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A coordinate or dative bond is any covalent bond that arose because one atom brought a pair of its electrons and donated them to another.
When the nitrogen donates a pair of electrons to share with the boron, the boron gains an octet. Both atoms are electronically saturated. In addition, a pair of non-bonding electrons becomes bonding; they are delocalized over two atoms and become lower in energy. This development acts as a driving force for formation of the dative bond.
• Formation of dative bonds can be driven by electronic saturation; that means octets become filled.
• Dative bond formation is also driven by the fact that bond formation allows nitrogen's lone pair to become delocalized.
However, sometimes bonds can be broken again. In the case of dative bonds, that means a pair of electrons that were donated from one atom to another can be taken back again. This event is called dissociation. In dissociation, two atoms that were connected to each other become disconnected.
Why might a dative bond be broken? From one point of view, maybe the nitrogen atom's positive charge in the Lewis acid-base complex is unfavorable. Nitrogen is a relatively electronegative element, and so it is easy to imagine the nitrogen pulling this pair of electrons back to itself, leaving the boron behind without an octet.
There is another reason Lewis-acid complexes can dissociate. That reason is entropy, or the distribution of energy. One of the fundamental laws of thermodynamics is that increased entropy is favorable. Looked at very loosely, that means it is favorable to have energy distributed in more states or packages. One of the ways of accomplishing this goal is to divide the available energy up between multiple molecules.
Even though it costs energy to break the B-N bond in the Lewis acid-base complex, doing so can still be energetically favorable if that increased amount of energy can be distributed better between two molecules than it was in one molecule. Exactly how that energy distribution is accomplished is the subject of statistical mechanics, and is beyond the scope of this course. However, energy distribution is generally related to the ability of atoms to move. In molecules, that means it is related to the ability of bonds to stretch and compress, of bond angles to squeeze down and widen out, and for molecules to tumble and zip around. There are more ways to tumble and zip around if two molecules are doing it, rather than one. That means energy may be better distributed by two molecules than by one.
• Entropy factors favor the dissociation, or break-up, of Lewis acid-base complexes.
The fact that there is a good reason for Lewis acid-base complexes to form, and also a good reason for them not to form, may be very unsatisfying. However, it simply means that there is a balance between a nucleophile and electrophile coming together (association) and going apart (dissociation). Which direction the system will go depends on the relative importance of these two factors. The relative importance of each factor will vary from one case to another. In many cases, there is a dynamic equilibrium between the two possible cases. That means that after complexes form, they will come apart again. After complexes come apart, they will form again. This process does not stop, but continues as long as there is enough energy around to keep things moving.
If it were possible to take a picture of a collection of molecules, the picture might show some of the molecules coordinated together and some of them wandering on their own. The ratio between the associated pairs and the dissociated molecules is related to the point of equilibrium. At the point of equilibrium, different factors driving the reaction to one side or the other balance each other out.
• Equilibrium is the balance between two possible states. For example, it is the balance between dissociated and associated ammonia and boron trifluoride.
In this case, the equilibrium heavily favors the Lewis acid-base complex. A picture of millions of ammonia and boron trifluoride molecules would show most of them coordinated together. A few molecules would be found on their own, however.
Exercise \(1\)
Use arrows and structures to show the dissociation of one nucleophile or Lewis base from each of the following polyatomic anions.
a) BF4- b) PF6- c) AlCl4- d) AlH4- e) Ag(NH3)2+
Answer a
Answer b
Answer c
Answer d
Answer e
Exercise \(2\)
Draw pictures to represent the following cases of equilibria. Use squares to represent Lewis acids and circle to represent Lewis bases.
1. The Lewis acid-base complex is heavily favored.
2. The Lewis acid-base complex is slightly favored.
3. The Lewis acid-base complex is heavily disfavored.
4. The Lewis acid-base complex does not form at all.
Answer a
Answer b
Answer c
Answer d
Sometimes, Lewis acid-base complexes are stabilized by chelation. Chelation (from the Greek word, chelos, meaning "crab") refers to the situation in which a Lewis base has more than one electron-donating site. That means it can bind to a Lewis acid through more than atom (just as a crab could grab something with both its claws).
The chelate effect refers to the fact that Lewis acid-base complexes are often more stable with respect to dissociation when they contain chelating electron donors. The equilibrium lies much farther towards the associated complex when chelation is a factor.
Chelation is seen in some very common biological molecules that incorporate Lewis acid-base complexes. For example, the heme subunit found in the oxygen-carrying proteins, hemoglobin and myoglobin, contains four electron-donating atoms that bind to iron. A very similar unit is found in a number of other molecules, including the important photosynthetic chromophore, chlorophyll. Related molecules are also found in the important biological cofactor, vitamin B12, and in cytochrome P450, which is involved in detoxification in the liver, among other tasks.
Exercise \(3\)
Using the concept of entropy, explain why an iron-heme complex is more stable than an iron ion complexed to four separate nitrogen donors, such as histidine or pyrrole anion.
Answer | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/14%3A_Concepts_of_Acidity/14.05%3A_Reversibility_of_the_Dative_Bond.txt |
The bond that forms between a Lewis base and a Lewis acid is sometimes called a dative bond or a coordinate bond. The term used for the donation of a Lewis base to a Lewis acid, without any other bonding changes, is coordination. Another term for Lewis acid-base complexes, especially used in the context of transition metal chemistry, is coordination complexes. Sometimes the Lewis base is referred to as a ligand; more generally, a ligand is just one molecule that binds to another.
An example of a coordination complex is hexaaquo cobalt dichloride, Co(H2O)6Cl2. This compound contains a Co2+ ion. This electrophilic metal ion is coordinated by six nucleophilic water ligands. Because the water molecules are neutral, the complex still has a 2+ charge overall. There are two chloride counterions to balance the charge, but the chlorides are not attached to the complex ion.
Notice that the usual Lewis conventions are usually abandoned in drawing coordination complexes of transition metals. With so many different nucleophiles sticking to the Lewis acid, the number of formal charges that must be drawn becomes very cumbersome. Usually the oxidation state of the metal cation is denoted. The oxidation state essentially means the charge on the metal cation and it is written in Roman numerals beside the metal atom. In addition, the overall charge on the Lewis acid-base complex is given, with square brackets indicating that the charge belongs to the entire complex within the brackets. Exactly where that charge resides is up to the reader to consider.
Coordination complexes are frequently useful in mining and metallurgy. For example, nickel can be extracted from nickel ore by converting the nickel into Ni(CO)4 via addition of carbon monoxide. Ni(CO)4, or tetracarbonyl nickel, is a gas that can be easily separated from the solid ore. When removed from the presence of carbon monoxide, the coordination complex decomposes back into Ni and CO.
Exercise \(1\)
These questions concern the formation of tetracarbonyl nickel.
a) Based on what you know about transition metals, is the nickel most likely a Lewis acid or a Lewis base?
b) Draw the Lewis structure for carbon monoxide, CO. Make sure each atom has an octet. Make sure you add formal charges.
c) Based on formal charge considerations, which atom in carbon monoxide will bind to the nickel?
d) Using arrow notation, show the coordination of CO to Ni. Do this one step at a time, showng each CO molecule binding to the nickel and the new complex that results after each step.
e) Why does nickel bind four carbon monoxide ligands, and not three or five?
Answer a
Lewis acid
Answer b
Answer c
Carbon will be nucleophilic due to the lone pair electrons and the negative charge.
Answer d
Answer e
Ni0 is d10 and with 4 CO ligands the nickel tetracarbonyl complex has 18 electrons.
Another example is seen in the liquid extraction of metals from ores. Often, a metal is coordinated by some kind of ligand that changes the solubility of the metal atom, so that it can be extracted from the ore with a solvent. The solvent is usually water, but other liquids could be used. One of the easiest cases to picture on paper is the leaching of gold from gold ores with cyanide.
Often, coordination of a ligand to a metal changes many properties of the metal, in addition to changing its solubility. Transition metals have different oxidation states, meaning they can give up different numbers of electrons and become cations with different charges. Sometimes, this event happens more easily upon coordination. In the case of gold, the gold atom can react with air and become a gold cation in the presence of cyanide.
Cyanide is an anion, so it would be added as a salt, such as sodium cyanide, NaCN, or potassium cyanide, KCN. The resulting complex, Au(CN)2, is actually a complex anion, and it would be associated with one of the counterions added, such as KAu(CN)2. It is not completely clear why the gold binds two cyanides, but binding two ligands is common in the chemistry of gold and silver, and is based on molecular orbital considerations that are beyond the level of this course.
This complex salt is water-soluble, so it can be removed from the ore with water. Later, it must be converted back into pure gold via electrochemical reactions. These reactions are not related to acid-base chemistry and will not be covered in this course.
Exercise \(2\)
These questions concern the formation of the potassium dicyano gold complex.
a) Based on what you know about transition metals, is the gold most likely a Lewis acid or a Lewis base?
b) Draw the Lewis structure for cyanide, CN-. Make sure each atom has an octet. Make sure you add formal charges.
c) Based on formal charge considerations, which atom in cyanide will bind to the gold?
d) Using arrow notation, show the coordination of cyanide to Au+. (We will simplify and assume the gold has already been oxidized). Do this one step at a time, showing each cyanide molecule binding to the gold ion and the new complex that results after each step.
Answer a
Lewis acid
Answer b
Answer c
Carbon will be nucleophilic due to the lone pair electrons and the negative charge.
Answer d
Perhaps some of the most interesting coordination complexes involve transition metals in biology. The most familiar of these compounds is haemoglobin. Haemoglobin is a complex protein that contains a crucial iron atom. The iron is present as a 2+ cation, but the haeme ring that binds to the iron has a 2- charge, so overall there is no charge on the complex. The iron is still electrophilic, however, and it can bind an oxygen molecule. Much more oxygen can be carried in the bloodstream this way than if the oxygen simply dissolved in the blood.
Exercise \(3\)
These questions concern the formation of the oxyhaemoglobin.
a) Based on what you know about transition metals, is the iron most likely a Lewis acid or a Lewis base?
b) Draw the Lewis structure for oxygen molecule, O2. Make sure each atom has an octet. Make sure you se if there are any formal charges.
N.B. This Lewis structure may not agree completely with the molecular orbital view of the molecule, but it is all we have to work with for our current method of understanding coordination.
c) Using arrow notation, show the coordination of oxygen to Fe2+.
d) There has been a great deal of research into the exact geometry of the dioxygen complex. Based on your Lewis structure of oxygen, show how you think the oxygen is attached to the iron. How would you describe the geometry?
Answer a
Lewis acid
Answer b
Answer c
Answer d
The bound oxygen would be predicted to have a trigonal planar electronic (bend molecular) geometry based on the Lewis structure.
There is another interesting event that happens when oxygen binds to the iron in haemoglobin. After binding O2, the iron actually transfers a single electron to the oxygen, becoming a Fe3+ cation. Although deoxyhaemoglobin, the Fe2+ or Fe(II) species, is purple, many Fe3+ or Fe(III) compounds are red. Thus, oxyhaemoglobin is red.
It is also interesting to note that we are used to seeing iron-containing materials turn a sort of red-brown color when they have been exposed to oxygen for a long time, especially in the presence of water and salts. Our cars, trains and bridges eventually rust as the iron in their steel turns to reddish iron oxide. William Tolman, a researcher in bioinorganic chemistry at the University of Minnesota, likes to raise the following question: we rely on iron complexes to carry oxygen through our salt-water bloodstream, so why don't we get rusty?
The answer is that, in a sense, we do. It has been estimated that after several passes through our bloodstream, a haemoglobin molecule meets its end when the oxygen fails to detach and the iron complex decomposes to an iron oxide. That's one reason why you need to eat a diet that contains iron, to replenish your iron stores in order to make new haemoglobin on a regular basis.
To see some additional problems that will help you get to know coordination compounds, go to the main coordination compound chapter. In particular, it is really useful to be able to count electrons in coordination chemistry. Counting electrons is just like learning Lewis structures, except that instead of going to a maximum count of eight electrons, we can go to a maximum of eighteen electrons. That's because the noble gas configuration for a transition metal would have eighteen electrons, not eight. To learn how to count electrons in coordination complexes, go here. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/14%3A_Concepts_of_Acidity/14.06%3A_Coordination_Complexes.txt |
Perhaps the most common example of a Lewis acid or electrophile is also the simplest. It is the hydrogen cation or proton. It is called a proton because, in most hydrogen atoms, the only particle in the nucleus is a proton. If an electron is removed to make a cation, a proton is all that is left.
• H+ is a very common Lewis acid or electrophile.
A proton is electrophilic for a couple of reasons. It has a positive charge, and so it will attract electrons, which are negative. Also, it lacks the electron configuration of its noble gas neighbour, helium. Helium has two electrons. If a Lewis base or nucleophile donates a pair of electrons to a proton, the proton will obtain a Noble gas configuration. That's part of the reason why, in some periodic tables, hydrogen is shown in two places: at the very left, illustrating its potential to lose an electron, like sodium and lithium; and at the right, illustrating its potential to take on helium's configuration.
There is something about hydrogen cations that is not so simple, however. They are actually not so common. Instead, protons are generally always bound to a Lewis base. Hydrogen is almost always covalently (or datively / coordinately) bonded to another atom.
Many of the other elements commonly found in compounds with hydrogen are more electronegative than hydrogen. As a result, hydrogen often has a partial positive charge. Remember, that is one of the reasons that atoms can act as Lewis acids: with a partial positive charge, an atom becomes electrophilic.
Our statement about protons might better be expressed as:
• Hd+ is a very common Lewis acid or electrophile.
If hydrogens are almost always bonded to other atoms, then the Lewis acid-base interactions we have looked at so far are slightly different here. Instead of two compounds coming together and forming a bond, we have one Lewis base replacing another at a proton.
• Protons are transferred from one basic site to another.
• Transfer occurs by donation of a lone pair to the proton.
Exercise \(1\)
Use arrows to show water acting as a Lewis base and donating electrons to the proton in the following compounds.
a) HBr b) HONO2 c) CH3(CO)OH
Answer a
Answer b
Answer c
Exercise \(2\)
Consider the following reaction:
• Draw an MO mixing diagram for the reaction above, using the following steps:
o Draw the orbital from the base that is likely to donate its electrons.
o Draw the orbital from the acid that is likely to accept electrons.
o Complete the MO mixing diagram of these two orbitals:
• Label the electron donating orbital
• Label the electron accepting orbital
• Populate the MO mixing diagram with electrons
• Draw a cartoon showing the mixing of these orbitals.
Answer
Exercise \(3\)
Histidine side-chains contain an imidazole ring, which are unique as they can behave as either a proton-donor or a proton-acceptor depending on surrounding conditions.
a) Draw in the hydrogen at the chiral center of histidine.
b) Draw the (R)- enantiomer of histidine.
c) Circle the side chain of histidine.
d) Using the energy axis below do the following:
Draw the Hückel MO diagram for imidazole.
Populate the Hückel MO diagram with electrons.
Draw the MO pictures for 3 of the 5 MOs and label each as π, π* or n.
e) Based on your MO diagram, the imidazole is (choose one):
aromatic anti-aromatic non-aromatic
f) Show what imidazole would look like if it gave away a proton.
g) Circle the atom in imidazole that could donate electrons to a proton.
h) Why can't the other nitrogen donate electrons to a proton?
i) Show what imidazole would look like if it did pick up a proton.
Histidine residues frequently hold metal ions in place within enzymes. Enzymes are proteins that catalyze chemical reactions in biological settings. For example, plastocyanins contain a copper(II) ion (Cu2+) which functions to store an electron as it is transported from photosystem II to photosystem I during photosynthesis.
j) The drawing below is a cartoon of an enzyme active site, the place in an enzyme that carries out a specific task. Complete the blank active site by adding thestructures of the missing amino acid side-chains, as indicated by their three-letter codes.
k) Explain why each of these chains is able to bind a Cu2+ ion.
l) Draw a Cu2+ ion in the active site.
m) What is the coordination number for the copper ion in the complex you drew?
n) What is the geometry name for the copper ion in the complex?
o) Let's think about how many valence electrons surround that copper in the complex.
What is the number of valence electrons for copper metal?
What is the number of valence electrons for copper (II) ion?
How many electrons will the donor atoms contribute?
What is total valence electron count for the copper (II) complex?
p) At low pH (when there are lots of protons around), one of the histidines binds a proton. What happens to its ability to bind copper ion?
q) What is the coordination geometry of the copper ion in the complex at low pH?
Answer e
aromatic
Answer g
Answer h
This nitrogen atom can donate a lone pair of electrons without disturbing the aromatic character of the molecule.
Answer k
Each side chain has a Lewis base with a lone pair of electrons to donate the copper ion.
Answer l
Answer m
CN = 4
Answer n
tetrahedral
Answer o
Cu: 11 e-
Cu(II): 9 e-
4 x 2e- = 8 e-
8 e- + 9 e- = 17 e-
Answer p
If the imidazole becomes occupied by a proton, then the imidazole no longer has the lone pair to donate to the copper ion.
Answer q
trigonal | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/14%3A_Concepts_of_Acidity/14.07%3A_Proton_as_a_Common_Lewis_Acid.txt |
Acidity involving protons is complicated because of the exchange of one base for another. The proton has attracted the electron pair from the base via its partial positive charge. Once the base arrives, however, the proton cannot bind to the new base and still retain its tie to the base to which it was already bound. A hydrogen can only have 2 electrons in its valence shell, not four.
What begins very much like the formation of an intermolecular hydrogen bond goes one step further. The proton forms a full bond to the incoming donor, and releases its bond to its former partner.
In hydrogen bonding, the lone pair is partially shared with a proton, without actually displacing the other partner from the proton. A true bond is not quite formed between the donor and the proton. However, the distinction between hydrogen bonding and proton transfer can be blurred. Often, both cases happen at the same time in a given system.
• Hydrogen atoms can normally have only one bond.
A conjugate base is any Lewis base that forms as a result of a proton transfer. This label is only relative. Any Lewis base could be called a conjugate base if, in the present circumstances, it has lost a proton. Also, a conjugate acid is any compound that has just gained a proton.
However, the terms conjugate base and conjugate acid imply that these compounds can potentially act as bases and acids, respectively. because the conjugate base has a lone pair and is a Lewis base, it could in fact take a proton again. Because a conjugate acid has a proton, it could conceivably give the proton up again. Whether or not this event will actually occur is another matter. Later, we will develop more of an understanding of when proton transfers will actually occur.
• The pair of electrons connecting the proton to the old Lewis base leave with the base. This base is called the "conjugate base".
• The proton is now bound only to the new Lewis base. This new compound is called the "conjugate acid".
We also need a new term to call the compound that provides the proton in these proton transfer reactions. The proton is the Lewis acid in these cases, but what do we call the entire compound that contains the acidic proton? This compound is commonly called a Brønsted acid. This term comes from studies of acids by Brønsted and Lowry, who independently observed that many reactions in chemistry involve proton transfers.
• A Brønsted acid is a compound that provides a proton for a base to take.
Again, the conjugate acid that forms after proton transfer is often a Brønsted acid; it could transfer the proton to another base. It could transfer the proton back to the conjugate base, reforming the starting materials. Alternatively, it could transfer the proton to a new base. In that case, the Lewis base in the first step has picked up a proton, turned into a Brønsted acid, delivered the proton to another base, and become a Lewis base again. It is like a taxicab that has picked up a passenger and dropped it off again. Sometimes in this situation, the Lewis base is called a proton shuttle.
Delivery of protons is an essential task in a wide array of chemical processes. In biological systems, histidine residues in proteins commonly function as proton shuttles (although they play many other roles as well). For example, the interconversion of aldose and ketose sugars is a basic step in a variety of biosynthetic pathways. In an aldose, a C=O group is found at the end of a carbon chain. In a ketose, this C=O group is not at the end of the chain.
An aldose can be converted into a ketose essentially by moving some protons.
When this transformation is carried out by the enzyme, glyoxalase, studies have shown that the proton marked with an asterisk in the aldose and ketose above are exactly the same proton, having moved from one position to another. The proton has simply been moved by a proton shuttle.
The evidence for the specific proton from the aldose ending up in the ketose comes from labeling studies. In a labeling study, a different kind of proton is placed in that specific position in the aldose. This task would involve a less-common isotope such as tritium or deuterium instead of the usual protium. Deuterium and tritium can both be distinguished from protium; for example, deuterium and protium both absorb radio waves in a technique similar to medical MRI, and by using this technique it is easy to distinguish one from the other. As a result, we can tell where a deuterium is found in a particular molecule. When the aldose is then exposed to glyoxalase, we can see that deuterium show up in a specific spot in the ketose.
The scheme above shows only the transfer of the marked proton from one carbon to another. In addition, a proton must be placed on the oxygen of the C=O group in the aldose, and another proton must be removed from the OH group on the next carbon in the chain. Additional proton transfer steps would accomplish these tasks.
Exercise \(1\)
Provide the conjugate bases of the following mineral acids. In each case, assume only one proton is lost.
a) HBr b) H3PO4, connected (HO)3PO c) HClO4, connected HOClO3.
Answer a
Answer b
Answer c
Exercise \(2\)
Provide the conjugate acids of the following bases.
a) H2O b) NH3 c) CH3NH2 d) CH3O- e) H-
Exercise \(3\)
Some of the more common mineral acids that you may be familiar with from high school chemistry are formed as a result of Lewis acid-base complexation, followed by proton transfer. Use arrow formalisms to show the elementary steps that occur when the following compounds are dissolved in water.
a) sulfur trioxide, SO3, to give sulfuric acid, H2SO4 (connected (HO)2SO2).
b) nitric oxide ion, NO2+, to give nitric acid, HNO3 (connected HONO2).
Answer a
Exercise \(4\)
Note that a number of common acids contain OH groups.
a) What is it about this group makes these compounds acidic?
b) Some compounds, such as NaOH or KOH, are not very acidic. In fact, they are termed Arhennius bases, meaning they are ionic compounds that dissociate in water to give hydroxide ions (OH-), rather than protons. Explain why these compounds behave differently from the acids containing the same group, and why the hydroxide ion is basic.
Answer a
The O-H bond is polar covalent. The oxygen is much more electronegative than the hydrogen, so the bond is easily ionized to give H+.
Answer b
In a case like NaOH, the electronegativity difference between the sodium and oxygen is much greater than the electronegativity between the oxygen and the hydrogen. The Na-O bond is ionic. The removal of a proton from hydroxide ion is harder than the removal from a proton from hydroxide. It would make an oxide anion, O2-; that buildup of negative charge is more difficult than the formation of -OH.
Exercise \(5\)
The following pairs of species will react with each other via an acid-base reaction. For each pair, complete the reaction by adding the products of the reaction, drawing a Lewis dot structure for all four reactants and products, labeling each species in the reaction as an acid, conjugate acid, base or conjugate base; and drawing arrows to show the electron flow.
A. H3O+ + OCl-
B. CH3CH2COOH + CH3-
C. HClO3 + OH-
D. CH3CH2CH2- + H2O
E. CH3CHO + (CH3)3CO-
F. NH4+ + NH2-
Answer A.
Answer B.
Answer C.
Answer D.
Answer E.
Answer F.
In Brønsted acidity, the Lewis acid is always a proton. However, protons aren't generally found by themselves. That's because they are such good Lewis acids; they are usually found sticking to a Lewis base already.
Earlier, we looked at how frontier orbitals are sometimes used to think about reactions. We think about the highest occupied molecular orbital on one reactant as the source of electrons. We imagine the lowest unoccupied orbital on the other reactant as the destination for the electrons.
Suppose a proton is transferred from a hydrogen chloride molecule, HCl, to a water molecule, H2O. That's what will happen if hydrogen chloride, a gas, is bubbled into some water, forming aqueous hydrochloric acid.
The electrons are donated from a non-bonding pair of electrons on the oxygen atom in the water molecule. That part seems pretty straightforward from the Lewis structure, and we see the same idea conveyed in the MO diagram. But where do these electrons go? There is no obvious acceptor in the Lewis structure, because the hydrogen atom has two electrons and the chlorine has eight, so both octets are satisfied. We need to bring in the idea of electronegativity to see that the two electrons in the H-Cl bond are not shared evenly; there is a parital negative charge on chlorine and a partial positive charge on hydrogen. That fact make hydrogen seem like the electron acceptor. Sure enough, an O-H bond has formed in the product, suggesting the oxygen atom has donated a pair of its electrons to the hydrogen.
If we look at the molecular orbital picture, we see that the LUMO on HCl is an antibonding orbital, the σ*HCl. The non-bonding electrons from oxygen will be donated into the empty antibonding orbital on HCl. Remember, putting electrons in an antibonding orbital always weakens or breaks the bond. That's because the energetic advantage of dropping electrons into a low-energy bonding orbital is negated by the energetic disadvantage of raising electrons into a high-energy antibonding orbital.
As a result, there is more going on in the molecular orbital interaction diagram than just bond formation. As in previous cases, the MO diagram for the product looks a little like the MO diagrams of the two reactants added together, but there are some changes. Once again, the HOMO from the base and the LUMO from the acid mix together to form a new low-energy bonding level and a new high-energy antibonding level in the product. Because the LUMO is an antibonding orbital, however, there is an additional consequence: the corresponding bondng orbital in the acid becomes a non-bonding orbital, because that bond must be broken.
In this case, the MO picture reminds us of something in addition to the bond formation, and that i the bond-breaking that accompanies proton transfers.
Exercise \(6\)
For the following reactions, draw (i) MO diagrams showing the HOMO-LUMO interaction and (ii) the molceular orbital interaction diagram showing the formation of the product.
1. Two molecules of water react to form H3O+-OH.
2. A molecule of methanal, H2CO, reacting with hydrogen chloride, HCl, to form H2COH+Cl-.
Answer a i
Answer a ii
Answer b i
Answer b ii | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/14%3A_Concepts_of_Acidity/14.08%3A_Proton_Transfer_from_One_Basic_Site_to_.txt |
It is helpful to have a way of comparing Brønsted-Lowry acidities of different compounds. If the chemistry of protons involves being passed from a more acidic site to a less acidic site, then the site that binds the proton more tightly will retain the proton, and the site that binds protons less tightly will lose the proton. If we know which sites bind protons more tightly, we can predict in which direction a proton will be transferred.
There is an experimentally-determined parameter that tells us how tightly protons are bound to different compounds. "Experimental" often implies to students "untested" or "unreliable", but here it means that someone has done the work to measure how tightly the proton is bound. Experimental in this sense means "based on physical evidence".
This experimental parameter is called "the pKa". The pKa measures how tightly a proton is held by a Brønsted acid. A pKa may be a small, negative number, such as -3 or -5. It may be a larger, positive number, such as 30 or 50. The lower the pKa of a Brønsted acid, the more easily it gives up its proton. The higher the pKa of a Brønsted acid, the more tightly the proton is held, and the less easily the proton is given up.
• Low pKa means a proton is not held tightly.
• pKa can sometimes be so low that it is a negative number!
• High pKa means a proton is held tightly.
For example, nitric acid and hydrochloric acid both give up their protons very easily. Nitric acid in water has a pKa of -1.3 and hydrobromic acid has a pKa of -9.0. On the other hand, acetic acid (found in vinegar) and formic acid (the irritant in ant and bee stings) will also give up protons, but hold them a little more tightly. Their pKas are reported as 4.76 and 3.77, respectively. Water can certainly give up a proton, but not very easily; it has a pKa of around 14. Methane is not really an acid at all, and it has an estimated pKa of about 50.
The pKa measures the "strength" of a Brønsted acid. A proton, H+, is a strong Lewis acid; it attracts electron pairs very effectively, so much so that it is almost always attached to an electron donor. A strong Brønsted acid is a compound that gives up its proton very easily.
A weak Brønsted acid is one that gives up its proton with more difficulty. Going to a farther extreme, a compound from which it is very, very difficult to remove a proton is not considered to be an acid at all.
When a compound gives up a proton, it retains the electron pair that it formerly shared with the proton. It becomes a conjugate base. Looked at another way, a strong Brønsted acid gives up a proton easily, becoming a weak Brønsted base. The Brønsted base does not easily form a bond to the proton. It is not good at donating its electron pair to a proton. It does so only weakly.
In a similar way, if a compound gives up a proton and becomes a strong base, the base will readily take the proton back again. Effectively, the strong base competes so well for the proton that the compound remains protonated. The compound remains a Brønsted acid rather than ionizing and becoming the strong conjugate base. It is a weak Brønsted acid.
• "Strong" Brønsted acids ionize easily to provide H+.
• This term is usually used to describe common acids such as sulfuric acid and hydrobromic acid.
• "Weak" Brønsted acids do not ionize as easily.
• This term is often used to describe common acids such as acetic acid and hydrofluoric acid.
However, the terms "strong" and "weak" are really relative. pKa values that we have seen range from -5 to 50. If something with a pKa of 4 is described as a weak acid, what is something with a pKa of 25? A very, very weak acid? It is certainly a better source of protons than something with a pKa of 35. Is that a very, very, very, very weak acid? How many "verys" are there in a pKa unit?
This idea is also true when considering the opposite: a base picking up a proton to form a conjugate acid. How tightly that conjugate acid holds a proton is related to how strongly the base can remove protons from other acids. The weaker something is as a source of protons, the stronger its conjugate is as a proton sponge.
• Be careful. The terms "strong acid" and "weak acid" can be used relatively, rather than absolutely.
• The same is true for "strong base" and "weak base".
• Sometimes, whether something is called "strong" or "weak" depends on what else it is being compared to.
Table \(1\). Approximate pKa values for selected compounds.
A more extensive pKa table can be found at Prof. David Evans' site at Harvard.
Exercise \(1\)
Find a pKa table. Use it to help you decide which of the following pairs is the most Brønsted acidic in water.
a) HNO3 or HNO2 b) H2Se or H2O c) HCl or H2SO4 d) Be(OH)2 or HSeO3
Answer a
Answer b
Answer c
Answer d
* A note on the pKa of water: physics and physical chemistry texts list 14 as the value of the pKa of water. Biochemistry and organic chemistry texts often list the value as 15.7. The latter texts have simply factored into the constant a molar value for the concentration of water; the former agree that this factor should be replaced by the activity of water, which has a value of 1. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/14%3A_Concepts_of_Acidity/14.09%3A_Proton_Donor_Strength-_pKa.txt |
Structure plays a key role in determining how easily a compound can provide protons. For example, note that a number of common acids contain OH groups, such as sulfuric acid and acetic acid. What is it about this group that makes these compounds acidic?
Probably it has something to do with the O-H bond being polar. However, water has an O-H bond, but it is not terribly acidic.
• A polar O-H bond may be one factor contributing to Brønsted acidity (the formation of H+).
• However, it cannot be the only factor.
In contrast, some compounds, such as NaOH or KOH, are not very acidic. In fact, they are termed Arhennius bases, which means they are ionic compounds that dissociate in water to give hydroxide ions (HO-). Remember, HO- is a good Lewis base, because it has lone pairs to donate, and it also has a negative charge that makes it especially nucleophilic. Why do these compounds ionize to form HO-, whereas other compounds containing the OH group ionize to form H+?
The difference has to do with which bond to the electronegative element, oxygen, is most polar. The most polar bond is the one most likely to ionize.
In alkali metal hydroxides, such as NaOH, the Na-O bond is most polar. The electronegativity difference between sodium and oxygen is larger than that between oxygen and hydrogen. In fact, sodium is on the very left hand side of the periodic table, whereas oxygen is in the upper right hand corner. A combination like that results in an ionic bond, not a covalent one, so sodium hydroxide should be thought of as Na+ and HO-.
In main group hydroxyls such as nitric acid or sulfuric acid, the most polar bond is the O-H bond. N-O bonds or S-O bonds are not nearly as polar as O-H bonds. The O-H bond ionizes more easily.
However, relative bond polarity is not the only factor here. Acidity is strongly influenced by the structure of the ions produced in each case. More stable ions are produced more easily.
If HO- dissociates to form a proton, a O2- or oxide anion will result. That oxygen has a high nuclear charge and a high electron affinity, but a 2- charge is a lot of negative charge on one small atom. This ion will not be very stable.
On the other hand, nitric acid dissociates to form a proton and nitrate ion, NO3-. This ion has a single negative charge on an element that has a high electron affinity, oxygen. In addition, there is an electronegative atom, nitrogen, attached to that oxygen, and another electronegative element, oxygen, attached to that one. Electronegativity is the ability to draw electronic charge through bonds. That means these other electronegative atoms will draw some negative charge away from the anionic atom. In this way, the negative charge is dispersed and stabilized.
• Dispersing or spreading charge out helps to stabilize charge.
• Concentrating more charge in one location is destabilizing.
• Nearby electronegative atoms can help stabilize negative charge.
There is another, related factor that stabilizes the nitrate anion by spreading out the negative charge. Notice that the negative charge could be drawn on two of the three oxygen atoms at once. The negative charge does not have to be on any particular oxygen. It can really be shared by all three. In Lewis terms, we show that fact by drawing all three resonance structures for the nitrate ion.
• Resonance or delocalization spreads out charge, stabilizing it.
In the subsequent sections, we can look at these and other factors in more detail.
Exercise \(1\)
Which of the following acids do you think has the lowest pKa?
a) HClO2 or HClO4 b) H3PO3 or H3PO4
Answer a
HClO4
Answer b
H3PO4
Attribution | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/14%3A_Concepts_of_Acidity/14.10%3A_The_Relationship_Between_Structure_and_.txt |
A Brønsted Acid provides a proton to an electron donor. In doing so, the former Brønsted acid becomes a conjugate base. We can understand a great deal about proton transfer by looking at that conjugate base. If the conjugate base is not very stable, then probably the proton will not be donated. If the conjugate base is very stable, then the proton may be given up more easily.
Electronegativity & Nuclear Charge
The first factor to consider is that atom attached to the proton in the Brønsted acid. That is the atom that will accept a pair of electrons from the covalent bond it shares with the proton. How easily can this atom accept a pair of electrons?
An obvious factor to consider is electronegativity. As the atom attached to the proton becomes more electronegative, the bonding pair of electrons becomes more strongly attracted to that atom, and less attracted to the proton. If the bond becomes more polarized away from the proton, it seems likely that the proton will more easily ionize. The molecule containing this bond will be a stronger Brønsted acid. It will not hold onto the proton as tightly. It will have a lower pKa.
Atoms with higher electronegativities are to the upper right in the periodic table. Moving to the right across a row, the nuclear core charge is increasing, so there is more attraction for electrons.
In addition, we should think about what happens after the proton has ionized. In most cases, a neutral (uncharged) Brønsted acid will give rise to an anionic conjugate base. Proton transfer is generally reversible, so it could always go back where it came from, unless something stabilizes the anion that forms. However, if an atom has a higher nuclear core charge, it will be more stable as an anion than would other atoms. That means a compound with a hydrogen attached to that atom will give up a proton more easily. When we consider anion stabilities, the trend across a row of the periodic table is exactly the same as the trend in bond polarity.
We can compare the pKa's of methane (CH4), ammonia (NH3), water (H2O) and hydrogen fluoride (HF) to examine these ideas. Carbon, nitrogen, oxygen and fluorine are all in a row in the periodic table. Fluorine, to the right, has the highest core charge and highest electronegativity. Carbon, to the left, has the lowest. Water should be a stronger acid than ammonia, which should be more acidic than methane.
In fact, the pKa of hydrogen fluoride is 4; that of water is 15; that of ammonia is 35; and methane's is about 50. Water is much more acidic than ammonia, which is much more acidic than methane. Hydroxide is a more stable ion than amide ion, NH2-, which is a more stable ion than methyl ion, CH3-.
• Electronegativity can reliably be used to compare acidities of two different X-H bonds in the same row.
• The higher the electronegativity of an atom, the more easily its X-H bond ionizes.
• Also, the higher the core charge of an atom, the more stable it will be as an anion, X-, after the proton is lost.
Clearly, none of the compounds illustrated above is highly acidic. If you have learned any chemistry before, you may be familiar with the idea that hydroxide ion is a strong base. Hydroxide has a very strong attraction for protons. It binds a proton to form water, and is much more stable in that form.
However, these comparisons are relative. Amide ion is an even stronger base than hydroxide ion; it binds protons very tightly. Methyl anion is an extremely strong base that binds protons extremely tightly.
Exercise \(1\)
In each case, choose the compound more likely to act as a proton donor.
a) H2S or SiH4 b) GaH3 or AsH3
c) PH3 or AlH3 d) H2Se or HBr
Answer a
H2S
Answer b
AsH3
Answer c
PH3
Answer d
HBr
Charge on the proton donor
The atom attached to the proton influences the acidity in other ways. Whether or not that atom has a formal charge becomes very important. The reason is simple: when a proton is given up, the proton will have a positive charge, and the atom releasing the proton will become more negative. Consequently, if the atom attached to the proton already has a negative charge, it is less likely to give up the proton. If it did, it would take on a charge of 2-. That charge build-up would not be very favorable.
On the other hand, if the same kind of atom had a positive charge, it would be much more likely to give up a proton. Once it did, it would have no charge at all. That would be pretty favorable; no energy would have to be expended in stabilizing a charge that is no longer there.
As an example, consider three related species: water (H2O), hydroxide ion (HO-), and hydronium ion (H3O+). Of the three, the hydronium ion would be the most likely to donate a proton, in order to relieve the oxygen atom of positive charge. Charge stabilisation costs energy, so having no charge at all is often better than having a charge.
The hydroxide ion would be least likely to give up a proton, because that would leave an oxide ion with a charge of 2-. That increase in negative charge on one atom costs energy.
The water ion would be intermediate. There would be an increase in charge on the atom, but it wouldn't be the same as a build up of a 2- charge on the atom.
In other words, hydronium ion is more acidic than water, and water is more acidic than hydroxide ion.
Of course, these general rules about acidity do not always apply in different situations. We know that halogens tend to be pretty stable as anions, for the most part, so their anionic form may be more stable than their neutral form. Conversely, positive metal ions may be more stable than the corresponding neutral atom. In the specific case of how likely an atom is to give up a proton, however, the development of charge or loss of charge does become an important factor. This factor is especially important when we are comparing two atoms of the same type, such as two oxygen atoms or two nitrogen atoms.
Exercise \(2\)
In each case, assess whether a formal charge is present. Choose the compound more likely to act as a proton donor.
a) +NH4 or NH3 b) PH2 or PH3 c) H2O or +NH4 d) H3O+ or CH4
Answer a
+NH4
Answer b
PH3
Answer c
+NH4
Answer d
H3O+
Polarizability
Let's look at another example, comparing the pKa's of hydrogen halides. These compounds are all much stronger acids than water. Hydrogen fluoride, HF, has a pKa of 4. Hydrogen chloride, HCl, has a pKa of -1. Hydrogen bromide, HBr, has a pKa of -5, and hydrogen iodide, HI, has a pKa of -7.
Fluoride has the highest electronegativity of the four conjugate bases here, and iodide has the lowest. However, fluoride binds its proton most strongly, and HF has the highest pKa. Something other than electronegativity is at work here.
• Electronegativity differences cannot explain the differences in acidity of H-X bonds in a column of the periodic table.
One way of discussing this trend is in terms of polarizability. A polarizable atom is generally a large atom that can distribute charge easily over a greater volume; charge is less concentrated than it would be in a smaller atom. The distribution of charge is stabilizing.
• When comparing anionic atoms from the same column of the periodic table, the polarizability of the atom (related to its size) can be used to explain different anion stabilities.
The reason polarizability dominates comparisons within a column, but not within a row, has to due with the relatively large change in size of atoms from one row to the next. As electrons occupy an additional energy level, the size of the atom increases greatly. Atoms also change size as we move across a row of the periodic table, getting a little smaller as the nuclear charge increases. However, this change is not as dramatic as the change in size from one row to the next.
Earlier, we looked at both sides of the equation, before and after ionization, in terms of core charge and electronegativity. Both considerations led to similar conclusions about which bonds would be most acidic. So far, we have looked at the anions formed when hydrogen halides ionize. Is there a factor (other than electronegativity) we can use in comparing the hydrogen halides directly?
Among these compounds, the bond strength increases from the bottom of the column to the top (from about 70 kcal/mol in HI to about 135 kcal/mol in HF). That means it is much easier to break a hydrogen atom away from an iodine atom in HI than to break a hydrogen atom away from a fluorine atom in HF. That isn't what we are doing when we ionize these bonds; we are breaking a proton away from an anion in each case, rather than a neutral hydrogen atom away from another neutral atom. Nevertheless, whatever factors influence bond strength may be affecting acidity as well.
Bond strengths actually vary in this way indirectly because of the relative electronegativities of the halogens. The valence electrons on fluorine, a more electronegative atom, are at a lower energy than those on either bromine or chlorine. When the valence electrons on a hydrogen atom and a halogen atom combine to form a bond, the bonding combination is at a lower energy than either of the originals.
Of course, the antibonding combination is at a higher energy than either of the originals, but since there are no electrons at that level we don't have to worry about that.
What is a bond strength? It's just the amount of stabilisation upon formation of the bond. Put in an allegorical way, it's how deep the gully is that the electrons have rolled into when the bond forms. To break the bond, the electrons would have to climb the hills back out of the gully again. For example, the bond strength of HBr involves the amount of energy by which the electron on hydrogen has fallen, ΔE1, plus the amount by which the hydrogen on bromine has fallen, ΔE2. To break the bond again, we would have to add in the sum of ΔE1 + ΔE2.
The analogous amount for HCl is greater, mostly because ΔE1 in that case is much larger, and it is even greater for HF. Thus, bond strengths vary in the order HF > HCl > HBr.
While we are looking at that picture, it's worth considering another common aspect of bond strengths. As a simple approximation, there are two important components that describe a bond. One component is overlap (how well the eletrons are shared; that is, how covalent is the bond). The other component is exchange (how much electrostatic attraction there is between the atoms; that is, how polar is the bond). In fact, bromine is better at sharing its electrons with hydrogen than is fluorine, but the H-F bond is more polar than the H-Br bond. The latter fact is another consequence of fluorine's electronegativity. Once again, the greater electronegativity of fluorine actually leads to a stronger bond in this case.
• In comparing two atoms in a column of the periodic table, bond strength with hydrogen is a good index of how acidic the bond will be.
Note that bond strength does not work as a comparison of H-X bonds in a row of the periodic table. An HF bond is stronger than an OH bond (about 110 kcal/mol) or CH bond (about 100 kcal/mol on average), yet it is much more acidic.
Now, in summary, we need to think about both sides of the equation. The H-F bond is the strongest; that is, it is lowest in energy. However, upon ionisation, bromide is the most stable anion, because it is the largest and most polarizable. Overall, the reaction from HBr to bromide is the most downhill in energy; that one will happen most easily. In contrast, the reaction from HF to fluoride is actually uphill in energy; that one will be most difficult to ionise.
Exercise \(3\)
In each case, choose the compound more likely to act as a proton donor.
a) H2S or H2Te b) GeH4 or SnH4 c) HCl or HBr d) NH3 or AsH3
Answer a
H2Te
Answer b
SnH4
Answer c
HBr
Answer d
AsH3
Exercise \(4\)
Rank the following groups of compounds from most acidic to least acidic.
a) PH3, NH3, AsH3 b) PH3, H2S, SiH4, HCl c) BH3, SiH4, SeH2, HI
Answer a
AsH3 > PH3 > NH3
Answer b
HCl > H2S > PH3 > SiH4
Answer c
HI > SeH2 > SiH4 > BH3
Exercise \(5\)
The relative acidities of compounds in some of the following pairs can be explained by polarizabilty. In others, they cannot. Explain why polarizability is or is not a useful comparison to make in each case.
a) CH3OH, (CH3)2CHOH b) SH2, SeH2 c) GeH4, SiH4 d) GeH4, GaH3
Answer a
Polarizability would not be useful as the atom that will be anionic in the conjugate base is the same (oxygen).
Answer b
Se is larger and therefore more polarizable leading to H2Se being more acidic compared to H2S.
Answer c
Ge is larger and therefore more polarizable leading to GeH4 being more acidic compared to SiH4.
Answer d
Polarizability would not be useful as the atoms are next to one another in the same row and therefore likely to have the same polarizability
Hybridization
There is a special case related to the electron affinity factor described above. It generally shows up in discussions of carbon or nitrogen compounds, and specifically deals with comparisons of two atoms of the same element having different geometry. For example, the C-H bond involving a tetrahedral carbon is extraordinarily non-acidic (pKa near 50), but a C-H bond involving a linear carbon is much more acidic (pKa of about 25).
In ethane, which has tetrahedral carbons, a similar argument is made. The difference is that a tetrahedron is a three-dimensional shape. The s orbital can still bond to neighbors in any direction because of its spherical symmetry. However, in order to bond with neighbors in all three dimensions, the entire set of p orbitals would be needed, because each p orbital lies along one dimension only.
The energy of the electrons in this bonding situation will be an average of the energy of an s electron and three p electrons. In terms of carbon's energetic contribution, the energy of the electrons in the C-H bond will be three-quarters of the way between what would be expected if the bond formed between an H atom and a carbon 2s electron and if the bond formed between an H atom and a carbon p orbital. That is because the energy should be an average of the energies of all four orbitals that could contribute to that bond.
Because s electrons are lower in energy than p electrons, the energy of the (s + p) combination will be lower than the energy of the (s + p + p + p) combination. An electron in an "sp orbital" at linear carbon is lower in energy than an electron in the "sp3 orbital" at tetrahedral carbon. If the electron is at lower energy on a linear carbon than a tetrahedral carbon, then a linear carbon has a higher electron affinity than a tetrahedral one.
This comparison is much like the comparison of electron affinity on an oxygen vs a carbon. Oxygen, with a higher electron affinity, generally forms more stable anion than does carbon. An O-H bond is thus more acidic than a C-H bond. A liner C-H bond is more acidic than a tetrahedral C-H bond.
Exercise \(6\)
Predict which of these two nitrogen compounds will be more basic.
Answer | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/14%3A_Concepts_of_Acidity/14.11%3A_Factors_Affecting_Bronsted-Lowry_Acidit.txt |
Sometimes there are factors farther away from the proton that affect Brønsted acidity. In addition to considering the atom to which the proton is directly attached, we may need to consider other parts of the molecule to understand anion stability.
A first factor to consider is delocalization due to conjugation. In Lewis structure terms, that means resonance. Delocalization of charge is stabilizing, so if a negative charge can be distributed across multiple atoms via resonance, a conjugate base will be more stable.
• resonance can delocalize negative charge and stabilize an anion.
• a more stable conjugate base forms more easily. A Brønsted acid will give up a proton more easily if it gives rise to a stable conjugate base.
Many organic compounds contain hydroxyl or OH groups, some of which are acidic and some of which are not. Three kinds of compounds containing this group are alcohols, phenols and carboxylic acids. Examples of these functional groups include cyclohexanol, phenol and benzoic acid.
Cyclohexanol is an alcohol. An alcohol contains an OH group connected to a tetrahedral carbon.
Phenol is slightly different; it contains an OH group connected to a trigonal planar carbon that is part of an aromatic ring. Aromatic rings are cyclic groups of atoms, usually carbons, with delocalized pi bonding all around the ring. Benzene is the most common example of an aromatic ring. It is a ring of six carbons and all of the carbons are trigonal planar. In the Lewis structure, it is drawn with alternating single and double bonds between the carbons.
Benzoic acid contains a very different functional group in which the OH is connected to a carbonyl. A carbonyl is a carbon-oxygen unit with a double bond, C=O. The carbon in a carbonyl is trigonal planar, as in a phenol, but the presence of the double bond to oxygen makes a big difference.
Cyclohexanol has a pKa of about 18. It is less acidic than water. It can give up a proton, but the proton is much more likely to be bound to the oxygen than disscociated. Phenol has a pKa of about 9 and is considered mildly acidic. Benzoic acid has a pKa of about 5; its acidity is similar to hydrofluoric acid, although not nearly as acidic as other hydrogen halides.
Comparing the conjugate bases of cyclohexanol, phenol and benzoic acid reveals some differences.
• In cyclohexanol, the anion formed by loss of a proton is localized on the oxygen. There is no resonance stabilization.
• In phenol, the anion formed by loss of a proton is delocalized. Resonance structures show the negative charge can be shared between the oxygen atom and three of the carbons in the benzene ring.
• In benzoic acid, the anion is also delocalized. This time the negative charge is shared between two different oxygen atoms.
Resonance delocalization plays a clear role in stabilizing the conjugate base formed after loss of a proton. Despite the similar bond polarity in cyclohexanol and phenol, the proton is much more tightly bound in cyclohexanol. The greater stability of the phenolate anion compared to the cyclohexanoxide anion makes it easier to remove a proton from phenol than cyclohexanol.
• The atoms onto which the charge is delocalized still plays a major role in determining stability. Delocalization onto an additional oxygen atom may be more stabilizing than delocalization onto several carbons.
Exercise \(1\)
Some compounds have acidic C-H bonds, despite the fact that many hydrocarbons have extremely high pKa's (such as methane, CH4, pKa = 50). Explain the reason for the trend in pKa's among the following compounds.
cyclopentadiene, pKa 25 cyclopentanone, pKa 18 2,4-hexanedione, pKa 12
Answer
Exercise \(2\)
Compare the acidity of the following pairs of compounds.
1. ethanol, CH3CH2OH, and vinyl alcohol, CH2=CHOH.
2. trimethylamine, (CH3)3N, and nitromethane, CH3-NO2.
3. acetonitrile, CH3-CN, and trimethylamine, (CH3)3N
Answer a
Vinyl alcohol would be more acidic as its anion is stabilized by resonance while the anionic charge on ethoxide would be localized on the oxygen atom.
Answer b
Nitromethane would be more acidic as its anion is stabilized by resonance while the anionic charge of the trimethylamine anion would be localized on the carbon atom.
Answer c
Acetonitrile would be more acidic as its anion is stabilized by resonance while the anionic charge of the trimethylamine would be localized on the carbon atom.
There is another way that distant atoms can influence anion stability. It has to do with electronegativity.
Consider the pKa of these three, similar, halogenated carboxylic acids: bromoacetic acid, chloroacetic acid and fluoroacetic acid. The pKa of fluoroacetic acid is lower than that of chloroacetic acid, which is lower than that of bromoacetic acid. Fluoroacetic acid is more acidic than chloroacetic acid, which is more acidic than bromoacetic acid.
When the proton is directly attached to these three halogens, the pKa runs in the opposite direction. HBr is more acidic than HCl, which is more acidic than HF. That was because of the greater polarizability of the bromide versus the chloride and fluoride.
In these haloacetic acids, the halogen can stabilize the conjugate anion via inductive delocalization. In an inductive effect, electronegative atoms can draw electron density toward themselves. That means the halogen shares the partial negative charge of the oxygen atoms in the carboxylate anion.
• electronegative atoms can draw negative charge toward themselves through bonds.
