structures from: Watt, I.N., Montgomery, M.G., Runswick, M.J., Leslie, A.G.W., Walker, J.E. Bioenergetic cost of making an adenosine triphosphate molecule in animal mitochondria. Proc.Natl.Acad.Sci.USA 2010, 107, 16823. Images from RCSB Protein Data Bank (2XND). | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/V%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_3/07%3A_Oxidative_Phosphorylation/7.06%3A_Complex_V.txt |
Exercise 7.2.1:
Exercise 7.2.2:
Exercise 6.2.3:
a) Iron charges: $Fe(II) + Fe(III) = 5^{+}$
Ligand charges: $2 \: sulfides \: = 2 \times 2^{-} = 4^{-} ; 4 \: cysteines \: = 4 \times 1^{-} = 4^{-} ; \: total \:= 8^{-}$
Overall: 3-
b) Iron charges: $2 \times Fe(II) + Fe(III) = 4^{+} + 3^{+} = 7^{+}$
Ligand charges: $4 \: sulfides \: = 4 \times 2^{-} = 8^{-}; 3 \: cysteines \: = 3 \times 1^{-} = 3^{-} ; \: total \: = 11^{-}$
Overall: 4-
c) Iron charges: $3 \times Fe(II) + Fe(III) = 6^{+} + 3^{+} = 9^{+}$
Ligand charges: $4 \: sulfides \: = 4 \times 2^{-} = 8^{-} ; 4 \: cysteines \: = 4 \times 1^{-} = 4^{-} ; \: total \: = 12^{-}$
Overall: 3-
Exercise 7.2.4:
Upon reduction, the charge on a 2Fe2S cluster will increase from 3- to 4-, assuming it starts in a mixed Fe(II)/(III) state (whereas if it starts in a Fe(III)/(III) state, the overall charge will increase from 2- to 3-). These anions would be stabilised by strong intermolecular interactions such as ion-dipole forces. Both states (oxidised and reduced) will be stabilised by a polar environment, but the more highly charged reduced state will depend even more strongly on stabilisation by the environment. As a result, we might expect the reduction potential to be lower when surrounded by nonpolar amino acid residues, and higher if surrounded by polar residues.
Exercise 7.2.5:
Exercise 7.2.6:
Exercise 7.2.7:
Exercise 7.2.8:
Exercise 7.2.9:
a) N1a and N1b are most likely not involved, because their reduction potentials are too negative.
b)
Exercise 7.2.10:
Assuming the reduction potentials are:
$N5(ox) + e^{-} \rightarrow N5(red)\) $E^{o}_{red}= -0.40V \nonumber$ $N6a(ox) + e^{-} \rightarrow N6a(red)$ $E^{o}_{red} = -0.30V \nonumber$ Then the potential difference for the reaction, \(\Delta E^{o} = -0.30 - (-0.40)V = 0.10V$
The Faraday relation $\Delta G= -n F \Delta E^{o}$ gives
$\Delta G = -1 \times 96485 \frac{J}{V \: mol} \times 0.10 V = 9649 \frac{J}{mol} = 9.7 \frac{kJ}{mol} \nonumber$
Exercise 7.3.1:
Heme b.
Exercise 7.3.2:
A porphyrin contains four pyrrole rings (five-membered, aromatic ring containing a nitrogen) arranged to form a 16-membered macrocycle.
Exercise 7.3.3:
Exercise 7.3.4:
Exercise 7.3.5:
Assuming the reduction potentials are:
$4Fe4s(ox) + e^{-} \rightarrow 4Fe4S(red)\) $E^{o}_{red} = -0.15V \nonumber$ $3Fe4S(ox) + e^{-} \rightarrow 3Fe4S(red)$ $E^{o}_{red} = 0.06V \nonumber$ Then the potential difference for the reaction, \(\Delta E^{o} = 0.06 -(-0.15)V = 0.21V$
The Faraday relation $\Delta G = -n F \Dleta E^{o}$ gives
$\Delta G = -1 \times 96485 \frac{J}{V \: mol} \times 0.21 V = 20262 \frac{J}{mol} = 20 \frac{kJ}{mol} \nonumber$
Exercise 7.3.6:
a) Iron charges: $2 \times Fe(III) = 6^{+}$
Ligand charges: $2 \: sulfides \: = 2 \times 2^{-}= 4^{-} ; 4 \: cysteines \: = 4 \times 1^{-} = 4^{-} ; \: total \: = 8^{-}$
Overall: 2-
b) Iron charges: $3 \times Fe(III) = 9^{+}$
Ligand charges: $4 \: sulfides \: 4 \times 2^{-} = 8^{-} ; 3 \: cysteines \: = 3 \times 1^{-} = 3^{-} ; \: total \: = 11^{-}$
Overall: 2-
c) Iron charges: $4 \times Fe(III) = 12^{+}$
Ligand charges: $4 \: sulfides = 4 \times 2^{-} = 8^{-} ; 4 \: cysteines \: = 4 \times 1^{-} = 4^{-} ; \: total \: = 12^{-}$
Overall: 0
Exercise 7.4.1:
Arninine and lysine are positively charged at neutral pH.
Exercise 7.4.2:
Exercise 7.4.3:
Assuming the reduction potentials are:
$2Fe2S(ox) + e^{-} \rightarrow 2Fe2S(red)\) $E^{o}_{red} = 0.10V \nonumber$ $cyt \: c_{1} (ox) + 3^{-} \rightarrow cyt \: c_{1} (red)$ $E^{o}_{red}= 0.230V \nonumber$ Then the potential difference for the reaction, \(\Delta E^{o} = 0.23- (0.10) V = 0.13V$
The Faraday relation $\Delta G = -nF \Delta E^{o}$ gives
$\Delta G = -1 \times 96485 \frac{J}{V \: mol} \times 0.13V = 12543 \frac{J}{mol} = 12.5 \frac{kJ}{mol} \nonumber$
Exercise 7.4.4:
The positive arginine residues would confer partial positive charge on the ubiquinone via hydrogen bonding; the ubiquinone would have a more positive reduction potential as a result.
Exercise 7.5.1:
$\ce{O2 -> H2O} \nonumber$
$\ce{O2 -> 2H2O}$ (O balanced)
$\ce{O2 + 4H^{+} -> 2H2O}$ (H balanced)
$\ce{O2 + 4e^{-} + 4H^{+} -> 2H2O}$ (charge balanced)
Exercise 7.5.2:
Exercise 7.5.3:
Exercise 7.5.4:
a)
b) tetrahedral
c) Cu(I) = d10
4 donors = 8 e-
total = 18e-
d) 2 x Cu(I) = 2+
2 x Cys-S- = 2-
All others neutral
Total = 0
Exercise 7.5.5:
a)
) trigonal planar
c) Cu(I) = d10
3 donors = 6 e-
total = 16 e-
d) Cu(I) = 1+
histidines neutral
Total = 1+
Exercise 7.5.6:
Exercise 7.5.7:
Assuming the reduction potentials are:
$heme \: a(ox) + e^{-} \rightarrow heme \: a(red)\) $E^{o}_{red} = 0.20V \nonumber$ $heme \: a_{3}(ox) + e^{-} \rightarrow heme \: a_{3}(red)$ $E^{o}_{red} = 0.38V \nonumber$ Then the potential difference for the reaction, \(\Delta E^{o} = 0.38 - (0.20)V = 0.18V$
The Faraday relation $\Delta G = -n F \Delta E^{o}$ gives
$\Delta G = -1 \times 96485 \frac{J}{V \: mol} \times 0.13V = 17367 \frac{J}{mol} = 17.4 \frac{kJ}{mol} \nonumber$
Exercise 7.6.1:
These amino acids would probably be non-polar: alanine, glycine, methionine, isoleucine, leucine, methionine, phenylalanine, tryptophan, valine.
Exercise 7.6.2:
There is always an equilibrium between the protonated state and the deprotonated state in a charged amino acid residue. For this position, an amino acid is needed that is more reliably in the protonated state; that is, the equilibrium lies more heavily to the protonated side of the equation. Because of the resonance-stabilised cation that results from protonation, arginine is much more likely to remain in a protonated state than lysine. That will make for a more efficient millwheel.
Exercise 7.6.3:
ADP and phosphate are both anions; they would repel normally each other. When bound in the active site, their charges are likely neutralized by complementary charges in the active site.
Exercise 7.6.4:
Assuming the reduction potentials are:
$\ce{NAD^{+} + 2e^{-} + 2H^{+} -> NADH}\) $E^{o}_{red} = -0.32V \nonumber$ $\ce{0.5 O2 + 2e^{-} + 2H^{+} -> H2O}$ $E^{o}_{red} = 0.816V \nonumber$ Then the potential difference for the reaction, \(\Delta E^{o} = 0.816 - (-0.32)V = 1.136V$
The Faraday relation $\Delta G = -NF \Delta E^{o}$ gives
$\Delta G = -2 \times 96475 \frac {J}{V \: mol} \times 1.136V = 219213 \frac{J}{mol} = 219 \frac{kJ}{mol} \nonumber$
so $\frac{219 \frac{kJ}{mol}} {30 \frac{kJ}{mol}} = 7.3$
With 100% efficiency, 7 moles of ATP could be produced per mole of NADH. In reality, about half that amount is produced (closer to 3 moles ATP per mole NADH). | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/V%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_3/07%3A_Oxidative_Phosphorylation/7.07%3A_Solutions.txt |
Let's think first about the interaction of light with matter.
We have all seen light shine on different objects. Some objects are shiny and some are matte or dull. Some objects are different colours. Light interacts with these objects in different ways. Sometimes, light goes straight through an object, such as a window or a piece of glass.
Because chemical reactions are frequently conducted in solution, we will think about light entering a solution.
Imagine sunlight shining through a glass of soda. Maybe it is orange or grape soda; it is definitely coloured. We can see that as sunlight shines through the glass, coloured light comes out the other side. Also, less light comes out than goes in.
Maybe some of the light just bounces off the glass, but some of it is definitely absorbed by the soda.
So, what is the soda made of? Molecules. Some of these molecules are principally responsible for the colour of the soda. There are others, such as the ones responsible for the flavour or the fizziness of the drink, as well as plain old water molecules. The soda is a solution; it has lots of molecules (the solute) dissolved in a solvent (the water).
Light is composed of photons. As photons shine through the solution, some of the molecules catch the photons. They absorb the light. Generally, something in the molecule changes as a result. The molecule absorbs energy from the photon and is left in an excited state.
The more of these molecules there are in the solution, the more photons will be absorbed. If there are twice as many molecules in the path of the light, twice as many photons will be absorbed. If we double the concentration, we double the absorbance.
• The amount of light absorbed depends on the concentration of the solution.
Alternatively, if we kept the concentration of molecules the same, but doubled the length of the vessel through which the light traveled, it would have the same effect as doubling the concentration. Twice as much light would be absorbed.
• The absorbance depends on the length of cell holding the solution.
These two factors together make up part of a mathematical relationship, called Beer's Law, describing the absorption of light by a material:
$A = \varepsilon c l \nonumber$
in which A = Absorbance, the percent of light absorbed; c = the concentration; l = the length of the light's path through the solution; ε = the "absorptivity" or "extinction coeficient" of the material, which is a measure of how easily it absorbs a photon that it encounters.
That last factor, ε, suggests that not all photons are absorbed easily, or that not all materials are able to absorb photons equally well. There are a couple of reasons for these differences.
Exercise $1$
Calculate the absorbance in the following cases.
1. A sample with a molar absorptivity ε = 60 L mol-1cm-1 is diluted to a 0.01 mol L-1 solution in water and placed in a 1 cm cell.
2. A sample with a molar absorptivity ε = 3,000 L mol-1cm-1 is diluted to a 3.5 x 10-5 mol L-1 solution in water and placed in a 1 cm cell.
3. A sample with a molar absorptivity ε = 1.4 L mol-1cm-1 is diluted to a 0.25 mol L-1 solution in water and placed in a 0.5 cm cell.
4. A sample with a molar absorptivity ε = 23,000 L mol-1cm-1 is diluted to a 2.5 x 10-6 mol L-1 solution in water and placed in a 1 cm cell.
5. A sample with a molar absorptivity ε = 14,000 L mol-1cm-1 is diluted to a 0.015 mmol L-1 solution in water and placed in a 1 cm cell.
Answer a
a) A = ε c l = 60 L mol-1 cm-1 x 0.01 mol L-1 x 1 cm = 0.60 = 60 %
Answer b
b) A = ε c l = 3,000 L mol-1 cm-1 x 3.5 x 10-5 mol L-1 x 1 cm = 0.105 = 10.5 %
Answer c
c) A = ε c l = 1.4 L mol-1 cm-1 x 0.25 mol L-1 x 0.5 cm = 0.175 = 17.5 %
Answer d
d) A = ε c l = 23,000 L mol-1 cm-1 x 2.5 x 10-6 mol L-1 x 1 cm = 0.0575 = 5.75 %
Answer e
e) A = ε c l = 14,000 L mol-1 cm-1 x 0.015 mmol L-1 x 1 cm = 14,000 L mol-1 cm-1 x 0.015 x 10-3 mol L-1 x 1 cm = 0.21 = 21 %
Exercise $2$
Calculate the extinction coefficient in the following cases.
1. 30% absorbance is observed with a 0.01 mol L-1 solution in a 1 cm cell.
2. 25% absorbance is observed with a 0.025 mol L-1 solution in a 1 cm cell.
3. 95% absorbance is observed with a 0.00175 mol L-1 solution in a 0.5 cm cell.
4. 66% absorbance is observed with a 0.025 mmol L-1 solution in a 1 cm cell.
Answer a
a) ε = A /c l = 0.30 / (0.01 mol L-1 x 1 cm) = 30 L mol-1 cm-1
Answer b
b) ε = A /c l = 0.25 / (0.025 mol L-1 x 1 cm) = 10 L mol-1 cm-1
Answer c
c) ε = A /c l = 0.30 / (0.01 mol L-1 x 1 cm) = 543 L mol-1 cm-1
Answer d
d) ε = A /c l = 0.66 / (0.025 m mol L-1 x 1 cm) = 0.66 / (0.025 x 10-3 mol L-1 x 1 cm) = 26,400 L mol-1 cm-1
Often a particular soda will absorb light of a particular colour. That means, only certain photons corresponding to a particular colour of light are absorbed by that particular soda.
How does that affect what we see? If the red light is being absorbed by the material, it isn't coming back out again. The blue and yellow light still are, though. That means the light coming out is less red, and more yellowy-blue. We see green light emerging from the glass.
A "colour wheel" or "colour star" can help us keep track of the idea of complementary colours. When a colour is absorbed on one side of the star, we see mostly the colour on the opposite side of the star.
Exercise $3$
1. What colour of photon is probably most strongly absorbed by a glass of orange soda?
2. What colour of photon is most strongly absorbed by a glass of lime soda?
3. What colour of photon is most strongly absorbed by a glass of blue raspberry kool-aid?
4. What colour of photon is most strongly absorbed by a glass of pineapple soda?
5. What colour of photon is most strongly absorbed by a glass of cherry soda?
Answer a
a) blue
Answer b
b) red
Answer c
c) orange
Answer d
d) violet
Answer e
e) green
Why do certain materials absorb only certain colours of light? That has to do with the properties of photons. Photons have particle-wave duality, just like electrons. They have wave properties, including a wavelength.
That wavelength corresponds to the energy of a photon, according to the Planck-Einstein equation:
$E = \frac{hc}{\lambda} \nonumber$
in which E = energy of the photon, h = Planck's constant (6.625 x 10-34 J s-1), c = speed of light (3.0 x 108 m s-1), λ = wavelength of light in m.
Alternatively, the Planck-Einstien equation can be thought of in terms of frequency of thr photon: as a photon passes through an object, how frequently does one of its "crests" or "troughs" encounter the object? How frequently does one full wavelength of the photon pass an object? That parameter is inversely proportional to the wavelength. The equation becomes:
$E = H \nu \nonumber$
in which ν = the frequency of the photon, in s-1.
Exercise $4$
1. Calculate the energy of a photon with a wavelength of 1 x 10-5 m.
2. Calculate the energy of a photon with a wavelength of 125 nm.
3. Calculate the energy of a photon with a wavelength of 1025 nm.
4. Calculate the energy of a photon with a wavelength of 450 µm.
5. Calculate the energy of a photon with a wavelength of 850 Å.
Answer a
a) E = hc/λ = (6.625 x 10-34 Js mol-1 x 3.0 x 108 m s-1)/1 x 10-5 m = 1.98 x 10-20 J
Answer b
b) E = hc/λ = (6.625 x 10-34 Js mol-1 x 3.0 x 108 m s-1)/125 x 10-9 m = 1.59 x 10-18 J
Answer c
c) E = hc/λ = (6.625 x 10-34 Js mol-1 x 3.0 x 108 m s-1)/1025 x 10-9 m = 1.94 x 10-19 J
Answer d
d) E = hc/λ = (6.625 x 10-34 Js mol-1 x 3.0 x 108 m s-1)/450 x 10-6 m = 4.42 x 10-22 J
Answer e
e) E = hc/λ = (6.625 x 10-34 Js mol-1 x 3.0 x 108 m s-1)/850 x 10--10 m = 2.3 x 10-18 J
Exercise $5$
1. Calculate the wavelength of a 1.36 x 10-17 J photon.
2. Calculate the wavelength of a 4.72 x 10-24 J photon.
3. Calculate the wavelength of a 9.26 x 10-7 J photon.
Answer a
a) λ = hc/E = (6.625 x 10-34 Js mol-1 x 3.0 x 108 m s-1)/1.46 x 10--17 J = 1.36 x 10-8 m
Answer b
b) λ = hc/E = (6.625 x 10-34 Js mol-1 x 3.0 x 108 m s-1)/4.72 x 10--24 J = 4.21 x 10-2 m
Answer c
c) λ = hc/E = (6.625 x 10-34 Js mol-1 x 3.0 x 108 m s-1)/9.26 x 10--17 J = 2.15 x 10-19 m
Exercise $6$
1. Calculate the wavelength of a photon with a frequency of 6.7 x 1010 s-1.
2. Calculate the wavelength of a photon with a frequency of 1500 MHz.
3. Calculate the frequency of a photon with a wavelength of 9.8 x 10-10 m.
4. Calculate the frequency of a photon with a wavelength of 4.3 x 10-12 m.
Answer a
a) E = hν = hc/λ ; so ν = c/λ; or λ = c/ν = 3.0 x 108 m s-1/6.7 x 1010 s-1 = 4.48 x 10-3 m
Answer b
b) λ = c/ν = 3.0 x 108 m s-1/1500 x 106 s-1 = 0.2 m
Answer c
c) ν = c/λ = 3.0 x 108 m s-1/9.8 x 10-10 m = 3.06 x 1015 s-1
Answer d
d) ν = c/λ = 3.0 x 108 m s-1/4.3 x 10-12 m = 7.0 x 1019 s-1
The visible spectrum, shown below, contains a very limited range of photon wavelengths, between about 400 and 700 nm.
The higher the frequency, the higher the energy of the photon. The longer the wavelength, the lower the energy of the photon.
As a result of this relationship, different photons have different amounts of energy, because different photons have different wavelengths.
Exercise $7$
Different portions of the electromagnetic spectrum interact with matter in different ways. Because of that, we can use different wavelengths of light to gain different kinds of information about a material. Calculate the amount of energy involved in the following kinds of interactions, in units of kJ/mol.
1. Molecular bond rotations, measured by microwave spectroscopy. Suppose the microwave has a wavelength 1 mm long.
2. Bond stretching and bending, measured by infrared spectroscopy. Suppose the IR wavelength is 1,000 nm long.
3. Nuclear magnetic moments, measured by radio waves in NMR. Assume the radio wave is 1 m long.
4. The excitation of an electron from one energy level to another, measured by ultraviolet and visible spectroscopy. Assume the visible light's wavelength is 500 nm long.
Answer a
a) E = h c / λ
E = (6.625 x 10-34 J s mol-1 x 3.0 x 108 m s-1 )/ (1 mm x 10-3 m mm-1) = 1.99 x 10-22 J
That's for one molecule. On a per mole basis, E = 1.99 x 10-22 J x 6.02 x 1023 mol-1 = 120 J mol-1 = 0.12 kJ mol-1.
Answer b
b) E = h c / λ
E = (6.625 x 10-34 J s mol-1 x 3.0 x 108 m s-1 )/ (1000 nm x 10-9 m nm-1) = 1.99 x 10-19 J
That's for one molecule. On a per mole basis, E = 1.99 x 10-19 J x 6.02 x 1023 mol-1 = 120,000 J mol-1 = 120 kJ mol-1.
Answer c
c) E = h c / λ
E = (6.625 x 10-34 J s mol-1 x 3.0 x 108 m s-1 )/ (1 m) = 1.99 x 10-25 J
That's for one molecule. On a per mole basis, E = 1.99 x 10-25 J x 6.02 x 1023 mol-1 = 0.120 J mol-1 = 1.2 x 10-4 kJ mol-1.
Answer d
d) E = h c / λ
E = (6.625 x 10-34 J s mol-1 x 3.0 x 108 m s-1 )/ (500 nm x 10-9 m nm-1) = 3.98 x 10-19 J
That's for one molecule. On a per mole basis, E = 3.98 x 10-19 J x 6.02 x 1023 mol-1 = 239,000 J mol-1 = 239 kJ mol-1.
Exercise $8$
The visible spectrum ranges from photons having wavelengths from about 400 nm to 700 nm. The former is the wavelength of violet light and the latter is the wavlength of red light. Which one has higher energy: a photon of blue light or a photon of red light?
Answer
Blue.
Exercise $9$
Ultraviolet light -- invisible to humans and with wavelengths beyond that of violet light -- is associated with damage to skin; these are the cancer-causing rays from the sun. Explain their danger in terms of their relative energy.
Answer
Ultraviolet light, with a shorter wavelength than visible light, is much higher in energy and potentially more damaging.
Different materials absorb photons of different wavelengths because absorption of a photon is an absorption of energy. Something must be done with that energy. In the case of ultraviolet and visible light, the energy is of the right general magnitude to excite an electron to a higher energy level.
However, we know that energy is quantized. That means photons will be absorbed only if they have exactly the right amount of energy to promote an electron from its starting energy level to a higher one (producing an "excited state"). Just like Goldilocks, a photon with too much energy won't do the trick. Neither will a photon with too little. It has to be just right.
If the absorption of a UV-visible photon is coupled to the excitation of an electron, what happens when the electron falls back down to the ground state? You might expect a photon to be released.
This phenomenon was observed during the late nineteenth century, when scientists studied the "emission spectra" of metal ions. In these studies, the metal ions would be heated in a flame, producing characteristic colours. In that event, the electron would be thermally promoted to a higher energy level, and when it relaxed, a photon would be emitted corresponding to the energy of relaxation.
By passing this light through a prism or grating, scientists could separate the observed colour into separate lines of different wavelengths. This evidence led directly to the idea of Niels Bohr and others that atoms had electrons in different energy levels, which is part of our view of electronic structure today. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/V%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_3/08%3A_Photochemical_Reactions/8.01%3A_Absorbance.txt |
So far, we have come across one big rule of photon absorbance. In order to be absorbed, a photon's energy has to match an energy difference within the compound that is absorbing it.
In the case of visible or ultraviolet light, the energy of a photon is roughly in the region that would be appropriate to promote an electron to a higher energy level. Different wavelengths would be able to promote different electrons, depending on the energy difference between an occupied electronic energy level and an unoccupied one.
Other types of electromagnetic radiation would not be able to promote an electron, but they would be coupled to other events. For example, absorption of infrared light is tied to vibrational energy levels. Microwave radiation is tied to rotational energy levels in molecules.
Thus, one reason a photon may or may not be absorbed has to do with whether its energy corresponds to the available energy differences within the molecule or ion that it encounters.
Franck-Condon: Electronic and Vibrational Coupling
Photons face other limitations. One of these is a moderate variation on our main rule. It is called the Frank Condon Principle. According to this idea, when an electron is excited from its normal position, the ground state, to a higher energy level, the optimal positions of atoms in the molecule may need to shift. Because electronic motion is much faster than nuclear motion, however, any shifting of atoms needed to optimize positions as they should be in the excited state will have to wait until after the electron gets excited. In that case, when the electron lands and the atoms aren't yet in their lowest energy positions for the excited state, the molecule will find itself in an excited vibrational state as well as an excited electronic state.
That means the required energy for excitation doesn't just correspond to the difference in electronic energy levels; it is fine-tuned to reach a vibrational energy level, which is quantized as well.
• The Franck Condon Principle states that electronic transitions are vertical.
• A vertical transition is one in which non of the nuclei move while the electron journeys from one state to another.
• A vertical transition may begin in a vibrational ground state of an electronic ground state and end in a vibrational excited state of an electronic excited state.
LaPorte: Orbital Symmetry
There are other restrictions on electronic excitation. Symmetry selection rules, for instance, state that the donor orbital (from which the electron comes) and the acceptor orbital (to which the electron is promoted) must have different symmetry. The reasons for this rule are based in the mathematics of quantum mechanics.
What constitutes the same symmetry vs. different symmetry is a little more complicated than we will get into here. Briefly, let's just look at one "symmetry element" and compare how two orbitals might differ with respect to that element.
If an orbital is centrosymmetric, one can imagine each point on the orbital reflecting through the very centre of the orbital to a point on the other side. At the end of the operation, the orbital appears unchanged. That means the orbital is symmetric with respect to a centre of inversion..
If we do the same thing with a sigma antibonding orbital, things turn out differently.
In the drawing, the locations of the atoms are labelled A and B, but the symmetry of the orbital itself doesn't depend on that. If we imagine sending each point on this orbital through the very centre to the other side, we arrive at a picture that looks exactly the opposite of what we started with.
These two orbitals have different symmetry. A transition from one to the other is allowed by symmetry.
Exercise \(1\)
Decide whether each of the following orbitals is centrosymmetric.
a) an s orbital b) a p orbital c) a d orbital d) a π orbital e) a π* orbital
Exercise \(2\)
Decide whether each of the following transitions would be allowed by symmetry.
a) π → π* b) p → π* c) p → σ* d) d → d
Symmetry selection rules are in reality more like "strong suggestions". They depend on the symmetry of the molecule remaining strictly static, but all kinds of distortions occur through molecular vibrations. Nevertheless, these rules influence the likelihood of a given transition. The likelihood of a transition, similarly, has an influence upon the extinction coefficient, ε.
transition ε, extinction coefficient
π → π* 3,000 - 25,000 M-1 cm-1
p → π* 20 - 150 M-1 cm-1
p → σ* 100 - 7,000
d → π* (MLCT) 10,000 - 50,000 M-1 cm-1
d → d 5 - 400 M-1 cm-1
Exercise \(1\)
Suggest the type of transition that is probably occurring in each case.
1. A Ti(III) complex absorbs at 377 nm with ε = 95 M-1 cm-1.
2. A terpenoid ketone absorbs at 538 nm with ε = 11,500 M-1 cm-1.
3. A Ru(II) compound absorbs at 444 nm with ε = 36,000 M-1 cm-1.
Answer a
a) d → d
Answer b
b) π → π*
Answer c
c) MLCT
Spin State
Let's take a quick look at one last rule about electronic emissions. This rule concerns the spin of the excited electron, or more correctly, the "spin state" of the excited species. The spin state describes the number of unpaired electrons in the molecule or ion.
number of unpaired electrons spin state
0 singlet
1 doublet
2 triplet
3 quartet
The rule says that in an electronic transition, the spin state of the molecule must be preserved. That means if there are no unpaired electrons before the transition, then the excited species must also have no unpaired electrons. If there are two unpaired electrons before the transition, the excited state must also have two unpaired electrons.
That allowed diagram should looks strange to you. We are used to filling in electrons with spin paired for a good reason: we are following Hund's rule, that says the lowest energy state has any unpaired spins parallel, not opposite. In fact, the state on the left really is lower in energy than the state on the right; it's just that the electrons can't get from the middle state (the ground state) to the left state easily via absorption of a photon. However, they can get from the ground state to the state on the right pretty easily via photon absorption.
The state on the left is called a triplet state (the "multipicity" of the state is the number of parallel spins plus one). The state on the right is called the singlet state. The ground state is also a singlet state. Overall, we have a ground singlet state, an excited singlet state and an excited triplet state.
As it happens, because the excited triplet state is a little lower in energy than the excited singlet state, the electron can eventually relax into the excited triplet state. It can flip, without absorbing or releasing a photon. This even is called intersystem crossing. It is related to fluorescence and phosphorescence phenomena that we will look at next.
Exercise \(4\)
Cis-trans isomerism of C=N bonds provides a path for rapid vibrational relaxation from excited states. Compounds containing this bond do not fluoresce because they are able to relax back to the ground state by transferring energy into rotational and vibrational states.
a) Illustrate this type of isomerism in the example below left.
b) Some imines (above right) are strongly fluorescent. Explain the difference.
c) Would you predict the following compound, Compound L, to absorb UV-Vis light?
d) Would you predict Compound L to be fluorescent?
Compound L shows the absorption spectrum below (conditions: [L] = 10-6 M; cell length = 1 cm).
e) What is its approximate molar absorptivity at its λmax (ε, M-1 cm-1)
f) Is this transition allowed or forbidden?
g) Identify the type of transition that is likely involved.
h) In the presence of zinc ion, compound L becomes strongly fluorescent (below). Explain why with a structure. (P.-F. Wang, Org. Lett. 2007, 9, 33-36)
i) Propose a practical application for compound L.
Answer b
b) The one on the left can relax by channelling some energy into molecular vibration, especially its cis-trans isomerisation. The one on the right can't do that because its rotation is restricted by the presence of the ring.
Answer c
c) Of course! It's crawling with pi bonds. A strong pi-pi* transition in the visible region seems likely.
Answer d
d) Probably not. It does not have restricted rotation, so the cis-trans isomerisation route is available for relaxation.
Answer e
e) A = ε b c, so ε = A / b c = 0.77 / (1 cm x 10-6 M) = 7.7 x 105 M-1 cm-1
Answer f
f) That's a large molar absorptivity constant. It's allowed.
Answer g
g) Undoubtedly this is that pi-pi* transition we were thinking about earlier.
Answer h
h) If the compound binds to zinc, it probably does so via bidentate coordination. The resulting ring restricts the degrees of freedom in the compound so it can't undergo cis-trans isomerisation, closing off a route to rapid relaxation.
Answer i
i) A compound like this could be used to detect metal ions such as Zn2+. Because the amount of fluorescence depends strongly on the Zn2+ concentration, it could be used to measure the amount of the ion present.
Exercise \(5\)
Electron transfer between metal ions is a crucial event in a number of biochemical processes, such as oxidative phosphorylation and photosynthesis. It is thought that amino acid residues may help conduct electrons over the long distances between metals.
Harry Gray and co-workers have conducted studies of electron transfer in blue copper (azurin) proteins (J. Am. Chem. Soc. 2013, 135, 15515-15525). In the study, the protein is first tagged with [(dmp)Re(CO)3]+, which binds to a histidine residue.
1. Draw the Re complex bound to histidine. It forms the fac isomer. Make sure you draw the ligand structures.
2. This complex is orange. What colour does it absorb?
3. If you wanted to use a laser to get the complex into an excited state, what wavelength would you choose?
4. When pulsed with a laser, the compound undergoes metal-to-ligand charge transfer (MLCT).
i) Draw the d orbital splitting diagram for this complex in the ground state.
ii) Add to your diagram a ligand orbital for MLCT at an appropriate energy level. Label it (σ, π, n etc).
iii) Show the corresponding orbital diagram for the excited state.
iv) The complex now contains a “hole” (where an electron used to be). Circle it.
e) The complex undergoes intersystem crossing to the triplet state within a picosecond (10-12 s). Show this change.
Electron transfer ensues involving the Cu(I)/(II) site (E0 = 0.3 V) and the Re complex (E0 = 1.4 V). This reaction occurs within several nanoseconds (10-9 s).
f) A portion of the X-ray crystal structure of the Cu(II) form of the protein is shown above; the structure is assumed to be similar when Cu(I) is present. Draw an arrow showing the direction of electron transfer on the structure.
g) This ET reaction is conveniently monitored by IR spectroscopy, observing the carbonyl ligands on the Re complex.
i) In what region of the IR spectrum would you observe these ligands?
ii) When electron transfer occurs, would the IR frequency shift to higher cm-1 or to lower cm-1? Explain why.
iii) The Cu-Re distance is approximately 20 Angstroms. Does electron transfer probably occur via an outer sphere mechanism or an inner sphere one?
h) In site-directed mutagenesis studies, electron transfer did not occur in a W122F mutant; i.e. with W122 (E0 = 1.37 V) replaced by F (E0 = 1.6 V). Draw a mechanism for electron transfer in the native, tagged protein that incorporates this information.
i) The reduction potential of the blue copper (II / I) site varies among different forms of the protein, from about 200 mV to 1000 mV. It has been suggested that this value is influenced mostly by amino acid residues beyond the coordination sphere. Draw in two amino acid residues that would decrease the reduction potential, and briefly state how they would do so.
Afterword: In the Re experiment, the electron eventually goes back to where it came from, but not until several microseconds (10-6 s) have passed.
Answer b
b) An orange complex would absorb blue light, its complementary colour.
Answer c
c) A blue laser would work. Maybe somewhere around 476 nm.
Answer d
d) Notice a Re(II) site results, because it has transferred an electron to the ligand.
Answer f
f) The reduction potentials (1.4 V > 0.3 V) suggest transfer from copper (I) to rhenium (II).
Answer g
g) i) The carbonyl stretches would be observed somewhere around 2000 cm-1.
ii) The electron transfer from the Cu(I) to the Re(II) site would result in a Re(I) site. The carbonyl stretch would shift to a lower wavenumber. That is because the lower oxidation state on the metal results in increased backbonding to the π-accepting carbonyls.
iii) At that distance, an outer sphere mechanism seems likely; it isn't clear how something could bridge that distance, other than the peptide chain itself, which probably lacks sufficient conjugation to conduct electrons.
Answer h
h) Tryptophan acts as a stepping stone between copper and rhenium. Phenylalanine, with a higher reduction potential than rhenium, does not. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/V%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_3/08%3A_Photochemical_Reactions/8.02%3A_Rules_of_Ab.txt |
Sometimes, when an excited state species relaxes, giving off a photon, the wavelength of the photon is different from the one that initially led to excitation. When this happens, the photon is invariably red-shifted; its wavelength is longer than the initial one. This situation is called "fluorescence".
How can that be? Isn't energy quantized? How is the molecule suddenly taking a commission out of the energy the original photon brought with it?
Relaxation and Fluorescence
This discrepancy is related to the Franck-Condon principle from the previous page. When an electron is promoted to an electronic excited state, it often ends up in an excited vibrational state as well. Thus, some of the energy put into electronic excitation is immediately passed into vibrational energy. Vibrational energy, however, doesn't just travel in photons. It can be gained or lost through molecular collisions and heat transfer.
The electron might simply drop down again immediately; a photon would be emitted of exactly the same wavelength as the one that was previously absorbed. On the other hand, if the molecule relaxes into a lower vibrational state, some of that initial energy will have been lost as heat. When the electron relaxes, the distance back to the ground state is a little shorter. The photon that is emitted will have lower energy and longer wavelength than the initial one.
Just how does a molecule undergo vibrational relaxation? Vibrational energy is the energy used to lengthen or shorten bonds, or to widen or squeeze bond angles. Given a big enough molecule, some of this vibrational energy could be transferred into bond lengths and angles further away from the electronic transition. Otherwise, if the molecule is small, it may transfer some of its energy in collisions with other molecules.
There are lots of examples of energy being transferred this way in everyday life. In a game of pool, one billiard ball can transfer its energy to another, sending it toward the pocket. Barry Bonds can transfer a considerable amount of energy through his bat into a baseball, sending it out of the park, just as Serena Williams can send a whole lot of energy whizzing back at her sister. In curling, one stone can transfer its energy to another, sending it out of the house and giving Canada the gold over Sweden.
Exercise $1$
How does the energy of an electronic absorption compare to other processes? To find out, you might consider the excitation of an entire mole of molecules, rather than a sinle molecule absorbing a single photon. Calculate the energy in kJ/mol for the following transitions.
1. absorbance at 180 nm (ultraviolet)
2. absorbance at 476 nm (blue)
3. absorbance at 645 nm (red)
Answer a
a) E = h c / λ
E = (6.625 x 10-34 J s mol-1 x 3.0 x 108 m s-1 )/ (180 nm x 10-9 m nm-1) = 1.10 x 10-18 J
That's for one molecule. On a per mole basis, E = 1.10 x 10-18 J x 6.02 x 1023 mol-1 = 665,000 J mol-1 = 665 kJ mol-1.
For comparison, the relatively strong and unreactive C-H bond in methane has a bond dissociation energy of only 440 kJ mol-1. (That's a thermodynamic value; to actually break the bond would cost more input of energy, to get over the kinetic energy barrier for bond cleavage.)
Answer b
b) E = h c / λ
E = (6.625 x 10-34 J s mol-1 x 3.0 x 108 m s-1 )/ (476 nm x 10-9 m nm-1) = 4.17 x 10-19 J
That's for one molecule. On a per mole basis, E = 4.42 x 10-19 J x 6.02 x 1023 mol-1 = 251,000 J mol-1 = 251 kJ mol-1.
Answer c
c) E = h c / λ
E = (6.625 x 10-34 J s mol-1 x 3.0 x 108 m s-1 )/ (645 nm x 10-9 m nm-1) = 3.08 x 10-19 J
That's for one molecule. On a per mole basis, E = 3.08 x 10-19 J x 6.02 x 1023 mol-1 = 138,000 J mol-1 = 138 kJ mol-1.
Exercise $2$
How does the energy of an excitation between vibrational states compare to that of an electronic excitation? Typically, infrared absorptions are reported in cm-1, which is simply what it looks like: the reciprocal of the wavelength in cm. Because wavelength and frequency are inversely related, wavenumbers are considered a frequency unit. Calculate the energy in kJ/mol for the following transitions.
1. absorbance at 3105 cm-1
2. absorbance at 1695 cm-1
3. absorbance at 963 cm-1
Answer a
a) λ = 1 / wavenumber = 1 / 3105 cm-1 = 3.22 x 10-4 cm x 0.01 m cm-1 = 3.22 x 10-6 m
E = h c / λ
E = (6.625 x 10-34 J s mol-1 x 3.0 x 108 m s-1 )/ (3.22 x 10-6 m) = 6.17 x 10-20 J
That's for one molecule. On a per mole basis, E = 6.17 x 10-20 J x 6.02 x 1023 mol-1 = 37,000 J mol-1 = 37 kJ mol-1.
Answer b
b) λ = 1 / wavenumber = 1 / 1695 cm-1 = 5.90 x 10-4 cm x 0.01 m cm-1 = 5.90 x 10-6 m
E = h c / λ
E = (6.625 x 10-34 J s mol-1 x 3.0 x 108 m s-1 )/ (3.22 x 10-6 m) = 3.37 x 10-20 J
That's for one molecule. On a per mole basis, E = 3.37 x 10-20 J x 6.02 x 1023 mol-1 = 20,000 J mol-1 = 20 kJ mol-1.
Answer c
c) λ = 1 / wavenumber = 1 / 963 cm-1 = 1.04 x 10-3 cm x 0.01 m cm-1 = 1.04 x 10-5 m
E = h c / λ
E = (6.625 x 10-34 J s mol-1 x 3.0 x 108 m s-1 )/ (1.04 x 10-5 m) = 1.91 x 10-20 J
That's for one molecule. On a per mole basis, E = 1.91 x 10-20 J x 6.02 x 1023 mol-1 = 11,500 J mol-1 = 11.5 kJ mol-1.
In molecules, as one molecule drops to a lower vibrational state, the other will hop up to a higher vibrational state with the energy it gains. In the drawing below, the red molecule is in an electronic excited and vibrational state. In a collision, it transfers some of its vibrational energy to the blue molecule.
Radiationless Transitions: Internal Conversion
If electrons can get to a lower energy state, and give off a little energy at a time, by hopping down to lower and lower vibrational levels, do they need to give off a giant photon at all? Maybe they can relax all the way down to the ground state via vibrational relaxation. That is certainly the case. Given lots of vibrational energy levels, and an excited state that is low enough in energy so that some of its lower vibrational levels overlap with some of the higher vibrational levels of the ground state, the electron can hop over from one state to the other, without releasing a photon.
This event is called a "radiationless transition", because it occurs without release of a photon. The electron simply slides over from a low vibrational state of the excited electronic state to a high vibrational state of the electronic ground state. We will see a couple of iinds of radiationless transitions. Specifically, if the electron simply keeps dropping a vibrational level at a time back to the ground state, the process is called "internal conversion".
Internal conversion has an important consequence. Because the absorption of UV and visible light can result in energy transfer into vibrational states, much of the energy that is absorbed from these sources is converted into heat. That can be a good thing if you happen to be a marine iguana trying to warm up in the sun after a plunge in the icy Pacific. It can also be a tricky thing if you are a process chemist trying to scale up a photochemical reaction for commercial production of a pharmaceutical, because you have to make sure the system has adequate cooling available.
Radiationless Transitions: Intersystem Crossing
There is a very similar event, called "intersystem crossing", that leads to the electron getting caught between the excited state and the ground state. Just as, little by little, vibrational relaxation can lead the electron back onto the ground state energy surface, it can also lead the electron into states that are intermediate in energy.
For example, suppose an organic molecule undergoes electronic excitation. Generally, organic molecules have no unpaired electrons. Their ground states are singlet states. According to one of our selection rules for electronic excitation, the excited state must also have no unpaired electrons. In other words, the spin on the electron that gets excited is the same after excitation as it was before excitation.
However, that's not the lowest possible energy state for that electron. When we think about atomic orbital filling, there is a rule that governs the spin on the electrons in degenerate orbitals: in the lowest energy state, spin is maximized. In other words, when we draw a picture of the valence electron configuration of nitrogen, we show nitrogen's three p electrons each in its own orbital, with their spins parallel.
The picture with three unpaired electrons, all with parallel spins, shows a nitrogen in the quartet spin state. Having one of those spins point the other way would result in a different spin state. One pair of electrons in the p level would be spin-paired, one up and one down, even though they are in different p orbitals. That would leave one electron without an opposite partner. The nitrogen would be in a doublet spin state.
That isn't what happens. The quartet spin state is lower in energy than the doublet state. That's just one of the rules of quantum mechanics: maximize spin when orbitals are singly occupied.
It's the same in a molecule. The triplet state is lower in energy than the singlet state. Why didn't the electron get excited to the triplet state in the first place? That's against the rules. But sliding down vibrationally onto the triplet state from the singlet excited state isn't, because it doesn't involve absorption of a photon.
Intersystem crossing can have important consequences in reaction chemistry because it allows access to triplet states that are not normally avaiable in many molecules. Because triplet states feature unpaired electrons, their reactivity is often typified by radical processes. That means an added suite of reactions can be accessed via this process.
Phosphorescence: A Radiationless Transition Followed by Emission
Intersystem crossing is one way a system can end up in a triplet excited state. Even though this state is lower in energy than a singlet excited state, it can't be accessed directly via electronic excitation because that would violate the spin selection rule.
That's where the electron gets stuck, though. The quick way back down to the bottom is by emitting a photon, but because that would involve a change in spin state, it isn't allowed. Realistically speaking, that means it takes a long time. By "a long time", we might mean a few seconds, several minutes, or possibly even hours. Eventually, the electron can drop back down, accompanied by the emission of a photon. This situation is called "phosphorescence".
Molecules that display phosphorescence are often incorporated into toys and shirts so that they will glow in the dark.
Photosensitization
We have already seen that an excited state molecule can transfer some vibrational energy to another molecule via a collision. What about the energy of the electroic excited state? Can a molecule transfer a large quantum of energy to another -- essentially a photon's worth, but without the photon? The answer is yes.
In a collision, one molecule in an electronic excited state can transfer its energy to another. In the process, the first molecule returns to the ground state and the second is excited. This process is called "photosensitization".
Photosensitization can occur in a couple of different ways. Because photosensitization does not involve absorption or emission of a photon, it can also lead to formation of a triplet excited state.
The significance of photosensitization is that compounds that do not have strong chromophores can still access electronic excited states if they come into contact with other molecules that do have strong chromophores. There are a number of compounds that are routinely used to induce excitation in other molecules; these photochemical enablers are referred to as photosensitizers.
Exercise $1$
Photoredox processes have experienced a great deal of attention from researchers. (For a nice overview, see Stephenson, J. Org. Chem. 2012, 77, 1617-1622).
1. Photochemically excited species are often good reducing agents (or photoredox agents). Show why with the help of a generic energy diagram showing a HOMO → LUMO excitation.
2. Ru(II) complexes such as Ru(bpy)32+ are good photoredox agents. Comment on the usefulness of Ru(bpy)32+ as a regular reducing agent (in its ground state), based on the following equation.
$\ce{Ru(bpy)3^{3+} + e^{-} -> Ru(bpy)3^{2+}}\) $E^{0} = 1.26V \nonumber$ c) Calculate the energy difference between states suggested by this standard reduction potential. d) Ru(bpy)32+ absorbs very strongly in the visible spectrum (450 nm). What colour is it? e) What type of transition do you think this is? f) Calculate the energy difference from the ground state to the initially formed excited state Ru(bpy)32+*. g) What is the multiplicity of this initially formed excited state (e.g. a singlet or a triplet etc)? Show why. h) The initially formed excited state Ru(bpy)32+* undergoes internal conversion (radiationless relaxation) and intersystem crossing to form triplet Ru(bpy)32+* before phosphorescence occurs at 615 nm. What colour is observed upon phosphorescence? i) Calculate the energy difference from the triplet Ru(bpy)32+* to the ground state. j) The difference between the wavelength absorbed and the wavelength emitted during fluorescence or phosphorescence is called the Stokes shift. What is the Stokes shift in this case? k) Calculate the amount of energy transferred to vibrational states during the internal conversion. l) Calculate the potential for the following reaction: \(\ce{Ru(bpy)3^{3+} + e^{-} -> Ru(bpy)3^{2+*}}$ (triplet state) $E^{0} = ?V$
m) Compare the usefulness of Ru(bpy)32+ +vs. Ru(bpy)32+*as a reducing agent.
Answer b
b) Ru(bpy)32+ would be a terrible reducing agent. The reduction potential of Ru(bpy)33+ is very high. That means Ru(bpy)32+ would not give up an electron very easily.
Answer c
c) ΔG = - n F E0
ΔG = - 1 x 97,485 J V-1 mol-1 x 1.26 V = -123, 000 J mol-1 = -123 kJ mol-1
Answer d
d) It absorbs in the blue part of the spectrum and appears orange.
Answer e
e) It absorbs very strongly, so probably not d-d. It is probably MLCT, from the ruthenium to the π* in the bpy ligand.
Answer f
f) E = h c / λ
E = (6.625 x 10-34 J s mol-1 x 3.0 x 108 m s-1 )/ (450 nm x 10-9 m nm-1) = 4.42 x 10-19 J
That's for one molecule. On a per mole basis, E = 4.42 x 10-19 J x 6.02 x 1023 mol-1 = 266,000 J mol-1 = 266 kJ mol-1.
Answer g
g) Ru2+ would be low spin d6, a singlet ground state. The excited state will also be a singlet state.
Answer h
h) 615 nm is in the orange region of the spectrum. This is light given off by the complex as it relaxes, so it is the colour we see.
Answer i
i) E = h c / λ
E = (6.625 x 10-34 J s mol-1 x 3.0 x 108 m s-1 )/ (615 nm x 10-9 m nm-1) = 3.23 x 10-19 J
That's for one molecule. On a per mole basis, E = 3.23 x 10-19 J x 6.02 x 1023 mol-1 = 195,000 J mol-1 = 195 kJ mol-1.
Answer j
j) Stokes shift = 615 nm - 450 nm = 165 nm
Answer k
k) ΔE = E1 - E2 = 266 - 195 kJ mol-1 = 71 kJ mol-1
However, it would not be lost all at once, but in small increments equivalent to the differences between vibrational states.
Answer l
l) If we were to reduce Ru(bpy)33+ directly into an excited state, we would arrive at a state much higher in energy than the ground state. That reduction would be harder to accomplish. In this case, the ending state would be 195 kJ mol-1 higher than Ru(bpy)32+.
ΔG = 195 - 123 kJ mol-1 = 72 kJ mol-1
E0 = - ΔG / n F = - (72 kJ mol-1 x 1000 kJ J-1) / (1 x 96,485 J V-1 mol-1) = -0.75 V
Answer m
m) The oxidation of Ru(bpy)32+* would be very favourable compared to the oxidation of Ru(bpy)32+. The former is a much better reducing agent. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/V%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_3/08%3A_Photochemical_Reactions/8.03%3A_Fluorescenc.txt |
We have seen that the absorption of photons (especially in the ultraviolet-visible spectrum) is connected to the excitation of electrons. After excitation, a number of different relaxation pathways lead back to the ground state. Sometimes, absorption of a photon leads to a vastly different outcome. Instead of just relaxing again, the molecules may undergo bond-breaking reactions, instead.
An example of this phenomenon is observed in the complex ion [Co(NH3)]63+. Addition of UV light to this complex results in loss of ammonia. In the absence of UV light, however, the complex ion is quite stable.
In many cases, loss of a ligand is followed by replacement by a new one. For example, if an aqueous solution of [Co(NH3)]63+ is photolysed, an ammonia ligand is easily replaced by water.
Exercise \(1\)
Draw a d orbital splitting diagram for [Co(NH3)]63+. Explain why this complex is normally inert toward substitution.
Exercise \(2\)
Use the d orbital splitting diagram for [Co(NH3)]63+ to explain why this complex undergoes substitution upon irradiation with UV light
Photolysis is the term used to describe the use of light to initiate bon-breaking events. Photolysis frequently involves the use of high-intensity ultraviolet lamps. The high intensity light is needed in order to provide enough photons to get higher conversion of reactant into a desired product.
Two very different things could happen as a result of photon absorption. In one case, the molecule absorbs the photon, then somehow relaxes again, remaining unchanged overall. In the other case, the absorption of the photon results in bond cleavage and the formation of a new product. As a result, for every photon absorbed, there is a certain chance that the molecule will actually undergo a reaction, and a certain chance that the molecule will just relax again.
"Quantum yield" is an expression used to define the efficiency of a photolytic reaction. The quantum yield is just the number of molecules of reactant formed per photon absorbed. On a macroscopic level, we might say it is the number of moles of reactant formed per mole of photons absorbed.
Quantum yield = number of molecules of product formed / number of photons absorbed
The higher the quantum yield, the more efficient the reaction, because it requires less light in order to successfully form the product.
Exercise \(3\)
Calculate the quantum yield in the following cases.
1. 6 mmoles of product results from absorption of 24 mmoles of photons.
2. 54 mmoles of photons are required to produce 3 mmoles of product.
3. 1.2 x 10-6 moles of product are formed after absorption of 4.2 x 10-5 moles of photons.
In practice, determination of quantum yield is complicated because of the need to calculate exactly how many photons have been absorbed, in addition to how much product has been formed.
8.05: Atmospheric
There are lots of ways that light plays an important role in the world. Life on earth is heavily dependent on photosynthesis. Energy from the sun can be harvested and used to make ATP, which can then be used for carbohydrate synthesis. The energy found in light, however, can also be very damaging. In fact, ultraviolet light would make life impossible on earth were it not for the intercession of the high-atmosphere ozone layer. By undergoing reactions that absorb UV light, the molecules involved in the ozone cycle block much of the most harmful sunlight from reaching the earth's surface.
The ozone cycle is a series of reactions involving allotropes of oxygen. Oxygen can be present as oxygen atom, O; as oxygen molecule, O2; or as ozone, O3. Of course, dioxygen, or molecular oxygen, is by far the most common of these forms, constituting about 20% of our atmosphere.
That's where we will start. There is plenty of dioxygen in our atmosphere, and so photons from the sun encounter these molecules pretty frequently. If the photons have sufficient energy, they can break the O-O bond, forming oxygen atoms.
$\ce{O2 + energy -> 2O} \nonumber$
That photolysis reaction (meaning, "broken by light") leads to a buildup of oxygen atoms in the upper atmosphere. Of course, these oxygen atoms are likely to encounter oxygen molecules, which are very abundant. When they do, they can undergo a bond-forming reaction, to make ozone. This reaction is purely bond-forming, and it is exothermic.
$\ce{O + O2 -> O3 + energy} \nonumber$
There are also a couple of reactions that cause removal of ozone from the atmosphere. The first is the reaction of ozone with ultraviolet light. The energy from the sunlight can break the O-O bond in ozone, converting it back into dioxygen and oxygen atom.
$\ce{O3 + energy -> O + O2} \nonumber$
In addition, as ozone and oxygen atom build up, collisions between these two species becomes increasingly likely. If that happens, an oxygen atom abstraction is an obvious outcome, in which two dioxygen molecules are the products. This reaction involves both bond-breaking and bond-making, but the reaction is overall exothermic.
$\ce{O3 + O -> 2O2 + energy} \nonumber$
Notice those two energy-consuming reactions. One of them occurs in the ozone creation phase and one in the ozone removal phase. Together, these reactions contribute to the consumption of ultraviolet light as it passes through the stratosphere, so that less ultraviolet light reaches the earth's surface below. However, the two reactions are on opposing sides of a cycle. Consumption of some sunlight leads to ozone formation, but consumption of sunlight also leads to ozone destruction.
The thermodynamics of the ozone cycle are illustrated below. Again, two of the reactions are endothermic, whereas two others are endothermic.
The endothermic steps actually require more energy than is implied in the picture above. As in any reaction, it usually isn't sufficient to supply enough energy to get from one side of the reaction to the other. There is also an energy barrier to overcome. That's part of the reason these reactions need ultraviolet light.
The first reaction requires light of wavelengths shorter than about 240 nm.
The second reaction requires light of wavelengths shorter than about 325 nm.
Exercise $1$
Why is there an upper limit to the wavelength of light capable of inducing these reactions?
Answer
The longer the wavelength, the lower the energy. Photons of wavelength longer than 240 nm would not have enough energy to overcome the barrier for the dioxygen-cleaving reaction.
Exercise $2$
Calculate the energy, in J, of the following photons:
1. 220 nm
2. 325 nm
Answer a
According to the Planck-Einstein relation:
E = hν
or, since ν = c / λ
E = hc / λ
in which h = Planck's constant = 6.625 x 10-34 Js,
c = speed of light = 3.0 x 108 m/s,
ν = frequency,
λ = wavelength.
a) E = hc / λ
E = (6.525 x 10-34 Js)(3.0 x 108 m/s) / (220 nm)(10-9 m/nm)
E = 9.03 x 10-19 J
Answer b
b) E = hc / λ
E = (6.525 x 10-34 Js)(3.0 x 108 m/s) / (325 nm)(10-9 m/nm)
E = 6.12 x 10-19 J
Exercise $3$
1. Given the wavelength of light needed for the photolysis of O2, calculate the barrier to the reaction, in kJ/mol.
2. Sketch a reaction progress diagram for the reaction.
Answer a
Remember, the mole is the conversion unit from the molecular scale to the macroscopic scale.
a) E = (8.28 x 10-19 J / photon)(6.02 x 1023 photons/mol)
E = 543,770 J/mol
E = 544 kJ/mol
Answer b
b)
Exercise $4$
1. Given the wavelength of light needed for the photolysis of O3, calculate the barrier to the reaction, in kJ/mol.
2. Sketch a reaction progress diagram for the reaction.
Answer a
a) E = (6.12 x 10-19 J / photon)(6.02 x 1023 photons/mol)
E = 368,146 J/mol
E = 368 kJ/mol
Answer b
b)
Exercise $5$
Given the principle of microscopic reversibility, calculate the barrier to the following reactions, in kJ/mol.
