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Orientational Interaction of Substituents
When a benzene ring has two substituent groups, each exerts an influence on subsequent substitution reactions. The activation or deactivation of the ring can be predicted more or less by the sum of the individual effects of these substituents. The site at which a new substituent is introduced depends on the orientation of the existing groups and their individual directing effects. We can identify two general behavior categories, as shown in the following table. Thus, the groups may be oriented in such a manner that their directing influences act in concert, reinforcing the outcome; or are opposed (antagonistic) to each other. Note that the orientations in each category change depending on whether the groups have similar or opposite individual directing effects.
Antagonistic or Non-Cooperative
Reinforcing or Cooperative
D = Electron Donating Group (ortho/para-directing)
W = Electron Withdrawing Group (meta-directing)
Reinforcing or Cooperative Substitutions
The products from substitution reactions of compounds having a reinforcing orientation of substituents are easier to predict than those having antagonistic substituents. For example, the six equations shown below are all examples of reinforcing or cooperative directing effects operating in the expected manner. Symmetry, as in the first two cases, makes it easy to predict the site at which substitution is likely to occur. Note that if two different sites are favored, substitution will usually occur at the one that is least hindered by ortho groups.
The first three examples have two similar directing groups in a meta-relationship to each other. In examples 4 through 6, oppositely directing groups have an ortho or para-relationship. The major products of electrophilic substitution, as shown, are the sum of the individual group effects. The strongly activating hydroxyl (–OH) and amino (–NH2) substituents favor dihalogenation in examples 5 and six.
Antagonistic or Non-Cooperative Substitutions
Substitution reactions of compounds having an antagonistic orientation of substituents require a more careful analysis. If the substituents are identical, as in example 1 below, the symmetry of the molecule will again simplify the decision. When one substituent has a pair of non-bonding electrons available for adjacent charge stabilization, it will normally exert the product determining influence, examples 2, 4 & 5, even though it may be overall deactivating (case 2). Case 3 reflects a combination of steric hindrance and the superior innate stabilizing ability of methyl groups relative to other alkyl substituents. Example 6 is interesting in that it demonstrates the conversion of an activating ortho/para-directing group into a deactivating meta-directing "onium" cation [–NH(CH3)2(+) ] in a strong acid environment.
Contributors
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry
Friedel-Crafts Reactions
Benzene was first suspected to exist way back in 1825 when British scientist Michael Faraday first isolated it from an oily mixture used in gaslights. This, and other compounds similar to it, formed a group called aromatic compounds. They were named aromatic due to their pleasing aroma, although not all smelled as such. The unusual stability of benzene makes it ideal for many reactions. Among these reactions is one known as the Friedel-Crafts Alkylation. However, the reaction suffers from a group of limitations making it a poor candidate to achieve desired results. Friedel-Crafts Alkylation was first discovered by French scientist Charles Friedel and his partner, American scientist James Crafts, in 1877. This reaction allowed for the formation of alkyl benzenes from alkyl halides, but was plagued with unwanted supplemental activity that reduced its effectively.
Limitations of Friedel-Crafts Alkylation
• Carbocation Rearrangement - Only certain alkylbenzenes can be made due to the tendency of cations to rearrange.
• Compound Limitations - Friedel-Crafts fails when used with compounds such as nitrobenzene and other strong deactivating systems.
• Polyalkylation - Products of Friedel-Crafts are even more reactive than starting material. Alkyl groups produced in Friedel-Crafts Alkylation are electron-donating substituents meaning that the products are more susceptible to electrophilic attack than what we began with. For synthetic purposes, this is a big disappointment.
To remedy these limitations, a new and improved reaction was devised: The Friedel-Crafts Acylation, also known as Friedel-Crafts Alkanoylation.
The goal of the reaction is the following:
The very first step involves the formation of the acylium ion which will later react with benzene:
The second step involves the attack of the acylium ion on benzene as a new electrophile to form one complex:
The third step involves the departure of the proton in order for aromaticity to return to benzene:
During the third step, AlCl4 returns to remove a proton from the benzene ring, which enables the ring to return to aromaticity. In doing so, the original AlCl3 is regenerated for use again, along with HCl. Most importantly, we have the first part of the final product of the reaction, which is a ketone. The first part of the product is the complex with aluminum chloride as shown:
The final step involves the addition of water to liberate the final product as the acylbenzene:
Because the acylium ion (as was shown in step one) is stabilized by resonance, no rearrangement occurs (Limitation 1). Also, because of of the deactivation of the product, it is no longer susceptible to electrophilic attack and hence, is no longer susceptible to electrophilic attack and hence, no longer goes into further reactions (Limitation 3). However, as not all is perfect, Limitation 2 still prevails where Friedel-Crafts Acylation fails with strong deactivating rings.
Problem 1:
Problem 2:
Problem 3:
Problem 4:
Problem 5:
Solutions
Solution to Problem 2:
Solution to Problem 3:
Solution to Problem 4:
Solution to Problem 5:
Contributors
• Mario Morataya (UCD)
Friedel-Crafts Acylation
This page gives details of the Friedel-Crafts reactions of benzene and methylbenzene (toluene).
Friedel-Crafts acylation of benzene
An acyl group is an alkyl group attached to a carbon-oxygen double bond. If "R" represents any alkyl group, then an acyl group has the formula RCO-. Acylation means substituting an acyl group into something - in this case, into a benzene ring.
The most commonly used acyl group is CH3CO-. This is called the ethanoyl group, and in this case the reaction is sometimes called "ethanoylation". In the example which follows we are substituting a CH3CO- group into the ring, but you could equally well use any other acyl group. The most reactive substance containing an acyl group is an acyl chloride (also known as an acid chloride). These have the general formula RCOCl. Benzene is treated with a mixture of ethanoyl chloride, CH3COCl, and aluminium chloride as the catalyst. The mixture is heated to about 60°C for about 30 minutes. A ketone called phenylethanone (old name: acetophenone) is formed.
or, if you want a more compact form:
$C_6H_6 + CH_3COCl \longrightarrow C_6H_5COCH_3 + HCl$
The aluminium chloride isn't written into these equations because it is acting as a catalyst. If you wanted to include it, you could write AlCl3 over the top of the arrow (see below).
Friedel-Crafts acylation of methylbenzene (toluene)
The reaction is just the same with methylbenzene except that you have to worry about where the acyl group attaches to the ring relative to the methyl group. Normally, the methyl group in methylbenzene directs new groups into the 2- and 4- positions (assuming the methyl group is in the 1- position). In acylation, though, virtually all the substitution happens in the 4- position.
Friedel-Crafts alkylation
Alkylation means substituting an alkyl group into something - in this case into a benzene ring. A hydrogen on the ring is replaced by a group like methyl or ethyl and so on.
Friedel-Crafts alkylation of benzene: Benzene reacts at room temperature with a chloroalkane (for example, chloromethane or chloroethane) in the presence of aluminium chloride as a catalyst. On this page, we will look at substituting a methyl group, but any other alkyl group could be used in the same way. Substituting a methyl group gives methylbenzene.
or:
$C_6H_6 + CH_3Cl \rightarrow C_6H_5CH_3 + HCl$
Friedel-Crafts alkylation of methylbenzene (toluene)
Again, the reaction is just the same with methylbenzene except that you have to worry about where the alkyl group attaches to the ring relative to the methyl group. Unfortunately this time there is a problem! Where the incoming alkyl group ends up depends to a large extent on the temperature of the reaction.
At 0°C, substituting methyl groups into methylbenzene, you get a mixture of the 2-.3- and 4- isomers in the proportion 54% / 17% / 29%. That's a higher proportion of the 3- isomer than you might expect. At 25°C, the proportions change to 3% / 69% / 28%. In other words the proportion of the 3- isomer has increased even more. Raise the temperature some more and the trend continues.
Friedel-Crafts alkylation industrially
The manufacture of ethylbenzene: Ethylbenzene is an important industrial chemical used to make styrene (phenylethene), which in turn is used to make polystyrene - poly(phenylethene). It is manufactured from benzene and ethene. There are several ways of doing this, some of which use a variation on Friedel-Crafts alkylation.
The reaction is done in the liquid state. Ethene is passed through a liquid mixture of benzene, aluminium chloride and a catalyst promoter which might be chloroethane or hydrogen chloride. We are going to assume it is HCl. Promoters are used to make catalysts work better.
There are two variants on the process. One (the Union Carbide / Badger process) uses a temperature no higher than 130°C and a pressure just high enough to keep everything liquid. The other (the Monsanto process) uses a slightly higher temperature of 160°C which needs less catalyst. (Presumably - although I haven't been able to confirm this - the pressure would also need to be higher to keep everything liquid at the higher temperature.)
or
$C_6H_6 + \ce{CH_2=CH_2} \rightarrow C_6H_5CH_2CH_3$
Again, the aluminium chloride and HCl aren't written into these equations because they are acting as catalysts. If you wanted to include them, you could write AlCl3 and HCl over the top of the arrow.
Contributors
Jim Clark (Chemguide.co.uk) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Arenes/Reactivity_of_Arenes/Electrophilic_Substitution_of_Disubstituted_Benzene_Rings.txt |
Halogenation is an example of electrophillic aromatic substitution. In electrophilic aromatic substitutions, a benzene is attacked by an electrophile which results in substition of hydrogens. However, halogens are not electrophillic enough to break the aromaticity of benzenes, which require a catalyst to activate.
Activation of Halogen
(where X= Br or Cl, we will discuss further in detail later why other members of the halogen family Flourine and Iodine are not used in halogenation of benzenes)
Hence, Halogen needs the help and aid of Lewis Acidic Catalysts to activate it to become a very strong electrophile. Examples of these activated halogens are Ferric Hallides (FeX3) Aluminum Halides (AlX3) where X= Br or Cl. In the following examples, the halogen we will look at is Bromine.
In the example of bromine, in order to make bromine electrophillic enough to react with benzene, we use the aid of an aluminum halide such as aluminum bromide.
With aluminum bromide as a Lewis acid, we can mix Br2 with AlBr3 to give us Br+. The presence of Br+ is a much better electrophile than Br2 alone. Bromination is acheived with the help of AlBr3 (Lewis acid catalysts) as it polarizes the Br-Br bond. The polarization causes polarization causes the bromine atoms within the Br-Br bond to become more electrophillic. The presence of Br+ compared to Br2 alone is a much better electrophille that can then react with benzene.
As the bromine has now become more electrophillic after activation of a catalyst, an electrophillic attack by the benzene occurs at the terminal bromine of Br-Br-AlBr3. This allows the other bromine atom to leave with the AlBr3 as a good leaving group, AlBr4-.
After the electrophilic attack of bromide to the benzene, the hydrogen on the same carbon as bromine substitutes the carbocation in which resulted from the attack. Hence it being an electrophilic aromatic SUBSTITUTION. Since the by-product aluminum tetrabromide is a strong nucleophile, it pulls of a proton from the Hydrogen on the same carbon as bromine.
In the end, AlBr3was not consumed by the reaction and is regenerated. It serves as our catalyst in the halogenation of benzenes.
Dissociation Energies of Halogens and its Effect on Halogenation of Benzenes
The electrophillic bromination of benzenes is an exothermic reaction. Considering the exothermic rates of aromatic halogenation decreasing down the periodic table in the Halogen family, Flourination is the most exothermic and Iodination would be the least. Being so exothermic, a reaction of flourine with benzene is explosive! For iodine, electrophillic iodination is generally endothermic, hence a reaction is often not possible. Similar to bromide, chlorination would require the aid of an activating presence such as Alumnium Chloride or Ferric Chloride. The mechanism of this reaction is the same as with Bromination of benzene.
Problems
1. What reagents would you need to get the given product?
What reagents would you need to gete given product
2. What product would result from the given reagents?
3. What is the major product given the reagents below?
4. Draw the formatin of Cl+ from AlCl3 and Cl2
5. Draw the mechanism of the reaction between Cl+ and a benzene.
Solutions
1. Cl2 and AlCl3 or Cl2 and FeCl3
2. No Reaction
3.
4.
5.
Contributors
• Catherine Nguyen | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Arenes/Reactivity_of_Arenes/Halogenation_of_Benzene-The_Need_for_a_Catalyst.txt |
This page looks at the reactions of benzene and methylbenzene (toluene) with chlorine and bromine under various conditions.
The halogenation of benzene
Substitution reactions
Benzene reacts with chlorine or bromine in the presence of a catalyst, replacing one of the hydrogen atoms on the ring by a chlorine or bromine atom. The reactions happen at room temperature. The catalyst is either aluminum chloride (or aluminum bromide if you are reacting benzene with bromine) or iron.
Strictly speaking iron is not a catalyst, because it gets permanently changed during the reaction. It reacts with some of the chlorine or bromine to form iron(III) chloride, $FeCl_3$, or iron(III) bromide, $FeBr_3$.
$2Fe + 3Cl_2 \rightarrow 2FeCl_3$
$2Fe + 3Br_2 \rightarrow 2FeBr_3$
These compounds act as the catalyst and behave exactly like aluminum chloride, $AlCl_3$, or aluminum bromide, $AlBr_3$, in these reactions.
The reaction with chlorine
The reaction between benzene and chlorine in the presence of either aluminum chloride or iron gives chlorobenzene.
or, written more compactly:
$C_6H_6 + Cl_2 \rightarrow C_6H_5Cl + HCl$
The reaction with bromine
The reaction between benzene and bromine in the presence of either aluminum bromide or iron gives bromobenzene. Iron is usually used because it is cheaper and is more readily available.
or
$C_6H_6 + Br_2 \rightarrow C_6H_5Br + HBr$
Addition reactions
In the presence of ultraviolet light (but without a catalyst present), hot benzene will also undergo an addition reaction with chlorine or bromine. The ring delocalization is permanently broken and a chlorine or bromine atom adds on to each carbon atom.
For example, if you bubble chlorine gas through hot benzene exposed to UV light for an hour, you get 1,2,3,4,5,6-hexachlorocyclohexane.
Bromine would behave similarly.
The chlorines and hydrogens can stick up and down at random above and below the ring and this leads to a number of geometric isomers. Although there aren't any carbon-carbon double bonds, the bonds are still "locked" and unable to rotate. One of these isomers was once commonly used as an insecticide known variously as BHC, HCH and Gammexane. This is one of the "chlorinated hydrocarbons" which caused so much environmental harm.
The halogenation of methylbenzene
Substitution reactions
It is possible to get two quite different substitution reactions between methylbenzene and chlorine or bromine depending on the conditions used. The chlorine or bromine can substitute into the ring or into the methyl group.
Substitution into the ring
Substitution in the ring happens in the presence of aluminum chloride (or aluminum bromide if you are using bromine) or iron, and in the absence of UV light. The reactions happen at room temperature. This is exactly the same as the reaction with benzene, except that you have to worry about where the halogen atom attaches to the ring relative to the position of the methyl group.
Methyl groups are 2,4-directing, which means that incoming groups will tend to go into the 2 or 4 positions on the ring - assuming the methyl group is in the 1 position. In other words, the new group will attach to the ring next door to the methyl group or opposite it. With chlorine, substitution into the ring gives a mixture of 2-chloromethylbenzene and 4-chloromethylbenzene.
With bromine, you would get the equivalent bromine compounds.
Substitution into the methyl group
If chlorine or bromine react with boiling methylbenzene in the absence of a catalyst but in the presence of UV light, substitution happens in the methyl group rather than the ring. For example, with chlorine (bromine would be similar):
The organic product is (chloromethyl)benzene. The brackets in the name emphasize that the chlorine is part of the attached methyl group, and isn't on the ring.
One of the hydrogen atoms in the methyl group has been replaced by a chlorine atom. However, the reaction doesn't stop there, and all three hydrogens in the methyl group can in turn be replaced by chlorine atoms. That means that you could also get (dichloromethyl)benzene and (trichloromethyl)benzene as the other hydrogen atoms in the methyl group are replaced one at a time.
If you use enough chlorine you will eventually get (trichloromethyl)benzene, but any other proportions will always lead to a mixture of products.
Addition reactions
I haven't been able to track down anything similar to the reaction between benzene and chlorine in which six chlorine atoms add around the ring.
That perhaps isn't surprising. Chlorine adds to benzene in the presence of ultraviolet light. With methylbenzene under those conditions, you get substitution in the methyl group. That is energetically easier because it doesn't involve breaking the delocalized electron system.
Whether you would get addition to the ring if you used a large excess of chlorine and did the reaction for a long time, I don't know. Once all the hydrogens in the methyl group had been substituted, perhaps you might then get addition to the ring as well.
Contributors
Jim Clark (Chemguide.co.uk) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Arenes/Reactivity_of_Arenes/Halogenation_of_Benzene_and_Methylbenzene.txt |
The strongest activating and ortho/para-directing substituents are the amino (-NH2) and hydroxyl (-OH) groups. Direct nitration of phenol (hydroxybenzene) by dilute nitric acid gives modest yields of nitrated phenols and considerable oxidative decomposition to tarry materials; aniline (aminobenzene) is largely destroyed. Bromination of both phenol and aniline is difficult to control, with di- and tri-bromo products forming readily. Because of their high nucleophilic reactivity, aniline and phenol undergo substitution reactions with iodine, a halogen that is normally unreactive with benzene derivatives. The mixed halogen iodine chloride (ICl) provides a more electrophilic iodine moiety, and is effective in iodinating aromatic rings having less powerful activating substituents.
C6H5–NH2 + I2 + NaHCO3 p-I–C6H4–NH2 + NaI + CO2 + H2O
By acetylating the heteroatom substituent on phenol and aniline, its activating influence can be substantially attenuated. For example, acetylation of aniline gives acetanilide (first step in the following equation), which undergoes nitration at low temperature, yielding the para-nitro product in high yield. The modifying acetyl group can then be removed by acid-catalyzed hydrolysis (last step), to yield para-nitroaniline. Although the activating influence of the amino group has been reduced by this procedure, the acetyl derivative remains an ortho/para-directing and activating substituent.
C6H5–NH2 + (CH3CO)2O
pyridine (a base)
C6H5–NHCOCH3
HNO3 , 5 ºC
p-O2N–C6H4–NHCOCH3
H3O(+) & heat
p-O2N–C6H4–NH
The following diagram illustrates how the acetyl group acts to attenuate the overall electron donating character of oxygen and nitrogen. The non-bonding valence electron pairs that are responsible for the high reactivity of these compounds (blue arrows) are diverted to the adjacent carbonyl group (green arrows). However, the overall influence of the modified substituent is still activating and ortho/para-directing.
Nitration and Sulfonation of Benzene
Nitration and sulfonation of benzene are two examples of electrophilic aromatic substitution. The nitronium ion (NO2+) and sulfur trioxide (SO3) are the electrophiles and individually react with benzene to give nitrobenzene and benzenesulfonic acid respectively.
Nitration of Benzene
The source of the nitronium ion is through the protonation of nitric acid by sulfuric acid, which causes the loss of a water molecule and formation of a nitronium ion.
Sulfuric Acid Activation of Nitric Acid
The first step in the nitration of benzene is to activate HNO3with sulfuric acid to produce a stronger electrophile, the nitronium ion.
Because the nitronium ion is a good electrophile, it is attacked by benzene to produce Nitrobenzene.
Mechanism
(Resonance forms of the intermediate can be seen in the generalized electrophilic aromatic substitution)
Sulfonation of Benzene
Sulfonation is a reversible reaction that produces benzenesulfonic acid by adding sulfur trioxide and fuming sulfuric acid. The reaction is reversed by adding hot aqueous acid to benzenesulfonic acid to produce benzene.
Mechanism
To produce benzenesulfonic acid from benzene, fuming sulfuric acid and sulfur trioxide are added. Fuming sulfuric acid, also refered to as oleum, is a concentrated solution of dissolved sulfur trioxide in sulfuric acid. The sulfur in sulfur trioxide is electrophilic because the oxygens pull electrons away from it because oxygen is very electronegative. The benzene attacks the sulfur (and subsequent proton transfers occur) to produce benzenesulfonic acid.
Reverse Sulfonation
Sulfonation of benzene is a reversible reaction. Sulfur trioxide readily reacts with water to produce sulfuric acid and heat. Therefore, by adding heat to benzenesulfonic acid in diluted aqueous sulfuric acid the reaction is reversed.
Further Applications of Nitration and Sulfonation
Nitration is used to add nitrogen to a benzene ring, which can be used further in substitution reactions. The nitro group acts as a ring deactivator. Having nitrogen present in a ring is very useful because it can be used as a directing group as well as a masked amino group. The products of aromatic nitrations are very important intermediates in industrial chemistry.
Because sulfonation is a reversible reaction, it can also be used in further substitution reactions in the form of a directing blocking group because it can be easily removed. The sulfonic group blocks the carbon from being attacked by other substituents and after the reaction is completed it can be removed by reverse sulfonation. Benzenesulfonic acids are also used in the synthesis of detergents, dyes, and sulfa drugs. Bezenesulfonyl Chloride is a precursor to sulfonamides, which are used in chemotherapy.
Outside Links
Aromatic Sulfonation
Aromatic Nitration
Problems
1. What is/are the required reagent(s)for the following reaction:
2. What is the product of the following reaction:
3. Why is it important that the nitration of benzene by nitric acid occurs in sulfuric acid?
4. Write a detailed mechanism for the sulfonation of benzene, including all resonance forms.
5. Draw an energy diagram for the nitration of benzene. Draw the intermediates, starting materials, and products. Label the transition states. (For questions 1 and 2 see Electrophilic Aromatic Substitution for hints)
For other problems involving Electrophilic Aromatic Substitution and similar reactions see:
Solutions
1. SO3 and H2SO4 (fuming)
2.
3. Sulfuric acid is needed in order for a good electrophile to form. Sulfuric acid protonates nitric acid to form the nitronium ion (water molecule is lost). The nitronium ion is a very good electrophile and is open to attack by benzene. Without sulfuric acid the reaction would not occur.
4.
5. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Arenes/Reactivity_of_Arenes/Modifying_the_Influence_of_Strong_Activating_Groups.txt |
This page looks at the facts about the nitration of benzene and methylbenzene. The mechanisms for these reactions are covered elsewhere on the site, and you will find links to these.
The nitration of benzene
Nitration happens when one (or more) of the hydrogen atoms on the benzene ring is replaced by a nitro group, NO2. Benzene is treated with a mixture of concentrated nitric acid and concentrated sulfuric acid at a temperature not exceeding 50°C. The mixture is held at this temperature for about half an hour. Yellow oily nitrobenzene is formed.
You could write this in a more condensed form as:
$C_6H_6 + HNO_3 \rightarrow C_6H_5NO_2 + H_2O$
The concentrated sulfuric acid is acting as a catalyst and so is not written into the equations. At higher temperatures there is a greater chance of getting more than one nitro group substituted onto the ring. You will get a certain amount of 1,3-dinitrobenzene formed even at 50°C. Some of the nitrobenzene formed reacts with the nitrating mixture of concentrated acids.
Notice that the new nitro group goes into the 3 position on the ring. Nitro groups "direct" new groups into the 3 and 5 positions. It is also possible to get a third nitro group attached to the ring (in the 5 position). However, nitro groups make the ring much less reactive than the original benzene ring. Two nitro groups on the ring make its reactions so slow that virtually no trinitrobenzene is produced under these conditions.
The nitration of methylbenzene (toluene)
Methylbenzene reacts rather faster than benzene - in nitration, the reaction is about 25 times faster. That means that you would use a lower temperature to prevent more than one nitro group being substituted - in this case, 30°C rather than 50°C. Apart from that, the reaction is just the same - using the same nitrating mixture of concentrated sulphuric and nitric acids.
You get a mixture of mainly two isomers formed: 2-nitromethylbenzene and 4-nitromethylbenzene. Only about 5% of the product is 3-nitromethylbenzene. Methyl groups are said to be 2,4-directing.
For 2-nitromethylbenzene:
and 4-nitromethylbenzene:
Just as with benzene, you will get a certain amount of dinitro compound formed under the conditions of the reaction, but virtually no trinitro product because the reactivity of the ring decreases for every nitro group added (bythe way, trinitromethylbenzene used to be called trinitrotoluene or TNT).
The reactivity of a benzene ring is governed by the electron density around the ring. Methyl groups tend to "push" electrons towards the ring - increasing the density, and so making the ring more attractive to attacking reagents. This is actually a simplification. In order to understand the rate effect properly you have to think about the stability of the intermediate ions formed during the reactions, because this affects the activation energy of the reactions.
Contributors
Jim Clark (Chemguide.co.uk)
Other Reactions of Benzene and Methylbenzene
This page gives details of some reactions of benzene and methylbenzene (toluene) not covered elsewhere in this section. It deals with the combustion, hydrogenation and sulfonation of benzene and methylbenzene (toluene), and with the oxidation of side chains attached to benzene rings.
Combustion
Like any other hydrocarbons, benzene and methylbenzene burn in a plentiful supply of oxygen to give carbon dioxide and water. For example:
For benzene:
$2C_6H_6 + 15O_2 \rightarrow 12CO_2 + 6H_2O$
. . . and methylbenzene:
$C_6H_5CH_3 + 9O_2 \rightarrow 7CO_2 + 4H_2O$
However, for these hydrocarbons, combustion is hardly ever complete, especially if they are burnt in air. The high proportion of carbon in the molecules means that you need a very high proportion of oxygen to hydrocarbon to get complete combustion. Look at the equations.
As a general rule, the hydrogen in a hydrocarbon tends to get what oxygen is available first, leaving the carbon to form carbon itself, or carbon monoxide, if there isn't enough oxygen to go round.
The arenes tend to burn in air with extremely smoky flames - full of carbon particles. You almost invariably get incomplete combustion, and the arenes can be recognised by the smokiness of their flames.
Hydrogenation
Hydrogenation is an addition reaction in which hydrogen atoms are added all the way around the benzene ring. A cycloalkane is formed. For example:
With benzene:
. . . and methylbenzene:
These reactions destroy the electron delocalisation in the original benzene ring, because those electrons are being used to form bonds with the new hydrogen atoms.
Although the reactions are exothermic overall because of the strengths of all the new carbon-hydrogen bonds being made, there is a high activation barrier to the reaction.
The reactions are done using the same finely divided nickel catalyst that is used in hydrogenating alkenes and at similar temperatures (around 150°C), but the pressures used tend to be higher.
sulfonation
sulfonation involves replacing one of the hydrogens on a benzene ring by the sulfonic acid group, -SO3H.
The sulfonation of benzene
There are two equivalent ways of sulfonating benzene:
• Heat benzene under reflux with concentrated sulfuric acid for several hours.
• Warm benzene under reflux at 40°C with fuming sulfuric acid for 20 to 30 minutes. Fuming sulfuric acid, H2S2O7, can usefully be thought of as a solution of sulfur trioxide in concentrated sulfuric acid.
Or:
$C_6H_6 + H_2SO_4 \rightarrow C_5H_5SO_3H + H_2O$
The product is benzenesulfonic acid.
The sulfonation of methylbenzene
Methylbenzene is more reactive than benzene because of the tendency of the methyl group to "push" electrons towards the ring.
The effect of this greater reactivity is that methylbenzene will react with fuming sulfuric acid at 0°C, and with concentrated sulfuric acid if they are heated under reflux for about 5 minutes.
As well as the effect on the rate of reaction, with methylbenzene you also have to think about where the sulfonic acid group ends up on the ring relative to the methyl group. Methyl groups have a tendency to "direct" new groups into the 2- and 4- positions on the ring (assuming the methyl group is in the 1- position). Methyl groups are said to be 2,4-directing.
So you get a mixture which mainly consists of two isomers. Only about 5 - 10% of the 3- isomer is formed. The main reactions are:
and:
In the case of sulfonation, the exact proportion of the isomers formed depends on the temperature of the reaction. As the temperature increases, you get increasing proportions of the 4- isomer and less of the 2- isomer. This is because sulfonation is reversible. The sulfonic acid group can fall off the ring again, and reattach somewhere else. This tends to favour the formation of the most thermodynamically stable isomer. This interchange happens more at higher temperatures.
The 4- isomer is more stable because there is no cluttering in the molecule as there would be if the methyl group and sulfonic acid group were next door to each other.
Side chain oxidation in alkylbenzenes
An alkylbenzene is simply a benzene ring with an alkyl group attached to it. Methylbenzene is the simplest alkylbenzene. Alkyl groups are usually fairly resistant to oxidation. However, when they are attached to a benzene ring, they are easily oxidised by an alkaline solution of potassium manganate(VII) (potassium permanganate).
Methylbenzene is heated under reflux with a solution of potassium manganate(VII) made alkaline with sodium carbonate. The purple colour of the potassium manganate(VII) is eventually replaced by a dark brown precipitate of manganese(IV) oxide. The mixture is finally acidified with dilute sulfuric acid.
Overall, the methylbenzene is oxidised to benzoic acid.
Interestingly, any alkyl group is oxidised back to a -COOH group on the ring under these conditions. So, for example, propylbenzene is also oxidised to benzoic acid.
Contributors
Jim Clark (Chemguide.co.uk) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Arenes/Reactivity_of_Arenes/Nitration_of_Benzene_and_Methylbenzene.txt |
Compounds in which two or more benzene rings are fused together were described in an earlier section, and they present interesting insights into aromaticity and reactivity. The smallest such hydrocarbon is naphthalene. Naphthalene is stabilized by resonance. Three canonical resonance contributors may be drawn, and are displayed in the following diagram.
The two structures on the left have one discrete benzene ring each, but may also be viewed as 10-pi-electron annulenes having a bridging single bond. The structure on the right has two benzene rings which share a common double bond. From heats of hydrogenation or combustion, the resonance energy of naphthalene is calculated to be 61 kcal/mole, 11 kcal/mole less than that of two benzene rings (2 * 36). As expected from an average of the three resonance contributors, the carbon-carbon bonds in naphthalene show variation in length, suggesting some localization of the double bonds. The C1–C2 bond is 1.36 Å long, whereas the C2–C3 bond length is 1.42 Å. This contrasts with the structure of benzene, in which all the C–C bonds have a common length, 1.39 Å.
Naphthalene is more reactive than benzene, both in substitution and addition reactions, and these reactions tend to proceed in a manner that maintains one intact benzene ring. The following diagram shows three oxidation and reduction reactions that illustrate this feature. In the last example, catalytic hydrogenation of one ring takes place under milder conditions than those required for complete saturation (the decalin product exists as cis/trans isomers). Electrophilic substitution reactions take place more rapidly at C1, although the C2 product is more stable and predominates at equilibrium. Examples of these reactions will be displayed by clicking on the diagram. The kinetically favored C1 orientation reflects a preference for generating a cationic intermediate that maintains one intact benzene ring. By clicking on the diagram a second time, the two naphthenonium intermediates created by attack at C1 and C2 will be displayed.
The structure and chemistry of more highly fused benzene ring compounds, such as anthracene and phenanthreneshow many of the same characteristics described above.
Contributors
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry
Reactions of Substituent Groups
Oxidation of Alkyl Side-Chains
The benzylic hydrogens of alkyl substituents on a benzene ring are activated toward free radical attack, as noted earlier. Furthermore, \(S_N1\), \(S_N2\) and E1 reactions ofbenzylic halides, show enhanced reactivity, due to the adjacent aromatic ring. The possibility that these observations reflect a general benzylic activation is supported by the susceptibility of alkyl side-chains to oxidative degradation, as shown in the following examples (the oxidized side chain is colored). Such oxidations are normally effected by hot acidic pemanganate solutions, but for large scale industrial operations catalyzed air-oxidations are preferred. Interestingly, if the benzylic position is completely substituted this oxidative degradation does not occur (second equation, the substituted benzylic carbon is colored blue).
C6H5CH2CH2CH2CH3 + KMnO4 + H3O(+) & heat
C6H5CO2H + CO2
p-(CH3)3C–C6H4CH3 + KMnO4 + H3O(+) & heat
p-(CH3)3C–C6H4CO2H
These equations are not balanced. The permanganate oxidant is reduced, usually to Mn(IV) or Mn(II). Two other examples of this reaction are given below, and illustrate its usefulness in preparing substituted benzoic acids.
Reduction of Nitro Groups and Aryl Ketones
Electrophilic nitration and Friedel-Crafts acylation reactions introduce deactivating, meta-directing substituents on an aromatic ring. The attached atoms are in a high oxidation state, and their reduction converts these electron withdrawing functions into electron donating amino and alkyl groups. Reduction is easily achieved either by catalytic hydrogenation (H2 + catalyst), or with reducing metals in acid. Examples of these reductions are shown here, equation 6 demonstrating the simultaneous reduction of both functions. Note that the butylbenzene product in equation 4 cannot be generated by direct Friedel-Crafts alkylation due to carbocation rearrangement. The zinc used in ketone reductions, such as 5, is usually activated by alloying with mercury (a process known as amalgamation).
Several alternative methods for reducing nitro groups to amines are known. These include zinc or tin in dilute mineral acid, and sodium sulfide in ammonium hydroxide solution. The procedures described above are sufficient for most cases.
Conversion of Halogens to Organometallic Reagents
The reaction of alkyl and aryl halides with reactive metals (usually Li & Mg) to give nucleophilic reagents has been noted. This provides a powerful tool for the conversion of chloro, bromo or iodo substituents into a variety of other groups. Many reactions of these aryl lithium and Grignard reagents will be discussed in later sections, and the following equations provide typical examples of carboxylation, protonation and Gilman coupling. Metal halogen exchange reactions take place at low temperature, and may be used to introduce iodine at designated locations. An example of this method is displayed below. In this example care must be taken to maintain a low temperature, because elimination to an aryne intermediate takes place on warming.
Hydrolysis of Sulfonic Acids
The potential reversibility of the aromatic sulfonation reaction was noted earlier. The following equation illustrates how this characteristic of the sulfonic acids may be used to prepare the 3-bromo derivative of ortho-xylene. Direct bromination would give the 4-bromo derivative.
Contributors
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Arenes/Reactivity_of_Arenes/Reactions_of_Fused_Benzene_Rings.txt |
Substituted rings are divided into two groups based on the type of the substituent that the ring carries:
• Activated rings: the substituents on the ring are groups that donate electrons.
• Deactivated rings: the substituents on the ring are groups that withdraw electrons.
Introduction
Examples of activating groups in the relative order from the most activating group to the least activating:
-NH2, -NR2 > -OH, -OR> -NHCOR> -CH3 and other alkyl groups
with R as alkyl groups (CnH2n+1)
Examples of deactivating groups in the relative order from the most deactivating to the least deactivating:
-NO2, -CF3> -COR, -CN, -CO2R, -SO3H > Halogens
with R as alkyl groups (CnH2n+1)
The order of reactivity among Halogens from the more reactive (least deactivating substituent) to the least reactive (most deactivating substituent) halogen is:
F> Cl > Br > I
The order of reactivity of the benzene rings toward the electrophilic substitution when it is substituted with a halogen groups, follows the order of electronegativity. The ring that is substituted with the most electronegative halogen is the most reactive ring ( less deactivating substituent ) and the ring that is substituted with the least electronegatvie halogen is the least reactive ring ( more deactivating substituent ), when we compare rings with halogen substituents. Also the size of the halogen effects the reactivity of the benzene ring that the halogen is attached to. As the size of the halogen increase, the reactivity of the ring decreases.
The direction of the reaction
The activating group directs the reaction to the ortho or para position, which means the electrophile substitute the hydrogen that is on carbon 2 or carbon 4. The deactivating group directs the reaction to the meta position, which means the electrophile substitute the hydrogen that is on carbon 3 with the exception of the halogens that is a deactivating group but directs the ortho or para substitution.
Substituents determine the reaction direction by resonance or inductive effect
Resonance effect is the conjugation between the ring and the substituent, which means the delocalizing of the $\pi$ electrons between the ring and the substituent. Inductive effect is the withdraw of the sigma ( the single bond ) electrons away from the ring toward the substituent, due to the higher electronegativity of the substituent compared to the carbon of the ring.
Activating groups (ortho or para directors)
When the substituents like -OH have an unshared pair of electrons, the resonance effect is stronger than the inductive effect which make these substituents stronger activators, since this resonance effect direct the electron toward the ring. In cases where the subtituents is esters or amides, they are less activating because they form resonance structure that pull the electron density away from the ring.
By looking at the mechanism above, we can see how groups donating electron direct the ortho, para electrophilic substition. Since the electrons locatinn transfer between the ortho and para carbons, then the electrophile prefer attacking the carbon that has the free electron.
Inductive effect of alkyl groups activates the direction of the ortho or para substitution, which is when s electrons gets pushed toward the ring.
Deactivating group (meta directors)
The deactivating groups deactivate the ring by the inductive effect in the presence of an electronegative atom that withdraws the electrons away from the ring.
we can see from the mechanism above that when there is an electron withdraw from the ring, that leaves the carbons at the ortho, para positions with a positive charge which is unfavorable for the electrophile, so the electrophile attacks the carbon at the meta positions.
Halogens are an exception of the deactivating group that directs the ortho or para substitution. The halogens deactivate the ring by inductive effect not by the resonance even though they have an unpaired pair of electrons. The unpaired pair of electrons gets donated to the ring, but the inductive effect pulls away the s electrons from the ring by the electronegativity of the halogens.
Substituents determine the reactivity of rings
The reaction of a substituted ring with an activating group is faster than benzene. On the other hand, a substituted ring with a deactivated group is slower than benzene.
Activating groups speed up the reaction because of the resonance effect. The presence of the unpaired electrons that can be donated to the ring, stabilize the carbocation in the transition state. Thus; stabilizing the intermediate step, speeds up the reaction; and this is due to the decrease of the activating energy. On the other hand, the deactivating groups, withdraw the electrons away from the carbocation formed in the intermediate step, thus; the activation energy is increased which slows down the reaction.
Problems
1. Predict the direction of the electrophile substition on these rings:
2. Which nitration product is going to form faster?
nitration of aniline or nitration of nitrobenzene?
3. Predict the product of the following two sulfonation reactions:
A.
4. Classify these two groups as activating or deactivating groups:
A. alcohol
B. ester
5. By which effect does trichloride effect a monosubstituted ring?
Answers
1. The first substitution is going to be ortho and/or para substitution since we have a halogen subtituent. The second substition is going to be ortho and/or para substitution also since we have an alkyl substituent.
2. The nitration of aniline is going to be faster than the nitration of nitrobenzene, since the aniline is a ring with NH2 substituent and nitrobenzene is a ring with NO2 substiuent. As described above NH2is an activating group which speeds up the reaction and NO2 is deactivating group that slows down the reaction.
3. A. the product is
B. the product is
4. A. alcohol is an activating group.
B. ester is a deactivating group.
5. Trichloride deactivate a monosubstitued ring by inductive effect.
Contributors
• Lana Alawwad (UCD)
Substitution Reactions of Benzene and Other Aromatic Compounds
The chemical reactivity of benzene contrasts with that of the alkenes in that substitution reactions occur in preference to addition reactions, as illustrated in the following diagram (some comparable reactions of cyclohexene are shown in the green box).
Many other substitution reactions of benzene have been observed, the five most useful are listed below (chlorination and bromination are the most common halogenation reactions). Since the reagents and conditions employed in these reactions are electrophilic, these reactions are commonly referred to as Electrophilic Aromatic Substitution. The catalysts and co-reagents serve to generate the strong electrophilic species needed to effect the initial step of the substitution. The specific electrophile believed to function in each type of reaction is listed in the right hand column.
Reaction Type
Typical Equation
Electrophile E(+)
Halogenation:
C6H6
+ Cl2 & heat
FeCl3 catalyst
——>
C6H5Cl + HCl
Chlorobenzene
Cl(+) or Br(+)
C6H6
+ HNO3 & heat
H2SO4 catalyst
——>
C6H5NO2 + H2O
Nitrobenzene
NO2(+)
Sulfonation:
C6H6
+ H2SO4 + SO3
& heat
——>
C6H5SO3H + H2O
Benzenesulfonic acid
SO3H(+)
Alkylation:
Friedel-Crafts
C6H6
+ R-Cl & heat
AlCl3 catalyst
——>
C6H5-R + HCl
An Arene
R(+)
Acylation:
Friedel-Crafts
C6H6
+ RCOCl & heat
AlCl3 catalyst
——>
C6H5COR + HCl
An Aryl Ketone
RCO(+)
Contributors
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Arenes/Reactivity_of_Arenes/Substitution_Reactions_of_Benzene_Derivatives.txt |
Many laboratory methods exist for the organic synthesis of arenes from non-arene precursors. Many methods rely on cycloaddition reactions. Alkyne trimerization describes the [2+2+2] cyclization of three alkynes, in the Dötz reaction an alkyne, carbon monoxide and a chromium carbene complex are the reactants. Another set of methods is the aromatization of cyclohexanes and other aliphatic rings: reagents are catalysts used in hydrogenation such as platinum, palladium and nickel (reverse hydrogenation), quinones and the elements sulfur and selenium.
Synthesis of Arenes
This section is on the general mechanism of how an electrophilic atom becomes a part of a benzene ring through the substitution of a hydrogen. Common reactions that proceed by electrophilic aromatic substitution include the nitration and sulfonation of benzene, hydration of benzene, friedel-crafts acylation and friedel-crafts alkylation.
Reactivity of benzene
Benzene is an aromatic compound that is greatly stabilized by its resonance forms. Stable compounds are much harder to react with, therefore a strong electrophile will be needed to attack the \(pi\) electrons in the benzene ring. Electrophiles used in alkene reactions will typically not be strong enough on their own, therefore a Lewis acid catalyst is required to help generate the electrophile. Although the general mechanism listed here starts with a non-substituted benzene ring, it should make sense that this same reaction could still occur even if there was already a constituent present on the ring, creating polysubstituted benzene rings.
Specific Reactions
Reaction Reagent Catalyst Product E+ or E
Halogenation X2 (X=Cl, Br) FeX3 ArCl, ArBr X+
Nitration HNO3 H2SO4 ArNO2 +NO2
Sulfonation H2SO4 or H2S2O7 None ArSO3H SO3
Friedel-Crafts alkylation RX, ArCH2X AlCl3 Ar-R, Ar-CH2Ar R+
ROH HF, H2SO4, or BF3 Ar-R R+
RCH=CH2 H3PO4 or HF Ar-CHRCH3 R+
Friedel-Crafts acylation RCOCl AlCl3 Ar-COR RC+=O
Potential Energy Diagram of Reaction
Below in a potential energy diagram showing the reaction course of benzene with an electrophile. Notice that the INTERMEDIATE IS NOT AROMATIC. Aromatic cyclic compounds are much more stable than cyclic alkenes, which is why the reaction will continue after the substitution until the aromaticity of the ring as been regained.
Step 1 has a high activation energy barrier because a non-aromatic intermediate is formed. Aromatic molecules are lower in energy than their non-aromatic counterparts, so this step is endothermic (product has higher energy than reactant). This step is slow and not thermodynamically favored because the product has higher energy than the reactant. From this intermediate there is a smaller energy of activation- this is because the reaction wants to proceed forward and regain the aromaticity lost in the first step, so this occurs quickly. The aromatic product has lower energy, so this step is exothermic. If you still have trouble interpreting this diagram, check out this site on reaction kinetics to review how to read potential energy diagrams.
Problems
1. Label the hybridization on all the carbons in a) reacting benzene ring, b) intermediate (including resonance forms), and c) product (monosubstituted benzene ring)
2. Is the energy of activation higher in the first step or second step of the mechanism? Explain your reasoning.
3. If you wanted to halogenate benzene, what sort of reagent and catalyst (if needed) would you use?
4. Which hydrogren is used in order to regain aromaticity after the electrophile has added to the ring?
5. Critical Thinking Question: Mentioned above was the fact that electrophilic aromatic substitution can and does happen when there are substituents already present on the ring. Already present substituents will determine where something adds onto the ring in relation to itself (ortho, meta, or para position). What sort of factors do you think influence where addition will occur?
Answers
1. a) All are sp2 b) Five carbons are sp2, the carbon with the attached E is sp3 hybridized c) All are sp2
2. The activation energy is higher for the first step- aromatic compounds are lower in energy ("happier") than their non-aromatic forms. Therefore it takes more energy to change a compound from being aromatic to non-aromatic, than it does in the reverse direction.
3. \(X_2 (X=Cl, Br) + FeX_3\)
4. The hydrogen used is the one attached to the same carbon that E added to
5. Critical Thinking: If you came up with a) Inductive Effects (correlates to electronegativity) b) Resonance effects c) Electrostatics d) Steric Hindrance then good job! All of these influence where a substituent will add to the ring. Go to Activating and Deactivating Benzene Rings to find out more about how position is determined.
Contributors
• Stevie Maxwell (UCD)
Synthesis of Arenes
This page looks at the manufacture of arenes such as benzene and methylbenzene (toluene) by the catalytic reforming of fractions from petroleum (crude oil).
Catalytic reforming
Reforming takes straight chain hydrocarbons in the C6 to C8 range from the gasoline or naphtha fractions and rearranges them into compounds containing benzene rings. Hydrogen is produced as a by-product of the reactions. For example, hexane, C6H14, loses hydrogen and turns into benzene. As long as you draw the hexane bent into a circle, it is easy to see what is happening.
Similarly, methylbenzene (toluene) is made from heptane:
The process
• The feedstock: The feedstock is a mixture of the naphtha or gasoline fractions and hydrogen. The hydrogen is there to help prevent the formation of carbon by decomposition of the hydrocarbons at the high temperatures used. The carbon would otherwise contaminate the catalyst.
• The catalyst: A typical catalyst is a mixture of platinum and aluminium oxide. With a platinum catalyst, the process is sometimes described as "platforming".
• Temperature and pressure: The temperature is about 500°C, and the pressure varies either side of 20 atmospheres.
Converting methylbenzene into benzene
Methylbenzene is much less commercially valuable than benzene. The methyl group can be removed from the ring by a process known as "dealkylation". The methylbenzene is mixed with hydrogen at a temperature of between 550 and 650°C, and a pressure of between 30 and 50 atmospheres, with a mixture of silicon dioxide and aluminium oxide as catalyst.
Contributors
Jim Clark (Chemguide.co.uk) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Arenes/Synthesis_of_Arenes/Electrophilic_Aromatic_Substitution.txt |
Aryl halides are termed Haloarenes in IUPAC nomenclature system. The prefix ‘halo” (bromo or chloro or iodo or fluoro) is placed before the name of the aromatic hydrocarbon. In case of disubstituted compounds, the relative positions are indicated by (1,2), (1,3) or (1,4). Ortho, meta and para are also used to indicate the positions (e.g, chlorobenzene,bromobenzene).
Figure 1: Structure of 1-Chlorobenzene
Properties of Aryl Halides
This page looks at the structure and physical properties of three simple aryl halides - chlorobenzene, bromobenzene and iodobenzene. An aryl halide has a halogen atom attached directly to a benzene ring.
The structure of chlorobenzene
We'll look in some detail at the structure of chlorobenzene. Bromobenzene and iodobenzene are just the same. The simplest way to draw the structure of chlorobenzene is:
To understand chlorobenzene properly, you need to dig a bit deeper than this. There is an interaction between the delocalised electrons in the benzene ring and one of the lone pairs on the chlorine atom. This overlaps with the delocalised ring electron system . . .
to giving a structure rather like this:
This delocalisation is by no means complete, but it does have a significant effect on the properties of both the carbon-chlorine bond and the polarity of the molecule. The delocalisation introduces some extra bonding between the carbon and the chlorine, making the bond stronger. This has a major effect on the reactions of compounds like chlorobenzene.
There is also some movement of electrons away from the chlorine towards the ring. Chlorine is quite electronegative and usually draws electrons in the carbon-chlorine bond towards itself. In this case, this is offset to some extent by the movement of electrons back towards the ring in the delocalisation. The molecule is less polar than you would otherwise have expected.
Physical properties
Boiling points
Chlorobenzene, bromobenzene and iodobenzene are all oily liquids. The boiling points increase as the halogen atom gets bigger.
boiling point (°C)
C6H5Cl 132
C6H5Br 156
C6H5I 189
The main attractions between the molecules will be van der Waals dispersion forces. These increase as the number of electrons in the molecule increases. This is the reason that the boiling points increase as the halogen atom gets bigger.
There will also be permanent dipole-dipole attractions involved in the chlorobenzene and bromobenzene, but very little in the iodobenzene. Iodine has much the same electronegativity as carbon. These dipole-dipole attractions must be very unimportant relative to the dispersion forces because the most polar molecule (the chlorobenzene) has the lowest boiling point of the three.
Solubility in water
The aryl halides are insoluble in water and are denser than water and form a separate lower layer. The molecules are quite large compared with a water molecule. For chlorobenzene to dissolve in water, it would have to break many existing hydrogen bonds between water molecules. Moreover, the quite strong van der Waals dispersion forces between chlorobenzene molecules would have to be broken. Both of these cost energy.
The only new forces between the chlorobenzene and the water would be van der Waals dispersion forces. These are not as strong as hydrogen bonds (or the original dispersion forces in the chlorobenzene), and so you wouldn't get much energy released when they form. It simply is not energetically profitable for chlorobenzene (and the others) to dissolve in water.
Contributors
Jim Clark (Chemguide.co.uk) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Aryl_Halides/Nomenclature_of_Aryl_Halides.txt |
This page only looks at one aspect of the chemistry of the aryl halides such as chlorobenzene - and that they are very unreactive compared with halogenoalkanes.
Nucleophilic substitution reactions
A nucleophile can be either a negative ion or a molecule which carries a partial negative charge somewhere on it. On this page, we are just going to be looking at a simple nucleophile - a hydroxide ion. A nucleophilic substitution reaction is one in which a part of a molecule is replaced after it has been attacked by a nucleophile.
Here is a quick summary of the two ways that halogenoalkanes can react with hydroxide ions. We'll compare these with the aryl halides afterwards. The two different ways in which these reactions can happen depends on what kind of halogenoalkane you are talking about.
Here is the mechanism for the reaction involving bromoethane - a primary halogenoalkane. A hydroxide ion attacks the slightly positive carbon atom and pushes off the bromine as a bromide ion.
A tertiary halogenoalkane reacts differently. The mechanism this time involves an initial ionisation of the halogenoalkane:
. . . followed by a very rapid attack by the hydroxide ion on the carbocation (carbonium ion) formed:
Nucleophilic substitution in the aryl halides
Simple aryl halides like chlorobenzene are very resistant to nucleophilic substitution. It is possible to replace the chlorine by -OH, but only under very severe industrial conditions - for example at 200°C and 200 atmospheres. In the lab, these reactions do not happen. There are two reasons for this - depending on which of the above mechanisms you are talking about.
1. The extra strength of the carbon-halogen bond in aryl halides: The carbon-chlorine bond in chlorobenzene is stronger than you might expect. There is an interaction between one of the lone pairs on the chlorine atom and the delocalized ring electrons, and this strengthens the bond. Both of the mechanisms above involve breaking the carbon-halogen bond at some stage. The more difficult it is to break, the slower the reaction will be.
2. Repulsion by the ring electrons: This will only apply if the hydroxide ion attacked the chlorobenzene by a mechanism like the first one described above. In that mechanism, the hydroxide ion attacks the slightly positive carbon that the halogen atom is attached to.If the halogen atom is attached to a benzene ring, the incoming hydroxide ion is going to be faced with the delocalized ring electrons above and below that carbon atom. The negative hydroxide ion will simply be repelled. That particular mechanism is simply a non-starter!
Contributors
Jim Clark (Chemguide.co.uk)
Synthesis of Aryl Halides
This page looks at the ways of making the aryl halides, chlorobenzene, bromobenzene and iodobenzene.
Making chlorobenzene
Benzene reacts with chlorine in the presence of a catalyst, replacing one of the hydrogen atoms on the ring by a chlorine atom. The reaction happens at room temperature. The catalyst is either aluminium chloride or iron. Strictly speaking iron is not a catalyst, because it gets permanently changed during the reaction. It reacts with some of the chlorine to form iron(III) chloride, FeCl3.
$2Fe + 3Cl_2 \rightarrow 2FeCl_3$
This compound acts as the catalyst and behaves exactly like aluminium chloride, $AlCl_3$, in this reaction. The reaction between benzene and chlorine in the presence of either aluminium chloride or iron gives chlorobenzene.
or, written more compactly:
$C_6H_6 + Cl_2 \rightarrow C_6H_5Cl + HCl$
Bromobenzene
The reaction between benzene and bromine in the presence of either aluminum bromide (rather than aluminum chloride) or iron gives bromobenzene. Iron is usually used because it is cheaper and more readily available. If you use iron, it is first converted into iron(III) bromide by the reaction between the iron and bromine.
or:
$C_6H_6 + Br_2 \rightarrow C_6H_5Br + HBr$
Iodobenzene
Iodobenzene can be made from the reaction of benzene with iodine if they are heated under reflux in the presence of concentrated nitric acid, but it is normally made from benzenediazonium chloride solution. If cold potassium iodide solution is added to ice-cold benzenediazonium chloride solution, nitrogen gas is released and oily droplets of iodobenzene are formed.
There is a simple reaction between the diazonium ions present in the benzenediazonium chloride solution and the iodide ions from the potassium iodide solution.
Contributors
Jim Clark (Chemguide.co.uk) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Aryl_Halides/Reactions_of_Aryl_Halides.txt |
Fischer's brilliant elucidation of the configuration of glucose did not remove all uncertainty concerning its structure. Two different crystalline forms of glucose were reported in 1895. Each of these gave all the characteristic reactions of glucose, and when dissolved in water equilibrated to the same mixture. This equilibration takes place over a period of many minutes, and the change in optical activity that occurs is called mutarotation. These facts are summarized in the diagram below.
When glucose was converted to its pentamethyl ether (reaction with excess CH3I & AgOH), two different isomers were isolated, and neither exhibited the expected aldehyde reactions. Acid-catalyzed hydrolysis of the pentamethyl ether derivatives, however, gave a tetramethyl derivative that was oxidized by Tollen's reagent and reduced by sodium borohydride, as expected for an aldehyde.
The search for scientific truth often proceeds in stages, and the structural elucidation of glucose serves as a good example. It should be clear from the new evidence presented above, that the open chain pentahydroxyhexanal structure drawn above must be modified. Somehow a new stereogenic center must be created, and the aldehyde must be deactivated in the pentamethyl derivative. A simple solution to this dilemma is achieved by converting the open aldehyde structure for glucose into a cyclic hemiacetal, called a glucopyranose, as shown in the following diagram. The linear aldehyde is tipped on its side, and rotation about the C4-C5 bond brings the C5-hydroxyl function close to the aldehyde carbon. For ease of viewing, the six-membered hemiacetal structure is drawn as a flat hexagon, but it actually assumes a chair conformation. The hemiacetal carbon atom (C-1) becomes a new stereogenic center, commonly referred to as the anomeric carbon, and the α and β-isomers are called anomers.
We can now consider how this modification of the glucose structure accounts for the puzzling facts noted above. First, we know that hemiacetals are in equilibrium with their carbonyl and alcohol components when in solution. Consequently, fresh solutions of either alpha or beta-glucose crystals in water should establish an equilibrium mixture of both anomers, plus the open chain chain form. Note that despite the very low concentration of the open chain aldehyde in this mixture, typical chemical reactions of aldehydes take place rapidly.
Second, a pentamethyl ether derivative of the pyranose structure converts the hemiacetal function to an acetal. Acetals are stable to base, so this product should not react with Tollen's reagent or be reduced by sodium borohydride. Acid hydrolysis of acetals regenerates the carbonyl and alcohol components, and in the case of the glucose derivative this will be a tetramethyl ether of the pyranose hemiacetal. This compound will, of course, undergo typical aldehyde reactions.
Contributors
Prof. Steven Farmer (Sonoma State University)
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry
Cyclic Forms of Monosaccharides
As noted above, the preferred structural form of many monosaccharides may be that of a cyclic hemiacetal. Five and six-membered rings are favored over other ring sizes because of their low angle and eclipsing strain. Cyclic structures of this kind are termed furanose (five-membered) or pyranose (six-membered), reflecting the ring size relationship to the common heterocyclic compounds furan and pyran shown on the right.
Ribose, an important aldopentose, commonly adopts a furanose structure, as shown in the following illustration. By convention for the D-family, the five-membered furanose ring is drawn in an edgewise projection with the ring oxygen positioned away from the viewer. The anomeric carbon atom (colored red here) is placed on the right. The upper bond to this carbon is defined as beta, the lower bond then is alpha.
The cyclic pyranose forms of various monosaccharides are often drawn in a flat projection known as a Haworth formula, after the British chemist, Norman Haworth. As with the furanose ring, the anomeric carbon is placed on the right with the ring oxygen to the back of the edgewise view. In the D-family, the alpha and beta bonds have the same orientation defined for the furanose ring (beta is up & alpha is down). These Haworth formulas are convenient for displaying stereochemical relationships, but do not represent the true shape of the molecules. We know that these molecules are actually puckered in a fashion we call a chair conformation. Examples of four typical pyranose structures are shown below, both as Section 3.8: Fischer and Haworth projections and as the more representative chair conformers. The anomeric carbons are colored red.
The size of the cyclic hemiacetal ring adopted by a given sugar is not constant, but may vary with substituents and other structural features. Aldolhexoses usually form pyranose rings and their pentose homologs tend to prefer the furanose form, but there are many counter examples. The formation of acetal derivatives illustrates how subtle changes may alter this selectivity. A pyranose structure for D-glucose is drawn in the rose-shaded box on the left. Acetal derivatives have been prepared by acid-catalyzed reactions with benzaldehyde and acetone. As a rule, benzaldehyde forms six-membered cyclic acetals, whereas acetone prefers to form five-membered acetals. The top equation shows the formation and some reactions of the 4,6-O-benzylidene acetal, a commonly employed protective group. A methyl glycoside derivative of this compound (see below) leaves the C-2 and C-3 hydroxyl groups exposed to reactions such as the periodic acid cleavage, shown as the last step. The formation of an isopropylidene acetal at C-1 and C-2, center structure, leaves the C-3 hydroxyl as the only unprotected function. Selective oxidation to a ketone is then possible. Finally, direct di-O-isopropylidene derivatization of glucose by reaction with excess acetone results in a change to a furanose structure in which the C-3 hydroxyl is again unprotected. However, the same reaction with D-galactose, shown in the blue-shaded box, produces a pyranose product in which the C-6 hydroxyl is unprotected. Both derivatives do not react with Tollens' reagent. This difference in behavior is attributed to the cis-orientation of the C-3 and C-4 hydroxyl groups in galactose, which permits formation of a less strained five-membered cyclic acetal, compared with the trans-C-3 and C-4 hydroxyl groups in glucose. Derivatizations of this kind permit selective reactions to be conducted at different locations in these highly functionalized molecules. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Carbohydrates/Anomeric_Forms_of_Glucose.txt |
When the alcohol component of a glycoside is provided by a hydroxyl function on another monosaccharide, the compound is called a disaccharide. Four examples of disaccharides composed of two glucose units are shown in the following diagram. The individual glucopyranose rings are labeled A and B, and the glycoside bonding is circled in light blue. Notice that the glycoside bond may be alpha, as in maltose and trehalose, or beta as in cellobiose and gentiobiose. Acid-catalyzed hydrolysis of these disaccharides yields glucose as the only product. Enzyme-catalyzed hydrolysis is selective for a specific glycoside bond, so an alpha-glycosidase cleaves maltose and trehalose to glucose, but does not cleave cellobiose or gentiobiose. A beta-glycosidase has the opposite activity.
In order to draw a representative structure for cellobiose, one of the glucopyranose rings must be rotated by 180º, but this feature is often omitted in favor of retaining the usual perspective for the individual rings. The bonding between the glucopyranose rings in cellobiose and maltose is from the anomeric carbon in ring A to the C-4 hydroxyl group on ring B. This leaves the anomeric carbon in ring B free, so cellobiose and maltose both may assume alpha and beta anomers at that site (the beta form is shown in the diagram). Gentiobiose has a beta-glycoside link, originating at C-1 in ring A and terminating at C-6 in ring B. Its alpha-anomer is drawn in the diagram. Because cellobiose, maltose and gentiobiose are hemiacetals they are all reducing sugars (oxidized by Tollen's reagent). Trehalose, a disaccharide found in certain mushrooms, is a bis-acetal, and is therefore a non-reducing sugar. A systematic nomenclature for disaccharides exists, but as the following examples illustrate, these are often lengthy.
• Cellobiose : 4-O-β-D-Glucopyranosyl-D-glucose (the beta-anomer is drawn)
• Maltose : 4-O-α-D-Glucopyranosyl-D-glucose (the beta-anomer is drawn)
• Gentiobiose : 6-O-β-D-Glucopyranosyl-D-glucose (the alpha-anomer is drawn)
• Trehalose : α-D-Glucopyranosyl-α-D-glucopyranoside
Although all the disaccharides shown here are made up of two glucopyranose rings, their properties differ in interesting ways. Maltose, sometimes called malt sugar, comes from the hydrolysis of starch. It is about one third as sweet as cane sugar (sucrose), is easily digested by humans, and is fermented by yeast. Cellobiose is obtained by the hydrolysis of cellulose. It has virtually no taste, is indigestible by humans, and is not fermented by yeast. Some bacteria have beta-glucosidase enzymes that hydrolyze the glycosidic bonds in cellobiose and cellulose. The presence of such bacteria in the digestive tracts of cows and termites permits these animals to use cellulose as a food. Finally, it may be noted that trehalose has a distinctly sweet taste, but gentiobiose is bitter.
Disaccharides made up of other sugars are known, but glucose is often one of the components. Two important examples of such mixed disaccharides are displayed above. Lactose, also known as milk sugar, is a galactose-glucose compound joined as a beta-glycoside. It is a reducing sugar because of the hemiacetal function remaining in the glucose moiety. Many adults, particularly those from regions where milk is not a dietary staple, have a metabolic intolerance for lactose. Infants have a digestive enzyme which cleaves the beta-glycoside bond in lactose, but production of this enzyme stops with weaning. Cheese is less subject to the lactose intolerance problem, since most of the lactose is removed with the whey. Sucrose, or cane sugar, is our most commonly used sweetening agent. It is a non-reducing disaccharide composed of glucose and fructose joined at the anomeric carbon of each by glycoside bonds (one alpha and one beta). In the formula shown here the fructose ring has been rotated 180º from its conventional perspective.
Contributors
Prof. Steven Farmer (Sonoma State University)
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry
Glycosides
Acetal derivatives formed when a monosaccharide reacts with an alcohol in the presence of an acid catalyst are called glycosides. This reaction is illustrated for glucose and methanol in the diagram below. In naming of glycosides, the "ose" suffix of the sugar name is replaced by "oside", and the alcohol group name is placed first. As is generally true for most acetals, glycoside formation involves the loss of an equivalent of water. The diether product is stable to base and alkaline oxidants such as Tollen's reagent. Since acid-catalyzed aldolization is reversible, glycosides may be hydrolyzed back to their alcohol and sugar components by aqueous acid.
The anomeric methyl glucosides are formed in an equilibrium ratio of 66% alpha to 34% beta. From the structures in the previous diagram, we see that pyranose rings prefer chair conformations in which the largest number of substituents are equatorial. In the case of glucose, the substituents on the beta-anomer are all equatorial, whereas the C-1 substituent in the alpha-anomer changes to axial. Since substituents on cyclohexane rings prefer an equatorial location over axial (methoxycyclohexane is 75% equatorial), the preference for alpha-glycopyranoside formation is unexpected, and is referred to as the anomeric effect.
Glycosides abound in biological systems. By attaching a sugar moiety to a lipid or benzenoid structure, the solubility and other properties of the compound may be changed substantially. Because of the important modifying influence of such derivatization, numerous enzyme systems, known as glycosidases, have evolved for the attachment and removal of sugars from alcohols, phenols and amines. Chemists refer to the sugar component of natural glycosides as the glycon and the alcohol component as the aglycon.
Two examples of naturally occurring glycosides and one example of an amino derivative are displayed above. Salicin, one of the oldest herbal remedies known, was the model for the synthetic analgesic aspirin. A large class of hydroxylated, aromatic oxonium cations called anthocyanins provide the red, purple and blue colors of many flowers, fruits and some vegetables. Peonin is one example of this class of natural pigments, which exhibit a pronounced pH color dependence. The oxonium moiety is only stable in acidic environments, and the color changes or disappears when base is added. The complex changes that occur when wine is fermented and stored are in part associated with glycosides of anthocyanins. Finally, amino derivatives of ribose, such as cytidine play important roles in biological phosphorylating agents, coenzymes and information transport and storage materials.
Contributors
Prof. Steven Farmer (Sonoma State University)
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Carbohydrates/Disaccharides.txt |
Overview
Emil Fischer made use of several key reactions in the course of his carbohydrate studies. These are described here, together with the information that each delivers.
Oxidation
As noted above, sugars may be classified as reducing or non-reducing based on their reactivity with Tollens', Benedict's or Fehling's reagents. If a sugar is oxidized by these reagents it is called reducing, since the oxidant (Ag(+) or Cu(+2)) is reduced in the reaction, as evidenced by formation of a silver mirror or precipitation of cuprous oxide. The Tollens' test is commonly used to detect aldehyde functions; and because of the facile interconversion of ketoses and aldoses under the basic conditions of this test, ketoses such as fructose also react and are classified as reducing sugars.
When the aldehyde function of an aldose is oxidized to a carboxylic acid the product is called an aldonic acid. Because of the 2º hydroxyl functions that are also present in these compounds, a mild oxidizing agent such as hypobromite must be used for this conversion (equation 1). If both ends of an aldose chain are oxidized to carboxylic acids the product is called an aldaric acid. By converting an aldose to its corresponding aldaric acid derivative, the ends of the chain become identical (this could also be accomplished by reducing the aldehyde to CH2OH, as noted below). Such an operation will disclose any latent symmetry in the remaining molecule. Thus, ribose, xylose, allose and galactose yield achiral aldaric acids which are, of course, not optically active. The ribose oxidation is shown in equation 2 below.
1.
2.
3.
Other aldose sugars may give identical chiral aldaric acid products, implying a unique configurational relationship. The examples of arabinose and lyxose shown in equation 3 above illustrate this result. Remember, a Fischer projection formula may be rotated by 180º in the plane of projection without changing its configuration.
Reduction
Sodium borohydride reduction of an aldose makes the ends of the resulting alditol chain identical, HOCH2(CHOH)nCH2OH, thereby accomplishing the same configurational change produced by oxidation to an aldaric acid. Thus, allitol and galactitol from reduction of allose and galactose are achiral, and altrose and talose are reduced to the same chiral alditol. A summary of these redox reactions, and derivative nomenclature is given in the following table.
Derivatives of HOCH2(CHOH)nCHO
HOBr Oxidation
——>
HOCH2(CHOH)nCO2H
an Aldonic Acid
HNO3 Oxidation
——>
H2OC(CHOH)nCO2H
an Aldaric Acid
NaBH4 Reduction
——>
HOCH2(CHOH)nCH2OH
an Alditol
Osazone Formation
1.
2.
The osazone reaction was developed and used by Emil Fischer to identify aldose sugars differing in configuration only at the alpha-carbon. The upper equation shows the general form of the osazone reaction, which effects an alpha-carbon oxidation with formation of a bis-phenylhydrazone, known as an osazone. Application of the osazone reaction to D-glucose and D-mannose demonstrates that these compounds differ in configuration only at C-2.
Chain Shortening and Lengthening
1.
2.
These two procedures permit an aldose of a given size to be related to homologous smaller and larger aldoses. The importance of these relationships may be seen in the array of aldose structures presented earlier, where the structural connections are given by the dashed blue lines. Thus Ruff degradation of the pentose arabinose gives the tetrose erythrose. Working in the opposite direction, a Kiliani-Fischer synthesis applied to arabinose gives a mixture of glucose and mannose. An alternative chain shortening procedure known as the Wohl degradation is essentially the reverse of the Kiliani-Fischer synthesis.
Fischer's train of logic in assigning the configuration of D-glucose
1. Ribose and arabinose (two well known pentoses) both gave erythrose on Ruff degradation. As expected, Kiliani-Fischer synthesis applied to erythrose gave a mixture of ribose and arabinose.
2. Oxidation of erythrose gave an achiral (optically inactive) aldaric acid. This defines the configuration of erythrose.
3. Oxidation of ribose gave an achiral (optically inactive) aldaric acid. This defines the configuration of both ribose and arabinose.
4. Ruff shortening of glucose gave arabinose, and Kiliani-Fischer synthesis applied to arabinose gave a mixture of glucose and mannose.
5. Glucose and mannose are therefore epimers at C-2, a fact confirmed by the common product from their osazone reactions.
6. A pair of structures for these epimers can be written, but which is glucose and which is mannose?
To determine which of these epimers was glucose, Fischer made use of the inherent C2 symmetry in the four-carbon dissymmetric core of one epimer (B). This is shown in the following diagram by a red dot where the symmetry axis passes through the projection formula. Because of this symmetry, if the aldehyde and 1º-alcohol functions at the ends of the chain are exchanged, epimer B would be unchanged; whereas A would be converted to a different compound.
Fischer looked for and discovered a second aldohexose that represented the end group exchange for the epimer lacking the latent C2 symmetry (A). This compound was L-(+)-gulose, and its exchange relationship to D-(+)-glucose was demonstrated by oxidation to a common aldaric acid product. The remaining epimer is therefore mannose.
Contributors
Prof. Steven Farmer (Sonoma State University)
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Carbohydrates/Important_Reactions.txt |
Carbohydrates are the most abundant class of organic compounds found in living organisms. They originate as products of photosynthesis, an endothermic reductive condensation of carbon dioxide requiring light energy and the pigment chlorophyll.
$nCO_2 + n H_2O + Energy \rightarrow C_nH_{2n}O_n + nO_2$
As noted here, the formulas of many carbohydrates can be written as carbon hydrates, $C_n(H_2O)_n$, hence their name. The carbohydrates are a major source of metabolic energy, both for plants and for animals that depend on plants for food. Aside from the sugars and starches that meet this vital nutritional role, carbohydrates also serve as a structural material (cellulose), a component of the energy transport compound ATP/ADP, recognition sites on cell surfaces, and one of three essential components of DNA and RNA.
Carbohydrates are called saccharides or, if they are relatively small, sugars. Several classifications of carbohydrates have proven useful, and are outlined in the following table.
Complexity
Simple Carbohydrates
monosaccharides
Complex Carbohydrates
disaccharides, oligosaccharides
& polysaccharides
Size
Tetrose
C4 sugars
Pentose
C5 sugars
Hexose
C6 sugars
Heptose
C7 sugars
etc.
C=O Function
Aldose
sugars having an aldehyde function or an acetal equivalent.
Ketose
sugars having a ketone function or an acetal equivalent.
Reactivity
Reducing
sugars oxidized by Tollens' reagent (or Benedict's or Fehling's reagents).
Non-reducing
sugars not oxidized by Tollens' or other reagents.
Contributors
Prof. Steven Farmer (Sonoma State University)
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry
Ketoses
If a monosaccharide has a carbonyl function on one of the inner atoms of the carbon chain it is classified as a ketose. Dihydroxyacetone may not be a sugar, but it is included as the ketose analog of glyceraldehyde. The carbonyl group is commonly found at C-2, as illustrated by the following examples (chiral centers are colored red). As expected, the carbonyl function of a ketose may be reduced by sodium borohydride, usually to a mixture of epimeric products. D-Fructose, the sweetest of the common natural sugars, is for example reduced to a mixture of D-glucitol (sorbitol) and D-mannitol, named after the aldohexoses from which they may also be obtained by analogous reduction. Mannitol is itself a common natural carbohydrate.
Although the ketoses are distinct isomers of the aldose monosaccharides, the chemistry of both classes is linked due to their facile interconversion in the presence of acid or base catalysts. This interconversion, and the corresponding epimerization at sites alpha to the carbonyl functions, occurs by way of an enediol tautomeric intermediate.
Because of base-catalyzed isomerizations of this kind, the Tollens' reagent is not useful for distinguishing aldoses from ketoses or for specific oxidation of aldoses to the corresponding aldonic acids. Oxidation by HOBr is preferred for the latter conversion.
Polysaccharides
As the name implies, polysaccharides are large high-molecular weight molecules constructed by joining monosaccharide units together by glycosidic bonds. They are sometimes called glycans. The most important compounds in this class, cellulose, starch and glycogen are all polymers of glucose. This is easily demonstrated by acid-catalyzed hydrolysis to the monosaccharide. Since partial hydrolysis of cellulose gives varying amounts of cellobiose, we conclude the glucose units in this macromolecule are joined by beta-glycoside bonds between C-1 and C-4 sites of adjacent sugars. Partial hydrolysis of starch and glycogen produces the disaccharide maltose together with low molecular weight dextrans, polysaccharides in which glucose molecules are joined by alpha-glycoside links between C-1 and C-6, as well as the alpha C-1 to C-4 links found in maltose. Polysaccharides built from other monosaccharides (e.g. mannose, galactose, xylose and arabinose) are also known, but will not be discussed here.
Over half of the total organic carbon in the earth's biosphere is in cellulose. Cotton fibers are essentially pure cellulose, and the wood of bushes and trees is about 50% cellulose. As a polymer of glucose, cellulose has the formula (C6H10O5)n where n ranges from 500 to 5,000, depending on the source of the polymer. The glucose units in cellulose are linked in a linear fashion, as shown in the drawing below. The beta-glycoside bonds permit these chains to stretch out, and this conformation is stabilized by intramolecular hydrogen bonds. A parallel orientation of adjacent chains is also favored by intermolecular hydrogen bonds. Although an individual hydrogen bond is relatively weak, many such bonds acting together can impart great stability to certain conformations of large molecules. Most animals cannot digest cellulose as a food, and in the diets of humans this part of our vegetable intake functions as roughage and is eliminated largely unchanged. Some animals (the cow and termites, for example) harbor intestinal microorganisms that breakdown cellulose into monosaccharide nutrients by the use of beta-glycosidase enzymes.
Cellulose is commonly accompanied by a lower molecular weight, branched, amorphous polymer called hemicellulose. In contrast to cellulose, hemicellulose is structurally weak and is easily hydrolyzed by dilute acid or base. Also, many enzymes catalyze its hydrolysis. Hemicelluloses are composed of many D-pentose sugars, with xylose being the major component. Mannose and mannuronic acid are often present, as well as galactose and galacturonic acid.
Starch is a polymer of glucose, found in roots, rhizomes, seeds, stems, tubers and corms of plants, as microscopic granules having characteristic shapes and sizes. Most animals, including humans, depend on these plant starches for nourishment. The structure of starch is more complex than that of cellulose. The intact granules are insoluble in cold water, but grinding or swelling them in warm water causes them to burst.
The released starch consists of two fractions. About 20% is a water soluble material called amylose. Molecules of amylose are linear chains of several thousand glucose units joined by alpha C-1 to C-4 glycoside bonds. Amylose solutions are actually dispersions of hydrated helical micelles. The majority of the starch is a much higher molecular weight substance, consisting of nearly a million glucose units, and called amylopectin. Molecules of amylopectin are branched networks built from C-1 to C-4 and C-1 to C-6 glycoside links, and are essentially water insoluble. Representative structural formulas for amylose and amylopectin are shown above. The branching in this diagram is exaggerated, since on average, branches only occur every twenty five glucose units.
Hydrolysis of starch, usually by enzymatic reactions, produces a syrupy liquid consisting largely of glucose. When cornstarch is the feedstock, this product is known as corn syrup. It is widely used to soften texture, add volume, prohibit crystallization and enhance the flavor of foods. Glycogen is the glucose storage polymer used by animals. It has a structure similar to amylopectin, but is even more highly branched (about every tenth glucose unit). The degree of branching in these polysaccharides may be measured by enzymatic or chemical analysis.
Contributors
Prof. Steven Farmer (Sonoma State University)
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Carbohydrates/Introduction_to_Carbohydrates.txt |
Cotton, probably the most useful natural fiber, is nearly pure cellulose. The manufacture of textiles from cotton involves physical manipulation of the raw material by carding, combing and spinning selected fibers. For fabrics the best cotton has long fibers, and short fibers or cotton dust are removed. Crude cellulose is also available from wood pulp by dissolving the lignan matrix surrounding it. These less desirable cellulose sources are widely used for making paper.
In order to expand the ways in which cellulose can be put to practical use, chemists have devised techniques for preparing solutions of cellulose derivatives that can be spun into fibers, spread into a film or cast in various solid forms. A key factor in these transformations are the three free hydroxyl groups on each glucose unit in the cellulose chain, --[C6H7O(OH)3]n--. Esterification of these functions leads to polymeric products having very different properties compared with cellulose itself.
• Cellulose Nitrate, first prepared over 150 years ago by treating cellulose with nitric acid, is the earliest synthetic polymer to see general use. The fully nitrated compound, --[C6H7O(ONO2)3]n--, called guncotton, is explosively flammable and is a component of smokeless powder. Partially nitrated cellulose is called pyroxylin. Pyroxylin is soluble in ether and at one time was used for photographic film and lacquers. The high flammability of pyroxylin caused many tragic cinema fires during its period of use. Furthermore, slow hydrolysis of pyroxylin yields nitric acid, a process that contributes to the deterioration of early motion picture films in storage.
• Cellulose Acetate, --[C6H7O(OAc)3]n--, is less flammable than pyroxylin, and has replaced it in most applications. It is prepared by reaction of cellulose with acetic anhydride and an acid catalyst. The properties of the product vary with the degree of acetylation. Some chain shortening occurs unavoidably in the preparations. An acetone solution of cellulose acetate may be forced through a spinneret to generate filaments, called acetate rayon, that can be woven into fabrics.
• Viscose Rayon, is prepared by formation of an alkali soluble xanthate derivative that can be spun into a fiber that reforms the cellulose polymer by acid quenching. The following general equation illustrates these transformations. The product fiber is called viscose rayon.
Many complex polysaccharides are found in nature. The galactomannans, consisting of a mannose backbone with galactose side groups, are an interesting and useful example. A (1-4)-linked beta-D-mannose chain is adorned with 1-6-linked alpha-D-galactose units, as shown in the diagram below. The ratio of galactose to mannose usually ranges from 1:2 to 1:4. An important source of this substance is the guar bean, grown principally in northwestern India, and Pakistan. Altough guar protein is not of nutritional value to humans (guar means 'cow food' in Hindi), the bean is important as a source of guar gum, a galactomannan which forms a gel in water. The food industry uses this material as a stabilizer in ice cream, cream cheese and salad dressings.
Recently, guar gum has becomes an essential ingredient for mining oil and natural gas in a process called hydraulic fracturing. The demand for guar has increased to such a degree that the poor farmers in northwestern India have had their lives transformed by unexpected wealth.
Contributors
Prof. Steven Farmer (Sonoma State University)
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry
The Configuration of Glucose and Other Sugars
The four chiral centers in glucose indicate there may be as many as sixteen (24) stereoisomers having this constitution. These would exist as eight diastereomeric pairs of enantiomers, and the initial challenge was to determine which of the eight corresponded to glucose. This challenge was accepted and met in 1891 by the German chemist Emil Fischer. His successful negotiation of the stereochemical maze presented by the aldohexoses was a logical tour de force, and it is fitting that he received the 1902 Nobel Prize for chemistry for this accomplishment. One of the first tasks faced by Fischer was to devise a method of representing the configuration of each chiral center in an unambiguous manner. To this end, he invented a simple technique for drawing chains of chiral centers, that we now call the Fischer projection formula.
At the time Fischer undertook the glucose project it was not possible to establish the absolute configuration of an enantiomer. Consequently, Fischer made an arbitrary choice for (+)-glucose and established a network of related aldose configurations that he called the D-family. The mirror images of these configurations were then designated the L-family of aldoses. To illustrate using present day knowledge, Fischer projection formulas and names for the D-aldose family (three to six-carbon atoms) are shown below, with the asymmetric carbon atoms (chiral centers) colored red.
The last chiral center in an aldose chain (farthest from the aldehyde group) was chosen by Fischer as the D / L designator site. If the hydroxyl group in the projection formula pointed to the right, it was defined as a member of the D-family. A left directed hydroxyl group (the mirror image) then represented the L-family. Fischer's initial assignment of the D-configuration had a 50:50 chance of being right, but all his subsequent conclusions concerning the relative configurations of various aldoses were soundly based. In 1951 x-ray fluorescence studies of (+)-tartaric acid, carried out in the Netherlands by Johannes Martin Bijvoet (pronounced "buy foot"), proved that Fischer's choice was correct.
It is important to recognize that the sign of a compound's specific rotation (an experimental number) does not correlate with its configuration (D or L). It is a simple matter to measure an optical rotation with a polarimeter. Determining an absolute configuration usually requires chemical interconversion with known compounds by stereospecific reaction paths.
Contributors
Prof. Steven Farmer (Sonoma State University)
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Carbohydrates/Synthetic_Modification_of_Cellulose.txt |
The IUPAC system of nomenclature assigns a characteristic suffix to these classes. The –e ending is removed from the name of the parent chain and is replaced -anoic acid. Since a carboxylic acid group must always lie at the end of a carbon chain, it is always is given the #1 location position in numbering and it is not necessary to include it in the name.
Many carboxylic acids are called by the common names. These names were chosen by chemists to usually describe a source of where the compound is found. In common names of aldehydes, carbon atoms near the carboxyl group are often designated by Greek letters. The atom adjacent to the carbonyl function is alpha, the next removed is beta and so on.
Formula Common Name Source IUPAC Name Melting Point Boiling Point
HCO2H formic acid ants (L. formica) methanoic acid 8.4 ºC 101 ºC
CH3CO2H acetic acid vinegar (L. acetum) ethanoic acid 16.6 ºC 118 ºC
CH3CH2CO2H propionic acid milk (Gk. protus prion) propanoic acid -20.8 ºC 141 ºC
CH3(CH2)2CO2H butyric acid butter (L. butyrum) butanoic acid -5.5 ºC 164 ºC
CH3(CH2)3CO2H valeric acid valerian root pentanoic acid -34.5 ºC 186 ºC
CH3(CH2)4CO2H caproic acid goats (L. caper) hexanoic acid -4.0 ºC 205 ºC
CH3(CH2)5CO2H enanthic acid vines (Gk. oenanthe) heptanoic acid -7.5 ºC 223 ºC
CH3(CH2)6CO2H caprylic acid goats (L. caper) octanoic acid 16.3 ºC 239 ºC
CH3(CH2)7CO2H pelargonic acid pelargonium (an herb) nonanoic acid 12.0 ºC 253 ºC
CH3(CH2)8CO2H capric acid goats (L. caper) decanoic acid 31.0 ºC 219 ºC
Example (Common Names Are in Red)
Naming carboxyl groups added to a ring
When a carboxyl group is added to a ring the suffix -carboxylic acid is added to the name of the cyclic compound. The ring carbon attached to the carboxyl group is given the #1 location number.
Naming carboxylates
Salts of carboxylic acids are named by writing the name of the cation followed by the name of the acid with the –ic acid ending replaced by an –ate ending. This is true for both the IUPAC and Common nomenclature systems.
Naming carboxylic acids which contain other functional groups
Carboxylic acids are given the highest nomenclature priority by the IUPAC system. This means that the carboxyl group is given the lowest possible location number and the appropriate nomenclature suffix is included. In the case of molecules containing carboxylic acid and alcohol functional groups the OH is named as a hydroxyl substituent. However, the l in hydroxyl is generally removed.
In the case of molecules containing a carboxylic acid and aldehydes and/or ketones functional groups the carbonyl is named as a "Oxo" substituent.
In the case of molecules containing a carboxylic acid an amine functional group the amine is named as an "amino" substituent.
When carboxylic acids are included with an alkene the following order is followed:
(Location number of the alkene)-(Prefix name for the longest carbon chain minus the -ane ending)-(an –enoic acid ending to indicate the presence of an alkene and carboxylic acid)
Remember that the carboxylic acid has priority so it should get the lowest possible location number. Also, remember that cis/tran or E/Z nomenclature for the alkene needs to be included if necessary.
Naming dicarboxylic acids
For dicarboxylic acids the location numbers for both carboxyl groups are omitted because both functional groups are expected to occupy the ends of the parent chain. The ending –dioic acid is added to the end of the parent chain. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Carboxylic_Acids/Nomenclature_of_Carboxylic_Acids.txt |
The important classes of organic compounds known as alcohols, phenols, ethers, amines and halides consist of alkyl and/or aryl groups bonded to hydroxyl, alkoxyl, amino and halo substituents respectively. If these same functional groups are attached to an acyl group (RCO–) their properties are substantially changed, and they are designated as carboxylic acid derivatives. Carboxylic acids have a hydroxyl group bonded to an acyl group, and their functional derivatives are prepared by replacement of the hydroxyl group with substituents, such as halo, alkoxyl, amino and acyloxy.
The following table lists some representative derivatives and their boiling points. An aldehyde and ketone of equivalent molecular weight are also listed for comparison. Boiling points are given for 760 torr (atmospheric pressure), and those listed as a range are estimated from values obtained at lower pressures. The relatively high boiling point of carboxylic acids is due to extensive hydrogen bonded dimerization. Similar hydrogen bonding occurs between molecules of 1º and 2º-amides (amides having at least one N–H bond), and the first three compounds in the table serve as hydrogen bonding examples.
Table 1: Physical Properties of Some Carboxylic Acid Derivatives
Formula
IUPAC Name
Molecular Weight
Boiling Point
Water Solubility
CH3(CH2)2CO2H butanoic acid 88 164 ºC very soluble
CH3(CH2)2CONH2 butanamide 87 216-220 ºC soluble
CH3CH2CONHCH3 N-methylpropanamide 87 205 -210 ºC soluble
CH3CON(CH3)2 N,N-dimethylethanamide 87 166 ºC very soluble
HCON(CH3)CH2CH3 N-ethyl, N-methylmethanamide 87 170-180 ºC very soluble
CH3(CH2)3CN pentanenitrile 83 141 ºC slightly soluble
CH3CO2CHO ethanoic methanoic
anhydride
88 105-112 ºC reacts with water
CH3CH2CO2CH3 methyl propanoate 88 80 ºC slightly soluble
CH3CO2C2H5 ethyl ethanoate 88 77 ºC moderately soluble
CH3CH2COCl propanoyl chloride 92.5 80 ºC reacts with water
CH3(CH2)3CHO pentanal 86 103 ºC slightly soluble
CH3(CH2)2COCH3 2-pentanone 86 102 ºC slightly soluble
The last nine entries in the above table cannot function as hydrogen bond donors, so hydrogen bonded dimers and aggregates are not possible. The relatively high boiling points of equivalent 3º-amides and nitriles are probably due to the high polarity of these functions. Indeed, if hydrogen bonding is not present, the boiling points of comparable sized compounds correlate reasonably well with their dipole moments.
Nomenclature
Three examples of acyl groups having specific names were noted earlier. These are often used in common names of compounds. In the following examples the IUPAC names are color coded, and common names are given in parentheses.
• Esters: The alkyl group is named first, followed by a derived name for the acyl group, the oic or ic suffix in the acid name is replaced by ate.
e.g. CH3(CH2)2CO2C2H5 is ethyl butanoate (or ethyl butyrate).
Cyclic esters are called lactones. A Greek letter identifies the location of the alkyl oxygen relative to the carboxyl carbonyl group.
• Acid Halides: The acyl group is named first, followed by the halogen name as a separate word.
e.g. CH3CH2COCl is propanoyl chloride (or propionyl chloride).
• Anhydrides: The name of the related acid(s) is used first, followed by the separate word "anhydride".
e.g. (CH3(CH2)2CO)2O is butanoic anhydride & CH3COOCOCH2CH3 is ethanoic propanoic anhydride (or acetic propionic anhydride).
• Amides: The name of the related acid is used first and the oic acid or ic acid suffix is replaced by amide (only for 1º-amides).
e.g. CH3CONH2 is ethanamide (or acetamide).
2º & 3º-amides have alkyl substituents on the nitrogen atom. These are designated by "N-alkyl" term(s) at the beginning of the name.
e.g. CH3(CH2)2CONHC2H5 is N-ethylbutanamide; & HCON(CH3)2 is N,N-dimethylmethanamide (or N,N-dimethylformamide).
Cyclic amides are called lactams. A Greek letter identifies the location of the nitrogen on the alkyl chain relative to the carboxyl carbonyl group.
• Nitriles: Simple acyclic nitriles are named by adding nitrile as a suffix to the name of the corresponding alkane (same number of carbon atoms).
Chain numbering begins with the nitrile carbon . Commonly, the oic acid or ic acid ending of the corresponding carboxylic acid is replaced by onitrile.
A nitrile substituent, e.g. on a ring, is named carbonitrile.
e.g. (CH3)2CHCH2C≡N is 3-methylbutanenitrile (or isovaleronitrile).
Acyl Group Substitution
This is probably the single most important reaction of carboxylic acid derivatives. The overall transformation is defined by the following equation, and may be classified either as nucleophilic substitution at an acyl group or as acylation of a nucleophile. For certain nucleophilic reagents the reaction may assume other names as well. If Nuc-H is water the reaction is often called hydrolysis, if Nuc–H is an alcohol the reaction is called alcoholysis, and for ammonia and amines it is called aminolysis.
Different carboxylic acid derivatives have very different reactivities, acyl chlorides and bromides being the most reactive and amides the least reactive, as noted in the following qualitatively ordered list. The change in reactivity is dramatic. In homogeneous solvent systems, reaction of acyl chlorides with water occurs rapidly, and does not require heating or catalysts. Amides, on the other hand, react with water only in the presence of strong acid or base catalysts and external heating.
Reactivity: acyl halides > anhydrides >> esters ≈ acids >> amides
Because of these differences, the conversion of one type of acid derivative into another is generally restricted to those outlined in the following diagram. Methods for converting carboxylic acids into these derivatives were shown in a previous section, but the amide and anhydride preparations were not general and required strong heating. A better and more general anhydride synthesis can be achieved from acyl chlorides, and amides are easily made from any of the more reactive derivatives. Specific examples of these conversions will be displayed by clicking on the product formula. The carboxylic acids themselves are not an essential part of this diagram, although all the derivatives shown can be hydrolyzed to the carboxylic acid state (light blue formulas and reaction arrows). Base catalyzed hydrolysis produces carboxylate salts.
Before proceeding further, it is important to review the general mechanism by means of which all these acyl transfer or acylation reactions take place. Indeed, an alert reader may well be puzzled by the facility of these nucleophilic substitution reactions. After all, it was previously noted that halogens bonded to sp2 or sp hybridized carbon atoms do not usually undergo substitution reactions with nucleophilic reagents. Furthermore, such substitution reactions of alcohols and ethers are rare, except in the presence of strong mineral acids. Clearly, the mechanism by which acylation reactions occur must be different from the SN1 and SN2 procedures described earlier.
In any substitution reaction two things must happen. The bond from the substrate to the leaving group must be broken, and a bond to the replacement group must be formed. The timing of these events may vary with the reacting system. In nucleophilic substitution reactions of alkyl compounds examples of bond-breaking preceding bond-making (the SN1 mechanism), and of bond-breaking and bond-making occurring simultaneously (the SN2 mechanism) were observed. On the other hand, for most cases of electrophilic aromatic substitution bond-making preceded bond-breaking.
As illustrated in the following diagram, acylation reactions generally take place by an addition-elimination process in which a nucleophilic reactant bonds to the electrophilic carbonyl carbon atom to create a tetrahedral intermediate. This tetrahedral intermediate then undergoes an elimination to yield the products. In this two-stage mechanism bond formation occurs before bond cleavage, and the carbonyl carbon atom undergoes a hybridization change from sp2 to sp3 and back again. The facility with which nucleophilic reagents add to a carbonyl group was noted earlier for aldehydes and ketones.
Acid and base-catalyzed variations of this mechanism will be displayed in turn as the "Mechanism Toggle" button is clicked. Also, a specific example of acyl chloride formation from the reaction of a carboxylic acid with thionyl chloride will be shown. The number of individual steps in these mechanisms vary, but the essential characteristic of the overall transformation is that of addition followed by elimination. Acid catalysts act to increase the electrophilicity of the acyl reactant; whereas, base catalysts act on the nucleophilic reactant to increase its reactivity. In principle all steps are reversible, but in practice many reactions of this kind are irreversible unless changes in the reactants and conditions are made. The acid-catalyzed formation of esters from carboxylic acids and alcohols, described earlier, is a good example of a reversible acylation reaction, the products being determined by the addition or removal of water from the system. The reaction of an acyl chloride with an alcohol also gives an ester, but this conversion cannot be reversed by adding HCl to the reaction mixture.
Mechanisms of Ester Cleavage
Esters are one of the most common carboxylic derivatives. Cleavage of the alkyl moiety in an ester may be effected in several different ways, the most common being the acyl transfer mechanism described above; however, other mechanisms have been observed. For examples and further discussion Click Here.
Thus far we have not explained the marked variation, noted above, in the reactivity of different carboxylic acid derivatives. The distinguishing carbonyl substituents in these compounds are: chloro (acyl chlorides), acyloxy (anhydrides), alkoxy (esters) and amino (amides). All of these substituents have bonds originating from atoms of relatively high electronegativity (Cl, O & N). They are therefore inductively electron withdrawing when bonded to carbon, as shown in the diagram on the right. The consequences of such inductive electron withdrawal on the acidity of carboxylic acids was previously noted.
Then these substituents are attached to an sp2 carbon that is part of a π-electron system, a similar inductive effect occurs, but n-π conjugation (p-π conjugation) moves electron density in the opposite direction. By clicking the "Toggle Effect" button the electron shift in both effects will be displayed sequentially. This competition between inductive electron withdrawal and conjugative electron donation was discussed earlier in the context of substituent effects on electrophilic aromatic substitution. Here, it was noted that amino groups were strongly electron donating (resonance effect >> inductive effect), alkoxy groups were slightly less activating, acyloxy groups still less activating (resonance effect > inductive effect) and chlorine was deactivating (inductive effect > resonance effect). In the illustration on the right, R and Z represent the remainder of a benzene ring.
This analysis also predicts the influence these substituent groups have on the reactivity of carboxylic acid derivatives toward nucleophiles (Z = O in the illustration). Inductive electron withdrawal by Y increases the electrophilic character of the carbonyl carbon, and increases its reactivity toward nucleophiles. Thus, acyl chlorides (Y = Cl) are the most reactive of the derivatives. Resonance electron donation by Y decreases the electrophilic character of the carbonyl carbon. The strongest resonance effect occurs in amides, which exhibit substantial carbon-nitrogen double bond character and are the least reactive of the derivatives. An interesting exception to the low reactivity of amides is found in beta-lactams such as penicillin G. The angle strain introduced by the four-membered ring reduces the importance of resonance, the non-bonding electron pair remaining localized on the pyramidally shaped nitrogen. Finally, anhydrides and esters have intermediate reactivities, with anhydrides being more reactive than esters.
Carbonyl Reactivity and IR Stretching Frequency
An interesting correlation between the reactivity of carboxylic acid derivatives and their carbonyl stretching frequencies exists. For a discussion of this topic Click Here
From the previous discussions you should be able to predict the favored product from each of the following reactions.
The acyl derivative is the reactant on the left, and the nucleophilic reactant is to its right. Click the "Show Products" button to display the answers.
The first three examples concern reactions of acyl chlorides, the most reactive acylating reagents discussed here. Although amines are among the most reactive nucleophiles, only 1º and 2º-amines give stable amide products. Reaction of 3º-amines with strong acylating reagents may generate acylammonium species reversibly (see below), but these are as reactive as acyl chlorides and will have only a very short existence. This explains why reactions #2 & 3 do not give amide products.
RCOCl + R'3N RCONR'3(+) Cl(–) (an acylammonium salt)
Reactions #4 & 5 display the acylating capability of anhydrides. Bear in mind that anhydrides may also be used as reagents in Friedel-Crafts acylation reactions. Esters are less reactive acylating reagents than anhydrides, and the ester exchange reaction (#6) requires a strong acid or base catalyst. The last example demonstrates that nitrogen is generally more nucleophilic than oxygen. Indeed, it is often possible to carry out reactions of amines with acyl chlorides and anhydrides in aqueous sodium hydroxide solution! Not only is the amine more nucleophilic than water, but the acylating reagent is generally not soluble in or miscible with water, reducing the rate of its hydrolysis.
No acylation reactions of amides were shown in these problems. The most important such reaction is hydrolysis, and this normally requires heat and strong acid or base catalysts. One example, illustrating both types of catalysis, is shown here. Mechanisms for catalyzed reactions of this kind were presented earlier.