• this phenomenon is called an inductive effect.
• inductive effects spread out negative charge and stabilize anions.
The effect is additive. If more than one halogen is nearby, there is more electron-withdrawing effect. There is more positive charge on the proton, and when the proton is released, the resulting anion has more charge delocalization.
Exercise \(3\)
Predict the order of acidity in the following compounds.
a) CF3CO2H, CFH2CO2H, CF2HCO2H b) CF3CHCl2, CF3CHClF, CF3CHF2
Answer a
CF3CO2H > CF2HCO2H > CFH2CO2H
Answer b
CF3CHF2 > CF3CHClF > CF3CHCl2
An important limitation on inductive effects is seen in comparing a series of chlorobutanoic acid derivatives. In 2-chlorobutanoic acid, the presence of a chlorine next to the carbonyl, four bonds from the acidic proton, renders this compound much more acidic than butanoic acid; the pKa's are about 2.9 and 4.8, respectively.
In 3-chlorobutanoic acid, the pKa is 4.0, whereas in 4-chlorobutanoic acid, with the fluorine all the way at the end of the chain, six bonds away from the acidic position, the pKa is 4.5. The fluorine in these last two cases has relatively little effect.
• Inductive effects fall off quickly with increasing distance from the acidic site.
However, in aromatic systems the effect of distance is slightly weaker. In 2-chlorobenzoic acid, the effect of the chlorine is substantial, despite its distance from the acidic proton. The pKa of 2-chlorobenzoic acid is 2.94, as compared to 4.2 for benzoic acid, despite the fact that the chlorine is 5 bonds away from the acidic proton. This increased influence of inductive effects is common in aromatic systems such as benzene.
• Inductive effects are readily transmitted across conjugated systems like benzene.
Aromatic systems contain delocalized, polarizable sets of electrons. In contrast to sigma bonds, in which electrons are delocalized between two atoms yet still located pretty reliably in between those two atoms, pi bonds hold electron density further from the nuclei, and are more easily distorted or polarized.
Exercise \(4\)
For each of the following compounds:
a. Determine the most acidic proton and provide a reason for your choice.
b. Draw the conjugate base.
Answer | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/14%3A_Concepts_of_Acidity/14.12%3A_Factors_affecting_Bronsted-Lowry_Acidit.txt |
All of the factors that we have discussed for Brønsted acidity, or the ability of a compound to provide a proton to its surroundings, have an effect on basicity as well. In other words, factors like nuclear charge / electron affinity influence how strongly a compound will attract or bind a proton.
In summary:
• the higher the electron affinity or core charge of an atom, the less likely it is to donate its electrons to a proton.
• the greater the delocalization of electrons that could potentially donate to a proton, the less able they are to donate.
• the greater the electron-withdrawing effects in another part of a molecule, the less likely the electrons on a particular atom are to donate.
These factors are generally complementary to the effects on acidity. A factor that makes a Brønsted acid more acidic usually makes the corresponding conjugate base less basic.
However, sometimes things can be more subtle.
• the higher the polarizability of an atom (i.e. the larger an atom), the more easily it can donate to a Lewis acid (its electrons are not held very tightly because they are far from the nucleus, and so they can be donated easily).
• except: a larger atom cannot donate easily to a proton. In this specific case, the Lewis acid (the proton) is too small to get good covalent overlap with the Lewis base, so it can't form a very strong bond
Exercise \(1\)
Rank the following in terms of base strength (1 = strongest base).
Answer a
Answer b
Answer c
Answer d
14.14: The Direction of Proton Transfer
Brønsted-Lowry acidity is a special case of Lewis acidity. In Lewis acidity, an electron donor shares electrons with an electron acceptor, forming a bond. In some cases, the electron acceptor is a proton.
If the proton accepts electrons from the donor, but does not relinquish its bond to its previous partner, the interaction is called a hydrogen bond.
If the proton exchanges a new bond with the donor for an old bond with its previous partner -- if it releases a pair of electrons to its partned as it accepts a pair of electrons from the new donor -- the event is called proton transfer.
The compound that provided the proton is called a Brønsted-Lowry acid. The compound that donated the new bond to the proton is called the Brønsted-Lowry base.
Because the proton allowed its former partner to take the pair of electrons from their former bond, that partner becomes a Lewis base. In a proton transfer, the proton moves from one Lewis base to another. The proton could conceivably move back to its original partner, however. The original partner could simply donate its pair of electrons to the proton again. It would displace the new partner and win the proton back.
This situation is called reversibility. Reactions can often move back and forth. In order to convey this idea, when illustrating or writing about these reactions, a pair of opposing arrows are used to show that the reaction can go from left to right as written, as well as from right to left.
• In a reversible reaction, the change that occurs in the reaction can be undone. The reaction can move forwards and backwards.
In most cases, the reaction settles out on one side or the other. Either the reaction goes mostly forward or it goes mostly backward. The point where the reaction settles is termed the equilibrium. At equilibrium, the reactants that come together may mostly be converted into new products. Looking at an equation or diagram of the reaction, , the equilibrium is said to "lie to the right", because the products of the reaction are usually written on the right hand side of the reaction arrow or equilibrium arrow. Some reactions "lie to the left", meaning very little of the original reactants are ultimately converted to the products show.
• The equilibrium is the balance established between the products formed and the original reactants in a reversible reaction.
In a proton transfer, the equilibrium is determined by how tightly the proton is held by each Brønsted acid. The proton will simply remain bonded to whichever compound binds it more tightly. If the difference in binding is great, the equilibrium will lie far to the left or far to the right. If the difference in proton binding is small, there will be a mixture, in which the proton could be in either position.
Remember, most reactions involve zillions of molecules. There is plenty of room for mixtures.
We can predict where the proton will end up by looking at pKas.
• A higher pKa means the proton is more tightly held.
• By comparing the pKa's of the Brønsted acids on both sides of the equation, we can determine which compound will retain the proton.
• The equilibrium will lie towards the compound with the higher pKa.
This idea is illustrated in the equilibrium between hydronium ion and ammonium ion.
As another example, if hydrogen chloride is dissolved in water, the HCl may give up its proton to the water. Water has a lone pair and can act as a base. However, in doing so, the water will form hydronium ion, H3O+. Hydronium ion is Brønsted acidic and can provide a proton to something else that has a lone pair, such as a chloride ion. This reaction could go back and forth. Where will it settle out?
HCl has a pKa of -8. Hydronium ion has a pKa of about -1.7. The equilibrium in the reaction described above lies to the right, towards the hydronium ion produced when the hydrogen chloride dissociates. The proton will remain on the oxygen.
Exercise \(1\)
Write an equation for the proton transfer reactions that could happen in each of the following mixtures. Use structures in your equations. Predict the position of the equilibrium in each case.
a) HF plus water b) CH3CO2H plus ammonia
c) phenol (C6H5OH) plus sodium carbonate d) HCN plus acetonitrile
Answer a
Answer b
Answer c
Answer d
Exercise \(2\)
Use curved arrows to show the proton transfer reaction between the following compounds.
Predict the products for these proton transfer reactions.
Use pKa to determine whether each reaction is reactant favored OR mixture OR product favored.
Answer | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/14%3A_Concepts_of_Acidity/14.13%3A_Effects_on_Basicity_%28Attraction_for_P.txt |
Knowing something about proton transfer changes how we look at some important biomolecules. Amino acids are the fundamental building blocks of peptides and proteins. Peptides, which are chains of amino acids, are frequently used as signaling molecules within the body (some hormones are peptides). Proteins, which are very large peptides, have a variety of uses. They form the key components of muscles, for instance, and they also form enzymes that carry out a multitude of chemical reactions necessary for life.
Amino acids are so called because they all contain two common components. One is an amine, or a tetrahedral nitrogen attached to a carbon. The other is a carboxylic acid, which is a carbon that is double bonded to an oxygen and also attached to an OH or hydroxyl group.
We have seen that carboxylic acids are moderately acidic. Most of them have pKa's of 3 to 5. That means a small fraction of the OH groups are ionized in a large group of carboxylic acids.
We have also seen that tetrahedral nitrogens are somewhat Lewis basic. The nitrogen can donate its lone pair to Lewis acidic atoms. Protons are good Lewis acids. Amines are easily protonated if protons are available.
Because the carboxylic acid is a pretty good source of protons and because protons bind to amines pretty well, it seems reasonable that a proton transfer may occur from one site to the other.
Does one of these forms dominate the equilibrium? Compare the pKa's. The pKa of the acid is near 5, and the pKa of the ammonium is near 9. The ammonium holds the proton more tightly than does the acid. The proton stays on the nitrogen.
Amino acids are zwitterionic. A zwitterion is a compound that has no overall charge but that has charge separation within it. The zwitterionic nature of amino acids has an effect on their properties. For example, they are usually pretty soluble in water and other polar solvents.
Amino acids are joined together via "amide linkages" to form peptides and proteins. In these structures, the individual amino acids no longer have the same acidic carboxylic acid group; the carbonyl (or C=O) no longer has a hydroxyl group attached. The amino acids no longer contain amines, either; a nitrogen attached to a C=O has very different properties than a regular nitrogen attached to carbon. Only the "N-terminus" and "C-terminus" are ionic. The nitrogens along the chain are not very basic and are not protonated.
Exercise \(1\)
Explain why an amide nitrogen is not very basic.
Answer
Exercise \(2\)
The favored site of protonation in an amide is the carbonyl oxygen. Show why.
Answer
The acidic and basic groups found in individual amino acids are masked in peptides and proteins. However, peptides and proteins do have basic and acidic sites. These sites are found on the side chains of the amino acids, the part that varies from one amino acid to another. In some cases, the side chain contains an acidic group. Examples are aspartic acid and glutamic acid.
There are other side chains that are weakly acidic. For example, tyrosine is sometimes able to supply protons, and so is cysteine. The proton comes from the OH and SH groups, respectively, on these two compounds. However, neither of these compounds can supply protons as easily as aspartic acid or glutamic acid. Furthermore, serine, which also has an OH group, is not really able to supply protons that easily.
Exercise \(3\)
Explain the difference in acidity between serine and tyrosine, which both contain OH groups.
Answer
Exercise \(4\)
Explain the difference in acidity between serine and cysteine, which have very similar structures but with a sulfur atom in place of an oxygen.
Answer
The thiol proton on cystine is more acidic due to the polarizability of the sulfur anion stabilizing the conjugate base's anionic charge. The oxygen anion is less polarizable and therefore less stable rendering the alcohol proton less acidic.
Sometimes the amino acid side chain contains a basic group. Examples are histidine, lysine and arginine. There is a big difference in basicity between these three compounds. The difference can be seen by looking at the pKa's of the conjugate acids in each case. The higher the pKa of the conjugate acid, the more tightly the proton is held, and so the more basic the nitrogen atom. Arginine is by far the most basic and histidine is the least basic.
Exercise \(5\)
Show why the indicated nitrogen in arginine's side chain is protonated, and not the others. Also, explain the difference in basicity between arginine and lysine.
Answer
Exercise \(6\)
Explain the difference in basicity between histidine and lysine.
Answer
In contrast to the basic amino acids shown above, other amino acids with nitrogen in their side chains are not considered basic. These compounds are not protonated easily on their side chains.
Exercise \(7\)
Explain why the three un-basic amino acids are not easily protonated on their side chain nitrogen atoms.
Answer | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/14%3A_Concepts_of_Acidity/14.15%3A_Amino_Acids_and_Peptides.txt |
Reactivity is strongly affected by the environment around the molecules that are reacting. Usually the environment is a solvent; a solvent is a liquid in which the molecules are dissolved. Solvent effects (how different solvents behave) and solvation (how solvents organize around a solute molecule) are very important to consider in thinking about acidity.
Solvation is very important. When molecules are dissolved in a liquid, they can move around easily and mix with other molecules. That ease of motion facilitates reactions. In contrast, molecules in the solid state hardly move at all. Solid state reactions are very, very slow because molecules can't easily come into contact with each other. If two solids are mixed, molecules on the surface of the solid grains may react, but the molecules buried inside will be left untouched.
In order for one compound to dissolve in another, some intermolecular attractions must be present. In order to maximize these interactions, the solvent molecules probably need to arrange themselves somehow.
This organization of solvent molecules becomes even more important when ions are involved than when neutral molecules are dissolved. When ions are dissolved, anions are separated from cations. Solvent molecules must be able to interact with the ions so as to mitigate against the energetic costs of charge separation.
In organizing around a solute molecule, the usual interactions between the solvent molecules are disrupted. The solute molecule occupies a gap in the solvent molecules. For example, if the solvent is water, there must be a break in hydrogen bonding between water molecules to allow for the solute to swim among the waters.
Earlier we saw that larger atoms can more easily accommodate charge. This rule does not extend to the size of molecules. A larger molecular ion is generally not as easy to solvate as a small one. Larger ions require much more organization of solvent molecules. In addition, interactions between the solvent molecules (generally very favorable in solvents such as water) must be given up so that solvent molecules can move away from each other to open up space for the guest ions.
• The size of a molecular ion being accommodated in a solvent matrix has an effect on the ion stability.
Different solvent effects arise partly because of different intermolecular attractions available to solvent molecules. Some solvents are hydrogen bonding. Water is a very common example, as are alcohols such as methanol. Others are only hydrogen bond-accepting, but not hydrogen bond-donating. Acetone and acetonitrile are examples.
Some solvents may more efficiently stabilize anions that form when Brønsted acids are deprotonated. As a result, pKa values may be different when measured in different solvents. For example, the pKa of water is reported as 15.7 in pure water, but when dissolved in DMSO it is reported as 32. Water is much less acidic in DMSO than when it is pure.
DMSO has something in common with water. Water is very polar, with a polar O-H bond. DMSO is very polar, having a strongly polarized S-O bond. Water can hydrogen bond. DMSO is capable of accepting hydrogen bonds. That means DMSO can use its lone pairs to donate to protons on other molecules. However, DMSO is aprotic. It does not have very positive hydrogens that can participate in hydrogen bonding. That means one DMSO molecule cannot hydrogen bond to another DMSO molecule like water can. As a result, it is not able to donate hydrogen bonds to anions, as water can. Water is therefore more able to stabilize anions, so molecules can ionize more easily in water than in DMSO.
• Hydrogen bonding can be important in anion stabilization.
There is another reason to be aware of solvent effects in proton transfer reactions. Sometimes, solvents can themselves become involved in acid-base reactions.
For example, water is a very weak acid, but it can give up a proton. When it does so, it forms a hydroxide ion. Water could give its proton up to another anion if that anion could bind a proton more tightly than could hydroxide.
An example of this situation would occur if sodium amide were dissolved in water. Ammonia binds its proton more tightly than water. Thus, if sodium amide were dissolved in water, it would immediately become ammonia, removing a proton from water and forming sodium hydroxide.
As a result, the pKa of ammonia could not easily be measured in water because its conjugate base does not really exist in water. Assessing a pKa requires comparing how much of a compound remains protonated and comparing it to how much of the compound ionizes. As it turns out, the pKa of ammonia is around 41. That is high enough that this value was not determined directly. Instead, it had to be extrapolated by comparison with other data.
• Sometimes, solvent participate in reactions. They are not always innocent bystanders.
Exercise \(1\)
Explain why ammonium bromide, NH4Br, is more soluble in water than is sodium bromide, NaBr.
Answer
Ammonium bromide is more soluble because in water it can participate in both ion-dipole and hydrogen-bonding, while sodium bromide only benefits from ion-dipole interactions.
Exercise \(2\)
The pKa of hydrogen cyanide, HCN, is about 13 in DMSO. Predict qualitatively how the pH would change if measured in a) water or b) pentane, CH3CH2CH2CH2CH3.
Answer a
Hydrogen cyanide will be have a lower pKa in water as the resulting cyanide anion will be stabilized by both ion-dipole and hydrogen-bonding interactions.
Answer b
Hydrogen cyanide would have a higher pKa in pentanes as the resulting cyanide anion would only experience ion-induced dipole interactions which are relatively weak. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/14%3A_Concepts_of_Acidity/14.16%3A_Relative_Conditions_and_pKa.txt |
The index we have used to assess Brønsted acidity, pKa, is a measurable quantity. It is determined by measuring the ratio of products to reactants in a proton transfer reaction.
For example, if HCl is dissolved in water, much of it ionizes, transferring a proton to water to form hydronium ion and chloride anions. The concentrations of all four of these species could be measured and compared. The ratio of products to reactants is called the equilibrium constant, or Keq:
$K_{eq} = \frac{[H_{3}O^{+}][Cl^{-}]}{[H_{2}O][HCl]} \nonumber$
The concentrations of these species are generally reported in moles per liter. A mole, you may know, is a unit used to count very large numbers of molecules. Since molecules are very small, we usually deal with very large numbers of them at a time.
By convention, the concentration of water in itself is defined as 1. That leads to a slightly different expression.
$K_{eq}' = \frac{[H_{3}O^{+}][Cl^{-}]}{[HCl]} \nonumber$
This ratio, dealing with proton transfer, is also called the acidity constant, Ka.
$K_{a}= \frac{[H_{3}O^{+}][Cl^{-}]}{[HCl]} \nonumber$
Exercise $1$
Write the expression for the Ka in each of the following mixtures.
a) HCN in water b) H2S in water c) NH3 in DMSO (DMSO = (CH3)2SO)
Answer a
Answer b
Answer c
The Ka is often a very, very small number or a very, very large one. In the case of HCl in water, the Ka is about 1 x 10-8. Dealing with exponents can be cumbersome. In order to simplify comparisons, the equilibrium constant is expressed logarithmically.
$pKa = -\log K_{a} \nonumber$
The Ka is often a very, very small number or a very, very large one. In the case of HCl in water, the Ka is about 1 x 10-8. Dealing with exponents can be cumbersome. In order to simplify comparisons, the equilibrium constant is expressed logarithmically.
Exercise $2$
Convert the following Ka's to pKa's.
a) 1 x 106 b) 1 x 10-9 c) 3.5 x 10-25 d) 8.5 x 10 -17
Answer a
pKa = -6
Answer b
pKa = 9
Answer c
pKa = 24
Answer d
pKa = 16
Exercise $3$
Convert the following pKa's to Ka's.
a) -3.5 b) 4.3 c) 9 d) 25
Answer a
Ka = 103.5
Answer b
Ka = 10-4.3
Answer c
Ka = 10-25
Answer d
Ka = 10-9
Equilibrium constants are not restricted to proton transfer. They can be used to describe the extent to which any reaction occurs. For example, they can be written for other, reversible processes involving acid-base chemistry.
Lewis acid-base interactions are very often reversible. For example, a Lewis base like ether can donate a pair of electrons to a Lewis acid such as BF3. The ether can take its electrons back again and leave the BF3 behind, too. How tightly the ether is held by the BF3 is termed the binding constant. In this case,
$K_{eq}= \frac{[BF_{3}][Et_{2}O]}{[BF_{3}OEt_{2}]} \nonumber$
In this case, the equilibrium constant has been reported to be about 0.25.
Exercise $4$
What does the equilibrium constant for formation of a complex between BF3 and diethyl ether (above) tell you about the position of the equilibrium?
Answer
At equilibrium formation of the Lewis acid-base complex is slightly favored.
Exercise $5$
The equilibrium constant for complex formation between dimethyl ether (CH3OCH3) and BF3 is 4.2. Compare this value with the one for diethyl ether and explain the difference.
Answer
At equilibrium formation of the Lewis acid-base complex of dimethylether and BF3 is more favored than the corresponding diethylether BF3 complex. This is likely due to the decreased steric bulk of the methy groups compared to the ethyl groups.
Exercise $6$
Write expressions for the binding constant in the following cases.
1. (NH3)2PtCl2 loses an ammonia
2. Mo(CO)6 loses a carbon monoxide
3. FeCl4- loses a chloride anion
Answer a
Answer b
Answer c | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/14%3A_Concepts_of_Acidity/14.17%3A_The_Meaning_of_pKa-_Product-to-Reactant.txt |
The term pH refers to the amount of readily available protons present in the environment. If the environment is water, pH is an index of the concentration of hydronium ion (\(\ce{H3O+}\)) in the water.
pH is related to pKa. Both indices work on a logarithmic scale to avoid carrying large amounts of decimal places in very small numbers. A pH of 6, typical in many biological environments, means that the hydronium ion concentration is about 10-6 moles/L (in which a mole is a unit used for convenient counting of nanoscopic things like molecules or ions, just like a dozen is used for counting eggs or doughnuts).
A low pH actually means there are lots of protons or hydronium ions around. At low pH, the environment is very acidic. Low pH is usually associated with the presence of strong Brønsted acids. The typical pH of about 3 in the stomach is produced by dilute hydrochloric acid, HCl.
A second factor that affects pH is the concentration of species in solution. For example, if a solution of HCl is more concentrated, then clearly more protons will be made available, so the pH will be lower. A solution in which the HCl concentration is 0.1 moles/liter (or 10-1 mol/ L) will have a pH of about 1, but a solution in which HCl concentration is 0.001 moles/liter (or 10-3 mol/L) will have pH of about 3. Note that pH is mathematically related to the exponent in the concentration of an acid when written in scientific notation.
Control of pH is very important in biological systems. Many biological processes operate at an optimum pH, and many biomoecules are stable only across a certain pH range. Proteins are particularly sensitive to conditions including pH. Changes in conditions can easily lead to proteins becoming denatured, which means the protein undergoes a shape change that severely affects its ability to function. This shape change is a conformational change, and it is brought about by changing interactions along the protein chain, including changing electrostatic interactions when different sites become protonated or deprotonated.
Different organisms can have different pH ranges over which they function best. Even different tissues within the same organism may work best at different pH values. In order to maintain pH balance at an optimum level, biological systems employ buffers. Buffers are compounds that can either absorb or provide protons in order to keep the pH of their environment from changing. Because they need to absorb or provide protons, buffers are weak Brønsted acids or weak bases, together with their conjugates.
Bicarbonate is an example of a buffer. When pH gets too high, bicarbonate can provide a proton, becoming carbonate.
When pH gets too low, the conjugate base, carbonate, can absorb a proton and become bicarbonate again.
• Buffers maintain pH balance by intercepting acids & bases.
• Adding a strong base such as hydroxide would deplete protons from the system, raising pH; the buffer provides protons.
• Adding a strong acid such as hydronium chloride would drop pH; the buffer picks up the protons.
Exercise \(1\)
For the buffer systems described below, use arrows to show how the buffer system would neutralize (i) a hydroxide ion and (ii) a hydronium ion.
1. Ammonia (\(\ce{NH3}\)) / ammonium chloride (\(\ce{NH4Cl}\))
2. Sodium dihydrogen phosphate (\(\ce{NaO2P(OH)2}\)) / sodium hydrogen phosphate (\(\ce{Na2O3POH}\))
3. Sodium hydrogen phosphate (\(\ce{Na2O3POH}\)) / sodium phosphate (\(\ce{Na3PO4}\))
4. Histidine (see above) / histidine hydrochloride (histidine.HCl)
Answer a
Answer b
Answer c
Answer d
Because the different Brønsted acids involved with these buffers have different pKa's, and so they hold protons more or less tightly, different buffers are able to operate at different pH ranges. This factor, together with control over the concentrations of the components of the buffer system, allows pH to be held fairly constant at almost any value necessary. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/14%3A_Concepts_of_Acidity/14.18%3A_pH_and_Buffers.txt |
Exercise \(1\)
Fraser Stoddart (Northwestern University) shared the 2016 Nobel Prize in chemistry for the development of “molecular machines”. The components of a simple molecular valve are shown here. (Adapted with permission from Nguyen, T. D.; Leung, K. C.-F.; Liong, M.; Pentecost, C. D.; Stoddart, J. F.; Zink, J. I. Org. Lett. 2006, 8, 3363-3366. Copyright 2006 American Chemical Society).
a) What structural feature do all bases have in common?
b) Which atoms in the molecules shown below could potentially be basic: F C N O Si H
Order these basic atoms from most basic to least basic.
State the reason for this order.
c) Circle the most basic site in each of the three molecules (one in the tether, one in the dye, one in the cap).
Box the most basic site of all.
State the reason for its superior basicity.
d) Modify the drawing above to show that site in its protonated state.
The tether is bonded to a silica surface. When the system is protonated, the cap remains bound to the tether (below, left).
e) What intermolecular force binds the cap and tether together?
f) Why does the dye get trapped?
When placed in water, the nanoparticles remain bright yellow and the water remains colorless. If triethylamine is added, the nanoparticles turn white and the water turns yellow.
g) After adding triethylamine, where is the dye?
h) Draw triethylamine.
Show, with curved arrows, what triethylamine does to the molecular valve. You don’t need the entire structures; you can abbreviate to just the part you are using.
i) Indicate any changes in the intermolecular forces after addition of triethylamine.
N,N-Diisopropylethylamine can also release the dye, but more slowly. Triethylamine causes the dye to be released with a half-life of 100 seconds; the half-life for release with N,N-diisopropylethylamine is 300 seconds.
j) Draw N,N-diisopropylethylamine.
k) Why is release so much slower with N,N-diisopropylethylamine?
This system has the potential to be used in a number of applications, such as drug delivery: the slow release of pharmaceuticals into the bloodstream from silica nanoparticles.
If possible, a nitrogen is circled in each molecule. The cap has only basic oxygen atoms. If there is a choice between two nitrogens (in the tether) or two oxygens (in the cap), the non-conjugated lone pair would be most basic. Conjugated lone pairs are held in place by their stable interaction with their neighbors.
The most basic of all is the amine (non-conjugated) nitrogen in the tether.
Answer a
all bases have lone pairs.
Answer b
F, N, and O have lone pairs. They can be bases.
Their order of basicity would be N > O > F.
These three elements come from the same row of the periodic table and so they are of similar sizes. However, N is the least electronegative and F is the most electronegative. N is most able to donate electrons and most able to support a positive charge, compared to the other two.
Answer c
Answer d
Answer e
The cap and tether would be bound by hydrogen bonding. However, hydrogen bonding becomes much stronger if there is an ionic component. It is not a full ionic bond because the cap does not contain an anion, but its strength is between that of a normal hydrogen bond and an ionic bond.
Answer f
Steric crowding traps the dye between the cap above and the silica surface below. The dye is too big to squeeze past the cap.
Answer g
The dye has been released, and is now in the water.
Answer h
There is an equilibrium between the protonated tether and the protonated triethylamine.
Answer i
Once the tether has been deprotonated, the tether-cap interaction is just a normal hydrogen bond. It's still strong, but not as strong as the ion-boosted hydrogen bonding that we had before.
Answer j
N,N-Diisopropylethylamine or Hunig's base:
Answer k):
The lone pair in Hunig's base is much more crowded than the one in triethylamine, so it cannot remove the proton as quickly.
Exercise \(2\)
The laboratories of Teresa Reineke and Tim Lodge at U MN collaborated to study the use of a polymer as a possible drug delivery device for gene therapy (Adapted with permission from Laaser, J. E.; Jiang, Y.; Sprouse, D.; Reineke, T. M.; Lodge, T. P. Macromolecules 2015, 48, 2677-2685. Copyright 2015 American Chemical Society).
Here is a section of their polymer:
Answer
These polymer chains coil up, forming spherical nanoparticles.
a) Add any lone pairs to the structure above.
b) Show what happens to the polymer structure when aqueous HCl is added.
After treatment with HCl, the polymer nanoparticles expand, from spheres with radii of about 20 nm to spheres with radii of about 40 nm.
c) Show cartoons of a long polymer chain coiled up to form (i) a sphere of radius 20 nm; (ii) a sphere of radius 40 nm.
d) Why do the nanoparticles expand when HCl is added?
After treatment with HCl, the polymer binds DNA molecules. Here is a short section of DNA. It has three repeating units.
e) Add any formal charges.
f) Circle one sugar. Put a square around one phosphate. Put a triangle around one base (as in “DNA base pair”). Put a dashed circle around an aromatic ring.
g) Show how one of these other base pairs binds to the DNA strand
h) One of the base pairs binds more tightly than the other one to the DNA strand. Which one? Why?
i) Why do the acid-treated polymer nanoparticles bind DNA?
Salt solutions (such as aqueous NaCl) were subsequently shown to trigger DNA release from the DNA-nanoparticle complexes.
j) Why would these conditions lead to DNA release?
Triethylamine solutions ([CH3CH2]3N) cause the nanoparticles to shrink and they also trigger DNA release.
k) Why would triethylamine inhibit DNA binding?
l) Why would triethylamine cause the nanoparticles to shrink?
Certain peptide solutions also trigger DNA release.
m) This is not the structure of a peptide at neutral pH. Modify the structure to reflect neutral pH.
n) Why would this peptide trigger DNA release?
o) This peptide would probably not trigger DNA release. Modify the structure to show why.
Answer a
Answer b
Answer c
Answer d
Repulsion between the positively-charged quaternary ammonium groups (poly-CH2-N(H)(CH3)2+) would cause the nanoparticle to uncoil.
Answer e
Answer f
Answer g
Answer h
The one on the right forms three hydrogen bonds with the DNA and binds more tightly than the one on the left, which forms only two hydrogen bonds with the DNA.
Answer i
The polymer is positively charged after acid treatment. It binds the negatively charged DNA via ion-ion interactions.
Answer j
If there are enough of them, the chloride anions could "wash out" the anionic DNA. These individual anions would replace the DNA anion previously bound to the nanoparticle.
Answer k
The triethylamine could remove the proton from the polymer chain. If the polymer chain were no longer charged, it would no longer bind the anionic DNA.
Answer l
Also, because the nanoparticles would no longer be charged, there would no longer be repulsive forces causing the polymer to uncoil.
Answer m
Answer n
This anionic (overall) peptide would bind to the cationic nanoparticles, replacing the DNA.
Answer o
This cationic (overall) peptide would not bind to the cationic nanoparticles, so it would not displace the DNA.
Exercise \(3\)
Frustrated Lewis pairs should react together but do not. An example of a frustrated Lewis pair is tri-t-butylphosphine, [(CH3)3C]3P, and tris(pentafluorophenyl)borane, B(C6F5)3.
1. Draw both structures.
2. Identify the Lewis acid and the Lewis base.
3. Draw a mechanism using curved arrow(s) to show how the acid and base would interact.
4. State why this interaction does not occur.
5. Stephan & Erker showed that the frustrated Lewis pair can work together to capture a molecule of carbon dioxide (Angewandte Chemie, 2015). Show a mechanism that explains how the Lewis acid would interact with the CO2, including curved arrows.
Exercise \(4\)
Researchers in the Rowan lab at University of Chicago have been experimenting with composite materials made from cellulose and plastic. (Adapted from Amanda E. Way, Lorraine Hsu, Kadhiravan Shanmuganathan, Christoph Weder, and Stuart J. Rowan, ACS Macro. Lett. 2012, 1, 1001-1006. Copyright 2012, American Chemical Society. Used with Permission.)
Cellulose is made of glucose building blocks or monomers.
a) Show the alpha-D-glucopyranose form of glucose in its most stable chair conformation.
When this form of glucose links with other molecules through its 1- and 4-positions, with loss of a water molecule, it forms cellulose, a polymer made of glucose building blocks.
b) Show a small cellulose molecule 3 glucose units long.
c) Cellulose is an example of what class of biomolecules: carbohydrates / lipids / nucleic acids / proteins?
d) Name one common household material that is made of cellulose.
e) Draw a cartoon of the cellulose polymer, using just a wiggly line for the long polymer chain. Add the parts of the structure that you would need in order to show why two cellulose chains would stick to each other. Name this intermolecular attraction.
The researchers used poly(vinyl acetate) (PVAc; a short section is shown below) as the plastic in their study. PVAc is also a polymer; the dashed lines here indicate the units keep repeating in a long chain. PVAc is mostly used in latex paint.
f) Name the type of functional group in PVAc.
g) Draw a cartoon of the PVAc polymer, using just a wiggly line for the long polymer chain. Add the parts of the structure that you would need in order to show why two PVAc chains would stick to each other. Name this intermolecular attraction.
Below 35°C, PVAc is “glassy”, or strong and stiff, like the plastic in an automobile bumper. Above 35°C, PVAc becomes much softer or “rubbery”, like an eraser.
h) Explain why this property prevents the use of PVAc for medical implants such as an artificial hip.
Blending cellulose nanocrystals (CNCs) in with the PVAc keeps the material stronger up until about 45°C.
i) Show why the CNCs make the PVAc stronger.
Clearly, composite materials, in which other materials are mixed in with plastics to provide extra strength, can sometimes be better than the plastics by themselves.
The Rowan lab is interested in something much more sophsticated than that, though. They want to make a material that can switch from rubbery to glassy, or vice versa, in an instant. They were inspired by tunicates, marine organisms that can change instantly from soft-shelled to hard-shelled if disturbed. Tunicate shells, it turns out, contains high levels of CNCs.
One possible stimulus is pH or acidity; this factor could be changed by adding HCl or NaOH.
j) HCl is (acidic / basic) because the H-Cl bond is polarized with electrons closer to (H / Cl).
k) NaOH is (acidic / basic) because the (Na-O / O-H) bond is more polarized, with electrons closer to (Na / O / H).
They modified the CNCs by adding different side groups. Only a portion of the cellulose chain is shown here.
The resulting materials respond to changes in pH or acidity levels; they display Brønsted acidity/basicity.
l) Which of these three materials would be most acidic? Explain why with a structure of the conjugate base.
m) Which of these three materials would be most basic? Explain why with a structure of the conjugate acid.
n) CNC-CO2H is more than twice as strong at pH 3 (i.e. when HCl is added) than at pH 11 (i.e. when NaOH is added). Why? Explain in terms of IMFs.
o) CNC-NH2 is more than twice as strong at pH 11 (i.e. when NaOH is added) than at pH 3 (i.e. when HCl is added). Why? Explain in terms of IMFs.
p) At neutral pH, a mixture of CNC-NH2 and CNC-CO2H also produced a very strong material. Show the structures of the two materials when mixed, and identify why the material is so strong.
Answer a
Answer b
Answer c
carbohydrates
Answer d
Cellulose is the major component of cotton (textiles and money) and paper.
Answer e
Answer f
ester
Answer g
Answer h
Body temperature is closer to 40 degrees Celsius, so the hip would be soft and rubbery. That would making walking a little unpredictable.
Answer i
Answer j
HCl is acidic because the H-Cl bond is polarized with electrons closer to Cl.
Answer k
NaOH is basic because the Na-O bond is more polarized, with electrons closer to O.
Answer l
CNC-CO2H is most acidic because of the resonance stabilized anion in the conjugate base.
Answer m
CNC-NH2 is most basic because, although all three compounds possess lone pairs, the lone pair on the nitrogen is on a less electronegative atom than the lone pairs on oxygens, so it is more readily donated to a proton.
Answer n
This one may seem counterintuitive. The intermolecular forces at pH 11 are ionic, which should be stronger than hydrogen bonding. However, interactions between two anionic chains are repulsive, which would decrease the attraction between neighbouring CNC chains.
Answer o
The intermolecular forces at pH 3 are ionic, which should be stronger than hydrogen bonding. However, interactions between two cationic chains are repulsive, which would decrease the attraction between neighboring CNC chains.
Answer p
A proton would be transferred from the acidic site to the basic site. The oppositely charged CNC chains would attract each other strongly. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/14%3A_Concepts_of_Acidity/14.19%3A_Application_Problems.txt |
With contributions from Nicholas Jones and Kate Graham, College of Saint Benedict / Saint John's University
Exercise 14.2.1:
A Lewis base must have lone pairs or non-bonding electron pairs so that it can donate them to a Lewis acid.
Alternatively, in some cases the electrons in a π-bond can be donated instead, so sometimes compounds with π-bonds can be Lewis basic.
Exercise 14.2.2:
a. not a Lewis base
b. not a Lewis base
c. Lewis base
d. Lewis base
e. Lewis base
Exercise 14.3.1:
A Lewis acid atom attracts electrons from a Lewis base.
The most common feature of a Lewis acid is an atom that is not "electronically saturated" or has not filled its octet. For example, an aluminum with only six electrons rather than eight is Lewis acidic.
Other atoms, like transition metals, have "octets" of eighteen electrons, so having fewer than eighteen electrons in their valence shell can make these atoms Lewis acidic.
Exercise 14.3.2:
a. Lewis acid
b. Lewis acid
c. not a Lewis acid
d. Lewis acid
e. not a Lewis acid
Exercise 14.3.3:
a)
b) cations
c) possibly ion-dipole forces; alternatively, the oxygen atoms could act as Lewis bases, donating lone pairs to a cation.
d)
e) anions
f)
g)
h) When the chloride ion binds to the sensor molecule it gives the overall complex a negative charge and addition of the potassium ion cancels out this charge.
Exercise 14.4.1:
Exercise 14.4.6:
a)
b)
• The benzene rings take up more space than a fluorine atom; they may get in the way, making it harder for the water molecule to approach.
Also, the aromatic rings may be able to form a conjugated system with the empty p orbital; if that p orbital is partially filled, the boron atom becomes less Lewis acidic.
• A larger group than fluorine may cause more steric hindrance. Replacing the two fluorines closest to the boron on each arene (aromatic ring) with a CH3 or CF3 group would slow down the formation of an adduct. Alternatively, a less electronegative group than fluorine would also make the boron seem less positive; a CH3 or OCH3 are two possibilities.
• An even more electron-withdrawing group than fluorine would make the Lewis acid more reactive. Examples include nitro (NO2) and carbonyl groups (such as CH3C=0); these groups are resonance-withdrawing. Alternatively, a smaller group such a hydrogen would lower steric resistance, but would also lead to lower electrophilicity at boron, owing to the lower electronegativity of hydrogen compared to fluorine.
Exercise 14.4.7:
a) (i)
(ii)
b) (i)
(ii)
Exercise 14.7.3:
a)
b)
c)
d)
e) aromatic
f)
g)
h) This nitrogen atom can donate a lone pair of electrons without disturbing the aromatic character of the molecule.
i)
j)
k) Each side chain has a Lewis base with a lone pair of electrons to donate the copper ion.
l)
m) CN = 4
n) tetrahedral
o) Cu: 11 e-
Cu(II): 9 e-
$4 \times 2e^{-} = 8 e^{-} \nonumber$
$8e^{-} + 9 e^{-} = 17 e^{-} \nonumber$
p) If the imidazole becomes occupied by a proton, then the imidazole no longer has the lone pair to donate to the copper ion.
q) trigonal
Exercise 14.8.1:
Exercise 14.8.2:
Exercise 14.8.3:
Exercise 14.8.4:
1. The O-H bond is polar covalent. The oxygen is much more electronegative than the hydrogen, so the bond is easily ionized to give H+.
2. In a case like NaOH, the electronegativity difference between the sodium and oxygen is much greater than the electronegativity between the oxygen and the hydrogen. The Na-O bond is ionic. The removal of a proton from hydroxide ion is harder than the removal from a proton from hydroxide. It would make an oxide anion, O2-; that buildup of negative charge is more difficult than the formation of -OH.
Exercise 14.8.5:
Exercise 14.8.6:
a) i)
ii)
b) i)
ii)
Exercise 14.9.1
a) HNO3 (pKa = -1.3); HNO2 (pKa = 3.3)
HNO3 is a stronger Brønsted acid compared to HNO2
b) H2Se (pKa = 3.9); H2O (pKa = 14)
H2Se is a stronger Brønsted acid compared to H2O
c) HCl (pKa = -8); H2SO4 (pKa = -3)
HCl is a stronger acid compared to H2SO4
d) Ba(OH)2 (pKa > 50); HSeO3 (pKa = 6.6)
HSeO3- is a stronger Brønsted acid compared to Ba(OH)2
Exercise 14.9.2:
a) NH4+ (pKa = 9.2); HN3 (pKa = 4.7)
NH3 is a stronger Brønsted base compared to N3-
b) HCN (pKa = 9.4); HSCN (pKa = 4)
-CN is a stronger Brønsted base compared to -SCN
c) NH3 (pKa = 35); H2O (pKa = 14)
-NH2 is a stronger Brønsted base compared to HO-
Exercise 14.10.1
1. HClO4
2. H3PO4
Exercise 14.11.1:
a) H2S or SiH4 b) GaH3 or AsH3
c) PH3 or AlH3 d) H2Se or HBr
Exercise 14.11.2:
a) +NH4 or NH3 b) -PH2 or PH3
c) H2O or +NH4 d) +H3O or CH4
Exercise 14.11.3:
a) H2S or H2Te b) GeH4 or SnH4
c) HCl or HBr d) NH3 or AsH3
Exercise 14.11.4:
Exercise 14.11.5:
a. Polarizability would not be useful as the atom that will be anionic in the conjugate base is the same (oxygen).
b. Se is larger and therefore more polarizable leading to H2Se being more acidic compared to H2S.
c. Ge is larger and therefore more polarizable leading to GeH4 being more acidic compared to SiH4.
d. Polarizability would not be useful as the atoms are next to one another in the same row and therefore likely to have the same polarizability.
Exercise 14.11.3:
Exercise 14.12.1:
Exercise 14.12.2:
a. Vinyl alcohol would be more acidic as its anion is stabilized by resonance while the anionic charge on ethoxide would be localized on the oxygen atom.
b. Nitromethane would be more acidic as its anion is stabilized by resonance while the anionic charge of the trimethylamine anion would be localized on the carbon atom.
c. Acetonitrile would be more acidic as its anion is stabilized by resonance while the anionic charge of the trimethylamine would be localized on the carbon atom.
Exercise 14.12.3:
Exercise 14.12.4:
Exercise 14.15.4:
The thiol proton on cystine is more acidic due to the polarizability of the sulfur anion stabilizing the conjugate base's anionic charge. The oxygen anion is less polarizable and therefore less stable rendering the alcohol proton less acidic.
Exercise 14.15.5:
Exercise 14.15.6:
Exercise 14.15.7
Exercise 14.16.1:
Ammonium bromide is more soluble because in water it can participate in both ion-dipole and hydrogen-bonding, while sodium bromide only benefits from ion-dipole interactions.
Exercise 14.16.2:
Hydrogen cyanide will be have a lower pKa in water as the resulting cyanide anion will be stabilized by both ion-dipole and hydrogen-bonding interactions. Hydrogen cyanide would have a higher pKa in pentanes as the resulting cyanide anion would only experience ion-induced dipole interactions which are relatively weak.
Exercise 14.17.1:
a. $K_{a} = \frac{[NC^{-}][H_{3}O^{+}]}{[HCN]}$
b. $K_{a} = \frac{[HS^{-}][H_{3}O^{+}]}{[H_{2}S]}$
c. $K_{a} = \frac{[H_{2}N^{-}][(CH_{3})_{2}SOH^{+}]}{[HCN]}$
Exercise 14.17.2:
a. pKa = -6
b. pKa = 9
c. pKa = 24
d. pKa = 16
Exercise 14.17.3:
a. Ka = 103.5
b. Ka = 10-4.3
c. Ka = 10-25
d. Ka = 10-9
Exercise 14.17.4:
At equilibrium formation of the Lewis acid-base complex is slightly favored.
Exercise 14.17.5:
At equilibrium formation of the Lewis acid-base complex of dimethylether and BF3 is more favored than the corresponding diethylether BF3 complex. This is likely due to the decreased steric bulk of the methy groups compared to the ethyl groups.
Exercise 14.17.6:
a. $K_{eq}= \frac{[(NH_{3})_{2}PtCl_{2}]}{[(NH_{3})PtCl_{2}][NH_{3}]}$
b. $K_{eq} = \frac{[Mo(CO)_{6}]}{[Mo(CO)_{5}][CO]}$
c. $K_{eq}= \frac{[FeCl_{4}^{-}]}{[FeCl_{3}][Cl^{-}]$
Exeercise 14.18.1:
Exercise 14.19.1:
1. all bases have lone pairs.
2. F, N, and O have lone pairs. They can be bases.
Their order of basicity would be N > O > F.
These three elements come from the same row of the periodic table and so they are of similar sizes. However, N is the least electronegative and F is the most electronegative. N is most able to donate electrons and most able to support a positive charge, compared to the other two.
c)
If possible, a nitrogen is circled in each molecule. The cap has only basic oxygen atoms. If there is a choice between two nitrogens (in the tether) or two oxygens (in the cap), the non-conjugated lone pair would be most basic. Conjugated lone pairs are held in place by their stable interaction with their neighbors.
The most basic of all is the amine (non-conjugated) nitrogen in the tether.
d)
e) The cap and tether would be bound by hydrogen bonding. However, hydrogen bonding becomes much stronger if there is an ionic component. It is not a full ionic bond because the cap does not contain an anion, but its strength is between that of a normal hydrogen bond and an ionic bond.
f) Steric crowding traps the dye between the cap above and the silica surface below. The dye is too big to squeeze past the cap.
g) The dye has been released, and is now in the water.
h) There is an equilibrium between the protonated tether and the protonated triethylamine.
i) Once the tether has been deprotonated, the tether-cap interaction is just a normal hydrogen bond. It's still strong, but not as strong as the ion-boosted hydrogen bonding that we had before.
j) N,N-Diisopropylethylamine or Hunig's base:
k) The lone pair in Hunig's base is much more crowded than the one in triethylamine, so it cannot remove the proton as quickly.
Exercise 14.19.2:
a)
b)
c)
d) Repulsion between the positively-charged quaternary ammonium groups (poly-CH2-N(H)(CH3)2+) would cause the nanoparticle to uncoil.
e)
f)
g)
h) The one on the right forms three hydrogen bonds with the DNA and binds more tightly than the one on the left, which forms only two hydrogen bonds with the DNA.
i) The polymer is positively charged after acid treatment. It binds the negatively charged DNA via ion-ion interactions.
j) If there are enough of them, the chloride anions could "wash out" the anionic DNA. These individual anions would replace the DNA anion previously bound to the nanoparticle.
k) The triethylamine could remove the proton from the polymer chain. If the polymer chain were no longer charged, it would no longer bind the anionic DNA.
l) Also, because the nanoparticles would no longer be charged, there would no longer be repulsive forces causing the polymer to uncoil.
m)
n) This anionic (overall) peptide would bind to the cationic nanoparticles, replacing the DNA.
o)
This cationic (overall) peptide would not bind to the cationic nanoparticles, so it would not displace the DNA.
Exercise 14.19.4:
a)
b)
c) carbohydrates
d) Cellulose is the major component of cotton (textiles and money) and paper.
e)
f) ester
g)
h) Body temperature is closer to 40%deg;C, so the hip would be soft and rubbery. That would making walking a little unpredictable.
i)
j) HCl is acidic because the H-Cl bond is polarized with electrons closer to Cl.
k) NaOH is basic because the Na-O bond is more polarized, with electrons closer to O.
l) CNC-CO2H is most acidic because of the resonance stabilized anion in the conjugate base.
m) CNC-NH2 is most basic because, although all three compounds possess lone pairs, the lone pair on the nitrogen is on a less electronegative atom than the lone pairs on oxygens, so it is more readily donated to a proton.
n) This one may seem counterintuitive. The intermolecular forces at pH 11 are ionic, which should be stronger than hydrogen bonding. However, interactions between two anionic chains are repulsive, which would decrease the attraction between neighbouring CNC chains.
o) The intermolecular forces at pH 3 are ionic, which should be stronger than hydrogen bonding. However, interactions between two cationic chains are repulsive, which would decrease the attraction between neighbouring CNC chains.
p) A proton would be transferred from the acidic site to the basic site. The oppositely charged CNC chains would attract each other strongly. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/14%3A_Concepts_of_Acidity/14.20%3A_Solutions_to_Selected_Problems.txt |
We use energy every day. We plug in our electronic devices, drawing energy that has been converted into electricity, but which originated in a waterfall on a spinning turbine, a wind that turned a windmill, the burning of coal to produce carbon dioxide, a nuclear reaction, or other sources. We drive cars, in which the energy released from a chemical reaction heats gas in a chamber, pushing a piston which drives a crankshaft that ultimately causes the wheels to turn.
Many of our sources of energy depend on chemical processes. Solar power, for example, depends on photochemistry. When a photon from the sun is absorbed by an atom of the right kind of material, an electron hops from one energy level to a higher one. That event leaves behind a "hole", a place where an electron used to be. An electron might move in from a neighbouring atom to occupy that hole. Now the original electron can't fall back to where it came from; it will instead need to drop into a hole on the next atom. We now have electrons moving from one atom to the next. We have electricity.
The burning of gasoline and coal depends on expanding gases which push against a turbine, like wind against a windmill. The gases expand because they get hotter when the combustion, or burning, reaction happens. But why do these reactions produce heat? That's related to the formation of chemical bonds. When chemical bonds are formed, energy is released.
• The formation of a chemical bond always releases energy.
Over the course of a chemical reaction, old bonds are frequently broken and new ones are frequently made. But if energy is always released when bonds are formed, what happens when a bond is broken? When a bond is broken, energy is consumed. It costs energy to break a bond, but the formation of a new bond pays some energy back.
• Breaking a chemical bond always costs energy.
So, if breaking bonds costs energy, and making bonds pays energy, and energy becomes heat, then a reaction will only produce heat if the energy released when the bonds are made is more than the energy consumed when bonds were broken. We need to replace weaker bonds with stronger ones.
Burning carbon-containing compounds, like wood, coal, or gasoline, is a fantastic way to release energy and make heat. Our ancestors have known that since the stone age. Carbon-containing compounds generally contains lots of carbon-carbon and carbon-hydrogen bonds (which are actually quite strong). When burned, they produce carbon dioxide and water, which contain carbon-oxygen and hydrogen-oxygen bonds, and those bonds are even stronger than the carbon-hydrogen and carbon-carbon bonds that were broke. Overall, energy is released.
We can use that energy to warm ourselves on a cold night under the stars, to cook our food, to drive a mill that manufactures steel, or to fly an airplane. We can use that energy to get work done.
Thermodynamics is the study of the relationship between heat (or energy) and work. In other words, thermodynamics looks at how we can put energy into a system (whether it is a machine or a molecule) and make it do work. Alternatively, we might be able to do some work on a system and make it produce energy (like spinning the turbines in a power station to produce electricity).
In chemistry, we sometimes speak more broadly about "energetics" of reactions (rather than thermodynamics), because energy given off during a reaction may simply be lost to the surroundings without doing useful work. Nevertheless, the ideas are the same: energy can be added to a set of molecules in order to produce a reaction, or a reaction can occur between a set of molecules in order to release energy.
A classic example of reaction energetics is the hydrolysis of ATP to ADP in biology. This reaction is used in the cell as a source of energy; the energy released from the reaction is frequently coupled to other processes that could not occur without the added energy.
The hydrolysis of ATP, or the addition of water to ATP in order to break ATP into two, smaller molecules, gives off energy. That energy can be used by the cell to carry out other processes that would cost energy. One molecule of ADP and one molecule of inorganic phosphate, sometimes abbreviated as Pi, are also produced.