1. $\ce{2O -> O2}$
2. $\ce{O + O2 -> O3}$
Answer a
The reactions must take the same pathway, and go over the same barrier, forward and back.
a) The reverse barrier: E = 544 - 498 kJ/mol
E = 46 kJ/mol
Answer b
b) The reverse barrier: E = 368 - 105 kJ/mol
E = 263 kJ/mol
Ozone is clearly at a higher energy than dioxygen. The fact that appreciable amounts exist in the upper atmosphere can be attributed to the constant input of energy from the sun. This is an example of an environmental steady state. The ratio of ozone to oxygen, although still pretty low even in the stratosphere, is elevated above its natural equilibrium value because energy is constantly being added, pushing the system uphill in energy.
The Problem with Chlorine
In 1974, Sherwood Rowland, a professor of chemistry at the University of California, Irvine, and Mario Molina, a Mexican post-doctoral associate in his laboratory, authored a paper in the journal Nature outlining how chlorofluorocarbons (CFCs) could play a role in interfering with the ozone cycle. CFCs at that time were an economically important compound because they were a non-toxic, highly effective refrigerant used in millions of homes and businesses worldwide. Rowland and Molina shared the 1995 Nobel Prize in Chemistry with Paul Crutzen, a Dutch citizen working at the Max Planck Institute in Germany, who similarly reported the effects of nitrogen oxides on the ozone cycle.
The problem with CFCs is really the fact that, in the upper atmosphere, they can break down to form chlorine atoms. For example, freon-12, or CF2Cl2, can absorb ultraviolet light, resulting in cleavage of the C-Cl bond.
Chlorine atom in the stratosphere interferes with the ozone cycle by consuming both ozone and, indirectly, a key intermediate in ozone formation, oxygen atom.
$\ce{O3 + Cl -> ClO + O2} \nonumber$
$\ce{ClO + O -> O2 + Cl} \nonumber$
What makes the role of chlorine particularly problematic is that, in addition to wiping out ozone and an intermediate needed to replace it, the chlorine is regenerated at the end, so it is able to go back into the system and do it all over again. It is destructive to ozone, and with catalytic efficiency.
After environmental regulations were introduced mandating an eventual industry-wide phase-out of CFCs, industry was able to respond with the development of chlorine-free substitutes that were still non-toxic and non-flammable (both important for household use). Hydrofluorocarbons are the major type of replacement that has been developed for use today. There are other compounds commonly used as refrigerants, but they have significant drawbacks. Ammonia is toxic, and hydrocarbons are very flammable.
There is an additional consideration that researchers must take into account in looking for CFC replacements, however: lifetime. If a molecule lasts too long in the atmosphere, there is the risk of a long-term buildup that may result in additional environmental consequences. The fact that chlorocarbons and fluorocarbons contain polar C-Cl and C-F bonds means that they will absorb infrared radiation strongly, so they will act as greenhouse gases. That's part of the reason hydrofluorocarbons are used. The C-H bond in a hydrofluorocarbon is easily broken by atmospheric hydroxy radical, because of the strength of the new O-H bond that is formed. That event provides for a decomposition pathway that removes the molecule from the atmosphere more quickly. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/V%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_3/08%3A_Photochemical_Reactions/8.04%3A_Photolysis.txt |
There are lots of ways that photochemistry can be harnessed to do useful things. Solar power is an obvious example: sunlight is used to excite an electron in a solar panel, leading to generation of electricity. That kind of power can potentially have a broad impact on other aspects of the energy industry. For example, a number of researchers have been trying to use sunlight as an energy source to make hydrogen from water.
Photoredox catalysis is the application of photochemistry to carry out redox reactions catalytically. For example, if sunlight were absorbed by a material such as a semiconductor, resulting in excitation of an electron, the resulting species may have the energy to reduce the protons in water to dihydrogen. If the semiconductor could be regenerated with a reducing agent, the process would be catalytic.
Perhaps the most powerful aspect of photoexcitation is that it results in both a powerful oxidant and a powerful reductant. Excitation of an electron results in electron/hole pair separation. After excitation, the electron is in a higher energy level, and it leaves behind a "hole", where the electron used to be.
Because the electron is at a higher energy level than it used to be, donation to an acceptor is more favorable than it was before. The compound has become a better reducing agent through photoexcitation. Also, because the hole is lower in energy than the LUMO of the ground state, it will more easily accept and electron from a donor. That means the compound has also become a better oxidising agent through photoexcitation.
We can estimate how much the reducing or oxidizing power of a compound changes upon excitation by a photon. For example, the reduction potential depends upon the relative energy level of the orbital into which an electron would be donated. Upon photoexcitation, that energy level changes. It goes from the ground state LUMO to the level of the hole introduced where an electron used to be. The difference in energy between those two levels is the same as the energy of the photon that was absorbed.
We can calculate the energy of the absorbed photon. That quantity is given by the Planck-Einstein relation:
$E = h \nu \: or \: E= \frac{hc}{\lambda} \nonumber$
in which E is energy, h is Planck's constant, ν is the frequency of the photon, c is the speed of light, and λ is the wavelngth of the photon.
That energy is the same as the distance between the ground state LUMO and the excited state hole. Now, that is a quantum-scale energy difference, found within a single molecule. When we talk about reduction potentials, we are discussing a macroscopic phenomenon. Reduction potentials are measured on the scale of human experience, the scale of the very large, the scale that we can easily observe. Macroscopic measurements involve vast numbers of atoms, and the convenient unit of a vast number of atoms is a mole. Remember, that's just the conversion factor between an easily measurable quantity of mass (a gram) and a quantum-scale quantity of mass (an atomic mass unit, also called an amu or a Dalton).
So we might think about how much energy it would take to get an entire mole of molecules into the excited state, which would simply require the use of Avogadro's number. Clearly this will be a much bigger amount of energy than it takes to excite a single molecule, and we would get there by multiplying by Avogadro's number.
$E = (h \nu ) N_{A} \: or \: E = (\frac{hc}{\lambda})N_{A} \nonumber$
in which NA = Avogadro's number.
(We could equally well be dealing with an individual atom or some other kind of compound rather than a molecule, but the argument would be the same.)
Because the excited state hole is that much lower in energy than the original LUMO, an electron falling into that hole would release that much more energy than an electron falling into the ground state LUMO. On the macroscopic scale, a mole of electrons falling into the excited state hole would release so much more energy than a mole of electrons falling into the ground state LUMO; that energy difference is the quentity that we have just calculated using Avogadro's number.
Of course, a similar argument holds for the excited state electron. It is much more likely to be donated to an acceptor after first having been excited.
We know that the free energy released when an electron falls to a lower level is related to the reduction potential. That difference is given by the relation
$\Delta G = -nF \Delta E^{o} \nonumber$
in which ΔG is the free energy change, n is the number of electrons transferred, F is the Faraday constant, and ΔEo is the cell reduction potential, or the difference in reduction potential between the donor and the acceptor.
Here we will make a slight approximation. The energy of the photon is roughly the difference in free energy upon electron transfer from a donor to the excited state hole compared to the ground state LUMO (although the energy calculated in the Planck-Einstein relation doesn't say anything about entropy, which is an implicit part of the free energy change).
$E_{excitation} \sim \Delta G_{reduction} \nonumber$
From there, it is a short step to figure out the change in reduction potential when the electron is excited.
$\Delta E^{o}_{new} = \Delta E^{o}_{original} + \Delta E^{o}_{photo} \nonumber$
The possibility for tremendous utility of an excited state species in catalysis arises from the fact that the excited state can be both a better reductant and a better oxidant than the ground state species. As a better oxidant, it can easily oxidize another compound; as a better reductant, it can then be reduced again, forming the original, ground state species and completing a catalytic cycle. Alternatively, it could act as a reductant first and complete the cycle by becoming reduced again; the order of the two steps does not really matter.
Let's think about the example of hydrogen production. Hydrogen is very appealling as a fuel because, when it burns, it provides a large amount of energy through the formation of strong O-H bonds. In addition, the product of its combustion is water, which does not pose any obvious problems as a pollutant. An economy based on hydrogen consumption is potentially more sustainable in environmental terms.
Where could we get hydrogen to drive this economy? Currently, we obtain hydrogen from coal and natural gas. In a process called steam reforming, hydrocarbon gases such as methane can be passed over a catalyst in the presence of steam, producing H2 and carbon oxygenates, mostly CO but also some CO2. A similar process can be carried out using coal instead of natural gas. The carbon monoxide is usually captured for other industrial processes, but it can also be used to make even more hydrogen via the water gas shift reaction. In the water gas shift reaction, the carbon monoxide is again treated with steam over a different catalyst, producing H2 and carbon dioxide.
Although the water gas shift reaction is exothermic, steam reforming is overall endothermic, so energy must be expended in order to produce hydrogen. That situation isn't optimal, because we end up having to spend a lot of energy in order to produce an energy source. In addition, we produce carbon dioxide, which is a greenhouse gas, the human production of which has been convincingly linked to global warming and climate change. Furthermore, coal and natural gas reserves may be plentiful right now, but the same was also once true of petroleum reserves, and although petroleum companies continue to locate and extract new deposits of petroleum, the extraction process becomes more difficult and expensive as these deposits become more and more remote or inaccessible.
Water is the most abundant source of hydrogen on earth. It would be much easier to produce hydrogen from water. People have been doing that for about 200 years, just by placing electrodes in water and applying a voltage. This is called electrolysis of water. The trouble is, the electrolysis of water requires massive amounts of electricity. Electricity production requires energy consumption. In the United States, electricity production still depends heavily on consumption of coal and natural gas, so instead of replacing our need for these commodities, we have simply removed them from sight.
Photoredox catalysis may offer a solution. By producing hydrogen from water catalytically, we take a roundabout approach that requires less energy than direct electrolysis of water. By driving the reaction photochemically, we can harness sunlight to provide that energy.
There are different possible routes that can convert protons to dihydrogen catalytically. One example occurs with coordination compounds. Some coordination compounds in low oxidation states can be protonated at the metal centre by water. Protonation of a metal centre is formally an oxidative addition. Once the proton is bound to the metal centre, it is considered a hydride ligand, so it has formally changed from H+ to H-. The two electrons to convert it to a hydride came from the metal.
Catalytic turnover results after the metal centre has been protonated twice, resulting in a pair of hydride ligands on the metal. Reductive elimination of dihydrogen returns two of the electrons to the metal centre; the other two leave with the dihydrogen. The metal still needs two more electrons from another source, because in undergoing two oxidative additions it has donated a total of four electrons to the protons.
One of the problems with this approach is that in order to make the process catalytic, the two electrons that were sent to the protons in water to make hydrogen must be replaced. Many researchers in this area have used sacrificial electron donors, compounds that can donate electrons to the catalyst to replace the ones given to the protons. Sacrificial donors could be biological cofactors such as ascorbate or succinate, or they could be simple amines.
Because these donors are consumed in this process, they represent a potential inefficiency in the form of a waste side product. Some researchers hope to circumvent this problem by tying in an additional catalytic cycle that regenerates the electron donors. Others hope to bypass this problem by using the oxygen atom in water as the sacrificial donor, producing dioxygen, but doing so with a low energy cost is a challenge.
Another problem is how to drive the system photochemically. A common approach in current research efforts involves the use of semiconducting materials that can absorb sunlight (that's what solar panels are, after all). The excited state electron is passed to the catalytic component of the system, and the hole is filled by the sacrificial electron donor.
Exercise $1$
Researchers at Colorado State have developed a new system for photolysis of water (Finke, ACS Appl. Mater. Interfaces, 2014, ASAP).
PDI (below), a type of dye used in automobile paint, was coated onto an indium tin oxide plate. The UV-Vis spectrum was obtained and the reduction potential was measured.
1. What colour is the original PDI-coated plate?
2. What is the longest wavelength absorbed?
3. Translate this wavelength into energy, in kJ/mol.
4. Translate this energy into a potential, in Volts.
5. Show a diagram of photon absorbance to explain why this energy is a more accurate reflection of the band gap in PDI than that of other wavelengths that are absorbed.
6. What type of transition is probably involved, given the structure of PDI?
$MLCT \: d-d \: n \rightarrow \pi * \pi \rightarrow \pi * \nonumber$
g) What happens to the energy of the transition after dipping the plate in acid?
It is thought that this transition can be intermolecular in this case:
$2 PDI + h \nu \rightarrow PDI + \: PDI- \nonumber$
h) Most molecules don't undergo intermolecular photonic excitation. What makes it easier in this case?
i) How is the idea of intermolecular photonic excitation consistent with the acid-dipped experiment?
j) Use the cyclic voltamogram to estimate the reduction potential of PDI vs Ag/AgCl.
Cobalt oxide was photodeposited on the PDI surface by dipping the plate in phosphate buffered (KH2PO4/ K2HPO4) cobalt nitrate solution (Co(NO3)2) and irradiating with a xenon arc lamp.
k) Show a d orbital splitting diagram for the octahedral aqueous cobalt ion before and after photon absorption.
l) Provide a mechanism for the formation of Co(IV)2O in that solution.
The following reduction potentials have been reported in the literature:
indium tin oxide: -0.10 V vs NHE
cobalt oxide: 1.23 V vs NHE
water: 0.83 V vs NHE
Now we have a problem, because some of our reduction potentials are compared to NHE; they tell us how easy it is to reduce something compared to reducing H+. Some are compared to Ag/AgCl; they tell us how easy it is to reduce something compared to reducing Ag+.
m) If the reduction potential of Ag+ is +0.20 V vs NHE, then what are the following potentials vs. Ag/AgCl?
i) indium tin oxide ii) cobalt oxide iii) water
n) Calculate the reduction potential of the excited state PDI*.
o) Fill in the appropriate energy levels for the species indicated.
p) Add curved arrows to indicate electron flow through the system during photolytic oxygen generation.
q) Studies indicate that the system does not work if the PDI does not have phosphates attached to it. Explain why.
Answer a
a) The original PDI has an absorption maximum at just under 500 nm. It absorbs blue-green, so it would be expected to appear red-orange.
Answer b
b) The longest wavelength absorbed is approximately 625 nm.
Answer c
c) E = hc/λ = (6.625 x 10-34 Js mol-1 x 3.0 x 108 m s-1)/625 x 10-9 m = 3.18 x 10-19 J
That's the energy of a photon. A mole of photons would have energy multiplied by Avogadro's number, NA.
E = 3.18 x 10-19 J x NA = 3.18 x 10-19 x 6.02 x 1023 mol-1 = 1.91 x 105 J mol-1 = 191 kJ mol-1
Answer d
d) If we approximate E = ΔG, and given that ΔG = - nFE0
Then for this single-electron excitation, E0 = - ΔG / n F = - 1.91 x 105 J mol-1 / (1 x 96,485 J V-1 mol-1) = -1.97 V
Answer f
f) Given the structure of PDI, the transition is probably π --> π*.
Answer g
g) The longest wavelength absorbed shifts toward the red, to about 700 nm, after dipping in acid. The energy of the transition is lowered to
E = hc/λ = (6.625 x 10-34 Js mol-1 x 3.0 x 108 m s-1)/700 x 10-9 m = 2.84 x 10-19 J
E = 2.84 x 10-19 J x NA = 3.18 x 10-19 x 6.02 x 1023 mol-1 = 1.71 x 105 J mol-1 = 171 kJ mol-1
Answer h
h) These very flat molecules can probably stack very tightly together. The distance between the π electrons on one molecule and the π* orbital on a neighbouring molecule is very small. An intermolecular transition is possible.
Answer i
i) What happens when the film is dipped in acid? It gets protonated. The anionic phosphate groups would become neutralized. The molecules would be able to stack even more closely together, lowering the energy required to excite an electron from one molecule to the other.
Answer j
j) The onset of the reduction wave appears to be about -0.40 V, as shown by the peak in the cyclic voltamogram.
Answer m
m) E0(vs Ag+/AgCl) = E0(vs NHE) + 0.20 V (i.e. the positive reduction potential of Ag+ vs. NHE indicates an electron is
i) E0(vs Ag+/AgCl) = - 0.10 V + 0.20 V = + 0.10 V
ii) E0(vs Ag+/AgCl) = + 1.23 V + 0.20 V = + 1.43 V
iii) E0(vs Ag+/AgCl) = + 0.83 V + 0.20 V = + 1.03 V
Answer n
n) The excited state can be reduced by placing an electron in the hole vacated by the excited electron. This level is 1.97 V below the conductance band in energy, which has a reduction potential of -0.40 V (its potential is 1.97 V more positive than the conductance band).
Thus, -0.40 V + 1.97 V = + 1.57 V.
Answer q
q) The phosphates probably coordinate to the cobalt oxide, forming a bridge for electron transfer.
Exercise $2$
Researchers in China have developed Ag-AgBr nanoparticles capable of catalysing the photochemical degradation of pollutants in sunlight. (Langmuir 2010, 26, 18723-18727)
1. The AgBr nanoparticles are first formed by autoclaving AgNO3 and tetraalkylammonium bromide. Show an equation for this reaction.
2. The AgBr nanoparticles absorb strongly between 250-450 nm. What colour are they?
3. The structure of AgBr can be described as face centered cubic bromide with silver in the octahedral holes. Draw a unit cell.
The AgBr nanoparticles are suspended in water and exposed to sunlight for several hours until Ag picoparticles form on the surfaces of the nanoparticles.
d) Compare the size of nanoparticles to picoparticles.
e) Photoreduction of silver is believed to involve electron transfer from solvent. Show a mechanism for Ag(0) formation.
f) The Ag-AgBr nanoparticles absorb strongly from 250-850 nm. What colour are they?
In remediation assays, methyl orange is used as a stand-in for water-borne pollutants. The authors speculate that photon absorption causes electron / hole pairs that lead to reduction of O2 and oxidation of water, respectively, generating reactive oxygen species.
h) Draw an MO or band gap picture to help illustrate these two processes (water oxidation and O2 reduction).
i) Show a mechanism for radical propagation using a reactive oxygen species and methyl orange.
Answer a
a) Ag+(aq) + NO3-(aq) + Bu4N+(aq) + Br -(aq) → AgBr(s) + Bu4N+ (aq) + NO3- (aq)
Answer b
b) They absorb in the violet-blue range and would appear yellow-orange.
Answer d
d) A nanometer is 10-9 m, whereas a picometer is 10-12 m. A picoparticle is 1,000 times smaller than a nanoparticle.
Answer f
f) They absorb across the visible spectrum and appear black. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/V%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_3/08%3A_Photochemical_Reactions/8.06%3A_Application.txt |
Exercise 8.1.1:
1. $A = \varepsilon c l = 6 \frac{L}{mol \: cm} \times 0.01 \frac{mol}{L} \times 1 cm = 0.60 = 60 \%$
2. $A = \varepsilon c l = 3000 \frac{L}{mol \: cm} \times 3.5 \times 10^{-5} \frac{mol}{L} \times 1 cm = 0.105 = 10.5 \%$
3. $A = \varepsilon c l = 1.4 \frac{L}{mol \: cm} \times 0.25 \frac{mol}{L} \times 0.5 cm = 0.175 = 17.5 \%$
4. $A = \varepsilon c l = 23000 \frac{L}{mol \: cm} \times 2.5 \times 10^{-6} \frac{mol}{L} \times 1 cm = 0.0575 = 5.75 \%$
5. $A = \varepsilon c l = 14000 \frac{L}{mol \: cm} \times 0.015 \frac{mmol}{L} \times 1 cm = 14000 \frac{L}{mol \: cm} 0.015 \times 10^{3-} \frac{mol}{L} \times 1 cm = 0.21 = 21 \%$
Exercise 8.1.2:
1. $\varepsilon= \frac{A}{cl} = \frac{0.30}{(0.01 mol \: L^{1-} \times 1 cm)} = 30 \frac{L}{mol \: cm}$
2. $\varepsilon = \frac{A}{cl} = \frac{0.25}{(0.025 mol \: L^{-1} \times 1cm)} = 10 \frac{L}{mol \: cm}$
3. $\varepsilon = \frac{A}{cl} = \frac{0.30}{(0.01 mol \: L^{-1} \times 1cm)} =543 \frac{L}{mol \: cm}$
4. $\varepsilon = \frac{A}{cl} = \frac{0.66}{(0.025 mmol \: L^{-1} \times 1cm)} = 26400 \frac{L}{mol \:cm}$
Exercise 8.1.3:
1. blue
2. red
3. orange
4. violet
5. green
Exercise 8.1.4:
1. $E = \frac{hc}{\lambda} = \frac{(6.625 \times 10^{-34} Js \: mol^{-1} \times 3.0 \times 10^{8} m \: s^{-1})}{1 \times 10^{-5}m} = 1.98 \times 10^{-20} J$
2. $E = \frac{hc}{\lambda} = \frac{(6.625 \times 10^{-34} Js \: mol^{-1} \times 3.0 \times 10^{-8} m \: s^{-1})}{125 \times 10^{-9}m} = 1.59 \times 10^{-18} J$
3. $E = \frac{hc}{\lambda} = \frac{(6.625 \times 10^{-34} Js \: mol^{-1} \times 3.0 \times 10^{-8} m \: s^{-1})}{1025 \times 10^{-9}m} = 1.94 \times 10^{-19}J$
4. $E = \frac{hc}{\lambda} = \frac{(6.625 \times 10^{-34} Js \: mol^{-1} \times 3.0 \times 10^{8} m \: s^{-1})}{450 \times 10^{-6}m} = 4.42 \times 10^{-22}J$
5. $E = \frac{hc}{\lambda} = \frac{(6.625 \times 10^{-34} Js \: mol^{-1} \times 3.0 \times 10^{-8} m \: s^{-1})}{850 \times 10^{-10}m} = 2.3 \times 10^{-18}J$
Exercise 8.1.5:
1. $\lambda = \frac{hc}{E} = \frac{(6.625 \times 10^{-34} Js \: mol^{-1} \times 3.0 \times 10^{-8} m \: s^{-1})}{1.46 \times 10^{-17} J} =1.36 \times 10^{-8} m$
2. $\lambda = \frac{hc}{E} = \frac{(6.625 \times 10^{-34} Js \: mol^{-1} \times 3.0 \times 10^{-8} m \: s^{-1})}{4.72 \times 10^{-24} J} =4.21 \times 10^{-2} m$
3. $\lambda = \frac{hc}{E} = \frac{(6.625 \times 10^{-34} Js \: mol^{-1} \times 3.0 \times 10^{-8} m \: s^{-1})}{9.26 \times 10^{-17} J} =2.15 \times 10^{-19} m$
Exercise 8.1.6:
1. $E = h \nu = \frac{hc}{\lambda}$; so $v = \frac {c}{\lambda}$; or $\lambda = \frac{c}{v} = \frac{3.0 \times 10^{8} m \: s^{-1}}{6.7 \times 10^{10} s^{-1}}=4.48 \times 10^{-3} m$
2. $\lambda = \frac{c}{v} = \frac{3.0 \times 10^{8} m \: s^{-1}}{1500 \times 10^{6} s^{-1}} = 0.2m$
3. $v = \frac{c}{\lambda} = \frac{3.0 \times 10^{8} m \: s^{-1}}{9.8 \times 10^{-10}m} = 3.06 \times 10^{15} s^{-1}$
4. $v \frac{c}{\lambda} = \frac{3.0 \times 10^{8} m \: s^{-1}}{4.3 \times 10^{-12}m}= 7.0 \times 10^{19} s^{-1}$
Exercise 8.1.7:
a) $E = \frac{hc}{\lambda}$
$E = \frac{(6.625 \times 10^{-34} Js \:mol^{-1} \times 3.0 \times 10^{8} m \: s^{-1})}{(1mm \times 10^{-3}m \: mm^{-1})} = 1.99 \times 10^{-22}J \nonumber$
That's for one molecule. On a per mole basis, $E = 1.99 \times 10^{-22} J \times 6.02 \times 10^{23} mol^{-1} = 120 J mol^{-1} = 0.12 kJ \: mol^{-1}$
b) $E = \frac{hc}{lambda}$
$E = \frac{(6.625 \times 10^{-34} Js \: mol^{-1} \times 3.0 times 10^{8} m \: s^{-1})}{(1000nm \times 10^{-9} m \: nm^{-1})} = 1.99 \times 10^{-19}J \nonumber$
That's for one molecule. On a per mole basis, $E = 1.99 \times 10^{-19} J \times 6.02 \times 10^{23} mol^{-1} = 120000 J \: mol^{-1} = 120 kJ \: mol^{-1}$
c) $E = \frac{hc}{\lambda}$
$E = \frac{(6.625 \times 10^{-34} Js \: mol^{-1} \times 3.0 \times 10^{8} m \: s^{-1})}{(1m)} = 1.99 \times 10^{-25} J \nonumber$
That's for one molecule. On a per mole basis, $E = 1.99 \times 10^{-25} J \times 6.02 \times 10^{23} mol^{-1} = 0.120 J \: mol^{-1} = 1.2 \times 10^{-4} kJ \: mol^{-1}$
d) $E = \frac{hc}{\lambda}$
$E = \frac{(6.625 \times 10^{-34} Js mol^{-1} \times 3.0 \times 10^{8} m s^{-1})} = {(500 nm \times 10^{-9} m \: nm^{-1})} = 3.98 \times 10^{-19} J \nonumber$
That's for one molecule. On a per mole basis, $E = 3.98 \times 10^{-19} J \times 6.02 \times 10^{23} mol^{-1} = 239000 J mol^{-1} = 239 kJ\: mol^{-1}$
Exercise 8.1.8:
Blue.
Exercise 8.1.9:
Ultraviolet light, with a shorter wavelength than visible light, is much higher in energy and potentially more damaging.
Exercise 8.2.3:
1. $d \rightarrow d$
2. $\pi \rightarrow \pi *$
3. MLCT
Exercise 8.2.4:
1. The one on the left can relax by channelling some energy into molecular vibration, especially its cis-trans isomerisation. The one on the right can't do that because its rotation is restricted by the presence of the ring.
2. Of course! It's crawling with pi bonds. A strong pi-pi* transition in the visible region seems likely.
3. Probably not. It does not have restricted rotation, so the cis-trans isomerisation route is available for relaxation.
4. $A = \varepsilon bc$ ; so $\varepsilon = \frac{A}{bc} = \frac{0.77}{1cm \times 10^{-6}M} = 7.7 \times 10^{5} M^{-1} cm^{-1}$
5. That's a large molar absorptivity constant. It's allowed.
6. Undoubtedly this is that pi-pi* transition we were thinking about earlier.
7. If the compound binds to zinc, it probably does so via bidentate coordination. The resulting ring restricts the degrees of freedom in the compound so it can't undergo cis-trans isomerisation, closing off a route to rapid relaxation.
i) A compound like this could be used to detect metal ions such as Zn2+. Because the amount of fluorescence depends strongly on the Zn2+ concentration, it could be used to measure the amount of the ion present.
Exercise 8.2.5:
a)
b) An orange complex would absorb blue light, its complementary colour.
c) A blue laser would work. Maybe somewhere around 476 nm.
d) Notice a Re(II) site results, because it has transferred an electron to the ligand.
e)
f) The reduction potentials (1.4 V > 0.3 V) suggest transfer from copper (I) to rhenium (II).
g) i) The carbonyl stretches would be observed somewhere around 2000 cm-1.
iii) The electron transfer from the Cu(I) to the Re(II) site would result in a Re(I) site. The carbonyl stretch would shift to a lower wavenumber. That is because the lower oxidation state on the metal results in increased backbonding to the π-accepting carbonyls.
iii) At that distance, an outer sphere mechanism seems likely; it isn't clear how something could bridge that distance, other than the peptide chain itself, which probably lacks sufficient conjugation to conduct electrons.
h) Tryptophan acts as a stepping stone between copper and rhenium. Phenylalanine, with a higher reduction potential than rhenium, does not.
i)
Exercise 8.3.1:
a) $E = \frac{hc}{\lambda}$
$E = \frac{(6.625 \times 10^{-34} Js mol^{-1} \times 3.0 \times 10^{8} m s^{-1})}{(180 nm \times 10^{-9} m \: nm^{-1})} = 1.10 \times 10^{-18}J \nonumber$
That's for one molecule. On a per mole basis, $E = 1.10 \times 10^{-18} J \times 6.02 \times 10^{23} mol^{-1} = 665000 J mol^{-1} = 665 kJ \: mol^{-1}$
For comparison, the relatively strong and unreactive C-H bond in methane has a bond dissociation energy of only 440 kJ mol-1. (That's a thermodynamic value; to actually break the bond would cost more input of energy, to get over the kinetic energy barrier for bond cleavage.)
b) $E = \frac{hc}{\lambda}$
$E = \frac{(6.625 \times 10^{-34} J \: s \: mol^{-1} \times 3.0 \times 10^{8} m \: s^{-1})}{(476 nm \times 10^{-9} m \: nm^{-1})} = 4.17 \times 10^{-19} J \nonumber$
That's for one molecule. On a per mole basis, $E = 4.42 \times 10^{-19} J \times 6.02 \times 10^{23} mol^{-1} = 251000 J \: mol^{-1} = 251 kJ \: mol^{-1}$
c) $E = \frac{hc}{\lambda}$
$E = \frac{(6.625 \times 10^{-34} J s \: mol^{-1} \times 3.0 \times 10^{8} ms^{-1})}{(645 nm \times 10^{-9} m \: nm^{-1})} = 3.08 \times 10^{-19} J \nonumber$
That's for one molecule. On a per mole basis, $E = 3.08 \times 10^{-19} J \times 6.02 \times 10^{23} mol^{-1} = 138000 J mol^{-1} = 138 kJ mol^{-1}$
Exercise 8.3.2:
a) $\lambda = \frac{1}{wavenumber} = \frac{1}{3105 cm^{-1}} = 3.22 \times 10^{-4} cm \times 0.01 m \: cm^{-1} = 3.22 \times 10^{-6} m$
$E = \frac{hc}{\lambda} \nonumber$
$E = \frac{(6.625 \times 10^{-34} J s \: mol^{-1} \times 3.0 \times 10^{8} m s^{-1})}{(3.22 \times 10^{-6}m)} = 6.17 \times 10^{-20} J \nonumber$
That's for one molecule. On a per mole basis, $E = 6.17 \times 10^{-20} J \times 6.02 \times 10^{23} mol^{-1} =37000 J \: mol^{-1} = 37 kJ mol^{-1}$
b) $\lambda = \frac{1}{wavenumber} = \frac{1}{1695 cm^{-1}} = 5.90 \times 10^{-4} cm \times 0.01 m cm^{-1} = 5.90 \times 10^{-6} m$
$E = \frac{hc}{\lambda} \nonumber$
$E = \frac{(6.625 \times 10^{-34} J \: s \: mol^{-1} \times 3.0 \times 10^{8} ms^{-1})} {(3.22 \times 10^{6} m)} = 3.37 \times 10^{-20} J \nonumber$
That's for one molecule. On a per mole basis, $E = 3.37 \times 10^{-20} J \times 6.02 \times 10^{23} mol^{-1} = 20000 J mol^{-1} = 20 kJ mol^{-1}$
c) $\lambda = \frac{1}{wavenumber} = \frac{1}{963 cm^{-1}} = 1.04 \times 10^{-3} cm \times 0.01 m \: cm^{-1} = 1.04 \times 10^{-5}$
$E = \frac{hc}{\lambda} \nonumber$
$E = \frac{(6.625 \times 10^{-34} J s mol^{-1} \times 3.0 \times 10^{8} m s^{-1})}{(1.04 \times 10^{-5} m )} = 1.91 \times 10^{-20} J \nonumber$
That's for one molecule. On a per mole basis, $E = 1.91 \times 10^{-20} J \times 6.02 \times 10^{23} mol^{-1} = 11500 J mol^{-1} = 11.5 kJ mol^{-1}$
Exercise 8.3.3:
a)
b) Ru(bpy)32+ would be a terrible reducing agent. The reduction potential of Ru(bpy)33+ is very high. That means Ru(bpy)32+ would not give up an electron very easily.
c) $\Delta G = -n F E^{0}$
$\Delta G = -1 \times 97485 J V^{-1}mol^{-1} \times 1.26 V = -123 kJ mol^{-1} \nonumber$
d) It absorbs in the blue part of the spectrum and appears orange.
e) It absorbs very strongly, so probably not d-d. It is probably MLCT, from the ruthenium to the π* in the bpy ligand
f) $E = \frac {hc}{\lambda}$
$E = \frac{(6.625 \times 10^{-34} Js \: mol^{-1} \times 3.0 \times 3.0 \times 10^{8} m s^{-1})}{(450 nm \times 10^{-9} m \: nm^{-1})} = 4.42 \times 10^{-19} J \nonumber$
That's for one molecule. On a per mole basis, $E = 4.42 \times 10^{-19} J \times 6.02 \times 10^{23} mol^{-1} = 266000 J mol^{-1} = 266 kJ mol^{-1}$
g) Ru2+ would be low spin d6, a singlet ground state. The excited state will also be a singlet state.
h) 615 nm is in the orange region of the spectrum. This is light given off by the complex as it relaxes, so it is the colour we see.
i) $E = \frac{hc}{\lambda}$
$E = \frac{(6.625 \times 10^{-34} Js mol^{-1} \times 3.0 \times 10^{8} m \: s^{-1})}{(615 nm \times 10^{-9} m \: nm^{-1})} = 3.23 \times 10^{-19} J \nonumber$
That's for one molecule. On a per mole basis, $E = 3.23 \times 10^{-19} J \times 6.02 \times 10^{23} mol^{-1} = 195000 J \: mol^{-1} = 195 kJ \: mol^{-1}$
j) Stokes shift $=615 nm - 450 nm = 165nm$
k) $\Delta E = E_{1} - E_{2} = 266 - 195 kJ \: mol^{-1} = 71 kJ \: mol^{-1}$
However, it would not be lost all at once, but in small increments equivalent to the differences between vibrational states.
l) If we were to reduce Ru(bpy)33+ directly into an excited state, we would arrive at a state much higher in energy than the ground state. That reduction would be harder to accomplish. In this case, the ending state would be 195 kJ mol-1 higher than Ru(bpy)32+.
$\Delta G = 195 - 123 kJ \: mol^{-1} = 72 kJ \: mol^{-1} \nonumber$
$E^{0} = \frac{- \Delta G}{nF} = -\frac{(72 kJ \: mol^{-1} \times 1000 kJ \: J^{-1})}{(1 \times 96485 J: V^{-1} mol^{-1})} = -0.75V \nonumber$
m) The oxidation of Ru(bpy)32+* would be very favorable compared to the oxidation of Ru(bpy)32+. The former is a much better reducing agent.
Exercise 8.5.1:
The longer the wavelength, the lower the energy. Photons of wavelength longer than 240 nm would not have enough energy to overcome the barrier for the dioxygen-cleaving reaction.
Exercise 8.5.2:
According to the Planck-Einstein relation:
$E = h \nu \nonumber$
or, since $V = \frac{c}{\lambda}$
$E = \frac{hc}{\lambda} \nonumber$
in which h = Planck's constant = 6.625 x 10-34 Js,
c = speed of light = 3.0 x 108 m/s,
ν = frequency,
λ = wavelength.
a) $E = \frac{hc}{\lambda}$
$E = \frac{(6.525 \times 10^{-34} Js)(3.0 \times 10^{8} m/s)}{(200 nm)(10^{-9} m/nm)} \nonumber$
$E = 9.03 \times 10^{-19} J \nonumber$
b) $E = \frac{hc}{\lambda}$
$E = \frac{(6.525 \times 10^{-34} Js)(3.0 \times 10^{8} m/s)}{(325nm)(10^{-9} m/nm)} \nonumber$
$E = 6.12 \times 10^{-19}J \nonumber$
Exercise 8.5.3:
Remember, the mole is the conversion unit from the molecular scale to the macroscopic scale.
a) $E = (8.28 \times 10^{-19} J/photon)(6.02 \times 10^{23} photons/mol)$
$E = 543770 \frac{J}{mol} \nonumber$
$E = 544 \frac{kJ}{mol} \nonumber$
b)
Exercise 8.5.4:
a) $E = (6.12 \times 10^{-19} J/photon)(6.02 \times 10^{23} photons/mol)$
$E = 368146 \frac{J}{mol} \nonumber$
$E = 368 \frac{kJ}{mol} \nonumber$
b)
Exercise 8.5.5:
The reactions must take the same pathway, and go over the same barrier, forward and back.
a) The reverse barrier: $E = 544-498 \frac{kJ}{mol}$
$E = 46 \frac{kJ}{mol} \nonumber$
b) The reverse barrier: $E = 368 - 105 \frac{kJ}{mol}$
$E = 263 \frac{kJ}{mol} \nonumber$
Exercise 8.6.1:
1. The original PDI has an absorption maximum at just under 500 nm. It absorbs blue-green, so it would be expected to appear red-orange.
2. The longest wavelength absorbed is approximately 625 nm.
3. $E = \frac{hc}{\lambda} = \frac{(6.625 \times 10^{-34} Js \: mol^{-1} \times 3.0 \times 10^{8} m s^{-1})}{625 \times 10^{-9} m} = 3.18 \times 10^{-19} J$
That's the energy of a photon. A mole of photons would have energy multiplied by Avogadro's number, NA.
$E = 3.18 \times 10^{-19} J \times NA = 3.18 \times 10^{-19} \times 6.02 \times 10^{23} mol^{-1} = 1.91 \times 10^{5} J \: mol^{-1} = 191 kJ mol^{-1} \nonumber$
d) If we approximate $E = \Delta G$, and given that $\Delta G = -NFE^{0}$
Then for this single-electron excitation, $E^{0} = \frac{- \Delta G}{nF} = \frac{-1.91 \times 10^{5} J \: mol^{-1}}{(1 \times 96485 J \: V^{-1} mol^{-1})} = -1.97 V$
e)
f) Given the structure of PDI, the transition is probably π --> π*.
g) The longest wavelength absorbed shifts toward the red, to about 700 nm, after dipping in acid. The energy of the transition is lowered to
$E = \frac{hc}{\lambda} = \frac{(6.625 \times 10^{-34} Js \: mol^{-1} \times 3.0 \times 10^{8} m \: s^{-1})}{300 \times 10^{-9} m} = 2.84 \times 10^{-19} J \nonumber$
$E = 2.84 \times 10^{-19} J \times NA = 3.18 \times 10^{-19} \times 6.02 \times 10^{23} mol^{-1} = 1.71 \times 10^{5} J \: mol^{-1} = 171 kJ \: mol^{-1} \nonumber$
h) These very flat molecules can probably stack very tightly together. The distance between the π electrons on one molecule and the π* orbital on a neighbouring molecule is very small. An intermolecular transition is possible.
i) What happens when the film is dipped in acid? It gets protonated. The anionic phosphate groups would become neutralized. The molecules would be able to stack even more closely together, lowering the energy required to excite an electron from one molecule to the other.
j) The onset of the reduction wave appears to be about -0.40 V, as shown by the peak in the cyclic voltamogram.
k)
l)
m) $E^{0}_{(_{vs}Ag+/AgCl)} = E^{0}_{(_{vs}NHE)} + 0.20 V$ (i.e. the positive reduction potential of Ag+ vs. NHE indicates an electron is
i) $E^{0}_{(_{vs} Ag+/AgCl)} = -0.10V + 0.20V = + 0.10 V$
ii) $E^{0}_{(_{vs}Ag+/AgCl)} = + 1.23 V + 0.20 V = + 1.43V$
iii) $E^{0}_{(_{vs}Ag+/AgCl)} = +0.83V + 0.20V = + 1.03V$
n) The excited state can be reduced by placing an electron in the hole vacated by the excited electron. This level is 1.97 V below the conductance band in energy, which has a reduction potential of -0.40 V (its potential is 1.97 V more positive than the conductance band).
Thus, $-0.40V + 1.97V = + 1.57V$.
o)
p)
q) The phosphates probably coordinate to the cobalt oxide, forming a bridge for electron transfer.
Exercise 8.6.2:
1. $\ce{Ag^{+}_{(aq)} + NO_{3}^{-}_{(aq)} + Bu_{4}N^{+}_{(aq)} + Br^{-}_{(aq)} -> AgBr_{(s)} + Bu_{4}N^{+}_{(aq)} + NO_{3}^{-}_{(aq)}}$
2. They absorb in the violet-blue range and would appear yellow-orange.
3. A nanometer is 10-9 m, whereas a picometer is 10-12 m. A picoparticle is 1,000 times smaller than a nanoparticle.
4. They absorb across the visible spectrum and appear black. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/V%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_3/08%3A_Photochemical_Reactions/8.08%3A_Solutions_t.txt |
Exercise 9.2.1:
1. blue and red; green
2. green; red
3. blue; orange
4. violet
Exercise 9.2.2:
Exercise 9.2.3:
Exercise 9.3.1:
a)
b)
Exercise 9.3.2:
Pheophytin b is probably more electrophilic because of the extra formyl (-HC=O) group. At first glance, it would be expected to have a more positive reduction potential.
Exercise 9.3.3:
Exercise 9.3.4:
$\Delta E = E^{o} (red) - E^{o} (ox) = 1.3 - 1.229 V = 0.07 V \nonumber$
Exercise 9.3.5:
Exercise 9.4.1:
Exercise 9.4.2:
a) Fe(II, III) active b) Mg(II) inactive c) Cu(I, II) active d) Li(I) inactive
e) Zn(II) inactive f) Sc(III) inactive g) Co(II, III) active h) Cr(II, III) active
Exercise 9.5.1:
The first pair of plastoquinols recycle two electrons between them, pumping two additional protons via a recycled quinol. The proton output is 1.5 times what it would be without the Q-loop (from 1 proton/electron + 0.5 proton/electron). Now remember that this recycled quinol also delivers two electrons, one of which is recycled. To get a second electron to generate a new recycled quinol, we would have to go through the entire, two-new-plastoquinol cycle again. We would get another two protons but it would have required eight electrons to get there, or 0.25 protons output per electron input. The proton output would rise to 1.75 times what it would be without the Q-loop (from 1 proton/electron + 0.5 proton/electron + 0.25 proton/electron). A series will result if we keep going. The series is 1 + 1/2 + 1/4 + 1/8 + 1/16 + ..... + 1/2n, and it will converge to 2. Therefore, the Q-loop doubles the number of protons pumped across the membrane.
Exercise 9.5.2:
The major factor appears to be charge stabilisation. Suppose both iron atoms start as Fe(III). They have an overall 6+ charge. The two sulfides are each 2-, for an overall Fe2S2 core charge of 2+. If ligated by four cysteines (Cys-S-), then the overall charge of the coordination compound is 2-. Addition of an electron will result in an increase in negative charge, to 3-. Now, that was starting from the fully oxidised state. If instead we start in a mixed Fe(II/III) oxidation state, then addition of an electron results in a 4- charge on the coordination complex. That charge buildup will be energetically difficult.
If ligated by two cysteines (Cys-S-) and two neutral histidines, then the overall charge of the coordination compound is 0. Addition of an electron will result in a negative charge, 1-. If instead we start in a mixed Fe(II/III) oxidation state, then addition of an electron results in a 2- charge on the coordination complex. That lower charge buildup required in this case will be energetically less difficult.
Exercise 9.5.3:
Copper is further to the right in the periodic table and is consequently more electronegative than iron. That is the major reason that copper is below iron in the activity series or metals. On that basis alone (and ignoring any differences in the coordination environment), we might expect a more positive reduction potential on a copper ion than an iron ion. Copper would therefore be a better electron acceptor, and would be useful in creating a greater driving force for electron transfer later in the electron transport chain.
Exercise 9.5.4:
1. With only one anionic donor, the charge of the complex that contains Cu(II) would be 1+.
2. With one anionic donor, the charge of the complex that contains Cu(I) would be 0.
3. Reduction of plastocyanin results in a decrease of charge, whereas reduction of the FeS cluster results in an increase of charge. That factor alone could make plastocyanin easier to reduce, with a more positive reduction potential.
Exercise 9.6.1:
Exercise 9.6.2:
Exercise 9.6.3:
Remember, NADP+, like NAD+, is solely a 2-electron oxidising agent. It can only receive a pair of electrons associated with a hydride. Ferredoxin, via an FeS cluster, can only give one electron at a time. An adapter is needed between these two carriers. FAD can accept either 1 or 2 electrons, proceeding to either FADH. radical (stabilised via extensive resonance delocalisation) or to FADH2.
Exercise 9.6.4:
1. Releasing antennae molecules from the light harvesting complex in photosystem II would decrease both the number of photons captured and the number of electrons sent into the electron transport chain. Conversely, in a "chromophore pool" model, releasing these molecules from photosystem II would make them more available for uptake by photosystem I, resulting in greater absorption of photons there, and greater capacity to accept incoming electrons. If regulated, these two systems should be able to level out at an identical rate of photon absorption.
2. Kinases phosphorylate the hydroxy side chains in serine residues (and sometimes threonines and tyrosines). That covalent modification results in a change in the charge of the side chain from neutral to anionic. Significant conformational changes in the protein can be expected to result. It is likely that this conformational change results in looser binding of the antennae molecules in the light harvesting complex of photosystem II.
Furthermore, kinase appears to promote tighter binding of antennae molecules at photosystem I, presumable through a conformational change there. The result is a shift of these molecules from the available pool, such that additional chromophores become available in photosystem I and fewer are retained in photosystem II, until photon absorption at the two sites is equal.
c) In fact, phosphatase, which is the complementary enzyme to kinase, is activated when photosystem I has high activity compared to photosystem I. Phosphatases remove phosphate groups from serines and related residues. The result would be tighter binding of antennae molecules at photosystem I and looser binding of antennae molecules at photosystem II, so that light absorption could be evened out between the two sites.
The use of phosphatases and kinases to effect opposite objectives in regulation is common in a number of biochemical systems.
Exercise 9.8.1:
Exercise 9.8.2:
Exercise 9.8.3:
Exercise 9.8.4:
Exercise 9.8.5:
a)
b)
Exercise 9.8.6:
a)
b)
Exercise 9.8.7:
The regular enol pathway is driven by pi-donation from an oxygen atom. The enamine pathway is driven by pi-donation from a nitrogen. Nitrogen is less electronegative and a better pi-donor than oxygen, so the enamine pathway is faster.
Exercise 9.8.8:
Exercise 9.8.9:
Exercise 9.8.10:
Reaction that are in equilibrium are often under substrate control. Because the enzyme can catalyse the reaction in both directions, the direction of the reaction driven by the enzyme is controlled by the relative amount of reactants on either side of the reaction. For example, if the xylulose concentration greatly increased, then the amount of xylulose bound by the enzyme would also increase, so the enzyme would shift xylulose into ribulose and thereby maintain equilibrium. If the ribulose concentration greatly increased, then more ribulose would be bound by the enzyme, and the ribulose would be converted into xylulose.
Exercise 9.9.1:
a)
Exercise 9.9.2: | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/V%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_3/09%3A_Photosynthesis/9.01%3A_Solutions_to_Selecte.txt |
In biology, energy is needed in order to drive all sorts of biochemical processes. Energy is needed to stay alive. There are plenty of energy sources on earth. Tectonic forces release massive amounts of heat and drive the conversion of some minerals into gaseous products; for example, metal sulfides such as zincblende can be converted to gaseous hydrogen sulfide, H2S. Sometimes, all that heat and gas finds its way to the earth's surface in the form of volcanoes. In the oceans, some marine organisms get their energy from gases released from volcanic vents, such as methane and hydrogen sulfide. However, sunlight is an even more abundant source of energy over most of the earth's surface. In photosynthesis, light energy is absorbed and used to make ATP. Remember, ATP is like a portable battery pack in biology; it can travel to different parts of a cell where it can be used to power uphill steps in biochemical reactions.
Plants, algae, and some bacteria are capable of carrying out photosynthesis. They could get the immediate benefit of portable ATP molecules to drive biochemical reactions. However, the production of ATP in photosynthesis is also connected to carbon capture. Carbon dioxide from the air is incorporated into carbohydrate molecules. This conversion happens in a series of reactions called "dark reactions", because they keep happening even without sunlight. The carbohydrates can be stored, long-term, and later they can be used as energy sources via glycolysis and the citric acid cycle.
Animals, of course, benefit from this process indirectly because they can also use carbohydrates as a source of energy. The breakdown of carbohydrates releases energy through the usual trade-off: slightly weaker C-H and C-C bonds are broken and slightly stronger O-H and C-O bonds are made, meaning there is an overall release of energy. By eating plants, we can immediately access these carbohydrates without all the fuss of standing in the sun all day making them ourselves.
If you remember some basic plant biology, you may be familar with another aspect of photosynthesis. The "balanced reaction" for photosynthesis also involves the conversion of water to molecular oxygen, as follows:
$\ce{6CO_{2 \: (g)} + 6H2O_{(l)} -> C6H12O6_{(s)} + 6O2_{(g)}} \nonumber$
Oxygen is a key player in oxidative phosphorylation, in which glycolysis and the TCA cycle are made even more efficient by boosting the amount of ATP produced for every glucose molecule broken down. Most organisms (including us) just can't survive without that extra ATP; we depend on plants for our survival in more ways than one.
But in contrast to what is suggested by the balanced reaction, the production of oxygen by plants is actually carried out separately from carbohydrate synthesis. Production of oxygen is actually part of the "light reaction", along with ATP synthesis.
In photosynthesis, water is oxidised to produce molecular oxygen. Plants take the electrons that they have stripped from the water molecules and divert them into an electron transport chain. Energy harnessed by that electron transport chain is used to convert ADP to ATP. In oxidative phosphorylation, organisms (including plants) take electrons from NADH and succinate and divert them into an electron transport chain, eventually depositing them onto an oxygen molecule to make water. Energy harnessed by that electron transport chain is used to convert ADP to ATP.
That means we have two opposite processes that both are harnessed to produce ATP. In one process, the electrons run downhill energetically and are deposited on dioxygen to make water. That's oxidative phosphorylation. Photosynthesis is really oxidative phosphorylation running in reverse: the electrons start on water and proceed through an electron chain from there. But if oxidative phosphorylation runs downhill, then photosynthesis must run uphill.
That's where the light comes in. The light absorbed in photosynthesis is used to lift the electrons uphill in energy; from there, they can start rolling downhill through the electron transport chain, releasing energy along the way that can be harnessed for ATP formation.
All of these events take place in a special organelle in the plant called the chloroplast. Chloroplasts are a little bit like mitochondria, where the important metabolic processes such the TCA cycle and oxidative phosphorylation take place. Like mitochondria, chloroplasts contain their own DNA and ribosomes for protein production and they are passed on directly from mother cell to daughter cell. Chloroplasts have a double membrane and are filled with an aqueous medium called the stroma.
Within the chloroplast there are structures called thylakoids. A thylakoid is like a complex water balloon; it has a membrane and is filled with an aqueous medium called the lumen. Unlike a simple water balloon, though, the thylakoid has portions that are deeply folded, so that they look like discs stacked in layers. These portions of the thylakoid are called the grana. The regular, non-folded portions are called the lamellae.
The thylakoid plays a very important role in phtosynthesis. A group of protein complexes bound to the thylakoid membrane carry out the absorption of light energy, the conversion of water to dioxygen, and the production of ATP, as well as an electron carrier, NADPH. The ATP and NADPH are released into the surrounding stroma. A soluble protein in the stroma, called ribulose bisphosphate carboxylase (RuBisCo) captures carbon dioxide and covalently attaches it to a carbohydrate molecule. Other proteins then use the ATP and NADPH to reduce the carboxylate group (from the CO2) into a regular part of the carbohydrate chain. In this way, rather than trying to knit six carbon dioxide molecules together into a glucose, the problem is simplified into just taking up one carbon dioxide at a time, adding it into a pre-existing sugar.
The ATP is produced by an ATP synthase, which is very similar to the complex used for the same purpose during oxidative phosphorylation. Just like the ATP synthase in the mitochondria, this one is driven by a proton gradient. The proton gradient is created through an electron transport pathway, just like the one in the mitochondria. In fact, many of the features of photosynthesis are pretty similar to oxidative phosphorylation. A crucial difference is that the electron transport chain in oxidative phosphorylation starts with NADH and ends with water, whereas in photosynthesis it is the reverse: the chain starts with water and ends with NADPH. The electron transport chain in oxidative phosphorylation is exothermic, running downhill in energy. The electron transport chain in photosynthesis would be endothermic, but can be sustained by the input of energy in the form of light.
See the section on photosynthesis at Henry Jakubowski's Biochemistry Online. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/V%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_3/09%3A_Photosynthesis/9.02%3A_Introduction_to_Phot.txt |
It all starts with Photosystem II. Photosystem I comes later; the two complexes are named that way because Photosystem I was discovered before Photosystem II.
The picture below is a "cartoon rendering" of an X-ray crystal structure of photosystem II. So, researchers first took a solution of protein and somehow coaxed the protein molecules out of solution in the form of a crystal. In a crystal, the molecules are oriented in an organised array, so they are all lined up in a regular way. The researchers took the crystal and placed it in a beam of X-rays, then measured how the X-rays scattered when they hit the crystal. That scattering pattern was used to construct a three-dimensional map of the atomic positions in the protein. Because that is an awful lot of atoms, biochemists often present the map in different ways that are less complicated.
The cartoon rendering doesn't show where the individual atoms are, but it shows the locations of any alpha helices (pink) and beta sheets (yellow) as well as the chain formed by the rest of the protein that isn't organized in either of those ways (white). The parallel helices are typical of a membrane-bound protein; the top part of the structure can easily fit in among the parallel tails of the lipid bilayer that forms the thylakoid membrane.
The structure below is shown from the same point of view, but with the protein removed. You can see that the protein is holding lots of smaller molecules in place; it is packed full of them. That's in sharp contrast to the complexes of the oxidative phosphorylation pathway, where there were at most a half-dozen metal clusters, which were directly attached to the protein, and a couple of ubiquinones, which were free.
We're going to take just a moment and focus on something interesting in the lower left part of that picture. The whole thing is interesting, of course, but this is something extraordinarily important to us as humans. Halfway between the big, tangled mass of molecules at the top and the single molecule at the bottom (a heme, in case you were wondering) there is a small blob of atoms. Some of the atoms are red (oxygen, in this case) and some are purple (manganese); there is also a grey one (calcium) and even a green one (probably a stray chloride ion that photobombed the structure). This is the oxygen-evolving complex. It's responsible for producing all of the molecular oxygen, O2, on earth. Your life depends on it.
We'll get back to the oxygen-evolving complex in a later unit. Right now let's figure out what all those other molecules are doing in that scrum at the top of the picture, and why they are important. This is the light harvesting complex. Its job is to collect energy from sunlight and convert that energy into forms that the plant can use. Of course, somewhere along the line you learned in biology class that chlorophyll is what makes photosynthesis possible, and there are several chlorophyll molecules in the light harvesting complex. There are many different members of this family of molecules in nature, found in different organisms that can survive on sunlight: plants, algae, some plankton, a few bacteria. The chlorophylls commonly found in plants are chlorophyll a and chlorophyll b, shown below.
A chlorophyll is an example of a porphyrin, like the one that forms the heme unit in hemoglobin. It has a flat, polycyclic structure composed of four nitrogen-containing pyrole rings. Those nitrogen atoms can bind metal ions such as iron (in hemoglobin, myoglobin, and various cytochromes) or, in this case, magnesium.
Why are there two different chlorophylls, though? There are even more if you count the variations that are common in organisms other than plants. It takes a great deal of energy for a cell to manufacture different molecules; wouldn't it be more efficient to just make one kind of chlorophyll?
The advantage of chlorophyll diversity can be seen in UV-Vis spectra of the different compounds. Chlorophyll a has a couple of absorption maxima around 430 and 660 nm; there are also some smaller absorption features near 410 and 600 nm.
A look at the UV-Vis spectrum of chlorophyll b shows some strong similarities. There are still two major absorption maxima, near those for chlorophyll a, but they have shifted to about 450 and 640 nm.
It probably isn't surprising that the two molecules have roughly similar UV-Vis spectra, since they have similar structures. Minor differences in the structures are responsible for subtle variations in energy levels of the donor and acceptor orbitals involved in the electronic transition, resulting in slightly different absorption maxima.