R–CO2(–) + CH3NH2 OH(–) & heat
R–CO–NH(CH3) + H2O
H(+) & heat
R–CO2H + CH3NH3(+)
Other Acylation Reagents and Techniques
Because acylation is such an important and widely used transformation, the general reactions described above have been supplemented by many novel procedures and reagents that accomplish similar overall change. These are normally beyond the scope of an introductory text, but a short description of some of these methods is provided for the interested reader by
Clicking Here.
Nitriles
Although they do not have a carbonyl group, nitriles are often treated as derivatives of carboxylic acids. Hydrolysis of nitriles to carboxylic acids was described earlier, and requires reaction conditions (catalysts and heat) similar to those needed to hydrolyze amides. This is not surprising, since addition of water to the carbon-nitrogen triple bond gives an imino intermediate which tautomerizes to an amide.
R–C≡N + H2O
acid or base
R–C(OH)=NH
R–CO–NH2
Reduction
Reductions of carboxylic acid derivatives might be expected to lead either to aldehydes or alcohols, functional groups having a lower oxidation state of the carboxyl carbon. Indeed, it was noted earlier that carboxylic acids themselves are reduced to alcohols by lithium aluminum hydride. At this point it will be useful to consider three kinds of reductions:
1. catalytic hydrogenation
2. complex metal hydride reductions
3. diborane reduction.
Catalytic Hydrogenation
As a rule, the carbonyl group does not add hydrogen as readily as do the carbon-carbon double and triple bonds. Thus, it is fairly easy to reduce an alkene or alkyne function without affecting any carbonyl functions in the same molecule. By using a platinum catalyst and increased temperature and pressure, it is possible to reduce aldehydes and ketones to alcohols, but carboxylic acids, esters and amides are comparatively unreactive. The exceptional reactivity of acyl halides, on the other hand, facilitates their reduction under mild conditions, by using a poisoned palladium catalyst similar to that used for the partial reduction of alkynes to alkenes. This reduction stops at the aldehyde stage, providing us with a useful two-step procedure for converting carboxylic acids to aldehydes, as reaction #1 below demonstrates. Equivalent reductions of anhydrides have not been reported, but we might speculate that they would be reduced more easily than esters. The only other reduction of a carboxylic acid derivative that is widely used is that of nitriles to 1º-amines. Examples of these reductions are provided in the following diagram.
The second and third equations illustrate the extreme difference in hydrogenation reactivity between esters and nitriles. This is further demonstrated by the last reaction, in which a nitrile is preferentially reduced in the presence of a carbonyl group and two benzene rings. The resulting 1º-amine immediately reacts with the carbonyl function to give a cyclic enamine product (colored light blue).
In most nitrile reductions ammonia is added to inhibit the formation of a 2º-amine by-product. This may occur by way of an intermediate aldehyde imine created by addition of the first equivalent of hydrogen. The following equations show how such an imine species might react with the 1º-amine product to give a substituted imine (2nd equation), which would then add hydrogen to generate a 2º-amine. Excess ammonia shifts the imine equilibrium to the left, as written below.
(1)
R–C≡N + H2
catalyst
RCH=NH
imine
H2
RCH2NH2
1º-amine
(2)
RCH=NH + RCH2NH2
imine 1º-amine
RCH=NCH2R + NH3
substituted imine
H2 & catalyst
RCH2NHCH2R
2º-amine
Complex Metal Hydride Reductions
The use of lithium aluminum hydride (LiAlH4) and sodium borohydride (NaBH4) as reagents for the reduction of aldehydes and ketones to 1º and 2º-alcohols respectively has been noted. Of these, lithium aluminum hydride, often abbreviated LAH, is the most useful for reducing carboxylic acid derivatives. Thanks to its high reactivity, LAH easily reduces all classes of carboxylic acid derivatives, generally to the –1 oxidation state. Acids, esters, anhydrides and acyl chlorides are all reduced to 1º-alcohols, and this method is superior to catalytic reduction in most cases. Since acyl chlorides and anhydrides are expensive and time consuming to prepare, acids and esters are the most commonly used reactants for this transformation.
Amides are reduced to amines by treatment with LAH, and this has proven to be one of the most general methods for preparing all classes of amines (1º, 2º & 3º). Because the outcome of LAH reduction is so different for esters and amides, we must examine plausible reaction mechanisms for these reactions to discover a reason for this divergent behavior. As in the reductions of aldehydes and ketones, the first step in each case is believed to be the irreversible addition of hydride to the electrophilic carbonyl carbon atom. This is shown in the following diagrams, with the hydride-donating moiety being written as AlH4(–). All four hydrogens are potentially available to the reduction, but when carboxylic acids are reduced, one of the hydrides reacts with the acidic O–H to generate hydrogen gas. Although the lithium is not shown, it will be present in the products as a cationic component of ionic salts.
One explanation of the different course taken by the reductions of esters and amides lies in the nature of the different hetero atom substituents on the carbonyl group (colored green in the diagram). Nitrogen is more basic than oxygen, and amide anions are poorer leaving groups than alkoxide anions. Furthermore, oxygen forms especially strong bonds to aluminum. Addition of hydride produces a tetrahedral intermediate, shown in brackets, which has a polar oxygen-aluminum bond. Neither the hydrogen nor the alkyl group (R) is a possible leaving group, so if this tetrahedral species is to undergo an elimination to reform a carbonyl group, one of the two remaining substituents must be lost. For the ester this is an easy choice (described by the curved arrows). By eliminating an aluminum alkoxide (R'O–Al), an aldehyde is formed, and this is quickly reduced to the salt of a 1º-alcohol by LAH. In the case of the amide, aldehyde formation requires the loss of an aluminum amide (R'2N–Al), an unlikely process. Alternatively, the more basic nitrogen may act to eject a metal oxide species (e.g. Al–O(–)), and the resulting iminium double bond would be reduced rapidly to an amine. This is the course followed in most amide reductions. In the case of 1º-amides, however, the acidity of the nitrogen hydrogens coupled with the basicity of hydride enables a facile elimination of the oxygen (as an oxide moiety), forming a nitrile intermediate. Nitriles are in fact a major product when less than a full equivalency of LiAlH4 is used. A mechanism will be shown above by clicking on the diagram.
Lithium aluminum hydride reduces nitriles to 1º-amines, as shown in the following equation. An initial hydride addition to the electrophilic nitrile carbon atom generates the salt of an imine intermediate. This is followed by a second hydride transfer, and the resulting metal amine salt is hydrolyzed to a 1º-amine. This method provides a useful alternative to the catalytic reduction of nitriles, described above, when alkene or alkyne functions are present.
In contrast to the usefulness of lithium aluminum hydride in reducing various carboxylic acid derivatives, sodium borohydride is seldom chosen for this purpose. First, NaBH4 is often used in hydroxylic solvents (water and alcohols), and these would react with acyl chlorides and anhydrides. Furthermore, it is sparingly soluble in relatively nonpolar solvents, particularly at low temperatures. Second, NaBH4 is much less reactive than LAH, failing to reduce amides and acids (they form carboxylate salts) at all, and reducing esters very slowly.
Since relatively few methods exist for the reduction of carboxylic acid derivatives to aldehydes, it would be useful to modify the reactivity and solubility of LAH to permit partial reductions of this kind to be achieved. The most fruitful approach to this end has been to attach alkoxy or alkyl groups on the aluminum. This not only modifies the reactivity of the reagent as a hydride donor, but also increases its solubility in nonpolar solvents. Two such reagents will be mentioned here; the reactive hydride atom is colored blue.
Lithium tri-tert-butoxyaluminohydride (LtBAH), LiAl[OC(CH3)3]3H : Soluble in THF, diglyme & ether.
Diisobutylaluminum hydride (DIBAH), [(CH3)2CHCH2]2AlH : Soluble in toluene, THF & ether.
Each of these reagents carries one equivalent of hydride. The first (LtBAH) is a complex metal hydride, but the second is simply an alkyl derivative of aluminum hydride. In practice, both reagents are used in equimolar amounts, and usually at temperatures well below 0 ºC. The following examples illustrate how aldehydes may be prepared from carboxylic acid derivatives by careful application of these reagents. A temperature of -78 ºC is easily maintained by using dry-ice as a coolant. The reduced intermediates that lead to aldehydes will be displayed on clicking the "Show Intermediates" button. With excess reagent at temperatures above 0 ºC most carboxylic acid derivatives are reduced to alcohols or amines.
Diborane, B2H6
The reducing characteristics of diborane (disassociated to BH3 in ether or THF solution) were first introduced as addition reactions to alkenes and alkynes. This remains a primary application of this reagent, but it also effects rapid and complete reduction of carboxylic acids, amides and nitriles. Other than LAH, this reagent provides one of the best methods for reducing carboxylic acids to 1º-alcohols.
1. (1)
R–CO2H + BH3
ether soln.
[RCH2O–B]
H2O2
RCH2–OH
(2)
R–C≡N + BH3
ether soln.
RCH2–NH–B
H2O
RCH2–NH
Overview of Reducing Agents
The following table summarizes the influence each of the reducing systems discussed above has on the different classes of carboxylic acid derivatives. Note that LAH is the strongest reducing agent listed, and it reduces all the substrates. In a similar sense, acyl chlorides are the most reactive substrate. They are reduced by all the reagents, but only a few of these provide synthetically useful transformations.
Function
Reagent
Aldehydes
& Ketones
Carboxylic
Acids
Carboxylic
Esters
Acyl
Chlorides
Amides
Nitriles
H2
& catalyst
alcohols
( slow, Pt, Pd )
(v. slow) (v. slow) aldehydes
( Pd/BaSO4 )
(v. slow) amines
( Ni cat. )
NaBH4
polar solvent
alcohols N.R. alcohols
(slow)
complex
mixture
N.R. N.R.
LiAlH4
ether or THF
alcohols 1º-alcohol alcohols 1º-alcohol amines 1º-amine
LiAlH(Ot-Bu)3
1 eq. in THF
alcohols
(slow at 0º)
N.R. v. slow aldehyde
(-78 º C)
aldehyde
(-78 º C)
aldehyde
(0 º C)
(iso-Bu)2AlH
1 eq. in toluene
alcohols 1º-alcohol aldehyde
(-78º C)
1º-alcohol aldehyde
(-78 º C)
aldehyde
(-78 º C)
B2H6
THF
alcohols
(slow)
1º-alcohol (v. slow) complex
mixture
1º-amine 1º-amine
Color Code
Reduction occurs readily under normal conditions of temperature and pressure.
Reduction occurs readily, but selectivity requires low temperature.
Slow reduction occurs. Heating and/or high pressures of hydrogen are needed for effective use.
Reduction occurs very slowly or not at all (N.R.).
3. Reactions with Organometallic Reagents
The facile addition of alkyl lithium reagents and Grignard reagents to aldehydes and ketones has been described. These reagents, which are prepared from alkyl and aryl halides, are powerful nucleophiles and very strong bases. Reaction of an excess of these reagents with acyl chlorides, anhydrides and esters leads to alcohol products, in the same fashion as the hydride reductions. As illustrated by the following equations (shaded box), this occurs by sequential addition-elimination-addition reactions, and finishes with hydrolysis of the resulting alkoxide salt. A common bonding pattern is found in all these carbonyl reactions. The organometallic reagent is a source of a nucleophilic alkyl or aryl group (colored purple), which bonds to the electrophilic carbon of the carbonyl group (colored orange). Substituent Y (colored green) is eliminated from the tetrahedral intermediate as its anion. The aldehyde or ketone product of this elimination then adds a second equivalent of the reagent.
Reactions of this kind are important synthetic transformations, because they permit simple starting compounds to be joined to form more complex structures. Esters are the most common carbonyl reactants, since they are cheaper and less hazardous to use than acyl chlorides and anhydrides. Most esters react with organometallic reagents to give 3º-alcohols; but formate esters (R=H) give 2º-alcohols. Some examples of these reactions are provided in the following diagram. As demonstrated by the last equation, lactones undergo ring opening and yield diol products.
The acidity of carboxylic acids and 1º & 2º-amides acts to convert Grignard and alkyl lithium reagents to hydrocarbons (see equations), so these functional groups should be avoided when these reagents are used.
R–CO2H + R'–MgBr
ether soln.
R–CO2(–) MgBr(+) + R'H
R–CONH2 + R'–Li
ether soln.
R–CONH(–) Li(+) + R'H
Since acyl chlorides are more reactive than esters, isolation of the ketone intermediate formed in their reactions with organometallic reagents becomes an attractive possibility. To achieve this selectivity we need to convert the highly reactive Grignard and lithium reagents to less nucleophilic species. Two such modifications that have proven effective are the Gilman reagent (R2CuLi) and organocadmium reagents (prepared in the manner shown).
2 R–MgBr + CdCl2
ether & benzene
R2Cd + MgBr2 + MgCl2
Specific examples of ketone synthesis using these reagents are presented in the following diagram. The second equation demonstrates the low reactivity of organocadmium reagents, inasmuch as the ester function is unchanged. Another related approach to this transformation is illustrated by the third equation. Grignard reagents add to nitriles, forming a relatively stable imino derivative which can be hydrolyzed to a ketone. Imines themselves do not react with Grignard reagents.
4. Other Reactions
Amides are very polar, thanks to the n-π conjugation of the nitrogen non-bonded electron pair with the carbonyl group. This delocalization substantially reduces the basicity of these compounds (pKa ca. –1) compared with amines (pKa ca. 11). When electrophiles bond to an amide, they do so at the oxygen atom in preference to the nitrogen. As shown below, the oxygen-bonded conjugate acid is stabilized by resonance charge delocalization; whereas, the nitrogen-bonded analog is not. One practical application of this behavior lies in the dehydration of 1º-amides to nitriles by treatment with thionyl chloride. This reaction is also illustrated in the following diagram. Other dehydrating agents such as P2O5 effect the same transformation.
\
Pyrolytic syn-Eliminations
Ester derivatives of alcohols may undergo unimolecular syn-elimination on heating.
To see examples of these Click Here
Practice Problems
The following problems review aspects of the chemistry of carboxylic acids and their derivatives. The first two questions concern their nomenclature. The third reviews three common reactions, applied to eight carbonyl compounds, including aldehydes and ketones. The fourth question asks you to draw the structural formulas for the products of more than fifty possible reactions of some carboxylic acids. The fifth problem concerns hydrolysis with aqueous acid or base, and requires drawing product structures for both conditions.
For a summary of the fundamental reactions of carboxylic acid derivatives Click Here
Reactions at the α-Carbon
Many aldehydes and ketones were found to undergo electrophilic substitution at an alpha carbon. These reactions, which included halogenation, isotope exchange and the aldol reaction, take place by way of enol tautomer or enolate anion intermediates, a characteristic that requires at least one hydrogen on the α-carbon atom. In this section similar reactions of carboxylic acid derivatives will be examined. Formulas for the corresponding enol and enolate anion species that may be generated from these derivatives are drawn in the following diagram.
Acid-catalyzed alpha-chlorination and bromination reactions proceed more slowly with carboxylic acids, esters and nitriles than with ketones. This may reflect the smaller equilibrium enol concentrations found in these carboxylic acid derivatives. Nevertheless, acid and base catalyzed isotope exchange occurs as expected; some examples are shown in equations #1 and #2 below. The chiral alpha-carbon in equation #2 is racemized in the course of this exchange, and a small amount of nitrile is hydrolyzed to the corresponding carboxylic acid.
Acyl halides and anhydrides are more easily halogenated than esters and nitriles, probably because of their higher enol concentration. This difference may be used to facilitate the alpha-halogenation of carboxylic acids. Thus, conversion of the acid to its acyl chloride derivative is followed by alpha-bromination or chlorination, and the resulting halogenated acyl chloride is then hydrolyzed to the carboxylic acid product. This three-step sequence can be reduced to a single step by using a catalytic amount of phosphorus tribromide or phosphorus trichloride, as shown in equation #3. This simple modification works well because carboxylic acids and acyl chlorides exchange functionality as the reaction progresses. The final product is the alpha-halogenated acid, accompanied by a trace of the acyl halide. This halogenation procedure is called the Hell-Volhardt-Zelinski reaction.
To see a mechanism for the acyl halide-carboxylic acid exchange click the "Show Mechanism" button.
In a similar fashion, acetic anhydride serves as a halogenation catalyst for acetic acid (first equation below). Carboxylic acids that have a higher equilibrium enol concentration do not need to be activated for alpha-halogenation to occur, as demonstrated by the substituted malonic acid compound in the second equation below. The enol concentration of malonic acid (about 0.01%) is roughly ten thousand times greater than that of acetic acid. This influence of a second activating carbonyl function on equilibrium enol concentrations had been noted earlier in the case of 2,4-pentanedione.
1. (i)
CH3-CO2H + Br2 & (CH3CO)2O catalyst
heat
BrCH2CO2H + HBr
(ii)
RCH(CO2H)2 + Br2
RCBr(CO2H)2 + HBr
1. Enolate Intermediates
Many of the most useful alpha-substitution reactions of ketones proceeded by way of enolate anion conjugate bases. Since simple ketones are weaker acids than water, their enolate anions are necessarily prepared by reaction with exceptionally strong bases in non-hydroxylic solvents. Esters and nitriles are even weaker alpha-carbon acids than ketones (by over ten thousand times), nevertheless their enolate anions may be prepared and used in a similar fashion. The presence of additional activating carbonyl functions increases the acidity of the alpha-hydrogens substantially, so that less stringent conditions may be used for enolate anion formation. The influence of various carbonyl and related functional groups on the equilibrium acidity of alpha-hydrogen atoms (colored red) is summarized in the following table. For common reference, these acidity values have all been extrapolated to water solution, even though the conjugate bases of those compounds having pKas greater than 18 will not have a significant concentration in water solution.
Acidity of α-Hydrogens in Mono- and Di-Activated Compounds
Mono-Activation Compound RCH2–NO2 RCH2–COR RCH2–CO2CH3 RCH2–C≡N RCH2–SO2R RCH2–CON(CH3)2
pKa 9 20 25 25 25 28
Di-Activation Compound CH2(NO2)2 (CH3CO)2CH2 CH3COCH2CO2C2H5 CH2(C≡N)2 CH2(CO2C2H5)2 CH2(SO2CH3)2
pKa 4 9 11 11 13 13
To illustrate the general nucleophilic reactivity of di-activated enolate anions, two examples of SN2 alkylation reactions are shown below. Malonic acid esters and acetoacetic acid esters are commonly used starting materials, and their usefulness in synthesis will be demonstrated later in this chapter. Note that each of these compounds has two acidic alpha-hydrogen atoms (colored red). In the equations written here only one of these hydrogens is substituted; however, the second is also acidic and a second alkyl substitution may be carried out in a similar fashion.
2. Claisen Condensation
The aldol reaction, is a remarkable and useful reaction of aldehydes and ketones in which the carbonyl group serves both as an electrophilic reactant and the source of a nucleophilic enol species. Esters undergo a similar transformation called the Claisen Condensation. Four examples of this base-induced reaction, which usually forms beta-ketoester products, are shown in the following diagram. Greek letter assignments for the ester products are given in blue. Equation #1 presents the synthesis of the important reagent ethyl acetoacetate, and #2 illustrates the general form of the Claisen condensation. Intramolecular reactions, such as #3, lead to rings (usually five or six-membered) and are referred to as Dieckmann Condensations. The last equation shows a mixed condensation between two esters, one of which has no alpha-hydrogens. The product in this case is a phenyl substituted malonic ester rather than a ketoester.
By clicking the "Structural Analysis" button below the diagram, a display showing the nucleophilic enolic donor molecule and the electrophilic acceptor molecule together with the newly formed carbon-carbon bond will be displayed. A stepwise mechanism for the reaction will be shown by clicking the "Reaction Mechanism" button. In a similar mode to the aldol reaction, the fundamental event in the Claisen condensation is a dimerization of two esters by an alpha C–H addition of one reactant to the carbonyl group of a second reactant. This bonding is followed by alcohol elimination from the resulting hemiacetal. The eventual formation of a resonance stabilized beta-ketoester enolate anion, as shown on the third row of the mechanism, provides a thermodynamic driving force for the condensation. Note that this stabilization is only possible if the donor has two reactive alpha-hydrogens.
The Claisen condensation differs from the aldol reaction in several important ways.
1. The aldol reaction may be catalyzed by acid or base, but most Claisen condensations require base.
2. In contrast to the catalytic base used for aldol reactions, a full equivalent of base (or more) must be used for the Claisen condensation. The extra base is needed because beta-ketoesters having acidic hydrogens at the alpha-carbon are stronger acids (by about 5 powers of ten) than the alcohol co-product. Consequently, the alkoxide base released after carbon-carbon bond formation (upper right structure in the mechanism diagram) immediately removes an alpha proton from the beta-ketoester product. As noted above, formation of this doubly-stabilized enolate anion provides a thermodynamic driving force for the condensation.
3. The aldol reaction may be catalyzed by hydroxide ion, but the Claisen condensation requires that alkoxide bases be used, in order to avoid ester hydrolysis. The specific alkoxide base used should match the alcohol component of the ester to avoid ester exchange reactions. Very strong bases such as LDA may also be used in this reaction.
4. The stabilized enolate product must be neutralized by aqueous acid in order to obtain the beta-ketoester product.
Transformations similar to the Claisen condensation may be effected with mixed carbonyl reactants, which may include ketones and nitriles as well as esters. Esters usually serve as the electrophilic acceptor component of the condensation. Acyl chlorides and anhydrides would also be good electrophilic acceptors, but they are more expensive than esters and do not tolerate the alcohol solvents often used for Claisen condensations.
In the case of mixed condensations, complex product mixtures are commonly avoided by using an acceptor ester that has no alpha-hydrogens. Examples of such reactants are: ethyl formate (HCO2C2H5), diethyl carbonate (C2H5OCO2C2H5), ethyl benzoate (C6H5CO2C2H5) and diethyl oxalate (C2H5O2C-CO2C2H5). Equations #2, 3 & 4 below illustrate the use of such acceptors with ester, ketone and nitrile donor compounds. The nucleophilic enol species from the nitrile in #4 may be written as: C6H5CH=C=N(–). The 2-formylcyclohexanone product from reaction #3 exists predominantly as its hydrogen-bonded enol. Most beta-ketoesters have significant enol concentrations, but the formyl group has an exceptional bias for this tautomer.
Equation #1 shows a condensation in which both reactants might serve either as donors or acceptors. The selective formation of one of the four possible condensation products is due to the reversibility of these reactions and the driving force provided by resonance stabilization of the enolate anion of 2,4-pentanedione (pKa=9). Protonation of this anion gives the product. The last equation (#5) presents an interesting example of selectivity. There are three ester functions, each of which has at least one alpha-hydrogen. Only one of these, that on the left, has two alpha-hydrogens and will yield an enolizable beta-ketoester by functioning as the donor in a Dieckmann cyclization. Strained four-membered rings are not favored by reversible condensation reactions, so ring closure to the ester drawn below the horizontal chain does not occur. The only reasonable product is the five-membered cyclic ketoester.
Although many Claisen condensations are carried out with a full equivalent of the alkoxide base, an effective alternative procedure, used in reaction #5, uses sodium hydride (NaH) together with a catalytic amount of alcohol. The catalytic alcohol reacts with NaH to produce alkoxide, this initiates a condensation reaction and the product alcohol then reacts with more NaH to give alkoxide.
Applications of Condensation Reactions to Synthesis
The construction of complex molecules by a series of suitable reactions carried out from simple starting compounds is called synthesis. Synthesis is not only of immense practical importance (aspirin and nylon are two examples of commercially valuable synthetic compounds), but it also allows us to prepare novel molecules with which to test our understanding of structure and reactivity. Three challenges must be met in devising a synthesis for a specific compound:
1. The carbon atom framework or skeleton that is found in the desired compound (the target) must be assembled.
2. The functional groups that characterize the target compound must be introduced or transformed from other groups at appropriate locations.
3. If centers of stereoisomerism are present, they must be fixed in a proper manner.
Recognition of these tasks does not imply that they are independent of each other, or should be approached and solved separately. A successful plan or strategy for a synthesis must correlate each step with all these goals, so that an efficient and practical solution to making the target molecule is achieved. Nevertheless, it is useful to classify the various reactions we have studied with respect to their ability to (i) enlarge or expand a given structure, (ii) transform or relocate existing functional groups, and (iii) do both of these in a stereoselective fashion. The organization of this text by functional group behavior partially satisfies the second point, and the following discussion focuses on the first.
1. Carbon-Carbon Bond Formation
A useful assortment of carbon-carbon bond forming reactions have been described in this and earlier chapters. These include:
(1) Friedel-Crafts alkylation and acylation.
(2) Diels-Alder cycloaddition.
(3) addition of organometallic reagents to aldehydes, ketones & carboxylic acid derivatives.
(4) alkylation of acetylide anions.
(5) alkylation of enolate anions.
(6) Claisen and aldol condensations.
With the exception of Friedel-Crafts alkylation these reactions all give products having one or more functional groups at or adjacent to the bonding sites. As a result, subsequent functional group introduction or modification may be carried out in a relatively straightforward manner. This will be illustrated for aldol and Claisen condensations in the following section.
2. Modification of Condensation Products
A. Reactions of Aldol Products
The aldol reaction produces beta-hydroxyaldehydes or ketones, and a number of subsequent reactions may be carried out with these products. As shown in the following diagram, they may be (i) reduced to 1,3-diols, (ii) a 2º-hydroxyl group may be oxidized to a carbonyl group, (iii) acid or base catalyzed beta-dehydration may produce an unsaturated aldehyde or ketone, and (iv) organometallic reagents may be added to the carbonyl group (assuming the hydroxyl group is protected as an ether or a second equivalent of reagent is used).
B. Reactions of Claisen Products
The Claisen condensation produces beta-ketoesters. These products may then be modified or enhanced by further reactions. Among these, the following diagram illustrates (i) partial reduction of the ketone with NaBH4, (ii) complete reduction to a 1,3-diol by LiAlH4, (iii) enolate anion alkylation, and (iv) ester hydrolysis followed by thermal decarboxylation of the resulting beta-ketoacid.
C. Synthesis Examples
To illustrate how the reaction sequences described above may be used to prepare a variety of different compounds, five examples are provided here. The first is a typical aldol reaction followed by reduction to a 1,3-diol (2-ethyl-1,3-hexanediol). In the second example, the absence of alpha-hydrogens on the aldehyde favors the mixed condensation, and conjugation of the double bond facilitates dehydration. The doubly-activated methylene group of malonic and acetoacetic acids or esters makes them good donors in any condensation, as is demonstrated by the third aldol-like reaction. A concerted dehydrative-decarboxylation (shown by the magenta arrows) leads to the unsaturated carboxylic acid product. Amine bases are often used as catalysts for aldol reactions, as in equations #2 & 3. The fourth reaction demonstrates that the conjugate base of the beta-ketoester products from Claisen or Dieckmann condensation may be alkylated directly. Thermal decarboxylation of the resulting beta-ketoacid gives a mono-alkylated cyclic ketone. Finally, both acidic methylene hydrogens in malonic ester or ethyl acetoacetate may be substituted, and the irreversible nature of such alkylations permits strained rings to be formed. In this case thermal decarboxylation of a substituted malonic acid generates a carboxylic acid. In all these examples the remaining functional groups could be used for additional synthetic operations.
Vinylagous Reactions
A large family of vinylagous reactions, related to the condensations, acylations and alkylations described here, increase the bond forming options available to the synthetic chemist. To learn more about these versatile reactions Click Here
Some Exercises
If you understand the previous discussion of reactions useful in synthesis you should try the following problems. Some of them are complex so don't be concerned if you don't solve them all immediately. Analyze each problem carefully, and try to learn from it. The solutions will be displayed by clicking the answer button under the diagram.
The following problems ask you to devise a synthesis for a given target molecule. The first two problems make use of the common starting materials, diethyl malonate and ethyl acetoacetate. The third problem leaves the choice of materials open. The nature of the target molecule suggests that an aldol condensation might be useful. The fourth problem must be solved by using diethyl succinate as the only reagent. Finally, other reactants composed of no more than five carbon atoms may be used in the last problem.
Practice Problemsss
The following problems review many aspects of the chemistry of carboxylic acids and their derivatives. The first question explores the relative acidity of various functional derivatives. The second tests your understanding of the Claisen condensation. The third is an introduction to multistep syntheses. The fourth and fifth questions ask you to draw the product structures for a number of multistep syntheses, involving other classes of compounds as well as carboxylic acids. The last two questions allow you to choose reagents for a multistep synthesis.
Carboxyl Derivatives
The important classes of organic compounds known as alcohols, phenols, ethers, amines and halides consist of alkyl and/or aryl groups bonded to hydroxyl, alkoxyl, amino and halo substituents respectively. If these same functional groups are attached to an acyl group (RCO–) their properties are substantially changed, and they are designated as carboxylic acid derivatives. Carboxylic acids have a hydroxyl group bonded to an acyl group, and their functional derivatives are prepared by replacement of the hydroxyl group with substituents, such as halo, alkoxyl, amino and acyloxy. Some examples of these functional derivatives were displayed earlier.
The following table lists some representative derivatives and their boiling points. An aldehyde and ketone of equivalent molecular weight are also listed for comparison. Boiling points are given for 760 torr (atmospheric pressure), and those listed as a range are estimated from values obtained at lower pressures. The relatively high boiling point of carboxylic acids is due to extensive hydrogen bonded dimerization. Similar hydrogen bonding occurs between molecules of 1º and 2º-amides (amides having at least one N–H bond), and the first three compounds in the table serve as hydrogen bonding examples.
Table 1: Physical Properties of Some Carboxylic Acid Derivatives
Formula
IUPAC Name
Molecular Weight
Boiling Point
Water Solubility
CH3(CH2)2CO2H butanoic acid 88 164 ºC very soluble
CH3(CH2)2CONH2 butanamide 87 216-220 ºC soluble
CH3CH2CONHCH3 N-methylpropanamide 87 205 -210 ºC soluble
CH3CON(CH3)2 N,N-dimethylethanamide 87 166 ºC very soluble
HCON(CH3)CH2CH3 N-ethyl, N-methylmethanamide 87 170-180 ºC very soluble
CH3(CH2)3CN pentanenitrile 83 141 ºC slightly soluble
CH3CO2CHO ethanoic methanoic
anhydride
88 105-112 ºC reacts with water
CH3CH2CO2CH3 methyl propanoate 88 80 ºC slightly soluble
CH3CO2C2H5 ethyl ethanoate 88 77 ºC moderately soluble
CH3CH2COCl propanoyl chloride 92.5 80 ºC reacts with water
CH3(CH2)3CHO pentanal 86 103 ºC slightly soluble
CH3(CH2)2COCH3 2-pentanone 86 102 ºC slightly soluble
The last nine entries in the above table cannot function as hydrogen bond donors, so hydrogen bonded dimers and aggregates are not possible. The relatively high boiling points of equivalent 3º-amides and nitriles are probably due to the high polarity of these functions. Indeed, if hydrogen bonding is not present, the boiling points of comparable sized compounds correlate reasonably well with their dipole moments.
Contributors
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry
Carboxylic Derivatives - Physical Properties
Nomenclature
Three examples of acyl groups having specific names were noted earlier. These are often used in common names of compounds. In the following examples the IUPAC names are color coded, and common names are given in parentheses.
• Esters: The alkyl group is named first, followed by a derived name for the acyl group, the oic or ic suffix in the acid name is replaced by ate.
e.g. CH3(CH2)2CO2C2H5 is ethyl butanoate (or ethyl butyrate).
Cyclic esters are called lactones. A Greek letter identifies the location of the alkyl oxygen relative to the carboxyl carbonyl group.
• Acid Halides: The acyl group is named first, followed by the halogen name as a separate word.
e.g. CH3CH2COCl is propanoyl chloride (or propionyl chloride).
• Anhydrides: The name of the related acid(s) is used first, followed by the separate word "anhydride".
e.g. (CH3(CH2)2CO)2O is butanoic anhydride & CH3COOCOCH2CH3 is ethanoic propanoic anhydride (or acetic propionic anhydride).
• Amides: The name of the related acid is used first and the oic acid or ic acid suffix is replaced by amide (only for 1º-amides).
e.g. CH3CONH2 is ethanamide (or acetamide).
2º & 3º-amides have alkyl substituents on the nitrogen atom. These are designated by "N-alkyl" term(s) at the beginning of the name.
e.g. CH3(CH2)2CONHC2H5 is N-ethylbutanamide; & HCON(CH3)2 is N,N-dimethylmethanamide (or N,N-dimethylformamide).
Cyclic amides are called lactams. A Greek letter identifies the location of the nitrogen on the alkyl chain relative to the carboxyl carbonyl group.
• Nitriles: Simple acyclic nitriles are named by adding nitrile as a suffix to the name of the corresponding alkane (same number of carbon atoms).
Chain numbering begins with the nitrile carbon . Commonly, the oic acid or ic acid ending of the corresponding carboxylic acid is replaced by onitrile.
A nitrile substituent, e.g. on a ring, is named carbonitrile.
e.g. (CH3)2CHCH2C≡N is 3-methylbutanenitrile (or isovaleronitrile). | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Carboxylic_Acids/Properties_of_Carboxylic_Acids/Carboxyl_Derivatives/Carboxylic_Derivatives_-_Physical_Properties/Carboxylic_Derivatives_-_Nomenclature.txt |
Acyl Group Substitution
This is probably the single most important reaction of carboxylic acid derivatives. The overall transformation is defined by the following equation, and may be classified either as nucleophilic substitution at an acyl group or as acylation of a nucleophile. For certain nucleophilic reagents the reaction may assume other names as well. If Nuc-H is water the reaction is often called hydrolysis, if Nuc–H is an alcohol the reaction is called alcoholysis, and for ammonia and amines it is called aminolysis.
Different carboxylic acid derivatives have very different reactivities, acyl chlorides and bromides being the most reactive and amides the least reactive, as noted in the following qualitatively ordered list. The change in reactivity is dramatic. In homogeneous solvent systems, reaction of acyl chlorides with water occurs rapidly, and does not require heating or catalysts. Amides, on the other hand, react with water only in the presence of strong acid or base catalysts and external heating.
Reactivity: acyl halides > anhydrides >> esters ≈ acids >> amides
Because of these differences, the conversion of one type of acid derivative into another is generally restricted to those outlined in the following diagram. Methods for converting carboxylic acids into these derivatives were shown in a previous section, but the amide and anhydride preparations were not general and required strong heating. A better and more general anhydride synthesis can be achieved from acyl chlorides, and amides are easily made from any of the more reactive derivatives. Specific examples of these conversions will be displayed by clicking on the product formula. The carboxylic acids themselves are not an essential part of this diagram, although all the derivatives shown can be hydrolyzed to the carboxylic acid state (light blue formulas and reaction arrows). Base catalyzed hydrolysis produces carboxylate salts.
Before proceeding further, it is important to review the general mechanism by means of which all these acyl transfer or acylation reactions take place. Indeed, an alert reader may well be puzzled by the facility of these nucleophilic substitution reactions. After all, it was previously noted that halogens bonded to sp2 or sp hybridized carbon atoms do not usually undergo substitution reactions with nucleophilic reagents. Furthermore, such substitution reactions of alcohols and ethers are rare, except in the presence of strong mineral acids. Clearly, the mechanism by which acylation reactions occur must be different from the SN1 and SN2 procedures described earlier.
In any substitution reaction two things must happen. The bond from the substrate to the leaving group must be broken, and a bond to the replacement group must be formed. The timing of these events may vary with the reacting system. In nucleophilic substitution reactions of alkyl compounds examples of bond-breaking preceding bond-making (the SN1 mechanism), and of bond-breaking and bond-making occurring simultaneously (the SN2 mechanism) were observed. On the other hand, for most cases of electrophilic aromatic substitution bond-making preceded bond-breaking.
As illustrated in the following diagram, acylation reactions generally take place by an addition-elimination process in which a nucleophilic reactant bonds to the electrophilic carbonyl carbon atom to create a tetrahedral intermediate. This tetrahedral intermediate then undergoes an elimination to yield the products. In this two-stage mechanism bond formation occurs before bond cleavage, and the carbonyl carbon atom undergoes a hybridization change from sp2 to sp3 and back again. The facility with which nucleophilic reagents add to a carbonyl group was noted earlier for aldehydes and ketones.
Acid and base-catalyzed variations of this mechanism are shown above. Also, a specific example of acyl chloride formation from the reaction of a carboxylic acid with thionyl chloride will be shown. The number of individual steps in these mechanisms vary, but the essential characteristic of the overall transformation is that of addition followed by elimination. Acid catalysts act to increase the electrophilicity of the acyl reactant; whereas, base catalysts act on the nucleophilic reactant to increase its reactivity. In principle all steps are reversible, but in practice many reactions of this kind are irreversible unless changes in the reactants and conditions are made. The acid-catalyzed formation of esters from carboxylic acids and alcohols, described earlier, is a good example of a reversible acylation reaction, the products being determined by the addition or removal of water from the system. The reaction of an acyl chloride with an alcohol also gives an ester, but this conversion cannot be reversed by adding HCl to the reaction mixture.
Mechanisms of Ester Cleavage
Esters are one of the most common carboxylic derivatives. Cleavage of the alkyl moiety in an ester may be effected in several different ways, the most common being the acyl transfer mechanism described above; however, other mechanisms have been observed.
Thus far we have not explained the marked variation, noted above, in the reactivity of different carboxylic acid derivatives. The distinguishing carbonyl substituents in these compounds are: chloro (acyl chlorides), acyloxy (anhydrides), alkoxy (esters) and amino (amides). All of these substituents have bonds originating from atoms of relatively high electronegativity (Cl, O & N). They are therefore inductively electron withdrawing when bonded to carbon, as shown in the diagram on the right. The consequences of such inductive electron withdrawal on the acidity of carboxylic acids was previously noted.
When these substituents are attached to an sp2 carbon that is part of a π-electron system, a similar inductive effect occurs, but n-π conjugation (p-π conjugation) moves electron density in the opposite direction. By clicking the "Toggle Effect" button the electron shift in both effects will be displayed sequentially. This competition between inductive electron withdrawal and conjugative electron donation was discussed earlier in the context of substituent effects on electrophilic aromatic substitution. Here, it was noted that amino groups were strongly electron donating (resonance effect >> inductive effect), alkoxy groups were slightly less activating, acyloxy groups still less activating (resonance effect > inductive effect) and chlorine was deactivating (inductive effect > resonance effect). In the illustration on the right, R and Z represent the remainder of a benzene ring.
This analysis also predicts the influence these substituent groups have on the reactivity of carboxylic acid derivatives toward nucleophiles (Z = O in the illustration). Inductive electron withdrawal by Y increases the electrophilic character of the carbonyl carbon, and increases its reactivity toward nucleophiles. Thus, acyl chlorides (Y = Cl) are the most reactive of the derivatives. Resonance electron donation by Y decreases the electrophilic character of the carbonyl carbon. The strongest resonance effect occurs in amides, which exhibit substantial carbon-nitrogen double bond character and are the least reactive of the derivatives. An interesting exception to the low reactivity of amides is found in beta-lactams such as penicillin G. The angle strain introduced by the four-membered ring reduces the importance of resonance, the non-bonding electron pair remaining localized on the pyramidally shaped nitrogen. Finally, anhydrides and esters have intermediate reactivities, with anhydrides being more reactive than esters.
Carbonyl Reactivity and IR Stretching Frequency
An interesting correlation between the reactivity of carboxylic acid derivatives and their carbonyl stretching frequencies exists.
From the previous discussions you should be able to predict the favored product from each of the following reactions.
The acyl derivative is the reactant on the left, and the nucleophilic reactant is to its right.
The first three examples concern reactions of acyl chlorides, the most reactive acylating reagents discussed here. Although amines are among the most reactive nucleophiles, only 1º and 2º-amines give stable amide products. Reaction of 3º-amines with strong acylating reagents may generate acylammonium species reversibly (see below), but these are as reactive as acyl chlorides and will have only a very short existence. This explains why reactions #2 & 3 do not give amide products.
RCOCl + R'3N RCONR'3(+) Cl(–) (an acylammonium salt)
Reactions #4 & 5 display the acylating capability of anhydrides. Bear in mind that anhydrides may also be used as reagents in Friedel-Crafts acylation reactions. Esters are less reactive acylating reagents than anhydrides, and the ester exchange reaction (#6) requires a strong acid or base catalyst. The last example demonstrates that nitrogen is generally more nucleophilic than oxygen. Indeed, it is often possible to carry out reactions of amines with acyl chlorides and anhydrides in aqueous sodium hydroxide solution! Not only is the amine more nucleophilic than water, but the acylating reagent is generally not soluble in or miscible with water, reducing the rate of its hydrolysis.
No acylation reactions of amides were shown in these problems. The most important such reaction is hydrolysis, and this normally requires heat and strong acid or base catalysts. One example, illustrating both types of catalysis, is shown here. Mechanisms for catalyzed reactions of this kind were presented earlier.
R–CO2(–) + CH3NH2 OH(–) & heat
R–CO–NH(CH3) + H2O
H(+) & heat
R–CO2H + CH3NH3(+)
Other Acylation Reagents and Techniques
Because acylation is such an important and widely used transformation, the general reactions described above have been supplemented by many novel procedures and reagents that accomplish similar overall change. These are normally beyond the scope of an introductory text, but a short description of some of these methods is provided for the interested reader by Clicking Here.
Nitriles
Although they do not have a carbonyl group, nitriles are often treated as derivatives of carboxylic acids. Hydrolysis of nitriles to carboxylic acids was described earlier, and requires reaction conditions (catalysts and heat) similar to those needed to hydrolyze amides. This is not surprising, since addition of water to the carbon-nitrogen triple bond gives an imino intermediate which tautomerizes to an amide.
R–C≡N + H2O
acid or base
R–C(OH)=NH
R–CO–NH2
Contributors
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Carboxylic_Acids/Properties_of_Carboxylic_Acids/Carboxyl_Derivatives/Carboxylic_Derivatives_-_Physical_Properties/Carboxylic_Derivatives_-_Reactions_%28Acyl_Group_.txt |
Many aldehydes and ketones were found to undergo electrophilic substitution at an alpha carbon. These reactions, which included halogenation, isotope exchange and the aldol reaction, take place by way of enol tautomer or enolate anion intermediates, a characteristic that requires at least one hydrogen on the α-carbon atom. In this section similar reactions of carboxylic acid derivatives will be examined. Formulas for the corresponding enol and enolate anion species that may be generated from these derivatives are drawn in the following diagram.
Acid-catalyzed alpha-chlorination and bromination reactions proceed more slowly with carboxylic acids, esters and nitriles than with ketones. This may reflect the smaller equilibrium enol concentrations found in these carboxylic acid derivatives. Nevertheless, acid and base catalyzed isotope exchange occurs as expected; some examples are shown in equations #1 and #2 below. The chiral alpha-carbon in equation #2 is racemized in the course of this exchange, and a small amount of nitrile is hydrolyzed to the corresponding carboxylic acid.
Acyl halides and anhydrides are more easily halogenated than esters and nitriles, probably because of their higher enol concentration. This difference may be used to facilitate the alpha-halogenation of carboxylic acids. Thus, conversion of the acid to its acyl chloride derivative is followed by alpha-bromination or chlorination, and the resulting halogenated acyl chloride is then hydrolyzed to the carboxylic acid product. This three-step sequence can be reduced to a single step by using a catalytic amount of phosphorus tribromide or phosphorus trichloride, as shown in equation #3. This simple modification works well because carboxylic acids and acyl chlorides exchange functionality as the reaction progresses. The final product is the alpha-halogenated acid, accompanied by a trace of the acyl halide. This halogenation procedure is called the Hell-Volhardt-Zelinski reaction.
To see a mechanism for the acyl halide-carboxylic acid exchange click the "Show Mechanism" button.
In a similar fashion, acetic anhydride serves as a halogenation catalyst for acetic acid (first equation below). Carboxylic acids that have a higher equilibrium enol concentration do not need to be activated for alpha-halogenation to occur, as demonstrated by the substituted malonic acid compound in the second equation below. The enol concentration of malonic acid (about 0.01%) is roughly ten thousand times greater than that of acetic acid. This influence of a second activating carbonyl function on equilibrium enol concentrations had been noted earlier in the case of 2,4-pentanedione.
1. (i)
CH3-CO2H + Br2 & (CH3CO)2O catalyst
heat
BrCH2CO2H + HBr
(ii)
RCH(CO2H)2 + Br2
RCBr(CO2H)2 + HBr
Contributors
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry
Carboxylic Derivatives - Reactions at the Alpha-Carbon
Enolate Intermediates
Many of the most useful alpha-substitution reactions of ketones proceeded by way of enolate anion conjugate bases. Since simple ketones are weaker acids than water, their enolate anions are necessarily prepared by reaction with exceptionally strong bases in non-hydroxylic solvents. Esters and nitriles are even weaker alpha-carbon acids than ketones (by over ten thousand times), nevertheless their enolate anions may be prepared and used in a similar fashion. The presence of additional activating carbonyl functions increases the acidity of the alpha-hydrogens substantially, so that less stringent conditions may be used for enolate anion formation. The influence of various carbonyl and related functional groups on the equilibrium acidity of alpha-hydrogen atoms (colored red) is summarized in the following table. For common reference, these acidity values have all been extrapolated to water solution, even though the conjugate bases of those compounds having pKas greater than 18 will not have a significant concentration in water solution.
Acidity of α-Hydrogens in Mono- and Di-Activated Compounds
Mono-Activation Compound RCH2–NO2 RCH2–COR RCH2–CO2CH3 RCH2–C≡N RCH2–SO2R RCH2–CON(CH3)2
pKa 9 20 25 25 25 28
Di-Activation Compound CH2(NO2)2 (CH3CO)2CH2 CH3COCH2CO2C2H5 CH2(C≡N)2 CH2(CO2C2H5)2 CH2(SO2CH3)2
pKa 4 9 11 11 13 13
To illustrate the general nucleophilic reactivity of di-activated enolate anions, two examples of SN2 alkylation reactions are shown below. Malonic acid esters and acetoacetic acid esters are commonly used starting materials, and their usefulness in synthesis will be demonstrated later in this chapter. Note that each of these compounds has two acidic alpha-hydrogen atoms (colored red). In the equations written here only one of these hydrogens is substituted; however, the second is also acidic and a second alkyl substitution may be carried out in a similar fashion. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Carboxylic_Acids/Properties_of_Carboxylic_Acids/Carboxyl_Derivatives/Carboxylic_Derivatives_-_Reactions_at_the_Alpha-Carbon/Carboxylic_Derivatives_-_Acidity_of_an_.txt |
Reductions of carboxylic acid derivatives might be expected to lead either to aldehydes or alcohols, functional groups having a lower oxidation state of the carboxyl carbon. Indeed, carboxylic acids themselves are reduced to alcohols by lithium aluminum hydride. At this point it will be useful to consider three kinds of reductions:
1. catalytic hydrogenation
2. complex metal hydride reductions
3. diborane reduction.
Catalytic Hydrogenation
As a rule, the carbonyl group does not add hydrogen as readily as do the carbon-carbon double and triple bonds. Thus, it is fairly easy to reduce an alkene or alkyne function without affecting any carbonyl functions in the same molecule. By using a platinum catalyst and increased temperature and pressure, it is possible to reduce aldehydes and ketones to alcohols, but carboxylic acids, esters and amides are comparatively unreactive. The exceptional reactivity of acyl halides, on the other hand, facilitates their reduction under mild conditions, by using a poisoned palladium catalyst similar to that used for the partial reduction of alkynes to alkenes. This reduction stops at the aldehyde stage, providing us with a useful two-step procedure for converting carboxylic acids to aldehydes, as reaction #1 below demonstrates. Equivalent reductions of anhydrides have not been reported, but we might speculate that they would be reduced more easily than esters. The only other reduction of a carboxylic acid derivative that is widely used is that of nitriles to 1º-amines. Examples of these reductions are provided in the following diagram.
The second and third equations illustrate the extreme difference in hydrogenation reactivity between esters and nitriles. This is further demonstrated by the last reaction, in which a nitrile is preferentially reduced in the presence of a carbonyl group and two benzene rings. The resulting 1º-amine immediately reacts with the carbonyl function to give a cyclic enamine product (colored light blue).
In most nitrile reductions ammonia is added to inhibit the formation of a 2º-amine by-product. This may occur by way of an intermediate aldehyde imine created by addition of the first equivalent of hydrogen. The following equations show how such an imine species might react with the 1º-amine product to give a substituted imine (2nd equation), which would then add hydrogen to generate a 2º-amine. Excess ammonia shifts the imine equilibrium to the left, as written below.
(1)
R–C≡N + H2
catalyst
RCH=NH
imine
H2
RCH2NH2
1º-amine
(2)
RCH=NH + RCH2NH2
imine 1º-amine
RCH=NCH2R + NH3
substituted imine
H2 & catalyst
RCH2NHCH2R
2º-amine
Contributors
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry
Carboxylic Derivatives - Reduction (Catalytic Reduction)
The use of lithium aluminum hydride (LiAlH4) and sodium borohydride (NaBH4) as reagents for the reduction of aldehydes and ketones to 1º and 2º-alcohols respectively has been noted. Of these, lithium aluminum hydride, often abbreviated LAH, is the most useful for reducing carboxylic acid derivatives. Thanks to its high reactivity, LAH easily reduces all classes of carboxylic acid derivatives, generally to the –1 oxidation state. Acids, esters, anhydrides and acyl chlorides are all reduced to 1º-alcohols, and this method is superior to catalytic reduction in most cases. Since acyl chlorides and anhydrides are expensive and time consuming to prepare, acids and esters are the most commonly used reactants for this transformation. As in the reductions of aldehydes and ketones, the first step in each case is believed to be the irreversible addition of hydride to the electrophilic carbonyl carbon atom. Coordinative bonding of the carbonyl oxygen to a Lewis acidic metal (Li or Al) undoubtedly enhances that carbon's electrophilic character. This hydride addition is shown in the following diagrams, with the hydride-donating moiety being written as AlH4(–). All four hydrogens are potentially available to the reduction, but when carboxylic acids are reduced, one of the hydrides reacts with the acidic O–H to generate hydrogen gas. Although the lithium is not shown, it will be present in the products as a cationic component of ionic salts.
Amides are reduced to amines by treatment with LAH, and this has proven to be one of the most general methods for preparing all classes of amines (1º, 2º & 3º). Because the outcome of LAH reduction is so different for esters and amides, we must examine plausible reaction mechanisms for these reactions to discover a reason for this divergent behavior.
One explanation of the different course taken by the reductions of esters and amides lies in the nature of the different hetero atom substituents on the carbonyl group (colored green in the diagram). Nitrogen is more basic than oxygen, and amide anions are poorer leaving groups than alkoxide anions. Furthermore, oxygen forms especially strong bonds to aluminum. Addition of hydride produces a tetrahedral intermediate, shown in brackets, which has a polar oxygen-aluminum bond. Neither the hydrogen nor the alkyl group (R) is a possible leaving group, so if this tetrahedral species is to undergo an elimination to reform a hetero atom double bond, one of the two remaining substituents must be lost. For the ester this is an easy choice (described by the curved arrows). By eliminating an aluminum alkoxide (R'O–Al), an aldehyde is formed, and this is quickly reduced to the salt of a 1º-alcohol by LAH. In the case of the amide, aldehyde formation requires the loss of an aluminum amide (R'2N–Al), an unlikely process. Alternatively, the more basic nitrogen may act to eject a metal oxide species (e.g. Al–O(–)), and the resulting iminium double bond would then be reduced to an amine. This is the course followed by most amide reductions; but in the case of 1º-amides, the acidity of the nitrogen hydrogens coupled with the basicity of hydride enables a facile elimination of the oxygen (as an oxide moiety). The resulting nitrile intermediate is then reduced to a 1º-amine. Nitriles are in fact a major product when less than a full equivalency of LiAlH4 is used. A mechanism will be shown above by clicking on the diagram.
Lithium aluminum hydride reduces nitriles to 1º-amines, as shown in the following equation. An initial hydride addition to the electrophilic nitrile carbon atom generates the salt of an imine intermediate. This is followed by a second hydride transfer, and the resulting metal amine salt is hydrolyzed to a 1º-amine. This method provides a useful alternative to the catalytic reduction of nitriles, described above, when alkene or alkyne functions are present.
In contrast to the usefulness of lithium aluminum hydride in reducing various carboxylic acid derivatives, sodium borohydride is seldom chosen for this purpose. First, NaBH4 is often used in hydroxylic solvents (water and alcohols), and these would react with acyl chlorides and anhydrides. Furthermore, it is sparingly soluble in relatively nonpolar solvents, particularly at low temperatures. Second, NaBH4 is much less reactive than LAH, failing to reduce amides and acids (they form carboxylate salts) at all, and reducing esters very slowly.
Since relatively few methods exist for the reduction of carboxylic acid derivatives to aldehydes, it would be useful to modify the reactivity and solubility of LAH to permit partial reductions of this kind to be achieved. The most fruitful approach to this end has been to attach alkoxy or alkyl groups on the aluminum. This not only modifies the reactivity of the reagent as a hydride donor, but also increases its solubility in nonpolar solvents. Two such reagents will be mentioned here; the reactive hydride atom is colored blue.
Lithium tri-tert-butoxyaluminohydride (LtBAH), LiAl[OC(CH3)3]3H : Soluble in THF, diglyme & ether.
Diisobutylaluminum hydride (DIBAH), [(CH3)2CHCH2]2AlH : Soluble in toluene, THF & ether.
Each of these reagents carries one equivalent of hydride. The first (LtBAH) is a complex metal hydride, but the second is simply an alkyl derivative of aluminum hydride. In practice, both reagents are used in equimolar amounts, and usually at temperatures well below 0 ºC. The following examples illustrate how aldehydes may be prepared from carboxylic acid derivatives by careful application of these reagents. A temperature of -78 ºC is easily maintained by using dry-ice as a coolant. The reduced intermediates that lead to aldehydes will be displayed on clicking the "Show Intermediates" button. With excess reagent at temperatures above 0 ºC most carboxylic acid derivatives are reduced to alcohols or amines. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Carboxylic_Acids/Properties_of_Carboxylic_Acids/Carboxyl_Derivatives/Carboxylic_Derivatives_-_Reduction_%28Catalytic_Reduction%29/Carboxylic_Derivatives_-_Reductio.txt |
This page explains what carboxylic acids are, and looks at the ions that they form in their salts. It also considers their simple physical properties such as solubility and boiling points.
Carboxylic acids contain a -COOH group
Carboxylic acids are compounds which contain a -COOH group. For the purposes of this page we shall just look at compounds where the -COOH group is attached either to a hydrogen atom or to an alkyl group.
Example
The name counts the total number of carbon atoms in the longest chain - including the one in the -COOH group. If you have side groups attached to the chain, notice that you always count from the carbon atom in the -COOH group as being number 1.
Salts of carboxylic acids
Carboxylic acids are acidic because of the hydrogen in the -COOH group. When the acids form salts, this is lost and replaced by a metal. Sodium ethanoate, for example, has the structure:
Depending on whether or not you wanted to stress the ionic nature of the compound, this would be simplified to CH3COO- Na+ or just CH3COONa. Notice:
• The bond between the sodium and the ethanoate is ionic. Do not draw a line between the two (implying a covalent bond). That is absolutely wrong!
• Although the name is written with the sodium first, the formula is always written in one of the ways shown. This is something you just have to get used to.
Physical properties of carboxylic acids
The physical properties (for example, boiling point and solubility) of the carboxylic acids are governed by their ability to form hydrogen bonds.
Boiling points
Before we look at carboxylic acids, a reminder about alcohols: The boiling points of alcohols are higher than those of alkanes of similar size because the alcohols can form hydrogen bonds with each other as well as van der Waals dispersion forces and dipole-dipole interactions. The boiling points of carboxylic acids of similar size are higher still. For example:
propan-1-ol CH3CH2CH2OH 97.2°C
ethanoic acid CH3COOH 118°C
These are chosen for comparison because they have identical relative molecular masses and almost the same number of electrons (which affects van der Waals dispersion forces). The higher boiling points of the carboxylic acids are still caused by hydrogen bonding, but operating in a different way. In a pure carboxylic acid, hydrogen bonding can occur between two molecules of acid to produce a dimer.
This immediately doubles the size of the molecule and so increases the van der Waals dispersion forces between one of these dimers and its neighbors - resulting in a high boiling point.
Solubility in water
In the presence of water, the carboxylic acids do not dimerize. Instead, hydrogen bonds are formed between water molecules and individual molecules of acid. The carboxylic acids with up to four carbon atoms will mix with water in any proportion. When you mix the two together, the energy released when the new hydrogen bonds form is much the same as is needed to break the hydrogen bonds in the pure liquids.
The solubility of the bigger acids decreases very rapidly with size. This is because the longer hydrocarbon "tails" of the molecules get between water molecules and break hydrogen bonds. In this case, these broken hydrogen bonds are only replaced by much weaker van der Waals dispersion forces. The energetics of dissolving carboxylic acids in water is made more complicated because some of the acid molecules actually react with the water rather than just dissolving in it.
Contributors
Jim Clark (Chemguide.co.uk)
Fatty Acids
Carboxylic acids are widespread in nature, often combined with other functional groups. Simple alkyl carboxylic acids, composed of four to ten carbon atoms, are liquids or low melting solids having very unpleasant odors. The fatty acids are important components of the biomolecules known as lipids, especially fats and oils. As shown in the following table, these long-chain carboxylic acids are usually referred to by their common names, which in most cases reflect their sources. A mnemonic phrase for the C10 to C20 natural fatty acids capric, lauric, myristic, palmitic, stearic and arachidic is: "Curly, Larry & Moe Perform Silly Antics" (note that the names of the three stooges are in alphabetical order).
Table: \(1\): Saturated Fatty Acids
Formula Common Name Melting Point
CH3(CH2)10CO2H lauric acid 45 ºC
CH3(CH2)12CO2H myristic acid 55 ºC
CH3(CH2)14CO2H palmitic acid 63 ºC
CH3(CH2)16CO2H stearic acid 69 ºC
CH3(CH2)18CO2H arachidic acid 76 ºC
Interestingly, the molecules of most natural fatty acids have an even number of carbon atoms. Analogous compounds composed of odd numbers of carbon atoms are perfectly stable and have been made synthetically. Since nature makes these long-chain acids by linking together acetate units, it is not surprising that the carbon atoms composing the natural products are multiples of two. The double bonds in the unsaturated compounds listed on the right are all cis (or Z).
Table \(2\): Unsaturated Fatty Acids
Formula Common Name Melting Point
CH3(CH2)5CH=CH(CH2)7CO2H palmitoleic acid 0 ºC
CH3(CH2)7CH=CH(CH2)7CO2H oleic acid 13 ºC
CH3(CH2)4CH=CHCH2CH=CH(CH2)7CO2H linoleic acid -5 ºC
CH3CH2CH=CHCH2CH=CHCH2CH=CH(CH2)7CO2H linolenic acid -11 ºC
CH3(CH2)4(CH=CHCH2)4(CH2)2CO2H arachidonic acid -49 ºC
The following formulas are examples of other naturally occurring carboxylic acids. The molecular structures range from simple to complex, often incorporate a variety of other functional groups, and many are chiral.
Contributors
• William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Carboxylic_Acids/Properties_of_Carboxylic_Acids/Carboxylic_Acids_Background.txt |
Structure of the carboxyl acid group
Carboxylic acids are organic compounds which incorporate a carboxyl functional group, CO2H. The name carboxyl comes from the fact that a carbonyl and a hydroxyl group are attached to the same carbon.
The carbon and oxygen in the carbonyl are both sp2 hybridized which give a carbonyl group a basic trigonal shape. The hydroxyl oxygen is also sp2 hybridized which allows one of its lone pair electrons to conjugate with the pi system of the carbonyl group. This make the carboxyl group planar an can represented with the following resonance structure.
Carboxylic acids are named such because they can donate a hydrogen to produce a carboxylate ion. The factors which affect the acidity of carboxylic acids will be discussed later.
Physical Properties of Some Carboxylic Acids
The table at the beginning of this page gave the melting and boiling points for a homologous group of carboxylic acids having from one to ten carbon atoms. The boiling points increased with size in a regular manner, but the melting points did not. Unbranched acids made up of an even number of carbon atoms have melting points higher than the odd numbered homologs having one more or one less carbon. This reflects differences in intermolecular attractive forces in the crystalline state. In the table of fatty acids we see that the presence of a cis-double bond significantly lowers the melting point of a compound. Thus, palmitoleic acid melts over 60º lower than palmitic acid, and similar decreases occur for the C18 and C20 compounds. Again, changes in crystal packing and intermolecular forces are responsible.
The factors that influence the relative boiling points and water solubilities of various types of compounds were discussed earlier. In general, dipolar attractive forces between molecules act to increase the boiling point of a given compound, with hydrogen bonds being an extreme example. Hydrogen bonding is also a major factor in the water solubility of covalent compounds To refresh your understanding of these principles Click Here. The following table lists a few examples of these properties for some similar sized polar compounds (the non-polar hydrocarbon hexane is provided for comparison).
Physical Properties of Some Organic Compounds
Formula
IUPAC Name
Molecular Weight
Boiling Point
Water Solubility
CH3(CH2)2CO2H butanoic acid 88 164 ºC very soluble
CH3(CH2)4OH 1-pentanol 88 138 ºC slightly soluble
CH3(CH2)3CHO pentanal 86 103 ºC slightly soluble
CH3CO2C2H5 ethyl ethanoate 88 77 ºC moderately soluble
CH3CH2CO2CH3 methyl propanoate 88 80 ºC slightly soluble
CH3(CH2)2CONH2 butanamide 87 216 ºC soluble
CH3CON(CH3)2 N,N-dimethylethanamide 87 165 ºC very soluble
CH3(CH2)4NH2 1-aminobutane 87 103 ºC very soluble
CH3(CH2)3CN pentanenitrile 83 140 ºC slightly soluble
CH3(CH2)4CH3 hexane 86 69 ºC insoluble
The first five entries all have oxygen functional groups, and the relatively high boiling points of the first two is clearly due to hydrogen bonding. Carboxylic acids have exceptionally high boiling points, due in large part to dimeric associations involving two hydrogen bonds. A structural formula for the dimer of acetic acid is shown here. When the mouse pointer passes over the drawing, an electron cloud diagram will appear. The high boiling points of the amides and nitriles are due in large part to strong dipole attractions, supplemented in some cases by hydrogen bonding.
Acidity of Carboxylic Acids
The pKa 's of some typical carboxylic acids are listed in the following table. When we compare these values with those of comparable alcohols, such as ethanol (pKa = 16) and 2-methyl-2-propanol (pKa = 19), it is clear that carboxylic acids are stronger acids by over ten powers of ten! Furthermore, electronegative substituents near the carboxyl group act to increase the acidity.
Compound
pKa
Compound
pKa
HCO2H 3.75 CH3CH2CH2CO2H 4.82
CH3CO2H 4.74 ClCH2CH2CH2CO2H 4.53
FCH2CO2H 2.65 CH3CHClCH2CO2H 4.05
ClCH2CO2H 2.85 CH3CH2CHClCO2H 2.89
BrCH2CO2H 2.90 C6H5CO2H 4.20
ICH2CO2H 3.10 p-O2NC6H4CO2H 3.45
Cl3CCO2H 0.77 p-CH3OC6H4CO2H 4.45
Why should the presence of a carbonyl group adjacent to a hydroxyl group have such a profound effect on the acidity of the hydroxyl proton? To answer this question we must return to the nature of acid-base equilibria and the definition of pKa , illustrated by the general equations given below. These relationships were described in an previous section of this text.
We know that an equilibrium favors the thermodynamically more stable side, and that the magnitude of the equilibrium constant reflects the energy difference between the components of each side. In an acid base equilibrium the equilibrium always favors the weaker acid and base (these are the more stable components). Water is the standard base used for pKa measurements; consequently, anything that stabilizes the conjugate base (A:(–)) of an acid will necessarily make that acid (H–A) stronger and shift the equilibrium to the right. Both the carboxyl group and the carboxylate anion are stabilized by resonance, but the stabilization of the anion is much greater than that of the neutral function, as shown in the following diagram. In the carboxylate anion the two contributing structures have equal weight in the hybrid, and the C–O bonds are of equal length (between a double and a single bond). This stabilization leads to a markedly increased acidity, as illustrated by the energy diagram displayed by clicking the "Toggle Display" button.
The resonance effect described here is undoubtedly the major contributor to the exceptional acidity of carboxylic acids. However, inductive effects also play a role. For example, alcohols have pKa's of 16 or greater but their acidity is increased by electron withdrawing substituents on the alkyl group. The following diagram illustrates this factor for several simple inorganic and organic compounds (row #1), and shows how inductive electron withdrawal may also increase the acidity of carboxylic acids (rows #2 & 3). The acidic hydrogen is colored red in all examples.
Water is less acidic than hydrogen peroxide because hydrogen is less electronegative than oxygen, and the covalent bond joining these atoms is polarized in the manner shown. Alcohols are slightly less acidic than water, due to the poor electronegativity of carbon, but chloral hydrate, Cl3CCH(OH)2, and 2,2,2,-trifluoroethanol are significantly more acidic than water, due to inductive electron withdrawal by the electronegative halogens (and the second oxygen in chloral hydrate). In the case of carboxylic acids, if the electrophilic character of the carbonyl carbon is decreased the acidity of the carboxylic acid will also decrease. Similarly, an increase in its electrophilicity will increase the acidity of the acid. Acetic acid is ten times weaker an acid than formic acid (first two entries in the second row), confirming the electron donating character of an alkyl group relative to hydrogen, as noted earlier in a discussion of carbocation stability. Electronegative substituents increase acidity by inductive electron withdrawal. As expected, the higher the electronegativity of the substituent the greater the increase in acidity (F > Cl > Br > I), and the closer the substituent is to the carboxyl group the greater is its effect (isomers in the 3rd row). Substituents also influence the acidity of benzoic acid derivatives, but resonance effects compete with inductive effects. The methoxy group is electron donating and the nitro group is electron withdrawing (last three entries in the table of pKa values).
For additional information about substituent effects on the acidity of carboxylic acids Click Here
Vinylagous Acids
Compounds in which an enolic hydroxyl group is conjugated with a carbonyl group also show enhanced acidity. To see examples of such compounds Click Here
Contributors
Prof. Steven Farmer (Sonoma State University)
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry
Physical Properties of Carboxylic Acids
The pKa 's of some typical carboxylic acids are listed in the following table. When we compare these values with those of comparable alcohols, such as ethanol (pKa = 16) and 2-methyl-2-propanol (pKa = 19), it is clear that carboxylic acids are stronger acids by over ten powers of ten! Furthermore, electronegative substituents near the carboxyl group act to increase the acidity.
Compound
pKa
Compound
pKa
HCO2H 3.75 CH3CH2CH2CO2H 4.82
CH3CO2H 4.74 ClCH2CH2CH2CO2H 4.53
FCH2CO2H 2.65 CH3CHClCH2CO2H 4.05
ClCH2CO2H 2.85 CH3CH2CHClCO2H 2.89
BrCH2CO2H 2.90 C6H5CO2H 4.20
ICH2CO2H 3.10 p-O2NC6H4CO2H 3.45
Cl3CCO2H 0.77 p-CH3OC6H4CO2H 4.45
Why should the presence of a carbonyl group adjacent to a hydroxyl group have such a profound effect on the acidity of the hydroxyl proton? To answer this question we must return to the nature of acid-base equilibria and the definition of pKa , illustrated by the general equations given below. These relationships were described in an previous section of this text.
We know that an equilibrium favors the thermodynamically more stable side, and that the magnitude of the equilibrium constant reflects the energy difference between the components of each side. In an acid base equilibrium the equilibrium always favors the weaker acid and base (these are the more stable components). Water is the standard base used for pKa measurements; consequently, anything that stabilizes the conjugate base (A:(–)) of an acid will necessarily make that acid (H–A) stronger and shift the equilibrium to the right. Both the carboxyl group and the carboxylate anion are stabilized by resonance, but the stabilization of the anion is much greater than that of the neutral function, as shown in the following diagram. In the carboxylate anion the two contributing structures have equal weight in the hybrid, and the C–O bonds are of equal length (between a double and a single bond). This stabilization leads to a markedly increased acidity, as illustrated by the energy diagram displayed by clicking the "Toggle Display" button.
Vinylagous Acids
Compounds in which an enolic hydroxyl group is conjugated with a carbonyl group also show enhanced acidity. To see examples of such compounds Click Here
The resonance effect described here is undoubtedly the major contributor to the exceptional acidity of carboxylic acids. However, inductive effects also play a role. For example, alcohols have pKa's of 16 or greater but their acidity is increased by electron withdrawing substituents on the alkyl group. The following diagram illustrates this factor for several simple inorganic and organic compounds (row #1), and shows how inductive electron withdrawal may also increase the acidity of carboxylic acids (rows #2 & 3). The acidic hydrogen is colored red in all examples.
Water is less acidic than hydrogen peroxide because hydrogen is less electronegative than oxygen, and the covalent bond joining these atoms is polarized in the manner shown. Alcohols are slightly less acidic than water, due to the poor electronegativity of carbon, but chloral hydrate, Cl3CCH(OH)2, and 2,2,2,-trifluoroethanol are significantly more acidic than water, due to inductive electron withdrawal by the electronegative halogens (and the second oxygen in chloral hydrate). In the case of carboxylic acids, if the electrophilic character of the carbonyl carbon is decreased the acidity of the carboxylic acid will also decrease. Similarly, an increase in its electrophilicity will increase the acidity of the acid. Acetic acid is ten times weaker an acid than formic acid (first two entries in the second row), confirming the electron donating character of an alkyl group relative to hydrogen, as noted earlier in a discussion of carbocation stability. Electronegative substituents increase acidity by inductive electron withdrawal. As expected, the higher the electronegativity of the substituent the greater the increase in acidity (F > Cl > Br > I), and the closer the substituent is to the carboxyl group the greater is its effect (isomers in the 3rd row). Substituents also influence the acidity of benzoic acid derivatives, but resonance effects compete with inductive effects. The methoxy group is electron donating and the nitro group is electron withdrawing (last three entries in the table of pKa values). For additional information about substituent effects on the acidity of carboxylic acids Click Here
Related Derivatives
Other functional group combinations with the carbonyl group can be prepared from carboxylic acids, and are usually treated as related derivatives. Five common classes of these carboxylic acid derivatives are listed in the following table. Although nitriles do not have a carbonyl group, they are included here because the functional carbon atoms all have the same oxidation state. The top row (yellow shaded) shows the general formula for each class, and the bottom row (light blue) gives a specific example of each. As in the case of amines, amides are classified as 1º, 2º or 3º, depending on the number of alkyl groups bonded to the nitrogen.
Functional groups of this kind are found in many kinds of natural products. Some examples are shown below with the functional group colored red. Most of the functions are amides or esters, cantharidin being a rare example of a natural anhydride. Cyclic esters are called lactones, and cyclic amides are referred to as lactams. Penicillin G has two amide functions, one of which is a β-lactam. The Greek letter locates the nitrogen relative to the carbonyl group of the amide.
Contributors
• William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Carboxylic_Acids/Properties_of_Carboxylic_Acids/Physical_Properties_of_Carboxylic_Acids/Acidity_of_Carboxylic_Acids.txt |
The direct conversion of a carboxylic acid to an amide is difficult because amines are basic and tend to convert carboxylic acids to their highly unreactive carboxylates. In this reaction the carboxylic acid adds to the DCC molecule to form a good leaving group which can then be displaced by an amine during nucleophilic substitution. DCC induced coupling to form an amide linkage is an important reaction in the synthesis of peptides.
Mechanism
1) Deprotonation
2) Nucleophilic attack by the carboxylate
3) Nucleophilic attack by the amine
4) Proton transfer
5) Leaving group removal
Conversion of a Carboxylic Acid to an Amide
The direct reaction of a carboxylic acid with an amine would be expected to be difficult because the basic amine would deprotonate the carboxylic acid to form a highly unreactive carboxylate. However when the ammonium carboxylate salt is heated to a temperature above 100 oC water is driven off and an amide is formed.
Contributors
Prof. Steven Farmer (Sonoma State University)
Conversion of carboxylic acids to acid chlorides
Carboxylic acids react with Thionyl Chloride (\(SOCl_2\)) to form acid chlorides. During the reaction the hydroxyl group of the carboxylic acid is converted to a chlorosulfite intermediate making it a better leaving group. The chloride anion produced during the reaction acts a nucleophile.
Mechanism
1) Nucleophilic attack on Thionyl Chloride
2) Removal of Cl leaving group
3) Nucleophilic attack on the carbonyl
4) Leaving group removal
5) Deprotonation
Contributors
Prof. Steven Farmer (Sonoma State University)
Conversion of carboxylic acids to alcohols using LiAlH4
Carboxylic acids can be converted to 1o alcohols using Lithium aluminum hydride (LiAlH4). Note that NaBH4 is not strong enough to convert carboxylic acids or esters to alcohols. An aldehyde is produced as an intermediate during this reaction, but it cannot be isolated because it is more reactive than the original carboxylic acid.
Possible Mechanism
1) Deprotonation
2) Nucleophilic attack by the hydride anion
3) Leaving group removal
4) Nucleophilic attack by the hydride anion
5) The alkoxide is protonated
Fischer Esterification
Fischer esterification is the esterification of a Carboxylic acid by heating it with an alcohol in the presence of a strong acid as the catalyst.
Mechanism
The overall reaction is reversible; to drive the reaction to completion, it is necessary to exploit Le Chateliers principle, which can be done either by continuously removing the water formed from the system or by using a large excess of the alcohol.
1) Protonation of the carbonyl by the acid. The carbonyl is now activated toward nucleophilic attack.
2) Nucleophilic attack on the carbonyl
3) Proton transfer
4) Water leaves
5) Deprotonation
Hell-Volhard-Zelinskii Reaction
Although the alpha bromination of some carbonyl compounds, such as aldehydes and ketones, can be accomplished with Br2 under acidic conditions, the reaction will generally not occur with acids, esters, and amides. This is because only aldehydes and ketones enolize to a sufficient extent to allow the reaction to occur. However, carboxylic acids, can be brominated in the alpha position with a mixture of Br2 and PBr3 in a reaction called the Hell-Volhard-Zelinskii reaction.
The mechanism of this reaction involves an acid bromide enol instead of the expected carboxylic acid enol. The reaction stats with the reaction of the carboxylic acid with PBr3 to form the acid bromide and HBr. The HBr then catalyzes the formation of the acid bromide enol which subsequently reacts with Br2 to give alpha bromination. Lastly, the acid bromide reacts with water to reform the carboxylic acid.
Contributors
Prof. Steven Farmer (Sonoma State University)
Making Acyl Chlorides (Acid Chlorides)
This page looks at ways of swapping the -OH group in the -COOH group of a carboxylic acid for a chlorine atom. This produces useful compounds called acyl chlorides (acid chlorides). It covers the use of phosphorus(V) chloride and phosphorus(III) chloride as well as sulfur dichloride oxide (thionyl chloride).
Replacing -OH by -Cl
We are going to be looking at converting a carboxylic acid, RCOOH, into an acyl chloride, RCOCl. Acyl chlorides are also known as acid chlorides.
By far the most commonly used example of the conversion of a carboxylic acid into an acyl chloride is ethanoic acid to ethanoyl chloride.
Acyl chlorides are very reactive, and can be used to make a wide range of other things. That's why they are important.
Replacing the -OH group using phosphorus(V) chloride, PCl5
Phosphorus(V) chloride is a solid which reacts with carboxylic acids in the cold to give steamy acidic fumes of hydrogen chloride. It leaves a liquid mixture of the acyl chloride and a phosphorus compound, phosphorus trichloride oxide (phosphorus oxychloride) - POCl3.
$CH_3COOH + PCl_5 \rightarrow CH_3COCl + POCl_3 + HCl$
The acyl chloride can be separated by fractional distillation.
Replacing the -OH group using phosphorus(III) chloride, PCl3
Phosphorus(III) chloride is a liquid at room temperature. Its reaction with a carboxylic acid is less dramatic than that of phosphorus(V) chloride because there is no hydrogen chloride produced.
You end up with a mixture of the acyl chloride and phosphoric(III) acid (old names: phosphorous acid or orthophosphorous acid), H3PO3. For example:
$CH_3COOH + PCl_3 \rightarrow 3CH_3COCl + H_3PO3$
Again, the ethanoyl chloride can be separated by fractional distillation.
Replacing the -OH group using thionyl chloride
Sulfur dichloride oxide (thionyl chloride) is a liquid at room temperature and has the formula SOCl2. Traditionally, the formula is written as shown, despite the fact that the modern name writes the chlorine before the oxygen (alphabetical order). The sulfur dichloride oxide reacts with carboxylic acids to produce an acyl chloride, and sulfur dioxide and hydrogen chloride gases are given off. For example:
$CH_3COOH + SOCl_2 \rightarrow CH_3COCl + SO_2 + HCl$
The separation is simplified to an extent because the by-products are both gases. You would obviously still have to fractionally distil the mixture to separate the acyl chloride from any excess acid or sulfur dichloride oxide.
Contributors
Jim Clark (Chemguide.co.uk) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Carboxylic_Acids/Reactivity_of_Carboxylic_Acids/Conversion_of_Carboxylic_acids_to_amides_using_DCC_as_an_activating_agent.txt |
This page looks at esterification - the reaction between alcohols and carboxylic acids to make esters.
Esters have a hydrocarbon group of some sort replacing the hydrogen in the -COOH group of a carboxylic acid. We shall just be looking at cases where it is replaced by an alkyl group, but it could equally well be an aryl group (one based on a benzene ring). The most commonly discussed ester is ethyl ethanoate. In this case, the hydrogen in the -COOH group has been replaced by an ethyl group. The formula for ethyl ethanoate is:
Notice that the ester is named the opposite way around from the way the formula is written. The "ethanoate" bit comes from ethanoic acid. The "ethyl" bit comes from the ethyl group on the end. In each case, be sure that you can see how the names and formulae relate to each other.
Remember that the acid is named by counting up the total number of carbon atoms in the chain - including the one in the -COOH group. So, for example, CH3CH2COOH is propanoic acid, and CH3CH2COO is the propanoate group.
Making esters
Esters are produced when carboxylic acids are heated with alcohols in the presence of an acid catalyst. The catalyst is usually concentrated sulphuric acid. Dry hydrogen chloride gas is used in some cases, but these tend to involve aromatic esters (ones containing a benzene ring). The esterification reaction is both slow and reversible. The equation for the reaction between an acid RCOOH and an alcohol R'OH (where R and R' can be the same or different) is:
So, for example, if you were making ethyl ethanoate from ethanoic acid and ethanol, the equation would be:
Doing the reactions
• Carboxylic acids and alcohols are often warmed together in the presence of a few drops of concentrated sulfuric acid in order to observe the smell of the esters formed.
• You would normally use small quantities of everything heated in a test tube stood in a hot water bath for a couple of minutes.
• Because the reactions are slow and reversible, you don't get a lot of ester produced in this time. The smell is often masked or distorted by the smell of the carboxylic acid. A simple way of detecting the smell of the ester is to pour the mixture into some water in a small beaker.
• Esters are virtually insoluble in water and tend to form a thin layer on the surface. Excess acid and alcohol both dissolve and are tucked safely away under the ester layer.
• Small esters like ethyl ethanoate smell like typical organic solvents (ethyl ethanoate is a common solvent in, for example, glues).
• As the esters get bigger, the smells tend towards artificial fruit flavoring - "pear drops", for example.
On a larger scale
If you want to make a reasonably large sample of an ester, the method used depends to some extent on the size of the ester. Small esters are formed faster than bigger ones. To make a small ester like ethyl ethanoate, you can gently heat a mixture of ethanoic acid and ethanol in the presence of concentrated sulfuric acid, and distil off the ester as soon as it is formed.
This prevents the reverse reaction happening. It works well because the ester has the lowest boiling point of anything present. The ester is the only thing in the mixture which doesn't form hydrogen bonds, and so it has the weakest intermolecular forces.
Larger esters tend to form more slowly. In these cases, it may be necessary to heat the reaction mixture under reflux for some time to produce an equilibrium mixture. The ester can be separated from the carboxylic acid, alcohol, water and sulfuric acid in the mixture by fractional distillation. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Carboxylic_Acids/Reactivity_of_Carboxylic_Acids/Making_Esters_From_Carboxylic_Acids.txt |
Salt Formation
Because of their enhanced acidity, carboxylic acids react with bases to form ionic salts, as shown in the following equations. In the case of alkali metal hydroxides and simple amines (or ammonia) the resulting salts have pronounced ionic character and are usually soluble in water. Heavy metals such as silver, mercury and lead form salts having more covalent character (3rd example), and the water solubility is reduced, especially for acids composed of four or more carbon atoms.
RCO2H + NaHCO3 RCO2(–) Na(+) + CO2 + H2O
RCO2H + (CH3)3N: RCO2(–) (CH3)3NH(+)
RCO2H + AgOH RCO2δ(-) Agδ(+) + H2O
Carboxylic acids and salts having alkyl chains longer than six carbons exhibit unusual behavior in water due to the presence of both hydrophilic (CO2) and hydrophobic (alkyl) regions in the same molecule. Such molecules are termed amphiphilic (Gk. amphi = both) or amphipathic. Depending on the nature of the hydrophilic portion these compounds may form monolayers on the water surface or sphere-like clusters, called micelles, in solution.
Substitution of the Hydroxyl Hydrogen
This reaction class could be termed electrophilic substitution at oxygen, and is defined as follows (E is an electrophile). Some examples of this substitution are provided in equations (1) through (4).
RCO2–H + E(+) RCO2E + H(+)
If E is a strong electrophile, as in the first equation, it will attack the nucleophilic oxygen of the carboxylic acid directly, giving a positively charged intermediate which then loses a proton. If E is a weak electrophile, such as an alkyl halide, it is necessary to convert the carboxylic acid to the more nucleophilic carboxylate anion to facilitate the substitution. This is the procedure used in reactions 2 and 3. Equation 4 illustrates the use of the reagent diazomethane (CH2N2) for the preparation of methyl esters. This toxic and explosive gas is always used as an ether solution (bright yellow in color). The reaction is easily followed by the evolution of nitrogen gas and the disappearance of the reagent's color. This reaction is believed to proceed by the rapid bonding of a strong electrophile to a carboxylate anion.
The nature of SN2 reactions, as in equations 2 & 3, has been described elsewhere. The mechanisms of reactions 1 & 4 will be displayed by clicking the "Toggle Mechanism" button below the diagram.
Alkynes may also serve as electrophiles in substitution reactions of this kind, as illustrated by the synthesis of vinyl acetate from acetylene. Intramolecular carboxyl group additions to alkenes generate cyclic esters known as lactones. Five-membered (gamma) and six-membered (delta) lactones are most commonly formed. Electrophilic species such as acids or halogens are necessary initiators of lactonizations. Even the weak electrophile iodine initiates iodolactonization of γ,δ- and δ,ε-unsaturated acids. Examples of these reactions will be displayed by clicking the "Other Examples" button.
Substitution of the Hydroxyl Group
Reactions in which the hydroxyl group of a carboxylic acid is replaced by another nucleophilic group are important for preparing functional derivatives of carboxylic acids. The alcohols provide a useful reference chemistry against which this class of transformations may be evaluated. In general, the hydroxyl group proved to be a poor leaving group, and virtually all alcohol reactions in which it was lost involved a prior conversion of –OH to a better leaving group. This has proven to be true for the carboxylic acids as well.
Four examples of these hydroxyl substitution reactions are presented by the following equations. In each example, the new bond to the carbonyl group is colored magenta and the nucleophilic atom that has replaced the hydroxyl oxygen is colored green. The hydroxyl moiety is often lost as water, but in reaction #1 the hydrogen is lost as HCl and the oxygen as SO2. This reaction parallels a similar transformation of alcohols to alkyl chlorides, although its mechanism is different. Other reagents that produce a similar conversion to acyl halides are PCl5 and SOBr2.
The amide and anhydride formations shown in equations #2 & 3 require strong heating, and milder procedures that accomplish these transformations will be described in the next chapter.
previous diagram illustrate the formation of tert-butyl and methyl esters respectively. The acid-catalyzed formation of ethyl acetate from acetic acid and ethanol shown here is reversible, with an equilibrium constant near 2. The reaction can be forced to completion by removing the water as it is formed. This type of esterification is often referred to as Fischer esterification. As expected, the reverse reaction, acid-catalyzed ester hydrolysis, can be carried out by adding excess water.
A thoughtful examination of this reaction (#4) leads one to question why it is classified as a hydroxyl substitution rather than a hydrogen substitution. The following equations, in which the hydroxyl oxygen atom of the carboxylic acid is colored red and that of the alcohol is colored blue, illustrate this distinction (note that the starting compounds are in the center).
H2O + CH3CO-OCH2CH3
H-substitution
CH3CO-OH + CH3CH2-OH
HO-substitution
CH3CO-OCH2CH3 + H2O
In order to classify this reaction correctly and establish a plausible mechanism, the oxygen atom of the alcohol was isotopically labeled as 18O (colored blue in our equation). Since this oxygen is found in the ester product and not the water, the hydroxyl group of the acid must have been replaced in the substitution. A mechanism for this general esterification reaction will be displayed on clicking the "Esterification Mechanism" button; also, once the mechanism diagram is displayed, a reaction coordinate for it can be seen by clicking the head of the green "energy diagram" arrow. Addition-elimination mechanisms of this kind proceed by way of tetrahedral intermediates (such as A and B in the mechanism diagram) and are common in acyl substitution reactions. Acid catalysis is necessary to increase the electrophilic character of the carboxyl carbon atom, so it will bond more rapidly to the nucleophilic oxygen of the alcohol. Base catalysis is not useful because base converts the acid to its carboxylate anion conjugate base, a species in which the electrophilic character of the carbon is reduced.
Since a tetrahedral intermediate occupies more space than a planar carbonyl group, we would expect the rate of this reaction to be retarded when bulky reactants are used. To test this prediction the esterification of acetic acid was compared with that of 2,2-dimethylpropanoic acid, (CH3)3CO2H. Here the relatively small methyl group of acetic acid is replaced by a larger tert-butyl group, and the bulkier acid reacted fifty times slower than acetic acid. Increasing the bulk of the alcohol reactant results in a similar rate reduction.
Reactions of Carboxylic Acids
Because of their enhanced acidity, carboxylic acids react with bases to form ionic salts, as shown in the following equations. In the case of alkali metal hydroxides and simple amines (or ammonia) the resulting salts have pronounced ionic character and are usually soluble in water. Heavy metals such as silver, mercury and lead form salts having more covalent character (3rd example), and the water solubility is reduced, especially for acids composed of four or more carbon atoms.
RCO2H + NaHCO3 RCO2(–) Na(+) + CO2 + H2O
RCO2H + (CH3)3N: RCO2(–) (CH3)3NH(+)
RCO2H + AgOH RCO2δ(-) Agδ(+) + H2O
Carboxylic acids and salts having alkyl chains longer than six carbons exhibit unusual behavior in water due to the presence of both hydrophilic (CO2) and hydrophobic (alkyl) regions in the same molecule. Such molecules are termed amphiphilic (Gk. amphi = both) or amphipathic. Depending on the nature of the hydrophilic portion these compounds may form monolayers on the water surface or sphere-like clusters, called micelles, in solution. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Carboxylic_Acids/Reactivity_of_Carboxylic_Acids/Reactions_of_Carboxylic_Acids/Carboxylate_formation_reactions.txt |
Reduction
The carbon atom of a carboxyl group is in a relatively high oxidation state. Reduction to a 1º-alcohol takes place rapidly on treatment with the powerful metal hydride reagent, lithium aluminum hydride, as shown by the following equation. One third of the hydride is lost as hydrogen gas, and the initial product consists of metal salts which must be hydrolyzed to generate the alcohol. These reductions take place by the addition of hydride to the carbonyl carbon, in the same manner noted earlier for aldehydes and ketones. The resulting salt of a carbonyl hydrate then breaks down to an aldehyde that undergoes further reduction.
4 RCO2H + 3 LiAlH4
ether
4 H2 + 4 RCH2OM + metal oxides
H2O
4 RCH2OH + metal hydroxides
Diborane, B2H6, reduces the carboxyl group in a similar fashion. Sodium borohydride, NaBH4, does not reduce carboxylic acids; however, hydrogen gas is liberated and salts of the acid are formed. Partial reduction of carboxylic acids directly to aldehydes is not possible, but such conversions have been achieved in two steps by way of certain carboxyl derivatives. These will be described later.
Oxidation
Because it is already in a high oxidation state, further oxidation removes the carboxyl carbon as carbon dioxide. Depending on the reaction conditions, the oxidation state of the remaining organic structure may be higher, lower or unchanged. The following reactions are all examples of decarboxylation (loss of CO2). In the first, bromine replaces the carboxyl group, so both the carboxyl carbon atom and the remaining organic moiety are oxidized. Silver salts have also been used to initiate this transformation, which is known as the Hunsdiecker reaction. The second reaction is an interesting bis-decarboxylation, in which the atoms of the organic residue retain their original oxidation states. Lead tetraacetate will also oxidize mono-carboxylic acids in a manner similar to reaction #1. Finally, the third example illustrates the general decarboxylation of β-keto acids, which leaves the organic residue in a reduced state (note that the CO2 carbon has increased its oxidation state.).
Three additional examples of the Hunsdiecker reaction and a proposed mechanism for the transformation will be shown above by clicking on the diagram. Note that the meta- dihalobenzene formed in reaction 4 could not be made by direct halogenation reactions, since chlorine and bromine are ortho/para-directing substituents. Also, various iodide derivatives may be prepared directly from the corresponding carboxylic acids. A heavy metal carboxylate salt is transformed into an acyl hypohalide by the action of a halogen. The weak oxygen-halogen bond in this intermediate cleaves homolytically when heated or exposed to light, and the resulting carboxy radical decarboxylates to an alkyl or aryl radical. A chain reaction then repeats these events. Since acyl hypohalites are a source of electrophilic halogen, this reaction takes a different course when double bonds and reactive benzene derivatives are present. In this respect remember the addition of hypohalous reagents to double bonds and the facile bromination of anisole.
For a summary of the basic reactions of carboxylic acids Click Here
Practice Problems
The following problems review many aspects of carboxylic acid chemistry. The first two questions concern nomenclature, including some carboxylic derivatives. The third and fourth questions focus on the relative acidity of selected compounds. The fifth asks you to draw the product of a reaction selected from 48 possible combinations of carboxylic acids and reagents.
Substitution of the Hydroxyl Hydrogen
1. Contributors
This reaction class could be termed electrophilic substitution at oxygen, and is defined as follows (E is an electrophile). Some examples of this substitution are provided in equations (1) through (4).
RCO2–H + E(+) RCO2E + H(+)
If E is a strong electrophile, as in the first equation, it will attack the nucleophilic oxygen of the carboxylic acid directly, giving a positively charged intermediate which then loses a proton. If E is a weak electrophile, such as an alkyl halide, it is necessary to convert the carboxylic acid to the more nucleophilic carboxylate anion to facilitate the substitution. This is the procedure used in reactions 2 and 3. Equation 4 illustrates the use of the reagent diazomethane (CH2N2) for the preparation of methyl esters. This toxic and explosive gas is always used as an ether solution (bright yellow in color). The reaction is easily followed by the evolution of nitrogen gas and the disappearance of the reagent's color. This reaction is believed to proceed by the rapid bonding of a strong electrophile to a carboxylate anion.
The nature of SN2 reactions, as in equations 2 & 3, has been described elsewhere. The mechanisms of reactions 1 & 4 will be displayed by clicking the "Toggle Mechanism" button below the diagram.
Alkynes may also serve as electrophiles in substitution reactions of this kind, as illustrated by the synthesis of vinyl acetate from acetylene. Intramolecular carboxyl group additions to alkenes generate cyclic esters known as lactones. Five-membered (gamma) and six-membered (delta) lactones are most commonly formed. Electrophilic species such as acids or halogens are necessary initiators of lactonizations. Even the weak electrophile iodine initiates iodolactonization of γ,δ- and δ,ε-unsaturated acids. Examples of these reactions will be displayed by clicking the "Other Examples" button. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Carboxylic_Acids/Reactivity_of_Carboxylic_Acids/Reactions_of_Carboxylic_Acids/Reduction_and_Oxidation_Reactions_of_Carboxylic_Acids.txt |
Reactions in which the hydroxyl group of a carboxylic acid is replaced by another nucleophilic group are important for preparing functional derivatives of carboxylic acids. The alcohols provide a useful reference chemistry against which this class of transformations may be evaluated. In general, the hydroxyl group proved to be a poor leaving group, and virtually all alcohol reactions in which it was lost involved a prior conversion of –OH to a better leaving group. This has proven to be true for the carboxylic acids as well.
Four examples of these hydroxyl substitution reactions are presented by the following equations. In each example, the new bond to the carbonyl group is colored magenta and the nucleophilic atom that has replaced the hydroxyl oxygen is colored green. The hydroxyl moiety is often lost as water, but in reaction #1 the hydrogen is lost as HCl and the oxygen as SO2. This reaction parallels a similar transformation of alcohols to alkyl chlorides, although its mechanism is different. Other reagents that produce a similar conversion to acyl halides are PCl5 and SOBr2.
The amide and anhydride formations shown in equations #2 & 3 require strong heating, and milder procedures that accomplish these transformations will be described in the next chapter.
Reaction #4 is called esterification, since it is commonly used to convert carboxylic acids to their ester derivatives. Esters may be prepared in many different ways; indeed, equations #1 and #4 in the previous diagram illustrate the formation of tert-butyl and methyl esters respectively. The acid-catalyzed formation of ethyl acetate from acetic acid and ethanol shown here is reversible, with an equilibrium constant near 2. The reaction can be forced to completion by removing the water as it is formed. This type of esterification is often referred to as Fischer esterification. As expected, the reverse reaction, acid-catalyzed ester hydrolysis, can be carried out by adding excess water.
A thoughtful examination of this reaction (#4) leads one to question why it is classified as a hydroxyl substitution rather than a hydrogen substitution. The following equations, in which the hydroxyl oxygen atom of the carboxylic acid is colored red and that of the alcohol is colored blue, illustrate this distinction (note that the starting compounds are in the center).
H2O + CH3CO-OCH2CH3
H-substitution
CH3CO-OH + CH3CH2-OH
HO-substitution
CH3CO-OCH2CH3 + H2O
In order to classify this reaction correctly and establish a plausible mechanism, the oxygen atom of the alcohol was isotopically labeled as 18O (colored blue in our equation). Since this oxygen is found in the ester product and not the water, the hydroxyl group of the acid must have been replaced in the substitution. A mechanism for this general esterification reaction will be displayed on clicking the "Esterification Mechanism" button; also, once the mechanism diagram is displayed, a reaction coordinate for it can be seen by clicking the head of the green "energy diagram" arrow. Addition-elimination mechanisms of this kind proceed by way of tetrahedral intermediates (such as A and B in the mechanism diagram) and are common in acyl substitution reactions. Acid catalysis is necessary to increase the electrophilic character of the carboxyl carbon atom, so it will bond more rapidly to the nucleophilic oxygen of the alcohol. Base catalysis is not useful because base converts the acid to its carboxylate anion conjugate base, a species in which the electrophilic character of the carbon is reduced.
Since a tetrahedral intermediate occupies more space than a planar carbonyl group, we would expect the rate of this reaction to be retarded when bulky reactants are used. To test this prediction the esterification of acetic acid was compared with that of 2,2-dimethylpropanoic acid, (CH3)3CO2H. Here the relatively small methyl group of acetic acid is replaced by a larger tert-butyl group, and the bulkier acid reacted fifty times slower than acetic acid. Increasing the bulk of the alcohol reactant results in a similar rate reduction.
Reduction of Carboxylic Acids with LiAlH 4
This page looks at the reduction of carboxylic acids to primary alcohols using lithium tetrahydridoaluminate(III) (lithium aluminium hydride), LiAlH4. The "(III)" is the oxidation state of the aluminium. Since aluminium only ever shows the +3 oxidation state in its compounds, the "(III)" is actually unnecessary.
The reaction
Lithium tetrahydridoaluminate has the structure:
In the negative ion, one of the bonds is a co-ordinate covalent (dative covalent) bond using the lone pair on a hydride ion (H-) to form a bond with an empty orbital on the aluminium.
The reduction of a carboxylic acid
The reaction happens in two stages - first to form an aldehyde and then a primary alcohol. Because lithium tetrahydridoaluminate reacts rapidly with aldehydes, it is impossible to stop at the halfway stage. Equations for these reactions are usually written in a simplified form with the "[H]" in the equations represents hydrogen from a reducing agent. Because of the impossibility of stopping at the aldehyde, there is not much point in giving an equation for the two separate stages. The overall reaction is:
$RCOOH + 4[H] \rightarrow RCH_2OH + H_2O$
"R" is hydrogen or a hydrocarbon group. For example, ethanoic acid will reduce to the primary alcohol, ethanol.