• Energy can be given off by a chemical reaction.
• That energy can be used to power other reactions that require energy.
In the cell, ATP is produced in high levels in the mitochondria. Because it is a relatively small molecule, it can be transported easily to other areas of the cell where energy may be needed. The ATP can be hydrolysed on site, providing energy for the cell to use for other reactions.
Note that the scheme above uses some thermodynamics jargon. The place where the reaction takes place, or the molecules participating in the reaction, are called "the system". Energy is supplied to "the surroundings", meaning places or molecules other than those directly involved in this reaction.
There are a couple of other ways in which energetics of reactions are commonly depicted. The energetic relationship between ATP plus water and ADP plus phosphate shown above is really a simplified graph of energy versus reaction progress (sometimes called reaction coordinate). This type of graph shows changes in energy over the course of a reaction. The energy of the system at the beginning of the reaction is shown on the left, and the energy at the end of the reaction is shown on the right. This type of graph is sometimes referred to as a reaction profile.
Another common way of discussing energetics is to include energy as a reactant or product in an equation describing the reaction. An equation for a reaction shows what the starting materials were for the reaction, and what they turned into after the reaction. The things that reacted together in the reaction are called the "reactants". They are written on the left hand side of the arrow that says a reaction took place. The things that the reactants turned into are called the "products". They show up on the right hand side of the arrow.
$ATP \:+ \: H_{2}O \rightarrow ADP \: + \: P_{i} \: + \:energy \nonumber$
For the hydrolysis of ATP, energy is simply included as one of the products of the reaction, since the reaction releases energy.
Alternatively, the energetic observation about ATP can be turned around, since there are evidently some reactions that cost energy. Probably the most well-know reaction of this type is the conversion of carbon dioxide to carbohydrates such as glucose. This conversion actually results from a long series of different reactions that happen one after another. Overall, the process requires a lot of energy. This energy is supplied in part by ATP, generated with assistance from photosystem I and II, which are arrays of molecules that interact with sunlight. A simplified reaction profile for carbohydrate synthesis is shown below.
• Energy can be consumed by a chemical reaction.
• Reactions that consume energy need an energy source in order to occur.
Again, this energetic relationship can be thought of in the form of a balanced reaction.
$energy \: + 6CO_{2} + 6H_{2}O \rightarrow \: C_{6}H_{12}O_{6} +O_{2} \nonumber$
In this case, energy is a reactant, not a product. It is one of the key ingredients needed to make the reaction happen.
Reactions that produce energy, like ATP hydrolysis, are referred to as exothermic reactions (or sometimes exergonic, meaning roughly the same thing). In reaction profiles, these reactions go downhill in energy as the reaction occurs from the left side of the diagram to the right. On the other hand, reactions that cost energy (the ones that go uphill on the reaction profile, like carbohydrate synthesis) are referred to as endothermic (or sometimes endergonic).
It is useful to think of reactions as "going downhill" or "going uphill" because one of these situations should seem inherently easier than the other (especially if you've ever been skiing). Exothermic reactions (the downhill ones) occur very easily; endothermic reactions do not (those are the uphill ones).
• Systems always go to lower energy if possible.
Reactions that are energetically "uphill" cannot happen easily by themselves. Those reactions must be powered by other reactions that are going downhill. The energy traded between these reactions keeps chemical reactions going, in cells and other important places. Sometimes, a process that is used to supply energy for another reaction is thought of as the "driving force" of the reaction. Without the driving force, the desired reaction would not be able to occur.
In general, a reaction will occur if more than enough energy is supplied. Excess energy does not hurt on the macroscopic scale. However, if not enough energy is supplied to make up for an endothermic reaction, the reaction is not likely to happen.
Energy is a lot like money. It can be passed from one set of hands to another. Doing so often helps get things done.
There is one problem with the use of chemical reactions as sources of energy. If ATP hydrolysis releases energy, and if the release of energy is always favored, why does not it happen spontaneously? In other words, why don't all the ATP molecules in all the cells in all the organisms in the whole world just slide downhill into ADP right now? What is stopping them?
Fortunately, all reactions have barriers that stop them from happening until they are ready to go. A reaction barrier is an initial investment of energy needed to get things started. Reaction barriers occur for a variety of physical reasons: two molecules may need to get oriented in the right direction to react with each other, or a bond may have to be broken to get the reaction going, costing an initial outlay of energy.
The reaction barriers of reactions influence how quickly reactions happen. High barriers slow reactions down a lot. Low barriers allow them to happen more easily. The study of reaction barriers, and how quickly reactions can occur, is called chemical kinetics.
Thermodynamics, on the other hand, is really concerned with the overall energy change from the beginning of a reaction to the end. It compares the energies of two sets of molecules to each other: the energies of the reactants and the energies of the products.
Exercise $1$
Which of the following reaction profiles describe reactions that will go forward, as opposed to ones that will probably not occur?
Answer
Reactions that go down in energy will proceed. Reactions that go up in energy will not proceed. If the reaction profile is higher on the left (reactant side) than the right, the reaction will go forward and it will form products. If the reverse is true, the reaction presumably will not occur. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/III%3A_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_1/01%3A_Thermodynamics/1.01%3A_Introduction.txt |
Thermodynamics is the study of the relationship between heat (or energy) and work. Enthalpy is a central factor in thermodynamics. It is the heat content of a system. The heat that passes into or out of the system during a reaction is the enthalpy change. Whether the enthalpy of the system increases (i.e. when energy is added) or decreases (because energy is given off) is a crucial factor that determines whether a reaction can happen.
Sometimes, we call the energy of the molecules undergoing change the "internal enthalpy". Sometimes, we call it the "enthalpy of the system". These two phrases refer to the same thing.
Similarly, the energy of the molecules that don't take part in the reaction is called the "external enthalpy" or the "enthalpy of the surroundings".
Roughly speaking, the energy changes that we looked at in the introduction to thermodynamics were changes in enthalpy. We will see in the next section that there is another energetic factor, entropy, that we also need to consider in reactions. For now, we will just look at enthalpy.
• Enthalpy is the heat content of a system.
• The enthalpy change of a reaction is roughly equivalent to the amount of energy lost or gained during the reaction.
• A reaction is favored if the enthalpy of the system decreases over the reaction.
That last statement is a lot like the description of energetics on the previous page. If a system undergoes a reaction and gives off energy, its own energy content decreases. It has less energy left over if it gave some away.
Why does the energy of a set of molecules change when a reaction occurs? To answer that, we need to think about what happens in a chemical reaction.
In a reaction, there is a change in chemical bonding. Some of the bonds in the reactants are broken, and new bonds are made to form the products. It costs energy to break bonds, but energy is released when new bonds are made.
Whether a reaction is able to go forward may depend on the balance between these bond-making and bond-breaking steps.
• A reaction is exothermic if more energy is released by formation of new bonds than is consumed by breaking old bonds.
• A reaction is exothermic if weaker bonds are traded for stronger ones.
• A reaction is endothermic if bond-breaking costs more energy than what is provided in bond-making.
one lots of work measuring bond strengths, and they have collected the information in tables, so if you need to know how strong a bond is, you can just look up the information you need.
Bond Bond Energy (kcal/mol) Bond Bond Energy (kcal/mol)
H-H 104 O-H 111
C-C 83 C-H 99
O=O 119 N-H 93
N=N 226 C=O 180
For example, suppose you wanted to know whether the combustion of methane were an exothermic or endothermic reaction. I am going to guess that it's exothermic, because this reaction (and others like it) is used to provide heat for lots of homes by burning natural gas in furnaces.
The "combustion" of methane means that it is burned in air, so that it reacts with oxygen. The products of burning hydrocarbons are mostly carbon dioxide and water. The carbon atom in methane (CH4) gets incorporated into a carbon dioxide molecule. The hydrogen atoms get incorporated into water molecules. There are four hydrogen atoms in methane, so that's enough to make two molecules of H2O.
• Four C-H bonds must be broken in the combustion of methane.
• Four new O-H bonds are made when the hydrogens from methane are added into new water molecules.
• Two new C=O bonds are made when the carbon from methane is added into a CO2 molecule.
The other piece of the puzzle is the oxygen source for the reaction. Oxygen is present in the atmosphere mostly as O2. Because we need two oxygen atoms in the CO2 molecule and two more oxygen atoms for the two water molecules, we need a total of four oxygen atoms for the reaction, which could be provided by two O2 molecules.
• Two O=O bonds must be broken to provide the oxygen atoms for the products.
Altogether, that's four C-H and two O=O bonds broken, plus two C=O and four O-H bonds made. That's 4 x 99 kcal/mol for the C-H bonds and 2 x 119 kcal/mol for the O=O bonds, a total of 634 kJ/mol added. The reaction releases 2 x 180 kcal/mol for the C=O bonds and 4 x 111 kcla/mol for the OH bonds, totaling 804 kcal/mol. Overall, there is 170 kcal/mol more released than is consumed.
That means the reaction is exothermic, so it produces heat. It's probably a good way to heat your home.
Exercise \(1\)
Compare the combustion of ethane to the combustion of methane.
1. Write a reaction for the combustion of ethane, CH3CH3, to carbon dioxide and water.
2. How many carbon dioxide molecules would be produced from one molecule of ethane?
3. How many water molecules would be produced from one molecule of ethane?
4. How many oxygen molecules would be needed to provide oxygen atoms to accomplish the steps in questions (b) and (c)?
5. How much energy is consumed / produced by the reaction? Compare this result to the one for methane.
Answer
Answer a
The reaction is given with structures below:
Answer b
Because there are two carbons in ethane, one molecule of ethane will give rise to two molecules of CO2.
Answer c
Because there are six hydrogens in ethane, one molecule of ethane will give rise to three molecules of H2O.
Answer d
In order to make two molecules of carbon dioxide (four oxygen atoms) and three water molecules (three oxygen atoms), we would need seven oxygen atoms total. Since oxygen molecules contain pairs of oxygen atoms, we would only need 3.5 oxygen molecules.
Answer e
The energy requirements are laid out in the following table. Overall, the reaction releases 375.5 kcal per mol of ethane burned. The negative sign in the table is often used to denote that this is excess energy released (whereas a positive sign would indicate that energy as consumed overall).
Bond Breaking Costs (kcal/mol) Sum of Cost Bond Making Releases (kcal/mol) Sum of Release Overall (kcal/mol)
6 x C-H 6 * 99 594 6 x O-H 6 * 111 666
3.5 x O=O 3.5 * 119 416.5 4 x C=O 4 * 180 720
1 x C-C 83 83
total breaking: 1093.5 making: 1386 -292.5
That's more energy than was produced from a molecule of methane (-170 kcal/mol).
Exercise \(2\)
The Haber-Bosch process is used to make ammonia for fertilizer. It employs the reaction of hydrogen gas (H2) with atmospheric nitrogen (N2) in a 3:1 ratio to produce ammonia (NH3).
1. Write a reaction for the Haber-Bosch process.
2. How many ammonia molecules would be produced from one molecule of nitrogen?
3. How much energy is consumed / produced by the reaction? | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/III%3A_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_1/01%3A_Thermodynamics/1.02%3A_Enthalpy.txt |
Entropy is another important aspect of thermodynamics. Enthalpy has something to do with the energetic content of a system or a molecule. Entropy has something to do with how that energy is stored.
We sometimes speak of the energy in a system as being "partitioned" or divided into various "states". How this energy is divided up is the concern of entropy.
By way of analogy, picture a set of mailboxes. You may have a wall of them in your dormitory or your apartment building. The mailboxes are of several different sizes: maybe there are a few rows of small ones, a couple of rows of medium sized ones, and a row of big mailboxes on the bottom.
Instead of putting mail in these boxes, we're going to use them to hold little packages of energy. Later on, you might take the energy packages out of your own mailbox and use them to take a trip to the mall or the gym. But how does the mail get to your mailbox in the first place?
The energy packages don't arrive in your molecular dormitory with addresses on them. The packages come in different sizes, because they contain different amounts of energy, but other than that there is no identifying information on them.
Some of the packages don't fit into some of the mailboxes, because some of the packages are too big and some of the mailboxes are smaller than the others. The energy packages need to go into mailboxes that they will fit into.
Still, there are an awful lot of mailboxes that most of the energy packages could still fit into. There needs to be some system of deciding where to put all of these packages. It turns out that, in the molecular world, there is such a system, and it follows a pretty simple rule. When a whole pile of energy packages arrive, the postmaster does her best to put one package into every mailbox. Then, when every mailbox has one, she starts putting a second one into each box, and so on.
It didn't have to be that way. It could have been the case that all the energy was simply put into the first couple of mailboxes and the rest were left empty. In other words, the rule could have been that all the energy must be sorted into the same place, instead of being spread around. But that's not how it is.
• Energy is always partitioned into the maximum number of states possible.
Entropy is the sorting of energy into different modes or states. When energy is partitioned or sorted into additional states, entropy is said to increase. When energy is bundled into a smaller number of states, entropy is said to decrease. Nature's bias is towards an increase in entropy.
This is a fundamental law of the universe; there is no reason that can be used to explain why nature prefers high entropy to low entropy. Instead, increasing entropy is itself the basic reason for a wide range of things that happen in the universe.
Entropy is popularly described in terms of "disorder". That can be a useful idea, although it does not really describe what is happening energetically.
A better picture of entropy can be built by looking at how a group of molecules might sort some energy that is added to them. In other words, what are some examples of "states" in which energy can be sorted?
If you get more energy -- maybe by eating breakfast -- one of the immediate benefits is being able to increase your physical activity. You have more energy to move around, to run, to jump. A similar situation is true with molecules.
Molecules have a variety of ways in which they can move, if they are given some energy. They can zip around; this kind of motion is usually called translation. They can tumble and roll; this kind of motion is referred to as rotation. Also, they can wiggle, letting their bonds get longer and shorter by moving individual atoms around a little bit. This type of motion is called vibration.
When molecules absorb extra energy, they may be able to sort the energy into rotational, vibrational and translational states. This only works with energy packages of a certain size; other packages would be sorted into other kinds of states. However, these are just a few examples of what we mean by states.
OK, so energy is stored in states, and it is sorted into the maximum possible number of states. But how does entropy change in a reaction? We know that enthalpy may change by breaking or forming certain bonds, but how does the energy get sorted again?
The changes in internal entropy during a reaction are often very small. In other words, the energy remaining at the end of the reaction gets sorted more or less the way it was before the reaction. However, there are some very common exceptions.
The most common case in which internal entropy changes a lot is when the number of molecules involved changes between the start of the reaction and the end of the reaction. Maybe two molecules react together to form one, new molecule. Maybe one molecule splits apart to make two, new molecules.
If one molecule splits apart in the reaction, entropy generally increases. Two molecules can rotate, vibrate and translate (or tumble, wiggle and zip around) independently of each other. That means the number of states available for partitioning energy increases when one molecule splits into two.
• Entropy generally increases when a reaction produces more molecules than it started with.
• Entropy generally decreases when a reaction produces fewer molecules than it started with.
Apart from a factor like a change in the number of molecules involved, internal entropy changes are often fairly subtle. They are not as easy to predict as enthalpy changes.
Nevertheless, there may sometimes be a trade-off between enthalpy and entropy. If a reaction splits a molecule into two, it seems likely that an increase in enthalpy will be involved, so that the bond that held the two pieces together can be broken. That's not favorable. However, when that happens, we've just seen that there will be an increase in entropy, because energy can then be sorted into additional modes in the two, independent molecules.
So we have two different factors to balance. There is a tool we often use to decide which factor wins out. It's called free energy, and we will look at it next.
Exercise \(1\)
In each of the following pairs, look at the distribution of energy packages (gray shapes) and decide which system has the highest entropy.
Answer
Entropy is higher if the energy is partitioned into more states. For example, in question (b), the same amount of energy is distributed into three states on the left hand side and only two states on the right. Entropy is higher in the left hand example than the right in that case.
Exercise \(2\)
In each of the following pairs, select the system (i or ii) that has the highest entropy.
Answer
Examples of "states" into which energy can be partitioned include molecular vibrational, rotational and translational states (which, loosely speaking, correspond to wiggling, spinning and zipping around). Entropy is higher if energy is distributed into more of these states. That might include a greater range of vibrational or rotational states used in (a) and (c), or similar states employed in a greater number of molecules in (b).
Exercise \(3\)
Based on what you know, would the following reactions be entropically favored or not?
1. The decarboxylation of one molecule of the anti-Parkinson's drug, L-DOPA, to produce one molecule of carbon dioxide and one molecule of the neurotransmitter, dopamine.
2. One molecule of nitrogen and three molecules of hydrogen react to produce two molecules of ammonia?
3. One molecule of methane reacts with two molecules of oxygen to produce two molecules of water and one molecule of carbon dioxide?
Answer
One general observation about internal entropy is that it increases if the number of molecules increases during a reaction and decreases if the number of molecules decreases during a reaction. It's just a matter of counting how many things on the left get turned into how many things on the right. For example, in question (a), one molecule produces two new molecules in the decarboxylation reaction, so the reaction is entropically favored. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/III%3A_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_1/01%3A_Thermodynamics/1.03%3A_Entropy.txt |
Entropy and enthalpy are two of the basic factors of thermodynamics. Enthalpy has something to do with the energetic content of a system or a molecule. Entropy has something to do with how that energy is stored.
There is a bias in nature toward decreasing enthalpy in a system. Reactions can happen when enthalpy is transferred to the surroundings.
• A reaction is favored if enthalpy decreases.
There is also a bias in nature toward increasing entropy in a system. Reactions can happen when entropy increases.
• A reaction is favored if entropy increases.
Consider the cartoon reaction below. Red squares are being converted to green circles, provided the reaction proceeds from left to right as shown.
Whether or not the reaction proceeds to the right depends on the balance between enthalpy and entropy. There are several combinations possible.
In one case, maybe entropy increases when the red squares turn into green circles, and the enthalpy decreases. If we think of the balance between these two factors, we come to a simple conlusion. Both factors tilt the balance of the reaction to the right. In this case, the red squares will be converted into green circles.
Alternatively, maybe entropy decreases when the red squares turn into green circles, and enthalpy increases. If we think of the balance between these two factors, we come to another simple conlusion. Both factors tilt the balance of the reaction to the left. In this case, the red squares will remain just as they are.
Having two factors may lead to complications. For example, what if enthalpy decreases, but so does entropy? Does the reaction happen, or does not it?
In that case, we may need quantitation to make a decision. How much does the enthalpy decrease? How much does the entropy decrease? If the effect of the enthalpy decrease is greater than that of the entropy decrease, the reaction may still go forward.
The combined effects of enthalpy and entropy are often combined in what is called "free energy". Free energy is just a way to keep track of the sum of the two effects. Mathematically, the symbol for the internal enthalpy change is " ΔH" and the symbol for the internal entropy change is "ΔS". Free energy is symbolized by "ΔG", and the relationship is given by the following expression:
$\Delta G = \Delta H - T \Delta S \nonumber$
(note: that may look like "?G" = "?H" - "?S" in some web browsers, rather than ΔG = ΔH - TΔS; the ? will show up as a Greek delta in Safari or Firefox)
The letter T in this expression stands for the temperature (in Kelvin, rather than Celsius or Fahrenheit). The temperature acts as a scaling factor in the expression, putting the entropy and enthalpy on equivalent footing so that their effects can be compared directly.
How do we use free energy? It works the same way we were using enthalpy earlier (that's why the free energy has the same sign as the enthalpy in the mathematical expression, whereas the entropy has an opposite sign). If free energy decreases, the reaction can proceed. If the free energy increases, the reaction can't proceed.
• A reaction is favored if the free energy of the system decreases.
• A reaction is not favored if the free energy of the system increases.
Because free energy takes into consideration both the enthalpy and entropy changes, we don't have to consider anything else to decide if the reaction occurs. Both factors have already been taken into account.
Remember the terms "endothermic" and "exothermic" from our discussion of enthalpy. Exothermic reactions were favores (in which enthalpy decreases). Endothermic ones were not. In free energy terms, we say that exergonic reactions are favored (in which free energy decreases). Endergonic ones (in which free energy increases) are not.
Exercise $1$
Imagine a reaction in which the effects of enthalpy and entropy are opposite and almost equally balanced, so that there is no preference for whether the reaction proceeds or not. Looking at the expression for free energy, how do you think the situation will change under the following conditions:
1. the temperature is very cold (0.09 K)
2. the temperature is very warm (500 K)
Answer
The expression ΔG = ΔH - TΔS includes both an enthalpy contribution and an enthalpy contribution and balances them against each other. However, the effect of entropy is multiplied by the temperature. The greater the temperature, the greater will be the influence of entropy (and therefore the smaller the influence of enthalpy). The lower the temperature, the smaller will be the influence of entropy (and therefore the greater the influence of enthalpy).
Exercise $2$
Which of the following reaction profiles describe reactions that will proceed? Which ones describe reactions that will not proceed?
Answer
See problem Exercise 1.1.1
How Entropy Rules Thermodynamics
Sometimes it is said that entropy governs the universe.
As it happens, enthalpy and entropy changes in a reaction are partly related to each other. The reason for this relationship is that if energy is added to or released from the system, it has to be partitioned into new states. Thus, an enthalpy change can also have an effect on entropy.
Specifically, the internal enthalpy change that we discussed earlier has an effect on the entropy of the surroundings. So far, we have just considered internal entropy changes.
• In an exothermic reaction, the external entropy (entropy of the surroundings) increases.
• In an endothermic reaction, the external entropy (entropy of the surroundings) decreases.
Free energy takes into account both the entropy of the system and the entropy changes that arise because of heat exchange with the surroundings. Together, the system and the surroundings are called "the universe". That's because the system is just everything involved in the reaction, and the surroundings are everything that isn't involved in the reaction.
Enthalpy changes in the system lead to additional partitioning of energy. We might visualize that with the mailbox analogy we used for entropy earlier. In this case, each molecule has its own set of mailboxes, into which it sorts incoming energy.
Looked at in this way, thermodynamics boils down to one major consideration, and that is the combined entropy of both the system and its surroundings (together known as the universe).
• For a reaction to proceed, the entropy of the universe must increase. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/III%3A_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_1/01%3A_Thermodynamics/1.04%3A_Free_Energy.txt |
Sometimes, there is not a big difference in energy between reactants and products of a reaction. What happens then? Does the reaction go forward, because it will not cost a lot of energy? Or does it not proceed, because there isn't enough driving force?
For example, one simple reaction that occurs all the time is the reaction of water with carbon dioxide. This is a reaction that happens when carbon dioxide dissolves in lakes, rivers and oceans. It even happens in your own bloodstream.
Water reacts with carbon dioxide to form carbonic acid.
However, carbonic acid also decomposes spontaneously in water. It reacts to form carbon dioxide and water.
In other words, this is a reaction that can go either direction. It can go forwards or backwards. It is an example of an equilibrium reaction. An equilibrium reaction is one that is energetically balanced, so that it really isn't favored to go in either direction.
Equilibrium reactions are extremely important in nature, partly because of the forward and reverse capabilities that they offer. In essence, they are reactions with an "undo" button. The reaction can proceed in one direction when needed, and it can proceed in the other direction when needed.
However, there are some inherent limitations involved. Frequently, equilibrium reactions only proceed "partway". That is, a group of molecules will start to produce products. However, at some point those products will begin reverting to the starting materials again. Eventually the system will settle out as a mixture of reactants and products.
What if it's really important that we have the products of the reaction at one point, with none of the reactants? And if later on we need the reactants, but not any of the products? It would be useful if there were a way to control the direction of an equilibrium reaction, so that we could "push" it to one side or the other.
Control of equilibrium reactions can be remarkably simple. It follows a rule that was observed by Henri le Chatelier (ah-REE luh shah-tell-YAY), a French industrial chemist, around 1900. Le Chaletelier noticed that equilibrium reactions often shift direction if the conditions of the reaction are changed.
In general, adding any product of the reaction shifts the balance back toward the reactants. If any product of the reaction is added, the reaction makes more starting materials. Thus, adding more carbonic acid to a carbon dioxide - water - carbonic acid mixture would result in reverse reaction, producing more water and carbon dioxide. Adding more carbon dioxide, on the other hand, would lead to production of more carbonic acid.
Here is a cartoon illustration of "le Chatelier's Principle" at work. Suppose red squares and blue ovals can react together to make black circles and green circles. Maybe there is a natural equilibrium in this reaction, so that the two piles of shapes are roughly equal in size.
What would happen if something knocked this system off balance? For example, maybe black circles are highly elusive, and they just wander away as soon as they are formed. The system won't be in equilibrium anymore, because without those black circles, the balance will be upset, with not enough things on the right side for the number of things on the left.
Le Chatelier noticed that nature automatically corrects for such changes. If some of the black circles disappear, the reaction will kick into action again, using up some red squares and blue ellipses to produce more green and black circles. The exact numbers of shapes won't return to exactly the same as before, because some of the black circles have still gone missing, but the system will have shifted to use up more reactants on the left and to produce more products on the right, so that the overall ratio between right and left is restored.
Alternatively, maybe we found a way to make the black circles stay where they are. Instead, we have dumped in a bunch of extra blue ellipses. Once again, the system is knocked off balance. This time, there is too much stuff on the left, compared to the amount on the right side.
The reaction goes into action again. It uses up some of those extra blue ellipses (and, at the same time, some of the red squares) to produce more black and green circles, bringing the system back to the original ratio of right side shapes to left side shapes.
In general, if molecules are added to a system, the reaction will shift to bring the system back into equilibrium. If molecules are removed from the system, the reaction will also shift to bring the system back into equilibrium.
Furthermore, because heat can be consumed by (or produced by) reactions, temperature can sometimes be used to shift equilibria. If a reaction is exothermic, heat is a product of the reaction. Adding more heat will result in the reaction shifting to produce more reactants. Cooling the reaction (removing heat) would do the opposite: the reaction would shift to produce more heat, and more products.
In the cartoon, we have a shape-shifting reaction again, but this time the reaction releases energy (those are orange flames, symbolic of the heat produced).
What happens if that energy is removed? For example, if heat is removed through addition of a pale blue ice cube, what will be the effect on the system?
Those orange energy shapes (the "flames") were a part of the system. If they are removed, the system will have to shift in order to restore them. If the reaction pushes to the right again, more energy will be released, bringing the system back into equilibrium.
Exercise \(1\)
The water-gas shift reaction involves the production of hydrogen gas from steam and carbon monoxide. It is important both for the commercial production of hydrogen gas and for its application in fuel cells. At 300 K, the reaction (and an approximate energy produced) is shown below:
Explain what would happen if this gas-phase reaction is already at equilibrium and the following changes take place:
1. The pressure of steam injected into the reaction is doubled.
2. The temperature is raised to 450 K.
3. The CO2 produced is "captured" and removed as carbonate.
4. The temperature is lowered to 250 K.
5. The pressure of CO added is cut in half.
Answer
Answer
The removal of any item produced on the right side of the reaction will shift the reaction to the right in order to restore equilibrium. On the other hand, adding any more of any of the items on the right will shift the reaction to the left.
Items on the left side will work in the opposite way. Adding more of anything on the left will shift the reaction to the right, to use up the newly added materials. Removing anything from the left will shift the reaction further left, to replace the items that were removed.
Answer a
The amount of water increases, moving the reaction to the right. More products are made.
Answer b
The amount of energy increases, moving the reaction to the left. Fewer products are made.
Answer c
The amount of carbon dioxide decreases, shifting the reaction to the right. More products are made.
Answer d
The amount of energy decreases, shifting the reaction to the right. More products are made.
Answer e
The amount of carbon monoxide decreases, shifting the reaction to the left. Fewer products are made.
Exercise \(2\)
Hydrochlorination of acetylene (ethyne) is another gas-phase reaction. It is used to produce vinyl chloride, the starting material for the polyvinyl chloride commonly used to make the pipes in household plumbing. At 300 K, the reaction (and an approximate energy produced) is shown below:
Explain what would happen if the system is at equilibrium and the following changes take place:
1. The temperature is raised to 350 K.
2. The pressure of HCl is doubled.
3. The pressure of acetylene is cut in half.
4. The temperature is dropped to 250 K.
5. The overall pressure in the system is increased from one atmosphere to two atmospheres.
Answer
Answer a
The amount of energy increases, moving the reaction to the left. Fewer products are made.
Answer b
The amount of hydrogen chloride increases, shifting the reaction to the right. More products are made.
Answer c
The amount of acetylene decreases, shifting the reaction to the left. Fewer products are made.
Answer d
The amount of energy decreases, shifting the reaction to the right. More products are made.
Answer e
This question doesn't follow the pattern. However, because the products and reactants are all gases, we can think about the effect they would have on pressure if the reaction moved one way or the other. Because fewer gas molecules are produced on the right than the left, pressure would decrease on going from left to right (and increase on going from right to left). Thus, we can pencil in "pressure" as an item on the left side of the reaction. That means increasing pressure will shift the reaction to the right, making more products.
Exercise \(3\)
Production of ATP in the cell proceeds according to the reaction below, with an approximate energy indicated at 310 K.
If the system is already at equilibrium, explain what happens when the following changes take place:
1. The temperature is raised to 320 K (It's OK. This organism is really hardy and it can handle the temperature change).
2. The temperature is lowered to 300 K.
3. The supply of inorganic phosphate is doubled.
Answer
Answer a
The amount of energy increases, shifting the reaction to the right. More products are made.
Answer b
The amount of energy decreases, moving the reaction to the left. Fewer products are made.
Answer c
The amount of phosphate increases, shifting the reaction to the right. More products are made.
Exercise \(4\)
Nitric acid, HNO3, is a common industrial chemical. For example, it is used to make azo dyes that are employed in paints. Nitric acid production involves the following reaction, with an approximate energy change indicated at 300 K.
Note that this is a multi-phase reaction: it involves gases (g), liquids (l) and aqueous solutions (aq, something dissolved in water). Explain what would happen if production were run under the following conditions:
1. The NO2 gas is introduced into a chamber that contains a tank of water. After reacting for a while, the gas is released and the water, containing the aqueous solution of nitric acid, is drained from the tank.
2. The NO2 gas is introduced into a chamber that contains a tank of water. Periodically, the water is drained from the tank, and new water is introduced, without releasing any gases.
3. The NO2 gas is continually introduced into a chamber, and there is a vent that slowly releases gases from the chamber at all times. There is also a constant flow of water into and out of the chamber.
Answer
Answer a
The nitric acid would build up in the water, and the NO gas would build up, until equilibrium is reached. The nitric acid in the water would be limited by that equilibrium point.
Answer b
Periodically removing the nitric acid solution and adding fresh water would help to shift the reaction further to the right, although the eventual buildup of NO gas might prevent the reaction from shifting too far.
Answer c
A constant source of both water and nitrogen dioxide (nitric oxide) would help to push the reaction to the right. Although allowing gases to vent would limit the amount of nitrogen dioxide in the system, it would also prevent a buildup of nitrogen monoxide (nitrous oxide), which would otherwise push the reaction to the left, eventually. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/III%3A_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_1/01%3A_Thermodynamics/1.05%3A_Reversibility_and_L.txt |
The balance between reactants and products in a reaction will be determined by the free energy difference between the two sides of the reaction. The greater the free energy difference, the more the reaction will favor one side or the other. The smaller the free energy difference, the closer the mixture will get to equal parts reactants and products (loosely speaking).
Exactly where the balance lies in an equilibrium reaction is described by the equilibrium constant. The equilibrium constant is just the ratio of products to reactants, once the reaction has settled out at equilibrium. That's the point at which the forward and reverse reactions are balanced, so that the ratio of products to reactants is stable.
• A reaction has reached equilibrium when the reaction has stopped progressing, so that the amount of reactants that have turned into products remains constant, and the amount of reactants left over stays constant.
• The equilibrium constant is the ratio of products to reactants when the reaction has reached equilibrium.
The equilibrium constant could be a large number (like a thousand). That means that there are much more products than reactants at equilibrium. It could also be a very small fraction (like one millionth). That would indicate that the reaction does not proceed very far, producing only a tiny amount of products at equilibrium.
• Every reaction has an equilibrium constant.
• A very large equilibrium constant (in the millions or bazillions) means the reaction goes "to completion", with all reactants essentially converted into products.
• A tiny equilibrium (very close to zero) constant means the reaction hardly moves forward at all.
• A modest equilibrium constant (close to one, or as close to one as numbers like 0.01 or 100) is considered to be a true equilibrium reaction, in which there is a significant amount of both products and reactants.
The equilibrium constant is related to the free energy change of the reaction by the expression:
$K = e^({\frac{- \Delta G}{RT}}) \nonumber$
or $ln K = \frac{-Delta G}{RT}$
in which T is the temperature in Kelvin and R is the "gas constant" (1.986 cal/K mol). Remember, e is just a number that occurs frequently in mathematical relationships in nature (sort of like π); it has a value of about 2.718. This expression for K does make some assumptions about the conditions that we won't worry about; we are using a slightly simplified model.
Exercise $1$
Arrange the following series of numbers from the largest quantity to the smallest, from left to right.
1. 105 104 106
2. 23 26 22
3. 33 30 32
4. e2 e1 e4
5. 10-1 10-5 10-3
6. 1 / 10 1 / 25 1 / 50
7. 20.5 20.1 20.9
Answer
The exponent is the number of times the base number is multiplied by itself. For example, 103 = 10 x 10 x 10. The higher the exponent, the larger the resulting mathematical product.
The same is true with the magnitude of a negative exponent, but the negative sign means that we are dealing with the inverse of the base number. For example, $10^{-2} = \frac{1}{10} \times \frac{1}{10} = \frac{1}{10 \times 10}$
Let's look at the form of this relationship between free energy and the equilibrium constant. First, we will see how we deal with endergonic versus exergonic reactions. The free energy changes in opposite directions in these two cases, and we usually deal with opposites by giving one quantity a positive sign and one quantity a negative sign. A reaction in which the free energy increases is given a positive value for its free energy. On the other hand, if free energy decreases over the course of the reaction, we show that by using a negative number for the value of the free energy.
If ΔG is negative, the exponent in the relationship becomes positive (because it is multiplied by -1 in the expression). Since e to a positive power will usually be a number greater than one, the relationship suggests there are more products than reactants. That's good, because the reaction is exergonic, and we expect the reaction to go forward. What's more, the larger the value of ΔG, the more product-favored the reaction will be.
• 10large number is a large number.
• 10small number is a smaller number.
However, if ΔG is a positive number, then the exponent in the relationship becomes negative. An number with a negative exponent, by the rules of exponents, is the same as the inverse of the number with a positive exponent of the same size.
In other words, 10-2 = 1 / 102.
• 10negative number is a fraction.
That means if ΔG is positive, the equilibrium constant becomes a fraction. That's because that positive value of ΔG is multiplied by -1 in the expression, becoming negative, and then it's placed in the exponent.
That's good, because a positive value of ΔG corresponds to an endergonic reaction, and that's not supposed to favor product formation.
Exercise $2$
Given the following free energy differences, arrange the corresponding equilibrium constants from largest to smallest.
1. 25 kcal/mol 17 kcal/mol 9 kcal/mol
2. 16 kcal/mol 19 kcal/mol 21 kcal/mol
3. 7 kcal/mol 22 kcal/mol 13 kcal/mol
4. -17 kcal/mol -3 kcal/mol -8 kcal/mol
5. -17 kcal/mol 3 kcal/mol -8 kcal/mol
Answer
The greater (and more positive) the free energy change, the smaller the equilibrium constant.
However, the greater (and more negative) the free energy change, the larger the equilibrium constant.
Equilibrium constants, from largest to smallest, would have associated free energies as follows:
(large K) big, negative ΔG > small, negative ΔG > small, positive ΔG > large, positive ΔG (small K)
There are other factors in the expression relating ΔG to the equilibrium constant. One of them, R, is just a "fudge factor"; it's the number that, when placed in the expression, makes the relationship agree with reality. And it's a constant, so it does not change.
The other factor is temperature. It does change. That means that the equilibrium constant may change with different temperatures.
Overall, the effect of temperature is to make the exponent in the expression a smaller number. That's because the free energy is divided by the temperature and the gas constant; the resulting number becomes the exponent in the relationship. At the extreme, a high temperature could make the exponent into a very, very small number, something close to zero. What happens then?
• 100 = 1
• e0 = 1
As the exponent gets smaller and smaller, the equilibrium constant could approach 1. That means there would be more or less equal amounts of products and reactants in our simplified approach.
However, the fact that there is a temperature factor in the expression for ΔG itself means that there is a limit to how small K will get as the temperature increases. At some point, the two values for temperature cancel out altogether and the expression becomes K = e (ΔS/R). At that point, the equilibrium constant is independent of temperature and is based only on internal entropy differences between the two sides of the reaction.
This relationship is useful because of its predictive value. Qualtitatively, it confirms ideas we had already developed about thermodynamics.
• Highly exergonic reactions (large, negative/decreasing ΔG) favor products.
• Highly endergonic reactions (large, positive/increasing ΔG) favor reactants.
• Reactions with small free energy changes lead to equilibrium mixtures of both products and reactants.
Exercise $3$
What is the value of the equilibrium constant at 300K in the following cases?
What is the value of the equilibrium constant in the following cases? (1 kcal = 1000 cal)
1. ΔG = 3 kcal /mol
2. ΔG = -2 kcal/mol
3. ΔG = -5 kcal /mol
4. ΔG = 15 kcal/mol
5. ΔG = -10 kcal/mol
6. The free energy increases by 8 kcal/mol over the reaction.
7. The free energy decreases by 1 kcal/mol over the reaction.
Answer
This is just an algorithm problem, but don't forget to convert kcal to cal.
For example, in (a), $K = e^ = e^{-5.035} = 0.0065$
Exercise $4$
In which of the cases in Exercise $3$ do you think there would be significant amounts of both products and reactants at equilibrium?
Answer
Remember, the closer K gets to 1, the closer the system gets to an equal mix of reactants and products. That's a slight approximation, because the value of K when there is an equal amount of reactants and products may be more or less than one depending on how many molecules (or moles) of each species are involved in the reaction.
Exercise $5$
The mathematical expression for the equilibrium constant says that K will get smaller at higher temperatures. Explain this phenomenon without the mathematical expression in terms of what you know about temperature and energy.
Answer
There are a couple of reasons, but one involves the enthalpy requirement compared to the available energy. Temperature is an index of how much energy is available in the surroundings. The more energy is available from the surroundings, the more likely energy can be supplied to overcome a deficit in enthalpy, for either the forward or the reverse reaction. Thus at high temperature, the equilibrium is just as likely to sit on the high energy side of the reaction as it is on the low energy side.
Another way of looking at things is that the external entropy change is relatively small at high temperature, because the additional distribution of energy resulting from the reaction is very small compared to the pre-existing distribution of external energy when there is already a lot of energy in the surroundings. That leaves only the internal entropy change to govern the equilibrium. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/III%3A_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_1/01%3A_Thermodynamics/1.06%3A_Free_Energy_and_Equ.txt |
Proton transfer reactions are very common. It's useful to be able to tell whether a proton is likely to be transferred from one position to another. Researchers have worked very hard to work out information that will give them some insight into that problem. The resulting data is often compiled into a pKa table, like the one below.
The table is organised into columns of compounds. On each compound, there is a hydrogen colored in blue. The pKa value listed next to the compound is an index of how tightly that hydrogen is held by the compound. The bigger the pKa, the more tightly that proton is held.
• A low pKa means the proton is given up easily. The compound is very acidic.
• A high pKa means the proton is held very tightly. The compound is not very acidic.
For example, methanesulfonic acid, CH3SO3H, is quite acidic. It has a low pKa. In this table, its approximate pKa is listed as -3. What does that mean exactly? Well, the equilibrium constant for the ionisation of any compound is 10-pKa. The equlibrium constant for the ionisation of methanesulfonic acid is 103, or 1000. That's the equilibrium constant for the reaction:
CH3SO2O-H = CH3SO2O- + H+
Of course, you may already know that a proton is not likely to wander around on its own. It is normally bound to a lone pair. In some studies, the lone pair is on a water molecule. Then the reaction would really look like this:
CH3SO2O-H + H2O = CH3SO2O- + H3O+
In other studies, the lone pair is on another solvent molecule, such as DMSO, (CH3)2SO. Then the reaction would really look like this:
CH3SO2O-H + (CH3)2SO = CH3SO2O- + (CH3)2SOH+
The actual value of the pKa varies depending on the conditions of the measurement. For example, a pKa measured in water is a little different from a pKa measured in DMSO, but the trends are generally the same.
We can use the pKa table to compare the acidity of different compounds. Methanesulfonic acid is much, much more acidic than butane, CH3CH2CH2CH3, which has a pKa of about 50. In addition, we can use the table to predict the direction of a reaction, or the equilibrium of a reaction. If a proton is being transferred from one position to another, a comparison of pKa values will tell us whether the reaction will proceed or not.
For example, consider the transfer of a proton from ethane thiol, CH3CH2SH, to pyridine, C5H5N (it looks like a benzene with one carbon replaced by a nitrogen). The pKa of the ethane thiol is 11. If the pyridine accepts the proton, it will form a pyridinium ion, C5H5NH+, with a pKa of 5. Qualitatively, we can already predict that the reaction will not proceed that well. The ethane thiol has a higher pKa than the pyridinium ion, so it will keep its proton, not give it away. However, the difference between the numbers isn't that large. Maybe there will be a measurable equilibrium, meaning an equilibrium in which there are measurable amounts of both products and reactants.
It can be shown that the equilibrium constant for a proton transfer reaction is K = 10(pKa2-pKa1), in which pKa2 is the pKa of the acid on the product side and pKa1 is the pKa of the acid on the reactant side. The acid is simply the species on either side that might give up its proton.
In this case,
K = 10(5-11) = 10-6.
The equilibrium constant is really defined as the ratio of the product concentrations to the reactant concentrations.
$K = \frac{[products]}{[reactants]} \nonumber$
K = [products]/[reactants]
In this case,
$K= \frac{[CH_{3}CH_{2}S^{-}][C_{5}H_{5}NH^{+}]}{[CH_{3}CH_{2}SH][C_{5}H_{5}N]} \nonumber$
We have two numbers multiplied together in the numerator and two numbers multiplied together in the denominator. Each time an ethane thiol and a pyridine react together and transfer a proton, we will get a thiolate anion and a pyridinium cation. In a simple case, we can think of an equal number of thiolate anions and pyridinium cations forming (since the same proton went from one to the other).
Then the equilibrium constant has a slightly simpler form:
$K= \frac{x^{2}}{y^{2}} \nonumber$
in which x = the concentration of either the thiolate or the pyridinium and y = the concentration of either the thiol or the pyridine.
In that case, the ratio of products to reactants, x/y, is equal to the square root of the equilibrium constant. In this case, the ratio is the square root of 0.000001. So the ratio of the products to the reactants is 0.001.
Exercise $1$
a) Predict the equilibrium constant for the reaction of pentane-2,4-dione, CH3COCH2COCH3, with sodium tert-butoxide, NaOC(CH3)3.
Answer
K = 1020-9 = 1011
The table can often be used even if you don't see the compound you are looking for. Often, you can see something that looks pretty similar to what you are interested in and you can estimate the pKa that you need.
Exercise $2$
a) Predict the equilibrium constant for the reaction of 2-pentanone, CH3CH2COCH2CH3, with sodium methoxide, NaOCH3.
Answer
K = 1017-19 = 10-2
Note that this pKa table is organised so that similar sorts of structures are placed in a box together. That arrangement is supposed to make it easier for you to look for trends and understand what kinds of factors make a proton more tightly held or more easily released.
Exercise $3$
1. Compare a pair of compounds in the box of nitrogen compounds and explain why one has a higher pKa than the other.
2. Compare a pair of compounds in the box of hydrocarbons and explain why one has a higher pKa than the other.
3. Compare a pair of compounds in the box of O-H containing compounds and explain why one has a higher pKa than the other. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/III%3A_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_1/01%3A_Thermodynamics/1.07%3A_Equilibrium_and_pKa.txt |
Sometimes we can use available information about the energetics of reactions to predict the energetics of a new reaction. To see how, we'll look at a relatively simple reaction: the combustion of carbon.
A couple of things could happen when we burn some carbon. Burning typically results in the combination of the elements in a material with oxygen. So we are just talking about combining carbon with oxygen to make a new compound. There are two possibilities. Either the reaction forms carbon monoxide, or it forms carbon dioxide.
The first reaction is:
$\ce{2C + O_{2} \rightarrow 2 CO}$
That reaction is exothermic. It releases about 25 kcal per mol of CO produced.
Note that the reaction has been balanced to keep track of the exact numbers of atoms involved in the transaction. Because only one oxygen atom is needed, and the oxygen molecules comes with two oxygen atoms, the oxygen molecule can actually convert two carbon atoms into carbon monoxide.
The second reaction is:
$\ce{C + O_{2} \rightarrow CO_{2}}$
That reaction is also exothermic. It releases about 90 kcal per mol of CO2 produced.
Because energy is released, or produced, by each reaction, we can think about the energy as another product of the reaction. We'll just list it on the product side of the equation with the other products.
$\ce{2C + O_{2} \rightarrow 2CO + 50 \frac{kcal}{mol}}$
and
$\ce{C + O_{2} \rightarrow CO_{2} + 90 \frac{kcal}{mol}}$
There's a third reaction that is related to these two reactions. It's the combustion of carbon monoxide.
$2CO + O_{2} \rightarrow 2CO_{2}$
Again, the reaction is exothermic, releasing about 65 kcal per mole of CO2 produced. It would release 130 kcal for the reaction as written, because we are showing the production of two moles of CO2. Rewriting the equation to include the energy produced:
$2CO + O_{2} \rightarrow 2 CO_{2} + 130 \frac{kcal}{mol}$
What if we conducted this reaction in stages? What if we combusted the carbon to carbon monoxide, then took the carbon monoxide and allowed it to react further to get carbon dioxide?
$2C + O_{2} \rightarrow 2 CO + 50 \frac{kcal}{mol}$
$2CO + O_{2} \rightarrow 2CO_{2} + 130\frac{kcal}{mol}$
Imagine this is a pair of algebraic equations. What would happen if we added them together?
$2C + O_{2} = 2CO + 50 \frac{kcal}{mol}$
$2CO + O_{2} = 2CO_{2} + 130 \frac{kcal}{mol}$
Sum: $2 C + 2 CO + 2 O_{2} = 2 CO + 2 CO_{2} + 180\frac{kcal}{mol}$
Note that the CO appears on both sides and would cancel.
$2C + 2O_{2} =2CO_{2} + 180 \frac{kcal}{mol}$
We can drop the factor of 2:
$C + O_{2} = CO_{2} + 90 \frac{kcal}{mol}$
So two reactions, one after the other, would add up to a third. In addition, the energies of those two reactions, added together, give the energy of the third.
This result is a pretty important aspect of thermodynamics. Enthalpy is a state function. that means it does not matter how a reaction is performed. Whether we convert carbon directly into carbon dioxide or we convert it to carbon monoxide, then continue, the e nergy involved is the same overall. That's because the energy of the reaction is a property of the products and the reactants only. It is independent of how we get from one to the other.
One more note on the reactions above. The enthalpies for these reactions, if measured in the correct way, are sometimes called the heats of formation of the compounds. The heat of formation refers to the energy change when the compounds are formed from the elements under standard conditions. We're not going too deeply into what those standard conditions are here. However, because C is the elemental form of carbon and O2 is the elemental form of oxygen, we would loosely consider the energies listed above to be heats of formation.
• When you hear the phrase "heat of formation", we're just talking about the formation of the compound from the elements.
Exercise $1$
1. If the heat of formation of potassium chloride, KCl, is -104 kcal/mol, and the heat of formation or potassium chlorite, KClO2, is -95 kcal/mol, then what is the heat of reaction when potassium chloride reacts with oxygen to produce potassium chlorite?
2. If the heat of formation of tantalum(IV) oxide, TaO2, is -40 kcal/mol, and the heat of formation of tantalum(V) oxide, Ta2O5, is -490 kcal/mol, then what is the heat of reaction for the combustion of TaO2 to Ta2O5?
3. If the heat of formation of carbon monoxide, CO, is -25 kcal/mol and the heat of formation of tetracarbonyl nickel, Ni(CO)4, is -145 kcal/mol, then what is the heat of reaction for the formation of tetracarbonyl nickel from nickel and carbon monoxide?
Answer
Answer a
We could write the equations for the reactions, including the energy change involved:
K + 0.5 Cl2 → KCl + 104 kcal/mol
There is a 0.5 in front of the Cl2. Although chlorine is diatomic in its elemental state, we only need half the number of Cl2 molecules as we need potassium atoms if we are to form potassium chloride.
K + 0.5 Cl2 + O2 → KClO2 + 95 kcal/mol
In both cases, the heat of formation is negative, so we are writing that energy as a product of the reaction.
If we write the first reaction in reverse,
104 kcal/mol + KCl → K + 0.5 Cl2
then we are saying that the reverse reaction would require the input of energy.
We will combine those two equations by adding them together:
K + 0.5 Cl2 + O2 → KClO2 + 95 kcal/mol
104 kcal/mol + KCl → K + 0.5 Cl2
Sum:
104 kcal/mol + KCl + K + 0.5 Cl2 + O2 → K + 0.5 Cl2 + KClO2 + 95 kcal/mol
Simplifying (the 0.5 chlorine molecules and potassium atoms appear on both sides, so they cancel):
104 kcal/mol + KCl + O2 → KClO2 + 95 kcal/mol
Now we use more algebra and combine the numerical part together:
104 kcal/mol - 95 kcal/mol + KCl + O2 → KClO2 + 95 kcal/mol - 95 kcal/mol
Which leaves:
9 kcal/mol + KCl + O2 → KClO2
The energy is added on the left. It is needed for the reaction. The heat of reaction, ΔHrxn = + 9 kcal/mol. It's an endothermic reaction. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/III%3A_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_1/01%3A_Thermodynamics/1.08%3A_Hess%27_Law.txt |
Exercise 1.1.1:
Reactions that go down in energy will proceed. Reactions that go up in energy will not proceed. If the reaction profile is higher on the left (reactant side) than the right, the reaction will go forward and it will form products. If the reverse is true, the reaction presumably will not occur.
Exercise 1.2.1:
b) Because there are two carbons in ethane, one molecule of ethane will give rise to two molecules of CO2.
c) Because there are six hydrogens in ethane, one molecule of ethane will give rise to three molecules of H2O.
d) In order to make two molecules of carbon dioxide (four oxygen atoms) and three water molecules (three oxygen atoms), we would need seven oxygen atoms total. Since oxygen molecules contain pairs of oxygen atoms, we would only need 3.5 oxygen molecules.