The combination of the two different chlorophylls would lead to a UV-Vis spectrum that looks a little like this one:
The effect is to harvest light from a slightly broader range of the visible spectrum. That makes for a more efficient process. Because sunlight contains wavelengths that range all across the visible spectrum (and beyond), the more of these wavelengths that can be captured, the more efficient photosynthesis can be.
For this reason, the light harvesting complex contains more than one kind of chlorophyll, but it also contains other molecules. The most important class of these non-chlorophyll molecules includes the carotenoids. Compounds such as lycopene, β-carotene, and zeaxanthin are examples of carotenoids.
These compounds help to further broaden absorption by the light harvesting complex. We can see how they might start to fill in the gap in the previous spectrum that was based only on chlorophyll content.
Overall, the result is a collection of molecules that can capture photons from across much of the visible spectrum, although not all. That's important, because solar radiation reaching the surface of the earth covers a broad range of wavelengths. Capturing as much light as possible increases the efficiency of the process. The fact that plants do reflect some visible light results in their green colour.
What next? All of that captured light energy needs to be converted into a different form; that's the whole point of photosynthesis. We aren't quite ready for that, though. The light harvesting complex just collects the light; the nearby reaction center is where the conversion process begins. You can think of the light harvesting complex as an antenna or as a funnel that sends energy to the reaction center, where the process of converting the energy into ATP (and, later, carbohydrates) will begin.
How does the energy get to the reaction center? It happens through a process of radiationless energy transfer. Once it absorbs the right amount of energy, the reaction centre will initiate an electron transport chain, much like the one that operated in oxidative phosphorylation. A photon doesn't need to strike the reaction centre itself, however, because it can strike any of the surrounding chromophores in the light harvesting complex. These molecules are all in close contact with each other, so it is relatively easy for molecules to collide and to transfer that entire amount of energy from one molecule to the next. During that transfer, the first molecule relaxes back to the ground state and the next molecule is excited into the excited state.
This process is very similar to photosensitization. In photosensitization, a molecule containing a chromophore, which can absorb a photon, transfers its energy to a second molecule, which has no chromophore and can't absorb light by itself. In the light harvesting complex, all of the molecules contain chromophores and could potentially absorb a photon. The point is that it doesn't matter which of them absorbs the photon; any of these absorptions will result in a cascade of radiationless transfers until the energy arrives at the right place. Once the energy reaches the reaction center, an electron transport chain will begin, and the energy released along this chain will be captured for the manufacture of ATP.
Conversely, it has been suggested that this array of molecules also affords protection to the cell. Under conditions in which the light intensity is too great, the reaction centre may become excited without an available outlet, as downstream participants in the pathway may not be ready to receive electrons. If that happens, a retro-mechanism may allow the reaction centre to relax once again, with energy dissipated to the surroundings through internal conversion.
Exercise \(1\)
In general, if a compound has one main absorption in the visible spectrum, we see the complementary colour. If a compound has two main absorptions in different parts of the spectrum, we generally see the colour of the "trough" or dip in the spectrum between these absorption maxima.
1. What colour(s) correspond to the UV-Vis absorption maxima of chlorophyll a? What colour do we see when we look at chlorophyll a?
2. What colour(s) correspond to the UV-Vis absorption maxima of lycopene? What colour do we see when we look at lycopene?
3. What colour(s) correspond to the UV-Vis absorption maxima of β-carotene? What colour do we see when we look at β-carotene?
4. Zeaxanthin is an example of a xanthophyll; xanthophylls get their name from the Greek roots "phyllon" (leaf) and "xanthos" (yellow). What colour probably corresponds to the UV-Vis absorption maximum of zeaxanthin?
Answer a
a) blue and red; green
Answer b
b) green; red
Answer c
c) blue; orange
Answer d
d) violet
Exercise \(2\)
The carotenoids are biosynthetically related to each other. Show how, in principle, lycopene could be converted to β-carotene via simple acid/base catalysis.
Answer
Exercise \(3\)
The carotenoids are examples of an enormous class of natural products called terpenes. Terpenes can be thought of as oligomers formed from 5-carbon isoprene units. Show the location of the five-carbon units in lycopene and β-carotene.
Answer
Exercise \(4\)
Some marine algae contain phycoerythrin, a protein that is covalently linked to a unique chromophore. The chromophore structure and an artist's rendering of the UV spectrum are shown below.
1. What colours are absorbed by the chromophore?
2. What colour does the chromophore appear?
3. What is the evolutionary advantage that drives the algae to make this chromophore?
1. X-ray crystal structures: Ferreira, K.N.; Iverson, T.M.; Maghlaoui, K.; Barber, J.; Iwata, S. Architecture of the photosynthetic oxygen-evolving center. Science 2004, 303, 1831-1838. Images obtained via RCSB Protein Data Bank (1S5L). | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/V%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_3/09%3A_Photosynthesis/9.03%3A_Photosystem_II-_Harv.txt |
The goal of photosynthesis is to capture light energy from the sun and convert it into forms that are useful to the plant. The process begins in Photosystem II, where the light harvesting complex absorbs photons and relays that energy to the reaction centre, which can refer to a specific protein within photosystem II or, more specifically, to a pair of chlorophylls within that protein. What happens there?
The first goal of photosynthesis is the production of ATP. As in oxidative phosphorylation, that task is accomplished by releasing energy through an electron transport chain. In general, electrons need to be transferred from a position of high energy (or low potential) to low energy (or high potential). Photoexcitation helps that process because it leads to the formation of a low-energy hole as well as a high-energy electron.
Once an electron has been excited, it finds itself at a much higher energy level. It can easily slide downhill to a lower energy acceptor orbital. Note that the energy level of the acceptor orbital could be anywhere below the higher electronic level in the excited state. It could even be above the original electronic level in the ground state. That means that an electron transfer that would have been uphill if it occurred from the ground state would now be downhill from the excited state.
Furthermore, any donor that is above the original electronic level in energy could drop an electron into the new hole. Without photoexcitation, electron donation would be much more endothermic.
The reaction centre is sometimes referred to as P680, for pigment 680, so called because it has a UV-Vis absorbance maximum at 680 nm. It could absorb visible light directly, but more likely it is excited by transfer of energy from the surrounding antennae molecules in the light-harvesting complex. The reaction centre is actually composed of two pairs of chlorophylls. One of these pairs sits very near together; they are parallel to each other and overlapping so closely that they are practically touching. This special pair is called the chlorophyll dimer, and each individual chlorphyll within the pair is sometimes given the abbreviation PD1 or PD2.
These two special chlorophylls form an excitonic dimer. That means that the two molecules behave as if they were only one molecule during a photochemical event. When a photon is absorbed, or the equivalent amount of energy transferred from another molecule, the excitonic dimer enters into an excited state in which an electron has been passed from one chlorophyll to the other.
$P680 \rightarrow P680* \nonumber$
or
$P_{D1}P_{D2} \rightarrow P_{D1}^{+}P_{D2}^{-}* \nonumber$
Why would that happen, from a biological perspective? Maybe the electronic separation simply makes one half a better reducing agent (it is negative) and the other half a better oxidizing agent (it is positive). Alternatively, that separation of the electron and the hole between two different molecules could lead to a more efficient photoredox process by making relaxation a little more difficult. Certainly the excited electron could recombine with the hole via the same pathway by which it was formed, with transfer of energy back to a surrounding molecule. Other relaxation pathways are less likely, however. It would be unlikely for the P680-* half to relax via a simple cascade through vibrational states since the electron would find no hole to drop into on that molecule.
Instead, the excited electron enters into an electron transport pathway. The electron is first transferred to a nearby chlorophyll a (ChlD1). From there, an immediate, rapid transfer to a pheophytin molecule follows. A pheophytin is really just a chlorophyll without a magnesium ion in the middle.
Exercise $1$
Reduction of pheophytin results in a resonance-delocalised anion. Show one resonance structure of the radical anion:
1. with an oxygen-centred anion.
2. with a nitrogen-centred anion.
Answer a
Exercise $2$
Based on the structures of the analogous chlorophyll a and chlorophyll b, which would be expected to have a more positive reduction potential: pheophytin a or pheophytin b?
Answer
Pheophytin b is probably more electrophilic because of the extra formyl (-HC=O) group. At first glance, it would be expected to have a more positive reduction potential.
The subsequent destination in the electron transport chain is a plastoquinone. Like the related ubiquinones found in oxidative phosphorylation, plastoquinones are mobile, two-electron carriers. Mobile electron carriers are needed in order to transport electrons from one complex to another. The plastoquinones are also quite lipophilic, so their range of motion is restricted to the thylakoid membrane. That restriction boosts the efficiency of photosynthesis by making it unlikely for the transported electrons to be lost elsewhere in the cell.
Exercise $3$
Show the product of the two-electron, two-proton reduction of plastoquinone (plastoquinol).
Answer
There are actually two plastoquinones in the electron transport chain, however, and only the second one is mobile. The first one is covalently bound to the protein. The two plastoquinones have different reduction potentials, probably because of their environments, and so the first plastoquinone not only acts as a stepping stone to the second, but also allows for a more gradual loss of energy as the electrons rolls downhill. At this point, the second plastoquinol leaves photosystem II behind and travels to complex b6f, which will play an important role in ATP production via proton pumping.
We are not quite finished with Photosystem II, though. What happened to that hole that was left behind in the excitonic dimer? PD2+* is pretty significant; it has been described as nature's most powerful oxidising agent. It needs to be: its job is ultimately to oxidise water to dioxygen. Remember, the opposite reaction, reduction of dioxygen to water, served as the final destination for electrons during oxidative phosphorylation. That exothermic half-reaction served as part of the driving force for the electron transport chain, the associated proton pumping, and ATP formation. To drive that reaction backward by stripping electrons from water will require a very strong oxidising agent.
In fact, the reduction potential of PD2+* has been estimated at +1.3 V (vs SHE); that is quite positive. For comparison, the reduction potential of dioxygen under acidic conditions is +1.229 V (vs SHE).
Exercise $4$
What is the reaction potential for the oxidation of water by PD2+*?
Answer
ΔE = Eo (red) - Eo (ox) = 1.3 - 1.229 V = 0.07 V
In a sense, the reaction centre does not stand at the start of the electron transport chain at all. It is partway along; the real start of the electron transport chain is water. Electrons from water drop into the hole on PD2+*, forming a complete P680. The P680 absorbs a photon, sending the electron all the way up into PD1-*, and down they fall from there along the rest of the pathway. The reaction centre is like the engine on a ski lift or roller coaster, pulling in electrons and then sending them past to their next destination.
There are actually a couple of intermediaries between PD2+* and the water, however. The nearest one is a tyrosine residue. It provides the electron that immediately replaces that which has been sent out of the reaction centre.
The next intermediary is the oxygen-evolving complex. The oxygen evolving complex is a manganese oxo cluster that strips electrons from water while, at the same time, combining them to make dioxygen.
We can summarise the events of photosystem II in a couple of ways. One way is to try to picture, roughly, how the different players involved so far are arranged in the protein complex. We can imagine how a photon is absorbed, and how that energy is passed along to the reaction centre. We can imagine a pathway for the electron through this system, too.
Alternatively, it is useful to display these components not in physical space, but in energy or potential space. By looking at the reduction potentials of the species, we can start to imagine how the electron transport chain works, and we can see more clearly the role played by light absorption. The absorbed photon lifts the electron up from a low energy level (corresponding roughly to the reduction potential of PD2+*) to a much higher one (corresponding to the reduction potential of PD1-*). Furthermore, we can understand another reason why the electron does not simply relax back to the ground state after the reaction centre becomes excited: the electron is quickly carried downhill through another pathway, involving transfer to other molecules.
Exercise $5$
The activity within photosystem II can be thought of as a catalytic cycle. Draw out the changes in electronic populations of the species below as a result of the sequence of events that starts with photon absorption.
1. X-ray crystal structures: Deisenhofer, J., Epp, O., Sinning, I., Michel, H. Crystallographic refinement at 2.3 A resolution and refined model of the photosynthetic reaction centre from Rhodopseudomonas viridis. J. Mol. Biol. 1995, 246, 429-457. Images obtained via RCSB Protein Data Bank (1PRC). | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/V%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_3/09%3A_Photosynthesis/9.04%3A_Photosystem_II-_Elec.txt |
You may recall that the oxygen-evolving complex is found deep in Photosystem II. It's found near that yellow-coded bundle of beta-sheets in the lower left of the picture below; that part of the complex is called the manganese-stabilizing protein. Remember, Photosystem II is embedded in the thylakoid membrane. The membrane surrounds the pink portion of alpha-helices in the top half of the picture. The stroma, the aqueous medium that surrounds the thylakoid, is at the very top of the picture. In the lower part of the picture, the protein is dipping its toes into the lumen; that's the watery medium inside the thylakoid.
When we strip away all of the proteins, we are left with a number of ligands that were embedded inside. That includes the light-harvesting complex that we have already looked at, as well as the small clutch of atoms called the oxygen-evolving complex (OEC). The OEC is found a little further down in the picture, at the edge of the lumen. You can see its manganese atoms, coloured purple in this picture, as well as red oxygens and grey calcium, mid-way down and a little to the left.
Let's zoom in for a closer look. A couple of stray atoms are visible in the picture, but focus on the atoms that are connected together. The "spokes" mean that the X-ray crystallography software has decided these atoms are close enough together that they are bonded to each other. The purple manganese atoms might not really be bonded to each other in terms of sharing a pair of electrons between the atoms, but they are magnetically coupled, sharing electrons in a different sense. We can see that the atoms almost form a cube, although it has one corner torn open. It's not unlike an FeS cluster, only with manganese and oxygen instead of iron and sulfur.
The task of this little cube is to carry out a reaction that is indispensible for our existence on earth. It needs to convert water into oxygen. From an evolutionary point of view, there is a distinct advantage to the use of water as a source of electrons. There is an almost endless supply: we have oceans full of it on this planet. The trouble is, removing an electron from water results in an oxygen radical; in biochemistry, this is called a reactive oxygen species. Reactive oxygen species can be chaotic, evil characters inside a cell, reacting indiscriminately with any other structures that they contact. For that reason, it would be safer to oxidise the water all the way to dioxygen.
Exercise \(1\)
Use the Lewis structures to confirm the number of electrons involved in the oxidation of water.
Answer
That four-electron oxidation of water is a pretty challenging task. The opposite event, the four-electron reduction of dioxygen to water, served as the thermodynamic sink for the electron transport chain in oxidative phosphorylation. Nevertheless, it was also pretty challenging, because it required the rapid introduction of four electrons into the system in order to fully reduce the dioxygen to water, without the formation of free reactive oxygen species that might wreak havoc in the cell.
In that case, the problem was solved through a large number of iron and copper atoms within Complex IV, serving as a reservoir of electrons for rapid reduction. Furthermore, the actual site of reduction was a close-set iron and copper pair, allowing the dioxygen to bridge between two metals to aid in that rapid reduction.
The oxygen-evolving complex uses a similar strategy, but with a reverse bias. The cluster of manganes ions provides a reservoir of holes rather than electrons, allowing for the rapid oxidation of water to dioxygen. The role of the calcium ion is not clear, since calcium is not redox-active; it is always Ca2+. Remember, metals in the first few columns of the periodic table are always found in their maximum oxidations states, corresponding to a noble gas configuration. However, it may play another role in binding water or stabilizing the cluster.
Exercise \(2\)
It can be useful to know which metal ions are typically redox-active and which are not (i.e. those metals that are always found in the same oxidation state). For the following metals, indicate whether the metal is redox-active and provide its most common oxidation state(s).
a) Fe b) Mg c) Cu d) Li
e) Zn f) Sc g) Co h) Cr
Answer a
a) Fe(II, III) active
Answer b
b) Mg(II) inactive
Answer c
c) Cu(I, II) active
Answer d
d) Li(I) inactive
Answer e
e) Zn(II) inactive
Answer f
f) Sc(III) inactive
Answer g
g) Co(II, III) active
Answer h
h) Cr(II, III) active
Remember that there is an intermediary between P680+* and the oxygen-evolving complex. There is a tyrosine residue, called Tyr161 because of its position in the primary structure of the protein, situated between the oxygen-evolving complex and P680+*. It conducts the electron from manganese to the chlorophyll in the reaction centre. An electron is first transferred from Tyr161 to P680+*. An electron from manganese then replaces the missing electron on Tyr161.
It takes the removal of four electrons to oxidise water to oxygen, so all of that has to happen four times. Four photons must be absorbed, exciting P680 four times. Each time, an electron from Tyr161 replaces the electron that was excited from the ground state in P680, leaving behind a hole. The electron in Tyr161 is replaced by an electron from manganese. These events are illustrated in an overall mechanism called the Kok cycle (pronounced "coke" cycle).
The Kok cycle illustrates the storage of holes on the manganese atoms. The holes are just the places where electrons used to be. We can think of them as a buildup of positive charge. The cycle also shows one of the steps in water oxidation that is evident in the equation for the reaction: the loss of four protons. The loss of protons is probably linked directly to the oxidation of the manganese ions. Any time an atom or molecule becomes more positively charged, it becomes more highly acidic.
From the point of view of the water molecule, just binding to a manganese makes it more likely to lose a proton, because it is donating electrons to another atom and thereby taking on a formal positive charge on oxygen. So, a water molecule becomes much more acidic when it binds to a metal atom. That's a common event in biochemistry. The buildup of additional positive charge on manganese makes this deprotonation even more likely.
In terms of how the O-O bond is formed, different pathways are conceivable, but the simplest possibility involves activation of the water as a nucleophile. Deprotonation results in a hydroxide ion. Although the hydroxide is bound to the metal atom, the polarity of the bond allows the hydroxide to behave as a nucleophile.
At this point, you have also seen water become activated as a nucleophilic hydroxide ion in other biochemical pathways, so this development should not be a big surprise. What about an electrophile? Well, one way to provide an electrophile is to deprotonate the hydroxide to make a metal oxo. That deprotonation could be linked to another oxidation at manganese.
Metal oxo compounds can develop electrophilic character at oxygen because of π-donation from oxygen to a metal ion in a high oxidation state. The donation of a lone pair to the metal would give oxygen a formal positive charge. The oxygen would be activated as an electrophile.
At that point, it is easy to imagine how one oxygen (a hydroxy ligand) could act as a nucleophile while a second (an oxo ligand) acts as an electrophile. An O-O bond would form, accompanied by a two-electron reduction at manganese.
Formation of a second O=O bond would result in a second two-electron reduction at manganese. In two easy steps, the entire four-electron reduction of manganese -- and four-electron oxidation of water -- has been accomplished. Because there are four manganese ions within the cluster, it is a simple matter to redistribute electrons so that the manganese ions have similar oxidation states.
Of course, because we are dealing with a manganese oxo cluster rather than an individual manganese ion, electrons can be reshuffled at any time. Oxidation states in clusters are complicated, because of the ease of transferring or even delocalising electrons within the system.
Again, there are other possible mechanisms, and you know enough chemistry at this point that you should be able to propose some alternatives.
Exercise \(3\)
1. Propose an alternative mechanism for oxygen evolution using classic organometallic mechanistic steps (oxidative addition, reductive elimination, insertion, beta-elimination).
2. What side products might be detected if this mechanism operated?
Exercise \(4\)
1. Propose an alternative mechanism for oxygen evolution using radical mechanistic steps.
2. What side products might be detected if this mechanism operated?
One last point about the oxygen-evolving complex. The goal of photosynthesis is to make ATP; that ATP will be used to convert CO2 to carbohydrates. The way to make ATP in this case is to use an electron transport chain to release enough energy to generate a proton gradient. The water is the starting point for that electron transport chain. But wait, there is a bonus: just by oxidising the water, we have released protons into the lumen. That's a head start on setting up a proton gradient.
1. X-ray crystal structures: Ferreira, K.N.; Iverson, T.M.; Maghlaoui, K.; Barber, J.; Iwata, S. Architecture of the photosynthetic oxygen-evolving center. Science 2004, 303, 1831-1838. Images obtained via RCSB Protein Data Bank (1S5L). | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/V%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_3/09%3A_Photosynthesis/9.05%3A_Photosystem_II_-_The.txt |
The goal of photosynthesis is to produce ATP, which is in turn used to produce carbohydrates through carbon capture. ATP is generated via a proton gradient, which in turn is maintained through an electron transport chain. The cytochrome b6f complex is closely analogous to complex III in oxidative phosphorylation. It plays a major role in generating a protein gradient during photosynthesis.
An X-ray crystal structure of cytochrome b6f is shown below. Once again, it is a membrane-bound protein. The stroma lies at the top of the picture and the lumen at the bottom.
The ligand-only view of the structure is provided below. We can see a dimeric structure here, with the left half mirroring the right half. Near the bottom of each dimer, there is a heme. Just above that, we can see an FeS cluster; it is actually a 2Fe2S cluster, and specifically a Rieske cluster. That is just like the FeS cluster in complex III. Rieske clusters have very positive reduction potentials because of their unique coordination environment.
Above the Rieske clusters, things get a little more cluttered. There are a couple of more hemes on each side with their red iron atoms, as well as a chlorophyll with its magnesium shown in yellow. In addition, there are lots of terpenoid structures, such as carotenoids. Not all of these ligands have roles that are clearly understood, and we will focus on the bare essentials.
Cytochrome b6f is shown in a simplified form in a diagram below. Onle one of the two units in the dimer is shown. Each unit is itself a relatively small complex, containing only four proteins: cyctchrome b6, cytochrome f, a Rieske protein, and "subunit IV", whose role is not well understood.
Photosystem II eventually produces a mobile plastoquinol which travels through the thylakoid membrane to cytochrome b6f, binding near the lumen side of the membrane. The plastoquinone releases its two electrons and its two protons and can return to photosystem II for more. Alternatively, the plastoquinone might just travel across to the stroma side of the mebrane, where it can pick up two new protons from the stroma as well as two more electrons.
Where do the two electrons come from? They are recycled in the "Q-loop". When it arrives at cyctochrome b6f, instead of sending both electrons along the electron transport chain, the plastoquinone only sends one electron in that direction. That first electron travels to the Rieske FeS cluster, then to a cytochrome f (that's a cytochrome b found in plants), and finally to a soluble, mobile electron carrier, plastocyanin.
The other electron travels in the other direction, back toward the stroma, via a pair of hemes. After two plastoquinols have delivered their electrons to cytochrome b6f, a waiting plastoquinone can be reduced again to plastoquinol, picking up two more protons from the stroma. That plastoquinol can then travel back across the membrane to deliver its two protons to the lumen. One of its electrons gets sent down the electron transport chain, and the other electron gets recycled (again).
The advantage of the Q-loop is explained in the following diagram. In the diagram, inputs to cytochrome b6f are shown in red, outputs are shown in blue, and recycled elements are in green. If one plastoquinol simply delivered its electrons and protons and was done, there would be two protons delivered per plastoquinol. That's one proton output per electron that was input.
A second plastoquinol would do exactly the same thing. There would be four protons output for four electrons initially input. That's still one proton output per electron that was input.
If, instead, one electron is recycled each time, then every second plastoquinol leads to the delivery of an extra pair of protons. That's because in picking up the recycled electrons, a plastoquinone has had to travel back to the stroma side of the membrane and pick up two more protons. Overall, that means six protons are delivered for four electron input, or 1.5 protons output per electron input. Since the proton gradient is what is generating the ATP, then by increasing the number of protons pumped per electron coming in, efficiency is increased.
Exercise \(1\)
The efficiency advantage is even greater than that described above. The Q-loop is thought to double the amount of protons pumped by cytochrome b6f. Explain why.
Answer
The first pair of plastoquinols recycle two electrons between them, pumping two additional protons via a recycled quinol. The proton output is 1.5 times what it would be without the Q-loop (from 1 proton/electron + 0.5 proton/electron). Now remember that this recycled quinol also delivers two electrons, one of which is recycled. To get a second electron to generate a new recycled quinol, we would have to go through the entire, two-new-plastoquinol cycle again. We would get another two protons but it would have required eight electrons to get there, or 0.25 protons output per electron input. The proton output would rise to 1.75 times what it would be without the Q-loop (from 1 proton/electron + 0.5 proton/electron + 0.25 proton/electron). A series will result if we keep going. The series is 1 + 1/2 + 1/4 + 1/8 + 1/16 + ..... + 1/2n, and it will converge to 2. Therefore, the Q-loop doubles the number of protons pumped across the membrane.
The electron transport chain through cytochrome b6f is rather short. The electron is first passed to a "high-potential FeS cluster", or Rieske cluster. It differs from most FeS clusters in that two of the amino acid residues that bind it in the protein are histidines, rather than the usual cysteines. Rieske clusters usually have higher reduction potentials than other FeS clusters. They were one of the last in a series of FeS clusters encountered in the electron transport chain in oxidative phosphorylation, and their more positive reduction potential was needed to keep the electron transport chain moving in the right direction.
Exercise \(2\)
The reduction potential of a Rieske FeS cluster is generally more positive than that of a regular FeS cluster, in which the Fe2S2 core is held in place by four cysteines. Explain why.
Answer
The major factor appears to be charge stabilisation. Suppose both iron atoms start as Fe(III). They have an overall 6+ charge. The two sulfides are each 2-, for an overall Fe2S2 core charge of 2+. If ligated by four cysteines (Cys-S-), then the overall charge of the coordination compound is 2-. Addition of an electron will result in an increase in negative charge, to 3-. Now, that was starting from the fully oxidised state. If instead we start in a mixed Fe(II/III) oxidation state, then addition of an electron results in a 4- charge on the coordination complex. That charge buildup will be energetically difficult.
If ligated by two cysteines (Cys-S-) and two neutral histidines, then the overall charge of the coordination compound is 0. Addition of an electron will result in a negative charge, 1-. If instead we start in a mixed Fe(II/III) oxidation state, then addition of an electron results in a 2- charge on the coordination complex. That lower charge buildup required in this case will be energetically less difficult.
The final stop for the electron is a small, soluble protein, plastocyanin. The electron carrier in plastocyanin is not an iron, but rather a copper atom. The copper ion, which can toggle between Cu+ and Cu2+, is held in place by two histidines, a methionine, and a cysteine. The plastocyanin transports the electron through the lumen to the next complex, photosystem I.
Exercise \(3\)
Use periodic trends to suggest a reason why a copper atom appears later in the electron transport chain, whereas several iron atoms appeared earlier in the chain.
Answer
Copper is further to the right in the periodic table and is consequently more electronegative than iron. That is the major reason that copper is below iron in the activity series or metals. On that basis alone (and ignoring any differences in the coordination environment), we might expect a more positive reduction potential on a copper ion than an iron ion. Copper would therefore be a better electron acceptor, and would be useful in creating a greater driving force for electron transfer later in the electron transport chain.
Exercise \(4\)
1. What would be the overall charge on the coordination complex in plastocyanin, assuming the Cu(II) state?
2. What would be the overall charge on the coordination complex in plastocyanin, assuming the Cu(I) state?
3. Based on charge considerations alone, how would you expect the reduction potential of plastocyanin to compare to the Rieske FeS cluster?
Answer a
a) With only one anionic donor, the charge of the complex that contains Cu(II) would be 1+.
Answer b
b) With one anionic donor, the charge of the complex that contains Cu(I) would be 0.
Answer c
c) Reduction of plastocyanin results in a decrease of charge, whereas reduction of the FeS cluster results in an increase of charge. That factor alone could make plastocyanin easier to reduce, with a more positive reduction potential.
1. X-ray crystal structures: Kurisu, G.; Zhang, H.; Smith, J.L.; Cramer, W.A. Structure of the cytochrome b6f complex of oxygenic photosynthesis: tuning the cavity. Science 2003, 302, 1009-1014. Images obtained via RCSB Protein Data Bank (1VF5). | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/V%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_3/09%3A_Photosynthesis/9.06%3A_Adding_a_Proton_Pump.txt |
Having already generated the proton gradient used for ATP generation, what is there left to do?
Remember back to that high school biology class when you first saw the equation for photosynthesis.
$\ce{6 CO2_{(g)} + 6H2O_{(l)} -> C6H12O6_{(s)} + 6O2_{(g)}} \nonumber$
We can break that into two half reactions. The first is the oxidation of water.
$\ce{2H2O_{(l)} -> O2_{(g)} + 4H^{+}_{(aq)} + 4e^{-}_{(aq)}} \nonumber$
The other half is the reduction of carbon dioxide.
$\ce{6CO2_{(g)} + 24H^{+}_{(aq)} -> 24e^{-}_{(aq)} -> C6H12O6_{(s)} + 6H2O_{(l)}} \nonumber$
Although these two half-reactions are physically separated from each other, the electrons must still get from one to the other in order to complete the full reaction. We have already accomplished the first reaction and now have electrons in hand to carry out the second. We are going to need a biological reducing agent, like NADH, to carry out the second. In this case, we will use a closely-related compound, NADPH.
In both of the drawings above, the purple portion of the compound is where the electrons are accepted. The black portion is sugar, the blue is a purine base, and the red is the phosphate part.
Exercise $1$
Because NADH and NADPH are electron carriers, it is important to recognise both reduced and oxidised forms. Given the reduced forms of NADH and NADPH shown above, draw the oxidised forms, and show where the :H- is incorporated in the molecule in the reduced form.
Answer
The problem is, NADPH is a relatively high-energy electron carrier -- remember, NADH was at the uphill end of the electron transport chain in oxidative phosphorylation -- and our electrons have just ridden their sled to the bottom of the hill. If we are going to make NADPH, we will need to tug the electrons back uphill again. That's essentially the job of photosystem I, shown below.
The ligand-only view of the crystal structure reveals a picture that should be somewhat familar now: a host of antennae molecules forming a light-harvesting complex. Once again, the antennae molecules include mostly chlorophylls, as well as carotenoids such as carotene and lycopene, plus a few xanthins like violaxanthin; the exact molecules vary depending on whether we are looking at plants, algae, or bacteria.
Also visible in the ligand-only view is a group of FeS clusters. These clusters play a role in the electron-transport chain of photosystem I, conducting the electron toward the stroma, where they will be used to reduce NADP+ to NADPH. That biological reducing agent, closely related to NADH, will be used to help reduce carbon dioxide.
Let's pause for a moment. Most of the X-ray crystal structures we have seen have come from bacteria. Of course, we are much more familiar with the idea of photosynthesis in plants. Are the proteins and processes involved in photosynthesis in one anything like the other? For comparison, here is photosystem I from a spinach plant.
It is roughly similar, although not identical, to the one from the bacterium. There is a broad swath of alpha helices through the middle, topped with a bundle of beta sheets. If anything, the protein from the spinach looks superficially a little simpler than the one from the bacterium. The ligand-only view, below, is strikingly similar, showing a spray of chromophores and a series of iron-sulfur clusters.
Let's pick up the story where we left off when we were discussing cytochrome b6f. A single electron was being transported by the soluble, mobile electron carrier, plastocyanin. That single electron is delivered directly to the reaction centre of photosystem I. That reaction centre is quite similar to the one in photosystem II, although it absorbs photons of a slightly longer wavelength, 700 nm. The P700 centre contains an excitonic dimer, a closely-spaced pair of chlorophyll molecules that absorb the photon and undergo electron / hole separation on the two different molecules.
Just as in P680, the P700* excited state can accept an electron onto the chlorphyll partner that has the hole, PD2+*. That's where the plastocyanin delivers its electron.
Meanwhile, the part of the dimer with the extra electron, PD1-*, passes the electron on to another station on the electron transport chain. Part of the goal here is to get the electrons back to the stroma where the NADP+ is waiting. If you think about it, the electrons started out in the lumen, on a water molecule. They traveled across the thylakoid membrane in photosystem II until they joined a plastoquinone and picked up a couple of protons; those protons came from the stroma. The plastoquinones travelled back across the membrane, a stealth migration of polar H+ in the form on nonpolar PQH2, and delivered their protons to the lumen. The side-kick electrons skipped along the edge of the lumen in a plastocyanin carrier, and are now traveling back toward the stroma to that waiting NADP+. The NADPH will be produced in the stroma because that's where it will be needed for CO2 capture.
Again, just as in P680, the first electron acceptor after P700 is a single chlorophyll. This chlorophyll anion that forms is slightly downhill in energy from the one in the dimer, PD1-*. From there, the electron passes to a phylloquinone, which is just another variation on the familar quinone family.
The phylloquinone sends the electron into that series of three FeS clusters visible in the crystal structure. From there, the electron leaves photosystem I and is deposited on a ferredoxin, a small protein that binds another iron sulfur cluster. The ferredoxin is located in the stroma, where it can bind with an enzyme called ferredoxin NADP+ reductase. Like most enzymes, the name says exactly what it does: it catalyses the reaction in which ferredoxin reduces NADP+, resulting in NADPH.
The crystal structure of a ferredoxin is shown below. This particular one is not from photosystem I (it is found in a nitrogen-fixing bacterium), but it gives you the idea. The protein is simple enough that you can see that it is a dimer; the top half is the same as the bottom half, flipped over. You can also clearly see a 2Fe2S cluster at the edge of each subunit. The location of that FeS cluster near the edge of the protein is probably important in facilitating electron transfer with another protein.
The ferredoxin is an electron carrier protein. Its job is to deliver electrons to the ferredoxin NADP+ reductase in the stroma, which will then use the electrons to make NADPH. NADPH is a biological source of hydride ion (formally, H-, or a proton plus two electrons) that in this context will be used to help reduce carbon dioxide into carbohydrate. Below, we can see an example. This is another dimeric protein with rotational symmetry: the left side is the same as the right, but turned over.
If we look inside, we can see the NADP+ molecules waiting at either end of the structure, at both the far left and the far right. In addition, there is a pair of molecules that look vaguely related to NADP+ toward the middle of the picture. Those are two FAD molecules. The FAD molecules are there to conduct electrons from the ferredoxin t the NADP+.
In the picture below, the purple part of the molecule is the redox-active region. It is able to accept electrons. The black part is sugar, the blue is a purine base, and the red is phosphate.
Exercise $2$
Show the reduced form of FAD. Remember, the reduced form is called FADH2.
Answer
Exercise $3$
What role do you think the FAD plays in ferredoxin NADP+ reductase? Why doesn't the ferredoxin reduce the NADP+ directly?
Answer
Remember, NADP+, like NAD+, is solely a 2-electron oxidising agent. It can only receive a pair of electrons associated with a hydride. Ferredoxin, via an FeS cluster, can only give one electron at a time. An adapter is needed between these two carriers. FAD can accept either 1 or 2 electrons, proceeding to either FADH. radical (stabilised via extensive resonance delocalisation) or to FADH2.
Exercise $4$
Because photosystem II and photosystem I are part of the same electron transport chain, it is important that the reaction centre in one is getting excited just as often as the reaction centre in the other. Regulation is partly accomplished via covalent modification. For example, insufficient activity at photosystem I results in activation of a kinase, which results in release of some of the antennae molecules from the light harvesting complex in photosystem II.
1. How would this situation help to remedy the lack of parity in activity between photosystem I and photosystem II?
2. Explain how the kinase would achieve this result.
3. Suppose the opposite situation occurred: insufficient activity at photosystem II. What enzyme would probably be activated to remedy the situation?
Answer a
a) Releasing antennae molecules from the light harvesting complex in photosystem II would decrease both the number of photons captured and the number of electrons sent into the electron transport chain. Conversely, in a "chromophore pool" model, releasing these molecules from photosystem II would make them more available for uptake by photosystem I, resulting in greater absorption of photons there, and greater capacity to accept incoming electrons. If regulated, these two systems should be able to level out at an identical rate of photon absorption.
Answer b
b) Kinases phosphorylate the hydroxy side chains in serine residues (and sometimes threonines and tyrosines). That covalent modification results in a change in the charge of the side chain from neutral to anionic. Significant conformational changes in the protein can be expected to result. It is likely that this conformational change results in looser binding of the antennae molecules in the light harvesting complex of photosystem II.
Furthermore, kinase appears to promote tighter binding of antennae molecules at photosystem I, presumable through a conformational change there. The result is a shift of these molecules from the available pool, such that additional chromophores become available in photosystem I and fewer are retained in photosystem II, until photon absorption at the two sites is equal.
Answer c
c) In fact, phosphatase, which is the complementary enzyme to kinase, is activated when photosystem I has high activity compared to photosystem I. Phosphatases remove phosphate groups from serines and related residues. The result would be tighter binding of antennae molecules at photosystem I and looser binding of antennae molecules at photosystem II, so that light absorption could be evened out between the two sites.
The use of phosphatases and kinases to effect opposite objectives in regulation is common in a number of biochemical systems.
We can summarize with a sketch of photosystem I:
1. X-ray crystal structures: Jordan, P.; Fromme, P.; Witt, H.T.; Klukas, O.; Saenger, W.; Krauss, N. Three-dimensional structure of cyanobacterial photosystem I at 2.5 A resolution. Nature 2001, 411, 909-917. Images obtained via RCSB Protein Data Bank (1JB0).
2. X-ray crystal structures: Amunts, A., Drory, O., Nelson, N. The structure of a plant photosystem I supercomplex at 3.4 A resolution. Nature 2007, 447, 58-63. Images obtained via RCSB Protein Data Bank (2o01).
3. X-ray crystal structures: Schlesier, J., Rohde, M., Gerhardt, S., Einsle, O. A Conformational Switch Triggers Nitrogenase Protection from Oxygen Damage by Shethna Protein II (FeSII). J. Am. Chem. Soc. 2016, 138, 239-247. Images obtained via RCSB Protein Data Bank (5ffi).
4. X-ray crystal structures: Komori, H., Seo, D., Sakurai, T., Higuchi, Y. Crystal structure analysis of Bacillus subtilis ferredoxin-NADP(+) oxidoreductase and the structural basis for its substrate selectivity. Protein Sci. 2010, 19, 2279-2290. Images obtained via RCSB Protein Data Bank (3Lzw).
9.08: ATP Synthase
One of the goals of photosynthesis is to make ATP. ATP is frequently employed in biology to drive uphill reactions. The hydrolysis of ATP to give ADP is a downhill reaction, so it releases energy.
Formation of ATP from ADP requires addition of phosphate. It's an uphill reaction, so the cell has to expend energy to produce ATP. ATP synthase harnesses a proton gradient to drive a molecular rotor, much like a millwheel, in order to bring ADP and phosphate in close proximity so that they can form ATP. You may already be familiar with ATP synthase from oxidative phosphorylation, where it works exactly the same way.
ATP synthase is thought of as having three states. In the "loose" state, ADP and phosphate (Pi) can enter the binding site easily. Once these two substrates are in the binding site, the enzyme enters the "tight" state, in which ADP and phosphate are constricted and form ATP. At that point, the enzyme goes through an "open" phase in which ATP can leave easily.
As in oxidative phosphorylation, the changes in states are coupled to rotation of one part of the multiprotein complex (the rotor) with respect to another (the stator). The principle is similar to that of an electric motor. If the stator is positively charged and certain positions of the rotor are negatively charged, then the opposite charges will attract and the rotor will begin to turn. Once the negatively charged part of the rotor reaches the positively charged part of the stator, however, everything should come to a stop. The reason it doesn't is because a proton drops into the the anionic position of the rotor just as it is approaching the stator. Now neutralized, this position is no longer attracted to the stator, and it just sails past. Meanwhile, another anionic position is approaching the stator, keeping the rotor turning. As soon as it is a safe distance away from the stator, a hole opens up below the rotor and the protons falls away, regenerating an anionic position that is once again attracted to the next stator.
In the case of photosynthesis, the proton is "dropping in" from the lumen and later "falling down" into the stroma; its motion in that direction is driven by the proton gradient. So that proton gradient in turn drives the motion of the rotor. The motion of the rotor, in turn, drives conformational changes within (and between) the protein subunits of ATP synthase. Those conformational changes promote ATP formation through an approximation effect. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/V%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_3/09%3A_Photosynthesis/9.07%3A_Harvesting_the_Photo.txt |
One of the main goals of photosynthesis is to take the ATP and NADPH produced through the electron transport chain and use them to store energy in the form of carbohydrates. The key event in carbohydrate storage is the capture of carbon dioxide from the atmosphere. That task is carried out by ribulose-1,5-bisphosphate carboxylase (usually abbreviated as rubisco). Rubisco is found inside the chloroplasts, in the stroma.
Rubisco is an aggregate of sixteen proteins, but it contains only two different kinds of proteins; it is composed of eight molecules of one protein plus eight molecules of the other. It is a huge protein, with a molecular weight of 50 or 60 kDa. Its primary function is to use a five-carbon sugar, ribulose, as a scaffold on which to attach a carbon dioxide molecule. The result, after reduction with NADH, is a six-carbon sugar.
In the picture below, you can see the ribulose-1,5-bisphosphate molecules awaiting their carbon dioxide. There are actually eight of them, but they are arranged in four stacks when viewed from this angle, with four molecules in the front and four behind them.
The capture of the carbon dioxide is accomplished through a simple aldol reaction. Under biological conditions, this reaction happens via the neutral, enol form of ribulose-1,5-bisphosphate, rather than the anionic enolate ion, because there is not a strong enough base to produce appreciable quantities of the enolate ion. To get to the enol form, a proton must be transferred from an alpha position to the oxygen of the ketone.
Exercise \(1\)
Provide a mechanism, with curved arrows, for the capture of the carbon dioxide by the enol of ribulose bisphosphate.
Answer
That intermediate, resulting from the aldol reaction of ribulose bisphosphate and carbon dioxide, does not last very long. It is quickly hydrated to form a geminal diol (twin OH groups on one carbon), and the geminal diol decomposes via a retro-aldol reaction. Remember, a retro-aldol reaction is an aldol reaction in reverse. Instead of an enolate nucleophile adding to a carbonyl electrophile, an alcoholic oxygen pi-donates to form a new carbonyl, and as a result an enolate is produced as a leaving group.
Exercise \(2\)
Provide a mechanism, with curved arrows, for the cleavage of the geminal diol into two 3-phosphoglycerate molecules.
Answer
At this point, the ATP and NADPH produced in the electron transport chain come into the picture. The NADPH converts the carboxylate group into an aldehyde; at that point, the CO2 has become fully incorporated into a normal sugar, albeit the smallest one there is. The ATP is first used in order to activate the 3-phosphoglycerate. The NADPH can't displace the oxide (O2-) from 3-phosphoglycerate; oxide is a terrible leaving group, and the carboxylate group isn't particularly electrophilic with that negative charge hanging around. The 3-phosphoglycerate would be a much better nucleophile. Well, instead of reacting directly with the NADPH, it reacts with the ATP forst. Phosphoglycerate kinase activates the ATP, which is then ready to behave as an electrophile. The negatively charged 3-phosphoglycerate donates to the terminal phosphate group on the ATP, displacing ADP. The kinase has done its job, forming a phosphoanhydride.
Let's think about phosphoanhydrides for a minute. They are a little like acid anhydrides in structure. An acid anhydride has an oxygen atom between two carbonyls. Normally, the excellent pi-donating ability of an oxygen atom "protects" an attached carbonyl from approaching nucleophiles; in a sense, the electrophilic site on the carbonyl is already filled by the lone pair from the oxygen. In an anhydride, however, that lone pair is in a dither. It is divided between two carbonyls and so it can only donate half as well to either one of them. Since it isn't doing much good as a pi-donor, the other half of oxygen's behaviour emerges: it is also an excellent sigma-acceptor, because it is a very electronegative atom. That leaves the acid anhydride as a very good electrophile.
Of course, phosphoanhydrides behave in exactly the same way, but the ostensibly pi-donating oxygen is stuck between one carbonyl and one phosphoryl group. It is unable to pi-donate effectively to the carbonyl, and so its electron-withdrawing ability takes over instead. A phosphoanhydride is an excellent electrophile, just like a regular anhydride. Both are "high on the ski hill", if you remember the analogy for carboxyloid reactivity.
Incidentally, polyphosphates tend to be pretty electrophilic, too, which is one of the factors driving the high reactivity of ATP.
The result is that the rather nucleophilic 3-phosphoglycerate has been transformed in one step into the highly electrophilic 1,3-bisphosphoglycerate, an example of a phosphoanhydride. That makes it much easier for the NADPH to donate a nucleophilic hydride to the carbonyl, forming an aldehyde. The exact mechanism, catalysed by glyceraldehyde phosphate dehydrogenase, involves a group transfer step. The carbonyl is transferred from the phosphate group to a cysteine residue in the enzyme. The advantage of group transfer is chiefly entropic: instead of the NADPH and the phosphoanhydride both binding in the enzyme and reacting together, at this point the NADPH must simply bind with the enzyme and react with the enzyme itself, because the glycerate group has become covalently bound to the enzyme.
Moreover, the new functional group formed during group transfer is a thioester. Thioesters, like phosphoanhydrides, are excellent electrophiles. Although the second-row pi-donors, oxygen and nitrogen, are exceptionally good at donating to carbonyls, the third-row pi-donors, such as chlorine and sulfur, are not. Instead, their relatively high electronegativity activates the carbonyl. That means group transfer occurs without any loss of reactivity, which is essential for catalysis.
Exercise \(3\)
Provide a mechanism for the NADPH reduction of 1,3-bisphosphoglycerate.
Answer
The glyceraldehyde formed via reduction by NADPH is the simplest of the carbohydrates; you can think of it as the grandmother of all sugars, which can be thought of as forming through extension of glyceraldehyde's carbon chain. In this case, we really have the phosphorylated form of the sugar, rather than the sugar itself. Glyceraldehyde, or glyceraldehyde-3-phosphate, can be transformed easily into another molecule via tautomerism. That is the movement of one proton from one position to another, and in this example it leads to dihydroxyacetonephosphate.
Exercise \(4\)
Show the enol intermediate between glyceraldehyde-3-phosphate and dihydroxyacetone phosphate.
Answer
Glyceraldehyde is, of course, an aldose, with its carbonyl at the end of the three-carbon chain, forming an aldehyde functional group. The other carbons have hydroxyl groups; that's what makes it a carbohydrate (a polyhydroxylated aldehyde or ketone). Dihydroxyacetone is a ketose, with its carbonyl in the middle of the chain, forming a ketone functional group. That group is flanked by two alcohols. The fact that glyceraldehyde can turn into dihydroxyacetone is partly significant because the biosynthetic potential of this small building block is opened up by allowing it to form twice as many structures.
There are a number of ensuing reactions that have the effect of quickly interconverting these sugars into different carbohydrates in a sort of carbon pool. The plant can draw on this carbon pool for various purposes, whether it is to send sugars elsewhere in the plant or to regenerate some more ribulose in order to grab more carbon dioxide. By grabbing more carbon dioxide, that carbon pool gets bigger and bigger. The series of reactions governing this carbon pool is called the Calvin cycle. It is really more of a network than a cycle, with molecules able to hop back and forth between various parts of the cycle, rather than everything following monotonously in lockstep.
For example, the aldol reaction of those two triose phosphate molecules -- dihydroxyacetone phosphate and glyceraldehyde-3-phosphate -- leads to fructose-1,6-bisphosphate. This reaction is accomplished through the help of an enzyme, aldolase. Fructose, or fruit sugar, is one of the few sugars that are readily absorbed through our digestive system.
Exercise \(5\)
Provide a mechanism for the formation of fructose-1,6-bisphosphate, above
1. under basic conditions
2. under acidic conditions
Answer a
a)
Answer b
b)
Exercise \(6\)
Many biological aldol reactions actually occur via an enamine intermediate.
1. Provide a mechanism for the formation of an enamine from dihydroxyacetone phosphate.
2. Provide a mechanism for formation of fructose-1,6-bisphosphate via the enamine of dihydroxyacetone phosphate.
Answer a
a)
Answer b
b)
Exercise \(7\)
Why is enamine catalysis used in biochemical pathways? What makes it faster than a regular enol pathway?
Answer
The regular enol pathway is driven by pi-donation from an oxygen atom. The enamine pathway is driven by pi-donation from a nitrogen. Nitrogen is less electronegative and a better pi-donor than oxygen, so the enamine pathway is faster.
The next step in the Calvin cycle is just a dephosphorylation, carried out with the help of a phosphatase. In this case, the enzyme is called fructose-1,6-bisphosphatase. Phosphorylations and dephosphorylations of substrates, like phosphorylation and dephosphorylation of enzymes, are often key regulatory steps, required to promote a reaction involving an enzyme.
On to the next step. This is when things get a little bit funky. Two carbohydrate molecules undergo a sort of metathesis together. In a metathesis reaction, compounds are split up and recombined in a new way. In this reaction, fructose-6-phosphate combines with another glyceraldehyde-3-phosphate. After the reaction, the two old sugars have been changed into two new ones: erythrose-4-phosphate and xylulose-5-phosphate. It appears as though the fructose has been split and two of its carbons have connected with the glyceraldehyde chain.
How does that happen? The main part of the fructose-6-phosphate molecule turns into erythrose-4-phosphate. You can imagine that much happening if the number three carbon's hydroxy group performs a pi donation to form the new aldehyde group. That's the topmost OH group, in black, in fructose-6-phosphate. That explains the erythrose-4-phosphate formation. The trouble is, that pi donation would require that something leaves carbon three. The leaving group would be an anionic carbonyl. You probably haven't seen such a leaving group before, for good reason.
Let's not abandon this idea, just yet. If that leaving group did form (and it doesn't), that would explain the formation of xylulose-5-phosphate, too. The carbonyl-based anion, if it were to act as a nucleophile, would form a new carbon-carbon bond, opening up the carbonyl of glyceraldehyde-3-phosphate to form the hydroxy group on what is now carbon three of xylulose-5-phosphate (uppermost on the left, in black).
There is more to this reaction than just one step. The problem is, we need to form a better leaving group. That will take some extra effort. The addition of the ylide of thiamine pyrophosphate (TPP) will help. By adding temporarily to the carbonyl of fructose-6-phosphate, we can get around the problem of having an anion form on the carbonyl carbon.
This same strategy comes up during the citric acid cycle (or TCA cycle), where you may have seen the TPP ylide before. In that cycle, the decarboxylation of pyruvate is impossible because there is no alpha carbon on which to stabilise the anion that forms via decarboxylation. The negative charge on the carbonyl carbon may seem like a good idea, because it is near the electronegative oxygen. Unlike an alpha position, though, the lone pair cannot fully delocalise onto the oxygen. That's because the lone pair associated with the anion -- the lone pair that forms from the broken bond -- must be orthogonal to the pi bond.
TPP comes to the rescue. Addition of TPP ylide to the ketone installs an iminium ion, turning the former carbonyl carbon into an alpha position (it is alpha to an imine instead of a carbonyl, but it works the same way). The required anion is now fully stabilised by resonance.
In the transketolase reaction, the anion also needs to be stabilised at the carbonyl position. The TPP ylide can add to the carbonyl of the sugar, installing an imine next to the former carbonyl carbon. After the reaction is done, the TPP can be displaced again to form a new carbonyl on xylulose-5-phosphate. All of this process is catalysed by an enzyme, transketolase.
Exercise \(8\)
Provide a mechanism for the TPP-mediated conversion of fructose-6-phosphate and 3-glyceraldehyde phosphate into erythrose-4-phosphate and xylylose-4-phosphate.
Answer
That isn't the only reaction catalysed by transketolase. Sedulose-7-phosphate also reacts with glyceraldehyde-3-phosphate, in this case forming ribulose-5-phosphate and xylulose-5-phosphate. Note that first product: ribose. Ribose is the ketose form of the aldose sugar, ribulose. Ribulose was the sugar we started with; its bisphosphate form was taken up by rubisco to trap CO2. We are very close to completing a cycle, and are almost ready to capture another carbon dioxide molecule, adding more stock to the carbon pool.
But where does that sedulose come from, anyway? That's the product of another aldol reaction, this time between dihydroxyacetone phosphate and erythrose-4-phosphate. Erythrose-4-phosphate was the other product of our first transketolase reaction.
We are almost finished the complete Calvin cycle. First, ribose-5-phosphate just needs to be converted to ribulose-5-phosphate. That entails a keto-enol mechanism, in which an alpha proton is transferred to the carbonyl oxygen. The reaction is accelerated in the presence of the enzyme, ribose phosphate isomerase.
Exercise \(9\)
Show the enol intermediate between ribose-5-phosphate and ribulose-5-phosphate.
Answer
Ribulose is produced in another reaction within the Calvin cycle, too. Ribulose-5-phosphate is just the epimer of xylulose-5-phosphate. In other words, it is the stereoisomer formed by removing a proton from the chiral alpha position of xylulose-5-phosphate and putting it back on the wrong side. The configuration at that position is changed from (S) to (R), producing the diastereomeric ribulose-5-phosphate instead. This reversible reaction is undertaken by the enzyme, ribulose phosphate epimerase.
Xylulose-5-phosphate was one of the products of both transketolase-catalysed reactions; one of those reactions also produced ribose-5-phosphate. That means the same reaction provides two precursors to ribulose-5-phosphate (xylulose-5-phosphate and ribose-5-phosphate) and another reaction, catalysed by the same enzyme, produces a third (another molecule of xylulose-5-phosphate).
Exercise \(10\)
The name, ribulose phosphate epimerase, suggests that the same enzyme actually carries out the reverse reaction: the conversion of ribulose-5-phosphate to xylulose-5-phosphate. How might such a reversible reaction be regulated?
Answer
Reaction that are in equilibrium are often under substrate control. Because the enzyme can catalyse the reaction in both directions, the direction of the reaction driven by the enzyme is controlled by the relative amount of reactants on either side of the reaction. For example, if the xylulose concentration greatly increased, then the amount of xylulose bound by the enzyme would also increase, so the enzyme would shift xylulose into ribulose and thereby maintain equilibrium. If the ribulose concentration greatly increased, then more ribulose would be bound by the enzyme, and the ribulose would be converted into xylulose.
The ribulose-5-phosphate will be ready to bind rubisco and take up a carbon dioxide molecule as soon as it is phosphorylated by ribulose phosphate kinase. This reaction is largely irreversible as written because of the expenditure of ATP.
X-ray crystal structures: Taylor, T.C.; Andersson, I. The structure of the complex between rubisco and its natural substrate ribulose 1,5-bisphosphate. J. Mol. Biol. 1997 265: 432-444. Images obtained via RCSB Protein Data Bank (1RCX).
9.10: Additional Problems
Exercise \(1\)
Provide arrows to illustrate the electron transport chain in the following scheme. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/V%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_3/09%3A_Photosynthesis/9.09%3A_Carbon_Capture_and_C.txt |
Pericyclic reactions differ from the ones we have looked at so far because they are not easily understood in Lewis acid- Lewis base terms. There is not always a clear nucleophile and electrophile in these reactions. In fact, they may appear to involve completely non-polar reactants.
The classic example of a pericyclic reaction is a Cope rearrangement. In a Cope rearrangement, a molecule reorganizes itself, trading in some sigma bonds for pi bonds, so that by the end of the reaction the atoms are connected in a slightly different way. At first, it doesn't seem like much has happened in this reaction, but if you look closely, you can see how specific bonds have changed places.
Highlighting specific bonds with colour is helpful here. Keeping track of electrons with arrows usually helps, too.
Cycloadditions are very closely related to pericyclic reactions. The most common example of a cycloaddition reaction is a Diels Alder reaction. A Diels Alder reaction takes place between two alkenes. Below, you can see a Diels Alder reaction between ethene and 1,3-butadiene.
Once again, keeping track of electrons can help us to establish what is going on here.
Normally, we think of both of these compounds as nucleophiles. It isn't easy to see why one would react with the other. It isn't easy to see how electrons would be attracted from one molecule to the other.
Instead, pericyclic reactions rely on weak attractions between (or within) molecules that can lead to electronic interactions that result in new bond formation. Normally, pericyclic reactions are studied using molecular orbital calculations to map out these electronic interactions. That analysis makes use of group theory, the mathematics of symmetry. However, they are also explained qualitatively using simplified molecular orbital pictures; that's the approach we'll take here.
Exercise \(1\)
Pericylic reactions depend on molecular orbital interactions. Draw the π-bonding molecular orbitals for butadiene, ranked in energy. In other words, draw a Huckel molecular orbital diagram for butadiene.
Answer
Exercise \(2\)
Draw a Huckel molecular orbital diagram for hexatriene.