$CH_3COOH + 4[H] \rightarrow CH_3CH_2OH + H_2O$
Sodium tetrahydridoborate (sodium borohydride) will not work! If you are familiar with the reduction of aldehydes and ketones using lithium tetrahydridoaluminate, you are probably aware that sodium tetrahydridoborate is often used as a safer alternative. It CAN'T be used with carboxylic acids. The sodium tetrahydridoborate isn't reactive enough to reduce carboxylic acids.
Reaction conditions
Lithium tetrahydridoaluminate reacts violently with water and so the reactions are carried out in solution in dry ethoxyethane (diethyl ether or just "ether"). The reaction happens at room temperature. At the end of the reaction, the product is a complex aluminum salt. This is converted into the alcohol by treatment with dilute sulfuric acid. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Carboxylic_Acids/Reactivity_of_Carboxylic_Acids/Reactions_of_Carboxylic_Acids/Substitution_of_the_hydroxyl_group.txt |
This page looks at the simple reactions of carboxylic acids as acids, including their reactions with metals, metal hydroxides, carbonates and hydrogencarbonates, ammonia and amines.
The acidity of the carboxylic acids
Using the definition of an acid as a "substance which donates protons (hydrogen ions) to other things", the carboxylic acids are acidic because of the hydrogen in the -COOH group. In solution in water, a hydrogen ion is transferred from the -COOH group to a water molecule. For example, with ethanoic acid, you get an ethanoate ion formed together with a hydroxonium ion, H3O+.
This reaction is reversible and, in the case of ethanoic acid, no more than about 1% of the acid has reacted to form ions at any one time. (This is a rough-and-ready figure and varies with the concentration of the solution.)
These are therefore weak acids.
$CH_3COOH + H_2O \rightleftharpoons CH_3COO^- + H_3O^+$
This equation is often simplified by removing the water to:
$CH_3COOH (aq) \rightleftharpoons CH_3COO^- (aq) + H^+$
However, if you are going to use this second equation, you must include state symbols. They imply that the hydrogen ion is actually attached to a water molecule.
The pH of carboxylic acid solutions
The pH depends on both the concentration of the acid and how easily it loses hydrogen ions from the -COOH group. Ethanoic acid is typical of the acids where the -COOH group is attached to a simple alkyl group. Typical lab solutions have pH's in the 2 - 3 range, depending on their concentrations.
Methanoic acid is rather stronger than the other simple acids, and solutions have pH's about 0.5 pH units less than ethanoic acid of the same concentration.
Reactions of carboxylic acids with metals
Carboxylic acids react with the more reactive metals to produce a salt and hydrogen. The reactions are just the same as with acids like hydrochloric acid, except they tend to be rather slower.
For example, dilute ethanoic acid reacts with magnesium. The magnesium reacts to produce a colorless solution of magnesium ethanoate, and hydrogen is given off. If you use magnesium ribbon, the reaction is less vigorous than the same reaction with hydrochloric acid, but with magnesium powder, both are so fast that you probably wouldn't notice much difference.
$2CH_3COOH + Mg \rightarrow (CH_3COO)_2Mg + H_2$
Reactions of carboxylic acids With metal hydroxides
These are simple neutralisation reactions and are just the same as any other reaction in which hydrogen ions from an acid react with hydroxide ions. They are most quickly and easily represented by the equation:
$H^+ (aq) + OH^- \rightarrow H_2O(l)$
If you mix dilute ethanoic acid with sodium hydroxide solution, for example, you simply get a colorless solution containing sodium ethanoate. The only sign that a change has happened is that the temperature of the mixture will have increased.
This change could well be represented by the ionic equation above, but if you want it, the full equation for this particular reaction is:
$CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O$
Reactions of carboxylic acids with carbonates and hydrogencarbonates
In both of these cases, a salt is formed together with carbon dioxide and water. Both are most easily represented by ionic equations.
For carbonates:
$2H^+ (aq) + CO_3^{2-} \rightarrow H_2O (l) + CO_2 (g)$
. . . and for hydrogencarbonates:
$H^+ (aq) + HCO_3^{-} \rightarrow H_2O (l) + CO_2 (g)$
If you pour some dilute ethanoic acid onto some white sodium carbonate or sodium hydrogencarbonate crystals, there is an immediate fizzing as carbon dioxide is produced. You end up with a colorless solution of sodium ethanoate.
With sodium carbonate, the full equation is:
$2CH_3COOH + Na_2CO_3 \rightarrow 2CH_3COONa + H_2O + CO_2$
. . . and for sodium hydrogencarbonate:
$CH_3COOH + NaHCO_3 \rightarrow CH_3COONa + H_2O + CO_2$
There is very little obvious difference in the vigor of these reactions compared with the same reactions with dilute hydrochloric acid. However, you would notice the difference if you used a slower reaction - for example with calcium carbonate in the form of a marble chip. With ethanoic acid, you would eventually produce a colorless solution of calcium ethanoate.
$2CH_3COOH + CaCO_3 \rightarrow (CH_3COO)_2Ca + H_2O + CO_2$
In this case, the marble chip would react noticeably more slowly with ethanoic acid than with hydrochloric acid.
Reactions of carboxylic acids with ammonia
Ethanoic acid reacts with ammonia in exactly the same way as any other acid does. It transfers a hydrogen ion to the lone pair on the nitrogen of the ammonia and forms an ammonium ion.
If you mix together a solution of ethanoic acid and a solution of ammonia, you will get a colorless solution of ammonium ethanoate.
$CH_3COOH + NH_3 \rightarrow CH_3COONH_4$
Reactions of carboxylic acids with amines
Amines are compounds in which one or more of the hydrogen atoms in an ammonia molecule have been replaced by a hydrocarbon group such as an alkyl group. For simplicity, we'll just look at compounds where only one of the hydrogen atoms has been replaced. These are called primary amines.
Simple primary amines include:
The small amines are very similar indeed to ammonia in many ways. For example, they smell very much like ammonia and are just as soluble in water. Because all you have done to an ammonia molecule is swap a hydrogen for an alkyl group, the lone pair is still there on the nitrogen atom.
That means that they will react with acids (including carboxylic acids) in just the same way as ammonia does. For example, ethanoic acid reacts with methylamine to produce a colorless solution of the salt methylammonium ethanoate.
However complicated the amine, because all of them have got a lone pair on the nitrogen atom, you would get the same sort of reaction.
Contributors
Jim Clark (Chemguide.co.uk)
The Decarboxylation of Carboxylic Acids and Their Salts
This page looks at the formation of hydrocarbons by the decarboxylation of the salts of carboxylic acids (and of certain acids themselves) by heating them with soda lime. It does NOT cover the decarboxylation of some acids by simply heating them.
Decarboxylation using soda lime
A carboxylic acid has the formula RCOOH where R can be hydrogen or a hydrocarbon group such as an alkyl group. The hydrocarbon group could equally well be based on a benzene ring. The sodium salt of a carboxylic acid will have the formula RCOONa. In decarboxylation, the -COOH or -COONa group is removed and replaced with a hydrogen atom.
Soda lime is manufactured by adding sodium hydroxide solution to solid calcium oxide (quicklime). It is essentially a mixture of sodium hydroxide, calcium oxide and calcium hydroxide. It comes as white granules. In equations, it is almost always written as if it were simply sodium hydroxide.
It is an easier material to handle than solid sodium hydroxide. Solid sodium hydroxide absorbs water from the atmosphere and you tend to end up with puddles of extremely concentrated (and corrosive) sodium hydroxide solution if you leave it exposed to the air. Soda lime has much less tendency to absorb water.
The reaction
The solid sodium salt of a carboxylic acid is mixed with solid soda lime, and the mixture is heated. For example, if you heat sodium ethanoate with soda lime, you get methane gas formed:
This reaction can be done with certain carboxylic acids themselves. For example, benzene can be made by heating soda lime with solid benzoic acid (benzenecarboxylic acid), C6H5COOH.
$C_6H_5COOH + 2NaOH \rightarrow C_6H_6 + Na_2CO_3 + H_2O$
You can think of this as first a reaction between the acid and the soda lime to make sodium benzoate, and then a decarboxylation as in the first example.
Contributors
Jim Clark (Chemguide.co.uk) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Carboxylic_Acids/Reactivity_of_Carboxylic_Acids/Simple_Reactions_of_Carboxylic_Acids_as_Acids.txt |
Primary alcohols and aldehydes are normally oxidized to carboxylic acids using potassium dichromate(VI) solution in the presence of dilute sulfuric acid. During the reaction, the potassium dichromate(VI) solution turns from orange to green. The potassium dichromate(VI) can just as well be replaced with sodium dichromate(VI). Because what matters is the dichromate(VI) ion, all the equations and color changes would be identical.
Primary alcohols are oxidized to carboxylic acids in two stages - first to an aldehyde and then to the acid. We often use simplified versions of these equations using "[O]" to represent oxygen from the oxidizing agent. The formation of the aldehyde is shown by the simplified equation:
"R" is a hydrogen atom or a hydrocarbon group such as an alkyl group. The aldehyde is then oxidised further to give the carboxylic acid:
If you start with an aldehyde, you are obviously just doing this second stage. Starting from the primary alcohol, you could combine these into one single equation to give:
$RCH_2OH + 2[O] \rightarrow RCOOH + H_2O$
For example, if you were converting ethanol into ethanoic acid, the simplified equation would be:
$CH_3CH_2OH + 2[O] \rightarrow CH_3COOH + H_2O$
It is possible that you might want to write proper equations for these reactions rather than these simplified ones. You can work these out from electron-half-equations. How you do this is described in detail elsewhere on the site.
The complete equation for the conversion of a primary alcohol to a carboxylic acid is:
$3RCH_2OH + 2Cr_2O_7^{2-} + 16H^+ \rightarrow 3RCOOH + 4Cr^{3+} + 11H_2O$
or if you were starting from an aldehyde is:
$3RCHO + Cr_2O_7^{2-} + 8H^+ \rightarrow 3RCOOH + 3Cr^{3+} + 4H_2O$
In the Lab
It would actually be quite uncommon to make an acid starting from an aldehyde, but very common to start from a primary alcohol. The conversion of ethanol into ethanoic acid would be a typical example. The alcohol is heated under reflux with an excess of a mixture of potassium dichromate(VI) solution and dilute sulfuric acid. Heating under reflux (heating in a flask with a condenser placed vertically in it) prevents any aldehyde formed escaping before it has time to be oxidized to the carboxylic acid.
Using an excess of oxidizing agent is to be sure that there is enough oxidizing agent present for the oxidation to go all the way to the carboxylic acid. When oxidation is complete, the mixture can be distilled. You end up with an aqueous solution of the acid.
Making Carboxylic Acids by the Hydrolysis of Nitriles
Nitriles are compounds which contain -CN attached to a hydrocarbon group. Some common examples include:
The name is based on the total number of carbons in the longest chain - including the one in the -CN group. Where you have things substituted into the chain (as in the third example), the -CN carbon counts as number 1. Nitriles are produced in two important reactions - both of which result in an increase in the length of the carbon chain because of the extra carbon in the -CN group. They are formed in the reaction between halogenoalkanes (haloalkanes or alkyl halides) and cyanide ions. For example:
$CH_3CH_2Br + CN^- \rightarrow CH_3CH_2CN + Br^-$
or during the reaction between aldehydes or ketones and hydrogen cyanide. For example, the reaction between ethanal and hydrogen cyanide to make 2-hydroxypropanenitrile is:
Converting the nitrile into a carboxylic acid
There are two ways of doing this, both of which involve reacting the carbon-nitrogen triple bond with water. This is described as hydrolysis. The two methods produce slightly different products - you just have to be careful to get this right.
Acid hydrolysis
The nitrile is heated under reflux with a dilute acid such as dilute hydrochloric acid. A carboxylic acid is formed. For example, starting from ethanenitrile you would get ethanoic acid. The ethanoic acid could be distilled off the mixture.
$CH_3CN + 2H_2O + H^+ \rightarrow CH_3COOH + NH_4^+$
Alkaline hydrolysis
The nitrile is heated under reflux with an alkali such as sodium hydroxide solution. This time you would not, of course, get a carboxylic acid produced - any acid formed would react with the sodium hydroxide present to give a salt. You also wouldn't get ammonium ions because they would react with sodium hydroxide to produce ammonia. Starting from ethanenitrile, you would therefore get a solution containing ethanoate ions (for example, sodium ethanoate if you used sodium hydroxide solution) and ammonia.
$CH_3CN + H_2O + OH^- \rightarrow CH_3COO^- + NH_3$
You have to remember to convert the ions into the free carboxylic acid, because that's what we are trying to make. To liberate the weak acid, ethanoic acid, you just have to supply hydrogen ions from a strong acid such as hydrochloric acid. You add enough hydrochloric acid to the mixture to make it acidic.
$CH_3COO^- + H^+ \rightarrow CH_3COOH$
Now you can distil off the carboxylic acid.
Preparation of carboxylic acids
The carbon atom of a carboxyl group has a high oxidation state. It is not surprising, therefore, that many of the chemical reactions used for their preparation are oxidations. Such reactions have been discussed in previous sections of this text, and the following diagram summarizes most of these:
Two other useful procedures for preparing carboxylic acids involve hydrolysis of nitriles and carboxylation of organometallic intermediates. As shown in the following diagram, both methods begin with an organic halogen compound and the carboxyl group eventually replaces the halogen. Both methods require two steps, but are complementary in that the nitrile intermediate in the first procedure is generated by a SN2 reaction, in which cyanide anion is a nucleophilic precursor of the carboxyl group. The hydrolysis may be either acid or base-catalyzed, but the latter give a carboxylate salt as the initial product.
In the second procedure the electrophilic halide is first transformed into a strongly nucleophilic metal derivative, and this adds to carbon dioxide (an electrophile). The initial product is a salt of the carboxylic acid, which must then be released by treatment with strong aqueous acid.
An existing carboxylic acid may be elongated by one methylene group, using a homologation procedure called the Arndt-Eistert reaction.
Contributors
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Carboxylic_Acids/Synthesis_of_Carboxylic_Acids/Making_Carboxylic_Acids_by_Oxidation_of_Primary_Alcohols_or_Aldehydes.txt |
Stereoisomers are isomers that differ in the spatial arrangement of atoms, rather than the order of atomic connectivity. One of the most interesting types of isomer is mirror-image stereoisomers, a non-superimposable set of two molecules that are mirror images of one another. The existence of these molecules is determined by concept known as chirality. The word "chiral" was derived from the Greek word for hand, because our hands are good example of chirality since they are non-superimposable mirror images of each other.
Chirality
To name the enantiomers of a compound unambiguously, their names must include the "handedness" of the molecule. The method for this is formally known as R/S nomenclature.
Introduction
The method of unambiguously assigning the handedness of molecules was originated by three chemists: R.S. Cahn, C. Ingold, and V. Prelog and is also often called the Cahn-Ingold-Prelog rules. In addition to the Cahn-Ingold system, there are two ways of experimentally determining the absolute configuration of an enantiomer:
1. X-ray diffraction analysis. Note that there is no correlation between the sign of rotation and the structure of a particular enantiomer.
2. Chemical correlation with a molecule whose structure has already been determined via X-ray diffraction.
However, for non-laboratory purposes, it is beneficial to focus on the R/S system. The sign of optical rotation, although different for the two enantiomers of a chiral molecule, at the same temperature, cannot be used to establish the absolute configuration of an enantiomer; this is because the sign of optical rotation for a particular enantiomer may change when the temperature changes.
Stereocenters are labeled R or S
The "right hand" and "left hand" nomenclature is used to name the enantiomers of a chiral compound. The stereocenters are labeled as R or S.
Consider the first picture: a curved arrow is drawn from the highest priority (1) substituent to the lowest priority (4) substituent. If the arrow points in a counterclockwise direction (left when leaving the 12 o' clock position), the configuration at stereocenter is considered S ("Sinister" → Latin= "left"). If, however, the arrow points clockwise,(Right when leaving the 12 o' clock position) then the stereocenter is labeled R ("Rectus" → Latin= "right"). The R or S is then added as a prefix, in parenthesis, to the name of the enantiomer of interest. For example: (R)-2-Bromobutane and (S)-2,3- Dihydroxypropanal.
Sequence rules to assign priorities to substituents
Before applying the R and S nomenclature to a stereocenter, the substituents must be prioritized according to the following rules:
Rule 1
First, examine at the atoms directly attached to the stereocenter of the compound. A substituent with a higher atomic number takes precedence over a substituent with a lower atomic number. Hydrogen is the lowest possible priority substituent, because it has the lowest atomic number.
1. When dealing with isotopes, the atom with the higher atomic mass receives higher priority.
2. When visualizing the molecule, the lowest priority substituent should always point away from the viewer (a dashed line indicates this). To understand how this works or looks, imagine that a clock and a pole. Attach the pole to the back of the clock, so that when when looking at the face of the clock the pole points away from the viewer in the same way the lowest priority substituent should point away.
3. Then, draw an arrow from the highest priority atom to the 2nd highest priority atom to the 3rd highest priority atom. Because the 4th highest priority atom is placed in the back, the arrow should appear like it is going across the face of a clock. If it is going clockwise, then it is an R-enantiomer; If it is going counterclockwise, it is an S-enantiomer.
When looking at a problem with wedges and dashes, if the lowest priority atom is not on the dashed line pointing away, the molecule must be rotated.
Remember that
• Wedges indicate coming towards the viewer.
• Dashes indicate pointing away from the viewer.
Rule 2
If there are two substituents with equal rank, proceed along the two substituent chains until there is a point of difference. First, determine which of the chains has the first connection to an atom with the highest priority (the highest atomic number). That chain has the higher priority.
If the chains are similar, proceed down the chain, until a point of difference.
For example: an ethyl substituent takes priority over a methyl substituent. At the connectivity of the stereocenter, both have a carbon atom, which are equal in rank. Going down the chains, a methyl has only has hydrogen atoms attached to it, whereas the ethyl has another carbon atom. The carbon atom on the ethyl is the first point of difference and has a higher atomic number than hydrogen; therefore the ethyl takes priority over the methyl.
Rule 3
If a chain is connected to the same kind of atom twice or three times, check to see if the atom it is connected to has a greater atomic number than any of the atoms that the competing chain is connected to.
• If none of the atoms connected to the competing chain(s) at the same point has a greater atomic number: the chain bonded to the same atom multiple times has the greater priority
• If however, one of the atoms connected to the competing chain has a higher atomic number: that chain has the higher priority.
Example 2
A 1-methylethyl substituent takes precedence over an ethyl substituent. Connected to the first carbon atom, ethyl only has one other carbon, whereas the 1-methylethyl has two carbon atoms attached to the first; this is the first point of difference. Therefore, 1-methylethyl ranks higher in priority than ethyl, as shown below:
However:
Remember that being double or triple bonded to an atom means that the atom is connected to the same atom twice. In such a case, follow the same method as above.
Caution!!
Keep in mind that priority is determined by the first point of difference along the two similar substituent chains. After the first point of difference, the rest of the chain is irrelevant.
When looking for the first point of difference on similar substituent chains, one may encounter branching. If there is branching, choose the branch that is higher in priority. If the two substituents have similar branches, rank the elements within the branches until a point of difference.
After all your substituents have been prioritized in the correct manner, you can now name/label the molecule R or S.
1. Put the lowest priority substituent in the back (dashed line).
2. Proceed from 1 to 2 to 3. (it is helpful to draw or imagine an arcing arrow that goes from 1--> 2-->3)
3. Determine if the direction from 1 to 2 to 3 clockwise or counterclockwise.
i) If it is clockwise it is R.
ii) if it is counterclockwise it is S.
USE YOUR MODELING KIT: Models assist in visualizing the structure. When using a model, make sure the lowest priority is pointing away from you. Then determine the direction from the highest priority substituent to the lowest: clockwise (R) or counterclockwise (S).
IF YOU DO NOT HAVE A MODELING KIT: remember that the dashes mean the bond is going into the screen and the wedges means that bond is coming out of the screen. If the lowest priority bond is not pointing to the back, mentally rotate it so that it is. However, it is very useful when learning organic chemistry to use models.
If you have a modeling kit use it to help you solve the following practice problems.
Exercise \(1\)
Are the following R or S?
Answer
1. S: I > Br > F > H. The lowest priority substituent, H, is already going towards the back. It turns left going from I to Br to F, so it's a S.
2. R: Br > Cl > CH3 > H. You have to switch the H and Br in order to place the H, the lowest priority, in the back. Then, going from Br to Cl, CH3 is turning to the right, giving you a R.
3. Neither R or S: This molecule is achiral. Only chiral molecules can be named R or S.
4. R: OH > CN > CH2NH2 > H. The H, the lowest priority, has to be switched to the back. Then, going from OH to CN to CH2NH2, you are turning right, giving you a R. (5)
5. S: \(\ce{-COOH}\) > \(\ce{-CH_2OH}\) > \(\ce{C#CH}\) > \(\ce{H}\). Then, going from \(\ce{-COOH}\) to \(\ce{-CH_2OH}\) to \(\ce{-C#CH}\) you are turning left, giving you a S configuration. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Chirality/Absolute_Configuration_R-S_Sequence_Rules.txt |
Stereoisomers are isomers that differ in spatial arrangement of atoms, rather than order of atomic connectivity. One of their most interesting type of isomer is the mirror-image stereoisomers, a non-superimposable set of two molecules that are mirror image of one another. The existence of these molecules are determined by concept known as chirality.
Introduction
Organic compounds, molecules created around a chain of carbon atom (more commonly known as carbon backbone), play an essential role in the chemistry of life. These molecules derive their importance from the energy they carry, mainly in a form of potential energy between atomic molecules. Since such potential force can be widely affected due to changes in atomic placement, it is important to understand the concept of an isomer, a molecule sharing same atomic make up as another but differing in structural arrangements. This article will be devoted to a specific isomers called stereoisomers and its property of chirality (Figure 1).
Figure 1: Two enantiomers of a tetrahedral complex.
The concepts of steroisomerism and chirality command great deal of importance in modern organic chemistry, as these ideas helps to understand the physical and theoretical reasons behind the formation and structures of numerous organic molecules, the main reason behind the energy embedded in these essential chemicals. In contrast to more well-known constitutional isomerism, which develops isotopic compounds simply by different atomic connectivity, stereoisomerism generally maintains equal atomic connections and orders of building blocks as well as having same numbers of atoms and types of elements.
What, then, makes stereoisomers so unique? To answer this question, the learner must be able to think and imagine in not just two-dimensional images, but also three-dimensional space. This is due to the fact that stereoisomers are isomers because their atoms are different from others in terms of spatial arrangement.
Spatial Arrangement
First and foremost, one must understand the concept of spatial arrangement in order to understand stereoisomerism and chirality. Spatial arrangement of atoms concern how different atomic particles and molecules are situated about in the space around the organic compound, namely its carbon chain. In this sense, spatial arrangement of an organic molecule are different another if an atom is shifted in any three-dimensional direction by even one degree. This opens up a very broad possibility of different molecules, each with their unique placement of atoms in three-dimensional space .
Stereoisomers
Stereoisomers are, as mentioned above, contain different types of isomers within itself, each with distinct characteristics that further separate each other as different chemical entities having different properties. Type called entaniomer are the previously-mentioned mirror-image stereoisomers, and will be explained in detail in this article. Another type, diastereomer, has different properties and will be introduced afterwards.
Enantiomers
This type of stereoisomer is the essential mirror-image, non-superimposable type of stereoisomer introduced in the beginning of the article. Figure 3 provides a perfect example; note that the gray plane in the middle demotes the mirror plane.
Figure 2: Comparison of Chiral and Achiral Molecules. (a) Bromochlorofluoromethane is a chiral molecule whose stereocenter is designated with an asterisk. Rotation of its mirror image does not generate the original structure. To superimpose the mirror images, bonds must be broken and reformed. (b) In contrast, dichlorofluoromethane and its mirror image can be rotated so they are superimposable.
Note that even if one were to flip over the left molecule over to the right, the atomic spatial arrangement will not be equal. This is equivalent to the left hand - right hand relationship, and is aptly referred to as 'handedness' in molecules. This can be somewhat counter-intuitive, so this article recommends the reader try the 'hand' example. Place both palm facing up, and hands next to each other. Now flip either side over to the other. One hand should be showing the back of the hand, while the other one is showing the palm. They are not same and non-superimposable.
This is where the concept of chirality comes in as one of the most essential and defining idea of stereoisomerism.
Chirality
Chirality essentially means 'mirror-image, non-superimposable molecules', and to say that a molecule is chiral is to say that its mirror image (it must have one) is not the same as it self. Whether a molecule is chiral or achiral depends upon a certain set of overlapping conditions. Figure 4 shows an example of two molecules, chiral and achiral, respectively. Notice the distinct characteristic of the achiral molecule: it possesses two atoms of same element. In theory and reality, if one were to create a plane that runs through the other two atoms, they will be able to create what is known as bisecting plane: The images on either side of the plan is the same as the other (Figure 4).
Figure 4.
In this case, the molecule is considered 'achiral'. In other words, to distinguish chiral molecule from an achiral molecule, one must search for the existence of the bisecting plane in a molecule. All chiral molecules are deprive of bisecting plane, whether simple or complex.
As a universal rule, no molecule with different surrounding atoms are achiral. Chirality is a simple but essential idea to support the concept of stereoisomerism, being used to explain one type of its kind. The chemical properties of the chiral molecule differs from its mirror image, and in this lies the significance of chilarity in relation to modern organic chemistry.
Compounds with Multiple Chiral Centers
We turn our attention next to molecules which have more than one stereocenter. We will start with a common four-carbon sugar called D-erythrose.
A note on sugar nomenclature: biochemists use a special system to refer to the stereochemistry of sugar molecules, employing names of historical origin in addition to the designators 'D' and 'L'. You will learn about this system if you take a biochemistry class. We will use the D/L designations here to refer to different sugars, but we won't worry about learning the system.
As you can see, D-erythrose is a chiral molecule: C2 and C3 are stereocenters, both of which have the R configuration. In addition, you should make a model to convince yourself that it is impossible to find a plane of symmetry through the molecule, regardless of the conformation. Does D-erythrose have an enantiomer? Of course it does – if it is a chiral molecule, it must. The enantiomer of erythrose is its mirror image, and is named L-erythrose (once again, you should use models to convince yourself that these mirror images of erythrose are not superimposable).
Notice that both chiral centers in L-erythrose both have the S configuration. In a pair of enantiomers, all of the chiral centers are of the opposite configuration.
What happens if we draw a stereoisomer of erythrose in which the configuration is S at C2 and R at C3? This stereoisomer, which is a sugar called D-threose, is not a mirror image of erythrose. D-threose is a diastereomer of both D-erythrose and L-erythrose.
The definition of diastereomers is simple: if two molecules are stereoisomers (same molecular formula, same connectivity, different arrangement of atoms in space) but are not enantiomers, then they are diastereomers by default. In practical terms, this means that at least one - but not all - of the chiral centers are opposite in a pair of diastereomers. By definition, two molecules that are diastereomers are not mirror images of each other.
L-threose, the enantiomer of D-threose, has the R configuration at C2 and the S configuration at C3. L-threose is a diastereomer of both erythrose enantiomers.
In general, a structure with n stereocenters will have 2n different stereoisomers. (We are not considering, for the time being, the stereochemistry of double bonds – that will come later). For example, let's consider the glucose molecule in its open-chain form (recall that many sugar molecules can exist in either an open-chain or a cyclic form). There are two enantiomers of glucose, called D-glucose and L-glucose. The D-enantiomer is the common sugar that our bodies use for energy. It has n = 4 stereocenters, so therefore there are 2n = 24 = 16 possible stereoisomers (including D-glucose itself).
In L-glucose, all of the stereocenters are inverted relative to D-glucose. That leaves 14 diastereomers of D-glucose: these are molecules in which at least one, but not all, of the stereocenters are inverted relative to D-glucose. One of these 14 diastereomers, a sugar called D-galactose, is shown above: in D-galactose, one of four stereocenters is inverted relative to D-glucose. Diastereomers which differ in only one stereocenter (out of two or more) are called epimers. D-glucose and D-galactose can therefore be refered to as epimers as well as diastereomers.
Example 3.10
Draw the structure of L-galactose, the enantiomer of D-galactose.
Solution
Example 3.11
Draw the structure of two more diastereomers of D-glucose. One should be an epimer.
Solution
Erythronolide B, a precursor to the 'macrocyclic' antibiotic erythromycin, has 10 stereocenters. It’s enantiomer is that molecule in which all 10 stereocenters are inverted.
In total, there are 210 = 1024 stereoisomers in the erythronolide B family: 1022 of these are diastereomers of the structure above, one is the enantiomer of the structure above, and the last is the structure above.
We know that enantiomers have identical physical properties and equal but opposite degrees of specific rotation. Diastereomers, in theory at least, have different physical properties – we stipulate ‘in theory’ because sometimes the physical properties of two or more diastereomers are so similar that it is very difficult to separate them. In addition, the specific rotations of diastereomers are unrelated – they could be the same sign or opposite signs, and similar in magnitude or very dissimilar.
Constitutional isomers
Constitutational isomers or structural isomers are molecules with the same chemical formula but different structures of atoms and bonds. For example, both 3-methylpentane and hexane have the same chemical formula, C6H14, yet they clearly have different structures:
3-methylpentane hexane
Another example involves functional groups. Methoxy methane, an ether, and ethanol, an alcohol, both have the chemical formula C2H6O:
Methoxy methane Ethanol
External Resources
1. Further Details of Stereochemistry: For those interested in the topic further!
2. MIT Online-Lecture including basic Organic Chemistry : Good background lecture to introduce Organic Chemistry.
3. Chirality Rap: Good way to understand the concept? Decide for yourself!
Problems
Identify the following as either a constitutional isomer or stereoisomer. If stereoisomer, determine if it is an enantiomer or diastereomer. Explain the reason behind the answer. Also mark chirality for each molecule.
1. 2. 3. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Chirality/Chirality_and_Stereoisomers.txt |
Diastereomers are stereoisomers that are not related as object and mirror image and are not enantiomers. Unlike enatiomers which are mirror images of each other and non-sumperimposable, diastereomers are not mirror images of each other and non-superimposable. Diastereomers can have different physical properties and reactivity. They have different melting points and boiling points and different densities. They have two or more stereocenters.
Introduction
It is easy to mistake between diasteromers and enantiomers. For example, we have four steroisomers of 3-bromo-2-butanol. The four possible combination are SS, RR, SR and RS (Figure 1). One of the molecule is the enantiomer of its mirror image molecule and diasteromer of each of the other two molecule (SS is enantiomer of RR and diasteromer of RS and SR). SS's mirror image is RR and they are not superimposable, so they are enantiomers. RS and SR are not mirror image of SS and are not superimposable to each other, so they are diasteromers.
Figure 1
Diastereomers vs. Enantiomers vs. Meso Compounds
Tartaric acid, C4H6O6, is an organic compound that can be found in grape, bananas, and in wine. The structures of tartaric acid itself is really interesting. Naturally, it is in the form of (R,R) stereocenters. Artificially, it can be in the meso form (R,S), which is achiral. R,R tartaric acid is enantiomer to is mirror image which is S,S tartaric acid and diasteromers to meso-tartaric acid (figure 2).
(R,R) and (S,S) tartaric acid have similar physical properties and reactivity. However, meso-tartaric acid have different physical properties and reactivity. For example, melting point of (R,R) & (S,S) tartaric is about 170 degree Celsius, and melting point of meso-tartaric acid is about 145 degree Celsius.
Figure 2
To identify meso, meso compound is superimposed on its mirror image, and has an internal plane that is symmetry (figure 3). Meso-tartaric acid is achiral and optically unactive.
Problems
Identify which of the following pair is enantiomers, diastereomers or meso compounds.
Answer
1. Diasteromers
2. Identical
3. Meso
4. Enantiomers
Fischer Projections
The Fischer Projections allow us to represent 3D molecular structures in a 2D environment without changing their properties and/or structural integrity.
Introduction
The Fischer Projection consists of both horizontal and vertical lines, where the horizontal lines represent the atoms that are pointed toward the viewer while the vertical line represents atoms that are pointed away from the viewer. The point of intersection between the horizontal and vertical lines represents the central carbon.
How to make Fischer Projections
To make a Fischer Projection, it is easier to show through examples than through words. Lets start with the first example, turning a 3D structure of ethane into a 2D fischer Projection.
Example \(1\):
Start by mentally converting a 3D structure into a Dashed-Wedged Line Structure. Remember, the atoms that are pointed toward the viewer would be designated with a wedged lines and the ones pointed away from the viewer are designated with dashed lines.
Figure A Figure B
Notice the red balls (atoms) in Figure A above are pointed away from the screen. These atoms will be designated with dashed lines like those in Figure B by number 2 and 6. The green balls (atoms) are pointed toward the screen. These atoms will be designated with wedged lines like those in Figure B by number 3 and 5. The blue atoms are in the plane of the screen so they are designated with straight lines.
Now that we have our Dashed- Wedged Line Structure, we can convert it to a Fischer Projection. However, before we can convert this Dashed-Wedged Line Structure into a Fischer Projection, we must first convert it to a “flat” Dashed-Wedged Line Structure. Then from there we can draw our Fischer Projection. Lets start with a more simpler example. Instead of using the ethane shown in Figure A and B, we will start with a methane. The reason being is that it allows us to only focus on one central carbon, which make things a little bit easier.
Figure C Figure D
Lets start with this 3D image and work our way to a dashed-wedged image. Start by imagining yourself looking directly at the central carbon from the left side as shown in Figure C. It should look something like Figure D. Now take this Figure D and flatten it out on the surface of the paper and you should get an image of a cross.
As a reminder, the horizontal line represents atoms that are coming out of the paper and the vertical line represents atoms that are going into the paper. The cross image to the right of the arrow is a Fischer projection. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Chirality/Diastereomers.txt |
Stereoisomers are isomers that differ in spatial arrangement of atoms, rather than order of atomic connectivity. One of their most interesting type of isomer is the mirror-image stereoisomers, a non-superimposable set of two molecules that are mirror image of one another. The existance of these molecules are determined by concept known as chirality. The word "chiral" was derived from the Greek word for hand, because our hands display a good example of chirality since they are non-superimposable mirror images of each other.
Introduction
The opposite of chiral is achiral. Achiral objects are superimposable with their mirror images. For example, two pieces of paper are achiral. In contrast, chiral molecules, like our hands, are non superimposable mirror images of each other.
Try to line up your left hand perfectly with your right hand, so that the palms are both facing in the same directions. Spend about a minute doing this. Do you see that they cannot line up exactly? The same thing applies to some molecules
A Chiral molecule has a mirror image that cannot line up with it perfectly- the mirror images are non superimposable. The mirror images are called enantiomers.
But why are chiral molecules so interesting? A chiral molecule and its enantiomer have the same chemical and physical properties(boiling point, melting point,polarity, density etc...). It turns out that many of our biological molecules such as our DNA, amino acids and sugars, are chiral molecules.
It is pretty interesting that our hands seem to serve the same purpose but most people are only able to use one of their hands to write. Similarily this is true with chiral biological molecules and interactions. Just like your left hand will not fit properly in your right glove, one of the enantiomers of a molecule may not work the same way in your body.
This must mean that enantiomers have properties that make them unique to their mirror images. One of these properties is that they cannot have a plane of symmetry or an internal mirror plane. So, a chiral molecule cannot be divided in two mirror image halves. Another property of chiral molecules is optical activity.
Optical Activity
As mentioned before, chiral molecules are very similar to each other since they have the same components to them. The only thing that obviously differs is their arrangement in space. As a result of this similarity, it is very hard to distinguish chiral molecules from each other when we try to compare their properties such as boiling points, melting points and densities.
However, we can differentiate them by their optical activity. When a plane-polarized light is passed through one of the 2 enantiomers of a chiral molecule that molecule rotates light in a certain direction. When the same plane polarized light is passed through the other enantiomer, that enantionmer rotates light by the same amount but in the opposite direction.
If one enantiomer rotates the light counterclockwise, the other would rotate it clockwise. Because chiral molecules are able to rotate the plane of polarization differently by interacting with the electric field differently, they are said to be optically active. In general molecules that rotate light in differen directions are called optical isomers.
• Dextrorotatory (+ enantiomer) – rotates the plane polarized light clockwise (when viewing towards the light source)
• Levorotatory (- enantiomer) – rotates the plane polarized light counterclockwise
Circular Dichroism
Another property of chiral molecules is called circular dichroism (CD). This pertains to their differential absorption of left and right circularly polarized light. When left and right circularly polarized light passes through chiral molecules, the absorption coefficients differ so that the change in absorption coefficients does not equal zero.
$\Delta A=A_L-A_R \,$
where ΔA is the difference between absorbance of left circularly polarized (LCP) and right circularly polarized (RCP) light (this is what is usually measured). Where [J] = molar concentration of the sample and l is the path length.
The CD signals of chiral molecules can give important information and this information can be used for visible and ultraviolet spectroscopy. Every chiral molecule shows a particular CD spectrum. By looking at the distinctive spectra of molecules such as proteins and DNA, we can obtain useful information about their secondary structures and see how they differ. An example of a CD spectrum comparing the differences between A-DNA and B-DNA.
Contributors
• Kristeen Pareja (UCD), Ifemayowa Aworanti (University Of Maryland Baltimore County) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Chirality/Fundamentals_of_Chirality.txt |
Meso compounds are achiral compounds that has multiple chiral centers. It is superimposed on its mirror image and is optically inactive despite its stereocenters.
Introduction
In general, a meso compound should contain two or more identical substituted stereocenters. Also, it has an internal symmetry plane that divides the compound in half. These two halves reflect each other by the internal mirror. The stereochemistry of stereocenters should "cancel out". What it means here is that when we have an internal plane that splits the compound into two symmetrical sides, the stereochemistry of both left and right side should be opposite to each other, and therefore, result in optically inactive. Cyclic compounds may also be meso.
Identification
If A is a meso compound, it should have two or more stereocenters, an internal plane, and the stereochemistry should be R and S.
1. Look for an internal plane, or internal mirror, that lies in between the compound.
2. The stereochemistry (e.g. R or S) is very crucial in determining whether it is a meso compound or not. As mentioned above, a meso compound is optically inactive, so their stereochemistry should cancel out. For instance, R cancels S out in a meso compound with two stereocenters.
trans-1,2-dichloro-1,2-ethanediol
(meso)-2,3-dibromobutane
Tips: An interesting thing about single bonds or sp3-orbitals is that we can rotate the substituted groups that attached to a stereocenter around to recognize the internal plane. As the molecule is rotated, its stereochemistry does not change. For example:
Another case is when we rotate the whole molecule by 180 degree. Both molecules below are still meso.
Remember the internal plane here is depicted on two dimensions. However, in reality, it is three dimensions, so be aware of it when we identify the internal mirror.
Example
This molecule has a plane of symmetry (the horizontal plane going through the red broken line) and, therefore, is achiral; However, it has two chiral carbons and is consequentially a meso compound.
Example 2
This molecules has a plane of symmetry (the vertical plane going through the red broken line perpendicular to the plane of the ring) and, therefore, is achiral, but has has two chiral centers. Thus, its is a meso compound.
Other Examples of meso compounds
Meso compounds can exist in many different forms such as pentane, butane, heptane, and even cyclobutane. They do not necessarily have to be two stereocenters, but can have more.
Optical Activity Analysis
When the optical activity of a meso compound is attempted to be determined with a polarimeter, the indicator will not show (+) or (-). It simply means there is no certain direction of rotation of the polarized light, neither levorotatory (-) and dexorotatory (+).
Problems
Beside meso, there are also other types of molecules: enantiomer, diastereomer, and identical. Determine if the following molecules are meso.
Answer key: A C, D, E are meso compounds.
• Duy Dang
Mixtures of Stereoisomers
Samples containing only a single stereoisomer are described as enantiomerically pure. However, many processes give mixtures of stereoisomers, at least to some extent.
Racemic Mixtures
A racemic mixture contains two enantiomers in equal amounts. As a result, a racemic mixture has no net optical activity.
Enantiomeric Excess
For non-racemic mixtures of enantiomers, one enantiomer is more abundant than the other. The composition of these mixtures is described by the enantiomeric excess, which is the difference between the relative abundance of the two enantiomers. Therefore, if a mixture contains 75% of the R enantiomer and 25% S, the enantiomeric excess if 50%. Similarly, a mixture that is 95% of one enantiomer, the enantiomeric excess is 90%, etc.
Enantiomeric excess is useful because it reflects the optical activity of the mixture. The standard optical rotation by the mixture ($[\alpha]_{mix}$) is equal to the product of the standard optical rotation of the major isomer ($[\alpha]_{major}$) and the enantiomeric excess ($EE$):
$[\alpha]_{mix} = EE \times [\alpha]_{major}$
In the same way, the enantiomeric excess in a mixture can be measured if the optical rotation of the pure enantiomer is known.
Diastereomeric Excess
A similar approach can be used to describe mixtures of diastereomers, resulting in the diastereomeric excess. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Chirality/Meso_Compounds.txt |
Optical activity is an effect of an optical isomer's interaction with plane-polarized light.
Introduction
Optical isomers, or enantiomers, have the same sequence of atoms and bonds but are different in their 3D shape. Two enantiomers are nonsuperimposible mirror images of one another (i.e., chiral), with the most common cited example being our hands. Our left hand is a mirror image of our right, yet there is no way our left thumb can be over our right thumb if our palms are facing the same way and placed over one another. Optical isomers also have no axis of symmetry, which means that there is no line that bisects the compound such that the left half is a mirror image of the right half.
Optical isomers have basically the same properties (melting points, boiling points, etc.) but there are a few exceptions (uses in biological mechanisms and optical activity). There are drugs, called enantiopure drugs, that have different effects based on whether the drug is a racemic mixture or purely one enantiomer. For example, d-ethambutol treats tuberculosis, while l-ethambutol causes blindness. Optical activity is the interaction of these enantiomers with plane-polarized light.
A Brief History
Optical activity was first observed by the French physicist Jean-Baptiste Biot. He concluded that the change in direction of plane-polarized light when it passed through certain substances was actually a rotation of light, and that it had a molecular basis. His work was supported by the experimentation of Louis Pasteur. Pasteur observed the existence of two crystals that were mirror images in tartaric acid, an acid found in wine. Through meticulous experimentation, he found that one set of molecules rotated polarized light clockwise while the other rotated light counterclockwise to the same extent. He also observed that a mixture of both, a racemic mixture (or racemic modification), did not rotate light because the optical activity of one molecule canceled the effects of the other molecule. Pasteur was the first to show the existence of chiral molecules.
Rotation of Light
An enantiomer that rotates plane-polarized light in the positive direction, or clockwise, is called dextrorotary [(+), or d-], while the enantiomer that rotates the light in the negative direction, or counterclockwise, is called levorotary [(-), or l-]. When both d- and l- isomers are present in equal amounts, the mixture is called a racemic mixture.
image source
In the picture above, you can see that unpolarized light passes through a filter so that only waves that oscillate in a certain direction can pass through. When these waves interact with an optically active material, they are rotated either clockwise or counterclockwise, depending on the enantiomer. In the case of the image above, the light is rotated clockwise so the substance is the dextrorotary enantiomer.
Measuring Optical Activity
Optical activity is measured by a polarimeter, and is dependent on several factors: concentration of the sample, temperature, length of the sample tube or cell, and wavelength of the light passing through the sample. Rotation is given in +/- degrees, depending on whether the sample has d- (positive) or l- (negative) enantiomers. The standard measurement for rotation for a specific chemical compound is called the specific rotation, defined as an angle measured at a path length of 1 decimeter and a concentration of 1g/ml. The specific rotation of a pure substance is an intrinsic property. In solution, the formula for specific rotation is:
$[\alpha]^T_\lambda = \dfrac{\alpha}{I\cdot c}$
where
• [α] is the specific rotation in degrees cm3 dm-1 g-1.
• λ is the wavelength in nanometers,
• α is the measured angle of rotation of a substance,
• T is the temperature in degrees,
• l is the path length in decimeters,
• c is the concentration in g/ml, and
Practice Questions
1. Why doesn’t a racemic mixture show optical activity?
2. Does the following image show optically active molecules (image source)?
1. What would happen if your feet were not optical isomers of each other?
2. Your friend has tuberculosis. Assuming you like your friend and you want him to live, which isomer of ethambutol do you give him and why?
3. Draw each compound's enantiomer.
Answers
1. In a racemic mixture, both dextrorotary and levorotary enantiomers are present in equal amounts, so the overall rotation of polarized light is zero. The clockwise rotation is canceled by the counterclockwise rotation.
2. Yes, because the molecules are mirror images of one another.
3. You'd be a terrible dancer. (Get it? Two left feet? Haha.)
4. Dextrorotary ethambutol, because l-ethambutol would not only fail to cure him, but it would also leave him blind.
5. Enantiomers
Contributors
• Nanki J Natt (UCD), Anna Zhu (UCD) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Chirality/Optical_Activity.txt |
As defined in an earlier introductory section, isomers are different compounds that have the same molecular formula. When the group of atoms that make up the molecules of different isomers are bonded together in fundamentally different ways, we refer to such compounds as constitutional isomers. For example, in the case of the C4H8 hydrocarbons, most of the isomers are constitutional. Shorthand structures for four of these isomers are shown below with their IUPAC names.
Note that the twelve atoms that make up these isomers are connected or bonded in very different ways. As is true for all constitutional isomers, each different compound has a different IUPAC name. Furthermore, the molecular formula provides information about some of the structural features that must be present in the isomers. Since the formula C4H8 has two fewer hydrogens than the four-carbon alkane butane (C4H10), all the isomers having this composition must incorporate either a ring or a double bond. A fifth possible isomer of formula C4H8 is CH3CH=CHCH3. This would be named 2-butene according to the IUPAC rules; however, a close inspection of this molecule indicates it has two possible structures. These isomers may be isolated as distinct compounds, having characteristic and different properties. They are shown here with the designations cis and trans.
The bonding patterns of the atoms in these two isomers are essentially equivalent, the only difference being the relative orientation or configuration of the two methyl groups (and the two associated hydrogen atoms) about the double bond. In the cis isomer the methyl groups are on the same side; whereas they are on opposite sides in the trans isomer. Isomers that differ only in the spatial orientation of their component atoms are called stereoisomers. Stereoisomers always require that an additional nomenclature prefix be added to the IUPAC name in order to indicate their spatial orientation, for example, cis (Latin, meaning on this side) and trans (Latin, meaning across) in the 2-butene case.
Stereoisomers
Configurational Stereoisomers of Alkenes
The carbon-carbon double bond is formed between two sp2 hybridized carbons, and consists of two occupied molecular orbitals, a sigma orbital and a pi orbital. Rotation of the end groups of a double bond relative to each other destroys the p-orbital overlap that creates the pi orbital or bond. Because the pi bond has a bond energy of roughly 60 kcal/mole, this resistance to rotation stabilizes the planar configuration of this functional group. As a result, certain disubstituted alkenes may exist as a pair of configurational stereoisomers, often designated cis and trans. The essential requirement for this stereoisomerism is that each carbon of the double bond must have two different substituent groups (one may be hydrogen). This is illustrated by the following general formulas. In the first example, the left-hand double bond carbon has two identical substituents (A) so stereoisomerism about the double bond is not possible (reversing substituents on the right-hand carbon gives the same configuration). In the next two examples, each double bond carbon atom has two different substituent groups and stereoisomerism exists, regardless of whether the two substituents on one carbon are the same as those on the other.
Some examples of this configurational stereoisomerism (sometimes called geometric isomerism) are shown below. Note that cycloalkenes smaller than eight carbons cannot exist in a stable trans configuration due to ring strain. A similar restriction holds against cycloalkynes smaller than ten carbons. Since alkynes are linear, there is no stereoisomerism associated with the carbon-carbon triple bond.
Nomenclature of Alkene Stereoisomers
Configurational stereoisomers of the kind shown above need an additional nomenclature prefix added to the IUPAC name, in order to specify the spatial orientations of the groups attached to the double bond. Thus far, the prefixes cis- and trans- have served to distinguish stereoisomers; however, it is not always clear which isomer should be called cis and which trans. For example, consider the two compounds on the right. Both compound A (1-bromo-1-chloropropene) and compound B ( 1-cyclobutyl-2-ethyl-3-methyl-1-butene) can exist as a pair of configurational stereoisomers (one is shown). How are we to name these stereoisomers so that the configuration of each is unambiguously specified? Assignment of a cis or trans prefix to any of these isomers can only be done in an arbitrary manner, so a more rigorous method is needed. A completely unambiguous system, based on a set of group priority rules, assigns a Z (German, zusammen for together) or E (German, entgegen for opposite) to designate the stereoisomers. In the isomers illustrated above, for which cis-trans notation was adequate, Z is equivalent to cis and E is equivalent to trans.
The Sequence Rule for Assignment of Alkene Configurations
Assign priorities to double bond substituents by looking at the atoms attached directly to the double bond carbons.
1. The higher the atomic number of the immediate substituent atom, the higher the priority.
For example, H– < C– < N– < O– < Cl–. (priority increases left to right)
(Different isotopes of the same element are assigned a priority according to their atomic mass.)
2. If two substituents have the same immediate substituent atom, move to the next atom (away from the double bond) until a difference is found.
For example, CH3– < C2H5– < ClCH2– < BrCH2– < CH3O–.
Once the relative priorities of the two substituents on each of the double bond carbons has been determined, a cis orientation of the higher priority pair is designated Z, and a trans orientation is termed E. Applying these rules to the isomers of compounds A and B shown above, we assign the configuration of the 1-bromo-1-chloropropene isomer as E (Br has higher priority than Cl, and CH3 a higher priority than H). The configuration of the 1-cyclobutyl-2-ethyl-3-methyl-1-butene isomer is determined to be Z (C4H7 has higher priority than H, and the isopropyl group has higher priority than an ethyl group). The following example elaborates the priority determination for a more complex case.
The line formula is expanded to give the structural formula in the center. The root name is heptene (the longest chain incorporating both carbons of the double bond), and the substituents (in red) are added to give the IUPAC name. In order to assign a configurational prefix the priority order of substituents at each double bond carbon must be determined. For carbon #3 the immediate substituent atoms are a chlorine and a carbon. The chlorine has a higher atomic number and therefore has higher priority (colored green and numbered 1). The more remote bromine atom does not figure in this choice. For carbon #4 the immediate substituent atoms are both carbons (colored orange). As a result, we must look at the next higher atomic number atoms in the substituent chain. These are also carbon, but the isopropyl group has two carbons (also orange) whereas the propyl group has only one. The priority order is therefore isopropyl (green) > propyl (magenta). Since the two higher priority groups (#1) are on the same side of the double bond, this configuration is (Z).
Butane Conformers
The hydrocarbon butane has a larger and more complex set of conformations associated with its constitution than does ethane. Of particular interest and importance are the conformations produced by rotation about the central carbon-carbon bond. Among these we shall focus on two staggered conformers (A & C) and two eclipsed conformers (B & D), shown below in several stereo-representations. As in the case of ethane, the staggered conformers are more stable than the eclipsed conformers by 2.8 to 4.5 kcal/mol. Since the staggered conformers represent the chief components of a butane sample they have been given the identifying prefix designations anti for A and gauche for C.
Four Conformers of Butane:
The following diagram illustrates the change in potential energy that occurs with rotation about the C2–C3 bond. The model on the right is shown in conformation D, and by clicking on any of the colored data points on the potential energy curve, it will change to the conformer corresponding to that point. The full rotation will be displayed by turning the animation on. This model may be manipulated by click-dragging the mouse for viewing from any perspective.
Summary of Conformational Stereoisomerism
1. Most conformational interconversions in simple molecules occur rapidly at room temperature. Consequently, isolation of pure conformers is usually not possible.
2. Specific conformers require special nomenclature terms such as staggered, eclipsed, gauche and anti when they are designated.
3. Specific conformers may also be designated by dihedral angles. In the butane conformers shown above, the dihedral angles formed by the two methyl groups about the central double bond are: A 180º, B 120º, C 60º & D 0º.
4. Staggered conformations about carbon-carbon single bonds are more stable (have a lower potential energy) than the corresponding eclipsed conformations. The higher energy of eclipsed bonds is known as eclipsing strain.
5. In butane the gauche-conformer is less stable than the anti-conformer by about 0.9 kcal/mol. This is due to a crowding of the two methyl groups in the gauche structure, and is called steric strain or steric hindrance.
6. Butane conformers B and C have non-identical mirror image structures in which the clockwise dihedral angles are 300º & 240º respectively. These pairs are energetically the same, and have not been distinguished in the potential energy diagram shown here.
For a more extensive discussion of rotamer analysis Click Here. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Chirality/Stereoisomers/Alkene_Stereoisomers.txt |
All objects may be classified with respect to a property we call chirality (from the Greek cheir meaning hand). A chiral object is not identical in all respects (i.e. superimposable) with its mirror image. An achiral object is identical with (superimposable on) its mirror image. Chiral objects have a "handedness", for example, golf clubs, scissors, shoes and a corkscrew. Thus, one can buy right or left-handed golf clubs and scissors. Likewise, gloves and shoes come in pairs, a right and a left. Achiral objects do not have a handedness, for example, a baseball bat (no writing or logos on it), a plain round ball, a pencil, a T-shirt and a nail. The chirality of an object is related to its symmetry, and to this end it is useful to recognize certain symmetry elements that may be associated with a given object. A symmetry element is a plane, a line or a point in or through an object, about which a rotation or reflection leaves the object in an orientation indistinguishable from the original. Some examples of symmetry elements are shown below.
The face playing card provides an example of a center or point of symmetry. Starting from such a point, a line drawn in any direction encounters the same structural features as the opposite (180º) line. Four random lines of this kind are shown in green. An example of a molecular configuration having a point of symmetry is (E)-1,2-dichloroethene. Another way of describing a point of symmetry is to note that any point in the object is reproduced by reflection through the center onto the other side. In these two cases the point of symmetry is colored magenta.
The boat conformation of cyclohexane shows an axis of symmetry (labeled C2 here) and two intersecting planes of symmetry (labeled σ). The notation for a symmetry axis is Cn, where n is an integer chosen so that rotation about the axis by 360/nº returns the object to a position indistinguishable from where it started. In this case the rotation is by 180º, so n=2. A plane of symmetry divides the object in such a way that the points on one side of the plane are equivalent to the points on the other side by reflection through the plane. In addition to the point of symmetry noted earlier, (E)-1,2-dichloroethene also has a plane of symmetry (the plane defined by the six atoms), and a C2 axis, passing through the center perpendicular to the plane. The existence of a reflective symmetry element (a point or plane of symmetry) is sufficient to assure that the object having that element is achiral. Chiral objects, therefore, do not have any reflective symmetry elements, but may have rotational symmetry axes, since these elements do not require reflection to operate. In addition to the chiral vs achiral distinction, there are two other terms often used to refer to the symmetry of an object. These are:
1. Dissymmetry: The absence of reflective symmetry elements. All dissymmetric objects are chiral.
2. Asymmetry: The absence of all symmetry elements. All asymmetric objects are chiral.
Models of some additional three-dimensional examples are provided on the interactive symmetry page. The symmetry elements of a structure provide insight concerning the structural equivalence or nonequivalence of similar component atoms or groups Examples of this symmetry analysis may be viewed by Clicking Here.
External Links
• George Hart has produced a nice treatment of symmetry in polyhedra that makes use of VRML (Click Here).
Chirality and Symmetry
Designating the Configuration of Chiral Centers
Although enantiomers may be identified by their characteristic specific rotations, the assignment of a unique configuration to each has not yet been discussed. We have referred to the mirror-image configurations of enantiomers as "right-handed" and "left-handed", but deciding which is which is not a trivial task. An early procedure assigned a D prefix to enantiomers chemically related to a right-handed reference compound and a L prefix to a similarly related left-handed group of enantiomers. Although this notation is still applied to carbohydrates and amino acids, it required chemical transformations to establish group relationships, and proved to be ambiguous in its general application. A final solution to the vexing problem of configuration assignment was devised by three European chemists: R. S. Cahn, C. K. Ingold and V. Prelog. The resulting nomenclature system is sometimes called the CIP system or the R-S system.
In the CIP system of nomenclature, each chiral center in a molecule is assigned a prefix (R or S), according to whether its configuration is right- or left-handed. No chemical reactions or interrelationship are required for this assignment. The symbol R comes from the Latin rectus for right, and S from the Latin sinister for left. The assignment of these prefixes depends on the application of two rules: The Sequence Rule and The Viewing Rule. The sequence rule is the same as that used for assigning E-Z prefixes to double bond stereoisomers. Since most of the chiral stereogenic centers we shall encounter are asymmetric carbons, all four different substituents must be ordered in this fashion.
The Sequence Rule for Assignment of Configurations to Chiral Centers
Assign sequence priorities to the four substituents by looking at the atoms attached directly to the chiral center.
1. The higher the atomic number of the immediate substituent atom, the higher the priority.
For example, H– < C– < N– < O– < Cl–. (Different isotopes of the same element are assigned a priority according to their atomic mass.)
2. If two substituents have the same immediate substituent atom,
evaluate atoms progressively further away from the chiral center until a difference is found.
For example, CH3– < C2H5– < ClCH2– < BrCH2– < CH3O–.
3. If double or triple bonded groups are encountered as substituents, they are treated as an equivalent set of single-bonded atoms.
For example, C2H5– < CH2=CH– < HC≡C–
The Viewing Rule
Once the relative priorities of the four substituents have been determined, the chiral center must be viewed from the side opposite the lowest priority group. If we number the substituent groups from 1 to 4, with 1 being the highest and 4 the lowest in priority sequence, the two enantiomeric configurations are shown in the following diagram along with a viewers eye on the side opposite substituent #4.
Remembering the geometric implication of wedge and hatched bonds, an observer (the eye) notes whether a curved arrow drawn from the # 1 position to the # 2 location and then to the # 3 position turns in a clockwise or counter-clockwise manner. If the turn is clockwise, as in the example on the right, the configuration is classified R. If it is counter-clockwise, as in the left illustration, the configuration is S. Another way of remembering the viewing rule, is to think of the asymmetric carbon as a steering wheel. The bond to the lowest priority group (# 4) is the steering column, and the other bonds are spokes on the wheel. If the wheel is turned from group # 1 toward group # 2, which in turn moves toward group # 3, this would either negotiate a right turn (R) or a left turn (S). This model is illustrated below for a right-handed turn, and the corresponding (R)-configurations of lactic acid and carvone are shown to its right. The stereogenic carbon atom is colored magenta in each case, and the sequence priorities are shown as light blue numbers. Note that if any two substituent groups on a stereogenic carbon are exchanged or switched, the configuration changes to its mirror image.
The sequence order of the substituent groups in lactic acid should be obvious, but the carvone example requires careful analysis. The hydrogen is clearly the lowest priority substituent, but the other three groups are all attached to the stereogenic carbon by bonds to carbon atoms (colored blue here). Two of the immediate substituent species are methylene groups (CH2), and the third is a doubly-bonded carbon. Rule # 3 of the sequence rules allows us to order these substituents. The carbon-carbon double bond is broken so as to give imaginary single-bonded carbon atoms (the phantom atoms are colored red in the equivalent structure). In this form the double bond assumes the priority of a 3º-alkyl group, which is greater than that of a methylene group. To establish the sequence priority of the two methylene substituents (both are part of the ring), we must move away from the chiral center until a point of difference is located. This occurs at the next carbon, which on one side is part of a carbonyl double bond (C=O), and on the other, part of a carbon-carbon double bond. Rule # 3 is again used to evaluate the two cases. The carbonyl group places two oxygens (one phantom) on the adjacent carbon atom, so this methylene side is ranked ahead of the other.
An interesting feature of the two examples shown here is that the R-configuration in both cases is levorotatory (-) in its optical activity. The mirror-image S-configurations are, of course, dextrorotatory (+). It is important to remember that there is no simple or obvious relationship between the R or S designation of a molecular configuration and the experimentally measured specific rotation of the compound it represents. In order to determine the true or "absolute" configuration of an enantiomer, as in the cases of lactic acid and carvone reported here, it is necessary either to relate the compound to a known reference structure, or to conduct a rather complex X-ray analysis on a single crystal of the sample.
The module on the right provides examples of chiral and achiral molecules for analysis. These are displayed as three-dimensional structures which may be moved about and examined from various points of view. By using this resource the reader's understanding of configurational notation may be tested.
This visualization makes use of the Jmol applet. With some browsers it may be necessary to click a button twice for action.
Select an Example
Click the Show Example Button
A three-dimensional molecular structure will be displayed here, and may be moved about with the mouse. Carbon is gray, hydrogen is cyan, oxygen is red, and nitrogen is dark blue. Other atoms are colored differently and are labeled.
Characterize the configuration of the molecule by selecting one of the three terms listed below. A response to your answer will be presented by clicking the Check Answer button.
Example 1
Example 2
Example 3
Example 4
Example 5
Example 6
Example 7
Example 8
Example 9
Example 10
Example 11
Example 12
S R Achiral
A response to your selection will appear here.
A sequence assignment will be shown above.
Configurational drawings of chiral molecules sometimes display structures in a way that does not permit an easy application of the viewing rule. In the example of carvone, shown above, the initial formula directed the lowest priority substituent (H) toward the viewer, requiring either a reorientation display or a very good sense of three-dimensional structure on the part of the reader. The Fischer projection formulas, described later, are another example of displays that challenge even experienced students. A useful mnemonic, suggested by Professor Michael Rathke, is illustrated below. Here a stereogenic tetrahedral carbon has four different substituents, designated 1, 2, 3 & 4. If we assume that these numbers represent the sequence priority of these substituents (1 > 2 > 3 > 4), then the R and S configurations are defined.
The viewing rule states that when the lowest priority substituent (4) is oriented behind the triangular face defined by the three higher priority substituents (shaded light gray here), a clockwise sequential arrangement of these substituents (1, 2 & 3) is defined as R, and a counter-clockwise sequence as S.
Now a tetrahedral structure may be viewed from any of the four triangular faces, and the symmetry of the system is such that a correct R/S assignment is made if the remote out-of plane group has an even number sequence priority (2 or 4), whereas the wrong assignment results when the out-of plane group has an odd priority (1 or 3). Once one recognizes this relationship, the viewing options are increased and a configurational assignment is more easily achieved. For an example, click on the diagram to see the 1:3:4-face, shaded light gray. oriented in front of substituent 2. Note that the R/S assignment is unchanged.
Configurational Nomenclature
Two or More Chiral Centers
The Chinese shrub Ma Huang (Ephedra vulgaris) contains two physiologically active compounds ephedrine and pseudoephedrine. Both compounds are stereoisomers of 2-methylamino-1-phenyl-1-propanol, and both are optically active, one being levorotatory and the other dextrorotatory. Since the properties of these compounds (see below) are significantly different, they cannot be enantiomers. How, then, are we to classify these isomers and others like them?
Ephedrine from Ma Huang: m.p. 35 - 40 º C, [α]D = –41º, moderate water solubility This isomer may be referred to as (–)-ephedrine
Pseudoephedrine from Ma Huang: m.p. 119 º C, [α]D = +52º, relatively insoluble in water This isomer may be referred to as (+)-pseudoephedrine
Since these two compounds are optically active, each must have an enantiomer. Although these missing stereoisomers were not present in the natural source, they have been prepared synthetically and have the expected identical physical properties and opposite-sign specific rotations with those listed above. The structural formula of 2-methylamino-1-phenylpropanol has two stereogenic carbons (#1 & #2). Each may assume an R or S configuration, so there are four stereoisomeric combinations possible. These are shown in the following illustration, together with the assignments that have been made on the basis of chemical interconversions.
As a general rule, a structure having n chiral centers will have 2n possible combinations of these centers. Depending on the overall symmetry of the molecular structure, some of these combinations may be identical, but in the absence of such identity, we would expect to find 2n stereoisomers. Some of these stereoisomers will have enantiomeric relationships, but enantiomers come in pairs, and non-enantiomeric stereoisomers will therefore be common. We refer to such stereoisomers as diastereomers. In the example above, either of the ephedrine enantiomers has a diastereomeric relationship with either of the pseudoephedrine enantiomers.
For an interesting example illustrating the distinction between a chiral center and an asymmetric carbon Click Here.
Stereogenic Nitrogen
A close examination of the ephedrine and pseudoephedrine isomers suggests that another stereogenic center, the nitrogen, is present. As noted earlier, single-bonded nitrogen is pyramidal in shape, with the non-bonding electron pair pointing to the unoccupied corner of a tetrahedral region. Since the nitrogen in these compounds is bonded to three different groups, its configuration is chiral. The non-identical mirror-image configurations are illustrated in the following diagram (the remainder of the molecule is represented by R, and the electron pair is colored yellow). If these configurations were stable, there would be four additional stereoisomers of ephedrine and pseudoephedrine. However, pyramidal nitrogen is normally not configurationally stable. It rapidly inverts its configuration (equilibrium arrows) by passing through a planar, sp2-hybridized transition state, leading to a mixture of interconverting R and S configurations. If the nitrogen atom were the only chiral center in the molecule, a 50:50 (racemic) mixture of R and S configurations would exist at equilibrium. If other chiral centers are present, as in the ephedrin isomers, a mixture of diastereomers will result. In any event, nitrogen groups such as this, if present in a compound, do not contribute to isolable stereoisomers. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Chirality/Stereoisomers/Chirality_and_Symmetry/Configurational_Nomenclature/Multiple_Stereogenic_Centers.txt |
A consideration of the chirality of molecular configurations explains the curious stereoisomerism observed for lactic acid, carvone and a multitude of other organic compounds. Tetravalent carbons have a tetrahedral configuration. If all four substituent groups are the same, as in methane or tetrachloromethane, the configuration is that of a highly symmetric "regular tetrahedron". A regular tetrahedron has six planes of symmetry and seven symmetry axes (four C3 & three C2) and is, of course, achiral. Examples of these Axes and Planes are found on George Hart's VRML site.
If one of the carbon substituents is different from the other three, the degree of symmetry is lowered to a C3 axis and three planes of symmetry, but the configuration remains achiral. The tetrahedral configuration in such compounds is no longer regular, since bond lengths and bond angles change as the bonded atoms or groups change. Further substitution may reduce the symmetry even more, but as long as two of the four substituents are the same there is always a plane of symmetry that bisects the angle linking those substituents, so these configurations are also achiral.
A carbon atom that is bonded to four different atoms or groups loses all symmetry, and is often referred to as an asymmetric carbon. The configuration of such a molecular unit is chiral, and the structure may exist in either a right-handed configuration or a left-handed configuration (one the mirror image of the other). This type of configurational stereoisomerism is termed enantiomorphism, and the non-identical, mirror-image pair of stereoisomers that result are called enantiomers. The structural formulas of lactic acid and carvone are drawn on the right with the asymmetric carbon colored red. Consequently, we expect, and find, these compounds to exist as pairs of enantiomers. The presence of a single asymmetrically substituted carbon atom in a molecule is sufficient to render the whole configuration chiral, and modern terminology refers to such asymmetric (or dissymmetric) groupings as chiral centers. Most of the chiral centers we shall discuss are asymmetric carbon atoms, but it should be recognized that other tetrahedral or pyramidal atoms may become chiral centers if appropriately substituted. When more than one chiral center is present in a molecular structure, care must be taken to analyze their relationship before concluding that a specific molecular configuration is chiral or achiral. This aspect of stereoisomerism will be treated later.
The identity or non-identity of mirror-image configurations of some substituted carbons may be examined as interactive models by .
A useful first step in examining structural formulas to determine whether stereoisomers may exist is to identify all stereogenic elements. A stereogenic element is a center, axis or plane that is a focus of stereoisomerism, such that an interchange of two groups attached to this feature leads to a stereoisomer. Stereogenic elements may be chiral or achiral. An asymmetric carbon is often a chiral stereogenic center, since interchanging any two substituent groups converts one enantiomer to the other. However, care must be taken when evaluating bridged structures in which bridgehead carbons are asymmetric. This caveat will be illustrated by Clicking Here.
Alkenes having two different groups on each double bond carbon (e.g. abC=Cab) constitute an achiral stereogenic element, since interchanging substituents at one of the carbons changes the cis/trans configuration of the double bond. Chiral stereogenic axes or planes may be present in a molecular configuration, as in the case of allenes, but these are less common than chiral centers and will not be discussed here.
For additional information about chiral axes and planes Click Here.
Structural formulas for eight organic compounds are displayed in the frame below. Some of these structures are chiral and some are achiral. First, try to identify all chiral stereogenic centers. Formulas having no chiral centers are necessarily achiral. Formulas having one chiral center are always chiral; and if two or more chiral centers are present in a given structure it is likely to be chiral, but in special cases, to be discussed later, may be achiral. Once you have made your selections of chiral centers, check them by pressing the "Show Stereogenic Centers" button. The chiral centers will be identified by red dots.
Structures F and G are achiral. The former has a plane of symmetry passing through the chlorine atom and bisecting the opposite carbon-carbon bond. The similar structure of compound E does not have such a symmetry plane, and the carbon bonded to the chlorine is a chiral center (the two ring segments connecting this carbon are not identical). Structure G is essentially flat. All the carbons except that of the methyl group are sp2 hybridized, and therefore trigonal-planar in configuration. Compounds C, D & H have more than one chiral center, and are also chiral. Remember, all chiral structures may exist as a pair of enantiomers. Other configurational stereoisomers are possible if more than one stereogenic center is present in a structure.
Enantiomorphism
The Fischer projection formula of meso-tartaric acid has a plane of symmetry bisecting the C2–C3 bond, as shown on the left in the diagram below, so this structure is clearly achiral. The eclipsed orientation of bonds that is assumed in the Fischer drawing is, however, an unstable conformation, and we should examine the staggered conformers that undoubtedly make up most of the sample molecules. The four structures that are shown to the right of the Fischer projection consist of the achiral Fischer conformation (A) and three staggered conformers, all displayed in both sawhorse and Newman projections. The second and fourth conformations (B & D) are dissymmetric, and are in fact enantiomeric structures. The third conformer (C) has a center of symmetry and is achiral.
Conformations of meso-Tartaric Acid
Fischer
Projection
A
eclipsed, achiral
B
staggered, chiral
C
staggered, achiral
D
staggered, chiral
Since a significant proportion of the meso-tartaric acid molecules in a sample will have chiral conformations, the achiral properties of the sample (e.g. optical inactivity) should not be attributed to the symmetry of the Fischer formula. Equilibria among the various conformations are rapidly established, and the proportion of each conformer present at equilibrium depends on its relative potential energy (the most stable conformers predominate). Since enantiomers have equal potential energies, they will be present in equal concentration, thus canceling their macroscopic optical activity and other chiral behavior. Simply put, any chiral species that are present are racemic.
It is interesting to note that chiral conformations are present in most conformationally mobile compounds, even in the absence of any chiral centers. The gauche conformers of butane, for example, are chiral and are present in equal concentration in any sample of this hydrocarbon. The following illustration shows the enantiomeric relationship of these conformers, which are an example of a chiral axis rather than a chiral center.
Conformational Enantiomorphism
Another class of compounds that display conformational enantiomorphism are the substituted biphenyls. As shown in the following diagram, biphenyl itself is not planar, one benzene ring being slightly twisted or canted in relation to the other as a consequence of steric crowding. This crowding will be demonstrated by clicking on the diagram. The resulting chiral conformation, having a dihedral angle of about 45º, equilibrates rapidly with its enantiomer by rotation about the connecting single bond. Note that a conformation having a 90º dihedral angle is achiral, as a consequence of a plane of symmetry.
If each of the phenyl rings of a biphenyl has two different ortho or meta substituents (one may be hydrogen), even the twisted 90º dihedral angle conformer becomes chiral. In order to interconvert such conformers with their mirror image structures, a rotation through the higher energy coplanar form must be made. The ease with which this interconversion occurs will depend on the size of the ortho substituents, since these groups must slide past each other. The 2,2'-dicarboxylic acid on the left below cannot be resolved at room temperature, since thermal (kinetic) energy is sufficient to provide the necessary activation energy for racemization. The two additionally substituted diacids to its right have a higher activation energy for racemization, and can be resolved if care is taken to avoid heating them. As a rule, an activation energy barrier of 16 to 19 kcal/mole is required to prevent spontaneous room temperature racemization of substituted biphenyls. Since fluorine is smaller than a nitro group, the center compound racemizes more rapidly on heating than does the nitro compound to its right. Conformational isomers that are isolable due to high energy barriers are called atropisomers.
The 2,2'-disulfonic acid (compound A) can be resolved with care, confirming the larger size of SO3H compared with CO2H. Compounds B and C provide additional insight into the racemization of biphenyls. Although these biphenyls have identical ortho substituents, the meta nitro substituent adjacent to the methoxyl group in C exerts a buttressing influence that increases the effective size of that ortho substituent.
Finally, by clicking on the diagram a second time two additional examples of substituted biphenyls will be shown. The left hand compound is held in a twisted conformation by the bridging carbon chain. Racemization requires passing through a planar configuration, and the increased angle and eclipsing strain in this structure contribute to a large activation energy. Consequently, this compound is easily resolved into enantiomeric stereoisomers. The right hand compound is heavily ortho-substituted and most certainly resists assuming a planar configuration. However, the right benzene ring has two identical ortho substituents, so the stable 90º dihedral angle conformer has a plane of symmetry. All chiral twisted conformers are present as racemates, so this compound cannot be resolved.
To see models of biphenyl and a chiral tetrasubstituted derivative . | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Chirality/Stereoisomers/Chirality_and_Symmetry/Enantiomorphism/Conformational_Enantiomorphism/Conformations_of_Biphenyls.txt |
Identifying and distinguishing enantiomers is inherently difficult, since their physical and chemical properties are largely identical. Fortunately, a nearly two hundred year old discovery by the French physicist Jean-Baptiste Biot has made this task much easier. This discovery disclosed that the right- and left-handed enantiomers of a chiral compound perturb plane-polarized light in opposite ways. This perturbation is unique to chiral molecules, and has been termed optical activity.
Polarimetry
Plane-polarized light is created by passing ordinary light through a polarizing device, which may be as simple as a lens taken from polarizing sun-glasses. Such devices transmit selectively only that component of a light beam having electrical and magnetic field vectors oscillating in a single plane. The plane of polarization can be determined by an instrument called a polarimeter, shown in the diagram below.
Figure \(1\): Schematic of a polarimeter showing the principles behind it's operation. Unpolarized light is passed through a polarizing filter before traveling through a sample. The degree of rotation of polarization is determined by a second, rotatable filter.
Monochromatic (single wavelength) light, is polarized by a fixed polarizer next to the light source. A sample cell holder is located in line with the light beam, followed by a movable polarizer (the analyzer) and an eyepiece through which the light intensity can be observed. In modern instruments an electronic light detector takes the place of the human eye. In the absence of a sample, the light intensity at the detector is at a maximum when the second (movable) polarizer is set parallel to the first polarizer (α = 0º). If the analyzer is turned 90º to the plane of initial polarization, all the light will be blocked from reaching the detector.
Chemists use polarimeters to investigate the influence of compounds (in the sample cell) on plane polarized light. Samples composed only of achiral molecules (e.g. water or hexane), have no effect on the polarized light beam. However, if a single enantiomer is examined (all sample molecules being right-handed, or all being left-handed), the plane of polarization is rotated in either a clockwise (positive) or counter-clockwise (negative) direction, and the analyzer must be turned an appropriate matching angle, α, if full light intensity is to reach the detector. In the above illustration, the sample has rotated the polarization plane clockwise by +90º, and the analyzer has been turned this amount to permit maximum light transmission.
The observed rotations (α) of enantiomers are opposite in direction. One enantiomer will rotate polarized light in a clockwise direction, termed dextrorotatory or (+), and its mirror-image partner in a counter-clockwise manner, termed levorotatory or (–). The prefixes dextro and levo come from the Latin dexter, meaning right, and laevus, for left, and are abbreviated d and l respectively. If equal quantities of each enantiomer are examined, using the same sample cell, then the magnitude of the rotations will be the same, with one being positive and the other negative. To be absolutely certain whether an observed rotation is positive or negative it is often necessary to make a second measurement using a different amount or concentration of the sample. In the above illustration, for example, α might be –90º or +270º rather than +90º. If the sample concentration is reduced by 10%, then the positive rotation would change to +81º (or +243º) while the negative rotation would change to –81º, and the correct α would be identified unambiguously.
Since it is not always possible to obtain or use samples of exactly the same size, the observed rotation is usually corrected to compensate for variations in sample quantity and cell length. Thus it is common practice to convert the observed rotation, α, to a specific rotation, [α], by the following formula:
Specific Rotation = where l = cell length in dm, c = concentration in g/ml
D is the 589 nm light from a sodium lamp
Compounds that rotate the plane of polarized light are termed optically active. Each enantiomer of a stereoisomeric pair is optically active and has an equal but opposite-in-sign specific rotation. Specific rotations are useful in that they are experimentally determined constants that characterize and identify pure enantiomers. For example, the lactic acid and carvone enantiomers discussed earlier have the following specific rotations.
Carvone from caraway: [α]D = +62.5º this isomer may be referred to as (+)-carvone or d-carvone
Carvone from spearmint: [α]D = –62.5º this isomer may be referred to as (–)-carvone or l-carvone
Lactic acid from muscle tissue: [α]D = +2.5º this isomer may be referred to as (+)-lactic acid or d-lactic acid
Lactic acid from sour milk: [α]D = –2.5º this isomer may be referred to as (–)-lactic acid or l-lactic acid
A 50:50 mixture of enantiomers has no observable optical activity. Such mixtures are called racemates or racemic modifications, and are designated (±). When chiral compounds are created from achiral compounds, the products are racemic unless a single enantiomer of a chiral co-reactant or catalyst is involved in the reaction. The addition of HBr to either cis- or trans-2-butene is an example of racemic product formation (the chiral center is colored red in the following equation).
CH3CH=CHCH3 + HBr (±) CH3CH2CHBrCH3
Chiral organic compounds isolated from living organisms are usually optically active, indicating that one of the enantiomers predominates (often it is the only isomer present). This is a result of the action of chiral catalysts we call enzymes, and reflects the inherently chiral nature of life itself. Chiral synthetic compounds, on the other hand, are commonly racemates, unless they have been prepared from enantiomerically pure starting materials.
There are two ways in which the condition of a chiral substance may be changed:
1. A racemate may be separated into its component enantiomers. This process is called resolution.
2. A pure enantiomer may be transformed into its racemate. This process is called racemization.
Practice Problems
Structural formulas for eight organic compounds are displayed in the frame below. Some of these structures are chiral and some are achiral. First, try to identify all chiral stereogenic centers. Formulas having no chiral centers are necessarily achiral. Formulas having one chiral center are always chiral; and if two or more chiral centers are present in a given structure it is likely to be chiral, but in special cases, to be discussed later, may be achiral.
AnswerS
Structures F and G are achiral. The former has a plane of symmetry passing through the chlorine atom and bisecting the opposite carbon-carbon bond. The similar structure of compound E does not have such a symmetry plane, and the carbon bonded to the chlorine is a chiral center (the two ring segments connecting this carbon are not identical). Structure G is essentially flat. All the carbons except that of the methyl group are sp2 hybridized, and therefore trigonal-planar in configuration. Compounds C, D & H have more than one chiral center, and are also chiral. Remember, all chiral structures may exist as a pair of enantiomers. Other configurational stereoisomers are possible if more than one stereogenic center is present in a structure.
Resolution of Racemates
As noted earlier, chiral compounds synthesized from achiral starting materials and reagents are generally racemic (i.e. a 50:50 mixture of enantiomers). Separation of racemates into their component enantiomers is a process called resolution. Since enantiomers have identical physical properties, such as solubility and melting point, resolution is extremely difficult. Diastereomers, on the other hand, have different physical properties, and this fact is used to achieve resolution of racemates. Reaction of a racemate with an enantiomerically pure chiral reagent gives a mixture of diastereomers, which can be separated. Reversing the first reaction then leads to the separated enantiomers plus the recovered reagent.
Many kinds of chemical and physical reactions, including salt formation, may be used to achieve the diastereomeric intermediates needed for separation. The following diagram illustrates this general principle by showing how a nut having a right-handed thread (R) could serve as a "reagent" to discriminate and separate a mixture of right- and left-handed bolts of identical size and weight. Only the two right-handed partners can interact to give a fully-threaded intermediate, so separation is fairly simple. The resolving moiety, i.e. the nut, is then removed, leaving the bolts separated into their right and left-handed forms. Chemical reactions of enantiomers are normally not so dramatically different, but a practical distinction is nevertheless possible.
To learn more about chemical procedures for achieving resolution Click Here. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Chirality/Stereoisomers/Chirality_and_Symmetry/Enantiomorphism/Optical_Activity.txt |
The problem of drawing three-dimensional configurations on a two-dimensional surface, such as a piece of paper, has been a long-standing concern of chemists. The wedge and hatched line notations we have been using are effective, but can be troublesome when applied to compounds having many chiral centers. As part of his Nobel Prize-winning research on carbohydrates, the great German chemist Emil Fischer, devised a simple notation that is still widely used. In a Fischer projection drawing, the four bonds to a chiral carbon make a cross with the carbon atom at the intersection of the horizontal and vertical lines. The two horizontal bonds are directed toward the viewer (forward of the stereogenic carbon). The two vertical bonds are directed behind the central carbon (away from the viewer). Since this is not the usual way in which we have viewed such structures, the following diagram shows how a stereogenic carbon positioned in the common two-bonds-in-a-plane orientation ( x–C–y define the reference plane ) is rotated into the Fischer projection orientation (the far right formula). When writing Fischer projection formulas it is important to remember these conventions. Since the vertical bonds extend away from the viewer and the horizontal bonds toward the viewer, a Fischer structure may only be turned by 180º within the plane, thus maintaining this relationship. The structure must not be flipped over or rotated by 90º.
In the above diagram, if x = CO2H, y = CH3, a = H & b = OH, the resulting formula describes (R)-(–)-lactic acid. The mirror-image formula, where x = CO2H, y = CH3, a = OH & b = H, would, of course, represent (S)-(+)-lactic acid.
Using the Fischer projection notation, the stereoisomers of 2-methylamino-1-phenylpropanol are drawn in the following manner. Note that it is customary to set the longest carbon chain as the vertical bond assembly.
The usefulness of this notation to Fischer, in his carbohydrate studies, is evident in the following diagram. There are eight stereoisomers of 2,3,4,5-tetrahydroxypentanal, a group of compounds referred to as the aldopentoses. Since there are three chiral centers in this constitution, we should expect a maximum of 23 stereoisomers. These eight stereoisomers consist of four sets of enantiomers. If the configuration at C-4 is kept constant (R in the examples shown here), the four stereoisomers that result will be diastereomers. Fischer formulas for these isomers, which Fischer designated as the "D"-family, are shown in the diagram. Each of these compounds has an enantiomer, which is a member of the "L"-family so, as expected, there are eight stereoisomers in all. Determining whether a chiral carbon is R or S may seem difficult when using Fischer projections, but it is actually quite simple. If the lowest priority group (often a hydrogen) is on a vertical bond, the configuration is given directly from the relative positions of the three higher-ranked substituents. If the lowest priority group is on a horizontal bond, the positions of the remaining groups give the wrong answer (you are in looking at the configuration from the wrong side), so you simply reverse it.
The aldopentose structures drawn above are all diastereomers. A more selective term, epimer, is used to designate diastereomers that differ in configuration at only one chiral center. Thus, ribose and arabinose are epimers at C-2, and arabinose and lyxose are epimers at C-3. However, arabinose and xylose are not epimers, since their configurations differ at both C-2 and C-3. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Chirality/Stereoisomers/Chirality_and_Symmetry/Fischer_Projections.txt |
Achiral Diastereomers (meso-Compounds)
The chiral centers in the preceding examples have all been different. In the case of 2,3-dihydroxybutanedioic acid, known as tartaric acid, the two chiral centers have the same four substituents and are equivalent. As a result, two of the four possible stereoisomers of this compound are identical due to a plane of symmetry, so there are only three stereoisomeric tartaric acids. Two of these stereoisomers are enantiomers and the third is an achiral diastereomer, called a meso compound. Meso compounds are achiral (optically inactive) diastereomers of chiral stereoisomers. Investigations of isomeric tartaric acid salts, carried out by Louis Pasteur in the mid 19th century, were instrumental in elucidating some of the subtleties of stereochemistry. Some physical properties of the isomers of tartaric acid are given in the following table.
(+)-tartaric acid: [α]D = +13º m.p. 172 ºC
(–)-tartaric acid: [α]D = –13º m.p. 172 ºC
meso-tartaric acid: [α]D = 0º m.p. 140 ºC
Fischer projection formulas provide a helpful view of the configurational relationships within the structures of these isomers. In the following illustration a mirror line is drawn between formulas that have a mirror-image relationship. In demonstrating the identity of the two meso-compound formulas, remember that a Fischer projection formula may be rotated 180º in the plane.
A model of meso-tartaric acid may be examined by
An additional example, consisting of two meso compounds, may be examined by
Other Configuration Notations
Fischer projection formulas are particularly useful for comparing configurational isomers within a family of related chiral compounds, such as the carbohydrates. However, the eclipsed conformations implied by these representations are unrealistic. When describing acyclic compounds incorporating two or more chiral centers, many chemists prefer to write zig-zag line formulas for the primary carbon chain. Here, the zig-zag carbon chain lies in a plane and the absolute or relative configurations at the chiral centers are then designated by wedge or hatched bonds to substituent groups. This is illustrated for D-(-)-ribose and the diastereoisomeric D-tetroses erythrose and threose in the following diagram.
These compounds are all chiral and only one enantiomer is drawn (the D-family member). Many times, however, we must refer to and name diastereoisomers that are racemic or achiral. For example, addition of chlorine to cis-2-butene yields a stereoisomer of 2,3-dichlorobutane different from the one obtained by chlorine addition to trans-2-butene. In cases having two adjacent chiral centers, such as this, the prefixes erythro and threo may be used to designate the relative configuration of the centers. These prefixes, taken from the names of the tetroses erythrose and threose (above), may be applied to racemic compounds, as well as pure enantiomers and meso compounds, as shown in the following diagram. In the commonly used zig-zag drawings substituents may lie on the same side of the carbon chain, a syn orientation, or on opposite sides, an anti orientation. For adjacent (vicinal) substituents this is opposite to their location in a Fischer formula. Thus, the substituents in the erythro isomer have an anti orientation, but are syn in the threo isomer.
The syn-anti nomenclature may be applied to acyclic compounds having more than two chiral centers, as illustrated by the example in the colored box. The stereogenic center nearest carbon #1 serves as a reference. At sites having two substituents, such as carbon #5, the terms refer to the relative orientation of the highest order substituent, as determined by the C.I.P. sequence rules. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Chirality/Stereoisomers/Chirality_and_Symmetry/Meso_Compounds.txt |
The symmetry of a molecule is determined by the existence of symmetry operations performed with respect to symmetry elements. A symmetry element is a line, a plane or a point in or through an object, about which a rotation or reflection leaves the object in an orientation indistinguishable from the original. A plane of symmetry is designated by the symbol σ (or sometimes s), and the reflection operation is the coincidence of atoms on one side of the plane with corresponding atoms on the other side, as though reflected in a mirror. A center or point of symmetry is labeled i, and the inversion operation demonstrates coincidence of each atom with an identical one on a line passing through and an equal distance from the inversion point (see chair cyclohexane). Finally, a rotational axis is designated Cn, where the degrees of rotation that restore the object is 360/n (C2= 180º rotation, C3= 120º rotation, C4= 90º rotation, C5= 72º rotation). C1 is called the identity operation E because it returns the original orientation.
An object having no symmetry elements other than E is called asymmetric. Such an object is necessarily chiral. Since a plane or point of symmetry involves a reflection operation, the presence of such an element makes an object achiral. One or more rotational axes of symmetry may exist in both chiral, dissymmetric, and achiral objects.
Three dimensional models illustrating these symmetry elements will be displayed on the right by clicking one of the following names. The forth and seventh of these are dissymmetric. The others are achiral.
Examples
• cis-1,2-Dichloroethene
• trans-1,2-Dichloroethene
• cis-1,2-Dimethylcyclopropane
• trans-1,2-Dimethylcyclopropane
• Cyclohexane (chair conformer)
• Cyclohexane (boat conformer)
• Cyclohexane (twist boat conformer)
• Allene
• 1,3,5,7-Tetrafluoro-1,3,5,7-Cyclooctatetraene
One more symmetry operation must be defined. Both trans-dimethylcyclopropane and 1,3,5,7-tetrafluoro-1,3,5,7-cyclooctatetraene have a C2 axis, and both lack a plane or center of symmetry. The former is chiral, but the latter is achiral because it has a S4 improper rotational axis (sometimes called an alternating axis). An improper axis, Sn, consists of a n-fold rotation followed by reflection through a mirror plane perpendicular to the rotation axis (n is always 3 or larger because S1 = σ and S2 = i). This is equivalent to saying that a n-fold rotation converts an object into its mirror image.
The S4 element in 1,3,5,7-tetrafluoro-1,3,5,7-cyclooctatetraene will be illustrated above by Clicking Here.
Symmetry Point Groups
An object may be classified with respect to its symmetry elements or lack thereof. This is done by assigning a symmetry point group, reflecting the combination of symmetry elements present in the structure. For example, bromochlorofluoromethane has no symmetry element other than C1 and is assigned to that point group. All C1 group objects are chiral. Other low symmetry point groups are Cs (only a single plane of symmetry) and Ci (only a point of symmetry). Objects in either of these point groups are achiral.
Some objects are highly symmetric and incorporate many symmetry elements. Methane is an example of a high symmetry molecule, having 4 C3 axes, 3 C2 axes and 6 σ (planes); it belongs to the tetrahedral point group Td. When combinations of rotational axes and planes are present, their relationship is designated by a v (vertical), h (horizontal) or d (diagonal). Thus, a plane containing the principle rotation axis is σv, a plane perpendicular to the principle rotation axis is σh, and a plane parallel to the principle rotation axis but bisecting the angle between two C2 axes is σd. By this notation, the six planes of the methane tetrahedron are all σd. Some of the symmetry elements of methane will be shown below by Clicking Here.
Objects of intermediate symmetry may be assigned to appropriate point groups by following the decision tree shown below. For example, trans-1,2-dichloroethene, which has a C2 axis perpendicular to its single plane of symmetry, belongs to the C2h point group. By clicking on any of the nine categories circled in light blue, further examples will be provided.
Compounds with Several Stereogenic Centers
The Chinese shrub Ma Huang (Ephedra vulgaris) contains two physiologically active compounds ephedrine and pseudoephedrine. Both compounds are stereoisomers of 2-methylamino-1-phenyl-1-propanol, and both are optically active, one being levorotatory and the other dextrorotatory. Since the properties of these compounds (see below) are significantly different, they cannot be enantiomers. How, then, are we to classify these isomers and others like them?
Ephedrine from Ma Huang: m.p. 35 - 40 º C, [α]D = –41º, moderate water solubility [this isomer may be referred to as (–)-ephedrine]
Pseudoephedrine from Ma Huang: m.p. 119 º C, [α]D = +52º, relatively insoluble in water [this isomer may be referred to as (+)-pseudoephedrine]
Since these two compounds are optically active, each must have an enantiomer. Although these missing stereoisomers were not present in the natural source, they have been prepared synthetically and have the expected identical physical properties and opposite-sign specific rotations with those listed above. The structural formula of 2-methylamino-1-phenylpropanol has two stereogenic carbons (#1 & #2). Each may assume an R or S configuration, so there are four stereoisomeric combinations possible. These are shown in the following illustration, together with the assignments that have been made on the basis of chemical interconversions.
As a general rule, a structure having n chiral centers will have 2n possible combinations of these centers. Depending on the overall symmetry of the molecular structure, some of these combinations may be identical, but in the absence of such identity, we would expect to find 2n stereoisomers. Some of these stereoisomers will have enantiomeric relationships, but enantiomers come in pairs, and non-enantiomeric stereoisomers will therefore be common. We refer to such stereoisomers as diastereomers. In the example above, either of the ephedrine enantiomers has a diastereomeric relationship with either of the pseudoephedrine enantiomers.
For an interesting example illustrating the distinction between a chiral center and an asymmetric carbon Click Here.
The configurations of ephedrine and pseudoephedrine enantiomers may be examined as interactive models by . | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Chirality/Stereoisomers/Chirality_and_Symmetry/Symmetry_and_Point_Groups.txt |
Designating the Configuration of Chiral Centers
Although enantiomers may be identified by their characteristic specific rotations, the assignment of a unique configuration to each has not yet been discussed. We have referred to the mirror-image configurations of enantiomers as "right-handed" and "left-handed", but deciding which is which is not a trivial task. An early procedure assigned a D prefix to enantiomers chemically related to a right-handed reference compound and a L prefix to a similarly related left-handed group of enantiomers. Although this notation is still applied to carbohydrates and amino acids, it required chemical transformations to establish group relationships, and proved to be ambiguous in its general application. A final solution to the vexing problem of configuration assignment was devised by three European chemists: R. S. Cahn,C. K. Ingold and V. Prelog. The resulting nomenclature system is sometimes called the CIP system or the R-S system.
In the CIP system of nomenclature, each chiral center in a molecule is assigned a prefix (R or S), according to whether its configuration is right- or left-handed. No chemical reactions or interrelationship are required for this assignment. The symbol R comes from the Latin rectus for right, and S from the Latin sinister for left. The assignment of these prefixes depends on the application of two rules: The Sequence Rule and The Viewing Rule.
The sequence rule is the same as that used for assigning E-Z prefixes to double bond stereoisomers. Since most of the chiral stereogenic centers we shall encounter are asymmetric carbons, all four different substituents must be ordered in this fashion.
The Sequence Rule for Assignment of Configurations to Chiral Centers
Assign sequence priorities to the four substituents by looking at the atoms attached directly to the chiral center.
1. The higher the atomic number of the immediate substituent atom, the higher the priority.
For example, H– < C– < N– < O– < Cl–. (Different isotopes of the same element are assigned a priority according to their atomic mass.)
2. If two substituents have the same immediate substituent atom,
evaluate atoms progressively further away from the chiral center until a difference is found.
For example, CH3– < C2H5– < ClCH2– < BrCH2– < CH3O–.
3. If double or triple bonded groups are encountered as substituents, they are treated as an equivalent set of single-bonded atoms.
For example, C2H5– < CH2=CH– < HC≡C–
The Viewing Rule
Once the relative priorities of the four substituents have been determined, the chiral center must be viewed from the side opposite the lowest priority group. If we number the substituent groups from 1 to 4, with 1 being the highest and 4 the lowest in priority sequence, the two enantiomeric configurations are shown in the following diagram along with a viewers eye on the side opposite substituent #4.
Remembering the geometric implication of wedge and hatched bonds, an observer (the eye) notes whether a curved arrow drawn from the # 1 position to the # 2 location and then to the # 3 position turns in a clockwise or counter-clockwise manner. If the turn is clockwise, as in the example on the right, the configuration is classified R. If it is counter-clockwise, as in the left illustration, the configuration is S. Another way of remembering the viewing rule, is to think of the asymmetric carbon as a steering wheel. The bond to the lowest priority group (# 4) is the steering column, and the other bonds are spokes on the wheel. If the wheel is turned from group # 1 toward group # 2, which in turn moves toward group # 3, this would either negotiate a right turn (R) or a left turn (S). This model is illustrated below for a right-handed turn, and the corresponding (R)-configurations of lactic acid and carvone are shown to its right. The stereogenic carbon atom is colored magenta in each case, and the sequence priorities are shown as light blue numbers. Note that if any two substituent groups on a stereogenic carbon are exchanged or switched, the configuration changes to its mirror image.
The sequence order of the substituent groups in lactic acid should be obvious, but the carvone example requires careful analysis. The hydrogen is clearly the lowest priority substituent, but the other three groups are all attached to the stereogenic carbon by bonds to carbon atoms (colored blue here). Two of the immediate substituent species are methylene groups (CH2), and the third is a doubly-bonded carbon. Rule # 3 of the sequence rules allows us to order these substituents. The carbon-carbon double bond is broken so as to give imaginary single-bonded carbon atoms (the phantom atoms are colored red in the equivalent structure). In this form the double bond assumes the priority of a 3º-alkyl group, which is greater than that of a methylene group. To establish the sequence priority of the two methylene substituents (both are part of the ring), we must move away from the chiral center until a point of difference is located. This occurs at the next carbon, which on one side is part of a carbonyl double bond (C=O), and on the other, part of a carbon-carbon double bond. Rule # 3 is again used to evaluate the two cases. The carbonyl group places two oxygens (one phantom) on the adjacent carbon atom, so this methylene side is ranked ahead of the other.