In principle, three and a half molecules is a problem. Where are we going to get a half of a molecule? In practice, it's nothing to worry about. We can't really do reactions with single molecules anyway. We are always working with vast numbers of molecules, but we have to make sure we keep them in the right ratio. Instead of using one molecule of ethane, we might use one billion ethane molecules, and 3.5 billion oxygen molecules.
a) The reaction is given with structures below:
e) The energy requirements are laid out in the following table. Overall, the reaction releases 375.5 kcal per mol of ethane burned. The negative sign in the table is often used to denote that this is excess energy released (whereas a positive sign would indicate that energy as consumed overall).
Bond Breaking Costs (kcal/mol) Sum of Cost Bond Making Releases (kcal/mol) Sum of Release Overall (kcal/mol)
6 x C-H 6 * 99 594 6 x O-H 6 * 111 666
3.5 x O=O 3.5 * 119 416.5 4 x C=O 4 * 180 720
1 x C-C 83 83
total breaking: 1093.5 making: 1386 -292.5
That's more energy than was produced from a molecule of methane (-170 kcal/mol).
Exercise 1.3.1:
Entropy is higher if the energy is partitioned into more states. For example, in question (b), the same amount of energy is distributed into three states on the left hand side and only two states on the right. Entropy is higher in the left hand example than the right in that case.
Exercise 1.3.2:
Examples of "states" into which energy can be partitioned include molecular vibrational, rotational and translational states (which, loosely speaking, correspond to wiggling, spinning and zipping around). Entropy is higher if energy is distributed into more of these states. That might include a greater range of vibrational or rotational states used in (a) and (c), or similar states employed in a greater number of molecules in (b).
Exercise 1.3.3:
One general observation about internal entropy is that it increases if the number of molecules increases during a reaction and decreases if the number of molecules decreases during a reaction. It's just a matter of counting how many things on the left get turned into how many things on the right. For example, in question (a), one molecule produces two new molecules in the decarboxylation reaction, so the reaction is entropically favored.
Exercise 1.4.1:
The expression ΔG = ΔH - TΔS includes both an enthalpy contribution and an enthalpy contribution and balances them against each other. However, the effect of entropy is multiplied by the temperature. The greater the temperature, the greater will be the influence of entropy (and therefore the smaller the influence of enthalpy). The lower the temperature, the smaller will be the influence of entropy (and therefore the greater the influence of enthalpy).
Exercise 1.4.2:
See problem 1.1.1
Exercise 1.5.1:
The removal of any item produced on the right side of the reaction will shift the reaction to the right in order to restore equilibrium. On the other hand, adding any more of any of the items on the right will shift the reaction to the left.
Items on the left side will work in the opposite way. Adding more of anything on the left will shift the reaction to the right, to use up the newly added materials. Removing anything from the left will shift the reaction further left, to replace the items that were removed.
1. The amount of water increases, moving the reaction to the right. More products are made.
2. The amount of energy increases, moving the reaction to the left. Fewer products are made.
3. The amount of carbon dioxide decreases, shifting the reaction to the right. More products are made.
4. The amount of energy decreases, shifting the reaction to the right. More products are made.
5. The amount of carbon monoxide decreases, shifting the reaction to the left. Fewer products are made.
Exercise 1.5.2:
1. The amount of energy increases, moving the reaction to the left. Fewer products are made.
2. The amount of hydrogen chloride increases, shifting the reaction to the right. More products are made.
3. The amount of acetylene decreases, shifting the reaction to the left. Fewer products are made.
4. The amount of energy decreases, shifting the reaction to the right. More products are made.
5. This question does not follow the pattern. However, because the products and reactants are all gases, we can think about the effect they would have on pressure if the reaction moved one way or the other. Because fewer gas molecules are produced on the right than the left, pressure would decrease on going from left to right (and increase on going from right to left). Thus, we can pencil in "pressure" as an item on the left side of the reaction. That means increasing pressure will shift the reaction to the right, making more products.
Exercise 1.5.3:
1. The amount of energy increases, shifting the reaction to the right. More products are made.
2. The amount of energy decreases, moving the reaction to the left. Fewer products are made.
3. The amount of phosphate increases, shifting the reaction to the right. More products are made.
Exercise 1.5.4:
1. The nitric acid would build up in the water, and the NO gas would build up, until equilibrium is reached. The nitric acid in the water would be limited by that equilibrium point.
2. Periodically removing the nitric acid solution and adding fresh water would help to shift the reaction further to the right, although the eventual buildup of NO gas might prevent the reaction from shifting too far.
3. A constant source of both water and nitrogen dioxide (nitric oxide) would help to push the reaction to the right. Although allowing gases to vent would limit the amount of nitrogen dioxide in the system, it would also prevent a buildup of nitrogen monoxide (nitrous oxide), which would otherwise push the reaction to the left, eventually.
Exercise 1.6.1:
The exponent is the number of times the base number is multiplied by itself. For example, 103 = 10 x 10 x 10. The higher the exponent, the larger the resulting mathematical product.
The same is true with the magnitude of a negative exponent, but the negative sign means that we are dealing with the inverse of the base number. For example, $10^{-2} = \frac{1}{10} \times \frac{1}{10} = \frac{1}{(10 \times 10)}$
Exercise 1.6.2:
The greater (and more positive) the free energy change, the smaller the equilibrium constant.
However, the greater (and more negative) the free energy change, the larger the equilibrium constant.
Equilibrium constants, from largest to smallest, would have associated free energies as follows:
(large K) big, negative ΔG > small, negative ΔG > small, positive ΔG > large, positive ΔG (small K)
Exercise 1.6.3:
This is just an algorithm problem, but don't forget to convert kcal to cal.
For example, in (a), $K = e^{-\frac{3000 cal \: mol^{-1}}{1.986 cal \: K^{-1} \: mol^{-1} \times 300K}} = e^{-5.035} = 0.0065$
Exercise 1.6.4:
Remember, the closer K gets to 1, the closer the system gets to an equal mix of reactants and products. That's a slight approximation, because the value of K when there is an equal amount of reactants and products may be more or less than one depending on how many molecules (or moles) of each species are involved in the reaction.
Exercise 1.6.5:
There are a couple of reasons, but one involves the enthalpy requirement compared to the available energy. Temperature is an index of how much energy is available in the surroundings. The more energy is available from the surroundings, the more likely energy can be supplied to overcome a deficit in enthalpy, for either the forward or the reverse reaction. Thus at high temperature, the equilibrium is just as likely to sit on the high energy side of the reaction as it is on the low energy side.
Another way of looking at things is that the external entropy change is relatively small at high temperature, because the additional distribution of energy resulting from the reaction is very small compared to the pre-existing distribution of external energy when there is already a lot of energy in the surroundings. That leaves only the internal entropy change to govern the equilibrium.
Exercise 1.7.1:
a) K = 1020-9 = 1011
Exercise 1.7.2:
a) K = 1017-19 = 10-2
Exercise 1.8.1:
a) We could write the equations for the reactions, including the energy change involved:
$K + 0.5 Cl_{2} \rightarrow KCl + 104 \frac{kcal}{mol} \nonumber$
There is a 0.5 in front of the Cl2. Although chlorine is diatomic in its elemental state, we only need half the number of Cl2 molecules as we need potassium atoms if we are to form potassium chloride.
$K + 0.5 Cl_{2} + O_{2} \rightarrow KClO_{2} + 95 \frac{kcal}{mol} \nonumber$
In both cases, the heat of formation is negative, so we are writing that energy as a product of the reaction.
If we write the first reaction in reverse,
$104 \frac{kcal}{mol} + KCl \rightarrow K + 0.5 Cl_{2} \nonumber$
then we are saying that the reverse reaction would require the input of energy.
We will combine those two equations by adding them together:
$K + 0.5Cl_{2} + O_{2} \rightarrow KClO_{2} + 95 \frac{kcal}{mol} \nonumber$
$104 \frac{kcal}{mol} + KCl \rightarrow K + 0.5Cl_{2} \nonumber$
Sum:
$104 \frac{kcal}{mol} + KCl + K + 0.5Cl_{2} + O_{2} \rightarrow K + 0.5Cl_{2} + KClO_{2} _ 95 \frac{kcal}{mol} \nonumber$
Simplifying (the 0.5 chlorine molecules and potassium atoms appear on both sides, so they cancel):
$104 \frac{kcal}{mol} + KCl + O_{2} \rightarrow KClO_{2} + 95 \frac{kcal}{mol} \nonumber$
Now we use more algebra and combine the numerical part together:
$104 \frac{kcal}{mol} -95 \frac{kcal}{mol} + KCl + O_{2} \rightarrow KClO_{2} + 95 \frac{kcal}{mol} - 95 \frac{kcal}{mol} \nonumber$
Which leaves:
$9 \frac{kcal}{mol} + KCl + O_{2} \rightarrow KClO_{2} \nonumber$
The energy is added on the left. It is needed for the reaction. The heat of reaction, ΔHrxn = + 9 kcal/mol. It's an endothermic reaction. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/III%3A_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_1/01%3A_Thermodynamics/1.09%3A_Solutions_to_Select.txt |
• 2.1: Introduction
• 2.2: How Tightly Do Ligands Bind?
• 2.3: Electron Counting in Transition Metal Complexes
• 2.4: Chelation
Monodentate ligands bind through only one donor atom. Bidentate ligands bind through two donor sites. Bidentate binding allows a ligand to bind more tightly. Tridentate ligands, which bind through three donors, can bind even more tightly, and so on. This phenomenon is generally called the "chelate effect".
• 2.5: Pi Coordination- Donation from Alkenes
• 2.6: Hapticity
The term used to describe the participation of multiple atoms simultaneously during pi coordination is hapticity.
• 2.7: Hard and Soft Acid and Base Concepts
Not all metals form coordination complexes with all possible ligands. Some metals are more likely to form compounds with certain ligands. This observation has eventually led to a classification system called Hard and Soft Acids and Bases (HSAB).
• 2.8: Ligand Field Theory
• 2.9: Ligand Field Stabilization Energy
We can use the relative energy levels of the d orbitals in a given complex to calculate whether the overall energy would be higher or lower in a high-spin vs. a low-spin case, for example. The calculation provides us with a value that is called the ligand field stabilization energy. Although we have been thinking of bonding in transition metal complexes in terms of molecular orbital ideas, ligand field stabilization energy actually has its roots in a separate approach called crystal field theory
• 2.10: Spectrochemical Series
The d orbital energy splitting is influenced by how strongly the ligand interacts with the metal. Ligands that interact only weakly produce little change in the d orbital energy levels, whereas ligands that interact strongly produce a larger change in d orbital energy levels. The spectrochemical series is a list of ligands based on the strength of their interaction with metal ions.
• 2.11: Ligand Lability
• 2.12: Jahn-Teller Distortion
In some cases, the d-electron count can have a subtle influence on the geometry of a complex. For example, an octahedral complex might be distorted, either stretched along one axis or else compressed. In Jahn-Teller Distortion, this effect arises from unequally-distributed electrons in the same level (degeneracy). Although this phenomenon is structural, it can sometimes exert an influence on the stability of complexes that translates into accelerated ligand substitutions.
• 2.13: Multiple Bonds in Coordination Complexes
• 2.14: Solutions to Selected Problems
• 2.15: More Solutions to Selected Problems
3.01: What is a So
From the student's point of view, the information presented in an equation of a reaction can be confusing. The starting material and the product are linked by a straight reaction arrow. The starting material is the compound at the beginning of the reaction; the product is the compound at the end. The reagent is usually shown above the arrow. The reagent is the compound needed to turn the starting material into the product.
However, something else is often listed along with the reagent: the solvent. That can make students wonder: what does this thing do? Is it a second reaction I should be worrying about?
Students sometimes make the assumption that the reagent is written above the arrow and the solvent written below the arrow. That's a good observation, because reactions are often written that way, although there is no rule that says they have to be. However, there are exceptions in which that typical way of writing things is abandoned. Some reactions require lots of different reagents, additives, and promoters, or else there is a need to report the temperature or the pressure. In these cases, additional items are written below the arrow, just because there isn't enough room on top.
In other cases, a series of reactions are run. For example, in the above reaction it is assumed that there was an aqueous workup to neutralise the product. We might write that reaction out explicitly. In that case, the two different steps are numbered, so that we know that they were done one step at a time, rather than throwing everything in all at once.
Chemists often list the solvent in the reaction because the solvent is, practically speaking, tremendously important. Performing a reaction without solvent is a little like washing your hands without water. You could take a bar of soap and run it between your fingers, but not much will happen without the power of the water. The water dissolves up the soap (or at least suspends it in micelles), moves it around, gets it into contact with the dirt and carries it away.
In fact, water is literally the solvent in the physical process of washing. It can be a solvent in many chemical reactions as well. The solvent has many roles to play in a reaction. Foremost, it dissolves the reactants. In that state, the reactants are very mobile. Without the solvent, the reactants may be solids, or if liquids, they may be too thick for molecules to move around very quickly; they may be more like oils. Depending on the nature of the solvent, intermediates may be stabilised, allowing them to form more easily and aiding the course of the reaction. Solvents also act like baths, moderating heat flow into or out of the reaction as needed.
In the cartoon below, nothing happens when the two reagents are dumped together. When a solvent is added, the two reagents start to dissolve, and as they move around in the solution the two reagents encounter each other and start to react.
At the beginning, you may not want to worry too much about the role of the solvent. However, you may still want to know what sorts of things are likely to be solvents, if only so that you can safely ignore them when trying to sort out how the reactant gets to the product.
The following table sums up a number of the most common solvents, displayed from most polar at the top to least polar at the bottom.
Note that, just because something acts as a solvent in one reaction does not mean it must be one in another. For example, acetone is a pretty common solvent, but it also happens to be a ketone. It's likely to undergo carbonyl addition reactions if presented with good nucleophiles. For that reason, carbonyl addition reactions wouldn't be carried out with acetone, because the nucleophile would just react with the solvent instead of the intended electrophile. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/III%3A_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_1/02%3A_Ligand_Binding_in_Coordination_Complexes_a.txt |
Some nucleophiles are added to carbonyls in the form of salts, such as sodium cyanide. In a salt, there is an anion and a cation. The anion can act as a nucleophile, donating a lone pair to the carbonyl. The cation is just a counterion; it is there to balance the charge but does not usually play an active role.
Some anions are too unstable and reactive to be used as salts. This is especially true with a number of carbon nucleophiles. C-H bonds are not usually acidic enough to deprotonate with a strong base. That makes it hard to make a simple salt containing such nucleophiles. There are exceptions, such as acetylide or alkynyl protons like CH3CCH. In that case, the resulting anion is relatively stable because the lone pair is in a lower-energy orbital with more s character, so it is held more tightly to the nucleus.
Less stable carbon anions can be be stabilized through a covalent bond. If the carbon is covalently attached to a less electronegative atom, the carbon has a partially negative charge. It can still act as though it were an anion. However, the covalent bond stabilizes the would-be "lone pair". Compounds like this can be considered to be "semi-anionic". Frequently, they are described as polar covalent compounds, although that is really a much more general term.
These polar covalent bonds can be found any time a carbon atom is bound to a metal. Remember, the metals are the aroms in the colored boxes in the periodic table below.
One of the most common classes of this type of compounds is the family of organomagnesium halides or Grignard reagents (Green-yard reagents). Victor Grignard was awarded the Nobel Prize in Chemistry for his development of these reagents. These compounds are made by reacting an alkyl halide (such as chloropropane, CH3CH2CH2Cl, or bromopropane, CH3CH2CH2Br) with magnesium metal. With bromopropane, the metal undergoes an insertion into the C-Br bond, forming CH3CH2CH2MgBr. (You don't need to worry about how this happens.) Because magnesium is less electronegative than carbon, the C-Mg bond acts as though it were a lone pair on the carbon and the magnesium acts as though it were a cation.
What's striking about Grignard formation is that polarity is reversed in this reaction. In the alkyl halide, the carbon attached to the halogen has a partially positive charge, because carbon is further to the left than halogens in the periodic table. After magnesium insertion, this same carbon has a partial negative charge, because carbon is farther to the right in the periodic table than magnesium. This sort of reversal in reactivity is sometimes called "umpolung chemistry".
At the extreme, we could think of propylmagnesium bromide as a propyl anion with a magnesium counterion. That picture really isn't very accurate; there really is a covalent bond between the carbon and the magnesium, which is what makes the compound more stable than if it were purely ionic. However, thinking of it as a propyl anion might help you understand its role as a nucleophile.
Propylmagnesium chloride and other Grignard reagents can deliver alkyl nucleophiles to carbonyls. Just like with simple anionic nucleophiles, an alkoxide ion results.
The relatively simple mechanism is shown below. The nucleophilic bond donates to the carbon of the carbonyl, breaking the π-bond, releasing a pair of electrons to the electronegative oxygen atom. Alternatively, you could still draw the bromine attached to a Mg+ at this point.
Of course, subsequent treatment of the alkoxide with acid provides a proton, resulting in an alcohol. We can see the two consecutive reactions in the following mechanism.
In that first step, there is no apparent lone pair involved. Instead, the polar magnesium carbon bond behaves just as if it were a lone pair on carbon.
Remember, the order of these two steps is very important. Adding the acid before the Grignard reagent would not work, because the Grignard reagent would become protonated at the carbon. Although the Mg-C bond is covalent, it is still polar enough so that the carbon can act as a nucleophile or as a base. Once propylmagnesium has become protonated, it forms propane, which isn't likely to act as a nucleophile.
Grignards can have a variety of structures, but they are almost always hydrocarbons, with no other functional groups in the structure. Other functional groups are frequently incompatible with the reactive metal-carbon bond. The same is true for the closely-related alkyllithium compounds, such a methyllithium,CH3Li. Other than that, Grignards and alkyllithiums can be saturated (containing only sp3 carbons) or they can contain double bonds.
Exercise \(1\)
Show the products of the following reactions. Assume workup with aqueous acid after each reaction.
Answer
Grignard reagents are very delicate. Solvents must be chosen very carefully for Grignard reactions. Grignard reagents are basic enough that they can't tolerate protic solvents. Protic solvents are solvents that are capable of hydrogen-bonding. Although they don't seem very acidic, they can still give up a proton to a strong enough base. A Grignard reagent is a strong enough base to take that proton from an O-H bond.
Exercise \(2\)
Show why Grignard reagents cannot be used with ethanol as a solvent.
Answer
In fact, Grignard reagents are even fussier than that. Not only do they not get along well with acidic or even semi-acidic protons, but they tend to need coordinating solvents to help support the magnesium atom and keep the complex stable. The most common solvents for this use are (diethyl) ether and tetrahydrofuran (THF). When two ethers bind to the magnesium, the magnesium has a full octet. Because coordination of magnesium by these weakly donating solvents is crucial, Grignard reagents can't generally be isolated. Instead, they are sold and used as solutions in ethereal solvents.
The THF complex, like other coordination complexes, can be drawn in a number of additional ways. The drawing above is probably the most common type, but a couple of other drawings are shown below. In the drawing on the left, we are distinguishing between two slightly different types of bonds on the magnesium. One bond, which we have already discussed, is a polar covalent bond. It is shown with a straight line. However, we know the carbon behaves as if it were an anion with a lone pair. The other is a dative bond from a neutral donor. The oxygen is simply sharing one of its lone pairs with the magnesium. If the oxygen took its lone pair back, it would be a neutral atom. If the bromine left, it would be an anion. A convention widely adopted in coordination chemistry shows bonds from anionic donors to metals as lines, whereas bonds from neutral donors to metals are shown as short, straight arrows. On the other hand, if we know anything about formal charge, we would look at the picture above and correct it, giving the version below, right.
Each drawing has its merits. The first one is probably the simplest; in this case, we could even afford the luxury of drawing in a wedge and a dash to show stereochemistry without getting too complicated. The second quickly conveys the idea of the charges on the ligands and the metal in the coordination complex; there are two neutral ligands and two anionic ones, and therefore the metal has a +2 charge. The third tells us something about how charge has been transferred in bonding. Because the oxygen has shared its electrons with magnesium, magnesium gets a little more electron density and oxygen's goes down a little bit. Despite being notably electronegative, oxygen frequently donates to other atoms if needed. (A generous act may not always be good for the donor, but it is frequently good for the universe.)
Eventually, it will be useful to know that coordination complexes (or Lewis acid-base adducts) frequently form reversibly. There is actually an equilibrium in which donors come on and off the metal, especially if the donors are pretty stable by themselves.
We can picture the propylmagnesium bromide swimming through the THF, swinging from one THF molecule to another. Of course, it is always possible that it bumps into another oxygen donor: an aldehyde or ketone. It would certainly coordinate to that oxygen as well, but the situation wouldn't last long as the nucleophilic addition would immediately ensue.
The carbon-magnesium bond is polar, and it allows carbon to act as a nucleophile, donating its bonding pair to an electrophile. Other polar bonds behave in a similar way. Common examples include carbon-lithium bonds (which you will see in a question below), as well as aluminum-hydrogen and boron-hydrogen bonds (in complexes such as Na+ BH4- and Li+ AlH4-, also in a question below). Perhaps surprisingly, because both aluminum and boron are less electronegative than hydrogen, these bonds are polarized toward the hydrogen (carbon, just to the right of boron, is slightly more electronegative than hydrogen, although not enough to make us think of a C-H bond as polar). That means that these compounds act as sources of nucleophilic hydride ion, H-.
Just like Grignard reagents, alkyllithium reagents and complex hydride anions are good nucleophiles for aldehydes and ketones. The nucleophilic part is donated to the carboonyl to make an alkoxide anion.
Like Grignard reactions, these reactions are usually followed by treatment with aqueous acid (such as HCl or H2SO4 in water). The alkoxide ion picks up a proton to make an alcohol. The salts (including sodium or lithium ions) are washed away in the water.
Exercise \(3\)
There are plenty of other semi-anionic nucleophiles. For example, alkyl lithium reagents are also very common, and they are prepared by treatment of alkyl halides with finely divided lithium metal. The reaction produces lithium chloride as a side product.
a) Show an equation, with structures, for the preparation of butyllithium from 1-bromobutane.
b) Explain what happens to polarity at carbon number one before and after this reaction.
c) Why would the amount of charge on carbon number one be somewhat similar in butyllithium and butylmagnesium bromide?
Answer
Exercise \(4\)
Another class of semi-anionic nucleophiles is the family of complex metal hydrides. Examples include sodium borohydride, NaBH4, and lithium aluminum hydride (LAH), LiAlH4. There are many other variations.
1. Draw a Lewis structure for lithium aluminum hydride.
2. Explain why LAH functions as a source of the hydride nucleophile, H-.
3. LAH is much more reactive that sodium borohydride; it can reduce compounds that sodium borohydride will not. For example, it can reduce a nitrile such as CH3CN to an amine such as CH3CH2NH2 (after an aqueous workup). Explain why LAH is so much more reactive than NaBH4.
4. Sodium borohydride is sometimes used in methanol, but care must be taken in dissolving the NaBH4. It does not just dissolve; it quickly reacts with the methanol to produce a flammable gas and NaBH3OCH3. Provide a mechanism for this reaction with arrows.
5. Although NaBH3OCH3 also reacts with methanol, it does so much more slowly than NaBH4, and so it is still able to reduce aldehydes and ketones in methanol. Explain the difference between sodium borohydride and sodium methoxyborohydride in terms of reactivity with methanol.
6. LAH cannot be used in protic solvents such as methanol. Explain why.
Answer a
Answer b & c
Answer d
Exercise \(5\)
Barbier reactions are a general class of reactions involving metal alkyls and carbonyls. Treatment of a halide such as propargyl bromide (HCCCH2Br) with zinc metal in the presence of an aldehyde such as benzaldehyde (C6H5CHO) results in nucleophilic addition of the propargyl group to the aldehyde.
1. Zinc can insert into a carbon-halogen bond, just like magnesium. Show the product of the insertion described above.
2. This reaction is usually performed in water with some ammonium chloride, NH4Cl, in solution. Show a mechanism, with curved arrows, for the reaction of the alkylzinc species with the aldehyde to yield an alcohol.
3. Explain why this alkylzinc reaction can be conducted in the presence of water, but a Grignard reaction cannot.
4. "Green chemistry" refers to the intentional use of processes that are better for the environment, by minimizing the use of toxic reagents and solvents. Compare the zinc-mediated Barbier reaction with the Grignard reaction in terms of "greenness".
Answer
Answer a
Exercise \(6\)
Semi-anionic nucleophiles do not just react with carbonyls. They are also frequently used to prepare organometallic compounds via "transmetallation". For example, treatment of tantalum pentachloride, TaCl5, with dimethylzinc, (CH3)2Zn, affords trimethyltantalum dichloride, (CH3)3TaCl2.
1. Assume for the moment that tantalum pentachloride and dimethylzinc are both covalently-bonded molecules. What would you say about bond polarity in each case?
2. What is the side-product of the reaction (i.e. what else must be produced given the production of trimethyltantalum dichloride from these reactants?)
3. In what ratio would you mix the two reactants to get these products?
4. Show a mechanism, with curved arrows, for the formation of trimethyltantalum dichloride.
5. Like some of the other compounds on this page, trimethyltantalum dichloride is remarkably pyrophoric: it catches fire upon contact with air. This behavior often depends on weather and humidity. Show a mechanism, with curved arrows, for what happens when this compound is exposed to air.
Answer
Answer
Exercise \(7\)
Answer
Exercise \(8\)
Provide IUPAC names for the following alcohols. For help, see the functional group section, simple heteroatomics.
Exercise \(9\)
Fill in the missing intermediates and add curved arrows to show electron movement.
Answer | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/III%3A_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_1/03%3A_Addition_to_Carbonyls/3.02%3A_Semi-Anionic.txt |
Enolate ions are just another example of anionic carbon nucleophiles. The reason they get a page to themselves is that they are especially important, especially in biological chemistry. They are also important in the synthesis of organic compounds, such as in the pharmaceutical industry.
Forming Enolate Ions
An enolate ion is the anion that forms when a proton is removed next to a carbonyl. The carbon next to the carbonyl is called the α-position (alpha position). The alpha position is acidic both because of the amount of positive charge on a proton in that position and because of the stability of the anion that results if that proton is removed.
You might have learned about metal hydroxides such as sodium hydroxide and lithium hydroxide. The metal-oxygen bond is ionic because of the large electronegativity difference between the metal and the oxygen. These compounds give rise to hydroxide ions. Those hydroxide ions are basic because they can easily pick up protons to become neutral water molecules. Metal hydroxides are commonly seen in chemistry, and they are thought of as strong bases.
Other oxygen anions are also able to act as strong bases, unless there is some resonance factor that delocalises the anion and makes it less reactive. Methoxide ion and butoxide ion are also common strong bases.
Exercise \(1\)
Show why an enolate ion, such as the one formed from 2-propanone, above, is particularly stable.
Answer
Exercise \(2\)
Show a mechanism, with curved arrows, for the formation of the enolate ion from 2-propanone, above.
Answer
In the example above, 2-propanone is deprotonated at the α position to form the corresponding enolate ion. Note that sodium hydroxide is not a strong enough base to convert all of the 2-propanone to its enolate. The resulting enolate is basic enough to pull a proton from a water molecule, so an equilibrium results.
That would be the case any time a strong base such as a hydroxide or an alkoxide was used to deprotonate a ketone or aldehyde. In the following example with pentanal, the reaction would also result in an equilibrium between the reactants and the products.
That means that all of those compounds on both the left hand and right hand side of the arrow would all be present as a mixture.
Negative charges are fairly stable on oxygen atoms. That allows this particular reaction to shift back to the left again, to form that hydroxide ion again. To make the reaction go all the way to the right, we would need a less stable anion on the left. That would make that anion more basic. Can you think of atoms that would be less stable as anions than oxygen?
The most commonly used very strong bases in synthetic chemistry involve anions of carbon, nitrogen or hydrogen. Some examples of compounds used as very strong bases are sodium hydride (NaH), sodium amide (NaNH2), lithium diisopropylamide (LiN[CH(CH3)2]2, also called LDA for short), and butyllithium (CH3CH2CH2CH2Li, abbreviated BuLi).
In all of these compounds, the negative charge is on a less electronegative element than the oxygen of the hydroxide ion. That means that they are less stable and more reaction than hydroxide.
If one of these bases were to react with an aldehyde or ketone, the proton would be removed irreversibly. The delocalised enolate ion is actually more stable than the original amide ion in sodium amide, for example.
As a consequence, adding a very strong base to an aldehyde or ketone results in complete conversion into products. At the end of the reaction, there are no reactants left.
In contrast, a "strong base" such as sodium hydroxide won't really do the job. If it did, we would be trading in an anion on a more electronegative atom (oxygen) for an anion on a less electronegative atom (carbon) in the same row of the periodic table. That's not possible. The enolate anion that forms would be more basic than the hydroxide we began with, and most of the time it would just snatch the proton back from the water again, making ketone and hydroxide again.
• Enolate ions form in equilibrium with their parent carbonyl compounds if a moderately strong base like sodium hydroxide is used.
• A very strong base, like sodium amide (NaNH2), butyllithium (CH3CH2CH2CH2Li, or BuLi) or sodium hydride (NaH), would result in complete enolate formation.
However, it is sometimes really useful to have an equilibrium between a carbonyl compound and its enolate. That situation allows both a ketone (the 2-propanone, left) and its enolate (right) to be present at the same time. That means there is both a nucleophile and an electrophile (the ketone and the enolate). They will be able to react together.
• Simple carbonyls are electrophiles.
• The enolate ions that form from simple carbonyls are nucleophiles.
• Carbonyls react with enolate ions.
While we are on the subject of bases, there is a third category of compounds that we would consider weak bases. The most common examples are amines (but not amides) and resonance-stabilised oxygen anions.
These compounds would be good at picking up excess protons that were floating around. However, in most cases they wouldn't be strong enough bases to provide an appreciable amount of enolate ion.
Exercise \(3\)
Identify the following compounds as weak, strong, or very strong bases.
Answer
Answer a
very strong
Answer b
very strong
Answer c
weak
Answer d
weak
Answer e
weak
Answer f
strong
Answer g
weak
Answer h
very strong
Answer i
weak
Aldol Reactions: Adding Enolates to Carbonyl Electrophiles
The reaction of an enolate nucleophile with another carbonyl compound is called an aldol reaction. A simple example of this reaction is shown here. This example involves the reaction of 2-propanone with its enolate.
Note the pattern in the product. The carbonyl of the enolate is connected to the enolate carbon which is connected to the alcohol carbon.
Exercise \(4\)
Provide a mechanism, with curved arrows, for the aldol reaction of 2-propanone, above.
Answer
The biosynthesis of sugars, such as fructose, involves coupling smaller sugars together. If one sugar is converted into a nucleophile, it can donate electrons to the carbonyl on the other sugar, forming a new C-C bond. The carbonyl on the second sugar becomes a hydroxyl group in the new, larger sugar.
In the cell, sugars are typically in a phosphorylated form when they react in this way. Phosphorylation is often an important step in activating molecules for biochemical reactions.
Exercise \(5\)
Show the mechanism for the formation of the phosphorylated fructose shown above.
Answer
Add texts here. Do not delete this text first.
A Variation: Aldol Condensation
Sometimes, aldol reactions are followed by a subsequent reaction, called an elimination reaction. That reaction formally produces a molecule of water. Early studies of this reaction would result in droplets of condensation on the glassware in which the reaction occured; hence, it is sometimes called a condensation reaction.
The term "condensation" comes from the fact that the reaction formally results in loss of a water molecule from the alpha and beta positions (H from alpha and OH from beta). Early reactions that were observed to result in loss of water were frequently described as condensations because of the water droplets that would appear on the glassware (literally, condensation) as the reaction proceeded. The term "dehydration" is also used to describe this loss of water.
However, don't get too tied to the descriptions of the reactions (condensation vs. addition). The terms are used loosely and sometimes interchanged. On this page, we try to use addition to describe the initial product and condensation to describe the product after loss of water, but sources elsewhere might not describe it that way.
Exercise \(6\)
Provide a mechanism for the dehydration step in the aldol condensation shown above.
Answer
It can be hard to predict the outcome of an aldol reaction because of the fact that there are two possible products from an aldol reaction (one with a new hydroxyl and one with a new double bond). A chemist might try to make one product in the laboratory, and end up with the other. This process can be difficult to control. However, in general, the elimination reaction is encouraged by heating the reaction. The reaction sometimes occurs without elimination if the reaction is kept cool. However, there are also other factors that may come into play.
Exercise \(7\)
Predict the products of the following aldol reactions.
Answer
Exercise \(8\)
The following compound would give multiple products through different aldol condensation reactions. Show the products.
Answer
Sometimes, two different compounds may react together in an aldol reaction. One compound acts as the nucleophile, and the other one acts as the electrophile. However, the reaction is really not much different that the aldol reactions we have already seen.
The only complication is that now there are two different compounds that could potentially be nucleophiles and two different compounds that could potentially be electrophiles. That makes predicting the outcome of the reaction a little more difficult.
Exercise \(9\)
The following compounds would give multiple products through different aldol condensation reactions. Show the products.
Answer
There are cases where it becomes a lot more obvious which compound would be the electrophile and which one would be the nucleophile. Maybe one of the compounds has a carbonyl that is much less crowded than the other. For example, maybe one compound is an aldehyde and the other one is a ketone. The less crowded carbonyl is much more likely to act as a good electrophile.
Maybe one of the compounds does not even have any alpha-protons. In that case, it can't be deprotonated and it can't form an enolate anion. It won't be able to act as the nucleophile.
Exercise \(10\)
Only some of the following compounds may undergo aldol reactions. Select which ones may not undergo the reaction, and explain what factor prevents them from reacting.
Answer
Some of these compounds do not have alpha-protons, so they cannot form enolate ions.
Exercise \(11\)
Fill in the products of the following aldol condensations.
Answer
Aldol reactions do not just occur with enolate anions, however. Enols are the neutral form of enolates, protonated on the oxygen instead of the alpha carbon. Enols are also good nucleophiles. In an enol nucleophile, the pi bond acts as the electron source, rather than the lone pair. However, the pi bond gets a boost from the lone pair on the oxygen.
Enols are always present in equilibrium with aldehydes and ketones. An enol is a simple tautomer of a carbonyl compound. To get from one to the other, a proton is simply transferred from one position to the other.
• Enamines and enols are also good nucleophiles for aldol reactions.
• Because either acid or base can catalyse keto-enol tautomerism, aldol reactions can be catalysed by either acid or base.
Exercise \(12\)
Show the subsequent protonation step in the reactions involving the enol and the enamine above.
Answer
Exercise \(13\)
An enamine reaction is usually followed by hydrolysis of the C=N bond in the iminium ion. Show the mechanism for conversion of the iminium ion to the carbonyl via addition of aqueous acid.
Answer
Exercise \(14\)
Which conditions would be most likely to form this aldol product (select a, b, c or d below)?
Answer
Exercise \(15\)
Provide products of the following acid-catalysed aldol reactions.
Answer
Exercise \(16\)
Draw the enol nucleophiles for the above question.
Answer
Exercise \(17\)
Provide the products of the following enamine additions, after hydrolysis.
Answer | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/III%3A_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_1/03%3A_Addition_to_Carbonyls/3.03%3A_Enolate_Addi.txt |
A secondary theme in carbonyl chemistry centers on the role played by the oxygen lone pairs. A compound with lone pairs can act as a Lewis base. Can carbonyl compounds also act as Lewis bases? The answer is yes, although it is most important to think about carbonyls primarily as Lewis acids.
One of the reasons the basicity of the lone pair matters is because of carbonyl activation. If a carbonyl donates a lone pair to a Lewis acid, forming a bond, the carbonyl gets a formal positive charge. If the carbonyl has a formal positive charge, it attracts electrons more strongly. In that case, nucleophiles react more easily with the carbonyl. The carbonyl is said to be activated.
A carbonyl can be activated by the addition of a proton donor, such as HCl or other common acids.
• An activated carbonyl has a positive charge.
• Carbonyls become activated by donating a lone pair to a Lewis acid (also called an electrophile).
• Once activated, carbonyls become more reactive.
• Activated carbonyls attract nucleophiles more strongly.
Most common mineral acids are used as aqueous solutions (the familiar HCl, H2SO4, HNO3, H3PO4 and so on). The acid is only found in the presence of water. Many of them are actually hydrates; if you take sulfuric acid,H2SO4, and set it to boil on a hotplate, eventually it reverts back to sulfur trioxide, SO3, as the water boils away, and a fog appears above the beaker. Sometimes, in a laboratory reaction, it isn't helpful to have all that water around (the reasons will become clear later). Other, organic acids are sometimes used instead, such as camphorsulfonic acid or toluenesulfonic acid; these are both solids that are easy to weight out and add to a reaction, and they don't add a bunch of water to the reaction.
Carbonyls are also activated by more general Lewis acids. Often, metal chloride salts are used. These may include main group metals, such as aluminum, bismuth or indium, or transition metals such as scandium, titanium or iron.
Once the carbonyl is activated, nucleophiles are more strongly attracted to the carbon. The carbon was already partially positive, but with a full positive charge on the molecule, electrons are attracted much more strongly.
It is tempting to donate electrons from a nucleophile to the positive oxygen. However, the oxygen already has three bonds and an octet. Remember, donating a lone pair from a nucleophile means the lone pair is becoming a bond between the nucleophile and the electrophile. Giving a pair of electrons directly to the oxygen would give it four bonds and more than an octet-- it would have 10 electrons. Instead, donation to the neighbouring carbon allows the C=O pi bond to move to the oxygen and become a lone pair. The positive charge on the oxygen disappears.
• The nucleophile donates to the activated carbonyl carbon
• That event lets the pi bond become a lone pair on oxygen
Figure \(4\): Donation of nucleophile to an activated carbonyl.
Exercise \(1\)
Show, with arrows, the activation of the following carbonyls.
Answer
Exercise \(2\)
Show, with arrows, the activation of the following carbonyls, followed by donation from the nucleophile.
Answer
Exercise \(3\)
Part a. Use curved arrows to denote electron flow in the following mechanism for the Mukaiyama Aldol Addition.
Part b. Using these starting molecules in Mukaiyama Aldol Addition, predict the product formed.
Answer
Add texts here. Do not delete this text first.
Exercise \(4\)
Provide Lewis-Kekule structures for the following commonly used acids.
a) hydrochloric acid, HCl b) hydrofluoric acid, HF c) hydrobromic acid, HBr
d) nitric acid, HNO3 e) perchloric acid, HClO4 f) phosphoric acid, H3PO4
g) formic acid, HCO2H h) acetic acid, CH3CO2H i) benzoic acid, C6H5CO2H
j) sulfuric acid, H2SO4 k) toluenesulfonic acid, CH3C6H4SO3H l) methanesulfonic acid, CH3SO3H | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/III%3A_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_1/03%3A_Addition_to_Carbonyls/3.04%3A_Activation_o.txt |
Nucleophiles do not have to be ionic, or even "semi-anionic". The basic requirement for a nucleophile is a lone pair. If a nucleophile has a lone pair, it can donate the lone pair to an electrophile such as a carbonyl. By donating a lone pair to a carbonyl, it can form a bond.
However, donation of a lone pair to a carbonyl is reversible. If there is something about the nucleophile/electrophile adduct that isn't very stable, the reaction may revert to reactants again. In the case of neutral nucleophiles, charge separation may destabilize the first-formed products of the reaction. Let's think about addition of water to propanone. Two neutral molecules, water and propanone, come together. The water donates a lone pair to the carbonyl carbon in propanone. That leaves the oxygen atom from the water with a positive charge, and the oxygen atom from the propanone with a negative charge.
One easy way to get rid of the charge separation is for the water to leave again. That step would just be the reverse of the first one.
On the other hand, another way to solve the charge problem is to move a proton (H+) from the positively charged oxygen to the negatively charged one. That turns out to be pretty easy to do. If that happens, a "hydrate" or a "geminal diol" forms. A geminal diol, or twin diol, has two hydroxy groups on one carbon.
Exercise \(1\)
In the following cases, a nucleophile donates to the carbonyl, followed by a proton transfer. Show mechanisms, with curved arrows, for each of the following reactions.
Answer
Answer a
Answer b
The immediate products that form from the addition of neutral nucleophiles to carbonyls turn out to be a little unstable. That's partly because they can easily revert back to reactants. It's also because there are other processes that carry the reaction away from the first-formed products and turn them into other things. For example, in most cases the reaction does not involve the simple addition of one nucleophile, but involves a second molecule of the nucleophile as well.
We will keep looking at these other processes and see where these reactions lead. In the meantime, it can be useful to know the patterns that different types of nucleophiles will usually follow. For example, addition of alcohols to aldehydes or ketones leads to the formation of ketals or acetals. Ketals or acetals have the specific chain of atoms C-O-C-O-C, in which each carbon is tetrahedral or sp3.
On the other hand, the addition of amines leads to a very different kind of structure. Instead of adding two amine molecules into the final structure, only one amine is incorporated, and a double bond appears again. The C=O of the aldehyde or ketone is replaced with the C=N of an imine. Imines form important linkages in biological chemistry, especially in connecting small molecules to lysine side chains in proteins.
If the amine is "secondary", meaning the nitrogen is connected to two carbons instead of just one, the double bond ends up one position over. It is between the former carbonyl carbon and the one next to it, which is called the alpha position.
Note that all three of these reactions involve ultimate loss of water from the original carbonyl compound. That's a very important feature of reactions with neutral nucleophiles.
• Addition of a neutral nucleophile to a carbonyl almost always leads to loss of the original carbonyl oxygen in a water molecule.
Exercise \(2\)
Fill in the missing reagents or products in the following reaction equations.
Answer
Exercise \(3\)
Answer | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/III%3A_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_1/03%3A_Addition_to_Carbonyls/3.05%3A_Addition_of_.txt |
Proton transfer is rapid, especially if it is transferred from a very acidic position. For example, a proton can easily be transferred from a positively charged oxygen atom to a neutral oxygen (resulting in a new, neutral oxygen and a new, positive oxygen). These species would be in equilibrium with each other.
• Proton transfer is rapid.
• Protons can be transferred from more acidic to less acidic position.
• Protons can be transferred from one acidic position to another of similar acidity, although the equilibrium may not be favored.
It would not be as easy to transfer a proton from a neutral oxygen to another neutral oxygen. Sometimes, a neutral oxygen can transfer a proton to a negatively charged one, but the equilibrium will depend on the relative pKa values of the two species. In the case below, the tert-butoxide is a less stable anion than the hydroxide because the tert-butoxide is larger and requires more organization of solvent molecules around it.
It is tempting to think that a proton could be transferred directly from a cationic position to an anionic position in the same molecule. That might not occur, however. In terms of conformational analysis, the two positions might not be able to twist around and reach each other. The usual rule applies: two atoms may need to be greater than five atoms away from each other along a chain before they can reach around and make contact.
If the solvent has a lone pair, it may pick up the proton from the acidic position and drop it off on the basic position. These events are made easy by the fact that the reacting molecules are usually surrounded by many solvent molecules.
• Solvent can often act as a proton shuttle.
3.07: Pi Donation
Some nucleophiles are added to carbonyls, lose a proton to drop a positive charge, and have a lone pair again. If the (former) carbonyl oxygen also has a lone pair, a potential lone-pair/lonepair repulsion problem exists. Partly for this reason, two heteroatoms bonded to one carbon often present an inherently unstable situation. These kinds of species often decompose readily via pi donation.
In pi donation, a lone pair on one heteroatom is donated to the carbon shared by both heteroatoms. As a result, the other heteroatom is pushed off the carbon. This event is helped if one of the heteroatoms is already protonated, so that it comes off as a neutral species.
Exercise \(1\)
Fill in any missing lone pairs, provide curved arrows to show pi donation, and show the resonance structures that result.
Answer
Exercise \(2\)
Show how each of the following species might break down into two molecules via pi donation.
Answer
As a result of pi donation, neutral, protic nucleophiles often replace the carbonyl oxygen entirely. In effect, the nucleophile adds twice. The lone pair that is revealed after a deprotonation step adds again to the same carbon, pushing that carbonyl oxygen out of the molecule entirely. In the case of alcohol nucleophiles, a ketal or acetal results. This kind of molecule looks like two ethers that meet at one carbon. A ketal is a "masked" carbonyl; it still contains a carbon with two bonds to oxygen. However, a ketone is no longer an electrophile like a carbonyl compound.
Imines also result from pi donation. If an primary amine donates to a carbonyl, it can lose its first proton to reveal a lone pair. Once that lone pair donates, pushing off the carbonyl oxygen, a second proton can be dropped to allow the nitrogen to lose its positive charge. Imines are similar to carbonyls in that they contain a carbon-heteroatom double bond and so they are still good electrophiles.
Exercise \(3\)
Fill in the missing reagents and products in the following reaction schemes.
Answer
Add texts here. Do not delete this text first. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/III%3A_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_1/03%3A_Addition_to_Carbonyls/3.06%3A_Proton_Trans.txt |
Carbohydrates are an important class of biological molecules. Although their best-known role is in energy storage in the form of glucose and starch, carbohydrates play a number of other roles. For example, they lend structural support in the backbone of DNA. They aid in homebody security and defense operations by forming molecular codes on the surfaces of cells that identify whether the cell is one of our own or an intruder. They carry specialized chemical reagents into enzymes where reactions essential to life are carried out. Some of them are sweet.
The term "sugar" is often applied to any simple carbohydrate. One strict definition of a carbohydrate is a polyhydroxylated aldehyde or ketone. In general, a sugar is a molecule that contains an aldehyde or a ketone, and every carbon other than the carbonyl carbon has a hydroxyl group attached to it.
This sort of structure presents lots of possibilities for reactions. In just one molecule, we have both an electrophile (the carbonyl) and a number of nucleophiles (the hydroxyls). Can sugars react with each other? Yes. Can a sugar react with itself? Of course.
In fact, if you have seen drawings of sugars before, you might not have noticed the carbonyl. That's because the carbonyl is usually "masked" as a hemiacetal. The hemiacetal forms when a hydroxyl group along the carbon chain reaches back and bonds to the electrophilic carbonyl carbon. As a result, five- and six-membered rings are very common in sugars. Five-membered rings are called "furanoses" and six-membered rings are called "pyranoses".
The most common way of drawing these rings are in "Haworth projections". Haworth projections don't reflect the real shape of the ring. For example, in a six-membered ring, the atoms in the ring adopt a zig-zag, up-and-down pattern in order to optimize bond angles. The chair drawing shows that relationship, but in a Haworth projection, the ring is drawn as though it were flat. Also, substituents on the atoms in the ring can be found above the ring, below the ring, or sticking out around the edge of the ring. The chair drawing or "diamond lattice projection" shows these relationships pretty well. A Haworth projection does not try to do that. Instead, it tries to depict stereochemical relationships: whether two substituents are on the same face of the ring or opposite faces of the ring. In a diamond lattice projection, we have to keep track of whether a given substituent is in the upper or lower position at its particular site on the ring, and that requires careful attention. In a Haworth projection, substituents are simply drawn straight up or straight down. Because there are many chiral canters in sugars, and becuase two sugars can differ by just one chiral center, Haworth projections make it easier to tell different sugars apart.
When an open-chain sugar cyclizes by forming a hemiacetal, it forms a new stereocenter. Because the carbonyl carbon is trigonal planar, the hydroxyl group can approach it from either face. There is nothing to distinguish one face from the other, and so approach from either face is equally likely. That means that two different stereochemical configurations can form at the hemiacetal carbon: R and S.
In sugar chemistry, these two isomers are named a different way: alpha and beta. To distinguish these two designations, you need to look at the Haworth projection. In a Haworth projection, the lower edge of the ring is read as being nearer to you. The upper edge is read as being farther away. Remember, that's how we usually read a chair structure, too. However, in a Haworth projection, we have to orient the ring in a specific way. The hemiacetal carbon is always placed at the right edge of the drawing. In addition, we always keep the oxygen atom on the back edge of the ring (i.e. the upper edge of the drawing). That means the ring oxygen in a Haworth projection is always found in the upper or upper right part of the drawing, with the hemiacetal carbon directly beside it to the right.
If we have the Haworth projection, we can designate whether we have the alpha or beta from by seeing whether the hydroxy part of the hemiacetal points up or down. If it is down like the ants, we have an alpha isomer. If it is up like the butterflies, we have a beta isomer.
Because the hemiacetal carbon can adopt either of two configurations in a ring, it is given a special name. It is called the anomeric position. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/III%3A_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_1/03%3A_Addition_to_Carbonyls/3.08%3A_Sugars-_Pyra.txt |
When a sugar cyclizes via donation of a hydroxy lone pair to the carbonyl, it forms a "hemiacetal". We have already seen that hemiacetals are unstable with respect to further substitution. Pi donation from an oxygen in the hemiacetal can displace the other oxygen. A second nucleophile can then donate to the pseudo-carbonyl that results.
The hemiacetal position in the sugar is called the anomeric center. The anomeric center is special for two reasons. First, as you have already seen, the anomeric center is a chiral center. This new center can form with either of two configurations. The sugar, which is already chiral, can become either of two diastereomers when it cyclizes. Second, the anomeric center is a site of enhanced reactivity in the sugar, in terms of substitution of the carbonyl.
Anomeric reactivity involves pi donation from one oxygen to push off the other oxygen. This mode of reaction should be familiar. The C=O+ unit that forms resembles a carbonyl. Furthermore, the positive charge on the oxygen brings to mind an activated carbonyl. This position is especially attractive for nucleophiles.
It isn't an accident that in many sugar-containing biomolecules, substituents are found at the anomeric center. For example, nucleosides sub-units found in DNA and RNA are all substituted at this position. A number of other biological agents contain this motif as well.
Exercise \(1\)
Substitution at the anomeric position can be accelerated if a proton source is available. Show why.
Answer
Exercise \(2\)
Show the stereochemical results of substitution at the anomeric center of glucose with methanol.
Answer
Exercise \(3\)
Nucleotides, which form DNA and RNA chains, are just like nucleosides, but they all have a phosphate at a specific position. Explain what is special about this position that could make it form a phosphate more easily than the other hydroxyl sites.
Answer
Add texts here. Do not delete this text first.
Exercise \(4\)
Show the products of the following substitution reactions.
Answer
Although sugars contain a number of chiral centers, characterizing them by polarimetry is complicated. Optical rotation measurements are done in solution, in a polarimetry cell. Polarimetry is always a little bit complicated, because the optical rotation varies with the concentration of the solution and the length of the polarimetry cell. When a pure enantiomer of a sugar such as alpha-D-glucose is dissolved, usually in water, its optical rotation also varies with time. In other words, the reading keeps changing, eventually settling out far from the initial value. That means care must be taken in measuring this information, and in interpreting the data.
Exercise \(5\)
Show why alpha-D-glucose would exhibit a changing optical rotation value after being dissolved.
Answer
Exercise \(6\)
Suppose a one gram sample of alpha-D-glucopyranose is dissolved in 1 mL of water and its optical rotation is measured in a a 1 dm cell. Initially, a value of 100 degrees is recorded. After several hours, the value has stopped changing, and is 48 degrees.