Answer
10.02: Co
A rearrangement is a reaction in which one molecule undergoes bonding changes, with the transfer of one atom or group from one position in the molecule to another.
Proton tautomerism is a kind of rearrangement. A proton is removed from one site in the molecule and put back in a different site nearby. Tautomerism generally requires a couple of proton transfer steps in a row. A proton is removed from one site and then a proton is placed on the new site. (In another variation, a proton is added at one site and then a proton is removed from the old site.)
However, rearrangements often involve the concerted transfer of a group from one site to another within the molecule. The group loses its bond to one site and gains its bond to the other site at the same time.
A Cope rearrangement happens that way.
At first it may not seem like much has happened in this reaction. The two pi bonds have changed position, however, and so has one of the sigma bonds. That means a total of six electrons have moved (two electrons per bond).
It does not matter which directions you draw the arrows moving in figure PR2.2. They could be shown going clockwise or counterclockwise. There is no electrophile or nucleophile. However, the arrows help with "electron book-keeping". The number of electrons is significant, however.
• Six electrons move in a circle of six atoms.
That patterns is reminiscent of benzene. This reaction may be related to the unusual stability of benzene. The transition state for this reaction is considered to be somewhat like benzene. Halfway between one structure and the other, the electrons are delocalized around the ring of atoms.
A Cope rearrangement can be considered to occur via a rearrangement of overlap between a group of orbitals around this ring. Two orbitals forming a sigma bond tilt away from each other while two orbitals that are pi bonding tilt toward each other.
Now there are p orbitals parallel to each other on the left, able to form new pi bonds.
Many concerted rearrangements can be thought of in terms of these orbital reorganizations.
Exercise \(1\)
A Claisen rearrangement is very similar to a Cope rearrangement, but oxygen is involved.
1. Draw curved arrows to keep track of electrons in this Claisen rearrangement.
2. Draw the aromatic transition state of the Claisen rearrangement.
3. Draw the orbital reorganization in the Claisen rearrangement.
Exercise \(2\)
Provide products for the following Cope rearrangements.
Exercise \(3\)
Provide products for the following Claisen rearrangements. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/V%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_3/10%3A_Reactions_Under_Orbital_Control/10.01%3A_In.txt |
As noted previously, the Diels Alder reaction is a classic example of a cycloaddition reaction. A cycloaddition reaction has a lot in common with a pericyclic reaction like a Cope rearrangement.
Unlike the Cope and Claisen rearrangements, this reaction often occurs intermolecularly (between two molecules). It requires an alkene on one molecule and a conjugated diene on the other molecule. The alkene is referred to as a "dienophile"; it reacts with the conjugated pair of double bonds. In the drawing below, the ethene and the 1,3-butadiene are labelled as diene and dienophile, respectively.
Exercise \(1\)
Draw curved arrows to keep track of electrons in the Diels Alder reaction.
Answer
Exercise \(2\)
Draw the aromatic transition state of the Diels Alder reaction.
Answer
Exercise \(3\)
Provide products for the Diels Alder reactions below.
Although Diels Alder reactions frequently occur between two molecules, they can happen intramolecularly as well. In these cases, there must be an alkene in one part of the molecule that is able to reach over and interact with a conjugated diene on another part. In some ways, intramolecular Diels Alder reactions can be easier than their intermolecular counterparts for reasons of internal entropy. Rather than bringing two molecules together and combining them into one, the intramolecular Diels Alder starts with one molecule to begin with.
Exercise \(4\)
Many pericyclic reactions are reversible. The reversible process is usually named the same way as the forward reaction, but with the prefix "retro". For example, a Diels Alder reaction and the corresponding retro-Diels Alder reaction are shown together below.
Typically, Diels Alder reactions occur at low or moderate temperatures (between 25 °C and 100 °C is a common range). In comparison, retro Diels Alder reactions require more elevated temperatures, often above 200 °C. The forward reaction is favored at low temperature, whereas the retro reaction is favored at high temperature.
Exercise \(5\)
Draw curved arrows for the retro-Diels Alder reaction to form ethene and 1,3-butadiene.
Exercise \(6\)
Explain why higher temperatures promote the retro-Diels Alder reaction using the expression for free energy of a reaction, \(\Delta G = \Delta H - T \Delta S\)
Answer
The retro-Diels Alder reaction is entropically favoured, because two molecules are made from one molecule. As a result, energy becomes partitioned into additional states because the degrees of freedom are increasing. Internal entropy increases during this reaction. Mathematically, the entropy term is negative in the expression for free energy, so as entropy increases the free energy becomes lower. Because that term is actually a product of entropy and temperature, an increase in temperature has the effect of amplifying the influence of entropy on the free energy of the reaction. Hence, this reaction is favoured by internal entropy factors, which come to dominate at elevated temperature.
Exercise \(7\)
Provide products for the following retro-Diels Alder reactions.
Exercise \(8\)
Sometimes, there are subsequent steps that occur after a cycloaddition reaction. For example, the following Diels Alder reaction involving naphthol produces a tautomerised product. Show the mechanism for the reaction sequence.
The reaction can be thought of in terms of a reorganization of electrons between these two molecules. In the Diels Alder reaction, we can think of an interaction between the LUMO on one molecule and the HOMO on the other. As it happens, the LUMO on one molecule has the correct symmetry such that it can overlap and form a bonding interaction with the HOMO on the other molecule.
Pay attention to the p orbital drawings on the carbons that will bond to each other to form the six-membered ring. It is important that those orbitals are able to overlap with each other to form an in-phase interaction. In that way, these carbon atoms at the ends of the diene and dienophile are able to bond with each other.
A Diels Alder reaction is sometimes called a [2+4] addition reaction. A 2-carbon unit on one molecule interacts with a 4-carbon unit on another molecule.
In contrast, the addition of one regular alkene to another regular alkene would be called a [2+2] addition reaction. If this reaction occurred, two alkenes would come together to form a four-membered ring.
However, [2+2] addition reactions don't occur without special circumstances. We will look at the requirements for that reaction later. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/V%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_3/10%3A_Reactions_Under_Orbital_Control/10.03%3A_Th.txt |
The Diels Alder reaction is the most common cycloaddition reaction. It allows the construction of six-membered rings, These rings are very common in biological small molecules. These compounds are often of interest in medicinal chemistry and other areas of chemical biology. As a result, they are frequently targeted in synthetic studies. Being able to make these compounds, as well as compounds that are structurally related, is a key part of carrying out studies that will help us understand how they work in nature.
When two compounds combine in a Diels Alder reaction, a new ring is formed. Each reactant has two ends: there are two ends on the double bond in the dienophiles, and two ends on the diene, as well. Which end of the diene connects to which end of the dienophile? That is a question of regiochemistry, because we are asking where the reaction happens, or where the new bonds form.
We have thought about pericyclic reactions as being very different from polar reactions involving electrophiles and nucleophiles. That isn't always the case. In fact, there can be a pronounced polar component to these reactions. The cycloaddition of polar molecules is often much faster than nonpolar ones, and can be accelerated in the presence of Lewis acids. The Lewis acids can act as catalysts that activate the reactants, making them even more polar so that the attraction between the two components is greater.
Exercise \(1\)
Maleic anhydride is frequently used as a dienophile in Diels Alder reactions. Explain how this component would enhance the interaction between the diene and the dienophile in the following case.
Exercise \(2\)
Show how, in the previous question, the interaction between the molecules would be enhanced through the addition of a Lewis acid, such as titanium tetrahloride, TiCl4.
Answer
Lewis acids frequently activate carbonyl compounds towards interaction with nucleophiles. They do so by binding to the lone pairs on the carbonyl oxygen, drawing electron density away from the oxygen and, inductively, away from the neighbouring portion of the molecule.
If the reactants are polar, they may not be very symmetric. There may be a polar group attached to a double bond, causing the double bond to become more reactive but also making the molecule more complicated.
As a result, regiochemistry matters. If each component, the diene and the dienophile, have two very distinctive ends, then it matters which end connects to which, because we could potentially have two different products with completely different properties.
Generally, an understanding of resonance structures can help us to predict how two compounds are going to combine in a Diels Alder reaction.
Exercise \(3\)
Identify each of the following compounds as electrophilic or nucleophilic at carbon. In each case, include an additional resonance structure to justify your choice.
If a diene and a dienophile are to react together, their interaction can be enhanced if one of the reactants is an electrophile and one is a nucleophile. Typically, in these cases the diene is designed to act as a nucleophile and the dienophile is chosen to act as an electrophile. However, there is no particular reason it has to be that way, other than the idea that the two double bonds of the diene already appear to be more electron-rich than the single double bond of the dienophile, and so it seems we only need to give them a little nudge in that direction.
In general, an electron-withdrawing group is installed on the dienophile. Remember, an electron-withdrawing group typically contains a multiple bond to an electronegative atom. If you think of electrophilic aromatic substitutions, these were the groups that typically acted as deactivating groups and meta-directors in that situation. Carbonyls, nitriles, and nitro groups are all examples of electron-withdrawing groups.
In a complementary manner, an electron-donating group is installed on the diene. Remember, π-donors are good examples of electron-donating groups. In electrophilic aromatic substitutions, electron-donating groups typically acted as ortho-para directors, and were usually activating groups in that situation. Alkoxyl groups and amines are good examples.
Exercise \(4\)
Illustrate this electrophile/nucleophile interaction using the diene, CH3OCH=CHCH=CH2 and the dienophile, CH2=CHNO2.
Answer
Exercise \(5\)
A similar interaction is possible using an isomer of the diene in the previous question, CH2=C(OCH3)CH=CH2. Illustrate the electrophile/nucleophile interaction between this diene and the dienophile, CH2=CHNO2.
Answer
Exercise \(6\)
Provide the products of Diels Alder reactions between the reactants below.
Perhaps the most widely used polar diene is Danishefsky's diene, developed in the laboratory of Samuel Danishefsky of Columbia University and Memorial Sloan-Kettering Cancer Center. Columbia University is a world leader in research in the field of organic chemistry. Danishefsky's lab specializes in the synthesis of complex molecules that may be of medicinal interest.
Danishefsky's diene, and others like it, allow for controlled regiochemistry in Diels Alder reactions because π-donation from the alkoxy and siloxy groups both push electron density toward the same location on the diene. It is then able to react more easily with a dienophile. If the dienophile is asymmetric, which in this case means one end is more positive than the other, then the reaction occurs in a predictable way.
Exercise \(7\)
Provide a mechanism for the formation of Danishefsky's diene under the following conditions.
Exercise \(8\)
Show the products of the following reactions, including preferred regiochemistry.
Exercise \(9\)
One of the reasons Danishefsky's is useful is that the silyl ether can be converted easily into a ketone. Provide mechanisms for the sequence below.
Sometimes, the Diels Alder reaction can work equally well with compounds other than alkenes. Of course, they work with alkynes instead of alkenes in the dienophile; the only difference in that case is that an extra double bond remains in the product. Sometimes, alkynes can react in the presence of alkenes, probably because the alkyne is less sterically hindered.
Carbonyls and imines also have double bonds. As such, they can also react sometimes. Because they are electrophilic by nature, they will react better with a more nucleophilic partner. They may also need to be activated.
Exercise \(10\)
Show the product of the following reaction.
Note that, so far, we have looked at polarity-assisted Diels Alder reactions in which the diene takes the role of nucleophile (it is more electron-rich to begin with) and the dienophile takes the role of eletrophile. It doesn't have to be that way. By placing an electron-withdrawing group on the diene and an electron-donating group on the dienophile, we would still get complementary reactivity. This type of approach is called an "inverse electron demand" DIels Alder reaction, because it is the opposite of the usual method.
Exercise \(11\)
Show a mechanism with curved arrows to make a prediction about regiochemistry in the following reaction.
Exercise \(12\)
Show the products of the following reactions, including preferred regiochemistry.
Exercise \(13\)
Show why the dienophile, 1,2-dimethoxybuta-1,3-diene, would be less successful as a regioselective nucleophile.
Answer
The drawings indicate the two electron-donating groups are operating at cross purposes. One directs electron density toward one end of the diene, whereas the other directs the electron desnity toward the other end. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/V%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_3/10%3A_Reactions_Under_Orbital_Control/10.04%3A_Re.txt |
The Diels Alder reaction is probably the most common cycloaddition. It allows the construction of six-membered rings, which are very common in biological small molecules which are frequently synthetic targets.
Often, there are already rings in the molecules undergoing reaction, and a new one is being added. When two cyclic structures combine in a Diels Alder reaction, a third ring is formed in between the original ones. There are different ways the two original rings can combine, leading to different stereochemical outcomes. These two outcomes are called "exo" and "endo" addition.
An exo addition looks something like this, schematically. The two molecules approach with minimal overal between their faces; they combine edge-on. The exo product that results has a sort of Z-shape.
An endo addition looks more like this, schematically. Two cyclic molecules approach each other so that there is maximum overlap between their faces. The endo product that results has a kind of C-shape.
Let's look at these two modes of addition with real molecules. Here we are adding furan, the diene, to maleic anhydride, the dienophile. The two reactants can approach each other such that one appears to be trailing behind the other, and in this case they appear to be facing the same direction, as far as the orientation of the oxygen atoms goes. This approach leads to the zig-zag exo product. In the other case, the two molecules can be directly on top of each other; one molecule appears to be folded underneath the other. This approach leads to the curled-up endo product.
In fact, as the diagram shows, the endo product is usually the favored one. Often, the margin is substantial; one might see 90% endo product or greater in some cases, although the ratio is sometimes much lower.
The reason for this difference has something to do with an interaction between the delocalised π-system of the diene and the substituent groups attached to the dienophile. A number of researchers attribute the prefernece to a "secondary molecular orbital interaction" between the diene and the dienophile, whereas others describe the interaction as a London dispersion interaction, in which the weak intermolecular attractions stabilise the transition state in one geometry.
The endo and exo products are really two different diastereomers. If you think about it, you can see that when two rings fuse together to make a third, four new stereocenters can be created.
Since each chiral centre could have two possible configurations, there are sixteen possible stereoisomers that could result in the reaction shown above. That's a lot of structures. Just eight of them are shown below. Note that they occur in pairs of enantiomers. However, most of those diastereomers don't really occur.
Exercise \(1\)
Draw the other stereisomers of the product formed from the reaction between furan and maleic anhydride.
Answer
If we look at the molecule in this way, with the hydrogens highlighted on the ends of the diene and the dienophile, it may be easier to see the stereochmical relationships in the exo and endo products. In the exo product, the pair of hydrogens on the diene ends up cis to the pair of hydrogens on the dienophile when the rings become fused. In the endo product, the opposite is true: the pair of hydrogens on the diene come out trans to the pair of hydrogens on the dienophile when the rings fuse together.
Note that the hydrogens on the ends of the diene come out cis to each other, as well. They would be cis to each other in the ring, and that relationship does not change over the course of the reaction.
The same thing is true with the hydrogens on the dienophile.
It would be very difficult for the two cis hydrogens on one ring to become trans in the product, because it would require that one ring react via two different faces at the same time. It would be difficult to get one ring twisted around to do that. As a result, we can ignore most of the diastereoisomers that could result from a Diels Alder reaction on paper, and focus on the endo and exo products.
Of course, Diels Alder reactions between rings like these always have the potential to produce pairs of enantiomers. Because two enantiomers are equal in energy to each other, there is no inherent preference for forming one over the other. In other words, there would actually be two endo products, and they would be enantiomers of each other. There would also be two exo products, and those would be anantiomers of each other, too.
Methods are available for influencing the formation of one enantiomer exclusively. These methods involve the use of chiral catalysts. These catalysts may contain Lewis acidic metals or they may contain some combination of hydrogen bond donors and acceptors. In either case, the role of the catalyst is to tether the reactants close together so that they will undergo the reaction. If the catalyst is itself chiral, the reactants will often fit together one way more easily than another, so that the formation of one enantiomer is preferred.
Exercise \(2\)
Show the products of the following reactions, including preferred stereochemistry. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/V%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_3/10%3A_Reactions_Under_Orbital_Control/10.05%3A_En.txt |
As noted previously, the Diels Alder reaction is a classic example of a pericyclic reaction.
A Diels Alder reaction is sometimes called a [2+4] addition reaction. A 2-carbon unit on one molecule interacts with a 4-carbon unit on another molecule.
In contrast, the addition of one regular alkene to another regular alkene would be called a [2+2] addition reaction. If this reaction occurred, two alkenes would come together to form a four-membered ring.
However, [2+2] addition reactions don't occur without special circumstances. There are a couple of reasons why, and you may be able to suggest some at this point.
You might say that the four-membered ring would be much more strained than the six-membered ring formed by the Diels Alder reaction. That is true, but it may not be reason enough to prevent the reaction from happening. Four-membered rings do occur in nature despite their strain energy.
You might also say that the benzene-like transition state that stabilizes the pathway through a Cope or Diels Alder reaction isn't possible in a [2+2] addition. In fact, the transition state would be more like antiaromatic cyclobutadiene. The transition state would be very high in energy.
Another problem shows up if we look at the orbital interactions in a [2+2] addition reaction. The HOMO on one alkene and LUMO on the other alkene do not overlap so that bonds can form between the two ends. If the p orbitals on one end are in phase, the p orbitals on the other end must be out of phase. The concerted reorganization of bonding possible for the Diels Alder reaction can't happen here.
In fact, there is a way around that problem. Irradiating an alkene with UV light leads to promotion of an electron from the LUMO to the HOMO. The alkene is now in an "excited state".
This does not happen with 100% efficiency, so only some of the alkenes will become excited. In the excited state alkene, the HOMO now resembles the LUMO of the ground state alkene. Because of the matching symmetry between these orbitals, the addition reaction can proceed.
A [4+2] reaction is sometimes referred to as "thermally-allowed", whereas a [2+2] addition is sometimes referred to as "photolytically-allowed." This distinction refers to the need for electronic excitation to accomplish the latter type of reaction.
10.07: Fa
In addition to the consequences of endo- vs. exo- additions in the Diels Alder reaction, pericyclic reactions are subject to additional stereochemical constraints. In this section we will look at more issues of topology, or how the surfaces of the molecules fit together. Although this topic applies to both cycloadditions and true pericyclic rearrangements, we will start by looking at cycloadditions.
So far, we have made the simple assumption that two molecules in a Diels Alder reaction would simply come together in a face-to-face fashion. It is easy to imagine that one molecule would sit above the other as they approach, and form bonds from one face of one molecule to one face of the other molecule. In fact, it often happens that way, and this relationship is called a suprafacial addition.
However, sometimes something else happens. Imagine one molecule is able to approach the other at a slight angle, such that it is able to slip in between the ends of the other molecule. A molecule would be interacting with its neighbour not just along one face, but along two. It would form bonds via both its top and its bottom face. This type of addition is called antarafacial. It's not exactly like the cartoon shown above, but to get additional detail we will need to look at molecular orbital pictures.
In a true pericyclic reaarrangement such as a Cope or Claisen rearrangement, similar things can happen. When two ends of a molecule fold in to bond with each other, you can imagine doing so in either of two ways. Maybe the two ends roll towards each other, so that the top face on one end of the molecule connects with the top face on the other. Maybe the top face on one end connects with the lower face on the other end.
Sometimes, the motions that the molecule would undergo to put these interactions in place are called "conrotatory" (rotating together) and "disrotatory" (rotating opposite ways). The interactions they produce, however, are just like the interactions in cycloadditions, with either an antarafacial or a suprafacial relationship, respectively.
Number of electron pairs suprafacial / disrotatory antarafacial / conrotatory
even photochemical thermal
odd thermal photochemical | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/V%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_3/10%3A_Reactions_Under_Orbital_Control/10.06%3A_Ph.txt |
There are a number of additions to alkenes that occur via concerted mechanisms. Alkene oxidations are among the most synthetically useful of these reactions because they are able to convert simple hydrocarbon starting materials into oxygen-containing compounds. The resulting heteroatomic functional groups may open up new avenues of synthetic utility or they may reflect aspects of a target natural product.
The three most common alkene oxidations are epoxidation, dihydroxylation and oxidative cleavage.
Epoxidation
Epoxidation is a method for converting an alkene into an epoxide. The reagent required is always a peroxo species. A peroxo species looks very much like a normal oxygen-containing compound, but with an extra oxygen in it. Historically, the most common such reagent was m-chloroperbenzoic acid (mCPBA).
However, other reagents can also be used, such as hydrogen peroxide (H2O2) or potassium hydrogen persulfate (KHSO5), marketed under the trade name Oxone. The latter methods are considered "greener" or more environmentally friendly, because the side poducts (water or sulfate, respectively) are pretty innocuous. These methods are generally slower and are often used with a catalyst. Catalysts used with hydrogen peroxide include Lewis acidic species such as sodium tungstate (Na2WO4) needed to activate the peroxide. A similar reaction using titanium (IV) and chiral ligands leads to an enantiomerically pure epoxide; this reaction is called "Sharpless epoxidation". With oxone, ketones are used as oxygen transfer catalysts in a method referred to as "Shi oxidation".
The electrophilicity of peroxy compounds continues a theme seen in halogens such as chlorine and bromine. When two oxygen atoms are connected to each other, one of the can act as an electrophile, just as when two halogens are connected together.
During the epoxidation, the peroxy compound simply delivers its extra oxygen to the double bond. The oxygen atom both accepts a pair of electrons from the double bond and donates an electron pair to the double bond at the same time.
The reaction has something in common with pericyclic reactions. In pericyclic reactions and other reactions that take place under control of orbital symmetry, it is common to see six electrons circulating in a ring as a central feature of the mechanism. This picture is reminiscent of the aromatic structure of benzene. In fact, that aromatic stabilization is thought to play a role in stabilizing the transition states of various reactions. In this case, the three electron pairs involve delivery of the oxygen, proton transfer and π donation to the carbonyl in mCPBA. However, a similar set of arrows might not be found in the reaction of hydrogen peroxide.
Exercise \(1\)
Show how the carbonyl in mCPBA may help activate the donor oxygen toward reaction with the alkene.
Exercise \(2\)
Show how a titanium (IV) ion may help activate the donor oxygen in hydrogen peroxide toward reaction with the alkene.
The electrophilic nature of the peroxy compound is seen in the selectivity of the epoxidation reaction. Alkenes that are more electron rich tend to react much more quickly than other ones. For example, more substituted alkenes, often regarded as being electron-rich, can be selectively epoxidized in the presence of other alkenes.
Furthermore, although enones can sometimes be epoxidized, the reaction is generally slower than with regular alkenes. Of course, the carbonyl attached to the alkene in an enone makes the alkene very electron-poor.
Part of the evidence for a concerted mechanism for epoxidation comes from the stereochemistry of the reaction. In general, if a cis-alkene is epoxidized, the two groups that were cis to each other in the alkene remain cis to each other in the epoxide. If the groups start out trans to each other, they remain trans in the epoxide. Just as in hydroboration, there is no opportunity for these stereochemical relationships to change.
Dihydroxylation
Dihydroxylation is the addition of an OH group to both sides of an alkene. Typically, when reagents such as osmium tetroxide are used, the hydroxyl groups are added to the same face of the double bond. This reaction is therefore called a syn-dihydroxylation.
Osmate esters can be isolated from this reaction, resulting from the concerted addition of osmium tetroxide to the alkene. Once again, this step can be compared to a pericyclic reaction. However, the osmate ester is usually decomposed in situ through the addition of a "reducing agent" such as sodium sulfite.
Once again, the concerted nature of the reaction is seen in the stereochemistry of the product. The fact that both oxygens, which come from the osmium, are delivered to the same face of the alkene suggests that they are added at the same time.
It isn't uncommon for oxygen atoms to form additional π bonds to transition metals such as osmium. In this case, we could think of the osmium as forming an 18 electron complex as a result. Whether or not that resonance contributor is an important representation of osmium tetroxide, it is a helpful device to think of how the oxygen might form an initial attraction to the alkene.
Because of osmium tetroxide's high cost, potent toxicity and alarming propensity to rapidly sublime, other reagents are preferred. It is quite common to still use a catalytic amount of osmium tetroxide, though, along with a co-oxidant. Co-oxidants can be things like Fe(III) salts and air, although hydrogen peroxide is often used.
Oxidative Cleavage: Ozonolysis
Ozonolysis results in the complete cleavage of a double bond into two parts. The resulting fragments are each capped by an oxygen atom.
Once again, this reaction starts out with a concerted addition of the ozone to the alkene.
However, the first-formed adduct, termed a molozonide, quickly rearranges to a second product, termed an ozonide. Both of these species can be isolated. However, in practice this is rarely done because of the appalling tendency of molozonides and ozonides to explode unexpectedly. The ozonide is instead decomposed through the addition of a reducing agent, such as dimethylsulfide or zinc, leaving two oxygen-containing fragments behind.
Exercise \(3\)
Fill in the reagents for the following alkene oxidations.
Exercise \(4\)
Fill in the products of the following alkene oxidations.
Exercise \(5\)
Fill in the blanks in the following synthesis.
Exercise \(6\)
Fill in the blanks in the following synthesis.
Exercise \(7\)
Fill in the blanks in the following synthesis.
10.09: De
Decarboxylation refers to the loss of carbon dioxide from a molecule. This event generally happens upon heating certain carboxylic acids. It can't be just any compound; most commonly, there must be a carbonyl group β- to the carboxylic acid functional group.
The presence of the second carbonyl in that position is crucial in stabilizing the transition state of the reaction. If we try to follow the reaction using curved arrows, we can see something similar to what we have already observed in pericyclic reactions such as the Cope rearrangement. Three pairs of electrons in a six-membered ring are responsible for the reaction.
The loss of carbon dioxide from β-keto acids and esters is a key event in some important processes. Some typical synthetic methods include the acetoacetic ester synthesis and the malonic ester synthesis. These are two methods of α-alkylation. In these approaches, the presence of a carbonyl β- to a second carbonyl makes it much easier to remove the α-proton in between. That means simple bases such as sodium ethoxide are good enough to remove the proton efficiently and allow an alkylation reaction. However, once it is no longer needed for this activating effect, one of the ester groups can then be removed via decarboxylation.
The Knoevenagel reaction is a third, related example of a synthetic transformation that relies on decarboxylation. It differs from the other two in that it involves a carbonyl condensation rather than an alylation. Once again, the reaction is followed by decarboxylation.
Decarboxylations are also common in biochemistry. For example, loss of CO2 from isocitrate provides α-ketoglutarate. This step is one of several exergonic events in the citric acid cycle.
Exercise \(1\)
Provide a mechanism for the conversion of isocitrate to α-ketoglutarate.
Many reactions in biochemistry involve decarboxylations, even without a stabilizing group in the β-position. For example, in the entry point to the citric acid cycle, pyruvate is decarboxylated during the formation of acetyl coenzyme A.
However, a closer look at the mechanistic pathways of these reactions reveals there is something more going on. Intermediate steps involve the introduction of these stabilizing groups, which are later removed again. In the case of pyruvate decarboxylation, the compound must be activated by the addition of a thiamine ylide. The iminium group in the resulting intermediate plays the same role as that of the β -keto group in the decarboxylations we have already seen.
Subsequent steps result in the loss of the thiamine group and its replacement with a thioester linkage.
Exercise \(2\)
Provide products for each of the following reactions.
Exercise \(3\)
Carbonyls are not the only groups that can promote decarboxylation. Provide a mechanism for each of the following reactions.
Exercise \(4\)
Provide a mechanism for the Carroll Reaction, which involves the initial formation of an enol intermediate. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/V%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_3/10%3A_Reactions_Under_Orbital_Control/10.08%3A_Al.txt |
Olefin metathesis, or alkene metathesis, is an important process in petroleum refining and in the synthesis of important compounds such as pharmaceuticals. The mechanism of olefin metathesis is related to pericyclic reactions like Diels Alder and [2+2] reactions. In other words, it occurs through the concerted interaction of one molecule with another.
In petroleum refining, heating alkenes over metal oxide surfaces results in the formation of longer-chain alkenes. In particular, terminal olefins (with the double bond at the end of the chain) are converted into internal olefins (with the double bond somewhere in the middle of the chain).
What does that reaction have to do with addition reactions involving double bonds?
Clearly, the alkenes have double bonds. In addition, so do the metal oxides. Metal atoms inside the metal oxides are bridged together by oxygen atoms. The surface of the metal oxides may be covered with a mixture of hydroxyl groups as well as terminal oxides (M=O groups). The terminal oxides on the surface are the important part of the catalyst.
When metal alkylidene complexes were developed in the 1970's, it was found that they, too, could catalyze this reaction.
In fact, scientists working in petroleum chemistry soon came to believe that metal oxides on the catalyst surface were converted to alkylidenes, which then carried out the work of olefin metathesis.
The reaction, it turns out, involves a [2+2] cycloaddition of an alkene to a metal alkylidene (to the metal-carbon double bond). This reaction results in a four-membered ring, called a metallacyclobutane.
The [2+2] cycloaddition is quickly followed by the reverse reaction, a retro-[2+2]. The metallacyclobutane pops open to form two new double bonds.
This mechanism is called the Chauvin mechanism, after its first proponent, Yves Chauvin of the French Petroleum Institute. Chauvin's proposal of this mechanism shortly after the discovery of metal alkylidenes by Dick Schrock at DuPont earned him a Nobel Prize in 2005. Chauvin and Schrock shared the prize with Bob Grubbs, who made it possible for the reaction to be adapted easily to the synthesis of complex molecules such as pharmaceuticals.
Why does olefin metathesis lead to the formation of internal alkenes? The [2+2] addition and retro-[2+2] reactions occur in equilibrium with each other. Each time the metallacyclobutane forms, it can form two different pairs of double bonds through the retro reaction. In the presence of terminal alkenes, one of those pairs of alkenes will eventually include ethene. Since ethene is a gas, it is easily lost from the system, and equilibrium shifts to the right in the equation below.
That leaves a longer-chain alkylidene on the metal, ready to be attached to another long chain through subsequent [2+2] addition and reversion reactions.
In most cases, a [2+2] addition won't work unless photochemistry is involved, but it does work with metal alkylidenes. The reason for this exception is thought to involve the nature of the metal-carbon double bond. In contrast to an orbital picture for an alkene, an orbital picture for an alkylidene pi bond suggests orbital symmetry that can easily interact with the LUMO on an alkene. That's because a metal-carbon pi bond likely involves a d orbital on the metal, and the d orbital has lobes alternating in phase like a pi antibonding orbital.
A variety of catalysts have been developed for olefin metathesis. The laboratories of Bob Grubbs at CalTech and Dick Schrock at MIT have been particularly important in this area, although other labs have contributed significantly, including that of Amir Hoveyda at Boston College. The Schrock catalysts are based on tungsten or, more commonly, molybdenum. These catalysts are among the most efficient available, operating at very high turnover frequencies.
There are other variations, including some designed for alkyne metathesis rather than alkene metathesis. There is also a commercially-available Schrock-Hoveyda catalyst; that one is chiral, and can be used to carry out olefin metathesis enantioselectively. When presented with a racemic mixture of substrates, this catalysts will select one substrate preferentially over the other. However, the Schrock catalysts can be intolerant of heteroatom functional groups. Molybdenum and tungsten are highly electrophilic, so lone pairs on atoms such as halogens or nitrogens can sometimes inhibit these metals. These compounds are also highly intolerant of air and moisture, for similar reasons.
Grubbs catalysts are more tolerant than Schrock catalysts, although they do not operate at nearly the speed that the molybdenum and tungsten ones can achieve. Because they are less sensitive to air and moisture, they are more commonly used for small-scale, benchtop reactions. In larger scale, industrial reactions, air- and moisture-sensitivity is usually a less compelling factor than speed.
There are several "generations" of Grubbs catalysts that are commercially available, but a range of others have also been developed. New generations tend to offer a much higher reaction rate.
In addition to the simple "partner-swapping" of olefin metathesis, which is very important to the petroelum industry, there are other useful variations on the reaction. For example, a cyclic alkene that undergoes olefin metathesis forms two new double bonds, but these parts of the molecule are still connected to each other. Thus, an olefin metathesis between a cyclic alkene a chain alkene might produce a diene.
But remember, these reactions occur in equilibrium. If a cyclic alkene can be converted into a diene, then under the right conditions, a diene can be converted to a cyclic alkene. This reaction has become very important in the synthesis of organic compounds by the agricultural and pharmaceutical industries. RIngs of many different sizes, even very large ones, can be made in this way.
On the other hand, if the cyclic alkene by itself is treated with an olefin metathesis catalyst, it may link to other cyclic alkenes. That's because each cyclic alkene forms two new double bonds, one on each end. If each molecule forms two double bonds, a long chain of dienes will form. The result would be a polymer.
Exercise \(1\)
Show the products of the following reaction.
Exercise \(2\)
Show the products of the following reaction.
Exercise \(3\)
Show the products of the following reaction.
Exercise \(4\)
Fill in the blanks in the following synthesis.
10.12: So
Exercise 10.1.1:
Exercise 10.1.2:
Exercise 10.2.2:
Exercise 10.2.3:
Exercise 10.3.1:
Exercise 10.3.2:
Exercise 10.3.3:
Exercise 10.3.4:
Exercise 10.3.6
The retro-Diels Alder reaction is entropically favored, because two molecules are made from one molecule. As a result, energy becomes partitioned into additional states because the degrees of freedom are increasing. Internal entropy increases during this reaction. Mathematically, the entropy term is negative in the expression for free energy, so as entropy increases the free energy becomes lower. Because that term is actually a product of entropy and temperature, an increase in temperature has the effect of amplifying the influence of entropy on the free energy of the reaction. Hence, this reaction is favored by internal entropy factors, which come to dominate at elevated temperature.
Exercise 10.3.7:
Exercise 10.3.8:
Exercise 10.4.1:
The maleic anhydride is polarized, with electron density drawn toward the anhydride functional group on one side of the ring. As a result, the other side of the ring is left more positive, and will attract the polarizable electron density from the diene.
Exercise 10.4.2:
Lewis acids frequently activate carbonyl compounds towards interaction with nucleophiles. They do so by binding to the lone pairs on the carbonyl oxygen, drawing electron density away from the oxygen and, inductively, away from the neighbouring portion of the molecule.
Exercise 10.4.3:
Exercise 10.4.4:
Exercise 10.4.5:
Exercise 10.4.6:
Exercise 10.4.7:
Exercise 10.4.8:
Exercise 10.4.9:
Exercise 10.4.10:
Exercise 10.4.11:
Exercise 10.4.12:
Exercise 10.4.13:
The drawings indicate the two electron-donating groups are operating at cross purposes. One directs electron density toward one end of the diene, whereas the other directs the electron desnity toward the other end.
Exercise 10.5.1:
Exercise 10.8.3:
a) OsO4, H2O
b) mCPBA
c) 1. O3; 2. Me2S or Zn
d) 1. O3; 2. Me2S or Zn
e) OsO4, H2O
f) mCPBA
Exercise 10.8.6:
Exercise 10.9.2:
Exercise 10.9.3:
Exercise 10.9.4:
Exercise 10.10.1:
Exercise 10.10.2:
Exercise 10.10.3:
Exercise 10.10.4: | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/V%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_3/10%3A_Reactions_Under_Orbital_Control/10.10%3A_Ol.txt |
The pinacol rearrangement is a reaction of 1,2-diols. It takes place under the influence of strong acids, including mineral acids like sulfuric acid. It can also be brought about via the use of Lewis acids. The reaction overall involves the loss of one of the hydroxyl groups, the conversion of the other hydroxyl group into a carbonyl, and the shift of an alkyl group.
The heart of the rearraction is the 1,2-shift of an alkyl group. This shift is assisted by π-donation from the oxygen, converting the C-O bond into a carbonyl at the same time.
The entire mechanism would look like this:
In many rearrangements, formation of the actual cation is not necessarily involved. The 1,2-shift could occur as the water molecule leaves. The protonated oxygen may draw enough electron density toward itself to induce the shift.
Frequently there is a fine line between alternative reaction pathways. In many cases, the stability of an intermediate determines whether a reaction happens in stages or in one step. If the intermediate is relatively stable, the reaction will happen in steps. The less stable the intermediate, the more likely the reaction will proceed in a concerted manner.
Exercise \(1\)
Fill in the products of the following pinacol rearrangements.
Exercise \(2\)
The Tiffeneau Demjanov rearrangement is superficially similar to a pinacol rearrangement. It occurs when a β-amino alcohol is treated with nitrous acid. The reaction results in loss of the amino group with rearrangement of the molecule. Predict the product of the reaction.
11.03: Baeyer
The Baeyer-Villiger rearrangement is the conversion of a ketone to an ester via the insertion of an oxygen atom next to the carbonyl.
The reaction involves initial addition of a peroxide to the carbonyl carbon. The resulting adduct undergoes rearrangement to form the ester.
Exercise \(1\)
Predict the products of the following Baeyer-Villiger reactions.
11.04: Beckma
The Beckmann rearrangement results when an oxime (an N-hydroxyimine) is treated with concentrated acid and heated.
The oxime, in turn, is generated by treatment of a ketone with hydroxylamine. A catalytic amount of acid can activate the carbonyl, accelerating the otherwise sluggish reaction.
The reaction, like the pinacol rearrangement, is triggered by the loss of water from the starting material. The incipient cation that results undergoes a 1,2-shift. Subsequently, the re-addition of water to the rearranged cation results in a new compound.
By "incipient", we describe a cation that is only on the brink of forming, but has not actually occurred yet. The 1,2-shift happens as soon as the partial positive charge on the nitrogen becomes great enough to draw the electrons from the neighbouring bond.
The complete mechanism is described pictorially below.
Exercise \(1\)
Predict the products of the following Beckmann rearrangements.
11.05: Wolff
The Wolff rearrangement is the conversion of a diazoketone to a ketene, usually under photolytic conditions.
The loss of dinitrogen from the diazonium compound would result in an electron-deficient carbene. Like a carbocation, the carbene would be susceptible to a 1,2-shift. If accompanied by π-donation from the carbene to the carbonyl, a ketene would result.
As with other rearrangements, the 1,2-shift could occur at the same time as the loss of the dinitrogen.
Ketenes are not terribly stable. In the presence of nucleophilic solvents such as water or alcohol, the ketene easily undergoes nucleophilic addition. Addition of water would result in a carboxylic acid.
The mechanism of that addition involves keto-enol tautomerism.
Exercise \(1\)
Predict the products of the following Wolff rearrangements.
11.06: Soluti
Exercise 11.1.1:
Exercise 11.1.2:
Exercise 11.2.1:
Exercise 11.3.1:
Exercise 11.4.1:
Exercise 11.5.1: | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/V%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_3/11%3A_Electrophilic_Rearrangement/11.02%3A_Pinaco.txt |
If, in some cataclysm, all of scientific knowledge were to be destroyed, and only one sentence passed on to the next generation of creatures, what statement would contain the most information in the fewest words? I believe it is the atomic hypothesis (or the atomic fact, or whatever you wish to call it) that all things are made of atoms—little particles that move around in perpetual motion, attracting each other when they are a little distance apart, but repelling on being squeezed into one another. In that one sentence, you will see, there is an enormous amount of information about the world, if just a little imagination and thinking are applied. – Nobel Laureate Richard Feynman, 1963
Most of us are quite familiar with the core principle of atomic theory—the idea that matter is composed of atoms—because we have been told that this is so since childhood. But how many of us really, and we mean really believe it, use it in our day-to-day life, understand its implications, or know the reasons why it is assumed to be true? It seems so completely and totally impossible and improbable because we do not experience atoms directly and it is easy to go through life quite successfully, at least for the vast majority of us, without having to take atoms seriously. The average person’s brain is simply not wired to believe in the reality of things like atoms in a concrete and day-to-day way. Yet most scientists, and certainly most chemists, would agree that Feynman’s deceptively simple statement contains the essence of chemistry.
Atomic theory is also critical for understanding a significant number of the underlying concepts of biology and physics, not to mention geology, astronomy, ecology, and engineering. How can one sentence contain so much information? Can we really explain such a vast and diverse set of scientific observations with so little to go on? In the next two chapters we will expand on Feynman’s sentence to see just what you can do with a little imagination and thinking. At the same time, it is worth remembering that the fact that atoms are so unreal from the perspective of our day-to-day experience means that the atomic theory poses a serious barrier to understanding modern chemistry. This is a barrier that can only be dealt with if you recognize it explicitly and try to address and adjust to it. You will be rewiring your brain in order to take atoms, and their implications, seriously. We are aware that this is not an easy task. It takes effort, and much of this effort will involve self-reflection, problem-solving, and question-answering. In an important sense, you do not have to believe in atoms, but you do have to understand them.
Thumbnail: Spinning Buckminsterfullerene (\(\ce{C60}\)). (CC BY-SA 3.0; unported; Sponk).
01: Atoms
You almost certainly have heard about atoms and it is very likely you have been taught about them. If asked you might profess to believe in their reality. You might accept that matter, in all its forms, is made up of atoms — particles that are the smallest entities that retain the identity of an element (we will discuss elements in much greater detail in the next few chapters.) It is very likely that you have been taught that atoms are made up of even smaller particles: positively charged protons, uncharged neutrons, and negatively charged electrons. You may even have heard, and perhaps even believe, that protons and neutrons can be further subdivided into quarks and gluons, while electrons are indivisible. Equally difficult to appreciate is that all atoms are organized in a very similar way, with a very tiny, but relatively heavy, positively charged nucleus surrounded by the much lighter, negatively charged electrons.
Part of the difficulty in really understanding atoms is the fact that the forces holding the atomic nucleus together, the so-called strong and weak forces, operate at such infinitesimal distances that we do not experience them directly. This is in contrast to electromagnetism and gravity, which we experience directly because they act over longer, macroscopic or visible distances. A second problem is associated with the fact that to experience the world we need to use energy; at the atomic scale the energy used to observe the system also perturbs it. This is the basis of the Heisenberg uncertainty principle, which you may have encountered or at least heard of before, and to which we will return. Finally, objects at the atomic and subatomic scales behave differently from the macroscopic objects with which we typically interact. A particle of light, a photon, an electron, a proton, or a neutron each behaves as both a particle and a wave. In terms of physics, these are neither particles nor waves; they are quantum mechanical particles. Luckily, the weirder behaviors of atomic and subatomic entities can often, but not always, be ignored in chemical and biological systems. We will touch on these topics as necessary.
Current theory holds that each atom contains a very, very small, but very dense nucleus, which contains protons and neutrons and is surrounded by electrons. These electrons are very light, relatively, but the space occupied by moving electrons accounts for the vast majority of the volume of an atom. Because the number of positively charged protons and negatively charged electrons are equal and the size of the charges are the same but opposite, atoms are electrically neutral when taken as a whole; that is, each positively-charged proton is counterbalanced by a negatively-charged electron.
Often the definition of an atom contains some language about how atoms are the smallest particle identifiable as that element. What do we mean by that? For example, can an atom have chemical properties? And how can ensembles of the same particles, that is protons, electrons, and neutrons, have different properties? This is the mystery of the atom and understanding it is the foundation of chemistry. In this first chapter, we hope to lead you to a basic understanding of atomic structure and inter-atomic interactions. Subsequent chapters will extend and deepen this understanding.
Questions
Questions to Ponder
• If you had to explain to a non-scientist why it is that scientists accept the idea that all material things are composed of atoms what evidence would you use?
• Does the ability of science to explain so much about the world influence your view about the reality of supernatural forces? | textbooks/chem/General_Chemistry/CLUE%3A_Chemistry_Life_the_Universe_and_Everything/01%3A_Atoms/1.1%3A_What_Do_You_Think_You_Know_About_Atoms.txt |
We assume that you have lots of ideas about atoms but did you ever stop to think how we came to accept this information as reasonable or what the reality of atoms implies about how the world we perceive behaves? Atoms are incredibly and unimaginably small. A gold atom with its full complement of electrons is less than a nanometer ($1 \times 10^{-9}$ meters) in diameter and its nucleus, which contains 79 protons and generally around 116 neutrons, has a radius of $\sim 1.5 \times 10^{-14}$ meters. While these sizes are actually unimaginable, there are a number of web-based activities that can help you come to terms with the scales of atoms.[1] There is no way you could see an atom with your eye or with a light microscope, although there are now techniques that allow us to view computer representations of individual atoms using various types of electron and force-probe microscopes. The smallest particle of matter that you can see with your naked eye contains more atoms than there are people in the world. Every cell in your body contains a huge number of atoms. Obviously, whatever we know about atoms is based on indirect evidence; we do not directly experience atoms.
The full story of how we know what we know about the existence and structure of atoms is fascinating, complex, and perhaps fortunately for you, too long to go into in detail. What we do want to do is to consider a number of key points that illustrate how our ideas of atoms arose and have changed over time. We will present the evidence that has made accepting the atomic theory unavoidable if you want to explain and manipulate chemical reactions and the behavior of matter.
Atomic theory is an example of a scientific theory that began as speculation and, through the constraints provided by careful observation, experimentation, and logical consistency, evolved over time into a detailed set of ideas that make accurate predictions and are able to explain an increasing number of diverse, and often previously unknown, phenomena. As scientists made new observations, atomic theory was adapted to accommodate and organize these observations.
A key feature of scientific ideas, as opposed to other types of ideas, is not whether they are right or wrong but whether they are logically coherent and make unambiguous, observable, and generally quantitative predictions. They tell us what to look for and predict what we will find if we look at or measure it. When we look, we may find the world acts as predicted or that something different occurs. If the world is different from what our scientific ideas suggest then we assume we are missing something important: either our ideas need altering or perhaps we are not looking at the world in the right way. As we will see, the types of observations and experimental evidence about matter have become increasingly accurate, complex, and often abstract, that is, not part of our immediate experience. Some of these observations can be quite difficult to understand, because matter behaves quite differently on the atomic and sub-atomic scale than it does in the normal, macroscopic world. It is the macroscopic world that evolutionary processes have adapted us to understand, or at least cope with, and with which we are familiar. Yet, if we are to be scientific, we have to go where the data lead us. If we obtain results that are not consistent with our intuitions and current theories, we have to revise those theories rather than ignore the data.
However, scientists tend to be conservative when it comes to revising well-established theories because new data can sometimes be misleading. This is one reason there is so much emphasis placed on reproducibility. A single report, no matter how careful it appears, can be wrong or misinterpreted and the ability of other scientists to reproduce the observation or experiment is key to its acceptance. This is why there are no miracles in science. Even so, the meaning of an observation is not always obvious or unambiguous; more often than not an observation that at first appears to be revolutionary turns out to have a simple and even boring explanation. Truly revolutionary observations are few and far between. This is one reason that the Carl Sagan (1934-1996) quote, “Extraordinary claims require extraordinary evidence” is so often quoted by scientists. In most cases where revolutionary data is reported, subsequent studies reveal that the results were due to poor experimental design, sloppiness, or some irrelevant factor. The fact that we do not all have cold fusion energy plants driving perpetual motion refrigerators in our homes is evidence that adopting a skeptical approach that waits for experimental confirmation is wise.
A common misconception about scientific theories is that they are simply ideas that someone came up with on the spur of the moment. In everyday use, the word theory may well mean an idea or even a guess, a hypothesis, or a working assumption, but in science the word theory is reserved for explanations that encompass and explain a broad range of observations. More than just an explanation, a theory must be well tested and make clear predictions relating to new observations or experiments. For example, the theory of evolution predicted that the fossil record would show evidence for animals that share many of the features of modern humans. This was a prediction made before any such fossils were found; many fossils of human-like organisms have since been and continue to be discovered. Based on these discoveries, and on comparative analyses of the structure of organisms, it is possible to propose plausible family trees, known as phylogenies, connecting different types of organisms. Modern molecular genetics methods, particularly genome ($\mathrm{DNA}$) sequencing, have confirmed these predictions and produced strong experimental support for the current view that all organisms now living on Earth are part of the same family–that is, they share a common ancestor that lived billions of years ago. The theory of evolution also predicts that the older the rocks, the more different the fossilized organisms found will be from modern organisms. In rocks dated to $\sim 410$ million years ago, we find fossils of various types of fish but not the fish that exist today. We do not find evidence of humans from that period; there are, in fact, no mammals, no reptiles, no insects, and no birds.
A scientific theory is also said to be falsifiable, which doesn’t mean that it is false but rather that it may be proven false by experimentation or observation. For example, it would be difficult to reconcile the current theory of evolution with the discovery of fossil rabbits from rocks older than 300 million years. Similarly, the atomic theory would require some serious revision if someone discovered an element that did not fit into the periodic table; the laws of thermodynamics would have to be reconsidered if someone developed a successful perpetual motion machine. A theory that can be too easily adapted to any new evidence has no real scientific value.
A second foundational premise of science is that all theories are restricted to natural phenomena; that is, phenomena that can be observed and measured, either directly or indirectly. Explanations that invoke the supernatural or the totally subjective are by definition not scientific, because there is no imaginable experiment that could be done that might provide evidence one way or another for their validity. In an important sense, it does not matter whether these supernatural explanations are true or not; they remain unscientific. Imagine an instrument that could detect the presence of angels. If such an instrument could be built, angels could be studied scientifically; their numbers and movements could be tracked and their structure and behaviors analyzed; it might even be possible to predict or control their behavior. Thus, they would cease to be supernatural and would become just another part of the natural world. Given these admitted arbitrary limitations on science as a discipline and an enterprise, it is rather surprising how well science works in explaining (and enabling us to manipulate) the world around us. At the same time, science has essentially nothing to say about the meaning of the world around us, although it is often difficult not to speculate on meaning based on current scientific ideas. Given that all theories are tentative, and may be revised or abandoned, perhaps it is wise not to use scientific ideas to decide what is good or bad, in any moral sense.
As we will see, the history of atomic theory is rife with examples of one theory being found to be inadequate, at which point it must be revised, extended, and occasionally totally replaced by a newer theory that provides testable explanations for both old and new experimental evidence. This does not mean that the original theory was necessarily completely false but rather that it was unable to fully capture the observable universe or to accurately predict newer observations. Older theories are generally subsumed as newer ones emerge; in fact, the newer theory must explain everything explained by the older one and more.
Questions
Questions to Answer: Scientific Questions and Theories:
• How would you decide whether a particular question was answerable scientifically?
• How would you decide whether an answer to a question was scientific?
• What is the difference between a scientific and a non-scientific question? Provide an example of each.
Questions to Ponder
• What things have atoms in them? Air, gold, cells, heat, light?
• How do you know atoms exist? | textbooks/chem/General_Chemistry/CLUE%3A_Chemistry_Life_the_Universe_and_Everything/01%3A_Atoms/1.2%3A_Atomic_Realities_and_Scientific_Theories.txt |
Modern atomic theories have their roots in the thinking of ancient peoples, in particular ancient Greek philosophers who lived over 2500 years ago. At that time the cultural, economic, and intellectual climate in Ancient Greece permitted a huge surge of philosophical and scientific development, the so-called Greek miracle. While most people of that time believed that the world was ruled by a cohort of semi-rational gods a series of philosophers, beginning with Thales of Miletus (died 546 bce),[2] were intent on developing rational and non-supernatural explanations for observable phenomena such as what we are made of and where we came from. As we know now, they could not possibly have understood the underlying nature of matter because they lacked the tools to observe and experiment at the atomic scale. However, this does not mean that their ideas were simple idle speculation. The ideas they produced, although not scientific as we understand the term today, contained remarkable insights – some of which appear to be true.
This era gave birth to a new way to approach and explore natural phenomena in order to gain understanding of their complexity and diversity in terms of natural explanations. It is worth considering that such a rational approach did not necessarily have to be productive; it could be that the world is really a totally irrational, erratic, and non-mechanistic place, constantly manipulated by supernatural forces; but given that science can not address these kinds of ideas, let us just leave them to fantasy authors. The assumption that the world is ruled solely by natural forces has been remarkably productive; that is, consistent with the way the world appears to behave when we look at it dispassionately.
The ancient Greeks developed complex ideas about the nature of the universe and the matter from which it was composed, some of which were accepted for a long time. However, in response to more careful observation and experimental analysis, these ideas were eventually superseded by more evidence-based theories. In large part this involved a process by which people took old ideas seriously, and tried to explain and manipulate the world based on them. When their observations and manipulations failed to produce the expected or desired outcomes, such as turning base metals into gold, curing diseases, or evading death altogether, they were more or less forced to revise their ideas, often abandoning older ideas for newer ideas that seemed to work.
The development of atomic theories is intertwined with ideas about the fundamental nature of matter, not to mention the origin of the universe and its evolution. Most Greek philosophers thought that matter was composed of some set of basic elements, for example, the familiar earth, air, fire, and water. Some philosophers proposed the presence of a fifth element, known as quintessence or aether.[3] These clearly inadequate ideas persist today as part of astrology and the signs of the Zodiac—a poor tribute to some very serious thinkers.
The original elements, that is, earth, air, fire, and water, were thought to be composed of tiny indestructible particles, called atoms by Leucippus and Democritus (who lived around 460 bce).[4] The atoms of different elements were assumed to be of different sizes and shapes, and their shapes directly gave rise to the properties of the particular element. For example, the atoms of earth were thought to be cubic; their close packing made earth solid and difficult to move. The idea that the structure of atoms determines the observable properties of the material is one that we will return to, in a somewhat different form, time and again. Although the particulars were not correct, the basic idea turns out to be sound.
In addition to their shapes, atoms were also thought to be in constant motion, based on watching the movement of dust motes in sunlight, with nothing, or a void, between them.[5] Many centuries later Einstein’s analysis of this type of motion, known as Brownian motion, provided strong experimental support for the physical reality of molecules, larger structures composed of atoms, and the relationship between molecular movement, temperature, and energy, which we will consider later on in this chapter.
All in all the combined notions of the Greek philosophers provided a self-consistent and satisfactory basis for an explanation of the behavior of matter, as far as they could tell. The trap here is one that is very easy to fall into, namely that a satisfying explanation for a phenomenon is not necessarily true. Even if it seems to be self-consistent, useful, or comforting, an explanation is not scientific unless it makes testable, quantitative predictions. For example, it was thought that different materials were made up of different proportions of the four ancient elements. Bones were made of water, earth, and fire in the proportions 1:1:2, whereas flesh was composed of these elements in a ratio of 2:1:1.[6] While these ideas are now considered strange, they contain a foreshadowing of the “law of multiple proportions”, which would come some 2300 years later and which we will deal with later in this chapter. Some philosophers even thought that the soul was composed of atoms or that atoms themselves had a form of consciousness, two ideas that seem quite foreign to (most of) us today.
Such ideas about atoms and elements provided logical and rational, that is, non-supernatural explanations for many of the properties of matter. But the Greeks were not the only ancient people to come up with explanations for the nature of matter and its behavior. In fact, it is thought that the root of the words alchemy and chemistry is the ancient Greek word Khem, the Greek name for Egypt, where alchemy and chemistry are thought to have originated.[7] Similar theories were being developed in India at about the same time, although it is the Greek ideas about atoms that were preserved and used by the people who eventually developed our modern atomic theories. With the passage of time ancient ideas about atoms and matter were kept alive by historians and chroniclers, in particular scholars in the Arab world. During the European Dark Ages and into medieval times, there were a few scattered revivals of ideas about atoms, but it was not until the Renaissance that the cultural and intellectual climate once again allowed the relatively free flowering of ideas. This included speculation on the nature of matter, atoms, and life. Experimental studies based on these ideas led to their revision and the eventual appearance of science, as we now know it. It is also worth remembering that this relative explosion of new ideas was occasionally and sometimes vigorously opposed by religious institutions, leading to torture, confinement, and executions.[8]
Questions:
Questions to Answer:
• What properties ascribed by the Greeks to atoms do we still consider to be valid?
Questions to Ponder:
• If earth had atoms that were cubic, what shape would you ascribe to the elements air, water, and fire?