An interesting feature of the two examples shown here is that the R-configuration in both cases is levorotatory (-) in its optical activity. The mirror-image S-configurations are, of course, dextrorotatory (+). It is important to remember that there is no simple or obvious relationship between the R or S designation of a molecular configuration and the experimentally measured specific rotation of the compound it represents. In order to determine the true or "absolute" configuration of an enantiomer, as in the cases of lactic acid and carvone reported here, it is necessary either to relate the compound to a known reference structure, or to conduct a rather complex X-ray analysis on a single crystal of the sample.
The configurations of lactic acid and carvone enantiomers may be examined as interactive models by .
The module on the right provides examples of chiral and achiral molecules for analysis. These are displayed as three-dimensional structures which may be moved about and examined from various points of view. By using this resource the reader's understanding of configurational notation may be tested. This visualization makes use of the Jmol applet. With some browsers it may be necessary to click a button twice for action.
Select an Example
Click the Show Example Button
A three-dimensional molecular structure will be displayed here, and may be moved about with the mouse. Carbon is gray, hydrogen is cyan, oxygen is red, and nitrogen is dark blue. Other atoms are colored differently and are labeled.
Characterize the configuration of the molecule by selecting one of the three terms listed below. A response to your answer will be presented by clicking the Check Answerbutton.
Example 1
Example 2
Example 3
Example 4
Example 5
Example 6
Example 7
Example 8
Example 9
Example 10
Example 11
Example 12
S R Achiral
A response to your selection will appear here.
A sequence assignment will be shown above.
Configurational drawings of chiral molecules sometimes display structures in a way that does not permit an easy application of the viewing rule. In the example of carvone, shown above, the initial formula directed the lowest priority substituent (H) toward the viewer, requiring either a reorientation display or a very good sense of three-dimensional structure on the part of the reader. The Fischer projection formulas, described later, are another example of displays that challenge even experienced students. A useful mnemonic, suggested by Professor Michael Rathke, is illustrated below. Here a stereogenic tetrahedral carbon has four different substituents, designated 1, 2, 3 & 4. If we assume that these numbers represent the sequence priority of these substituents (1 > 2 > 3 > 4), then the R and Sconfigurations are defined.
The viewing rule states that when the lowest priority substituent (4) is oriented behind the triangular face defined by the three higher priority substituents (shaded light gray here), a clockwise sequential arrangement of these substituents (1, 2 & 3) is defined as R, and a counter-clockwise sequence as S.
Now a tetrahedral structure may be viewed from any of the four triangular faces, and the symmetry of the system is such that a correct R/S assignment is made if the remote out-of plane group has an even number sequence priority (2 or 4), whereas the wrong assignment results when the out-of plane group has an odd priority (1 or 3). Once one recognizes this relationship, the viewing options are increased and a configurational assignment is more easily achieved. For an example, click on the diagram to see the 1:3:4-face, shaded light gray. oriented in front of substituent 2. Note that the R/S assignment is unchanged. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Chirality/Stereoisomers/Configurational_Nomenclature.txt |
Structural formulas show the manner in which the atoms of a molecule are bonded together (its constitution), but do not generally describe the three-dimensional shape of a molecule, unless special bonding notations (e.g. wedge and hatched lines) are used. The importance of such three-dimensional descriptive formulas became clear in discussing configurational stereoisomerism, where the relative orientation of atoms in space is fixed by a molecule's bonding constitution (e.g. double-bonds and rings). Here too it was noted that nomenclature prefixes must be used when naming specific stereoisomers. In this section we shall extend our three-dimensional view of molecular structure to include compounds that normally assume an array of equilibrating three-dimensional spatial orientations, which together characterize the same isolable compound. We call these different spatial orientations of the atoms of a molecule that result from rotations or twisting about single bonds conformations.
In the case of hexane, we have an unbranched chain of six carbons which is often written as a linear formula: CH3CH2CH2CH2CH2CH3. We know this is not strictly true, since the carbon atoms all have a tetrahedral configuration. The actual shape of the extended chain is therefore zig-zag in nature. However, there is facile rotation about the carbon-carbon bonds, and the six-carbon chain easily coils up to assume a rather different shape. Many conformations of hexane are possible and two are illustrated below.
Extended Chain Coiled Chain
For an animation of conformational motion in hexane .
Cycloalkane Stereoisomer
Configurational Stereoisomers of Cycloalkanes
Stereoisomers are also observed in certain disubstituted (and higher substituted) cyclic compounds. Unlike the relatively flat molecules of alkenes, substituted cycloalkanes must be viewed as three-dimensional configurations in order to appreciate the spatial orientations of the substituents. By agreement, chemists use heavy, wedge-shaped bonds to indicate a substituent located above the average plane of the ring (note that cycloalkanes larger than three carbons are not planar), and a hatched line for bonds to atoms or groups located below the ring. As in the case of the 2-butene stereoisomers, disubstituted cycloalkane stereoisomers may be designated by nomenclature prefixes such as cis and trans. The stereoisomeric 1,2-dibromocyclopentanes shown to the right are an example.
In general, if any two sp3 carbons in a ring have two different substituent groups (not counting other ring atoms) stereoisomerism is possible. This is similar to the substitution pattern that gives rise to stereoisomers in alkenes; indeed, one might view a double bond as a two-membered ring. Four other examples of this kind of stereoisomerism in cyclic compounds are shown below.
If more than two ring carbons have different substituents (not counting other ring atoms) the stereochemical notation distinguishing the various isomers becomes more complex.
Examples of the IUPAC Rules in Practice
When one substituent and one hydrogen atom are attached at each of more than two positions of a monocycle, the steric relations of the substituents may be expressed by first identifying a reference substituent (labeled r) followed by a hyphen and the substituent locator number and name. The relative configuration of other substituents are then reported as cis (c) or trans (t) to the reference substituent.
When two different substituents are attached at the same position of a monocycle, then the lowest-numbered substituent named as a suffix is selected as reference group. If none of the substituents is named as a suffix, then that substituent of the pair of substituents having the lowest number, and which is preferred by the sequence rule, is chosen as the reference group. The relationship of the sequence-rule-preferred substituent at geminally substituted positions, relative to the reference group, is cited as c- or t-, as appropriate.
An alternative system which specifies the absolute configuration of substituted carbon atoms may also be used. This system, known as the Cahn-Ingold-Prelog rules, uses and elaborates the priority rules developed earlier.
Practice Problems
Three problems concerning the naming of alkene stereoisomers.
Ethane Conformers
The simple alkane ethane provides a good introduction to conformational analysis. Here there is only one carbon-carbon bond, and the rotational structures (rotamers) that it may assume fall between two extremes, staggered and eclipsed. In the following description of these conformers, several structural notations are used. The first views the ethane molecule from the side, with the carbon-carbon bond being horizontal to the viewer. The hydrogens are then located in the surrounding space by wedge (in front of the plane) and hatched (behind the plane) bonds. If this structure is rotated so that carbon #1 is canted down and brought closer to the viewer, the "sawhorse" projection is presented. Finally, if the viewer looks down the carbon-carbon bond with carbon #1 in front of #2, the Newman projection is seen.
Name of Conformer Wedge-Hatched Bond Structure Sawhorse Structure Newman Projection
Figure 1: Extreme Conformations of Ethane
As a result of bond-electron repulsions, illustrated in Figure 2, the eclipsed conformation is less stable than the staggered conformation by roughly 3 kcal / mol (eclipsing strain). The most severe repulsions in the eclipsed conformation are depicted by the red arrows. There are six other less strong repulsions that are not shown. In the staggered conformation there are six equal bond repulsions, four of which are shown by the blue arrows, and these are all substantially less severe than the three strongest eclipsed repulsions.
Figure 2: Bond Repulsions in Ethane
Consequently, the potential energy associated with the various conformations of ethane varies with the dihedral angle of the bonds, as shown below. Although the conformers of ethane are in rapid equilibrium with each other, the 3 kcal/mol energy difference leads to a substantial preponderance of staggered conformers (> 99.9%) at any given time.
Dihedral Angle
Potential Energy Profile for Ethane Conformers: The above animation illustrates the relationship between ethane's potential energy and its dihedral angle | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Chirality/Stereoisomers/Conformational_Stereoisomers.txt |
Resolution of Racemates
As noted earlier, chiral compounds synthesized from achiral starting materials and reagents are generally racemic (i.e. a 50:50 mixture of enantiomers). Separation of racemates into their component enantiomers is a process called resolution. Since enantiomers have identical physical properties, such as solubility and melting point, resolution is extremely difficult. Diastereomers, on the other hand, have different physical properties, and this fact is used to achieve resolution of racemates. Reaction of a racemate with an enantiomerically pure chiral reagent gives a mixture of diastereomers, which can be separated. Reversing the first reaction then leads to the separated enantiomers plus the recovered reagent.
Many kinds of chemical and physical reactions, including salt formation, may be used to achieve the diastereomeric intermediates needed for separation. The following diagram illustrates this general principle by showing how a nut having a right-handed thread (R) could serve as a "reagent" to discriminate and separate a mixture of right- and left-handed bolts of identical size and weight. Only the two right-handed partners can interact to give a fully-threaded intermediate, so separation is fairly simple. The resolving moiety, i.e. the nut, is then removed, leaving the bolts separated into their right and left-handed forms. Chemical reactions of enantiomers are normally not so dramatically different, but a practical distinction is nevertheless possible.
To learn more about chemical procedures for achieving resolution Click Here.
Stereogenic Nitrogen
A close examination of the ephedrine and pseudoephedrine isomers suggests that another stereogenic center, the nitrogen, is present. As noted earlier, single-bonded nitrogen is pyramidal in shape, with the non-bonding electron pair pointing to the unoccupied corner of a tetrahedral region. Since the nitrogen in these compounds is bonded to three different groups, its configuration is chiral. The non-identical mirror-image configurations are illustrated in the following diagram (the remainder of the molecule is represented by R, and the electron pair is colored yellow). If these configurations were stable, there would be four additional stereoisomers of ephedrine and pseudoephedrine. However, pyramidal nitrogen is normally not configurationally stable. It rapidly inverts its configuration (equilibrium arrows) by passing through a planar, sp2-hybridized transition state, leading to a mixture of interconverting R and S configurations. If the nitrogen atom were the only chiral center in the molecule, a 50:50 (racemic) mixture of R and S configurations would exist at equilibrium. If other chiral centers are present, as in the ephedrin isomers, a mixture of diastereomers will result. In any event, nitrogen groups such as this, if present in a compound, do not contribute to isolable stereoisomers.
Figure 1: An Example of a stereogenic nitrogen in a chiral molecule.
Contributors
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry
Stereoisomerism in Disubstituted Cyclohexanes
The distinction between configurational stereoisomers and the conformers they may assume is well-illustrated by the disubstituted cyclohexanes. The following discussion uses the various isomers of dichlorocyclohexane as examples. The 1,1-dichloro isomer is omitted because it is an unexceptional constitutional isomer of the others, and has no centers of chirality (asymmetric carbon atoms). The 1,2- and 1,3-dichlorocyclohexanes each have two centers of chirality, bearing the same set of substituents. The cis & trans-1,4-dichlorocyclohexanes do not have any chiral centers, since the two ring groups on the substituted carbons are identical.
There are three configurational isomers of 1,2-dichlorocyclohexane and three configurational isomers of 1,3-dichlorocyclohexane. These are shown in the following table.
The 1,2-Dichlorocyclohexanes The 1,3-Dichlorocyclohexanes
All the 1,2-dichloro isomers are constitutional isomers of the 1,3-dichloro isomers. In each category (1,2- & 1,3-), the (R,R)-trans isomer and the (S,S)-trans isomer are enantiomers. The cis isomer is a diastereomer of the trans isomers. Finally, all of these isomers may exist as a mixture of two (or more) conformational isomers, as shown in the table.
The chair conformer of the cis 1,2-dichloro isomer is chiral. It exists as a 50:50 mixture of enantiomeric conformations, which interconvert so rapidly they cannot be resolved (ie. separated). Since the cis isomer has two centers of chirality (asymmetric carbons) and is optically inactive, it is a meso-compound. The corresponding trans isomers also exist as rapidly interconverting chiral conformations. The diequatorial conformer predominates in each case, the (R,R) conformations being mirror images of the (S,S) conformations. All these conformations are diastereomeric with the cis conformations.
The diequatorial chair conformer of the cis 1,3-dichloro isomer is achiral. It is the major component of a fast equilibrium with the diaxial conformer, which is also achiral. This isomer is also a meso compound. The corresponding trans isomers also undergo a rapid conformational interconversion. For these isomers, however, this interconversion produces an identical conformer, so each enantiomer (R,R) and (S,S) has predominately a single chiral conformation. These enantiomeric conformations are diastereomeric with the cis conformations.
The 1,4-dichlorocyclohexanes may exist as cis or trans stereoisomers. Both are achiral, since the disubstituted six-membered ring has a plane of symmetry. These isomers are diastereomers of each other, and are constitutional isomers of the 1,2- and 1,3- isomers.
The 1,4-Dichlorocyclohexanes
All the chair conformers of these isomers are achiral, and the diequatorial conformer of the trans isomer is the predominate species at equilibrium. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Chirality/Stereoisomers/Resolution.txt |
Ring Conformations
Although the customary line drawings of simple cycloalkanes are geometrical polygons, the actual shape of these compounds in most cases is very different.
Cyclopropane is necessarily planar (flat), with the carbon atoms at the corners of an equilateral triangle. The 60º bond angles are much smaller than the optimum 109.5º angles of a normal tetrahedral carbon atom, and the resulting angle strain dramatically influences the chemical behavior of this cycloalkane. Cyclopropane also suffers substantial eclipsing strain, since all the carbon-carbon bonds are fully eclipsed. Cyclobutane reduces some bond-eclipsing strain by folding (the out-of-plane dihedral angle is about 25º), but the total eclipsing and angle strain remains high. Cyclopentane has very little angle strain (the angles of a pentagon are 108º), but its eclipsing strain would be large (about 10 kcal/mol) if it remained planar. Consequently, the five-membered ring adopts non-planar puckered conformations whenever possible. Rings larger than cyclopentane would have angle strain if they were planar. However, this strain, together with the eclipsing strain inherent in a planar structure, can be relieved by puckering the ring. Cyclohexane is a good example of a carbocyclic system that virtually eliminates eclipsing and angle strain by adopting non-planar conformations, such as those shown below. Cycloheptane and cyclooctane have greater strain than cyclohexane, in large part due to transannular crowding (steric hindrance by groups on opposite sides of the ring).
Some Conformations of Cyclohexane Rings
A planar structure for cyclohexane is clearly improbable. The bond angles would necessarily be 120º, 10.5º larger than the ideal tetrahedral angle. Also, every carbon-carbon bond in such a structure would be eclipsed. The resulting angle and eclipsing strains would severely destabilize this structure. If two carbon atoms on opposite sides of the six-membered ring are lifted out of the plane of the ring, much of the angle strain can be eliminated. This boat structure still has two eclipsed bonds (colored magenta in the drawing) and severe steric crowding of two hydrogen atoms on the "bow" and "stern" of the boat. Thissteric crowding is often called steric hindrance. By twisting the boat conformation, the steric hindrance can be partially relieved, but the twist-boat conformer still retains some of the strains that characterize the boat conformer. Finally, by lifting one carbon above the ring plane and the other below the plane, a relatively strain-free chair conformer is formed. This is the predominant structure adopted by molecules of cyclohexane.
An energy diagram for these conformational interconversions is drawn below. The activation energy for the chair-chair conversion is due chiefly to a high energy twist-chair form (TC), in which significant angle and eclipsing strain are present. A facile twist-boat (TB)-boat (B) equilibrium intervenes as one chair conformer (C) changes to the other.
Conformational Energy Profile of Cyclohexane
TC = twist chair
B = boat
TB = twist boat
C = chair
These conformations may be examined as interactive models by .
Investigations concerning the conformations of cyclohexane were initiated by H. Sachse (1890) and E. Mohr (1918), but it was not until 1950 that a full treatment of the manifold consequences of interconverting chair conformers and the different orientations of pendent bonds was elucidated by D. H. R. Barton (Nobel Prize 1969 together with O. Hassel). The following discussion presents some of the essential features of this conformational analysis.
On careful examination of a chair conformation of cyclohexane, we find that the twelve hydrogens are not structurally equivalent. Six of them are located about the periphery of the carbon ring, and are termedequatorial. The other six are oriented above and below the approximate plane of the ring (three in each location), and are termed axial because they are aligned parallel to the symmetry axis of the ring. In the stick model shown on the left below, the equatorial hydrogens are colored blue, and the axial hydrogens are red. Since there are two equivalent chair conformations of cyclohexane in rapid equilibrium, all twelve hydrogens have 50% equatorial and 50% axial character.
Because axial bonds are parallel to each other, substituents larger than hydrogen generally suffer greater steric crowding when they are oriented axial rather than equatorial. Consequently, substituted cyclohexanes will preferentially adopt conformations in which large substituents assume equatorial orientation. In the two methylcyclohexane conformers shown above, the methyl carbon is colored blue. When the methyl group occupies an axial position it suffers steric crowding by the two axial hydrogens located on the same side of the ring. This crowding or steric hindrance is associated with the red-colored hydrogens in the structure. A careful examination of the axial conformer shows that this steric hindrance is due to two gauche-like orientations of the methyl group with ring carbons #3 and #5. The use of models is particularly helpful in recognizing and evaluating these relationships.
These conformations may be examined as interactive models by
To view an animation of the interconversion of cyclohexane chair conformers
The relative steric hindrance experienced by different substituent groups oriented in an axial versus equatorial location on cyclohexane may be determined by the conformational equilibrium of the compound. The corresponding equilibrium constant is related to the energy difference between the conformers, and collecting such data allows us to evaluate the relative tendency of substituents to exist in an equatorial or axial location. A table of these free energy values (sometimes referred to as A values) may be examined by .
Clearly the apparent "size" of a substituent is influenced by its width and bond length to cyclohexane, as evidenced by the fact that an axial vinyl group is less hindered than ethyl, and iodine slightly less than chlorine. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Chirality/Stereoisomers/Stereoisomers%3A_Ring_Conformations.txt |
Because it is so common among natural and synthetic compounds, and because its conformational features are rather well understood, we shall focus on the six-membered cyclohexane ring in this discussion. In a sample of cyclohexane, the two identical chair conformers are present in equal concentration, and the hydrogens are all equivalent (50% equatorial & 50% axial) due to rapid interconversion of the conformers. When the cyclohexane ring bears a substituent, the two chair conformers are not the same. In one conformer the substituent is axial, in the other it is equatorial. Due to steric hindrance in the axial location, substituent groups prefer to be equatorial and that chair conformer predominates in the equilibrium.
We noted earlier that cycloalkanes having two or more substituents on different ring carbon atoms exist as a pair (sometimes more) of configurational stereoisomers. Now we must examine the way in which favorable ring conformations influence the properties of the configurational isomers. Remember, configurational stereoisomers are stable and do not easily interconvert, whereas, conformational isomers normally interconvert rapidly. In examining possible structures for substituted cyclohexanes, it is useful to follow two principles.
1. Chair conformations are generally more stable than other possibilities.
2. Substituents on chair conformers prefer to occupy equatorial positions due to the increased steric hindrance of axial locations.
The following equations and formulas illustrate how the presence of two or more substituents on a cyclohexane ring perturbs the interconversion of the two chair conformers in ways that can be predicted.
Conformational Structures of Disubstituted Cyclohexanes
1,1-dimethylcyclohexane
1-t-butyl-1-methylcyclohexane
cis-1,2-dimethylcyclohexane
trans-1,2-dimethylcyclohexane
cis-1,3-dimethylcyclohexane
trans-1,3-dimethylcyclohexane
cis-1,4-dimethylcyclohexane
trans-1,4-dimethylcyclohexane
In the case of 1,1-disubstituted cyclohexanes, one of the substituents must necessarily be axial and the other equatorial, regardless of which chair conformer is considered. Since the substituents are the same in 1,1-dimethylcyclohexane, the two conformers are identical and present in equal concentration. In 1-t-butyl-1-methylcyclohexane the t-butyl group is much larger than the methyl, and that chair conformer in which the larger group is equatorial will be favored in the equilibrium( > 99%). Consequently, the methyl group in this compound is almost exclusively axial in its orientation.
In the cases of 1,2-, 1,3- and 1,4-disubstituted compounds the analysis is a bit more complex. It is always possible to have both groups equatorial, but whether this requires a cis-relationship or a trans-relationship depends on the relative location of the substituents. As we count around the ring from carbon #1 to #6, the uppermost bond on each carbon changes its orientation from equatorial (or axial) to axial (or equatorial) and back. It is important to remember that the bonds on a given side of a chair ring-conformation always alternate in this fashion. Therefore, it should be clear that for cis-1,2-disubstitution, one of the substituents must be equatorial and the other axial; in the trans-isomer both may be equatorial. Because of the alternating nature of equatorial and axial bonds, the opposite relationship is true for 1,3-disubstitution (cis is all equatorial, trans is equatorial/axial). Finally, 1,4-disubstitution reverts to the 1,2-pattern.
The conformations of some substituted cyclohexanes may be examined as interactive models by .
For additional information about six-membered ring conformations Click Here. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Chirality/Stereoisomers/Substituted_Cyclohexanes.txt |
Like nonconjugated dienes, conjugated dienes are subject to attack by electrophiles. In fact, conjugated electrophiles experience relatively greater kinetic reactivity when reacted with electrophiles than nonconjugated dienes do. Upon electrophilic addition, the conjugated diene forms a mixture of two products—the kinetic product and the thermodynamic product—whose ratio is determined by the conditions of reaction. A reaction yielding more thermodynamic product is under thermodynamic control, and likewise, a reaction that yields more kinetic product is under kinetic control.
The reaction of one equivalent of hydrogen bromide with 1,3-butadiene gives different products at under different conditions
This is a classic example of the concept of thermodynamic versus kinetic control of a reaction. Take a look at this energy profile diagram in Figure 1.
Figure 1: Energy profile diagram for $A \rightarrow B$ (left) and $A \rightarrow C$ (rght). The horizontal axis is a reaction coordinate, and the vertical axis represents Gibbs energy. IB and IC are intermediates in the two reactions and TB1, TB2, TC1, and TC2 are transitions states .
In this scenario, the starting material $\ce{A}$ can react to form either $\ce{B}$ (to the left) or $\ce{C}$ (to the right). The formation of the product $\ce{C}$ involves evolution over barrier (assuming a two step process) with lower activation energy, which means that it will form faster (ignoring the pre-exponential constant effects):
$E_{\mathrm{a},\ce{AB}} > E_{\mathrm{a},\ce{AC}} \qquad \Longrightarrow \qquad k_\ce{AB} < k_\ce{AC}$
If we keep the temperature sufficiently low, the molecules of $\ce{C}$, which are inevitably formed faster, will probably not have enough energy to overcome the reverse activation barrier (i.e., $\ce{C} \rightarrow \ce{A}$) to regenerate $\ce{A}$. The forward reactions $\ce{A->B}$ and $\ce{A->C}$ are, under such conditions, effectively irreversible. Since the formation of $\ce{C}$ is faster, it will predominate, and the major product formed will be $\ce{C}$. This is known as kinetic control and $\ce{C}$ is the kinetic product.
At elevated temperatures, $\ce{C}$ is still going to be the product that is formed faster. However, it also means that all the reactions will be reversible. This means that molecules of $\ce{C}$ can revert back to $\ce{A}$. Since the system is no longer limited by temperature, the system will minimize its Gibbs free energy, which is the thermodynamic criterion for chemical equilibrium. This means that, as the most thermodynamically stable molecule, $\ce{B}$ will be predominantly formed.2 The reaction is said to be under thermodynamic control and $\ce{B}$ is the thermodynamic product.
A simple definition is that the kinetic product is the product that is formed faster, and the thermodynamic product is the product that is more stable. This is precisely what is happening here. The kinetic product is 3-bromobut-1-ene, and the thermodynamic product is 1-bromobut-2-ene (specifically, the trans isomer).
A Warning: Not every reaction has different thermodynamic and kinetic products!
Note that not every reaction has an energy profile diagram like Figure 1, and not every reaction has different thermodynamic and kinetic products! If the transition states leading to the formation of $\ce{C}$ (e.g., TC1, and TC2) were to be higher in energy than that leading to $\ce{B}$ (e.g., TB1, and TB2), then $\ce{B}$ would simultaneously be both the thermodynamic and kinetic product. There are plenty of reactions in which the more stable product (thermodynamic) is also formed faster (kinetic).
The Reaction Mechanism
The first step is the protonation of one of the $\ce{C=C}$ double bonds. In butadiene (1), both double bonds are the same, so it does not matter which one you protonate. The protonation occurs regioselectively to give the more stable carbocation:
The more stable cation is not only secondary, but also allylic, and therefore enjoys stabilization via resonance (or conjugation). This is depicted in the resonance forms 2a and 2b above. This allylic carbocation, more properly denoted as the resonance hybrid 2, has two carbons which have significant positive charge, and the bromide ion (here denoted as $\ce{X-}$) can attack either carbon. Attacking the central carbon, adjacent to the site of protonation, leads to the kinetic product 3 (called the 1,2-adduct); attacking the terminal carbon, distant from the site of protonation, leads to the thermodynamic product 4 (called the 1,4-adduct).
A Common Mistake: Resonance Structures do not Independently Exist
There are some people who write that 3 results from attack of $\ce{X-}$ on resonance form 2a, and 4 from attack of $\ce{X-}$ on resonance form 2b. This is not correct! Resonance forms do not separately exist, and they are not distinct species that rapidly interconvert. As such, one cannot speak of one single resonance form undergoing a reaction.
Now, why 4 is the thermodynamic product, and why 3 is the kinetic product for this reaction?
The thermodynamic product: trans-1-bromobut-2-ene
It is perhaps simple enough to see why 4 is more stable than 3. It has an internal, disubstituted double bond, and we know that as a general rule of thumb, the thermodynamic stability of an alkene increases with increasing substitution. So, compared to the terminal, monosubstituted alkene 3, 4 is more stable.
Both the trans isomer 4 as well as the cis isomer 5 can be formed via attack of the nucleophile at the terminal carbon, and both are disubstituted alkenes. However, the trans isomer 4 is more stable than the cis isomer 5, because there is less steric repulsion between the two substituents on the double bond. As such, 4 is the thermodynamic product.
The kinetic product: 3-bromobut-1-ene
Several explanations may be proposed to explain the nature of the kinetic product.
The worst possible argument argues that the resonance form 2a, being an allylic secondary carbocation, is more stable than resonance form 2b, which is an allylic primary carbocation. Therefore, resonance form 2a exists in greater relative proportion (i.e., more molecules will look like 2a than 2b), and the nucleophile preferentially reacts with this specific carbocation, leading to the formation of 3. However, this is incorrect, since individual resonance forms do not exist. Moreover, such an argument suggests that we are looking for the more stable intermediate (IB or IC in Figure 1). In fact, we should be looking for the more stable transition states (TB1, TB2, TC1, and TC2 in Figure 1). The carbocation is an intermediate, and not a transition state.
The most common argument is since resonance form 2a is more stable than 2b, is that it contributes more towards the resonance hybrid 2. As such, the positive charge on the internal carbon is greater than the positive charge on the terminal carbon. The nucleophile, being negatively charged, is more strongly attracted to the more positively charged or more electrophilic carbon, and therefore attack there occurs faster (the transition state being stabilized by greater electrostatic interactions). That's actually a very sensible explanation; with only the data that has been presented so far, we would not be able to disprove it, and it was indeed the accepted answer for quite a while.
Experimental Results
In 1979, Nordlander et al. carried out a similar investigation on the addition of $\ce{DCl}$ to a different substrate, 1,3-pentadiene.7 This experiment was ingenious, because it was designed to proceed via an almost symmetrical intermediate:
Resonance forms 7a and 7b are both allylic and secondary. There is a very minor difference in their stabilities arising from the different hyperconjugative ability of $\ce{C-D}$ vs $\ce{C-H}$ bonds, but in any case, it is not very large. Therefore, if we adopt the explanation in the previous section, one would expect there not to be any major kinetic pathway, and both 1,2- and 1,4-addition products (8 and 9) would theoretically be formed roughly equally.
Instead, it was found that the 1,2-addition product was favored over the 1,4-addition product. For example, at $-78\ ^\circ\mathrm{C}$ in the absence of solvent, there was a roughly $75:25$ ratio of 1,2- to 1,4-addition products. Clearly, there is a factor that favors 1,2-addition that does not depend on the electrophilicity of the carbon being attacked! The authors attributed this effect to an ion pair mechanism. This means that, after the double bond is protonated (deuterated in this case), the chloride counterion remains in close proximity to the carbocation generated. Immediately following dissociation of $\ce{DCl}$, the chloride ion is going to be much closer to $\ce{C-2}$ than it is to $\ce{C-4}$, and therefore attack at $\ce{C-2}$ is much faster. In fact, normal electrophilic addition of $\ce{HX}$ to conjugated alkenes in polar solvents can also proceed via similar ion pair mechanisms. This is reflected by the greater proportion of syn addition products to such substrates.11
The mechanism that favors 1,2-addition clearly does not depend on the electrophilicity of the carbon being attacked. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Conjugation/Electrophilic_Attack_on_Conjugated_Dienes_-_Kinetic_and_Thermodynamic_Control.txt |
Esters are known for their distinctive odors and are commonly used for food aroma and fragrances. The general formula of an ester is RCOOR'.
Introduction
Esters are formed through reactions between an acid and an alcohol with the elimination of water. An example of this is the reaction of acetic acid with an alcohol, which yields an acetic ester and water.
The part enclosed by the red circle represents the ethyl group from the alcohol and the part enclosed by the green rectangle represents the acetate group from the acid.
Esters can be named using a few steps
Esters are named as if the alkyl chain from the alcohol is a substituent. No number is assigned to this alkyl chain. This is followed by the name of the parent chain from the carboxylic acid part of the ester with an –e remove and replaced with the ending –oate.
Example
Esters are formed through reactions between an acid and an alcohol with the elimination of water. An example
1. First, identify the oxygen that is part of the continuous chain and bonded to carbon on both sides. (On one side of this oxygen there will be a carbonyl present but on the other side there won't be.)
2. Second, begin numbering the carbon chains on either side of the oxygen identified in step 1.
3. Next, use this format: [alkyl on side further from the carbonyl] (space) [alkane on the side with the carbonyl] - (In this case: [methyl] [methane])
4. Finally, change the ending of the alkane on the same side as the carbonyl from -e to -oate. (In this case: methyl methanoate)
When an ester group is attached to a ring, the ester is named as a substituent on the ring.
Benzenecarboxylic acid (Benzoic acid)
Other substituents that exist on either side of the ester are named in the same way as they are on regular alkane chains. The only thing you must make sure of is placing the substituent name on the part of the name that corresponds to the side of the ester that it is on.
CH3COOCH2CH3
methyl propanoate
CH3COOCH2CH2CH2CH2CH2CH2CH2CH3
Problems
Name the following:
Answer: propyl ethanoate
Answer: 3-bromopentyl 2-chlorobutanoate
Answer: ethyl hexanoate
Answer: ethyl 3-bromopentanoate
Answer: 4-nitrobenzenecarboxylic acid or
4-nitrobenzoic acid
Properties of Esters
Esters are derived from carboxylic acids. A carboxylic acid contains the -COOH group, and in an ester the hydrogen in this group is replaced by a hydrocarbon group. This could be an alkyl group like methyl or ethyl, or one containing a benzene ring such as a phenyl or benzyl group. The most commonly discussed ester is ethyl ethanoate. In this case, the hydrogen in the -COOH group has been replaced by an ethyl group. The formula for ethyl ethanoate is:
Notice that the ester is named the opposite way around from the way the formula is written. The "ethanoate" bit comes from ethanoic acid. The "ethyl" bit comes from the ethyl group replacing the hydrogen.
Example
In each case, be sure that you can see how the names and formulae relate to each other.
Notice that the acid is named by counting up the total number of carbon atoms in the chain - including the carbonyl carbon. So, for example, \(CH_3CH_2COOH\) is propanoic acid, and \(CH_3CH_2COO\) is the propanoate group.
Fats and oils
Animal and vegetable fats and oils are composed of long-chain, complicated esters. The physical differences observed between a fat (like butter) and an oil (like sunflower oil) are due to differences in melting points of the mixture of esters they contain. If the melting point of the substance is below room temperature, it will be a liquid - an oil. If the melting point is above room temperature, it will be a solid - a fat. The causes of the differences in melting points are discussed below.
Fats and oils as big esters
Esters can be made from carboxylic acids and alcohols. This is discussed in detail on another page; in general terms, the two combine together, losing a molecule of water in the process. Consider a very simple ester such as ethyl ethanoate. The figure below shows its formation from ethanoic acid and ethanol.
Figure: The diagram shows the relationship between the ethanoic acid, the ethanol and the ester. This is not intended to be a full equation. Water, of course, is also produced.
The same process can be carried out for more complicated alcohols. The diagram below shows the structure of propane-1,2,3-triol (commonly known as glycerol).
Just as with the ethanol in the previous equation, I've drawn this back-to-front to make the next diagrams clearer. Normally, it is drawn with the -OH groups on the right-hand side. By the esterification process shown above, three ethanoate groups can be formed.
Lengthening each carbon chain creates a triglyceride, otherwise known as a fat.
The acid CH3(CH2)16COOH is called octadecanoic acid, frequently referred to by its common name, stearic acid. The full name for the ester of this with propane-1,2,3-triol is propane-1,2,3-triyl trioctadecanoate, unsurprisingly known by by its common name of glyceryl tristearate.
Saturated and unsaturated fats and oils
If the fat or oil is saturated, the acid from which it is derived has no carbon-carbon double bonds in its chain. Stearic acid is a saturated acid; therefore glyceryl tristearate is a saturated fat. If the acid has just one carbon-carbon double bond somewhere in the chain, it is called mono-unsaturated. If it has more than one carbon-carbon double bond, it is polyunsaturated.
The acids below are saturated acids, and will therefore form saturated fats and oils:
Oleic acid is a typical mono-unsaturated acid:
Linoleic and linolenic acids are typical polyunsaturated acids.
The terms "omega-6" and "omega-3" have become popular in the context of fats and oils. Linoleic acid is an omega-6 acid. This indicates that the first carbon-carbon double bond starts on the sixth carbon from the CH3 end. Linolenic acid is an omega-3 acid for the same reason. Because of their relationship with fats and oils, all of the acids above are sometimes described as fatty acids.
Physical properties
Boiling points
Small esters have boiling points which are similar to those of aldehydes and ketones with the same number of carbon atoms. Esters, like aldehydes and ketones, are polar molecules and so have dipole-dipole interactions as well as van der Waals dispersion forces. However, they do not form ester-ester hydrogen bonds, so their boiling points are significantly lower than those of an acid with the same number of carbon atoms.
molecule type boiling point (°C)
CH3COOCH2CH3 ester 77.1
CH3CH2CH2COOH carboxylic acid 164
Solubility in water
Small esters are fairly soluble in water but solubility decreases with increasing chain length, as shown below:
ester formula solubility (g per 100 g of water)
ethyl methanoate HCOOCH2CH3 10.5
ethyl ethanoate CH3COOCH2CH3 8.7
ethyl propanoate CH3CH2COOCH2CH3 1.7
The reason for this trend in solubility is that although esters cannot hydrogen bond with each other, they can hydrogen bond with water molecules. One of the partially-positive hydrogen atoms in a water molecule can be sufficiently attracted to one of the lone pairs on one of the oxygen atoms in an ester, forming a hydrogen bond. Dispersion forces and dipole-dipole attractions are also present.
Forming these intermolecular attractions releases some of the energy needed to solvate the ester. As chain length increases, the hydrocarbon portion forces itself between water molecules, breaking the relatively strong hydrogen bonds between water molecules without offering an energetic compensation; furthermore, the water molecules are forced into an ordered alignment along the chain, decreasing the entropy in the system. This makes the process thermodynamically less favorable, and so solubility decreases.
Solubility in water
Fats and oils are not water soluble. The chain lengths are so great that far too many hydrogen bonds between water molecules must be broken; this is not an energetically profitable arrangement.
Melting point
Melting points determine whether the substance is a fat (a solid at room temperature) or an oil (a liquid at room temperature). Fats normally contain saturated chains. These allow more effective van der Waals dispersion forces between the molecules: more energy is required to separate the chains, increasing the melting point.
A greater number of double bonds, or degree of unsaturation, in the molecules results in a lower melting point, because the van der Waals forces are less effective. The efficacy of van derWaals forces depends on the ability of molecules to pack closely together. The presence of carbon-carbon double bonds in the chains disrupts otherwise tidy packing. Here is a simplified diagram of a saturated fat:
The hydrocarbon chains are, of course, in constant motion in the liquid, but it is possible for them to lie tidily when the substance solidifies. If the chains in one molecule can lie tidily, that means that neighboring molecules can get close. That increases the attractions between one molecule and its neighbors and so increases the melting point.
Unsaturated fats and oils have at least one carbon-carbon double bond in at least one chain. Rotation about a carbon-carbon double bond is constrained, locking a permanent kink into the chain. This makes packing molecules close together more difficult. If the chains cannot pack well, the van der Waals forces will be less effective. This effect is much stronger for molecules in which the hydrocarbon chains at either end of the double bond are arranged cis- to each other, as shown in the figure below:
In the trans- conformation, the effect is not as marked. It is, however, rather more than the diagram below suggests because of the changes in bond angles around the double bond compared with the rest of the chain.
Trans-fats and oils have higher melting points than their corresponding cis- conformations because the packing is not interrupted to the same degree. Naturally occurring unsaturated fats and oils tend to adopt the cis- conformations. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Esters/Nomenclature_of_Esters.txt |
This page looks in detail at the mechanism for the hydrolysis of esters in the presence of a dilute acid (such as hydrochloric acid or sulfuric acid) acting as the catalyst. It uses ethyl ethanoate as a typical ester.
The mechanism for the hydrolysis of ethyl ethanoate
When ethyl ethanoate is heated under reflux with a dilute acid such as dilute hydrochloric acid or dilute sulfuric acid, the ester reacts with the water present to produce ethanoic acid and ethanol. Because the reaction is reversible, an equilibrium mixture is produced containing all four of the substances in the equation. In order to get as much hydrolysis as possible, a large excess of water can be used. The dilute acid provides both the acid catalyst and the water.
The mechanism
All the steps in the mechanism below are shown as one-way reactions because it makes the mechanism look less confusing. The reverse reaction is actually done sufficiently differently that it affects the way the mechanism is written. You will find a link to the esterification reaction further down the page if you are interested.
Note
The explanation assumes that you know about the use of curly arrows in organic reaction mechanisms. If you aren't happy about these follow this link before you go any further.
Step 1
The actual catalyst in this case is the hydroxonium ion, H3O+, present in all solutions of acids in water. In the first step, the ester takes a proton (a hydrogen ion) from the hydroxonium ion. The proton becomes attached to one of the lone pairs on the oxygen which is double-bonded to the carbon.
The transfer of the proton to the oxygen gives it a positive charge, but the charge is actually delocalized (spread around) much more widely than this shows.
One way of representing this delocalization is to draw a number of structures called resonance structures or canonical forms joined by double-headed arrows. You could, if you wished, put in some curly arrows to show the movements of electrons which change one of these structures into the next.
None of these formulae represents the true structure of the ion - but each gives you some information about it. For example, notice where the positive charge is in the three structures. What this means in reality is that the positive charge is spread around over those three atoms - the two oxygens and the carbon.
Note
If you haven't come across canonical forms as a way of representing delocalization, don't worry about it particularly. It is important, however, that you do not imagine that the molecule is rapidly flipping from one structure to another. The double-headed arrows mean something different.
A mule is a hybrid of a donkey and a horse. In this notation, you could represent a mule by writing donkey and horse connected by a double-headed arrow. Neitherdonkey nor horse accurately represents what a mule looks like, but with a bit of imagination you could build up a fairly good picture of a mule by combining together the characteristics of both donkey and horse. But a mule obviously doesn't spend its time rapidly changing back and forth between being a donkey and a horse!
The next stage of the mechanism involves an attack on the carbon, and so it is convenient to use the structure showing the positive charge on that carbon in the next step.
Step 2
The positive charge on the carbon atom is attacked by one of the lone pairs on the oxygen of a water molecule.
Note
You could work out precisely why that particular oxygen carries the positive charge on the right-hand side. On the other hand, you could realise that there has to be a positive charge somewhere (because you started with one), and that particular oxygen doesn't look right - it has too many bonds. Put the charge on there!
That's a quick rough-and-ready reasoning which works every time I use it!
Step 3
What happens next is that a proton (a hydrogen ion) gets transferred from the bottom oxygen atom to one of the others. It gets picked off by one of the other substances in the mixture (for example, by attaching to a lone pair on a water molecule), and then dumped back onto one of the oxygens more or less at random. Eventually, by chance, it will join to the oxygen with the ethyl group attached. When that happens, the net effect is:
Step 4
Now a molecule of ethanol is lost from the ion. That's one of the products of the reaction.
The structure for the latest ion is just like the one we discussed at length back in step 1. The positive charge is actually delocalized all over that end of the ion. The real structure will be a hybrid of these:
It is easier to follow what is happening if we keep going with the structure with the charge on the carbon.
Step 5
The hydrogen is removed from the oxygen by reaction with a water molecule.
And there we are! We have produced the ethanoic acid (the other product of the reaction) and the hydroxonium ion catalyst has been regenerated.
Contributors
Jim Clark (Chemguide.co.uk) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Esters/Reactivity_of_Esters/Acid_Catalyzed_Hydrolysis_of_Esters.txt |
Introduction
Esters can be cleaved back into a carboxylic acid and an alcohol by reaction with water and a catalytic amount of acid.
Example 1:
Mechanism
1) Protonation of the Carbonyl
2) Nucleophilic attack by water
3) Proton transfer
4) Leaving group removal
Contributors
Prof. Steven Farmer (Sonoma State University)
Esters can be converted aldehydes using diisobutylaluminum hydride (DIBAH).
Introduction
Esters can be converted to aldehydes using diisobutylaluminum hydride (DIBAH). The reaction is usually carried out at -78 oC to prevent reaction with the aldehyde product.
Example 1:
Contributors
Prof. Steven Farmer (Sonoma State University)
Esters can be reduced to 1 alcohols using (LiAlH 4)
Esters can be converted to 1o alcohols using LiAlH4, while sodium borohydride (\(NaBH_4\)) is not a strong enough reducing agent to perform this reaction.
Example 1:
Mechanism
1) Nucleophilic attack by the hydride
2) Leaving group removal
3) Nucleopilic attack by the hydride anion
4) The alkoxide is protonated
Contributors
Prof. Steven Farmer (Sonoma State University)
General mechanism of ester reactions
Introduction
Carboxylic acid derivatives are a group of functional groups whose chemistry is closely related. The main difference is the presence of an electronegative substituent that can act as a leaving group during nucleophile substitution reactions. Although there are many types of carboxylic acid derivatives known we will be focusing on just four: Acid halides, Acid anhydrides, Esters, and Amides.
General mechanism
1) Nucleophilic attack on the carbonyl
2) Leaving group is removed
Although aldehydes and ketones also contain a carbonyl their chemistry is distinctly different because they do not contain a suitable leaving group. Once the tetrahedral intermediate is formed aldehydes and ketones cannot reform the carbonyl. Because of this aldehydes and ketones typically undergo nucleophilic additions and not substitutions.
The relative reactivity of carboxylic acid derivatives toward nucleophile substitutions is related to the electronegative leaving group’s ability to activate the carbonyl. The more electronegative leaving groups withdrawn electron density from the carbonyl, thereby, increasing its electrophilicity.
Contributors
Prof. Steven Farmer (Sonoma State University)
Grignard Reagents Convert Esters into Tertiary Alcohols
Addition of Grignard reagents convert esters to 3o alcohols. In effect the Grignard reagent adds twice.
Mechanism
1) Nucleophilic attack
2) Leaving group removal
3) Nucleophilic attack
4) Protonation
Contributors
Prof. Steven Farmer (Sonoma State University)
Polyesters
This page looks at the formation, structure and uses of a common polyester sometimes known as Terylene if it is used as a fibre, or PET if it used in, for example, plastic drinks bottles. A polyester is a polymer (a chain of repeating units) where the individual units are held together by ester linkages.
The diagram shows a very small bit of the polymer chain and looks pretty complicated. But it is not very difficult to work out - and that's the best thing to do: work it out, not try to remember it. You will see how to do that in a moment.
The usual name of this common polyester is poly(ethylene terephthalate). The everyday name depends on whether it is being used as a fibre or as a material for making things like bottles for soft drinks. When it is being used as a fiber to make clothes, it is often just called polyester. It may sometimes be known by a brand name like Terylene. When it is being used to make bottles, for example, it is usually called PET.
Making polyesters as an example of condensation polymerisation
In condensation polymerisation, when the monomers join together a small molecule gets lost. That's different from addition polymerisation which produces polymers like poly(ethene) - in that case, nothing is lost when the monomers join together. A polyester is made by a reaction involving an acid with two -COOH groups, and an alcohol with two -OH groups. In the common polyester drawn below.
Figure: The acid is benzene-1,4-dicarboxylic acid (old name: terephthalic acid) and the alcohol is ethane-1,2-diol (old name: ethylene glycol).
Now imagine lining these up alternately and making esters with each acid group and each alcohol group, losing a molecule of water every time an ester linkage is made.
That would produce the chain shown above (although this time written without separating out the carbon-oxygen double bond - write it whichever way you like).
Manufacturing poly(ethylene terephthalate)
The reaction takes place in two main stages: a pre-polymerisation stage and the actual polymerisation. In the first stage, before polymerization happens, you get a fairly simple ester formed between the acid and two molecules of ethane-1,2-diol.
In the polymerisation stage, this is heated to a temperature of about 260°C and at a low pressure. A catalyst is needed - there are several possibilities including antimony compounds like antimony(III) oxide. The polyester forms and half of the ethane-1,2-diol is regenerated. This is removed and recycled.
Hydrolysis of polyesters
Simple esters are easily hydrolyzed by reaction with dilute acids or alkalis. Polyesters are attacked readily by alkalis, but much more slowly by dilute acids. Hydrolysis by water alone is so slow as to be completely unimportant. (You wouldn't expect your polyester fleece to fall to pieces if you went out in the rain!). If you spill dilute alkali on a fabric made from polyester, the ester linkages are broken. Ethane-1,2-diol is formed together with the salt of the carboxylic acid. Because you produce small molecules rather than the original polymer, the fibers are destroyed, and you end up with a hole! For example, if you react the polyester with sodium hydroxide solution:
Saponification
Esters can be cleaved back into a carboxylic acid and an alcohol by reaction with water and a base. The reaction is called a saponification from the Latin sapo which means soap. The name comes from the fact that soap used to be made by the ester hydrolysis of fats. Due to the basic conditions a carboxylate ion is made rather than a carboxylic acid.
Mechanism
Step 1: Nucleophilic attack by hydroxide
Step 2: Leaving group removal
Step 3: Deprotonation | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Esters/Reactivity_of_Esters/Acid_Catalyzed_Hydrolysis_of_Esters_%28II%29.txt |
This page describes ways of hydrolyzing esters - splitting them into carboxylic acids (or their salts) and alcohols by the action of water, dilute acid or dilute alkali. It starts by looking at the hydrolysis of simple esters like ethyl ethanoate, and goes on to look at hydrolyzing bigger, more complicated ones to make soap.
Technically, hydrolysis is a reaction with water. That is exactly what happens when esters are hydrolyzed by water or by dilute acids such as dilute hydrochloric acid. The alkaline hydrolysis of esters actually involves reaction with hydroxide ions, but the overall result is so similar that it is lumped together with the other two.
Hydrolysis using water or dilute acid
The reaction with pure water is so slow that it is never used. The reaction is catalyzed by dilute acid, and so the ester is heated under reflux with a dilute acid like dilute hydrochloric acid or dilute sulfuric acid. Here are two simple examples of hydrolysis using an acid catalyst.
First, hydrolyzing ethyl ethanoate:
. . . and then hydrolyzing methyl propanoate:
Notice that the reactions are reversible. To make the hydrolysis as complete as possible, you would have to use an excess of water. The water comes from the dilute acid, and so you would mix the ester with an excess of dilute acid.
Hydrolysis using dilute alkali
This is the usual way of hydrolyzing esters. The ester is heated under reflux with a dilute alkali like sodium hydroxide solution. There are two advantages of doing this rather than using a dilute acid. The reactions are one-way rather than reversible, and the products are easier to separate. Taking the same esters as above, but using sodium hydroxide solution rather than a dilute acid:
First, hydrolyzing ethyl ethanoate using sodium hydroxide solution:
and then hydrolyzing methyl propanoate in the same way:
Notice that you get the sodium salt formed rather than the carboxylic acid itself. This mixture is relatively easy to separate. Provided you use an excess of sodium hydroxide solution, there will not be any ester left - so you don't have to worry about that. The alcohol formed can be distilled off. That's easy!
If you want the acid rather than its salt, all you have to do is to add an excess of a strong acid like dilute hydrochloric acid or dilute sulfuric acid to the solution left after the first distillation. If you do this, the mixture is flooded with hydrogen ions. These are picked up by the ethanoate ions (or propanoate ions or whatever) present in the salts to make ethanoic acid (or propanoic acid, etc). Because these are weak acids, once they combine with the hydrogen ions, they tend to stay combined. The carboxylic acid can now be distilled off.
Hydrolyzing complicated esters to make soap
This next bit deals with the alkaline hydrolysis (using sodium hydroxide solution) of the big esters found in animal and vegetable fats and oils. If the large esters present in animal or vegetable fats and oils are heated with concentrated sodium hydroxide solution exactly the same reaction happens as with the simple esters. A salt of a carboxylic acid is formed - in this case, the sodium salt of a big acid such as octadecanoic acid (stearic acid). These salts are the important ingredients of soap - the ones that do the cleaning. An alcohol is also produced - in this case, the more complicated alcohol, propane-1,2,3-triol (glycerol).
Because of its relationship with soap making, the alkaline hydrolysis of esters is sometimes known as saponification.
Transesterification
Transesterification is the conversion of a carboxylic acid ester into a different carboxylic acid ester.
Introduction
When in ester is placed in a large excess of an alcohol along with presence of either an acid or a base there can be an exchange of alkoxy groups. The large excess of alcohol is used to drive the reaction forward. The most common method of transesterification is the reaction of the ester with an alcohol in the presence of an acid catalyst eg:
This reaction has the following mechanism:
Since both the reactants and the products are an ester and an alcohol, the reaction is reversible and the equilibrium constant is close to one. Consequently, the Le Chatelier’s principle has to be exploited to drive the reaction to completion. The simplest way to do so is to use the alcohol as the solvent as well.
Mechanism in basic conditions
Nucleophilic attack by an alkoxide
2) Leaving group removal
Mechanism in acidic conditions
1) Protonation of the carbonyl by the acid. The carbonyl is now activated toward nucleophilic attack.
2) Nucleophilic attack on the carbonyl
3) Proton transfer
4) Removal of the leaving group
5) Deprotonation
Contributors
Prof. Steven Farmer (Sonoma State University) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Esters/Reactivity_of_Esters/The_Hydrolysis_of_Esters.txt |
• Acetoacetic Ester Synthesis
Acetoacetic ester (ethyl acetoacetate) is an extremely useful molecule that can be used to make ketones and other molecules. You’ll even use this later on in amino acid synthesis, so let’s break down the way it reacts.
• Esterification
The π bond of the carbonyl group can act as a base to a strong inorganic acid due to the distortion of the electrons from the electronegativity difference between the oxygen atom and the carbon atom and also the resonance dipole. The cation produced in the reaction with sulfuric acid will have resonance stabilization.
• Making Esters From Alcohols
This page looks at esterification - mainly the reaction between alcohols and carboxylic acids to make esters. It also looks briefly at making esters from the reactions between acyl chlorides (acid chlorides) and alcohols, and between acid anhydrides and alcohols.
• Preparation of Esters
This page describes ways of making esters in the lab from alcohols and phenols using carboxylic acids, acyl chlorides (acid chlorides) or acid anhydrides as appropriate.
• Synthesis of Esters
Carboxylic acids can react with alcohols to form esters.
Synthesis of Esters
Acetoacetic ester (ethyl acetoacetate) is an extremely useful molecule that can be used to make ketones and other molecules. You’ll even use this later on in amino acid synthesis, so let’s break down the way it reacts.
From Beta-Ketoester to Ketone:
How do we accomplish this transformation
Enolate Formation
Labeled alpha-carbons
See those two carbonyls there? Each carbonyl has something called an alpha-carbon, and each alpha-carbon has hydrogens that are easily abstracted. The pKa of the green alpha-hydrogen is about 20, and the pKa of the blue alpha-hydrogen is actually about 10. Why? Because of the resonance structures the anions can form!
Green enolate resonance structures
Blue enolate resonance structures
Whenever you have a beta-dicarbonyl like this one, the enolate will preferentially form on the shared alpha-carbon. The anion on the blue alpha-carbon above can form more resonance structures than the anion on the green alpha-carbon can, so the blue hydrogen’s pKa will be lower (more acidic).
That all sounds cool, but can we just use any ol’ base to form our enolate? Definitely not! Let’s say we were to try using NaOH. Instead of forming the enolate, we’d actually wind up with a competing reaction: saponification, a type of nucleophilic acyl substitution. Notice that the hydroxide replaces the ethoxy group.
abridged saponification mechanism
So, how can we specifically avoid that type of acyl substitution? We can use a bulky base like LDA or the anionic version of our alkoxy group! See how we’ve got an ethoxy group (—OEt) in our starting material? In order to prevent any substitutions of that group, we can actually use NaOEt. Those ethoxy groups are totally exchanging, but the same molecule is produced.
Fischer esterification with sodium ethoxide
Methylene enolate formation
Enolate Alkylation
Okay, cool! These enolates are pretty good at SN2 reactions. They can act as nucleophiles on alkyl halides, acyl (acid) chlorides, and more! Let’s try adding a propyl group.
Enolate alkylation
Decarboxylation
Once we’ve got our alkyl group on there, we can actually get rid of the ester entirely through a mechanism called decarboxylation if we want to. All it takes is some heat and a little bit of aqueous acid. It could be written a ton of different ways—H2SO4 (aq), HCl (aq), or even generically as H3O+.
Acid-catalyzed ester hydrolysis
First we hydrolyze the ester to make a beta ketoacid, and then we heat things up to lose CO2. After acidic hydrolysis, the enol (vinyl alcohol) that results will tautomerize back into a substituted ketone.
Tautomerization and decarboxylation
Boom! There’s our product, a substituted ketone, in the green box! Not so bad, right? Of course, there are tons of different ways to use this molecule. We’ve just walked through the steps for a single alkylation, but there’s nothing stopping us from adding different groups.
Adding Two Alkyl Groups
We’ve added one alkyl group, but what if we want to add another one? Well, we just have to follow the same steps! So, let’s start from the beginning. Let’s first add a propyl group and then an ethyl group. Once we’ve added the propyl group, all we need to do is add another equivalent of base and then the ethyl group. Here’s what the order of reagents looks like:
Double alkylation reagents
And here’s what the mechanism would look like:
Double alkylation mechanism
Adding Cyclic Alkyl Groups
Okay, but what if we want to add a cyclic group to our molecule? Well, luckily that’s not so bad either. We just need a molecule that has two leaving groups at terminal positions. Basically, it’s going to be very similar to the double alkylation but with just one equivalent of our alkyl molecule being added. Here’s what the reagents look like: Cyclic alkylation reagents
And here’s what the mechanism looks like:
Cyclic alkylation mechanism
Adding Acyl Groups
Let’s take a step back and use the same enolate we used in the alkylation, but let’s use an acyl chloride instead of an alkyl halide this time. This follows basically the same pattern as the alkylation, but I’m going to rotate the acetoacetic ester a little bit and highlight the acid chloride so that it’s easier to follow.
Enolate acylation mechanism
See how we just followed the same pattern? Form the enolate, provide an electrophile, and cleave off the ester by adding acid and heating it up!
Contributors
• Johnny Betancourt, Clutchprep. Source page can be accessed here. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Esters/Synthesis_of_Esters/Acetoacetic_Ester_Synthesis.txt |
The π bond of the carbonyl group can act as a base to a strong inorganic acid due to the distortion of the electrons from the electronegativity difference between the oxygen atom and the carbon atom and also the resonance dipole. The cation produced in the reaction with sulfuric acid will have resonance stabilization.
Mechanism for Acid Catalyzed Esterification
Step 1: Formation of cation
Step 2: The methanol can act as a nucleophile to a carbocation. Remember that there are many methanol molecules in the solution...it is always in excess in this reaction.
Step 3: The protonated ether can leave as methanol but that will not accomplish anything. A proton can be transferred to one of the hydroxyl groups and thus make it a good leaving group.
Step 4: The alcohol oxygen atom from the hydroxy group can donate a pair of electrons to the carbon atom making a π bond and eliminating water. The water will not be a viable nucleophile that will reverse the reaction because its concentration will be low compared to the concentration of the methanol.
Step 5: The water will be in too low a concentration to reverse the reaction.
Making Esters From Alcohols
This page looks at esterification - mainly the reaction between alcohols and carboxylic acids to make esters. It also looks briefly at making esters from the reactions between acyl chlorides (acid chlorides) and alcohols, and between acid anhydrides and alcohols. Esters are derived from carboxylic acids. A carboxylic acid contains the -COOH group, and in an ester the hydrogen in this group is replaced by a hydrocarbon group of some kind. We shall just be looking at cases where it is replaced by an alkyl group, but it could equally well be an aryl group (one based on a benzene ring).
A common ester - ethyl ethanoate
The most commonly discussed ester is ethyl ethanoate. In this case, the hydrogen in the -COOH group has been replaced by an ethyl group. The formula for ethyl ethanoate is:
Notice that the ester is named the opposite way around from the way the formula is written. The "ethanoate" bit comes from ethanoic acid. The "ethyl" bit comes from the ethyl group on the end.
A few more esters
In each case, be sure that you can see how the names and formulae relate to each other.
Notice that the acid is named by counting up the total number of carbon atoms in the chain - including the one in the -COOH group. So, for example, CH3CH2COOH is propanoic acid, and CH3CH2COO is the propanoate group.
Making esters from carboxylic acids and alcohols
Esters are produced when carboxylic acids are heated with alcohols in the presence of an acid catalyst. The catalyst is usually concentrated sulphuric acid. Dry hydrogen chloride gas is used in some cases, but these tend to involve aromatic esters (ones containing a benzene ring). If you are a UK A level student you won't have to worry about these.
The esterification reaction is both slow and reversible. The equation for the reaction between an acid RCOOH and an alcohol R'OH (where R and R' can be the same or different) is:
So, for example, if you were making ethyl ethanoate from ethanoic acid and ethanol, the equation would be:
Doing the reactions
On a test tube scale
Carboxylic acids and alcohols are often warmed together in the presence of a few drops of concentrated sulfuric acid in order to observe the smell of the esters formed. You would normally use small quantities of everything heated in a test tube stood in a hot water bath for a couple of minutes.
Because the reactions are slow and reversible, you don't get a lot of ester produced in this time. The smell is often masked or distorted by the smell of the carboxylic acid. A simple way of detecting the smell of the ester is to pour the mixture into some water in a small beaker. Apart from the very small ones, esters are fairly insoluble in water and tend to form a thin layer on the surface. Excess acid and alcohol both dissolve and are tucked safely away under the ester layer. Small esters like ethyl ethanoate smell like typical organic solvents (ethyl ethanoate is a common solvent in, for example, glues). As the esters get bigger, the smells tend towards artificial fruit flavoring - "pear drops", for example.
On a larger scale
If you want to make a reasonably large sample of an ester, the method used depends to some extent on the size of the ester. Small esters are formed faster than bigger ones. To make a small ester like ethyl ethanoate, you can gently heat a mixture of ethanoic acid and ethanol in the presence of concentrated sulfuric acid, and distil off the ester as soon as it is formed. This prevents the reverse reaction happening. It works well because the ester has the lowest boiling point of anything present. The ester is the only thing in the mixture which doesn't form hydrogen bonds, and so it has the weakest intermolecular forces.
Larger esters tend to form more slowly. In these cases, it may be necessary to heat the reaction mixture under reflux for some time to produce an equilibrium mixture. The ester can be separated from the carboxylic acid, alcohol, water and sulfuric acid in the mixture by fractional distillation.
Making esters from alcohols and acyl chlorides
Esters can also be made from the reactions between alcohols and either acyl chlorides (acid chlorides) or acid anhydrides.If you add an acyl chloride to an alcohol, you get a vigorous (even violent) reaction at room temperature producing an ester and clouds of steamy acidic fumes of hydrogen chloride. For example, if you add the liquid ethanoyl chloride to ethanol, you get a burst of hydrogen chloride produced together with the liquid ester ethyl ethanoate.
$CH_3COCl + CH_3CH_2OH \rightarrow CH_3COOCH_2CH_3 + HCl$
Making esters from alcohols and acid anhydrides
The reactions of acid anhydrides are slower than the corresponding reactions with acyl chlorides, and you usually need to warm the mixture. Taking ethanol reacting with ethanoic anhydride as a typical reaction involving an alcohol. There is a slow reaction at room temperature (or faster on warming). There is no visible change in the colorless liquids, but a mixture of ethyl ethanoate and ethanoic acid is formed.
$(CH_3CO)_2O + CH_3CH_2OH \rightarrow CH_3COOCH_2CH_3 + CH_3COOH$ | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Esters/Synthesis_of_Esters/Esterification.txt |
This page describes ways of making esters in the lab from alcohols and phenols using carboxylic acids, acyl chlorides (acid chlorides) or acid anhydrides as appropriate.
Making esters from carboxylic acids
This method can be used for converting alcohols into esters, but it doesn't work with phenols - compounds where the -OH group is attached directly to a benzene ring. Phenols react with carboxylic acids so slowly that the reaction is unusable for preparation purposes.
Esters are produced when carboxylic acids are heated with alcohols in the presence of an acid catalyst. The catalyst is usually concentrated sulphuric acid. Dry hydrogen chloride gas is used in some cases, but these tend to involve aromatic esters (ones where the carboxylic acid contains a benzene ring). If you are a UK A level student you won't have to worry about these.
The esterification reaction is both slow and reversible. The equation for the reaction between an acid RCOOH and an alcohol R'OH (where R and R' can be the same or different) is:
So, for example, if you were making ethyl ethanoate from ethanoic acid and ethanol, the equation would be:
Doing the reactions
On a test tube scale: Carboxylic acids and alcohols are often warmed together in the presence of a few drops of concentrated sulfuric acid to observe the smell of the esters formed. You would normally use small quantities of everything heated in a test tube stood in a hot water bath for a couple of minutes.
Because the reactions are slow and reversible, you don't get a lot of ester produced in this time. The smell is often masked or distorted by the smell of the carboxylic acid. A simple way of detecting the smell of the ester is to pour the mixture into some water in a small beaker.
Apart from the very small ones, esters are fairly insoluble in water and tend to form a thin layer on the surface. Excess acid and alcohol both dissolve and are tucked safely away under the ester layer. Small esters like ethyl ethanoate smell like typical organic solvents (ethyl ethanoate is a common solvent in, for example, glues). As the esters get bigger, the smells tend towards artificial fruit flavoring - "pear drops", for example.
On a larger scale: If you want to make a reasonably large sample of an ester, the method used depends to some extent on the size of the ester. Small esters are formed faster than bigger ones. To make a small ester like ethyl ethanoate, you can gently heat a mixture of ethanoic acid and ethanol in the presence of concentrated sulphuric acid, and distil off the ester as soon as it is formed. This prevents the reverse reaction happening. It works well because the ester has the lowest boiling point of anything present. The ester is the only thing in the mixture which doesn't form hydrogen bonds, and so it has the weakest intermolecular forces.
Larger esters tend to form more slowly. In these cases, it may be necessary to heat the reaction mixture under reflux for some time to produce an equilibrium mixture. The ester can be separated from the carboxylic acid, alcohol, water and sulphuric acid in the mixture by fractional distillation.
Making esters using acyl chlorides (acid chlorides)
This method will work for alcohols and phenols. In the case of phenols, the reaction is sometimes improved by first converting the phenol into a more reactive form.
If you add an acyl chloride to an alcohol, you get a vigorous (even violent) reaction at room temperature producing an ester and clouds of steamy acidic fumes of hydrogen chloride. For example, if you add the liquid ethanoyl chloride to ethanol, you get a burst of hydrogen chloride produced together with the liquid ester ethyl ethanoate.
\[ CH_3OCl + CH_3CH_2OH \longrightarrow CH_3COOCH_2CH_3 + HCl \]
The substance normally called "phenol" is the simplest of the family of phenols. Phenol has an -OH group attached to a benzene ring - and nothing else. The reaction between ethanoyl chloride and phenol is similar to the ethanol reaction although not so vigorous. Phenyl ethanoate is formed together with hydrogen chloride gas.
Improving the reactions between phenols and some less reactive acyl chlorides
Benzoyl chloride has the formula C6H5COCl. The -COCl group is attached directly to a benzene ring. It is much less reactive than simple acyl chlorides like ethanoyl chloride. The phenol is first converted into the ionic compound sodium phenoxide (sodium phenate) by dissolving it in sodium hydroxide solution.
The phenoxide ion reacts more rapidly with benzoyl chloride than the original phenol does, but even so you have to shake it with benzoyl chloride for about 15 minutes. Solid phenyl benzoate is formed.
Making esters with acid anhydrides
This reaction can again be used to make esters from both alcohols and phenols. The reactions are slower than the corresponding reactions with acyl chlorides, and you usually need to warm the mixture. In the case of a phenol, you can react the phenol with sodium hydroxide solution first, producing the more reactive phenoxide ion.
Taking ethanol reacting with ethanoic anhydride as a typical reaction involving an alcohol: There is a slow reaction at room temperature (or faster on warming). There is no visible change in the colorless liquids, but a mixture of ethyl ethanoate and ethanoic acid is formed.
\[ (CH_3CO)_2O + CH_3CH_2OH \longrightarrow CH_3COOCH_2CH_3 + CH_3COOH \]
The reaction with phenol is similar, but will be slower. Phenyl ethanoate is formed together with ethanoic acid.
This reaction is not important itself, but a very similar reaction is involved in the manufacture of aspirin (covered in detail on another page - link below). If the phenol is first converted into sodium phenoxide by adding sodium hydroxide solution, the reaction is faster. Phenyl ethanoate is again formed, but this time the other product is sodium ethanoate rather than ethanoic acid.
Contributors
Jim Clark (Chemguide.co.uk)
Synthesis of Esters
Contributors
Prof. Steven Farmer (Sonoma State University) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Esters/Synthesis_of_Esters/Preparation_of_Esters.txt |
Ethers are compounds having two alkyl or aryl groups bonded to an oxygen atom, as in the formula R1–O–R2. The ether functional group does not have a characteristic IUPAC nomenclature suffix, so it is necessary to designate it as a substituent. To do so the common alkoxy substituents are given names derived from their alkyl component (below):
Alkyl Group Name Alkoxy Group Name
CH3 Methyl CH3O– Methoxy
CH3CH2 Ethyl CH3CH2O– Ethoxy
(CH3)2CH– Isopropyl (CH3)2CHO– Isopropoxy
(CH3)3C– tert-Butyl (CH3)3CO– tert-Butoxy
C6H5 Phenyl C6H5O– Phenoxy
Ethers can be named by naming each of the two carbon groups as a separate word followed by a space and the word ether. The -OR group can also be named as a substituent using the group name, alkox
Example \(1\)
CH3-CH2-O-CH3 is called ethyl methyl ether or methoxyethane.
The smaller, shorter alkyl group becomes the alkoxy substituent. The larger, longer alkyl group side becomes the alkane base name. Each alkyl group on each side of the oxygen is numbered separately. The numbering priority is given to the carbon closest to the oxgen. The alkoxy side (shorter side) has an "-oxy" ending with its corresponding alkyl group. For example, CH3CH2CH2CH2CH2-O-CH2CH2CH3 is 1-propoxypentane. If there is cis or trans stereochemistry, the same rule still applies.
Examples \(2\)
• \(CH_3CH_2OCH_2CH_3\), diethyl ether (sometimes referred to as just ether)
• \(CH_3OCH_2CH_2OCH_3\), ethylene glycol dimethyl ether (glyme).
Exercises \(2\)
Try to name the following compounds using these conventions:
Try to draw structures for the following compounds:
• 2-pentyl 1-propyl ether J
• 1-(2-propoxy)cyclopentene J
Common names
Simple ethers are given common names in which the alkyl groups bonded to the oxygen are named in alphabetical order followed by the word "ether". The top left example shows the common name in blue under the IUPAC name. Many simple ethers are symmetrical, in that the two alkyl substituents are the same. These are named as "dialkyl ethers".
• anisole (try naming anisole by the other two conventions. J )
• oxirane
1,2-epoxyethane, ethylene oxide, dimethylene oxide, oxacyclopropane,
• furan (this compound is aromatic)
tetrahydrofuran
oxacyclopentane, 1,4-epoxybutane, tetramethylene oxide,
• dioxane
1,4-dioxacyclohexane
Exercise \(2\)
Try to draw structures for the following compounds-
• 3-bromoanisole J
• 2-methyloxirane J
• 3-ethylfuran J
Heterocycles
In cyclic ethers (heterocycles), one or more carbons are replaced with oxygen. Often, it's called heteroatoms, when carbon is replaced by an oxygen or any atom other than carbon or hydrogen. In this case, the stem is called the oxacycloalkane, where the prefix "oxa-" is an indicator of the replacement of the carbon by an oxygen in the ring. These compounds are numbered starting at the oxygen and continues around the ring. For example,
If a substituent is an alcohol, the alcohol has higher priority. However, if a substituent is a halide, ether has higher priority. If there is both an alcohol group and a halide, alcohol has higher priority. The numbering begins with the end that is closest to the higher priority substituent. There are ethers that are contain multiple ether groups that are called cyclic polyethers or crown ethers. These are also named using the IUPAC system.
Sulfides
Sulfur analogs of ethers (R–S–R') are called sulfides, e.g., (CH3)3C–S–CH3 is tert-butyl methyl sulfide. Sulfides are chemically more reactive than ethers, reflecting the greater nucleophilicity of sulfur relative to oxygen.
Problems
Name the following ethers:
(Answers to problems above: 1. diethyl ether; 2. 2-ethoxy-2-methyl-1-propane; 3. cis-1-ethoxy-2-methoxycyclopentane; 4. 1-ethoxy-1-methylcyclohexane; 5. oxacyclopropane; 6. 2,2-Dimethyloxacyclopropane)
Exercise \(3\)
Answer
A one-eyed one-horned flying propyl people ether | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Ethers/Nomenclature_of_Ethers.txt |
Ethers are a class of organic compounds that contain an oxygen between two alkyl groups. They have the formula R-O-R', with R's being the alkyl groups. these compounds are used in dye, perfumes, oils, waxes and industrial use.
Properties of Ethers
Dimethylether and ethyl methyl ether are gases at ordinary temperature. The other lower homologues are colorless, pleasant smelling, volatile liquids with typical ether smell.
Boiling points
The C - O bonds in ether are polar and thus ethers have a net dipole moment. The weak polarity of ethers do not appreciably affect their boiling points which are comparable to those of the alkenes of comparable molecular mass. Ethers have much lower boiling points as compared to isomeric alcohols. This is because alcohols molecules are associated by hydrogen bonds while ether molecules are not.
Solubility
Ethers containing up to 3 carbon atoms are soluble in water, due to their hydrogen bond formation with water molecules.
The solubility decreases with increase in the number of carbon atoms. The relative increase in the hydrocarbon portion of the molecule decreases the tendency of H-bond formation. Ethers are appreciably soluble in organic solvents like alcohol, benzene, acetone etc.
Structure of Ethers
Ethers are a class of organic compounds that contain an sp3 hybridized oxygen between two alkyl groups and have the formula R-O-R'. these compounds are used in dyes, perfumes, oils, waxes and other industrial uses. Aliphatic ethers have no aryl groups directly attached to the ether oxygen.
Examples of Aliphatic Ethers
Aromatic ethers have at least one aryl ring directly attached to the ether oxygen. In aryl ethers, the lone pair elections on oxygen are conjugated with the aromatic ring which significantly changes the properties of the ether.
Example of Aromatic Ethers
The sp3 hybridization of oxygen gives ethers roughly the same geometry as alcohols and water. The R-O-R' bond angle is close to what is expected in a tetrahedral geometry. The bond angle of dimethyl ether is 112o which is larger than the H-O-H bond angle in water (104.5o) due to the steric repulsion of the methyl groups.
The presence of an electronegative oxygen atom gives ethers a small dipole moment.
Comparisons of Physical Properties of Alcohols and Ethers
Ethers, unlike alcohols, have no hydrogen atom on the oxygen atom (that is, no OH group). Therefore, there is no intermolecular hydrogen bonding between ether molecules, which makes their boiling points much lower than an alcohol with similar mass. Despite the presence of a small dipole moment, ethers have boiling points that about the same alkanes of comparable molar mass. (Table 18.1.2).
Table 18.1.2 Comparison of Boiling Points of Alkanes, Alcohols, and Ethers
Condensed Structural Formula Name Molar Mass Boiling Point (°C) Intermolecular Hydrogen Bonding in Pure Liquid?
CH3CH2CH3 propane 44 –42 no
CH3OCH3 dimethyl ether 46 –25 no
CH3CH2OH ethyl alcohol 46 78 yes
CH3CH2CH2CH2CH3 pentane 72 36 no
CH3CH2OCH2CH3 diethyl ether 74 35 no
CH3CH2CH2CH2OH butyl alcohol 74 117 yes
Ether molecules do have an oxygen atom, however, and engage in hydrogen bonding with water molecules. Consequently, an ether has about the same solubility in water as the alcohol that is isomeric with it. For example, dimethyl ether and ethanol (both having the molecular formula C2H6O) are completely soluble in water, whereas diethyl ether and 1-butanol (both C4H10O) are barely soluble in water (8 g/100 mL of water).
Peroxide Formation
Many ethers can react with oxygen to form explosive peroxide compounds n a free radical process called autoxidation. For this reason ethers should not be stored for long periods of time and should not be stored in glass bottles. The danger is particularly acute when ether solutions are distilled to near dryness. The hydroperoxides can become more concentrated during a distillation because they tend to have a slightly higher boiling point than the corresponding ether. Before performing an ether distillation great care should be taken to test for the presence of peroxides.
Contributors
Binod Shrestha (University of Lorraine) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Ethers/Properties_of_Ethers/Physical_Properties_of_Ether.txt |
Ethers are widely used as solvents for a variety of organic compounds and reactions, suggesting that they are relatively unreactive themselves. Indeed, with the exception of the alkanes, cycloalkanes and fluorocarbons, ethers are probably the least reactive, common class of organic compounds. The inert nature of the ethers relative to the alcohols is undoubtedly due to the absence of the reactive O–H bond.
Reactivity of Ethers
The most common reaction of ethers is cleavage of the C–O bond by strong acids. This may occur by SN1 or E1 mechanisms for 3º-alkyl groups or by an SN2 mechanism for 1º-alkyl groups. Some examples are shown in the following diagram. The conjugate acid of the ether is an intermediate in all these reactions, just as conjugate acids were intermediates in certain alcohol reactions.
The first two reactions proceed by a sequence of SN2 steps in which the iodide or bromide anion displaces an alcohol in the first step, and then converts the conjugate acid of that alcohol to an alkyl halide in the second. Since SN2 reactions are favored at least hindered sites, the methyl group in example #1 is cleaved first. The 2º-alkyl group in example #3 is probably cleaved by an SN2 mechanism, but the SN1 alternative cannot be ruled out. The phenol formed in this reaction does not react further, since SN2, SN1 and E1 reactions do not take place on aromatic rings. The last example shows the cleavage of a 3º-alkyl group by a strong acid. Acids having poorly nucleophilic conjugate bases are often chosen for this purpose so that E1 products are favored. The reaction shown here (#4) is the reverse of the tert-butyl ether preparation described earlier.
Ethers in which oxygen is bonded to 1º- and 2º-alkyl groups are subject to peroxide formation in the presence of air (gaseous oxygen). This reaction presents an additional hazard to the use of these flammable solvents, since peroxides decompose explosively when heated or struck. The mechanism of peroxide formation is believed to be free radical in nature (note that molecular oxygen has two unpaired electrons).
$\ce{R–O–CH(CH3)2 + O2 }\rightarrow \underset{\text{a peroxide}}{\ce{R–O–C(CH3)2–O–O–H }}$
Ethers as Protective Groups
Because of their chemical stability, ethers may be used to protect hydroxyl functions from undergoing unwanted reactions.
Reactions of Epoxides
Epoxides (oxiranes) are three-membered cyclic ethers that are easily prepared from alkenes by reaction with peracids. Because of the large angle strain in this small ring, epoxides undergo acid and base-catalyzed C–O bond cleavage more easily than do larger ring ethers. Among the following examples, the first is unexceptional except for the fact that it occurs under milder conditions and more rapidly than other ether cleavages. The second and third examples clearly show the exceptional reactivity of epoxides, since unstrained ethers present in the same reactant or as solvent do not react. The aqueous acid used to work up the third reaction, following the Grignard reagent cleavage of the ethylene oxide, simply neutralizes the magnesium salt of the alcohol product.
Sulfur Analogs of Alcohols and Ethers
Sulfur is below oxygen in the periodic table. To see examples of organosulfur compounds and their chemistry Click Here
Contributors
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Ethers/Reactivity_of_Ethers/Cleaving_Ethers.txt |
There are two primary reactions to generate ethers: either by Dehydration of Alcohols or by the Williamson Synthesis. Acyclic ethers can be prepared using Williamson's synthesis, which involves reacting an alkoxide with a haloalkane. As stated previously, alkoxides are created by reacting an alcohol with metallic sodium or potassium, or a metal hydride, such as sodium hydride (NaH). To minimize steric hindrance and achieve a good yield, the haloalkane must be a primary haloalkane. This is because the mechanism is SN2, where the oxygen atom does a backside attack on the carbon atom with the halogen atom, causing the halogen atom to leave with its electrons.
Synthesis of Ethers
The alkoxymercuration reaction is similar to oxymercuration reaction, except alcohol is used instead of water.
Reactions #3 and #4 are examples of this two-step procedure. Note that the alcohol reactant is used as the solvent, and a trifluoroacetate mercury (II) salt is used in preference to the acetate (trifluoroacetate anion is a poorer nucleophile than acetate). The mechanism of alkoxymercuration is similar to that of oxymercuration, with an initial anti-addition of the mercuric species and alcohol being followed by reductive demercuration.
Dehydration of Alcohols to Make Ethers
Acid-catalyzed dehydration of small 1º-alcohols constitutes a specialized method of preparing symmetrical ethers.
Introduction
As shown in the following two equations, the success of this procedure depends on the temperature. At 110º to 130 ºC an SN2 reaction of the alcohol conjugate acid leads to an ether product. At higher temperatures (over 150 ºC) an E2 elimination takes place.
\[\ce{2 CH3CH2-OH + H2SO4 ->[130 \,^oC] CH3CH2-O-CH2CH3 + H2O}\]
\[\ce{CH3CH2-OH + H2SO4 ->[150 \,^oC] CH2=CH2 + H2O}\]
In this reaction alcohol has to be used in excess and the temperature has to be maintained around 413 K. If alcohol is not used in excess or the temperature is higher, the alcohol will preferably undergo dehydration to yield alkene.
If ethanol is dehydrated to ethene in presence of sulfuric acid at 433 K, but as 410 K, ethoxyethane is the main product. The dehydration of secondary and tertiary alcohols to get corresponding ethers is unsuccessful as alkenes are formed easily in these reactions.
This reaction cannot be employed to prepare unsymmetrical ethers. It is because a mixture of products is likely to be obtained. The Williamson Ether synthesis or alkoxymercuration/demercuration apporach can be used to prepare unsymmetrical ethers.
Ether Synthesis
Ethers are usually prepared from alcohols or their conjugate bases. One important procedure, known as the Williamson Ether Synthesis, proceeds by an SN2 reaction of an alkoxide nucleophile with an alkyl halide. Reactions #1 and #2 below are two examples of this procedure. When applied to an unsymmetrical ether, as in this case, there are two different combinations of reactants are possible. Of these one is usually better than the other. Since alkoxide anions are strong bases, the possibility of a competing E2 elimination must always be considered. Bearing in mind the factors that favor substitution over elimination, a 1º-alkyl halide should be selected as a preferred reactant whenever possible. Thus, reaction #1 gives a better and cleaner yield of benzyl isopropyl ether than does reaction #2, which generates considerable elimination product.
A second general ether synthesis, alkoxymercuration, is patterned after the oxymercuration reaction. Reactions #3 and #4 are examples of this two-step procedure. Note that the alcohol reactant is used as the solvent, and a trifluoroacetate mercury (II) salt is used in preference to the acetate (trifluoroacetate anion is a poorer nucleophile than acetate). The mechanism of alkoxymercuration is similar to that of oxymercuration, with an initial anti-addition of the mercuric species and alcohol being followed by reductive demercuration.
Acid-catalyzed dehydration of small 1º-alcohols constitutes a specialized method of preparing symmetrical ethers. As shown in the following two equations, the success of this procedure depends on the temperature. At 110º to 130 ºC an SN2 reaction of the alcohol conjugate acid leads to an ether product. At higher temperatures (over 150 ºC) an E2 elimination takes place.
2 CH3CH2-OH + H2SO4 130 ºC
CH3CH2-O-CH2CH3 + H2O
CH3CH2-OH + H2SO4 150 ºC
CH2=CH2 + H2O | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Ethers/Synthesis_of_Ethers/Alkoxy-mercuration_of_Alkenes.txt |
This page looks at the manufacture of epoxyethane from ethene, and then at some of the products that are made from epoxyethane.
The manufacture of epoxyethane
Conditions
Temperature: about 250 - 300°C
Pressure: about 15 atmospheres
Catalyst: silver
Problems and hazards during manufacture
The main problem comes in controlling the temperature. The reaction is exothermic and so the temperature will tend to rise unless it is carefully controlled. At higher temperatures the ethene burns in the oxygen to produce carbon dioxide and water which means that the temperature would increase even more - and the whole thing get completely out of hand! Two hazards during manufacture come from the nature of epoxyethane. It is
1. poisonous and carcinogenic (cancer producing);
2. highly flammable or explosive in contact with air.
The reactivity of epoxyethane
The reason that epoxyethane is so reactive is that bonding pairs in the ring of atoms in the molecule are forced very close together. The bond angles are about 60° rather than about 109.5° when carbon atoms normally form single bonds.
The overlap between the atomic orbitals in forming the carbon-carbon and carbon-oxygen bonds is less good than it is normally, and there is considerable repulsion between the bonding pairs. The system becomes more stable if the ring is broken. When epoxyethane reacts a carbon-oxygen bond is always broken and the ring opens up.
Uses of epoxyethane
Acid catalyzed hydrolysis of epoxyethane
Epoxyethane reacts with water in the presence of an acid catalyst (very dilute sulphuric acid) at a temperature of about 60°C. Ethane-1,2-diol is produced.
A large excess of water is used to try to prevent the product from reacting with the original epoxyethane. Ethane-1,2-diol is an alcohol (because it contains simple -OH groups), and alcohols react with epoxyethane (see below). Even in the presence of a large excess of water, this reaction happens as well:
The product is still an alcohol, and similar reactions can also lead to quite long chains.
Uses of ethane-1,2-diol
Ethane-1,2-diol is used as an antifreeze in car engines. It is added to the cooling water to prevent it from freezing under very cold conditions. Ethane-1,2-diol is also used in the manufacture of polyesters such as poly(ethylene terephthalate). You may have come across this as a fibre used to make clothes (perhaps under the brand name Terylene), or as a clear material used to make plastic drinks bottles (PET).
The reaction of epoxyethane with alcohols
This is a reaction which students at this level often find difficulty remembering. It is actually probably easier to work out than remember. Think of it as an extension of the reaction with water.
Alcohols have the formula R-OH, where R is an alkyl group. Water can be thought of as H-OH. The reaction of epoxyethane with water can be color-coded like this:
Now do the same thing with the alcohol:
Product molecules of this type are used as solvents.
Notice that the product is still an alcohol. It has an -OH group at the right-hand end of the molecule. If the epoxyethane is in excess, the reaction can continue. (In fact, it continues to some extent even if the epoxyethane isn't in excess.)
The product from this reaction is again an alcohol, and can go on to react with even more epoxyethane! What you get eventually is a chain with a structure:
Compounds of this type are used as plasticisers (added, for example, to PVC to make it more flexible) or as non-ionic surfactants (detergents). To make the surfactant, you would start with a fairly long chain alcohol to produce a molecule such as:
Williamson Ether Synthesis
The Williamson Ether synthesis is the easiest, and perhaps the fastest, way to create ethers.
Introduction
Williamson Ether Reactions involve an alkoxide that reacts with a primary haloalkane or a sulfonate ester. Alkoxides consist of the conjugate base of an alcohol and are comprised of an R group bonded to an oxygen atom. They are often written as RO, where R is the organic substituent.
Sn2 reactions are characterized by the inversion of stereochemistry at the site of the leaving group. Williamson Ether synthesis is not an exception to this rule and the reaction is set in motion by the backside attack of the nucleophile. This requires that the nucleophile and the electrophile are in anti-configuration.
Ethers are prepared by SN2 reactions
Ethers can be synthesized in standard SN2 conditions by coupling an alkoxide with a haloalkane/sulfonate ester. The alcohol that supplies the electron rich alkoxide can be used as the solvent, as well as dimethyl sulfoxide (DMSO) or hexamethylphosphoric triamide (HMPA).
For example
Intramolecular Williamson Ethers
You can also use the Williamson synthesis to produce cyclic ethers. You need a molecule that has a hydroxyl group on one carbon and a halogen atom attached to another carbon. This molecule will then undergo an SN2 reaction with itself, creating a cyclic ether and a halogen anion. Another way of deriving ethers is by converting halo alcohols into cyclic ethers. This reaction is prompted by the deprotonation of the hydrogen attached to the oxygen by an OH- anion. This leads to the departure of the halogen, forming a cyclic ether and halogen radical.
Another factor in determining whether a cyclic ether will be formed is ring size. Three-membered rings along with five membered rings form the fastest, followed by six, four, seven, and lastly eight membered rings. The relative speeds of ring formation are influenced by both enthalpic and entropic contributions.
Enthalpy and Entropy Contributions
Ring strain is the primary enthalpy effect on ring formation however it is not the only thing that effects formation. If this were the case, rings with the most strain would be formed the slowest. The reason why this is not the trend for ring formation is because of entropy conditions. Smaller rings have less entropy making them more favorable because of less ordering of the molecule.
However, the reason why ring formation does not follow this trend is because of another factor called the proximity effect. The proximity effect states that the nucleophilic part of the carbon chain is so close to the electrophilic carbon that a small amount of ring strain is evident in the ground state of the molecule. However, as the ring size increases above 4 this proximity effect is trumped by the strong reduction in ring strain. Five and six membered rings have less strain allowing them to form faster. However, as rings get larger (8,9,10 etc. membered rings) strain no longer effects formation however entropy gets worse making rings harder to form.
Contributors
• Kirtan Patel (UCD) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Ethers/Synthesis_of_Ethers/Making_Epoxyethane_from_Ethene.txt |
Just what is organic? The term "organic" often conjures up the image of life; something very separate from inanimate or not possessing any traits commonly attributed to being alive. That is, when something is alive and it therefore is organic. In the supermarket, we are used to seeing the "organic" label to designate something special, something more connected to nature. However, the term organic in chemistry has a very specific definition involving chemicals and chemistry that involves carbon. As discussed below, there a strong diversity of chemistry associated with organic compounds.
Fundamentals
Includes basic electronic structure, bonding in methane, ethene, benzene and carbonyl compounds, and ideas about electronegativity and bond polarity.
Bonding in Organic Compounds
Kekulé was the first to suggest a sensible structure for benzene. The carbons are arranged in a hexagon, and he suggested alternating double and single bonds between them. Each carbon atom has a hydrogen attached to it.
This diagram is often simplified by leaving out all the carbon and hydrogen atoms! In diagrams of this sort, there is a carbon atom at each corner. You have to count the bonds leaving each carbon to work out how many hydrogens there are attached to it. In this case, each carbon has three bonds leaving it. Because carbon atoms form four bonds, that means you are a bond missing - and that must be attached to a hydrogen atom.
Problems with the Kekulé structure
Although the Kekulé structure was a good attempt in its time, there are serious problems with it regarding chemistry, structure and stability.
The Kekulé structure has problems with the chemistry. Because of the three double bonds, you might expect benzene to have reactions like ethene - only more so! Ethene undergoes addition reactions in which one of the two bonds joining the carbon atoms breaks, and the electrons are used to bond with additional atoms. Benzene rarely does this. Instead, it usually undergoes substitution reactions in which one of the hydrogen atoms is replaced by something new.
The Kekulé structure has problems with the shape. Benzene is a planar molecule (all the atoms lie in one plane), and that would also be true of the Kekulé structure. The problem is that C-C single and double bonds are different lengths.
• C-C (0.154 nm)
• C=C (0.134 nm)
That would mean that the hexagon would be irregular if it had the Kekulé structure, with alternating shorter and longer sides. In real benzene all the bonds are exactly the same - intermediate in length between C-C and C=C at 0.139 nm. Real benzene is a perfectly regular hexagon.
The Kekulé structure has problems with the stability of benzene. Real benzene is a lot more stable than the Kekulé structure would give it credit for. Every time you do a thermochemistry calculation based on the Kekulé structure, you get an answer which is wrong by about 150 kJ mol-1. This is most easily shown using enthalpy changes of hydrogenation. Hydrogenation is the addition of hydrogen to something. If, for example, you hydrogenate ethene you get ethane:
$\ce{CH_2=CH_2 + H_2 \rightarrow CH_3CH_3} \tag{1}$
In order to do a fair comparison with benzene (a ring structure) we're going to compare it with cyclohexene. Cyclohexene, C6H10, is a ring of six carbon atoms containing just one C=C.
When hydrogen is added to this, cyclohexane, C6H12, is formed. The "CH" groups become CH2 and the double bond is replaced by a single one. The structures of cyclohexene and cyclohexane are usually simplified in the same way that the Kekulé structure for benzene is simplified - by leaving out all the carbons and hydrogens.
In the cyclohexane case, for example, there is a carbon atom at each corner, and enough hydrogens to make the total bonds on each carbon atom up to four. In this case, then, each corner represents CH2. The hydrogenation equation could be written:
The enthalpy change during this reaction is -120 kJ mol-1. In other words, when 1 mole of cyclohexene reacts, 120 kJ of heat energy is evolved. Where does this heat energy come from? When the reaction happens, bonds are broken (C=C and H-H) and this costs energy. Other bonds have to be made, and this releases energy. Because the bonds made are stronger than those broken, more energy is released than was used to break the original bonds and so there is a net evolution of heat energy.
If the ring had two double bonds in it initially (cyclohexa-1,3-diene), exactly twice as many bonds would have to be broken and exactly twice as many made. In other words, you would expect the enthalpy change of hydrogenation of cyclohexa-1,3-diene to be exactly twice that of cyclohexene - that is, -240 kJ mol-1.
In fact, the enthalpy change is -232 kJ mol-1 - which isn't far off what we are predicting. Applying the same argument to the Kekulé structure for benzene (what might be called cyclohexa-1,3,5-triene), you would expect an enthalpy change of -360 kJ mol-1, because there are exactly three times as many bonds being broken and made as in the cyclohexene case.
In fact what you get is -208 kJ mol-1 - not even within distance of the predicted value! This is very much easier to see on an enthalpy diagram. Notice that in each case heat energy is released, and in each case the product is the same (cyclohexane). That means that all the reactions "fall down" to the same end point.
Heavy lines, solid arrows and bold numbers represent real changes. Predicted changes are shown by dotted lines and italics. The most important point to notice is that real benzene is much lower down the diagram than the Kekulé form predicts. The lower down a substance is, the more energetically stable it is. This means that real benzene is about 150 kJ mol-1 more stable than the Kekulé structure gives it credit for. This increase in stability of benzene is known as the delocalization energy or resonance energy of benzene.
Contributors
Jim Clark (Chemguide.co.uk) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Fundamentals/Bonding_in_Organic_Compounds/Bonding_in_Benzene%3A_the_Kekule_Structure.txt |
Building the orbital model
Benzene is built from hydrogen atoms (1s1) and carbon atoms (1s22s22px12py1). Each carbon atom has to join to three other atoms (one hydrogen and two carbons) and doesn't have enough unpaired electrons to form the required number of bonds, so it needs to promote one of the 2s2 pair into the empty 2pz orbital.
Figure 1.1: Step 1: Promotion of an electron
There is only a small energy gap between the 2s and 2p orbitals, and an electron is promoted from the 2s to the empty 2p to give 4 unpaired electrons. The extra energy released when these electrons are used for bonding more than compensates for the initial input. The carbon atom is now said to be in an excited state.
Because each carbon is only joining to three other atoms, when the carbon atoms hybridize their outer orbitals before forming bonds, they only need to hybridise three of the orbitals rather than all four. They use the 2s electron and two of the 2p electrons, but leave the other 2p electron unchanged.'
Figure 1.2: Step 2: Hydribidization of the atomic orbitals
The new orbitals formed are called sp2 hybrids, because they are made by an s orbital and two p orbitals reorganizing themselves. The three sp2 hybrid orbitals arrange themselves as far apart as possible - which is at 120° to each other in a plane. The remaining p orbital is at right angles to them. Each carbon atom now looks like the diagram above. This is all exactly the same as happens in ethene.
The difference in benzene is that each carbon atom is joined to two other similar carbon atoms instead of just one. Each carbon atom uses the sp2 hybrids to form sigma bonds with two other carbons and one hydrogen atom. The next diagram shows the sigma bonds formed, but for the moment leaves the p orbitals alone.
Only a part of the ring is shown because the diagram gets extremely cluttered if you try to draw any more. Notice that the p electron on each carbon atom is overlapping with those on both sides of it. This extensive sideways overlap produces a system of pi bonds which are spread out over the whole carbon ring. Because the electrons are no longer held between just two carbon atoms, but are spread over the whole ring, the electrons are said to be delocalized. The six delocalized electrons go into three molecular orbitals - two in each.
In common with the great majority of descriptions of the bonding in benzene, we are only going to show one of these delocalized molecular orbitals for simplicity.
In the diagram, the sigma bonds have been shown as simple lines to make the diagram less confusing. The two rings above and below the plane of the molecule represent one molecular orbital. The two delocalized electrons can be found anywhere within those rings. The other four delocalized electrons live in two similar (but not identical) molecular orbitals.\
The Symbol for benzene
Although you will still come across the Kekulé structure for benzene, for most purposes we use the structure with a hexagon showing the ring of six carbon atoms, each of which has one hydrogen attached. (You have to know that - counting bonds to find out how many hydrogens to add doesn't work in this particular case.)
The circle represents the delocalized electrons. It is essential that you include the circle. If you miss it out, you are drawing cyclohexane and not benzene.
Relating Electronic Structrure to Properties of Benzene
• The shape of benzene: Benzene is a planar regular hexagon, with bond angles of 120°. This is easily explained. It is a regular hexagon because all the bonds are identical. The delocalization of the electrons means that there aren't alternating double and single bonds. It is planar because that is the only way that the p orbitals can overlap sideways to give the delocalized $\pi$ system.
• The energetic stability of benzene: This is accounted for by the delocalization. As a general principle, the more you can spread electrons around - in other words, the more they are delocalized - the more stable the molecule becomes. The extra stability of benzene is often referred to as "delocalization energy".
• The reluctance of benzene to undergo addition reactions: With the delocalized electrons in place, benzene is about 150 kJ mol-1 more stable than it would otherwise be. If you added other atoms to a benzene ring you would have to use some of the delocalized electrons to join the new atoms to the ring. That would disrupt the delocalization and the system would become less stable. Since about 150 kJ per mole of benzene would have to be supplied to break up the delocalization, this isn't going to be an easy thing to do.