The experiment is repeated with beta-D-glucopyranose.
1. What can you predict about the initial value with beta-D-glucopyranose?
2. What can you predict after several hours?
Exercise \(7\)
In water, alpha-D-glucopyranose predominates over the beta form in solution. Explain why.
Answer
Exercise \(8\)
In less polar solvents (compared to water) such as dichloromethane, beta-D-glucopyranose predominates over the alpha form. This phenomenon is thought to result from the influence of lone pair-lone pair repulsion.
1. Show why this factor might favor one isomer over the other.
2. Show why this factor is less important in water.
Answer
3.10: Biological R
Addition to a carbonyl by a semi-anionic hydride, such as NaBH4, results in conversion of the carbonyl compound to an alcohol. The hydride from the BH4- anion acts as a nucleophile, adding H- to the carbonyl carbon. A proton source can then protonate the oxygen of the resulting alkoxide ion, forming an alcohol.
Formally, that process is referred to as a reduction. Reduction generally means a reaction in which electrons are added to a compound; the compound that gains electrons is said to be reduced. Because hydride can be thought of as a proton plus two electrons, we can think of conversion of a ketone or an aldehyde to an alcohol as a two-electron reduction. An aldehyde plus two electrons and two protons becomes an alcohol.
Aldehydes, ketones and alcohols are very common features in biological molecules. Converting between these compounds is a frequent event in many biological pathways. However, semi-anionic compounds like sodium borohydride don't exist in the cell. Instead, a number of biological hydride donors play a similar role.
NADH is a common biological reducing agent. NADH is an acronym for nicotinamide adenine dinucleotide hydride. Insetad of an anionic donor that provides a hydride to a carbonyl, NADH is actually a neutral donor. It supplies a hydride to the carbonyl under very specific circumstances. In doing so, it forms a cation, NAD+. However, NAD+ is stabilized by the fact that its nicotinamide ring is aromatic; it was not aromatic in NADH. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/III%3A_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_1/03%3A_Addition_to_Carbonyls/3.09%3A_The_Anomeric.txt |
Organic compounds are a broad class of compounds that got their name because they were originally found in living, or organic, matter. Physicians during the Renaissance and afterward strove to develop new medicines by extracting "active principles" from plants that were known to have medicinal properties. The same practice remains the most common method by which pharmaceutical companies develop medicines today.
Carbonyls are organic compounds that contain C=O bonds; that is, they contain double bonds between carbon and oxygen. (The word is pronounced car-bow-KNEEL.) This group of compounds is probably the most important single class of organic molecules. They can be divided into two classes. There are simple carbonyls, in which the carbon of the C=O bond is attached to other carbons, or possibly to hydrogens. There are also carboxylic acid derivatives, or carboxyloids, in which the carbony carbon is attached to a "heteroatom": an atom other than carbon or hydrogen, such as oxygen, nitrogen, sulfur, or a halogen.
Carbonyl compounds are very common in biological chemistry. Classes of compounds that contain the C=O bond include amides (found in proteins & peptides, used as signaling molecules and to help catalyze and guide reactions), aldehydes and ketones (found in carbohydrates, which play structural roles in cellulose, starch and DNA, for example) and esters (found in fats that form cell membranes, among other things). Understanding the reactivity of these bonds will help you to learn about many biological processes, as well as other transformations that are important in human society.
Common clusters of atoms, such as the H-C=O group in an aldehyde or the HO-C=O group in a carboxylic acid, are called "functional groups".
The amino acids are some of the fundamental building blocks of life. Fundamentally, they are carbonyl compounds.
Figure \(2\): An alphabet of amino acids illustrates how common carbonyls are in biological chemistry.
Exercise \(1\)
What functional groups contain carbonyls in the family of amino acids illustrated above?
Answer
The HO-C=O or CO2H group present in all of the amino acids is called a carboxylic acid.
An additional carboxylic acid is present in aspartic acid and glutamic acid.
The H2N-C=O or CONH2 group present in asparagine and glutamine is called an amide.
Amino acids are connected together to form peptides. Peptides are "polymers", which means they are very large molecules made up of small, repeating units.
Carbohydrates are another important class of biomolecules. They also contain carbonyl groups. Like amino acids, there are many variations of carbohydrates, and they are sometimes found bound together to form polymers. Of course, carbohydrates are important in energy storage. In plants, they also provide structural strength, helping to form cell walls.
Figure \(4\): A simple carbohydrate, glyceraldehyde.
Exercise \(2\)
Identify which group is contained in each of the following carbohydrates: an aldehyde or a ketone?
Answer
Answer a
aldehyde
Answer b
ketone
Answer c
ketone
Answer d
aldehyde
Fatty acids, triglycerides and phospholipids are a third class of biomolecules that contain carbonyls. These compounds are also used in energy storage. In addition, phopholipids form a major part of cell membranes, so these compounds also play an important structural role in biology.
Figure \(5\): A triglyceride.
Exercise \(3\)
Locate the carbonyls in the following biological molecules and the functional groups that contain carbonyls in each case.
Answer
Ginkgolides are biologically active terpenoids from Ginkgo trees. They are thought to have medicinal properties.
Okundoperoxide is isolated from a type of sedge in Cameroon. It has modest anti-malarial properties.
D-erythrose is a typical carbohydrate.
Exercise \(4\)
Frequently, the carbonyls in carbohydrates are "masked", as in deoxyribose (below).
Answer
By moving one proton from one position to another, and then breaking a single C-O bond, discover where the carbonyl is hiding. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/III%3A_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_1/03%3A_Addition_to_Carbonyls/3.11%3A_Carbonyls_in.txt |
You may recall that conversion of an aldehyde or ketone to an alcohol is referred to as a reduction. The hydride from an NADH molecule or a BH4- anion acts as a nucleophile, adding H- to the carbonyl carbon. A proton source can then protonate the oxygen of the resulting alkoxide ion, forming an alcohol.
In this reduction, two electrons and two protons are donated to the carbonyl compound to produce an alcohol.
The opposite process, the loss of two protons and two electrons from an alcohol to form a ketone or aldehyde, is an oxidation.
In biological pathways, oxidation is often the microscopic reverse of reduction. That means that the products of a reduction, NAD+ and an alcohol, could react together under the right circumstances to form NADH and a carbonyl. The reduction of NAD+ by a hydride donor is possible because, although the NAD+ loses the aromaticity of its nicotinamide ring upon becoming NADH, it also loses its positive charge. Charge stabilization is frequently an energetic problem for molecules.
This general type of reaction, hydride transfer reduction, has been adapted by Ryoji Noyori, of Nagoya University in Japan, to produce a single enantiomer of a chiral alcohol product. Noyori's work on this reaction, and others, led to him being awarded the Nobel Prize in Chemistry in 2001.
Exercise \(1\)
Show the two enantiomers that could be produced from reduction of acetophenone, CH3(CO)C6H5
Exercise \(2\)
Provide a mechanism with arrows for the Oppenauer oxidation of benzyl alcohol, C6H5CH2OH.
Exercise \(3\)
Explain why acetone is used as the solvent in an Oppenauer oxidation.
Answer
Add texts here. Do not delete this text first.
Exercise \(4\)
Meerwein-Ponndorf-Verley reduction of a ketone is carried out with aluminum tris(isopropoxide) in isopropanol as solvent. Provide a mechanism for the reduction of acetophenone, CH3(CO)C6H5, via this reaction.
Exercise \(5\)
Explain why isopropanol is used as the solvent in a Meerwein-Pondorf-Verley reduction
A second general method for alcohol oxidation employs a "redox-active" transition metal to accept a pair of electrons from from an alcohol during the oxidation. Because oxidation of an alcohol formally involves the loss of two electrons and two protons, a proton acceptor is also involved in this oxidation. There are many redox-active metals, but one of the most commonly used is Cr(VI). When Cr(VI) accepts a pair of electrons, it becomes Cr(IV).
In order to look at how chromium oxidation works, we'll use chromium oxide, CrO3, as an oxidant and water as a solvent. Note that water could also act as a proton acceptor or proton shuttle, moving protons from one place to another as needed. To carry out an oxidation, a number of events need to happen.
• The alcohol needs to bind to the chromium.
• A proton needs to be removed. This event is helped by the formal positive charge on the alcohol after it donates a lone pair to the chromium.
• A second proton must be removed and a pair of electrons given to the chromium for good.
In reality, CrO3 isn't used that often as an oxidant. It tends to catch fire when mixed with organic compounds. Instead, a variety of other chromium compounds are used.
Exercise \(6\)
In determining an oxidation state, we imagine giving both electrons in a bond to the more electronegative atom and looking at the resulting charges on the ions that result. Assuming all of the oxygens in chromium oxide can be thought of as dianions, confirm that the chromium can be thought of as a Cr6+ cation (in other words, in oxidation state Cr(VI)).
Answer
Exercise \(7\)
By the reasoning used in the previous question, determine the oxidation state of the transition metal in the following compounds. Note that in some cases, there is an anion and cation in the compound.
a) KMnO4 b) NaIO4 c) Ag2O d) OsO4 e) (CH3CH2CH2)4N RuO4
Answer
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Sometimes, nucleophiles adding to a carbonyl do not follow the normal reactivity patterns that have been common so far. This is often the case with the addition of ylides.
Ylides are compounds that are often depicted with a positive charge one one atom and a negative charge on the next atom. They are examples of zwitterions, compounds that contain both positive and negative charges within the same molecule. What distinguishes them from other zwitterions is the proximity of the opposite charges.
The classic example of an ylide addition to a carbonyl is the Wittig reaction. The Wittig reaction involves the addition of a phosphorus ylide to an aldehyde or ketone. Rather than producing an alcohol, the reaction produces and alkene. The reaction is driven by formation of a phosphorus oxide "side product".
This is a special case. The phosphorus-oxygen bond is strong enough to change the course of this reaction away from the normal pattern, and it isn't something you would have been able to predict based on related reactions.
An ylide is an example of a molecular compound that contains both a positive and a negative formal charge on two adjacent atoms. The charges are right beside each other: in this case, there is a positive charge on the phosphorus and a negative charge on the carbon.
Ylides are specific examples of zwitterions, which are molecules that contain positive and negative charges. The most common example of a zwitterion is probably an amino acid, which contains a positive ammonium ion and a negative carboxylate ion, within the same molecule.
Phosphorus ylides are made one charge at a time. A phosphonium ion must first be assembled, containing the positive charge on phosphorus. This event occurs via a nucleophilic substitution reaction, in which a phosphorus nucleophile displaces a halogen from an alkyl halide.
Exercise \(1\)
Show, with reaction arrows, formation of the three alkyltriphenyl phosphonium bromide salts shown below.
Answer
In most cases, the source of the phosphorus is triphenylphosphine. Triphenylphosphine is used for several practical reasons. First of all, it is a solid, so it is easy to weigh out the right amount of it and add it to a reaction. Secondly, organophosphorus compounds are often very toxic and smelly, but triphenylphosphine is less offensive. Thirdly, in the Wittig reaction, the original phosphorus compound is eventually discarded as waste, and the more useful alkene is kept. Since the phosphorus part does not matter that much, the most convenient possible phosphine is generally used. However, there are other variations of this reaction that use other phosphorus compounds.
Once the phosphonium salt has been made, the phosphorus ylide can then be obtained via deprotonation of a phosphonium ion. The hydrogens on a carbon next to a phosphorus cation are a little bit acidic because of the positive charge on the phosphorus. One of these hydrogens is easily removed via adddition of a very strong base such as sodium hydride.
Exercise \(2\)
Show, with reaction arrows, formation of ylides from the three alkyltriphenyl phosphonium bromide salts shown above in Exercise \(1\)
Answer
The reaction of a phosphorus ylide with a carbonyl compound does begin like other nucleophilic additions. The ylide donates its nucleophilic lone pair to the carbonyl and the carbonyl pi bond breaks. However, the strong P-O bond then takes over the reaction. To begin, a lone pair on the resulting alkoxide ion is donated to the positively charged phosphonium ion.
Wait! That violates one of our mechanistic rules. Usually, we don't have an atom donate to a positively charged atom that already has an octet; if we do so, the atom will have too many electrons. However, the octet rule does not strictly apply to sulfur and phosphorus. These atoms are larger than second-row atoms like nitrogen and oxygen, and they are often observed to "exceed the octet rule". Sulfur and phosphorus are frequently observed with trigonal bipyramidal or octahedral molecular geometries, meaning they may have up to 12 electrons in their valence shells.
So go ahead! Donate a pair of electrons to the phosphorus. It can't help itself, because of the strength of the P-O bond that forms.
Exercise \(3\)
Show the products of the reactions of each of the ylides you made in Exercise \(2\) (Problem CO18.2.) with the following electrophiles:
a) butanal b) benzaldehyde c) 4-methylpentanal
Answer
This is when things really get interesting. It turns out that one P-O bond just isn't enough. The phosphorus is so oxophilic that it takes the oxygen atom all to itself, pulling it right out of the molecule. It probably does not hurt that the four-membered ring is pretty strained, so it is motivated to decompose (but be careful: there are plenty of stable four- and even three-membered rings in nature).
The arrows shown in the decomposition of the four-membered ring (called a betaine) are just meant to keep track of electrons; there isn't a true nucleophile and electrophile in this step. Instead this step may resemble a pericyclic reaction, which is covered in another section.
Exactly how to draw the P=O bond is debatable. There isn't much doubt that it is a double bond; it is stronger and shorter than a P-O single bond. However, quantum mechanical calculations indicate that the phosphorus can't form a pi bond. This double bond is different than other double bonds you have seen. For that reason, some people prefer to draw this compound as an ylide, too, with a positive charge on the phosphorus, a single bond, and a negative charge on the oxygen.
The phosphorus oxide compound forms, leaving behind an alkene. Alkenes are very common in nature, and this reaction has frequently been used to make interesting alkene-containing compounds for further use or study.
Exercise \(4\)
The juvenile hormone of the cecropia moth caterpillar (JH-1, below) is a regulatory hormone used to control the organism's development by preventing it from pupating until conditions are right.
Synthesis of insect hormones is often undertaken in order to control insect populations. The following synthesis of JH-1 was developed by Barry Trost (Stanford) in the 1960's. Fill in the missing reagents and reaction products.
Answer
One of the keys in this problem is recognizing that in some steps, two different reactions are involved. For example, in the first box, there is an addition of a diol to a carbonyl followed by an ylide addition.
Sulfur ylides are also good nucleophiles for aldehydes and ketones. However, the unusual stability of the phosphorus-oxygen bond does not have a similar analogue in sulfur chemistry.
Sulfur ylides are formed in a manner very similar to phosphorus ylides.
Exercise \(5\)
Show, with arrows, the mechanism for formation of the sulfur ylide above.
Answer
Once formed, sulfur ylides react with aldehydes or ketones. Like phosphorus ylides, the reaction starts out just like any other nucleophile, but a second step takes a very different direction. Epoxides are formed in these reactions, and the original sulfur compound (a thioether) is regenerated.
Exercise \(6\)
Show, with arrows, the mechanism for the epoxide-forming reaction above.
Answer
Exercise \(7\)
Fill in the product or reagent for each of the following transformations. Remember there is always an acidic workup assumed.
Answer
Exercise \(8\)
Identify the starting material as nucleophile or electrophile in the following reactions (from the synthesis of an epothilone analogue by K.H. Altmann at ETH). In the product, box the part of the structure that came from the compound on the left; circle the new part.
Answer
Exercise \(9\)
Fill in the blanks in the following synthesis. Requires knowledge of aldol addition, Grignard additions, and Wittig / Horner-Wadsworth-Emmons reactions.
Answer
Exercise \(10\)
Altmann's epothilone analogue synthesized above (shown on the right, below) clearly bears some resemblance to natural epothilone A (below, left). However, the analogue is effective at much lower concentrations than the natural product. That information may actually reveal something about how the natural product interacts with its target.
1. Parts of the epothilone A are highlighted. Circle and box the corresponding parts of the analog.
2. Describe the differences between the highlighted portions of the natural and synthetic versions.
3. One could imagine the boxed alcohol group forming part of a pharmacophore -- the part of the compound that binds with its target. What intermolecular attractions seem likely with this group?
4. Comment on the apparent importance of this group in binding to the target, based on the evidence.
5. One could imagine the circled part of the natural compound adopting different conformations via changes in the dihedral angle. What dihedral angle appears to be shown in the natural epothilone?
6. What dihedral angle appears to be shown in the analogue?
7. Based on the evidence, which dihedral angle is preferred in order to bind to the target?
Answer
Answer a
Answer b
The circled part changed from a single bond to a double bond. The boxed part changed from an alcohol to unadorned hydrocarbon chain.
Answer c
Hydrogen bonding is the most obvious.
Answer d
However, the analog works better without this group; this particular alcohol group is probably not an important part of the pharmacophore. It is probably not needed in order to bind to the target.
Answer e
In epothilone A, as drawn, the dihedral angle appears to be 0 degrees.
Answer f
In the new analog, the dihedral angle is 180 degrees.
Answer g
Based on the superior activity of the analog, the active conformation of the ring is probably more like the one on the right than the one on the left. The circled bond probably adopts a dihedral angle closer to 180 degrees, with the rest of the ring twisting into a shape more like the one shown on the right, in order to bind to the target. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/III%3A_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_1/03%3A_Addition_to_Carbonyls/3.13%3A_Addition_of_.txt |
Conjugated systems are structures that contain alternating double and single bonds (or, in some cases, a double bond that is next to an atom with either a lone pair or a vacant orbital). Conjugated systems are usually at lower energy than regular double bonds because the electrons involved in bonding are delocalized; they are spread out over a greater area and thus can have a longer wavelength.
For example, the -bonding system for 3-butene-2-one (or methyl vinyl ketone) is described by orbitals involving both the carbonyl group and the alkene group. These two groups become linked together so that there is not longer an independent carbonyl nor an independent alkene, but one "enone" (a term taken from the words alkene and ketone).
Because of that extra stability, it might not be surprising that conjugated carbonyls are often a little slower to react than regular carbonyls. The surprise is that conjugated carbonyls can sometimes give additional products in which addition does not take place at the carbonyl.
The product shown above is called a conjugate addition product, or a 1,4-addition product. In conjugate addition, the nucleophile does not donate to the carbonyl, but instead donates to an atom that is involved in conjugation with the carbonyl. This additional electrophilic position is sometimes called a "vinylogous" position (from the word vinyl, which refers to that CH=CH2 unit next to the carbonyl).
• Conjugate additions (or 1,4-additions) can occur when a carbonyl is attached to a C=C bond.
Exercise $1$
Draw a mechanism with curved arrows for the conjugate addition shown above.
Answer
Exercise $2$
Regular additions to carbonyls are sometimes called 1,2-additions, whereas conjugate additions are called 1,4-additions. Show why.
Answer
Remember that we can look at another resonance structure of a carbonyl, one that emphasizes the electron-poverty of the carbonyl carbon. It's not a good Lewis structure because of the lack of an octet on carbon, but it does reinforce the idea that there is at least some positive charge at that carbon because it is less electronegative than oxygen. Extending that idea, we can draw an additional resonance structure in a conjugated system. That third structure suggests there may be some positive charge two carbons away from the carbonyl, on the β position on the double bond.
The idea that there are two electrophilic positions in an enone is reinforced by the picture of the LUMO (the lowest-energy "empty" frontier orbital, the virtual place where an additional electron would probably go). When a lone pair is donated to an electrophile, the electrons are most likely to be donated into the LUMO.
Although it isn't obvious from the cartoons we often draw for molecular orbitals, quantum mechanical calculations suggest that the LUMO is "larger" at the carbonyl position as well as the β-position on the vinyl group.
How can it be larger on some atoms than others? A molecular orbital is an algebraic combination of atomic orbitals. In this case,
$LUMO = ap_{C1} + bp_{C2} + cp_{C3} + d_{pO} \nonumber$
in which pC1 is the p orbital on the carbon on the left, pO is the p orbital on the oxygen, and so on. The letters a, b, c and d are just numbers; they are the coefficients in the equation. The result of the molecular orbital calculation in this case suggests that the numbers a and c are a little bigger than b and d. Incidentally, it also suggests that a and d have opposite sign from b and c (maybe a and d are positive numbers whereas b and c are negative numbers), meaning that a and d are out of phase with b and c.
In any case, we sometimes think of the large LUMO on particular atoms as being an easier "target", an easier place to throw the incoming electrons. These mathematical results really just reflect what we would expect from the resonance structures.
Exercise $3$
Indicate whether the following systems are capable of undergoing conjugate addition, and show why or why not.
Answer
Having two possible products of a reaction can be confusing. How do you know which one will result? Often, you don't know. Frequently, both products result, so there is a mixture of compounds. However, one product often predominates. In conjugate addition, there are a few different factors that may tilt the reaction in one direction or another.
Possibly the simplest reason is steric effects. Maybe one of the electrophilic positions is more crowded than the other, and the nucleophile can access that position more easily.
• Addition of a nucleophile often occurs at the least crowed electrophile.
•
Exercise $4$
In each of the following cases, indicate whether the addition of a nucleophile will be via 1,2-addition, via 1,4-addition, or an equal mixture.
Answer
The hard/soft acid characteristics of the two electrophilic positions also influence the reaction. The carbonyl position is closer to the oxygen, of course, and it makes sense that the oxygen would have a greater influence on this carbon. The carbonyl position, with its more concentrated positive charge, is a harder electrophile. The vinylogous position, with less positive charge, is a softer electrophile.
• Carbonyls are hard electrophiles.
• Vinylogous positions are soft electrophiles.
Soft nucleophiles are more likely to react with soft electrophiles, and hard nucleophiles are more likely to react with hard electrophiles. The amount of negative charge concentrated at the nucleophilic atom is the biggest factor determining hardness. The lower the charge, or the more spread out the charge, the softer the nucleophile.
Exercise $5$
Indicate whether the following nucleophiles are more likely to undergo 1,2-addition or 1,4-addition.
There are other factors that play roles in influencing the course of these reactions. Sometimes the mechanism of reaction is slightly different under different circumstances. The presence of Lewis acid catalysts can also influence reactivity in these systems, but not always in a predictable way.
Lewis acid catalysts can influence whether a reaction proceeds via 1,4-addition.
Exercise $6$
Fill in the products of the following reactions.
Answer
The Michael addition is one of the most important examples of conjugate addition. In a Michael addition, an enolate nucleophile undergoes 1.4-addition to an enone. Typically, the nucleophile is diactivated; that is, there is a carbonyl on either side of the alpha position.
Exercise $7$
Provide the product of the following Michael additions.
Answer
Robinson Annulation
Robinson annnulation allows for formation of a six-membered ring via a Michael addition and subsequent aldol condensation.
The first reaction, a Michael addition, links two molecules together via a 1,4-addition of an enolate to an enone.
After that, a regular aldol reaction ensues, closing the ring.
Typically, the aldol reaction continues through a condensation. That's the variation of the aldol reaction in which a water molecule is lost, forming a double bond in conjugation with the carbonyl.
Exercise $8$
Provide mechanism for the following steps in the Robinson annulation above:
1. the Michael addition
2. the aldol reaction to form the six membered ring
3. the dehydration of the aldol product to form the enone product
Answer
Answer a
Answer b
Answer c
Exercise $9$
Provide products of the following Robinson annulations.
Answer
Sometimes, it is useful to think "backwards" about a reaction. Given a particular structure, can we picture the reaction that may have taken place to form that structure? This type of analysis is very useful to bio-organic chemists, who often seek to find out how different compounds formed in nature. By imagining different ways in which a natural product may have formed, they can design different experiments that may shed light on how the process really happened.
Of course, synthetic chemists find this way of thinking about things is very helpful, too. Given the task of making a particular compound, they must imagine the most efficient ways in which the compound could be made.
Let's take a look at how that approach to thinking about reactions would work with a Robinson Annulation. There are two reactions in a Robinson Annulation. One is a Michael addition and another is an aldol condensation. They always happen in that order. To analyse the product, we need to work backward from the product and “disconnect” the molecule. That means we need to find where the aldol reaction happened, then where the Michael reaction happened.
A diagnostic aldol fragment can look either like R-CO-CH2-CHOH-R (carbonyl-carbon-alcohol) or, if there is a condensation/dehydration step, RCO-CH=CHR (carbonyl-alkene). That’s the first part of the molecule you need to find. It forms when the enolate fragment RCO-CH2- adds to the carbonyl RCHO. We must break the product at the double bond to uncover the aldol reactant.
A Michael fragment looks like R-CO-CH2-CH2-CH2-CO-R. There are two carbonyls, with three carbons in between them. That’s the part of the molecule we need to find next, but we won’t see it until we have stepped back to the situation we had before the aldol step. The Michael fragment forms from the electrophile RCO-CH=CH2 and the enolate nucleophile -CH2–CO-R.
Overall, the Robinson annulation is the sum of these two steps.
Exercise $10$
Provide the starting materials needed to make the following compounds via Robinson annulation.
Answer | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/III%3A_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_1/03%3A_Addition_to_Carbonyls/3.14%3A_Conjugate_Ad.txt |
In conjugate addition, a carbonyl group turns a neighbouring alkene into an electrophilic site. An enone, such as the one below, has two electrophilic positions.
A similar situation happens when pi-acceptors such as nitro groups are attached to aromatic rings.
The key step in the mechanism is the loss of the halide ion, which allows the aromaticity to be restored.
Exercise \(1\)
The location of the halogen and the electron-withdrawing group matters. Explain why the reaction occurs if the groups are in the 1 and 2 positions (ortho to each other) or the 1 and 4 positions (para to each other), but not if they are in the 1 and 3 positions (meta to each other).
Exercise \(2\)
Explain why the reaction is faster if additional electron-withdrawing groups are present.
Answer
Add texts here. Do not delete this text first.
3.16: Carbonyl Add
As you are probably aware, most of our medicines ultimately come from natural sources. They are derived from natural products, although they have often been modified in some way in order to limit side effects or maximise their effectiveness. You probably also know that the contents of a pill you get from the pharmacy does not come directly from nature. More likely, it is a synthetic product. Pharmaceutical companies make these compounds from smaller, more readily available starting materials.
The total synthesis of natural products provides an interesting context for reactions of organic molecules. We can see where familar reactions have been used to make something important. Whether or not the natural product is ever developed as a commercial drug is beside the point. Often, the exercise of making one compound provides answers that allow researchers to find a better way to make another compound.
Looking at reactions within an actual synthesis also forces us to develop some spatial reasoning skills. We need to be able to recognise familiar reactions in unfamiliar contexts.
For example, in the following sequence of steps, a set of reagents has been left out. How would we get from one compound to the other?
The first thing we need to do is block out the infromation we don't need to worry about right now. We'll circle the one reaction that is missing information, and focus only on that one.
The compound on the right still looks pretty similar to the one on the right. We might try circling the part of the molecule where we can see a difference. That way, we can focus on what reagents would make that difference. What kind of reaction is happening here?
It looks like an aldol reaction. We'll put some typical reagents in the box. First, we'll add a base, to remove a proton from the alpha-position (the acidic hydrogen next to the carbonyl, in the blue ellipse on the left). Second, we'll add the electrophile that would lead to the product on the right.
Those might not be exactly the same conditions used by the researchers in this synthesis, but it is a reasonable idea of what needs to be added.
In the following example, an intermediate product has been left out. It is the product of the first reaction, but the starting material for the second reaction.
Once again, we might do an exercise to focus on the important part of the question. What is the electrophile? What is the nucleophile?
On the left, there is an aldehyde. It's a carbonyl compound, so maybe it is an electrophile. We'll circle it in blue. Above the arrow, there is a phosphorus compound. It does not have a positive charge as shown, but there is a very positive phosphorus atom because of the attached oxygens. There is also a base present, although it's a relatively weak one. Maybe we will remove a proton next to the phosphorus to make an ylide. Ylides are good nucleophiles.
If we are correct, the ylide would combine with the aldehyde to make a C=C double bond. The phosphorus would remove the oxygen atom from the aldehyde. The resulting compound is shown in the box.
Sometimes, our guesses can be confirmed by looking further ahead in the synthesis to see evidence for the compound we think has formed. In this case, the product of the second reaction may be complicated enough to make that process difficult (even if we already know something about alkene oxidations).
Exercise \(1\)
Practice the methods described above to fill in the missing reagents.
Answer
Exercise \(2\)
Practice the methods described above to fill in the missing products.
Answer
Because chemists are generally working toward a specific synthetic target, they frequently need to work backwards to plan out a series of reactions that will produce the desired compound. A good memory of specific patterns that occur in reaction products can help to identify what reaction is needed to get there. For example, if the product contains a cyanohydrin, a CN attached to the same carbon as an OH, then it may be possible to construct that part of the compound using addition of cyanide to a carbonyl.
This type of thinking is called "retrosynthetic analysis". The open arrow shown above is called a retrosynthetic arrow, and it roughly translates as "can be made from". That is, the cyanohydrin on the left can be made from the carbonyl compound on the right. By working backwards through a series of such reactions, we could eventually arrive at simple starting materials that could be used to make a very complicated molecule. This approach to thinking about how to make organic molecules is generally attributed to E. J. Corey of Harvard University. Corey was awarded the Nobel Prize in chemistry for his transformative work in organic synthesis.
Other retrosynthetic patterns may be familiar to you at this point. For example, an alcohol could always be made from reduction of a carbonyl.
On the other hand, it may be more convenient to make the alcohol via addition of a Grignard reagent (or something similar) to a carbonyl. That approach may be especially appealing if the alcohol carbon is attached to a common hydrocarbon fragment such as a vinyl, a phenyl or an alkyl group.
So, there are a number of ways in which an alcohol could be incorporated into a molecule. Other functional groups are also commonly targeted in this approach. For example, alkenes are very frequently added via Wittig or Horner-Wadsworth-Emmonds reactions, which make use of phosphorus ylides.
Epoxides are often added via the Corey-Chaykovsky epoxidation, just one of many reactions named after E. J. Corey. This reaction uses a sulfur ylide.
Sometimes a wider collection of atoms indicates a particular route to the target. For example, the pattern carbonyl-carbon-alcohol -- or C=O-C-COH -- would result from an aldol reaction.
An enone -- that is, a C=O-C=C pair in conjugation -- also suggests origins in an aldol reaction, although the aldol in this case resulted in a condensation or dehydration.
These relationships between products and the starting materials that they can be made from are useful in planning out syntheses. The ones we have looked at here involve only the addition to carbonyls of anionic and semianionic nucleophiles, as well as ylides. There are plenty of other such relationships in other areas of organic reactivity.
Exercise \(3\)
Perform retrosynthetic analysis in each of the following cases, proposing something that the compound could be made from.
Answer
Exercise \(4\)
Practice the methods described above to fill in the missing reagents and products in the following synthesis. Requires knowledge of anionic and semi-anionic nucleophiles, aldols and ylides. A list of possible reagents is provided below the roadmap.
Answer
Exercise \(5\)
Practice the methods described above to fill in the missing reagents and products in the following synthesis. Requires knowledge of anionic and semi-anionic nucleophiles, aldols and ylides. A list of possible reagents is provided below the roadmap.
Use the reagents below to finish the tulearin A roadmap.
Answer
Exercise \(6\)
Practice the methods described above to fill in the missing reagents and products in the following synthesis. Requires knowledge of anionic and semi-anionic nucleophiles, aldols and ylides.
Answer | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/III%3A_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_1/03%3A_Addition_to_Carbonyls/3.15%3A_Conjugate_Ad.txt |
The reactions of carbonyls can become very complicated, involving many steps. In essence, though, the steps involve only a few, different elementary reactions.
• Proton transfer.
• Protons are most often transferred from a positively charged atom to a neutral atom with a lone pair.
• Protons are also easily transferred from a positively charged atom to a negatively charged atom.
• Sometimes, a proton might be transfered from a neutral atom to a negatively charged one.
In many of the reactions of anionic and semi-anionic nucleophiles, these two steps complete the entire reaction mechanism. However, if an additional lone pair can be revealed at the nucleophilic atom (often by transferring a proron away from this site), additional steps occur.
• Pi donation.
• In pi donation, two heteroatoms, both with lone pairs, are attached to the same carbon. A lone pair is donated to the carbon, and one of the heteroatoms is pushed off.
Many mechanisms involve a number of proton transfers and pi donations. These steps occur over and over, inching the molecule along step by step towards the product. Usually, each proton transfer helps to prepare an atom for eventual removal via pi donation.
Occasionally, if the nucleophile is neutral, these steps are preceded by an initial activation step.
• Carbonyl activation.
• Usually makes the reaction faster.
• Is especially helpful when the nucleophile is uncharged, and hence less reactive.
• The carbonyl is often activated by a proton (from a protic acid) but it can also be activated by a Lewis acid (such as a metal ion).
3.18: Additional P
Exercise \(1\)
Synthesize the following compounds starting from acetone.
Answer
Exercise \(2\)
Researchers are investigating cyclohexenone derivatives as potential inhibitors for esterases. Below is a scheme for the synthesis of several of these derivatives. Fill in the boxes with the appropriate intermediates or reagents.
Below is another derivative that they made. Provide a mechanism (make sure to draw arrows) for the following reaction:
Answer
Exercise \(3\)
Fill in the product or reagent for each of the following transformations. Remember there is always an acidic workup assumed.
Answer
Exercise \(4\)
Fill in the blanks in the following synthesis. Includes addition to carbonyls (anionic nucleophiles).
Answer
Exercise \(5\)
Fill in the blanks in the following synthesis. Includes addition of nucleophiles to carbonyls (anionic nucleophiles, enolates, ylides).
Answer
Exercise \(6\)
Fill in the blanks in the following synthesis. Includes addition of nucleophiles to carbonyls (anionic nucleophiles, enolates).
Answer
Exercise \(7\)
Fill in the blanks in the following synthesis. Includes addition of nucleophiles to carbonyls (anionic nucleophiles, enolates, ylides).
Answer
Exercise \(8\)
Fill in the blanks in the following synthesis. Includes addition of nucleophiles to carbonyls (anionic nucleophiles, enolates, conjugate additions, carboxyloid substitutions).
Answer
Exercise \(9\)
Fill in the blanks in the following synthesis. Includes addition of nucleophiles to carbonyls (anionic nucleophiles, enolates, conjugate addition).
Answer
3.19: Solutions to
Problem CO1.1.
The HO-C=O or CO2H group present in all of the amino acids is called a carboxylic acid.
An additional carboxylic acid is present in aspartic acid and glutamic acid.
The H2N-C=O or CONH2 group present in asparagine and glutamine is called an amide.
Problem CO1.2.
a) aldehyde b) ketone c) ketone d) aldehyde
Problem CO1.3.
a) The double bond means two pairs of electrons are shared between the carbon and oxygen, instead of just one. As a result, the oxygen is able to pull more electron density away from the carbon. The carbon becomes much more positive in this case than in the case of a double bond.
Not only that, but the second bond between the carbon and the oxygen is a pi bond. Those electrons are farther from the nucleus than a sigma bond, in which the electrons are tightly held between the atoms. That means the pi electrons are more easily drawn toward the oxygen, so the bond becomes even more polarized.
b) A C=N bond would be very similar to a C=O bond, because nitrogen is the third most electronegative element after oxygen and fluorine (ignoring noble gases).
a) 2-hexanone or hexan-2-one b) 5-methylhexan-2-one c) 3-methylhexan-2-one
d) hexan-3-one e) heptan-4-one f) 4-methylhexan-3-one
g) hexanal h) 3-methylbutanal i) 2-methylbutanal
j) 2-ethylpentanal k) 4,4-dimethylpentanal l) 2,3-dimethylpentanal
m) 3-ethoxypropanal n) 4-chloropentan-2-one o) 4-aminohexan-3-one
p) hex-4-en-2-one (in the solution above, (E)-hex-4-en-2-one is shown first and (Z)-hex-4-en-2-one is shown second).
q) hept-4-yn-2-one r) benzaldehyde (this is a common name adopted for formal naming. Benz means a carbon attached to a benzene ring.)
The LUMO in this case is the π*, an antibonding level. If electrons populate this level, the π bond will break.
Both the imine (b) and nitrile (c) have a low-lying pi antibonding level (π*), similar to a carbonyl.
1. On the basis of steric crowding, the first one is most reactive, then the last one, then the middle.
2. On the basis of steric crowding, the last one is most reactive, then the first one, and then the middle.
Although there is a large group on the nitrogen in that last compound, the site of reactivity is the carbon, which is less crowded.
c) On the basis of electronics, the middle one is most reactive, then the last, and then the first. The fluorine atom is very electronegative and pulls electron density towards itself. That leaves more positive charge on the nearby carbonyl carbon. The more fluorines on that nearby carbon, the more positive the carbon. The more positive the carbon, the more it attracts electrons from a nucleophile.
a) propanal b) butanal c) propanal, again!
d) pentanal e) hexanal f) heptanal
a) 3-pentanone b) 3-hexanone c) 4-heptanone
d) 2-butanone e) 3-octanone f) 5-decanone
a) 2-methyl-3-pentanone or 2-methylpentan-3-one b) 4-ethyl-3-hexanone or 4-ethylhexan-3-one
c) 3,3-dimethyl-2-butanone or 3,3-dimethylbutan-2-one d) 2,5,5-trimethyl-4-heptanone or 2,5,5-trimethylheptan-4-one
e) 6-ethyl-4-methyl-3-octanone or 6-ethyl-4-methyloctan-3-one f) 6-ethyl-4,5-dimethyl-3-octanone or 6-ethyl-4,5-dimethyloctan-3-one
Ammonium chloride has an N-H bond, normally less polar than the O-H bond of water. However, the positive charge makes this compound give up a proton more easily, because it results in a neutral (uncharged) ammonia molecule.
Sodium carbonate has a polar O-H bond, just like water. However, the anion that results from loss of a proton is resonance-stabilised. That makes this compound more acidic than water.
a) In acetylide, the lone pair is on a linear carbon or sp carbon. In methyl anion, the lone pair is on a tetrahedral carbon or sp3 carbon. The description "sp" indicates that sigma bonding to neighbors involves a 2s orbital and a 2p orbital on carbon; there is a 50% contribution from the s orbital.
The description "sp3", on the other hand, indicates that sigma bonding to neighbors involves a 2s orbital and three 2p orbitals; there is a 25% contribution from the s orbital.
The 2s orbital is lower in energy than the 2p orbital. The greater the s orbital contribution to the bond (or in this case to the lone pair), the lower it is in energy. Thus, a lone pair on an sp carbon is lower in enrgy than a lone pair on an sp3 carbon.
b) In cyanide, the same argument outlined in pary (a) hold true. In addition, the nearby electronegative nitrogen stabilizes the charge by drawing electron density toward itself.
1. CH3OK, because of the ionic O-K bond. This is an anionic nucleophile. It is more reactive and nucleophilic than the corresponding neutral nucleophile.
2. CH3NH2, because nitrogen is less electronegative than oxygen. Its lone pair is held less tightly and is more easily donated to the electrophile.
3. NaCCH, because the neighbouring carbon in this case does not have the inductive electron-withdrawing effect that the nitrogen does in the case of NaCN. In that case, the lone pair is stabilized and made less reactive.
4. c-C6H11ONa, because the negative charge is localized on one atom. In c-C6H5ONa, the negative charge is delocalized over four different positions in the molecule. Delocalization of charge stabilizes the anion and makes it less reactive.
The resulting anion is stabilised by resonance.
The case with more steric crowding is more likely to result in deprotonation.
Problem CO11.5.
a) pentanol b) 2-butanol c) 4-octanol
d) 2-methylpropan-2-ol or 2-methyl-2-pronanol e) 3-methylhexan-2-ol or 3-methyl-3-hexanol
f) 5,6-dimethylheptan-1-ol or 5,6-dimethyl-1-heptanol
Some of these compounds do not have alpha-protons, so they cannot form enolate ions.
Problem CO12.11. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/III%3A_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_1/03%3A_Addition_to_Carbonyls/3.17%3A_Summary_of_E.txt |
Part a.
Part b.
One of the keys in this problem is recognizing that in some steps, two different reactions are involved. For example, in the first box, there is an addition of a diol to a carbonyl followed by an ylide addition.
a)
b) The circled part changed from a single bond to a doube bond. The boxed part changed from an alcohol to unadorned hydrocarbon chain.
c) Hydrogen bonding is the most obvious.
d) However, the analog works better without this group; this particular alcohol group is probably not an important part of the pharmacophore. It is probably not needed in order to bind to the target.
e) In epothilone A, as drawn, the dihedral angle appears to be 0 degrees.
f) In the new analog, the dihedral angle is 180 degrees.
g) Based on the superior activity of the analog, the active conformation of the ring is probably more like the one on the right than the one on the left. The circled bond probably adopts a dihedral angle closer to 180 degrees, with the rest of the ring twisting into a shape more like the one shown on the right, in order to bind to the target.
a)
Problem CO26.8.
3.21: Carbonyls ar
The carbonyl bond is very polar. There is a partial positive charge on the carbon and a partial negative charge on the oxygen, because oxygen is more electronegative than carbon. This charge separation is intensified because of the double bond between the carbon and oxygen. Rather than just pulling one pair of bonding electrons towards itself, the oxygen pulls two pairs of electrons towards itself.
• The C=O bond is very polar.
• The carbonyl carbon is very positive.
Sometimes, a resonance structure is drawn to emphasize the charge separation in the carbonyl. The structure has only one bond between the carbon and oxygen. In this structure, oxygen has an octet but carbon does not. This is not really a good Lewis structure, because the other resonance structure satisfies octets on all the atoms. However, this Lewis structure emphasizes the polarity of the bond and is sometimes drawn to reinforce that idea.
Because of the positive charge on the carbonyl carbon, the most important theme in carbonyl chemistry is reaction of the carbonyl as a Lewis acid. Reactions of carbonyls almost always involve addition of an electron donor to the carbonyl carbon.
• Electrophile is another term for Lewis acid.
• Lewis acids attract electrons.
• Lewis acids have a positive charge on an atom, a partial positive charge on an atom, or an atom lacking an octet.
• Carbonyl compounds are good electrophiles.
The electrophilicity of carbonyls is very important in their reactivity. The goal of this chapter is to develop an understanding of how carbonyls react. We will learn about a few key factors that will be used in different combinations under different circumstances. Eventually, you will build an understanding that will allow you to follow both biological reactions and modern synthetic reactions.
It is important to realize that biological reactions, such as carbohydrate synthesis, are very complex and can involve many, many steps. For example, the carbohydrate synthesis shown above involves additional acid-base steps as well as a reaction of a carbonyl. The additional acid base steps may involve proton donors and acceptors as well as more general Lewis acids.
Exercise \(1\)
Problem CO2.1.
a) Explain why the carbon in a C=O unit is very electrophilic, but the carbon in a C-O unit is much less so.
b) Propose other carbon-heteroatom bonds that may make the carbon electrophilic (heteroatom means not carbon or hydrogen).
Answer a
The double bond means two pairs of electrons are shared between the carbon and oxygen, instead of just one. As a result, the oxygen is able to pull more electron density away from the carbon. The carbon becomes much more positive in this case than in the case of a double bond.
Not only that, but the second bond between the carbon and the oxygen is a pi bond. Those electrons are farther from the nucleus than a sigma bond, in which the electrons are tightly held between the atoms. That means the pi electrons are more easily drawn toward the oxygen, so the bond becomes even more polarized.
Answer b
A C=N bond would be very similar to a C=O bond, because nitrogen is the third most electronegative element after oxygen and fluorine (ignoring noble gases).
Exercise \(2\)
Problem CO2.2.
Provide line structures for the following compounds.
Exercise \(3\)
Problem CO2.3.
Translate the following condensed formulae into Lewis-Kekule structures (i.e. like Lewis structures, but use lines for bonds). All of the compounds are adehydes or ketones.
a) CH3CH2CH2CH2COCH3 b) ((CH3)2CH2CH2COCH3 c) CH3CH2CH2CH(CH3)COCH3
d) CH3CH2CH2COCH2CH3 e) CH3CH2CH2COCH2CH2CH3 f) CH3CH2CC(CH3)COCH2CH3
g) CH3CH2CH2CH2CHO h) ((CH3)2CH2CHO i) CH3CH2CH(CH3)CHO
j) CH3CH2CH2CH(CH2CH3)CHO k) ((CH3)3CH2CH2CHO l) CH3CH2CH(CH3)CH(CH3)CHO
m) CH3CH2OCH2CH2CHO n) CH3CHClCH2COCH3 o) CH3CH2CH(NH2)COCH2CH3
p) CH3CHCHCH2COCH3 q) CH3CH2CCCH2COCH3 r) C6H5CHO
Answer
Exercise \(4\)
Problem CO2.4.
Translate the condensed formulae in the previous problem into line structures.
Answer
Exercise \(5\)
Provide IUPAC names for the compounds in the previous problem. For help, see the functional group section, simple carbonyls.
Answer a
2-hexanone or hexan-2-one
Answer b
5-methylhexan-2-one
Answer c
3-methylhexan-2-one
Answer d
hexan-3-one
Answer e
heptan-4-one
Answer f
4-methylhexan-3-one
Answer g
hexanal
Answer h
3-methylbutanal
Answer i
2-methylbutanal
Answer j
2-ethylpentanal
Answer k
4,4-dimethylpentanal
Answer l
2,3-dimethylpentanal
Answer m
3-ethoxypropanal
Answer n
4-chloropentan-2-one
Answer o
4-aminohexan-3-one
Answer p
hex-4-en-2-one (in the solution above, (E)-hex-4-en-2-one is shown first and (Z)-hex-4-en-2-one is shown second).
Answer q
hept-4-yn-2-one
Answer r
benzaldehyde (this is a common name adopted for formal naming. Benz means a carbon attached to a benzene ring.)
Exercise \(6\)
Fill in any missing lone pairs in the following structures. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/III%3A_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_1/03%3A_Addition_to_Carbonyls/3.20%3A_Solutions_to.txt |
In the following pictures, a number of anions are added to a simple carbonyl compound, a ketone (2-propanone, or acetone). In each case, addition of the nucleophile is followed by addition of a proton source. Note that, overall, the reaction involves addition of the nucleophile to the carbonyl carbon and addition of the proton to the carbonyl oxygen.
• Addition of anionic nucleophiles to ketones or aldehydes transforms the carbonyl into an alcohol.
Look at the way the reaction is presented in each case. The organic (carbon-based) starting material is presented on the left hand side of the reaction arrow. The reagent added to this starting material is often shown over the arrow. This reagent transforms the starting material into something else. That something else, the product, is shown to the right of the arrow.
Very often, the solvent for the reaction is shown underneath the arrow. The solvent is the liquid that is used to dissolve the starting material and reagents. This is done for a number of reasons. First, reactions generally happen much more quickly in solution than they do without a solvent. When dissolved, the reactants can move around more easily and bump into each other, as if they are swimming. Also, most useful reactions generate heat, and the solvent acts as a heat sink, carrying the excess heat away. (People who have not thought about the importance of solvent sometimes accidentally start fires as a result.) However, there are exceptions, and not all reactions need solvent.
These reactions shown above do need solvent, but the solvent is not shown for other reasons. There is something else to focus on, and the solvent would have just cluttered up the picture. Instead, the focus of the picture is that these reagents must be added in a particular order: first the nucleophile and then the acid. The nucleophile and acid cannot be allowed to mix before the nucleophile has a chance to react with the carbonyl. If they did, they would just react with each other, and leave the carbonyl electrophile alone.
Exercise \(1\)
1. For each of the cases shown above, use curved arrows to show the movement of electrons in the reaction between the anion and the carbonyl.
2. Show the intermediate that results.
3. Use curved arrows to show the movement of electrons in the reaction of the intermediate with acid to form the product.
Answer
Exercise \(2\)
Problem CO3.2.
a) For each of the cases shown above, used curved arrows to show what would happen if acid were mixed with the nucleophile.
b) Why would the nucleophile no longer be able to react with the carbonyl?
Answer
Previously, we saw that nucleophiles add to carbonyl electrophiles, breaking the pi bond of the carbonyl and converting it into an OH group.
The pattern of reactivity is very different with another class of nucleophile. These could be called neutral nucleophiles (as opposed to anionic ones). Neutral nucleophiles do not have a negative charge like the previous ones. However, they still have a lone pair, and that fact still makes them nucleophiles.
However, the outcome of the reaction looks a little different from what we saw with the earlier anionic nucleophiles. In this case, the carbonyl is not converted into an OH group. (At some point we will see that it can be under some circumstances, but that is the exception rather than the rule.) Instead, the osygen is completely displaced from the carbonyl. It is lost as water. It is replaced by two nucleophiles, instead.
The same thing happens in the following case, involving an oxygen nucleophile instead of a sulfur. Oxygen is in the same column of the periodic table as sulfur, so similar behavior is not really a surprise.
With nitrogen nucleophiles, the oxygen of the carbonyl is still displaced as water, but the nucleophile does not appear to add twice. Instead, the C=O is replaced with a C=N.
Other nitrogen compounds give a similar product, with the double bond in a slightly different place. We will see that the difference has to do with how many hydrogens there are on the original nitrogen atom. If there are two, it is an even trade: the nitrogen replaces the carbonyl oxygen, and the oxygen takes the two hydrogens to form water. If there is only one NH, the oxygen needs to pick up a second hydrogen from elsewhere, and it takes it from the alpha position, next to the former carbonyl.
These nucleophiles are a little less likely to react with the acid, so we have not taken care to add them in a particular order. They are less likely to react with the acid because they are not anions. They are neutral, so they will not attract the proton as strongly as anions would. In reality, you have to be a little bit careful about conducting reactions like these. With the amine nucleophiles in particular, if you add too much acid, the nucleophiles will react with the acid after all.
Take another look at the general pattern of reactivity for the anionic nucleophiles and the neutral nucleophiles. In the case of the anionic nucleophiles, the pattern is relatively easy to discover. The product has incorporated the nucleophile into its structure (or at least the anionic part of the nucleophile, which you will soon learn about). The nucleophile has attached at the carbonyl carbon. The carbonyl oxygen has become part of a hydroxyl group. These are very common patterns in the addition of nucleophiles to carbonyls.
In the case of the neutral nucleophiles, there are some similarities and some differences. The nucleophile is still incorporated into the product structure. It has added at the carbonyl position in the electrophile. However, the fate of the carbonyl oxygen is a little bit different with neutral nucleophiles. Generally, this atom is lost as a water molecule in these cases. If you look closely, you will be able to tell where the two protons come from in each case in order to form the water molecule. It's not really the HCl, which is only added in very tiny amounts and acts catalytically. The protons come from other positions in the nucleophile, and sometimes from the electrophile, too.