Questions for Later
• If atoms are in constant motion, what do you think keeps them moving? | textbooks/chem/General_Chemistry/CLUE%3A_Chemistry_Life_the_Universe_and_Everything/01%3A_Atoms/1.3%3A_Some_History_of_Atomic_Theory.txt |
The Greek notion of atoms and elements survived for many centuries and it was eventually fleshed out with the addition of a few more elements, mostly through the efforts of the alchemists. Some elements such as gold were discovered much earlier – mainly because they exist as elements rather than compounds. By the late eighteenth century, the idea of an element as a substance that cannot be broken down into more fundamental substances had begun to be accepted. In 1789 Antoine Lavoisier (1743–1794) produced a list of 33 elements. His list did not include earth, air, fire, and water, but it did contain light and heat, along with a number of modern elements including cobalt, mercury, zinc, and copper. It had already been established that oxygen and hydrogen were elements, while water was not. The stage was set for a rapid growth in our knowledge about the underlying structure of matter. We now know of 91 naturally occurring elements, and quite a number of unnatural, that is, human-made ones which are not found in nature because they unstable. These human-made elements are heavier in atomic terms than the naturally occurring elements and are typically generated by smashing atoms of natural elements into one another; they break down, or decay, rapidly into atoms of other elements. As examples of how science can remove some of the mystery from the universe: our understanding of atoms and elements means that no new natural, light elements are theoretically possible. We know of all the light elements that can possibly exist anywhere in the universe, a pretty amazing fact. Similarly, our current understanding of the theory of general relativity and the laws of thermodynamics make faster-than-light travel and perpetual motion machines impossible, although it does not stop people from speculating about them.
The first modern chemical isolation of an element is attributed to the alchemist Hennig Brand (c. 1630–c. 1710).[9] He isolated phosphorus from urine while in pursuit of the philosopher’s stone.[10] While this may seem like an odd thing to do, people have done much stranger things in pursuit of gold or cures for diseases like syphilis. Imagine his surprise when, after boiling off all the water from the urine, the residue burst into flames and gave off a gas that, when condensed, produced a solid that glowed green in the dark. It was for this reason that he named it phosphorus, from the Greek for light-bearer. Similarly, mercury was originally isolated by roasting the mineral cinnabar. Despite being quite toxic, mercury was used as a treatment for syphilis prior to the discovery of effective antibiotics.
Questions
Questions to Answer
• Given what you know, how would you explain the difference between an atom and an element?
• What differentiates one element from another?
• What is the difference between an atom and a molecule?
• What is the difference between an element and a compound?
Questions to Ponder
• What types of evidence might be used to prove you had isolated a new element?
• When can unproven/unsubstantiated assumptions be scientific?
• Under what conditions are such assumptions useful?
• Why do you think gold was recognized as an element earlier than many others? | textbooks/chem/General_Chemistry/CLUE%3A_Chemistry_Life_the_Universe_and_Everything/01%3A_Atoms/1.4%3A_Identifying_and_Isolating_Elements.txt |
It is important to note that from the time that the first ideas of atoms arose, and for thousands of years thereafter, there was not one shred of evidence for the particulate nature of matter or the physical existence of atoms. The idea of atoms was purely a product of imagination, and while there was vigorous debate about the nature of matter, this debate could not be settled scientifically until there was objective empirical evidence one way or another.
So the question arises, how did scientists in the nineteenth century eventually produce clear evidence for the existence of atoms? We have already said atoms are much too small to be seen by any direct method. So what would lead scientists to the unavoidable conclusion that matter is composed of discrete atoms? It is often the case that a huge intuitive leap must be made to explain the results of scientific observations. For example, the story about Isaac Newton (1643–1727) and the falling apple captures this truism, namely the remarkable assumption that the movement of Earth around the Sun, the trajectory of a cannon ball, and the falling of an apple to Earth are all due to a common underlying factor, the force of gravity, which acts at a distance and obeys an inverse square relationship, $\frac{1}{r^{2}}$ where $r$ is the distance between two objects. This seems like a pretty weird and rather over-blown speculation; how does this “action at a distance” between two objects work? Yet, followed scientifically, it appeared to be very powerful and remarkably accurate. The point is that Newton was able to make sense of the data, something that is in no way trivial. It requires a capacity for deep, original, and complex thought. That said, it was not until Albert Einstein (1874-1955) proposed his general theory of relativity in 1915 that there was a coherent, mechanistic explanation for gravitational forces.
The first scientific theory of atomic structure was proposed by John Dalton (1766–1844), a self-taught Quaker[11] living in Manchester, England.[12] In 1805 Dalton published his atomic theory to explain the observed law of multiple, or definite, proportions, which stated briefly is “when elements combine, they do so in the ratio of small whole numbers”, we will return to this idea later on, in much greater detail.[13] Rather surprisingly, Dalton never really explained what led him to propose his atomic theory, although he certainly used it to explain existing rules about how different elements combine. Among these rules was the observation that the total matter present in a system does not change during a chemical reaction, although a reaction might lead to a change from a solid to a gas or vice versa. Dalton’s atomic theory (1805) had a number of important components:
• Elements are composed of small indivisible, indestructible particles called atoms.
• All atoms of an element are identical and have the same mass and properties.
• Atoms of a given element are different from atoms of other elements.
• Compounds are formed by combinations of atoms of two or more elements.
• Chemical reactions are due to the rearrangements of atoms, and atoms (matter) are neither created nor destroyed during a reaction.
Based on these tenets he was able to explain many of the observations that had been made, by himself and others, about how matter behaves and reacts. More modern atomic theories have made some modifications, for example to include the existence of atomic isotopes, that is, atoms with different numbers of neutrons, but the same number of protons and electrons, and the conversion of energy into matter and vice versa, but Dalton’s core ideas remain valid.
Questions
Questions to Answer
• In what ways is Dalton’s atomic theory different from the ideas of the Greek philosophers?
• Which tenets of Dalton’s theory still hold up today?
• Design an experiment to investigate whether there is a change in mass when water changes phase. What data would you collect? How would you analyze it?
Questions to Ponder
• How did Dalton conclude that there were no half-atoms?
• Which parts of Dalton’s theory were unfounded speculation and which parts were based on direct observation? | textbooks/chem/General_Chemistry/CLUE%3A_Chemistry_Life_the_Universe_and_Everything/01%3A_Atoms/1.5%3A_Evidence_for_Atoms.txt |
The opposite of a correct statement is a false statement. But the opposite of a profound truth may well be another profound truth. Neils Bohr (1865–1962)
Dalton’s theory of atoms as indivisible, indestructible, objects of different sizes, weights, and perhaps shapes, depending on the element, held up for almost 100 years, although there was considerable dissent about whether atoms really existed, particularly among philosophers. By 1900 the atomic theory was almost universally accepted by chemists. More evidence began to accumulate, more elements were discovered, and it even became possible to calculate the number of atoms in a particular sample. The first step, along this direction, was made by Amedeo Avogadro (1776–1856). In 1811 he proposed that, under conditions of equal temperature and pressure, equal volumes of gases contained equal numbers of particles (molecules) and that the densities of the gases, that is their weight divided by their volume, were proportional to the weight of the individual molecules. This was expanded on by the Austrian high school teacher Josef Loschmidt (1821–1895) who, in 1865, combined Avogadro’s conclusion with the assumption that atoms and molecules move very much as elastic objects, think billiard balls. This enabled him to calculate the force a molecule would exert when traveling at a particular speed, something difficult to measure, and relate that to the pressure, something easily measured. In fact, this assumption enabled physicists to deduce that the temperature of a gas is related to the average kinetic energy of the molecules within it, a concept we will return to shortly.
Probing the Substructure of Atoms
The initial Greek assumption was that atoms were indivisible, essentially unchangeable from their initial creation. However, gradually evidence began to accumulate that atoms were neither indivisible nor indestructible. Evidence for the existence of particles smaller than atoms had been building up for some time, although it was not recognized as such. For example, the well-recognized phenomenon of static electricity had been known since the ancient Greeks. The name electricity comes from the Latin electricus, meaning amber-like. Rubbing amber with fur generates static electricity—the same type of spark that jumps from your finger to a doorknob or another person under dry conditions. In the late 1700s Luigi Galvani (1737–1798) discovered that animals can produce and respond to electricity, perhaps the most dramatic example being the electric eels and rays that stun their prey through electrical shocks. The discovery of bioelectricity was exploited in many novels and movies, beginning with Mary Shelly’s (1797–1851) novel Frankenstein and continuing through Mel Brook’s (b. 1926) comedy film, Young Frankenstein. Galvani discovered that a dead frog’s leg would twitch in response to exposure to static electricity; it appeared to come back to life, just like Frankenstein’s monster. He assumed, correctly it turns out, that electrical activity was involved in the normal movement of animals. He thought that a specific form of electricity, bioelectricity, was carried in the fluid within the muscles and was a unique product of biological systems, a type of life-specific force. We now recognize that a number of biological phenomena, such as muscle contraction and brain activity, are initiated by changes in electric fields (across membranes) and that the underlying physicochemical principles are similar to those taking place in non-biological systems.
The excitement about electricity and its possible uses prompted Alessandro Volta (1745–1827) to develop the first modern battery, now known as a voltaic pile. He alternated sheets of two different metals, such as zinc and copper, with discs soaked in salt water (brine). It produced the first steady electrical current that, when applied to frog muscles, caused them to contract. Such observations indicated that biological systems can both generate and respond to electrical currents, suggesting that bioelectricity was no different than any other form of electricity. What neither Volta nor Galvani knew was the nature of electricity. What was it, exactly, and how did it flow from place to place? What was in the spark that jumped from finger to metal doorknob, or from Benjamin Franklin’s (1705–1790) kite string to his finger? What was this “electrical fluid” made of?
Progress in the understanding of the nature and behavior of electricity continued throughout the 19th century and the power of electricity was harnessed to produce dramatic changes in the way people lived and worked, powering factories, lighting houses and streets, and so on. Yet there was no deep of understanding as to the physical nature of electricity. It was known that electric charge came in two forms, positive and negative, and that these charges were conserved; that is, they could not be created or destroyed, ideas first proposed by Franklin. The electrical (charged) nature of matter was well established, but not where those charges came from or what they were.
A key step to understanding electricity involved unraveling the idea of the indivisible atom and involved a series of experiments by J. J. Thompson (1856–1940), another Mancunian.[14] Although the idea of electricity was now well appreciated, Thompson and other scientists wanted to study it in a more controlled manner. They used what were, and are now, known as cathode ray tubes (CRTs). Once common in televisions, these have now been replaced by various flat screen devices. CRTs are glass tubes with wires embedded in them; these wires are connected to metal discs. The inside of the tube is coated with a chemical that glows (fluoresces) in response to electricity. They generally have ports in the walls that can be connected to a vacuum pump, so that most of the air within the tube can be removed, typically the ports are then sealed. When connected to a source of electricity, such as a voltaic pile, the fluorescent material at one end of the tube glows. In a series of experiments (1897) Thomson was able to show that:
• Rays emerged from one disc (the cathode) and moved to the other (the anode).
• “Cathode” rays were deflected by electrical fields in a direction that indicated that they were negatively charged.
• The rays could also be deflected by magnetic fields.[15]
• The rays carried the electrical charge; that is, if the ray was bent, for example by a magnetic field, the charge went with it.
• The metal that the cathode was made of did not affect the behavior of the ray; so whatever the composition of the ray, it appeared to be independent of the element that it came from.
In all of these experiments, it needs to be stressed that “positive” and “negative” are meant to indicate opposite and are assigned by convention. That means that we could decide tomorrow that positive was negative, and negative positive, and nothing would change, as long as we were consistent. From these experiments, Thompson concluded that “cathode” rays were carried by discrete charged particles, he called them corpuscles, and he assigned these particles a negative charge. But the truly stunning conclusion he reached was that these particles must come from within the atoms of the metal cathode. Because the type of metal did not affect the nature or behavior of the cathode rays, he assumed that these particles were not newly created but must pre-exist within the atoms of the cathode. Moreover, he hypothesized that identical particles must be present in all atoms, not just in the atoms of one particular metal. Do you see how he jumps from experimental results using a few metals to all elements and all atoms? Of course, we now know these particles as electrons but it is difficult to imagine what a huge impact this new theory had on scientists at the time.
Since electrons can be produced by all chemical elements, we must conclude that they enter the constitution of all atoms. We have thus taken our first step in understanding the structure of the atom. —J. J. Thomson, The Atomic Theory, 1914[16]
The discovery of the electron made the old idea of an atom as a little indestructible billiard ball-like objects obsolete and necessitated a new model. It is an example of a paradigm shift[17]—a fundamental change in scientific thinking driven by new evidence. Thompson’s first version of this new model became known as the plum pudding model.[18] His basic idea was that the atom is a ball of positively charged, but apparently amorphous, matter with electrons studded here and there, like the raisins in a pudding. Because it contained equal numbers of positive and negative charges, the overall structure was electrically neutral. Subsequent work by Thompson and Robert A. Millikan (1868–1953) established that all electrons are identical, each with the same, very small mass and negative charge. The mass of an electron is less than $\frac{1}{1000\mathrm{th}}$ of the mass of a hydrogen atom.
Thompson’s proposed plum pudding model of the atom spurred much experimental and theoretical work and led to a remarkable number of subsequent discoveries. For example, it was soon recognized that the $\beta$ particles emitted by some radioactive minerals and elements, were, in fact, electrons. Other studies found that the number of electrons present in the atoms of a particular element was roughly proportional to half the element’s atomic weight, although why this should be the case was unclear.
However, as more and more data began to accumulate, the plum pudding model had to be abandoned because it just could not explain what was being observed. The key experiment that led to a new model of the atom was carried out in 1908 by Ernest Rutherford (1871–1937). As you may have already guessed, he was working at the University of Manchester. In this experiment, he examined how alpha ($\alpha$) particles, which he knew to be positively charged particles made of the element helium without it’s electrons, behaved when they were fired at a very thin sheet of metal, such as gold or platinum. In the experiment a narrow parallel beam of α particles was directed at a thin sheet of gold foil and the angles at which the deflected particles scattered were detected. The observed result was completely unexpected. Instead of passing straight through the thin sheet of foil, he found that a few particles were deflected, some of them at large angles. Rutherford wrote, “It is as if I had fired a cannon ball at a piece of tissue paper, and it bounced right back.” Here again, we see a particular aspect of the scientific enterprise, namely that even though only a few alpha particles bounced back, we still need to explain how this could possibly occur. We could not just say, “Only a few particles were bounced so it doesn’t matter”; we have to provide a plausible scenario to explain the observation. Often it is paying attention to, and taking seriously, the unexpected result that leads to the most profound discoveries.
Based on these experimental results Rutherford reasoned that the positively charged α particles were being repelled by positive parts of the atom. Because only a very small percentage of alpha particles were deflected, only a very small region of each atom could be positively charged. That is, the positive charge in an atom could not be spread out more or less uniformly, as the plum pudding model assumed; instead it must be concentrated in a very small region. This implied that most of the atom is empty (remember the void of the ancient Greeks?) or occupied by something that poses little or no resistance to the passage of the α particles. What it left unexplained was why positively charged particles (which we now know as protons) concentrated in such a small volume, did not repel one another – the answer to which had to wait to discovery of the strong nuclear force (see below). Again we see a scientist making a huge intuitive leap from the experimental observation to a hypothesis that was consistent with that evidence and that makes specific predictions that can be confirmed or falsified by further experiment and observation. Rutherford’s model, which became known as the planetary model, postulated a very, very small nucleus where all of the positive charge and nearly all of the mass of the atom was located; this nucleus was encircled by electrons. In 1920 Rutherford went on to identify the unit of positive charge and called it the proton. In 1932 James Chadwick (1891–1974)(who co-incidentally studied at the University of Manchester) identified a second component of the nucleus, the neutron. Neutrons are heavy, like protons. In fact they are slightly heavier than protons, but have no charge. The identity of the element depends on the number of protons, however the number of neutrons may be different in different atoms of the same element. For example an atom of carbon always has six protons, but it can have different numbers of neutrons. Most carbon atoms have six neutrons ($\mathrm{C-} 12$), but some have seven ($\mathrm{C-} 13$) and some have eight ($\mathrm{C-} 14$).
Questions
Questions for Later
• If atoms are mostly empty space, why can’t we walk through walls?
• What is radiation?
• How does an atom change when it emits an alpha particle? Or a beta particle/electron?
Questions to Ponder
• If the original discoverers of electricity had decided that electrons have a positive charge, would that have made a difference in our understanding of electricity?
• Why do you think electrons were the first sub-atomic particles to be discovered?
• How exactly did Rutherford detect alpha particles?
• Can you think of an alternative model of the atom based on Rutherford’s observations?
• How would the experiment change if he had used electrons or neutrons? | textbooks/chem/General_Chemistry/CLUE%3A_Chemistry_Life_the_Universe_and_Everything/01%3A_Atoms/1.6%3A_The_Divisible_Atom.txt |
At this point we have arrived at a relatively simple model of the atom. Do not to worry, we will move to more complex and realistic models in the next chapter. In this simple model the atom has a very small but heavy nucleus that contains both protons and neutrons. As we talk about biology now and again, take care not to confuse the nucleus of an atom with the nucleus of a cell; they are completely different – besides the fact that they are of very different sizes. For example, there is no barrier round the nucleus of an atom—an atomic nucleus is a clump of protons and neutrons. Surrounding the atomic nucleus are electrons, in the same number as there are protons. The atom has no net electrical charge since the number of electrons is equal to the number of protons.
Where the electrons actually are in an atom, however, is a trickier question to answer, because of quantum mechanical considerations, specifically the Heisenberg uncertainty principle, which we will return to in the next chapter. For now we are going to assume the electrons are outside the nucleus and moving. We can think of them as if they were a cloud of electron density rather than particles whizzing around (don’t worry we will provide evidence for this model soon). This simple model captures important features and enables us to begin to consider how atoms interact with one another to form molecules and how those molecules can be rearranged—real chemistry!
There are four fundamental forces that we know about at the moment: gravity, the electromagnetic force, the strong nuclear force, and the weak nuclear force. For now we can largely ignore the strong nuclear force that is involved in holding the nucleus together: .it is an attractive force between neutrons and protons and is the strongest of all known forces in the universe, $\sim 137$ times stronger than the electromagnetic force. The strong nuclear force, acts at very short ranges, $\sim 10^{-15}$ m, or about the diameter of the nucleus. The other force involved in nuclear behavior, the weak force, plays a role in nuclear stability, specifically the stability of neutrons, but it has an even shorter range of action ($10^{-18}$ m). Because the nucleus is much smaller than the atom itself we can (and will) ignore the weak and strong nuclear forces when we consider chemical interactions. The force we are probably most familiar with is gravity, which is the weakest force, more than $10^{-37}$ times weaker than the electromagnetic force, and we can ignore it from the perspective of chemistry, although it does have relevance for the biology of dinosaurs, elephants, whales, and astronauts. The electromagnetic force is responsible for almost all the phenomena that we encounter in our everyday lives. While we remain grounded on the Earth because of the gravitational interaction between our body and the Earth, the fact that we don’t fall through to the center of the earth is entirely due to electromagnetic interactions. One obvious feature of the world that we experience is that it is full of solid things—things that get in each other’s way. If atoms and molecules did not interact with one another, one might expect to be able to walk through walls, given that atoms are mostly empty space, but clearly this is not the case. Similarly, your own body would not hold together if your atoms, and the molecules they form, failed to interact. As we will see, all atoms and molecules attract one another—a fact that follows directly from what we know about the structure of atoms and molecules.
Questions
Questions to Ponder
• What would a modern diagram of an atom look like and what could it be used to explain?
• Why don’t the protons within a nucleus repel one another?
• Why don’t the electrons and protons attract each other and end up in the nucleus?
• How the electrons within an atom interact?
Questions for Later
• Can an atom have chemical and/or physical properties; if so, what are they?
• What are chemical and physical properties? Can you give some examples?
• What distinguishes one element from another?
Interactions Between Atoms: A Range of Effects
The attractions and repulsions between charged particles and magnets are both manifestations of the electromagnetic force. Our model of the interactions between atoms will involve only electric forces; that is, interactions between electrically charged particles, electrons and protons. In order to understand this we need to recall from physics that when charged particles come close to each other they interact. You probably recall that “like charges repel and unlike charges attract”, and that this interaction, which is known as a Coulombic interaction, depends on the sizes and signs of the charges, and is inversely proportional to the square of the distance between them (this interaction can be modeled by the equation: $F = \alpha \frac{(q_1 x q_2 )}{r^2} \text { (Coulomb's Law).}$
where $q_{1}$ and $q_{2}$ are the charges on the particles and $r$ is the distance between them. That is: there is a force of attraction (or repulsion if the two charges are of the same sign) that operates between any two charged particles. This mathematical description of the electromagnetic interaction is similar to the interaction due to gravity. That is, for a gravitational interaction there must be at least two particles (e.g. you and the Earth) and the force of the attraction depends on both masses, and is inversely proportional to the square of the distance between them: $F = \alpha \frac{(m_1 x m_2 )}{r^2}$
The difference between the two forces are:
1. gravitational interactions are much weaker than electromagnetic interactions and
2. gravity is solely an attractive interaction while electromagnetic interactions can be either attractive or repulsive.
Now, let us consider how atoms interact with one another. Taken as a whole, atoms are electrically neutral, but they are composed of discrete electrically charged particles. Moreover, their electrons behave as moving objects.[19] When averaged over time the probability of finding an electron is spread uniformly around an atom, the atom is neutral. At any one instant, however, there is a non-zero probability that the electrons are more on one side of the atom than the other. This results in momentary fluctuations in the charge density around the atom and leads to a momentary charge build up; for a instant one side of the atom is slightly positive ($delta +$) and the other side is slightly negative ($delta -$). This produces what is known as an instantaneous and transient electrical dipole – that is a charge separation. As one distorted atom nears another atom it affects the second atom’s electron density distribution and leads to what is known as an “induced dipole”. So, for example, if the slightly positive end of the atom is located next to another atom, it will attract the electron(s) in the other atom. This results in an overall attraction between the atoms that varies as $\frac{1}{r^{6}}$ – where $r$ is the distance between the atoms. Note that this is different than the attraction between fully charged species, the Coulombic attraction, which varies as $\frac{1}{r^{2}}$. What does that mean in practical terms? Well, most importantly it means that the effects of the interaction will be felt only when the two atoms are quite close to one another.
As two atoms approach, they will be increasingly attracted to one another. But this attraction has its limit – when the atoms get close enough, the interactions between the negatively charged electrons (and positively charged nuclei) of each atom increase very rapidly, which leads to an overall repulsion,which will stop the two atoms approaching so closely.
A similar effect was in also seen in Rutherford’s experiment. Recall that he accelerated positively charged alpha particles toward a sheet of gold atoms. As an alpha particle approaches a gold atom’s nucleus, the positively (+2) charged alpha particle and the gold atom’s positively (+79) nucleus begin to repel each other. If no other factors were involved, the repulsive force would approach infinity as the distance between the nuclei ($r$) approached 0. (You should be able to explain why.) But infinite forces are not something that happens in the macroscopic, atomic, or subatomic worlds, if only because the total energy in the universe is not infinite. As the distance between the alpha particle and gold nucleus approaches zero, the repulsive interaction grows strong enough to slow the incoming alpha particle and then push it away from the target particle. If the target particle is heavy compared to the incoming particle, as it was in Rutherford’s experiments, the target, composed of gold atoms that weigh about 50 times as much as the alpha particle, will not move much while the incoming alpha particle will be reflected away. But, if the target and incoming particle are of similar mass, then both will be affected by the interaction and both will move. Interestingly, if the incoming particle had enough initial energy to get close enough (within $\sim 10^{-15}$ m) to the target nucleus, then the strong nuclear force of attraction would come into play and start to stabilize the system. The result would be the fusion of the two nuclei and the creation of a different element, a process that occurs only in very high-energy systems such as the center of stars or during a stellar explosion, a supernova. We return to this idea in Chapter $3$.
Questions
Questions to Answer
• How does the discovery that atoms have parts alter Dalton’s atomic theory?
• What would the distribution of alpha particles, relative to the incident beam, look like if the positive nucleus took up the whole atom (sort of like the plum pudding)? What if it took up 50% of the atom?
• What does the distribution of alpha particles actually look like (recall that 1 in every 8000 particles were deflected)?
Forces and Energy: an overview.
We would like to take some time to help you think about the interactions (forces) between atoms and molecules, and how these interactions lead to energy changes. These energy changes are responsible for the formation of molecules, their reorganization through chemical reactions, and the macroscopic properties of chemical substances (i.e. everything). While you may have learned about forces and energy in your physics classes, most likely these concepts were not explicitly related to how things behave at the atomic-molecular level. We are going to begin with a discussion of the interactions and energy changes that result from the force of gravity, because these ideas are almost certainly something you are familiar with, certainly more familiar with than electromagnetic interactions – but the purpose of this section is to help you make the connections between what you already know (at the macroscopic level), and how these ideas are transferred to the molecular level, including similarities and differences. For example, Newton’s Laws of Motion describe how objects behave when they come into contact, say when a baseball comes in contact with a bat. But often objects interact with one another at a distance. After the ball is hit, its movements are determined primarily by its gravitational interactions with all other objects in the Universe, although because of the nature of the gravitational interaction, by far the most important interaction is between the ball and the Earth (see below).
A force is an interaction between objects that causes a pull (attraction) or a push (repulsion) between those objects. When such an interaction occurs, there is a change in energy of the objects. As noted above, there are four fundamental forces: gravitational, electromagnetic, the strong and the weak nuclear forces. We will have more to say about the electromagnetic force that is relevant for understanding chemical interactions, that is how atoms and molecules behave. Many of the phenomena you are familiar with are based on electromagnetic forces. For example, electromagnetic forces stop the ball from going through the bat – or you from falling down to the center of the Earth.
Now let us consider what happens when you throw a ball straight up into the air. You apply a force to the ball (through the action of your muscles), and once it leaves your hand the only force acting on the ball is gravity (we are, of course, ignoring friction due to interactions with the molecules in the air). The ball, initially at rest, starts moving upward. Over time, you observe the velocity of the ball changes, as the ball slows, stops and falls back to earth. So what forces cause these changes? The answer is the force of gravity, which is a function of the masses of the ball and the Earth, which do not change over time, and the distance ($r$) between the Earth and the ball, which does. This gravitational force $F$, can be modeled by an equation that shows it is proportional to the product of the masses of the ball ($M_{1}$) and the Earth ($M_{2}$) divided by the square of the distance between the objects ($r$).[20]
In gravitational interactions, the force decreases as the distance between the objects increases (the decrease is proportional to $\frac{1}{r^{2}}$), which means the further away you get from the Earth the smaller is the attractive force between you and the Earth. If you get far enough away, and you are moving away from the Earth, the interaction will not be enough to keep you attracted to the Earth and you will continue to move away forever.
Of course, why objects with mass attract each other is a subject for physics – beyond the scope of this course.[21] What we can say is that the force is mediated by a gravitational field. Any object with mass will interact with other objects with mass through this field. The field can also be said to transfer energy through space between two (or more) objects. That is, the interaction leads to an energy change in the system of interacting objects. In chemistry we are concerned with both the forces that cause interactions and the energy changes that result.
How do forces influence energy?
If we take our macroscopic example of your throwing a ball upwards, we know that you transfer some energy to the ball. Of course this begs the question “what do we mean by energy?” and unfortunately we do not have an easy answer, in fact Richard Feynman once famously said “in physics we have no idea of what energy is”. Physicists might say energy is the capacity to do work, and then define work as force times distance, which does not really get us anywhere, especially in chemistry where the notion of work is often not helpful. What we can say is that any changes are accompanied by energy changes, and that we can calculate or measure these energy changes.[22]
You may be familiar with what are often referred to as “forms of energy”, such as mechanical, or elastic, or chemical, but at the most basic level all forms of energy we will be concerned with can be described either as kinetic energy, potential energy, or electromagnetic energy (e.g. light). Kinetic energy is often called the energy of motion ($\mathrm{KE}=1 / 2 mv^{2}$, where $m$ is the mass and $v$ the velocity of the object), and potential energy the energy of position, or stored energy (it is calculated in various ways as we will see). Changes between kinetic and potential forms of energy involve forces. The ball that you throw straight up and then comes down has changing amounts of kinetic energy (it changes as the velocity of the ball changes) and potential energy (which changes as the distance between the Earth and the ball changes.) As the ball rises, you can observe that the velocity of the ball decreases, and therefore the $\mathrm{KE}$ decreases. At the same time the $\mathrm{PE}$ increases since the distance between the Earth and ball is increasing. On the way down the opposite is true, the ball starts moving faster – the $\mathrm{KE}$ increases and the $\mathrm{PE}$ decreases. Recall the principle of the conservation of energy; after the ball leaves your hand, no energy is added or taken away as the ball is traveling, if one form of energy increases, the other must decrease.
Another important point about energy is that it is a property of a system, rather than of an object. Although it may be tempting to consider that a ball in motion has a certain amount of kinetic energy it is important to remember the frame of reference from which you are considering the ball. Certainly the ball’s velocity is related to the $\mathrm{KE}$, but that velocity depends upon where you are viewing the ball from. Usually (almost always) we consider the velocity from the point of view of an observer who is stationary, but if we changed the system we were considering, and viewed the ball while we were also moving, then the velocity of the ball would be different. This may seem quite an abstract point, but it is an important one.
Similarly it is quite tempting to say that the ball has potential energy, but in fact this is also not entirely accurate. It is more accurate – and more useful – to say that the system of the ball and the Earth has potential energy – again we are taking a systems perspective here. Unlike kinetic energy, the potential energy in a system also depends on the force that is acting on it, and that force is a function of the position of the objects that are interacting within the gravitational field. For example, a “frictionless” object traveling through a space free of fields (gravitational or otherwise) at a constant velocity has a constant kinetic energy, but no potential energy.
Potential energy (often called stored energy) or the energy of position, raises the question – where is the energy “stored”? A useful way to think about this is that for the example of the ball and the Earth, this energy is stored in the gravitational field. In this way we can accommodate the idea that the $\mathrm{PE}$ depends on the distance between the two interacting objects. It will also allow us to generate a more overarching concept of potential energy that will be useful in chemistry, as we extend these ideas to interactions of atoms and molecules. You might ask why then is it OK to say an object has kinetic energy (as long as you specify the frame of reference), and the difference here is that any object in motion can have energy associated with it (for example, you, an atom or a car), but potential energy must be associated with objects that are interacting via a field, be it gravitational or electromagnetic. That said, fields are everywhere – there is no place in the universe where there are no fields (although they can be balanced, leaving the net force zero). What is important here is that
1. you understand that objects interact,
2. that these interactions cause a change in energy of the system, and
3. that the interacting forces depend on the distance between the interacting objects (as well as other factors, such as mass, which are constant).
The electromagnetic force:
While gravitational interactions are, for all intents and purposes, irrelevant in chemistry (except to hold the beaker down on the lab bench!) they do provide a familiar example of the relationship between the kinetic and potential energies of a system that we can use to explore the electromagnetic interactions that are responsible for the behavior of atoms and molecules. There are some important similarities between gravitational and electromagnetic interactions; both act at a distance, both are mediated by fields, and both display the same relationship between force and distance. There are also important differences. In the context of chemistry, electromagnetic interactions are much stronger and while gravity is always attractive, electromagnetic interactions can be either attractive or repulsive.[23]
All electrically charged objects interact via electromagnetic forces. As we have already seen (and will to return to again) atoms and molecules are made up of charged particles (electrons and protons) and these produce unequal charge distributions that lead to the same kinds of interactions. The strength of these interactions between charged particles can be modeled using an equation, Coulomb’s Law. You will note that its form is similar to Newton’s Law of Gravitation. Instead of the masses of the two interacting objects, however, the electromagnetic force depends on the charges on the two particles ($q_{1}$ and $q_{2}$). The electromagnetic force typically acts over much shorter distances than gravitation, but is much stronger. It is the force that affects interactions of atoms and molecules.
As with the gravitational force as the charged particles get closer together, the interaction (whether attractive or repulsive) gets stronger. Just like gravity, the interaction between charged particles is mediated by a field, which transfers energy between interacting objects. We can identify (and calculate) the types of energy changes that are occurring as the particles interact. For example two oppositely charged particles are attracted to each other. As they approach one another, the force of attraction becomes stronger, the particles will move faster – and their kinetic energies increase. Given the fact that energy is conserved, the potential energy of the system of particles must decrease to a similar extent.[24] If, on the other hand the two charges are of the same sign, then the force between them is repulsive. So if two particles of the same charge are moving toward each other, this repulsive force will decrease their velocity (and kinetic energy), and increase their potential energy. As the distance between the particles decreases, the repulsion will eventually lead to the two particles moving away from one another.
Of course you may have noticed that there is a little problem with the equations that describe both gravitation and electromagnetic forces. If the forces change as $r$ decreases, what happens as the distance between the interacting objects approaches zero? If we were to rely on the equations we have used so far, as $r$ approaches 0, the force (whether repulsive or attractive) would approach infinity. Clearly something is wrong here since infinite forces are not possible (do you know why?). The ball is stopped by the surface of the Earth – it does not plummet to the center of the Earth, and charged particles do not merge into each other (or fly away at infinite speed). What is it that we are missing? Well, the problem lies in the idea that these equations are really dealing with idealized situations such as point charges or masses, rather than taking into account the fact that matter is made up of atoms, molecules and ions. When two atoms, or two molecules (or two particles made up of atoms or molecules) approach each other, they will eventually get close enough that the repulsions between like charges will become stronger than the attractive forces between unlike charges. As we will see, when two macroscopic objects appear to touch, they do not really – what stops them is the electron-electron repulsions of the atoms on the surface of the objects.[25] We will revisit all these ideas as we discuss how atoms and molecules interact at the atomic-molecular level, and how electrons behave (quantum mechanically).
Interacting Atoms: Forces, Energy Conservation and Conversion
Let us step back, collect our thoughts, and reflect on the physics of the situation. First, remember that the total matter and energy of an isolated system are conserved; that is the first law of thermodynamics. As we mentioned above, while energy and matter can, under special circumstances, be interconverted, typically they remain distinct. That means in most systems the total amount of matter is conserved and the total amount of energy is conserved, and that these are separate.
So let us consider the situation of atoms or molecules in a gas. These atoms and molecules are moving randomly in a container, colliding with one another and the container’s walls. We can think of the atoms/molecules as a population. Population thinking is useful for a number of phenomena, ranging from radioactive decay to biological evolution. For the population of atoms/molecules as a whole, there is an average speed and this average speed is a function of the temperature of the system.[26] If we were to look closely at the population of molecules, however, we would find that some molecules are moving very fast and some are moving very slowly; there is a distribution of speeds and velocities (speed + direction).
As two atoms/molecules approach each other they will feel the force of attraction caused by the electron density distortions, these are known as London dispersion forces, which we will abbreviate as LDFs. The effects of these LDFs depend on the strength of the interaction (that is the magnitude of the charges and the distance between them) and on the kinetic energies of the atoms and molecules. LDF are one of a number of intermolecular forces (IMFs), which we will consider later. LDFs are the basis of van der Waals interactions in biological systems.
To simplify things we are going to imagine a very simple system: assume for the moment that there are just two isolated atoms, $\text{atom}_{1}$ and $\text{atom}_{2}$. The atoms are at rest with respect to one another, but close enough that the LDF-based attractive interactions between them are significant. For this to occur they have to be quite close, since such attractive interactions decreases rapidly, as $\frac{1}{r^{6}}$ where $r$ is the distance between the two atoms. At this point, the system, which we will define as the two atoms, has a certain amount of energy. The exact amount does not matter, but as long as these two atoms remain isolated, and do not interact with anything else, the energy will remain constant.
So what does all this have to do with atoms approaching one another? We can use the same kinds of reasoning to understand the changes in energy that occur as the atoms approach each other. Initially, the system will have a certain amount of energy (kinetic + potential). If the atoms are close enough to feel the effects of the attractive LDFs, they begin to move toward each other, think of a ball falling towards the Earth, and some of the potential energy associated with the atoms’ initial state is converted into kinetic energy ($\mathrm{E}_{\mathrm{K}}=1 / 2 mv^{2}$).
As they approach each other the LDFs grow stronger, the atoms are more strongly attracted to each other; the system’s potential energy decreases and is converted into kinetic energy, the atoms move faster.[27] The total energy remains the same as long as there are no other atoms around. This continues until the atoms get close enough that repulsive interactions between the electrons become stronger and as they approach even more closely the repulsive interactions between the positively charged nuclei also come into play, causing the potential energy in the system to rise. As the atoms begin to slow down their kinetic energy is converted back into potential energy. They will eventually stop and then be repelled from one another. At this point potential energy will be converted back into kinetic energy. As they move away, however, repulsion will be replaced by attraction and they will slow; their kinetic energy will be converted back into potential energy.[28] With no other factors acting within the system, the two atoms will oscillate forever. In the graph showing potential energy versus the distance between the atoms, we see that the potential energy of the system reaches a minimum at some distance. Closer than that and the repulsive electromagnetic forces come into play, further away and the attractive electromagnetic forces (LDF’s) are dominant. The distance between the two atoms is a function of the relative strengths of the attractive and repulsive interactions. However, even at the minimum, there is some potential energy in the system, stored in the electromagnetic field between the two atoms. At temperatures above absolute zero ($0 \mathrm{ K}$), the pair of atoms will also have kinetic energy – as they oscillate back and forth.
Here we have a core principle that we will return to time and again: a stabilizing interaction always lowers the potential energy of the system, and conversely a destabilizing interaction always raises the potential energy of the system. In an isolated system with only two atoms, this oscillation would continue forever because there is no way to change the energy of the system. This situation doesn’t occur in real life because two-atom systems do not occur. For example, even in a gas, where the atoms are far apart, there are typically large numbers of atoms that have a range of speeds and kinetic energies present in the system. These atoms frequently collide and transfer energy between one another. Therefore, when two atoms collide and start to oscillate, some energy may be transferred to other particles by collisions. If this happens, a stable interaction can form between the two particles; they will “stick” together. If more particles approach, they can also become attracted, and if their extra energy is transferred by collisions, the particles can form a bigger and bigger clump.
As we discussed earlier, LDFs arise due to the fluctuations of electron density around nuclei and are a feature common to all atoms; all atoms/molecules attract one another in this manner. The distance between atoms/molecules where this attraction is greatest is known as the van der Waals radius of the atom/molecule. If atoms/molecules move closer to one another than their van der Waals radii they repel one another. The van der Waals radius of an atom is characteristic for each type of atom/element. As mentioned earlier, it is only under conditions of extreme temperature and pressure that the nuclei of two atoms can fuse together to form a new type of atom; such a nuclear/atomic fusion event results in the interconversion of matter into energy.[29]
Questions
Questions to Answer
• What is potential energy? Can you provide an example?
• What is kinetic energy? Can you provide an example?
• At the atomic level, what do you think potential energy is?
• At the atomic level, what do you think kinetic energy is?
• Why does raising the temperature affect the speed of a gas molecule?
Questions to Ponder
• What is energy (have your ideas changed from before)?
Questions for Later:
• When we talk about potential energy of a system, what does system mean?
• Helium liquefies at around 4K. What makes the helium atoms stick together? (Why don’t they turn into a gas?)
• Consider two atoms separated by 1 spatial unit versus 4 spatial units. How much weaker is the interaction between the more distant atoms? How does that compared to the behavior of simple charges (rather than atoms)? | textbooks/chem/General_Chemistry/CLUE%3A_Chemistry_Life_the_Universe_and_Everything/01%3A_Atoms/1.7%3A_Interactions_Between_Atoms_and_Molecules.txt |
Now let’s take a look at a couple of real systems. We begin by considering interactions between the simplest atoms, hydrogen ($\mathrm{H}$) and helium ($\mathrm{He}$), and the simplest molecule, molecular hydrogen ($\mathrm{H}_{2}$). A typical hydrogen atom consist of one proton and one electron, although some contain one or two neutrons and form “isotopes” known as deuterium and tritium, respectively. A hydrogen molecule is a completely different chemical entity: it contains two hydrogen atoms, but its properties and behavior are quite different. Helium atoms have 2 protons and 2 neutrons in their nuclei, and 2 electrons in their electron clouds. We will consider more complicated atoms and molecules after we discuss atomic structure in greater detail in the next chapter. One advantage of focusing on molecular hydrogen and helium is that it also allows us to introduce, compare, and briefly consider both van der Waals interactions (due to IMFs) and covalent bonds; we will do much more considering later on.
When two atoms of helium approach each other LDFs come into play and a attractive interaction develops. In the case of $\mathrm{He}$ the drop in potential energy due to the interaction is quite small, that is, the stabilization due to the interaction, and it does not take much energy to knock the two atoms apart. This energy is delivered by collisions with other $\mathrm{He}$ atoms. In fact at atmospheric pressures, Helium is never a solid and liquid $\mathrm{He}$ boils at $\sim 4 \mathrm{~K}\left(-268.93^{\circ} \mathrm{C}\right)$, only a few degrees above absolute zero or $\sim 0 \mathrm{~K}\left(-273.15^{\circ} \mathrm{C}\right)$.[30] This means that at all temperatures above $\sim 4 \mathrm{~K}$ there is enough kinetic energy in the atoms of the system to disrupt the interactions between He atoms. The weakness of these interactions means that at higher temperatures, above $4 \mathrm{~K}$, helium atoms do not “stick together”. Helium is a gas at temperatures above $4 \mathrm{~K}$.
Now let us contrast the behavior of helium with that of hydrogen ($\mathrm{H}$). As two hydrogen atoms approach one another they form a much more stable interaction, about 1000 times stronger than the $\mathrm{He-He}$ London dispersion forces. In an $\mathrm{H-H}$ interaction the atoms are held together by the attraction of each nucleus for both electrons. The attractive force is much stronger and as the atoms get closer this leads to a larger drop in potential energy and a minimum for the two interacting hydrogen atoms that is much deeper than that for $\mathrm{He-He}$. Because of its radically different stability the $\mathrm{H-H}$ system gets a new name; it is known as molecular hydrogen or $\mathrm{H}_{2}$ and the interaction between the $\mathrm{H}$ atoms is known as a covalent bond. In order to separate a hydrogen molecule back into two hydrogen atoms, that is, to break the covalent bond, we have to supply energy.[31] This energy can take several forms: for example, energy delivered by molecular collisions with surrounding molecules or by the absorption of light both lead to the breaking of the bond.
Each $\mathrm{H}$ can form only a single covalent bond, leading to the formation of $\mathrm{H-H}$ molecules, which are often also written as $\mathrm{H}_{2}$ molecules. These $\mathrm{H-H}$ molecules are themselves attracted to one another through LDFs. We can compare energy associated with the $\mathrm{H-H}$ covalent bond and the $\mathrm{H}_{2} - \mathrm{H}_{2}$ IMF. To break a $\mathrm{H-H}$ covalent bond one needs to heat the system to approximately $5000 \mathrm{~K}$. On the other hand to break the intermolecular forces between separate $\mathrm{H}_{2}$ molecules, the system temperature only needs to rise to $\sim 20 \mathrm{~K}$; above this temperature $\mathrm{H}_{2}$ is a gas. At this temperature the IMFs between individual $\mathrm{H}_{2}$ molecules are not strong enough to resist the kinetic energy of colliding molecules. Now you may ask yourself, why does $\mathrm{H}_{2}$ boil at a higher temperature than $\mathrm{He}$? Good question! It turns out that the strengths of LDFs depend on several factors including shape of the molecule, surface area, and number of electrons. For example the greater the surface areas shared between interacting atoms or molecules the greater the LDFs experienced and the stronger the resulting interaction. Another factor is the ability of the electron cloud to become charged, a property known as polarizability. You can think of polarizability as the floppiness of the electron cloud. As a rough guide, the further away from the nucleus the electrons are, the more polarizable (floppy) the electron cloud becomes. We will return to this and related topics later on. As we will see, larger molecules with more complex geometries, such as biological macromolecules (proteins and nucleic acids), can interact through more surface area and polarizable regions, leading to correspondingly stronger interactions.
At this point, you are probably (or should be) asking yourself some serious questions, such as, why don’t helium atoms form covalent bonds with one another? Why does a hydrogen atom form only one covalent bond? What happens when other kinds of atoms interact? To understand the answers to these questions, we need to consider how the structure of atoms differs between the different elements, which is the subject of the next chapter.
Questions
Questions to Answer
• Can you draw a picture (with about 20 helium atoms, represented as circles) of what solid helium would look like if you could see it?
• How would that differ from representations of liquid helium or gaseous helium?
• Now make a similar drawing of $\mathrm{H}_{2}$. Does this help explain the higher melting point of $\mathrm{H}_{2}$?
Question to Ponder
• How do the properties of solids, liquids, and gases differ? | textbooks/chem/General_Chemistry/CLUE%3A_Chemistry_Life_the_Universe_and_Everything/01%3A_Atoms/1.8%3A_Interactions_Between_Helium_Atoms_and_Hydrogen_Molecules.txt |
1. Scale of the universe: http://htwins.net/scale2/
2. http://www.iep.utm.edu/t/thales.htm
3. Of course if you know your movies, you know that the “Fifth Element” is love.
4. http://plato.stanford.edu/entries/democritus/
5. First description of Brownian motion - Epicurus
6. A History of Greek Philosophy by William Keith Chambers Guthrie. p. 212.
7. http://www.chm.bris.ac.uk/webproject...b/history.html and http://en.Wikipedia.org/wiki/Chemistry_(etymology)
8. An important event was the rediscovery by Poggio of Lucretius’s “On the Nature of Things,” a poem centered on the atomic nature of the universe (see The Swerve by Stephen Greenblatt). One reason Giordano Bruno was burnt at the stake was the fact that he took these ideas seriously.
9. http://elements.vanderkrogt.net/elem/p.html
10. The Philosopher’s stone was thought to be able to turn base (common) metals into gold, and perhaps even be the key to everlasting life. It was the ultimate goal of the alchemists. Interestingly the first Harry Potter book was titled Harry Potter and the Philosopher’s Stone in England but was re-titled in America, because the publishers thought that American children would not be interested in a book with this title, perhaps due a failure to appreciate the importance of philosophy.
11. Religious dissenters, that is, non-Anglicans, were not allowed access to English universities at that time.
12. An extraordinary number of discoveries related to the structure of the atom were made by scientists in or from Manchester. There must be something in the air there. It is, of course, completely fortuitous that one of the authors was also born and bred in Manchester!
13. https://en.Wikipedia.org/wiki/Law_of...le_proportions
14. That is, a person from Manchester, England.
15. This works because the electrons are spinning.
16. http://www.aip.org/history/electron/jjsound.htm
17. A term made popular (although often misunderstood) by T. S. Kuhn, The Structure of Scientific Revolutions, 1st. ed., Chicago: Univ. of Chicago Pr., 1962
18. This can be a little confusing to those not familiar with plum pudding – a “delicious” English delicacy composed of dried fruit (raisins) in a spongy base, usually prepared by boiling for several days and often served with rum sauce.
19. Yes we did tell you to think of electrons as a cloud - because this is a helpful model - but electrons are both particles and “clouds” as we will discuss later, in fact in some instances they appear to be quite close to perfect spheres in shape, In fact “The experiment, which spanned more than a decade, suggests that the electron differs from being perfectly round by less than $0.000000000000000000000000001$ cm. This means that if the electron was magnified to the size of the solar system, it would still appear spherical to within the width of a human hair. (Hudson et al "Improved measurement of the shape of the electron" DOI: 10.1038/nature10104).
20. See http://www.youtube.com/watch?v=p_o4aY7xkXg for an excellent explanation of this phenomenon.
21. That said, we recommend the discription given in Einstein and Infeld’s Evolution of Physics: https://archive.org/details/evolutionofphysi033254mbp
22. The trouble with chemical energy: why understanding bond energies requires an interdisciplinary systems approach. CBE Life Sci. Education,12:306-12.
23. Magnetic, like electrical force can also be attractive or repulsive. Most of us have played with magnets and felt the force of attraction between a north and south pole of a set of magnets, which gets stronger as the magnets get closer together, and the repulsion between two north poles which also gets stronger as the magnets get closer together.
24. A point we have not considered is why the atoms or molecules stop moving toward each other, which will return to shortly.
25. http://www.youtube.com/watch?v=BksyMWSygnc
26. Remember speed is a directionless value, while velocity involves both speed and direction.
27. Imagine, as an analogy that the two atoms are balls rolling down opposite sides of a hill towards a valley, their potential energy falls as they move down - but their kinetic energy rises and they speed up.
28. To continue our analogy as the balls get to the bottom of the hill, they collide and bounce back - rolling back up the hill, until once again the force of gravity takes over and they start to roll back down. In an ideal (unreal) situation with no friction, this situation would simply continue, until some other factor is introduced.
29. It is these factors that made the report of cold fusion so strange and so exciting to physicists. The temperatures and pressures required for fusion are so high that they are extremely difficult to achieve under controlled conditions. The failure to reproduce the original cold fusion report reinforces our understanding of how atoms interact. That scientists around the world attempted to reproduce the original observation (and failed), illustrates the open-mindedness of the scientific community. The fact that badly controlled and irreproducible observations were published, illustrates how scientific effort and resources (that is, research funds) can be wasted by inadequate pre-publication review. But science, like all human activities, is imperfect. The price for open-mindedness may be be wasted time and effort, yet it remains critical to scientific process and progress. At the same time, once the replication efforts failed, it became a waste of time (or a delusional obsession) to pursue cold fusion.
30. According to Robert Parson, “At $1$ atmosphere pressure, Helium does not melt at ANY temperature - it stays liquid down to absolute zero. (If you want to be picky, it is a liquid down to the lowest temperatures that anyone has ever achieved, which are orders of magnitude less than $1 \mathrm{~K}$ (http://en.Wikipedia.org/wiki/Dilution_refrigerator), and our best theories predict that it will remain a liquid no matter how low the temperature.) To get solid helium you have to increase the pressure to $25$ atmospheres or above. This is one of the most dramatic consequences of zero-point energy: the intermolecular forces in He are so weak that it melts under its own zero point energy. (This leads to the peculiar consequence that Helium at zero Kelvin is a liquid with zero entropy.)
31. In fact this is known as the bond energy – the energy required to break the bond – which in the case of $\mathrm{H}_{2}$ is $432 \mathrm{~kJ} / \mathrm{mol}$. | textbooks/chem/General_Chemistry/CLUE%3A_Chemistry_Life_the_Universe_and_Everything/01%3A_Atoms/1.9%3A_In-Text_References.txt |
Even as he articulated his planetary model of the atom, Rutherford was aware that there were serious problems with it. For example because like charges repel and unlike charges attract, it was not at all clear why the multiple protons in the nuclei of elements heavier than hydrogen did not repel each other and cause the nuclei to fragment. What enables them to stay so close to each other? On the other hand, if electrons are orbiting the nucleus like planets around the Sun, why don’t they repel each other, leading to quite complex and presumably unstable orbits? Why aren’t they ejected spontaneously and why doesn’t the electrostatic attraction between the positively-charged nucleus and the negatively-charged electrons result in the negatively-charged electrons falling into the positively charged nucleus? Assuming that the electrons are moving around the nucleus, they are constantly accelerating (changing direction). If you know your physics, you will recognize that (as established by J.C. Maxwell – see below) a charged object emits radiation when accelerating.[1] As the electron orbits the nucleus this loss of energy will lead it to spiral into the nucleus – such an atom would not be stable. But, as we know, most atoms are generally quite stable.
So many questions and so few answers! Clearly Rutherford’s model was missing something important and assumed something that cannot be true with regard to forces within the nucleus, the orbital properties of electrons, and the attractions between electrons and protons. To complete this picture leads us into the weird world of quantum mechanics.
Thumbnail: Covalently bonded hydrogen and carbon in a w:molecule of methane. (CC BY-SA 2.5; DynaBlast via Wikipedia)
02: Electrons and Orbitals
While Rutherford and his colleagues worked on the nature of atoms, other scientists were making significant progress in understanding the nature of electromagnetic radiation, that is, light. Historically, there had been a long controversy about the nature of light, with one side arguing that light is a type of wave, like sound or water waves, traveling through a medium like air or the surface of water and the other side taking the position that light is composed of particles. Isaac Newton called them corpuscles. There was compelling evidence to support both points of view, which seemed to be mutually exclusive, and the attempt to reconcile these observations into a single model proved difficult.
By the end of the 1800s, most scientists had come to accept a wave model for light because it better explained behaviors such as interference[2] and diffraction,[3] the phenomena that gives rise to patterns when waves pass through or around objects that are of similar size to the wave itself. James Clerk Maxwell (1831–1879) developed the electromagnetic theory of light, in which visible light and other forms of radiation, such as microwaves, radio waves, X-rays, and gamma rays, were viewed in terms of perpendicular electric and magnetic fields. A light wave can be described by defining its frequency ($ν$) and its wavelength ($\lambda$). For all waves, the frequency times its wavelength equals the velocity of the wave. In the case of electromagnetic waves, $\lambda ν = c$, where $c$ is the velocity of light.
Although the wave theory explained many of the properties of light, it did not explain them all. Two types of experiments in particular gave results that did not appear to be compatible with the wave theory. The first arose during investigations by the German physicist Max Planck (1858–1947) of what is known as black body radiation. In these studies, an object heated to a particular temperature emits radiation. Consider your own body, which typically has a temperature of approximately $98.6^{\circ} \mathrm{F}$ or $36^{\circ} \mathrm{C}$. Your body emits infrared radiation that can be detected by some cameras.[4] Some animals, like snakes, have infrared detectors that enable them to locate their prey—typically small, warm-blooded, infrared-light-emitting mammals.[5] Because mammals tend to be warmer than their surroundings, infrared vision can be used to find them in the dark or when they are camouflaged.
Planck had been commissioned by an electric power company to produce a light bulb that emitted the maximum amount of light using the minimum amount of energy. In the course of this project he studied how the color of the light emitted (a function of its wavelength) changed as a function of an object’s (such as a light bulb filament) temperature. We can write this relationship as $\lambda \text { (wavelength) } = f(t)$ where $t =$ temperature and $f$ indicates “function of.” To fit his data Planck had to invoke a rather strange and non-intuitive idea, namely that matter absorbs and emits energy only in discrete chunks, which he called quanta. These quanta occurred in multiples of $E \text{ (energy) } = hν$, where $h$ is a constant, now known as Planck’s constant, and $ν$ is the frequency of light. Planck’s constant is considered one of the fundamental numbers that describes our universe.[6] The physics that uses the idea of quanta is known as quantum mechanics.
One problem with Planck’s model, however, is that it disagreed with predictions of classical physics; in fact as the frequency of the light increased, his measurements diverged more and more from the predictions of the then current, wave-based theory.[7] This divergence between classical theory and observation became known, perhaps over-dramatically, as the ultraviolet catastrophe. It was a catastrophe for the conventional theory because there was no obvious way to modify classical theories to explain Planck’s observations; this was important because Planck’s observations were reproducible and accurate. Once again, we see an example of the rules of science: a reproducible discrepancy, even if it seems minor, must be addressed or the theory must be considered either incomplete or just plain wrong.
The idea that atoms emit and absorb energy only in discrete packets is one of the most profound and revolutionary discoveries in all of science, and set the stage for a radical rethinking of the behavior of energy and matter on the atomic and subatomic scales. Planck himself proposed the idea with great reluctance and spent a great deal of time trying to reconcile it with classical theories of light. In the next section we will see how this property can be used to identify specific types of atoms, both in the laboratory and in outer space.
Questions
Questions to Answer
• What is a constant? What is a function?
• What happens to the energy of a photon of light as the frequency increases? What about as the wavelength increases? (remember: $\lambda ν = c$)
• Why is it difficult to detect cold-blooded animals using infrared detectors?
Questions to Ponder
• How can the phenomena of diffraction and interference be used as evidence that light behaves like it a wave?
• How can light be both a wave and a particle?