Contributors
Jim Clark (Chemguide.co.uk) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Fundamentals/Bonding_in_Organic_Compounds/Bonding_in_Benzene_-_a_Modern_Orbital_View.txt |
The simple view of the bonding in carbon - oxygen double bonds
Where the carbon-oxygen double bond, C=O, occurs in organic compounds it is called a carbonyl group. The simplest compound containing this group is methanal. We are going to look at the bonding in methanal, but it would equally apply to any other compound containing C=O. The interesting thing is the nature of the carbon-oxygen double bond - not what it's attached to.
Note
Methanal is normally written as HCHO. If you wrote it as HCOH, it looks as if it contains an -O-H group - and it doesn't. Methanal is an aldehyde. All aldehydes contain the CHO group.
An orbital view of the bonding in carbon - oxygen double bonds
The carbon atom
Just as in ethene or benzene, the carbon atom is joined to three other atoms. The carbon's electrons rearrange themselves, and promotion and hybridisation give sp2 hybrid orbitals. Promotion gives:
Hybridization of the 2s orbital and two of the 2p orbitals means that the carbon atom now looks like the diagram below. Three sp2 hybrid orbitals are formed and these arrange themselves as far apart in space as they can - at 120° to each other. The remaining p orbital is at right angles to them. This is exactly the same as in ethene or in benzene.
The oxygen atom
Oxygen's electronic structure is 1s22s22px22py12pz1. The 1s electrons are too deep inside the atom to be concerned with the bonding and so we'll ignore them from now on. Hybridisation occurs in the oxygen as well. It is easier to see this using "electrons-in-boxes".
This time two of the sp2 hybrid orbitals contain lone pairs of electrons.
The carbon atom and oxygen atom then bond in much the same way as the two carbons do in ethene. In the next diagram, we are assuming that the carbon will also bond to two hydrogens to make methanal - but it could equally well bond to anything else.
End-to-end overlap between the atomic orbitals that are pointing towards each other produce sigma bonds. Notice that the p orbitals are overlapping sideways.
This sideways overlap produces a pi bond. So just like C=C, C=O is made up of a sigma bond and a pi bond. Does that mean that the bonding is exactly the same as in ethene? No! The distribution of electrons in the pi bond is heavily distorted towards the oxygen end of the bond, because oxygen is much more electronegative than carbon.
This distortion in the pi bond causes major differences in the reactions of compounds containing carbon-oxygen double bonds like methanal compared with compounds containing carbon-carbon double bonds like ethene.
Contributors
Jim Clark (Chemguide.co.uk)
Bonding in Ethene
At a simple level, you will have drawn ethene showing two bonds between the carbon atoms. Each line in this diagram represents one pair of shared electrons. Ethene is actually much more interesting than this.
Ethene is built from hydrogen atoms (1s1) and carbon atoms (1s22s22px12py1). The carbon atom doesn't have enough unpaired electrons to form the required number of bonds, so it needs to promote one of the 2s2 pair into the empty 2pz orbital. This is exactly the same as happens whenever carbon forms bonds - whatever else it ends up joined to.
Promotion of an electron
There is only a small energy gap between the 2s and 2p orbitals, and an electron is promoted from the 2s to the empty 2p to give 4 unpaired electrons. The extra energy released when these electrons are used for bonding more than compensates for the initial input. The carbon atom is now said to be in an excited state.
Hybridization
In the case of ethene, there is a difference from, say, methane or ethane, because each carbon is only joining to three other atoms rather than four. When the carbon atoms hybridize their outer orbitals before forming bonds, this time they only hybridize three of the orbitals rather than all four. They use the 2s electron and two of the 2p electrons, but leave the other 2p electron unchanged.
The new orbitals formed are called sp2 hybrids, because they are made by an s orbital and two p orbitals reorganizing themselves. sp2 orbitals look rather like sp3 orbitals that you have already come across in the bonding in methane, except that they are shorter and fatter. The three sp2 hybrid orbitals arrange themselves as far apart as possible - which is at 120° to each other in a plane. The remaining p orbital is at right angles to them.
The two carbon atoms and four hydrogen atoms would look like this before they joined together:
The various atomic orbitals which are pointing towards each other now merge to give molecular orbitals, each containing a bonding pair of electrons. These are sigma bonds - just like those formed by end-to-end overlap of atomic orbitals in, say, ethane.
The p orbitals on each carbon are not pointing towards each other, and so we'll leave those for a moment. In the diagram, the black dots represent the nuclei of the atoms. Notice that the p orbitals are so close that they are overlapping sideways.
This sideways overlap also creates a molecular orbital, but of a different kind. In this one the electrons aren't held on the line between the two nuclei, but above and below the plane of the molecule. A bond formed in this way is called a $\pi$ bond.
For clarity, the sigma bonds are shown using lines - each line representing one pair of shared electrons. The various sorts of line show the directions the bonds point in. An ordinary line represents a bond in the plane of the screen (or the paper if you've printed it), a broken line is a bond going back away from you, and a wedge shows a bond coming out towards you.
Be clear about what a $\pi$ bond is. It is a region of space in which you can find the two electrons which make up the bond. Those two electrons can live anywhere within that space. It would be quite misleading to think of one living in the top and the other in the bottom. The $\pi$ bond dominates the chemistry of ethene. It is very vulnerable to attack - a very negative region of space above and below the plane of the molecule. It is also somewhat distant from the control of the nuclei and so is a weaker bond than the sigma bond joining the two carbons. All double bonds (whatever atoms they might be joining) will consist of a sigma bond and a pi bond.
The shape of Ethene
The shape of ethene is controlled by the arrangement of the sp2 orbitals. Notice two things about them:
• They all lie in the same plane, with the other p orbital at right angles to it. When the bonds are made, all of the sigma bonds in the molecule must also lie in the same plane. Any twist in the molecule would mean that the p orbitals wouldn't be parallel and touching any more, and you would be breaking the $\pi$ bond.
There is no free rotation about a carbon-carbon double bond. Ethene is a planar molecule.
• The sp2 orbitals are at 120° to each other. When the molecule is constructed, the bond angles will also be 120°. That is approximate though; there will be a slight distortion because you are joining 2 hydrogens and a carbon atom to each carbon, rather than 3 identical groups.
Contributors
Jim Clark (Chemguide.co.uk) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Fundamentals/Bonding_in_Organic_Compounds/Bonding_in_Carbonyl_Compounds.txt |
Ethyne, C2H2, has a triple bond between the two carbon atoms. In the diagram each line represents one pair of shared electrons.
If you have read the ethene page, you will expect that ethyne is going to be more complicated than this simple structure suggests.
An orbital view of the bonding in ethyne
Ethyne is built from hydrogen atoms (1s1) and carbon atoms (1s22s22px12py1). The carbon atom does not have enough unpaired electrons to form four bonds (1 to the hydrogen and three to the other carbon), so it needs to promote one of the 2s2 pair into the empty 2pz orbital. This is exactly the same as happens whenever carbon forms bonds - whatever else it ends up joined to.
Each carbon is only joining to two other atoms rather than four (as in methane or ethane) or three (as in ethene) and so when the carbon atoms hybridise their outer orbitals before forming bonds, this time they only hybridise two of the orbitals. They use the 2s electron and one of the 2p electrons, but leave the other 2p electrons unchanged. The new hybrid orbitals formed are called sp1 hybrids (sometimes just sp hybrids), because they are made by an s orbital and a single p orbital reorganizing themselves.
What these look like in the atom (using the same color coding) is:
Notice that the two green lobes are two different hybrid orbitals - arranged as far apart from each other as possible. Do not confuse them with the shape of a p orbital. The two carbon atoms and two hydrogen atoms would look like this before they joined together:
The various atomic orbitals which are pointing towards each other now merge to give molecular orbitals, each containing a bonding pair of electrons. These are sigma bonds - just like those formed by end-to-end overlap of atomic orbitals in, say, ethane. The sigma bonds are shown as orange in the next diagram. The various p orbitals (now shown in slightly different reds to avoid confusion) are now close enough together that they overlap sideways.
Sideways overlap between the two sets of p orbitals produces two pi bonds - each similar to the pi bond found in, say, ethene. These pi bonds are at 90° to each other - one above and below the molecule, and the other in front of and behind the molecule. Notice the different shades of red for the two different pi bonds.
Contributors
Jim Clark (Chemguide.co.uk)
Bonding in Methane
You will be familiar with drawing methane, CH4, using dots and crosses diagrams, but it is worth looking at its structure a bit more closely.
There is a serious mis-match between this structure and the modern electronic structure of carbon, 1s22s22px12py1. The modern structure shows that there are only 2 unpaired electrons to share with hydrogens, instead of the 4 which the simple view requires.
You can see this more readily using the electrons-in-boxes notation. Only the 2-level electrons are shown. The 1s2 electrons are too deep inside the atom to be involved in bonding. The only electrons directly available for sharing are the 2p electrons. Why then isn't methane CH2?
Promotion of an electron
When bonds are formed, energy is released and the system becomes more stable. If carbon forms 4 bonds rather than 2, twice as much energy is released and so the resulting molecule becomes even more stable.
There is only a small energy gap between the 2s and 2p orbitals, and so it pays the carbon to provide a small amount of energy to promote an electron from the 2s to the empty 2p to give 4 unpaired electrons. The extra energy released when the bonds form more than compensates for the initial input.
The carbon atom is now said to be in an excited state.
Now that we've got 4 unpaired electrons ready for bonding, another problem arises. In methane all the carbon-hydrogen bonds are identical, but our electrons are in two different kinds of orbitals. You aren't going to get four identical bonds unless you start from four identical orbitals.
Hybridization
The electrons rearrange themselves again in a process called hybridization. This reorganizes the electrons into four identical hybrid orbitals called sp3 hybrids (because they are made from one s orbital and three p orbitals). You should read "sp3" as "s p three" - not as "s p cubed".
sp3 hybrid orbitals look a bit like half a p orbital, and they arrange themselves in space so that they are as far apart as possible. You can picture the nucleus as being at the centre of a tetrahedron (a triangularly based pyramid) with the orbitals pointing to the corners. For clarity, the nucleus is drawn far larger than it really is.
What happens when the bonds are formed?
Remember that hydrogen's electron is in a 1s orbital - a spherically symmetric region of space surrounding the nucleus where there is some fixed chance (say 95%) of finding the electron. When a covalent bond is formed, the atomic orbitals (the orbitals in the individual atoms) merge to produce a new molecular orbital which contains the electron pair which creates the bond.
Four molecular orbitals are formed, looking rather like the original sp3 hybrids, but with a hydrogen nucleus embedded in each lobe. Each orbital holds the 2 electrons that we've previously drawn as a dot and a cross.
The principles involved - promotion of electrons if necessary, then hybridisation, followed by the formation of molecular orbitals - can be applied to any covalently-bound molecule.
The shape of methane
When sp3 orbitals are formed, they arrange themselves so that they are as far apart as possible. That is a tetrahedral arrangement, with an angle of 109.5°. Nothing changes in terms of the shape when the hydrogen atoms combine with the carbon, and so the methane molecule is also tetrahedral with 109.5° bond angles.
Ethane, \(C_2H_6\)
Ethane isn't particularly important in its own right, but is included because it is a simple example of how a carbon-carbon single bond is formed. Each carbon atom in the ethane promotes an electron and then forms sp3hybrids exactly as we've described in methane. So just before bonding, the atoms look like this:
The hydrogens bond with the two carbons to produce molecular orbitals just as they did with methane. The two carbon atoms bond by merging their remaining sp3 hybrid orbitals end-to-end to make a new molecular orbital. The bond formed by this end-to-end overlap is called a sigma bond. The bonds between the carbons and hydrogens are also sigma bonds.
In any sigma bond, the most likely place to find the pair of electrons is on a line between the two nuclei.
The shape of ethane around each carbon atom
The shape is again determined by the way the sp3 orbitals are arranged around each carbon atom. That is a tetrahedral arrangement, with an angle of 109.5°. When the ethane molecule is put together, the arrangement around each carbon atom is again tetrahedral with approximately 109.5° bond angles. Why only "approximately"? This time, each carbon atoms doesn't have four identical things attached. There will be a small amount of distortion because of the attachment of 3 hydrogens and 1 carbon, rather than 4 hydrogens.
Free rotation about the carbon-carbon single bond
The two ends of this molecule can spin quite freely about the sigma bond so that there are, in a sense, an infinite number of possibilities for the shape of an ethane molecule. Some possible shapes are:
In each case, the left hand CH3 group has been kept in a constant position so that you can see the effect of spinning the right hand one.
Other alkanes
All other alkanes will be bonded in the same way:
• The carbon atoms will each promote an electron and then hybridize to give sp3 hybrid orbitals.
• The carbon atoms will join to each other by forming sigma bonds by the end-to-end overlap of their sp3 hybrid orbitals.
• Hydrogen atoms will join on wherever they are needed by overlapping their 1s1 orbitals with sp3 hybrid orbitals on the carbon atoms. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Fundamentals/Bonding_in_Organic_Compounds/Bonding_in_Ethyne_%28Acetylene%29.txt |
The molecular formula which defines a very large number of chemical structure, in this particular case, it is a Herculean task to calculate the nature and number of bonds. Earlier Badertscher et al. studied a novel formalism to characterize the degree of unsaturation of organic molecules.1 But no such work has not been taken till now to calculate the number and types of bonds in open chain olefinic system having complex molecular formulae like C176H250, C2000H2000. Keeping this in view, a rapid method has been proposed2,3,4 for the calculation of number of π-bonds, σ-bonds, single and double bonds with the help of following formulae for certain aliphatic unsaturated open chain and cyclic olefinic hydrocarbons.
Open Chain Olefinic Hydrocarbons
Calculation of π-bonds and double bonds (P):
In the first case, we have to count the number of carbon atoms (X) and the number of hydrogen atoms (Y) in a given unsaturated hydrocarbon containing double bonds. The formula to calculate the number of π bonds or double bonds for an aliphatic straight chain olefin is
\[P= \dfrac{2X-Y}{2} + 1 \tag{1}\]
where, X = number of carbon atoms; Y = number of hydrogen atoms and P = number of π bonds/double bonds. E.g.: In C176H250, X = 176, Y = 250, therefore P = (2 x 176 – 250)/2 +1 = 51 + 1 = 52 number of π bonds or double bonds.
Calculation of σ-bonds (S):
In this case, first we have to count the number of carbon atoms (X) and the number of hydrogen atoms (Y) in the given unsaturated hydrocarbon containing double bonds. The formula to calculate the number of σ bonds for an aliphatic straight chain olefin is
\[S = X + Y - 1 \tag{2}\]
where, X = number of carbon atoms; Y = number of hydrogen atoms and S = number of sigma bonds (σ-bonds). E.g.: In C176H250, X = 176, Y = 250, therefore P = 176 + 250 -1 = 425 σ bonds.
Calculation of Single bonds (A):
The total number of single bond for an aliphatic straight chain olefin is
\[A = \dfrac{3Y}{2}-2 \tag{3}\]
where A = number of single bonds and Y is number of hydrogen atoms. E.g.: In C176H250, Y = 250, therefore A =[(3 x 250)/2] = 375 -2 = 373 single bonds. Examples have been illustrated in Table 1.
Table 1: Calculation of π-bonds, σ-bonds, single and double bonds in open chain olefinic hydrocarbons
Example
(CxHy)
Straight-chain Structure
π bond/ bonds
[(2X-Y)/2+1]
σ bonds
[X+Y-1]
Single bonds
[(3Y/2)-2]
Double bond/bonds
[(2X-Y)/2 + 1]
C2H4
H2C=CH2
1
5
4
1
C3H6
H2C=CH-CH3
1
8
7
1
C3H4
H2C=C=CH2
2
6
4
2
C4H8
H2C=CH-CH2-CH3 or
H3C-HC=CH-CH3
1
11
10
1
C4H6
H2C=C=CH-CH3 or
H2C=CH-CH=CH2
2
9
7
2
C4H4
H2C=C=C=CH2
3
7
4
3
C176H250
-
52
425
373
52
C2000H2000
-
1001
3999
2998
1001
C99H4
-
98
102
4
98
Cyclic Olefinic Hydrocarbons
Calculation of π-bonds and double bonds (Pc):
In the first case, we have to count the number of carbon atoms (X) and the number of hydrogen atoms (Y) in the given unsaturated cyclic olefinic hydrocarbons. The formula to calculate the number of π bonds or double bonds for an aliphatic cyclic olefin is
\[P_c= \dfrac{2X-Y}{2} \tag{4}\]
where, X = number of carbon atoms; Y = number of hydrogen atoms and Pc = number of π bonds or double bonds in the cyclic olefinic system. E.g.: In cyclooctatetraene (C8H8), X = Y = 8, therefore Pc = 16-8/2 = 4 number of π bonds or double bonds.
Calculation of σ-bonds (Sc):
In the first case, we have to count the number of carbon atoms (X) and the number of hydrogen atoms (Y) in the given unsaturated cyclic olefinic hydrocarbons. The formula to calculate the number of σ bonds for an aliphatic cyclic olefin is
\[S_c = X + Y \tag{5}\]
where, X = number of carbon atoms; Y = number of hydrogen atoms and Sc = number of sigma bonds (σ-bonds) in cyclic olefinic system. Eg: In cyclooctatetraene (C8H8), X = Y = 8, therefore Sc = 8+8 = 16 number of σ bonds.
Calculation of Single bonds (Ac):
The total number of single bonds in aliphatic cyclic olefin can be calculated by using the formula
\[A_c = \dfrac{3Y}{2} \tag{6}\]
where Ac = number of single bonds and y is number of hydrogen atoms in aliphatic cyclic olefin. E.g.: In cyclooctatetraene (C8H8), Y = 8, therefore Ac = 24/2 = 12 number of single bonds. Examples have been illustrated in Table 2.
Table 2: Calculation of π-bonds, σ-bonds, single and double bonds in Cyclo Alkene system
Example
(CxHy)
Cycloalkene
π bond / bonds (Pc) =
[(2X-Y)/2]
σ bonds (Sc)
[X+Y]
Single bonds (Ac)
[(3Y/2)]
Double bond/bonds
[(2X-Y)/2]
C3H4
Cyclopropene
1
7
6
1
C4H4
Cyclobutadiene
2
8
6
2
C5H6
Cyclopentadiene
2
11
9
2
C6H8
Cyclohexadiene
2
14
12
2
C7H8
Cycloheptatriene
3
15
12
3
C8H8
Cyclooctatetraene
4
16
12
4 | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Fundamentals/Bonding_in_Organic_Compounds/Calculating_of_-bonds_-bonds_single_and_double_bonds_in_Straight_Chain_and_Cycloalkene_Systems.txt |
It was first devised by Huckel in 1931. The present study will be an innovative method1-8 involving two formulae by just manipulating the no of π bonds within the ring system and delocalized electron pair (excluding π electron pair within the ring system) with one (01).
Predicting Aromatic behavior
In the first case, the compound must be cyclic, planar (i.e. all the carbon atoms having same state of hybridization) with even number of A value, where
$A = \pi b + e^-p + 1(constant) \tag{1}$
, here πb = number of π bonds with in the ring system and e-p = number of electron pair outside or adjacent to the ring system i.e. if the ring contains hetero atoms (atoms containing lone pair of electrons) which can undergo delocalization and each negative charge if present may be treated as one pair of electrons.
If the value of ‘A’, for a certain organic compound comes out as even number then this compound will be treated as aromatic compound.
Predicting Anti-aromatic behavior
In the second case, the compound must be cyclic, planar (i.e. all the carbon atoms having same state of hybridization) with odd number of A value, where
$A = \pi b + e^-p + 1(constant)$
here πb = number of π bonds with in the ring system and e-p = number of electron pair outside or adjacent to the ring system i.e. if the ring contains hetero atoms which can undergo delocalization and each negative charge if present, may be treated as one pair of electrons.
If the value of ‘A’, for a certain organic compound comes out as odd number then this compound is anti-aromatic.
General Conditions for Non-Aromatic Behavior
Any compound that lacks one or more of the above features i.e. it may be acyclic / non-planar, is to be treated as non aromatic. But in this case, ‘A’ value may be even or odd number. It is always to be noted that if the ring contains hetero atom like N, O, S etc, in this case we must count that electron pair in the evaluation of ‘A’ value which can undergo delocalization. We never count localized electron pair. Examples have been illustrated in Table 1.
Table 1: Aromatic, anti-aromatic and non-aromatic behavior of organic compounds
Organic Compound
(Cyclic, Planar/Cyclic, non-planar)
πb value
[number of π bonds with in the ring system]
e-p value
[ number of delocalized electron pair outside or adjacent to the ring system]
A value
[A = πb + e-p + 1(constant)]
(even no/odd no)
Nature
of compound
( aromatic/anti-aromatic/non aromatic)
Benzene or [6] annulene
(Cyclic, Planar)
3 π bonds
0
3 + 0+1 = 4
(even no)
Aromatic
Naphthalene
(Cyclic, Planar)
5 π bonds
0
5 + 0 +1 = 6
(even no)
Aromatic
Anthracene
(Cyclic, Planar)
7 π bonds
0
7 + 0 + 1 = 8
(even no)
Aromatic
Cyclopropene
(Cyclic, non planar due to one sp3 hybridized carbon atom)
1 π bond
0
1 + 0 + 1 = 2
(even no)
Non-aromatic
Cyclopropenyl cation
(Cyclic, Planar)
1 π bond
0
1 + 0 + 1 = 2
(even no)
Aromatic
Cyclopropenyl anion
(Cyclic, Planar)
1 π bond
1
(For one negative charge on carbon which undergoes delocalization)
1 + 1+ 1 = 3
(odd no)
Anti-aromatic
Cyclobutadiene or
[4] annulene
(Cyclic, Planar)
2 π bonds
0
2 + 0 + 1 = 3 (odd no)
Anti aromatic
Cyclopentadiene
(Cyclic, non planar due to one sp3 hybridised carbon atom)
2 π bonds
0
2 + 0 + 1 = 3 (odd no)
Non-aromatic
Cyclopentadienyl cation
(Cyclic, Planar)
2 π bonds
0
2 + 0 + 1 = 3 (odd no)
Anti-aromatic
Cyclopentadienyl anion
(Cyclic, Planar)
2 π bonds
01(For one negative charge on carbon which undergo delocalization)
2 + 1 + 1 = 4
(even no)
Aromatic
Cyclooctatetraene or
[8] annulene
(Cyclic, Planar)
4 π bonds
0
4 + 0 + 1 = 5
(odd no)
Anti-aromatic
Cyclooctatrienyl cation
(Cyclic, non-planar due to one sp3 hybridized carbon atom adjacent to positive charge)
3 π bonds
0
3 + 0 + 1 = 4
(even no)
Non aromatic
Pyridine
(Cyclic, Planar)
3 π bonds
0
( Here lone pair on N does not take part in delocalization)
3 + 0 + 1 = 4
(even no)
Aromatic
Pyrrole
2 π bonds
1
( Here lone pair on N take part in delocalization)
2 + 1 + 1 = 4
(even no)
Aromatic
Furan
2 π bonds
1
( Here out of two lone pairs on O only one LP take part in delocalization)
2 + 1 + 1 = 4
(even no)
Aromatic
There are some compounds which do not follow the above rule. Huckel’s also cannot explain the aromatic or non aromatic behavior of these compounds. These compounds have been represented in the Table 2.
Table 2 : Omission behavior of aromatic and non aromatic organic compounds
Organic Compound
(Cyclic, Planar/Cyclic, non-planar)
πb value
[number of π bonds with in the ring system]
e-p value
[ number of delocalized electron pair outside or adjacent to the ring system]
A value
[A = πb + e-p + 1(constant)]
Nature
of compound
Remarks on Exception Behavior
Cyclodecapentaene
or [10]annulene
(C10H10)
5 π bonds
0
5 + 0 + 1 = 6 (even no.)
Not aromatic
Due to the interaction of the hydrogen of 1 and 6 compound become non planar.
(combination of steric and angular strain)
Pyrene
(C16H10)
8 π bonds
0
8 + 0 + 1 = 9 (odd no.)
Aromatic
Because double bonded C15-C16 do not take part in resonance.
If we easily predict the nature of organic compound i.e. aromatic, anti aromatic or non aromatic then we can resolve different kind of problems regarding stability, reactivity, acidity etc. by using the following supposition.
1. Order of stability is aromatic > non aromatic > anti aromatic
2. Order of reactivity just follows the reverse order of stability as Anti-aromatic > non aromatic > aromatic
3. Acidity: Stability of Conjugate base α acidity
eg: cyclopentadienyl anion(aromatic) > cyclopentadiene (non-aromatic) > cyclopentadienyl cation (anti aromatic). Hence, cyclopentadiene (its conjugate base i.e. Cyclopentadienyl anion is aromatic in nature) is much more acidic than cycloheptatriene (its conjugate base i.e. Cycloheptatrienyl anion is anti-aromatic in nature).
Predicting the Hybridization of Heterocyclic Compounds
The state of s-p hybridization of hetero atoms in heterocyclic compounds have been empirically calculated1 from the number of bonds and delocalized ion pair of electrons associated with it in the following way:
\[X = TNBS + DLP \tag{1}\]
where
• \(X\) is the total power on Hybridization State,
• \(TNBS\) is the total Number of bonds directly attached with hetero atom excluding H bond if any attached with hetero atom, and
• \(DLP\) is the delocalized ion pair electrons through resonance
• For sp ; Power on s = 1 and Power on p = 1, Hence Total power = (1+1) = 2
• For sp2 ; Power on s =1 and Power on p = 2, Hence Total power =3 = (1+2) = 3
• For sp3 ; Power on s = 1 and Power on p = 3,Hence Total power = (1+3) = 4
Examples have been illustrated in Table 1
Contributor
• Dr. Arijit Das, Ph.D. (Inorganic Chemistry), MACS ( Invited,USA ), SFICS, MISC, MIAFS (India), Assistant Professor, Department of Chemistry, Ramthakur College, Agartala, Tripura(W), Tripura, India, Pin-799003. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Fundamentals/Bonding_in_Organic_Compounds/Identifing_Aromatic_and_Anti-Aromatic_Compounds.txt |
If you scan any organic textbook you will encounter what appears to be a very large, often intimidating, number of reactions. These are the "tools" of a chemist, and to use these tools effectively, we must organize them in a sensible manner and look for patterns of reactivity that permit us make plausible predictions. Most of these reactions occur at special sites of reactivity known as functional groups, and these constitute one organizational scheme that helps us catalog and remember reactions. Ultimately, the best way to achieve proficiency in organic chemistry is to understand how reactions take place, and to recognize the various factors that influence their course. This is best accomplished by perceiving the reaction pathway or mechanism of a reaction.
Introduction
Organic chemistry encompasses a very large number of compounds (many millions), and our previous discussion and illustrations have focused on their structural characteristics. Now that we can recognize these actors ( compounds ), we turn to the roles they are inclined to play in the scientific drama staged by the multitude of chemical reactions that define organic chemistry. We begin by defining some basic terms that will be used frequently as this subject is elaborated.
• Chemical Reaction: A transformation resulting in a change of composition, constitution and/or configuration of a compound ( referred to as the reactant or substrate ).
• Reactant or Substrate: The organic compound undergoing change in a chemical reaction. Other compounds may also be involved, and common reactive partners ( reagents ) may be identified. The reactant is often ( but not always ) the larger and more complex molecule in the reacting system. Most ( or all ) of the reactant molecule is normally incorporated as part of the product molecule.
• Reagent: A common partner of the reactant in many chemical reactions. It may be organic or inorganic; small or large; gas, liquid or solid. The portion of a reagent that ends up being incorporated in the product may range from all to very little or none.
• Product(s) The final form taken by the major reactant(s) of a reaction.
• Reaction Conditions The environmental conditions, such as temperature, pressure, catalysts & solvent, under which a reaction progresses optimally. Catalysts are substances that accelerate the rate ( velocity ) of a chemical reaction without themselves being consumed or appearing as part of the reaction product. Catalysts do not change equilibria positions.
Chemical reactions are commonly written as equations:
Classification by Structural Change
First, we identify four broad classes of reactions based solely on the structural change occurring in the reactant molecules. This classification does not require knowledge or speculation concerning reaction paths or mechanisms. The letter R in the following illustrations is widely used as a symbol for a generic group. It may stand for simple substituents such as H– or CH3–, or for complex groups composed of many atoms of carbon and other elements.
Four General Reaction Classes
Addition Elimination
Substitution Rearrangement
In an addition reaction the number of σ-bonds in the substrate molecule increases, usually at the expense of one or more π-bonds. The reverse is true of elimination reactions, i.e.the number of σ-bonds in the substrate decreases, and new π-bonds are often formed. Substitution reactions, as the name implies, are characterized by replacement of an atom or group (Y) by another atom or group (Z). Aside from these groups, the number of bonds does not change. A rearrangement reaction generates an isomer, and again the number of bonds normally does not change.
The examples illustrated above involve simple alkyl and alkene systems, but these reaction types are general for most functional groups, including those incorporating carbon-oxygen double bonds and carbon-nitrogen double and triple bonds. Some common reactions may actually be a combination of reaction types. The reaction of an ester with ammonia to give an amide, as shown below, appears to be a substitution reaction ( Y = CH3O & Z = NH2 ); however, it is actually two reactions, an addition followed by an elimination.
The addition of water to a nitrile does not seem to fit any of the above reaction types, but it is simply a slow addition reaction followed by a rapid rearrangement, as shown in the following equation. Rapid rearrangements of this kind are called tautomerizations.
Additional examples illustrating these classes of reaction may be examined by Clicking Here
Classification by Reaction Type
At the beginning, it is helpful to identify some common reaction types that will surface repeatedly as the chemical behavior of different compounds is examined. This is not intended to be a complete and comprehensive list, but should set the stage for future elaborations.
Acidity and Basicity
It is useful to begin a discussion of organic chemical reactions with a review of acid-base chemistry and terminology for several reasons. First, acid-base reactions are among the simplest to recognize and understand. Second, some classes of organic compounds have distinctly acidic properties, and some other classes behave as bases, so we need to identify these aspects of their chemistry. Finally, many organic reactions are catalyzed by acids and/or bases, and although such transformations may seem complex, our understanding of how they occur often begins with the functioning of the catalyst. Organic chemists use two acid-base theories for interpreting and planning their work: the Brønsted theory and the Lewis theory.
Brønsted Theory
According to the Brønsted theory, an acid is a proton donor, and a base is a proton acceptor. In an acid-base reaction, each side of the equilibrium has an acid and a base reactant or product, and these may be neutral species or ions.
H-A + B:(–) A:(–) + B-H
(acid1) (base1) (base2) (acid2)
Structurally related acid-base pairs, such as {H-A and A:(–)} or {B:(–) and B-H} are called conjugate pairs. Substances that can serve as both acids and bases, such as water, are termed amphoteric.
H-Cl + H2O Cl:(–) + H3O(+)
(acid) (base) (base) (acid)
H3N: + H2O NH4(+) + HO(–)
(base) (acid) (acid) (base)
The relative strength of a group of acids (or bases) may be evaluated by measuring the extent of reaction that each group member undergoes with a common base (or acid). Water serves nicely as the common base or acid for such determinations. Thus, for an acid H-A, its strength is proportional to the extent of its reaction with the base water, which is given by the equilibrium constant Keq.
H-A + H2O
H3O(+) + A:(–)
Since these studies are generally extrapolated to high dilution, the molar concentration of water (55.5) is constant and may be eliminated from the denominator. The resulting K value is called the acidity constant, Ka. Clearly, strong acids have larger Ka's than do weaker acids. Because of the very large range of acid strengths (greater than 1040), a logarithmic scale of acidity (pKa) is normally employed. Stronger acids have smaller or more negative pKa values than do weaker acids.
Some useful principles of acid-base reactions are:
Strong acids have weak conjugate bases, and weak acids have strong conjugate bases.
Acid-base equilibria always favor the weakest acid and the weakest base.
Examples of Brønsted Acid-Base Equilibria
Acid-Base Reaction Conjugate
Acids
Conjugate
Bases
Ka pKa
HBr + H2O H3O(+) + Br(–) HBr
H3O(+)
Br(–)
H2O
105 -5
CH3CO2H + H2O H3O(+) + CH3CO2(–) CH3CO2H
H3O(+)
CH3CO2(–)
H2O
1.77*10-5 4.75
C2H5OH + H2O H3O(+) + C2H5O(–) C2H5OH
H3O(+)
C2H5O(–)
H2O
10-16 16
NH3 + H2O H3O(+) + NH2(–) NH3
H3O(+)
NH2(–)
H2O
10-34 34
In all the above examples water acts as a common base. The last example ( NH3 ) cannot be measured directly in water, since the strongest base that can exist in this solvent is hydroxide ion. Consequently, the value reported here is extrapolated from measurements in much less acidic solvents, such as acetonitrile.
Since many organic reactions either take place in aqueous environments ( living cells ), or are quenched or worked-up in water, it is important to consider how a conjugate acid-base equilibrium mixture changes with pH. A simple relationship known as the Henderson-Hasselbach equation provides this information.
When the pH of an aqueous solution or mixture is equal to the pKa of an acidic component, the concentrations of the acid and base conjugate forms must be equal ( the log of 1 is 0 ). If the pH is lowered by two or more units relative to the pKa, the acid concentration will be greater than 99%. On the other hand, if the pH ( relative to pKa ) is raised by two or more units the conjugate base concentration will be over 99%. Consequently, mixtures of acidic and non-acidic compounds are easily separated by adjusting the pH of the water component in a two phase solvent extraction.
For example, if a solution of benzoic acid ( pKa = 4.2 ) in benzyl alcohol ( pKa = 15 ) is dissolved in ether and shaken with an excess of 0.1 N sodium hydroxide ( pH = 13 ), the acid is completely converted to its water soluble ( ether insoluble ) sodium salt, while the alcohol is unaffected. The ether solution of the alcohol may then be separated from the water layer, and pure alcohol recovered by distillation of the volatile ether solvent. The pH of the water solution of sodium benzoate may then be lowered to 1.0 by addition of hydrochloric acid, at which point pure benzoic acid crystallizes, and may be isolated by filtration.
For a discussion of how acidity is influenced by molecular structure Click Here.
Basicity
The basicity of oxygen, nitrogen, sulfur and phosphorus compounds or ions may be treated in an analogous fashion. Thus, we may write base-acid equilibria, which define a Kb and a corresponding pKb. However, a more common procedure is to report the acidities of the conjugate acids of the bases ( these conjugate acids are often "onium" cations ). The pKa's reported for bases in this system are proportional to the base strength of the base. A useful rule here is: pKa + pKb = 14.
We see this relationship in the following two equilibria:
Acid-Base Reaction Conjugate
Acids
Conjugate
Bases
K pK
NH3 + H2O NH4(+) + OH(–) NH4(+)
H2O
NH3
OH(–)
Kb = 1.8*10-5 pKb = 4.74
NH4(+) + H2O H3O(+) + NH3 NH4(+)
H3O(+)
NH3
H2O
Ka = 5.5*10-10 pKa = 9.25
Tables of pKa values for inorganic and organic acids ( and bases) are available in many reference books, and may be examined here by clicking on the appropriate link:
• Inorganic Acidity Constants
• Organic Acidity Constants
• Basicity Constants
Although it is convenient and informative to express pKa values for a common solvent system (usually water), there are serious limitations for very strong and very weak acids. Thus acids that are stronger than the hydronium cation, H3O(+), and weak acids having conjugate bases stronger than hydroxide anion, OH(–), cannot be measured directly in water solution. Solvents such as acetic acid, acetonitrile and nitromethane are often used for studying very strong acids. Relative acidity measurements in these solvents may be extrapolated to water. Likewise, very weakly acidic solvents such as DMSO, acetonitrile, toluene, amines and ammonia may be used to study the acidities of very weak acids. For both these groups, the reported pKa values extrapolated to water are approximate, and many have large uncertainties. A useful table of pKa values in DMSO solution has been compiled from the work of F.G. Bordwell, and may be reached by Clicking Here
Lewis Theory
According to the Lewis theory, an acid is an electron pair acceptor, and a base is an electron pair donor. Lewis bases are also Brønsted bases; however, many Lewis acids, such as BF3, AlCl3 and Mg2+, are not Brønsted acids. The product of a Lewis acid-base reaction, is a neutral, dipolar or charged complex, which may be a stable covalent molecule. Two examples of Lewis acid-base equilibria are shown in equations 1 & 2 below.
In the first example, an electron deficient aluminum atom bonds to a covalent chlorine atom be sharing one of its non-bonding valence electron pairs, and thus achieves an argon-like valence shell octet. Because this sharing is unilateral (chlorine contributes both electrons), both the aluminum and the chlorine have formal charges, as shown. If the carbon chlorine bond in this complex breaks with both the bonding electrons remaining with the more electronegative atom (chlorine), the carbon assumes a positive charge. We refer to such carbon species as carbocations. Carbocations are also Lewis acids, as the reverse reaction demonstrates.
Many carbocations (but not all) may also function as Brønsted acids. Equation 3 illustrates this dual behavior; the Lewis acidic site is colored red and three of the nine acidic hydrogen atoms are colored orange. In its Brønsted acid role the carbocation donates a proton to the base (hydroxide anion), and is converted to a stable neutral molecule having a carbon-carbon double bond.
A terminology related to the Lewis acid-base nomenclature is often used by organic chemists. Here the term electrophile corresponds to a Lewis acid, and nucleophile corresponds to a Lewis base.
Electrophile: An electron deficient atom, ion or molecule that has an affinity for an electron pair, and will bond to a base or nucleophile.
Nucleophile: An atom, ion or molecule that has an electron pair that may be donated in bonding to an electrophile (or Lewis acid).
To learn more about the relationship of basicity and nucleophilicity,
and for examples of acid/base catalysis of organic reactions
Click Here.
Oxidation and Reduction Reactions
A parallel and independent method of characterizing organic reactions is by oxidation-reduction terminology. Carbon atoms may have any oxidation state from –4 (e.g. CH4 ) to +4 (e.g. CO2 ), depending upon their substituents. Fortunately, we need not determine the absolute oxidation state of each carbon atom in a molecule, but only the change in oxidation state of those carbons involved in a chemical transformation. To determine whether a carbon atom has undergone a redox change during a reaction we simply note any changes in the number of bonds to hydrogen and the number of bonds to more electronegative atoms such as O, N, F, Cl, Br, I, & S that has occurred. Bonds to other carbon atoms are ignored. This count should be conducted for each carbon atom undergoing any change during a reaction.
1. If the number of hydrogen atoms bonded to a carbon increases, and/or if the number of bonds to more electronegative atoms decreases, the carbon in question has been reduced (i.e. it is in a lower oxidation state).
2. If the number of hydrogen atoms bonded to a carbon decreases, and/or if the number of bonds to more electronegative atoms increases, the carbon in question has been oxidized (i.e. it is in a higher oxidation state).
3. If there has been no change in the number of such bonds, then the carbon in question has not changed its oxidation state. In the hydrolysis reaction of a nitrile shown above, the blue colored carbon has not changed its oxidation state.
These rules are illustrated by the following four addition reactions involving the same starting material, cyclohexene. Carbon atoms colored blue are reduced, and those colored red are oxidized. In the addition of hydrogen both carbon atoms are reduced, and the overall reaction is termed a reduction. Peracid epoxidation and addition of bromine oxidize both carbon atoms, so these are termed oxidation reactions. Addition of HBr reduces one of the double bond carbon atoms and oxidizes the other; consequently, there is no overall redox change in the substrate molecule.
For a discussion of how oxidation state numbers may be assigned to carbon atoms Click Here.
Since metals such as lithium and magnesium are less electronegative than hydrogen, their covalent bonds to carbon are polarized so that the carbon is negative (reduced) and the metal is positive (oxidized). Thus, Grignard reagent formation from an alkyl halide reduces the substituted carbon atom. In the following equation and half-reactions the carbon atom (blue) is reduced and the magnesium (magenta) is oxidized.
Classification by Functional Group
Functional groups are atoms or small groups of atoms (usually two to four) that exhibit a characteristic reactivity when treated with certain reagents. To view a table of the common functional groups and their class names Click Here. A particular functional group will almost always display its characteristic chemical behavior when it is present in a compound. Because of this, the discussion of organic reactions is often organized according to functional groups. The following table summarizes the general chemical behavior of the common functional groups. For reference, the alkanes provide a background of behavior in the absence of more localized functional groups.
Functional Class Formula Characteristic Reactions
Alkanes C–C, C–H Substitution (of H, commonly by Cl or Br)
Combustion (conversion to CO2 & H2O)
Alkenes C=C–C–H Addition
Substitution (of H)
Alkynes C≡C–H Addition
Substitution (of H)
Alkyl Halides H–C–C–X Substitution (of X)
Elimination (of HX)
Alcohols H–C–C–O–H Substitution (of H); Substitution (of OH)
Elimination (of HOH); Oxidation (elimination of 2H)
Ethers (α)C–O–R Substitution (of OR); Substitution (of α–H)
Amines C–NRH Substitution (of H);
Addition (to N); Oxidation (of N)
Benzene Ring C6H6 Substitution (of H)
Aldehydes (α)C–CH=O Addition
Substitution (of H or α–H)
Ketones (α)C–CR=O Addition
Substitution (of α–H)
Carboxylic Acids (α)C–CO2H Substitution (of H); Substitution (of OH)
Substitution (of α–H); Addition (to C=O)
Carboxylic Derivatives (α)C–CZ=O
(Z = OR, Cl, NHR, etc.)
Substitution (of Z); Substitution (of α–H)
Addition (to C=O)
This table does not include any reference to rearrangement, due to the fact that such reactions are found in all functional classes, and are highly dependent on the structure of the reactant. Furthermore, a review of the overall reaction patterns presented in this table discloses only a broad and rather non-specific set of reactivity trends. This is not surprising, since the three remaining categories provide only a coarse discrimination (comparable to identifying an object as animal, vegetable or mineral). Consequently, apparent similarities may fail to reflect important differences. For example, addition reactions to C=C are significantly different from additions to C=O, and substitution reactions of C-X proceed in very different ways, depending on the hybridization state of carbon.
The Variables of Organic Reactions
In an effort to understand how and why reactions of functional groups take place in the way they do, chemists try to discover just how different molecules and ions interact with each other as they come together. To this end, it is important to consider the various properties and characteristics of a reaction that may be observed and/or measured as the reaction proceeds . The most common and useful of these are listed below:
1. Reactants and Reagents
A. Reactant Structure: Variations in the structure of the reactant may have a marked influence on the course of a reaction, even though the functional group is unchanged. Thus, reaction of 1-bromopropane with sodium cyanide proceeds smoothly to yield butanenitrile, whereas 1-bromo-2,2-dimethylpropane fails to give any product and is recovered unchanged. In contrast, both alkyl bromides form Grignard reagents (RMgBr) on reaction with magnesium.
B. Reagent Characteristics: Apparently minor changes in a reagent may lead to a significant change in the course of a reaction. For example, 2-bromopropane gives a substitution reaction with sodium methylthiolate but undergoes predominant elimination on treatment with sodium methoxide.
2. Product Selectivity
A. Regioselectivity: It is often the case that addition and elimination reactions may, in principle, proceed to more than one product. Thus 1-butene might add HBr to give either 1-bromobutane or 2-bromobutane, depending on which carbon of the double bond receives the hydrogen and which the bromine. If one possible product out of two or more is formed preferentially, the reaction is said to be regioselective.
Simple substitution reactions are not normally considered regioselective, since by definition only one constitutional product is possible. However, rearrangements are known to occur during some reactions.
B. Stereoselectivity: If the reaction products are such that stereoisomers may be formed, a reaction that yields one stereoisomer preferentially is said to be stereoselective. In the addition of bromine to cyclohexene, for example, cis and trans-1,2-dibromocyclohexane are both possible products of the addition. Since the trans-isomer is the only isolated product, this reaction is stereoselective.
C. Stereospecificity: This term is applied to cases in which stereoisomeric reactants behave differently in a given reaction. Examples include:
Here, the (R)-reactant gives the configurationally inverted (S)-product, and (S)-reactant produces (R)-product. The (R) and (S) notations for configuration are described in a later section of this text.
(ii) Different rates of reaction, as in the base-induced eleimination of cis & trans-4-tert-butylcyclohexyl bromide (equation 1 below).
(iii) Different reaction paths leading to different products, as in the base-induced eleimination of cis & trans-2-methylcyclohexyl bromide (equation 2 below).
The mechanisms of these substitution and elimination reactions are discussed in the alkyl halide section of this text.
1. (i) Formation of different stereoisomeric products, as in the reaction of enantiomeric 2-bromobutane isomers with sodium methylthiolate, shown in the following diagram.
3. Reaction Characteristics
1. Reaction Rates: Some reactions proceed very rapidly, and some so slowly that they are not normally observed. Among the variables that influence reaction rates are temperature (reactions are usually faster at a higher temperature), solvent, and reactant / reagent concentrations. Useful information about reaction mechanisms may be obtained by studying the manner in which the rate of a reaction changes as the concentrations of the reactant and reagents are varied. This field of study is called kinetics.
2. Intermediates: Many reactions proceed in a stepwise fashion. This can be convincingly demonstrated if an intermediate species can be isolated and shown to proceed to the same products under the reaction conditions. Some intermediates are stable compounds in their own right; however, some are so reactive that isolation is not possible. Nevertheless, evidence for their existence may be obtained by other means, including spectroscopic observation or inference from kinetic results.
4. Factors that Influence Reactions
It is helpful to identify some general features of a reaction that have a significant influence on its facility. Some of the most important of these are:
1. Energetics: The potential energy of a reacting system changes as the reaction progresses. The overall change may be exothermic ( energy is released ) or endothermic ( energy must be added ), and there is usually an activation energy requirement as well. Tables of Standard Bond Energies are widely used by chemists for estimating the energy change in a proposed reaction. As a rule, compounds constructed of strong covalent bonds are more stable than compounds incorporating one or more relatively weak bonds.
2. Electronic Effects: The distribution of electrons at sites of reaction (functional groups) is a particularly important factor. Electron deficient species or groups, which may or may not be positively charged, are attracted to electron rich species or groups, which may or may not be negatively charged. We refer to these species as electrophiles & nucleophiles respectively. In general, opposites attract and like repel. The charge distribution in a molecule is usually discussed with respect to two interacting effects: An inductive effect, which is a function of the electronegativity differences that exist between atoms (and groups); and a resonance effect, in which electrons move in a discontinuous fashion between parts of a molecule.
3. Steric Effects: Atoms occupy space. When they are crowded together, van der Waals repulsions produce an unfavorable steric hindrance. Steric hindrance may influence conformational equilibria, as well as destabilizing transition states of reactions.
4. Stereoelectronic Effects: In many reactions atomic or molecular orbitals interact in a manner that has an optimal configurational or geometrical alignment. Departure from this alignment inhibits the reaction.
5. Solvent Effects: Most reactions are conducted in solution, not in a gaseous state. The solvent selected for a given reaction may exert a strong influence on its course. Remember, solvents are chemicals, and most undergo chemical reaction under the right conditions.
Mechanisms of Organic Reactions
A detailed description of the changes in structure and bonding that take place in the course of a reaction, and the sequence of such events is called the reaction mechanism. A reaction mechanism should include a representation of plausible electron reorganization, as well as the identification of any intermediate species that may be formed as the reaction progresses. These features are elaborated in the following sections.
The Arrow Notation in Mechanisms
Since chemical reactions involve the breaking and making of bonds, a consideration of the movement of bonding ( and non-bonding ) valence shell electrons is essential to this understanding. It is now common practice to show the movement of electrons with curved arrows, and a sequence of equations depicting the consequences of such electron shifts is termed a mechanism. In general, two kinds of curved arrows are used in drawing mechanisms:
A full head on the arrow indicates the movement or shift of an electron pair:
A partial head (fishhook) on the arrow indicates the shift of a single electron:
The use of these symbols in bond-breaking and bond-making reactions is illustrated below. If a covalent single bond is broken so that one electron of the shared pair remains with each fragment, as in the first example, this bond-breaking is called homolysis. If the bond breaks with both electrons of the shared pair remaining with one fragment, as in the second and third examples, this is called heterolysis.
Bond-Breaking Bond-Making
Other Arrow Symbols
Chemists also use arrow symbols for other purposes, and it is essential to use them correctly.
The Reaction Arrow
The Equilibrium Arrow
The Resonance Arrow
The following equations illustrate the proper use of these symbols:
For further information about the use of curved arrows in reaction mechanisms Click Here.
Reactive Intermediates
The products of bond breaking, shown above, are not stable in the usual sense, and cannot be isolated for prolonged study. Such species are referred to as reactive intermediates, and are believed to be transient intermediates in many reactions. The general structures and names of four such intermediates are given below.
A pair of widely used terms, related to the Lewis acid-base notation, should also be introduced here.
Electrophile: An electron deficient atom, ion or molecule that has an affinity for an electron pair, and will bond to a base or nucleophile.
Nucleophile: An atom, ion or molecule that has an electron pair that may be donated in bonding to an electrophile (or Lewis acid).
Using these definitions, it is clear that carbocations ( called carbonium ions in the older literature ) are electrophiles and carbanions are nucleophiles. Carbenes have only a valence shell sextet of electrons and are therefore electron deficient. In this sense they are electrophiles, but the non-bonding electron pair also gives carbenes nucleophilic character. As a rule, the electrophilic character dominates carbene reactivity. Carbon radicals have only seven valence electrons, and may be considered electron deficient; however, they do not in general bond to nucleophilic electron pairs, so their chemistry exhibits unique differences from that of conventional electrophiles. Radical intermediates are often called free radicals.
The importance of electrophile / nucleophile terminology comes from the fact that many organic reactions involve at some stage the bonding of a nucleophile to an electrophile, a process that generally leads to a stable intermediate or product. Reactions of this kind are sometimes called ionic reactions, since ionic reactants or products are often involved. Some common examples of ionic reactions and their mechanisms may be examined by Clicking Here
The shapes ideally assumed by these intermediates becomes important when considering the stereochemistry of reactions in which they play a role. A simple tetravalent compound like methane, CH4, has a tetrahedral configuration. Carbocations have only three bonds to the charge bearing carbon, so it adopts a planar trigonal configuration. Carbanions are pyramidal in shape ( tetrahedral if the electron pair is viewed as a substituent ), but these species invert rapidly at room temperature, passing through a higher energy planar form in which the electron pair occupies a p-orbital. Radicals are intermediate in configuration, the energy difference between pyramidal and planar forms being very small. Since three points determine a plane, the shape of carbenes must be planar; however, the valence electron distribution varies.
Practice Problems
The following problems include acid base relationships, recognition of different functional groups, recognition of nucleophiles and electrophiles, classification of reactions by structural change and oxidation/reduction change, and the use of curved arrow notation.
Chemical Reactivity
The general idea is that there is a “natural” way to write each half-cell reaction, somehow showing the electron transfer that is characteristic of redox reactions. With inorganic reactions, it is common to show free electrons. With organic reactions, it is common to show species such as free hydrogen atoms or oxygen atoms. Importantly, these various ways to show electron transfer are easily related to each other.
A. Introduction
Abbreviations and conventions:
• ON = oxidation number(s).
• Equations that are incomplete or not balanced are marked with ??? in front of the Eqn number.
• The formal species “hydrogen atom” and “oxygen atom” are shown as H and O, respectively. (This point is discussed further in Notes 2 and 3.) In general chemistry, with its usual emphasis on inorganic reactions, one learns methods for balancing redox equations by focusing on oxidation numbers (ON). However, with organic reactions, the use of ON may be unwieldy. The purpose of this document is to present an alternative approach to balancing organic reactions, avoiding direct use of ON. The approach is chemically logical, using mainstream ideas from organic chemistry. It is also consistent with the use of ON.
B. The reaction: oxidation of ethanol with dichromate
We focus here on one chemical reaction, the oxidation of ethanol by dichromate. We will explore various ways of balancing it -- especially the organic part. The reaction is the oxidation of ethanol by dichromate, to yield ethanoic acid and Cr3+. The reaction occurs in acidic aqueous medium. The final, balanced equation is:
$3 CH_3CH_2OH + 2 Cr_2O_7^{2-} + 16H^+ \rightarrow 3 CH_3COOH + 4Cr^{3+} + 11\,H_2O \tag{1}$
So, how do we get it? The first step is to break the reaction into two parts: the oxidation half cell and the reduction half cell. We understand that the dichromate is oxidizing the organic compound -- and that the Cr itself is being reduced. We thus consider the two half cells separately, and later combine them.
The reduction half cell
The inorganic half cell, the reduction of dichromate to Cr3+, is straightforward, using the usual methods learned in general chemistry. The Cr starts at ON = +6, and ends at ON = +3. Thus each Cr gains 3 electrons. We then balance out the H and O by using species present in the aqueous medium. The result is:
$Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O \tag{2}$
This equation meets all the criteria for a balanced equation. It is balanced for each kind of atom, and for charge. The only thing about the equation that is “unusual” is showing the free electrons. But this is a half-cell reaction. The free electrons reflect a simple view of what happens to the Cr, and we understand that they will not appear in the final overall equation.
The oxidation half cell
The oxidation half reaction is the oxidation of ethanol to ethanoic acid.
Let’s look at this half reaction more carefully, trying to use good organic chemistry. Dealing with ON for carbon often is difficult. (An alert reader may counter that the ON in this particular example are actually not difficult at all. True, but they often are difficult with complex organic compounds. For the record, I show how to balance this equation using ON in Note 1.) Thus in organic chemistry, we often deal with oxidation and reduction of C by looking at changes in the number of H and O. Oxidation is the addition of O or removal of H.
That statement refers to O and H atoms, just as the symbols indicate, not to ions. It is important to understand that, because oxidation and reduction deal with electron transfers. If we are considering a C atom getting oxidized (losing electrons), showing loss of an H atom is chemically logical, but showing loss of an H+ ion is not. The reaction being considered is actually the sum of two distinct reactions. I think it will be useful here to look at the two steps separately, since they are different in the H and O issues. (In practice, many people will be comfortable balancing the complete oxidation half-cell equation without considering the steps. This is addressed below, in Section B.6.)
B.3. Oxidation step 1. Alcohol to aldehyde
$CH_3CH_2OH \rightarrow CH_3CHO + \, ??? \tag{3}$
(Equations that are incomplete or not balanced are marked with ???.) Inspection of Eqn 3 shows that this oxidation reaction involves loss of two hydrogen atoms. Thus we write:
$CH_3CH_2OH \rightarrow CH_3CHO + 2H \tag{4}$
This is a complete and balanced equation for this half reaction. It is also simple and chemically logical. The oxidation involves removing two H atoms, so we show 2 H atoms.
B.4. Oxidation step 2. Aldehyde to acid
$CH_3CHO \rightarrow CH_3COOH ??? \tag{5}$
In this case, the oxidation involves a gain of one oxygen atom. Thus we write:
$CH_3CHO + O \rightarrow CH_3COOH \tag{6}$
This is a complete and balanced equation for this half reaction. Again, it is simple and chemically logical. We can also write this equation in another form. For the moment, this is “for fun” -- just to show two equivalent equations. However, it turns out that the form below will be convenient in the next section.
Eqn 6 uses an O atom to reflect the electrons of this redox reaction. We can transform the equation to use H, instead of O. To do this, we recognize that:
$2H + O = H_2O \tag{7}$
Substituting Eqn 7 in Eqn 6 for O gives:
$CH_3CHO + H_2O \rightarrow CH_3COOH + 2H \tag{8}$
Both Eqn 6 and Eqn 8 are complete and balanced for this second oxidation step. One is no “better” than the other.
B.5. Oxidation: Combining steps 1 and 2
The desired equation is the sum of Eqn 4 (first oxidation step) and Eqn 6 or 8 (second oxidation step). We choose Eqn 8 here for the second step, because the equation for the first step, Eqn 4, is in terms of H. The desired summed equation is:
$CH_3CH_2OH + H_2O \rightarrow CH_3COOH + 4 H \tag{9}$
Again, this is fully balanced for this double-oxidation half reaction. It is balanced for each kind of atom, and it is balanced for charge. Charge was not a difficulty, since all species were written with proper charge at the start. Eqn 9 is chemically logical, chemically instructive, and fully balanced. It is a good chemical equation -- for a half cell. It contains an unrealistic product, the H atom -- just as equations for common inorganic half cells typically contain electrons. The H is not the final product, but we can deal with that separately, when we combine half cells (Section B.7).
B.6. Balancing the overall oxidation half-cell equation without considering the two steps separately
Although I worked through the two oxidation steps above one by one, many will find it easy enough to balance the overall reaction in one step. The oxidation of CH3CH2OH to CH3COOH requires one O atom and releases two H atoms. Just count the atoms. That leads to:
CH3CH2OH + O → CH3COOH + 2 H (10)
That equation is logical and balanced. We then put it in a more convenient form by using Eqn 7; this leads to Eqn 9, as before.
This section combines what was done above in Sections B.3, B.4 and B.5. It is fine to do it either way: going through the individual steps and then adding them, as in the earlier sections, or doing it all at once, as in this section. The purpose of presenting the individual steps earlier was to present one small step at a time, to make the logic clear. It is fine to do multiple steps together, if you
are comfortable doing so.
B.7. Combining the half cells
We now have equations for two half-cell reactions. The reduction half cell is described in Eqn 2. The oxidation half cell is described in Eqn 9. The former shows electron transfer in terms of free electrons; the latter shows electron transfer in terms of hydrogen atoms. To combine these two half-cell equations, we need to get them in the same electron language. To do that, we re-write the H atom, H, as the sum of its parts:
H = H+ + e- (11)
Substituting Eqn 11 in Eqn 9 gives
CH3CH2OH + H2O → CH3COOH + 4 H+ + 4 e- (12)
To obtain the final equation for the overall reaction, we combine the reduction half cell and the oxidation half cell, as usual with redox equations. The reduction half cell is Eqn 2, and shows 6 e-. The form of the oxidation half cell that is convenient here is Eqn 12; it shows 4 e-. To get the electrons to balance out when the two are combined, we take 2 times Eqn 2, and add 3 times Eqn 12. The result is Eqn 1 -- the final desired equation.
3 CH3CH2OH + 2 Cr2O72- + 16H+ → 3 CH3COOH + 4 Cr3+ + 11 H2O (1)
This final equation includes the H+ ion. That is fine here, since the reaction occurs in acidic aqueous medium. The choice of which species are acceptable depends on the specifics of the case at hand. However, free electrons and free hydrogen atoms generally do not belong in final equations for real processes.
C. Perspective
Redox reactions involve electron transfer
To help us visualize and balance redox reactions, we commonly divide them into two parts: one for oxidation and one for reduction. We recognize an oxidation half cell because there is a loss of electrons; equivalently, there may be a loss of hydrogen or gain of oxygen. We recognize a reduction half cell because there is a gain of electrons -- or a gain of hydrogen or loss of oxygen. These “half-cell reactions” are hypothetical, and we often write them with unrealistic species, such as free electrons or hydrogen atoms. These unrealistic species must disappear when we combine the two half cells into an equation for the complete redox reaction. However, the half-cell reactions are instructive to us, as well as useful in working out the complete equation.
For simple chemicals, we may write the redox half cell with free electrons, making use of the oxidation numbers. For example:
$Cr(VI) + 3 e^- \rightarrow Cr(III) \tag{13}$
For more complex chemicals, and particularly for organic chemicals, we may write the redox half cell with free hydrogen or oxygen atoms. For example:
$CH_2=CH_2 + 2 H \rightarrow CH_3CH_3 \tag{14}$
That is, redox half cells may show the formal species e-, H or O. In combining the half cells, we need to eliminate these formal intermediates. To do that, we recognize various relationships between them, all of which follow good chemical logic. We used two such relationships above:
Eqn 7 is a relationship between H and O atoms:
$2 H + O = H_2O \tag{7}$
We used this relationship in Section B.4. In that case, we had two partial reactions, one showing H and one showing O. Eqn 7 allowed us to combine those into an equation showing only one electron form.
Eqn 11 is a relationship between H+ and e-.
$H = H^+ + e^- \tag{11}$
We used this relationship in Section B.7. In that case, we had two partial reactions, one written with $e^-$ and one written with H. Eqn 11 allowed us to combine those into a single equation -- in this case, one with all electron forms canceling out. Another relationship we might have used is:
$O + 2 e^- = O^{2-} \tag{15}$
Overall, Equations 7, 11 and 15 allow us to interconvert among three formal ways of representing electron intermediates: free electrons or hydrogen or oxygen atoms. Thus, we are free to write whichever of them seems most chemically appropriate for a particular half-cell; we can then combine half-cell equations no matter which electron form they show by using these relationships. Some additional relationships that might come up are briefly noted in Section F.
D. Shortcuts
From time to time the question comes up of whether we should teach -- or encourage -- a “shortcut”. This is a gray area, and needs to be considered in each particular case. However, as a general philosophical point, I prefer that we start by teaching things that are correct and logical. If shortcuts follow, that may be fine. Certainly, when students discover a shortcut on their own, they deserve a good explanation of why it works. And there may be times when we teach a shortcut or simplified method because it is not practical to be more rigorous. Examples here where this might be relevant include the breakdown of the oxidation to two steps vs one, and considering the half cells separately, rather than trying to balance the whole equation at once. Whether these are good examples is a matter of taste. As noted, this section is gray. The main point is to encourage the use of good chemical logic as much as possible. Shortcuts should be identified as such.
E. A possible shortcut for balancing the organic half cell
Here is a possible alternative approach that was offered to me for balancing the organic half cell; the approach might be considered a shortcut. The presentation of this shortcut here is loosely based on a real discussion. The purpose here is to discuss the merits of one or another approach. Discussing alternative approaches should serve to highlight their strengths and weaknesses.
For simplicity, let’s consider step 1 of the oxidation reaction discussed above: Eqn 3.
CH3CH2OH → CH3CHO ??? (3)
The suggested approach starts by noting that H+ is a likely product. Since 2 H are removed, we start by writing the following as our “trial equation”:
CH3CH2OH → CH3CHO + 2 H+ ??? (16)
That achieves balance for H, but it is unbalanced for charge. To solve that problem, we now add electrons as needed to achieve charge balance:
CH3CH2OH → CH3CHO + 2 H+ + 2 e- (17)
This is a complete and balanced equation for this half reaction. It is fully equivalent to Eqn 4 -- because
2 H = 2 H+ + 2 e- (by Eqn 11).
Thus the first point is that this approach works; it gives a correct answer. The purpose of discussing it further, comparing it to what was done in Section B.3, is to explore the logic of the two approaches. Both approaches ultimately make use of electrons, in some form. In the approach of Section B.3, presented earlier, the electrons are considered from the start -- because this is a redox Balancing Organic Redox Reactions Page 8 equation, fundamentally involving electron transfer. The first equation written, Eqn 4, is chemically logical and balanced. Further manipulation of the equation may be done for convenience. In contrast, in the shortcut approach, the electrons that are so fundamental are ignored in the initial step, and added back only later to get the equation to work out. The first equation written, Eqn 16, is neither chemically logical nor balanced; further manipulation is required simply to get a meaningful equation. In the end, this comes out fine, but why not just do the chemistry logically in the first place? It is not obvious that the “shortcut” is any shorter, and it is certainly less logical. What, then, is its merit?
In discussing the shortcut with one person, they seemed unsure how to explain what they had done, and unsure what to do in a case that did not ultimately produce H+. That discussion suggested that the shortcut was being used as a blind mechanical trick, with poor understanding of the chemistry. In contrast, the approach in Section B.3 is chemically logical right from the start. Then, it is meshed with whatever specific constraints are relevant to the problem at hand, such as relating H atoms to free electrons, using Eqn 11. Further, the approach is easily generalized to other cases -- simply by looking at what the electrons are doing.
If anyone has an example where they think the proposed shortcut is particularly appropriate, please contribute. Discussing such examples could be instructive.
F. Other relationships
We briefly note some additional cases. These were not relevant in our example above, but are relevant to some reactions. They can be seen as simple extensions to the story above. The first two reflect simple chemistry, and the final group considers biochemical cofactors that carry electrons (hydrogen).
Hydrogen gas
$2 H = H_2 \tag{7}$
Hydride ion
The hydride ion is $H^-$ -- and a good source of electrons, as a reducing agent. For example, the reducing agent $LiAlH_4$ can be thought of as formally containing $Li^+ + Al^{3+} + 4 H^-$. The hydride ion is easily related to other hydrogen species, as desired. One such relationship is:
$H^- = H^+ + 2e^- \tag{18}$
F.3. Biochemical cofactors
In biochemistry, electrons are carried by cofactors, such as flavin adenine dinucleotide (FAD) or nicotinamide adenine dinucleotide (NAD). (The details of these cofactors are not of concern here.) We often show the electrons in the form of hydrogen atoms. Thus we commonly write simply FADH2 and FAD for the two forms of FAD -- with and without the two electrons (two hydrogen atoms).
FAD + 2H = FADH2 (19)
NAD is a bit more complicated, because NAD itself actually is an ion. The idea is the same, however:
NAD+ + 2H = NADH + H+ (20)
There is also a phosphorylated form of NAD, called NADP. However, the phosphate group has no direct role in carrying H, and the relationship is quite the same as for NAD:
NADP+ + 2H = NADPH + H+ (21)
This approach is easily extended to other electron (hydrogen) carriers, such as pyrroloquinoline quinone (PQQ):
PQQ + 2H = PQQH2 (22)
1. Using oxidation numbers (ON)
The particular equation at hand is actually easy enough to balance by the usual procedures using ON, as taught in general chemistry. For the record, we do that here. The basic description of the reaction is
CH3CH2OH + Cr2O72- → CH3COOH + 2 Cr3+ ??? (23)
I have included a coefficient 2 for the Cr3+, since there must be 2 Cr3+ for a single Cr2O72-.
For Cr: The Cr is reduced from +6 to +3, and there are 2 Cr; therefore, the equation as written above involves a 6 electron gain, total, for the two Cr atoms. For C: We assign the usual +1 for H and -2 for O. For CH3CH2OH , there are 6 H = 6+ and 1 O = 2-. Total of H and O is 4+, therefore the total ON for the two C atoms is 4-. Similarly, for CH3COOH, the total ON for the two C atoms is 0. Thus the C atoms, together, lose 4 electrons.
To balance the 6 e- gain of the Cr and the 4 e- loss of the C, we multiply the former by 2 and the latter by 3:
3 CH3CH2OH + 2 Cr2O72- → 3 CH3COOH + 4 Cr3+ ??? (24)
That takes care of the redox (electron transfer) part of the balancing. We now complete the balancing, using water and its common ions. There are 17 O on the left, 6 on the right. So we add 11 H2O to the right to achieve O balance:
3 CH3CH2OH + 2 Cr2O72- → 3 CH3COOH + 4 Cr3+ + 11 H2O ??? (25)
There are now 18 H on the left and 34 on the right. So we add 16 H + to the left to achieve H balance:
3 CH3CH2OH + 2 Cr2O72- + 16 H+ → 3 CH3COOH + 4 Cr3+ + 11 H2O (1)
The usual checking shows that the equation is fully balanced, for each element and for charge. In fact, it is the correct final equation, Eqn 1.
On my web page of practice quizzes for intro organic/biochem is one called “Quiz: Oxidation and reduction”. This is an exercise to help students see the equivalence of the two methods for dealing with redox of carbon compounds: assigning ON and counting H and O. Both methods focus on counting the electrons, the heart of the redox reaction.
2. The hydrogen atom: how to show it in an equation.
There are various ways to show the hydrogen atom. These include H, as I have done, or H· or [H]. It does not matter, so long as we understand what is meant -- in particular, that it is the neutral species, with one electron. (The dot on H· serves to remind us of that one electron. It also relates to the common electron dot formula for hydrogen gas, H:H.) The H atom is not a stable final product (or reactant) in such equations, and will not appear in the final equation for a complete real process (under ordinary conditions). However, it lets us focus on one step of a complex reaction, without worrying about the detail of where the hydrogen comes from or goes.
Sometimes we use [H] to denote “reducing power”. The term may be used qualitatively or quantitatively. For example, the symbol [H] over the reaction arrow means to use a reducing agent, typically some source of H. In biochemistry, 2 H may well end up reducing NAD+ to NADH + H+. This was introduced in
Section F.3
But it is also common that we simply want to be able to count the amount of reducing power, without worrying about its form. My Metabolism handout, posted at the website for Intro Organic/Biochem, has a section on “Electron carriers (H carriers)”. It includes a discussion of how we show hydrogen atoms, with an emphasis on the biochemical usage. Similarly, the hydride ion, H-, might be shown as H:-. (The hydride ion was introduced in Section F.2.)
3. The oxygen atom: how to show it in an equation
This note follows from the preceding note, on the hydrogen atom. We might show the oxygen atom as O or [O]. I choose to show it simply as O. I suppose we might show it with dots, but that seems unwieldy in this case.
Contributors
The discussion here started with M. Farooq Wahab, then of the Chemistry Dept, University of Karachi, noting the difficulty of using ON with organic reactions. He also emphasized the importance of balancing such reactions in analytical chemistry, even though organic chemists are often rather casual about balancing equations. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Fundamentals/Chemical_Reactivity/Balancing_Organic_Redox_Reactions.txt |
This page deals with electronegativity in an organic chemistry context. If you want a wider view of electronegativity.
What is electronegativity?
Electronegativity is a measure of the tendency of an atom to attract a bonding pair of electrons. The Pauling scale is the most commonly used. Fluorine (the most electronegative element) is given a value of 4.0, and values range down to caesium and francium which are the least electronegative at 0.7.
What happens if two atoms of equal electronegativity bond together?
The most obvious example of this is the bond between two carbon atoms. Both atoms will attract the bonding pair to exactly the same extent. That means that on average the electron pair will be found half way between the two nuclei, and you could draw a picture of the bond like this:
It is important to realize that this is an average picture. The electrons are actually in a sigma orbital, and are moving constantly within that orbital.
The carbon-fluorine bond
Fluorine is much more electronegative than carbon. The actual values on the Pauling scale are
carbon 2.5
fluorine 4.0
That means that fluorine attracts the bonding pair much more strongly than carbon does. The bond - on average - will look like this:
Why is fluorine more electronegative than carbon?
A simple dots-and-crosses diagram of a C-F bond is perfectly adequate to explain it.
The bonding pair is in the second energy level of both carbon and fluorine, so in the absence of any other effect, the distance of the pair from both nuclei would be the same. The electron pair is shielded from the full force of both nuclei by the 1s electrons - again there is nothing to pull it closer to one atom than the other. BUT, the fluorine nucleus has 9 protons whereas the carbon nucleus has only 6.
Allowing for the shielding effect of the 1s electrons, the bonding pair feels a net pull of about 4+ from the carbon, but about 7+ from the fluorine. It is this extra nuclear charge which pulls the bonding pair (on average) closer to the fluorine than the carbon.
The carbon-chlorine bond
The electronegativities are:
carbon 2.5
chlorine 3.0
The bonding pair of electrons will be dragged towards the chlorine but not as much as in the fluorine case. Chlorine isn't as electronegative as fluorine.
Why isn't chlorine as electronegative as fluorine?
Chlorine is a bigger atom than fluorine.
• fluorine: 1s22s22px22py22pz1
• chlorine: 1s22s22px22py22pz23s23px23py23pz1
In the chlorine case, the bonding pair will be shielded by all the 1-level and 2-level electrons. The 17 protons on the nucleus will be shielded by a total of 10 electrons, giving a net pull from the chlorine of about 7+.
That is the same as the pull from the fluorine, but with chlorine the bonding pair starts off further away from the nucleus because it is in the 3-level. Since it is further away, it feels the pull from the nucleus less strongly.
Bond polarity and inductive effects
Polar bonds
Think about the carbon-fluorine bond again. Because the bonding pair is pulled towards the fluorine end of the bond, that end is left rather more negative than it would otherwise be. The carbon end is left rather short of electrons and so becomes slightly positive.
The symbols + and - mean "slightly positive" and "slightly negative". You read + as "delta plus" or "delta positive". We describe a bond having one end slightly positive and the other end slightly negative as being polar.
Inductive effects
An atom like fluorine which can pull the bonding pair away from the atom it is attached to is said to have a negative inductive effect.
Most atoms that you will come across have a negative inductive effect when they are attached to a carbon atom, because they are mostly more electronegative than carbon. You will come across some groups of atoms which have a slight positive inductive effect - they "push" electrons towards the carbon they are attached to, making it slightly negative. Inductive effects are sometimes given symbols: -I (a negative inductive effect) and +I (a positive inductive effect).
Hydrogen bromide (and other hydrogen halides)
Bromine (and the other halogens) are all more electronegative than hydrogen, and so all the hydrogen halides have polar bonds with the hydrogen end slightly positive and the halogen end slightly negative. The polarity of these molecules is important in their reactions with alkenes.
The carbon-bromine bond in halogenoalkanes
Bromine is more electronegative than carbon and so the bond is polarised in the way that we have already described with C-F and C-Cl.
The polarity of the carbon-halogen bonds is important in the reactions of the halogenoalkanes.
The carbon-oxygen double bond
An orbital model of the C=O bond in methanal, HCHO, looks like this:
The very electronegative oxygen atom pulls both bonding pairs towards itself - in the sigma bond and the pi bond. That leaves the oxygen fairly negative and the carbon fairly positive.
Contributors
Jim Clark (Chemguide.co.uk) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Fundamentals/Electronegativity.txt |
Compound Class General Structure Functional Group Example
Alcohols
Aldehydes
Alkanes None
Alkenes
Alkynes
Amides
Amines
Anhydrides
Aromatics
Carboxylic Acids
Esters
Ethers
Haloalkanes
Ketones
Nitriles
Thiols
Functional groups A
In organic chemistry, functional groups are specific groups of atoms within molecules arranged in a specific manner. The following tables list common functional groups arranged by heteroatom. The symbols R, R', R'' and R''' usually denote a hydrocarbon chain or a hydrogen but can sometimes be any group of atoms. The symbol X denotes a halide such as F, Br, Cl or I.
Carbon based
Chemical class Group Formula Structural Formula Prefix Suffix Example
Alkane Alkyl RH alkyl- -ane
Butane
Alkene Alkenyl R2C=CR2 alkenyl- -ene
(E)-But-2-ene
Alkyne Alkynyl RC≡CR' alkynyl- -yne
But-2-yne
Allene Allenyl R2C=C=CR2 allenyl- -diene
Penta-2,3-diene
Cumulene Cumulenyl R2C=C=C=CR2 cumulenyl- -triene
Hexa-2,3,4-triene
Benzene derivative Phenyl RC6H5
RPh
phenyl- -benzene
Propylbenzene
(Phenylpropane)
Toluene derivative Benzyl RCH2C6H5
RBn
benzyl- -toluene
Benzylbromide
(α-bromotoluene)
Halide Halo RX halo- -halide
Butyl iodide
(1-Iodobutane)
Oxygen based
Chemical class Group Formula Structural Formula Prefix Suffix Example
Ether Ether ROR' alkoxy- -ether
Diethyl ether
(Ethoxyethane)
Alcohol Hydroxyl ROH hydroxy- -ol
n-Butanol
(Butan-1-ol)
Aldehyde Carbonyl RCHO aldo- -al
Butyraldehyde
(Butanal)
Ketone Carbonyl RCOR' keto- -one
2-Butanone
(Methyl ethyl ketone)
Carboxylic Acid Carboxyl RCO2H carboxy- -oic acid
Butanoic acid
(Butyric acid)
Acyl halide Haloformyl RCOX haloformyl- -oyl halide
Butanoyl bromide
(Butyryl bromide)
Ester Ester RCO2R' / -(o)ate
Ethyl acetate
(Ethyl ethanoate)
Lactone Lactone / / -lactone
γ-Butyrolactone
Peroxy acid Peracid RCO3H per-
peroxy-
-peroxoic acid
Peracetic acid
Peroxyacetic acid
(Ethaneperoxoic acid)
Anhydride Anhydride RCOOCOR' -(o)ic anhydride
Acetic anhydride
(Ethanoic anhydride)
Carbonate Carbonate ROCOOR' / -carbonate
Dimethyl carbonate
Epoxide Epoxide/Oxirane / epoxy- -oxirane / -ene oxide
2-Methyloxirane
1,2-Epoxypropane
1,2-Propylene oxide
Hydrate Diol RC(OH)(OH)R' / -diol
Butane-2,2-diol
Hemiacetal Hemiacetal RCH(OR')(OH) / /
1-Methoxypropan-1-ol
Acetal Acetal RCH(OR')(OR'') dialkoxy- -dialkyl acetal
1,1-Dimethoxypropane
Butyraldehyde dimethyl acetal
Hemiketal Hemiketal RC(OR'')(OH)R' / /
2-Methoxybutan-2-ol
Ketal Ketal RC(OR'')(OR''')R' dialkoxy- -dialkyl acetal
2,2-Dimethoxybutane
2-Butanone dimethyl acetal
Orthoester Orthoester RC(OR')(OR'')(OR''') trimethoxy- -orthoacetate
Trimethylorthoacetate
(1,1,1-Trimethoxyethane)
Orthocarbonate Orthocarbonate ROC(OR')(OR'')(OR''') tetraalkoxy- -orthocarbonate
Tetramethylorthocarbonate
(1,1,1,1-Tetramethoxymethane)
Phenol Phenol PhOH / -phenol
3-Methylphenol
m-Cresol
Enol Enol RC(OH)=CR'2 / -enol
But-2-en-2-ol
Enol ether Enol ether RC(OR')=CR''2 / alkoxy -ene
2-methoxybut-2-ene
Hydroperoxide Hydroperoxide ROOH hydroperoxy- -hydroperoxide
1-Hydroperoxypropane
n-Propylhydroperoxide
Peroxide Peroxide ROOR' peroxy- -peroxide
Di-tert-butyl peroxide
Ketene Ketene RC(=C=O)R' / -ketene
Methyl ethyl ketene
(2-Methylbut-1-en-1-one)
Nitrogen based
Chemical class Group Formula Structural Formula Prefix Suffix Example
Amine Amino RNH2 amino- -amine
Butan-1-amine
1-Aminobutane)
Aldimine / Imine RCHNH2
Butan-1-imine
Ketimine / Imine
Butan-2-imine
Ketenimine
2-Methylbut-1-en-1-imine
Nitrile / Cyanide
Butyronitrile
1-Cyanobutane
Isocyanide / Isonitrile
1-Isocyanopropane
1-Propylisocyanide
Azide
1-Azidopropane
1-Propylazide
Azo compound / Diimide / Diazene
(E)-1,2-diethyldiazene
(Azoethane)
Diazonium salt
Butane-2-diazonium
(2-Diazobutane)
Carbodiimide
N,N'-Diisopropylcarbodiimide
Amidine
Propylamidine
Amidrazone
Propylamidrazone
(Propionimidohydrazide)
Guanidine
1,3-Diethyl-2-methylguanidine
Hydrazine
1-Ethyl-2-methylhydrazine
Hydrazone
Acetone N,N-dimethylhydrazone
Aminal
Propane-2,2-diamine
Hemiaminal
2-Aminopropan-2-ol
Enamine
(Z)-But-2-en-2-amine
Aniline
Hydroxylamine
Amine oxide / N-oxide
Aldoxime
Propionaldehyde oxime
Ketoxime
Aldoxime ether
Propionaldehyde O-methyl oxime
Ketoxime ether
Amide
Lactam
Carbamic acid
Carbamate
Urea
Isourea
Carbamic acyl halide
Hydroxamic acid
Imide
Imidoester
Imidoyl halide
Hydrazide
Cyanate
Isocyanate
Nitro compound
Nitroso
Nitrite
Nitrate
Acyl cyanide
Propionyl cyanide
Semicarbazide
Semicarbazone
Sulfur based
Chemical class Group Formula Structural Formula Prefix Suffix Example
Thioether Thioether RSR' thioalkoxy- -thioether File:.png
Thiol RSH thioxy- -thiol
n-Butanol
(Butan-1-ol)
Thial / Thioaldehyde RCHS File:Thioaldehyde.png File:.png
Thione / Thioketone RCSR' File:Thioketone.png File:.png
[[Thioic O-acid]] RC(S)OH File:Thioic O-acid.png File:.png
[[Thioic S-acid]] RC(O)SH File:Thioic S-acid.png File:.png
Dithioic acid RCS2H File:Dithioic acid.png File:.png
Thioyl_halide RCSX File:Thioyl halide.png File:.png
[[Thioic O-ester]] RC(S)OR' File:Thioic O-ester.png File:.png
[[Thioic S-ester]] RC(O)SR' File:Thioic S-ester.png File:.png
Dithioic ester RCS2R' File:Dithioic acid.png File:.png | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Fundamentals/Functional_Groups.txt |
The homolytic bond dissociation energy is the amount of energy needed to break apart one mole of covalently bonded gases into a pair of radicals. The SI units used to describe bond energy are kiloJoules per mole of bonds (kJ/Mol). It indicates how strongly the atoms are bonded to each other.
Introduction
Breaking a covalent bond between two partners, A-B, can occur either heterolytically, where the shared pair of electron goes with one partner or another
$A-B \rightarrow A^+ + B:^-$
or
$A-B \rightarrow A:^- + B^+ $
or homolytically, where one electron stays with each partner.
$A-B \rightarrow A^• + B^•$
The products of homolytic cleavage are radicals and the energy that is required to break the bond homolytically is called the Bond Dissociation Energy (BDE) and is a measure of the strength of the bond.
Calculation of the BDE
The BDE for a molecule A-B is calculated as the difference in the enthalpies of formation of the products and reactants for homolysis
$BDE = \Delta_fH(A^•) + \Delta_fH(B^•) - \Delta_fH(A-B)$
Officially, the IUPAC definition of bond dissociation energy refers to the energy change that occurs at 0 K, and the symbol is $D_o$. However, it is commonly referred to as BDE, the bond dissociation energy, and it is generally used, albeit imprecisely, interchangeably with the bond dissociation enthalpy, which generally refers to the enthalpy change at room temperature (298K). Although there are technically differences between BDEs at 0 K and 298 K, those difference are not large and generally do not affect interpretations of chemical processes.
Bond Breakage/Formation
Bond dissociation energy (or enthalpy) is a state function and consequently does not depend on the path by which it occurs. Therefore, the specific mechanism in how a bond breaks or is formed does not affect the BDE. Bond dissociation energies are useful in assessing the energetics of chemical processes. For chemical reactions, combining bond dissociation energies for bonds formed and bonds broken in a chemical reaction using Hess's Law can be used to estimate reaction enthalpies.
Example 1: Chlorination of Methane
Consider the chlorination of methane
$CH_4 + Cl_2 \rightarrow CH_3Cl + HCl$
the overall reaction thermochemistry can be calculated exactly by combining the BDEs for the bonds broken and bonds formed
CH4 → CH3• + H• BDE(CH3-H)
Cl2 → 2Cl• BDE(Cl2)\]
H• + Cl• → HCl -BDE(HCl)
CH3• + Cl• → CH3Cl -BDE(CH3-Cl)
---------------------------------------------------
$CH_4 + Cl_2 \rightarrow CH_3Cl + HCl$
$\Delta H = BDE(R-H) + BDE(Cl_2) - BDE(HCl) - BDE(CH_3-Cl)$
Because reaction enthalpy is a state function, it does not matter what reactions are combined to make up the overall process using Hess's Law. However, BDEs are convenient to use because they are readily available.
Alternatively, BDEs can be used to assess individual steps of a mechanism. For example, an important step in free radical chlorination of alkanes is the abstraction of hydrogen from the alkane to form a free radical.
RH + Cl• → R• + HCl
The energy change for this step is equal to the difference in the BDEs in RH and HCl
$\Delta H = BDE(R-H) - BDE(HCl)$
This relationship shows that the hydrogen abstraction step is more favorable when BDE(R-H) is smaller. The difference in energies accounts for the selectivity in the halogenation of hydrocarbons with different types of C-H bonds.
Representative C-H BDEs in Organic Molecules
R-H Do, kJ/mol D298, kJ/mol R-H Do, kJ/mol D298, kJ/mol
CH3-H 432.7±0.1 439.3±0.4 H2C=CH-H 456.7±2.7 463.2±2.9
CH3CH2-H 423.0±1.7 C6H5-H 465.8±1.9 472.4±2.5
(CH3)2CH-H 412.5±1.7 HCCH 551.2±0.1 557.8±0.3
(CH3)3C-H 403.8±1.7
H2C=CHCH2-H 371.5±1.7
HC(O)-H 368.6±0.8 C6H5CH2-H 375.3±2.5
CH3C(O)-H 374.0±1.2
Trends in C-H BDEs
It is important to remember that C-H BDEs refer to the energy it takes to break the bond, and is the difference in energy between the reactants and the products. Therefore, it is not appropriate to interpret BDEs solely in terms of the "stability of the radical products" as is often done.
Analysis of the BDEs shown in the table above shows that there are some systematic trends:
1. BDEs vary with hybridization: Bonds with sp3 hybridized carbons are weakest and bonds with sp hybridized carbons are much stronger. The vinyl and phenyl C-H bonds are similar, reflecting their sp2 hybridization. The correlation with hybridization can be viewed as a reflection of the C-H bond lengths. Longer bonds formed with sp3 orbitals are consequently weaker. Shorter bonds formed with orbitals that have more s-character are similarly stronger.
2. C-H BDEs vary with substitution: Among sp3 hybridized systems, methane has the strongest C-H bond. C-H bonds on primary carbons are stronger than those on secondary carbons, which are stronger than those on tertiary carbons.
Interpretation of C-H BDEs for sp3 Hybridized Carbons
The interpretation of the BDEs in saturated molecules has been subject of recent controversy. As indicated above, the variation in BDEs with substitution has traditionally been interpreted as reflecting the stabilities of the alkyl radicals, with the assessment that more highly substituted radicals are more stable, as with carbocations. Although this is a popular explanation, it fails to account fo the fact the bonds to groups other than H do not show the same types of variation.
R BDE(R-CH3) BDE(R-Cl) BDE(R-Br) BDE(R-OH)
CH3- 377.0±0.4 350.2±0.4 301.7±1.3 385.3±0.4
CH3CH2-
372.4±1.7
354.8±2.1 302.9±2.5 393.3±1.7
(CH3)2CH- 370.7±1.7 356.5±2.1 309.2±2.9 399.6±1.7
(CH3)3C- 366.1±1.7 355.2±2.9 303.8±2.5 400.8±1.7
Therefore, although C-CH3 bonds get weaker with more substitution, the effect is not nearly as large as that observed with C-H bonds. The strengths of C-Cl and C-Br bonds are not affected by substitution, despite the fact that the same radicals are formed as when breaking C-H bonds, and the C-OH bonds in alcohols actually increase with more substitution.
Gronert has proposed that the variation in BDEs is alternately explained as resulting from destabilization of the reactants due to steric repulsion of the substituents, which is released in the nearly planar radicals.1 Considering that BDEs reflect the relative energies of reactants and products, either explanation can account for the trend in BDEs.
Another factor that needs to be considered is the electronegativity. The Pauling definition of electronegativity says that the bond dissociation energy between unequal partners is going to be dependent on the difference in electrongativities, according to the expression
$D_o(A-B) = \dfrac{D_o(A-A) + D_o(B-B)}{2} + (X_A - X_B)^2$
where $X_A$ and $X_B$ are the electronegativities and the bond energies are in eV. Therefore, the variation in BDEs can be interpreted as reflecting variation in the electronegativities of the different types of alkyl fragments.
There is likely some merit in all three interpretations. Since Gronert's original publication of his alternate explanation, there have been many desperate attempts to defend the radical stability explanation.
.
Further Reading
MasterOrganicChemistry
Bond Strengths And Radical Stability | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Fundamentals/Homolytic_C-H_Bond_Dissociation_Energies_of_Organic_Molecules.txt |
This page explains the various ways that organic molecules can be represented on paper or on screen - including molecular formulae, and various forms of structural formulae.
Molecular formulae
A molecular formula simply counts the numbers of each sort of atom present in the molecule, but tells you nothing about the way they are joined together. For example, the molecular formula of butane is $C_4H_{10}$, and the molecular formula of ethanol is $C_2H_6O$.
Molecular formulae are very rarely used in organic chemistry, because they do not give useful information about the bonding in the molecule. About the only place where you might come across them is in equations for the combustion of simple hydrocarbons, for example:
$C_5H_{12} + 8O_2 \rightarrow 5CO_2 + 6H_2O$
In cases like this, the bonding in the organic molecule isn't important.
Structural formulae
A structural formula shows how the various atoms are bonded. There are various ways of drawing this and you will need to be familiar with all of them.
Displayed formulae
A displayed formula shows all the bonds in the molecule as individual lines. You need to remember that each line represents a pair of shared electrons. For example, this is a model of methane together with its displayed formula:
Notice that the way the methane is drawn bears no resemblance to the actual shape of the molecule. Methane isn't flat with 90° bond angles. This mismatch between what you draw and what the molecule actually looks like can lead to problems if you aren't careful. For example, consider the simple molecule with the molecular formula CH2Cl2. You might think that there were two different ways of arranging these atoms if you drew a displayed formula.
The chlorines could be opposite each other or at right angles to each other. But these two structures are actually exactly the same. Look at how they appear as models.
One structure is in reality a simple rotation of the other one. Consider a slightly more complicated molecule, C2H5Cl. The displayed formula could be written as either of these:
But, again these are exactly the same. Look at the models.
The commonest way to draw structural formulae
For anything other than the most simple molecules, drawing a fully displayed formula is a bit of a bother - especially all the carbon-hydrogen bonds. You can simplify the formula by writing, for example, CH3 or CH2 instead of showing all these bonds. For example, ethanoic acid would be shown in a fully displayed form and a simplified form as:
You could even condense it further to CH3COOH, and would probably do this if you had to write a simple chemical equation involving ethanoic acid. You do, however, lose something by condensing the acid group in this way, because you can't immediately see how the bonding works. You still have to be careful in drawing structures in this way. Remember from above that these two structures both represent the same molecule:
The next three structures all represent butane.
All of these are just versions of four carbon atoms joined up in a line. The only difference is that there has been some rotation about some of the carbon-carbon bonds. You can see this in a couple of models.
Not one of the structural formulae accurately represents the shape of butane. The convention is that we draw it with all the carbon atoms in a straight line - as in the first of the structures above. This is even more important when you start to have branched chains of carbon atoms. The following structures again all represent the same molecule - 2-methylbutane.
The two structures on the left are fairly obviously the same - all we've done is flip the molecule over. The other one isn't so obvious until you look at the structure in detail. There are four carbons joined up in a row, with a CH3 group attached to the next-to-end one. That's exactly the same as the other two structures. If you had a model, the only difference between these three diagrams is that you have rotated some of the bonds and turned the model around a bit.
To overcome this possible confusion, the convention is that you always look for the longest possible chain of carbon atoms, and then draw it horizontally. Anything else is simply hung off that chain. It does not matter in the least whether you draw any side groups pointing up or down. All of the following represent exactly the same molecule.
If you made a model of one of them, you could turn it into any other one simply by rotating one or more of the carbon-carbon bonds.
How to draw structural formulae in 3-dimensions
There are occasions when it is important to be able to show the precise 3-D arrangement in parts of some molecules. To do this, the bonds are shown using conventional symbols:
For example, you might want to show the 3-D arrangement of the groups around the carbon which has the -OH group in butan-2-ol.
Example 1: butan-2-ol
Butan-2-ol has the structural formula:
Using conventional bond notation, you could draw it as, for example:
The only difference between these is a slight rotation of the bond between the centre two carbon atoms. This is shown in the two models below. Look carefully at them - particularly at what has happened to the lone hydrogen atom. In the left-hand model, it is tucked behind the carbon atom. In the right-hand model, it is in the same plane. The change is very slight.
It doesn't matter in the least which of the two arrangements you draw. You could easily invent other ones as well. Choose one of them and get into the habit of drawing 3-dimensional structures that way. My own habit (used elsewhere on this site) is to draw two bonds going back into the paper and one coming out - as in the left-hand diagram above.
Notice that no attempt was made to show the whole molecule in 3-dimensions in the structural formula diagrams. The CH2CH3 group was left in a simple form. Keep diagrams simple - trying to show too much detail makes the whole thing amazingly difficult to understand!
Skeletal formulae
In a skeletal formula, all the hydrogen atoms are removed from carbon chains, leaving just a carbon skeleton with functional groups attached to it. For example, we've just been talking about butan-2-ol. The normal structural formula and the skeletal formula look like this:
In a skeletal diagram of this sort
• there is a carbon atom at each junction between bonds in a chain and at the end of each bond (unless there is something else there already - like the -OH group in the example);
• there are enough hydrogen atoms attached to each carbon to make the total number of bonds on that carbon up to 4.
Beware! Diagrams of this sort take practice to interpret correctly - and may well not be acceptable to your examiners (see below).
There are, however, some very common cases where they are frequently used. These cases involve rings of carbon atoms which are surprisingly awkward to draw tidily in a normal structural formula. Cyclohexane, C6H12, is a ring of carbon atoms each with two hydrogens attached. This is what it looks like in both a structural formula and a skeletal formula.
And this is cyclohexene, which is similar but contains a double bond:
But the commonest of all is the benzene ring, C6H6, which has a special symbol of its own.
Deciding which sort of formula to use
There's no easy, all-embracing answer to this problem. It depends more than anything else on experience - a feeling that a particular way of writing a formula is best for the situation you are dealing with.
Don't worry about this - as you do more and more organic chemistry, you will probably find it will come naturally. You'll get so used to writing formulae in reaction mechanisms, or for the structures for isomers, or in simple chemical equations, that you won't even think about it.
Contributors
Jim Clark (Chemguide.co.uk) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Fundamentals/How_to_Draw_Organic_Molecules.txt |
Hybridization was introduced to explain molecular structure when the valence bond theory failed to correctly predict them. It is experimentally observed that bond angles in organic compounds are close to 109o, 120o, or 180o. According to Valence Shell Electron Pair Repulsion (VSEPR) theory, electron pairs repel each other and the bonds and lone pairs around a central atom are generally separated by the largest possible angles.
Introduction
Carbon is a perfect example showing the value of hybrid orbitals. Carbon's ground state configuration is:
According to Valence Bond Theory, carbon should form two covalent bonds, resulting in a CH2, because it has two unpaired electrons in its electronic configuration.However, experiments have shown that \(CH_2\) is highly reactive and cannot exist outside of a reaction. Therefore, this does not explain how CH4 can exist. To form four bonds the configuration of carbon must have four unpaired electrons.
One way CH4 can be explained is, the 2s and the 3 2p orbitals combine to make four, equal energy sp3 hybrid orbitals. That would give us the following configuration:
Now that carbon has four unpaired electrons it can have four equal energy bonds. The hybridization of orbitals is favored because hybridized orbitals are more directional which leads to greater overlap when forming bonds, therefore the bonds formed are stronger. This results in more stable compounds when hybridization occurs.
The next section will explain the various types of hybridization and how each type helps explain the structure of certain molecules.
sp3 hybridization
sp3 hybridization can explain the tetrahedral structure of molecules. In it, the 2s orbitals and all three of the 2p orbitals hybridize to form four sp3 orbitals, each consisting of 75% p character and 25% s character. The frontal lobes align themselves in the manner shown below. In this structure, electron repulsion is minimized.
Energy changes occurring in hybridization
Hybridization of an s orbital with all three p orbitals (px , py, and pz) results in four sp3 hybrid orbitals. sp3 hybrid orbitals are oriented at bond angle of 109.5o from each other. This 109.5o arrangement gives tetrahedral geometry (Figure 4).
Example: sp3 Hybridization in Methane
Because carbon plays such a significant role in organic chemistry, we will be using it as an example here. Carbon's 2s and all three of its 2p orbitals hybridize to form four sp3 orbitals. These orbitals then bond with four hydrogen atoms through sp3-s orbital overlap, creating methane. The resulting shape is tetrahedral, since that minimizes electron repulsion.
Hybridization
Lone Pairs: Remember to take into account lone pairs of electrons. These lone pairs cannot double bond so they are placed in their own hybrid orbital. This is why H2O is tetrahedral. We can also build sp3d and sp3d2 hybrid orbitals if we go beyond s and p subshells.
sp2 hybridization
sp2 hybridization can explain the trigonal planar structure of molecules. In it, the 2s orbitals and two of the 2p orbitals hybridize to form three sp orbitals, each consisting of 67% p and 33% s character. The frontal lobes align themselves in the trigonal planar structure, pointing to the corners of a triangle in order to minimize electron repulsion and to improve overlap. The remaining p orbital remains unchanged and is perpendicular to the plane of the three sp2 orbitals.