This chapter will help you to develop skills so that you can recognize where nucleophilic additions have taken place in reactions. You will also be able to predict what products may result from a nucleophilic addition. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/III%3A_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_1/03%3A_Addition_to_Carbonyls/3.22%3A_General_Reac.txt |
Much of the chapter will focus on mechanisms of reaction. A reaction mechanism is, at the very least, the series of elementary steps needed to accomplish an overall reaction, and all of the intermediate structures that would be formed on the way from the reactants to the products.
• A reaction mechanism shows the structures of intermediates that occur after each elementary step.
Consider the following reaction. It's called a keto-enol tautomerism. A tautomerism is just a reaction in which, overall, a proton or hydrogen atom has changed positions. The structure on the left is a ketone. The structure on the right is called an enol, because it has a hydroxyl group (OH) attached directly to an alkene carbon (C=C). Just by moving one hydrogen atom, we go from one structure to the other.
Exercise \(1\)
Try drawing the reaction above using skeletal drawings instead of full Lewis structures.
Answer
An elementary reaction is typically a bond-forming or a bond-breaking step. In a bond-forming step, a pair of electrons are donated from one atom to another. In a bond-breaking step, a pair of electrons that were shared between two atoms are drawn to one end of the bond or the other, so that the bond breaks and the electrons end up on one atom only.
What are the elementary steps in a keto-enol tautomerism? What sorts of stops do we make along the way?
These reactions can actually occur in a couple of different ways, depending on whether the compounds are in acidic conditions or basic conditions. In acidic conditions, there are extra protons floating around. They aren't all by themselves; remember, protons tend to stick to things that have lone pairs to share.
Under basic conditions, there aren't a significant amount of extra protons around. There might be hydroxide ions or other nucleophilic species around. Nucleophile species are electron-donating compounds that are attracted to positive charges or electrophiles.
We're going to look at this reaction under acidic conditions. If there are protons around, maybe some mineral acid has been added, such as hydrochloric acid or sulfuric acid. Those things are typically used in water, so we'll assume there is some water around. We have hydronium ions (H3O)+ in solution. There must be some counterion, too, but we'll ignore it.
Under those conditions, what will the first step look like? Maybe a proton is transferred from the hydronium ion to the oxygen atom on the ketone. That would get us halfway there. Remember, the keto-enol tautomerism involves addition of a proton to that oxygen.
The bond-making event involves the carbonyl oxygen. What differences do you see at that atom before and after the transfer? Certainly a proton has appeared, and a positive charge, but there is also a lone pair missing. Where did it go?
On the hydronium ion, meanwhile, a lone pair has appeared along with the departure of the proton. Where did that come from?
Of course, a covalent bond is a pair of electrons shared between two atoms. If we are making and breaking bonds, electrons are playing a prominent role. It may be useful to illustrate the role they are playing.
Very often, curved arrows are used to show the path that electrons take in these elementary steps. These arrows are always drawn from the source of the electrons to the place to which the electrons are attracted. These arrows help to illustrate bond-making and bond-breaking steps and also serve a book-keeping function, helping us to keep track of electrons over the course of the reaction.
Notice that, in the elementary step shown above, a bond forms between the carbonyl oxygen and one of the protons on the hydronium ion (H3O+). A covalent bond is a pair of electrons shared by two atoms. Where do the electrons come from to form that bond? They used to be a lone pair on the carboyl oxygen. A curved arrow is used to show that.
At the same time, the bond breaks between that hydrogen and the oxygen in the hydronium ion. Where do those electrons go? They become a lone pair on the oxygen. Another curved arrow shows that event.
We're not finished, yet. What happens after that initial transfer? It seems reasonable that we might just take a proton off the carbon next to the carbonyl. That position, right next to the carbonyl carbon, is called the alpha position. We are taking a proton that was attached to an alpha carbon.
Is it OK to take protons away and break C-H bonds? Only sometimes, but this is one of those cases. Removal of a proton from an alpha position happens all the time in organic and biochemical reactions (those involving carbon-based molecules, and those involved in living systems).
Filling in curved arrows shows the bonds have been made or broken.
• Curved arrows from the nucleophile to the electrophile show the path of electrons in the reaction.
• Curved arrows illustrate bond-making and bond-breaking events.
Sometimes, only one arrow is required in showing an elementary step, but not always. Often, a bond-making step can happen at the same time as a bond-breaking step. This usually happens when an atom isn't large enough to accommodate the electrons from the new bond and sill keep the electrons from an old bond. In this case, two pairs of electrons move in the same elementary step, so two curved arrows are shown. Very rareley, more than two curved arrows are needed to show the events in one elementary step.
This is how chemists have thought about reactions, on paper, for about a hundred years. Always they try to draw a sequence of reasonable intermediates along the course of a reaction. Reactions rarely happen in one step, especially if multiple bonds are formed and broken, although you will eventually learn about some that happen that way. Usually, especially in organic and biochemical reactions, curved arrows are used in an attempt to map out the movement of electrons.
Exercise \(2\)
Draw the entire keto-enol tauomerism mechanism shown above using skeletal drawings rather than full Lewis structures. Remember, it is important that you still show the lone pairs, for electron accounting purposes.
Sometimes other information is displayed in a reaction mechanism. Computational chemists will often leave out the curved arrow notation but will instead indicate the relative energy differences between all the intermediate structures along the reaction pathway. These energies may be experimentally determined (i.e. they may be based on the measurement of real reactions) or they may be calculated using an appropriate level of quantum theory. The energies may be displayed numerically, possibly in a table, or they may be illustrated using a picture, such as a reaction profile.
Exercise \(3\)
Fill in curved arrows on the following mechanisms.
Exercise \(4\)
In the following overall reactions, identify where bonds have been broken and where bonds have been made.
Exercise \(5\)
In the following reactions, specific atoms have moved to specific places. Propose elementary mechanisms for these transformations.
Exercise \(6\)
Propose a mechanism, with arrows, for the keto-enol tautomerism above, but this time under basic conditions. Assume there is some sodium hydroxide dissolved in aqueous solution. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/III%3A_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_1/03%3A_Addition_to_Carbonyls/3.23%3A_Reaction_Mec.txt |
Often it is useful to look at the molecular orbital picture of a molecule to learn something about its reactivity.
In the case of carbonyls, frontier orbital ideas tell us to look at the lowest unoccupied molecular orbital (LUMO) and the highest occupied molecular orbital (HOMO).
When two different atoms bond together, the molecular orbitals that they form are not evenly distributed between the atoms. Instead, the new molecular orbital is closest in space to the atom to which it is closest in energy.
In the case of carbon and oxygen, oxygen is more electronegative than carbon. That means its electrons are more tightly held than carbon’s. That means its electrons are lower in energy than carbon’s.
When carbon and oxygen combine, a bonding orbital and an antibonding orbital result. The bonding orbital is lower in energy than the orbitals on either carbon or oxygen. However, it is closer in energy to oxygen. Thus, the orbital itself is more centered on oxygen. In other words, the electrons in the bond are closer to oxygen than to carbon.
The antibonding orbital, on the other hand, is closer to carbon in energy, although it is higher in energy than either carbon or oxygen. It is more centered on the carbon than the oxygen. That means the “target” for the electron donation is mostly found on the carbon. The carbonyl carbon is the electrophilic position.
If electrons are going to be donated to the molecule, the lowest energy position available for electrons in the molecule is described by the LUMO. The LUMO in this case is the C=O pi* or pi antibonding orbital.
If the carbonyl is going to donate electrons, the electrons will come from the HOMO. In this case, that refers to the non-bonding electrons. These electrons are found on the oxygen, and are equivalent to the lone pairs in the Lewis structure.
Exercise \(1\)
Explain what happens to bonding in the molecule if an electron pair is donated to the LUMO of H2C=O.
Answer
The LUMO in this case is the π*, an antibonding level. If electrons populate this level, the π bond will break.
Exercise \(2\)
Draw the frontier-orbital diagram (including orbital pictures) for the following compounds.
a) CH3OH b) H2C=NH c) CH3CN
Answer
Exercise \(3\)
Problem CO5.3.
In the previous problem, which MO pictures most closely resemble the one shown for H2C=O? Which compounds will behave most like a carbonyl compound? Explain.
Answer
Both the imine (b) and nitrile (c) have a low-lying pi antibonding level (π*), similar to a carbonyl.
3.25: Relative Rea
Sterics
Steric hindrance, or crowdedness around the electrophile, is an important factor that influences reactivity.
• The less crowded the electrophile, the more easily it will react.
• Aldehydes are more reactive than ketones.
Crowdedness affects reactivity simply by preventing nucleophiles from easily approaching the electrophilic site in the carbonyl. If the nucleophile hits something other than the carbonyl carbon, it will probably just bounce off. It needs to collide with the carbonyl carbon in order to deliver its electrons to the right place.
Charge
Amount of positive charge on the electrophile is an important factor that influences reactivity.
• The more positive the electrophile, the more easily it will react.
Factors that place more positive charge on the carbonyl (electron withdrawing groups nearby) make the carbonyl more positive and more reactive. Factors that place additional electron density on the carbonyl (electron donors nearby) make the carbonyl less reactive.
There is another resonance structure that we can think about that illustrates the electrophilicity of a carbonyl. That structure places a full negative charge on the oxygen and a full positive charge on the carbon. This isn’t a good Lewis structure because the carbon does not have an octet. Nevertheless, when taken together with the regular Lewis structure, it suggests something real about the nature of the carbonyl: there is partial positive charge on the carbon and partial negative charge on the oxygen.
There is a general rule about cation stability on carbon atoms: a carbocation with more carbons attached to it is more stable than a carbocation with more hydrogens attached to it.
This observation is sometimes explained as an inductive effect. The positively charged carbon is more electronegative than the uncharged carbons, so it draws electrons away from them. It can polarize the neighbouring carbons, drawing some negative charge towards itself and leaving some positive charge on the other carbons. In that way, it s charge is delocalized and stabilized.
In a more sophisticated explanation, the cation becomes stabilized by a molecular orbital interaction involving the empty p orbital on the carbocation and C-H bonds on the neighbouring carbons.
A similar situation results in the partially positive carbon in the carbonyl. The carbonyl carbon in the ketone is a little more stable than the carbonyl carbon in the aldehyde.
• The partial positive charge on an aldehyde carbonyl carbon is less stable than the partial positive charge on a ketone carbonyl carbon.
• Again, aldehydes are more reactive than ketones.
Exercise \(1\)
Rank the following carbonyl compounds from most reactive to least reactive towards nucleophilic addition. Explain your reasoning.
Exercise \(2\)
Provide names for the following aldehydes.
Exercise \(3\)
Problem CO6.3.
Provide names for the following ketones.
Exercise \(1\)
Problem CO6.4.
Provide names for the following ketones. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/III%3A_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_1/03%3A_Addition_to_Carbonyls/3.24%3A_Molecular_Or.txt |
Carbonyls act most importantly as electrophiles. They attract a pair of electrons from a nucleophile. When that happens, a bond forms between the nucleophile and the carbonyl carbon.
At the same time, the carbon-oxygen bond breaks. We think of that as a consequence of donating a pair of electrons into the LUMO of the carbonyl. The LUMO on the carbonyl is the C-O pi antibonding orbital. When that orbital is populated, there is no longer a net lowering in energy due to the pi interaction between the carbon and oxygen. The pi bond breaks. The electron pair from the pi bond goes to the oxygen, the more electronegative of the two atoms in the original bond. It becomes a lone pair.
After the pi bond breaks, the reaction reaches a branching point or decision point. The reaction may go forwards or backwards.
In other words, this reaction can occur in equilibrium.
To go backwards, the reaction simply slides into reverse. A lone pair on the oxygen donates to the carbon, forming a pi bond again, and pushes the nucleophile off. Whether the reaction ends up going forward or sliding backward depends partly on the relative stability of those two ends of the reaction. That’s often very difficult to assess qualitatively, because there are too many factors involved. However, one factor that plays a role is charge stability. Because an “O minus” or alkoxide is produced in this reaction, if the original nucleophile was a more reactive ion than an alkoxide, the reaction probably goes to the right.
For that reason, many of the best nucleophiles for these reactions involve carbon anions or hydrogen anions. Those anions are less stable than oxygen anions.
If the nucleophile were less reactive than alkoxide, the reaction could easily go to the left again. For that reason, stable halide ions (fluorides, chlorides, bromides, iodides) are not very good nucleophiles for these reactions. They have lone pairs, they even have negative charges, but the anion that would be produced would generally be less stable than the original halide ion.
Exercise \(1\)
Provide curved arrows and predict the direction of equilibrium in the following cases.
• More reactive nucleophiles push the equilibrium to the right
• Less reactive nucleophiles do not push the reaction to the right; the reaction remains on the left
What happens after the initial equilibrium? In most cases, the alkoxide that is formed will become protonated. It will pick up a proton to become an alcohol. The source of the proton may be an acid, deliberately added to provide the H+. Alternatively, it may just be a very slightly acidic molecule such as water or another alcohol.
3.27: Protonation
After the addition of an anionic nucleophile, reaction mixtures are usually treated with or water or dilute aqueous acid. The acid provides a proton that can be picked up by the alkoxide ion formed in the nucleophilic addition. An alcohol is formed as a result. Overall, carbonyl addition reactions usually involve addition of a nucleophile to the carbonyl carbon and addition of a proton to the carbonyl oxygen.
The order of these steps can be very important. On paper, a carbonyl could be turned into an alcohol by adding a proton to the carbonyl oxygen, and then adding a nucleophile to the carbonyl carbon. However, things might not work out that way in reality. The potential problem lies in the fact that anionic nucleophiles can be pretty basic. If protons are added first, the anionic nucleophile is likely to pick up a proton rather than donate to the carbonyl. Once the nucleophile has picked up a proton, it is no longer anionic, and is less attracted to the partially positive carbonyl. Furthermore, it may have donated its only lone pair to the proton, leaving it completely unable to donate to the carbonyl.
What is the source of protons? In most cases, a dilute aqueous solution of a strong acid is used. That is, a little bit of strong acid is dissolved in water to provide plenty of protons to form the alcohol. The most common strong acids are hydrochloric acid, HCl, and sulfuric acid, H2SO4.
Actually, just the water itself would be enough to donate protons. Water has a polar O-H bond and so it can transfer a positive hydrogen ion to an anion. However, that would result in the exchange of one oxygen anion for another. Energetically, there wouldn't be much difference between having the negative charge on the reaction product, the alkoxide, or on the hydroxide ion from the water. That's why a strong acid is usually added.
In terms of le Chatelier, if you added enough water, you should be able to drive that equilibrium to the right. However, there would always be a little unprotonated alkoxide left over at the end.
Exercise \(1\)
Problem CO8.1.
Provide curved arrows and predict the direction of equilibrium in the following cases.
Exercise \(2\)
Problem CO8.2.
Suppose nucleophilic addition was performed in methanol (10 mL) as a solvent, using 25 microliters of cyclopentanone and an equimolar amount as the of sodium cyanide (that means the same number of moles of sodium cyanide as cyclopentanone).
1. Show the mechanism for the reaction, using curved arrows in each step.
2. Comment on how the use of methanol as a solvent (rather than just adding an equimolar amount of methanol) would influence the direction of equilibrium for the final protonation step.
Answer
Exercise \(3\)
Provide structures for the following acids. They are moderately strong acids, but not as commonly used in this context as hydrochloric and sulfuric acid. Note that, like sulfuric acid, they all contain polar OH groups.
a) phosphoric acid, H3PO4 b) nitric acid, HNO3 c) acetic acid, CH3CO2H
Answer
Sometimes, weak acids are used to protonate the alkoxide intermediate. These acids are more acidic than water, so the equilibrium in the protonation reaction lies more towards the alcohol side. However, because they are only weak acids, they are safer and easier to work with than strong acids. Commone examples include ammonium chloride and sodium bicarbonate.
Exercise \(4\)
What factor makes each of these compounds mildly acidic?
Answer
Ammonium chloride has an N-H bond, normally less polar than the O-H bond of water. However, the positive charge makes this compound give up a proton more easily, because it results in a neutral (uncharged) ammonia molecule.
Sodium carbonate has a polar O-H bond, just like water. However, the anion that results from loss of a proton is resonance-stabilised. That makes this compound more acidic than water.
Exercise \(5\)
Problem CO8.5.
Fill in the product or reagent for each of the following transformations. Remember there is always an acidic workup assumed.
Exercise \(6\)
Draw a mechanism with curved arrows for each of the transformations listed below. Show all intermediates. Assume an acidic workup. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/III%3A_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_1/03%3A_Addition_to_Carbonyls/3.26%3A_The_Mechanis.txt |
Lots of things can be nucleophiles. In principle, a nucleophile only needs a lone pair. However, some nucleophiles are better than others.
You already know something about nucleophiles if you know something about acidity and basicity. Nucleophiles are really Lewis bases. Some of the factors that account for basicity also account for nucleophilicity.
Halides are not very good nucleophiles for carbonyls. The negative charge on a halide is pretty stable, either because of electronegativity or polarizability. If a halide donates to a carbonyl, producing an oxygen anion, the reaction is uphill.
Hydroxide and alkoxide anions (such as CH3O-) are more reactive than halides. They are better nucleophiles. The sulfur analogues are similarly good nucleophiles (such as CH3S-). In addition, water, alcohols and thiols are nucleophilic, because they all have lone pairs that could be donated to an electrophile.
Nitrogen also has a lone pair in most compounds. That means amines are good nucleophiles, too.
Carbon does not normally have a lone pair, unless it is a carbanion. Carbanions are usually not very stable. As a result, they are not very common, except for cyanide (CN-) and acetylides (RCC-, in which R is a hydrogen or an alkyl group). However, when carbon does have a lone pair (and a negative charge), it is a good nucleophile. Because carbon is less electronegative than other elements with lone pairs, it is able to donate its lone pair easily.
Carbon nucleophiles add to carbonyls because that less stable carbon anion is traded for a more stable alkoxide anion. The reaction is downhill energetically.
Exercise \(1\)
Carbanions such as CH3- (methyl anion) are very unstable and highly reactive. Explain why the following anions are more stable than a methyl anion.
1. Acetylide, HCC-
2. cyanide, CN-
Answer a
In acetylide, the lone pair is on a linear carbon or sp carbon. In methyl anion, the lone pair is on a tetrahedral carbon or sp3 carbon. The description "sp" indicates that sigma bonding to neighbors involves a 2s orbital and a 2p orbital on carbon; there is a 50% contribution from the s orbital.
The description "sp3", on the other hand, indicates that sigma bonding to neighbors involves a 2s orbital and three 2p orbitals; there is a 25% contribution from the s orbital.
The 2s orbital is lower in energy than the 2p orbital. The greater the s orbital contribution to the bond (or in this case to the lone pair), the lower it is in energy. Thus, a lone pair on an sp carbon is lower in energy than a lone pair on an sp3 carbon.
Answer b
In cyanide, the same argument outlined in part (a) hold true. In addition, the nearby electronegative nitrogen stabilizes the charge by drawing electron density toward itself.
Other nucleophiles, such as halides, do not proceed. They are going uphill, from a more stable halide ion to a less stable alkoxide ion.
If the nucleophilic atom were an oxygen anion, there might be an equilibrium. The reaction would be neither uphill nor downhill. It would result in a mixture of the original reactants and the new products.
Exercise \(2\)
Nucleophilicity is the degree of attraction of a nucleophile to a positive charge (or partial positive charge). It is related to basicity. Choose the most nucleophilic item from each of the following pairs, and explain your answer.
1. CH3OK or CH3OH
2. CH3OH or CH3NH2
3. NaCN or NaCCH
4. c-C6H11ONa or c-C6H5ONa (c- in this case means "cyclo")
Answer a
CH3OK, because of the ionic O-K bond. This is an anionic nucleophile. It is more reactive and nucleophilic than the corresponding neutral nucleophile.
Answer b
CH3NH2, because nitrogen is less electronegative than oxygen. Its lone pair is held less tightly and is more easily donated to the electrophile.
Answer c
NaCCH, because the neighbouring carbon in this case does not have the inductive electron-withdrawing effect that the nitrogen does in the case of NaCN. In that case, the lone pair is stabilized and made less reactive.
Answer d
c-C6H11ONa, because the negative charge is localized on one atom. In c-C6H5ONa, the negative charge is delocalized over four different positions in the molecule. Delocalization of charge stabilizes the anion and makes it less reactive.
Exercise \(3\)
Carbonyl compounds such as aldehydes and ketones contain a very slightly acidic hydrogen next to the carbonyl. Some nucleophiles are basic enough to remove that proton instead of donating to the carbonyl. Show why the resulting anion is stable, using cyclopentanone as an example.
Answer
The resulting anion is stabilised by resonance.
Exercise \(4\)
Accidental deprotonation (proton removal) alpha to a carbonyl (one carbon away from the carbonyl) can occur when a nucleophile is added to a ketone. One reason the proton might be taken instead is if the carbonyl is too crowded for the nucleophile to reach.
In the following cases, explain which nucleophile is more likely to add to the carbonyl in cyclohexanone and which is more likely to deprotonate it. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/III%3A_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_1/03%3A_Addition_to_Carbonyls/3.28%3A_What_is_a_Nu.txt |
Insertions constitute a class of reactions involving transition metals. These reactions are important in some industrial processes that are catalyzed by transition metals. One of the most important processes that involves these reactions is hydroformylation. Hydroformylation is used to convert unsaturated hydrocarbons into aldehydes. These long-chain aldehydes are then converted into important commodities, such as detergents and fragrances.
There are two basic kinds of insertions: migratory insertions (or 1,1-insertions, more generally) and beta-insertions (or 1,2-insertions).
• Migratory insertions are related to the addition of nucleophiles to carbonyls.
• Migratory insertions involve transition metal compounds that bind to a carbon monoxide.
• The transition metal is bound to another group, such as an alkyl group or a hydride.
• The alkyl or hydride is transferred from the metal to the carbon of the bound carbon monoxide (figure \(1\)).
The other type of insertion, 1,2-insertion, often involves alkenes, or other ligands that can bind to a metal through two atoms instead of just one (Figure \(2\)).
Note that, in transition metal chemistry, formalisms are often used differently than in simple main group compounds involving carbon, oxygen and nitrogen..
• Sometimes, formal charges are not shown.
• Non-bonding electrons on metals are rarely shown.
Structures are often complicated enough that it becomes difficult to draw all the non-bonding electrons and formal charges and still have a clear picture. People who work in this area will keep count of electrons in their head, although they will frequently jot down the electron count on the metal beside the structure in order to keep track.
4.02: CO Binding
Migratory insertion involves the transfer of a hydride or alkyl group from a metal to a bound carbon monoxide. Because this reaction specifically involves bound carbon monoxide, we should take a look at how CO binds to transition metals. We should begin by reviewing the Lewis structure of carbon monoxide.
Note the lone pair on carbon monoxide. It is a potential Lewis base or nucleophile. For a number of reasons, transition metals are almost always electrophiles: they are often positively charged ions, but in general they have an 18-electron octet that is difficult to fill, so they frequently need more electrons.
• Lewis: CO is a two electron donor
• Transition metals are electrophiles
• CO binds to metal atoms or ions
• The carbon is the usual donor atom; it has a lone pair and a negative formal charge
The donation of an electron pair to a metal cation is shown in figure \(2\).
Remember, because the 18-electron rule for transition metals makes them electrophilic, the electron pair does not need a positive charge to attract it (figure \(3\)).
Frequently, the formal charges and lone pairs are not even shown in the transition metal compound, because of the complexity of the picture.
Note that CO in the context of metal complexes is often referred to as carbonyl. For example, Cr(CO)6 is called hexacarbonyl chromium.
Exercise \(1\)
Draw structures for the following metal carbonyl compounds. For each compound, indicate
i) the electron count at the metal in the complex (show your work)
ii) the geometry at the metal
1. tetracarbonyl nickel, Ni(CO)4
2. pentacarbonyl iron, Fe(CO)5
3. hexacarbonyl chromium, Cr(CO)6
4. tetracarbonyl cobalt anion, Co(CO)4-
5. tetracarbonyl cobalt hydride, Co(CO)4H
6. octacarbonyl cobalt, Co2(CO)9 (there is a bond between the two cobalt atoms)
Answer
Exercise \(2\)
Draw, with structures and arrows, the equilibrium between pentacarbonyl iron and tetracarbonyl iron, Fe(CO)4 plus carbon monooxide.
Answer
An important aspect of CO binding is called "back-donation". In back-donation, not only does the ligand donate electrons to the metal, but the metal also donates to the ligand. We can think of the CO as donating a pair of electrons from a carbon-based orbital into an empty orbital on the metal, such as a p orbital (figure MI2.5). The metal has d orbitals that have good symmetry overlap with the pi antibonding orbitals in the CO. Electron density can be donated from a metal d orbital into the pi* level (figure MI2.6). Thus, binding to a metal actually weakens the CO bond because a pi* orbital receives electron density from the metal.
• MO picture: donation from carbon-based orbital into vacant metal p orbital
• MO picture, part 2: donation from occupied metal d orbital into CO pi* orbital
• this interaction weakens the C-O multiple bond
Exercise \(3\)
Draw a Lewis structure that takes into account the effect of metal-to-carbonyl electron donation in tetracarbonyl nickel.
Answer
Exercise \(4\)
Infrared spectroscopy is often used to assess bond order between specific atoms within a molecule. Because stretching frequencies are proportional to bond strength, a comparison of frequencies from a bond one molecule to a similar bond in another can give insight into the bond orders in each case.
1. A C-O bond in organic compounds shows up between 1000-1200 cm-1 in most cases. A C=O bond normally shows up near 1600-1700 cm-1. Explain why these two bonds give rise to two different stretching frequencies.
2. Based on the Lewis structure, what do you predict for this bond frequency in a CO molecule?
3. What will happen to the CO stretching frequency in carbon monoxide if the molecule binds to a palladium atom?
Exercise \(5\)
Explain the differences in CO stretching frequencies in the following pairs of compounds. (Note: the number of peaks is related to molecular symmetry and group theory; focus only on the difference in magnitude of the frequency.)
1. CO at 2143 cm-1 vs. Mo(CO)6 at 2004 cm-1.
2. Ni(CO)4 at 2060 cm-1 vs. Fe(CO)42- at 1790 cm-1.
3. Cr(CO)6 at 2000 cm-1 vs. Mn(CO)6+ at 2090 cm-1.
4. (PF3)Mo(CO)3 at 2055 & 2090 cm-1 vs (PPh3)Mo(CO)3 at 1835 & 1934 cm-1.
5. Cr(CO)6 at 2000 cm-1 vs. (C6H5CH3)Cr(CO)3 at 1963 and 1869 cm-1.
Exercise \(6\)
Sometimes, carbonyls can bridge between two metals. For example, the iron cluster Fe2(CO)9 contains six "terminal" carbonyls, bound to only one iron each, and three "bridging" carbonyls, each of which is bound to both iron atoms. This complex also features a metal-metal bond.
1. draw a structure for this compound.
2. explain why the terminal carbonyls display stretching frequencies of 2082 and 2019 cm-1, but the bridging carbonyls display a stretching frequency of 1829 cm-1.
Answer
"Organic" carbonyls, such as aldehydes and ketones, can also bind to transition metals, as you may have seen before. These compounds bind to transition metals in a very different way than carbon monoxide. Normally, we think of them as simple lone pair donors. The oxygen lone pair donates to the metal atom or ion. The resulting complexes are important because the carbonyl becomes "activated" or ready to accept nucleophiles.
• Organic carbonyls can also bind to metal atoms or ions.
• Binding is usually via the heteroatom.
Exercise \(7\)
Draw, with arrows, the coordination of the following organic carbonyl compounds to metal complexes.
1. 2-pentanone with TiCl4.
2. ethyl acetate with ScCl3.
3. propanal with Cp2ZrH2 (Cp = cyclopentadienyl anion, C5H5-).
Answer | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/III%3A_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_1/04%3A_Insertion/4.01%3A_Introduction.txt |
Migratory insertion is a term to describe the transfer of a ligand from a metal to a carbon monoxide that is also bound to the metal. It is a special case of a 1,1-insertion. In a 1,1-insertion, a group is transferred from a metal to an atom attached to the same metal. A general 1,1-insertion is shown in figure \(1\).
At the end of the 1,1-insertion, the ligand Z has attached to the 1st atom in the next ligand attached to the metal.
Carbonyl groups in organic compounds are electrophilic. The polar carbon-oxygen double bond places positive charge on the carbon, so the carbon atom attracts nucleophiles. One of the nucleophiles that can react with a carbonyl is a complex hydride, such as a borohydride ion or an aluminum hydride ion. Sometimes, these complex hydrides are anionic, making them more nucleophilic. An example is sodium borohydride, NaBH4. Sometimes, the hydride compound is neutral, as in BH3. However, the hydride is still nucleophilic even if the compound is not negatively charged, because of the electronegativity difference between the hydrogen and the boron (or the aluminum). A hydride ion is donated as a nucleophile to the electrophilic carbonyl. Transition metal hydrides, like boron and aluminum hydrides, are frequently nucleophilic. They can donate hydrides to electrophiles.
• A hydrogen attached to a metal atom frequently acts like a hydride.
• A nucleophilic hydride can donate to a carbonyl carbon.
"Inorganic carbonyls", or metal-bound CO compounds, behave in many ways like organic carbonyl compounds. In one sense, the bound CO can be thought of as having a positive formal charge on the oxygen, so it is easy to imagine it as an electrophile. It looks like an "activated" organic carbonyl (for example, a ketone that has been protonated, and has a positively charged oxygen). If a metal has a hydride attached to it (a nucleophile) as well as a CO (an electrophile), then a reaction can occur between them. The hydride can add to the carbonyl. This is one of the most useful things about transition metal chemistry: by binding different, reactive ligands, metals can organize reactants so that they react together.
• This reaction can occur intramolecularly; i.e. from a hydride to a carbonyl on the same metal.
• This event is called a migratory insertion.
Note that the metal does not have to be anionic for the hydride to act as a nucleophile. The electronegativities of transition metals vary from about 1.0 to 1.75, but the electronegativity of hydrogen is about 2.2. Hydrogen is more electronegative than the transition metals, and so a hydrogen attached to a transition metal is usually nucleophilic.
This reaction, migratory insertion of a hydride to a carbonyl, forms a metal "formyl" compound. The "formyl" is the CH=O group attached to the metal. Migratpry insertions can also take place with metal alkyls. Metal alkyls are also nucleophilic, just like metal hydrides. The alkyl carbon is usually more electronegative than the attached transition metal, so it has a partial negative charge.
• Nucleophilic alkyl groups can also undergo migratory insertion reactions.
In fact, migratory insertion to a carbonyl is actually much more common with alkyl ligands than it is with hydrides. Conceptually, the reactions are very similar. The reason alkyl migrations are much more common has to do with relative bonds strengths in the reactants compared to the products of the reaction.
Migratory insertion is industrially important. It is used in the manufacture of things many of us use every day. For example, the main component of body wash. sodium laureth sulfate and related compounds, is synthesized via a key hydroformylation process. The catalyst used for hydroformylation varies from one application to another and from one company to another. In the case of soap synthesis, a rhodium catalyst is usually employed.
Hydroformylation results in an alkene chain being lengthened by one carbon, with an aldehyde at the end of the chain. Overall, it results in the addition of a aldehyde fragment (CHO) to one end of the alkene double bond and a hydrogen atom the the other. A few additional steps are needed to extend the resulting aldehyde and cap the chain with a polar sulfonate group. Of course, the compound works by forming micelles when suspended in water. Those long tails gather together, with the polar heads interacting with the water. Stuff that won't wash off with plain water because it isn't polar enough to dissolve in water may instead dissolve among the soapy tails of the sodium laureth sulfate molecules.
Exercise \(1\)
Explain why a coordinated carbon monoxide acts as an electrophile.
Exercise \(2\)
Draw, with arrows, the reaction to give the 1,1- insertion products in the following cases.
Exercise \(3\)
Binding or coordination to a metal may also occur with organic carbonyl compounds such as ketones, aldehydes, esters and so on.
1. Explain why a coordinated carbonyl compound, such as propanal, can be especially electrophilic.
2. Compare and contrast the electrophilicity of a coordinated organic carbonyl with the electrophilicity of coordinated carbon monoxide. What atom is the electrophile in each case? What will happen if a nucleophile donates electrons in each case?
Exercise \(4\)
Provide products of the following migration reactions.
Exercise \(5\)
Provide the products of the following migration reactions. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/III%3A_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_1/04%3A_Insertion/4.03%3A_Hydride_and_Alkyl_Migrat.txt |
Like a coordinated carbon monoxide molecule, a coordinated organic carbonyl is also electrophilic. This time, the oxygen can be thought of as having a formal positive charge because it is donating a lone pair to a metal.
Consequently, the carbonyl is activated. It becomes a better electrophile and is more likely to undergo an addition reaction, if there are any nucleophiles nearby that are able to donate a pair of electrons.
A hydride attached to the metal can donate to the carbonyl. Remember, the hydride is nucleophilic, because the hydrogen atom is always more electronegative than a transition metal. Even the most electronegative transition metal, mercury, only has a Pauling electronegativity of 1.9, compared to 2.2 for hydrogen. When a hydride moves to the carbonyl from the metal, the carbonyl is turned into an alkoxide ligand.
Notice that there is a difference between migratory insertion, or 1,1-insertion, and this beta-insertion, which is sometimes called a 1,2-insertion. Unlike migratory insertion, the nucleophile does not move to the atom attached to the metal. The nucleophile moves to the second atom away from the metal: the 2 position or the beta position.
Furthermore, beta-insertions of hydride ligands are easily reversible. An alkoxide ligand attached to a metal can easily lose a beta hydrogen and become a ketone or aldehyde.
• In this case, the insertion can be reversible.
• The reverse of an insertion is called a 1,2-elimination or a beta-elimination.
The Greek lettering refers to the number of atoms away from the metal. The first atom attached to the metal is called the alpha position. A hydrogen on that atom is called an alpha hydrogen. The next atom along the chain is called the beta position. The third atom along the chain is the gamma position. A hydrogen attached to the beta position can undergo 1,2-elimination or beta-elimination.
• elimination leads to formation of a double bond.
• The double bond forms between the alpha and the beta position.
This nomenclature can be confusing because a carbonyl compound already has an alpha position and a beta position. The position is the carbonyl carbon and the position is the carbon next to the carbonyl. This Greek lettering system is a general way of designating positions and it is used in a number of different contexts; you need to be able to decide which context fits. If elimination is occurring, then the term, position, may mean one thing. If enolate formation is occurring, then the term, position, means something else.
Exercise \(1\)
Problem MI4.1.
In the following structures, identify any alpha, beta or gamma positions on the groups attached to the metal. In each case, show the potential products of 1,2-elimination of hydrogen.
Exercise \(2\)
Problem MI4.2.
Hydride transfer reduction is a method of converting aldehydes or ketones to alcohols. This catalytic reaction involves the use of a sacrificial alcohol as the solvent. As the aldehyde or ketone substrate is converted into alcohol, the alcohol solvent is converted into ketone. The mechanism is quite complicated, but at a simple level it could be imagined as taking place through a series of binding steps, eliminations and insertions.
Provide mechanisms based on the following descriptions for the reduction of benzaldehyde to benzyl alcohol.
1. The reaction involves binding of isopropanol or 2-propanol to a metal ion, such as Ru(II) or Ru2+. A deprotonation ensues, followed by elimination, and 2-propanone dissociates from the metal ion.
2. The ruthenium hydride formed in step (a) binds to the substrate, benzaldehyde. An insertion occurs, forming a metal alkoxide.
3. The metal alkoxide is protonated, and the resulting alcohol dissociates from the metal.
Answer
Just like aldehydes and ketones, alkenes can undergo 1,2-insertions (also called beta hydride insertions). In terms of electrophiles and nucleophiles, this reaction is a little harder to imagine. However, we can still think of the hydride as a nucleophile. Maybe the alkene is an electrophile. Given that it is donating its pi electrons to the metal, we can think of it as "activated", a little bit like an activated carbonyl.
The formalisms of drawing a beta alkene insertion are tricky. If we use the metallacycle drawing of a bound alkene, it might look like this:
More often, bound alkenes are drawn as shown as in the picture below. In that case, we could try to show the pi bond forming a new carbon-metal bond. The bond between the metal and alkene on the picture to the left does not really stand for a separate pair of electrons in this case; it just stands for the pi bond donating to the metal.
• The reverse of elimination is insertion.
• A coordinated alkene is sometimes considered electrophilic because it is giving electrons to the metal.
• A coordinated alkene is activated, like a coordinated carbonyl compound.
So far, we have just looked at the 1,2-insertion of hydride ligands, but 1,2-insertions of alkyl ligands are also possible. In this case, an alkyl ligand would relocate onto the second atom in the former alkene. As a result, the alkyl chain attached to the metal would become two carbons longer.
However, the opposite reaction, 1,2-elimination of an alkyl, is not very common. That reaction is usually restricted to movement of a hydrogen atom.
Beta-insertion of hydride or alkyl ligands across alkenes is very important commercially. Many catalytic processes that are used to convert hydrocarbons into useful products depend on this reaction. Alkenes are readily made from petroleum or natural gas, so they are convenient starting materials to make more useful compounds. For example, ethene can be grown into longer carbon chains via a series of 1,2-alkyl insertions into coordinated ethene. This process is called "olefin oliomerization".
That sequence of steps is crucial in the conversion of ethene into more valuable materials. It can be grown into hydrocarbon chains long enough to be used as fuels. However, more valuable products can be made such as soaps, detergents, and surfactants (via hydroformylation and other steps). In addition, the alkene may be functionalised in a variety of different ways if a long carbon chain needs to be incorporated into the synthesis of a more valuable commodity, such as a paint or a pharmacuetical.
Exercise \(3\)
Provide the products of beta-elimination in each of the following cases. In each case, assume the elimination product is still bound to the metal.
Exercise \(4\)
Provide products of the following beta-eliminations. In each case, assume the elimination product has already dissociated from the metal.
Exercise \(5\)
Provide products of the following insertion reactions.
Exercise \(6\)
Predict whether the following compounds will undergo 1,1-insertion, 1,2-insertion, or beta-elimination.
Exercise \(7\)
Insertion and elimination reactions can also occur on solid metal surfaces. Often, solid metals are used as catalysts in industrial processes.
a) Copper adopts a face-centered cubic structure. Show a unit cell of copper on the following template.
b) Alcohols such as ethanol can bind to the surface of the copper metal. Show an ethanol molecule binding to a copper in the unit cell.
c) The ethanol can undergo beta-elimination. Show the product of the reaction.
Answer a
Answer b
Answer c
4.05: Solutions for Selected P
Note: In some cases, formal charge has been shown only on a particular atom to emphasize an aspect of its reactivity, but is left off other atoms.
Exercise 4.2.1:
Exercise 4.2.2:
Exercise 4.2.3:
Exercise 4.2.6:
Exercise 4.2.7:
Exercise 4.3.2:
Exercise 4.3.4:
Exercise 4.3.5:
Exercise 4.4.1:
Exercise 4.4.2:
Exercise 4.4.3:
Exercise 4.4.4:
Exercise 4.4.5:
Exercise 4.4.6:
Exercise 4.4.7:
a)
b)
c) | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/III%3A_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_1/04%3A_Insertion/4.04%3A_Beta-Hydride_Insertion_a.txt |
n/a
5.02: Prote
After proteins are synthesized via transcription of DNA, their structures may be modified in a number of ways. This process, in general, is referred to as "post-translational modification". It might involve the attachment of small organic groups to the protein, attachment of other biomolecules, or even the complete alteration of an amino acid residue into another form, so that the protein actually contains a residue that does not exist in the basic beginner's amino acid alphabet.
Minor modifications of proteins can be very important in regulating enzyme activity. Modest changes in structure can have a big impact on the conformation of the protein, which in turn might open or close access into the enzyme's interior or alter the shape of the active site. Alternatively, these changes might affect the location of a protein by moving it from the cell membrane into the cell interior or vice versa. They might also result in the formation or dissociation of supramolecular assemblies, by increasing or decreasing the attraction between two separate biomolecules.
The physical reasons for these structure-property changes are based in simple intermolecular attractions or, in the case of a large, complicated protein, intramolecular attractions between different parts of the same protein. A hydrogen bonding interaction or ion-dipole interaction that holds the protein in one conformation may be disrupted if a key player in that interaction is suddenly masked. For example, a cationic ammonium side chain may react to become an amide group; the absence of a full positive charge on this group significantly alters its intermolecular attractions.
There are probably hundreds of ways in which proteins are modified. We will look at just a few different modifications that occur in proteins, some of which are closly tied to carboxylate substitution. Note that any of these modifications might act to turn an enzyme "off" or "on"; the details depend on the individual case.
Acetylation
Acetylation is the attachment of an acetyl group, CH3C=O, to another compound such as an amino acid residue. Serine and threonine groups might be acetylated to make esters, cysteine side chains might be acetylated to make thioesters, or amino groups. Frequently, acetylation refers specifically to reaction at an amino group, such as the N-terminus of a protein or in a lysine side chain, forming an amide.
The source of the acetyl group is acetyl coenzyme A (AcSCoA, below), the thioester workhorse of the cell. This structure may seem a little bit complicated, but at the most basic level it is just a carboxyl electrophile (CH3C=O) attached to a very large thiolate leaving group.
Of course, lysine side chains and the N-termini of proteins are usually in a protonated state under biological conditions. That means that there must be a deprotonation step along the reaction pathway.
One example of the role of acetylation is seen in histones. Histones are proteins found in chromatin, the mixture of DNA and proteins in the cell nucleus. The already-coiled DNA helix is further wrapped around histones, bundling it up into a smaller package. DNA has many, many negative charges all along it because the phosphate units in its phosphate-sugar copolymer backbone are negatively charged at typical biological pH. The DNA interacts easily with the histones because they are rich in positively charged lysine residues.
Storing DNA in smaller bundles lets you keep more junk in your nucleus, but you don't want to just let the DNA sit there. Once in a while you want to take it out and do something with it, but how can you do that when it's stuck to those darned histones? The answer is, you just have to turn off the histones' force field. Get rid of that positive charge, and the DNA won't be stuck anymore. It's easy to do that by acetylating the histones.
This change in charge results because, although amines are easily protonated, amides are not. Thus, amines are likely to carry positive charge under biological conditions, whereas amides are likely to be neutral.
Exercise \(1\)
Explain why an amine is easily protonated but an amide is not.
Answer
This change in charge results because, although amines are easily protonated, amides are not. Protonation of an amide would result in a cation adjacent to the very positive carbonyl carbon, leading to a buildup of localized positive charge. That wouldn't be easy. Furthermore, the amide nitrogen is not very likely to donate its electrons to a proton in the first place. Its protons are too busy. They are tied up in conjugation with the carbonyl, so they really aren't available to act as the lone pair of a base.
Exercise \(2\)
Provide a mechanism for the acetylation of a lysine side chain. Assume the presence of histidine and its conjugate acid, histidinium ion; these species are common proton shuttles in biological reactions.
Phosphorylation
Serine, tyrosine and threonine residues are frequently modified by phosphorylation, which is the formation of a phosphate ester. The phosphorylation of these groups results in a change from a neutral side chain to an anionic side chain. The phosphate donor in these cases is another ubiquitous cellular performer, ATP.
The complete structure of ATP is shown here. It is a little less complicated than AcSCoA. Once again, at the most basic level of reactivity, it provides a phosphate electrophile, attached to a phosphate leaving group.
The anionic nature of the phosphate may not be apparent in the above drawing of ATP, but in reality, at biological pH, a phosphate would be deprotonated. Phosphates actually undergo multiple equilibria involving the loss of two protons.
Exercise \(3\)
Provide a mechanism for phosphorylation of serine in the presence of an appropriate biological proton shuttle.
Alkylation (Farnesylation)
There are other modifications that occur via different mechanisms. In a farnesylation, reaction does not even take place at a carbonyl. It does not take place at a phosphonate or a sulfonate.
Exercise \(4\)
Sulfur is the nucleophile in the farnesylation shown above. Identify the electrophile and show a mechanism with curved arrows.
Exercise \(5\)
Palmitoylation is closely related to acetylation, but the electrophile in this case is a palmitoyl group instead of an acetyl.
1. Provide a mechanism for palmitoylation of cysteine in the presence of an appropriate biological proton shuttle.
2. Explain how palmitoylation might influence the interaction of a protein with a cell membrane, composed of phospholipids like this one:
5.03: Addit
Exercise \(1\)
Fill in the blanks in the following syntheses.
Exercise \(2\)
Fill in the blanks in the following synthesis involving anionic nucleophilic addition to carbonyl and nucleophilic substitution at carboxyl.
Exercise \(3\)
Fill in the blanks in the following synthesis involving anionic nucleophilic addition to carbonyl and nucleophilic substitution at carboxyl.
Exercise \(4\)
Fill in the blanks in the following synthesis involving anionic nucleophilic addition to carbonyl and nucleophilic substitution at carboxyl (including Wittig and aldol reactions).
Exercise \(5\)
Fill in the blanks in the following synthesis involving nucleophilic substitution at carboxyl.
Exercise \(6\)
Fill in the blanks in the following synthesis involving anionic nucleophilic addition to carbonyl, nucleophilic substitution at carboxyl and conjugate addition.
Exercise \(7\)
Fill in the blanks in the following synthesis involving anionic nucleophilic addition to carbonyl, conjugate addition and nucleophilic substitution at carboxyl.
Exercise \(8\)
Fill in the blanks in the following synthesis involving anionic nucleophilic addition to carbonyl, nucleophilic substitution at carboxyl and neutral nucleophilic addition to carbonyl.
Exercise \(9\)
Fill in the blanks in the following synthesis involving anionic nucleophilic addition to carbonyl/em>, conjugate addition, nucleophilic substitution at carboxyl and transition metal-catalysed coupling.
Exercise \(10\)
Fill in the blanks in the following synthesis involving anionic nucleophilic addition to carbonyl, conjugate addition, nucleophilic substitution at carboxyl and transition metal-catalysed coupling.
Exercise \(11\)
Fill in the blanks in the following synthesis involving anionic nucleophilic addition to carbonyl, conjugate addition, nucleophilic substitution at carboxyl, transition metal-catalysed coupling and neutral nucleophilic addition to carbonyl.
Exercise \(12\)
Fill in the blanks in the following synthesis involving anionic nucleophilic addition to carbonyl, conjugate addition, nucleophilic substitution at carboxyl, transition metal-catalysed coupling and neutral nucleophilic addition to carbonyl.
Exercise \(13\)
Fill in the blanks in the following synthesis involving anionic nucleophilic addition to carbonyl, conjugate addition, nucleophilic substitution at carboxyl, transition metal-catalysed coupling and neutral nucleophilic addition to carbonyl.
Exercise \(14\)
Fill in the blanks. Requires knowledge of formation of esters and amides as well as Wittig or Horner-Wadworth-Emmons reactions.
Exercise \(15\)
In each case, analyse whether the starting material is a nucleophile or electrophile. Fill in the product in part (c). This question requires knowledge of carbonyl conjugate addition.
These strategies are used in Fukumoto's synthesis of atisine (below).
Answer
Exercise \(16\)
Fill in the blanks in the following synthesis of atisine (Keiichiro Fukumoto, Tohoku University). Atisine is a natural product of Aconitum sp., the family of poisonous plants that includes wolfsbane.
Requires knowledge of conjugate addition / Michael reaction, ester reductions, formation of esters and amides, and Wittig or Horner-Wadworth-Emmons reactions. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/III%3A_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_1/05%3A_Substitution_at_Carboxyloids/5.01%3A_Biosy.txt |
Exercise 5.1.1:
These heteroatoms are all found in the upper right-hand corner of the periodic table. They are all pretty electronegative and they all have lone pairs.
Exercise 5.2.2:
1. The electronegativity of the heteroatom attached to the carbonyl group in a carboxyloid is one factor that allows it to leave and form its own stable anion.
2. Although carbon and hydrogen are more electronegative than many of the elements in the periodic table, they are not stable enough as anions to form easily on their own.
Exercise 5.2.3:
Exercise 5.2.4:
Exercise 5.3.1:
We might expect carboxyloids with the most electronegative elements attached to the carbonyl to be the most reactive and least stable towards substitution (in other words, carboxyloids with the most electronegative heteroatoms would become substituted the most easily).
In that case, we would predict that the carboxyloids with the most electronegative substituent (oxygen) would be the most reactive. There are a number of different kinds and we will think about how they relate to each other shortly.
After the oxygen derivatives we would predict either the nitrogen derivatives or the chloride, depending on what electronegativity scale we happen to use (remember, electronegativity is not an experimentally pure property, but the result of a calculation that can be performed in different ways). The sulfur derivative would be least reactive.
There are still several different oxygen derivatives to compare: carboxylic acids (OH), carboxylates (O-), esters (OR, in which R is an alkyl or carbon chain) and acid anhydrides (OC=O). The easiest to differentiate is the carboxylate, because of its negative charge. It must be less attractive to a nucleophile than the other oxygen derivatives, because it would offer more repulsion to an incoming lone pair.
However, we can't really predict whether it would be any less reactive than the nitrogen, chlorine or sulfur analogues, because who knows whether the charge or the nature of the atom matters more?
As it happens, the charge probably matters more. We learn that simply by looking at the experimental trend and seeing that the carboxylate is the least reactive of all the carboxyloids.
Turning to the other three oxygen derivatives, it would be difficult to differentiate between the effect of a remote hydrogen atom versus an alkyl chain in the ester versus the carboxylic acid, so we'll say those two are about the same. On the other hand, the additional electron-withdrawing carbonyl group in the acid anhydride probably has a profound effect, so we would expect that compound to attract nucleophiles more strongly.
Of course, the series we have produced above is not the "right answer". It does not match the experimentally observed series of carboxyloid reactivities. Nevertheless, it is very useful in terms of building an understanding of carboxyloids. It tells us that electronegativity may play a role here, but that it can't be the only factor.
Some other factor is putting some of the derivatives out of order. In particular, the acid chloride (C=OCl) and the thioester (C=OSR) do not fit.
Exercise 5.3.2:
Electronegativity is an abvious factor that could influence an atom's ability to π-donate, but we just looked at that factor in the previous section, so let's look at another atomic property instead. Of course, different atoms have different sizes. In particular, if we look at the atoms involved in carboxyloid substituents, we can divide them into 2nd row atoms and 3rd row atoms.
It's actually well-documented that the degree of overlap between two orbitals influences how well they bond together. Since carbon is in the second row, it is about the same size as, and overlaps pretty well with, other second row atoms. Third row atoms are a little too big, on the other hand.
That factor breaks the carboxyloids into two different groups. Assuming π-donation is a major factor, sulfur and chlorine may be placed above the others in tems of reactivity. They cannot donate as well as oxygen or nitrogen can.
From there, differences among the atoms from the same row may be sorted out based on electronegativity differences.