• Is light energy? | textbooks/chem/General_Chemistry/CLUE%3A_Chemistry_Life_the_Universe_and_Everything/02%3A_Electrons_and_Orbitals/2.1%3A_Light_and_Getting_Quantum_Mechanical.txt |
In 1905, Albert Einstein used the idea of quanta to explain the photoelectric effect, which was described by Philipp Lenard (1862-1947). The photoelectric effect occurs when light shines on a metal plate and electrons are ejected, creating a current.[8] Scientists had established that there is a relationship between the wavelength of the light used, the type of metal the plate is made of, and whether or not electrons are ejected. It turns out that there is a threshold wavelength (energy) of light that is characteristic for the metal used, beyond which no electrons are ejected. The only way to explain this is to invoke the idea that light comes in the form of particles, known as photons, that also have a wavelength and frequency (we know: this doesn’t make sense, but bear with us for now). The intensity of the light is related to the number of photons that pass by us per second, whereas the energy per photon is dependent upon its frequency or wavelength, because wavelength and frequency of light are related by the formula $\lambda ν = c$ where $c$ is the speed of light in a vacuum, is a constant and equal to $\sim 3.0 \times 10^{8} \mathrm{ m/s}$. The higher the frequency $ν$ (cycles per second, or Hertz), the shorter the wavelength $\lambda$ (length per cycle) and the greater the energy per photon. Because wavelength and frequency are inversely related–that is, as one goes up the other goes down– energy is directly related to frequency by the relationship $\mathrm{E} = hν$ or inversely related to the wavelength $\mathrm{E}= \frac{hc}{\lambda}$, where $h$ is Planck’s constant. So radiation with a very short wavelength, such as x rays ($\lambda = \sim 10^{-10} \mathrm{ m}$) and ultraviolet light (between $10^{-7} \text{ to } 10^{-8} \mathrm{ m}$), have much more energy per particle than long wavelength radiation like radio and microwaves ($\lambda = \sim 10^{3} \mathrm{ m}$). This is why we (or at least most of us) do not mind being surrounded by radio waves essentially all the time yet we closely guard our exposure to gamma rays, X-rays, and UV light; their much higher energies cause all kinds of problems with our chemistry, as we will see later.
Because of the relationship between energy and wavelength ($\lambda ν = c$), when you shine long-wavelength, low energy, such as infrared, but high intensity (many photons per second) light on a metal plate, no electrons are ejected. But when you shine short-wavelength, high energy (such as ultraviolet or x rays) but low intensity (few photons per second) light on the plate, electrons are ejected. Once the wavelength is short enough (or the energy is high enough) to eject electrons, increasing the intensity of the light now increases the number of electrons emitted. An analogy is with a vending machine that can only accept quarters; you could put nickels or dimes into the machine all day and nothing will come out. The surprising result is that the same total amount of energy can produce very different effects. Einstein explained this observation (the photoelectric effect) by assuming that only photons with “enough energy” could eject an electron from an atom. If photons with lower energy hit the atom no electrons are ejected – no matter how many photons there are.[9] You might ask: Enough energy for what? The answer is enough energy to overcome the attraction between an electron and the nucleus. In the photoelectric effect, each photon ejects an electron from an atom on the surface of the metal. These electrons exist somewhere within the atoms that make up the metal (we have not yet specified where) but it takes energy to remove them and the energy is used to overcome the force of attraction between the negative electron and the positive nucleus.
Now you should be really confused, and that is a normal reaction! On one hand we were fairly convinced that light acted as a wave but now we see some of its behaviors can be best explained in terms of particles. This dual nature of light is conceptually difficult for most normal people because it is completely counter intuitive. In our macroscopic world things are either particles, such as bullets, balls, coconuts, or waves (in water); they are not—no, not ever—both. As we will see, electromagnetic radiation is not the only example of something that has the properties of both a wave and a particle; this mix of properties is known as wave–particle duality. Electrons, protons, and neutrons also display wavelike properties. In fact, all matter has a wavelength, defined by Louis de Broglie (1892–1987), by the equation $\mathrm{E}= \frac{h}{mv}$ where $mv$ is the object’s momentum (mass $\times$ velocity) and $h$ is Planck’s constant. For heavy objects, moving at slow speeds, the wavelength is very, very small, but it becomes a significant factor for light objects moving fast, such as electrons. Although light and electrons can act as both waves and as particles, it is perhaps better to refer to them as quantum mechanical particles, a term that captures all features of their behavior and reminds us that they are weird! Their behavior will be determined by the context in which we study (and think of) them. | textbooks/chem/General_Chemistry/CLUE%3A_Chemistry_Life_the_Universe_and_Everything/02%3A_Electrons_and_Orbitals/2.2_Taking_Quanta_Seriously.txt |
As we will often see, there are times when an old observation suddenly fits into and helps clarify a new way of thinking about a problem or process. In order to understand the behavior of electrons within atoms scientists brought together a number of such observations. The first observation has its roots in understanding the cause of rainbows. The scientific explanation of the rainbow is based on the fact that light of different wavelengths is bent through different angles (refracted) when it passes through an air–water interface. When sunlight passes through approximately spherical water droplets, it is refracted at the air–water interface, partially reflected (note the difference) from the backside of the water droplet, and then refracted again as it leaves the droplet. The underlying fact that makes rainbows possible is that sunlight is composed of photons with an essentially continuous distribution of visible wavelengths. Isaac Newton illustrated this nicely by using a pair of prisms to show that white light could be separated into light of many different colors by passing it through a prism and then recombined back into white light by passing it through a second prism. On the other hand, light of a single color remained that color, even after it passed through a second prism.
When a dense body, like the Sun or the filament of an incandescent light bulb, is heated, it emits light of many wavelengths (colors)—essentially all wavelengths in the visible range. However, when a sample of an element or mixture of elements is heated, for example in a flame provided by a Bunsen burner, it emits light of only very particular wavelengths. The different wavelengths present in the emitted light can be separated from one another using a prism to produce what is known as an emission spectra. When projected on a screen these appear as distinct, bright-colored lines, known emission lines. In a complementary manner, if white light, which consists of a continuous distribution of wavelengths of light, is passed through a cold gaseous element the same wavelengths that were previously emitted by the heated element will be absorbed, while all other wavelengths will pass through unaltered. By passing the light through a prism we can see which wavelengths of light have been absorbed by the gas. We call these dark areas “absorption” lines within the otherwise continuous spectrum. The emission and absorption wavelengths for each element the same and unique for each element. Emission and absorption phenomenon provide a method (spectroscopy) by which the absorbance or emission of specific wavelengths of light by can be is used to study the composition and properties of matter. Scientists used spectroscopic methods to identify helium, from the Greek “sun”, in the Sun before it was isolated on Earth.
In the 1800s, it became increasingly clear that each element, even the simplest, hydrogen, has a distinctive and often quite complex emission/absorption spectra. In 1855 Johann Balmer (1825-1898) calculated the position of the lines in the visible region. In 1888 Johannes Rydberg (1854-1919) extended those calculations to the entire spectrum. These calculations, however, were based on an empirical formula and it was unclear why this formula worked or what features of the atom it was based on—this made the calculations rather unsatisfying. Although useful, they provided no insight into the workings of atoms.
Making sense of spectra:
How do we make sense of these observations? Perhaps the most important clue is again the photoelectric effect; that is, the observation that illuminating materials with light can in some circumstances lead to the ejection of electrons. This suggests that it is the interactions between light and the electrons in atoms that are important. Using this idea and the evidence from the hydrogen spectra Niels Bohr (1885-1962) proposed a new model for the atom His first hypothesis was that the electrons within an atom can only travel along certain orbits at a fixed distance from the nucleus, each orbit corresponding to a specific energy. The second idea was that electrons can jump from one orbit to another, but this jump requires either the capture (absorption) or release (emission) of energy, in the form of a photon. An electron can move between orbits only if a photon of exactly the right amount of energy is absorbed (lower to higher) or emitted (higher to lower). Lower (more stable) orbits are often visualized as being closer to the nucleus whereas higher, less stable and more energetic orbits are further away. Only when enough energy is added in a single packet is the electron removed completely from the atom, leaving a positively-charged ion (an ion is an atom or molecule that has a different number of protons and electrons) and a free electron. Because the difference in energy between orbits is different in different types of atoms, moving electrons between different orbits requires photons carrying different amounts of energy (different wavelengths).
Bohr’s model worked well for hydrogen atoms; in fact, he could account for and accurately calculate the wavelengths for all of hydrogen’s observed emission/absorption lines. These calculations involved an integer quantum number that corresponded to the different energy levels of the orbits.[11] Unfortunately, this model was not able to predict the emission/absorption spectrum for any other element, including helium and certainly not for any molecule. Apparently Bohr was on the right track—because every element does have a unique spectrum and therefore electrons must be transitioning from one energy level to another—but his model was missing something important. It was not at all clear what restricted electrons to specific energy levels. What happens in atoms with more than one electron? Where are those electrons situated and what governs their behavior and interactions? It is worth remembering that even though the Bohr model of electrons orbiting the nucleus is often used as a visual representation of an atom, it is not correct. Electrons do not circle the nucleus in defined orbits. Bohr’s model only serves as an approximate visual model for appearance of an atom–it is not how electrons actually behave!
Questions
Questions to Answer
• If the intensity of a beam of light is related to the number of photons passing per second, how would you explain intensity using the model of light as a wave? What would change and what would stay the same?
• Why do we not worry about being constantly bombarded by radio waves (we are), but yet we guard our exposure to x rays?
• Draw a picture of what you imagine is happening during the photoelectric effect.
• Is the energy required to eject an electron the same for every metal?
Questions to Ponder
• Can you think of other scientific ideas that you find nonsensical? Be honest.
• How does the idea of an electron as a wave fit with your mental image of an atom?
• Where is the electron if it is a wave?
Questions for Later
• What trends might you expect in the energies required to eject an electron?
• Why do you think this phenomenon (the photoelectric effect) is most often seen with metals? What property of metals is being exploited here?
• What other kinds of materials might produce a similar effect? | textbooks/chem/General_Chemistry/CLUE%3A_Chemistry_Life_the_Universe_and_Everything/02%3A_Electrons_and_Orbitals/2.3%3A_Exploring_Atomic_Organization_Using_Spectroscopy.txt |
Eventually, as they considered the problems with the Bohr model, scientists came back to the idea of the wave–particle duality as exemplified by the photon. If light (electromagnetic radiation), which was classically considered to be a wave, could have the properties of a particle, then perhaps matter, classically considered as composed of particles, could have the properties of waves, at least under conditions such as those that exist within an atom. Louis De Broglie (1892–1987) considered this totally counterintuitive idea in his Ph.D. thesis. De Broglie used Planck’s relationship between energy and frequency ($\mathrm{E} = hn$), the relationship between frequency and wavelength ($c = \lambda n$), and Einstein’s relationship between energy and mass ($\mathrm{E} = mc^{2}$) to derive a relationship between the mass and wavelength for any particle (including photons).[12] You can do this yourself by substituting into these equations, to come up with $\lambda = \frac{h}{mv}$, where $mv$ is the momentum of a particle with mass $m$ and velocity $v$. In the case of photons, $v = c$, the velocity of light.
Although the math involved in deriving the relationship between momentum ($mv$) of a particle and its wavelength $\lambda$ is simple, the ideas behind it are most certainly not. It is even more difficult to conceptualize the idea that matter, such as ourselves, can behave like waves, and yet this is consistent with a broad range of observations. We never notice the wavelike properties of matter because on the macroscopic scale, the wavelength associated with a particular object is so small that it is negligible. For example, the wavelength of a baseball moving at $100 \mathrm{~m/s}$ is much smaller than the baseball itself. It is worth thinking about what you would need to know to calculate it. At the atomic scale, however, the wavelengths associated with particles are similar to their size, meaning that the wave nature of particles such as electrons cannot be ignored; their behavior cannot be described accurately by models and equations that treat them as simple particles. The fact that a beam of electrons can undergo diffraction, a wave-like behavior provides evidence of this idea.
Certainty and Uncertainty
Where is a wave located? The answer is not completely obvious. You might think it would be easier to determine where a particle is, but things get complicated as they get smaller and smaller. Imagine that we wanted to view an electron within an atom using some type of microscope, in this case, an imaginary one with unlimited resolution. To see something, photons have to bounce or reflect off it and then enter our eye, be absorbed by a molecule in a retinal cell, and start a signal to our brain where that signal is processed and interpreted. When we look at macroscopic objects, their interactions with light have little effect on them. For example, objects in a dark room do not begin to move just because you turn the lights on! Obviously the same cannot be said for atomic-scale objects; we already know that a photon of light can knock an electron completely out of an atom (the photoelectric effect). Now we come to another factor: the shorter the wavelength of light we use, the more accurately we can locate an object.[13] Remember, however, that wavelength and energy are related: the shorter the wavelength the greater its energy. To look at something as small as an atom or an electron we have to use electromagnetic radiation of a wavelength similar to the size of the electron. We already know that an atom is about $10^{-10}$ m in diameter, so electrons are presumably much smaller. Let us say that we use gamma rays, a form of electromagnetic radiation, whose wavelength is $\sim 10^{-12}$ m. But radiation of such short wavelength carries lots of energy, so these are high-energy photons. When such a high-energy photon interacts with an electron, it dramatically perturbs the electron’s position and motion. That is, if we try to measure where an electron is, we perturb it by the very act of measurement. The act of measurement introduces uncertainty and this uncertainty increases the closer we get to the atomic molecular scale.
This idea was first put forward explicitly by Werner Heisenberg (1901-1976) and is known as the Heisenberg Uncertainty Principle. According to the uncertainty principle, we can estimate the uncertainty in a measurement using the formula $\Delta m v \times \Delta x> \frac{h}{2 \pi}$, where $\Delta mv$ is the uncertainty in the momentum of the particle (mass times velocity or where it is going and how fast), $\Delta x$ is the uncertainty in its position in space (where it is at a particular moment), and $h$ is Planck’s constant now divided by $2 \pi$. If we know exactly where the particle is ($\Delta x = 0$) then we have absolutely no information about its velocity, which means we do not know how fast or in what direction it is going. Alternatively, if we know its momentum exactly ($\Delta mv = 0$), that is, we know exactly how fast and in which direction it is going, we have no idea whatsoever where it is! The end result is that we cannot know exactly where an electron is without losing information on its momentum, and vice versa. This has lots of strange implications. For example, if we know the electron is within the nucleus ($\Delta x \sim 1.5 \times 10^{-14} \mathrm{~m}$), then we have very little idea of its momentum (how fast and where it is going). These inherent uncertainties in the properties of atomic-level systems are one of their key features. For example, we can estimate some properties very accurately but we cannot know everything about an atomic/molecular-level system at one point in time. This is a very different perspective from the one it replaced, which was famously summed up by Pierre-Simon Laplace (1749–1827), who stated that if the positions and velocities of every object in the universe were known, the future would be set:
We may regard the present state of the universe as the effect of its past and the cause of its future. An intellect which at a certain moment would know all forces that set nature in motion, and all positions of all items of which nature is composed, if this intellect were also vast enough to submit these data to analysis, it would embrace in a single formula the movements of the greatest bodies of the universe and those of the tiniest atom; for such an intellect nothing would be uncertain and the future just like the past would be present before its eyes. — Pierre-Simon Laplace (1745-1827)
It turns out that a major flaw in Bohr’s model of the atom was that he attempted to define both the position of an electron (a defined orbit) and its energy, or at least the energy difference between orbits, at the same time. Although such a goal seems quite reasonable and would be possible at the macroscopic level, it simply is not possible at the atomic level. The wave nature of the electron makes it impossible to predict exactly where that electron is if we also know its energy level. In fact, we do know the energies of electrons very accurately because of the evidence from spectroscopy. We will consider this point again later in this chapter.
Questions
Questions to Answer
• How does the wavelength of a particle change as the mass increases?
• Planck’s constant is $h=6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}$. What are the implications for particles of macroscopic size? ($1 \mathrm{~J} =$ the kinetic energy of a two-kilogram mass moving at the speed of $1 \mathrm{~m/s}$.)
• What would be the wavelength of the world-record holder for the $100 \mathrm{~-m}$ sprint? What assumptions do you have to make to answer this question?
• What is the wavelength of a protein of $60,000$ daltons? (That is, if the protein has a molar mass of $60,000 \mathrm{~g/M}$, what is the mass of one molecule of the protein?)
Questions to Ponder
• What is the uncertainty in your momentum, if the error in your position is $0.01 \mathrm{~m}$ (remembering
that Planck’s constant $h=6.626068 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}$)?
• How is it that we experience objects as having very definite velocities and positions?
• Does it take energy to determine your position?
• How is the emission and absorption behavior of atoms related to electron energies? | textbooks/chem/General_Chemistry/CLUE%3A_Chemistry_Life_the_Universe_and_Everything/02%3A_Electrons_and_Orbitals/2.4%3A_Beyond_Bohr.txt |
Up to this point we have made a number of unjustified assumptions. We have talked about elements but we have not explicitly specified how they are different, so let us do that now. If we start with hydrogen we characterized it by the presence of one proton in the nucleus and one electron surrounding it. Atoms are always neutral, which means that the number of positively-charged particles is equal to the number of negatively-charged particles, and charges come in discrete, equal, and opposite units. The presence of one proton and one electron defines a hydrogen atom but the world is a little more complex than that. A hydrogen atom may also contain one or two neutrons in its nucleus. A neutron can be considered, with the forgiveness of physicists, a proton, an electron, and an uncharged neutrino, and so it is electrically neutral. Neutrons are involved in the strong nuclear force and become increasingly important as the element increases in atomic number. In hydrogen, the neutrons (if they are present) have rather little to do, but in heavier elements the strong nuclear force is critical in holding the nucleus together, because at short distances this force is $\sim 100$ times stronger than the electrostatic repulsion between positively charged protons, which is why nuclei do not simply disintegrate. At the same time, the strong force acts over a very limited range, so when particles are separated by more than about $2 \times 10^{-15} \mathrm{~m}$ ($2$ femtometers or fm), we can ignore it.
As we add one proton after another to an atom, which we can do in our minds, and which occurs within stars and supernova, in a rather more complex manner, we generate the various elements. The number of protons determines the elemental identity of an atom, whereas the number of neutrons can vary. Atoms of the same element with different numbers of neutrons are known as isotopes of that element. Each element is characterized by a distinct, whole number (1, 2, 3, …) of protons and the same whole number of electrons. An interesting question emerges here: is the number of possible elements infinite? And if not, why not? Theoretically, it might seem possible to keep adding protons (and neutrons and electrons) to produce a huge number of different types of atoms. However, as Rutherford established, the nucleus is quite small compared to the atom as a whole, typically between one and ten femtometers in diameter. As we add more and more protons (and neutrons) the size of the nucleus exceeds the effective range of the strong nuclear force ($< 2 \mathrm{~fm}$), and the nucleus becomes unstable. As you might expect, unstable nuclei break apart (nuclear fission), producing different elements with smaller numbers of protons, a process that also releases tremendous amounts of energy. Some isotopes are more stable than others, which is why the rate of their decay, together with a knowledge of the elements that they decay into can be used to calculate the age of rocks and other types of artifacts.[14]
Each element is defined by the number of protons in the nucleus, and as such is different from every other element. In fact, careful analysis of different elements reveals that there are periodicities (repeating patterns) in the properties of elements. Although John Dalton produced a table of elements with their atomic weights in 1805, it was only when Dimitri Mendeleev (1834–1907) tried to organize the elements in terms of their chemical and physical properties that some semblance of order began to emerge. Mendeleev, a Russian chemistry professor, was frustrated by the lack of organization of chemical information, so he decided to write his own textbook (not unlike your current authors). At the time, scientists had identified about 60 elements and established their masses relative to hydrogen. Scientists had already noticed that the elements display repeating patterns of behavior: and that some elements have very similar properties. It was Mendeleev’s insight that these patterns could be used as a guide for arranging the elements in a systematic way. In his periodic table, published in 1869, he placed elements in order of increasing atomic weight in repeating rows from left to right; elements with similar chemical properties were placed in vertical columns (known as groups).
Although several other scientists were working on schemes to show patterns in elemental behavior, it was Mendeleev’s arrangement that emerged as the basis for the modern periodic table, not only because of the way he arranged the elements but also for what he left out and what he changed. For example he was so sure about the underlying logic of his table that where certain elements seemed out of place, based, for example, on their reported atomic weights, such as tellurium and iodine, he reversed them and he turned out to be correct. Where Mendeleev predicted elements should be, he left gaps in his table to accommodate them. Subsequently, scientists discovered these missing elements (for example germanium, gallium, and scandium). In fact, we now know that it is not atomic weight (that is the number of protons and neutrons) but rather atomic number, $Z$, (the number of protons and electrons) that increases periodically. This explains why tellurium ($\text{atomic mass } 127.6, Z = 52$) must come before iodine ($\text{atomic mass } 126.9, Z = 53$). The important point to note is that although the modern periodic table is arranged in order of increasing number of protons and electrons, the repetition and patterns that emerge are the property of the electrons, their arrangements, and energies. This is our next subject.
Questions
Question to Answer
• Science fiction authors like weird elements. Provide a short answer for why no new elements with atomic numbers below 92 are possible.
• Isotopes of the same element are very similar chemically. What does that imply about what determines chemical behavior?
Questions to Ponder
• Why do you think there were no noble gases in Mendeleev’s periodic table?
• Why aren’t the atomic weights in Mendeleev’s periodic table whole numbers?
• Why would you expect different isotopes of the same element to differ in stability?
• You discover a new element. How would you know where would it should go in the periodic table? | textbooks/chem/General_Chemistry/CLUE%3A_Chemistry_Life_the_Universe_and_Everything/02%3A_Electrons_and_Orbitals/2.5%3A_Organizing_Elements%3A_Introduction_to_the_Periodic_Table.txt |
Our current working model of the atom is based on quantum mechanics that incorporate the ideas of quantized energy levels, the wave properties of electrons, and the uncertainties associated with electron location and momentum. If we know their energies, which we do, then the best we can do is to calculate a probability distribution that describes the likelihood of where a specific electron might be found, if we were to look for it. If we were to find it, we would know next to nothing about its energy, which implies we would not know where it would be in the next moment. We refer to these probability distributions by the anachronistic, misleading, and Bohrian term orbitals. Why misleading? Because to a normal person, the term orbital implies that the electron actually has a defined and observable orbit, something that is simply impossible to know (can you explain why?)
Another common and often useful way to describe where the electron is in an atom is to talk about the electron probability density or electron density for short. In this terminology, electron density represents the probability of an electron being within a particular volume of space; the higher the probability the more likely it is to be in a particular region at a particular moment. Of course you can’t really tell if the electron is in that region at any particular moment because if you did you would have no idea of where the electron would be in the next moment.
Erwin Schrödinger (1887–1961) developed, and Max Born (1882–1970) extended, a mathematical description of the behavior of electrons in atoms. Schrödinger used the idea of electrons as waves and described each atom in an element by a mathematical wave function using the famous Schrödinger equation ($H \Psi=E \Psi$). We assume that you have absolutely no idea what either $H \Psi$ or $E \Psi$ are but don’t worry—you don’t really need to. The solutions to the Schrödinger equation are a set of equations (wave functions) that describe the energies and probabilities of finding electrons in a region of space. They can be described in terms of a set of quantum numbers; recall that Bohr’s model also invoked the idea of quantum numbers. One way to think about this is that almost every aspect of an electron within an atom or a molecule is quantized, which means that only defined values are allowed for its energy, probability distribution, orientation, and spin. It is far beyond the scope of this book to present the mathematical and physical basis for these calculations, so we won’t pretend to try. However, we can use the results of these calculations to provide a model for the arrangements of electrons in an atom using orbitals, which are mathematical descriptions of the probability of finding electrons in space and determining their energies. Another way of thinking about the electron energy levels is that they are the energies needed to remove that electron from the atom or to move an electron to a “higher” orbital. Conversely, this is the same amount of energy released when an electron moves from a higher energy to a lower energy orbital. Thinking back to spectroscopy, these energies are also related to the wavelengths of light that an atom will absorb or release. Let us take a look at some orbitals, their quantum numbers, energies, shapes, and how we can used them to explain atomic behavior.
Examining Atomic Structure Using Light: On the Road to Quantum Numbers
J.J. Thompson’s studies (remember them?) suggested that all atoms contained electrons. We can use the same basic strategy in a more sophisticated way to begin to explore the organization of electrons in particular atoms. This approach involves measuring the amount of energy it takes to remove electrons from atoms. This is known as the element’s ionization energy which in turn relates directly back to the photoelectric effect.
All atoms are by definition electrically neutral, which means they contain equal numbers of positively- and negatively-charged particles (protons and electrons). We cannot remove a proton from an atom without changing the identity of the element because the number of protons is how we define elements, but it is possible to add or remove an electron, leaving the atom’s nucleus unchanged. When an electron is removed or added to an atom the result is that the atom has a net charge. Atoms (or molecules) with a net charge are known as ions, and this process (atom/molecule to ion) is called ionization. A positively charged ion (called a cation) results when we remove an electron; a negatively charged ion (called an anion) results when we add an electron. Remember that this added or removed electron becomes part of, or is removed from, the atom’s electron system.
Now consider the amount of energy required to remove an electron. Clearly energy is required to move the electron away from the nucleus that attracts it. We are perturbing a stable system that exists at a potential energy minimum – that is the attractive and repulsive forces are equal at this point. We might naively predict that the energy required to move an electron away from an atom will be the same for each element. We can test this assumption experimentally by measuring what is called the ionization potential. In such an experiment, we would determine the amount of energy (in kilojoules per mole of molecules) required to remove an electron from an atom. Let us consider the situation for hydrogen ($\mathrm{H}$). We can write the ionization reaction as: $\mathrm{H} \text { (gas) }+\text { energy } \rightarrow \mathrm{H}^{+} \text {(gas) }+\mathrm{e}^{-}.$[15]
What we discover is that it takes $1312 \mathrm{~kJ}$ to remove a mole of electrons from a mole of hydrogen atoms. As we move to the next element, helium (He) with two electrons, we find that the energy required to remove an electron from helium is $2373 \mathrm{~kJ/mol}$, which is almost twice that required to remove an electron from hydrogen!
Let us return to our model of the atom. Each electron in an atom is attracted to all the protons, which are located in essentially the same place, the nucleus, and at the same time the electrons repel each other. The potential energy of the system is modeled by an equation where the potential energy is proportional to the product of the charges divided by the distance between them. Therefore the energy to remove an electron from an atom should depend on the net positive charge on the nucleus that is attracting the electron and the electron’s average distance from the nucleus. Because it is more difficult to remove an electron from a helium atom than from a hydrogen atom, our tentative conclusion is that the electrons in helium must be attracted more strongly to the nucleus. In fact this makes sense: the helium nucleus contains two protons, and each electron is attracted by both protons, making them more difficult to remove. They are not attracted exactly twice as strongly because there are also some repulsive forces between the two electrons.
The size of an atom depends on the size of its electron cloud, which depends on the balance between the attractions between the protons and electrons, making it smaller, and the repulsions between electrons, which makes the electron cloud larger.[16] The system is most stable when the repulsions balance the attractions, giving the lowest potential energy. If the electrons in helium are attracted more strongly to the nucleus, we might predict that the size of the helium atom would be smaller than that of hydrogen. There are several different ways to measure the size of an atom and they do indeed indicate that helium is smaller than hydrogen. Here we have yet another counterintuitive idea: apparently, as atoms get heavier (more protons and neutrons), their volume gets smaller!
Given that
1. helium has a higher ionization energy than hydrogen and
2. that helium atoms are smaller than hydrogen atoms, we infer that the electrons in helium are attracted more strongly to the nucleus than the single electron in hydrogen.
Let us see if this trend continues as we move to the next heaviest element, lithium ($\mathrm{Li}$). Its ionization energy is $520 \mathrm{~kJ/mol}$. Oh, no! This is much lower than either hydrogen ($1312 \mathrm{~kJ/mol}$) or helium ($2373 \mathrm{~kJ/mol}$). So what do we conclude? First, it is much easier (that is, requires less energy) to remove an electron from $\mathrm{Li}$ than from either $\mathrm{H}$ or $\mathrm{He}$. This means that the most easily removed electron in $\mathrm{Li}$ is somehow different than are the most easily removed electrons of either $\mathrm{H}$ or $\mathrm{He}$. Following our previous logic we deduce that the “most easily removable” electron in $\mathrm{Li}$ must be further away (most of the time) from the nucleus, which means we would predict that a $\mathrm{Li}$ atom has a larger radius than either $\mathrm{H}$ or $\mathrm{He}$ atoms. So what do we predict for the next element, beryllium ($\mathrm{Be}$)? We might guess that it is smaller than lithium and has a larger ionization energy because the electrons are attracted more strongly by the four positive charges in the nucleus. Again, this is the case. The ionization energy of $\mathrm{Be}$ is $899 \mathrm{~kJ/mol}$, larger than $\mathrm{Li}$, but much smaller than that of either $\mathrm{H}$ or $\mathrm{He}$. Following this trend the atomic radius of $\mathrm{Be}$ is smaller than $\mathrm{Li}$ but larger than $\mathrm{H}$ or $\mathrm{He}$. We could continue this way, empirically measuring ionization energies for each element (see figure), but how do we make sense of the pattern observed, with its irregular repeating character that implies complications to a simple model of atomic structure?
Questions
Questions to Answer
• Why are helium atoms smaller than hydrogen atoms?
• What factors govern the size of an atom? List all that you can. Which factors are the most important?
Questions to Ponder
• What would a graph of the potential energy of a hydrogen atom look like as a function of distance of the electron from the proton?
• What would a graph of the kinetic energy of an electron in a hydrogen atom look like as a function of distance of the electron from the nucleus?
• What would a graph of the total energy of a hydrogen atom look like as a function of distance of the electron from the proton? | textbooks/chem/General_Chemistry/CLUE%3A_Chemistry_Life_the_Universe_and_Everything/02%3A_Electrons_and_Orbitals/2.6%3A_Orbitals_Electron_Clouds_Probabilities_and_Energies.txt |
Quantum numbers (whose derivation we will not consider here) provide the answer to our dilemma. Basically we can describe the wave function for each individual electron in an atom by a distinct set of three quantum numbers, known as $n$, $l$, and $ml$. The principal quantum number, $n$, is a non-zero positive integer ($n=1,2,3,4$, etc.). These are often referred to as electron shells or orbitals, even though they are not very shell- or orbital-like. The higher the value of $n$, the higher the overall energy level of the electron shell. For each value of $n$ there are only certain allowable values of $l$, and for each value of $l$, only certain allowable values of $ml$. Table $2.7.1$ (next page) shows the allowable values of $l$ and $ml$ for each value of $n$ are shown. There are a few generalizations we can make. Three quantum numbers, $n$, $l$, and $ml$, describe each orbital in an atom and each orbital can contain a maximum of two electrons. As they are typically drawn, each orbital defines the space in which the probability of finding an electron is 90%. Because each electron is described by a unique set of quantum numbers, the two electrons within a particular orbital must be different in some way.[18] But because they are in the same orbital they must have the same energy and the same probability distribution. So what property is different? This property is called spin. The spin quantum number, ms, can have values of either $+ \frac{1}{2}$ or $- \frac{1}{2}$. Spin is responsible for a number of properties of matter including magnetism.
Hydrogen has one electron in a $1\mathrm{s}$ orbital and we write its electron configuration as $1\mathrm{s}^{1}$. Helium has both of its electrons in the $1\mathrm{s}$ orbital ($1\mathrm{s}^{2}$). In lithium, the electron configuration is $1\mathrm{s}^{2} 2\mathrm{s}^{1}$, which tells us that during ionization, an electron is being removed from a $2\mathrm{s}$ orbital. Quantum mechanical calculations tell us that in $2\mathrm{s}$ orbital there is a higher probability of finding electrons farther out from the nucleus than the $1\mathrm{s}$ orbital, so we might well predict that it takes less energy to remove an electron from a $2\mathrm{s}$ orbital (found in $\mathrm{Li}$) than from a $1\mathrm{s}$ orbital (found in $\mathrm{H}$). Moreover, the two $1\mathrm{s}$ electrons act as a sort of shield between the nucleus and the $2\mathrm{s}$ electrons. The $2\mathrm{s}$ electrons feel what is called the effective nuclear charge, which is smaller than the real charge because of shielding by the $1\mathrm{s}$ electrons. In essence, two of the three protons in the lithium nucleus are counterbalanced by the two $1\mathrm{s}$ electrons. The effective nuclear charge in lithium is $+1$. The theoretical calculations are borne out by the experimental evidence—always a good test of a theory.
Table $2.7.1$ Elemental electron shell organization
$s (l = 0)$ $p (l = 1)$ $d (l = 2)$
$m = 0$ $m = 0$ $m=\pm 1$ $m = 0$ $m=\pm 1$ $m=\pm 2$
$s$ $p_{z}$ $p_{x}$ $p_{y}$ $d_{z^{2}}$ $d_{xz}$ $d_{yz}$ $d_{xy}$ $d_{ x^{2}-y^{2}}$
$m = 1$
$m = 2$
$m = 3$
At this point, you might start getting cocky; you may even be ready to predict that ionization energies across the periodic table from lithium to neon ($\mathrm{Ne}$) will increase, with a concomitant decrease in atomic radius. In the case of atomic radius, this is exactly what we see in the figure – as you go across any row in the periodic table the atomic radius decreases. Again, the reason for both these trends is that same: that is, each electron is attracted by an increasing number of protons as you go from $\mathrm{Li}$ to $\mathrm{Ne}$, which is to say that the effective nuclear charge is increasing. Electrons that are in the same electron shell do not interact much and each electron is attracted by all the unshielded charge on the nucleus. By the time we get to fluorine ($\mathrm{F}$), which has an effective nuclear charge of $9 – 2 = +7$, and neon ($10 – 2 = +8$), each of the electrons are very strongly attracted to the nucleus, and very difficult to dislodge. Meaning that the size of the atom gets smaller, and the ionization energy gets larger[19]
As you have undoubtedly noted from considering the graph, the increase in ionization energy from lithium to neon is not uniform: there is a drop in ionization energy from beryllium to boron and from nitrogen to oxygen. This arises from the fact that as the number of electrons in an atom increases the situation becomes increasingly complicated. Electrons in the various orbitals influence one another and some of these effects are quite complex and chemically significant. We will return to this in a little more detail in Chapter $3$ and at various points through the rest of the book.
If we use the ideas of orbital organization of electrons, we can make some sense of patterns observed in ionization energies. Let us go back to the electron configuration. Beryllium ($\mathrm{Be}$) is $1\mathrm{s}^{2} 2\mathrm{s}^{2}$ whereas Boron ($\mathrm{B}$) is $1\mathrm{s}^{2} 2\mathrm{s}^{2} 2\mathrm{p}^{1}$. When electrons are removed from $\mathrm{Be}$ and $\mathrm{B}$ they are removed from the same quantum shell ($n = 2$) but, in the case of $\mathrm{Be}$, one is removed from the $2\mathrm{s}$ orbital whereas in $\mathrm{B}$, the electron is removed from a $2\mathrm{p}$ orbital. $\mathrm{s}$ orbitals are spherically symmetric, $\mathrm{p}$ orbitals have a dumbbell shape and a distinct orientation. Electrons in a $2\mathrm{p}$ orbital have lower ionization energies because they are on average a little further from the nucleus and so a little more easily removed compared to $2\mathrm{s}$ electrons. That said, the overall average atomic radius of boron is smaller than beryllium, because on average all its electrons spend more time closer to the nucleus.
The slight drop in ionization potential between nitrogen and oxygen has a different explanation. The electron configuration of nitrogen is typically written as $1\mathrm{s}^{2} 2\mathrm{s}^{2} 2\mathrm{p}^{3}$, but this is misleading: it might be better written as $1\mathrm{s}^{2} 2\mathrm{s}^{2} 2\mathrm{px}^{1} 2\mathrm{py}^{1} 2\mathrm{pz}^{1}$, with each $2\mathrm{p}$ electron located in a separate $\mathrm{p}$ orbital. These $\mathrm{p}$ orbitals have the same energy but are oriented at right angles (orthogonally) to one another. This captures another general principle: electrons do not pair up into an orbital until they have to do so.[20] Because the $\mathrm{p}$ orbitals are all of equal energy, each of them can hold one electron before pairing is necessary. When electrons occupy the same orbital there is a slight repulsive and destabilizing interaction; when multiple orbitals of the same energy are available, the lowest energy state is the one with a single electron in an orbital. Nitrogen has all three $2\mathrm{p}$ orbitals singly occupied and therefore the next electron, which corresponds to oxygen, has to pair up in one of the p orbitals. Thus it is slightly easier to remove a single electron from oxygen than it is to remove a single electron from nitrogen, as measured by the ionization energy.
To pull together a set of seriously obscure ideas, the trends in ionization energies and atomic radii indicate that electrons are not uniformly distributed around an atom’s nucleus but rather have distinct distributions described by the rules of quantum mechanics. Although we derive the details of these rules from rather complex calculations and the wave behavior of electrons, we can cope with them through the use of quantum numbers and electron probability distributions. Typically electrons in unfilled shells are more easily removed or reorganized than those in filled shells because atoms with unfilled shells have higher effective nuclear charges. Once the shell is filled, the set of orbitals acts like a shield and cancels out an equal amount of nuclear charge. The next electron goes into a new quantum shell and the cycle begins again. This has profound implications for how these atoms react with one another to form new materials because, as we will see, chemical reactions involve those electrons that are energetically accessible: the valence electrons.
We could spend the rest of this book (and probably one or two more) discussing how electrons are arranged in atoms but in fact your average chemist is not much concerned with atoms as entities in themselves. As we have said before, naked atoms are not at all common. What is common is combinations of atoms linked together to form molecules. From a chemist’s perspective, we need to understand how, when, and where atoms interact. The electrons within inner and filled quantum shells are “relatively inert” which can be translated into English to mean that it takes quite a lot of energy (from the outside world) to move them around. Chemists often refer to these electrons as core electrons, which generally play no part in chemical reactions; we really do not need to think about them much more except to remember that they form a shield between the nucleus and the outer electrons. The results of their shielding does, however, have effects on the strong interactions, commonly known as bonds, between atoms of different types, which we will discuss in Chapters $4$ and $5$. Reflecting back on Chapter $1$, we can think about the distinction between the London Dispersion Forces acting between He atoms and between $\mathrm{H}_{2}$ molecules versus the bonds between the two $\mathrm{H}$ atoms in a $\mathrm{H}_{2}$ molecule.
Bonds between atoms involve the valence electrons found in outer, and usually partially filled, orbitals. Because of the repeating nature of electron orbitals, it turns out that there are patterns in the nature of interactions atoms make—a fact that underlies the organization of elements in the periodic table. We will come back to the periodic table once we have considered how atomic electronic structure influences the chemical properties of the different elements.
Questions
Questions to Answer
• Try to explain the changes in ionization potential as a function of atomic number by drawing your impression of what each atom looks like as you go across a row of the periodic table, and down a group.
Questions to Ponder
• How does the number of valence electrons change as you go down a group in the periodic table? How does is change as you go across a row?
• How do you think the changes in effective nuclear charge affect the properties of elements as you go across a row in the periodic table?
2.8: In-Text References
1. This may (or may not) be helpful: http://www.cv.nrao.edu/course/astr53.../LarmorRad.pdf
2. http://phet.colorado.edu/en/simulati...e-interference
3. Link to “Dr. Quantum” double slit experiment: http://www.youtube.com/watch?v=DfPeprQ7oGc
4. In fact, there is lots of light within your eyeball, even in the dark, due to black body radiation. You do not see it because it is not energetic enough to activate your photosensing cells. See: http://blogs.discovermagazine.com/co...ose-your-eyes/
5. http://www.physorg.com/news76249412.html
6. $h= 6.626068 \times 10^{-34} \mathrm{~m}^{2} \mathrm{kg/s}$ (or joule-seconds, where a joule is the kinetic energy of a $2 \mathrm{~kg}$ mass moving at a velocity of $1 \mathrm{~m/s}$)
7. This is known as the Rayleigh-Jeans law.
8. http://phet.colorado.edu/simulations...lectric_Effect
9. One type of semi-exception is illustrated by what are known as two- and multi-photon microscopes, in which two lower energy photons hit a molecule at almost the same moment, allowing their energies to be combined; see http://en.Wikipedia.org/wiki/Two-pho...ion_microscopy.
10. For a more complex explanation, see: http://www.coffeeshopphysics.com/art...y_of_rainbows/
11. Bohr model applet particle and wave views: http://www.walter-fendt.de/ph11e/bohrh.htm
12. Although the resting mass of a photon is zero, a moving photon does has an effective mass because it has energy.
13. Good reference: http://ww2010.atmos.uiuc.edu/(Gl)/gu...asics/wvl.rxml
14. see https://www.youtube.com/watch?v=6SxzfZ8bRO4 and https://www.youtube.com/watch?v=1920gi3swe4
15. These experiments are carried out using atoms in the gas phase in order to simplify the measurement.
16. There are a number different ways of defining the size of an atom, and in fact the size depends on the atom’s chemical environment (for example, whether it is bonded to another atom or not). In fact, we can only measure the positions of atomic nuclei, and it is impossible to see where the electron cloud actually ends; remember that orbitals are defined as the surface within which there is a 90% probability of finding an electron. Therefore, we often use the van der Waals radius, which is half the distance between the nuclei of two adjacent unbonded atoms.
17. For more information see: http://winter.group.shef.ac.uk/orbit.../1s/index.html http://www.uark.edu/misc/julio/orbitals/index.html
18. This is called the Pauli exclusion principle, which states that no two electrons may occupy the same quantum state; that is, no two electrons can have the same value for all four quantum numbers.
19. We should note that this model for calculating the effective nuclear charge is just that – a model. It provides us with an easy way to predict the relative attractions between the nuclei and electrons, but there are of course more accurate ways of calculating the attraction which take into account the fact that the nuclei is only partial shielded by the core electrons.
20. This is often called Hund’s rule. Just as passengers on a bus do not sit together until they have to, neither do electrons. | textbooks/chem/General_Chemistry/CLUE%3A_Chemistry_Life_the_Universe_and_Everything/02%3A_Electrons_and_Orbitals/2.7%3A_Quantum_Numbers.txt |
Up to now we have been concerned mainly with isolated atoms, an extremely abstract topic. We now move on to consider the macroscopic behavior of atoms, that is, the behaviors of very, very large numbers of atoms that form the materials that we touch, feel, smell, and observe with our own eyes. Before we do that, it is important to understand and be explicit about what properties atoms, molecules, and their aggregates can and cannot exhibit.
03: Elements Bonding and Physical Properties
When atoms interact with one another to form molecules or larger structures, the molecules have different properties than their component atoms; they display what are often referred to as emergent properties, where the whole is more than, or different from the sum of its parts. In a similar way groups of atoms or molecules have different properties from isolated atoms/molecules. For example while groups of atoms/molecules exist in solid, liquid, or gaseous states, and often have distinct colors and other properties, isolated atoms/molecules do not; there are no solid or liquid isolated atoms and they do not have a color or a boiling point. So the obvious question is, how many atoms or molecules need to aggregate before they display these emergent properties, before they have a color, before they have a melting point, boiling point, heat capacity, and other properties that isolated atoms do not? The answer is not completely simple, as you are probably slowly coming to expect. As we add more and more atoms or molecules together their properties change but not all at once. You have probably heard about nanoscience and nanotechnologies, which have been the focus of a great deal of research and economic interest in the past decade or so. Nanoparticles are generally classified as being between $1$ and $100 \mathrm{~nm}$ in diameter (a nanometer is one billionth of a meter or $1 \times 10^{-9} \mathrm{~m}$). Such particles often have properties that are different from those of bulk (macroscopic) materials. Nanomaterials can be thought of as a bridge between the atomic-molecular and macroscopic scales.
Assuming that they are pure, macroscopic materials have predictable properties and it doesn’t really matter the size of the sample. A macroscopic sample of pure gold behaves the same regardless of its size and if Archimedes (ca. 287–212 bce) were alive today, he could tell you whether it was pure or not based on its properties, for example, its density. But gold nanoparticles have different properties depending upon their exact size. For example, when suspended in water, they produce colors ranging from orange to purple, depending on their diameter (see Figure). Often the differences in the properties displayed are due to differences in the ratio of surface area to volume, which implies that intermolecular forces (forces between molecules) are more important for nanomaterials. As we cluster more and more particles together, the properties of the particles change. Biomolecules generally fall into the size range of nanomaterials, and as we will see their surface properties are very important in determining their behavior.
Unfortunately when we are talking about the properties of atoms and molecules versus substances and compounds, it can be difficult, even for experienced chemists, to keep the differences clear. In addition different representations are often used for different organizational levels; it is an important skill to be able to recognize and translate between levels. We will be using a range of representations to picture atoms and molecules; chemists (and we) typically use various shorthand rules, methods, and chemical equations to represent molecular composition, shape, and behaviors. But just knowing the equations, often the only thing learned in introductory chemistry courses, is not sufficient to understand chemistry and the behavior of atoms and molecules. Much of the information implied by even the simplest chemical equations can easily be missed—or misunderstood—if the reader does not also have a mental picture of what the diagram or equation represents, how a molecule is organized and its shape, and how it is reorganized during a particular reaction. We will be trying to help you get these broader pictures, which should you make sense of the diagrams and equations used here. That said, it is always important to try to explicitly identify what you are assuming when you approach a particular chemical system; that way you can go back and check whether your assumptions are correct.
Where Do Atoms Come From?
“We are stardust, we are golden, We are billion-year-old carbon.”
– Woodstock, Joni Mitchell
“Sometimes I’ve believed as many as six impossible things before breakfast.”
Alice in Wonderland, Lewis Carroll
Did you ever stop to ask yourself where the atoms in your body came from? Common answers might be that the atoms in our bodies come from food, water, or air. But these are not the ultimate answers, because we then need to ask, where did the atoms in food, water, and air come from? Where did the atoms in the Earth come from? There are really two general possibilities: either the atoms that make up the Earth and the rest of the universe are eternal or they were generated/created by some process. How do we decide which is true? What is the evidence favoring one model over the other? The answers come not from chemistry, but from astrophysics.
Given that we are thinking scientifically what kinds of evidence can we look for to decide whether atoms (or the universe) are eternal or recently created? Clearly we must be able to observe the evidence here and now and use it to formulate logical ideas that make clear and unambiguous predictions. As we will see we will be called upon once again to believe many apparently unbelievable things. The current organizing theory in astrophysics and cosmology, known as the Big Bang theory, holds that the universe is $\sim 13,820,000,000 \pm 120,000,000$ years old or $13.82 \pm 0.12$ billion years – an unimaginable length of time. The Sun and Earth are $\sim 5,000,000,000$ years old, and the universe as a whole is $\sim 156$ billion light-years in diameter.[1]
The Big Bang theory was put forward in a response to the observation that galaxies in the universe appear to be moving away from one another. Because the galaxies that are further away from us are moving away more rapidly than those that are closer, it appears that space itself is expanding, another seriously weird idea.[2] Based on this observation, we can carry out what scientists call a thought experiment. What happens if we run time backwards, so that the universe is contracting rather than expanding? Taken to its logical conclusion, the universe would shrink until, at some point, all of the universe would be in a single place, at a single point, which would be unimaginably dense. Based on a range of astronomical measurements, this so-called singularity existed $\sim 13.73 \times 10^{9}$ years ago, which means the universe is about $13.73$ billion years old. The Big Bang theory tells us nothing about what happened before $13.73 \times 10^{9}$ years ago, and although there is no shortage of ideas, nothing scientific can be said about it, because it is theoretically unobservable, or at least that is what we have been led to believe by astrophysicists!
Thinking About Atomic Origins
The current model of the universe begins with a period of very rapid expansion, from what was essentially a dimensionless point, a process known as inflation. As you might well imagine there is some debate over exactly what was going on during the first $10^{-43}$ seconds (known as the Planck time) after the universe’s origin. Surprisingly, there is a remarkable level of agreement on what has happened since then.[3] This is because there is lots of observable evidence that makes it relatively easy to compare hypotheses, accepting some and ruling out others. Initially remarkably hot (about $10^{23} \mathrm{~K}$), over time the temperature (local energy levels) of the universe dropped to those that are reachable in modern particle accelerators, so we have actual experimental evidence of how matter behaves under these conditions. At $1$ picosecond after the Big Bang, there were no atoms, protons, or neutrons, because the temperature was simply too high. There were only elementary particles such as photons, quarks, and leptons (electrons are leptons) – particles that appear to have no substructure. By the time the universe was $\sim 0.000001$ seconds old (a microsecond or $1 \times 10^{-6}$ second), the temperature had dropped sufficiently to allow quarks and gluons to form stable structures, and protons and neutrons appeared. A few minutes later the temperature dropped to about $1,000,000,000 \mathrm{~K}$ ($1 \times 10^{9} \mathrm{~K}$), which is low enough for some protons and neutrons to stick together and stay together without flying apart again. That is, the kinetic energy of the particles colliding with them was less than the forces (the weak and strong nuclear forces) holding the protons, neutrons, and nuclei together. At this point the density of particles in the universe was about that of our air.
By the time the universe was a few minutes old it contained mostly hydrogen (${ }^{1} \mathrm{H}^{1}=$ one proton, no neutrons) and deuterium (${ }^{2} \mathrm{H}^{1}=$ one proton and one neutron) nuclei, with some helium (${ }^{3} \mathrm{He}^{2}=$ and ${ }^{4} \mathrm{He}^{2}=$ two protons and either one or two neutrons, respectively), and a few lithium (${ }^{7} \mathrm{Li}^{3}=$ = three protons and four neutrons).[4] These nuclei are all formed by nuclear fusion reactions such as ${ }^{1} \mathrm{p}^{+}+{ }^{1} \mathrm{n}^{0} \rightarrow{ }^{2} \mathrm{H}^{+}+\text {gamma radiation and }{ }^{2} \mathrm{H}^{+}+{ }^{2} \mathrm{H}^{+} \rightarrow{ }^{3} \mathrm{He}^{2+}+{ }^{1} \mathrm{n}^{0} .$
These fusion reactions take place in a temperature range where the nuclei have enough kinetic energy to overcome the electrostatic repulsion associated with the positively charged protons but less than that needed to disrupt the nuclei once formed. After a few minutes the temperature of the universe fell below $\sim 10,000,000 (10^{7}) \mathrm{~K}$. At these temperatures, the kinetic energy of protons and nuclei was no longer sufficient to overcome the electrostatic repulsion between their positive charges. The end result was that there was a short window of time following the Big Bang when a certain small set of nuclei (including ${ }^{1} \mathrm{H}^{+}$, ${ }^{2} \mathrm{H}^{+}$, ${ }^{3} \mathrm{He}^{2+}$, ${ }^{4} \mathrm{He}^{2+}$, and ${ }^{7} \mathrm{Li}^{3+}$) could be formed. After $\sim 400,000$ years the temperature of the universe had dropped sufficiently for electrons to begin to associate in a stable manner with these nuclei and the first atoms (as opposed to bare nuclei) were formed. This early universe was made up of mostly ($> 95%$) hydrogen atoms with a small percentage each of deuterium, helium, and lithium, which is chemically not very interesting.
The primary evidence upon which these conclusions are based comes in the form of the cosmic microwave background radiation ($\mathrm{CMBR}$), which is the faint glow of radiation that permeates the universe. The $\mathrm{CMBR}$ is almost perfectly uniform which means that no matter where you look in the sky the intensity of the $\mathrm{CMBR}$ is (essentially) the same. To explain the $\mathrm{CMBR}$, scientists assume that the unimaginably hot and dense early universe consisted almost entirely of a plasma of hydrogen nuclei that produced vast amounts of electromagnetic radiation, meaning that the early universe glowed. The $\mathrm{CMBR}$ is what is left of this radiation, it is a relic of that early universe. As the universe expanded it cooled but those photons continued to whiz around. Now that they have to fill a much larger universe individual photons have less energy, although the total energy remains the same! The current background temperature of the universe is $\sim 2.27 \mathrm{~K}$, which corresponds to a radiation wavelength of $\sim 1.9 \mathrm{~mm}$ (radiation in the microwave region); hence the name cosmic microwave background radiation.
After a billion years or so things began to heat up again literally (albeit locally). As in any randomly generated object the matter in the universe was not distributed in a perfectly uniform manner and as time passed this unevenness became more pronounced as the atoms began to be gravitationally attracted to each other. The more massive the initial aggregates the more matter was attracted to them. As the clumps of (primarily) hydrogen became denser the atoms banged into each other and these systems, protostars, began to heat up. At the same time the gravitational attraction resulting from the overall mass of the system caused the matter to condense into an even smaller volume and draw in more (mostly) hydrogen. As this matter condensed its temperature increased, as gravitational potential energy was converted into kinetic energy. At a temperature of $\sim 10,000,000 (10^{7}) \mathrm{~K}$ the atoms (which had lost their electrons again because of the higher temperature) began to undergo nuclear fusion. At this point we would probably call such an aggregate of matter a star. This process of hydrogen fusion produced a range of new types of nuclei. Hydrogen fusion, or hydrogen burning as it is sometimes called, is exemplified by reactions such as the formation of helium nuclei: $4^{1} $\mathrm{H}^{+} \rightarrow^{4} \(\mathrm{He}^{2+}+2 e^{+}+\text {energy. }$ When four protons are fused together they produce one helium-4 nucleus, containing two protons and two neutrons, plus two positrons (\(e^{+}$ – the antiparticle of the electron), and a great deal of energy. As the number of particles decreases ($4^{1} \(\mathrm{H}^{+}$ into $1 { }^{4} \mathrm{He}^{2+}$), the volume decreases. Gravity produces an increase in the density of the star (fewer particles in a smaller volume). The star’s core, where fusion occurs, gets smaller and smaller. The core does not usually collapse totally into a black hole, because the particles have a huge amount of kinetic energy, which keeps them in motion and moving on average away from one another.[5]
As the star’s inner temperature reaches $\sim 108 \mathrm{~K}$ there is enough kinetic energy available to drive other fusion reactions. For example three helium nuclei could fuse to form a carbon nuclei: $3 { }^{4} \mathrm{He}^{2+} \rightarrow { }^{12} \mathrm{C}^{6+}+\text { lots of energy (note again, the result is fewer atoms). }$
If the star is massive enough, a further collapse of its core would increase temperatures so that carbon nuclei could fuse, leading to a wide range of new types of nuclei, including those of elements up to iron (${ }^{56} \mathrm{Fe}^{26+}$) and nickel (${ }^{58} \mathrm{Ni}^{28+}$), as well as many of the most common elements found in living systems, such as nitrogen ($\mathrm{Ni}^{7}$), oxygen ($\mathrm{O}^{8}$), sodium ($\mathrm{Na}^{11}$), magnesium ($\mathrm{Mg}^{12}$), phosphorus ($\mathrm{P}^{15}$), sulfur ($\mathrm{S}^{16}$), chlorine ($\mathrm{Cl}^{17}$), potassium ($\mathrm{K}^{19}$), calcium ($\mathrm{Ca}^{20}$), manganese ($\mathrm{Mn}^{25}$), cobalt ($\mathrm{Co}^{27}$), copper ($\mathrm{Cu}^{29}$), and zinc ($\mathrm{Zn}^{30}$).
In some instances these nuclear reactions cause a rapid and catastrophic contraction of the star’s core followed by a vast explosion called a supernova. Supernovae can be observed today, often by amateur astronomers, in part because seeing one is a matter of luck. They are characterized by a sudden burst of electromagnetic radiation, as the supernova expels most of its matter into interstellar dust clouds. The huge energies involved in such stellar explosions are required to produce the naturally occurring elements heavier than iron and nickel, up to and including Uranium ($\mathrm{Ur}^{82+}$). The material from a supernova is ejected out into the interstellar regions, only to reform into new stars and planets and so begin the process all over. So the song is correct, many of the atoms in our bodies were produced by nuclear fusion reactions in the cores of stars that, at one point or another, must have blown up; we are literally stardust, except for the hydrogen formed before there were stars!