Energy changes occurring in hybridization
Hybridization of an s orbital with two p orbitals (px and py) results in three sp2 hybrid orbitals that are oriented at 120o angle to each other (Figure 3). Sp2 hybridization results in trigonal geometry.
Example: sp2 Hybridization in Aluminum Trihydride
In aluminum trihydride, one 2s orbital and two 2p orbitals hybridize to form three sp2 orbitals that align themselves in the trigonal planar structure. The three Al sp2 orbitals bond with with 1s orbitals from the three hydrogens through sp2-s orbital overlap.
Example: sp2 Hybridization in Ethene
Similar hybridization occurs in each carbon of ethene. For each carbon, one 2s orbital and two 2p orbitals hybridize to form three sp2 orbitals. These hybridized orbitals align themselves in the trigonal planar structure. For each carbon, two of these sp orbitals bond with two 1s hydrogen orbitals through s-sp orbital overlap. The remaining sp2 orbitals on each carbon are bonded with each other, forming a bond between each carbon through sp2-sp2 orbital overlap. This leaves us with the two p orbitals on each carbon that have a single carbon in them. These orbitals form a ? bonds through p-p orbital overlap, creating a double bond between the two carbons. Because a double bond was created, the overall structure of the ethene compound is linear. However, the structure of each molecule in ethene, the two carbons, is still trigonal planar.
sp Hybridization
sp Hybridization can explain the linear structure in molecules. In it, the 2s orbital and one of the 2p orbitals hybridize to form two sp orbitals, each consisting of 50% s and 50% p character. The front lobes face away from each other and form a straight line leaving a 180° angle between the two orbitals. This formation minimizes electron repulsion. Because only one p orbital was used, we are left with two unaltered 2p orbitals that the atom can use. These p orbitals are at right angles to one another and to the line formed by the two sp orbitals.
Energy changes occurring in hybridization
Figure 1: Notice how the energy of the electrons lowers when hybridized.
These p orbitals come into play in compounds such as ethyne where they form two addition? bonds, resulting in in a triple bond. This only happens when two atoms, such as two carbons, both have two p orbitals that each contain an electron. An sp hybrid orbital results when an s orbital is combined with p orbital (Figure 2). We will get two sp hybrid orbitals since we started with two orbitals (s and p). sp hybridization results in a pair of directional sp hybrid orbitals pointed in opposite directions. These hybridized orbitals result in higher electron density in the bonding region for a sigma bond toward the left of the atom and for another sigma bond toward the right. In addition, sp hybridization provides linear geometry with a bond angle of 180o.
Example: sp Hybridization in Magnesium Hydride
In magnesium hydride, the 3s orbital and one of the 3p orbitals from magnesium hybridize to form two sp orbitals. The two frontal lobes of the sp orbitals face away from each other forming a straight line leading to a linear structure. These two sp orbitals bond with the two 1s orbitals of the two hydrogen atoms through sp-s orbital overlap.
Hybridization
Example: sp Hybridization in Ethyne
The hybridization in ethyne is similar to the hybridization in magnesium hydride. For each carbon, the 2s orbital hybridizes with one of the 2p orbitals to form two sp hybridized orbitals. The frontal lobes of these orbitals face away from each other forming a straight line. The first bond consists of sp-sp orbital overlap between the two carbons. Another two bonds consist of s-sp orbital overlap between the sp hybridized orbitals of the carbons and the 1s orbitals of the hydrogens. This leaves us with two p orbitals on each carbon that have a single carbon in them. This allows for the formation of two ? bonds through p-p orbital overlap. The linear shape, or 180° angle, is formed because electron repulsion is minimized the greatest in this position.
Hybridization
Problems
Using the Lewis Structures, try to figure out the hybridization (sp, sp2, sp3) of the indicated atom and indicate the atom's shape.
1. The carbon.
2. The oxygen.
3. The carbon on the right.
Answers
1. sp2- Trigonal Planar
The carbon has no lone pairs and is bonded to three hydrogens so we just need three hybrid orbitals, aka sp2.
2. sp3 - Tetrahedral
Don't forget to take into account all the lone pairs. Every lone pair needs it own hybrid orbital. That makes three hybrid orbitals for lone pairs and the oxygen is bonded to one hydrogen which requires another sp3 orbital. That makes 4 orbitals, aka sp3.
3. sp - Linear
The carbon is bonded to two other atoms, that means it needs two hybrid orbitals, aka sp.
An easy way to figure out what hybridization an atom has is to just count the number of atoms bonded to it and the number of lone pairs. Double and triple bonds still count as being only bonded to one atom. Use this method to go over the above problems again and make sure you understand it. It's a lot easier to figure out the hybridization this way.
Contributors
• Harpreet Chima (UCD), Farah Yasmeen | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Fundamentals/Hybrid_Orbitals.txt |
There is no simple way of predicting how many isomers a given molecular formula will yield, (it can range from one to many). Structures are different if they cannot be superimposed upon one another. Keep in mind that there is rotation about all single bonds not involved in a ring, but not about double bonds. Because all of the formulas that you will be dealing with are based on the \(C\) atom, it may be useful to review the ways that \(C\) can bond to itself and to other atoms. We will limit ourselves, for now, to the \(C\) atom with four bonds. Below are the possible combinations of C having a total of four bonds.
In a hydrocarbon where all the C atoms have only single bonds and no rings are involved, the compound would have the maximum number of H atoms. If any of the bonds are replaced with double or triple bonds, or if rings are involved, there would be a “deficiency” of H atoms. By calculating the index of hydrogen deficiency (IHD), we can tell from the molecular formula whether and how many multiple bonds and rings are involved. IHD is also called the Degree of Unstaturation. This will help cut down the possibilities one has to consider in trying to come up with all the isomers of a given formula.
Here is a summary of how the index of hydrogen deficiency (IHD) works.
• A double bond and ring each counts as one IHD.
• A triple bond counts as two IHD.
Hydrocarbons (\(C_xH_y\)):
\[IHD = \dfrac{2x + 2 - y}{2} \]
(where \(x\) and \(y\) stand for # of C and H respectively.)
Example 1
IHD for \(C_2H_4\) is
\[ \dfrac{2(2) + 2 - 4}{2} = 1\]
This means it can have either one double bond or one ring, but it cannot have a triple bond. Since you cannot form a ring with only two C’s, it must have a double bond.
Example 2
IHD for \(C_4H_6\) is
\[\dfrac{2(4) + 2 - 6}{2} = 2\]
This means it can have either one double bond and a ring such as or two double bonds such as CH2=CH−CH=CH2 or CH2=C=CH−CH3 or two rings , or one triple bond, such as CH3C≡CCH3.
Compounds Containing Elements Other than C and H
O and S atoms do not affect the IHD.
• Halogens (F, Cl, Br, I) are treated like H atoms (CH2Cl2 has the same IHD as CH4).
• For each N, add one to the number of C and one to the number H (CH5N is treated as C2H6. CH4N2O is treated as C3H6 by adding 2 to # of C and 2 to # of H).
Do not forget that when double bonds and rings are involved, geometric isomers are possible.
Practice problems
Calculate the IHD for each of the following and see whether it corresponds to the structure shown. (Obviously it should!) Don’t peek until you’ve worked it out yourself, but answers are provided at the bottom.
a)
b) CH3CHCHCH2CHCH2
c)
d)
e) CH3C≡CCOCH3
a) IHD = 3
b) IHD = 2
c) IHD = 5
d) IHD = 1
e) IHD = 3 | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Fundamentals/Index_of_Hydrogen_Deficiency_%28IHD%29.txt |
The molecule is the smallest observable group of uniquely bonded atoms that represent the composition, configuration and characteristics of a pure compound. Our chief focus up to this point has been to discover and describe the ways in which atoms bond together to form molecules. Since all observable samples of compounds and mixtures contain a very large number of molecules (~1020), we must also concern ourselves with interactions between molecules, as well as with their individual structures. Indeed, many of the physical characteristics of compounds that are used to identify them (e.g. boiling points, melting points and solubilities) are due to intermolecular interactions.
All atoms and molecules have a weak attraction for one another, known as van der Waals attraction. This attractive force has its origin in the electrostatic attraction of the electrons of one molecule or atom for the nuclei of another. If there were no van der Waals forces, all matter would exist in a gaseous state, and life as we know it would not be possible. It should be noted that there are also smaller repulsive forces between molecules that increase rapidly at very small intermolecular distances.
Boiling Points
For general purposes it is useful to consider temperature to be a measure of the kinetic energy of all the atoms and molecules in a given system. As temperature is increased, there is a corresponding increase in the vigor of translational and rotation motions of all molecules, as well as the vibrations of atoms and groups of atoms within molecules. Experience shows that many compounds exist normally as liquids and solids; and that even low-density gases, such as hydrogen and helium, can be liquefied at sufficiently low temperature and high pressure. A clear conclusion to be drawn from this fact is that intermolecular attractive forces vary considerably, and that the boiling point of a compound is a measure of the strength of these forces. Thus, in order to break the intermolecular attractions that hold the molecules of a compound in the condensed liquid state, it is necessary to increase their kinetic energy by raising the sample temperature to the characteristic boiling point of the compound.
The following table illustrates some of the factors that influence the strength of intermolecular attractions. The formula of each entry is followed by its formula weight in parentheses and the boiling point in degrees Celsius. First there is molecular size. Large molecules have more electrons and nuclei that create van der Waals attractive forces, so their compounds usually have higher boiling points than similar compounds made up of smaller molecules. It is very important to apply this rule only to like compounds. The examples given in the first two rows are similar in that the molecules or atoms are spherical in shape and do not have permanent dipoles. Molecular shape is also important, as the second group of compounds illustrate. The upper row consists of roughly spherical molecules, whereas the isomers in the lower row have cylindrical or linear shaped molecules. The attractive forces between the latter group are generally greater. Finally, permanent molecular dipoles generated by polar covalent bonds result in even greater attractive forces between molecules, provided they have the mobility to line up in appropriate orientations. The last entries in the table compare non-polar hydrocarbons with equal-sized compounds having polar bonds to oxygen and nitrogen. Halogens also form polar bonds to carbon, but they also increase the molecular mass, making it difficult to distinguish among these factors.
Table 1: Boiling Points (ºC) of Selected Elements and Compounds
Increasing Size
Atomic Ar (40) -186 Kr (83) -153 Xe (131) -109
Molecular CH4 (16) -161 (CH3)4C (72) 9.5 (CH3)4Si (88) 27 CCl4 (154) 77
Molecular Shape
Spherical: (CH3)4C (72) 9.5 (CH3)2CCl2 (113) 69 (CH3)3CC(CH3)3 (114) 106
Linear: CH3(CH2)3CH3 (72) 36 Cl(CH2)3Cl (113) 121 CH3(CH2)6CH3 (114) 126
Molecular Polarity
Non-polar: H2C=CH2 (28) -104 F2 (38) -188 CH3C≡CCH3 (54) -32 CF4 (88) -130
Polar: H2C=O (30) -21 CH3CH=O (44) 20 (CH3)3N (59) 3.5 (CH3)2C=O (58) 56
HC≡N (27) 26 CH3C≡N (41) 82 (CH2)3O (58) 50 CH3NO2 (61) 101
The melting points of crystalline solids cannot be categorized in as simple a fashion as boiling points. The distance between molecules in a crystal lattice is small and regular, with intermolecular forces serving to constrain the motion of the molecules more severely than in the liquid state. Molecular size is important, but shape is also critical, since individual molecules need to fit together cooperatively for the attractive lattice forces to be large. Spherically shaped molecules generally have relatively high melting points, which in some cases approach the boiling point. This reflects the fact that spheres can pack together more closely than other shapes. This structure or shape sensitivity is one of the reasons that melting points are widely used to identify specific compounds. The data in the following table serves to illustrate these points.
Table 2: Alkane Properties
Compound Formula Boiling Point Melting Point
pentane CH3(CH2)3CH3 36ºC –130ºC
hexane CH3(CH2)4CH3 69ºC –95ºC
heptane CH3(CH2)5CH3 98ºC –91ºC
octane CH3(CH2)6CH3 126ºC –57ºC
nonane CH3(CH2)7CH3 151ºC –54ºC
decane CH3(CH2)8CH3 174ºC –30ºC
tetramethylbutane (CH3)3C-C(CH3)3 106ºC +100ºC
Notice that the boiling points of the unbranched alkanes (pentane through decane) increase rather smoothly with molecular weight, but the melting points of the even-carbon chains increase more than those of the odd-carbon chains. Even-membered chains pack together in a uniform fashion more compactly than do odd-membered chains. The last compound, an isomer of octane, is nearly spherical and has an exceptionally high melting point (only 6º below the boiling point).
Hydrogen Bonding
The most powerful intermolecular force influencing neutral (uncharged) molecules is the hydrogen bond. If we compare the boiling points of methane (CH4) -161ºC, ammonia (NH3) -33ºC, water (H2O) 100ºC and hydrogen fluoride (HF) 19ºC, we see a greater variation for these similar sized molecules than expected from the data presented above for polar compounds. This is shown graphically in the following chart. Most of the simple hydrides of group IV, V, VI & VII elements display the expected rise in boiling point with molecular mass, but the hydrides of the most electronegative elements (nitrogen, oxygen and fluorine) have abnormally high boiling points for their mass.
The exceptionally strong dipole-dipole attractions that cause this behavior are called the hydrogen bond. Hydrogen forms polar covalent bonds to more electronegative atoms such as oxygen, and because a hydrogen atom is quite small, the positive end of the bond dipole (the hydrogen) can approach neighboring nucleophilic or basic sites more closely than can other polar bonds. Coulombic forces are inversely proportional to the sixth power of the distance between dipoles, making these interactions relatively strong, although they are still weak (ca. 4 to 5 kcal per mole) compared with most covalent bonds. The unique properties of water are largely due to the strong hydrogen bonding that occurs between its molecules. In the following diagram the hydrogen bonds are depicted as magenta dashed lines.
The molecule providing a polar hydrogen for a hydrogen bond is called a donor. The molecule that provides the electron rich site to which the hydrogen is attracted is called an acceptor. Water and alcohols may serve as both donors and acceptors, whereas ethers, aldehydes, ketones and esters can function only as acceptors. Similarly, primary and secondary amines are both donors and acceptors, but tertiary amines function only as acceptors. Once you are able to recognize compounds that can exhibit intermolecular hydrogen bonding, the relatively high boiling points they exhibit become understandable. The data in the following table serve to illustrate this point.
Table: Boiling Points vs. Intermolecular Forces
Compound Formula Mol. Wt. Boiling Point Melting Point
dimethyl ether CH3OCH3 46 –24ºC –138ºC
ethanol CH3CH2OH 46 78ºC –130ºC
propanol CH3(CH2)2OH 60 98ºC –127ºC
diethyl ether (CH3CH2)2O 74 34ºC –116ºC
propyl amine CH3(CH2)2NH2 59 48ºC –83ºC
methylaminoethane CH3CH2NHCH3 59 37ºC
trimethylamine (CH3)3N 59 3ºC –117ºC
ethylene glycol HOCH2CH2OH 62 197ºC –13ºC
acetic acid CH3CO2H 60 118ºC 17ºC
ethylene diamine H2NCH2CH2NH2 60 118ºC 8.5ºC
Alcohols boil cosiderably higher than comparably sized ethers (first two entries), and isomeric 1º, 2º & 3º-amines, respectively, show decreasing boiling points, with the two hydrogen bonding isomers being substantially higher boiling than the 3º-amine (entries 5 to 7). Also, O–H---O hydrogen bonds are clearly stronger than N–H---N hydrogen bonds, as we see by comparing propanol with the amines. As expected, the presence of two hydrogen bonding functions in a compound raises the boiling point even further.
Acetic acid (the ninth entry) is an interesting case. A dimeric species, shown above, held together by two hydrogen bonds is a major component of the liquid state. If this is an accurate representation of the composition of this compound then we would expect its boiling point to be equivalent to that of a C4H8O4 compound (formula weight = 120). A suitable approximation of such a compound is found in tetramethoxymethane, (CH3O)4C, which is actually a bit larger (formula weight = 136) and has a boiling point of 114ºC. Thus, the dimeric hydrogen bonded structure appears to be a good representation of acetic acid in the condensed state.
A related principle is worth noting at this point. Although the hydrogen bond is relatively weak (ca. 4 to 5 kcal per mole), when several such bonds exist the resulting structure can be quite robust. The hydrogen bonds between cellulose fibers confer great strength to wood and related materials.
Properties of Crystalline Solids
Melting Points
Most organic compounds have melting points below 200 ºC. Some decompose before melting, a few sublime, but a majority undergo repeated melting and crystallization without any change in molecular structure. When a pure crystalline compound is heated, or a liquid cooled, the change in sample temperature with time is roughly uniform. However, if the solid melts, or the liquid freezes, a discontinuity occurs and the temperature of the sample remains constant until the phase change is complete. This behavior is shown in the diagram on the right, with the green segment representing the solid phase, light blue the liquid, and red the temperature invariant liquid/solid equilibrium. For a given compound, this temperature represents its melting point (or freezing point), and is a reproducible constant as long as the external pressure does not change. The length of the horizontal portion depends on the size of the sample, since a quantity of heat proportional to the heat of fusion must be added (or removed) before the phase change is complete
Now it is well known that the freezing point of a solvent is lowered by a dissolved solute, e.g. brine compared with water. If two crystalline compounds (A & B) are thoroughly mixed, the melting point of that mixture is normally depressed and broadened, relative to the characteristic sharp melting point of each pure component. This provides a useful means for establishing the identity or non-identity of two or more compounds, since the melting points of numerous solid organic compounds are documented and commonly used as a test of purity.
The phase diagram below shows the melting point behavior of mixtures ranging from pure A on the left to pure B on the right. A small amount of compound B in a sample of compound A lowers (and broadens) its melting point; and the same is true for a sample of B containing a litle A. The lowest mixture melting point, e, is called the eutectic point. For example, if A is cinnamic acid, m.p. 137 ºC, and B is benzoic acid, m.p. 122 ºC, the eutectic point is 82 ºC.
Below the temperature of the isothermal line ced, the mixture is entirely solid, consisting of a conglomerate of solid A and solid B. Above this temperature the mixture is either a liquid or a liquid solid mixture, the composition of which varies. In some rare cases of nonpolar compounds of similar size and crystal structure, a true solid solution of one in the other, rather than a conglomerate, is formed. Melting or freezing takes place over a broad temperature range and there is no true eutectic point.
An interesting but less common mixed system involves molecular components that form a tight complex or molecular compound, capable of existing as a discrete species in equilibrium with a liquid of the same composition. Such a species usually has a sharp congruent melting point and produces a phase diagram having the appearance of two adjacent eutectic diagrams. An example of such a system is shown on the right, the molecular compound being represented as A:B or C. One such mixture consists of α-naphthol, m.p. 94 ºC, and p-toluidine, m.p. 43 ºC. The A:B complex has a melting point of 54 ºC, and the phase diagram displays two eutectic points, the first at 50 ºC, the second at 30 ºC. Molecular complexes of this kind commonly have a 50:50 stoichiometry, as shown, but other integral ratios are known.
In addition to the potential complications noted above, the simple process of taking a melting point may also be influenced by changes in crystal structure, either before or after an initial melt. The existence of more than one crystal form for a given compound is called polymorphism.
Polymorphism
Polymorphs of a compound are different crystal forms in which the lattice arrangement of molecules are dissimilar. These distinct solids usually have different melting points, solubilities, densities and optical properties. Many polymorphic compounds have flexible molecules that may assume different conformations, and X-ray examination of these solids shows that their crystal lattices impose certain conformational constraints. When melted or in solution, different polymorphic crystals of this kind produce the same rapidly equilibrating mixture of molecular species. Polymorphism is similar to, but distinct from, hydrated or solvated crystalline forms. It has been estimated that over 50% of known organic compounds may be capable of polymorphism.
The ribofuranose tetraacetate, shown at the upper left below, was the source of an early puzzle involving polymorphism. The compound was first prepared in England in 1946, and had a melting point of 58 ºC. Several years later the same material, having the same melting point, was prepared independently in Germany and the United States. The American chemists then found that the melting points of their early preparations had risen to 85 ºC. Eventually, it became apparent that any laboratory into which the higher melting form had been introduced was no longer able to make the lower melting form. Microscopic seeds of the stable polymorph in the environment inevitably directed crystallization to that end. X-ray diffraction data showed the lower melting polymorph to be monoclinic, space group P2. The higher melting form was orthorhombic, space group P212121.
Polymorphism has proven to be a critical factor in pharmaceuticals, solid state pigments and polymer manufacture. Some examples are described below.
Example 1: Acetaminophen
Acetaminophen is a common analgesic (e.g. Tylenol).
It is usually obtained as monoclinic prisms (right)) on crystallization from water. A less stable orthorhombic polymorph, having better physical properties for pressing into tablets, is shown on left.
Two polymorphs of Acetaminophen.
Example 2: Quinacridone
Quinacridone is an important pigment used in paints and inks. It has a rigid flat molecular structure, and in dilute solution has a light yellow color. Three polymorphs have been identified. Intermolecular hydrogen bonds are an important feature in all off these. The crystal colors range from bright red to violet.
Example 3: Ranitidine
The anti-ulcer drug ranitidine (Zantac) was first patented by Glaxo-Wellcome in 1978. Seven years later a second polymorph of ranitidine was patented by the same company. This extended the licensing coverage until 2002, and efforts to market a generic form were thwarted, because it was not possible to prepare the first polymorph uncontaminated by the second.
Example 4: EL1
The relatively simple aryl thiophene, designated EL1, was prepared and studied by chemists at the Eli Lilly Company. It displayed six polymorphic crystal forms.
Polymorphs of EL1
Polymorph Color/Shape Space Group Melting Point
I yellow prisms monoclinic
P21/n[14]
110 ºC
II reddish plates monoclinic
P21/n[14]
113 ºC
III orange needles monoclinic
P21/c[14]
115 ºC
IV yellow needles triclinic
P1[2]
rearranges
to VI
V orange plates orthorhombic
Pbca[61]
rearranges
to I
VI red prisms triclinic
P1[2]
106 ºC
A common example of changes in polymorphism is shown by chocolate that has suffered heating and/or long storage. Over time, or when it resets after softening, it may have white patches on it, no longer melts in your mouth, and doesn't taste as good as it should. This is because chocolate has more than six polymorphs, and only one is ideal as a confection. It is created under carefully-controlled factory conditions. Improper storage or transport conditions cause chocolate to transform into other polymorphs.
Chocolate is in essence cocoa mass and sugar particles suspended in a cocoa butter matrix. Cocoa butter is a mixture of triglycerides in which stearoyl, oleoyl and palmitoyl groups predominate. It is the polymorphs of this matrix that influence the quality of chocolate. Low melting polymorphs feel too sticky or thick in the mouth. Form V, the best tasting polymorph of cocoa butter, has a melting point of 34 to 36 ºC, slightly less than the interior of the human body, which is one reason it melts in the mouth. Unfortunately, the higher melting form VI is more stable and is produced over time.
Table 3: Polymorphs of Chocolate
Polymorph Melting Point Comments
I 17.4 ºC Produced by rapid cooling of a melt.
II 23.4 ºC Produced by cooling the melt at 2 ºC/min.
III 26 ºC Produced by transformation of form II at 5-10 ºC.
IV 27 ºC Produced by transformation of form III by storing at 16-21 ºC.
V 34 ºC Produced by tempering (cooling then reheating slightly while mixing).
VI 36-37 ºC Produced from V after spending 4 months at room temperature.
Solubility in Water
Water has been referred to as the "universal solvent", and its widespread distribution on this planet and essential role in life make it the benchmark for discussions of solubility. Water dissolves many ionic salts thanks to its high dielectric constant and ability to solvate ions. The former reduces the attraction between oppositely charged ions and the latter stabilizes the ions by binding to them and delocalizing charge density. Many organic compounds, especially alkanes and other hydrocarbons, are nearly insoluble in water. Organic compounds that are water soluble, such as most of those listed in the above table, generally have hydrogen bond acceptor and donor groups. The least soluble of the listed compounds is diethyl ether, which can serve only as a hydrogen bond acceptor and is 75% hydrocarbon in nature. Even so, diethyl ether is about two hundred times more soluble in water than is pentane.
The chief characteristic of water that influences these solubilities is the extensive hydrogen bonded association of its molecules with each other. This hydrogen bonded network is stabilized by the sum of all the hydrogen bond energies, and if nonpolar molecules such as hexane were inserted into the network they would destroy local structure without contributing any hydrogen bonds of their own. Of course, hexane molecules experience significant van der Waals attraction to neighboring molecules, but these attractive forces are much weaker than the hydrogen bond. Consequently, when hexane or other nonpolar compounds are mixed with water, the strong association forces of the water network exclude the nonpolar molecules, which must then exist in a separate phase. This is shown in the following illustration, and since hexane is less dense than water, the hexane phase floats on the water phase.
It is important to remember this tendency of water to exclude nonpolar molecules and groups, since it is a factor in the structure and behavior of many complex molecular systems. A common nomenclature used to describe molecules and regions within molecules is hydrophilic for polar, hydrogen bonding moieties and hydrophobic for nonpolar species.
Intermolecular Forces and Physical Properties
The attractive forces that exist between molecules are responsible for many of the bulk physical properties exhibited by substances. Some compounds are gases, some are liquids, and others are solids. The melting and boiling points of pure substances reflect these intermolecular forces, and are commonly used for identification. Of these two, the boiling point is considered the most representative measure of general intermolecular attractions. Thus, a melting point reflects the thermal energy needed to convert the highly ordered array of molecules in a crystal lattice to the randomness of a liquid. The distance between molecules in a crystal lattice is small and regular, with intermolecular forces serving to constrain the motion of the molecules more severely than in the liquid state. Molecular size is important, but shape is also critical, since individual molecules need to fit together cooperatively for the attractive lattice forces to be large. Spherically shaped molecules generally have relatively high melting points, which in some cases approach the boiling point, reflecting the fact that spheres can pack together more closely than other shapes. This structure or shape sensitivity is one of the reasons that melting points are widely used to identify specific compounds.
Boiling points, on the other hand, essentially reflect the kinetic energy needed to release a molecule from the cooperative attractions of the liquid state so that it becomes an unincumbered and relative independent gaseous state species. All atoms and molecules have a weak attraction for one another, known as van der Waals attraction. This attractive force has its origin in the electrostatic attraction of the electrons of one molecule or atom for the nuclei of another, and has been called London dispersion force.
The following animation illustrates how close approach of two neon atoms may perturb their electron distributions in a manner that induces dipole attraction. The induced dipoles are transient, but are sufficient to permit liquifaction of neon at low temperature and high pressure.
In general, larger molecules have higher boiling points than smaller molecules of the same kind, indicating that dispersion forces increase with mass, number of electrons, number of atoms or some combination thereof. The following table lists the boiling points of an assortment of elements and covalent compounds composed of molecules lacking a permanent dipole. The number of electrons in each species is noted in the first column, and the mass of each is given as a superscript number preceding the formula.
# Electrons Molecules & Boiling Points ºC
10 20Ne –246 ; 16CH4 –162
18 40Ar –186 ; 32SiH4 –112 ; 30C2H6 –89 ; 38F2 –187
34-44 84Kr –152 ; 58C4H10 –0.5 ; 72(CH3)4C 10 ; 71Cl2 –35 ; 88CF4 –130
66-76 114[(CH3)3C]2 106 ; 126(CH2)9 174 ; 160Br2 59 ; 154CCl4 77 ; 138C2F6 –78
Two ten electron molecules are shown in the first row. Neon is heavier than methane, but it boils 84º lower. Methane is composed of five atoms, and the additional nuclei may provide greater opportunity for induced dipole formation as other molecules approach. The ease with which the electrons of a molecule, atom or ion are displaced by a neighboring charge is called polarizability, so we may conclude that methane is more polarizable than neon. In the second row, four eighteen electron molecules are listed. Most of their boiling points are higher than the ten electron compounds neon and methane, but fluorine is an exception, boiling 25º below methane. The remaining examples in the table conform to the correlation of boiling point with total electrons and number of nuclei, but fluorine containing molecules remain an exception.
The anomalous behavior of fluorine may be attributed to its very high electronegativity. The fluorine nucleus exerts such a strong attraction for its electrons that they are much less polarizable than the electrons of most other atoms.
Of course, boiling point relationships may be dominated by even stronger attractive forces, such as those involving electrostatic attraction between oppositely charged ionic species, and between the partial charge separations of molecular dipoles. Molecules having a permanent dipole moment should therefore have higher boiling points than equivalent nonpolar compounds, as illustrated by the data in the following table.
# Electrons Molecules & Boiling Points ºC
14-18 30C2H6 –89 ; 28H2C=CH2 –104 ; 26HC≡CH –84 30H2C=O –21 27HC≡N 26 34CH3-F –78
22-26 42CH3CH=CH2 –48 ; 40CH3C≡CH –23 ; 44CH3CH=O 21 41CH3C≡N 81 46(CH3)2O –24 50.5CH3-Cl –24 52CH2F2 –52
32-44 58(CH3)3CH –12 56(CH3)2C=CH2 –7 58(CH3)2C=O 56 59(CH3)3N 3 95CH3-Br 45 85CH2Cl2 40 70CHF3 –84
In the first row of compounds, ethane, ethene and ethyne have no molecular dipole, and serve as useful references for single, double and triple bonded derivatives that do. Formaldehyde and hydrogen cyanide clearly show the enhanced intermolecular attraction resulting from a permanent dipole. Methyl fluoride is anomalous, as are most organofluorine compounds. In the second and third rows, all the compounds have permanent dipoles, but those associated with the hydrocarbons (first two compounds in each case) are very small. Large molecular dipoles come chiefly from bonds to high-electronegative atoms (relative to carbon and hydrogen), especially if they are double or triple bonds. Thus, aldehydes, ketones and nitriles tend to be higher boiling than equivalently sized hydrocarbons and alkyl halides. The atypical behavior of fluorine compounds is unexpected in view of the large electronegativity difference between carbon and fluorine.
Hydrogen Bonding
Most of the simple hydrides of group IV, V, VI & VII elements display the expected rise in boiling point with number of electrons and molecular mass, but the hydrides of the most electronegative elements (nitrogen, oxygen and fluorine) have abnormally high boiling points (Table 4).
Table 4:
Group Molecules & Boiling Points ºC
VII HF 19 ; HCl –85 ; HBr –67 ; HI –36
VI H2O 100 ; H2S –60 ; H2Se –41 ; H2Te –2
V NH3 –33 ; PH3 –88 ; AsH3 –62 ; SbH3 –18
The exceptionally strong dipole-dipole attractions that are responsible for this behavior are called hydrogen bonds. When a hydrogen atom is part of a polar covalent bond to a more electronegative atom such as oxygen, its small size allows the positive end of the bond dipole (the hydrogen) to approach neighboring nucleophilic or basic sites more closely than can components of other polar bonds. Coulombic forces are inversely proportional to the sixth power of the distance between dipoles, making these interactions relatively strong, although they are still weak (ca. 4 to 5 kcal per mole) compared with most covalent bonds. The table of data on the right provides convincing evidence for hydrogen bonding. In each row the first compound listed has the fewest total electrons and lowest mass, yet its boiling point is the highest due to hydrogen bonding. Other compounds in each row have molecular dipoles, the interactions of which might be called hydrogen bonding, but the attractions are clearly much weaker. The first two hydrides of group IV elements, methane and silane, are listed in the first table above, and do not display any significant hydrogen bonding.
Organic compounds incorporating O-H and N-H bonds will also exhibit enhanced intermolecular attraction due to hydrogen bonding. Some examples are given below.
Table 3: Organic compounds incorporating O-H and N-H
Class Molecules & Boiling Points ºC
Oxygen
Compounds
C2H5OH 78 ; (CH3)2O –24 ; (CH2)2O 11
ethanol dimethyl ether ethylene oxide
(CH2)3CHOH 124 & (CH2)4O 66
cyclobutanol tetrahydrofuran
Nitrogen
Compounds
C3H7NH2 50 ; C2H5NH(CH3) 37 ; (CH3)3N 3
propyl amine ethyl methyl amine trimethyl amine
(CH2)4CHNH2 107 & (CH2)4NCH3 80
cyclopentyl amine N-methylpyrrolidine
Complex
Functions
C2H5CO2H 141 & CH3CO2CH3 57
propanoic acid methyl acetate
C3H7CONH2 218 & CH3CON(CH3)2 165
butyramide N,N-dimethylacetamide
Water Solubility
Water is the single most abundant and important liquid on this planet. The miscibility of other liquids in water, and the solubility of solids in water, must be considered when isolating and purifying compounds. To this end, the following table lists the water miscibility (or solubility) of an assortment of low molecular weight organic compounds. The influence of the important hydrogen bonding atoms, oxygen and nitrogen is immediately apparent. The first row lists a few hydrocarbon and chlorinated solvents. Without exception these are all immiscible with water, although it is interesting to note that the π-electrons of benzene and the nonbonding valence electrons of chlorine act to slightly increase their solubility relative to the saturated hydrocarbons. When compared with hydrocarbons, the oxygen and nitrogen compounds listed in the second, third and fourth rows are over a hundred times more soluble in water, and many are completely miscible with water.
Table 4: Water Solubility of Characteristic Compounds
Compound Type Specific Compounds Grams/100mL Moles/Liter Specific Compounds Grams/100mL Moles/Liter
Hydrocarbons &
Alkyl Halides
butane
hexane
cyclohexane
0.007
0.0009
0.006
0.0012
0.0001
0.0007
benzene
methylene chloride
chloroform
0.07
1.50
0.8
0.009
0.180
0.07
Compounds
Having
One Oxygen
1-butanol
tert-butanol
cyclohexanol
phenol
9.0
complete
3.6
8.7
1.2
complete
0.36
0.90
ethyl ether
THF
furan
anisole
6.0
complete
1.0
1.0
0.80
complete
0.15
0.09
Compounds
Having
Two Oxygens
1,3-propanediol
2-butoxyethanol
butanoic acid
benzoic acid
complete
complete
complete
complete
complete
complete
complete
complete
1,2-dimethoxyethane
1,4-dioxane
ethyl acetate
γ-butyrolactone
complete
complete
8.0
complete
complete
complete
0.91
complete
Nitrogen
Containing
Compounds
1-aminobutane
cyclohexylamine
aniline
pyrrolidine
pyrrole
complete
complete
3.4
complete
6.0
complete
complete
0.37
complete
0.9
triethylamine
pyridine
propionitrile
1-nitropropane
DMF
5.5
complete
10.3
1.5
complete
0.54
complete
2.0
0.17
complete
Some general trends are worth noting from the data above. First, alcohols (second row left column) are usually more soluble than equivalently sized ethers (second row right column). This reflects the fact that the hydroxyl group may function as both a hydrogen bond donor and acceptor; whereas, an ether oxygen may serve only as an acceptor. The increased solubility of phenol relative to cyclohexanol may be due to its greater acidity as well as the pi-electron effect noted in the first row.
The cyclic ether THF (tetrahydrofuran) is more soluble than its open chain analog, possibly because the oxygen atom is more accessible for hydrogen bonding to water molecules. Due to the decreased basicity of the oxygen in the aromatic compound furan, it is much less soluble. The oxygen atom in anisole is likewise deactivated by conjugation with the benzene ring (note, it activates the ring in electrophilic substitution reactions). A second oxygen atom dramatically increases water solubility, as demonstrated by the compounds listed in the third row. Again hydroxyl compounds are listed on the left.
Nitrogen exerts a solubilizing influence similar to oxygen, as shown by the compounds in the fourth row. The primary and secondary amines listed in the left hand column may function as both hydrogen bond donors and acceptors. Aromaticity decreases the basicity of pyrrole, but increases its acidity. The compounds in the right column are only capable of an acceptor role. The low solubility of the nitro compound is surprising.
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Intermolecular Forces
For general purposes it is useful to consider temperature to be a measure of the kinetic energy of all the atoms and molecules in a given system. As temperature is increased, there is a corresponding increase in the vigor of translational and rotation motions of all molecules, as well as the vibrations of atoms and groups of atoms within molecules. Experience shows that many compounds exist normally as liquids and solids; and that even low-density gases, such as hydrogen and helium, can be liquified at sufficiently low temperature and high pressure. A clear conclusion to be drawn from this fact is that intermolecular attractive forces vary considerably, and that the boiling point of a compound is a measure of the strength of these forces. Thus, in order to break the intermolecular attractions that hold the molecules of a compound in the condensed liquid state, it is necessary to increase their kinetic energy by raising the sample temperature to the characteristic boiling point of the compound.
The following table illustrates some of the factors that influence the strength of intermolecular attractions. The formula of each entry is followed by its formula weight in parentheses and the boiling point in degrees Celsius. First there is molecular size. Large molecules have more electrons and nuclei that create van der Waals attractive forces, so their compounds usually have higher boiling points than similar compounds made up of smaller molecules. It is very important to apply this rule only to like compounds. The examples given in the first two rows are similar in that the molecules or atoms are spherical in shape and do not have permanent dipoles. Molecular shape is also important, as the second group of compounds illustrate. The upper row consists of roughly spherical molecules, whereas the isomers in the lower row have cylindrical or linear shaped molecules. The attractive forces between the latter group are generally greater. Finally, permanent molecular dipoles generated by polar covalent bonds result in even greater attractive forces between molecules, provided they have the mobility to line up in appropriate orientations. The last entries in the table compare non-polar hydrocarbons with equal-sized compounds having polar bonds to oxygen and nitrogen. Halogens also form polar bonds to carbon, but they also increase the molecular mass, making it difficult to distinguish among these factors.
Table 1: Boiling Points (ºC) of Selected Elements and Compounds
Increasing Size
Atomic Ar (40) -186 Kr (83) -153 Xe (131) -109
Molecular CH4 (16) -161 (CH3)4C (72) 9.5 (CH3)4Si (88) 27 CCl4 (154) 77
Molecular Shape
Spherical: (CH3)4C (72) 9.5 (CH3)2CCl2 (113) 69 (CH3)3CC(CH3)3 (114) 106
Linear: CH3(CH2)3CH3 (72) 36 Cl(CH2)3Cl (113) 121 CH3(CH2)6CH3 (114) 126
Molecular Polarity
Non-polar: H2C=CH2 (28) -104 F2 (38) -188 CH3C≡CCH3 (54) -32 CF4 (88) -130
Polar: H2C=O (30) -21 CH3CH=O (44) 20 (CH3)3N (59) 3.5 (CH3)2C=O (58) 56
HC≡N (27) 26 CH3C≡N (41) 82 (CH2)3O (58) 50 CH3NO2 (61) 101
The melting points of crystalline solids cannot be categorized in as simple a fashion as boiling points. The distance between molecules in a crystal lattice is small and regular, with intermolecular forces serving to constrain the motion of the molecules more severely than in the liquid state. Molecular size is important, but shape is also critical, since individual molecules need to fit together cooperatively for the attractive lattice forces to be large. Spherically shaped molecules generally have relatively high melting points, which in some cases approach the boiling point. This reflects the fact that spheres can pack together more closely than other shapes. This structure or shape sensitivity is one of the reasons that melting points are widely used to identify specific compounds.
Table 2: Melting and Boiling points of Select Compounds
Compound Formula Boiling Point Melting Point
pentane CH3(CH2)3CH3 36ºC –130ºC
hexane CH3(CH2)4CH3 69ºC –95ºC
heptane CH3(CH2)5CH3 98ºC –91ºC
octane CH3(CH2)6CH3 126ºC –57ºC
nonane CH3(CH2)7CH3 151ºC –54ºC
decane CH3(CH2)8CH3 174ºC –30ºC
tetramethylbutane (CH3)3C-C(CH3)3 106ºC +100ºC
Notice that the boiling points of the unbranched alkanes (pentane through decane) increase rather smoothly with molecular weight, but the melting points of the even-carbon chains increase more than those of the odd-carbon chains. Even-membered chains pack together in a uniform fashion more compactly than do odd-membered chains. The last compound, an isomer of octane, is nearly spherical and has an exceptionally high melting point (only 6º below the boiling point). | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Fundamentals/Intermolecular_Forces/Boiling_Points.txt |
The most powerful intermolecular force influencing neutral (uncharged) molecules is the hydrogen bond. If we compare the boiling points of methane (CH4) -161ºC, ammonia (NH3) -33ºC, water (H2O) 100ºC and hydrogen fluoride (HF) 19ºC, we see a greater variation for these similar sized molecules than expected from the data presented above for polar compounds. This is shown graphically in the following chart. Most of the simple hydrides of group IV, V, VI & VII elements display the expected rise in boiling point with molecular mass, but the hydrides of the most electronegative elements (nitrogen, oxygen and fluorine) have abnormally high boiling points for their mass.
The exceptionally strong dipole-dipole attractions that cause this behavior are called the hydrogen bond. Hydrogen forms polar covalent bonds to more electronegative atoms such as oxygen, and because a hydrogen atom is quite small, the positive end of the bond dipole (the hydrogen) can approach neighboring nucleophilic or basic sites more closely than can other polar bonds. Coulombic forces are inversely proportional to the sixth power of the distance between dipoles, making these interactions relatively strong, although they are still weak (ca. 4 to 5 kcal per mole) compared with most covalent bonds. The unique properties of water are largely due to the strong hydrogen bonding that occurs between its molecules. In the following diagram the hydrogen bonds are depicted as magenta dashed lines.
The molecule providing a polar hydrogen for a hydrogen bond is called a donor. The molecule that provides the electron rich site to which the hydrogen is attracted is called an acceptor. Water and alcohols may serve as both donors and acceptors, whereas ethers, aldehydes, ketones and esters can function only as acceptors. Similarly, primary and secondary amines are both donors and acceptors, but tertiary amines function only as acceptors. Once you are able to recognize compounds that can exhibit intermolecular hydrogen bonding, the relatively high boiling points they exhibit become understandable. The data in the following table serve to illustrate this point.
Compound Formula Mol. Wt. Boiling Point Melting Point
dimethyl ether CH3OCH3 46 –24ºC –138ºC
ethanol CH3CH2OH 46 78ºC –130ºC
propanol CH3(CH2)2OH 60 98ºC –127ºC
diethyl ether (CH3CH2)2O 74 34ºC –116ºC
propyl amine CH3(CH2)2NH2 59 48ºC –83ºC
methylaminoethane CH3CH2NHCH3 59 37ºC
trimethylamine (CH3)3N 59 3ºC –117ºC
ethylene glycol HOCH2CH2OH 62 197ºC –13ºC
acetic acid CH3CO2H 60 118ºC 17ºC
ethylene diamine H2NCH2CH2NH2 60 118ºC 8.5ºC
Alcohols boil cosiderably higher than comparably sized ethers (first two entries), and isomeric 1º, 2º & 3º-amines, respectively, show decreasing boiling points, with the two hydrogen bonding isomers being substantially higher boiling than the 3º-amine (entries 5 to 7). Also, O–H---O hydrogen bonds are clearly stronger than N–H---N hydrogen bonds, as we see by comparing propanol with the amines.
As expected, the presence of two hydrogen bonding functions in a compound raises the boiling point even further. Acetic acid (the ninth entry) is an interesting case. A dimeric species, shown on the right, held together by two hydrogen bonds is a major component of the liquid state. If this is an accurate representation of the composition of this compound then we would expect its boiling point to be equivalent to that of a C4H8O4 compound (formula weight = 120). A suitable approximation of such a compound is found in tetramethoxymethane, (CH3O)4C, which is actually a bit larger (formula weight = 136) and has a boiling point of 114ºC. Thus, the dimeric hydrogen bonded structure appears to be a good representation of acetic acid in the condensed state.
A related principle is worth noting at this point. Although the hydrogen bond is relatively weak (ca. 4 to 5 kcal per mole), when several such bonds exist the resulting structure can be quite robust. The hydrogen bonds between cellulose fibers confer great strength to wood and related materials. For additional information on this subject Click Here.
Polar Protic and Aprotic Solvents
Solvents used in organic chemistry are characterized by their physical characteristics. Among the most important are whether the solvents are polar or non-polar, and whether they are protic or aprotic. Because non-polar solvents tend to be aprotic,the focus is upon polar solvents and their structures.
Solvent Polarity
Solvents are generally classified by the polarity, and considered either polar or non-polar, as indicated by the dielectric constant. However, as with many properties, the polarity is a continuous scale, and the correct question is not "is it polar or non-polar" but "how polar is it." Nonetheless, guidelines have been created to make it easier. Generally, solvents with dielectric constants greater than about 5 are considered "polar" and those with dielectric constants less than 5 are considered "non-polar."
Table 1: Examples of a few common solvents used in organic chemistry
Solvent Boiling Point, Celsius Dielectric Constant
NON-POLAR SOLVENTS
Pentane, C5H12 36 1.8
Hexane, C6H14 69 1.9
Benzene, C6H6 80 2.3
Chloroform, CHCl3 61 4.8
Diethyl ether, (CH3CH2)2O 35 4.3
1,40-Dioxane, cyc-(CH2CH2OCH2CH2O) 101 2.3
POLAR PROTIC SOLVENTS
Water, H2O 100 78.5
methanol, CH3OH 65 32.6
ethanol, CH3CH2OH 78.5 24.3
isopropyl alcohol, CH3CH(OH)CH3 82 18
acetic acid, CH3COOH 118 6
POLAR APROTIC SOLVENTS
dichloromethane, CH2Cl2 40 9.1
tetrahydrofuran (THF), cyc-(CH2)4O 66 7.5
ethyl acetate, CH3C(O)OCH2CH3 77 6
acetonitrile, CH3CN 81.6 37.5
dimethylformamide (DMF), HCON(CH3)2 153 38
dimethyl sulfoxide (DMSO), CH3SOCH3 189 47
acetone, CH3COCH3 56.5 21
hexamethylphosphoric triamde (HMPT), [(CH3)2N]3PO 232 30
Protic vs Aprotic Solvents
The table above distinguishes between protic and aprotic solvents. For the solvents included in the table, the distinguishing feature is the presence of an -OH group, and that is the most common characteristic of a protic solvent. However, there are exceptions, such as nitromethane, CH3NO2, which is also considered a protic solvent. That might suggest that Bronsted acidity is the most important feature, because nitromethane is very acidic, with a pKa of about 10. However, acetone is still considered a polar aprotic solvent, despite the fact that it is relatively acidic, and not significantly less acidic than alcohols. Then again, acetone (and other carbonyl containing solvents) are, indeed, poor solvents when using strong bases due to their relatively high acidity.
Significance
Solvent properties are in important consideration in many chemical reactions, including nucleophilic substitution reactions. As strong hydrogen-bond donors, protic solvents are very effective at stabilizing ions. Therefore, they favor reactions in which ions are formed, such as the SN1 reaction, and disfavor reactions where ions are reactants, such as the SN2 reaction. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Fundamentals/Intermolecular_Forces/Hydrogen_Bonding.txt |
The study of organic chemistry must at some point extend to the molecular level, for the physical and chemical properties of a substance are ultimately explained in terms of the structure and bonding of molecules. This module introduces some basic facts and principles that are needed for a discussion of organic molecules.
Electronic Configurations
Four elements, hydrogen, carbon, oxygen and nitrogen, are the major components of most organic compounds. Consequently, our understanding of organic chemistry must have, as a foundation, an appreciation of the electronic structure and properties of these elements. The truncated periodic table shown above provides the orbital electronic structure for the first eighteen elements (hydrogen through argon). According to the Aufbau principle, the electrons of an atom occupy quantum levels or orbitals starting from the lowest energy level, and proceeding to the highest, with each orbital holding a maximum of two paired electrons (opposite spins).
The periodic table shown here is severely truncated. There are, of course, over eighty other elements.
1A 2A 3A 4A 5A 6A 7A 8A
1
H
1s1
2
He
1s2
3
Li
1s2
2s1
4
Be
1s2
2s2
5
B
1s2
2s22p1
6
C
1s2
2s22p2
7
N
1s2
2s22p3
8
O
1s2
2s22p4
9
F
1s2
2s22p5
10
Ne
1s2
2s22p6
11
Na
[Ne]
3s1
12
Mg
[Ne]
3s2
13
Al
[Ne]
3s23p1
14
Si
[Ne]
3s23p2
15
P
[Ne]
3s23p3
16
S
[Ne]
3s23p4
17
Cl
[Ne]
3s23p5
18
Ar
[Ne]
3s23p6
Electron shell #1 has the lowest energy and its s-orbital is the first to be filled. Shell #2 has four higher energy orbitals, the 2s-orbital being lower in energy than the three 2p-orbitals. (x, y & z). As we progress from lithium (atomic number=3) to neon (atomic number=10) across the second row or period of the table, all these atoms start with a filled 1s-orbital, and the 2s-orbital is occupied with an electron pair before the 2p-orbitals are filled. In the third period of the table, the atoms all have a neon-like core of 10 electrons, and shell #3 is occupied progressively with eight electrons, starting with the 3s-orbital. The highest occupied electron shell is called the valence shell, and the electrons occupying this shell are called valence electrons.
The chemical properties of the elements reflect their electron configurations. For example, helium, neon and argon are exceptionally stable and unreactive monoatomic gases. Helium is unique since its valence shell consists of a single s-orbital. The other members of group 8 have a characteristic valence shell electron octet (ns2 + npx2 + npy2 + npz2). This group of inert (or noble) gases also includes krypton (Kr: 4s2, 4p6), xenon (Xe: 5s2, 5p6) and radon (Rn: 6s2, 6p6). In the periodic table above these elements are colored beige.
The halogens (F, Cl, Br etc.) are one electron short of a valence shell octet, and are among the most reactive of the elements (they are colored red in this periodic table). In their chemical reactions halogen atoms achieve a valence shell octet by capturing or borrowing the eighth electron from another atom or molecule. The alkali metals Li, Na, K etc. (colored violet above) are also exceptionally reactive, but for the opposite reason. These atoms have only one electron in the valence shell, and on losing this electron arrive at the lower shell valence octet. As a consequence of this electron loss, these elements are commonly encountered as cations (positively charged atoms).
The elements in groups 2 through 7 all exhibit characteristic reactivities and bonding patterns that can in large part be rationalized by their electron configurations. It should be noted that hydrogen is unique. Its location in the periodic table should not suggest a kinship to the chemistry of the alkali metals, and its role in the structure and properties of organic compounds is unlike that of any other element.
Bonding & Valence
As noted earlier, the inert gas elements of group 8 exist as monoatomic gases, and do not in general react with other elements. In contrast, other gaseous elements exist as diatomic molecules (H2, N2, O2, F2 & Cl2), and all but nitrogen are quite reactive. Some dramatic examples of this reactivity are shown in the following equations.
2Na + Cl2 2NaCl
2H2 + O2 2H2O
C + O2 CO2
C + 2F2 CF4
Why do the atoms of many elements interact with each other and with other elements to give stable molecules? In addressing this question it is instructive to begin with a very simple model for the attraction or bonding of atoms to each other, and then progress to more sophisticated explanations.
Ionic Bonding
When sodium is burned in a chlorine atmosphere, it produces the compound sodium chloride. This has a high melting point (800 ºC) and dissolves in water to to give a conducting solution. Sodium chloride is an ionic compound, and the crystalline solid has the structure shown on the right. Transfer of the lone 3s electron of a sodium atom to the half-filled 3p orbital of a chlorine atom generates a sodium cation (neon valence shell) and a chloride anion (argon valence shell). Electrostatic attraction results in these oppositely charged ions packing together in a lattice. The attractive forces holding the ions in place can be referred to as ionic bonds.
Covalent Bonding
The other three reactions shown above give products that are very different from sodium chloride. Water is a liquid at room temperature; carbon dioxide and carbon tetrafluoride are gases. None of these compounds is composed of ions. A different attractive interaction between atoms, called covalent bonding, is involved here. Covalent bonding occurs by a sharing of valence electrons, rather than an outright electron transfer. Similarities in physical properties (they are all gases) suggest that the diatomic elements H2, N2, O2, F2 & Cl2 also have covalent bonds.
Examples of covalent bonding shown below include hydrogen, fluorine, carbon dioxide and carbon tetrafluoride. These illustrations use a simple Bohr notation, with valence electrons designated by colored dots. Note that in the first case both hydrogen atoms achieve a helium-like pair of 1s-electrons by sharing. In the other examples carbon, oxygen and fluorine achieve neon-like valence octets by a similar sharing of electron pairs. Carbon dioxide is notable because it is a case in which two pairs of electrons (four in all) are shared by the same two atoms. This is an example of a double covalent bond.
These electron sharing diagrams (Lewis formulas) are a useful first step in understanding covalent bonding, but it is quicker and easier to draw Couper-Kekulé formulas in which each shared electron pair is represented by a line between the atom symbols. Non-bonding valence electrons are shown as dots. These formulas are derived from the graphic notations suggested by A. Couper and A. Kekulé, and are not identical to their original drawings. Some examples of such structural formulas are given in the following table.
Common Name Molecular Formula Lewis Formula Kekulé Formula
Methane CH4
Ammonia NH3
Ethane C2H6
Methyl Alcohol CH4O
Ethylene C2H4
Formaldehyde CH2O
Acetylene C2H2
Hydrogen Cyanide CHN
Charge Distribution
If the electron pairs in covalent bonds were donated and shared absolutely evenly there would be no fixed local charges within a molecule. Although this is true for diatomic elements such as H2, N2 and O2, most covalent compounds show some degree of local charge separation, resulting in bond and / or molecular dipoles. A dipole exists when the centers of positive and negative charge distribution do not coincide.
Formal Charges
A large local charge separation usually results when a shared electron pair is donated unilaterally. The three Kekulé formulas shown here illustrate this condition.
In the formula for ozone the central oxygen atom has three bonds and a full positive charge while the right hand oxygen has a single bond and is negatively charged. The overall charge of the ozone molecule is therefore zero. Similarly, nitromethane has a positive-charged nitrogen and a negative-charged oxygen, the total molecular charge again being zero. Finally, azide anion has two negative-charged nitrogens and one positive-charged nitrogen, the total charge being minus one.
In general, for covalently bonded atoms having valence shell electron octets, if the number of covalent bonds to an atom is greater than its normal valence it will carry a positive charge. If the number of covalent bonds to an atom is less than its normal valence it will carry a negative charge. The formal charge on an atom may also be calculated by the following formula:
Polar Covalent Bonds
H
2.20
Electronegativity Values
for Some Elements
Li
0.98
Be
1.57
B
2.04
C
2.55
N
3.04
O
3.44
F
3.98
Na
0.90
Mg
1.31
Al
1.61
Si
1.90
P
2.19
S
2.58
Cl
3.16
K
0.82
Ca
1.00
Ga
1.81
Ge
2.01
As
2.18
Se
2.55
Br
2.96
Because of their differing nuclear charges, and as a result of shielding by inner electron shells, the different atoms of the periodic table have different affinities for nearby electrons. The ability of an element to attract or hold onto electrons is called electronegativity. A rough quantitative scale of electronegativity values was established by Linus Pauling, and some of these are given in the table to the right. A larger number on this scale signifies a greater affinity for electrons. Fluorine has the greatest electronegativity of all the elements, and the heavier alkali metals such as potassium, rubidium and cesium have the lowest electronegativities. It should be noted that carbon is about in the middle of the electronegativity range, and is slightly more electronegative than hydrogen.
When two different atoms are bonded covalently, the shared electrons are attracted to the more electronegative atom of the bond, resulting in a shift of electron density toward the more electronegative atom. Such a covalent bond is polar, and will have a dipole (one end is positive and the other end negative). The degree of polarity and the magnitude of the bond dipole will be proportional to the difference in electronegativity of the bonded atoms. Thus a O–H bond is more polar than a C–H bond, with the hydrogen atom of the former being more positive than the hydrogen bonded to carbon. Likewise, C–Cl and C–Li bonds are both polar, but the carbon end is positive in the former and negative in the latter. The dipolar nature of these bonds is often indicated by a partial charge notation (δ+/–) or by an arrow pointing to the negative end of the bond.
Although there is a small electronegativity difference between carbon and hydrogen, the C–H bond is regarded as weakly polar at best, and hydrocarbons in general are considered to be non-polar compounds.
The shift of electron density in a covalent bond toward the more electronegative atom or group can be observed in several ways. For bonds to hydrogen, acidity is one criterion. If the bonding electron pair moves away from the hydrogen nucleus the proton will be more easily transferred to a base (it will be more acidic). A comparison of the acidities of methane, water and hydrofluoric acid is instructive. Methane is essentially non-acidic, since the C–H bond is nearly non-polar. As noted above, the O–H bond of water is polar, and it is at least 25 powers of ten more acidic than methane. H–F is over 12 powers of ten more acidic than water as a consequence of the greater electronegativity difference in its atoms.
Electronegativity differences may be transmitted through connecting covalent bonds by an inductive effect. Replacing one of the hydrogens of water by a more electronegative atom increases the acidity of the remaining O–H bond. Thus hydrogen peroxide, HO–O–H, is ten thousand times more acidic than water, and hypochlorous acid, Cl–O–H is one hundred million times more acidic. This inductive transfer of polarity tapers off as the number of transmitting bonds increases, and the presence of more than one highly electronegative atom has a cumulative effect. For example, trifluoro ethanol, CF3CH2–O–H is about ten thousand times more acidic than ethanol, CH3CH2–O–H.
Excellent physical evidence for the inductive effect is found in the influence of electronegative atoms on the NMR chemical shifts of nearby hydrogen atoms.
Functional Groups
Functional groups are atoms or small groups of atoms (two to four) that exhibit a characteristic reactivity when treated with certain reagents. A particular functional group will almost always display its characteristic chemical behavior when it is present in a compound. Because of their importance in understanding organic chemistry, functional groups have characteristic names that often carry over in the naming of individual compounds incorporating specific groups. In the following table the atoms of each functional group are colored red and the characteristic IUPAC nomenclature suffix that denotes some (but not all) functional groups is also colored.
Functional Group Tables
Exclusively Carbon Functional Groups
Group Formula
Class Name Specific Example IUPAC Name Common Name
Alkene H2C=CH2 Ethene Ethylene
Alkyne HC≡CH Ethyne Acetylene
Arene C6H6 Benzene Benzene
Functional Groups with Single Bonds to Heteroatoms
Group Formula Class Name Specific Example IUPAC Name Common Name
Halide H3C-I Iodomethane Methyl iodide
Alcohol CH3CH2OH Ethanol Ethyl alcohol
Ether CH3CH2OCH2CH3 Diethyl ether Ether
Amine H3C-NH2 Aminomethane Methylamine
Nitro Compound H3C-NO2 Nitromethane
Thiol H3C-SH Methanethiol Methyl mercaptan
Sulfide H3C-S-CH3 Dimethyl sulfide
Functional Groups with Multiple Bonds to Heteroatoms
Group Formula Class Name Specific Example IUPAC Name Common Name
Nitrile H3C-CN Ethanenitrile Acetonitrile
Aldehyde H3CCHO Ethanal Acetaldehyde
Ketone H3CCOCH3 Propanone Acetone
Carboxylic Acid H3CCO2H Ethanoic Acid Acetic acid
Ester H3CCO2CH2CH3 Ethyl ethanoate Ethyl acetate
Acid Halide H3CCOCl Ethanoyl chloride Acetyl chloride
Amide H3CCON(CH3)2 N,N-Dimethylethanamide N,N-Dimethylacetamide
Acid Anhydride (H3CCO)2O Ethanoic anhydride Acetic anhydride
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William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry
Introduction to Organic Chemistry
The three dimensional shape or configuration of a molecule is an important characteristic. This shape is dependent on the preferred spatial orientation of covalent bonds to atoms having two or more bonding partners. Three dimensional configurations are best viewed with the aid of models. In order to represent such configurations on a two-dimensional surface (paper, blackboard or screen), we often use perspective drawings in which the direction of a bond is specified by the line connecting the bonded atoms.
Introduction
In most cases the focus of configuration is a carbon atom so the lines specifying bond directions will originate there. As defined in the diagram on the right, a simple straight line represents a bond lying approximately in the surface plane. The two bonds to substituents A in the structure on the left are of this kind. A wedge shaped bond is directed in front of this plane (thick end toward the viewer), as shown by the bond to substituent B; and a hatched bond is directed in back of the plane (away from the viewer), as shown by the bond to substituent D. Some texts and other sources may use a dashed bond in the same manner as we have defined the hatched bond, but this can be confusing because the dashed bond is often used to represent a partial bond (i.e. a covalent bond that is partially formed or partially broken).
The following examples make use of this notation, and also illustrate the importance of including non-bonding valence shell electron pairs (colored blue) when viewing such configurations.
Methane Ammonia Water
Bonding configurations are readily predicted by valence-shell electron-pair repulsion theory, commonly referred to as VSEPR in most introductory chemistry texts. This simple model is based on the fact that electrons repel each other, and that it is reasonable to expect that the bonds and non-bonding valence electron pairs associated with a given atom will prefer to be as far apart as possible. The bonding configurations of carbon are easy to remember, since there are only three categories.
Configuration Bonding Partners Bond Angles Example
Tetrahedral 4 109.5º
Trigonal 3 120º
Linear 2 180º
In the three examples shown above, the central atom (carbon) does not have any non-bonding valence electrons; consequently the configuration may be estimated from the number of bonding partners alone. For molecules of water and ammonia, however, the non-bonding electrons must be included in the calculation. In each case there are four regions of electron density associated with the valence shell so that a tetrahedral bond angle is expected. The measured bond angles of these compounds (H2O 104.5º & NH3 107.3º) show that they are closer to being tetrahedral than trigonal or linear. Of course, it is the configuration of atoms (not electrons) that defines the the shape of a molecule, and in this sense ammonia is said to be pyramidal (not tetrahedral). The compound boron trifluoride, BF3, does not have non-bonding valence electrons and the configuration of its atoms is trigonal. Nice treatments of VSEPR theory have been provided by Oxford and Purdue. The best way to study the three-dimensional shapes of molecules is by using molecular models. Many kinds of model kits are available to students and professional chemists.
One way in which the shapes of molecules manifest themselves experimentally is through molecular dipole moments. A molecule which has one or more polar covalent bonds may have a dipole moment as a result of the accumulated bond dipoles. In the case of water, we know that the O-H covalent bond is polar, due to the different electronegativities of hydrogen and oxygen. Since there are two O-H bonds in water, their bond dipoles will interact and may result in a molecular dipole which can be measured. The following diagram shows four possible orientations of the O-H bonds.
The bond dipoles are colored magenta and the resulting molecular dipole is colored blue. In the linear configuration (bond angle 180º) the bond dipoles cancel, and the molecular dipole is zero. For other bond angles (120 to 90º) the molecular dipole would vary in size, being largest for the 90º configuration. In a similar manner the configurations of methane (CH4) and carbon dioxide (CO2) may be deduced from their zero molecular dipole moments. Since the bond dipoles have canceled, the configurations of these molecules must be tetrahedral (or square-planar) and linear respectively.
The case of methane provides insight to other arguments that have been used to confirm its tetrahedral configuration. For purposes of discussion we shall consider three other configurations for CH4, square-planar, square-pyramidal and triangular-pyramidal.
Substitution of one hydrogen by a chlorine atom gives a CH3Cl compound. Since the tetrahedral, square-planar and square-pyramidal configurations have structurally equivalent hydrogen atoms, they would each give a single substitution product. However, in the trigonal-pyramidal configuration one hydrogen (the apex) is structurally different from the other three (the pyramid base). Substitution in this case should give two different CH3Cl compounds if all the hydrogens react. In the case of disubstitution, the tetrahedral configuration of methane would lead to a single CH2Cl2 product, but the other configurations would give two different CH2Cl2 compounds.
Isomers
Structural Formulas
It is necessary to draw structural formulas for organic compounds because in most cases a molecular formula does not uniquely represent a single compound. Different compounds having the same molecular formula are called isomers, and the prevalence of organic isomers reflects the extraordinary versatility of carbon in forming strong bonds to itself and to other elements. When the group of atoms that make up the molecules of different isomers are bonded together in fundamentally different ways, we refer to such compounds as constitutional isomers. There are seven constitutional isomers of C4H10O, and structural formulas for these are drawn in the following table. These formulas represent all known and possible C4H10O compounds, and display a common structural feature. There are no double or triple bonds and no rings in any of these structures.
Structural Formulas for C4H10O isomers
Kekulé Formula Condensed Formula Shorthand Formula
Simplification of structural formulas may be achieved without any loss of the information they convey. In condensed structural formulas the bonds to each carbon are omitted, but each distinct structural unit (group) is written with subscript numbers designating multiple substituents, including the hydrogens. Shorthand (line) formulas omit the symbols for carbon and hydrogen entirely. Each straight line segment represents a bond, the ends and intersections of the lines are carbon atoms, and the correct number of hydrogens is calculated from the tetravalency of carbon. Non-bonding valence shell electrons are omitted in these formulas.
Developing the ability to visualize a three-dimensional structure from two-dimensional formulas requires practice, and in most cases the aid of molecular models. As noted earlier, many kinds of model kits are available to students and professional chemists, and the beginning student is encouraged to obtain one.
Constitutional isomers have the same molecular formula, but their physical and chemical properties may be very different. For an example Click Here.
Distinguishing Carbon Atoms
When discussing structural formulas, it is often useful to distinguish different groups of carbon atoms by their structural characteristics. A primary carbon (1º) is one that is bonded to no more than one other carbon atom. A secondary carbon (2º) is bonded to two other carbon atoms, and tertiary (3º) and quaternary (4º) carbon atoms are bonded respectively to three and four other carbons. The three C5H12 isomers shown below illustrate these terms.
Structural differences may occur within these four groups, depending on the molecular constitution. In the formula on the right all four 1º-carbons are structurally equivalent (remember the tetrahedral configuration of tetravalent carbon); however the central formula has two equivalent 1º-carbons (bonded to the 3º carbon on the left end) and a single, structurally different 1º-carbon (bonded to the 2º-carbon) at the right end. Similarly, the left-most formula has two structurally equivalent 2º-carbons (next to the ends of the chain), and a structurally different 2º-carbon in the middle of the chain. A consideration of molecular symmetry helps to distinguish structurally equivalent from nonequivalent atoms and groups. The ability to distinguish structural differences of this kind is an essential part of mastering organic chemistry. It will come with practice and experience.
Our ability to draw structural formulas for molecules is remarkable. To see how this is done Click Here.
Analysis of Molecular Formulas
Although structural formulas are essential to the unique description of organic compounds, it is interesting and instructive to evaluate the information that may be obtained from a molecular formula alone. Three useful rules may be listed:
1. The number of hydrogen atoms that can be bonded to a given number of carbon atoms is limited by the valence of carbon. For compounds of carbon and hydrogen (hydrocarbons) the maximum number of hydrogen atoms that can be bonded to n carbons is 2n + 2 (n is an integer). In the case of methane, CH4, n=1 & 2n + 2 = 4. The origin of this formula is evident by considering a hydrocarbon made up of a chain of carbon atoms. Here the middle carbons will each have two hydrogens and the two end carbons have three hydrogens each. Thus, a six-carbon chain (n = 6) may be written H-(CH2)6-H, and the total hydrogen count is (2 x 6) + 2 = 14. The presence of oxygen (valence = 2) does not change this relationship, so the previously described C4H10O isomers follow the rule, n=4 & 2n + 2 = 10. Halogen atoms (valence = 1) should be counted equivalent to hydrogen, as illustrated by C3H5Cl3, n = 3 & 2n + 2 = 8 = (5 + 3). If nitrogen is present, each nitrogen atom (valence = 3) increases the maximum number of hydrogens by one.
Some Plausible
Molecular Formulas
C7H16O3, C9H18, C15H28O3, C6H16N2
Some Impossible
Molecular Formulas
C8H20O6, C23H50, C5H10Cl4, C4H12NO
1. For stable organic compounds the total number of odd-valenced atoms is even. Thus, when even-valenced atoms such as carbon and oxygen are bonded together in any number and in any manner, the number of remaining unoccupied bonding sites must be even. If these sites are occupied by univalent atoms such as H, F, Cl, etc. their total number will necessarily be even. Nitrogen is also an odd-valenced atom (3), and if it occupies a bonding site on carbon it adds two additional bonding sites, thus maintaining the even/odd parity.
Some Plausible
Molecular Formulas
C4H4Cl2, C5H9OBr, C5H11NO2, C12H18N2FCl
Some Impossible
Molecular Formulas
C5H9O2, C4H5ClBr, C6H11N2O, C10H18NCl2
1. The number of hydrogen atoms in stable compounds of carbon, hydrogen & oxygen reflects the number of double bonds and rings in their structural formulas. Consider a hydrocarbon with a molecular structure consisting of a simple chain of four carbon atoms, CH3CH2CH2CH3. The molecular formula is C4H10 (the maximum number of bonded hydrogens by the 2n + 2 rule). If the four carbon atoms form a ring, two hydrogens must be lost. Similarly, the introduction of a double bond entails the loss of two hydrogens, and a triple bond the loss of four hydrogens.
From the above discussion and examples it should be clear that the molecular formula of a hydrocarbon (CnHm) provides information about the number of rings and/or double bonds that must be present in its structural formula. By rule #2 must be an even number, so if m < (2n + 2) the difference is also an even number that reflects any rings and double bonds. A triple bond is counted as two double bonds.
The presence of one or more nitrogen atoms or halogen substituents requires a modified analysis. The above formula may be extended to such compounds by a few simple principles:
• The presence of oxygen does not alter the relationship.
• All halogens present in the molecular formula must be replaced by hydrogen.
• Each nitrogen in the formula must be replaced by a CH moiety.
Resonance
Kekulé structural formulas are essential tools for understanding organic chemistry. However, the structures of some compounds and ions cannot be represented by a single formula. For example, sulfur dioxide (SO2) and nitric acid (HNO3) may each be described by two equivalent formulas (equations 1 & 2). For clarity the two ambiguous bonds to oxygen are given different colors in these formulas.
sulfur dioxide
nitric acid
If only one formula for sulfur dioxide was correct and accurate, then the double bond to oxygen would be shorter and stronger than the single bond. Since experimental evidence indicates that this molecule is bent (bond angle 120º) and has equal length sulfur : oxygen bonds (1.432 Å), a single formula is inadequate, and the actual structure resembles an average of the two formulas. This averaging of electron distribution over two or more hypothetical contributing structures (canonical forms) to produce a hybrid electronic structure is called resonance. Likewise, the structure of nitric acid is best described as a resonance hybrid of two structures, the double headed arrow being the unique symbol for resonance.
The above examples represent one extreme in the application of resonance. Here, two structurally and energetically equivalent electronic structures for a stable compound can be written, but no single structure provides an accurate or even an adequate representation of the true molecule. In cases such as these, the electron delocalization described by resonance enhances the stability of the molecules, and compounds or ions composed of such molecules often show exceptional stability.
formaldehyde
The electronic structures of most covalent compounds do not suffer the inadequacy noted above. Thus, completely satisfactory Kekulé formulas may be drawn for water (H2O), methane (CH4) and acetylene C2H2). Nevertheless, the principles of resonance are very useful in rationalizing the chemical behavior of many such compounds. For example, the carbonyl group of formaldehyde (the carbon-oxygen double bond) reacts readily to give addition products. The course of these reactions can be explained by a small contribution of a dipolar resonance contributor, as shown in equation 3. Here, the first contributor (on the left) is clearly the best representation of this molecular unit, since there is no charge separation and both the carbon and oxygen atoms have achieved valence shell neon-like configurations by covalent electron sharing. If the double bond is broken heterolytically, formal charge pairs result, as shown in the other two structures. The preferred charge distribution will have the positive charge on the less electronegative atom (carbon) and the negative charge on the more electronegative atom (oxygen). Therefore the middle formula represents a more reasonable and stable structure than the one on the right. The application of resonance to this case requires a weighted averaging of these canonical structures. The double bonded structure is regarded as the major contributor, the middle structure a minor contributor and the right hand structure a non-contributor. Since the middle, charge-separated contributor has an electron deficient carbon atom, this explains the tendency of electron donors (nucleophiles) to bond at this site.
The basic principles of the resonance method may now be summarized.
For a given compound, a set of Lewis / Kekulé structures are written, keeping the relative positions of all the component atoms the same. These are the canonical forms to be considered, and all must have the same number of paired and unpaired electrons.
The following factors are important in evaluating the contribution each of these canonical structures makes to the actual molecule.
1. The number of covalent bonds in a structure. (The greater the bonding, the more important and stable the contributing structure.)
2. Formal charge separation. (Other factors aside, charge separation decreases the stability and importance of the contributing structure.)
3. Electronegativity of charge bearing atoms and charge density. (High charge density is destabilizing. Positive charge is best accommodated on atoms of low electronegativity, and negative charge on high electronegative atoms.)
The stability of a resonance hybrid is always greater than the stability of any canonical contributor. Consequently, if one canonical form has a much greater stability than all others, the hybrid will closely resemble it electronically and energetically. This is the case for the carbonyl group (eq.3). The left hand C=O structure has much greater total bonding than either charge-separated structure, so it describes this functional group rather well. On the other hand, if two or more canonical forms have identical low energy structures, the resonance hybrid will have exceptional stabilization and unique properties. This is the case for sulfur dioxide (eq.1) and nitric acid (eq.2).
carbon monoxide
azide anion
To illustrate these principles we shall consider carbon monoxide (eq.4) and azide anion (eq.5). In each case the most stable canonical form is on the left. For carbon monoxide, the additional bonding is more important than charge separation. Furthermore, the double bonded structure has an electron deficient carbon atom (valence shell sextet). A similar destabilizing factor is present in the two azide canonical forms on the top row of the bracket (three bonds vs. four bonds in the left most structure). The bottom row pair of structures have four bonds, but are destabilized by the high charge density on a single nitrogen atom.
All the examples on this page demonstrate an important restriction that must be remembered when using resonance. No atoms change their positions within the common structural framework. Only electrons are moved.
Atomic and Molecular Orbitals
A more detailed model of covalent bonding requires a consideration of valence shell atomic orbitals. For second period elements such as carbon, nitrogen and oxygen, these orbitals have been designated 2s, 2px, 2py & 2pz. The spatial distribution of electrons occupying each of these orbitals is shown in the diagram below.
The valence shell electron configuration of carbon is 2s2, 2px1, 2py1 & 2pz0. If this were the configuration used in covalent bonding, carbon would only be able to form two bonds. Very nice displays of orbitals may be found at the following sites: J. Gutow, Univ. Wisconsin Oshkosh, R. Spinney, Ohio State and M. Winter, Sheffield University
Hybrid Orbitals
In order to explain the structure of methane (CH4), the 2s and three 2p orbitals must be converted to four equivalent hybrid atomic orbitals, each having 25% s and 75% p character, and designated sp3. These hybrid orbitals have a specific orientation, and the four are naturally oriented in a tetrahedral fashion.
Molecular Orbitals
Just as the valence electrons of atoms occupy atomic orbitals (AO), the shared electron pairs of covalently bonded atoms may be thought of as occupying molecular orbitals (MO). It is convenient to approximate molecular orbitals by combining or mixing two or more atomic orbitals. In general, this mixing of n atomic orbitals always generates n molecular orbitals. The hydrogen molecule provides a simple example of MO formation. In the following diagram, two 1s atomic orbitals combine to give a sigma (σ) bonding (low energy) molecular orbital and a second higher energy MO referred to as an antibonding orbital. The bonding MO is occupied by two electrons of opposite spin, the result being a covalent bond.
The notation used for molecular orbitals parallels that used for atomic orbitals. Thus, s-orbitals have a spherical symmetry surrounding a single nucleus, whereas σ-orbitals have a cylindrical symmetry and encompass two (or more) nuclei. In the case of bonds between second period elements, p-orbitals or hybrid atomic orbitals having p-orbital character are used to form molecular orbitals. For example, the sigma molecular orbital that serves to bond two fluorine atoms together is generated by the overlap of p-orbitals (part A below), and two sp3 hybrid orbitals of carbon may combine to give a similar sigma orbital. When these bonding orbitals are occupied by a pair of electrons a covalent bond, the sigma bond results. Although we have ignored the remaining p-orbitals, their inclusion in a molecular orbital treatment does not lead to any additional bonding, as may be shown by activating the fluorine correlation diagram below.
Another type of MO (the π orbital) may be formed from two p-orbitals by a lateral overlap, as shown in part A of the following diagram. Since bonds consisting of occupied π-orbitals (pi-bonds) are weaker than sigma bonds, pi-bonding between two atoms occurs only when a sigma bond has already been established. Thus, pi-bonding is generally found only as a component of double and triple covalent bonds. Since carbon atoms involved in double bonds have only three bonding partners, they require only three hybrid orbitals to contribute to three sigma bonds. A mixing of the 2s-orbital with two of the 2p orbitals gives three sp2 hybrid orbitals, leaving one of the p-orbitals unused. Two sp2 hybridized carbon atoms are then joined together by sigma and pi-bonds (a double bond), as shown in part B.
The manner in which atomic orbitals overlap to form molecular orbitals is commonly illustrated by a correlation diagram. Two examples of such diagrams for the simple diatomic elements F2 and N2 will be drawn above when the appropriate button is clicked. The 1s and 2s atomic orbitals do not provide any overall bonding, since orbital overlap is minimal, and the resulting sigma bonding and antibonding components would cancel. In both these cases three 2p atomic orbitals combine to form a sigma and two pi-molecular orbitals, each as a bonding and antibonding pair. The overall bonding order depends on the number of antibonding orbitals that are occupied.
The subtle change in the energy of the σ2p bonding orbital, relative to the two degenerate π-bonding orbitals, is due to s-p hybridization that is unimportant to the present discussion. An impressive example of the advantages offered by the molecular orbital approach to bonding is found in the oxygen molecule. Click Here to see this application. The p-orbitals in this model are represented by red and blue colored spheres, which represent different phases, defined by the mathematical wave equations for such orbitals.
Finally, in the case of carbon atoms with only two bonding partners only two hybrid orbitals are needed for the sigma bonds, and these sp hybrid orbitals are directed 180º from each other. Two p-orbitals remain unused on each sp hybridized atom, and these overlap to give two pi-bonds following the formation of a sigma bond (a triple bond), as shown below.
Contributors
William Reusch, Professor Emeritus (Michigan State U.), Virtual Textbook of Organic Chemistry | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Fundamentals/Introduction_to_Organic_Chemistry/Molecular_Shape.txt |
There are many types of chemical bonds and forces that bind molecules together. The two most basic types of bonds are characterized as either ionic or covalent. In ionic bonding, atoms transfer electrons to each other. Ionic bonds require at least one electron donor and one electron acceptor. In contrast, atoms with the same electronegativity share electrons in covalent bonds, because neither atom preferentially attracts or repels the shared electrons.
Introduction
Ionic bonding is the complete transfer of valence electron(s) between atoms. It is a type of chemical bond that generates two oppositely charged ions. In ionic bonds, the metal loses electrons to become a positively charged cation, whereas the nonmetal accepts those electrons to become a negatively charged anion. Ionic bonds require an electron donor, often a metal, and an electron acceptor, a nonmetal.
Ionic bonding is observed because metals have few electrons in their outer-most orbitals. By losing those electrons, these metals can achieve noble gas configuration and satisfy the octet rule. Similarly, nonmetals that have close to 8 electrons in their valence shells tend to readily accept electrons to achieve noble gas configuration. In ionic bonding, more than 1 electron can be donated or received to satisfy the octet rule. The charges on the anion and cation correspond to the number of electrons donated or received. In ionic bonds, the net charge of the compound must be zero.
This sodium molecule donates the lone electron in its valence orbital in order to achieve octet configuration. This creates a positively charged cation due to the loss of electron.
This chlorine atom receives one electron to achieve its octet configuration, which creates a negatively charged anion.
The predicted overall energy of the ionic bonding process, which includes the ionization energy of the metal and electron affinity of the nonmetal, is usually positive, indicating that the reaction is endothermic and unfavorable. However, this reaction is highly favorable because of the electrostatic attraction between the particles. At the ideal interatomic distance, attraction between these particles releases enough energy to facilitate the reaction. Most ionic compounds tend to dissociate in polar solvents because they are often polar. This phenomenon is due to the opposite charges on each ion.
Example \(1\): Chloride Salts
In this example, the sodium atom is donating its 1 valence electron to the chlorine atom. This creates a sodium cation and a chlorine anion. Notice that the net charge of the resulting compound is 0.
In this example, the magnesium atom is donating both of its valence electrons to chlorine atoms. Each chlorine atom can only accept 1 electron before it can achieve its noble gas configuration; therefore, 2 atoms of chlorine are required to accept the 2 electrons donated by the magnesium. Notice that the net charge of the compound is 0.
Covalent Bonding
Covalent bonding is the sharing of electrons between atoms. This type of bonding occurs between two atoms of the same element or of elements close to each other in the periodic table. This bonding occurs primarily between nonmetals; however, it can also be observed between nonmetals and metals.
If atoms have similar electronegativities (the same affinity for electrons), covalent bonds are most likely to occur. Because both atoms have the same affinity for electrons and neither has a tendency to donate them, they share electrons in order to achieve octet configuration and become more stable. In addition, the ionization energy of the atom is too large and the electron affinity of the atom is too small for ionic bonding to occur. For example: carbon does not form ionic bonds because it has 4 valence electrons, half of an octet. To form ionic bonds, Carbon molecules must either gain or lose 4 electrons. This is highly unfavorable; therefore, carbon molecules share their 4 valence electrons through single, double, and triple bonds so that each atom can achieve noble gas configurations. Covalent bonds include interactions of the sigma and pi orbitals; therefore, covalent bonds lead to formation of single, double, triple, and quadruple bonds.
Example \(2\): \(PCl_3\)
In this example, a phosphorous atom is sharing its three unpaired electrons with three chlorine atoms. In the end product, all four of these molecules have 8 valence electrons and satisfy the octet rule.
Bonding in Organic Chemistry
Ionic and covalent bonds are the two extremes of bonding. Polar covalent is the intermediate type of bonding between the two extremes. Some ionic bonds contain covalent characteristics and some covalent bonds are partially ionic. For example, most carbon-based compounds are covalently bonded but can also be partially ionic. Polarity is a measure of the separation of charge in a compound. A compound's polarity is dependent on the symmetry of the compound and on differences in electronegativity between atoms. Polarity occurs when the electron pushing elements, found on the left side of the periodic table, exchanges electrons with the electron pulling elements, on the right side of the table. This creates a spectrum of polarity, with ionic (polar) at one extreme, covalent (nonpolar) at another, and polar covalent in the middle.
Both of these bonds are important in organic chemistry. Ionic bonds are important because they allow the synthesis of specific organic compounds. Scientists can manipulate ionic properties and these interactions in order to form desired products. Covalent bonds are especially important since most carbon molecules interact primarily through covalent bonding. Covalent bonding allows molecules to share electrons with other molecules, creating long chains of compounds and allowing more complexity in life.
Problems
1. Are these compounds ionic or covalent?
2. In the following reactions, indicate whether the reactants and products are ionic or covalently bonded.
a)
b) Clarification: What is the nature of the bond between sodium and amide? What kind of bond forms between the anion carbon chain and sodium?
c)
Solutions
• 1) From left to right: Covalent, Ionic, Ionic, Covalent, Covalent, Covalent, Ionic.
• 2a) All products and reactants are ionic.
• 2b) From left to right: Covalent, Ionic, Ionic, Covalent, Ionic, Covalent, Covalent, Ionic.
• 2c) All products and reactants are covalent. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Fundamentals/Ionic_and_Covalent_Bonds.txt |
Includes structural isomerism and stereoisomerism (both geometric and optical).
Isomerism in Organic Compounds
• Two enantiomers have the same physical properties. They cannot be separated easily by standard laboratory techniques.
• Two diastereomers have different physical properties and can often be separated via standard laboratory techniques.
• If an additional chiral center can be incorporated into a pair of enantiomers so that they become diastereomers, they can be separated.
An example of this method of obtaining one isomer of a compound involves the formation of a diastereomeric salt. The salt has multiple chiral centers, and so diastereomers are possible.
If a racemic mixture of phenylsuccinic acid is mixed with a pure sample of (-)-proline (a naturally available amino acid), a proton transfer (or Bronsted acid-base) reaction occurs. Two protons are transferred from the phenylsuccinic acid to the proline. One proline has a position available for one extra proton, so the two protons end up on two different prolines. Because protons have +1 charge, each proline is cationic. The phenylsuccinic acid gave up two positives, so it is a dianion. Together, these three ions form a salt.
Figure SC14.1. Formation of a salt containing three chiral centers.
Figure SC14.2. Formation of a salt containing three chiral centers. This one is a diastereomer of the salt in the previous picture.
If a racemic mixture of phenylsuccinic acid is used, but pure (-)-proline is added, two possible diastereomers result. One chiral center, in the proline, is always the same. The other chiral center, in the phenylsuccinic acid, can be in two different configurations. As a result, cations and anions may pack differently together in each case, so different melting points and solubilities result. One salt precipitates or forms a solid from the solution, but the other stays dissolved. The two diastereomers can be separated by filtration.
Problem SC14.1.
Suppose you have a pure sample of L-phenylalanine. Write equations for reactions, using structures, that show how you could use it to obtain a sample of (S)-naproxen, an analgesic.
Enantiomers
An atom with four groups attached to it can also adopt a tetrahedral geometry. This geometry often occurs when the central atom is a little smaller. A tetrahedral geometry allows neighbouring groups to get a little farther from each other.
Unlike square planar compounds, simple tetrahedral compounds do not have the same kind of cis- and trans- isomers. That is, two groups can't be placed on a tetrahedron so that they are opposite each other or beside each other. The relationship between any two groups on a tetrahedron is the same as the relationship between any other two groups on a tetrahedron.
Dichlorodimethylsilane is a compound that can be used to make silicone polymers. Like platin, it has two each of two groups attached to the central atom. However, the central tom is tetrahedral. There is only one way to arrange these four groups.
Figure SC3.1. A tetrahedral atom with two different types of groups attached, (CH3)2SiCl2.
However, if four different groups are attached to a tetrahedral atom, the four groups can be arranged in two possible ways. The two compounds that result are mirror images of each other. These two isomers are called enantiomers.
Figure SC3.2. A pair of enantiomers. The (-) enantiomer is on the left and the (+) enantiomer is on the right. Note that the tetrahedral silicon atom has four different groups attached.
• Enantiomers are pairs of compounds with exactly the same connectivity but opposite three-dimensional shapes.
• Enantiomers are not the same as each other; one enantiomer cannot be superimposed on the other.
• Enantiomers are mirror images of each other.
Two compounds with the exact same connectivity, that are mirror images of each other but that are not identical to each other are called enantiomers. The more common definition of an enantiomer is that it is not superimposable on its mirror image. It can be distinguished easily from its mirror image, just as a right hand can easily be identified and distinguished from a left hand.
• Compounds that occur in these pairs are called "chiral".
• "Chiral" comes from the Greek word for "hand".
It can be shown using group theory, the mathematics of symmetry, that an enantiomer may also be defined as a molecule that does not contain a mirror plane, meaning it cannot be divided into two identical and opposite halves.
• Enantiomers contain no mirror planes.
• Enantiomers do not contain two equal and opposite halves.
Unlike cis- and trans-isomers, two enantiomers have the same physical properties. they have the same melting point, the same solubility, and so on. Two compounds that are almost identical, but mirror images of each other, have exactly the same kinds of intermolecular attraction, so it may not be a surprise that their physical properties are identical.
• Enantiomers are another example of a type of stereoisomers.
• Two enantiomers have identical physical properties, except for optical rotation.
Optical rotation involves the interaction of plane-polarized light with a material. If a material is not symmetric, the light that passes through it will be rotated. That means if the waves making up the light are oscillating in one direction as they enter the material, they will have tilted slightly to oscillate in another direction when they emerge from the material. We will look at this phenomenon later.
• Two enantiomers have an equal but opposite rotational effect on plane-polarized light.
• (+) enantiomers rotate light in a clockwise direction.
• (-) enantiomers rotate light in a counterclockwise direction.
For example, in the chiral silicon compound shown above, the (+) enantiomer rotates plane-polarized light in a clockwise direction. It has a "standard optical rotation" of [a] = +12 (+/-2)o. The (-) enantiomer rotates plane-polarized light in a counterclockwise direction. It has a "standard optical rotation" of [a] = -9.9 (+/-2)o.
Problem SC3.1.
A certain compound exists in two forms; enantiomer A and enantiomer B. Enantiomer A has a molecular weight of 126 g/mol, a density of 0.995 g/mL, an optical rotation of [a] = 26o, a melting point of 65 oC, a boiling point of 225 oC, and an odour of citrus fruit. What can you say about the corresponding properties of enantiomer B? | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Fundamentals/Isomerism_in_Organic_Compounds/Diastereomers_and_Optical_Resolution.txt |
Geometric isomerism (also known as cis-trans isomerism or E-Z isomerism) is a form of stereoisomerism. This page explains what stereoisomers are and how you recognise the possibility of geometric isomers in a molecule.
What are isomers?
Isomers are molecules that have the same molecular formula, but have a different arrangement of the atoms in space. That excludes any different arrangements which are simply due to the molecule rotating as a whole, or rotating about particular bonds. Where the atoms making up the various isomers are joined up in a different order, this is known as structural isomerism. Structural isomerism is not a form of stereoisomerism, and is dealt with on a separate page.
In stereoisomerism, the atoms making up the isomers are joined up in the same order, but still manage to have a different spatial arrangement. Geometric isomerism is one form of stereoisomerism.
Geometric (cis / trans) isomerism
These isomers occur where you have restricted rotation somewhere in a molecule. At an introductory level in organic chemistry, examples usually just involve the carbon-carbon double bond - and that's what this page will concentrate on. Think about what happens in molecules where there is unrestricted rotation about carbon bonds - in other words where the carbon-carbon bonds are all single. The next diagram shows two possible configurations of 1,2-dichloroethane.
These two models represent exactly the same molecule. You can get from one to the other just by twisting around the carbon-carbon single bond. These molecules are not isomers.
If you draw a structural formula instead of using models, you have to bear in mind the possibility of this free rotation about single bonds. You must accept that these two structures represent the same molecule:
But what happens if you have a carbon-carbon double bond - as in 1,2-dichloroethene?
These two molecules are not the same. The carbon-carbon double bond won't rotate and so you would have to take the models to pieces in order to convert one structure into the other one. That is a simple test for isomers. If you have to take a model to pieces to convert it into another one, then you've got isomers. If you merely have to twist it a bit, then you haven't!
Drawing structural formulae for the last pair of models gives two possible isomers:
1. In one, the two chlorine atoms are locked on opposite sides of the double bond. This is known as the trans isomer. (trans : from latin meaning "across" - as in transatlantic).
2. In the other, the two chlorine atoms are locked on the same side of the double bond. This is know as the cis isomer. (cis : from latin meaning "on this side")
The most likely example of geometric isomerism you will meet at an introductory level is but-2-ene. In one case, the CH3 groups are on opposite sides of the double bond, and in the other case they are on the same side.
The importance of drawing geometric isomers properly
It's very easy to miss geometric isomers in exams if you take short-cuts in drawing the structural formulae. For example, it is very tempting to draw but-2-ene as
CH3CH=CHCH3
If you write it like this, you will almost certainly miss the fact that there are geometric isomers. If there is even the slightest hint in a question that isomers might be involved, always draw compounds containing carbon-carbon double bonds showing the correct bond angles (120°) around the carbon atoms at the ends of the bond. In other words, use the format shown in the last diagrams above.
How to recognize the possibility of geometric isomerism
You obviously need to have restricted rotation somewhere in the molecule. Compounds containing a carbon-carbon double bond have this restricted rotation. (Other sorts of compounds may have restricted rotation as well, but we are concentrating on the case you are most likely to meet when you first come across geometric isomers.) If you have a carbon-carbon double bond, you need to think carefully about the possibility of geometric isomers.
What needs to be attached to the carbon-carbon double bond?
Think about this case:
Although we've swapped the right-hand groups around, these are still the same molecule. To get from one to the other, all you would have to do is to turn the whole model over. You won't have geometric isomers if there are two groups the same on one end of the bond - in this case, the two pink groups on the left-hand end. So there must be two different groups on the left-hand carbon and two different groups on the right-hand one. The cases we've been exploring earlier are like this:
But you could make things even more different and still have geometric isomers:
Here, the blue and green groups are either on the same side of the bond or the opposite side. Or you could go the whole hog and make everything different. You still get geometric isomers, but by now the words cis and trans are meaningless. This is where the more sophisticated E-Z notation comes in.
Summary
To get geometric isomers you must have:
• restricted rotation (often involving a carbon-carbon double bond for introductory purposes);
• two different groups on the left-hand end of the bond and two different groups on the right-hand end. It doesn't matter whether the left-hand groups are the same as the right-hand ones or not.
Contributors
Jim Clark (Chemguide.co.uk) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Fundamentals/Isomerism_in_Organic_Compounds/Geometric_Isomerism_in_Organic_Molecules.txt |
Optical isomerism is a form of stereoisomerism. This page explains what stereoisomers are and how you recognize the possibility of optical isomers in a molecule.
What are stereoisomers?
Isomers are molecules that have the same molecular formula, but have a different arrangement of the atoms in space. That excludes any different arrangements which are simply due to the molecule rotating as a whole, or rotating about particular bonds. Where the atoms making up the various isomers are joined up in a different order, this is known as structural isomerism. Structural isomerism is not a form of stereoisomerism, which involve the atoms of the complex bonded in the same order, but in different spatial arrangements. Optical isomerism is one form of stereoisomerism; geometric isomers are a second type.
Optical isomerism
Optical isomers are named like this because of their effect on plane polarized light. Simple substances which show optical isomerism exist as two isomers known as enantiomers.
• A solution of one enantiomer rotates the plane of polarisation in a clockwise direction. This enantiomer is known as the (+) form.
• For example, one of the optical isomers (enantiomers) of the amino acid alanine is known as (+)alanine.
• A solution of the other enantiomer rotates the plane of polarisation in an anti-clockwise direction. This enantiomer is known as the (-) form. So the other enantiomer of alanine is known as or (-)alanine.
• If the solutions are equally concentrated the amount of rotation caused by the two isomers is exactly the same - but in opposite directions.
• When optically active substances are made in the lab, they often occur as a 50/50 mixture of the two enantiomers. This is known as a racemic mixture or racemate. It has no effect on plane polarised light.
Origin of Optical Isomers
The examples of organic optical isomers contain a carbon atom joined to four different groups. These two models each have the same groups joined to the central carbon atom, but still manage to be different:
Obviously as they are drawn, the orange and blue groups are not aligned the same way. Could you get them to align by rotating one of the molecules? The next diagram shows what happens if you rotate molecule B.
They still are not the same - and there is no way that you can rotate them so that they look exactly the same. These are isomers of each other. They are described as being non-superimposable in the sense that (if you imagine molecule B being turned into a ghostly version of itself) you couldn't slide one molecule exactly over the other one. Something would always be pointing in the wrong direction.
What happens if two of the groups attached to the central carbon atom are the same? The next diagram shows this possibility.
The two models are aligned exactly as before, but the orange group has been replaced by another pink one. Rotating molecule B this time shows that it is exactly the same as molecule A. You only get optical isomers if all four groups attached to the central carbon are different.
Chiral and achiral molecules
The essential difference between the two examples we've looked at lies in the symmetry of the molecules. If there are two groups the same attached to the central carbon atom, the molecule has a plane of symmetry. If you imagine slicing through the molecule, the left-hand side is an exact reflection of the right-hand side.
Where there are four groups attached, there is no symmetry anywhere in the molecule
A molecule which has no plane of symmetry is described as chiral. The carbon atom with the four different groups attached which causes this lack of symmetry is described as a chiral center or as an asymmetric carbon atom. The molecule on the left above (with a plane of symmetry) is described as achiral. Only chiral molecules have optical isomers.
The relationship between the enantiomers
One of the enantiomers is simply a non-superimposable mirror image of the other one. In other words, if one isomer looked in a mirror, what it would see is the other one. The two isomers (the original one and its mirror image) have a different spatial arrangement, and so cannot be superimposed on each other.
If an achiral molecule (one with a plane of symmetry) looked in a mirror, you would always find that by rotating the image in space, you could make the two look identical. It would be possible to superimpose the original molecule and its mirror image.
Example 1: Isobutanol
The asymmetric carbon atom in a compound (the one with four different groups attached) is often shown by a star.
It's extremely important to draw the isomers correctly. Draw one of them using standard bond notation to show the 3-dimensional arrangement around the asymmetric carbon atom. Then draw the mirror to show the examiner that you know what you are doing, and then the mirror image.