Exercise 5.3.3:
Amide bonds are among the most stable carboxyloids possible. That stability makes them well-suited to form useful structures that will not decompose easily. Remember, any change that occurs in matter occurs through chemical reactions, including the formation and decomposition of biomaterials. Shutting down a potential chemical reaction means a material will be more durable.
Exercise 5.3.4:
Exercise 5.3.5:
One possibility is the presence of an additional electronegative substituent. In the oxalyl chloride, the presence of an additional carbonyl next to the electrophilic acid chloride group would make each carbonyl even more electrophilic. In the thionyl chloride, the presence of two chlorines, instead of just one, could make this compound much less stable and more electrophilic.
Exercise 5.4.1:
Exercise 5.4.2:
Exercise 5.4.3:
a)
b)
Exercise 5.4.4
Exercise 5.4.5:
Exercise 5.4.6:
Exercise 5.4.7:
Exercise 5.5.4:
Exercise 5.5.8:
Exercise 5.6.1:
Because acid chlorides are at the top of the carboxyloid reactivity diagram (the ski hill), and other halides are likely to be similar in reactivity to the chloride, this reaction would be uphill from the other carboxyloids.
Exercise 5.6.4:
Exercise 5.6.5:
Amides and carboxylates are the least reactive carboxyloids, so it might not be too surprising that they do not react with these nucleophiles.
Exercise 5.6.6:
Acid chlorides typically react with these cuprate reagents.
Exercise 5.6.7:
Borohydrides could presumably react with acid chlorides, anhydrides and thioesters, which are the most reactive carboxyloids. They probably can't react with amids or carboxylate ions, which are even farther downhill than esters.
Exercise 5.6.11:
Exercise 5.7.1:
That combination would give thw following dipeptides:
ala-ala ala-gly ala-val
gly-gly gly-ala gly-val
val-val val-ala val-gly
Of course, we might also get tripeptides, such as ala-ala-gly-val, and so on.
Carboxylic acids usually require activation before they can act as nucleophiles. That problem is actually complicated here because the carboxylic acid is in equilibrium with a carboxylate salt (read further on the page).
The relative reactivity of carboxyloids results from a balance between sigma electron withdrawing effects and pi-donation. An electronegative atom attached to a carbonyl tends to withdraw electron density, making the carbonyl even more positive and electrophilic. On the other hand, pi-donation from a neighbouring atom with a lone pair actually lowers electrophilicity by forming a stable, conjugated system.
In a carbamate, an additional electronegative atom is added to the carbonyl: it has an oxygen as well as a nitrogen adjacent to the C=O group. That atom draws electron density away from the carbonyl, making it more electrophilic. However, pi-donation from the additional oxygen does not result in a more stable conjugated system. The maximum pi system is still just three atoms long; it either involves conjugation of the O-C=O or the N-C=O unit. It does not, for example, lead to an even more stable conjugated system that is four atoms long.
As a result, the added oxygen probably contributes more to the electron-withdrawing effect than it does to stabilisation of the pi system.
The more polar hydrochloride salt of EDCI is often used, as pictured. This more polar compound dissolves well in polar solvents, such as water, that also dissolve amino acids.
One can also imagine the amino group in EDCI acting as a site for catalysis, shuttling protons from one place to another.
Problem CX10.1.
This change in charge results because, although amines are easily protonated, amides are not. Protonation of an amide would result in a cation adjacent to the very positive carbonyl carbon, leading to a buildup of localized positive charge. That wouldn't be easy. Furthermore, the amide nitrogen is not very likely to donate its electrons to a proton in the first place. Its protons are too busy. They are tied up in conjugation with the carbonyl, so they really aren't available to act as the lone pair of a base. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/III%3A_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_1/05%3A_Substitution_at_Carboxyloids/5.04%3A_Solut.txt |
Aldehydes and ketones are carbonyl compounds in which the carbonyl carbon is connected only to carbons atoms (in the case of a ketone) or to one carbon and one hydrogen atom (in the case of an aldehyde).
Carboxyloids, or carboxylic acid derivatives, are carbonyl compounds in which the carbonyl carbon is attached to one carbon and one heteroatom. Most commonly, the heteroatom is an oxygen, nitrogen, sulfur or chlorine.
Carboxylic acid derivatives were historically thought of as being made from carboxylic acids; hence the name. That name is a mouthful; we will use the term "carboxyloids", a term coined by the 20th century physical organic chemist, Christopher Ingold.
The reactivity of carboxyloids is typically different from aldehydes and ketones. Because of this difference it is useful to study these compounds separately from the simple carbonyls.
Exercise \(1\)
What are some things that the heteroatoms involved in carboxyloids have in common?
Answer
These heteroatoms are all found in the upper right-hand corner of the periodic table. They are all pretty electronegative and they all have lone pairs.
Like simple carbonyls, carboxyloids react with nucleophiles. Just like simple carbonyls, the LUMO of a carboxyloid is usually the C=O pi antibonding orbital (the π*). Populating this orbital with a pair of electrons from the nucleophile results in breaking the C=O pi bond, leaving only a C-O sigma bond.
Remember that this addition to the C=O bond is often reversible. In the case of carboxyloids, however, re-forming the pi bond can go through two pathways. In one pathway, the nucleophile can be displaced, returning to starting materials. In another pathway, the group attached to the carbonyl can be displaced instead. In that pathway, a new product results.
5.06: Gener
Just as a simple carbonyl is an electrophile, so is a carboxyloid. Carboxyloids react with many of the same nucleophiles that react with aldehydes and ketones. The overall result of the reaction is very different. However, the mechanism of the reaction is really quite similar.
Nucleophiles have the overall effect of adding across the carbonyl group of an aldehyde or ketone. Addition of a proton produces an alcohol. The C=O bond is parlty broken to a C-O bond. In some cases, the C=O bond is completely broken and after several steps the carbonyl oxygen is replaced by some other heteroatom. On the other hand, nucleophiles add to carboxyloids and end up replacing the heteroatom next to the carbonyl. The carbonyl itself remains intact after the reaction is complete.
The general pattern in carboxyloid chemistry is for nucleophiles to substitute for the heteroatomic group next to the carbonyl. In these reactions, the group next to the carbonyl is sometimes referred to as a "leaving group". That term simply means that this group has been replaced by the end of the reaction.
Note that there is no reason to believe that the initial elementary reaction between a carboxyloid and a nucleophie is any different than that of a simple carbonyl with a nucleophile. An examination of the frontier orbitals in a carboxyloid suggests the pi antibonding level would be the site of population by a nucleophilic lone pair. The carbonyl pi bond would break as the nucleophile approaches.
However, an important feature of carbonyl chemistry is that two heteroatoms on one tetrahedral carbon cannot last. One always pushes the other off. If the former carbonyl oxygen pushes off the nucleophile, the system returns to the starting materials. However, if the former carbonyl oxygen displaces the heteroatomic group next to it, there is an overall change in bonding and a different product is formed. The net result is replacement of the group next to the carbonyl.
Exercise \(1\)
The overall result of reaction with a carboxyloid is to displace the heteroatom group next to the carbonyl. This group is typically liberated as an anion.
1. What feature of the heteroatoms typically found attached to the carbonyl in carboxyloids allows them to be displaced from the molecule as anions?
2. Why doesn't this same reaction happen with aldehydes and ketones?
Answer a
The electronegativity of the heteroatom attached to the carbonyl group in a carboxyloid is one factor that allows it to leave and form its own stable anion.
Answer b
Although carbon and hydrogen are more electronegative than many of the elements in the periodic table, they are not stable enough as anions to form easily on their own.
Exercise \(2\)
Suggest an order of reactivity for the carboxyloids: rank them from most reactive to least reactive. Provide a reason for your trend.
Exercise \(3\)
For the following reactions
a. Draw mechanistic arrows for the nucleophilic addition. Clearly show the tetrahedral structure formed immediately after the nucleophile adds.
b. Determine if a leaving group is present. If so, show the pi donation step and resulting product. If not, draw the product in the neutral form.
Exercise \(4\)
When acidic compounds such as HCl are produced in a reaction, the reaction is often followed by treatment with a base that will neutralise the acid. Provide Lewis-Kekule structures of the following bases.
a) lithium hydroxide, LiOH b) sodium hydroxide, NaOH c) potassium hydroxide, KOH
d) sodium bicarbonate, NaHCO3 e) sodium carbonate, Na2CO3 f) sodium acetate CH3CO2Na
g) sodium dihydrogen phosphate, NaH2PO4 h) sodium hydrogen phosphate, Na2HPO4 i) sodium phosphate, Na3PO4
j) pyridine, C5H5N k) triethylamine, (CH3CH2)3N
Answer | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/III%3A_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_1/05%3A_Substitution_at_Carboxyloids/5.05%3A_Intro.txt |
In most cases, the reactivity of carboxyloids involves converting one carboxyloid into another. This is done by replacing one heteroatom substituent with another. For example, the chloride in an acid chloride may be replaced by an alcohol or alkoxide ion to make an ester.
The fact that one carboxyloid can be converted into another suggests that there would be an equilibrium between them. The ratio of two carboxyloids at equilibrium would be determined by their relative stability, as well as the stability of other associated species in solution.
We can map out the stability of carboxyloids on a potential energy surface, as shown below. The higher energy, less stable, more reactive carboxyloids are shown at the top of the potential energy curve. The lower energy, more stable, less reactive carboxyloids are found lower down on the potential energy curve.
The heteroatom attached to the carbonyl in a carboxyloid is always an electronegative atom with a lone pair. Either of those two features might be useful in understanding the reactivity trend illustrated above. For example, an electronegative atom would make the carbonyl carbon more positive. That carbon is already very positive because of the double bond to oxygen. Adding an additional electronegative atom should make it even more so. The amount of positive charge on the carbonyl carbon would be even greater if the atom attached to it were exceptionally electronegative.
On the other hand, a nearby lone pair might counteract the electron-attracting power of the carbonyl carbon. In a sense, we might think about that lone pair as competing with donation from a potential nucleophile.
The ability of an atom to π-donate, then, might have an influence on how strongly the carbonyl will attract nucleophiles. Of course, there is some trade-off involved in π-donation. Usually the atom that donates must take on a positive charge, since it is lending a pair of its own electrons to another atom. Factors that influence how easily this may happen could be important in determining carboxyloid reactivity.
Exercise \(1\)
Based on electronegativity of the atom attached to the carbonyl carbon, we might expect a specific trend in carboxyloid reactivity. Explain how this factor would affect electrophilicity at the carbonyl carbon and predict the corresponding trend in reactivity. Compare this trend with the information in Figure \(1\) (CX3.1).
Answer
We might expect carboxyloids with the most electronegative elements attached to the carbonyl to be the most reactive and least stable towards substitution (in other words, carboxyloids with the most electronegative heteroatoms would become substituted the most easily).
In that case, we would predict that the carboxyloids with the most electronegative substituent (oxygen) would be the most reactive. There are a number of different kinds and we will think about how they relate to each other shortly.
After the oxygen derivatives we would predict either the nitrogen derivatives or the chloride, depending on what electronegativity scale we happen to use (remember, electronegativity is not an experimentally pure property, but the result of a calculation that can be performed in different ways). The sulfur derivative would be least reactive.
There are still several different oxygen derivatives to compare: carboxylic acids (OH), carboxylates (O-), esters (OR, in which R is an alkyl or carbon chain) and acid anhydrides (OC=O). The easiest to differentiate is the carboxylate, because of its negative charge. It must be less attractive to a nucleophile than the other oxygen derivatives, because it would offer more repulsion to an incoming lone pair.
However, we can't really predict whether it would be any less reactive than the nitrogen, chlorine or sulfur analogues, because who knows whether the charge or the nature of the atom matters more?
As it happens, the charge probably matters more. We learn that simply by looking at the experimental trend and seeing that the carboxylate is the least reactive of all the carboxyloids.
Turning to the other three oxygen derivatives, it would be difficult to differentiate between the effect of a remote hydrogen atom versus an alkyl chain in the ester versus the carboxylic acid, so we'll say those two are about the same. On the other hand, the additional electron-withdrawing carbonyl group in the acid anhydride probably has a profound effect, so we would expect that compound to attract nucleophiles more strongly.
Of course, the series we have produced above is not the "right answer". It does not match the experimentally observed series of carboxyloid reactivities. Nevertheless, it is very useful in terms of building an understanding of carboxyloids. It tells us that electronegativity may play a role here, but that it can't be the only factor.
Some other factor is putting some of the derivatives out of order. In particular, the acid chloride (C=OCl) and the thioester (C=OSR) do not fit.
Exercise \(2\)
Lone pair donation from the atom attached to the carbonyl carbon could also influence carboxyloid reactivity. Explain how this factor would affect electrophilicity at the carbonyl carbon and predict the corresponding trend in reactivity. Compare this trend with the information in Figure \(1\).
Answer
Electronegativity is an abvious factor that could influence an atom's ability to π-donate, but we just looked at that factor in the previous section, so let's look at another atomic property instead. Of course, different atoms have different sizes. In particular, if we look at the atoms involved in carboxyloid substituents, we can divide them into 2nd row atoms and 3rd row atoms.
It's actually well-documented that the degree of overlap between two orbitals influences how well they bond together. Since carbon is in the second row, it is about the same size as, and overlaps pretty well with, other second row atoms. Third row atoms are a little too big, on the other hand.
That factor breaks the carboxyloids into two different groups. Assuming π-donation is a major factor, sulfur and chlorine may be placed above the others in tems of reactivity. They cannot donate as well as oxygen or nitrogen can.
From there, differences among the atoms from the same row may be sorted out based on electronegativity differences.
Exercise \(3\)
Using the information in Figure \(1\), explain why peptides (containing a number of amide bonds, R(C=O)N) are such a common structural feature in biology.
Answer
Amide bonds are among the most stable carboxyloids possible. That stability makes them well-suited to form useful structures that will not decompose easily. Remember, any change that occurs in matter occurs through chemical reactions, including the formation and decomposition of biomaterials. Shutting down a potential chemical reaction means a material will be more durable.
The potential energy curve in Figure \(1\) (CX3.1) is a useful index for the interconversion of carboxyloids. In general, it is easy to go downhill on the curve, but more difficult to go uphill. That means that compounds lower down on the ski hill can be made easily from compounds farther up the ski hill.
In general, pi donation from the heteroatom attached to the carbonyl is a primary factor that determines carboxyloid reactivity. The more able the heteroatom is to donate its pi electrons, the less electrophilic is the carbonyl. Nitrogen is very good at donating its lone pair. It is about the same size as the carbon atom it needs to donate to, and it only a little more electronegative than the carbon.
Oxygen (in esters and carboxylic acids) is next in line, since oxygen is more electronegative than nitrogen.
Chlorine and sulfur are a little too large to donate very well to a carbon atom. The size and energy mismatch between these atoms leads to poor pi bonding, and poor pi donation.
Exercise \(4\)
Place the following compounds in their relative positions on the ski hill.
Exercise \(5\)
Oxalyl chloride (left) and thionyl chloride (right) are even higher on the ski hill than regular acid chlorides. Propose a reason that explains this relative instability as electrophiles. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/III%3A_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_1/05%3A_Substitution_at_Carboxyloids/5.07%3A_Compa.txt |
The potential energy curve linking the carboxyloids can be used as a guide to how these compounds can be interconverted. In general, it is possible to take a compound that is higher on the ski hill and convert it to a compound that is lower on the ski hill.
For example, acid chlorides are widely used to make other carboxyloids. By choosing the correct nucleophile, an acid chloride could be converted to any of the other derivatives. This is really the whole point of an acid chloride; it has no other function other than to provide an easy way to make other derivatives.
Acid anhydrides, also high on the potential energy curve, are also used in the same way. They could be used to make any of the derivatives lower than they are on the ski hill. In turn, acid anhydrides could conceivably be made from acid chlorides. However, because an acid anhydride plays the same role as an acid chloride -- providing a source with which to make the other derivatives -- we wouldn't normally make an acid anhydride from an acid chloride. We would either make one of the other derivatives directly from the acid chloride, or else make it from an acid anhydride that was obtained in another way.
Exercise \(1\)
Show all of the ways that you could make the following compounds by going down the ski hill. (Show the starting material, the reagent added and a reaction arrow going to the product.) For help with names, see the appendix.
1. ethyl propanoate
2. butanoic acid
3. N-methylhexanamide
4. thioethyl pentanoate
5. potassium octanoate
Answer
Exercise \(2\)
Converting a carboxylic acid into an amide is complicated by a side reaction, as a result of which amide formation becomes an uphill process.
1. Show the mechanism for the conversion of butanoic acid into N-ethylbutanamide via the addition of ethylamine.
2. Show the side reaction that would easily occur between these two reactants.
3. Explain why amide formation becomes an uphill process as a result of this reaction.
Answer
Sometimes, two carboxyloids are close enough on the ski hill that it may be possible to convert in either direction between them. In other words, there is an equilibrium between these two compounds, and the equilibrium constant is close enough to unity (K = 1) that the equilibrium can be pushed in either direction.
For selected cases, we can choose which direction the equilibrium will go by changing the reaction conditions. To do so, we can use an important concept of equilibrium: le Chatelier's principle (luh sha-TELL-yay). According to le Chatelier, if conditions in the reaction cause the reaction to move away from equilibrium, the reaction will shift direction until it is back at equilibrium again. In other words, if the actual ratio of reactants to products strays from what it ought to be, the correct reaction will occur so that the ratio returns to normal.
le Chatelier's principle should be partly intuitive. If a reaction is occurring in both reactions, there are really two reactions, one going in each direction. The ability of these reactions to occur depends partly on the amount of reactants available for the forward reaction or the reverse reaction. If extra starting materials are added (on the left side of the reaction), there is too much reactant as defined by the equilibrium constant. The denominator gets bigger and the ratio of products to reactants goes down. However, because there is extra starting material for the forward reaction, more product is quickly made, until the ratio returns to normal.
The opposite situation applies if too much product is made. From the point of view of the reverse reaction, those products of the forward reaction are really the reactants needed to go in the opposite direction. The reverse reaction has more material to work with, and this material can quickly be converted into the stuff on the left hand side of the reaction.
Exercise \(3\)
Draw the mechanism for the conversion of:
1. hexyl propanoate to propanoic acid.
2. propanoic acid to hexyl propanoate.
Answer a
Answer b
In the interconversion of carboxyloids, equilibrium can be influenced in different ways. The conversion between esters and carboxylic acids can be influenced by the solvent used for the reaction. For example, an ester might be converted to a carboxylic acid under aqueous conditions, but a carboxylic acid might be converted to an ester using the appropriate alcohol as the solvent. In other words, because water and alcohol can be viewed as reactants in these cases, adding more can shift the reaction to one side. Solvent is usually present in concentrations many tens or hundreds or thousands of times higher than the reactants and products. Changing the solvent thus has a big impact on the direction of equilibrium.
Another strategy used to influence the equilibrium involves removal of product. For example, if water is a product of the reaction, a drying agent can be added to absorb the water. Drying agents include compounds such as MgSO4 or zeolites (mixed aluminosilicates containing Al, Si, O and other metal ions such as Mg2+, Ca2+ or Ti4+).
If a carboxylic acid is a product of a reaction, its concentration can be lowered by converting it to a carboxylate salt; this would happen easily in the presence of base. In either case, the disappearance of a product of the reaction (or a side product) would draw the reaction to the right in order to replace those products an re-establish the correct equilibrium ratio.
Exercise \(4\)
Show, with curved arrows, how a magnesium ion could remove water from solution.
Answer
Exercise \(5\)
Show, with curved arrows, how sodium hydroxide would remove butanoic acid from solution.
Answer
Exercise \(6\)
For each of the following reactions:
a. Predict the nucleophile that would be need to make this product.
b. Based on the location on the hill, would this reaction be thermodynamically favored?
Exercise \(7\)
Fill in the missing intermediates and add curved arrows to show electron movement.
5.09: Getti
The uphill carboxyloids are useful materials in terms of being able to make other compounds. For example, a thioester such as acetyl coenzyme A may be able to make a variety of acyl esters. In the laboratory, acid chlorides are very common starting materials to make other carboxyloids. However, if they are so far uphill, how are they formed in the first place?
The two most common methods of making acid chlorides are treatment with thionyl chloride or with oxalyl chloride.
The key part of making the uphill acid chloride out of the downhill carboxylic acid is the reagent used. Structurally, the reagents can be compared to acid chlorides themselves. They can be though of as being a little bit like uphill carboxyloids themselves. Thus, as one compound gives its chloride and gets oxygenated on its way downhill, it provides the energy needed to drive the carboxylic acid uphill.
Exercise \(1\)
Reaction of a carboxylic acid with oxalyl chloride starts with a carboxyloid substitution using the oxalyl chloride as electrophile and carboxylic acid as nucleophile. The chloride ion that is liberated then acts as a nucleophile in a cascade reaction that releases CO2 and CO as an acid chloride forms. Draw the mechanism.
Exercise \(2\)
Reaction of a carboxylic acid with thionyl chloride is very similar to reaction with oxalyl chloride, described above. Draw the mechanism of the reaction.
Exercise \(3\)
Provide additional factors (such as energetics, equilibrium concepts) that explain why oxalyl chloride and thionyl chloride can drive the conversion of a carboxylic acid to an acid chloride.
Acid anhydrides are usually made from the corresponding carboxylic acids. One molecule of carboxylic acid acts as a nucleophile and a second acts as an electrophile. Because the same kind of molecule acts as nucleophile and electrophile, acid anhydrides are typically symmetric: they havae a (C=O)O(C=O) unit in the middle, with the same alkyl groups on either side of it.
To aid formation of acid anhydrides, carboxylic acids are often heated strongly (well above 100 oC). Otherwise, they are sometimes heated in the presence of a strong drying agent, such as phosphorus pentoxide (empirically, P2O5). In the presence of water, phosphorus pentoxide is converted to phosphoric acid, H3PO4.
Exercise \(4\)
Draw a mechanism for the conversion of ethanoic acid to ethanoic anhydride.
Answer
Exercise \(5\)
Show the reverse reaction to the conversion of ethanoic acid to ethanoic anhydride (the same reaction, in the other direction). Which of the two directions do you think is favorable, based on the ski hill?
Exercise \(6\)
Draw a mechanism for the conversion of phosphorus pentoxide to phosphoric acid.
Exercise \(7\)
Explain why the conditions outlined above lead to acid anhydride formation.
Exercise \(8\)
Fill in the blanks in the following problem. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/III%3A_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_1/05%3A_Substitution_at_Carboxyloids/5.08%3A_Inter.txt |
Carboxyloids can be interconverted through addition of typical heteroatomic nucleophiles: amines, alcohols, thiols and water. In addition, other nucleophiles can displace the "leaving group" on a carboxyloid, provided the nucleophile is reactive enough.
Exercise \(1\)
Could a halide, such as bromide or chloride, replace a carboxyloid leaving group easily? Explain.
Answer
Because acid chlorides are at the top of the carboxyloid reactivity diagram (the ski hill), and other halides are likely to be similar in reactivity to the chloride, this reaction would be uphill from the other carboxyloids.
Carbon and hydrogen anions (or "semianions") are very good nucleophiles. Earlier, we saw how they can react with simple carbonyls. Although the lone pair on a carbon or a hydrogen is often masked in a covalent bond with a moderately electropositive metal such as aluminum or magnesium, that bonding pair of electrons is still nucleophilic enough to donate to a good electrophile. These nucleophiles can often react with carboxylic acid derivatives.
Exercise \(2\)
Draw a mechanism for the replacement of the chloride in propanoyl chloride with a hydride from lithium aluminum hydride.
Exercise \(3\)
Draw a mechanism for the replacement of the chloride in propanoyl chloride with a methyl from methylmagnesium chloride.
Addition of a Grignard reagent (alkylmagnesium halide) to a carboxyloid results in the formation of a ketone. Addition of a complex hydride reagent (such as lithium aluminum hydride) to a carboxyloid results in the formation of an aldehyde. We have already seen that these reagents can add to aldehydes and ketones to afford alcohols.
So, what happens if a Grignard reagent is added to an acid chloride or an acid anhydride? The carboxyloid would be converted to a ketone. If there are still more Grignard molecules around, they would probably convert the ketone into an alkoxide ion (and ultimately an alcohol via protonation). The thing is, it is very likely that there will be more Grignard molecules around. A reaction tends to involve millions of reactant molecules, so by the time the first thousand or so molecules of carboxyloid have been converted to ketone, hundreds of those ketone molecules have already been converted to alkoxide.
In most cases, alkyl reagents and hydride reagents will add twice to carboxyloids. They will convert the carboxyloid into an aldehyde or ketone. Because aldehydes and ketones also react with hydride and alkyl nucleophiles, they will react a second time.
Exercise \(4\)
Grignard reagents will not effect leaving group replacement in carboxylic acids. Show why that particular reaction does not occur, with the help of a mechanism.
Answer
Exercise \(5\)
Grignard reagents generally do not react with either amides or carboxylate ions. Explain why.
Answer
Amides and carboxylates are the least reactive carboxyloids, so it might not be too surprising that they do not react with these nucleophiles.
When a series of compounds varies from more reactive to less reactive in a particular reaction type, the possibility for selectivity arises. Some reagents may react with a few carboxyloids, but not with others. For example, organocuprates such as (CH3)2CuLi, which do not generally react well with aldehydes and ketones, are very selective in terms of which carboxyloids they will react with.
Exercise \(6\)
Which carboxyloids do you think will react with (CH3)2CuLi? Show the reaction in each case (the reactant, reagent, and a reaction arrow going to the product). Explain your reasoning.
Answer
Acid chlorides typically react with these cuprate reagents.
Exercise \(7\)
Complex hydride reagents can be very selective towards carboxyloids. For example, sodium borohydride is not powerful enough to react with esters.
1. Which of the carboxyloids can sodium borohydride react with? Explain.
2. What other carboxyloids can sodium borohydride NOT react with? Explain.
Answer
Borohydrides could presumably react with acid chlorides, anhydrides and thioesters, which are the most reactive carboxyloids. They probably can't react with amids or carboxylate ions, which are even farther downhill than esters.
Exercise \(8\)
Lithium aluminum hydride can induce carboxylic substitution with carboxylate salts such as sodium octanoate.
1. What would be the ultimate product of this reaction? Explain.
2. What other carboxyloids can lithium aluminum hydride react with? Explain.
Exercise \(9\)
a) Which is more reactive: lithium aluminum hydride or sodium borohydride?
1. Which is more selective: lithium aluminum hydride or sodium borohydride?
2. How can you explain the difference in reactivity between lithium aluminum hydride and sodium borohydride?
Exercise \(10\)
The reaction of lithium aluminum hydride with amides is unusual in that the final product of the reaction is generally an amine.
1. Why does this reaction seem to be different from other carboxyloid reactions?
2. Draw a mechanism for this reaction. (Hint: at some point, an oxygen atom donates a pair of electrons to aluminum.)
3. Propose a reason why this hydride reaction follows a different path than other reactions of hydrides with carboxyloids.
Exercise \(11\)
Fill in the products of the following reactions.
5.11: Enola
Alkylmagnesium reagents, alkylcuprates and complex hydrides can all react with carboxyloids. When they do, a carbon or hydrogen nucleophile bonds to the carbonyl carbon, usually replacing the leaving group at that position.
Another common carbon nucleophile is an enolate ion. Enolate ions can also react with carboxyloids, although not typically with amides.
Probably the most common enolate reaction involving carboxyloids is the reaction of esters. If a strong base is added to solution of ester, some of the esters will become deprotonated, forming enolate anions. These ions will be nucleophiles.
Some esters will remain protonated. These esters will be electrophiles. Donation of the enolate to the ester, with subsequent loss of the leaving group, leads to a beta-ketoester.
You are familiar with the term "alpha-position". That's the position next to a carbonyl. The "beta-position" is the next one after the alpha position. In a beta-ketoester, there is a ketone in the beta position of the ester. The formation of a beta-ketoester from two esters is called a "Claisen condensation".
In principle, this reaction could conceivably go backwards. The enolate ion could potentially be displaced by an alkoxide to get back to an ester and an enolate ion. That's because the enolate ion is a relatively stable ion, and a moderately good leaving group. However, that generally doesn't happen.
Under basic conditions, the beta-ketoester is usually deprotonated, forming a particularly stable ion. This ion formation acts as a "thermodynamic sink" for the reaction, pulling it forward until all of the ester has been consumed.
Exercise \(1\)
Show why the ion that results from deprotonation of the beta-ketoester is particularly stable.
Answer
Exercise \(2\)
Fill in the products of the following reactions.
Exercise \(3\)
Predict the reactants needed to make these products via a Claisen, Aldol or Crossed Aldol reaction.
The formation of a beta-ketoester from two esters is called a "Claisen condensation".
It is often followed by another important reaction: decarboxylation. If a beta-ketoester is treated with aqueous acid and heated, a couple of reactions take place. First, the ester portion of the molecule is converted into a carboxylic acid.
Second, the carboxylic acid is decarboxylated. Carbon dioxide is formed, and the organic molecule becomes a ketone. The carboxyl group is lost completely from the original molecule, and is converted into CO2.
Decarboxylation is related to the retro-aldol reaction; formally, it can be thought of as leading to an enolate leaving group. Decarboxylation most commonly occurs in beta-ketoacids, rather than in other carboxylic acids. Otherwise, that leaving group could not occur. The ease of decarboxylation in beta-ketoacids is related to the stability of the enolate anion.
Under acidic conditions, of course, an enolate anion does not occur; instead, an enol is formed. However, enols are rapidly converted into the keto tautomers.
Exercise \(4\)
Draw a mechanism for:
1. Conversion of ethyl-3-oxyhexanoate into 3-oxyhexanoic acid. (Oxy is a prefix meaning a ketone or aldehyde is foundalong the chain).
2. Decarboxylation of the resulting 3-oxyhexanoic acid.
Exercise \(5\)
Fill in the products of the following reactions. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/III%3A_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_1/05%3A_Substitution_at_Carboxyloids/5.10%3A_Semi-.txt |
Carboxyloids, such as esters, can interconvert with each other in the presence of the appropriate nucleophile. In the case of esters, an equilibrium will result. If an alcohol is added to an ester, under the right conditions we might get a new ester. A new alcohol would also appear, originating from the old OR group of the original ester. Two different esters and two different alcohols would be in equilibrium. This equilibrium might be perturbed in one direction or another, for example, by the addition of an excess of one nucleophile.
In the jargon of synthetic organic chemistry, an ester is a functional group. It is a site on the molecule at which reactions take place. It is also a site on the molecule that is easily subject to synthetic transformations. In other words, one functional group might easily be converted into another. In the case of a trans-esterification reaction, one ester simply gets converted into another.
Some compounds have more than one functional group. A compound could be both an ester and an alcohol, for instance. We might call this compound a hydroxyester. But wait -- if an alcohol can react with an ester, to make a new ester, doesn't this compound have both components of a reaction built in? Could it react with itself?
There are a couple of ways that could happen. If the chain between the carbonyl and the hydroxyl group is long enough (remember, a six-atom interaction between the hydroxyl oxygen and the carbonyl carbon may be optimal), the hydroxyl could wrap around and form a cyclic ester. That's an intramolecular reaction -- a reaction within one molecule.
Alternatively, if there is another one of these molecules around someplace, an intermolecular reaction might occur. That's a reaction between two different molecules. The hydroxyl group on one molecule can reach out and react with the carbonyl on another molecule.
Now there are two molecules of the same kind bonded to each other. This double molecule is called a dimer. The individual molecules that have been linked togethr to make the dimer are called monomers.
That dimer has two esters in it, not just one. Of course, it still has a hydroxyl group on one end. That hydroxyl group can still react with another carbonyl on another molecule.
Now there are three molecules bonded together. This molecule is called a trimer.
This process could keep going on indefinitely, of course. We might end up with a very large molecule, composed of many individual (former) molecules that have bonded together. This very large molecule made up of repeating units is called a polymer. A polymer is built up from many monomers linked together. This particular kind of polymer is called a polyester.
This polymer is frequently drawn in a way that emphasizes the repeating pattern of monomers that have been incorporated into the chain.
Polyesters can also be made by co-polymerizing two different monomers together. One monomer could be a diester, for example. The other monomer could be a diol. These two molecules are ready to react together, with one molecule acting as an electrophile and the other molecule acting as the nucleophile.
Together, the diester and the diol could be polymerized. The result would be an alternating copolymer, in which diester and diol monomer units alternate all along the polymer chain.
Polymers make up an important class of materials with many uses. Many polymers are lightweight, strong materials used to make parts for automobiles and other products. Polymers can also be very flexible or elastic. The physical properties of polymers are very different from the properties of other molecular compounds. These differences are a direct result of the very large size of polymer molecules. A polymer molecule might be thousands of monomers long, with a molecular weight in the millions.
Exercise \(1\)
Problem CX8.1.
Express the following polymer structures in abbreviated structures showing n repeating units in parentheses.
Exercise \(2\)
Show the structures of the polymers that would result from the following monomers. In each case, show a drawing with several enchained monomers.
Exercise \(3\)
Ring-opening polymerization involves a multi-step reaction in which a cyclic compound, such as a lactone (below) is opened into a chain through the addition of a nucleophile (called the "initiator"). The resulting chain is able to act as a nucleophile and open the next lactone, and so on, until a polymer has formed. Show a mechanism for formation of the oligomer in which n = 3.
Exercise \(4\)
Ring-opening polymerizations are frequently accelerated through the addition of small amounts of metal compounds, such as diethylzinc (Et2Zn) or tin octoate (Sn(O2CCH(CH2CH3)CH2CH2CH2CH3)2. Explain the role of these compounds in the reaction mechanism.
Exercise \(5\)
Karen Wooley at Texas A&M recently reported the following synthesis of a polyphosphoramidate for use as a pharmaceutical delivery agent. The goal is to use a benign delivery agent that is easily broken down and excreted by the body, resulting in low toxicity and minimal side effects.
1. Provide a mechanism for the synthesis.
2. Explain why the polyphosphoramidate is expected to be broken down and excreted easily by the body.
Difunctional molecules can sometimes polymerise, provided they have appropriate partners with which they can react on other molecules. For example, hydroxyesters might react with other hydoxyesters, with the hydroxyl group on one molecule reacting with the ester group on another, forming a polyester. Alternatively, there might be two different kinds of molecules that react with each other. For example, a diamine might react with a diacid chloride to form an amide.
In ring-opening polymerisation, the monomer is not difunctional. Instead, it is embedded in a ring. Ideally, there is a little ring strain in the molecule, bumping it up in energy just a little so that it will react more easily. Common examples include caprolactone and lactide, used to make biodegradable yard waste bags and produce containers, respectively (among many other applications). These cyclic esters are sometimes referred to as "lactones".
If an alcohol is added, it can act as an "initiator" in a "chain reaction". The alcohol is a nucleophile, and it donates to the carbonyl, eventually cleaving the carboxyl C-O bond and popping open the ring.
At some point, a proton gets transferred to the oxygen that used to be embedded in the ring. Now we have a new alcohol. What does it do? It reacts with another cyclic ester, popping it open and forming a new alcohol. The cycle repeats itself.
• A chain reaction keeps happenning over and over again.
• A chain reaction must be started by an initiator.
• A chain reaction leads to a product that looks just like an earlier reactant, so the product reacts again.
In reality, ring opening polymerisations don't really work if you just add an alcohol to a lactone. Typically, a catalyst is also added. Catalysts most commonly are Lewis acids, such as aluminum, iron or tin compounds. One of the most common catalysts is tin octoate, more properly called tin(II) 2-ethylhexanoate.
Exercise \(6\)
Provide a mechanism, with arrows, for the ring-opening polymerisation of caprolactone with tin octoate.
Exercise \(7\)
Perform end-group analysis in the following cases to determine
1. the degree of polymerisation (what is the value of "n"?).
2. the molecular weight.
i. The ratio of the integrals for the 1H NMR peaks representing positions b:a is 50:1.
ii. The ratio of the integrals for the 1H NMR peaks representing positions b:a is 80:1. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/III%3A_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_1/05%3A_Substitution_at_Carboxyloids/5.12%3A_Conde.txt |
Peptides and proteins are very important in biology. As a result, synthesis of these molecules has become very important, allowing for the laboratory study of model compounds that can give us insight into how proteins work, as well as pharmaceutically important compounds.
Insulin and glucagon are two important peptides (or small proteins) that regulate blood sugar. Insulin signals that blood sugar levels are high and that the body should begin storing this excess sugar. Glucagon signals that blood sugar levels are low and so the body may need to access its long-term energy stores. Both compounds are important in medicine.
Glucagon has a relatively simple structure, for a protein. It is a string of just 29 amino acids, connected together in the following order:
His-Ser-Gln-Gly-Thr-Phe-Thr-Ser-Asp-Tyr-Ser-Lys-Tyr-Leu-Asp-Ser-Arg-Arg-Ala-Gln-Asp-Phe-Val-Gln-Trp-Leu-Met-Asn-Thr.
The structure is composed of readily-available starting materials. The amino acids are simple units that connect together to form a chain. Conceptually, it seems easy to put two amino acids together to make a side chain. When two amino acids come together, they lose a molecule of water together, and the remaining pieces are able to bond to each other to make the dipeptide. It ought to be just as easy to add a third, and so on.
Structurally, amide or peptide bonds are very stable and resistant to carboxyl substitution. That stability makes optimal structures from which to construct proteins. Despite being composed of very long chains of linked amino acids, proteins actually have some limits on their conformational flexibility (their "floppiness"). That allows proteins to more reliably hold a particular shape. The shape of proteins is crucial to their function as enzymes and other
Both the stability and the structural rigidity of peptides arises from the nature of the peptide bond. The pi donation that hinders nucleophiles from substituting at the carbonyl is pronounced enough that it can be considered to form an additional bond. Thus, peptides behave as though they contain C=N bonds rather than C-N bonds. X-ray structure determinations show that the peptide nitrogens in proteins are trigonal planar, not pyramidal. In addition, many peptides exhibit cis-trans isomerism. For every peptide bond, two different isomers can occur, depending on whether a substituent attached to nitrogen is on the same side of the C=N bond as the carbonyl oxygen or the opposite side.
The great stability of these structures does not mean they are easy to make. Part of the difficulty stems from the fact that amino acids are difunctional. In order to form long chain structures, amino acids must be able to react twice: once with an amine, to grow in one direction, and once with a carboxylic acid to grow in the other direction. In other words, an amino acid contains both a nucleophile and an electrophile.
Suppose we were to try to make the dipeptide, ala-phe. This peptide contains an alanine connected to a phenylalanine through a peptide bond. The peptide bond is formed between the carboxylic acid of alanine and the amine of phenylalanine.
Assuming the amino acids do react together to form the peptide, combining these two reactants would likely produce a mixture of four dipeptides:
Ala-Phe Ala-Ala Phe-Phe Phe-Ala
In other words, peptide formation from amino acids is non-selective.
Exercise \(1\)
Draw structures for the four peptides formed by combining glycine and leucine.
Answer
Exercise \(2\)
What tripeptides would be produced by mixing Ala, Gly and Val?
Answer
That combination would give thw following dipeptides:
ala-ala ala-gly ala-val
gly-gly gly-ala gly-val
val-val val-ala val-gly
Of course, we might also get tripeptides, such as ala-ala-gly-val, and so on.
Exercise \(3\)
Simply combining these peptides might not result in any peptide formation at all. Why not?
Answer
Carboxylic acids usually require activation before they can act as nucleophiles. That problem is actually complicated here because the carboxylic acid is in equilibrium with a carboxylate salt (read further on the page).
An additional complication in peptide synthesis is that amines and carboxylic acids do not really exist together. Instead, a proton is transferred from the carboxylic acid to the amine, forming a salt. The carboxylate is no longer very electrophilic, due to its negative charge. Because of its positive charge, the ammonium ion is no longer very nucleophilic.
As a result, there are actually two distinct problems in peptide synthesis. There is a selectivity problem, because each amino acid has a nucleophilic part and an electrophilic part. There is no way to ask one compound to react only using its electrophile and another compound to react only using its nucleophile. There is also a reactivity problem: the carboxyl group in this case is a terrible electrophile, and the amine is a terrible nucleophile.
In laboratory syntheses, a number of techniques have been used to make peptide synthesis selective. Most frequently, protecting groups are used. A protecting group "masks" one of the two functional groups on an amino acid, but leaves the other one open. If one amino acid has its amine protected, it can only react via its carboxylic acid. If the other amino acid has its carboxylic acid protected, it can only react via its amino group. Only one combination will result.
The key to protecting groups is that the reaction used to mask one of the functional groups must be reversible. You must be able to take the protecting group back off when it is no longer needed.
Carboxylic acids are normally protected as esters. Esters can be removed via acid- or base-catalyzed hydrolysis (as can amides, but esters are more reactive, being farther up the ski hill).
Amines are normally protected as amides. However, we need to be able to remove specific amides her: the ones that mask the amines, not the ones that we have formed to link two amino acids together. As a result, in peptide synthesis, amines are usually protected as carbamates. Carbamates can be cleaved more easily than amides.
Exercise \(4\)
Propose a reason for the relatively higher reactivity of carbamates compared to amides.
Answer
The relative reactivity of carboxyloids results from a balance between sigma electron withdrawing effects and pi-donation. An electronegative atom attached to a carbonyl tends to withdraw electron density, making the carbonyl even more positive and electrophilic. On the other hand, pi-donation from a neighbouring atom with a lone pair actually lowers electrophilicity by forming a stable, conjugated system.
In a carbamate, an additional electronegative atom is added to the carbonyl: it has an oxygen as well as a nitrogen adjacent to the C=O group. That atom draws electron density away from the carbonyl, making it more electrophilic. However, pi-donation from the additional oxygen does not result in a more stable conjugated system. The maximum pi system is still just three atoms long; it either involves conjugation of the O-C=O or the N-C=O unit. It does not, for example, lead to an even more stable conjugated system that is four atoms long.
As a result, the added oxygen probably contributes more to the electron-withdrawing effect than it does to stabilisation of the pi system.
Exercise \(5\)
Fill in the blanks in the following peptide synthesis.
To get around the problem of low electrophilicity of the carboxylic acid, a number of coupling agents have been developed. A coupling agent can temporarily convert the carboxylate anion into a more reactive electrophile. To do so, it exploits the nucleophilicity of the carboxylate anion. After donating to the coupling agent, the carbonyl compound becomes more electrophilic.
Thionyl chloride (SOCl2) can accomplish this goal, of course. It converts relatively non-electrophilic carboxylic acids into much more electrophilic acid chlorides. Thionyl chloride can be a little harsh, however, so chemists have sought to develop milder conditions that can knit two amino acids together without the necessity of forming a reactive acid chloride.
Some of the most commonly-used coupling agents for peptide synthesis are carbodiimides. These compounds contain an electrophilic N=C=N unit to act as an initial electrophile. Diisopropylcarbodiimide (DIC) and dicyclohexylacarbodiimide (DCC) were some of the earliest and simplest examples of these compounds developed for peptide synthesis.
Once a carboxylate has donated to the electrophilic carbon of the carbodiimide, a better leaving group is formed. The adjacent nitrogen atoms act as basic sites, picking up a proton from the carboxylic acid on one amino acid and from the ammonium ion intermediate formed by the other amino acid.
Exercise \(6\)
Propose an advantage of EDCI as a coupling agent for amino acids. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/III%3A_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_1/05%3A_Substitution_at_Carboxyloids/5.13%3A_Pepti.txt |
If you look around the laundry room cupboard, you might see some different household products that contain enzymes. That might seem a little strange at first; enzymes are found in the body, and have no business being in stain remover or drain cleaner. But enzymes are just molecules that are really, really good at performing specific tasks. In these household applications, enzymes are really good at breaking specific bonds and replacing them with new ones. In doing so, they are breaking down larger molecules, which might be causing a stain or a slow drain. Those bigger molecules turn into smaller molecules that can be washed away more easily.
What kind of molecules are enzymes? They are special proteins. Proteins play lots of roles in biology; they may provide structure, for example, or they might catalyse reactions. Enzymes are proteins that catalyse reactions. Proteins belong to a class of molecules called biopoylmers. Biopolymers are large molecules composed of smaller molecules that have been bonded together. The small building blocks of proteins are the amino acids. One of the important features of amino acids is the fact that they are chiral; they have very specific three-dimensional shapes, like a left-handed glove or a right-handed glove.
• Enzymes are proteins
• Proteins are composed of chiral amino acids
• Enzymes are great, big, chiral molecules
How big are these proteins, exactly? The biggest protein is titin, which is partly responsible for the elastic properties of muscle. Titin is composed of about 35,000 individual amino acids. The molecular weight of an amino acid is roughly 100 Da (100 Daltons or 100 amu), so that means titin has a molecular weight of about 35 million Da.
But titin is one of those structural proteins. Let's look at an enzyme. Trypsin is a really common enzyme, found in the gut. The picture below comes from a structure determination experiment involving pig trypsin. It's made of about 450 amino acids, so its molecular weight is about 45,000 Da.
That picture shows every atom other than the hydrogens in trypsin; the hydrogens are too small to bother with. Even so, it's a bit busy. More often we look at enzymes and other proteins in a simplified, cartoon form. The cartoon model of trypsin below highlights the beta sheets and alpha helices, the secondary structures within trypsin. In the cartoon model, you can get a better sense of the shape of things. You can see gaps and grooves in the protein, and you can see the chirality in those helices.
Trypsin belongs to a class of enzymes called hydrolases. Hydrolases break bonds with the help of a water molecule. Trypsin breaks specific amide or peptide bonds in other proteins (yes, it's a cannibal), converting the carbonyl side of the bond into a carboxylic acid. That starts breaking down proteins in food; it's an essential part of digestion.
But how does trypsin do that?
At the most basic level, enzymes bind their targets, called substrates, perform chemical reactions on them, and let them go. Binding the substrate is a key step, and it's one of the reasons the chirality of enzymes can be important. Some enzymes are extremely specific in what substrates they bind. For example, they might bind one enantiomer and not another.
Trypsin is also very specific; it specializes in breaking the amide bond next to either arginine or lysine residues. The reaction that it catalyses is not necessarily specific to that particular peptide bond, but the trypsin selectively binds at that site.
This role of specific binding in enzymes is often referred to as the "lock and key" mechanism. We think of enzymes as being unlocked or turned on when the proper key is inserted, causing the enzyme to spring into action. In the following sections, we'll take a closer look at how enzymes bind their substrates and carry out their reactions.
Exercise \(1\)
Problem EZ1.1.
Enzymes often have two-part names: the first part identifies the substrate, and the second part describes what the enzyme does with that substrate. See if you can tell what an enzyme would do if one of the following words was in its name:
Match each item in one column to an item in the other column.
1. isomerase i) add oxygen from O2 into a molecule
2. hydrolase ii) add both oxygens from O2 into a molecule
3. oxygenase iii) oxidize or remove electrons from the substrate
4. dioxygenase iv) reduce or add electrons to the substrate
5. reductase v) reorganize atoms from one isomer into another
6. transferase vi) cleave a phosphate group off a protein
7. phosphatase vii) add water into a molecule, helping to break it down
8. oxidase viii) cause two molecules to be bound together
9. ligase ix) transfer a functional group to or from a molecule
Answer
a) isomerase v) reorganize atoms from one isomer into another
b) hydrolase vii) add water into a molecule, helping to break it down
c) oxygenase i) add oxygen from O2 into a molecule
d) dioxygenase ii) add both oxygens from O2 into a molecule
e) reductase iv) reduce or add electrons to the substrate
f) transferase ix) transfer a functional group to or from a molecule
g) phosphatase vi) cleave a phosphate group off a protein
h) oxidase iii) oxidize or remove electrons from the substrate
i) ligase viii) cause two molecules to be bound together
Trypsin Structure: PDB ID: 4AN7
Patil, D.N., Chaudhary, A., Sharma, A.K., Tomar, S., Kumar, P. Structural Basis for Dual Inhibitory Role of Tamarind Kunitz Inhibitor (Tki) Against Factor Xa and Trypsin. FEBS J. 2012, 279, 4547. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/III%3A_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_1/06%3A_Enzyme_Catalysis/6.01%3A_Introduction_to_E.txt |
The lock-and-key model of enzymes says that enzymes bind specific molecules and carry out reactions on those molecules. The enzyme recognizes the shape of its substrate and it is able to hold it in position in what is called the active site. The active site is the part of the enzyme that binds the substrate and carries out the reaction.
Enzyme specificity means that the enzyme only binds certain molecules that have the right shape. It will leave other molecules alone.
Once the molecule is bound, the enzyme carries out some change in the molecule. It might add new pieces on to the molecule or else take pieces off. After it carries out its work on the substrate, the products will be released. The enzyme is then regenerated into its original form and is ready to bind another substrate.
An enzyme is an example of a catalyst. It provides an alternative pathway for a reaction. It gives the substrate the tools to undergo a reaction that might not be possible otherwise. As a result, the reaction occurs much more quickly in the presence of the enzyme. However, one of the key features of a catalyst is that it remains unchanged at the end of the reaction (or, strictly speaking, it is regenerated into its original form). When the enzyme is finished with the substrate, it is ready to bind another one and repeat the process.
Exercise \(1\)
Identify the substrate that fits in each enzyme below.
So far, we have been looking at some cartoons to get the idea of enzyme-substrate specificity. If you know anything about small, organic molecules, you may have some ideas about how their shapes might vary in reality. Maybe the substrate is chiral; the enzyme might bind the substrate but not its enantiomer or diastereomer, because those molecules do not have exactly the same shape. Fitting the wrong enantiomer into an enzyme might be like fitting a left hand into a right-handed glove.
Otherwise, functional groups on the substrate play a key role in binding to the enzyme. The functional groups are responsible for the intermolecular attractions between the substrate and the active site. We have been thinking of the active site as a very specific shape, like a circle or a square. In reality, it is often a fold or an opening in the enzyme, a space where a small molecule might settle down. The amino acid side chains in that active site are situated to hold the substrate in place. Once everything is in place, the enzyme can get to work.
Exercise \(2\)
The following are some hypothetical small molecules and their potential binding sites. In each case, orient the molecule within the binding site in order to maximize binding interactions.
Things may be more subtle, still. Sometimes, binding a molecule causes a change in the shape of the enzyme. Enzymes are very large molecules. They are composed of long chains of amino acids; you can pictures those chains sliding past each other to make room for a guest molecule or to bind it more tightly. Frequently, enzymes are composed of more than one protein, stuck together, adding to the complexity of their shape. As a result, when a molecule binds to an enzyme, various changes might occur to trigger a reaction.
These changes in shape of the enzyme occur via conformational changes in the protein. Simple rotations around bonds cause changes in shapes of even small molecules. In large molecules such as proteins, these changes in shape can be dramatic.