Looking at Stars
At this point you may still be unclear as to how we know all this. How can we know about processes and events that took place billions of years ago? Part of the answer lies in the fact that all the processes involved in the formation of new elements are still occurring today in the centers of stars. Our own Sun is an example of a fairly typical star; it is composed of $\sim 74%$ (by mass) and $\sim 92%$ (by volume) hydrogen, $\sim 24%$ helium, and trace amounts of heavier elements. There are many other stars (billions) just like it. How do we know? Analysis of the emission spectra of the light emitted by the Sun or the light emitted from any other celestial object enables us to deduce which elements are present.[6] Similarly, we can deduce which elements and molecules are present in the clouds between stars by looking at which wavelengths of light are absorbed! Remember that emission/absorption spectra are a result of the interaction between the atoms of a particular element and electromagnetic radiation (light). They serve as a fingerprint of that element (or molecule). The spectrum of a star reveals which elements are present. No matter where an element is found in the universe it appears to have the same spectroscopic properties.
Astrophysicists have concluded that our Sun (Sol) is a third generation star, which means that the material in it has already been through two cycles of condensation and explosive redistribution. This conclusion is based on the fact that the Sun contains materials (heavy elements) that it could not have formed itself, and so must have been generated previously within larger and/or exploding stars. Various types of data indicate that the Sun and its planetary system were formed by the rapid collapse of a molecular (mostly hydrogen) cloud $\sim 4.59$ billion years ago. It is possible that this collapse was triggered by a shock wave from a nearby supernova. The gas condensed in response to gravitational attraction and the conservation of angular momentum; most of this gas ($> 98%$) became the Sun, and the rest formed a flattened disc, known as a planetary nebulae. The planets were formed from this disc, with the small rocky/metallic planets closer to the Sun, gas giants further out, and remnants of the dust cloud distributed in the Oort cloud.[7] As we will see, living systems as we know them depend upon elements produced by second and third generation stars. This process of planet formation appears to be relatively common and more and more planetary systems are being discovered every year.[8]
Stars have a life cycle from birth to death; our Sun is currently about half way through this life cycle. There is not enough matter in the Sun for it to become a supernova, so when most of its hydrogen has undergone fusion, $\sim 5$ billion years from now, the Sun’s core will collapse and helium fusion will begin. This will lead to the formation of heavier elements. At this point, scientists predict that the Sun’s outer layer will expand and the Sun will be transformed into a red giant. Its radius will grow to be larger than the Earth’s current orbit. That will be it for life on Earth, although humans are likely to become extinct much sooner than that. Eventually the Sun will lose its outer layers of gas and they may become a part of other stars elsewhere in the galaxy. The remaining core will shrink, grow hotter and hotter, and eventually form a white dwarf star. Over (a very long) time, the Sun will cool down, stop emitting light, and fade away.
Questions
Questions to Answer
• How do the properties of isolated atoms or molecules give rise to the world we observe? Why are objects different colors, or have different melting points?
• Do isolated atoms/molecules exist in a state such as solid, liquid, or gas?
• Where do the atoms in your body come from? (Trace their origin back as far as you can.)
• How does the size of the universe influence the density of particles?
• How many protons, neutrons, and electrons does ${ }^{4} \mathrm{He}$ have? How about ${ }^{4} \mathrm{He}^{2+}$?
• Generate a graph that estimates the number of atoms in the universe as a function of time, beginning with the Big Bang and continuing up to the present day.
• Draw another graph to illustrate the number of elements in the universe as a function of time. Explain your reasoning behind both graphs.
Questions to Ponder
• Can an atom of one element change into an atom of another element?
• Is the number of atoms in the universe constant?
• How does the big bang theory constrain the time that life could have first arisen in the universe? | textbooks/chem/General_Chemistry/CLUE%3A_Chemistry_Life_the_Universe_and_Everything/03%3A_Elements_Bonding_and_Physical_Properties/3.1%3A_Elements_and_Bonding.txt |
“From the 115 elements you can build a near infinity of molecules, of any type you need, to get all the structural and functional diversity you can ask for. There are at least 100,000 different molecules in the human body. Some 900 volatile aroma components have been found in wine. Chemistry is molecules. We are molecules. Chemistry is a truly anthropic science.”
Roald Hoffman (1937) quoted by Natalie Angier in The Canon
By this point, you probably have some idea or mental image of the basic (low resolution) structure of atoms. We will therefore return to some questions left unanswered at the end of Chapter $1$. Why is it that two hydrogen atoms form a strong stable (covalent) bond, a bond that requires temperatures above $5000 \mathrm{~K}$ to break, whereas two helium atoms and two hydrogen molecules have only a weak (London dispersion forces) attraction for each other that is broken at very low temperatures ($4–20 \mathrm{~K}$)? Because temperature is a measure of the kinetic energy present in the system we have to ask, what gives rise to this huge difference? Is there something fundamentally different going on in these situations? Other potentially troubling questions may also come to mind, in particular, how can pure samples of different elements be so different? Why is carbon either black (graphite) or transparent (diamond)? Why is gold shiny and yellow, while sulfur is dull and yellow? Why are most metals, but not gold or copper, shiny, solid, and colorless, while mercury is shiny, colorless, and liquid? Why are some elements more or less inert (such as gold, which does not tarnish) while others, such as iron (which rusts) and phosphorus (which bursts into flames) are highly reactive? To answer these questions will lead to an understanding of the basics of chemistry or how atoms interact with one another under various conditions. We will approach the answers in a step-by-step manner. In some cases where the answers are very complex (as is the case for why gold is yellow and mercury is a liquid), we will sketch out the answer but probably not provide a satisfying explanation. Luckily, most of the chemistry we need to address is not nearly so arcane!
Before we consider these and other questions, let us recap what we think we know about atoms and electrons. Most of an atom’s mass is localized in a very small region, the nucleus, surrounded by electrons that occupy most of the volume of the atom. Electrons have a number of strange properties because they are quantum mechanical particles. This means that under some conditions their behavior is best described by considering them as negatively charged particles, and under other conditions it is more helpful to consider them as waves; they are really both (and neither). Because of the uncertainty principle when we know the energy of each electron rather accurately we do not (and cannot) know where, exactly, a particular electron is at a particular moment in time. In fact, because all electrons are identical, if we had two electrons and turned away from them we could not tell which was which when we turned back. Within an atom each electron has a discrete energy and is characterized by its set of quantum numbers; no two electrons in an atom have the same set of quantum numbers. Perhaps you will be disappointed (or perhaps pleased) to know that a rigorous quantum mechanical (and relativistic) treatment of atoms and their interactions is beyond the scope of this book.[9] That said, we can give a reasonable overview of how the behavior of atoms can be explained in terms of atomic and molecular electron orbitals. We will also indicate where our description is an over-simplification.
It is worth remembering that there are very few (if any) instances when we come across isolated atoms. Although we often describe matter as being composed of atoms, that is a bit of an abstraction; most atoms are stuck to other atoms by bonds and interactions. As mentioned previously, this leads to emergent properties that are quite distinct from those of the isolated atoms of which they are composed. It is the interactions between atoms and molecules that makes surfaces solid.
If isolated atoms are rare, the obvious question is, why are they rare? What determines when and how atoms interact? The answer is simple really and based on a principle we have already encountered (and that we will return to time and again): systems will adopt the lowest energy state accessible to them. The reason is that at the lowest accessible energy state, the forces of attraction and repulsion are equilibrated. It would take more energy to move the components of the system (that is atoms in molecules, or electrons in atoms) because the forces acting on them would increase. Interactions and bond formation lead to lower potential energy. Whether the bonded system is stable will then depend upon the strength of the interaction/bond and the forces that impact the molecule. For example, surrounding molecules/atoms with a range of kinetic energies may collide with the molecule. If this kinetic energy of the impacting particle is larger than the interaction/bond energy, the collision can disrupt the interaction or break the bond(s) between them; if not, the interaction/bond will be stable. At the same time, there must be overarching principles governing which interactions occur and which do not; otherwise everything would clump together, which would be messy and not particularly interesting. These principles arise from the way electrons are organized in different types of atoms.
Thinking about the nature of the chemical bond
There is no single explanation that captures all the properties observed when atoms interact to form a bond.[10] Instead we use a range of models of bonding. Now, what do we mean by model? Models are much more limited than theories, which have global application and can be proven wrong through observation and experimental data. Models are more like strategies that simplify working with and making predictions about complex systems. A model often applies to only very specific situations. For example the Bohr model of the atom applies only to hydrogen and then only under quite specific circumstances. We are going to consider a variety of bonding models, some of which you may already be familiar with, but it is important that you remember that different models are used depending upon which properties you want to predict and explain.
So back to our original dilemma, namely why is it that the interaction between two hydrogen atoms is so much stronger than that between two helium atoms? One useful model of bonding uses the idea that electrons can be described in terms of orbitals.[11] Each orbital can contain a maximum of two electrons (with opposite spins). Recall that in an isolated atom, the electrons are described by atomic orbitals; therefore when in molecules, they are described by molecular orbitals ($\mathrm{MOs}$). When atoms approach each other, the atomic orbitals containing their outermost electrons, known as the valence electrons, begin to interact. Because of the wavelike nature of the electron, these interactions can be either constructive or destructive. If they interact in a constructive manner, the interaction is stabilizing, which means that potential energy decreases and (if that energy is released into the surrounding system) the two atoms adopt a more stable configuration; they form a bond that holds them together. If the interaction is destructive, there is no stabilizing interaction. In the case of hydrogen each atom has a single ($1\mathrm{s}$) orbital occupied by a single electron. As the atoms approach one another these $1\mathrm{s}$ atomic orbitals interact to form two possible $\mathrm{MOs}$: a lower energy, constructive or bonding $\mathrm{MO}$, and a higher energy, destructive or anti-bonding $\mathrm{MO}$. Notice that the bonding $\mathrm{MOs}$, a so-called $\sigma 1\mathrm{s}$ (sigma) orbital, has electron density (that is a high probability that the electrons would be found there if we looked) between the two hydrogen nuclei. In the anti-bonding $\mathrm{MO}$, known as $\sigma^{\star} 1 \mathrm{s}$, the electrons are mostly not between the nuclei. One way to think about this is that in the bonding orbital the protons in the hydrogen nuclei are attracting both electrons (one from each atom) and it is this common attractive force between electrons and nuclei that holds the two hydrogen atoms together. In contrast in the anti-bonding orbital there is little electron density between the two nuclei and any electrons in that orbital are actually destabilizing the system by enhancing the repulsive interactions between the nuclei. (Can you provide a short reason why this would be the case?)
Just like an atomic orbital each $\mathrm{MO}$, both bonding and anti-bonding, can hold two electrons. In the case of two approaching hydrogens there are only two electrons present in the system and the lowest energy state would have them both in the bonding orbital. Typically, both electrons in a $\mathrm{H–H}$ molecule are found in the lower energy (more stable) $\sigma 1\mathrm{s}$ bonding orbital. This arrangement of electrons is referred to as a covalent bond; this is the arrangement that requires temperatures of $\sim 5000 \mathrm{~K}$ to break, which means it requires a lot of energy to break a covalent bond.
Now let us take a look at what happens when two helium atoms approach. Each He atom has two electrons in its $1\mathrm{s}$ orbital. As the orbitals approach they interact and again produce two $\mathrm{MOs}$, the bonding $\sigma 1\mathrm{s}$ orbital and the anti-bonding $\sigma^{\star} 1 \mathrm{s}$ orbital. The $\sigma^{\star} 1 \mathrm{s MO}$ has no electron density between the two He nuclei and has considerably higher energy than the atomic orbitals of the isolated atoms. Since there are 4 electrons present in the two He atoms and only two can occupy the $\sigma 1 \mathrm{s}$ bonding orbital; the other two have to go into the $\sigma^{\star} 1 \mathrm{s}$ anti-bonding orbital. The end result is that the decrease in potential energy (increased stability) associated with occupying the bonding orbital is more than off-set by the increased energy associated with occupying the $\sigma^{\star} 1 \mathrm{s}$ anti-bonding orbital. So, the end result is no overall stabilization and no decrease in energy associated with bond formation; no covalent bond is formed. The only interactions between helium atoms are the van der Waals interactions that occur between the two atoms that depend exclusively on London dispersion forces, as discussed in Chapter $1$.
The interaction between two helium atoms is very similar to that between two $\mathrm{H}_{2}$ molecules. There is no possibility of stabilizing $\mathrm{MOs}$ forming and, as in the case of the helium atoms, hydrogen molecules ($\mathrm{H–H}$ or $\mathrm{H}_{2}$) interact exclusively through London dispersion forces ($\mathrm{LDFs}$). The $\mathrm{LDFs}$ will be somewhat stronger between hydrogen molecules than between helium atoms, however, because there is a larger surface area over which they can interact.
The idea that—all other things being equal—a system will move to the lowest accessible energy state (losing the excess energy to their surroundings), where the forces of attraction and repulsion are equal, is applicable to a wide range of situations. The potential energy of the system falls as the distance between the atoms decreases until the system reaches a balance between the stabilizing interaction of bond formation and the destabilizing repulsion of the two nuclei. The energy difference between the separated atoms and the minimum energy is called the bond energy and this amount of energy must be supplied to the system to break the two atoms apart again. The distance between the nuclei when the bond energy is at its minimum is the bond length. When a bond is formed between two atoms energy is always released to the surroundings and the new material is always more stable than the two separate atoms. Because energy is conserved a bond cannot form unless this bond energy is transferred from the interacting atoms to the rest of the system (usually by colliding with other atoms and transferring energy). Making bonds is always exothermic (meaning that energy is released not absorbed). This implies that energy (from the surrounding system) is always needed to break a bond. To break a bond energy must be transferred from the surroundings. Bond breaking is endothermic meaning it requires energy from the external world, normally delivered through collisions with other molecules.
When we consider more complex chemical reactions we will find that these generally involve both bond breaking and bond formation; the overall reaction will be exothermic when more energy is released from bond formation than is used for bond breaking. Conversely a reaction is endothermic (that is, uses energy) if more energy is required to break bonds than is released in bond formation. The important point is that we have to consider the system as a whole, including all of the bonds formed and broken. We will come back to this topic (in much greater depth) in Chapters $5$ and $7$.
Discrete Versus Continuous Molecules
Having considered the bonding situation with hydrogen and helium, the simplest two elements, we can now move on to consider other elements and the types of molecules that they form. In this discussion, we begin with molecules made up of a single type of atom. More complex molecules made of atoms of multiple elements will be considered in the next and subsequent chapters. As the number of protons in the nucleus of an element’s atoms increases, from $1$ in hydrogen to $10$ in neon, we find dramatic changes in physical properties that correlate with whether the elemental form is discrete or continuous. The discrete forms are either monoatomic—meaning that they exist as separate atoms (such as $\mathrm{He}$ and $\mathrm{Ne}$) with no covalent bonds between them (although they do interact via van der Waals interactions)—or diatomic molecules (such as $\mathrm{H}_{2}$, $\mathrm{N}_{2}$, $\mathrm{O}_{2}$, and $\mathrm{F}_{2}$), meaning that they exist as molecules that have only two atoms. The elements that exist as small molecules have very low melting points (the temperatures at which they change from a solid to a liquid) and low boiling points (the temperatures at which they change from a liquid to a gas). But don’t confuse these phase transitions with the breaking of a diatomic molecule into separate atoms. Phase transitions, which we will discuss in greater detail later, involve disruption of interactions between molecules (intermolecular forces), such as London dispersion forces, rather than interactions within molecules, that is, covalent bonds.
Table $3.2.1$ The First 10 Elements in Their Naturally Occurring Elemental State
Elemental Form $\mathrm{H}_{2}$
molecular
$\mathrm{He}$
atomic
$\mathrm{Li}$
continuous
$\mathrm{Be}$
continuous
$\mathrm{B}$
continuous
$\mathrm{C}$
continuous
$\mathrm{N}_{2}$
molecular
$\mathrm{O}_{2}$
molecular
$\mathrm{F}_{2}$
molecular
$\mathrm{Ne}$
atomic
Melting Point $13.81 \mathrm{~K}$ $0.00 \mathrm{~K}$ $453.65 \mathrm{~K}$ $1560 \mathrm{~K}$ $2348 \mathrm{~K}$ $3823 \mathrm{~K}$ $63.15 \mathrm{~K}$ $54.36 \mathrm{~K}$ $53.53 \mathrm{~K}$ $24.56 \mathrm{~K}$
Boiling Point $20.28 \mathrm{~K}$ $4.22 \mathrm{~K}$ $1615 \mathrm{~K}$ $2744 \mathrm{~K}$ $4273 \mathrm{~K}$ $4098 \mathrm{~K}$ $77.36 \mathrm{~K}$ $90.20 \mathrm{~K}$ $85.03 \mathrm{~K}$ $27.07 \mathrm{~K}$
Bp-Mp (*) $6.47 \mathrm{~K}$ $3.27 \mathrm{~K}$ $1161 \mathrm{~K}$ $1184 \mathrm{~K}$ $1925 \mathrm{~K}$ $275 \mathrm{~K}$ $14.21 \mathrm{~K}$ $35.84 \mathrm{~K}$ $31.5 \mathrm{~K}$ $2.51 \mathrm{~K}$
Name hydrogen helium Lithium beryllium boron carbon nitrogen oxygen fluorine neon
* boiling point (Bp) minus melting point (Mp).
In contrast to the elements that form discrete molecules, the atoms of the other elements we are considering (that is $\mathrm{Li}$, $\mathrm{Be}$, $\mathrm{B}$, $\mathrm{C}$) interact with one another in a continuous manner. Rather than forming discrete molecules, these elements can form ensembles of atoms in which the number of atoms can range from the small (a few billion) to the astronomical (very, very large). Whether the materials are at the nano- or the macroscopic levels, the atoms in these ensembles are held together by bonds that are very difficult to break, like the bond in $\mathrm{H-H}$. That is, a lot of energy must be put into the system to separate the component atoms. However, unlike hydrogen, the atoms that form these structures must form bonds with more than one other atom.
A consequence of this difference in organization is a dramatic increase in both the melting and boiling points compared to atomic ($\mathrm{He}$, $\mathrm{Ne}$) and molecular ($\mathrm{H}_{2}$, $\mathrm{N}_{2}$, $\mathrm{O}_{2}$, and $\mathrm{F}_{2}$) species (Table $3.2.1$). The reason is that when a substance changes from solid to liquid (at the melting point) the component particles have to be able to move relative to one another. When the substance changes from a liquid to a gas (at the boiling point) the particles have to separate entirely. Consequently the magnitude of the melting and boiling points gives us a relative estimate of how strongly the particles are held together in the solid and liquid states. As we have already seen temperature is a measure of the average kinetic energy of the molecules in a system. For elements that exist as discrete atoms or molecules the only forces that are holding these particles together are London dispersion forces, which are relatively weak compared to covalent bonds. In contrast, the elements that exist as extensive networks of atoms joined by bonds require much more energy to break as the material goes from solid to liquid to gas.
Questions
Questions to ponder
• Are all bonds the same?
• What factors might influence bond strength?
• Why are the properties of atoms and molecules different?
Questions to Answer
• Where are the electrons in $\mathrm{H}_{2}$ when the temperature is greater than $5000 \mathrm{~K}$?
• What would happen if you could form a $\mathrm{He-He}$ system with 3 electrons (instead of 4)?
• What would a molecular-level picture of $\mathrm{H}_{2}$ (g) look like?
• What would a molecular-level picture of $\mathrm{H}$ (g) look like?
• Where does the energy to break a bond come from?
• Where does the energy released upon bond formation go?
• The melting point of molecular hydrogen ($\mathrm{H}_{2}$) is $\sim 14 \mathrm{~K} (-259^{\circ} \mathrm{~C})$. Draw a molecular level picture of what molecular hydrogen looks like below this temperature (as a solid). Why are the molecules of hydrogen sticking together?
• The boiling point of molecular hydrogen ($\mathrm{H}_{2}$) is $\sim 20 \mathrm{~K} (-253^{\circ} \mathrm{~C})$. Draw a molecular level picture of what molecular hydrogen looks like above this temperature (as a gas).
• Molecular hydrogen dissociates at high temperatures ($> 6000 \mathrm{~K}$). Draw a picture of what you imagine this might look like. Why do you think it takes such a high temperature to bring about this change? | textbooks/chem/General_Chemistry/CLUE%3A_Chemistry_Life_the_Universe_and_Everything/03%3A_Elements_Bonding_and_Physical_Properties/3.2%3A_Elements_and_Their_Interactions.txt |
To give you an idea of some of the different types of bonds that form between elements, we are going to consider several representative elements from different areas of the periodic table. For starters, let us take a look at carbon. Carbon ($\mathrm{C}$) belongs to the family of elements known as non-metals. The bonding between $\mathrm{C}$ atoms (and to other types of atoms) is typically described as covalent bonding where each bond involves two electrons (one from each of the bonded atoms). Although this is the most common model we will see that it is not the only possible one; we will introduce other models as they are needed. Diamond is the name given to one of the naturally occurring forms (known as allotropes) of pure $\mathrm{C}$; the other allotropes of carbon are graphite, graphene, and various fullerenes ($\downarrow$), which we will return to later. The important question is: How can these substances have such different properties, yet be composed of the same types of atoms? For example: diamond is hard, colorless, sparkly and do not conduct electricity, but graphite is soft, grey, shiny, and does conduct electricity. The answer lies in the molecular level structure of these allotropes of carbon.
Diamonds form from carbon-rich materials subjected to very high pressure ($45,000–60,000$ atmospheres) but relatively low temperatures ($900–1300 { }^{\circ}\mathrm{C}$).[12] Such conditions can be found about $100$ miles under the Earth’s crust, the region known as the lithosphere. Diamonds have also been found in asteroids, which originate from outside of the Earth. Diamonds are so valued because they are rare, sparkly, hard, and almost completely inert. It is very hard to make diamonds do anything at all except sit there and sparkle; they don’t dissolve in water and they melt only at very high temperatures ($\mathrm{mp } = 3330^{\circ}\mathrm{C}$). Diamond has the highest melting point of any known substance, so high that these measurements are actually done under high pressure and then calculated to estimate what the value would be at atmospheric pressure. In addition, when diamond is melted it decomposes. When it is heated in an atmosphere of oxygen it reacts to produce carbon dioxide; when oxygen is absent it transforms into graphite. There is no such thing as molten diamond. Diamonds are extremely hard (the hardest naturally occurring substance) and do not conduct electricity at all (as we will see, to conduct electricity, electrons must be able to flow through the material). This suggests that the molecular-level structure of diamond is quite different from that found in metals (which we will see are malleable and conduct electricity). Any useful model of diamond’s structure must explain how these properties arise from atomic interactions.
Let us step back and look at the properties of diamond and see if we can make sense of them. To be so stable (chemically inert) diamond must have very strong bonds that take a lot of energy to break. The fact that it does not conduct electricity indicates that the electrons must not be free to move around within a diamond. A polished diamond is sparkly because some light is reflected from the surface and some light passes through it, making it transparent. If the diamond were not cut with so many facets it would allow most light to pass through it.
When we look at an X-ray diffraction-based structure[13] of diamond we find that each carbon atom is surrounded by four other carbon atoms situated at equal distances and equal angles from each other. In this context, the most useful model of bonding involves thinking of each carbon atom as forming four covalent (electron-sharing) bonds, all arranged so that the electron pairs are as far apart as possible. This places the four bonded atoms at the corners of a tetrahedron, with a central carbon atom. Each of these corner atoms is itself at the center of a similar tetrahedron of carbons (see Figure). Experimental evidence indicates that all the bonds, bond lengths, and bond angles in diamond are identical; the $\mathrm{C–C–C}$ bond angles are $\sim 109^{\circ}$. A diamond can be considered as one huge molecule connected by a network of carbon–carbon bonds.
How do we explain this arrangement in terms of what we know about the electronic structure of carbon atoms? The answer is that the electronic structure of the carbon atoms is reorganized to form bonding orbitals. In the case of carbon, each atom can form four bonding orbitals that are oriented as far apart as possible. There are several models to explain how this occurs, but it is important to remember that they are all models, designed to help us understand the properties of diamond.
The Hybrid Orbital Model
In this model the orbitals involved in carbon–carbon bonding are considered to be hybrids or mixtures of atomic orbitals. If carbon forms four bonds (and it does) then four bonding orbitals are needed. Carbon has available orbitals in the second ($n = 2$) quantum shell: the $2\mathrm{s}$, $2\mathrm{p}_{\mathrm{x}}$, $2\mathrm{p}_{\mathrm{y}}$, and $2\mathrm{p}_{\mathrm{z}}$ In an isolated carbon atom there is a full $2\mathrm{s}$ orbital and two half-filled $\mathrm{p}$ orbitals. When the carbon atoms form a bond, these orbitals are somehow mysteriously transformed into four new bonding orbitals, which are called $\mathrm{sp}^{3}$ hybrid orbitals because they are a mixture of an $\mathrm{s}$ and three $\mathrm{p}$ orbitals. These $\mathrm{sp}^{3}$ orbitals exist only in the context of bonded carbon; they are not present in isolated carbon atoms. They spring into existence when one carbon atom interacts with another atom to form a bond; they are generated through the interaction. In the case of carbon the four electron clouds (bonds) move as far apart as possible to minimize the repulsions between them, adopting a tetrahedral configuration (\rightarrow).
The Molecular Orbital Model
Another way to consider how these bonds form is similar to the way we approached molecular hydrogen. That is, we consider that when carbon–carbon ($\mathrm{C–C}$) bonds form, atomic orbitals are transformed into molecular orbitals ($\mathrm{MOs}$). For each stabilizing bonding orbital, a destabilizing antibonding orbital is also formed. Using the molecular orbital approach, we can model the bonding in diamond as carbon atoms forming a three-dimensional network held together by these molecular bonding orbitals. $\mathrm{C–C}$ bonds are very stable because there is a large energy gap between the bonding orbitals and the high-energy antibonding orbitals. The bonding molecular orbitals are occupied while the antibonding molecular orbitals are unoccupied. Because of this large gap between the filled bonding and empty antibonding orbitals it is hard to remove an electron from a $\mathrm{C–C}$ bonding $\mathrm{MO}$. The electrons are not free to move between energy levels. Given that electrical conduction depends upon the relatively free movement of electrons it is not surprising that diamonds do not conduct electricity. But why, you might ask, is a diamond transparent, rather than opaque, like a block of graphite, which is also composed of only carbon atoms? For an object to be transparent most of the light that hits it must pass through it; the light can be neither reflected or absorbed. For a diamond to absorb light a photon would need to move an electron from a low-energy bonding $\mathrm{MO}$ to a high-energy antibonding orbital. However, visible light does not have enough energy to bridge the energy gap between the bonding and antibonding orbitals. Based on this thinking we conclude that there is something different between bonds holding C atoms together in diamond from the bonds holding $\mathrm{C}$ atoms together in graphite even though we do not know, at this point, what it could be.
An important point to consider here is that we have described the bonding in carbon using two different models: the hybrid orbital (valence bond) and molecular orbital models. Although this may be (a bit!) confusing, and may take some getting used to, it is quite common to describe chemical and physical phenomena using different models. Typically we use the simplest model that will allow us to explain and predict the phenomenon we are interested in. Usually the bonding in carbon is described using the hybrid orbital model, because it is highly predictive and easier to use in practice.
Graphite
As we have already mentioned, different allotropes (different forms of the same element) can have quite different properties. The carbon allotrope graphite is soft, grey/black, opaque, conducts electricity, and slippery – it makes a good lubricant.[14] Diamond is hard, transparent, and does not conduct electricity. How can this be possible if both are pure carbon? The answer lies in how the carbon atoms are organized with respect to one another. Whereas the carbon atoms in diamond form a three-dimensional network, in graphite, the atoms are organized in two-dimensional sheets that stack one on top of the other. Within each two-dimensional sheet the carbon atoms are linked by covalent bonds in an extended array of six-membered rings. This means that the carbon sheets are very strongly bonded, but the interactions between sheets are much weaker. Although there are no covalent bonds between the sheets, the atoms of the sheets do interact through London dispersion forces, very much like the interactions that hold helium atoms together. Because the sheets interact over very much larger surface areas ($\rightarrow), however, these interactions are much stronger than those in helium. Yet another allotrope of carbon, graphene, consists of a single sheet of carbon atoms.[15] These sheets can be rolled into tubes to form nanotubes that are the subject of intense research interest because of their inherently high tensile strength. Carbon atoms can also form spherical molecules, known as buckminsterfullerenes or buckyballs.[16] The obvious question is, why don’t covalent bonds form between graphite sheets? Why are the patterns of covalent bonding so different: three-dimensional (tetrahedral) in diamond, with each carbon bonded to four others, and two-dimensional (planar) in graphite and graphene, with each carbon atom bonded to only three others? One way to describe the molecular structure is to use the hybrid orbital bonding model. As we discussed previously, to form the four bonds attached to each carbon atom in diamond, we needed to hybridize four atomic orbitals to form four bonding orbitals. We might think we only need three bonds in graphite/graphene because each carbon is only connected to three others. This is not exactly true. In graphite and graphene we use a model in which only three atomic orbitals are hybridized—an s and two \(2\mathrm{p}$ orbitals in order to form three $\mathrm{sp}^{2}$ bonding orbitals. These orbitals attach each carbon atom to three other atoms. Just like in diamond the three bonds associated with each carbon atom in graphite/graphene move as far apart as possible to minimize electron pair repulsion; they lie at the points of a triangle (rather than a tetrahedron). This geometry is called trigonal planar and the $\mathrm{C–C–C}$ bond angle is $120^{\circ}$ ($\rightarrow$).
All well and good, but this does not really explain why the carbons in graphite/graphene are attached to three other carbon atoms, whereas in diamond each carbon is attached to four others. Perhaps surprisingly there is no good answer for why carbon takes up different forms—except that it can. But in fact carbon does form four bonds in graphite (carbon almost always forms four bonds—a central principle of organic chemistry). The trick is that the four bonds are not always equivalent; in graphite the fourth bond is not formed by the $\mathrm{sp}^{2}$ bonding orbitals but rather involves an unhybridized $2\mathrm{p}$ atomic orbital. These p orbitals stick out at right angles to the sheet and can overlap with $\mathrm{p}$ orbitals from adjacent carbons in the same sheet (see Figure). Remember that $\mathrm{p}$ orbitals have two regions of electron density. To explain the fact that graphite conducts electricity, we use an idea from molecular orbital ($\mathrm{MO}$) theory, namely that bonding and antibonding $\mathrm{MOs}$ are formed from the adjacent $\mathrm{p}$ orbitals that extend over the sheet surface. The energy different between these orbitals is not large and electrons can move from one to the other, allowing the movement of electrons throughout the whole sheet of graphite, which gives it many of the properties that we associate with metals. Note that we use both the hybridization model, which explains the planar framework of $\mathrm{C-C}$ bonds in graphite, and molecular orbital theory, which explains graphite’s electrical conductivity. So before we delve further into the properties associated with graphite, let us take a look at bonding in metals.
Questions
Questions to Answer
• Diamond and graphite appear to be quite different substances, yet both contain only carbon atoms. Why are the observable properties of diamond and graphite so different when they are made of the same substance?
• The electron configuration of $C$ is $1\mathrm{s}^{2}2\mathrm{s}^{2}2\mathrm{p}^{2}$. Using the idea that each atom provides one electron to a bond, if carbon used atomic orbitals to bond, how many bonds would it form? Would they all be the same? What would be the bond angles if this were to happen? (Draw a picture of what this might look like.)
• The electron configuration of $C$ is $1\mathrm{s}^{2}2\mathrm{s}^{2}2\mathrm{p}^{2}$ this means that carbon has 6 electrons. Why doesn’t it form 6 bonds?
• We have seen that carbon can form materials in which it bonds to 4 other atoms ($\mathrm{sp}^{3}$ hybridization) or three other atoms ($\mathrm{sp}^{2}$ hybridization). What would be the hybridization for a carbon that was only bonded to two atoms? How would the other (unhybridized) $\mathrm{p}$ orbitals influence the behavior of such material (assuming that it could form)?
Questions to Ponder
• Could carbon form a three-dimensional structure by linking to two other carbon atoms?
• Do you think diamonds are transparent to all forms of light, such as X-rays?
• What does the color of graphite imply about the energies of the photons it absorbs? | textbooks/chem/General_Chemistry/CLUE%3A_Chemistry_Life_the_Universe_and_Everything/03%3A_Elements_Bonding_and_Physical_Properties/3.3%3A_Carbon%3A_An_Amazingly_Allotropic_Element.txt |
Metals have quite a wide range of properties at normal temperatures, from liquid (like mercury) to extremely hard (like tungsten). Most are shiny but not all are colorless. For example gold and copper have distinct colors. All metals conduct electricity but not all equally. How can we explain all these properties? Let us use aluminum ($\mathrm{Al}$) as an example because most of us have something made of aluminum such as a pan or aluminum foil. With modern instrumentation it is quite easy to visualize atoms and a variety of techniques have been used to image where the aluminum atoms are in the solid structure. What emerges is a picture of aluminum nuclei and their core electrons, packed like spheres where one layer of spheres rests in the interstices of the underlying and overlying layers; where the positions of the electrons are within this structure not well defined.
In $\mathrm{H-H}$ or diamond the electrons involved in bonding are located (most probably) between the two nuclei. In contrast in aluminum and other metals the valence electrons are not closely associated with each nucleus. Instead they are dispersed over the whole macroscopic piece of metal. Imagine that instead of two or three or four atomic orbitals combining to form $\mathrm{MOs}$, a mole ($6 \times 10^{23}$) of atomic orbitals were combined to produce a mole of $\mathrm{MOs}$. As more and more $\mathrm{MOs}$ are formed the energies between them gets smaller and smaller. For a macroscopic piece of metal (one you can see) the energy gap between the individual bonding $\mathrm{MOs}$ will be negligible for all intents and purposes. These orbitals produce what is essentially a continuous band of (low-energy) bonding $\mathrm{MOs}$ and a continuous band of (higher-energy) anti-bonding $\mathrm{MOs}$. The energy gap between the bonding and anti-bonding orbitals is called the band-gap and in a metal this band-gap is quite small (recall that the gap between the bonding and anti-bonding $\mathrm{MOs}$ in diamond is very large). Moreover in metals the bonding $\mathrm{MOs}$ (known as the valence band) are able to accommodate more electrons. This is because in metals there are typically fewer electrons than there are atomic orbitals. Consider aluminum: it has three valence electrons and in the ground (lowest energy) state has an electron configuration of $3\mathrm{s}^{2} 3\mathrm{p}^{1}$. This suggests that it has two unoccupied $3\mathrm{p}$ orbitals. We can consider the bonding $\mathrm{MOs}$ in aluminum to be formed from all the available atomic orbitals, which means that there are many bonding $\mathrm{MOs}$ that are not occupied by electrons. The physical consequences of this are that the valence electrons can move relatively easily from one $\mathrm{MO}$ to another because their energies are very close together. Whereas nuclei and core electrons remain more or less locked in position the valence electrons can spread out to form a kind of electron sea within the metal. When an electrical potential is applied across the metal, electrons from an external source can easily enter the valence band and electrons can just as easily leave the metal. Electrical conductivity is essentially a measure of how easily electrons can flow through a substance. Metals typically have high conductivity due to the ease with which electrons can move from one $\mathrm{MO}$ to another and the fact that each $\mathrm{MO}$ extends throughout the whole piece of metal. Because the numbers of electrons entering and leaving are the same, the piece of metal remains uncharged.
In this model the atomic cores are packed together and surrounded by a cloud of electrons that serve as the “glue” that binds them together. There are no discrete bonds in this type of structure. When a piece of metal is put under physical stress (for example it is stretched or deformed) the atoms can move relative to one another but the electrons remain spread throughout the structure. Metals can often be slowly deformed into different shapes without losing their structural integrity or electrical conductivity—they are malleable! They can be melted (increased atomic movement), become liquid, and then allowed to cool until they solidify; throughout this process they retain their integrity and their metallic properties and so continue to conduct electricity.[17] This is quite different from how other substances (such as diamond or water) behave. The hardness of a solid metal depends on how well its atoms packed together and how many electrons are contributed to the valence band of orbitals.
So why do some elements behave as metals and others do not? For example graphite conducts electricity but it is not malleable and can’t be heated and molded into other shapes. The answer lies in the behavior of the $\mathrm{MOs}$ and the resulting bonds they can produce. Graphite has a rigid backbone of carbon–carbon bonds that makes it strong and stable but overlaying those bonds is the set of delocalized $\mathrm{MOs}$ that spread out over the whole sheet. As a result graphite has some properties that are similar to diamond (stability and strength), some that are similar to metals (electrical conductivity), and some that are a consequence of its unique sheet structure (slipperiness).
Why Are Metals Shiny?
We see things because photons hit the back of our retinas and are absorbed by specialized molecules (proteins and associated pigment molecules). This leads to changes in protein structure and initiates a cascade of neuron-based cellular events that alters brain activity. So where do these photons come from? First and foremost they can be emitted from a source (the Sun, a light bulb, etc.) that appears to shine and can be seen in the dark. Alternatively, photons can be reflected off a surface; in fact most of the things we see do not emit light, but rather reflect it. A red T-shirt appears red because it absorbs other colors and reflects red light. Photons can also be refracted when they pass through a substance. A cut diamond sparkles because light is refracted as it passes through the material and exits from the many facets. Refraction is caused when photons bump into electrons, are absorbed, and then (very shortly thereafter) are re-emitted as they travel through a material. These processes take time, so the apparent speed of light slows down. It can take a photon many thousands of years to move from the core to the surface of the Sun because of all the collisions that it makes during the journey.[18]
To explain why metals (and graphite) are shiny, we invoke a combination of reflection, refraction, and the energy levels of $\mathrm{MOs}$. When a photon of light is absorbed and reemitted, the electron moves from one orbital to another. Let us consider a piece of metal at room temperature. When a photon arrives at the metal’s surface it encounters the almost continuous band of $\mathrm{MOs}$. Most photons, regardless of their wavelength, can be absorbed because there is an energy gap between orbitals corresponding to the energy of the photon. This process promotes electrons up to a higher energy level. As the electrons drop back down to a lower energy level, the photons are re-emitted, resulting in the characteristic metallic luster. Metals actually emit light, although this does not mean metals glow in the dark (like a light bulb or the Sun). Instead, metals absorb and re-emit photons, even at room temperature.
The color of a particular metal depends upon the range of wavelengths that are re-emitted. For most metals the photons re-emitted have a wide range of wavelengths which makes the metallic surface silvery. A few metals, such as copper and gold, absorb light in the blue region and re-emit light with wavelengths that are biased toward the red end region of the spectrum ($400–700 \mathrm{~nm}$) and therefore they appear yellowish. This is due to relativistic effects way beyond the scope of this book, but something to look forward to in your future physical chemistry studies!
Now we can also understand why metals emit light when they are heated. The kinetic energy of the atoms increases with temperature which promotes electrons from low to higher energy orbitals. When these electrons lose that energy by returning to the ground state, it is emitted as light. The higher the temperature the shorter the wavelength of the emitted light. As a filament heats up, it first glows red and then increasing whiter as photons of more and more wavelengths are emitted.
Questions
Questions to Answer
• What properties indicate that a substance is metallic?
• Why are metals shiny?
• How can metallic properties be explained by the atomic-molecular structure of $\mathrm{Al}$ (for example)?
• Why can we see through diamond but not aluminum? How about graphite?
• Why does aluminum (and for that matter all metals) conduct electricity? What must be happening at the atomic-molecular scale for this to occur?
• What does the fact that diamond doesn’t conduct electricity tell you about the bonding in diamond?
• How do the bonding models for diamond and graphite explain the differences in properties between diamond, graphite, and a metal like aluminum?
• Why is it OK to use different models to describe bonding in different species?
This chapter has brought us to a point where we should have a fairly good idea of the kinds of interactions that can occur among atoms of the same element. We have seen that the properties of different elements can be explained by considering the structure of their atoms and in particular the way their electrons behave as the atoms interact to form molecules or large assemblies of atoms (like diamond.) What we have not considered yet is how atoms of different elements interact to form compounds (substances that have more than one element). In Chapter $4$ we will take up this subject and much more. | textbooks/chem/General_Chemistry/CLUE%3A_Chemistry_Life_the_Universe_and_Everything/03%3A_Elements_Bonding_and_Physical_Properties/3.4%3A_Metals.txt |
1. http://www.space.com/scienceastronom...ay_040524.html
2. http://www.youtube.com/watch?v=th_9Z...W7hkUMVaZz4eDg
3. http://www.astro.ucla.edu/~wright/BBhistory.html
4. In this nomenclature (described more on the web), the first superscript number is the number of proton and neutrons, while the second superscript number is the number of protons; both numbers are always integers. The letter is the symbol of the element, e.g. $\mathrm{He}$ for helium or $\mathrm{Li}$ for lithium.
5. Nuclear fusion releases huge amounts of energy (some of the mass is transformed into energy). On Earth, controllable nuclear fusion has long been a potential target in the search for new energy sources, but so far the energy required to bring about the initial fusion has not been replaced when the fusion occurs – i.e., nuclear fusion reactors have yet to break even. Uncontrolled nuclear fusion takes place in hydrogen bombs – clearly not a viable option for a useful energy source at the moment. Nuclear fusion does however take place in stars, and is self-sustaining. The reason you can see and feel the energy from the Sun is that is it undergoing nuclear fusion reactions, which supply us with almost all the energy that is used on Earth today.
6. It has been estimated that it takes between 10,000 to 170,000 years for a photon released during a fusion reaction at the Sun’s core to reach its surface. http://sunearthday.nasa.gov/2007/loc...t_sunlight.php
7. More physics that we will conveniently pass over, but it is worth noting that this is why the planets all move around the Sun in the same direction.
8. You may want to search the web for “extrasolar planets.”
9. For those who want more, rest assured that you will find out if you take more advanced classes either in physics or physical chemistry.
10. This study shows images of bonds forming http://www.sciencemag.org/content/34.../1434.abstract
11. Although perhaps the word orbital is confusing because it implies a circular or elliptical motion, what we mean is the volume in which there is a 90% probability of finding an electron. That said, orbitals are the way chemists (and the occasional physicist) talk, so we have to use it.
12. How much pressure is that exactly in real world terms?
13. http://en.Wikipedia.org/wiki/X-ray_crystallography
14. In fact the sheets in graphite do not slip relative to each other very readily. On Earth graphite is a lubricant, but in space in the absence of small molecules like $\mathrm{O}_{2}$, $\mathrm{N}_{2}$ and $\mathrm{H}_{2}\mathrm{O}$, graphite does not lubricate. It is thought that the sheets slip relative to each other as if they were rolling on ball bearings (the small molecules). As you might imagine, this discovery caused some consternation in high-flying airplanes where the engines began to fail because of lack of lubrication.
15. The Nobel Prize in Physics was awarded in 2010 for the discovery of graphene. http://nobelprize.org/nobel_prizes/p...aureates/2010/
16. In 1996 Smalley, Kroto, and Curl were awarded the Nobel Prize in Chemistry for the discovery of fullerenes. http://nobelprize.org/nobel_prizes/c...aureates/1996/
17. We need to mentions (at least) what electricity is, i.e. the flow of electrons.
18. Robert Naeye (1998). Through the Eyes of Hubble: Birth, Life and Violent Death of Stars. CRC Press. ISBN 0750304847. Of course this raises the question, is it the same photon? | textbooks/chem/General_Chemistry/CLUE%3A_Chemistry_Life_the_Universe_and_Everything/03%3A_Elements_Bonding_and_Physical_Properties/3.5%3A_In-Text_References.txt |
Up until this point we have considered only bonds between atoms of the same element. While this makes things simpler (although you might not agree after thinking about the many forms of carbon), it leaves out the vast majority of the compounds that exist in the world and their chemistries. Moreover, pure elements are rare in nature. Much of the efforts of alchemists, early chemists, and the modern refining industry involve determining how to (economically) separate specific types of atoms (elements) away from others. Modern chemistry is concerned (largely) with putting atoms together to form new and useful molecules. Both involve understanding the concepts underlying how atoms interact.
04: Heterogeneous Compounds
To extend our discussion to the wider world of what we might call heterogenous molecules, that is, molecules made up of atoms of more than one element, we will begin with carbon. Why carbon? Well, here are some reasons. Carbon is the fourth most abundant element in the universe ($\sim 3,032$ atoms per million), after hydrogen ($\sim 705,700$ atoms per million), helium ($\sim 275,200$ atoms per million), and oxygen ($\sim 5,920$ atoms per million). Carbon is distinguished from most other elements in its ability to form a vast array of diverse compounds by bonding with itself and other elements with bonds that are not too strong and not too weak. Under the conditions that persist on the surface of the Earth carbon compounds are stable enough to hang around but not stable enough to persist forever, so they are not dead ends. Carbon is a key building block of the major molecules of life: proteins, nucleic acids, lipids, and carbohydrates. We are carbon-based life forms! Carbon compounds are also used in a wide range of synthetic materials, such as pharmaceuticals, polymers, and high-tech materials; we also consume a lot of carbon compounds by burning them for fuel.
Carbon: Always Tetravalent and Often Tetrahedral
Atoms combine in many different ways. We have already seen an example of how a covalent bond can form between two hydrogen atoms producing molecular ($\mathrm{H}_{2}$) as opposed to the atomic form of hydrogen. Similarly atoms of carbon can be linked together in various ways to form diamond, graphite, and graphene (see Chapter $3$). Now we move on to molecules involving atoms of carbon and other elements. In keeping with our ongoing attempt to keep things simple (or better put, as simple as possible), let us start by examining the types of molecules that can be formed by combining carbon with hydrogen. There are many such molecules, and collectively they are known as hydrocarbons. The simplest such compound is methane $\mathrm{CH}_{4}$, a major component of natural gas. As in all its compounds and its elemental forms, carbon is tetravalent, which means that it always forms four bonds. We will now consider in greater detail why this is so, what forms the bonds can take, and what are the consequences of this fact. In this discussion, we will be building on the ideas introduced when we talked about diamond, graphite, and graphene.
To answer these questions we need to return to the ideas (introduced in Chapter $2$) about the quantization of electron energy levels. Carbon has a total of six electrons, two of which are in a filled ($1\mathrm{s}$) quantum shell, and four valence electrons; it is these valence electrons that can take part in bonding. Remember that the formation of a bond always lowers the energy of a system. It therefore makes sense that a carbon atom would form as many bonds as possible, resulting in the most stable possible molecular species.
What happens if we combine hydrogen with carbon? Do we get a compound with properties intermediate between the two? Absolutely not, as you might have expected when considering the differences between diamonds and graphite. As previously we use the hybridization model to explain the behaviors we observe. We begin with what we know: in methane the carbon atoms make four bonds, one to each of four hydrogen atoms. We also know, from experiment, that the shape of the methane molecule is tetrahedral; there is a carbon at the center and the four $\mathrm{C-H}$ bonds pointing towards the corners of a four-sided figure. Since each $\mathrm{C-H}$ bond is formed from bonding orbitals we can use the model for bonding where these four bonding orbitals arise from the “hybridization” of the pre-existing $2\mathrm{s}$ and three $2\mathrm{p}$ atomic orbitals. The electrons in the $1\mathrm{s}$ orbital are not used because the amount of energy needed to use those electrons is greater than the energy that would be released upon bond formation (they are held tightly to the nucleus by the electromagnetic force). It turns out to be a general rule that electrons in the core of the atom—in filled shells—tend not to take part in bonding. This means we need only consider the valence electrons when thinking about bonding.
The hybridization of the $2\mathrm{s}$ and the three $2\mathrm{p}$ orbitals results in four $\mathrm{sp}^{3}$ molecular orbitals, each of which can interact with the $\mathrm{H}$ atom’s $1\mathrm{s}$ orbital to form a bond. When a bonding orbital is formed it contains two electrons. Because carbon has four valence electrons and each of the four hydrogens has one electron the result is a total of eight electrons distributed in four bonding orbitals.
Recall that we say the hybridization of carbon is $\mathrm{sp}^{3}$ and the arrangement of the bonds is tetrahedral, which means the angle between orbitals (and the C–H bonds) is 109.5º. Another way to say this is that the $\mathrm{H-C-H}$ bond angle is $109.5^{\circ}$. We can predict that this will be the case based on theoretical calculations; these have been confirmed by experimental observations. But why should this be true? How many different arrangements are there for four hydrogens bonded to a single carbon? Why aren’t the hydrogens all arranged in a single plane (around a central $\mathrm{C}$ with $90^{\circ}$ bond angles) rather than in the tetrahedral arrangement? The planar arrangement, which is known as a square planar geometry, is actually possible and is sometimes observed under some special conditions, usually in molecules involving transition metals as we will see later). The square planar arrangement is not as stable as the tetrahedral arrangement for carbon because each $\mathrm{C-H}$ bond can be considered as a region of high electron (negative charge) density. Given that like repels like, each bond repels the others and moves as far away from the other bonds as possible. The optimum bond angle turns out to be $109.5^{\circ}$ away from each of their neighbors. At that point, if they moved away from one orbital they would move closer to another. You may want to convince yourself of this geometric fact by using a marshmallow, toothpicks, and gumdrops! This principle goes by the unwieldy name of valence shell electron pair repulsion (VSEPR) and can be used to predict (once you get the hang of it) the three-dimensional (3D) structure of simple molecules—assuming that you know how the atoms within a molecule are connected. For example, using VSEPR logic, you should be able to present a compelling argument for why the $\mathrm{C-H}$ bonds in methane do not adopt a square planar orientation, as well as the general shape of many other types of molecules. You can even go further, in methane all four atoms attached to the central carbon are the same but what if they are different? You should be able to make plausible predictions about how bond angles would change if one of the attached groups is larger than the others – how would that influence bond angles?
One problem for many people is that 3D visualization of molecular structures is not easy. It is particularly tricky when one is called upon to translate the more or less abstract two-dimensional (2D) representations (Lewis and dot stuctures $\downarrow$) that you find printed on the page of a book, into a 3D model you can manipulate with your hands or in your mind. In addition, chemists (and molecular biologists) have an annoying tendency of representing complex 3D structures using various 2D representations, which can be confusing if you don’t know what you are looking at (or for). You have probably already seen some of these different structures, and we will consider a number of them below. Each provides specific kinds of information about the molecule. Note that actual 3D physical models and web activities can be very helpful in solidifying your ideas about structure.
If we were able to see a methane molecule, what we observe would probably be closest to the electrostatic potential map. This visualization provides a picture of the surface of the molecule, generally color coded to represent fluctuations in electron density. Notice that there are no color fluctuations on this model of methane indicating that there are no (permanent) electron cloud distortions in the molecule—the surface of the molecule is uniformly electrically neutral. What is not so easy to discern from this representation is the fact that the methane is tetrahedral or that the central carbon atom is bonded to four hydrogen atoms, a fact that is much easier to appreciate in the other representations. The electrostatic potential representation is very useful for large biological molecules for several reasons: it is much simpler than the other kinds of models because individual atoms are not represented; it shows the molecule’s shape; and it shows where charges and partial charges are located.
The space-filling or van der Waals model gives more structural information in that the individual atoms that make up the molecule are distinguished by color (black for carbon, white for hydrogen, red for oxygen, and blue for nitrogen.) The surface of the model represents the molecule’s van der Waals radius, which is the distance where attraction turns to repulsion when two molecules approach one another. As its name implies, such models represent the space occupied by each atom.
The ball-and-stick model of methane shows the central carbon (black ball) attached to four hydrogens (white balls) by sticks that represent the bonds between the atoms. Although this model is probably the easiest to visualize, it is misleading because it could give the impression that bonds are like sticks holding the atoms together. It also does not represent either the actual volume occupied by the molecule or its electrostatic surface features. Another problem with all three of the preceding types of models is that you need a computer and specialized software (or some artistic ability) to draw them, which may not always be convenient or possible.
One strategy to address this problem is through what is known as a perspective formula. In a perspective formula the atoms are represented by their atomic symbols (for example, $\mathrm{C}$ or $\mathrm{H}$) and bonds are represented by various kinds of lines. A normal line is meant to indicate a bond that is in the plane of the paper, a wedged line represents a bond that is coming out of the plane toward you (the reader), and a hatched line represents a bond that is coming out of the plane, but away from you. This convention makes it easier to draw 3D perspective structures by hand without specialized software (or graphical talent.) We can, in fact, go one step further and draw methane without indicating its 3D structure at all. Structures that show all the bonds, atoms, and any valence electrons that are not in bonds, but do not attempt to accurately represent the 3D shape of a molecule are called Lewis structures. The Lewis structure for methane (see above) and the molecular formula $\mathrm{CH}_{4}$ represent a chemical shorthand that can provide a huge amount of information; we will see even more extreme examples as we go on. However, to be able to understand these representations, you must already know that the methane molecule is tetrahedral and the rules that apply to the geometry of carbon bonds, because neither is shown explicitly. If you didn’t know these things, you might even be tempted to assume that methane is organized with a square planar geometry or that the hydrogens are all located to one side of the carbon atom, neither of which is true!
Why, you might ask, would one want to draw structures with so much information missing? Perhaps, like medieval alchemists, modern chemists want to keep their secrets from the average person. Perhaps they just like secret codes and mystical symbols. Or perhaps it is because these shorthand representations of molecules are just much more compact and easy to draw, particularly when we get to large molecules with lots of atoms.[1] Drawing Lewis structures is an important and useful chemistry skill and we will return to it in more detail shortly. Once you have mastered it you will be able to look at a molecular formula such as $\mathrm{CH}_{4}$ (or $\mathrm{C}_{5}\mathrm{H}_{12}$) and (together with other information) be able to visualize the 3D structure of the molecule represented and predict many of the substance’s physical and chemical properties.
For example, models of the methane molecule predict that it is symmetrical. Again, this might not be entirely obvious just by looking at the structure, but if you make a model, or look at a rotatable interactive 3D model on the web you will see that it does not matter which way you look at the structure—all the $\mathrm{C-H}$ bonds are the same, and all the bond angles are the same. A little more information (which we will discuss later on) will let you deduce that there are no permanent electron density distortions in the molecule—just as is shown by the electrostatic potential map. Together these enable you to deduce that methane molecules are attracted to one another solely through London dispersion forces (like helium atoms or hydrogen molecules). Given how weak these interactions between molecules are we might be brave enough to predict that the melting and boiling points of methane are low (melting and boiling occur at relatively low temperatures) and we would be right! Methane melts at $91 \mathrm{~K}$ and boils at $112 \mathrm{~K}$.[2]
Questions
Question to Answer
• Why (when present) are the four bonds formed by carbon usually arranged so that they point towards the corners of a tetrahedron?
Questions to Ponder
• If bond formation is stabilizing, why doesn’t carbon form six bonds, given that it has six electrons?
• Why doesn’t helium bond with carbon?
• What would be the consequences if carbon bonds with other atoms were very weak?
• What would be the consequences if carbon bonds with other atoms were very strong?
Building Increasingly Complex Molecules
You will soon realize that it is possible to build a rather amazing number of compounds using just hydrogen and carbon. For example imagine that we remove one hydrogen from a methane molecule; this leaves us with what is known as a methyl ($\mathrm{-CH}_{3}$) group. We can combine two methyl groups by forming a $\mathrm{C-C}$ bond between them (you might want to convince yourself that each carbon atom is still making four bonds with neighboring atoms). The resulting molecule is known as ethane ($\rightarrow$). The structure of ethane can be written in a number of ways, for example $\mathrm{H}_{3}\mathrm{C-CH}_{3}$, $\mathrm{CH}_{3}\mathrm{-CH}_{3}$ or $\mathrm{C}_{2}\mathrm{H}_{6}$. As the number of atoms increases so does the number of different ways a molecule can be represented. It is for this reason that chemists have developed a number of rules that are rather strictly adhered to; these rules make it possible to unambiguously communicate the structure of a molecule to others.[3] We will not spend much time on all of these various rules but there are web activities that you can do if you want to get an introduction and to practice them. These naming conventions are controlled by the International Union of Pure and Applied Chemistry, known as IUPAC and these rules can be found in the Compendium of Chemical Terminology.[4]
The process of removing hydrogens and adding methyl groups can continue, essentially without limit, to generate a family of hydrocarbons[5] known as the alkanes; the rules that govern these molecules are simple: each hydrogen makes one and only one bond; each carbon must make four discrete bonds; and these four bonds are tetrahedral in orientation. The number of carbons is in theory unlimited and how they are linked together determines the number of hydrogens. (Can you see how two hydrocarbons with the same number of carbon atoms could have different numbers of hydrogens?)