Notice that you don't literally draw the mirror images of all the letters and numbers! It is, however, quite useful to reverse large groups - look, for example, at the ethyl group at the top of the diagram. It doesn't matter in the least in what order you draw the four groups around the central carbon. As long as your mirror image is drawn accurately, you will automatically have drawn the two isomers.
So which of these two isomers is (+)butan-2-ol and which is (-)butan-2-ol? There is no simple way of telling that. For A'level purposes, you can just ignore that problem - all you need to be able to do is to draw the two isomers correctly.
Example 2: 2-hydroxypropanoic acid (lactic acid)
Once again the chiral center is shown by a star.
The two enantiomers are:
It is important this time to draw the COOH group backwards in the mirror image. If you don't there is a good chance of you joining it on to the central carbon wrongly.
If you draw it like this in an exam, you will not get the mark for that isomer even if you have drawn everything else perfectly.
Example 3: 2-aminopropanoic acid (alanine)
This is typical of naturally-occurring amino acids. Structurally, it is just like the last example, except that the -OH group is replaced by -NH2
The two enantiomers are:
Only one of these isomers occurs naturally: the (+) form. You cannot tell just by looking at the structures which this is.
It has, however, been possible to work out which of these structures is which. Naturally occurring alanine is the right-hand structure, and the way the groups are arranged around the central carbon atom is known as an L- configuration. Notice the use of the capital L. The other configuration is known as D-.
So you may well find alanine described as L-(+)alanine. That means that it has this particular structure and rotates the plane of polarization clockwise.
Even if you know that a different compound has an arrangement of groups similar to alanine, you still cannot say which way it will rotate the plane of polarization. The other amino acids, for example, have the same arrangement of groups as alanine does (all that changes is the CH3 group), but some are (+) forms and others are (-) forms.
It's quite common for natural systems to only work with one of the enantiomers of an optically active substance. It is not too difficult to see why that might be. Because the molecules have different spatial arrangements of their various groups, only one of them is likely to fit properly into the active sites on the enzymes they work with.
In the lab, it is quite common to produce equal amounts of both forms of a compound when it is synthesized. This happens just by chance, and you tend to get racemic mixtures.
Identifying Chiral Centers
A skeletal formula is the most stripped-down formula possible. Look at the structural formula and skeletal formula for butan-2-ol.
Notice that in the skeletal formula all of the carbon atoms have been left out, as well as all of the hydrogen atoms attached to carbons. In a skeletal diagram of this sort:
• there is a carbon atom at each junction between bonds in a chain and at the end of each bond (unless there is something else there already - like the -OH group in the example);
• there are enough hydrogen atoms attached to each carbon to make the total number of bonds on that carbon up to 4.
We have already discussed the butan-2-ol case further up the page, and you know that it has optical isomers. The second carbon atom (the one with the -OH attached) has four different groups around it, and so is a chiral center.
Is this obvious from the skeletal formula? Well, it is, provided you remember that each carbon atom has to have 4 bonds going away from it. Since the second carbon here only seems to have 3, there must also be a hydrogen attached to that carbon. So it has a hydrogen, an -OH group, and two different hydrocarbon groups (methyl and ethyl).
Four different groups around a carbon atom means that it is a chiral center.
Example 4: A slightly more complicated case: 2,3-dimethylpentane
The diagrams show an uncluttered skeletal formula, and a repeat of it with two of the carbons labeled.
Look first at the carbon atom labeled 2. Is this a chiral center? No, it is not. Two bonds (one vertical and one to the left) are both attached to methyl groups. In addition, of course, there is a hydrogen atom and the more complicated hydrocarbon group to the right. It doesn't have 4 different groups attached, and so is not a chiral center.
What about the number 3 carbon atom? This has a methyl group below it, an ethyl group to the right, and a more complicated hydrocarbon group to the left. Plus, of course, a hydrogen atom to make up the 4 bonds that have to be formed by the carbon. That means that it is attached to 4 different things, and so is a chiral center.
Introducing Rings
We will start with a fairly simple ring compound:
When you are looking at rings like this, as far as optical isomerism is concerned, you don't need to look at any carbon in a double bond. You also don't need to look at any junction which only has two bonds going away from it. In that case, there must be 2 hydrogens attached, and so there cannot possibly be 4 different groups attached.
In this case, that means that you only need to look at the carbon with the -OH group attached. It has an -OH group, a hydrogen (to make up the total number of bonds to four), and links to two carbon atoms. How does the fact that these carbon atoms are part of a ring affect things?
You just need to trace back around the ring from both sides of the carbon you are looking at. Is the arrangement in both directions exactly the same? In this case, it is not. Going in one direction, you come immediately to a carbon with a double bond. In the other direction, you meet two singly bonded carbon atoms, and then one with a double bond. That means that you haven't got two identical hydrocarbon groups attached to the carbon you are interested in, and so it has 4 different groups in total around it. It is asymmetric - a chiral center.
What about this near-relative of the last molecule?
In this case, everything is as before, except that if you trace around the ring clockwise and counter-clockwise from the carbon at the bottom of the ring, there is an identical pattern in both directions. You can think of the bottom carbon being attached to a hydrogen, an -OH group, and two identical hydrocarbon groups. It therefore is not a chiral center.
The other thing which is very noticeable about this molecule is that there is a plane of symmetry through the carbon atom we are interested in. If you chopped it in half through this carbon, one side of the molecule would be an exact reflection of the other. In the first ring molecule above, that is not the case.
If you can see a plane of symmetry through the carbon atom it will not be a chiral center. If there is not a plane of symmetry, it will be a chiral center.
Example 5: Cholesterol
The skeletal diagram shows the structure of cholesterol. Some of the carbon atoms have been numbered for discussion purposes below. These are not part of the normal system for numbering the carbon atoms in cholesterol.
Before you read on, look carefully at each of the numbered carbon atoms, and decide which of them are chiral centers. The other carbon atoms in the structure cannot be chiral centers, because they are either parts of double bonds, or are joined to either two or three hydrogen atoms.
So . . . how many chiral centers did you find? In fact, there are 8 chiral centers out of the total of 9 carbons marked. If you didn't find all eight, go back and have another look before you read any further. It might help to sketch the structure on a piece of paper and draw in any missing hydrogens attached to the numbered carbons, and write in the methyl groups at the end of the branches as well.
This is done for you below, but it would be a lot better if you did it yourself and then checked your sketch afterwards .
Starting with the easy one - it is obvious that carbon 9 has two methyl groups attached. It doesn't have 4 different groups, and so cannot be chiral. If you take a general look at the rest, it is fairly clear that none of them has a plane of symmetry through the numbered carbons. Therefore they are all likely to be chiral centers. But it's worth checking to see what is attached to each of them.
• Carbon 1 has a hydrogen, an -OH and two different hydrocarbon chains (actually bits of rings) attached. Check clockwise and anticlockwise, and you will see that the arrangement is not identical in each direction. Four different groups means a chiral center.
• Carbon 2 has a methyl and three other different hydrocarbon groups. If you check along all three bits of rings , they are all different - another chiral center. This is also true of carbon 6.
• Carbons 3, 4, 5 and 7 are all basically the same. Each is attached to a hydrogen and three different bits of rings. All of these are chiral centers.
• Finally, carbon 8 has a hydrogen, a methyl group, and two different hydrocarbon groups attached. Again, this is a chiral center.
This all looks difficult at first glance, but it is not. You do, however, have to take a great deal of care in working through it - it is amazingly easy to miss one out. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Fundamentals/Isomerism_in_Organic_Compounds/Optical_Isomerism_in_Organic_Molecules.txt |
What do you notice about these three pictures? Count the number of left gloves and right gloves.
6 left and 6 right gloves, correct?
What about this one:
I count 8 right gloves, 4 left gloves. So there’s a slight excess of right gloves here.
Finally, this figure:
ONLY right hand gloves here. 12 right gloves, zero left gloves.
Application to organic chemistry?
Gloves are chiral objects. That is, they lack an internal plane of symmetry. Left gloves and right gloves are mirror images of each other, but they can’t be superimposed. In chemistry, there’s a word we have to describe a pair of non-superimposable mirror images – they’re called enantiomers.
Tying it back to the drawings, we can have three types of situations.
1. Racemic Mixture: In the first drawing, we have an equal number of left and right gloves (i.e. enantiomers). This is called a racemic mixture of enantiomers.
2. Enantiomeric excess: In the second drawing, we have an excess of right gloves compared to left gloves. In a situtation like this we can say we have an “enantiomeric excess” of gloves, or alternatively, the mixture is “enantioenriched” in the right-hand glove. [We can also calculate the "excess" here: the mixture is 66% right and 33% left - so we have a 33% "excess" of the right-hand enantiomer].
3. Enantiomeric pure: In the third drawing, we have only right-hand gloves. This is said to be an “enantiomerically pure” mixture of gloves, since we have only one enantiomer present.
To tie it back to chemistry, let’s say we have a solution of a chiral molecule, like 2-butanol, which can exist as either the (R)-enantiomer or the (S)-enantiomer.
• A solution containing equal amounts of (R)-2-butanol and (S)-2-butanol is a racemic mixture.
• A solution containing an excess of either the (R)-enantiomer or the (S)-enantiomer would be enantioenriched.
• A solution containing only the (R)-enantiomer or the (S)-enantiomer will be enantiomerically pure.
Contributors
James Ashenhurst (MasterOrganicChemistry.com)
• A big thanks to Agnieszka at IlluScientia for the glove drawings.
Structural Isomerism in Organic Molecules
This page explains what structural isomerism is, and looks at some of the various ways that structural isomers can arise.
What is structural isomerism?
Isomers are molecules that have the same molecular formula, but have a different arrangement of the atoms in space. That excludes any different arrangements which are simply due to the molecule rotating as a whole, or rotating about particular bonds. For example, both of the following are the same molecule. They are not isomers. Both are butane.
There are also endless other possible ways that this molecule could twist itself. There is completely free rotation around all the carbon-carbon single bonds. If you had a model of a molecule in front of you, you would have to take it to pieces and rebuild it if you wanted to make an isomer of that molecule. If you can make an apparently different molecule just by rotating single bonds, it's not different - it's still the same molecule.
In structural isomerism, the atoms are arranged in a completely different order. This is easier to see with specific examples. What follows looks at some of the ways that structural isomers can arise. The names of the various forms of structural isomerism probably don't matter all that much, but you must be aware of the different possibilities when you come to draw isomers.
Chain Isomerism
These isomers arise because of the possibility of branching in carbon chains. For example, there are two isomers of butane, \(C_4H_{10}\). In one of them, the carbon atoms lie in a "straight chain" whereas in the other the chain is branched.
Be careful not to draw "false" isomers which are just twisted versions of the original molecule. For example, this structure is just the straight chain version of butane rotated about the central carbon-carbon bond.
You could easily see this with a model. This is the example we've already used at the top of this page.
Example 1: Chain Isomers in Pentane
Pentane, C5H12, has three chain isomers. If you think you can find any others, they are simply twisted versions of the ones below. If in doubt make some models.
Position isomerism
In position isomerism, the basic carbon skeleton remains unchanged, but important groups are moved around on that skeleton.
Example 2: Positional Isomers in C5H12
For example, there are two structural isomers with the molecular formula C3H7Br. In one of them the bromine atom is on the end of the chain, whereas in the other it's attached in the middle.
If you made a model, there is no way that you could twist one molecule to turn it into the other one. You would have to break the bromine off the end and re-attach it in the middle. At the same time, you would have to move a hydrogen from the middle to the end.
Another similar example occurs in alcohols such as \(C_4H_9OH\)
These are the only two possibilities provided you keep to a four carbon chain, but there is no reason why you should do that. You can easily have a mixture of chain isomerism and position isomerism - you aren't restricted to one or the other.
So two other isomers of butanol are:
You can also get position isomers on benzene rings. Consider the molecular formula \(C_7H_7Cl\). There are four different isomers you could make depending on the position of the chlorine atom. In one case it is attached to the side-group carbon atom, and then there are three other possible positions it could have around the ring - next to the \(CH_3\) group, next-but-one to the \(CH_3\) group, or opposite the \(CH_3\) group.
Functional group isomerism
In this variety of structural isomerism, the isomers contain different functional groups - that is, they belong to different families of compounds (different homologous series).
Example 3: Isomers in C3H6O
A molecular formula \(C_3H_6O\) could be either propanal (an aldehyde) or propanone (a ketone).
There are other possibilities as well for this same molecular formula - for example, you could have a carbon-carbon double bond (an alkene) and an -OH group (an alcohol) in the same molecule.
Another common example is illustrated by the molecular formula \(C_3H_6O_2\). Amongst the several structural isomers of this are propanoic acid (a carboxylic acid) and methyl ethanoate (an ester).
Contributors
Jim Clark (Chemguide.co.uk) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Fundamentals/Isomerism_in_Organic_Compounds/Racemic_Mixtures.txt |
Lewis structures, also known as Lewis-dot diagrams, show the bonding relationship between atoms of a molecule and the lone pairs of electrons in the molecule. Lewis structures can also be useful in predicting molecular geometry in conjuntion with hybrid orbitals. A compound may have multiple resonance forms that are also all correct Lewis structures. This section will discuss the rules for writing out Lewis structures correctly.
Introduction
Writing out Lewis structures can be at times, tricky and somewhat difficult. A compound can have multiple Lewis Structures that contribute to the shape of the overall compound, so one Lewis structure of a compound may not necessarily be exactly what the compound looks like. But before we begin, there are a few things to know. An electron is represnted as a dot. A bond, which is made up of 2 shared electrons, is represented by two dots between the bonded atoms or a line (Kekulé structures lines).Double bonds and triple bonds are represented as two and three lines/(pairs of electrons), respectively. Lone pairs on the outer rims of an atom are represented as two dots. The electrons represented in a lewis stucture are the outer-shell electrons, which are called valence electrons. This is because they are the ones involved in chemical reactions. The examples listed below may or may not be consistent for each step for the sake of showing multiple examples, since all steps apply to every compound. Now we are ready to begin.
Step 1: Draw out the molecular backbone of the compound. This is probably the most difficult step in drawing Lewis structures because we must pick one or multiple atoms to be the central connector to terminal (outside) atoms. Usually, the central atom is already known as in the case with many organic compounds containing Carbon as the central atom. But at the basics, the atom with the lowest electronegativity is the central atom. Other factors come into play such as the octet rule, which will be discussed in this section. One can find the electronegativity values of atoms by looking at a periodic table of elements that contains the electronegativity values for each element. Electronegativity typically increases from left to right and decreases from top to bottom in a periodic table.
Example 1
Correct Correct Wrong
CCl4 (A refrigerant) C2H6 (Ethane) CHClCH (Bad layout, hydrogen and chlorine are central atoms)
Note: Electronegativity values: C = 2.55; Cl = 3.16; H = 2.20
Step 2: Add up the valence electrons for each atom in the molecule.
For example,
H2O 2 H: 2 x 1 electron = 2 electrons
1 O: 1 x 6 electrons = 6 electrons
--------------------------------------------------
Total: 8 electrons
Step 3: (Octet Rule) All covalent bonds are shown by two shared electrons. Place a pair of electrons between two elements that are connected to each other with a single bond. Next, give as many atoms as possible 8 electrons, except for Hydrogen, which has a duet (2-shared electrons). Do this by placing lone electron pairs on the outer atoms until each outer atom fulfills the octet. Place any remaining lone electron pairs on the central atom. Usually, electronegative atoms will have lone pairs on them. The number of electrons placed on the molecule should correspond to the total number of valence electrons of the atoms in the molecule.
Example 2
Hydrochloric Acid
Note: Hydrogen has 2 electrons, meeting its duet and Chlorine has 8 electrons, meeting its octet. All electrons are placed in correct spots.
Many times, satisfying the octet rule and total number of valence electrons in the molecule with only single bonds is not the case, meaning that there could be less valence electrons than needed to satisfy all octets with only single bond connections. In these situations, double or triple bonds may be needed to obtain octets. Rectify these situations by moving a lone pair of electrons from one atom to the bond between those two atoms. By doing this, one atom that is deficient of an electron pair of its octet can be filled, while having no effect on the other atom’s octet. Each compound is a case-by-case scenario and thus must be thought out carefully to solve the octet/total # of valence electron problems.
Example 3
Ethene Ethyne
****A simple way to find the need for double or triple bonds is by counting the number of electrons to fill all octets, then finding the difference with the total number of valence electrons available in that compound. The difference will show exactly how much electrons that need to be shared to meet the octet (duet for hydrogen) for all atoms.*****
-Note: Ethene has 2 C + 4 H amounting to 12 available electrons. If every bond was single electron paired, there would be 10 electrons. The hydrogen would have its duet met but both Carbons will not have their octet met. Placing the extra pair of electrons in between the two carbons, making a double bond amends the problem. Ethyne has 2 C + 2 H amounting to 10 available electrons. Placing the two extra pair of electrons between both carbons, making a triple bond fixes the problem with the octet rule and allows for all available electrons to be placed on the compound.
Step 4: Assign Formal Charges (FC) to each atom in the molecule.The formal charge of an atom is pretty much the difference between the number of valence electrons that a neutral atom would have and the number of electrons that is contained on the atom of the Lewis structure. If a charge on an atom is zero, you don't need to put any symbols on that atom. The total of the formal charges on a neutral molecule should be equal to zero. An atom is charged if the total charge of the atom is different from the valence electrons of the free, non-bonded atom.
Formal Charge
FC = # of valence electrons - # of lone pair electrons - ½(number of bonding electrons)
Charge is distributed over the surroundings and thus called, a formal charge.
Example 4: Formal Charge
Hydroxide Anion Methyl Cation
Exceptions
There are some exceptions to the Lewis structure rules that one always needs to keep in mind. Regardless, these exceptions are still correct Lewis structures.
Exception 1: Some compounds have odd # of electrons.
For example,
Ethyl Radical
Exception 2: Some compounds don't have enough valence electrons for an octet. The octet rule only works for elements in the second row given enough valence electrons and usually, some of these compounds are very reactive.
For example,
Lithium Hydride
-Note: Lithium and hydrogen each have 1 electron to share. Hydrogen meets its duet, but Lithium is still 6 electrons deficient from meeting octet.
Exception 3: After the second row of table of elements, it is not uncommon to see an "expanded valence shell," where atoms have more than 8 valence electrons. The usual explanation is that the bonding involves d-orbital electrons, and the resulting Lewis structures look like that shown below.
For example,
Sulfuric Acid
Notice that sulfur atom has 12 electrons around it.
However, expanded octets are not required to draw Lewis structures for these types of molecules. For example, another Lewis structure of sulfuric acid is shown here, and does not include any expanded octets.
By the rules of drawing Lewis structures, both of these structures have undesirable features. The top structure has expanded octets, whereas the bottom structure contains separated formal charge. However, quantum mechanical electronic structure calculations find little d-orbital participation in the bonding, as would be expected for expanded octet bonding, suggesting that the formal charge separation is preferable to expanding the octets. Therefore, there is generally no reason to draw structures with expanded octets, because they usually don't contribute significantly to the overall structure of the molecule. Even molecules such as XeF6 can be drawn without expanded octets.
Example 5: Ozone
Consider Ozone (O3)
Step 1: Build the backbone of ozone.
Step 2: Total up the valence electrons.
O3 3 O: 3 x 6 electrons = 18 electrons
---------------------------------------------
Total: 18 electrons
Step 3: Fill up each oxygen atom’s octet.
There are 20 electrons in the drawing but only 18 available electrons.
20 - 18 electrons = 2 electrons that need to be shared.
So, drag one pair of electrons from one of the exterior oxygens and make a double bond with the central oxygen to the respective side, while maintaining the octet rule.
Step 4: Assign formal charges to each oxygen atom.
Oxygen 1: 6 – 4 – ½(4) = 0
Oxygen 2: 6 – 2 – ½(6) = +1
Oxygen 3: 6 – 6 – ½(2) = -1
We found a Lewis structure for O3. But wait, there are other valid Lewis structures, just by moving electrons around!
For example,
These are the resonance forms of ozone.
Problems
Draw out a correct Lewis Structure for the following compounds.
1. HCN
2. LiF
3. C3H6 (two possibilities)
4. CO32-
5. CH3NO2
Answers:
1.
2.
3. or
4.
5.
Another simple and general procedure to draw Lewis structures has been proposed by A.B.P. Lever (see reference 5). Before beginning this procedure it is necessary to know the basic geometry of the molecule, i.e. whether it is cyclic or noncyclic, and which atoms are connected to which. Several worked examples for the determination of the Lewis structures of simple and more complicated species using the above described method can be found in the original paper and in Chemistry Net.
Contributors
• Daniel S. Kim, Paul Wenthold
Lewis Structures
Most descriptions of the bonding in \(XeF_6\) invoke an "expanded octet" model. However, an expanded octet is not required to draw \XeF_6\). Alternatively, the structure of \(XeF_6\) can be described as a resonance hybrid of octet-conforming structures, as shown below.
The calculated Mullkin charge at the xenon in \(XeF_6\) is approximately +2.3, which is very similar to the charge of +3 that is predicted based on this resonance model, and suggests that "expanded octet" is not an entirely valid model. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Fundamentals/Lewis_Structures/Drawing_XeF6_without_an_expanded_octet.txt |
The increasingly large number of organic compounds identified with each passing day, together with the fact that many of these compounds are isomers of other compounds, requires that a systematic nomenclature system be developed. Just as each distinct compound has a unique molecular structure which can be designated by a structural formula, each compound must be given a characteristic and unique name. As organic chemistry grew and developed, many compounds were given trivial names, which are now commonly used and recognized. Some examples are:
Name Methane Butane Acetone Toluene Acetylene Ethyl Alcohol
Formula CH4 C4H10 CH3COCH3 CH3C6H5 C2H2 C2H5OH
Such common names often have their origin in the history of the science and the natural sources of specific compounds, but the relationship of these names to each other is arbitrary, and no rational or systematic principles underly their assignments.
The IUPAC Nomenclature System
A rational nomenclature system should do at least two things. First, it should indicate how the carbon atoms of a given compound are bonded together in a characteristic lattice of chains and rings. Second, it should identify and locate any functional groups present in the compound. Since hydrogen is such a common component of organic compounds, its amount and locations can be assumed from the tetravalency of carbon, and need not be specified in most cases.
The IUPAC nomenclature system is a set of logical rules devised and used by organic chemists to circumvent problems caused by arbitrary nomenclature. Knowing these rules and given a structural formula, one should be able to write a unique name for every distinct compound. Likewise, given a IUPAC name, one should be able to write a structural formula. In general, an IUPAC name will have three essential features:
• A root or base indicating a major chain or ring of carbon atoms found in the molecular structure.
• A suffix or other element(s) designating functional groups that may be present in the compound.
• Names of substituent groups, other than hydrogen, that complete the molecular structure.
As an introduction to the IUPAC nomenclature system, we shall first consider compounds that have no specific functional groups. Such compounds are composed only of carbon and hydrogen atoms bonded together by sigma bonds (all carbons are sp3 hybridized).
Alkanes
Hydrocarbons having no double or triple bond functional groups are classified as alkanes or cycloalkanes, depending on whether the carbon atoms of the molecule are arranged only in chains or also in rings. Although these hydrocarbons have no functional groups, they constitute the framework on which functional groups are located in other classes of compounds, and provide an ideal starting point for studying and naming organic compounds. The alkanes and cycloalkanes are also members of a larger class of compounds referred to as aliphatic. Simply put, aliphatic compounds are compounds that do not incorporate any aromatic rings in their molecular structure.
The following table lists the IUPAC names assigned to simple continuous-chain alkanes from C-1 to C-10. A common "ane" suffix identifies these compounds as alkanes. Longer chain alkanes are well known, and their names may be found in many reference and text books. The names methane through decane should be memorized, since they constitute the root of many IUPAC names. Fortunately, common numerical prefixes are used in naming chains of five or more carbon atoms.
Table: Simple Unbranched Alkanes
Name Molecular
Formula
Structural
Formula
Isomers Name Molecular
Formula
Structural
Formula
Isomers
methane CH4 CH4 1 hexane C6H14 CH3(CH2)4CH3 5
ethane C2H6 CH3CH3 1 heptane C7H16 CH3(CH2)5CH3 9
propane C3H8 CH3CH2CH3 1 octane C8H18 CH3(CH2)6CH3 18
butane C4H10 CH3CH2CH2CH3 2 nonane C9H20 CH3(CH2)7CH3 35
pentane C5H12 CH3(CH2)3CH3 3 decane C10H22 CH3(CH2)8CH3 75
Some important behavior trends and terminologies
1. The formulas and structures of these alkanes increase uniformly by a \(CH_2\) increment.
2. A uniform variation of this kind in a series of compounds is called homologous.
3. These formulas all fit the \(C_nH_{2n+2}\) rule. This is also the highest possible H/C ratio for a stable hydrocarbon.
4. Since the H/C ratio in these compounds is at a maximum, we call them saturated (with hydrogen).
Beginning with butane (C4H10), and becoming more numerous with larger alkanes, we note the existence of alkane isomers. For example, there are five C6H14 isomers, shown below as abbreviated line formulas (A through E):
Although these distinct compounds all have the same molecular formula, only one (A) can be called hexane. How then are we to name the others?
The IUPAC system requires first that we have names for simple unbranched chains, as noted above, and second that we have names for simple alkyl groups that may be attached to the chains. Examples of some common alkyl groups are given in the following table. Note that the "ane" suffix is replaced by "yl" in naming groups. The symbol R is used to designate a generic (unspecified) alkyl group.
Group CH3 C2H5 CH3CH2CH2 (CH3)2CH– CH3CH2CH2CH2 (CH3)2CHCH2 CH3CH2CH(CH3)– (CH3)3C– R–
Name Methyl Ethyl Propyl Isopropyl Butyl Isobutyl sec-Butyl tert-Butyl Alkyl
IUPAC Rules for Alkane Nomenclature
1. Find and name the longest continuous carbon chain.
2. Identify and name groups attached to this chain.
3. Number the chain consecutively, starting at the end nearest a substituent group.
4. Designate the location of each substituent group by an appropriate number and name.
5. Assemble the name, listing groups in alphabetical order.
The prefixes di, tri, tetra etc., used to designate several groups of the same kind, are not considered when alphabetizing.
Halogen substituents are easily accommodated, using the names: fluoro (F-), chloro (Cl-), bromo (Br-) and iodo (I-).
Example 1: Halogen Substitution
For example, (CH3)2CHCH2CH2Br would be named 1-bromo-3-methylbutane. If the halogen is bonded to a simple alkyl group an alternative "alkyl halide" name may be used. Thus, C2H5Cl may be named chloroethane (no locator number is needed for a two carbon chain) or ethyl chloride.
For the above isomers of hexane the IUPAC names are: B 2-methylpentane C 3-methylpentane D 2,2-dimethylbutane E 2,3-dimethylbutane
Cycloalkanes
Cycloalkanes have one or more rings of carbon atoms. The simplest examples of this class consist of a single, unsubstituted carbon ring, and these form a homologous series similar to the unbranched alkanes. The IUPAC names of the first five members of this series are given in the following table. The last (yellow shaded) column gives the general formula for a cycloalkane of any size. If a simple unbranched alkane is converted to a cycloalkane two hydrogen atoms, one from each end of the chain, must be lost. Hence the general formula for a cycloalkane composed of n carbons is CnH2n. Although a cycloalkane has two fewer hydrogens than the equivalent alkane, each carbon is bonded to four other atoms so such compounds are still considered to be saturated with hydrogen.
Table 3: Examples of Simple Cycloalkanes
Name Cyclopropane Cyclobutane Cyclopentane Cyclohexane Cycloheptane Cycloalkane
Molecular
Formula
C3H6 C4H8 C5H10 C6H12 C7H14 CnH2n
Structural
Formula
(CH2)n
Line
Formula
Substituted cycloalkanes are named in a fashion very similar to that used for naming branched alkanes. The chief difference in the rules and procedures occurs in the numbering system. Since all the carbons of a ring are equivalent (a ring has no ends like a chain does), the numbering starts at a substituted ring atom.
IUPAC Rules for Cycloalkane Nomenclature
1. For a monosubstituted cycloalkane the ring supplies the root name (table above) and the substituent group is named as usual. A location number is unnecessary.
2. If the alkyl substituent is large and/or complex, the ring may be named as a substituent group on an alkane.
3. If two different substituents are present on the ring, they are listed in alphabetical order, and the first cited substituent is assigned to carbon #1. The numbering of ring carbons then continues in a direction (clockwise or counter-clockwise) that affords the second substituent the lower possible location number.
4. If several substituents are present on the ring, they are listed in alphabetical order. Location numbers are assigned to the substituents so that one of them is at carbon #1 and the other locations have the lowest possible numbers, counting in either a clockwise or counter-clockwise direction.
5. The name is assembled, listing groups in alphabetical order and giving each group (if there are two or more) a location number. The prefixes di, tri, tetra etc., used to designate several groups of the same kind, are not considered when alphabetizing.
Small rings, such as three and four membered rings, have significant angle strain resulting from the distortion of the sp3 carbon bond angles from the ideal 109.5º to 60º and 90º respectively. This angle strain often enhances the chemical reactivity of such compounds, leading to ring cleavage products. It is also important to recognize that, with the exception of cyclopropane, cycloalkyl rings are not planar (flat). The three dimensional shapes assumed by the common rings (especially cyclohexane and larger rings) are described and discussed in the Conformational Analysis Section.
Hydrocarbons having more than one ring are common, and are referred to as bicyclic (two rings), tricyclic (three rings) and in general, polycyclic compounds. The molecular formulas of such compounds have H/C ratios that decrease with the number of rings. In general, for a hydrocarbon composed of n carbon atoms associated with m rings the formula is: CnH(2n + 2 - 2m). The structural relationship of rings in a polycyclic compound can vary. They may be separate and independent, or they may share one or two common atoms. Some examples of these possible arrangements are shown in the following table.
Table 4: Examples of Isomeric \(C_8H_{14}\) Bicycloalkanes
Isolated Rings Spiro Rings Fused Rings Bridged Rings
No common atoms One common atom One common bond Two common atoms
Alkenes and Alkynes
Alkenes and alkynes are hydrocarbons which respectively have carbon-carbon double bond and carbon-carbon triple bond functional groups. The molecular formulas of these unsaturated hydrocarbons reflect the multiple bonding of the functional groups:
Alkane R–CH2–CH2–R CnH2n+2 This is the maximum H:C ratio for a given number of carbon atoms.
Alkene R–CH=CH–R CnH2n Each double bond reduces the number of hydrogen atoms by 2.
Alkyne R–C≡C–R CnH2n-2 Each triple bond reduces the number of hydrogen atoms by 4.
As noted earlier in the Analysis of Molecular Formulas section, the molecular formula of a hydrocarbon provides information about the possible structural types it may represent. For example, consider compounds having the formula C5H8. The formula of the five-carbon alkane pentane is C5H12 so the difference in hydrogen content is 4. This difference suggests such compounds may have a triple bond, two double bonds, a ring plus a double bond, or two rings. Some examples are shown here, and there are at least fourteen others!
IUPAC Rules for Alkene and Cycloalkene Nomenclature
1. The ene suffix (ending) indicates an alkene or cycloalkene.
2. The longest chain chosen for the root name must include both carbon atoms of the double bond.
3. The root chain must be numbered from the end nearest a double bond carbon atom. If the double bond is in the center of the chain, the nearest substituent rule is used to determine the end where numbering starts.
4. The smaller of the two numbers designating the carbon atoms of the double bond is used as the double bond locator. If more than one double bond is present the compound is named as a diene, triene or equivalent prefix indicating the number of double bonds, and each double bond is assigned a locator number.
5. In cycloalkenes the double bond carbons are assigned ring locations #1 and #2. Which of the two is #1 may be determined by the nearest substituent rule.
6. Substituent groups containing double bonds are:
H2C=CH– Vinyl group
H2C=CH–CH2 Allyl group
IUPAC Rules for Alkyne Nomenclature
1. The yne suffix (ending) indicates an alkyne or cycloalkyne.
2. The longest chain chosen for the root name must include both carbon atoms of the triple bond.
3. The root chain must be numbered from the end nearest a triple bond carbon atom. If the triple bond is in the center of the chain, the nearest substituent rule is used to determine the end where numbering starts.
4. The smaller of the two numbers designating the carbon atoms of the triple bond is used as the triple bond locator.
5. If several multiple bonds are present, each must be assigned a locator number. Double bonds precede triple bonds in the IUPAC name, but the chain is numbered from the end nearest a multiple bond, regardless of its nature.
6. Because the triple bond is linear, it can only be accommodated in rings larger than ten carbons. In simple cycloalkynes the triple bond carbons are assigned ring locations #1 and #2. Which of the two is #1 may be determined by the nearest substituent rule.
7. Substituent groups containing triple bonds are:
HC≡C– Ethynyl group
HC≡CH–CH2 Propargyl group
Benzene Derivatives
The nomenclature of substituted benzene ring compounds is less systematic than that of the alkanes, alkenes and alkynes. A few mono-substituted compounds are named by using a group name as a prefix to "benzene", as shown by the combined names listed below. A majority of these compounds, however, are referred to by singular names that are unique. There is no simple alternative to memorization in mastering these names.
Two commonly encountered substituent groups that incorporate a benzene ring are phenyl, abbreviated Ph-, and benzyl, abbreviated Bn-. These are shown here with examples of their use. Be careful not to confuse a phenyl (pronounced fenyl) group with the compound phenol (pronounced feenol). A general and useful generic notation that complements the use of R- for an alkyl group is Ar- for an aryl group (any aromatic ring).
When more than one substituent is present on a benzene ring, the relative locations of the substituents must be designated by numbering the ring carbons or by some other notation. In the case of disubstituted benzenes, the prefixes ortho, meta & para are commonly used to indicate a 1,2- or 1,3- or 1,4- relationship respectively. In the following examples, the first row of compounds show this usage in red. Some disubstituted toluenes have singular names (e.g. xylene, cresol & toluidine) and their isomers are normally designated by the ortho, meta or para prefix. A few disubstituted benzenes have singular names given to specific isomers (e.g. salicylic acid & resorcinol). Finally, if there are three or more substituent groups, the ring is numbered in such a way as to assign the substituents the lowest possible numbers, as illustrated by the last row of examples. The substituents are listed alphabetically in the final name. If the substitution is symmetrical (third example from the left) the numbering corresponds to the alphabetical order.
Alkyl halides
One of the simplest functional groups is the alkyl halide. In an alkyl halide, one of the hydrogen atoms in an alkane has been replaced by a halogen. Alkyl halides are easy to name. The name of the alkane is preceded by the number of the carbon on which the halogen is substituted and the name of the halogen, modified so that -ine is replaced by -o (e.g. 2-bromopropane). Alkyl halides are treated the same as alkyl substituents; if a molecule contains both, they are ranked to give the lowest possible number to a substituent and then listed in alphabetical order. If a molecule contains an alkyl halide and a multiple bond, numbers are assigned to give the lowest number to the first functional group. In the event of a tie, the lowest number goes to the multiple bond.
Alcohols
The alcohol is a very common functional group and a very easy one to name. The molecule is named as if it were an alkane (or alkene or alkyne), except that the suffix -ane is replaced by -ol and the number of the carbon atom on which the -OH group is located is placed before the name of the compound (e.g. 2-butanol). The alcohol functional group takes precedence over alkyl substituents, multiple bonds, and halides and always gets the lowest number.
Ethers
An ether is a molecule consisting of two alkyl groups connected to an oxygen atom. Ethers are named by considering one alkyl group (the shorter one) plus the oxygen atom to be a substituent and the other alkyl group (the longer one) to be an alkane. The alkyl group plus oxygen atom is called an "alkoxy" substituent and is named by replacing -ane suffix from the alkane with -oxy (e.g. methane becomes methoxy). The allkoxy substituent gets priority over alkyl and halide substituents, but not over alcohols, which will get the lower number.
Amines
An amine is a derivatives of the molecule ammonia, NH3, in which one or more of the hydrogens has been replaced by an alkyl substituent (R group). Amines are named by replacing the suffix -ane with -amine, like in alcohols. If there are further substituents attached to the nitrogen atom, they are preceded by N-. The amine gets the lowest number.
Nomenclature
This page explains how to write the formula for an organic compound given its name - and vice versa. It covers alkanes, cycloalkanes, alkenes, simple compounds containing halogens, alcohols, aldehydes and ketones.
Background
There are two skills you have to develop in this area:
• You need to be able to translate the name of an organic compound into its structural formula.
• You need to be able to name a compound from its given formula.
The first of these is more important (and also easier!) than the second. In an exam, if you can't write a formula for a given compound, you aren't going to know what the examiner is talking about and could lose lots of marks. However, you might only be asked to write a name for a given formula once in a whole exam - in which case you only risk 1 mark.
So, we're going to look mainly at how you decode names and turn them into formulae. In the process you will also pick up tips about how to produce names yourself.
In the early stages of an organic chemistry course people frequently get confused and daunted by the names because they try to do too much at once. Don't try to read all these pages in one go. Just go as far as the compounds you are interested in at the moment and ignore the rest. Come back to them as they arise during the natural flow of your course.
Cracking the code
A modern organic name is simply a code. Each part of the name gives you some useful information about the compound. For example, to understand the name 2-methylpropan-1-ol you need to take the name to pieces.
The prop in the middle tells you how many carbon atoms there are in the longest chain (in this case, 3). The an which follows the "prop" tells you that there aren't any carbon-carbon double bonds.
The other two parts of the name tell you about interesting things which are happening on the first and second carbon atom in the chain. Any name you are likely to come across can be broken up in this same way.
Counting the carbon atoms
You will need to remember the codes for the number of carbon atoms in a chain up to 6 carbons. There is no easy way around this - you have got to learn them. If you don't do this properly, you won't be able to name anything!
code no of carbons
meth 1
eth 2
prop 3
but 4
pent 5
hex 6
Types of carbon-carbon bonds
Whether or not the compound contains a carbon-carbon double bond is shown by the two letters immediately after the code for the chain length.
code means
an only carbon-carbon single bonds
en contains a carbon-carbon double bond
For example, butane means four carbons in a chain with no double bond.
Propene means three carbons in a chain with a double bond between two of the carbons.
Alkyl groups
Compounds like methane, CH4, and ethane, CH3CH3, are members of a family of compounds called alkanes. If you remove a hydrogen atom from one of these you get an alkyl group. For example:
• A methyl group is CH3.
• An ethyl group is CH3CH2.
These groups must, of course, always be attached to something else.
The alkanes
Example \(1\): 2-methylpentane
Write the structural formula for 2-methylpentane.
Solution
Start decoding the name from the bit that counts the number of carbon atoms in the longest chain - pent counts 5 carbons.
Are there any carbon-carbon double bonds? No - an tells you there aren't any.
Now draw this carbon skeleton:
Put a methyl group on the number 2 carbon atom:
Does it matter which end you start counting from? No - if you counted from the other end, you would draw the next structure. That's exactly the same as the first one, except that it has been flipped over.
Finally, all you have to do is to put in the correct number of hydrogen atoms on each carbon so that each carbon is forming four bonds.
If you had to name this yourself:
• Count the longest chain of carbons that you can find. Don't assume that you have necessarily drawn that chain horizontally. 5 carbons means pent.
• Are there any carbon-carbon double bonds? No - therefore pentane.
• There's a methyl group on the number 2 carbon - therefore 2-methylpentane. Why the number 2 as opposed to the number 4 carbon? In other words, why do we choose to number from this particular end? The convention is that you number from the end which produces the lowest numbers in the name - hence 2- rather than 4-.
Example \(2\): 2,3-dimethylbutane
Write the structural formula for 2,3-dimethylbutane.
Solution
Start with the carbon backbone. There are 4 carbons in the longest chain (but) with no carbon-carbon double bonds (an).
This time there are two methyl groups (di) on the number 2 and number 3 carbon atoms.
Completing the formula by filling in the hydrogen atoms gives:
Example \(3\): 2,2-dimethylbutane
Write the structural formula for 2,2-dimethylbutane.
Solution
This is exactly like the last example, except that both methyl groups are on the same carbon atom. Notice that the name shows this by using 2,2- as well as di. The structure is worked out as before:
Example \(4\): 3-ethyl-2-methylhexane
Write the structural formula for 3-ethyl-2-methylhexane.
Solution
hexan shows a 6 carbon chain with no carbon-carbon double bonds.
This time there are two different alkyl groups attached - an ethyl group on the number 3 carbon atom and a methyl group on number 2.
Filling in the hydrogen atoms gives:
If you had to name this yourself:
How do you know what order to write the different alkyl groups at the beginning of the name? The convention is that you write them in alphabetical order - hence ethyl comes before methyl which in turn comes before propyl.
The cycloalkanes
In a cycloalkane the carbon atoms are joined up in a ring - hence cyclo.
Example \(5\)
Write the structural formula for cyclohexane.
Solution
hexan shows 6 carbons with no carbon-carbon double bonds. cyclo shows that they are in a ring. Drawing the ring and putting in the correct number of hydrogens to satisfy the bonding requirements of the carbons gives:
The alkenes
Example \(6\): propene
Write the structural formula for propene.
Solution
prop counts 3 carbon atoms in the longest chain. en tells you that there is a carbon-carbon double bond. That means that the carbon skeleton looks like this:
Putting in the hydrogens gives you:
Example \(7\): but-1-ene
Write the structural formula for but-1-ene.
Solution
"but" counts 4 carbon atoms in the longest chain and en tells you that there is a carbon-carbon double bond. The number in the name tells you where the double bond starts.
No number was necessary in the propene example above because the double bond has to start on one of the end carbon atoms. In the case of butene, though, the double bond could either be at the end of the chain or in the middle - and so the name has to code for the its position.
The carbon skeleton is:
And the full structure is:
Incidentally, you might equally well have decided that the right-hand carbon was the number 1 carbon, and drawn the structure as:
Example \(8\)
Write the structural formula for 3-methylhex-2-ene.
Solution
The longest chain has got 6 carbon atoms (hex) with a double bond starting on the second one (-2-en).
But this time there is a methyl group attached to the chain on the number 3 carbon atom, giving you the underlying structure:
Adding the hydrogens gives the final structure:
Be very careful to count the bonds around each carbon atom when you put the hydrogens in. It would be very easy this time to make the mistake of writing an H after the third carbon - but that would give that carbon a total of 5 bonds.
Compounds containing halogens
Example \(1\): 1,1,1-trichloroethane
Write the structural formula for 1,1,1-trichloroethane.
Solution
This is a two carbon chain (eth) with no double bonds (an). There are three chlorine atoms all on the first carbon atom.
Example \(1\): 2-bromo-2-methylpropane
Write the structural formula for 2-bromo-2-methylpropane.
Solution
First sort out the carbon skeleton. It's a three carbon chain with no double bonds and a methyl group on the second carbon atom.
Draw the bromine atom which is also on the second carbon.
And finally put the hydrogen atoms in.
If you had to name this yourself:
Notice that the whole of the hydrocarbon part of the name is written together - as methylpropane - before you start adding anything else on to the name.
Example \(1\): 1-iodo-3-methylpent-2-ene
Write the structural formula for 1-iodo-3-methylpent-2-ene.
Solution
This time the longest chain has 5 carbons (pent), but has a double bond starting on the number 2 carbon. There is also a methyl group on the number 3 carbon.
Now draw the iodine on the number 1 carbon.
Giving a final structure:
Alcohols
All alcohols contain an -OH group. This is shown in a name by the ending ol.
Example \(1\): methanol
Write the structural formula for methanol.
Solution
This is a one carbon chain with no carbon-carbon double bond (obviously!). The ol ending shows it's an alcohol and so contains an -OH group.
Example \(1\): 2-methylpropan-1-ol
Add example text here.
Solution
Write the structural formula for 2-methylpropan-1-ol.
The carbon skeleton is a 3 carbon chain with no carbon-carbon double bonds, but a methyl group on the number 2 carbon.
The -OH group is attached to the number 1 carbon.
The structure is therefore:
Example \(1\): ethane-1,2-diol
Write the structural formula for ethane-1,2-diol.
Solution
This is a two carbon chain with no double bond. The diol shows 2 -OH groups, one on each carbon atom.
Aldehydes
All aldehydes contain the group:
If you are going to write this in a condensed form, you write it as -CHO - never as -COH, because that looks like an alcohol. The names of aldehydes end in al.
Example: propanal
Write the structural formula for propanal.
This is a 3 carbon chain with no carbon-carbon double bonds. The al ending shows the presence of the -CHO group. The carbon in that group counts as one of the chain.
Example: 2-methylpentanal
Write the structural formula for 2-methylpentanal.
This time there are 5 carbons in the longest chain, including the one in the -CHO group. There aren't any carbon-carbon double bonds. A methyl group is attached to the number 2 carbon. Notice that in aldehydes, the carbon in the -CHO group is always counted as the number 1 carbon.
Ketones
Ketones contain a carbon-oxygen double bond just like aldehydes, but this time it's in the middle of a carbon chain. There isn't a hydrogen atom attached to the group as there is in aldehydes. Ketones are shown by the ending one.
Example: propanone
Write the structural formula for propanone.
This is a 3 carbon chain with no carbon-carbon double bond. The carbon-oxygen double bond has to be in the middle of the chain and so must be on the number 2 carbon.
Ketones are often written in this way to emphasize the carbon-oxygen double bond.
Example: pentan-3-one
Write the structural formula for pentan-3-one.
This time the position of the carbon-oxygen double bond has to be stated because there is more than one possibility. It's on the third carbon of a 5 carbon chain with no carbon-carbon double bonds. If it was on the second carbon, it would be pentan-2-one.
This could equally well be written: | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Fundamentals/Nomenclature/An_Overview_of_Naming_Organic_Molecules.txt |
This page continues looking at the names of organic compounds containing chains of carbon atoms.
Carboxylic acids
Carboxylic acids contain the -COOH group, which is better written out in full as:
Carboxylic acids are shown by the ending oic acid. When you count the carbon chain, you have to remember to include the carbon in the -COOH group. That carbon is always thought of as number 1 in the chain.
Example: 3-methylbutanoic acid
Write the structural formula for 3-methylbutanoic acid.
This is a four carbon acid with no carbon-carbon double bonds. There is a methyl group on the third carbon (counting the -COOH carbon as number 1).
Example: 2-hydroxypropanoic acid
Write the structural formula for 2-hydroxypropanoic acid.
The hydroxy part of the name shows the presence of an -OH group. Normally, you would show that by the ending ol, but this time you can't because you've already got another ending. You are forced into this alternative way of describing it.
The old name for 2-hydroxypropanoic acid is lactic acid. That name sounds more friendly, but is utterly useless when it comes to writing a formula for it. In the old days, you would have had to learn the formula rather than just working it out should you need it.
Example: 2-chlorobut-3-enoic acid
Write the structural formula for 2-chlorobut-3-enoic acid.
This time, not only is there a chlorine attached to the chain, but the chain also contains a carbon-carbon double bond (en) starting on the number 3 carbon (counting the -COOH carbon as number 1).
Salts of carboxylic acids
Example: sodium propanoate
Write the structural formula for sodium propanoate.
This is the sodium salt of propanoic acid - so start from that. Propanoic acid is a three carbon acid with no carbon-carbon double bonds.
When the carboxylic acids form salts, the hydrogen in the -COOH group is replaced by a metal. Sodium propanoate is therefore:
Notice that there is an ionic bond between the sodium and the propanoate group. Whatever you do, do not draw a line between the sodium and the oxygen. That would represent a covalent bond, but it is wrong!
In a shortened version, sodium propanoate would be written CH3CH2COONa or, if you wanted to emphasize the ionic nature, as CH3CH2COO- Na+.
Esters
Esters are one of a number of compounds known collectively as acid derivatives. In these the acid group is modified in some way. In an ester, the hydrogen in the -COOH group is replaced by an alkyl group (or possibly some more complex hydrocarbon group).
Example:
Example 1: Write the structural formula for methyl propanoate.
An ester name has two parts - the part that comes from the acid (propanoate) and the part that shows the alkyl group (methyl).
Start by thinking about propanoic acid - a 3 carbon acid with no carbon-carbon double bonds.
The hydrogen in the -COOH group is replaced by an alkyl group - in this case, a methyl group.
Ester names are confusing because the name is written backwards from the way the structure is drawn. There's no way round this - you just have to get used to it! In the shortened version, this formula would be written CH3CH2COOCH3.
Example: ethyl ethanoate
Write the structural formula for ethyl ethanoate.
This is probably the most commonly used example of an ester. It is based on ethanoic acid ( hence, ethanoate) - a 2 carbon acid. The hydrogen in the -COOH group is replaced by an ethyl group.
Make sure that you draw the ethyl group the right way round. A fairly common mistake is to try to join the CH3 group to the oxygen. If you count the bonds if you do that, you will find that both the CH3 carbon and the CH2 carbon have the wrong number of bonds.
Acyl chlorides (acid chlorides)
An acyl chloride is another acid derivative. In this case, the -OH group of the acid is replaced by -Cl. All acyl chlorides contain the -COCl group:
Example: ethanoyl chloride
Write the structural formula for ethanoyl chloride.
Acyl chlorides are shown by the ending oyl chloride. So ethanoyl chloride is based on a 2 carbon chain with no carbon-carbon double bonds and a -COCl group. The carbon in that group counts as part of the chain. In a longer chain, with side groups attached, the -COCl carbon is given the number 1 position.
Acid anhydrides
Another acid derivative! An acid anhydride is what you get if you dehydrate an acid - that is, remove water from it.
Example: propanoic anhydride
Write the structural formula for propanoic anhydride.
These are most easily worked out by writing it down on a scrap of paper in the following way:
Draw two molecules of acid arranged so that the -OH groups are next to each other. Tweak out a molecule of water - and then join up what's left. In this case, because you want propanoic anhydride, you draw two molecules of propanoic acid.
Amides
Yet another acid derivative! Amides contain the group -CONH2 where the -OH of an acid is replaced by -NH2.
Example: propanamide
Write the structural formula for propanamide.
This is based on a 3 carbon chain with no carbon-carbon double bonds. At the end of the chain is a -CONH2 group. The carbon in that group counts as part of the chain.
Nitriles
Nitriles contain a -CN group, and used to be called cyanides.
Example: ethanenitrile
Example 1: Write the structural formula for ethanenitrile.
The name shows a 2 carbon chain with no carbon-carbon double bond. nitrile shows a -CN group at the end of the chain. As with the previous examples involving acids and acid derivatives, don't forget that the carbon in the -CN group counts as part of the chain.
The old name for this would have been methyl cyanide. You might think that that's easier, but as soon as the chain gets more complicated, it doesn't work - as the next example shows.
Example: 2-hydroxypropanenitrile
Write the structural formula for 2-hydroxypropanenitrile.
Here we've got a 3 carbon chain, no carbon-carbon double bonds, and a -CN group on the end of the chain. The carbon in the -CN group counts as the number 1 carbon. On the number 2 carbon there is an -OH group (hydroxy). Notice that you can't use the ol ending because you've already got a nitrile ending.
Primary amines
A primary amine contains the group -NH2 attached to a hydrocarbon chain or ring. You can think of amines in general as being derived from ammonia, NH3. In a primary amine, one of the hydrogens has been replaced by a hydrocarbon group.
Example: ethylamine
Write the structural formula for ethylamine.
In this case, an ethyl group is attached to the -NH2 group.
This name (ethylamine) is fine as long as you've only got a short chain where there isn't any ambiguity about where the -NH2 group is found. But suppose you had a 3 carbon chain - in this case, the -NH2 group could be on an end carbon or on the middle carbon. How you get around that problem is illustrated in the next example.
Example: 2-aminopropane
Write the structural formula for 2-aminopropane.
The name shows a 3 carbon chain with an amino group attached to the second carbon. amino shows the -NH2 group.
Ethylamine (example 1 above) could equally well have been called aminoethane.
Secondary and tertiary amines
You are only likely to come across simple examples of these. In a secondary amine, two of the hydrogen atoms in an ammonia molecule have been replaced by hydrocarbon groups. In a tertiary amine, all three hydrogens have been replaced.
Example: dimethylamine
Write the structural formula for dimethylamine.
In this case, two of the hydrogens in ammonia have been replaced by methyl groups.
Example: trimethylamine
Write the structural formula for trimethylamine.
Here, all three hydrogens in ammonia have been replaced by methyl groups.
Amino acids
An amino acid contains both an amino group, -NH2, and a carboxylic acid group, -COOH, in the same molecule. As with all acids the carbon chain is numbered so that the carbon in the -COOH group is counted as number 1.
Example: 2-aminopropanoic acid
Write the structural formula for 2-aminopropanoic acid.
This has a 3 carbon chain with no carbon-carbon double bonds. On the second carbon (counting the -COOH carbon as number 1) there is an amino group, -NH2. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Fundamentals/Nomenclature/More_Organic_Names.txt |
This page looks at the names of some simple aromatic compounds. An aromatic compound is one which contains a benzene ring. Naming aromatic compounds isn't quite so straightforward as naming chain compounds. Often, more than one name is acceptable and it is not uncommon to find the old names still in use as well.
The Benzene Ring
Most, but not all, aromatic compounds are based on benzene, C6H6, which has a ring of six carbon atoms and has the symbol:
Each corner of the hexagon has a carbon atom with a hydrogen attached.
The Phenyl Group
Remember that you get a methyl group, CH3, by removing a hydrogen from methane, CH4. You get a phenyl group, C6H5, by removing a hydrogen from a benzene ring, C6H6. Like a methyl or an ethyl group, a phenyl group is always attached to something else.
Aromatic compounds with only one group attached to the benzene ring
Cases where the name is based on benzene
chlorobenzene
This is a simple example of a halogen attached to the benzene ring. The name is self-obvious.
The simplified formula for this is C6H5Cl. You could therefore (although you never do!) call it phenyl chloride. Whenever you draw a benzene ring with one other thing attached to it, you are in fact drawing a phenyl group. In order to attach something else, you have to remove one of the existing hydrogen atoms, and so automatically make a phenyl group.
nitrobenzene
The nitro group, NO2, is attached to a benzene ring.
The simplified formula for this is C6H5NO2.
methylbenzene
Another obvious name - the benzene ring has a methyl group attached. Other alkyl side-chains would be named similarly - for example, ethylbenzene. The old name for methylbenzene is toluene, and you may still meet that.
The simplified formula for this is C6H5CH3.
(chloromethyl)benzene
A variant on this which you may need to know about is where one of the hydrogens on the CH3 group is replaced by a chlorine atom. Notice the brackets around the (chloromethyl) in the name. This is so that you are sure that the chlorine is part of the methyl group and not somewhere else on the ring.
If more than one of the hydrogens had been replaced by chlorine, the names would be (dichloromethyl)benzene or (trichloromethyl)benzene. Again, notice the importance of the brackets in showing that the chlorines are part of the side group and not directly attached to the ring.
benzoic acid (benzenecarboxylic acid)
Benzoic acid is the older name, but is still in common use - it's a lot easier to say and write than the modern alternative! Whatever you call it, it has a carboxylic acid group, -COOH, attached to the benzene ring.
Cases where the name is based on phenyl
Remember that the phenyl group is a benzene ring minus a hydrogen atom - C6H5. If you draw a benzene ring with one group attached, you have drawn a phenyl group.
phenylamine
Phenylamine is a primary amine and contains the -NH2 group attached to a benzene ring.
The old name for phenylamine is aniline, and you could also reasonably call it aminobenzene.
phenylethene
This is an ethene molecule with a phenyl group attached. Ethene is a two carbon chain with a carbon-carbon double bond. Phenylethene is therefore:
The old name for phenylethene is styrene - the monomer from which polystyrene is made.
phenylethanone
This is a slightly awkward name - take it to pieces. It consists of a two carbon chain with no carbon-carbon double bond. The one ending shows that it is a ketone, and so has a C=O group somewhere in the middle. Attached to the carbon chain is a phenyl group. Putting that together gives:
phenyl ethanoate
This is an ester based on ethanoic acid. The hydrogen atom in the -COOH group has been replaced by a phenyl group.
phenol
Phenol has an -OH group attached to a benzene ring and so has a formula C6H5OH.
Aromatic compounds with more than one group attached to the benzene ring
Any group already attached to the ring is given the number 1 position. Where you draw it on the ring (at the top or in any other position) doesn't matter - that's just a question of rotating the molecule a bit. It's much easier, though, to get in the habit of drawing your main group at the top.
The other ring positions are then numbered from 2 to 6. You can number them either clockwise or anti-clockwise. As with chain compounds, you number the ring so that the name you end up with has the smallest possible numbers in it. Examples will make this clear.
Example 1: Chlorine atoms on the Ring
Look at these compounds:
All of these are based on methylbenzene and so the methyl group is given the number 1 position on the ring. Why is it 2-chloromethylbenzene rather than 6-chloromethylbenzene? The ring is numbered clockwise in this case because that produces a 2- in the name rather than a 6-. 2 is smaller than 6.
2-hydroxybenzoic acid
This might also be called 2-hydroxybenzenecarboxylic acid. There is a -COOH group attached to the ring and, because the name is based on benzoic acid, that group is assigned the number 1 position. Next door to it in the 2 position is a hydroxy group, -OH.
benzene-1,4-dicarboxylic acid
The di shows that there are two carboxylic acid groups, -COOH, one of them in the 1 position and the other opposite it in the 4 position.
2,4,6-trichlorophenol
This is based on phenol - with an -OH group attached in the number 1 position on the ring. There are 3 chlorine atoms substituted onto the ring in the 2, 4 and 6 positions.
methyl 3-nitrobenzoate
This is a name you might come across as a part of a practical exercise in nitrating benzene rings. It's included partly for that reason, and partly because it is a relatively complicated name to finish with! The structure of the name shows that it is an ester. You can tell that from the oate ending, and the methyl group floating separately from the rest of the name at the beginning. The ester is based on the acid, 3-nitrobenzoic acid - so start with that.
There will be a benzene ring with a -COOH group in the number 1 position and a nitro group, NO2, in the 3 position. The -COOH group is modified to make an ester by replacing the hydrogen of the -COOH group by a methyl group. Methyl 3-nitrobenzoate is therefore: | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Fundamentals/Nomenclature/Naming_Aromatic_Compounds.txt |
In naming organic chemicals, the nomenclature must specify the location of all special features. This is commonly done using "locants" such as numbers, as in 2-bromobutane. There are some cases where the number is omitted. However, students are sometimes confused about when they can omit a number. This page summarizes the cases where the number, most often a 1, can be omitted, and discusses a common misconception about when it can be omitted.
Introduction
The general rule is that the locant (number) can be omitted when there is no ambiguity; that is, when only one location is possible. It cannot be omitted by making any assumption that a certain location is more likely.
The examples below are presented without showing the structures. All are simple compounds, and you should be able to draw them. In fact, this page will probably work best if you draw the structures for yourself as you go, so you can think about whether or not a number is required.
The cases where the number can be omitted can be considered in three classes:
• Certain functional groups can only occur at a particular position, so simply stating the group defines the position unambiguously. The most common examples are the aldehyde and carboxylic acid groups, which can only be at the terminal C.
• With (mono)-cyclic compounds carrying a single substituent, all positions are equivalent. The position of the substituent is defined as the 1-position, and is omitted from the name.
• Some special cases with small molecules. These are actually special examples of one of the underlying ideas in the previous two cases, but are not as obvious.
Functional groups that define the position
The aldehyde and carboxylic acid functional groups can occur only at the end of a chain. When one of these is the main functional group, its position is defined as "1". Thus butanal (aldehyde) and butanoic acid (carboxylic acid) have the indicated functional group at the 1-position.
Similarly, butanedial and butanedioic acid have one functional group at each end; there is no other possibility, and the numbers are omitted.
Cyclic compounds with one substituent
All positions on a ring are equivalent. If there is one substituent, that substituent defines the "1" position. Thus bromocyclobutane and bromobenzene do not need numbers. There is only one possible compound of each name, and the bromo position is defined as "1".
This applies to single ring compounds, where all positions on the ring are equivalent.
Special cases
There are several chemicals, generally small ones, where the number for a substituent is commonly omitted, even though they do not quite fit one of the specific exceptions discussed above. However, they do fit the general criterion that the number is not needed, because there is no other choice. A list of some of those special cases follows. I encourage you to draw the structures, and be sure you agree that only one structure is possible in each case:
• Methanol. Alcohols usually need a number, but there is only one C here, so no number is needed.
• Ethanol. There are 2 C, but they are equivalent. There is only one alcohol possible based on ethane.
• Ethene and propene. Double bonds usually need a number. But these cases are unambiguous simply because the molecule is small, and there is only one possible structure.
• Propanone and butanone. Normally one would need to give a number for a ketone group. However, these two ketones are so small that they have only one possible ketone; the number is usually omitted. Note that with butanone, there are 4 C. However, two are terminal, so cannot be ketone C, and the other two are equivalent.
• Methylpropane. Draw propane and then put on a methyl group. The only possible position is at 2, so the number in 2-methylpropane is often omitted. (If you added the methyl group at a terminal C, the compound would be butane, not methylpropane.)
• Octachloropropane. There are only eight positions possible on propane, and "octachloro" means that all of them are Cl. One might write 1,1,1,2,2,3,3,3-octachloropropane, but the numbers are quite unnecessary, and usually omitted.
Misconception
Sometimes people think that the 1 is a "default" number. That is, they think that the number 1 is implied if no number is given. Using this logic, they might give octanol or octene as (incorrect) names for 1-octanol or 1-octene. As you read the cases above where the number is omitted, it is important to realize that it is omitted because there is no other possibility, not because there is a default number. That is, one never "assumes" that a group is at "1".
Perspective
A chemical name conveys information about the structure of the chemical; a name can be thought of as an "instruction manual" for how to draw the structure. When we first started discussing the nomenclature of organic compounds, we made the point that there are two types of wrong names. One type of incorrect name fails to provide the correct information. Another type of incorrect name may provide the information, but not follow the official (IUPAC) "rules" for doing so.
As an example, consider the simple chemical 1-chloropropane. If someone names this chloropropane, they have failed to communicate what compound they mean (since the chloro could be at either the 1-position or the 2-position). If someone names it 3-chloropropane, it is quite clear what they mean, even though they have violated the naming rules (give the substituent the lowest possible number).
The first type of error is much more important, as it means the primary goal of naming -- communication of the structure -- has failed.
How is this relevant to the discussion of omitting numbers? Well, you can do more harm by omitting a number that is needed than by giving one that is not needed. If it doubt, include the number. Omit the number only when it is perfectly clear what the meaning is. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Fundamentals/Nomenclature/Omitting_numbers_in_Nomenclature.txt |
Includes the acid strengths of carboxylic acids, phenols and alcohols, and the base strengths of primary amines.
Organic Acids and Bases
This page explains the acidity of simple organic acids and looks at the factors which affect their relative strengths.
Organic acids as weak acids
For the purposes of this topic, we are going to take the definition of an acid as "a substance which donates hydrogen ions (protons) to other things". We are going to get a measure of this by looking at how easily the acids release hydrogen ions to water molecules when they are in solution in water.
An acid in solution sets up this equilibrium:
A hydroxonium ion is formed together with the anion (negative ion) from the acid.
This equilibrium is sometimes simplified by leaving out the water to emphasise the ionisation of the acid.
If you write it like this, you must include the state symbols - "(aq)". Writing H+(aq) implies that the hydrogen ion is attached to a water molecule as H3O+. Hydrogen ions are always attached to something during chemical reactions.
The organic acids are weak in the sense that this ionisation is very incomplete. At any one time, most of the acid will be present in the solution as un-ionised molecules. For example, in the case of dilute ethanoic acid, the solution contains about 99% of ethanoic acid molecules - at any instant, only about 1% have actually ionised. The position of equilibrium therefore lies well to the left.
Comparing the strengths of weak acids
The strengths of weak acids are measured on the pKa scale. The smaller the number on this scale, the stronger the acid is.
Three of the compounds we shall be looking at, together with their pKa values are:
Remember - the smaller the number the stronger the acid. Comparing the other two to ethanoic acid, you will see that phenol is very much weaker with a pKa of 10.00, and ethanol is so weak with a pKa of about 16 that it hardly counts as acidic at all!
Why are these acids acidic?
In each case, the same bond gets broken - the bond between the hydrogen and oxygen in an -OH group. Writing the rest of the molecule as "X":
So . . . if the same bond is being broken in each case, why do these three compounds have such widely different acid strengths?
Differences in acid strengths between carboxylic acids, phenols and alcohols
Two of the factors which influence the ionization of an acid are:
• the strength of the bond being broken,
• the stability of the ions being formed.
In these cases, you seem to be breaking the same oxygen-hydrogen bond each time, and so you might expect the strengths to be similar. The most important factor in determining the relative acid strengths of these molecules is the nature of the ions formed. You always get a hydroxonium ion - so that's constant - but the nature of the anion (the negative ion) varies markedly from case to case.
Example \(1\): Ethanoic Acid
Ethanoic acid has the structure:
The acidic hydrogen is the one attached to the oxygen. When ethanoic acid ionises it forms the ethanoate ion, CH3COO-.
You might reasonably suppose that the structure of the ethanoate ion was as below, but measurements of bond lengths show that the two carbon-oxygen bonds are identical and somewhere in length between a single and a double bond.
To understand why this is, you have to look in some detail at the bonding in the ethanoate ion. Like any other double bond, a carbon-oxygen double bond is made up of two different parts. One electron pair is found on the line between the two nuclei - this is known as a sigma bond. The other electron pair is found above and below the plane of the molecule in a pi bond. Pi bonds are made by sideways overlap between p orbitals on the carbon and the oxygen.
In an ethanoate ion, one of the lone pairs on the negative oxygen ends up almost parallel to these p orbitals, and overlaps with them
This leads to a delocalised pi system over the whole of the -COO- group, rather like that in benzene.
All the oxygen lone pairs have been left out of this diagram to avoid confusion. Because the oxygens are more electronegative than the carbon, the delocalised system is heavily distorted so that the electrons spend much more time in the region of the oxygen atoms.
So where is the negative charge in all this? It has been spread around over the whole of the -COO- group, but with the greatest chance of finding it in the region of the two oxygen atoms. Ethanoate ions can be drawn simply as:
The dotted line represents the delocalisation. The negative charge is written centrally on that end of the molecule to show that it isn't localised on one of the oxygen atoms. The more you can spread charge around, the more stable an ion becomes. In this case, if you delocalise the negative charge over several atoms, it is going to be much less attractive to hydrogen ions - and so you are less likely to re-form the ethanoic acid.
Example \(2\): Phenol
Phenols have an -OH group attached directly to a benzene ring. Phenol itself is the simplest of these with nothing else attached to the ring apart from the -OH group.
When the hydrogen-oxygen bond in phenol breaks, you get a phenoxide ion, C6H5O-.
Delocalisation also occurs in this ion. This time, one of the lone pairs on the oxygen atom overlaps with the delocalised electrons on the benzene ring.
This overlap leads to a delocalisation which extends from the ring out over the oxygen atom. As a result, the negative charge is no longer entirely localised on the oxygen, but is spread out around the whole ion.
Why then is phenol a much weaker acid than ethanoic acid?
Think about the ethanoate ion again. If there wasn't any delocalisation, the charge would all be on one of the oxygen atoms, like this:
But the delocalisation spreads this charge over the whole of the COO group. Because oxygen is more electronegative than carbon, you can think of most of the charge being shared between the two oxygens (shown by the heavy red shading in this diagram).
If there wasn't any delocalisation, one of the oxygens would have a full charge which would be very attractive towards hydrogen ions. With delocalisation, that charge is spread over two oxygen atoms, and neither will be as attractive to a hydrogen ion as if one of the oxygens carried the whole charge.
That means that the ethanoate ion won't take up a hydrogen ion as easily as it would if there wasn't any delocalisation. Because some of it stays ionised, the formation of the hydrogen ions means that it is acidic.
In the phenoxide ion, the single oxygen atom is still the most electronegative thing present, and the delocalised system will be heavily distorted towards it. That still leaves the oxygen atom with most of its negative charge.
What delocalisation there is makes the phenoxide ion more stable than it would otherwise be, and so phenol is acidic to an extent.
However, the delocalisation hasn't shared the charge around very effectively. There is still lots of negative charge around the oxygen to which hydrogen ions will be attracted - and so the phenol will readily re-form. Phenol is therefore only very weakly acidic.
Example \(3\): Ethanol
Ethanol, CH3CH2OH, is so weakly acidic that you would hardly count it as acidic at all. If the hydrogen-oxygen bond breaks to release a hydrogen ion, an ethoxide ion is formed:
This has nothing at all going for it. There is no way of delocalising the negative charge, which remains firmly on the oxygen atom. That intense negative charge will be highly attractive towards hydrogen ions, and so the ethanol will instantly re-form.
Since ethanol is very poor at losing hydrogen ions, it is hardly acidic at all.
Variations in acid strengths between different carboxylic acids
You might think that all carboxylic acids would have the same strength because each depends on the delocalization of the negative charge around the -COO- group to make the anion more stable, and so more reluctant to re-combine with a hydrogen ion.
In fact, the carboxylic acids have widely different acidities. One obvious difference is between methanoic acid, HCOOH, and the other simple carboxylic acids:
pKa
HCOOH 3.75
CH3COOH 4.76
CH3CH2COOH 4.87
CH3CH2CH2COOH 4.82
Remember that the higher the value for pKa, the weaker the acid is.
Why is ethanoic acid weaker than methanoic acid? It again depends on the stability of the anions formed - on how much it is possible to delocalise the negative charge. The less the charge is delocalised, the less stable the ion, and the weaker the acid.
The methanoate ion (from methanoic acid) is:
The only difference between this and the ethanoate ion is the presence of the CH3 group in the ethanoate.
But that's important! Alkyl groups have a tendency to "push" electrons away from themselves. That means that there will be a small amount of extra negative charge built up on the -COO- group. Any build-up of charge will make the ion less stable, and more attractive to hydrogen ions.
Ethanoic acid is therefore weaker than methanoic acid, because it will re-form more easily from its ions.
The other alkyl groups have "electron-pushing" effects very similar to the methyl group, and so the strengths of propanoic acid and butanoic acid are very similar to ethanoic acid. The acids can be strengthened by pulling charge away from the -COO- end. You can do this by attaching electronegative atoms like chlorine to the chain.
As the next table shows, the more chlorines you can attach the better:
pKa
CH3COOH 4.76
CH2ClCOOH 2.86
CHCl2COOH 1.29
CCl3COOH 0.65
Trichloroethanoic acid is quite a strong acid.
Attaching different halogens also makes a difference. Fluorine is the most electronegative and so you would expect it to be most successful at pulling charge away from the -COO- end and so strengthening the acid.
pKa
CH2FCOOH 2.66
CH2ClCOOH 2.86
CH2BrCOOH 2.90
CH2ICOOH 3.17
The effect is there, but isn't as great as you might expect.
Finally, notice that the effect falls off quite quickly as the attached halogen gets further away from the -COO- end. Here is what happens if you move a chlorine atom along the chain in butanoic acid.
pKa
CH3CH2CH2COOH 4.82
CH3CH2CHClCOOH 2.84
CH3CHClCH2COOH 4.06
CH2ClCH2CH2COOH 4.52
The chlorine is effective at withdrawing charge when it is next-door to the -COO- group, and much less so as it gets even one carbon further away. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Fundamentals/Organic_Acids_and_Bases/Organic_Acids.txt |
This page explains why simple organic bases are basic and looks at the factors which affect their relative strengths. For A'level purposes, all the bases we are concerned with are primary amines - compounds in which one of the hydrogens in an ammonia molecule, NH3, is replaced either by an alkyl group or a benzene ring.
Ammonia as a weak base
All of the compounds we are concerned with are derived from ammonia and so we'll start by looking at the reason for its basic properties. For the purposes of this topic, we are going to take the definition of a base as "a substance which combines with hydrogen ions (protons)". We are going to get a measure of this by looking at how easily the bases take hydrogen ions from water molecules when they are in solution in water.
Ammonia in solution sets up this equilibrium:
An ammonium ion is formed together with hydroxide ions. Because the ammonia is only a weak base, it doesn't hang on to the extra hydrogen ion very effectively and so the reaction is reversible. At any one time, about 99% of the ammonia is present as unreacted molecules. The position of equilibrium lies well to the left.
The ammonia reacts as a base because of the active lone pair on the nitrogen. Nitrogen is more electronegative than hydrogen and so attracts the bonding electrons in the ammonia molecule towards itself. That means that in addition to the lone pair, there is a build-up of negative charge around the nitrogen atom. That combination of extra negativity and active lone pair attracts the new hydrogen from the water.
Comparing the strengths of weak bases
The strengths of weak bases are measured on the pKb scale. The smaller the number on this scale, the stronger the base is. Three of the compounds we shall be looking at, together with their pKb values are:
Remember - the smaller the number the stronger the base. Comparing the other two to ammonia, you will see that methylamine is a stronger base, whereas phenylamine is very much weaker.
• Methylamine is typical of aliphatic primary amines - where the -NH2 group is attached to a carbon chain. All aliphatic primary amines are stronger bases than ammonia.
• Phenylamine is typical of aromatic primary amines - where the -NH2 group is attached directly to a benzene ring. These are very much weaker bases than ammonia.
Explaining the differences in base strengths
Two of the factors which influence the strength of a base are:
• the ease with which the lone pair picks up a hydrogen ion,
• the stability of the ions being formed.
Why are aliphatic primary amines stronger bases than ammonia?
Methylamine
Methylamine has the structure:
The only difference between this and ammonia is the presence of the CH3 group in the methylamine. But that's important! Alkyl groups have a tendency to "push" electrons away from themselves. That means that there will be a small amount of extra negative charge built up on the nitrogen atom. That extra negativity around the nitrogen makes the lone pair even more attractive towards hydrogen ions.
Making the nitrogen more negative helps the lone pair to pick up a hydrogen ion. What about the effect on the positive methylammonium ion formed? Is this more stable than a simple ammonium ion? Compare the methylammonium ion with an ammonium ion:
In the methylammonium ion, the positive charge is spread around the ion by the "electron-pushing" effect of the methyl group. The more you can spread charge around, the more stable an ion becomes. In the ammonium ion there isn't any way of spreading the charge.
To summarize:
• The nitrogen is more negative in methylamine than in ammonia, and so it picks up a hydrogen ion more readily.
• The ion formed from methylamine is more stable than the one formed from ammonia, and so is less likely to shed the hydrogen ion again.
Taken together, these mean that methylamine is a stronger base than ammonia.
The other aliphatic primary amines
The other alkyl groups have "electron-pushing" effects very similar to the methyl group, and so the strengths of the other aliphatic primary amines are very similar to methylamine. For example:
pKb
CH3NH2 3.36
CH3CH2NH2 3.27
CH3CH2CH2NH2 3.16
CH3CH2CH2CH2NH2 3.39
Why are aromatic primary amines much weaker bases than ammonia?
An aromatic primary amine is one in which the -NH2 group is attached directly to a benzene ring. The only one you are likely to come across is phenylamine. Phenylamine has the structure:
The lone pair on the nitrogen touches the delocalized ring electrons . . .
. . . and becomes delocalized with them:
That means that the lone pair is no longer fully available to combine with hydrogen ions. The nitrogen is still the most electronegative atom in the molecule, and so the delocalized electrons will be attracted towards it, but the intensity of charge around the nitrogen is nothing like what it is in, say, an ammonia molecule.
The other problem is that if the lone pair is used to join to a hydrogen ion, it is no longer available to contribute to the delocalisation. That means that the delocalization would have to be disrupted if the phenylamine acts as a base. Delocalization makes molecules more stable, and so disrupting the delocalization costs energy and will not happen easily.
Taken together - the lack of intense charge around the nitrogen, and the need to break some delocalization - this means that phenylamine is a very weak base indeed.
Oxidation States of Organic Molecules
By the end of gen chemistry, calculating oxidation states of different metals should be pretty familiar. Here’s what you do. Take a typical compound – \(FeCl_3\), for instance. Treat every bond between the metal and a different atom as if it were an ionic bond. That means the more electronegative elements (like chlorine, say, or oxygen) bear negative charges, and the less electronegative element (such as the metal) bears the positive charge.
If the compound is neutral, the sum of the oxidation states also has to be neutral. (If the compound has a charge, you adjust the oxidation states accordingly so that their sum equals the charge).
Now here’s a fun exercise. Try applying the same rules to carbon. It’s going to feel a little bit weird. Why? Because there are two key differences:
• First, carbon is often more electronegative (2.5) than some of the atoms it’s bound to (such as H, 2.2). So what do you do in this case?
• Secondly, unlike metal-metal bonds, carbon-carbon bonds are ubiquitous. So how do you deal with them?
Two answers.
1. In a C-H bond, the H is treated as if it has an oxidation state of +1. This means that every C-H bond will decrease the oxidation state of carbon by 1.
2. Any two bonds between the same atom do not affect the oxidation state (recall that the oxidation state of Cl in Cl-Cl (and that of H in H-H) is zero. So a carbon attached to 4 carbons has an oxidation state of zero.
So unlike metals, which are almost always in a positive oxidation state, the oxidation state of carbon can vary widely, from -4 (in CH4) to +4 (such as in CO2). Here are some examples.
(Don’t forget that this is called a “formalism” for a reason. The charge on the carbon is not really +4 or –4. But the oxidation state formalism helps us keep track of where the electrons are going, which will come in handy very soon).
With an understanding of how to calculate oxidation states on carbon, we’re ready for the next step: understanding changes in the oxidation state at carbon, through reactions known as oxidations (where the oxidation state is increased), and reductions (where the oxidation state is reduced). More on that next time.
Carbanions
a reactive intermediate is a short-lived, high-energy, highly reactive molecule. When generated in a chemical reaction, it will quickly convert into a more stable molecule. Only in exceptional cases can these compounds be isolated and stored (e.g., low temperatures, matrix isolation). When their existence is indicated, reactive intermediates can help explain how a chemical reaction takes place
Reactive Intermediates
The removal of a hydrogen atom on a carbon atom in an organic molecule as a proton results in a species that has a formal charge of -1 on a carbon atom. It is called a carbanion. eg:
Some carbanions are resonance stabilized, in which case some or all resonance forms fit the above definition.
eg. 1:
eg. 2: | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Fundamentals/Organic_Acids_and_Bases/Organic_Bases.txt |
A carbanion is an anion in which carbon has an unshared pair of electrons and bears a negative charge usually with three substituents for a total of eight valence electrons.[1] The carbanion exists in a trigonal pyramidal geometry. Formally, a carbanion is the conjugate base of a carbon acid.
$\ce{R_3C-H + B^- \rightarrow R_3C^- + H-B}$
where B stands for the base. A carbanion is one of several reactive intermediates in organic chemistry.
Theory
A carbanion is a nucleophile, which stability and reactivity determined by several factors:
1. The inductive effect. Electronegative atoms adjacent to the charge will stabilize the charge;
2. Hybridization of the charge-bearing atom. The greater the s-character of the charge-bearing atom, the more stable the anion;
3. The extent of conjugation of the anion. Resonance effects can stabilize the anion. This is especially true when the anion is stabilized as a result of aromaticity.
A carbanion is a reactive intermediate and is encountered in organic chemistry for instance in the E1cB elimination reaction and in organometallic chemistry in for instance a Grignard reaction or in alkyl lithium chemistry. Stable carbanions do however exist. In 1984 Olmstead presented the lithium crown ether salt of the triphenylmethyl carbanion from triphenylmethane, n-butyllithium and 12-crown-4 at low temperatures:[2]
Adding n-butyllithium to triphenylmethane in THF at low temperatures followed by 12-crown-4 results in a red solution and the salt complex precipitates at −20 °C. The central C-C bond lengths are 145 pm with the phenyl ring propelled at an average angle of 31.2°. This propeller shape is less pronounced with a tetramethylammonium counterion.[3] One tool for the detection of carbanions in solution is proton NMR.[4] A spectrum of cyclopentadiene in DMSO shows four vinylic protons at 6.5 ppm and two methylene bridge protons at 3 ppm whereas the cyclopentadienyl anion has a single resonance at 5.50 ppm.
Carbon acids
Any molecule containing a C-H can lose a proton forming the carbanion. Hence any hydrocarbon containing C-H bonds can be considered an acid with a corresponding pKa value. Methane is certainly not an acid in its classical meaning yet its estimated pKa is 56. Compare this to acetic acid with pKa 4.76. The same factors that determine the stability of the carbanion also determine the order in pKa in carbon acids. These values are determined for the compounds either in water in order to compare them to ordinary acids, indimethyl sulfoxide in which the majority of carbon acids and their anions are soluble or in the gas phase. With DMSO the acidity window for solutes is limited to its own pKa of 35.5.
Table 1. Carbon acid acidities in pKa in DMSO [5]. Reference acids in bold.
name formula structural formula pKa
Methane CH4 ~ 56
Ethane C2H6 ~ 50
Anisole C7H8O ~ 49
Cyclopentane C5H10 ~ 45
Propene C3H6 ~ 44
Benzene C6H6 ~ 43
Toluene C6H5CH3 ~ 43
Dimethyl sulfoxide (CH3)2SO 35.5
Diphenylmethane C13H12 32.3
Aniline C6H5NH2 30.6
Triphenylmethane C19H16 30.6
Xanthene C13H10O 30
Ethanol C2H5OH 29.8
Phenylacetylene C8H6 28.8
Thioxanthene C13H10S 28.6
Acetone C3H6O 26.5
Acetylene C2H2 25
Benzoxazole C7H5NO 24.4
Fluorene C13H10 22.6
Indene C9H8 20.1
Cyclopentadiene C5H6 18
Malononitrile C3H2N2 11.2
Hydrogen cyanide HCN 9.2
Acetylacetone C5H8O2 8.95
Dimedone C8H12O2 5.23
Meldrum's acid C6H8O4 4.97
Acetic acid CH3COOH 4.76
Barbituric acid C4H2O3(NH)2 4.01
Trinitromethane HC(NO2)3 0.17
Fulminic acid HCNO -1.07
Carborane superacid HCHB11Cl11 -9
Note that the anions formed by ionization of acetic acid, ethanol or aniline are not carbanions.
Starting from methane in Table 1, the acidity increases:
• when the anion is aromatic, either because the added electron causes the anion to become aromatic (as in indene and cyclopentadiene), or because the negative charge on carbon can be delocalized over several already-aromatic rings (as in triphenylmethane or the carborane superacid).
• when the carbanion is surrounded by strongly electronegative groups, through the partial neutralisation of the negative charge (as in malononitrile).
• when the carbanion is immediately next to a carbonyl group. The α-protons of carbonyl groups are acidic because the negative charge in the enolate can be partially distributed in the oxygen atom. Meldrum's acid and barbituric acid, historically named acids, are in fact a lactone and a lactam respectively, but their acidic carbon protons make them acidic. The acidity of carbonyl compounds is an important driving force in many organic reactions such as the aldol reaction.
Chiral carbanions
With the molecular geometry for a carbanion described as a trigonal pyramid the question is whether or not carbanions can display chirality, because if the activation barrier for inversion of this geometry is too low any attempt at introducing chirality will end inracemization, similar to the nitrogen inversion. However, solid evidence exists that carbanions can indeed be chiral for example in research carried out with certain organolithium compounds.
The first ever evidence for the existence of chiral organolithium compounds was obtained in 1950. Reaction of chiral 2-iodooctane with sec-butyllithium in petroleum ether at −70 °C followed by reaction with dry ice yielded mostly racemic 2-methylbutyric acid but also an amount of optically active 2-methyloctanoic acid which could only have formed from likewise optical active 2-methylheptyllithium with the carbon atom linked to lithium the carbanion:[6]
On heating the reaction to 0 °C the optical activity is lost. More evidence followed in the 1960s. A reaction of the cis isomer of 2-methylcyclopropyl bromide with sec-butyllithium again followed by carboxylation with dry ice yielded cis-2-methylcyclopropylcarboxylic acid. The formation of the trans isomer would have indicated that the intermediate carbanion was unstable.[7]
In the same manner the reaction of (+)-(S)-l-bromo-l-methyl-2,2-diphenylcyclopropane with n-butyllithium followed by quench with methanol resulted in product with retention of configuration:[8]
Of recent date are chiral methyllithium compounds:[9]
The phosphate 1 contains a chiral group with a hydrogen and a deuterium substituent. The stannyl group is replaced by lithium to intermediate 2 which undergoes a phosphate-phosphorane rearrangement to phosphorane 3 which on reaction with acetic acid givesalcohol 4. Once again in the range of −78 °C to 0 °C the chirality is preserved in this reaction sequence.[10]
History
A carbanionic structure first made an appearance in the reaction mechanism for the benzoin condensation as correctly proposed by Clarke and Lapworth in 1907.[11] In 1904 Schlenk prepared Ph3C-NMe4+ in a quest for pentavalent nitrogen (fromTetramethylammonium chloride and Ph3CNa) [12] and in 1914 he demonstrated how triarylmethyl radicals could be reduced to carbonions by alkali metals.[13] The phrase carbanion was introduced by Wallis and Adams in 1933 as the negatively charged counterpart of the carbonium ion. [14][15]
External links
• Large database of Bordwell pKa values at www.chem.wisc.edu Link
• Large database of Bordwell pKa values at daecr1.harvard.edu Link | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Fundamentals/Reactive_Intermediates/Carbanions_II.txt |
A carbene is a molecule containing a neutral carbon atom with a valence of two and two unshared valence electrons. The general formula is R-(C:)-R' or R=C:. The term "carbene" may also refer to the specific compound H2C:, also called methylene, the parent hydride from which all other carbene compounds are formally derived. Carbenes are classified as either singlets or triplets depending upon their electronic structure. Most carbenes are very short lived, although persistent carbenes are known. One well studied carbene is Cl2C:, or dichlorocarbene, which can be generated in situ from chloroform and a strong base.
Singlet and triplet carbenes
The two classes of carbenes are singlet and triplet carbenes. Singlet carbenes are spin-paired. In the language of valence bond theory, the molecule adopts an sp2 hybrid structure. Triplet carbenes have two unpaired electrons. They may be either linear or bent, i.e. sp or sp2 hybridized, respectively. Most carbenes have a nonlinear triplet ground state, except for those with nitrogen, oxygen, or sulfur atoms, and halides directly bonded to the divalent carbon.
Figure 1: from Wikipedia
Carbenes are called singlet or triplet depending on the electronic spins they possess. Triplet carbenes are paramagnetic and may be observed by electron spin resonance spectroscopy if they persist long enough. The total spin of singlet carbenes is zero while that of triplet carbenes is one (in units of [\hbar] ). Bond angles are 125-140° for triplet methylene and 102° for singlet methylene (as determined by EPR). Triplet carbenes are generally stable in the gaseous state, while singlet carbenes occur more often in aqueous media.
For simple hydrocarbons, triplet carbenes usually have energies 8 kcal/mol (33 kJ/mol) lower than singlet carbenes (see also Hund's rule of maximum multiplicity), thus, in general, triplet is the more stable state (the ground state) and singlet is the excited state species.Substituents that can donate electron pairs may stabilize the singlet state by delocalizing the pair into an empty p-orbital. If the energy of the singlet state is sufficiently reduced it will actually become the ground state. No viable strategies exist for triplet stabilization. The carbene called 9-fluorenylidene has been shown to be a rapidly equilibrating mixture of singlet and triplet states with an approximately 1.1 kcal/mol (4.6 kJ/mol) energy difference.[3] It is, however, debatable whether diaryl carbenes such as the fluorene carbene are true carbenes because the electrons can delocalize to such an extent that they become in fact biradicals. In silico experiments suggest that triplet carbenes can be thermodynamically stabilized with electropositive heteroatoms such as in silyl and silyloxy carbenes, especially trifluorosilyl carbenes.[4]
Reactivity
Carbene addition to alkenes
Singlet and triplet carbenes exhibit divergent reactivity. Singlet carbenes generally participate in cheletropic reactions as either electrophiles or nucleophiles. Singlet carbenes with unfilled p-orbital should be electrophilic. Triplet carbenes can be considered to be diradicals, and participate in stepwise radical additions. Triplet carbenes have to go through an intermediate with two unpaired electrons whereas singlet carbene can react in a single concerted step.
Due to these two modes of reactivity, reactions of singlet methylene are stereospecific whereas those of triplet methylene are stereoselective. This difference can be used to probe the nature of a carbene. For example, the reaction of methylene generated from photolysis of diazomethane with cis-2-butene or with trans-2-butene each give a single diastereomer of the 1,2-dimethylcyclopropane product: cis from cis and trans from trans, which proves that the methylene is a singlet.[5] If the methylene were a triplet, one would not expect the product to depend upon the starting alkene geometry, but rather a nearly identical mixture in each case.
Reactivity of a particular carbene depends on the substituent groups. Their reactivity can be affected by metals. Some of the reactions carbenes can do are insertions into C-H bonds, skeletal rearrangements, and additions to double bonds. Carbenes can be classified as nucleophilic, electrophilic, or ambiphilic. For example, if a substituent is able to donate a pair of electrons, most likely carbene will not be electrophilic. Alkyl carbenes insert much more selectively than methylene, which does not differentiate between primary, secondary, and tertiary C-H bonds.
Cyclopropanation
Carbenes add to double bonds to form cyclopropanes. A concerted mechanism is available for singlet carbenes. Triplet carbenes do not retain stereochemistry in the product molecule. Addition reactions are commonly very fast and exothermic. The slow step in most instances is generation of carbene. A well-known reagent employed for alkene-to-cyclopropane reactions is Simmons-Smith reagent. This reagent is a system of copper, zinc, and iodine, where the active reagent is believed to be iodomethylzinc iodide. Reagent is complexed by hydroxy groups such that addition commonly happens syn to such group.
Carbene cyclopropanation
C—H insertion
Carbene insertion
Insertions are another common type of carbene reactions. The carbene basically interposes itself into an existing bond. The order of preference is commonly: 1. X–H bonds where X is not carbon 2. C–H bond 3. C–C bond. Insertions may or may not occur in single step.
Intramolecular insertion reactions present new synthetic solutions. Generally, rigid structures favor such insertions to happen. When an intramolecular insertion is possible, no intermolecular insertions are seen. In flexible structures, five-membered ring formation is preferred to six-membered ring formation. Both inter- and intramolecular insertions are amendable to asymmetric induction by choosing chiral ligands on metal centers.
Carbene intramolecular reaction
Carbene intermolecular reaction
Alkylidene carbenes are alluring in that they offer formation of cyclopentene moieties. To generate an alkylidene carbene a ketone can be exposed to trimethylsilyl diazomethane.
Alkylidene carbene
Carbene dimerization
Carbenes and carbenoid precursors can undergo dimerization reactions to form alkenes. While this is often an unwanted side reaction, it can be employed as a synthetic tool and a direct metal carbene dimerization has been used in the synthesis of polyalkynylethenes. Persistent carbenes exist in equilibrium with their respective dimers. This is known as the Wanzlick equilibrium.
Figure : Wanzlick equilibrium
Carbene ligands in organometallic chemistry
In organometallic species, metal complexes with the formulae LnMCRR' are often described as carbene complexes. Such species do not however react like free carbenes and are rarely generated from carbene precursors, except for the persistent carbenes. The transition metal carbene complexes can be classified according to their reactivity, with the first two classes being the most clearly defined:
• Fischer carbenes, in which the carbene is bonded to a metal that bears an electron-withdrawing group (usually a carbonyl). In such cases the carbenoid carbon is mildly electrophilic.
• Schrock carbenes, in which the carbene is bonded to a metal that bears an electron-donating group. In such cases the carbenoid carbon is nucleophilic and resembles Wittig reagent (which are not considered carbene derivatives).
• Persistent carbenes, also known as Arduengo or Wanzlick carbenes. These include the class of N-heterocyclic carbenes (NHCs) and are often are used as ancillary ligands in organometallic chemistry. Such carbenes are spectator ligands of low reactivity.
Generation of carbenes
• A method that is broadly applicable to organic synthesis is induced elimination of halides from gem-dihalides employing organolithium reagents. It remains uncertain if under these conditions free carbenes are formed or metal-carbene complex. Nevertheless, these metallocarbenes (or carbenoids) give the expected organic products.
R2CBr2 + BuLi → R2CLi(Br) +
BuBr
R
2CLi(Br) → R2C + LiBr
• For cyclopropanations, zinc is employed in the Simmons–Smith reaction. In a specialized but instructive case, alpha-halomercury compounds can be isolated and separately thermolyzed. For example, the "Seyferth reagent" releases CCl2 upon heating.
C6H5HgCCl3 → CCl2 + C6H5HgCl
• Most commonly, carbenes are generated from diazoalkanes, via photolytic, thermal, or transition metal-catalyzed routes. Catalysts typically feature rhodium and copper. The Bamford-Stevens reaction gives carbenes in aprotic solvents and carbenium ions in protic solvents.
• Base-induced elimination HX from haloforms (CHX3) with under phase-transfer conditions.
• Photolysis of diazirines and epoxides can also be employed. Diazirines are cyclic forms of diazoalkanes. The strain of the small ring makes photoexcitation easy. Photolysis of epoxides gives carbonyl compounds as side products. With asymmetric epoxides, two different carbonyl compounds can potentially form. The nature of substituents usually favors formation of one over the other. One of the C-O bonds will have a greater double bond character and thus will be stronger and less likely to break. Resonance structures can be drawn to determine which part will contribute more to the formation of carbonyl. When one substituent is alkyl and another aryl, the aryl-substituted carbon is usually released as a carbene fragment.
• Carbenes are intermediates in the Wolff rearrangement
References
1. Hoffmann, Roald (2005). Molecular Orbitals of Transition Metal Complexes. Oxford. p. 7. ISBN 0-19-853093-5.
2. IUPAC, Compendium of Chemical Terminology, 2nd ed. (the "Gold Book") (1997). Online corrected version: (2006–) "carbenes".
3. Grasse, P. B.; Brauer, B. E.; Zupancic, J. J.; Kaufmann, K. J.; Schuster, G. B. (1983). "Chemical and physical properties of fluorenylidene: equilibration of the singlet and triplet carbenes". Journal of the American Chemical Society 105 (23): 6833.doi:10.1021/ja00361a014. edit
4. Nemirowski, A; Schreiner, P. R. (November 2007). "Electronic Stabilization of Ground State Triplet Carbenes". J. Org. Chem. 72 (25): 9533–9540. doi:10.1021/jo701615x.PMID 17994760.
5. Skell, P. S.; Woodworth, R. C. (1956). Journal of the American Chemical Society 78(17): 4496. doi:10.1021/ja01598a087. edit
6. Bajzer, W. X. (2004). "Fluorine Compounds, Organic". Kirk-Othmer Encyclopedia of Chemical Technology. John Wiley & Sons.doi:10.1002/0471238961.0914201802011026.a01.pub2. edit
7. Buchner, E.; Feldmann, L. (1903). "Diazoessigester und Toluol". Berichte der deutschen chemischen Gesellschaft 36 (3): 3509. doi:10.1002/cber.190303603139.edit
8. Staudinger, H.; Kupfer, O. (1912). "Über Reaktionen des Methylens. III. Diazomethan".Berichte der deutschen chemischen Gesellschaft 45: 501.doi:10.1002/cber.19120450174. edit
9. Von E. Doering, W.; Hoffmann, A. K. (1954). "The Addition of Dichlorocarbene to Olefins". Journal of the American Chemical Society 76 (23): 6162.doi:10.1021/ja01652a087. edit | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Fundamentals/Reactive_Intermediates/Carbenes.txt |
A carbocation is an ion with a positively-charged carbon atom. Among the simplest examples are methenium CH3+, methanium CH5+, and ethanium C2H7+. Some carbocations may have two or more positive charges, on the same carbon atom or on different atoms; such as the ethylene dication C2H42+.[1]
Definitions
Until the early 1970s, all carbocations were called carbonium ions.[2] In present-day chemistry, a carbocation is any positively charged carbon atom, classified in two main categories according to the valence of the charged carbon:
• +3 in carbenium ions (protonated carbenes),
• +5 or +6 in the carbonium ions (protonated alkanes, named by analogy to ammonium).
This nomenclature was proposed by G. A. Olah.[3] University-level textbooks discuss carbocations only as if they are carbenium ions,[4] or discuss carbocations with a fleeting reference to the older phrase of carbonium ion[5] or carbenium and carbonium ions.[6]
History
The history of carbocations dates back to 1891 when G. Merling[8] reported that he added bromine to tropylidene (cycloheptatriene) and then heated the product to obtain a crystalline, water-soluble material, \(C_7H_7Br\). He did not suggest a structure for it; however, Doering and Knox[9] convincingly showed that it was tropylium (cycloheptatrienylium) bromide. This ion is predicted to be aromatic by Hückel's rule.
In 1902, Norris and Kehrman independently discovered that colorless triphenylmethanol gives deep-yellow solutions in concentrated sulfuric acid. Triphenylmethyl chloride similarly formed orange complexes with aluminium and tin chlorides. In 1902, Adolf von Baeyer recognized the salt-like character of the compounds formed.
He dubbed the relationship between color and salt formation halochromy, of which malachite green is a prime example.
Carbocations are reactive intermediates in many organic reactions. This idea, first proposed by Julius Stieglitz in 1899,[10] was further developed by Hans Meerwein in his 1922 study[11][12] of the Wagner-Meerwein rearrangement. Carbocations were also found to be involved in the \(S_N1\) reaction, the \(E1\0 reaction, and in rearrangement reactions such as the Whitmore 1,2 shift. The chemical establishment was reluctant to accept the notion of a carbocation and for a long time the Journal of the American Chemical Society refused articles that mentioned them.
The first NMR spectrum of a stable carbocation in solution was published by Doering et al.[13] in 1958. It was the heptamethylbenzenium ion, made by treating hexamethylbenzene with methyl chloride and aluminium chloride. The stable 7-norbornadienyl cation was prepared by Story et al. in 1960[14] by reacting norbornadienyl chloride with silver tetrafluoroborate in sulfur dioxide at −80 °C. The NMR spectrum established that it was non-classically bridged (the first stable non-classical ion observed).
In 1962, Olah directly observed the tert-butyl carbocation by nuclear magnetic resonance as a stable species on dissolving tert-butyl fluoride in magic acid. The NMR of the norbornyl cation was first reported by Schleyer et al.[15] and it was shown to undergo proton-scrambling over a barrier by Saunders et al.[16]
Structure and properties
The charged carbon atom in a carbocation is a "sextet", i.e. it has only six electrons in its outer valence shell instead of the eight valence electrons that ensures maximum stability (octet rule). Therefore, carbocations are often reactive, seeking to fill the octet of valence electrons as well as regain a neutral charge. One could reasonably assume a carbocation to have \(sp^3\) hybridization with an empty \(sp_3\) orbital giving positive charge. However, the reactivity of a carbocation more closely resembles \(sp^2\) hybridization with atrigonal planar molecular geometry. An example is the methyl cation, \(CH_3^+\).
Order of stability of examples of tertiary (III), secondary (II), and primary (I) alkylcarbenium ions, as well as the methyl cation (far right).
Carbocations are often the target of nucleophilic attack by nucleophiles like hydroxide (OH) ions or halogen ions.
Carbocations typically undergo rearrangement reactions from less stable structures to equally stable or more stable ones with rate constants in excess of 109/sec. This fact complicates synthetic pathways to many compounds. For example, when 3-pentanol is heated with aqueous HCl, the initially formed 3-pentyl carbocation rearranges to a statistical mixture of the 3-pentyl and 2-pentyl. These cations react with chloride ion to produce about 1/3 3-chloropentane and 2/3 2-chloropentane.
A carbocation may be stabilized by resonance by a carbon-carbon double bond next to the ionized carbon. Such cations as allyl cation CH2=CH–CH2+ and benzyl cation C6H5–CH2+ are more stable than most other carbocations. Molecules that can form allyl or benzyl carbocations are especially reactive. These carbocations where the C+ is adjacent to another carbon atom that has a double or triple bond have extra stability because of the overlap of the empty p orbital of the carbocation with the p orbitals of the π bond. This overlap of the orbitals allows the charge to be shared between multiple atoms – delocalization of the charge - and, therefore, stabilizes the carbocation.
Non-classical ions
Some carbocations such as the norbornyl cation exhibit more or less symmetrical three centre bonding. Cations of this sort have been referred to as non-classical ions. The energy difference between "classical" carbocations and "non-classical" isomers is often very small, and in general there is little, if any, activation energy involved in the transition between "classical" and "non-classical" structures. In essence, the "non-classical" form of the 2-butyl carbocation is 2-butene with a proton directly above the centre of what would be the carbon-carbon double bond. "Non-classical" carbocations were once the subject of great controversy. One of George Olah's greatest contributions to chemistry was resolving this controversy.[17]
Specific carbocations
Cyclopropylcarbinyl cations can be studied by NMR:[18][19]
In the NMR spectrum of a dimethyl derivative, two nonequivalent signals are found for the two methyl groups, indicating that the molecular conformation of this cation not perpendicular (as in A) but is bisected (as in B) with the empty p-orbital and the cyclopropyl ring system in the same plane:
In terms of bent bond theory, this preference is explained by assuming favorable orbital overlap between the filled cyclopropane bent bonds and the empty p-orbital.[20]
References
1. Hansjörg Grützmacher, Christina M. Marchand (1997), "Heteroatom stabilized carbenium ions", Coordination Chemistry Reviews, volume 163, pages 287-344. doi:10.1016/S0010-8545(97)00043-X
2. Gold Book definition carbonium ion HTML
3. George A. Olah (1972), "Stable carbocations. CXVIII. General concept and structure of carbocations based on differentiation of trivalent (classical) carbenium ions from three-center bound penta- of tetracoordinated (nonclassical) carbonium ions. Role of carbocations in electrophilic reactions." Journal of the American Chemical Society, volume 94, issue 3, pages 808–820 doi:10.1021/ja00758a020
4. Organic chemistry 5th Ed. John McMurry ISBN 0-534-37617-7
5. Organic Chemistry, Fourth Edition Paula Yurkanis Bruice ISBN 0-13-140748-1
6. Clayden, Jonathan; Greeves, Nick; Warren, Stuart; Wothers, Peter (2001). Organic Chemistry (1st ed.). Oxford University Press. ISBN 978-0-19-850346-0.
7. Organic Chemistry by Marye Anne Fox and James K. Whitesell ISBN 0-7637-0413-X
8. Chem. Ber. 24, 3108 1891
9. The Cycloheptatrienylium (Tropylium) Ion W. Von E. Doering and L. H. Knox J. Am. Chem. Soc.; 1954; 76(12) pp 3203 - 3206; doi:10.1021/ja01641a027
10. On the Constitution of the Salts of Imido-Ethers and other Carbimide Derivatives; Am. Chem. J. 21, 101; ISSN: 0096-4085
11. H. Meerwein and K. van Emster, Berichte, 1922, 55, 2500.
12. Rzepa, H. S.; Allan, C. S. M. (2010). "Racemization of Isobornyl Chloride via Carbocations: A Nonclassical Look at a Classic Mechanism". Journal of Chemical Education 87 (2): 221. Bibcode:2010JChEd..87..221R. doi:10.1021/ed800058c. edit
13. The 1,1,2,3,4,5,6-heptamethylbenzenonium ion W. von E. Doering and M. Saunders H. G. Boyton, H. W. Earhart, E. F. Wadley and W. R. Edwards G. Laber Tetrahedron Volume 4, Issues 1-2 , 1958, Pages 178-185 doi:10.1016/0040-4020(58)88016-3
14. The 7-norbornadienyl carbonium ion Paul R. Story and Martin Saunders J. Am. Chem. Soc.; 1960; 82(23) pp 6199 - 6199; doi:10.1021/ja01508a058
15. Stable Carbonium Ions. X.1 Direct Nuclear Magnetic Resonance Observation of the 2-Norbornyl Cation Paul von R. Schleyer, William E. Watts, Raymond C. Fort, Melvin B. Comisarow, and George A. Olah J. Am. Chem. Soc.; 1964; 86(24) pp 5679 - 5680; doi:10.1021/ja01078a056
16. Stable Carbonium Ions. XI.1 The Rate of Hydride Shifts in the 2-Norbornyl Cation Martin Saunders, Paul von R. Schleyer, and George A. Olah J. Am. Chem. Soc.; 1964; 86(24) pp 5680 - 5681; doi:10.1021/ja01078a057
17. George A. Olah - Nobel Lecture
18. Nuclear magnetic double resonance studies of the dimethylcyclopropylcarbinyl cation. Measurement of the rotation barrier David S. Kabakoff, , Eli. Namanworth J. Am. Chem. Soc. 1970, 92 (10), pp 3234–3235 doi:10.1021/ja00713a080
19. Stable Carbonium Ions. XVII.1a Cyclopropyl Carbonium Ions and Protonated Cyclopropyl Ketones Charles U. Pittman Jr., George A. Olah J. Am. Chem. Soc., 1965, 87 (22), pp 5123–5132 doi:10.1021/ja00950a026
20. F.A. Carey, R.J. Sundberg Advanced Organic Chemistry Part A 2nd Ed.
Carbynes
Carbyne is often a general term for any compound whose molecular structure includes an electrically neutral carbon atom with three non-bonded electrons, connected to another atom by a single bond. A carbyne has the general formula R-C3•, where R is any monovalent group and the superscript 3• indicates the three unbounded valences. Carbynes are named after the simplest such compound, HC3•, the methylidyne radical or (unsubstituted) carbyne.
Electronic configuration
Carbyne molecules are generally found to be in electronic doublet states: the nonbonding electrons on carbon are arranged as one radical (unpaired electron) and one electron pair, leaving a vacant atomic orbital, rather than being a tri-radical (the quartet state). The simplest case is the CH radical, which has an electron configuration 1σ222 1π. Here the 1σ molecular orbital is essentially the carbon 1s atomic orbital, and the 2σ is the C-H bonding orbital formed by overlap of a carbon s-p hybrid orbital with the hydrogen 1s orbital. The 3σ is a carbon non-bonding orbital pointing along the C-H axis away from the hydrogen, while there are two non-bonding 1π orbitals perpendicular to the C-H axis. However the 3σ is an s-p hybrid which has lower energy than the 1π orbital which is pure p, so the 3σ is filled before the 1π.
Figure: doublet (1 radical, 1 pair, 1 vacant orbital) (lefT) quartet (3 radicals)
The CH radical is in fact isoelectronic with the nitrogen atom which does have three unpaired electrons in accordance with Hund's rule of maximum multiplicity. However the N atom has three degenerate p orbitals, in contrast to the CH radical where hybridization of one orbital (the 3σ) leads to an energy difference.
• Wikipedia | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Fundamentals/Reactive_Intermediates/Carbocations.txt |
In chemistry, a radical (more precisely, a free radical) is an atom, molecule, or ion that has unpaired valence electrons or an open electron shell, and therefore may be seen as having one or more "dangling" covalent bonds.
With some exceptions, these "dangling" bonds make free radicals highly chemically reactive towards other substances, or even towards themselves: their molecules will often spontaneously dimerize or polymerize if they come in contact with each other. Most radicals are reasonably stable only at very low concentrations in inert media or in a vacuum.
A notable example of a free radical is the hydroxyl radical (HO•), a molecule that is one hydrogen atom short of a water molecule and thus has one bond "dangling" from the oxygen. Two other examples are the carbene molecule (:CH2), which has two dangling bonds; and the superoxide anion (•O−2), the oxygen molecule O2 with one extra electron, which has one dangling bond. In contrast, the hydroxyl anion (HO), the oxide anion (O2−) and thecarbenium cation (CH+3) are not radicals, since the bonds that may appear to be dangling are in fact resolved by the addition or removal of electrons.
Free radicals may be created in a number of ways, including synthesis with very dilute or rarefied reagents, reactions at very low temperatures, or breakup of larger molecules. The latter can be affected by any process that puts enough energy into the parent molecule, such as ionizing radiation, heat, electrical discharges, electrolysis, and chemical reactions. Indeed, radicals are intermediate stages in many chemical reactions.
Free radicals play an important role in combustion, atmospheric chemistry, polymerization, plasma chemistry, biochemistry, and many other chemical processes. In living organisms, the free radicals superoxide and nitric oxideand their reaction products regulate many processes, such as control of vascular tone and thus blood pressure. They also play a key role in the intermediary metabolism of various biological compounds. Such radicals can even be messengers in a process dubbed redox signaling. A radical may be trapped within a solvent cage or be otherwise bound.
Until late in the 20th century the word "radical" was used in chemistry to indicate any connected group of atoms, such as a methyl group or a carboxyl, whether it was part of a larger molecule or a molecule on its own. The qualifier "free" was then needed to specify the unbound case. Following recent nomenclature revisions, a part of a larger molecule is now called a functional group or substituent, and "radical" now implies "free". However, the old nomenclature may still occur in the literature.
History
The first organic free radical identified was triphenylmethyl radical. This species was discovered by Moses Gomberg in 1900 at the University of Michigan USA. Historically, the term radical in radical theory was also used for bound parts of the molecule, especially when they remain unchanged in reactions. These are now called functional groups. For example, methyl alcohol was described as consisting of a methyl "radical" and a hydroxyl "radical". Neither are radicals in the modern chemical sense, as they are permanently bound to each other, and have no unpaired, reactive electrons; however, they can be observed as radicals in mass spectrometry when broken apart by irradiation with energetic electrons.
Depiction in chemical reactions
In chemical equations, free radicals are frequently denoted by a dot placed immediately to the right of the atomic symbol or molecular formula as follows:
$\mathrm{Cl}_2 \; \xrightarrow{UV} \; {\mathrm{Cl} \cdot} + {\mathrm{Cl} \cdot}$
Chlorine gas can be broken down by ultraviolet light to form atomic chlorine radicals.
Radical reaction mechanisms use single-headed arrows to depict the movement of single electrons:
The homolytic cleavage of the breaking bond is drawn with a 'fish-hook' arrow to distinguish from the usual movement of two electrons depicted by a standard curly arrow. It should be noted that the second electron of the breaking bond also moves to pair up with the attacking radical electron; this is not explicitly indicated in this case.
Free radicals also take part in radical addition and radical substitution as reactive intermediates. Chain reactions involving free radicals can usually be divided into three distinct processes. These are initiation, propagation, and termination.
• Initiation reactions are those that result in a net increase in the number of free radicals. They may involve the formation of free radicals from stable species as in Reaction 1 above or they may involve reactions of free radicals with stable species to form more free radicals.
• Propagation reactions are those reactions involving free radicals in which the total number of free radicals remains the same.
• Termination reactions are those reactions resulting in a net decrease in the number of free radicals. Typically two free radicals combine to form a more stable species, for example: 2Cl·→ Cl2
Formation
The formation of radicals may involve breaking of covalent bonds homolytically, a process that requires significant amounts of energy. For example, splitting H2 into 2H· has a ΔH° of +435 kJ/mol, and Cl2 into 2Cl· has a ΔH° of +243 kJ/mol. This is known as the homolytic bond dissociation energy, and is usually abbreviated as the symbol ΔH°. The bond energy between two covalently bonded atoms is affected by the structure of the molecule as a whole, not just the identity of the two atoms. Likewise, radicals requiring more energy to form are less stable than those requiring less energy. Homolytic bond cleavage most often happens between two atoms of similar electronegativity. In organic chemistry this is often the O-O bond in peroxide species or O-N bonds. Sometimes radical formation is spin-forbidden, presenting an additional barrier. However, propagation is a very exothermic reaction. Likewise, although radical ions do exist, most species are electrically neutral. Radicals may also be formed by single electron oxidation or reduction of an atom or molecule. An example is the production of superoxide by the electron transport chain. Early studies of organometallic chemistry, especially tetra-alkyl lead species by F.A. Paneth and K. Hahnfeld in the 1930s supported heterolytic fission of bonds and a radical based mechanism.
Persistence and stability
The radical derived from α-tocopherol
Although radicals are generally short-lived due to their reactivity, there are long-lived radicals. These are categorized as follows:
Stable radicals
The prime example of a stable radical is molecular dioxygen (O2). Another common example is nitric oxide (NO). Organic radicals can be long lived if they occur in a conjugated π system, such as the radical derived from α-tocopherol (vitamin E). There are also hundreds of examples of thiazyl radicals, which show low reactivity and remarkable thermodynamic stability with only a very limited extent of π resonance stabilization.[1][2]
Persistent radicals
Persistent radical compounds are those whose longevity is due to steric crowding around the radical center, which makes it physically difficult for the radical to react with another molecule.[3] Examples of these include Gomberg's triphenylmethyl radical, Fremy's salt(Potassium nitrosodisulfonate, (KSO3)2NO·), nitroxides, (general formula R2NO·) such as TEMPO, TEMPOL, nitronyl nitroxides, and azephenylenyls and radicals derived from PTM (perchlorophenylmethyl radical) and TTM (tris(2,4,6-trichlorophenyl)methyl radical). Persistent radicals are generated in great quantity during combustion, and "may be responsible for the oxidative stress resulting in cardiopulmonary disease and probably cancer that has been attributed to exposure to airborne fine particles."[4]
Diradicals
Diradicals are molecules containing two radical centers. Multiple radical centers can exist in a molecule. Atmospheric oxygen naturally exists as a diradical in its ground state as triplet oxygen. The low reactivity of atmospheric oxygen is due to its diradical state. Non-radical states of dioxygen are actually less stable than the diradical. The relative stability of the oxygen diradical is primarily due to the spin-forbidden nature of the triplet-singlet transition required for it to grab electrons, i.e., "oxidize". The diradical state of oxygen also results in its paramagnetic character, which is demonstrated by its attraction to an external magnet.[5]
Reactivity
Radical alkyl intermediates are stabilized by similar physical processes to carbocations: as a general rule, the more substituted the radical center is, the more stable it is. This directs their reactions. Thus, formation of a tertiary radical (R3C·) is favored over secondary (R2HC·), which is favored over primary (RH2C·). Likewise, radicals next to functional groups such as carbonyl, nitrile, and ether are more stable than tertiary alkyl radicals.
Radicals attack double bonds. However, unlike similar ions, such radical reactions are not as much directed by electrostatic interactions. For example, the reactivity of nucleophilic ions with α,β-unsaturated compounds (C=C–C=O) is directed by the electron-withdrawing effect of the oxygen, resulting in a partial positive charge on the carbonyl carbon. There are two reactions that are observed in the ionic case: the carbonyl is attacked in a direct addition to carbonyl, or the vinyl is attacked in conjugate addition, and in either case, the charge on the nucleophile is taken by the oxygen. Radicals add rapidly to the double bond, and the resulting α-radical carbonyl is relatively stable; it can couple with another molecule or be oxidized. Nonetheless, the electrophilic/neutrophilic character of radicals has been shown in a variety of instances. One example is the alternating tendency of the copolymerization of maleic anhydride (electrophilic) and styrene (slightly nucleophilic).
In intramolecular reactions, precise control can be achieved despite the extreme reactivity of radicals. In general, radicals attack the closest reactive site the most readily. Therefore, when there is a choice, a preference for five-membered rings is observed: four-membered rings are too strained, and collisions with carbons six or more atoms away in the chain are infrequent.
Carbenes and nitrenes, which are diradicals, have distinctive chemistry.
Combustion
Spectrum of the blue flame from a butane torch showing excited molecular radical band emission and Swan bands
A familiar free-radical reaction is combustion. The oxygen molecule is a stable diradical, best represented by ·O-O·. Because spins of the electrons are parallel, this molecule is stable. While the ground stateof oxygen is this unreactive spin-unpaired (triplet) diradical, an extremely reactive spin-paired (singlet) state is available. For combustion to occur, the energy barrier between these must be overcome. This barrier can be overcome by heat, requiring high temperatures. The triplet-singlet transition is also "forbidden". This presents an additional barrier to the reaction. It also means molecular oxygen is relatively unreactive at room temperature except in the presence of a catalytic heavy atom such as iron or copper.
Combustion consists of various radical chain reactions that the singlet radical can initiate. The flammability of a given material strongly depends on the concentration of free radicals that must be obtained before initiation and propagation reactions dominate leading to combustion of the material. Once the combustible material has been consumed, termination reactions again dominate and the flame dies out. As indicated, promotion of propagation or termination reactions alters flammability. For example, because lead itself deactivates free radicals in the gasoline-air mixture, tetraethyl lead was once commonly added to gasoline. This prevents the combustion from initiating in an uncontrolled manner or in unburnt residues (engine knocking) or premature ignition (preignition).
When a hydrocarbon is burned, a large number of different oxygen radicals are involved. Initially, hydroperoxyl radical (HOO·) are formed. These then react further to give organic hydroperoxides that break up into hydroxyl radicals (HO·).
Polymerization
In addition to combustion, many polymerization reactions involve free radicals. As a result many plastics, enamels, and other polymers are formed through radical polymerization. For instance, drying oils and alkyd paints harden due to radical crosslinking by oxygen from the atmosphere.
Recent advances in radical polymerization methods, known as living radical polymerization, include:
• Reversible addition-fragmentation chain transfer (RAFT)
• Atom transfer radical polymerization (ATRP)
• Nitroxide mediated polymerization (NMP)
These methods produce polymers with a much narrower distribution of molecular weights.
Atmospheric radicals
The most common radical in the lower atmosphere is molecular dioxygen. Photodissociation of source molecules produces other free radicals. In the lower atmosphere, the most important examples of free radical production are the photodissociation of nitrogen dioxide to give an oxygen atom and nitric oxide (see eq. 1 below), which plays a key role in smog formation—and the photodissociation of ozone to give the excited oxygen atom O(1D) (see eq. 2 below). The net and return reactions are also shown (eq. 3 and 4, respectively).
$1. \; \; \mathrm{NO}_2 \; \xrightarrow{h \nu } \; \mathrm{NO + O}$
$2. \; \; \mathrm{O} + \mathrm{O}_2 \; \xrightarrow{} \; \mathrm{O}_3$
$3. \; \; \mathrm{NO}_2 + \mathrm{O}_2 \; \xrightarrow{h \nu } \; \mathrm{NO} + \mathrm{O}_3 \;$
$4. \; \; \mathrm{NO} + \mathrm{O}_3 \; \xrightarrow{} \; \mathrm{NO}_2 + \mathrm{O}_2 \;$
In the upper atmosphere, a particularly important source of radicals is the photodissociation of normally unreactive chlorofluorocarbons (CFCs) by solar ultraviolet radiation, or by reactions with other stratospheric constituents (see eq. 1 below). These reactions give off the chlorine radical, Cl•, which reacts with ozone in a catalytic chain reaction ending in Ozone depletion and regeneration of the chlorine radical, allowing it to reparticipate in the reaction (see eq. 2–4 below). Such reactions are believed to be the primary cause of depletion of the ozone layer (the net result is shown in eq. 5 below), and this is why the use of chlorofluorocarbons as refrigerants has been restricted.
$1. \; \; \mathrm{CFCS} \; \xrightarrow{h \nu} \; {\mathrm{Cl} \cdot}$
$2. \; \; {\mathrm{Cl} \cdot} + \mathrm{O}_3 \; \xrightarrow{} \; {\mathrm{ClO} \cdot} + \mathrm{O}_2 \;$
$3. \; \; \mathrm{O}_3 \; \xrightarrow{h \nu} \; \mathrm{O} + \mathrm{O}_2$
$4. \; \; \mathrm{O} + {\mathrm{ClO} \cdot} \; \xrightarrow \; {\mathrm{Cl} \cdot} + \mathrm{O}_2$
$5. \; \; 2\mathrm{O}_3 \; \xrightarrow{h \nu } \; 3\mathrm{O}_2$
In biology
Free radicals play an important role in a number of biological processes. Many of these are necessary for life, such as the intracellular killing of bacteria by phagocytic cells such as granulocytes and macrophages. Researchers have also implicated free radicals in certain cell signalling processes,[6] known as redox signaling.
The two most important oxygen-centered free radicals are superoxide and hydroxyl radical. They derive from molecular oxygen under reducing conditions. However, because of their reactivity, these same free radicals can participate in unwanted side reactions resulting in cell damage. Excessive amounts of these free radicals can lead to cell injury and death, which may contribute to many diseases such as cancer, stroke, myocardial infarction, diabetes and major disorders.[7] Many forms of cancer are thought to be the result of reactions between free radicals and DNA, potentially resulting in mutations that can adversely affect the cell cycle and potentially lead to malignancy.[8] Some of the symptoms of aging such as atherosclerosis are also attributed to free-radical induced oxidation of cholesterol to 7-ketocholesterol.[9] In addition free radicals contribute to alcohol-induced liver damage, perhaps more than alcohol itself. Free radicals produced by cigarette smoke are implicated in inactivation of alpha 1-antitrypsin in the lung. This process promotes the development of emphysema.
Free radicals may also be involved in Parkinson's disease, senile and drug-induced deafness, schizophrenia, and Alzheimer's.[10] The classic free-radical syndrome, the iron-storage disease hemochromatosis, is typically associated with a constellation of free-radical-related symptoms including movement disorder, psychosis, skin pigmentary melanin abnormalities, deafness, arthritis, and diabetes mellitus. The free-radical theory of aging proposes that free radicals underlie the aging process itself. Similarly, the process of mitohormesis suggests that repeated exposure to free radicals may extend life span.
Because free radicals are necessary for life, the body has a number of mechanisms to minimize free-radical-induced damage and to repair damage that occurs, such as the enzymes superoxide dismutase, catalase, glutathione peroxidase and glutathione reductase. In addition, antioxidants play a key role in these defense mechanisms. These are often the three vitamins, vitamin A, vitamin C and vitamin E and polyphenol antioxidants. Furthermore, there is good evidence indicating that bilirubin and uric acid can act as antioxidants to help neutralize certain free radicals. Bilirubin comes from the breakdown of red blood cells' contents, while uric acid is a breakdown product of purines. Too much bilirubin, though, can lead to jaundice, which could eventually damage the central nervous system, while too much uric acid causes gout.[11]
Reactive oxygen species
Reactive oxygen species or ROS are species such as superoxide, hydrogen peroxide, and hydroxyl radical and are associated with cell damage. ROS form as a natural by-product of the normal metabolism of oxygen and have important roles in cell signaling.
Oxybenzone has been found to form free radicals in sunlight, and therefore may be associated with cell damage as well. This only occurred when it was combined with other ingredients commonly found in sunscreens, like titanium oxide and octyl methoxycinnamate.[12]
Loose definition of radicals
In most fields of chemistry, the historical definition of radicals contends that the molecules have nonzero spin. However in fields including spectroscopy, chemical reaction, and astrochemistry, the definition is slightly different. Gerhard Herzberg, who won the Nobel prize for his research into the electron structure and geometry of radicals, suggested a looser definition of free radicals: "any transient (chemically unstable) species (atom, molecule, or ion)".[13] The main point of his suggestion is that there are many chemically unstable molecules that have zero spin, such as C2, C3, CH2 and so on. This definition is more convenient for discussions of transient chemical processes and astrochemistry; therefore researchers in these fields prefer to use this loose definition.[14]
Diagnostics
Free radical diagnostic techniques include:
• Electron spin resonance
A widely used technique for studying free radicals, and other paramagnetic species, is electron spin resonance spectroscopy (ESR). This is alternately referred to as "electron paramagnetic resonance" (EPR) spectroscopy. It is conceptually related to nuclear magnetic resonance, though electrons resonate with higher-frequency fields at a given fixed magnetic field than do most nuclei.
• Nuclear magnetic resonance using a phenomenon called CIDNP
• Chemical labelling
Chemical labelling by quenching with free radicals, e.g. with nitric oxide (NO) or DPPH (2,2-diphenyl-1-picrylhydrazyl), followed by spectroscopic methods like X-ray photoelectron spectroscopy (XPS) or absorption spectroscopy, respectively.
• Use of free radical markers
Stable, specific or non-specific derivates of physiological substances can be measured e.g. lipid peroxidation products (isoprostanes, TBARS), amino acid oxidation products (meta-tyrosine, ortho-tyrosine, hydroxy-Leu, dityrosine etc.), peptide oxidation products (oxidized glutathione – GSSG)
2,2'-Azobis(2-amidinopropane) dihydrochloride (AAPH) is a chemical compound used to study the chemistry of the oxidation of drugs.[15] It is a free radical-generating azo compound. It is gaining prominence as a model oxidant in small molecule and proteintherapeutics for its ability to initiate oxidation reactions via both nucleophilic and free radical mechanisms.[16]
• Indirect method
Measurement of the decrease in the amount of antioxidants (e.g. TAS, reduced glutathione – GSH)
• Trapping agents
Using a chemical species that reacts with free radicals to form a stable product that can then be readily measured (Hydroxyl radical and salicylic acid)
See also
• -yl
• Electron pair
• Globally Harmonized System of Classification and Labelling of Chemicals
• Hofmann–Löffler reaction
References
1. Jump up^ Oakley, Richard T. (1988). "Cyclic and Heterocyclic Thiazenes" (PDF). Progress in Inorganic Chemistry. Progress in Inorganic Chemistry 36. pp. 299–391.doi:10.1002/9780470166376.ch4. ISBN 978-0-470-16637-6.
2. Jump up^ Rawson, J; Banister, A; Lavender, I (1995). "The Chemistry of Dithiadiazolylium and Dithiadiazolyl Rings". Advances in Heterocyclic Chemistry Volume 62. Advances in Heterocyclic Chemistry 62. pp. 137–247. doi:10.1016/S0065-2725(08)60422-5. ISBN 978-0-12-020762-6.
3. Jump up^ Griller, David; Ingold, Keith U. (1976). "Persistent carbon-centered radicals". Accounts of Chemical Research 9: 13.doi:10.1021/ar50097a003.
4. Jump up^ Lomnicki S.; Truong H.; Vejerano E.; Dellinger B. (2008). "Copper oxide-based model of persistent free radical formation on combustion-derived particulate matter". Environ. Sci. Technol. 42 (13): 4982–4988. doi:10.1021/es071708h.PMID 18678037.
5. Jump up^ However, paramagnetism does not necessarily imply radical character.
6. Jump up^ Pacher P, Beckman JS, Liaudet L (2007). "Nitric oxide and peroxynitrite in health and disease". Physiol. Rev. 87 (1): 315–424. doi:10.1152/physrev.00029.2006. PMC 2248324.PMID 17237348.
7. Jump up^ Rajamani Karthikeyan, Manivasagam T, Anantharaman P, Balasubramanian T, Somasundaram ST (2011). "Chemopreventive effect of Padina boergesenii extracts on ferric nitrilotriacetate (Fe-NTA)-induced oxidative damage in Wistar rats". J. Appl. Phycol. 23, Issue 2, Page 257 (2): 257–263.doi:10.1007/s10811-010-9564-0.
8. Jump up^ Mukherjee, P. K., Marcheselli, V. L., Serhan, C. N., & Bazan, N. G. (2004). Neuroprotecin D1: A docosahexanoic acid-derived docosatriene protects human retinal pigment epithelial cells from oxidative stress. Proceedings of the National Academy of Sciences of the USA, 101(22), 8491–8496.doi:10.1073/pnas.0402531101
9. Jump up^ http://www.ncbi.nlm.nih.gov/pubmed/10224662
10. Jump up^ Floyd, R. A., (1999). Neuroinflammatory processes are important in neurodegenerative diseases: An hypothesis to explain the increased formation of reactive oxygen and nitrogen species as major factors involved in neurodegenerative disease development. Free Radical Biology and Medicine, 26(9–10), 1346–1355. doi:10.1016/S0891-5849(98)002937
11. Jump up^ An overview of the role of free radicals in biology and of the use of electron spin resonance in their detection may be found in Rhodes C.J. (2000). Toxicology of the Human Environment – the critical role of free radicals. London: Taylor and Francis.ISBN 0-7484-0916-5.
12. Jump up^ Serpone N, Salinaro A, Emeline AV, Horikoshi S, Hidaka H, Zhao JC. 2002. An in vitro systematic spectroscopic examination of the photostabilities of a random set of commercial sunscreen lotions and their chemical UVB/UVA active agents. Photochemical & Photobiological Sciences 1(12): 970-981.
13. Jump up^ G. Herzberg (1971), "The spectra and structures of simple free radicals", ISBN 0-486-65821-X.
14. Jump up^ 28th International Symposium on Free Radicals.
15. Jump up^ Betigeri, Seema; Thakur, Ajit; Raghavan, Krishnaswamy (2005). "Use of 2,2?-Azobis(2-Amidinopropane) Dihydrochloride as a Reagent Tool for Evaluation of Oxidative Stability of Drugs".Pharmaceutical Research 22 (2): 310–7. doi:10.1007/s11095-004-1199-x. PMID 15783080.
16. Jump up^ Werber, Jay; Wang, Y. John; Milligan, Michael; Li, Xiaohua; Ji, Junyan A. (2011). "Analysis of 2,2′-azobis (2-amidinopropane) dihydrochloride degradation and hydrolysis in aqueous solutions". Journal of Pharmaceutical Sciences 100(8): 3307–15. doi:10.1002/jps.22578. PMID 21560126. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Fundamentals/Reactive_Intermediates/Free_Radicals.txt |
A resonance form is another way of drawing a Lewis dot structure for a given compound. Equivalent Lewis structures are called resonance forms. They are used when there is more than one way to place double bonds and lone pairs on atoms. Resonance structures arise when there are more than one way to draw a Lewis dot diagram that satisfies the octet rule. Remember the octet rule is where the atom gains, loses, or shares electrons so that the outer electron shell has eight electrons. We draw them when one structure does not accurately show the real structure.
Introduction
There are some basic principle on the resonance theory. First resonance structures are not real, they just show possible structures for a compound. Resonance structures are not in equilibrium with each other. Resonance structures are not isomers. Isomers have different arrangement of both atoms and electrons. Resonance forms differ only in arrangement of electrons.
Resonance structures are a better depiction of a Lewis dot structure because they clearly show bonding in molecules. Not all resonance structures are equal there are some that are better than others. The better ones have minimal formal charges, negative formal charges are the most electronegative atoms, and bond is maximized in the structure. The more resonance forms a molecule has makes the molecule more stable. They are drawn with a double-headed arrow between them to show the actual structure is somewhere between the resonance structures. These structures used curved arrow notation to show the movement of the electrons in one resonance form to the next.
Formal charges are used in Chemistry to determine the location of a charge in a molecule and determine how good of a Lewis structure it will be. Remember, the best resonance structure is the one with the least formal charge. This is why formal charges are very important. Atoms that are missing one or more electrons will have a positive charge. An atom with many electrons will have a negative charge. Assigning formal charges to an atom is very useful in resonance forms.
Formal charge is calculated using this format:
# of valence electrons- (#non bonding electrons + 1/2 #bonding electrons)
Curved arrow notation is used in showing the placement of electrons between atoms. The tail of the arrow begins at the electron source and the head points to where the electron will be. Make sure the arrows are clear including the single and half headed arrow. The reader must know the flow of the electrons.
Common Examples
Benzene is commonly seen in Organic Chemistry and it has a resonance form. Benzene has two resonance structures, showing the placements of the bonds.
Another example of resonance is ozone. Ozone is represented by two different Lewis structures. The difference between the two structures is the location of double bond.
Drawing Resonance Forms
There are several things that should be checked before and after drawing the resonance forms. First know where the nonbonding electrons are, keep track of formal charges on atoms, and do not break sigma bonds. Finally, after drawing the resonance form make sure all the atoms have eight electrons in the outer shell. Checking these will make drawing resonance forms easier.
When drawing a resonance structure there are three rules that need to be followed for the structures to be correct:
1. Only electrons move and the nuclei of the atoms never move.
2. Only electrons that can move are pi electrons, single unpaired electrons, and lone pair electrons.
3. The total number of electrons in the molecule do not change and neither do the number of paired and unpaired electrons.
Approaches for moving electrons are move pi electrons toward a positive charge or toward an another pi bond. Move a single nonbonding electron towards a pi bond. Move lone pair electrons toward a pi bond and when electrons can be moved in more than one direction, move them to the more electronegative atom.
Helpful hints
1. Two resonance structures differ in the position of multiple bonds and non bonding electron. The placement of atoms and single bonds always stays the same.
2. They must make sense and agree to the rules. Hydrogens must have two electrons and elements in the second row cannot have more than 8 electrons. If so, the resonance structure is not valid. Always look at the placement of arrows to make sure they agree.
3. Electrons move toward a sp2 hybridized atom. The sp2 hybridized atom is either a double-bonded carbon, or a carbon with a positive charge, or it is an unpaired electron. Electrons do not move toward a sp3 hybridized carbon because there is no room for the electrons.
After drawing resonance structures check the net charge of all the structures. For example, if a structure has a net charge of +1 then all other structures must also have a net charge of +1. If not, the structure is not correct. Always check the net charge after each structure. These important details can ensure success in drawing any Resonance structure.
Problems
1)
2)
Fill in any lone pair electrons and identify any pi bond electrons.
3)
Fill in any lone pair electrons and identify any pi bond electrons.
4)
5) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Fundamentals/Resonance_Forms.txt |
Rotation in substituted ethanes is an example of general rotation about single, sigma, carbon-carbon bonds. The different conformations resulting from these single bond rotations are energetically very similar, making such rotations both common and important in understanding how alkanes act. Often a particular conformation of the molecule is required for a reaction to proceed, and can explain the product stereochemistry, so understanding how to visualize the different forms and how they interconvert is essential to understanding complicated reaction mechanisms.
Introduction
While the carbon-carbon single bonds of ethane must adhere to certain physical limitations (both carbons are sp3 hybridized, all bond angles are approximately 109.5°, etc), these criteria still allow a lot movement within the molecule. With a small amount of activation energy, substituents are free to rotate around the bond, like the blades on a fan rotate about the hub.
Figure 1: Internal configurations of the groups in ethane. Image used withpermission from Wikipedia: [email protected]
The most common and simplest way of depicting rotations about substituted ethanes is a Newman Projection. This simplifies the molecule into a 2-D sketch that can readily show the differences between conformations. (A conformation is a particular arrangement of the substituents of one carbon in a bond, with respect to the substituents of the other carbon in the bond)
There are two different classifications for the conformations:
Staggered - the substituents, as drawn in the Newman Projection, all have 60 degrees between them. These conformers are always the lowest points in relative energy diagrams.
• anti conformations are the lowest possible energy conformations; all substituents have a maximum amount of free space.
• gauche conformations are higher energy than anti conformations, but lower energy than any eclipsed forms. This is due to proximity of the substituents. In the gauche conformation, unlike any eclipsed forms, the substituents are not directly aligned. However, they are close enough to influence each other and raise the energy of the molecule (see attached "Energy Diagram" for relative energies of Newman Projections of the C3-C4 bond in 3,4-dimethylhexane).
Eclipsed - the substituents are overlapping, when viewed as in the Newman projection. With a model kit, this is the form where all the substituents are aligned and the ones on the carbon farthest from your eyes are hidden behind the substituents of the foremost carbon. Depending on which groups are overlapping, eclipsed forms can have varying energies. They do not have specific names, like anti vs gauche. The eclipsed conformer with the largest group on each carbon overlapping with each other will always be the highest energy conformer of that bond.
To convert between eclipsed and staggered conformations (or vice versa), one carbon, with all of its substituents, is rotated 60 degrees. You can imagine the projection like a combination lock. The back carbon is like the lock itself and the front carbon like the knob you turn to enter your combination. "grasp" the front carbon, and turn in 60 degrees. The placement of the back substituents will not change. (In reality, all the substituents are rotating simultaneously.) Repeating this process 6 times will produce all staggered and eclipsed conformations of the bond in question, and bring you back to the conformation you began with.
Any conformations between eclipsed and staggered are called skew conformations, but as there are an infinite number of these for each bond, (depending on how much or little the bond is rotated) skew conformations are rarely represented. (See attached "Newman Projections" to view Newman Projections of 3,4-dimethylhexane)
Unsubstituted Ethane: Only has two conformers! Because all the substituents are exactly the same (all hydrogens), there is only one eclipsed and one anti conformation, each. The eclipsed formation is 3 kcal/mol higher in energy - 1 kcal/mol for each eclipsed hydrogen.
Substituted Ethane: Substituted ethanes have slightly greater differences in energy between conformations because of steric hindrance. Essentially, the larger groups bump up against each other during the transition state, so it takes more energy to move them around eachother. The large substituents are like beach balls that knock into each other while moving around, as opposed to hydrogens, which are smaller, and more compact (imagine trying the move two volleyballs past each other in a 16 inch box, versus two golf balls in the same volume). This hindrance raises the energy of all conformations of the molecule (when compared to unsubstituted ethane, or less-substituted bonds). (See bottom of attached "Newman Projections" for Ethane Conformations)
Problems
1. Doing these problems with a model kit, even if it seems absurd to do so, is invaluable to understanding and visualizing what you are drawing. These are generally easiest to convert from molecule to projection if you begin with one of the staggered conformations and work your way through 360 degrees of rotation.
2. Draw and label all Newman projections of propane. Are any of these conformations equal energy?
3. Draw and label all Newman Projections of the C2-C3 bond in 3-ethyl-3-methylbutane. Which is the highest energy conformation? The lowest?
4. Make up your own molecule(s)! Draw all the Newman Projections and predict their relative energies.
Contributors
• Megan Maurano (UCD) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Fundamentals/Rotation_in_Substituted_Ethanes.txt |
Solutions with Liquid Solvents
A solution is a homogeneous mixture in which the solute particles are so small that they cannot be filtered out and they do not scatter light. For a solution to form, there needs to be the correct combination of enthalpic (intermolecular forces) and entropic effects to enable the solvent and solute particles to interact strongly with each other without decreasing the entropy of the system. In this section we will closely examine the effects of enthalpy, but we will not spend too much time discussing the effects of entropy. To understand the effects of intermolecular forces, we need to go all the way back to bonds.
How Does Bond Polarity Affect/Determine Bond Type?
A bond is defined as the sharing of electrons by two atoms. Bond polarity occurs when two atoms share electrons unequally because of a difference in the strength of attraction for the shared electrons by the nuclei of the two different atoms. The strength of attraction scale that is most commonly used is electronegativity (EN). There are three common categories of bonds, based on the difference in electronegativity (∆EN) values of the two bonding atoms:
1. If two atoms have identical, or similar EN values, their nuclei both pull on the shared electrons nearly equally. This equal sharing is called a nonpolar covalent bond – neither nucleus has a greater electron density after forming the bond than it did before forming the bond. Examples: H-H, F-F
2. If two atoms have a modest difference in EN values, the nucleus of the atom with the higher EN value has a greater pull on the shared electrons, so that these electrons are more likely to be found closer to that nucleus than the nucleus with the lower EN value. This unequal sharing is called a polar covalent bond – there is a partial negative charge around the nucleus of the atom with the higher EN value because it is “using/borrowing” the electron(s) of the atom with the lower EN value. This leaves a partial positive charge around the nucleus of the atom with the lower EN value because its electron(s) are less likely to be found near it. Examples: H-Cl, C-F
3. If two atoms have a large difference in EN values, the shared electrons are much more likely to be found closer to the nucleus of the atom with the higher EN value. There is always some small likelihood of finding the shared electron(s) near the nucleus of the atom with the smaller EN value, so there are never really separate, fully-charged ions in a sample of a pure substance. Nevertheless, it is common practice to designate these bonds as ionic bonds. This practice is reasonable because of the formation of separate, fully-charged ions when compounds involving these extremely polar bonds are placed in water. (See ion-dipole IMFs below.) Examples: Na-Cl, K-F
There are many approaches to the task of determining into which category a given bond should be placed. Some methods are more rigorous than others, but often the rigorous methods are too complicated to be used quickly and simply. You should find a method that you like and that serves your needs sufficiently and stick with it until you are asked to use a more rigorous method. The simplest method is that a bond between a metal and a nonmetal is ionic, whereas a bond between two nonmetals is covalent.
There are two common conventions used to show the polarity of a bond:
1. lower case deltas + and δ-): These two symbols must always be used together, because they show the partial positive pole and the partial negative pole of the polar bond. Example:
δ+ δ-
B – Cl
1. a dipole arrow : This symbol is used to show the positive end of the dipole and the negative end at which the electron density is greatest. Example:
B – Cl
What is Molecular Polarity?
The concept of molecular polarity is most easily applied to small molecules, but it can be applied to larger molecules with the understanding that larger molecules may have multiple “identities.” The first task in determining molecular polarity is to identify whether a substance is comprised of ions or of molecules. Individual small ions are not polar, because they do not have a charge separation that leads to the existence of a positive pole and a negative pole. Thus, if you identify a compound as consisting of ions (either monatomic or small polyatomic), you are no longer concerned about molecular polarity of that substance. To determine the molecular polarity of a compound that contains only covalent bonds, you need to consider three aspects of the molecule:
1. the shape of the molecule
2. the polarity of all of the bonds in the molecule
3. the presence of atoms with lone pairs.
Taking these three aspects into account, you should be able to determine if a small molecule is polar or nonpolar. Some helpful shortcuts that are almost always true:
a) If the central atom(s) in a molecule has at least one lone electron pair, the molecule will be polar.
NH3 is polar, and so is O3
b) If the molecule contains only nonpolar bonds and there are no lone pairs on the central atom(s), the molecule will be nonpolar.
All CxHy molecules are nonpolar
c) If all of the atoms attached to the central atom are of the same element and there are no lone pairs on the central atom, the molecule is nonpolar.
CCl4 and CO2 are nonpolar
d) If the atoms attached to the central atom are of different elements, the molecule is polar.
CH2Cl2 and OCS are polar
e) If a molecule of a compound contains at least one N, O, F, or Cl atom, the molecule is usually polar, unless c) is true.
CH3CH2OH, HF, CH3NH2, and CH3Cl are all polar
f) When molecules get large enough, they might have polar sections and nonpolar sections. A general rule of thumb for carbon-containing compounds is that molecules with 4/5 or fewer C atoms per polar bond can be considered to act as if they are polar. Thus, molecules with 4/5 or more C atoms per polar bond can be considered to act as if they were nonpolar.
CH3CH2OH is considered polar; CH3CH2CH2CH2CH2CH2CH2OH is considered nonpolar; CH3CH2CH2CH2OH is right on the polar/nonpolar divide
g) All single atoms are nonpolar. Most, but not all, molecules of elements are nonpolar.
Polar molecules have a dipole moment greater than zero, and thus are considered dipoles.
What is an Intermolecular Force?
If you have a collection of independent atoms, molecules, or ions, these particles will always exert forces of attraction and repulsion among themselves as they more closely approach each other. The repulsive forces tend to be ignored, but they do exist and come about mainly because of the repulsion of the negatively-charged electron clouds surrounding each particle (even for cations, although the positive charges do lead to repulsion with other cations and the nucleus of other atoms.)
A great deal of time is spent studying the attractive forces among independent particles. These forces are what enable molecules and atoms to gather together and remain in a condensed phase as liquids and solids. Collectively, these attractive forces are commonly known as intermolecular forces (IMFs). They are the forces of electrostatic attraction that exist between separate particles.
How is an IMF different from a bond?
Along with the common definition that bonds occur within molecules and IMFs occur between molecules, there are two closely related factors that distinguish bonds and IMFs:
1. The atomic nuclei are closer together in a bond between two atoms than in an IMF between two particles.
2. The potential energy released when a bond forms is greater than the potential energy released when an IMF forms.
These statements are generalizations, and there are certainly exceptions, but they hold true for most common substances.
What are the Types of IMFs?
The simplest way to categorize IMFs is to classify the particles that are interacting as one of the following:
1. ions
2. polar molecules
3. nonpolar molecules
From these three particle types, you can describe five different types of interparticle interactions:
1. dipole/dipole forces
2. ion/dipole forces
3. dipole/induced dipole forces.
4. ion/induced dipole forces
5. dispersion or London forces
(All particles exert and experience London forces, but the only forces among nonpolar particles are London forces)
The only remaining interaction would be ions interacting with ions, but that is an ionic bond. There is, however, a sixth IMF. It is a case of extreme dipole/dipole interactions:
1. molecules with very polar bonds interacting with molecules with very polar bonds – known as hydrogen bonding (but it is NOT a bond!)
The only bonds that are polar enough to result in a large enough partial positive charge on a small atom are:
δ- δ+ δ- δ+ δ- δ+
F—H, O—H, and N—H bonds.
The only atoms that have a large enough EN value to result in a large enough partial negative charge on a small atom are:
δ- δ- δ-
F, O, and N
What Substances Will Mix with Each Other to Create a Solution?
This discussion ignores entropy and focuses on enthalpy, but entropy definitely plays a role. That discussion is for later.
A solution is a homogeneous mixture in which the solute particles are so small that they cannot be filtered out and so dispersed throughout the mixture that they do not scatter a beam of light that travels through the mixture. The probability of one substance dissolving/mixing in/with another substance to form a solution depends on whether or not the newly formed IMFs are as energetically favorable as the currently existing IMFs. Thus, the mutual solubility of two substances depends mainly on the substance with the IMFs that involve the largest charge separation. In other words, there is a hierarchy of what will mix:
a) Ionically-bonded solids will form a solution (dissolve) in a solvent only if the solvent can provide very strong ion/dipole interactions.
- Usually water is the only material that can dissolve inorganic salts to any great extent, although some inorganic salts dissolve slightly in methanol and DMSO. You can find tables of solubility rules for common inorganic salts in water. One of these rules is that almost all compounds containing Group IA cations are soluble.
- Organic salts have a reasonable solubility in many polar organic solvents, but are generally much more soluble in water than in these organic solvents. The non-organic counter cations are usually Na+, K+, or NH4+, all of which generally form soluble salts. The non-organic counter anions are usually Cl-, NO3-, and SO42-, all of which generally form soluble salts.
b) Water will form a solution (mix or dissolve) with another covalently-bonded substance only if:
i) the molecules of that substance can form hydrogen bonds with the water molecules.
-The molecules of the other substance do not need to be able to form hydrogen bonds among themselves when in the pure form, but they must be polar molecules and contain an N, O, or F atom that has a partial negative charge. Acetone is an example of such a substance.
The ability of a molecule to form hydrogen bonds with water molecules does not guarantee that the molecule will dissolve in water. If there is a portion of the molecule that is nonpolar, this nonpolar section interferes with the water molecules’ ability to form hydrogen bonds among themselves. Thus, the larger the nonpolar section, the less soluble the substance is in water. A general rule of thumb is the 4-5 carbon atom rule: Molecules with fewer than 4-5 C atoms per hydrogen bonding group will likely be soluble in water. Molecules with more than 4-5 C atoms per hydrogen bonding group will likely be insoluble in water. The water solubility of molecules with 4-5 C atoms per hydrogen bonding group is best determined by experiment or by a search of the literature. For instance, the following compounds, all of which are polar and can form hydrogen bonds with water, contain four C atoms per molecule: butanoic acid is soluble with water in all ratios; 1-butanol is reasonably soluble in water; and diethyl ether is so slightly soluble in water that it is considered insoluble.
or
ii) the molecules of that substance react with water molecules to form ions, which then dissolve in the water because of ion/dipole interactions.
- The six strong acids (HCl, HBr, HI, HNO3, HClO4, and H2SO4) are the most common examples. All of these acids have Ka values greater than 10. (Remember, the Ka value tells you how likely it is for the acid to react with water, and a Ka greater than 1 means that the reaction is product-favored at equilibrium)
All molecular weak acids and molecular weak bases react with water to some extent to create ions, but these weak acids all have Ka values that are less than 1 and these weak bases all have Kb values that are less than 1. Thus, all these reactions are considered reactant-favored, and for purposes of predicting water solubility, we ignore the fact that some ions are formed. Remember that the intact weak acid or weak base molecules may themselves dissolve in water through hydrogen bonding. (See section b) (i) above.
c) The common adage “Like dissolves like” works well for most organic compounds. Polar compounds tend to mix well with other polar compounds. The IMFs involved can be hydrogen boding, dipole/dipole, and London forces. Nonpolar compounds tend to mix well with other nonpolar compounds, with London forces being the only IMF involved.
- With a few notable exceptions (CCl4, for instance), organic compounds containing chlorine and bromine and iodine do have a dipole moment, and are therefore polar molecules. These molecules cannot, however, form hydrogen bonds with water, nor are the halogen atoms present as ions. Thus, halogenated organic compounds do not dissolve well in water.
- Because most organic compounds have at least one small nonpolar section on their molecules, some polar organic compounds mix well with nonpolar organic compounds.
d) Solubility in aqueous acid solutions and/or aqueous base solutions is a confusing concept, mainly because of the way that chemists talk about the process. The statement, “Check the solubility of benzoic acid in 3 M NaOH” may be clear to a practicing chemist, but it is misleading to most beginning students. The statement would be much clearer if it were written, “Check the water solubility of the organic anion formed when benzoic acid reacts with the OH- ion in an aqueous 3M NaOH solution.” The same clarity, or lack thereof, is true for statements about solubility in aqueous acid solutions.
The order of events when an aqueous solution of a base is added to an organic compound that can act as an acid is:
i) The organic compound reacts with the OH- ion, donating an H+ ion to the OH- ion.
ii) The products of this reaction are water and the conjugate base of the organic compound. If the original organic compound was a neutral molecule, the conjugate base will be a water-soluble anion.
The order of events when an aqueous solution of an acid is added to an organic compound that can act as a base is:
i) The organic compound reacts with the acid, accepting an H+ ion.
ii) The products of this reaction are the conjugate base of the acid, and the conjugate acid of the organic compound. If the original organic compound was a neutral molecule, the conjugate acid will be a water-soluble cation.
If no acid/base reaction occurs between the aqueous acid/base and the organic compound, then the organic compound remains as a neutral substance, and maintains its original solubility (or lack thereof ) in water.
If knowing the solubility of two compounds is an essential piece of information for your work, it is always best to not only look up the solubility data, but also perform careful solubility tests yourself in the lab. | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Fundamentals/Solubility_-_What_dissolves_in_What%3F.txt |
Here you will learn how to understand, write, draw, and talk-the-talk of organic molecules. Why were different drawing techniques developed? Organic molecules can get complicated and large. It is a tedious to have to constantly draw out every detail, especially when not necessary, so the o-chemist of the past developed these techniques to make it more convenient and easy. In addition, some of these shorthand ways of drawing molecules give us insight into the bond angles, relative positions of atoms in the molecule, and some eliminate the numerous hydrogens that can get in the way of looking at the backbone of the structure.
Introduction
Observe the following drawings of the structure of Retinol, the most common form of vitamin A. The first drawing follows the straight-line (a.k.a. Kekulé) structure which is helpful when you want to look at every single atom; however, showing all of the hydrogen atoms makes it difficult to compare the overall structure with other similar molecules and makes it difficult to focus in on the double bonds and OH group.
Retinol: Kekulé straight-line drawing
The following is a bond-line (a.k.a. zig-zag) formula for retinol. With this simiplified representation, one can easily see the carbon-carbon bonds, double bonds, OH group, and CH3 groups sticking off of the the main ring and chain. Also, it is much quicker to draw this than the one above. You will learn to appreciate this type of formula writing after drawing a countless number of organic molecules.
Retinol: Bond-line or zig-zag formula
Importance of Structure
Learning and practicing the basics of Organic Chemistry will help you immensely in the long run as you learn new concepts and reactions. Some people say that Organic Chemistry is like another language, and in some aspects, it is. At first it may seem difficult or overwhelming, but the more you practice looking at and drawing organic molecules, the more familiar you will become with the structures and formulas. Another good idea is to get a model kit and physically make the molecules that you have trouble picturing in your head.
Through general chemistry, you may have already experienced looking at molecular structure. The different ways to draw organic molecules include Kekulé (straight-line), Condensed Formulas, and Bond-Line Formulas (zig-zag). It will be more helpful if you become comfortable going from one style of drawing to another, and look at drawings and understanding what they mean, than knowing which kind of drawing is named what.
An example of a drawing that incorporates all three ways to draw organic molecules would be the following additional drawing of Retinol. The majority of the drawing is Bond-line (zig-zag) formula, but the -CH3 are written as condensed formulas, and the -OH group is written in Kekulé form.
A widely used way of showing the 3D structure of molecules is the use of dashes, wedges, and straight lines. This drawing method is essential because the placement of different atoms could yield different molecules even if the molecular formulas were exactly the same. Below are two drawings of a 4-carbon molecule with two chlorines and two bromines attached.
4-carbon molecule with 2 chlorines and 2 bromines 4-carbon molecule with 2 chlorines and 2 bromines
Both drawings look like they represent the same molecule; however, if we add dashes and wedged we will see that two different molecules could be depicted:
The two molecules above are different, prove this to yourself by building a model. An easier way to compare the two molecules is to rotate one of the bonds (here, it is the bond on the right):
Notice how the molecule on the right has both bromines on the same side and chlorines on the same side, whereas the first molecule is different. Read about Dashed-Wedged Line structures, bottom of page, to understand what has been introduced above. You will learn more about the importance of atomic connectivity in molecules as you continue on to learn about Stereochemistry.
Drawing the Structure of Organic Molecules
Although larger molecules may look complicated, they can be easily understood by breaking them down and looking at their smaller components.
All atoms want to have their valence shell full, a "closed shell." Hydrogen wants to have 2 e- whereas carbon, oxygen, and nitrogen want to have 8 e-. When looking at the different representations of molecules, keep in mind the Octet Rule. Also remember that hydrogen can bond one time, oxygen can bond up to two times, nitrogen can bond up to three times, and carbon can bond up to four times.
Kekulé (a.k.a. Straight-Line Structures)
Kekulé structures are similar to Lewis Structures, but instead of covalent bonds being represented by electron dots, the two shared electrons are shown by a line.
(A) (B)(C)
Lone pairs remain as two electron dots, or are sometimes left out even though they are still there. Notice how the three lone pairs of electrons were not draw in around chlorine in example B.
Condensed Formulas
A condensed formula is made up of the elemental symbols. The order of the atoms suggests the connectivity. Condensed formulas can be read from either direction and H3C is the same as CH3, although the latter is more common because Look at the examples below and match them with their identical molecule under Kekulé structures and bond-line formulas.
(A) CH3CH2OH (B) ClCH2CH2CH(OCH3)CH3 (C) H3CNHCH2COOH
Let's look closely at example B. As you go through a condensed formula, you want to focus on the carbons and other elements that aren't hydrogen. The hydrogen's are important, but are usually there to complete octets. Also, notice the -OCH3 is in written in parentheses which tell you that it not part of the main chain of carbons. As you read through a a condensed formula, if you reach an atom that doesn't have a complete octet by the time you reach the next hydrogen, then it's possible that there are double or triple bonds. In example C, the carbon is double bonded to oxygen and single bonded to another oxygen. Notice how COOH means C(=O)-O-H instead of CH3-C-O-O-H because carbon does not have a complete octet and oxygens.
Bond-Line (a.k.a. zig-zag) Formulas
The name gives away how this formula works. This formula is full of bonds and lines, and because of the typical (more stable) bonds that atoms tend to make in molecules, they often end up looking like zig-zag lines. If you work with a molecular model kit you will find it difficult to make stick straight molecules (unless they contain sp triple bonds) whereas zig-zag molecules and bonds are much more feasible.
(A) (B) (C)
These molecules correspond to the exact same molecules depicted for Kekulé structures and condensed formulas. Notice how the carbons are no longer drawn in and are replaced by the ends and bends of a lines. In addition, the hydrogens have been omitted, but could be easily drawn in (see practice problems). Although we do not usually draw in the H's that are bonded to carbon, we do draw them in if they are connected to other atoms besides carbon (example is the OH group above in example A) . This is done because it is not always clear if the non-carbon atom is surrounded by lone pairs or hydrogens. Also in example A, notice how the OH is drawn with a bond to the second carbon, but it does not mean that there is a third carbon at the end of that bond/ line.
Dashed-Wedged Line Structure
As you may have guessed, the Dashed-Wedged Line structure is all about lines, dashes, and wedges. At first it may seem confusing, but with practice, understanding dash-wedged line structures will become like second nature. The following are examples of each, and how they can be used together.
Above are 4-carbon chains with attached OH groups or Cl and Br atoms. Remember that each line represents a bond and that the carbons and hydrogens have been omitted. When you look at or draw these structures, the straight lines illustrate atoms and bonds that are in the same plane, the plane of the paper (in this case, computer screen). Dashed lines show atoms and bonds that go into the page, behind the plane, away from you. In the above example, the OH group is going into the plane, while at the same time a hydrogen comes out (wedged).
Blue bead= OH group; White bead=H
Wedged lines illustrate bonds and atoms that come out of the page, in front of the plane, toward you. In the 2D diagram above, the OH group is coming out of the plane of the paper, while a hydrogen goes in (dashed).
Blue bead= OH group; White bead=H
As stated before, straight lines illustrate atoms and bonds that are in the same plane as the paper, but in the 2D example, the straight line bond for OH means that it it unsure or irrelevant whether OH is going away or toward you. It is also assumed that hydrogen is also connected to the same carbon that OH is on.
Blue bead= OH group; H is not shown
Try using your model kit to see that the OH group cannot lie in the same plane at the carbon chain (don't forget your hydrogens!). In the final 2Dexample, both dashed and wedged lines are used because the attached atoms are not hydrogens (although dashed and wedged lines can be used for hydrogens).The chlorine is coming out the page while bromine is going into the page.
Blue bead=Cl; Red bead=Br
Practice Problems
1. How many carbons are in the following drawing? How many hydrogens?
2. How many carbons are in the following drawing? How many hydrogens?
3. How many carbons are in the following drawing? How many hydrogens?
4. Look at the following molecule of vitamin A and draw in the hidden hydrogens and electron pairs.
(hint: Do all of the carbons have 4 bonds? Do all the oxygens have a full octet?)
5. How many bonds can hydrogen make?
6. How many bonds can chlorine make?
7. Dashed lines means the atomic bond goes ___________(away/toward) you.
8. Draw ClCH2CH2CH(OCH3)CH3 in Kekuléand zig-zag form.
9. Extra practice problems can be found ______?
Answers
1. Remember the octet rule and how many times carbons and hydrogens are able to bond to other atoms.
2. Electron pairs drawn in blue and hydrogens draw in red.
3. Hygrogen can make one bond.
4. Chlorine can make one bond.
5. Away
6. See (B) under Kekulé and Bond-line (zig-zag) formulas.
7. Extra practice problems can be found: in your textbook, homework, lecture notes, online, reference books, and more. Try making up some of your own molecules, they may exist!
Contributors
• Choo, Ezen (2009, UCD '11)
Structure of Organic Molecules
Several organic compounds may have identical compositions but will have widely different physical and chemical properties because the arrangement of the atoms is different. Isomers and identical compounds both have the same number of each kind of element in a formula. A simple count will establish this fact.
Introduction
As a consequence of the double bond, some alkene compounds exhibit a unique type of isomerism. Rotation around a single bond occurs readily, while rotation around a double bond is restricted. The pi bond prevents rotation because of the electron overlap both above and below the plane of the atoms.
A single bond is analogous to two boards nailed together with one nail. A double bond is analogous to two boards nailed together with two nails. In the first case you can twist the boards, while in the second case you cannot twist them.
Geometric Isomers are compounds with different spatial arrangements of groups attached to the carbons of a double bond. In alkenes, the carbon-carbon double bond is rigidly fixed. Even though the attachment of atoms is the same, the geometry (the way the atoms "see" each other) is different.
When looking for geometric isomers, a guiding principle is that there MUST BE TWO DIFFERENT "GROUPS" ON EACH CARBON OF THE DOUBLE BOND. A "group" can be hydrogen, alkyls, halogens, etc.
Identical compounds may appear to have different arrangements as written, but closer examination by rotation or turning will result in the molecules being superimposed. If they are super impossible or if they have identical names, then the two compounds are in fact identical.
Isomers of compounds have a different arrangement of the atoms. Isomer compounds will differ from identical compounds by the arrangement of the atoms. See example below.
Both compounds have the same number of atoms, C5H12. They are isomers because in the left molecule the root is 4 carbons with one branch. In the right molecule, the root is 3 carbons with 2 branches. They are isomers because they have the same number of atoms but different arrangements of those atoms.
Completely different compounds: If the number of each element is different, the two compounds are merely completely different. A simple count of the atoms will reveal them as different.
1,2-dichlorethene
In the example on the left, the chlorine atoms can be opposite or across from each other in which case it is called the "trans" isomer. If the the chlorine atoms are next to or adjacent each other, the isomer is called " cis".
If one carbon of the double bond has two identical groups such as 2 H's or 2 Cl's or 2 CH3 etc. there cannot be any geometric isomers.
2-butene
Consider the longest chain containing the double bond: If two groups (attached to the carbons of the double bond) are on the same side of the double bond, the isomer is a cis alkene. If the two groups lie on opposite sides of the double bond, the isomer is a trans alkene. One or more of the "groups" may or may not be part of the longest chain. In the case on the left, the "group" is a methyl - but is actually part of the longest chain.
A common mistake is to name this compound as 1,2-dimethylethene. Look at all carbons for the longest continuous chain - the root is 4 carbons - butene.
Problems
For the structures below:
a. Draw both cis/trans isomers, if any, of the structure based upon the name.
b. Look at the graphic and state whether the compound is cis, trans, or not cis/trans isomers.
• 2-methyl-2-butene
• 3-methyl-2-pentene
• 1-pentene
Contributors
• Charles Ophardt, Professor Emeritus, Elmhurst College; Virtual Chembook | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Fundamentals/Structure_of_Organic_Molecules/Cis_and_Trans_Isomers_of_Alkanes.txt |
The traditional system for naming the geometric isomers of an alkene, in which the same groups are arranged differently, is to name them as cis or trans. However, it is easy to find examples where the cis-trans system is not easily applied. IUPAC has a more complete system for naming alkene isomers. The R-S system is based on a set of "priority rules", which allow you to rank any groups. The rigorous IUPAC system for naming alkene isomers, called the E-Z system, is based on the same priority rules.These priority rules are often called the Cahn-Ingold-Prelog (CIP) rules, after the chemists who developed the system
The general strategy of the E-Z system is to analyze the two groups at each end of the double bond. At each end, rank the two groups, using the CIP priority rules, discussed in Ch 15. Then, see whether the higher priority group at one end of the double bond and the higher priority group at the other end of the double bond are on the same side (Z, from German zusammen = together) or on opposite sides (E, from German entgegen = opposite) of the double bond.
Example 1: Butene
The Figure below shows the two isomers of 2-butene. You should recognize them as cis and trans. Let's analyze them to see whether they are E or Z.
Start with the left hand structure (the cis isomer). On C2 (the left end of the double bond), the two atoms attached to the double bond are C and H. By the CIP priority rules, C is higher priority than H (higher atomic number). Now look at C3 (the right end of the double bond). Similarly, the atoms are C and H, with C being higher priority. We see that the higher priority group is "down" at C2 and "down" at C3. Since the two priority groups are both on the same side of the double bond ("down", in this case), they are zusammen = together. Therefore, this is (Z)-2-butene.
Now look at the right hand structure (the trans isomer). In this case, the priority group is "down" on the left end of the double bond and "up" on the right end of the double bond. Since the two priority groups are on opposite sides of the double bond, they are entgegen = opposite. Therefore, this is (E)-2-butene.
E,Z will always work, even when cis,trans fails
In simple cases, such as 2-butene, Z corresponds to cis and E to trans. However, that is not a rule. This section and the following one illustrate some idiosyncrasies that happen when you try to compare the two systems. The real advantage of the E-Z system is that it will always work. In contrast, the cis-trans system breaks down with many ambiguous cases.
Example 2
The following figure shows two isomers of an alkene with four different groups on the double bond, 1-bromo-2-chloro-2-fluoro-1-iodoethene.
It should be apparent that the two structures shown are distinct chemicals. However, it is impossible to name them as cis or trans. On the other hand, the E-Z system works fine... Consider the left hand structure. On C1 (the left end of the double bond), the two atoms attached to the double bond are Br and I. By the CIP priority rules, I is higher priority than Br (higher atomic number). Now look at C2. The atoms are Cl and F, with Cl being higher priority. We see that the higher priority group is "down" at C1 and "down" at C2. Since the two priority groups are both on the same side of the double bond ("down", in this case), they are zusammen = together. Therefore, this is the (Z) isomer. Similarly, the right hand structure is (E).
E,Z will work, but may not agree with cis,trans
Consider the molecule shown at the left.
This is 2-bromo-2-butene -- ignoring the geometric isomerism for now. Cis or trans? This molecule is clearly cis. The two methyl groups are on the same side. More rigorously, the "parent chain" is cis.
E or Z? There is a methyl at each end of the double bond. On the left, the methyl is the high priority group -- because the other group is -H. On the right, the methyl is the low priority group -- because the other group is -Br. That is, the high priority groups are -CH3 (left) and -Br (right). Thus the two priority groups are on opposite sides = entgegen = E.
Note
This example should convince you that cis and Z are not synonyms. Cis/trans and E,Z are determined by distinct criteria. There may seem to be a simple correspondence, but it is not a rule. Be sure to determine cis,trans or E,Z separately, as needed.
Multiple double bonds
If the compound contains more than one double bond, then each one is analyzed and declared to be E or Z.
Example 3
The configuration at the left hand double bond is E; at the right hand double bond it is Z. Thus this compound is (1E,4Z)-1,5-dichloro-1,4-hexadiene.
The double-bond rule in determining priorities
Example 4
Consider the compound below:
This is 1-chloro-2-ethyl-1,3-butadiene -- ignoring, for the moment, the geometric isomerism. There is no geometric isomerism at the second double bond, at 3-4, because it has 2 H at its far end.
What about the first double bond, at 1-2? On the left hand end, there is H and Cl; Cl is higher priority (by atomic number). On the right hand end, there is -CH2-CH3 (an ethyl group) and -CH=CH2 (a vinyl or ethenyl group). Both of these groups have C as the first atom, so we have a tie so far and must look further. What is attached to this first C? For the ethyl group, the first C is attached to C, H, and H. For the ethenyl group, the first C is attached to a C twice, so we count it twice; therefore that C is attached to C, C, H. CCH is higher than CHH; therefore, the ethenyl group is higher priority. Since the priority groups, Cl and ethenyl, are on the same side of the double bond, this is the Z-isomer; the compound is (Z)-1-chloro-2-ethyl-1,3-butadiene.
The "first point of difference" rule
Which is higher priority, by the CIP rules: a C with an O and 2 H attached to it or a C with three C? The first C has one atom of high priority but also two atoms of low priority. How do these "balance out"? Answering this requires a clear understanding of how the ranking is done. The simple answer is that the first point of difference is what matters; the O wins.
To illustrate this, consider the molecule at the left. Is the double bond here E or Z? At the left end of the double bond, Br > H. But the right end of the double bond requires a careful analysis.
At the right hand end, the first atom attached to the double bond is a C at each position. A tie, so we look at what is attached to this first C. For the upper C, it is CCC (since the triple bond counts three times). For the lower C, it is OHH -- listed in order from high priority atom to low. OHH is higher priority than CCC, because of the first atom in the list. That is, the O of the lower group beats the C of the upper group. In other words, the O is the highest priority atom of any in this comparison; thus the O "wins".
Therefore, the high priority groups are "up" on the left end (the -Br) and "down" on the right end (the -CH2-O-CH3). This means that the isomer shown is opposite = entgegen = E. And what is the name? The "name" feature of ChemSketch says it is (2E)-2-(1-bromoethylidene)pent-3-ynyl methyl ether.
Contributors
>Robert Bruner (http://bbruner.org) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Fundamentals/Structure_of_Organic_Molecules/The_E-Z_system_for_naming_alkenes.txt |
Your goal should be to understand, not memorize organic chemistry. The following 7 Golden Rules should be learned at the beginning of this semester. These simple ideas explain a very large number of things about the way organic molecules interact. Thus, understanding the 7 Golden Rules will allow you to develop an intuitive feel for organic chemistry, and things will make sense!
1. Atoms prefer filled valence shells: This rule explains why atoms make bonds, and the type of bonds created. A corollary is that centers of electron density (bonds and lone pairs of electrons) repel each other so they stay as far apart as possible. This latter rule, the basis for the so-called VSEPR model, explains 3-dimensional molecular structure.
2. The most important question in chemistry is "Where are the electrons?": The answer is that electrons are generally in higher amounts around the more electronegative atoms (e.g. F, Cl, O, N). The electronegative atoms pull electron density away from the less electronegative atoms (e.g. C, H) to which they are bonded. Thus, understanding electronegativities provides a simple method of deciding which portions of a molecule have a relatively high electron density, and which portions have a relatively low electron density.
3. Nature hates unpaired electrons: If a molecule must have an unpaired electron (a.k.a. radical), it is better to have the unpaired electron distributed over as many atoms as possible through resonance, inductive effects, and hyperconjugation.
4. Nature hates localized charges: If a molecule must have a charge, it is better to have the charge distributed over as many atoms as possible through resonance, inductive effects, and hyperconjugation. In addition, when given the choice, it is better to have more negative charge on a more electronegative atom (e.g. O), and more positive charge on a less electronegative atom (e.g. C).
5. Most reactions involve nucleophiles (molecules with a location of particularly high electron density) attacking electrophiles (molecules with a location of particularly low electron density). When in doubt, transfer a proton! Thus, simply understanding where electrons are provides you with the best way of analyzing new molecules so that you will be able to PREDICT how they will react.
6. Steric interactions (atoms bumping into each other) can prevent reactions by keeping the reactive atoms away from each other.
7. $\pi$ electrons prefer to be delocalized over as many adjacent sp2 hybridized atoms (or sp hybridized atoms in some cases) as possible, and aromaticity is the most stable form of $\pi$ electron delocalization. $\pi$ electrons cannot delocalize onto or through sp3 hybridized atoms since an sp3 atom has no 2p orbital available.
There are two Golden Subrules:
1. In mechanisms, proton transfers are generally the fastest possible reaction, so this usually happens before other possible processes such as nucleophilic attack: The exception is deprotonation of non-acidic carbon atoms such as in alkanes, these can be slow.
2. If a five or six-membered ring can be formed, intramolecular reaction will predominate if a molecule has two functional groups that can react with each other: Other rings can sometimes be formed, but when five or six-membered rings are not possible, intermolecular reactions become important competing reactions.
How to Think About Reactions
A good way to think about chemical reactions is that they are like crimes. Both crimes and chemical reactions need motive and opportunity to take place.
Motive
For reactions, the motive refers to the thermodynamic driving force. In other words, a reaction can be thought of as having a motive (thermodynamic driving force) if the products are more stable than the reactants. If the reaction does have a motive (thermodynamic driving force), it is said to be thermodynamically favorable and it will occur if given the opportunity. Reactions will have a favorable motive (thermodynamic driving force) if $\Delta{G}$ is negative ($\Delta{G} = \Delta{H} - T\Delta{S}$). This equation can be hard to apply to new situations, but the following rules of thumb can be helpful.
1. Reactions will usually have a motive (thermodynamic driving force) if stronger bonds are made than are broken in going from starting materials to products. This is primarily a $\Delta{H}$ effect and the reaction is enthalpically driven.
2. In reactions involving proton transfers, the reaction will generally have a motive (thermodynamic driving force) if the products represent the weaker acid and/or weaker base. Recall that equilibrium favors formation of the weaker acid/weaker base in an acid-base reaction. This is primarily a $\Delta{H}$ effect and the reaction is enthalpically driven.
3. Reactions will usually have a motive (thermodynamic driving force) if a greater number of smaller molecules are created from fewer larger molecules, especially if a small gaseous molecule such as $CO_2$, $N_2$ or $HCl$ is produced as a product. This is primarily a $\Delta{S}$ effect and the reaction is entropically driven.
Of course, the above rules of thumb also predict when reactions are not likely to have a favorable motive (thermodynamic driving force) as well. For example, reactions will usually not have a favorable motive (thermodynamic driving force) if weaker bonds are made than are broken in going from starting materials to products. This is primarily a $\Delta{H}$ effect.
Opportunity
Even if reactions have a motive (thermodynamic driving force), they can only occur if given the opportunity for the atoms and electrons to rearrange into the product. This rearrangement of atoms and electrons is what we refer to as the mechanism of the reaction. For a reaction to have an opportunity to react, the reaction cannot have an energy barrier that is too large. In other words, the mechanism cannot have any species (i.e. transition state) in it that is too high in energy (too unstable) to be formed at a given temperature. The Golden Rules of Chemistry are used to help predict the relative stabilities of proposed transition states. An obvious corollary to all of this is that reactions find the lowest energy opportunity (mechanism) to react out of all the possibilities, that is why reactions can usually be thought of as having a single mechanism. Thus, predicting mechanisms comes down to predicting the relative stabilities of potential transition states using the Golden Rules of Chemistry as a guide.* Great rule of thumb for most mechanisms: Each step involves a nucleophile attacking an electrophile, and when in doubt as to what to do, transfer a proton!
*The emphasis in this class is on qualitative thinking. Even though modern computers can usually calculate exact motives (thermodynamic driving forces) and exact transition state energies with a high degree of quantitative accuracy, that will not help you unless you have a suitable computer handy. The Golden Rules of Chemistry presented here are intended to give you the qualitative tools you need to think about chemistry without the aid of a computer calculation.
The Golden Rules of Organic Chemistry
http://iverson.cm.utexas.edu/courses...quivalents.pdf | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Fundamentals/The_Golden_Rules_of_Organic_Chemistry/Base_Equivalents_in_enolate_reactions.txt |
The Correct Use of Arrows to Indicate Electron Movement
The ability to write an organic reaction mechanism properly is key to success in organic chemistry classes. Organic chemists use a technique called arrow pushing to depict the flow or movement of electrons during chemical reactions. Arrow pushing helps chemists keep track of the way in which electrons and their associated atoms redistribute as bonds are made and broken. The first essential rule to keep in mind is the following:
First rule: Arrows are used to indicate movement of electrons
A regular arrow (double-sided arrowhead) is used to indicate the movement of two electrons, while a line with a single-sided arrowhead (sometimes called a “fish hook arrow”) is used for single electron movement involved with radical reactions that are first described in Chapter 8.
The great majority of reactions that will be discussed in this book involve movement of pairs of electrons, so they are represented by double-sided arrowheads.
Arrow pushing was first introduced in Section 1.8A in the discussion of resonance contributing structures. Recall that when comparing two or more contributing structures, an arrow was used to show how two electrons (lines representing bonds or pairs of dots representing lone pairs) could be redistributed within a single chemical structure to create an alternative Lewis line structure representation of the bonding. By convention, arrows are used to keep track of all pairs of electrons that are in different locations in the two different contributing Lewis line structures, shown here for the acetate anion and benzene molecule.
Keep in mind that in the case of resonance, 1) the atoms do not move between contributing structures, and 2) the electrons are not actually moving. The true chemical structure should be thought of as a hybrid of the contributing Lewis line structures. It is worth pointing out that when used with contributing structures, arrows generally indicate only the interconversion of p bonds and lone pairs (acetate ions) or just p bonds (benzene), not the formation or breaking of s bonds.
In chemical reactions, both electrons and atoms change positions as both p and s bonds are formed and broken. Arrow pushing is used to keep track of the movement of all electrons involved with each step of the overall transformation. Because electrons are located in orbitals surrounding atoms, when bonds are formed or broken, the movement of electrons between orbitals is necessarily accompanied by the movement of the associated atoms, which leads to the second rule of arrow pushing when depicting chemical reaction mechanisms:
Second Rule: Arrows are never used to indicate the movement of atoms directly. The arrows only show atom movement indirectly as a consequence of electron movement when covalent bonds are made and broken.
We have already used arrow pushing to show proton transfer several times in Chapter 4. The example below shows the transfer of a proton from the relatively acidic acetic acid molecule to the relatively basic hydroxide anion. We show this process with one arrow (labeled “a” in the diagram) that starts at a lone pair of electrons on the basic oxygen atom of the hydroxide anion, then points to the acidic H atom of acetic acid to indicate formation of the new bond being made. A second arrow originates at the line representing the breaking O-H bond and points to the O atom to denote creation of a lone pair (arrow “b”). In this reaction, the proton is being transferred between molecules, and the arrows indicate movement of the electrons involved.
A common mistake beginning students make is that they will erroneously write an arrow pointing from the H of the acetic acid to the O atom of the hydroxide anion. This is wrong, because such an arrow would be indicating the H atom movement directly, not electron movement! Other common mistakes in arrow pushing are given at the end.
Electron Sources and Sinks: How to Predict What Will Occur in an Organic Reaction Mechanism
Combined with the arrows shown for the contributing structures shown previously, we have now seen all three of the situations illustrated by arrows with double-sided arrowheads, namely the redistribution of p bonds and/or lone pairs, formation of a new s bond (generally from a lone pair or sometimes a new p bond), and breaking of a s bond (generally to form a new lone pair or sometimes a new p bond). Often, as in the case of the acetate-hydroxide ion reaction, more than one arrow is used in a given mechanism step. Now that you have seen all of the important types of arrows, we can point out the most important common feature between them:
Third Rule: Arrows always start at an electron source and end at an electron sink.
An electronsource is a bond or a lone pair of electrons. It is either a p bond or a lone pair on an atom of relatively high electron density in a molecule or ion, or a bond that must break during a reaction. An electron sink is an atom on a molecule or ion that can accept a new bond or lone pair of electrons.
Learning to identify the characteristic sources and sinks in different functional groups is the key to learning organic chemistry reaction mechanisms. For example, for arrows that depict the formation of new s bonds, the electron source is often readily identified as being a lone pair on the most electron rich atom of a molecule or ion, and the electron sink is readily identified as the most electron poor atom of a molecule or ion. Thus, the prediction of many of the most important electron sources and sinks comes down to lessons concerning the differences in electronegativity between atoms that were presented in Section 1.2, which allow you to identify partial and formal negative and positive charges in molecules. As an aid to your analysis, the red and blue colors of the various electrostatic surface maps given throughout this book indicate the negative and positive regions of molecules. We will have more to say about this reactivity pattern a little bit later.
This leads us to another commonly encountered type of process that deserves mention. As you will see in this and many later chapters, making a new bond to an electron sink often requires the simultaneous breaking of one of the bonds present at the sink atom to avoid overfilling its valence orbitals, a situation referred to as hypervalence.
Fourth rule: Breaking a bond will occur to avoid overfilling valence (hypervalence) on an atom serving as an electron sink.
In these cases, the electron source for the arrow is the bond being broken, and the sink is an atom able to accommodate the electrons as a lone pair, generally an electronegative atom such as an O atom or a halogen. If an ion is created, that ion is often stabilized by resonance delocalization or other stabilizing interactions.
Returning to the proton transfer reaction between acetic acid and hydroxide, we can now summarize our analysis of this simple one-step mechanism.
Viewed in the context of the third rule, when considering the arrow used to make a new s bond (arrow a), the hydroxide O atom is the electron source (most negatively charged atom) and the acetic acid H atom is the electron sink (atom with highest partial positive charge). This is illustrated using the electrostatic molecular surfaces shown below the reaction equation. The O atom of hydroxide ion has the greatest localized negative charge as indicated by the most intense red color and the acetic acid proton being transferred has the most intense positive charge character indicated by the most intense blue color. In order to avoid overfilling the valence of the H atom during the reaction (fourth rule), the O-H bond of acetic acid must be broken (arrow “b”). In so doing, the acetate ion is formed. Note that the acetate ion is stabilized by resonance delocalization.
Based on our analysis of the reaction between acetic acid and the hydroxide anion, you should now appreciate that the transfer of a proton (a so-called Brønsted acid-base reaction) is really just a special case of the common pattern of reactivity between an electron source (the base) and the proton as an electron sink, combined with breaking a bond to satisfy valence and create a relatively stable ion.
The addition or removal of protons during chemical reactions is so common that proton transfer steps are referred to by name directly, and we will use phrases such as “add a proton” or “take a proton away” when referring to them. However, proton transfer reactions are not the only case in which we use special names to describe a particular type of common reaction that involves arrows between electron sources and electron sinks.
As briefly mentioned in Section 4.7, a broader terminology is applied to the very common case of reactions in which new s bonds form between electron rich and electron poor regions of molecules. Nucleophiles (meaning nucleus seeking) are molecules that have relatively electron rich p bonds or lone pairs that act as electron sources for arrows making new bonds. Electrophiles (meaning electron seeking) are molecules with relatively electron poor atoms that serve as sinks for these arrows. Analogously, a molecule, or region of a molecule, that is a source for such an arrow is called nucleophilic, while a molecule or region of a molecule that is a sink for these arrows is referred to as being electrophilic. Based on this description, it should be clear that nucleophiles are analogous to Lewis bases and electrophiles are analogous to Lewis acids. Chemists use these terms interchangeably, although nucleophile and electrophile are more commonly used in kinetics discussions while Lewis acid and Lewis base are more commonly used in discussions about reaction thermodynamics. We will use all of these terms throughout the rest of the book.
It is helpful to summarize the appropriate use of key terms associated with arrow pushing and reaction mechanisms. The terms “source” and “sink” are used to identify the start and end of each reaction mechanism arrow, which is indicating the change in location of electron pairs. The terms “nucleophile” and “electrophile” (as well as “Lewis base” and “Lewis acid”) are used to describe molecules based on their chemical reactivity and propensity to either donate or receive electrons when they interact. Protons can be thought of as a specific type of electrophile, and for reactions in which a proton is transferred, the nucleophile is called a base.
Example 1: I see you
The following two sets of reactions (A and B) show possibilities for arrow pushing in individual reaction steps. Identify which is wrong and explain why. Next, using arrow pushing correctly, label which molecule is the nucleophile and which is the electrophile.
Solution
In each case the first arrow pushing scenario is wrong. The arrows shown below with stars over them do not start at a source of electrons, but rather they start at positions of relative positive charge, which is incorrect.
In the correct arrow pushing, the arrow labeled “a” depicts the interaction of a region of relative high negative charge (a p-bond or lone pair) with an atom of relatively high partial positive charge on the other reactant. Therefore, the molecule acting as the source for arrow the s bond-forming arrow “a” is the nucleophile while the molecule containing the sink atom is the electrophile. The arrow labeled “b” is needed to satisfy valence, and is not considered when defining the nucleophile and electrophile.
Putting it All Together: It Comes Down to a Multiple Choice Situation
In the sections and chapters that follow, many different reaction mechanisms will be described in a stepwise fashion. Each arrow can be classified according to one of the three overall situations we have already encountered (redistribution of $\pi$ bonds and/or lone pairs, formation of a new s bond from a lone pair or $\pi$ bond, breaking a s bond to give a new lone pair or $\pi$ bond).
When learning new mechanisms, first focus on the overall transformation that takes place. It might be a reaction in which atoms or groups are added (an addition reaction), a reaction in which atoms or groups are removed (an elimination reaction), a reaction in which atoms or groups replace an atom or group (a substitution reaction), or other processes we will encounter. Often, the overall process is composed of multiple steps. Once you have the overall process in mind, it is time to think about the individual steps that convert starting material(s) into product(s). Predicting complete multi-step mechanisms, then, comes down to learning how to predict the individual steps.
Understanding, as opposed to memorizing, mechanisms is critical to mastering organic chemistry. Although the mechanisms you encounter throughout the course may seem entirely different, they are actually related in fundamental ways. In fact, almost all of the organic reaction mechanisms you will learn are composed of only a few different individual elements (steps) that are put together in various combinations. Your job is to learn these individual mechanism elements, and then understand how to assemble them into the steps of the correct mechanism for the overall reaction.
Fortunately, there are a surprisingly small number of different types of characteristic mechanism elements (patterns of arrows) to be considered when trying to predict individual steps of even complex chemical reactions. For this reason, you should view the prediction of each step in an organic mechanism as essentially a multiple choice situation in which your most common choices are the following:
1. Make a new bond between a nucleophile (source for an arrow) and an electrophile (sink for an arrow). Use this element when there is a nucleophile present in the solution as well as an electrophile suitable for reaction to occur.
1. Break a bond so that relatively stable molecules or ions are created Use this element when there is no suitable nucleophile-electrophile or proton transfer reaction, but breaking a bond can create neutral molecules or relatively stable ions, or both.
1. Add a proton Use this element when there is no suitable nucleophile-electrophile reaction, but the molecule has a strongly basic functional group or there is a strong acid present.
1. Take a proton away Use this element when there is no suitable nucleophile-electrophile reaction, but the molecule has a strongly acidic proton or there is a strong base present.
The situation is even simpler than you might expect because 1. and 2. are the functional reverse of each other, as are 3. and 4. in many cases.
Many times, more than one of the four choices occurs simultaneously in the same mechanism step and there are some special situations in which unique or different processes such as electrophilic addition or 1,2 shifts occur. These different processes are described in detail as they are encountered.
In the following sections and chapters of the book, you will learn important properties of the different functional groups that allow you to deduce the appropriate choices for the individual steps in reaction mechanisms. To help you accomplish this, as new mechanisms are introduced throughout the rest of the book, we will label each mechanistic step as one of the four mentioned here when appropriate, emphasizing the common features between even complex mechanisms. When you are able to predict which of the above choices is(are) the most appropriate for a given step in a mechanism, you will then be able to push electrons correctly without relying on memorization. At that point, you will have taken a major step toward mastering organic chemistry!
Common Mistakes in Arrow Pushing
Throughout this book arrow pushing is used to indicate the flow of electrons in the various organic reaction mechanisms that are discussed. A few simple rules for properly performing arrow pushing were introduced in Section 6.2. In this Appendix we examine some of the most common mistakes that students make when first learning arrow-pushing methods and tell you how to avoid them. The mistakes given below are the ones seen most often by the authors during their cumulative dozens of year of experience in teaching Introductory Organic Chemistry.
Backwards Arrows
Reversing the direction of one or more arrows during a chemical step is the most common mistake made by students when writing organic reaction mechanisms. Backwards arrow pushing usually derives from a student thinking about the movement of atoms, not the movement of electrons. Hence, to avoid this mistake it is important to remember that arrows depict how electrons move, not where atoms move, within or between chemical structures. Further, one can avoid this mistake by remembering that every arrow must start at an electron source (a bond or lone pair) and terminate at an electron sink (an atom that can accept a new bond or lone pair).
Not Enough Arrows
A second common mistake in writing arrow-pushing schemes is to not use enough arrows. This usually results from not keeping track of all lone pairs, bonds made, or bonds broken in a mechanism step. In other words, if you analyze exactly the new position of electrons resulting from each arrow, missing arrows will become evident. In the following example we compare two arrow-pushing scenarios, one of which is missing an arrow. In the incorrect scheme there is no arrow that indicates breaking of the C-H bond of the reactant and formation of the p-bond in the alkene product. Note that when an arrow is missing, the result is commonly too many bonds and/or lone pairs on one atom (see the next section on hypervalency) and not enough bonds or lone pairs on another.
Hypervalency
Another frequent mistake when writing arrow-pushing schemes is to expand the valency of an atom to more electrons than an atom can accommodate, a situation referred to as hypervalency. An overarching principle of organic chemistry is that carbon has eight electrons in its valence shell when present in stable organic molecules (the Octet Rule, Section 1.2). Analogously, many of the other most common elements in organic molecules, such as nitrogen, oxygen, and chlorine, also obey the Octet Rule. There are three common ways in which students incorrectly draw hypervalent atoms: 1) Too many bonds to an atom, 2) Forgetting the presence of hydrogens, and 3) Forgetting the presence of lone pairs.
In the following case an arrow is used to depict a potential resonance structure of nitromethane. However, the result is a nitrogen atoms with 10 electrons in its valence shell because there are too many bonds to N. Such mistakes can be avoided by remembering to draw all bonds and lone pairs on an atom so that the total number of electrons in each atoms valence shell is apparent.
Another common way students mistakenly end up with a hypervalent atom is to forget the presence of hydrogens that are not explicitly written. When using stick diagrams to write organic chemical structures not all the hydrogens are drawn, and hence it is common to forget them during an arrow pushing exercise. The following example shows two proposed resonance contributing structures of an amide anion. The arrow drawn on the molecule to the left is incorrect because it depicts the formation of a new bond to a carbon that already has four bonds. When both bonds to hydrogen are drawn explicitly as on the structure farthest to the right, it is clear there are now five bonds around the indicated carbon atom. Notice also that the negative charge was lost upon drawing the contributing structures on the right, providing another clear signal that something was wrong because overall charge is always conserved when arrows are drawn correctly.
Another common way to make a hypervalency mistake is by forgetting to count all lone pairs of electrons. The following example shows a negatively charged nucleophile incorrectly adding to the formal positive charge on an alkylated ketone. This may look correct because atoms with positive and negative charges are being directly combined, but when counting bonds and lone pairs of electrons, it is found that the oxygen ends up with 10 electrons overall. Hence, this is a mistake.
Mixed Media Errors
Acids and bases are catalysts, reactants, products, and intermediates in many organic chemistry transformations. When writing mechanisms for reactions involving acids and bases, there are three general rules that will help guide you in depicting the correct mechanism.
Do not show the creation of a strong acid for a mechanism of a reaction that is performed in strongly basic media.
Do not show the creation of a strong base for a mechanism of a reaction that is performed in strongly acidic media.
In strongly acidic media, all the intermediates and products will be either neutral or positively charged, while in strongly basic media, all the products and intermediates will be neutral or negatively charged.
The reason for these rules is that significant extents of strong acids and bases cannot co-exist simultaneously in the same medium because they would rapidly undergo a proton transfer reaction before anything else would happen in the solution.
An example of a mixed media error is given below. The first example shows a strong base being created although the reaction is performed under acidic conditions (see conditions over the first equilibrium arrows). Not shown are the three steps that lead to the intermediate drawn. A mistake is made in the arrow pushing because a strong base (methoxide) is generated as the leaving group even though the reaction is run in strong acid. In the correct mechanism, the next step would be protonation of the ether oxygen atom followed by loss of methanol in the last step (not shown) to give a carboxylic acid product.
Failing to conserve charge
Overall charge must be conserved in all mechanism steps. Failure to conserve overall charge could be caused by some of the preceding errors (hypervalency, failure to draw arrows, mixed media errors), but we mention it by itself because it is always helpful to check that your arrow pushing is consistent by confirming that overall charge conservation is obeyed. In the example shown below, an arrow is missing leading to a neutral intermediate even thought the overall charge on the left side of the equation was minus one. Notice there are five bonds to carbon on the intermediate (hypervalency), providing another obvious indication that something was incorrect in the mechanism step as drawn.
Keys to Carbonyl Chemistry Mechanisms
http://iverson.cm.utexas.edu/courses...rbMechKeys.pdf
Mechanism Handout
http://iverson.cm.utexas.edu/courses...echWebPage.pdf | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Fundamentals/The_Golden_Rules_of_Organic_Chemistry/Everything_You_Need_to_Know_About_Mechanisms.txt |
This page explains the use of curly arrows to show the movement both of electron pairs and of single electrons during organic reaction mechanisms.
Using curly arrows to show the movement of electron pairs
Curly arrows (and that's exactly what they are called!) are used in mechanisms to show the various electron pairs moving around.
• The arrow tail is where the electron pair starts from. That's always fairly obvious, but you must show the electron pair either as a bond or, if it is a lone pair, as a pair of dots. Remember that a lone pair is a pair of electrons at the bonding level which isn't currently being used to join on to anything else.
• The arrow head is where you want the electron pair to end up.
For example, in the reaction between ethene and hydrogen bromide, one of the two bonds between the two carbon atoms breaks. That bond is simply a pair of electrons.
Those electrons move to form a new bond with the hydrogen from the HBr. At the same time the pair of electrons in the hydrogen-bromine bond moves down on to the bromine atom.
There's no need to draw the pairs of electrons in the bonds as two dots. Drawing the bond as a line is enough, but you could put two dots in as well if you wanted to.
Notice that the arrow head points between the C and H because that's where the electron pair ends up. Notice also that the electron movement between the H and Br is shown as a curly arrow even though the electron pair moves straight down. You have to show electron pair movements as curly arrows - not as straight ones.
The second stage of this reaction nicely illustrates how you use a curly arrow if a lone pair of electrons is involved.
The first stage leaves you with a positive charge on the right hand carbon atom and a negative bromide ion. You can think of the electrons shown on the bromide ion as being the ones which originally made up the hydrogen-bromine bond.
The lone pair on the bromide ion moves to form a new bond between the bromine and the right hand carbon atom. That movement is again shown by a curly arrow. Notice again, that the curly arrow points between the carbon and the bromine because that's where the electron pair ends up.
That leaves you with the product of this reaction, bromoethane:
Using curly arrows to show the movement of single electrons
The most common use of "curly arrows" is to show the movement of pairs of electrons. You can also use similar arrows to show the movement of single electrons - except that the heads of these arrows only have a single line rather than two lines.
shows the movement of an electron pair
shows the movement of a single electron
The first stage of the polymerization of ethene, for example, could be shown as:
You should draw the dots showing the interesting electrons. The half arrows show where they go. This is very much a "belt-and-braces" job, and the arrows don't add much. Whether you choose to use these half arrows to show the movement of a single electron should be governed by what your syllabus says. If your syllabus encourages the use of these arrows, then it makes sense to use them. If not - if the syllabus says that they "may" be used, or just ignores them altogether - then they are as well avoided.
There is some danger of confusing them with the arrows showing electron pair movements, which you will use all the time. If, by mistake, you use an ordinary full arrow to show the movement of a single electron you run the risk of losing marks.
Contributors
Jim Clark (Chemguide.co.uk) | textbooks/chem/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Fundamentals/The_Use_of_Curly_Arrows.txt |
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