Depending on how the carbons are connected it is possible to generate a wide variety of molecules with dramatically different shapes. For example there are cage-like, spherical, and long, string-like alkanes. Consider the four-carbon alkanes. There are butane and isobutane that have the formula $\mathrm{C}_{4}\mathrm{H}_{10}$ as well as others with four carbons but different numbers of hydrogens, for example: cyclobutane, methylcyclopropane, and tetrahedrane. Butane has a boiling point of $-0.5^{\circ}\mathrm{C}$, and isobutane has a boiling point of $-11.7^{\circ}\mathrm{C}$. Why are the boiling points of butane and isobutane, which have the same atomic composition ($\mathrm{C}_{4}\mathrm{H}_{10}$), different? The answer lies in the fact that they have different shapes. The roughly linear carbon chain of butane has a larger surface area than isobutane, which gives it more surface area through which to interact with other molecules via London dispersion forces. This idea, that the shape of a molecule and its composition, determine the compound’s macroscopic properties is one that we will return to repeatedly.
Questions
Question to Answer
• Why are the melting and boiling points of methane higher than the melting and boiling points of $\mathrm{H}_{2}$?
• How many different compounds can you draw for the formula $\mathrm{C}_{5}\mathrm{H}_{12}$?
• What structures could you imagine for hydrocarbons containing five carbon atoms?
• Is there a generic formula for an alkane containing n carbon atoms? How does forming a ring of carbons change your formula?
• Which has the higher boiling point, a spherical or a linear alkane?
• How do boiling points and melting points change as molecular weight increases?
Question to Ponder
• Make a prediction as to the melting and boiling points of ethane, compared to methane. What assumptions are you making? How would you test whether those assumptions are valid?
• Why does the shape of a molecule influence its behavior and its macroscopic properties? | textbooks/chem/General_Chemistry/CLUE%3A_Chemistry_Life_the_Universe_and_Everything/04%3A_Heterogeneous_Compounds/4.1%3A_3D_and_2D_Representations.txt |
$\mathrm{C–C}$ and $\mathrm{C–H}$ bonds are described by molecular orbitals; calculations indicate that most of the electron density associated with these orbitals lies between the two nuclei. The $\mathrm{C–H}$ bonds have a length of $109 \times 10^{-12} \mathrm{~m}$ ($109 \mathrm{~pm}$) while the $\mathrm{C–C}$ bond is approximately 50% longer, $154 \times 10^{-12} \mathrm{~m}$ ($154 \mathrm{~pm}$). This is because the $\mathrm{C–C}$ bonding orbital is made from $\mathrm{sp}^{3}$ hybrid orbitals, which are larger than the $1\mathrm{s}$ orbital that hydrogen uses to form bonds. These so-called σ (sigma) bonds have an interesting property; the atoms that they link can spin relative to each other without breaking the bond between them. For a $\mathrm{C–H}$ bond, if the $\mathrm{H}$ spins it would be impossible to tell, since the $\mathrm{H}$ atom is radially symmetric around the $\mathrm{C–H}$ bond axis. But if the carbons in the $\mathrm{C–C}$ bond of ethane spin relative to each other, then it is possible to observe different arrangements by looking down the $\mathrm{C–C}$ bond axis. For example:
and
are both representations of ethane (the $\mathrm{C–C}$ bond is not seen in this depiction because you are looking straight down the $\mathrm{C–C}$ bond). They appear different because the arrangement of the atoms is different in space, but in fact at room temperature these two arrangements can easily interconvert by rotating around the $\mathrm{C–C}$ bond.
This raises another point to consider, namely that starting (and stopping) bond rotations requires energy. Similarly, there can be vibrations along the length of a bond, which again involves the absorption or release of energy. We will consider this further later on.[6] In the case of the rotating bond it turns out that as the bulk of the groups attached to the carbons increases the energy required for the rotation around the $\mathrm{C–C}$ bond also increases. Big, bulky groups can bump into each other, occupying each other’s space causing electron-electron repulsions and raising the energy of any shape where the groups are too close. This tends to lock the molecule into specific orientations that can influence the compound’s physical properties. An example of how structure interferes with the formation of a molecule is a molecule containing 17 carbon atoms and 36 hydrogen atoms ($\rightarrow$); although it is possible to draw this molecule it has never been synthesized because the atoms crowd each other, and intrude on each other’s space. It is possible, however, to synthesize molecules with the same number of carbon atoms but fewer hydrogen atoms.[7] Can you produce a plausible explanation for why?
Collapsing Real Structures Down to 2-Dimensional Representations
Now, an obvious problem with complex three-dimensional molecules, even those made up only of hydrogen and carbon, is how to convey their structure when they must be depicted in two dimensions, like when you are writing on paper. Research indicates that students (that is, most people) have a tough time with this task, which is why we will describe various approaches here.
Before we begin, we need to have some rules. Let us use the set of possible molecules that contain 5 carbon atoms and 12 hydrogen atoms; these are generically known as pentanes. You can begin with a piece of paper and a pencil; how many different molecules can you draw with the composition of $\mathrm{C}_{5}\mathrm{H}_{12}$? Clearly $\mathrm{C}_{5}\mathrm{H}_{12}$ does not uniquely define the structure of the molecule; it is better to use their distinct names: pentane, isopentane, and neopentane ($\rightarrow$). Each of the different molecules you have drawn has the same molecular formula but a different shape and, it turns out, different properties. For example, pentane has a boiling point of $308 \mathrm{~K}$, whereas the boiling points of isopentane and neopentane are $301 \mathrm{~K}$ and $283 \mathrm{~K}$, respectively. Their shape, rather than their elemental composition, influences the strength of the attractions between the individual molecules, which in turn influences their boiling points. We call these kinds of related compounds structural isomers, which means they have the same composition (for example $\mathrm{C}_{5}\mathrm{H}_{12}$) but their constituent atoms are connected differently to give different structures and shapes.
It is common to use a number of different types of representations to picture molecules. One way is through what are known as text formulas (or linear formulae). In this scheme, pentane is written $\mathrm{CH}_{3}\mathrm{–CH}_{2}\(\mathrm{–CH}_{2}\mathrm{–CH}_{2}\(\mathrm{–CH}_{3}$, which can also be written as $\mathrm{CH}_{3}-\left[\mathrm{CH}_{2}\right]_{3}-\mathrm{CH}_{3}$. This captures some of the structural subtleties of pentane, but not all. For example, it does not illustrate the fact that the molecule is not strictly linear. Nevertheless, we can already anticipate complications. How would we write isopentane? The most obvious way would be $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCH}_{2} \mathrm{CH}_{3}$. Neopentane is written as $\left(\mathrm{CH}_{3}\right)_{4} \mathrm{C}$. Does that make sense? Try deciphering them. We will return to this point later on in this chapter.
If we followed the logic of this approach we could draw a more complete representation of pentane, isopentane, and neopentane as Lewis structures, but again, we are missing the three-dimensionality. You might even be led to think that the molecules are actually flat when they are much more like balls. Although it is possible to make the representation a little more realistic by trying to indicate three-dimensionality using the wedge and dash symbols, these structures become very complicated very fast. It is not really practical to draw out full 3D structures for larger, complex molecules. One important skill you will need to master is the ways that short-hand structures (such as Lewis structures) can provide information about the 3D structure of a molecule that allows us to predict chemical and physical properties.
There is one more representation you will often see used that leaves out even more information. In the line structure the only things that are shown are the bonds between carbons! So for example for the pentanes ($\mathrm{C}_{5}\mathrm{H}_{12}$) we can draw structures such as those shown in the figure that omit all the symbols for atoms and all the $\mathrm{C–H}$ bonds. These structures should be used with caution because it is very easy to forget atoms or bonds when they are not in the representation. But what these line structures do show clearly is how the carbon atoms are connected, which can be very helpful at times.
Questions
Questions to Answer
• How many different compounds can you draw for $\mathrm{C}_{6}\mathrm{H}_{14}$? Draw out the full Lewis structure, the condensed formula, and the line formula.
• What are the advantages and disadvantages of each type of structure?
Questions for Later:
• When you think about rotating around a $\mathrm{C–C}$ bond (say in ethane), there are more and less stable orientations. Which orientation do you think is the most stable and why?
• Now imagine a butane molecule ($\mathrm{C}_{4}\mathrm{H}_{10}$) looking along the $\mathrm{C}_{2-3}$ bond. You would see one methyl group and two hydrogen atoms bonded to the two carbon atoms. How would that influence rotation around the $\mathrm{C–C}$ bond we have been considering? | textbooks/chem/General_Chemistry/CLUE%3A_Chemistry_Life_the_Universe_and_Everything/04%3A_Heterogeneous_Compounds/4.2%3A_Single_Bonds_and_Molecular_Shape.txt |
So far, we have considered what are known as single bonds; that is, all the $\mathrm{C–C}$ and $\mathrm{C–H}$ bonds in alkanes, and all the bonds in diamond. Each single bond involves two (and only two) electrons that are described by a bonding molecular orbital. In such a bonding orbital, most of the electron density is located between the two bonded atoms in a linear sigma ($\sigma$) bond. We have, however, already discussed albeit briefly bonds that involve more than one pair of electrons, namely those found in graphite. Recall that for graphite and graphene the bonds between carbon atoms in the sheet plane involve hybridized orbitals that are mixtures of the $2\mathrm{s}^{2}$ and $2\mathrm{p}_{x}$ and $2\mathrm{p}_{y}$ (that is $\mathrm{sp}^{2}$ hybrid orbitals) leaving an unhybridized $2\mathrm{p}_{z}$ orbital. On bonding, these unhybridized $2\mathrm{p}_{z}$ orbitals reorganize to form what is known as a pi ($\pi$) bonding orbital. In $\pi$ orbitals, the electron density lies above and below the axis connecting the bonded atoms. The combination of $\simga$ and $\pi$ bonding orbitals produces a double bond. Double bonds are indicated by two lines, for example as in $\mathrm{CH}_{2}\mathrm{=CH}_{2}$ (ethene).
Shapes of Molecules with Double (and Triple) Bonds
We can apply the same thinking about the arrangement of bonds around the carbon atoms in $\mathrm{CH}_{2}\mathrm{=CH}_{2}$ in much the same manner as we did for $\mathrm{CH}_{3}\mathrm{-CH}_{3}$. In ethene each carbon atom is surrounded by three centers of electron density, two $\mathrm{H}$s and one $\mathrm{C}$. Note that the double bond counts as a single center of electron density ($\rightarrow$). There are a number of important points to keep in mind when considering the effects of double bonds on a molecule and its properties. First, a $\mathrm{C=C}$ double bond is typically less stable (that is more reactive) than two separate single bonds. When we come to thinking about reactions we will find that replacing a double bond by two single bonds typically produces a more stable system. Second, although there is more or less free rotation around the axis of a single bond at room temperature, rotation is blocked by the presence of a double bond. For a rotation to occur, the π bond (in which there is electron density above and below the axis between the two carbon atoms) must be broken and then reformed. The presence of a double bond has distinct effects on molecular shape. The minimum energy arrangement for three centers is a two-dimensional arrangement in which the groups are oriented at about $120^{\circ}$ to one another; an arrangement known as trigonal planar geometry.
There is one more common type of bond that carbon can form, which is a triple bond. For example each carbon in $\mathrm{C}_{2}\mathrm{H}_{2}$ (ethyne) is surrounded by only two centers of electron density: a single sp hybrid orbital bonds between a carbon atom and a hydrogen atom and a triple bond, which can be thought of as a $\sigma$ bond and two $\pi$ bonds between the carbons, shown in the figure ($\rightarrow$). The lowest energy arrangement around each carbon is a line in which the angle between the bonds is $180^{\circ}$. As before, a triple bond is less stable than three single bonds, and reactions can be expected!
We see that under most conditions, a carbon atom can participate in a maximum of four bonds; either four single bonds, two single bonds and a double bond, or one single bond and a triple bond.
Questions
Questions to Answer
• Given a particular hydrocarbon, what factors would influence your prediction of its melting and boiling points? Can you generate some tentative rules?
• How does the presence of a double bond influence the structure of a hydrocarbon?
• How is the presence a triple bond different from that of a double bond?
• Why do you think there is no tetrabonded from of carbon (that is $\mathrm{C}$ four bonds $\mathrm{C}$).
Questions to Ponder
• What limits the size and shape of a hydrocarbon? | textbooks/chem/General_Chemistry/CLUE%3A_Chemistry_Life_the_Universe_and_Everything/04%3A_Heterogeneous_Compounds/4.3%3A_Double_and_Triple_Bonds.txt |
Even though the bonding of hydrogen and carbon atoms can generate a remarkable array of molecules, the hydrocarbons are really rather boring (chemically). They take part in a rather limited range of reactions and would not, on their own, be expected to produce anything like life. Of course there are many other elements, and their properties add chemical complexity to molecular behavior. From the perspective of living systems two of the most interesting elements are nitrogen and oxygen. Carbon has six electrons (two core and four valence) and can form four bonds with neighboring atoms. Nitrogen has seven electrons: two core and five valence: $1 \mathrm{s}^{2}, 2 \mathrm{s}^{2}, 2 \mathrm{p}_{\mathrm{x}} {}^{1}, 2 \mathrm{p}_{\mathrm{y}} {}^{1}, 2 \mathrm{p}_{\mathrm{z}} {}^{1}$. So if you are following the rules, you might well assume that nitrogen would be able to form five bonds (after all, it has five valence electrons). But when we look carefully, we never see a nitrogen atom making five bonds, and in all stable compounds it makes only three bonds. We can explain this observation in several ways. One factor is that nitrogen atoms are too small to support five centers of electron density around themselves because the bonds begin to overlap, which is destabilizing, just like we saw with bulky groups around a carbon. Another factor is that there are only four orbitals available in nitrogen in the second quantum shell. If nitrogen were to form five bonds it would have to use orbitals from the next quantum shell (3), but these orbitals are so high in energy that the energy required would not be offset by the energy released upon on bond formation. Together these factors mean that nitrogen, and in fact all elements in the second row of the periodic table, are limited to bonding arrangements with no more than four centers of electron density. As we will see later on, elements in the next row, such as phosphorus ($\mathrm{P}$) and sulfur ($\mathrm{S}$), are larger and have more available orbitals for bonding. These elements can form up to six centers of electron density.
The simplest compound of nitrogen is molecular nitrogen, $\mathrm{N}_{2}$. The two nitrogen atoms are bonded together by a triple bond, consisting of a $\sigma$ and two $\pi$ bonds. Molecular nitrogen, $\mathrm{N}_{2}$[8]A common nitrogen-containing molecule is ammonia ($\mathrm{NH}_{3}$), which is analogous to methane ($\mathrm{CH}_{3}$). In ammonia the nitrogen atom is bonded to three hydrogen atoms. These three bonds involve three of nitrogen’s valence electrons; the remaining two valence electrons occupy a non-bonding orbital and are referred to as a lone pair. Given the molecular hybridization orbital model that we are using this implies that four $\mathrm{sp}^{3}$ orbitals are formed from the nitrogen atom’s $2\mathrm{s}$ and $2\mathrm{s}$ orbitals leading to four electron density centers around the nitrogen. The figure shows three representations of ammonia. The first indicates the $\mathrm{N–H}$ bonds but fails to show the lone pair orbital. The second uses the wedge and dash convention and dots to illustrate the geometry of both bonds and the lone pair. The actual shape of the molecule is determined by the arrangements of electron clouds and the bonded atoms. In $\mathrm{NH}_{3}$ all three bonds are equivalent ($\mathrm{N–H}$) and so must be symmetrical, but the lone pair orbital is different because it takes up more space than bonding pairs, can you imagine why? This has a subtle effect on the shape of the molecule. The angles between the $\mathrm{C–H}$ bonds in $\mathrm{CH}_{3}$ are equal and $109^{\circ}$ while the angles between the $\mathrm{N–H}$ bonds in $\mathrm{NH}_{3}$ are slightly smaller, $107.8^{\circ}$. The shape of the molecule itself (as outlined by the atoms) is a triangle-based pyramid rather than a tetrahedron. Finally the Lewis structure (the most abstract representation), indicates the bonds and lone pair electrons but gives an unrealistic depiction of the molecule’s geometry. It is up to the reader to supply the implicit information contained in the structure about bond angles and overall shape.
Bonding of Oxygen and Fluorine
Let us now consider oxygen ($\mathrm{O}$) which has eight electrons, two in the core and six valence ($1 \mathrm{s}^{2}, 2 \mathrm{s}^{2}, 2 \mathrm{p}_{\mathrm{x}} {}^{1}, 2 \mathrm{p}_{\mathrm{y}} {}^{1}, 2 \mathrm{p}_{\mathrm{z}} {}^{1}$). As with nitrogen, oxygen does not use all its electrons to form six bonds because it is too small and the orbitals that would need to be used to make six bonds are too high in energy to be energetically accessible; that is, not enough energy would be released upon bond formation to “pay for” that energy.
The simplest oxygen-containing molecule is molecular oxygen, $\mathrm{O}_{2}$. On our simple covalent bond model the two oxygen atoms are connected by a $\sigma$ and a $\pi$ bond, forming a double bond.[9] The next simplest, stable, most common, and by far the most important compound of oxygen at least from the perspective of living organisms, is water ($\mathrm{H}_{2}\mathrm{O}$). In water there are two $\mathrm{O-H}$ bonds and two lone pair non-bonding orbitals. As in the case of nitrogen, the orbitals are $\mathrm{sp}^{3}$ hybrids and the oxygen atom is surrounded by four centers of electron density (see a pattern here?), two bonds, and two lone pairs. Again, the lone pair orbitals are larger than the $\mathrm{O-H}$ bonding orbitals, which distorts the tetrahedral symmetry of the molecule. Instead of equal angles of $109^{\circ}$ between the orbitals, the angle between the $\mathrm{O-H}$ bonds is $104.5^{\circ}$. When we use a Lewis structure to represent the structure of $\mathrm{H}_{2}\mathrm{O}$, it is critical to include all valence shell electrons.
Continuing on across the periodic table we see that fluorine is the next element after oxygen. It has nine electrons: two core and seven valence. Rather than forming seven bonds fluorine only forms a single bond for basically the same reasons that oxygen only forms two bonds. Hydrogen fluoride, $\mathrm{HF}$, has one bond, but four centers of electron density around the fluorine. Because $\mathrm{HF}$ has only two atoms, they must by definition lie on a line and therefore we do not need to discuss its shape.
Compound Molar mass
(g/mole)
Boiling point Bond type Bond length (pm) Atomic radius
(pm)
$\mathrm{CH}_{4}$ $16$ $–161^{\circ}\mathrm{C}$ $\mathrm{C–H}$ (in $\mathrm{CH}_{4}$) $109$ $\mathrm{C} – 70$
$\mathrm{NH}_{3}$ $17$ $–33^{\circ}\mathrm{C}$ $\mathrm{N–H}$ in ($\mathrm{NH}_{3}$) $101$ $\mathrm{N} – 65$
$\mathrm{H}_{2}\mathrm{O}$ $18$ $100^{\circ}\mathrm{C}$ $\mathrm{O–H}$ (in $\mathrm{H}_{2}\mathrm{O}$) $96$ $\mathrm{O} – 60$
$\mathrm{HF}$ $20$ $19.5^{\circ}\mathrm{C}$ $\mathrm{F–H}$ in ($\mathrm{HF}$) $92$ $\mathrm{F} – 50$
$\mathrm{Ne}$ $20$ $–246.08^{\circ}\mathrm{C}$ not applicable not applicable $\mathrm{Ne} – 38$
As we will see, a valid Lewis structure makes it possible to extrapolate a significant amount of information about a molecule’s chemical and physical properties. A confusing point is that the Lewis structure can be written in a number of apparently different ways, which are actually equivalent. The key to remember is that the Lewis structure does not attempt to depict a molecule’s actual three-dimensional structure. It is a shorthand (a “cartoon” if you like) that assumes you already know the arrangement of orbitals. No matter how it is drawn, the actual structure of a $\mathrm{H}_{2}\mathrm{O}$ molecule is the same with a $104.5^{\circ}$ bond angle between the $\mathrm{O–H}$ bonds
$\mathrm{CH}_{4}$ $\mathrm{NH}_{3}$ $\mathrm{H}_{2}\mathrm{O}$ $\mathrm{HF}$ $\mathrm{Ne}$
$-258.7^{\circ} \mathrm{F}\left(-161.5^{\circ} \mathrm{C}\right)$ $-28.01^{\circ} \mathrm{F}\left(-33.34^{\circ} \mathrm{C}\right)$ $212^{\circ} \mathrm{F}\left(100^{\circ} \mathrm{C}\right)$ $67.1^{\circ} \mathrm{F}\left(19.5^{\circ} \mathrm{C}\right)$ $-410.9^{\circ} \mathrm{F}\left(-246.1^{\circ} \mathrm{C}\right)$
The tendency to form four centers (bonds or non-bonding pairs) has led to the rather misleading “octet rule”, which states that some elements tend to form molecules that have eight electrons around any atom (except for hydrogen). Unfortunately, the octet rule is far from being a rule because there are many exceptions, as we will see later. For example many of the elements past the second row of the periodic table are capable of bonding to more than four other atoms and some elements form stable compounds with less than eight electrons. It is important to remember that the octet rule is not the reason why atoms bond with each other, but it is a useful heuristic when constructing Lewis structures for the second row elements ($\mathrm{C}$, $\mathrm{N}$, $\mathrm{O}$, $\mathrm{F}$).
Polarized Bonds and Electronegativity
Earlier we saw that the boiling points of hydrocarbons tend to increase as the number of carbons in the compound increases and that molecules with similar molecular weights have similar but not identical boiling points, with the shapes of the molecules having an effect, although a relatively small one. The attractions between hydrocarbons are due to London dispersion forces that depend on the size, surface area, and shape of the molecule. The larger these forces, the more strongly molecules will stick together and the more energy (higher temperature) will be needed to overcome these attractions.
Let us consider the boiling points of some common second row compounds involving bonds with hydrogen, that is, $\mathrm{CH}_{4}$, $\mathrm{NH}_{3}$, $\mathrm{H}_{2}\mathrm{O}$ and $\mathrm{HF}$, and neon ($\mathrm{Ne}$), which does not form bonds with hydrogen (the compounds of lithium, beryllium, and boron with hydrogen are much less common.) These compounds all have about the same molecular weight but different shapes. Based on our experiences with hydrocarbons, we would be well justified in predicting that they would have somewhat similar boiling points. Unfortunately, this prediction is not supported by experimental evidence (see Table). There is no clear trend, so something is going on that we have not yet considered. To explain this data we have to return to an idea that we discussed in Chapter $3$, namely that the size of atoms decreases as you go across a row of the periodic table. Not only does the size (radius) of the atoms decrease (from $70 \mathrm{~pm}$ for carbon to $38 \mathrm{~pm}$ for neon) but so does the length of the bonds between the atoms and hydrogen (from $109 \mathrm{~pm}$ to $92 \mathrm{~pm}$). This is both surprising and counterintuitive (which is why we are reminding you about it!)
Table $4.4.1$ Electronegativities of Selected Elements
$\mathrm{H}$ $\mathrm{C}$ $\mathrm{N}$ $\mathrm{O}$ $\mathrm{F}$
$2.2$ $2.55$ $3.04$ $3.44$ $3.98$
Remember that the size of the atom is based on a balance between the attraction between the negatively charged electrons to the positively charged protons in the nucleus, the repulsions between the electrons as they get close to each other, and of course the arcane, but highly accurate rules of quantum mechanics. The reason that the atom’s size is decreasing as the number of protons increases is that each electron in the valence shell is attracted by an increasing number of protons in the nucleus. The more protons, the larger this attractive force. At the same time, the electrons in the same valence shell do not tend to repel each other as much as you might suspect because they are in different orbitals. Therefore the effective nuclear charge increases from left to right across the periodic table. This increase in effective nuclear charge doesn’t just affect the electrons in isolated atoms; it also affects the electrons in bonds. The ability to attract the electrons in bonds is called electronegativity, and because it derives from the same effect as that that determines effective nuclear charge and atomic radius, electronegativity also tends to increase from left to right across a row in the periodic table. It also decreases from top to bottom in a group of the periodic table. This makes sense because the further electrons are from the nucleus, the less they will be attracted to it. The exceptions to this rule are the noble gases (helium, neon, argon, etc.); because they do not form bonds with other elements (under normal circumstances) their electronegativities are usually not reported.
Fluorine is the most electronegative element and the Lewis structure of $\mathrm{HF}$ shows one $\mathrm{H–F}$ bond and three lone pairs. Fluorine attracts electrons very strongly—even the ones in the $\mathrm{H–F}$ bond so that the fluorine atom ends up with more than its fair share of electrons and the hydrogen atom ends up with less. One way to think about this is that the electron density in the $\mathrm{H–F}$ bond is shifted closer to the fluorine atom and away from the hydrogen atom ($\rightarrow$). The result of this is that the fluorine atom has more negative charge than positive charge and the hydrogen atom has more positive than negative charge. We indicate this by writing a $\delta –$ charge on the fluorine atom and a $\delta +$ charge on the hydrogen atom ($\delta$ is often used to denote a small increment, that is less than 1). That means that there is an unequal distribution of charge in the molecule. The $\mathrm{HF}$ molecule has a permanent dipole, that is, a separation of charge; the $\mathrm{H–F}$ bond is said to be polarized and the molecule is considered polar. Permanent dipoles are different from the transient dipoles associated with London dispersion forces. Because of their permanent dipoles molecules of $\mathrm{HF}$ interact with one another both attractively and repulsively, more strongly in some orientations than in others. $\mathrm{HF}$ molecules are attracted to each other much more strongly than neon atoms, for example, because of the presence of these permanent dipoles. This results in a much higher boiling point for $\mathrm{HF}$ than for neon (see above). That is, much more energy has to be supplied to the system to overcome the force of attraction and to separate $\mathrm{HF}$ molecules from each other than is needed to separate neon atoms. An important point to note is that HF only has one bond, and the polarity of the bond is the same as the polarity of the whole molecule. As we will see, this is not the case in molecules with more complex structures.
It is relatively easy to predict whether a particular bond is polar by looking at the electronegativity differences between the atoms in that bond. Typically, elements on the left-hand side of the periodic table (metals) have rather low electronegativities and elements over toward the right-hand side (non-metals) have higher electronegativities. There are several ways to calculate electronegativities but in general it is not very useful to memorize specific numbers. It is helpful, however, to understand the trends and to be able to predict bond polarities. Because fluorine is the most electronegative element it can be expected to make the most polarized bonds with hydrogen.[10] So let us take this logic a bit further. If HF has the most polar bonds then HF molecules should stick together with the strongest attractions and HF should have the highest boiling point. But oh no! Water’s boiling point is significantly higher ($100 { }^{\circ}\mathrm{C}$ compared to $19 { }^{\circ}\mathrm{C}$ for $\mathrm{HF}$). What is going on? Oxygen is not as electronegative as fluorine and so the $\mathrm{O-H}$ bond is not as polar as the $\mathrm{H–F}$ bond. Why then is the boiling point of $\mathrm{H}_{2}\mathrm{O}$ $81 { }^{\circ}\mathrm{C}$ higher than $\mathrm{HF}$? To answer this question we need to consider another factor that affects the polarity of a molecule – and that is molecular shape.
Questions
Questions to Answer
• Why do you think that the trends in effective nuclear charge, ionization energy, and electronegativity are correlated?
• What does correlated mean?
• Can you draw a picture of (say) four $\mathrm{H-F}$ molecules sticking together?
• Is there any arrangement that they might take up or would they stick together in a totally random way?
Questions to Ponder
• Why would you not expect polymeric oxygen, that is molecules similar to hydrocarbon chains (or perhaps you would)? | textbooks/chem/General_Chemistry/CLUE%3A_Chemistry_Life_the_Universe_and_Everything/04%3A_Heterogeneous_Compounds/4.4%3A_Bonding_in_Nitrogen%2C_Oxygen%2C_and_Fluorine.txt |
Now we really have to begin to use our 3D thinking and consider several additional factors: the shape of the molecules and how they interact. Much of this thinking is best done hands on with molecular models but we will outline the logic involved here. The HF molecule has a simple shape; it is linear with (partially) positively and (partially) negatively charged ends. In contrast, the $\mathrm{H}_{2}\mathrm{O}$ molecule has a more complex shape; it has two polar $\mathrm{O–H}$ bonds. To understand how this affects the polarity of the molecule we have to take into account the number of bonds, their polarization, and the overall shape of the molecule. Bond polarity is a vector quantity, which means it has both a magnitude and a direction. This is where an understanding of the 3D structure of the molecule becomes critical. Each $\mathrm{O–H}$ bond is polarized and the overall polarity of the molecule is determined by the vector sum of these bond polarities (that is you have to take into account both the magnitude and the direction of the bond dipoles). This may sound a bit complicated but in practice it is relatively easy to predict qualitatively what the overall polarity of the molecule is as long as you keep in mind its 3D structure. In water the two $\mathrm{O–H}$ bonds are at an angle of about $107^{\circ}$ to each other ($\rightarrow$). If we add the bond dipole moments up you can see that the overall direction of the dipole for the molecule bisects that angle, as shown in the figure. Now you might think that this exercise is a bit of a waste of time—surely it would make sense that if a molecule has polar bonds, then the molecule itself should be polar. However, as we will see shortly this is not always the case.
If we apply a similar analysis to ammonia ($\mathrm{CH}_{4}$) we see that the $\mathrm{N–H}$ bond is polar with a $\delta +$s on the hydrogen atoms and a $\delta –$ on the nitrogen atom. Remembering that the actual shape of $\mathrm{NH}_{3}$ is a triangular based pyramid, with an $\mathrm{H-N-H}$ bond angle of $\sim 105^{\circ}$, we can see that there is an overall dipole moment in ammonia. Therefore ammonia is a polar molecule.
If we contrast this with methane, however, we see two differences. The first is that carbon is not nearly as electronegative as nitrogen, oxygen, or fluorine, so the $\mathrm{C–H}$ bond is not as polar. That said, there is an electronegativity difference and so the electron density in the $\mathrm{C–H}$ bond is distorted towards the carbon atom (because it is a little more electronegative than the hydrogen atom.) At the same time, $\mathrm{CH}_{4}$ is symmetrical (tetrahedral.) If we add up all the bond dipoles they cancel each other out giving a molecular dipole moment of zero. Even if we were to replace the hydrogen atoms in methane with fluorine atoms to give $\mathrm{CF}_{4}$ (carbon tetrafluoride) the resulting molecule would still be non-polar, despite the fact that the electronegativity difference between carbon and fluorine is greater than that between hydrogen and oxygen! This is another example of something counterintuitive: something made up of polar parts that is not polar.
The Famous Hydrogen “Bond”
Now that we have a better idea of how the shape and types of bonds in a molecule can affect its polarity, let us look a little more closely at how molecules interact with each other. The first thing to note is that globally non-polar molecules interact solely via London dispersion forces just like atoms of neon or helium. The boiling point of neon is $–246 { }^{\circ}\mathrm{C}$ while the boiling point of $\mathrm{CH}_{4}$ is $–161 { }^{\circ}\mathrm{C}$. This means that methane molecules are more strongly attracted to each other than are neon atoms. We can explain this based on the fact that a methane molecule is larger than a neon atom. Because the electrons in methane molecules are dispersed over a larger area and their distribution (in space) is easier to distort, we say methane molecules are more polarizable. At the same time because methane molecules are non-polar, the boiling point of methane is much lower than that of substances made of polar molecules of similar size.[11]
Let us consider three such molecules: $\mathrm{HF}$ (bp $19.5 { }^{\circ}\mathrm{C}$), $\mathrm{H}_{2}\mathrm{O}$ (bp $100 { }^{\circ}\mathrm{C}$), and $\mathrm{NH}_{3}$ (bp $-33 { }^{\circ}\mathrm{C}$). All three are polar so they stick together but why are there such large differences in their boiling points? The answer lies in the fact that the molecules interact with one another in multiple ways. They all interact via London dispersion forces and dipole–dipole interactions. In addition, a new type of interaction, known as a hydrogen bond (or $\mathrm{H}$-bond) is also possible. The term $\mathrm{H}$-bond is somewhat misleading because these are much weaker than covalent bonds and do not involve shared electrons; the energy required to break a typical hydrogen bond is between $5$ and $30 \mathrm{~kJ/mole$, whereas it requires over $400 \mathrm{~kJ/mole$ to break a $\mathrm{C–C}$ bond.[12] In biological systems and in liquid water, $\mathrm{H}$-bonds are continuously breaking and reforming. Hydrogen bonds are formed between two separate molecules.[13] In contrast to London dispersion forces, but like covalent bonds, $\mathrm{H}$-bonds have a direction; they form when the hydrogen of one molecule, which is covalently bonded to an $\mathrm{O}$, $\mathrm{N}$ or $\mathrm{F}$, is attracted by the lone pair on an $\mathrm{O}$, $\mathrm{N}$ of $\mathrm{F}$ of a neighboring molecule.
$\mathrm{H}$-bonds are a special case of an electrostatic interaction involving a hydrogen atom that is bonded to a very electronegative atom (typically oxygen or fluorine) and an electronegative atom that has lone pairs of electrons. When a hydrogen is bonded in this way most of the electron density moves toward the electronegative atom, leaving a relatively large $\delta +$ on the hydrogen. Water is a particularly important example of a molecule able to engage in hydrogen bonding, because each molecule of water has the possibility of forming four $\mathrm{H}$-bonds ($\rightarrow$). Each of the hydrogen atoms within a water molecule can bond to another water molecule, while each oxygen atom has two lone pairs that can interact with the electron-deficient hydrogen atoms of two different neighboring water molecules, shown in the figure. The ability to form large numbers and networks of hydrogen bonds is responsible for many of the unique properties of water including its relatively high melting point, boiling point, heat capacity, viscosity, and low vapor pressure. In contrast, $\mathrm{HF}$ and $\mathrm{NH}_{3}$ can form, on average, only two $\mathrm{H}$-bonds per molecule. Can you figure out why this is so? Because there are fewer $\mathrm{H}$-bonds to break, they have lower boiling points. $\mathrm{HF}$ has a higher boiling point than $\mathrm{NH}_{3}$ because the $\mathrm{H}$-bonds in $\mathrm{HF}$ are stronger than those in $\mathrm{NH}_{3}$. (Can you figure out why?) In addition to their role in the bulk properties of substances like water, we will see that $\mathrm{H}$-bonds play a critical role in the organization of biological systems, from the structure of DNA and proteins, to the organization of lipid membranes and catalytic mechanisms (but more about that later).
Other Polar Bonds
We have seen that when hydrogen is covalently bonded to oxygen, nitrogen, or fluorine, the result is that the covalent bond is highly polarized and the majority of the electron density is located on the most electronegative atom. This means that the hydrogen atom has very little electron density remaining around it. Because hydrogen is such a small atom, the resulting positive charge density on the hydrogen atom is high. This leads to unusually strong attractions ($\mathrm{H}$-bonds) with atoms that have lone pairs with which the positively charged hydrogen atom can interact. H-bonding is unique to molecules in which a hydrogen atom is covalently bonded to an oxygen, nitrogen, or fluorine atom. However, there are uneven charge distributions possible whenever two atoms with different electronegativities form a bond. Consider, for example, methanol ($\mathrm{CH}_{3}\mathrm{OH}$). It has several different types of bonds with different distributions of charge in them. The familiar $\mathrm{O-H}$ bond in methanol is very much like the $\mathrm{O-H}$ bond found in water. That is, it is highly polarized and the hydrogen atom is a small, dense region of highly positive charge that can attract and will be attracted to regions of high electron density such as the lone pairs on oxygen. The methanol molecule also has a $\mathrm{C-O}$ bond and three $\mathrm{C-H}$ bonds. If we consider the differences in electronegativity we can predict the polarization of these bonds. Remember that carbon and hydrogen have quite similar electronegativities, and so the $\mathrm{C-H}$ bond is not very polarized. Carbon and oxygen, in contrast, are quite different in their electronegativities and the result is that the $\mathrm{C-O}$ bond is strongly polarized, with the $\delta +$ located on the carbon atom and the negative end of the bond dipole on the oxygen atom. As we will see later this has implications for how methanol (and all $\mathrm{C-O}$ containing compounds) interact (and react) with other substances.
An inspection of the Lewis structure can reveal (to the trained mind!) a huge amount about the structure and polarity of a molecule and taking that one step further we can make predictions about the properties of the compound. For example if we compare the relative boiling points of methanol ($\mathrm{CH}_{3}\mathrm{OH}$, bp $65 { }^{\circ}\mathrm{C}$) and ethane ($\mathrm{CH}_{3}\mathrm{CH}_{3}$, bp $-88.6 { }^{\circ}\mathrm{C}$) we see (just as you already predicted no doubt) that methanol has a much higher boiling point because it takes more energy to separate molecules of methanol. The question arises: is this because methanol can form an $\mathrm{H}$-bond with itself? Can you draw a picture of how this happens? Or is it because of the $\mathrm{C-O}$ dipole? We can look at this idea a little more closely by comparing the boiling points of three compounds that have similar molecular weights (so that they experience similar London dispersion forces), but different types of bonds in them.
If we classify the kinds of bonds as before we see that dimethyl ether has non-polar $\mathrm{C-H}$ bonds and polar $\mathrm{C-O}$ bonds. The $\mathrm{C-O-C}$ bond angle is about $104^{\circ}$. Because each atom (except for $\mathrm{H}$) is surrounded by four centers of electron density, the molecule is not linear as pictured. (Why not?) The molecule as whole is polar but cannot form hydrogen bonds with itself because none of the hydrogen atoms have a significant $\delta +$ as they would if they were bonded to an oxygen atom. We call the type of forces between dimethyl ether molecules, dipole–dipole forces. On the other hand, an ethanol molecule—which has exactly the same molecular weight and formula—can form hydrogen bonds with itself because it has an $\mathrm{O-H}$ bond, and so has a small partially positively charged hydrogen atom. This minor difference has a huge effect on boiling point: ethanol boils at $78 { }^{\circ}\mathrm{C}$ whereas dimethyl ether boils at $-23 { }^{\circ}\mathrm{C}$. Both of them are considerably higher than propane at $-44 { }^{\circ}\mathrm{C}$ (remembering that absolute zero is $-273.15 { }^{\circ}\mathrm{C}$). From comparing these three similar compounds we can see that a simple dipole–dipole attraction increases the boiling point by $21 { }^{\circ}\mathrm{C}$, and on top of that the $\mathrm{H}$-bonding attraction in ethanol is worth another $99 { }^{\circ}\mathrm{C}$, bringing the boiling point of ethanol to $78 { }^{\circ}\mathrm{C}$.
Intermolecular Forces
Taken together, London dispersion forces, dipole-dipole interactions, and hydrogen bonds comprise a set of attractive forces that make separate molecules stick together. These are collectively named intermolecular forces, IMFs. These forces are caused by either permanent or temporary distortions of the electron cloud in a molecule – which leads to electrostatic attractions between separate molecules. For small molecules, the typical order for strengths of IMFs is:
$\mathrm{H}$-bonding (where available) > dipole–dipole interactions > London dispersion forces. At the same time, because London dispersion forces increase with molecular size and the extent of surface-surface interactions, they are often the predominant intermolecular force between large biological macromolecules.
The Importance of Shape
While we are on the subject of carbon and oxygen containing compounds, let us take a look at one of the most common compounds of carbon and oxygen, carbon dioxide. You can draw the structure of $\mathrm{CO}_{2}$ with the carbon atom in the middle, double bonded to each of the oxygen atoms. That is, $\mathrm{CO}_{2}$ has two quite polar bonds in it, and so we might reasonably predict that its boiling point might lie somewhere between dimethyl ether and ethanol. But, as you probably already know, this is not the case. $\mathrm{CO}_{2}$ exists as a gas at room temperature. In fact $\mathrm{CO}_{2}$ does not have a liquid phase at standard atmospheric pressure; it changes directly from a solid to a gas, a process called sublimation, at $-78 { }^{\circ}\mathrm{C}$. How is such behavior to be understood, particularly given that $\mathrm{CO}_{2}$ has about the same molar mass as ethanol ($\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{OH}$), which is a liquid at room temperature? Once again we have to make sure we have considered all the factors that affect molecular polarity including bond polarity and shape. If you reflect back to the ideas about bond polarity and structure you will see that we have another case here of a molecule with polar bonds, but no overall polarity. $\mathrm{CO}_{2}$ has a linear structure so the bond polarities cancel each other out (they are at $180^{\circ}$ from each other) ($\rightarrow$). $\mathrm{CO}_{2}$ has no overall molecular polarity, even though it has polar bonds. Therefore the molecules do not stick together very well and it is a gas at room temperature.
Questions
Questions to Answer
• What is the direction of the molecular dipole moment in ammonia? Draw out a picture showing how you came up with the answer. Does it matter which way you draw the molecule? What if you draw it upside down? Will that affect the direction of the dipole (in the real world)?
• Why are the interactions between $\mathrm{H}_{2}\mathrm{O}$ molecules stronger than those between HF molecules even though the polarity of the $\mathrm{HF}$ bond is larger than the polarity of the $\mathrm{OH}$ bond?
• Why don’t more than four water molecules interact with a central water molecule?
• What would you predict would be the relative boiling points of methanol ($\mathrm{CH}_{3}\mathrm{OH}$) and ethane ($\mathrm{CH}_{3}\mathrm{CH}_{3}$), which have similar molecular weights?
• What would you predict would be the relative boiling points of methanol ($\mathrm{CH}_{3}\mathrm{OH}$) and ethanol ($\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{OH}$)?
• What kind of compound (or what structural feature) would you expect might be attracted to the $\delta +$ located on the carbon atom in methanol?
Questions to Ponder
• What would be the consequences (for life, the universe, and everything) if water molecules were linear? | textbooks/chem/General_Chemistry/CLUE%3A_Chemistry_Life_the_Universe_and_Everything/04%3A_Heterogeneous_Compounds/4.5%3A_Molecular_Shapes%2C_Polarity%2C_and_Molecular_Interactions.txt |
Our discussion up to now has centered on types of bonds that involve valence electrons being shared between (or more correctly being fought over – attracted to the opposite nuclei) different atoms. We have seen that we can consider electron density to be equally distributed between the bonding atoms, or that it may be distorted by being attracted to the more electronegative atom. What we have not looked at yet is the extreme case of this kind of distortion, in which the valence electrons are attracted so much by the electronegative atom that they are transferred completely. This kind of bonding is called ionic bonding (as you are almost certainly already aware).
Let us take a look at some common ionic compounds and see if we can make some sense of their properties from a consideration of their atomic-molecular structure. For the sake of simplicity we will confine ourselves (for the moment) to binary compounds (compounds with only two elements in them.) The most familiar of these compounds is sodium chloride ($\mathrm{NaCl}$), common table salt. $\mathrm{NaCl}$ is a continuous compound that extends in three-dimensional array much like diamond (see Chapter $3$.) $\mathrm{NaCl}$ is a solid at room temperature, with a very high melting point ($801 { }^{\circ}\mathrm{C}$), similar to the melting points of silver ($961.78 { }^{\circ}\mathrm{C}$) and gold ($1064.18 { }^{\circ}\mathrm{C}$), although much lower than the decomposition temperature of diamond ($3550 { }^{\circ}\mathrm{C}$). An interesting difference between diamond and sodium chloride occurs on heating. Remember diamond does not melt; it decomposes once enough energy is added to the system to break the $\mathrm{C–C}$ bonds. Under normal circumstances, the carbon atoms react with oxygen ($\mathrm{O}_{2}$) in the air to form carbon dioxide—a process that requires the addition of lots of energy to reverse (as we will see later). On the other hand $\mathrm{NaCl}$ melts (solid $\rightarrow$ liquid) and freezes (liquid $\rightarrow$ solid) at $801 { }^{\circ}\mathrm{C}$, much like water, just at a higher temperature. Based on this difference, we might be tempted to conclude that covalent bonds are not broken when salt melts but that something stronger than the $\mathrm{H}$-bonds that hold water molecules together are broken. What could that be?
A hint comes from studies first carried out by the English chemist Humphrey Davy.[14] Davy used a voltaic pile to study the effects of passing electricity through a range of substances.[15] Solid table salt did not conduct electricity, but liquid (molten) salt did. Not only did it conduct electricity, but when electricity (electrons) was passed through it, it decomposed to produce globules of a shiny, highly reactive metal (sodium, $\mathrm{Na}$) and a pale green gas (chlorine, $\mathrm{Cl}_{2}$). Davy correctly (as it turned out) deduced that the elements in table salt (what we now know as sodium and chlorine) are held together by what he termed electrical forces. Just what caused those electrical forces was not discovered until the atomic nature of matter was elucidated over 100 years later.
It takes a great deal of energy to change table salt into its constituent elements. First the salt has to be heated to its melting point, and then electrical energy must be added to release the elements sodium and chlorine. The reverse reaction, combining the elements sodium and chlorine (don’t do this at home), produces sodium chloride and releases a great deal of energy ($411 \mathrm{~kJ/mol}$). Given the release of energy, we suspect that bonds are being formed during this reaction.
One of the important principles of chemistry is that structure on the atomic-molecular level is reflected in the behavior of materials in the real world. So, let us review some of the real-world properties of sodium chloride:
• It forms colorless crystals that are often cubical in shape and are hard and brittle.
• It has a high melting point and conducts electricity when melted, but not in the solid state.
Based on these properties, and what we know about interactions, bonds, and electricity, we can begin to make hypotheses about how atoms are organized in $\mathrm{NaCl}$. For example, the fact that $\mathrm{NaCl}$ is a stable, crystalline solid at room temperature and that it melts at a high temperature implies that forces holding the atoms together are strong. The regular shape of salt crystals implies that bonds holding the atoms together extend in three dimensions with some regular pattern. If you take a large salt crystal and give it a sharp knock, it breaks cleanly along a flat surface. Diamond also behaves in this way. The ability of molten, but not solid, salt to conduct electricity suggests that melting leads to the appearance of moveable, electrically charged particles. The current interpretation of all these observations and experiments is that in the solid state salt ($\mathrm{NaCl}$) is held together by the coulombic (electrical) attractions between sodium ($\mathrm{Na}^{+}$) and chloride ($\mathrm{Cl}^{-}$) ions. So when sodium metal ($\mathrm{Na}$) reacts with chlorine ($\mathrm{Cl}_{2}$) gas, sodium and chloride ions are produced. In the solid state, these ions are strongly attracted to each other and cannot move, but they can move in the molten (liquid) state, and their movement is what conducts electricity (electrons).
One way to think of ionic bonding is that it is the extreme limit of a polar covalent bond. Typically, simple ionic compounds are formed from elements on the left-hand side of the periodic table (metals, such as sodium) and elements on the right-hand side (non-metals, such as chlorine). The non-metals tend to have a high electronegativity as a result of their high effective nuclear charge, whereas the metals have low electronegativity because their valence electrons are not very strongly attracted to their nuclei. When a metal atom meets a non-metal atom the non-metal attracts the valence electrons from the metal, so that for all intents and purposes electrons move from the metal atom (which then has a net positive charge) to the non-metal atom (which now has a net negative charge). This effect, however, applies only to the electrons in the unfilled valence shells. Electrons in a metal atom’s filled core orbitals require a lot more energy to remove. Why? Because they are closer to the positively charged nucleus (recall the jump in ionization energy when an electron is removed from the core). If there is a single outer-shell electron (as is the case with $\mathrm{Na}$ and other Group I metals) that electron is often lost, and the resulting atom (now called an ion) has a single positive charge (for example, $\mathrm{Na}^{+}$). If there are two outer-shell electrons, as in the case of the Group II metals, such as calcium and magnesium, both can be lost to produce doubly charged ions, such as $\mathrm{Ca}^{++}$ and $\mathrm{Mg}^{++}$ (usually written as $\mathrm{Ca}^{2+}$ and $\mathrm{Mg}^{2+}$). At the other side of the periodic table, the non-metals show exactly the opposite pattern, gaining electrons to become negatively charged ions.[16]
Questions
Questions to Answer
• The melting point of table salt is over $800 { }^{\circ}\mathrm{C}$. Why is it so high?
• What properties do you associate with a solid?
• What happens on the atomic-molecular level when a solid melts?
• Why don’t metals tend to gain electrons? Why don’t non-metals lose electrons?
• What happens to the size of a sodium atom when it loses an electron to become $\mathrm{Na}^{+}$?
• What happens to the size of a chlorine atom when it gains an electron and becomes $\mathrm{Cl}^{-}$?
Questions to Ponder
• Why doesn’t solid table salt conduct electricity?
• Why does molten table salt conduct electricity?
Back to Sodium Chloride
By this point, we have concluded that $\mathrm{NaCl}$ is composed of $\mathrm{Na}^{+}$ ions (cations) and $\mathrm{Cl}^{-}$ ions (anions), but we have not yet discussed how these ions are arranged with respect to one another in space. As you may have come to expect, there is usually more than one way to represent a chemical structure. Different models emphasize different features of a substance but none of them are real in the sense that if we could look at the molecular-level structure, these models are not what we would see. At the same time, visible cubes of salt crystals provide a clue to atomic-molecular structure. If we follow the structure down from the macroscopic to the molecular, this cubic/rectangular structure is retained. A diagram of sodium chloride showing the relative positions of the ions, shown here, illustrates this cubic organization.
Another way to look at $\mathrm{NaCl}$ is to think of each $\mathrm{Na}^{+}$ ion as being surrounded by six $\mathrm{Cl}^{-}$ ions, and each $\mathrm{Cl}^{-}$ ion is surrounded by six $\mathrm{Na}^{+}$ ions. Such an arrangement is possible because of the relative sizes of the sodium and chloride ions; the smaller $\mathrm{Na}^{+}$ ions can sit in the holes between the larger $\mathrm{Cl}^{-}$ ions (why are the chloride ions bigger than the sodium ions?). One consequence of this arrangement is that there is not an “ionic” bond that is analogous to a covalent bond. Our model of bonding here is best understood as this three-dimensional lattice of interacting ions. The alternating network of positive and negative ions makes for a very stable structure that is difficult to disrupt. The implication? Lots of energy is required to break these interactions and allow the ions to move with respect to one another. Many ionic compounds are organized in similar kinds of crystalline structures. A complexity (to which we will return in Chapter $6$) is that many ionic compounds, including $\mathrm{NaCl}$, are highly soluble in water, which means they interact strongly with water molecules. Often salts crystallize together with water molecules and form hydrated (with water) forms, as opposed to anhydrous (without water) forms.
How Ionic Bonding Explains the Properties of Ionic Compounds
Let us return to the properties of ionic compounds and see how this molecular-level (microscopic) model of bonding explains their properties. First, their high melting points arise from the fact that enough energy must be supplied so that multiple (strong) coulombic interactions (recall each cation is surrounded by six anions and vice versa) between the ions must be overcome. In contrast for water, it is only the intermolecular forces between molecules that must be overcome to melt ice; IMFs are significantly weaker than full ionic interactions. Similarly it takes even more energy to vaporize (liquid $\rightarrow$ gas) $\mathrm{NaCl}$.
Now let us predict the melting points of different ionic compounds. Remember that the force between the ions is a Coulombic attraction: $F_{\alpha} = \frac{q^{+} \times q^{-}}{r^{2}}$, where $q^{+}$ and $q^{-}$ are the charges on the ions, and $r$ is the distance between them. This equation tells us that as the charge on the ions increases, so does the force of attraction, but as the distance between them increases, the force of attraction decreases. That is, the coulombic attraction should be larger for small, highly charged ions, and this should be reflected in the melting points of ionic compounds. Even when we don’t factor in the size of the ions, $q_{1} \times q_{2} = 4$ which means that the attractive forces for $\mathrm{CaO}$ should be on the order of 4 times those for $\mathrm{NaCl}$. Indeed, the melting point of calcium oxide ($\mathrm{CaO}$) which has $q_{1} = 2^{+}$ and $q_{2} = 2^{-}$ is $2,572^{\circ}\mathrm{C}$.
Questions
Questions to Answer
• Draw a molecular level picture of liquid water, and a molecular level picture of liquid sodium chloride. Use this picture to explain why it takes more energy to melt solid salt than it does to melt solid water.
• Arrange these ionic compounds in order of increasing melting point: $\mathrm{NaCl}$, $\mathrm{KBr}$, $\mathrm{CaO}$, $\mathrm{Al}_{2}\mathrm{O}_{3}$. Look up your answers and see if your predictions were correct.
• Arrange these materials in order of increasing melting point: $\mathrm{CH}_{4}$, $\mathrm{MgBr}_{2}$, $\mathrm{HF}$, $\mathrm{C}$(diamond). Look up your answers and see if your predictions were correct.
• What do you think happens to the size of the particle when a chlorine atom gains an electron to become a chloride ion? (hint recall that the size of an atom depends on the balance between the attractions between the electrons and the nucleus, and the repulsions between the electrons)
• What do you think happens to the size of the particle when a sodium atom loses an electron to become a sodium ion?
4.7: In-Text References
1. We do not believe that their intent is to torment students, and perhaps they have just forgotten that not every student knows or remembers all of the rules.
2. Unless otherwise noted, we always consider melting and boiling points at atmospheric pressure.
3. Chemists are not being unnecessarily difficult; anatomists also have a very strict set of names for the various bones and nerves in the body, in part to avoid confusion during medical procedures.
4. http://www.iupac.org/ and http://goldbook.iupac.org/index.html
5. Hydrocarbons contain only hydrogen and carbon. Be careful not to confuse them with carbohydrates, which contain carbon, hydrogen, and oxygen and include sugars. We will consider carbohydrates in more detail later on.
6. In fact, we will see that these rotations and vibrations are quantized!
7. http://pubs.acs.org/doi/full/10.1021/ci0497657
8. However, a nitrogen compound with some structural similarities to diamond has been identified. It was synthesized from $\mathrm{N}_{2}$ at high pressure and temperatures. In this polymeric nitrogen, each nitrogen is connected to three neighbors via single bonds, in a similar way that diamond has carbons connected to four neighbors. However this polymeric nitrogen is highly unstable and reactive – unlike diamond. http://www.nature.com/nmat/journal/v.../nmat1146.html
9. Interestingly $\mathrm{O}_{2}$ cannot be well described by a simple valence bond model, because it can be shown that molecular oxygen has two unpaired electrons (it is a di-radical). The bonding is best explained by using molecular orbital theory.
10. Another way to talk about polarity is to say the bond (or molecule) has a dipole moment (unit Debye)- that is the magnitude of the charges $\times$ distance separating them.
11. It is worth keeping in mind the distinction between the molecules a substance is composed of, and the substance itself. Molecules do not have a boiling point, substances do.
12. Remember what a mole is, and that a kilojoule ($\mathrm{kJ}$) is a unit of energy.
13. In larger molecules, such as proteins and nucleic acids, $\mathrm{H}$-bonds can also form between distinct regions of a single molecule.
14. While Davy is well known now for his experiments on the nature of salts, he began his chemical career in his early twenties researching medical uses of gases. He apparently became very fond of nitrous oxide ($\mathrm{N}_{2}\mathrm{O}$, laughing gas), which he reported was an enjoyable recreational drug and a cure for hangovers (ref SALT).
15. In 1800 the first electric battery, the Voltaic Pile, was developed. It was promptly put to use by a growing number of scientists. For example, molecular hydrogen and oxygen could be produced by passing electricity through water.
16. Positively charged ions are known as cations and negatively charged ions are known as anions. | textbooks/chem/General_Chemistry/CLUE%3A_Chemistry_Life_the_Universe_and_Everything/04%3A_Heterogeneous_Compounds/4.6%3A_Ionic_Bonding.txt |
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