Exercise \(3\)
Binding the substrate in the active site (red) will cause one of the gates to close. Which one? | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/III%3A_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_1/06%3A_Enzyme_Catalysis/6.02%3A_Enzyme_Binding.txt |
Once a substrate has been bound, it is the enzyme's job to quickly transform the substrate into product. The enzyme does so by carrying the substrate over a catalytic pathway. In a catalytic pathway, the reaction takes a different course than it would on its own. Sometimes the catalytic pathway is longer, involving additional steps, but because the energetic terrain is easier to traverse than in the un-catalyzed process, the catalytic reaction actually takes much less time.
Enzymes have a range of structures and reaction properties, so there are a wide number of different reactions they can catalyze. Nevertheless, there are a few common strategies displayed in catalytic reactions that are useful to know.
Approximation
When we make an approximation, we are getting close to the answer. If someone asks us what time it is, and it is 3:02 pm, we probably tell them it's about three o'clock. That's close enough.
In enzyme catalysis, approximation means getting things close to each other. If we have a substrate that is going to react with something, the enzyme can bind the substrate in such a way as to get the substrate in close proximity to the reactant. Sometimes, but not always, that might mean binding two substrates, so that they can more easily react with each other. If you want to meet your friend, it's a much better idea to say, let's meet at Cafe Santropol at 7:00, rather than wander the streets of Montreal, along with four million other people, hoping to bump into your friend. The reactant and substrate are much more likely to encounter each other in the protected confines of the enzyme than they are floating around in the wilderness of the cell.
• In approximation, two substrates are held close together within the enzyme
• This proximity makes them react together more easily
Sometimes, this idea can be a little more subtle. Imagine that the reactant is the enzyme itself, so that just by binding the single substrate, the reaction is already much more likely to occur. The substrate may be held in such a way that it is already in close proximity to amino acid side chains that will work on it and transform it into a new molecule.
• In approximation, the substrate is held in position so that subsequent reaction is much more likely
Approximation is really an entropy factor. By binding the substrate, we can limit its degrees of freedom, restricting its location or even its orientation so that there is no way the reactant can miss its intended target.
Of course, the substrates still have to find the active site of the enzyme. Sometimes, that task is aided by sticky surfaces on the enzyme; there can be groups on the surface of the enzyme that can interact with substrates, so that the substrates are less likely to drift away after a random collision with the enzyme. With its movement thus restricted, the substrate is more likely to move into the active site than drift off through the cell once more.
Although we are discussing enzymes in the cell, other kinds of catalysis make use of approximation as well. For example, transition metal catalysis often makes use of solid chunks of metal to catalyze the reactions of gaseous vapors. The surface of the metal gives the gas-phase molecules a place to bind, giving them a place to gather, rather than wandering around in three dimensions in the gas phase. Once they are together in one place, they are more likely to react with each other or with additional species on the surface of the metal.
Acid-Base Catalysis
Acid-base catalysis is a very common phenomenon. So many reactions involve the addition or removal of protons, especially the carbonyl reactions that are so prevalent in biochemical pathways, that proton donors and acceptors become key players. Acidic and basic side chains of the amino acids in the protein naturally fill these roles.
Exercise \(1\)
Which amino acids are thought of as having "acidic" side chains?
Answer
Aspartic acid (abbreviations Asp or D) and glutamic acid (abbreviations Glu or E).
Exercise \(2\)
Which amino acids are thought of as having "basic" side chains?
Answer
Histidine (abbreviations His or H), lysine (abbreviations Lys or K) and arginine (abbreviations Arg or R).
Exercise \(3\)
Draw the acidic amino acid side chains in both their acidic and conjugate basic forms.
Answer
Exercise \(4\)
Draw the basic amino acid side chains in both their basic and conjugate acidic forms.
Answer
Acid-base catalysis can provide mechanistic advantages by rapidly enhancing the electrophilicity of a molecule. Any carbonyl compound is electrophilic, but if it gets protonated, the overall positive charge makes the carbonyl even more electrophilic.
Alternatively, acid-base catalysis might increase the nucleophilicity of a molecule. Any alcohol is nucleophilic, because it has lone pairs. However, if its proton is removed, it becomes even more nucleophilic, because of the overall negative charge.
So, in the most straightforward case, adding a proton might accelerate a reaction involving an electrophile. Removing a proton might accelerate a reaction involving a nucleophile.
• Acid/base catalysis involves rapid proton shuttling
• Acidic side chains can activate electrophiles
• Basic side chains can activate nucleophiles
There are other variations on this approach. For example, consider a keto-enol tautomerism. We think of ketones classically as electrophiles, but their enol isomer is easily accessible in general, and the enol form is an excellent nucleophile. As part of a series of reaction, rapid conversion of a ketone into an enol might be a key step.
That's really a couple of different steps; you have to add a proton to the carbonyl oxygen, and then you have to take a proton away from the alpha position. You could do it the other way around, but it would still be two steps. Fortunately, enzymes have lots of acidic and basic sites, so there may be sources and sinks of protons that are readily available for a molecule bound in an enzyme. Furthermore, the molecule may be bound in such a way that these proton transfer steps are essentially simultaneous.
Under normal circumstances, this looks like three different molecules reacting together; that's basically impossible under normal circumstances. But if the lysine and the aspartate are part of the same enzyme, that reduces the problem to an interaction between two molecules; that happens all the time. If the ketone is already bound in the enzyme, the difficulty is reduced even further.
In order to make the reaction catalytic, the enzyme has to be regenerated. Again, that isn't a problem for enzymes, because of the number of acidic and basic sites available. A key site that is missing a crucial proton can pick one up from another amino acid side chain sitting nearby. Protonation states get reshuffled pretty quickly, and soon the enzyme is ready to go again.
Acid-Base Catalysis: Metal Ion Catalysis
Metal ion catalysis can often be thought of as a special case of acid-base catalysis. With metal ions, we get Lewis acid catalysis. Lewis acid catalysis can accelerate reactions in a couple of different ways, in close analogy with general acid/base catalysis.
It sometimes helps to think of a metal as a great, big proton. That's an oversimplification, as we'll see in a moment. Nevertheless, it can be useful to keep the analogy in mind. When a compound binds to a metal ion, the effect can be similar to binding a proton. The compound has just donated a pair of electrons elsewhere (to the metal ion or to the proton), and so the compound suddenly looks electron-deficitient. It has enhanced electrophilicity.
On the other hand, we don't usually think of protonation as causing increased basicity; that would be completely backwards. With metals, though, that can happen, indirectly. If an alcohol, for example, donates a lone pair to a metal, the oxygen becomes positively charged. It becomes much more acidic. Suddenly, the alcohol can be deprotonated by a very, very weak base, such as water. That leads to formation of an alkoxide ion, which is much more nucleophilic than the original alcohol. Water would really never take a proton from an alcohol, but it can do it once the oxygen has a positive charge.
Metal ions can play a number of other roles in catalysis, but that's enough to get an idea of just some of the ways in which they might be useful. To learn more, we would have to explore more transition metal reactivity, including the ability of metals to donate and accept individual electrons.
Group Transfer
Group transfer, also called covalent catalysis, has something in common with approximation. It limits the degrees of freedom involved in a key step of the pathway to product. Rather than just binding two substrates near each other in the active site, or holding a substrate in a specific orientation, group transfer involves a reaction between the substrate and the enzyme such that the substrate becomes attached to the enzyme. It is attached via a covalent bond.
In a sense, like approximation, this mechanism limits the problem of two substrates bumping into each other -- an unlikely prospect in the roliing sea of the cell -- to two things coming together within the limited space of the active site. Of course, any time a molecule wanders into an active site, there is a chance it could wander off again; but that isn't true if you fasten it down. That's what group transfer does. It fastens one substrate to the enzyme, so that a second substrate is more likely to react with it.
That's not all. Sometimes when the substrate, or a group forming the crucial part of the substrate, is transferred to a side chain on the enzyme, it becomes modified in such a way as to become more reactive. For example, an aldehyde may be electrophilic to begin with. However, if it becomes attached to a lysine residue, it may be transformed into an imine. Now, an imine may seem less electrophilic than an aldehyde, because nitrogen is less electronegative than oxygen. However that same feature also makes nitrogen much more basic than oxygen. As a result, imines are very likely to rest in a protonated state; aldehydes are much less likely to be protonated. That protonated iminium ion, of course, is much more electrophilic than an aldehyde, because the iminium has a full positive charge.
Group transfer may even result in complete reversal of reactivity in the substrate. Again, in the case in which an aldehyde becomes tethered to a lysine residue, the aldehyde may be transformed into an enamine. The difference between forming an enamine and an imine is just the difference in which of the protons is removed. In an enzyme, that decision is easily settled by the positioning of the amino acid side chain that removes the proton. In one enzyme, this side chain removes the nearest proton, leading to an imine. In another enzyme, another side chain, situated in a different place, removes the nearest proton, leading to an enamine.
Of course, the consequence of forming an enamine rather than an imine is much like having an enolate or enol rather than an aldehyde. The former is nucleophilic, even though the latter is electrophilic. So, group transfer can even flip something from an electrophile into a nucleophile.
A very different example that is also pretty common involves a set of amino acid side chains called "the catalytic triad". These three groups -- aspartate, histidine, and serine -- are frequently observed in group transfer catalysis. With the catalytic triad, it is the serine that acts as the nucleophile, rather than a lysine. As a result, it is the serine that becomes bonded to the substrate. The role of the aspartate and histidine is to act as a proton relay, helping to activate the serine so that it becomes a more effective nucleophile.
Transition State Stabilization
We think of enzymes as being perfectly suited to bind a particular substrate, but that may not be the case at all. If you think about it, an enzyme that perfectly matches its substrate would pick up its substrate, snap it into place and... that would be all. Nothing else would ever happen. The substrate would sit there forever. Why mess with perfection?
Instead, it is thought that, at least in some cases, the enzyme is ideally suited to bind something a little further down the path. The idea is that the substrate will change shape as it goes through the reaction, and if the enzyme actually better fits the shape of something else to come, it will exert pressure on the substrate to go ahead and react. Once the substrate gets to the right shape, it will be rewarded with perfect binding.
In the extreme case, we might think about the enzyme being a perfect match for the product. That way, the enzyme could coax the reaction along, and the strong binding interaction between enzyme and product would pull the reaction through its equilibrium all the way to the product side.
Well, if you think about it, that wouldn't work very well, either. Once the product forms, it would bind perfectly with the enzyme, which would never let it go. What good is it if the product is made but never gets released into the cell? The whole point of enzymatic catalysis is to quickly make things that the cell desperately needs; without these things, the cell will die.
But there is one kind of structure that could bind the enzyme perfectly, and we wouldn't have to worry about it getting stuck. It's a transition state. A transition state is an inherently unnatural structure; the molecule can't stay that way for long. It's just passing through on its way from one structure to another, and the transition state is the awkward, gangly phase in between, that horrid thing we glimpse in the dark when it thinks we aren't looking.
So if we stabilize that thing, then the substrate is pushed onward, a little further, and just when it reaches the perfect point, it finds it's gone too far. It must collapse. It's like Wile E. Coyote sublimely achieving his goal just as he passes over the cliff's edge.
Transition State Stabilization: Distortion
If transition state stabilization is the carrot, then distortion is the stick. Transition state stabilization coaxes the substrate further on; distortion pushes and prods until the substrate can take it no longer.
In distortion, the substrate finds itself in a binding site in which its shape is not quite right. Its geometry is subtly squeezed. The substrate itself is destabilized, whereas forming the product, or the transition state leading to the product, actually relieves that strain.
A good example is an atom that is supposed to be trigonal planar, but finds itself sitting in a place where it can't quite lay flat. Instead, it finds intermolecular interactions that are pushing it into another shape; maybe tetrahedral would be a better fit.
If a tetrahedral intermediate lies ahead, then by the time we reach the transition state, the geometry of that atom may actually look more like a tetrahedron than a trigonal plane. As a result, those intermolecular forces are distorting the otherwise stable structure of the substrate and bending towards the transition state. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/III%3A_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_1/06%3A_Enzyme_Catalysis/6.03%3A_Strategies_in_Enz.txt |
Compounds other than substrates can play important roles in interacting with enzymes. There are compounds that can either activate or inhibit enzymes, speeding up their activity or slowing them down, respectively. These compounds can be very important in naturally regulating enzymes in the cell. We can also exploit these compounds as medicines, treating illness by tuning the activity of enzymes until the body is back to normal.
Knowing for sure whether a compound is going to turn an enzyme on or off can get pretty complicated, so we are just going to start with a couple of simple ways in which enzymes can be turned off.
Reversible Inhibitors
There are several different ways in which compounds can inhibit enzymes. The simplest way is simply by taking up space in the active site. If something is blocking the active site, the true substrate can't get in, so the enzyme can't do its job.
To block the active site, the inhibitor must also be a pretty good fit. It must have a similar shape, or in some way be able to interact with amino acid residues in the active site. That means it might not have a shape that is obviously familiar, but must be capable of intermolecular interactions that are similar to those of the substrate in the active site.
Exercise \(1\)
If the substrate is the compound in blue, indicate which of the red compounds would be a better competitive inhibitor.
In drug design, the part of the enzyme that binds to the substrate is called the pharmacophore. That's the feature that we will try to exploit in designing an inhibitor. The inhibitor will be designed so that it can also bind to the pharmacophore, either by fitting in terms of shape or by the intermolecular attractions it can supply.
Exercise \(2\)
Select the compound that would more efficiently bind to the pharmacophore.
Most of the time, if we just want to moderate the effects of an over-active enzyme, we use what is called a reversible inhibitor. We don't want to block the active site of the enzyme permanently; we just want to slow things down, to get things back to normal. After all, the enzyme probably does something useful for us, and we may need it later. So, we use an inhibitor that will slow it down for a while, but whose effects can be reversed.
A competitive inhibitor simply competes with the substrate for the active site. Both the substrate and the inhibitor occupy the same niche, but they can't both sit there at the same time. In general, there is some equilibrium between bound and unbound states. That means that after a while, one molecule wanders back out of the enzyme, and the other one has a chance to enter.
We have mechanisms in the cell that scrub out molecules that shouldn't be there, so eventually, that's what happens to the inhibitor. While the concentration of inhibitor remains high, it can effective compete for the active site of the enzyme. As more and more of it gets removed, the enzyme gets back to its normal levels of activity. Hopefully, by that point, the disease state has eased off.
Irreversible inhibitors
Maybe a disease state is serious enough that we want to shut an enzyme down completely. That might be the case if we are fighting a bacterial or fungal infection, for example; we don't care whether the fungus needs that enzyme later. Causing big problems for the fungus is kind of the whole point at this stage.
In a case like this, we might use what is called an irreversible inhibitor. Once it goes into the enzyme, it never comes back out. It forms a permanent bond, hopefully in a place that blocks the active site or some other important feature of the enzyme. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/III%3A_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_1/06%3A_Enzyme_Catalysis/6.04%3A_Enzyme_Inhibition.txt |
Reversible inhibitors are extremely important in regulating enzyme activity. Unlike irreversible inhibitors, they do no shut down an enzyme completely by permanently disabling it. They are much more subtle, just slowing it down temporarily. There are a number of different ways that the inhibitor could do that, however, and so we will take a look at those possibilities here.
The simplest idea is that an inhibitor can bind in the active site of the enzyme. If the inhibitor is already bound there, the substrate cannot. The inhibitor is in the way. This is called competitive inhibition.
This case is a true competition. If there are more inhibitor molecules than substrate molecules, the inhibitors will probably win out, blocking the substrate from entering the active site. But if there are more substrate molecules than inhibitor molecules, then chances are the substrate will be able to bind, and the subsequent reaction will proceed.
There are a number of cases, however, in which the inhibitor does not bind at the active site at all. It binds someplaces else on the enzyme, at a place called an allosteric site. When the inhibitor is bound at the allosteric site, it somehow interferes with the function of the enzyme. An inhibitor that binds at a site other than the active site is generally called an allosteric inhibitor.
Allosteric inhibitors can work in a few different ways. In perhaps the simplest case, when the allosteric inhibitor binds to the enzyme, it causes some sort of conformational change that prevents the enzyme from carrying out reactions. It doesn't interfere with substrate binding, so the substrate can still complex with the enzyme, but nothing will happen after that. This type of inhibition is called noncompetitive inhibition, or sometimes pure noncompetitive inhibition, for the simple reason that the inhibitor is not interfering directly with the substrate; it's temporarily disabling the enzyme in some other way.
Noncompetitive inhibition interferes with the machinery of the enzymatic reaction, but leaves substrate binding alone. But if allosteric inhibitors cause some sort of conformational change in the enzyme, then it's easy to imagine they could somehow mess up the binding site. As a result, they might interfere with substrate binding, even without being in direct competition with the substrate. This mode is called mixed noncompetitive inhibition. Although the inhibitor is not directly competing for the same binding site as the substrate, it ends up preventing the substrate from binding anyway.
Well, that game can go both ways. If an inhibitor can change the binding site of the substrate, maybe the substrate can change the binding site of the inhibitor. In uncompetitive inhibition, the inhibitor is not able to bind to the free enzyme. However, when the substrate binds, it induces a conformational change in the allosteric site, allowing the inhibitor to bind. If the inhibitor binds, it interferes with the machinery of the enzyme, so the enzyme can't do its job, even though the substrate is bound.
There is one more case that is a complemetary idea; it isn't about inhibition at all. If a compound can bind at an allosteric site, changing the conformation of the active site so that a substrate can no longer bind, then maybe the reverse is true. Maybe a compound can bind at an allosteric site, allowing a substrate to bind that previously could not.
Instead of inhibiting the enzyme, this compound would be activating it. It would make binding possible, and suddenly the enzyme could do ts job. This compound is an allosteric activator. As in inhibition, there may be different modes through which the allosteric activator could turn the enzyme on. The important thing is to know that in order to regulate an enzyme fully, there nood to be both on and off switches. Activators turn on enzymes that are waiting to be used. Inhibitors turn off enzymes that we don't need right now. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/III%3A_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_1/06%3A_Enzyme_Catalysis/6.05%3A_Types_of_Reversib.txt |
Reversible inhibitors are extremely important in regulating enzyme activity. They can turn enzymes on or off, acting as activators or inhibitors, respectively. In addition, enzymes can be regulated via covalent modification or post-translational modification. That means that, after the enzyme has been assembled in the cell, its structure can be modified further by adding special groups to specific locations. In the case of regulation, these groups are added reversibly. Even though the group is added covalently -- it is bonded to the protein -- a reaction path exists for the removal of the group again.
There are three examples of these modifications that we will look at here. In each case, the behaviour of the protein is modified because of changes in the intermolecular attractions within the protein (or between the protein and another molecule).
Phosphorylation is a very common modification. In phosphorylation, a phosphate group is attached to an amino acid side chain. The most commonly phosphorylated side chain is a serine. Tyrosine is often phosphorylated, too. Those sites are preferred because of the strong P-O bond formed during the reaction.
Phosphorylation is typically carried out under the control of another enzyme called a kinase. That's right, one enzyme will bind another, tying a phosphate group onto it before releasing it again. A phosphate group can be removed again via another enzyme called a phosphatase. The fact that these modifications are carried out by specific enzymes helps to explain their specificity. A particular kinase may bind to its target protein in a well-defined position, phosphorylating it only at that position (although some kinases may be less selective).
Of course, the key result of phosphorylation is that a neutral serine is suddenly masked by an anionic phosphate group. That negative charge alters intermlecular attractions quite starkly, because suddenly attractive and repulsive forces pop up where there were none before.
Acetylation is also quite common. Acetylation is the addition of an acetyl or ethanoyl group; usually, the group is added to a lysine. The forward reaction is driven by the strong amide bond that results.
Acetylation is carried out by an acetylase or an acyltransferase. Like phosphatases, these are enzymes that bind their target proteins in order to modify their structures. Because lysines are normally positively charged at biological pH, acetylation results in the sudden disappearance of charged species because the are masked with neutral acyl groups. Intermolecular attractions can be dramatically affected as a result.
Prenylation is the addition of a hydrocarbon side chain, most often to a cysteine side chain. Sulfur is a particularly good nucleophile for carbon chains of this sort, enhancing selectivity for cysteine. As in the other cases, however, the reaction is carried out under the control of an enzyme (a geranyltrasnferase, for example, or a farnesyltransferase), and so other side chains might be targeted, instead.
Those enzyme types highlight two common groups that are added in these cases: the geranyl group and the related farnesyl group. These groups are structurally related, and are both in the terpenoid family of natural products. Terpenes are based on five-carbon building blocks, which you can easily trace out in the geranyl and farnesyl structures.
Changes in intermolecular attractions are more subtle in these cases, because prenylation is not accompanied by a full change in the charge of a side chain as it was in the other cases. Instead, addition of this group enhances the hydrophobicity of that part of the protein. A hydrophobic group is one that is completely incapable of interacting with water molecules, lacking ions, dipoles, or hydrogen bonding sites. As a result, it tends to get displaced or pushed out of the way by surrounding water molecules. That "pushing away" by water has the net effect of piling hydrophobic groups together.
What we tend to get, then, is hydrophobic groups sticking to each other, even though their attraction for each other (London forces) is actually pretty weak. Of course, because London forces are relatively weak, the amount of London forces, or the relative area of a molecule that can interact with another via London forces, becomes very important. Geranyl and farnesyl groups are moderately long, and so they exert significant amounts of London forces. As a result, two prenylated parts of a protein might stick together, or one newly prenylated part might huddle together with some hydrophobic side chains, such as valines, leucines, and isoleucines. Also, prenylation sometimes causes a protein that was formerly more water-soluble to migrate over to the cell membrane, where it could interact with the lipid bilayer. So, prenylation, despite causing seemingly modest changes in intermolecular attractions, can elicit dramatic changes in behaviour in the protein.
Exercise \(1\)
Show the products of the following modifications.
Exercise \(2\)
Indicate whether attraction between the strands would increase or decrease after modification.
Exercise \(3\)
In the following drawings, the object's interaction with one of the surfaces will change upon modification. This change in interaction will result in a net shift of the object to the left or to the right. The shift may result because of attractive or repulsive forces. Indicate the direction of shift in each case.
Exercise \(4\)
In the following drawings, the active site is closed. Show how the regulator will bind, causing the active site to open, and also show how the substrate will bind.
Exercise \(5\)
The following drawings depict an allosteric site that will be modified on the left and a binding site on the right. Imagine the middle bar is a see-saw that can tilt left or right. In each case, decide whether the modification will lead to tighter or to looser binding of the substrate.
6.07: Enzyme Solutions
Exercise 6.1.1:
1. isomerase v) reorganize atoms from one isomer into another
2. hydrolase vii) add water into a molecule, helping to break it down
3. oxygenase i) add oxygen from O2 into a molecule
4. dioxygenase ii) add both oxygens from O2 into a molecule
5. reductase iv) reduce or add electrons to the substrate
6. transferase ix) transfer a functional group to or from a molecule
7. phosphatase vi) cleave a phosphate group off a protein
8. oxidase iii) oxidize or remove electrons from the substrate
i) ligase viii) cause two molecules to be bound together
Exercise 6.2.1:
Exercise 6.2.2:
Exercise 6.2.3:
Exercise 6.3.1:
Aspartic acid (abbreviations Asp or D) and glutamic acid (abbreviations Glu or E).
Exercise 6.3.2:
Histidine (abbreviations His or H), lysine (abbreviations Lys or K) and arginine (abbreviations Arg or R).
Exercise 6.3.3:
Exercise 6.3.4:
Exercise 6.4.1:
Exercise 6.4.2:
Exercise 6.6.1:
Exercise 6.6.2:
Exercise 6.6.3:
Exercise 6.6.4:
Exercise 6.6.5: | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/III%3A_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_1/06%3A_Enzyme_Catalysis/6.06%3A_Covalent_Modifica.txt |
Biological cells have a daunting task. They must carry out 1000s of different chemical reactions required to carry out cell function. These reactions can include opposing goals such as energy production and energy storage, macromolecule degradation and synthesis, and breakdown and synthesis of small molecules. All of these reactions are catalyzed by proteins and RNAs enzymes whose activities must be regulated, again through chemical reactions, to avoid a futile and energy wasting scenario of having opposing pathways functioning simultaneously in a cell.
Metabolism can be divided into two main parts, catabolism, the degradation of molecules, usually to produce energy or small molecules useful for cell function, and anabolism, the synthesis of larger biomolecules from small precursors.
CATBOLISM: Catabolic reactions involve the breakdown of carbohydrates, lipids, proteins, and nucleic acids to produce smaller molecules and biological energy in the form of heat or small thermodynamically reactive molecules like ATP whose further degradation can drive endergonic process such as biosynthesis. Our whole world is reliant on the oxidation of organic hydrocarbons to water and carbon dioxide to produce energy (at the expense of releasing a potent greenhouse gas, \(\ce{CO2}\)). In the biological world, reduced molecules like fatty acids and partially oxidized molecules such as glucose polymers (glycogen, starch), as well as simple sugars, can be partially or fully oxidized to ultimately produce \(\ce{CO2}\) as well. Energy released from oxidative reactions is used to produce molecules like ATP as well as heat. Oxidative pathways include glycolysis, the tricarboxylic acid cycle (aka Kreb's cycle) and mitochondrial oxidative phosphorylation/electron transport. To fully oxidize carbon in glucose and fatty acids to carbon dioxide requires splitting C-C bonds and the availability of series of oxidizing agents that can perform controlled, step-wise oxidation reactions, analogous to the sequential oxidation of methane, CH4 to methanol (CH3OH), formaldehyde (CH2O) and carbon dixoxide.
• Glycolysis: This most primitive of metabolic pathways is found in perhaps all organisms. In glycolysis, glucose (\(\ce{C6H12O6}\)), a 6C molecule, is split (or lysed) into two, 3C carbon molecules, glyceraldehyde-3-phosphate, which are then partially oxidized under anaerobic conditions (without \(\ce{O2}\)) to form two molecules of pyruvate (\(\ce{CH3COCO2^{-}}\)). Instead of the very strong oxidizing agent, \(\ce{O2}\), a weaker one, NAD+ is used, which is reduced in the process to form NADH. Since none of the carbon atoms is oxidized to the state of \(\ce{CO2}\), little energy is released compared to the complete oxidation to \(\ce{CO2}\). This pathway comes to a screeching halt if all cellular NAD+ is converted to NADH as NAD+ is not replenished by the simple act of breathing as is the case with \(\ce{O2}\) in aerobic oxidation. To prevent the depletion of NAD+ from inhibiting the cycle and to allow the cycle to continue under anaerobic conditions, excess NADH is reconverted to NAD+ when the other product of glycolysis, pyruvate is converted to lactate by the enzyme lactate dehydrogenase. Glycolysis occurs in the cytoplasm of the cell.
• Tricarboxylic Acid (Kreb's) Cycle: The TCA cycle is an aerobic pathway which takes place in an intracellular organelle called the mitochondria. It takes pyruvate, the incompletely oxidized product from glycolysis, and finishes the job of oxidizing the 3C atoms all the way to \(\ce{CO2}\). First the pyruvate moves into the mitochondria where is is oxidized to the 2C molecule acetylCoA with the release of one \(\ce{CO2}\) by the enzyme pyruvate dehydrogenase. The acetyl-CoA then enters the TCA cycle where two more \(\ce{CO2}\) are released. As in glycolysis, C-C bonds are cleaved and C is oxidized by NAD+ and another related oxidizing agent, FAD. What is very different about this pathway is that instead of being a series of linear, sequential reactions with one reactant (glucose) and one product (two pryuvates), it is a cyclic pathway. This has significant consequences since if any of the reactants within the pathways becomes depleted, the whole cyclic pathway can slow down and stop. To see how this happens consider the molecule oxaloacetate (OAA) which condenses with acetyl-CoA to form citrate (see diagram below). In this reaction, one OAA is consumed. However, when the cycle returns, one malate is converted to OAA so there is no net loss of OAA, unless OAA is pulled out of the TCA cycle for other reactions, which happens.
• Mitochondrial Oxidative Phosphorylation/Electron Transport: The TCA cycle accomplishes what glycolysis didn't, that is the cleavage of all C-C bonds in glucose (in the form of pyruvate and acetyl-CoA, and the complete oxidation of all C atoms to \(\ce{CO2}\). Yet two problem remains. The pool of oxidizing molecules, NAD+ and FAD get converted to their reduced forms, NADH and FADH2. Unless NAD+ and FAD are regenerated, as was the case in anaerobic conditions when pyruvate gets converted to lacate, the pathway would again come to a grinding halt. In addition, not much ATP is made in the cycle (in the form of a related molecule GTP). Both these problems are resolved as the resulting NADH and FADH2 formed are reoxidized by mitochondrial membrane enzyme complexes which pass electrons from the oxidized NADH and FADH2 to increasingly potent oxidizing agents until they are accepted by the powerful oxidant \(\ce{O2}\),which is converted reduced to water. The net oxidation of NADH and FADH2 by dioxygen is greatly exergonic, and the energy released by the process drives the synthesis of ATP from ADP and Pi by an mitochondrial enzyme complex, the F0F1ATPase.
Feeder Pathways: Other catabolic pathways produce products that can enter glycolysis or the TCA cycle. Two examples are given below.
• Complex carbohydrates: In mammals, the major carbohydrate storage molecule is glycogen, a polymer of glucose linked a1-4 with a1-6 branches. The terminal acetal linkages in this highly branched polymer is cleaved sequentially at the ends not through hydrolysis but through phosphorolysis to produce lots of glucose-1-phosphate which can enter glycolysis.
• Lipids: Lipids are stored mostly as triacylglycerides in fat cells (adipocytes). When needed for energy, fatty acids are hydrolyzed from the glycerol backbone of the triacylglyceride, and send into cells where they broken down in an oxidative process to form acetyl-CoA with the concomitant production of lots of NADH and FADH2. These can then enter the mitochondrial oxidative phosphorylation/electrons transport system, which produces, under aerobic conditions, lots of ATP.
• Proteins: When intracellular proteins get degraded, they from individual amino acids. The amine N is lost as it enters the urea cycle. The rest of some amino acid structures can be ultimately converted to acetyl-CoA or keto acids (like alpha-ketoglutarate- a-KG) that are TCA intermediate. These amino acids are called ketogenic. Alternatively, some amino acids, after deamination, are coveted to pyruvate which can either enter the TCA cycle or in the liver be used to synthesize glucose in an anabolic process. These amino acids are called glucogenic. Chemical reactions such as these can be used to replenish intermediates in the TCA cycle which can become depleted as they are withdraw for other reactions. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/III%3A_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_1/07%3A_Metabolic_Pathways/7.01%3A_An_Overview_of_.txt |
Anabolism: Anabolic reactions are those that lead to the synthesis of biomolecules. In contrast to the catabolic reactions just discussed (glycolysis, TCA cycle and electron transport/oxidative phosphorylation) which lead to the oxidative degradation of carbohydrates and fatty acids and energy release, anabolic reactions lead to the synthesis of more complex biomolecules including biopolymers (glycogen, proteins, nucleic acids) and complex lipids. Many biosynthetic reactions, including those for fatty acid synthesis, are reductive and hence require reducing agents. Reductive biosynthesis and complex polymer formation require energy input, usually in the form of ATP whose exergonic cleavage is coupled to endergonic biosynthesis.
Cells have evolved interesting mechanism so as not to have oxidative degradation reactions (which release energy) proceed at the same time and in the same cell as reductive biosynthesis (which requires energy input). Consider this scenario. You dive into a liver cell and find palmitic acid, a 16C fatty acid. From where did it come? Was it just synthesized by the liver cell or did it just enter the cell from a distant location such as adipocytes (fat cells). Should it be oxidized, which should happen if there is a demand for energy production by the cell, or should the liver cell export it, perhaps to adipocytes, which might happen if there is an excess of energy storage molecules? Cells have devised many ways to distinguish these opposing needs. One is by using a slightly different pool of redox reagents for anabolic and catabolic reactions. Oxidative degradation reactions typically use the redox pair NAD+/NADH (or FAD/FADH2) while reductive biosynthesis often uses phosphorylated variants of NAD+, NADP+/NADPH. In addition, cells often carry out competing reactions in different cellular compartments. Fatty acid oxidation of our example molecule (palmitic acid) occurs in the mitochondrial matrix, while reductive fatty acid synthesis occurs in the cytoplasm of the cell. Fatty acids entering the cell destined for oxidative degradation are transported into the mitochondria by the carnitine transport system. This transport system is inhibited under conditions when fatty acid synthesis is favored. We will discuss the regulation of metabolic pathways in a subsequent section. One of the main methods, as we will see, is to activate or inhibit key enzymes in the pathways under a given set of cellular conditions. The key enzyme in fatty acid synthesis, acetyl-CoA carboxylase, is inhibited when cellular conditions require fatty acid oxidation.
The following examples give short descriptions of anabolic pathways. Compare them to the catabolic pathways from the previous section.
• Glucose synthesis, better known as Gluconeogenesis: In glycolysis, glucose (C6H12O6), a 6C molecule, is converted to two, 3C molecules (pyruvate) in an oxidative process that requires NAD+ and makes two net ATP molecules. In a few organs, most predominately in the liver, the reverse pathway can take place. The liver does this to provide glucose to the brain when the body is deficient in circulating glucose, for example, under fasting and starving conditions. (The liver under these conditions can get its energy from oxidation of fatty acids). The reactions in gluconeogenesis are the same reactions in glycolysis but run in reverse, with the exception of three glycolytic steps which are essentially irreversible. These three steps have bypass enzymes in the gluconeogenesis pathway. Although the synthesis of glucose is a reductive pathway, it uses NADH instead of NADPH as the redundant as the same enzyme used in glycolysis is simply run in reverse. Gluconeogenesis, which also occurs in the cortex of the kidney, is more than just a simple reversal of glycolysis, however. It can be thought of as the net synthesis of glucose from non-carbohydrate precursors. Pyruvate, as seen in the section on catabolism, can be formed from protein degradation to glucogenic amino acids which can be converted to pyruvate. It can also be formed from triacylglycerides from the 3C molecule glycerol formed and released from adipocytes after hydrolysis of three fatty acids from triacylglycerides. However, in humans, glucose can not be made in net fashion from fatty acids. Fatty acids can be converted to acetyl-CoA by fatty acid oxidation. The resulting acetyl-CoA can not form pyruvate since the enzyme that catalyzes the formation for acetyl-CoA from pyruvate, pyruvate dehydrogenase, is irreversible and there is no bypass reaction known. The acetyl-CoA can enter the TCA cycle but since the pathway is cyclic and proceeds in one direction, it can not form in net fashion oxaloacetate. Although oxaloacetate can be remove from the TCA cycle and be use to form phosphoenolpyuvate, a glycolytic intermediate, one acetyl-CoA condenses with one oxaloacetate to form citrate which leads back to one oxaloacetate. Hence fatty acids can not be converted to glucose and other sugars in a net fashion.
• Pentose Phosphate Shunt: This two-part pathway doesn't appear to start as a reductive biosynthetic pathway as the first part is the oxidative conversion of a glycolytic intermediate, glucose-6-phosphate, to ribulose-5-phosphate. The next, nonoxidative branch leads to the formation of ribose-5-phosphate, a key biosynthetic intermediate in nucleic acid synthesis as well as erthyrose-4-phosphate used for biosynthesis of aromatic amino acids . The oxidative branch is important in reductive biosynthesis as it is a major source of the reductant NADPH used in biosynthetic reactions.
• Fatty acid and isoprenoid/sterol biosynthesis: Acetyl-CoA is the source of carbon atoms for the synthesis of more complex lipids such as fatty acids, isoprenoids, and sterols. When energy needs in a cell are not high, citrate, the condensation product of oxaloacetate and acetyl-CoA in the TCA cycle, builds up in the mitochondrial matrix. It is then transported by the citrate transporter (an inner mitochondrial membrane protein) to the cytoplasm, where it is cleaved back to oxaloacetate and acetyl-CoA by the cytoplasmic enzyme citrate lyase. The oxaloacetate is returned to the mitochondria by conversion first to malate (reduction reaction using NADH), which can move back into the mitochondria through the malate transporter, or further conversion to pyruate, using the cytosolic malic enzyme, which uses NADP+ to oxidize malate to pyruvate which then enters the mitochondria. The acetyl-CoA formed in the cytoplasm can then be used in reductive biosynthesis using NADPH as the reductant to form fatty acids, isoprenoids, and sterols. The NADPH for the reduction comes from the oxidative branch of the pentose phosphate pathway and from the reaction catalyzed by malic enzyme. The liver cells can still run the glycolytic pathway as the NADH/NAD+ ratio is low in the cytoplasm while NADPH/NADP+ ratio is high.
Now its time to see how the various pathways fit together to form an integrated set of pathways. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/III%3A_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_1/07%3A_Metabolic_Pathways/7.02%3A_An_Overview_of_.txt |
Now its time to see how the various pathways fit together to form an integrated set of pathways. Metabolic map pathways are by nature very messy and complex. I've created a series of maps below which display some important anabolic and catabolic pathways and how they connect. Many pathways have been omitted. These maps will evolve with time as more relevant information is added. The first maps show the interconnected pathways without much detail. Subsequent maps give increasingly amount of detail. Eventually, the most detailed maps will contain web links to show how given reactions are regulated and will display interactive Jmol molecular models of key enzymes.
These maps are tailored to support foundational courses in chemical reactivity that highlight specific metabolic pathways to illustrate how enzyme-catalyzed reactions can be explain using the language of organic and inorganic chemistry. Of course they are also useful for foundational level biochemistry courses which seek to give an overview of metabolic pathways and their connections. More detailed and comprehensive sites are available on the web. Once such site is the Kyoto Encylopedia of Gene and Genomes (KEGG) Pathway Data Base.
INTEGRATED METABOLIC PATHWAYS MAPS (pdf version)
INDIVIDUAL METABOLIC PATHWAY MAPS (pdf version)
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Catabolic
Anabolic
Both
ORGAN PROFILES
• Liver
• Brain
• Heart
• Muscle
• Adipose
7.04: Regulation of
Exquisite mechanisms have evolved that control the flux of metabolites through metabolic pathways to insure that the output of the pathways meets biological demand and that energy in the form of ATP is not wasted by having opposing pathways run concomitantly in the same cell.
Enzymes can be regulated by changing the activity of a preexisting enzyme or changing the amount of an enzyme.
A. Changing the activity of a pre-existing enzyme: The quickest way to modulate the activity of an enzyme is to alter the activity of an enzyme that already exists in the cell. The list below, illustrated in the following figure, gives common ways to regulate enzyme activity
1. Substrate availability: Substrates (reactants) bind to enzymes with a characteristic affinity (characterized by a dissociation constant) and a kinetic parameter called Km (units of molarity). If the actual concentration of a substrate in a cell is much less than the Km, the activity of the enzyme is very low. If the substrate concentration is much greater than Km, the enzyme active site is saturated with substrate and the enzyme is maximally active.
2. Product inhibition: A product of an enzyme-catalyzed reaction often resembles a starting reactant, so it should be clear that the product should also bind to the activity site, albeit probably with lower affinity. Under conditions in which the product of a reaction is present in high concentration, it would be energetically advantageous to the cell if no more product was synthesized. Product inhibition is hence commonly observed. Likewise it be energetically advantageous to a cell if the end product of an entire pathway could likewise bind to the initial enzyme in the pathways and inhibit it, allowing the whole pathway to be inhibited. This type of feedback inhibition is commonly oberved
1. Allosteric regulation: As many pathways are interconnected, it would be optimal if the molecules of one pathway affected the activity of enzymes in another interconnected pathway, even if the molecules in the first pathway are structurally dissimilar to reactants or products in a second pathway. Molecules that bind to sites on target enzymes other than the active site (allosteric sites) can regulate the activity of the target enzyme. These molecules can be structurally dissimilar to those that bind at the active site. They do so my conformational changes which can either activate or inhibit the target enzyme's activity.
2. pH and enzyme conformation: Changes in pH which can accompany metabolic process such as respiration (aerobic glycolysis for example) can alter the conformation of an enzyme and hence enzyme activity. The initial changes are covalent (change in protonation state of the protein) which can lead to an alteration in the delicate balance of forces that affect protein structure.
3. pH and active site protonation state: Changes in pH can affect the protonation state of key amino acid side chains in the active site of proteins without affecting the local or global conformation of the protein. Catalysis may be affected if the mechanism of catalysis involves an active site nucleophile (for example), that must be deprotonated for activity.
4. Covalent modification: Many if not most proteins are subjected to post-translational modifications which can affect enzyme activity through local or global shape changes, by promoting or inhibiting binding interaction of substrates and allosteric regulators, and even by changing the location of the protein within the cell. Proteins may be phosphorylated, acetylated, methylated, sulfated, glycosylated, amidated, hydroxylated, prenylated, myristolated, often in a reversible fashion. Some of these modifications are reversible. Regulation by phosphorylation through the action of kinases, and dephosphorylation by phosphates is extremely common. Control of phosphorylation state is mediated through signal transduction process starting at the cell membrane, leading to the activation or inhibition of protein kinases and phosphatases within the cell.
Extracellular regulated kinase 2 (ERK2), also known as mitogen activate protein kinase 2 (MAPK2) is a protein the plays a vital role in cell signaling across the cell membrane. Phosphoryation of ERK2 on Threonine 183 (Thr153) and Tyrosine 185 (Tyr185) leads to a structural change in the protein and the regulation of its activity.
B. Changing the amount of an enzyme: Another and less immediate but longer duration method to modulate the activity of an enzyme is to alter the activity of an enzyme that already exists in the cell. The list below, illustrated in the following figure, shows way in which enzyme concentration is regulated.
1. Alternation in transcription of enzyme's gene: Extracellular signal (hormones, neurotransmitters, etc) can lead to signal transductions responses and ultimate activation or inhibition of the transcription of the gene for a protein enzyme. These changes result from recruitment of transcription factors (proteins) to DNA sequences that regulate transcription of the enzyme gene.
2. Degradation of messenger RNA for the enzyme: The levels of messenger RNA for a protein will directly determin the amount of that protein synthesized. Small inhibitor RNAs, derived from microRNA molecules transcribed from cellular DNA, can bind to specific sequences in the mRNA of a target enzyme. The resulting double-stranded RNA complex recruits an enzyme (Dicer) that cleaves the complex with the effect of decreasing translation of the protein enzyme from its mRNA.
3. Co/Post translational changes: Once a protein enzymes is translated from its mRNA, it can undergo a changes to affect enzyme levels. Some proteins are synthesized in a "pre"form which must be cleaved in a targeted and limited fashion by proteases to active the protein enzyme. Some proteins are not fully folded and must bind to other factors in the cell to adopted a catalytically active form. Finally, fully active protein can be fully proteolyzed by the proteasome, a complex within cells, or in lysosomes, which are organelles within cells containing proteolytic enzymes.
Next we will consider which enzymes in pathways make the best target for regulation. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/III%3A_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_1/07%3A_Metabolic_Pathways/7.03%3A_Metabolic_Maps_.txt |
All proteins are ultimately regulated, if only by modulating the rates of their synthesis and degradation. However, some enzymes positioned at key points in metabolic pathways are ideal candidates for regulation, as their activity can affect the output of entire pathways. These enzymes typically have two common characteristics, they catalyze reactions far from equilibrium and they catalyze early committed steps in pathways.
A. Regulation of enzymes for reactions not at equilibrium
The optimal enzymes for regulation are those at the beginning of pathways and that carry out thermodynamically favored reactions. Why is the latter so important? These enzymes control the flux of metabolites through pathway, so to understand their regulation we can use the analogy of flow (or flux) of water from one container to another. Let's say you wish to fill a swimming pool at any desired height you wish and you have two ways to do so (see figure below). You could open a valve that controls the flow from your towns water tower to the pool. In this the reaction (flow of water) is energetically (thermodynamically) favored given the difference in height of the water levels and the potential energy difference between the two. Even though flow (or flux) is cleared flavored, you can regulate it, from no flow, to maximal flow, by opening and closing the valve (analogous to activating and inhibiting an enzyme). Your choices in the other scenario, filling the pool from a lake, are not so great. It would be hard to fill the water to the desired level (especially if it was an above ground pool). It would be hard to regulate the flow.
By analogy, the best candidates for regulation are those enzymes whose reactions are thermodynamically favored (not at equilibrium) but which can be controlled by the mechanisms discussed in the previous section.
Which reactions are commonly not at equilibrium (i.e. \(\Delta G^o<0\) and usually also \(\Delta G^o<0\) if the ratio of products to reactants is not too high)? The answer is those that have reactants that are thermodynamically unstable compared to their reaction products. There are several types of reactions that often fit these criteria:
Hydrolysis (or similar reactions) of anhydride or analogous motifs: The figure below shows molecules with similar "anhydride" motifs and the \(\Delta G^o\) for hydrolysis of the molecules. Those with more negative \(\Delta G^o\) values can transfer their phosphate group to ADP to make ATP, which is necessary to drive unfavorable biological reactions. Metabolic reactions that involve hydrolysis (or other type of transfer reaction of these groups) usually proceed with a negative \(\Delta G^o\) and \(\Delta G\), making them prime candidates for pathway regulation. Many textbooks label these types of molecules as having "high energy" bonds. This is confusing to many student as bonds between atoms lower the energy compare to when the atoms are not bonded. It takes energy to break the "high" energy phosphoanhydride covalent bond. What make hydrolysis of the molecules below so exergonic is that more energy is released on bond formation within the new products than was required to break the bonds in the reactants. In addition, other effects such as preferential hydration of the products, lower charge density in the products, and less competing resonances in the products all contribute to the thermodynamically favorable hydrolysis of the reactants.
Thioesters (such as Acetyl-SCoA) are also included since that have the same negative \(\Delta G^o\) of hydrolysis as ATP, even though the lack an "anhydride" motif. Thioesters are destabilized compared to their hydrolysis products and in comparison to esters made with alcohol since the C-S bond is weaker. Why?
Redox reactions: Everyone knows that redox reactions are thermodynamically favored if the oxidizing agent deployed is strong enough. The oxidation reactions of hydrocarbons, sugars, and fats by dioxygen are clearly exergonic (we do call these combustion reactions after all). What about redox reactions with less powerful oxidants? NAD+ is used frequently as a biological oxidizing agent. Are all these reactions as favored as combustion? Hardly so. Remember that in every redox reaction, an oxidizing and reducing agent react to form another oxidizing and reducing agent. Consider the following reaction:
\[\ce{Pyruvate + NADH -> Lactate + NAD^{+}}.\]
This reaction can go either way and is reversible. In the above form, it is written in the favored direction in aeroboic metabolism when both Pyr and NADH level are high. Although the \(\Delta G^o\) actually favors the oxidation of lactate, given the high concentration of Pyr and NADH, the reaction is driven in the opposite direction and proceeds as shown. To determine if a redox reaction is favored and likely to occur, and possibly be regulated, the \(\Delta G^o\) for a redox reaction should be calculated from standard reduction potentials, using the formula
\[\Delta G^o = -nFE^o.\]
B. Regulation of enzymes catalyzing committed steps in pathways:
The best enzymes to regulate are those that catalyze the first committed step in the reaction pathway. The committed step proceeds with a DG < 0 and is essentially irreversible. These reactions often occur from key metabolic intermediates that are immediately before or proximal to branches in reaction pathways. Two examples are shown below. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/III%3A_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_1/07%3A_Metabolic_Pathways/7.05%3A_Regulation_of_M.txt |
Glycolysis is a biochemical pathway in which glucose is consumed and ATP is produced. This pathway is an example of catabolism, in which larger molecules are broken down in the cell to make smaller ones. The opposite kind of pathway is anabolism, in which larger molecules are synthesized from smaller ones in the cell.
From the biologist's perspective, catabolism is associated with the breakdown of larger molecules to release energy. For example, in grade school science, you may have learned that most organisms derive their energy from the breakdown of carbohydrates. You may have seen the process of respiration expressed through the following equation of reaction:
$\ce{C6H12O6_{(s)} + 6O2_{(g)} -> 6CO2_{(g)} + 6H2O_{(l)} + energy} \nonumber$
That idea gives rise to the slightly misleading paradigm that energy is stored in chemical bonds. The idea goes that, for example, when the single sugar molecule represented by the formula, C6H12O6 , is broken down to make six carbon dioxide molecules, the energy from all of those broken bonds is released for the benefit of the organism.
You may also have learned about another important energy-storage molecule, ATP. Like the breakdown of sugar, the breakdown of ATP is used to power other processes in the cell. That process might be expressed in the following expression:
$\ce{ATP_{(aq)} + H2O_{(l)} -> ADP_{(aq)} + P_{i(aq)} + energy} \nonumber$
Once again, this can be considered a breaking-down process, in which an ATP molecule is split into a smaller ADP molecule and an inorganic phosphate.
From the chemist's perspective, it is wrong to suggest that energy is stored in chemical bonds. Instead, energy is released when bonds are formed. This chemical perspective is more than an idea; it represents physical reality. It can be demonstrated in a number of ways that energy is released when bonds are made, and energy must be used up in order to break bonds; apparently, this situation is the opposite of the biological viewpoint.
Some authors have suggested that this apparent disagreement is something like a difference of perspective. Think of an observer standing on the shore of the ocean, watching a ship sail away. From the observer's viewpoint, the ship eventually sinks below the ocean. After a while its hull is no longer visible; only its masts remain, and finally they, too, slip down and are gone. To a passenger on the ship, however, the ship is still sailing along on the surface of the ocean. Biologists and chemists think about bonding differently because they are looking at it from a different viewpoint.
Biologists say that energy is stored in chemical bonds because thinking about things that way is useful to them. It is useful to think of catabolic processes, such as the breakdown of sugars, as energy-releasing. It is useful to think of anabolic processes, such as photosynthesis or the synthesis of complex natural products, as energy-intensive.
Biologists are looking at things purely from the point of view of the biomolecule. Either it is breaking down into smaller pieces (its bonds are breaking), releasing energy, or else it is getting built up into something bigger (its bonds are being made), costing energy.
In a very loose sense, it is as if the reaction of carbohydrate breakdown is pared down to:
$\ce{C6H12O6_{(s)} -> 6CO2_{(g)} + energy} \nonumber$
And the reaction of ATP breakdown is abbreviated to:
$\ce{ATP_{(aq)} -> ADP_{(aq)} + P_{i(aq)} + energy} \nonumber$
In other words, part of the reaction is ignored. That viewpoint allows a focus on the biomolecule, but it neglects some important things. For example, in the breakdown of carbohydrates, it isn't the C-C bond breaking of the carbohydrates that is the source of energy. It is the formation of strong, new O-H and C=O bonds, and other, more subtle changes, that release the energy.
As always, we get more insight into a reaction by looking at the structural formulae in the equation, rather than condensed formulae. This way, we can actually see what bonds are being made and broken.
The case of ATP is a little different. The bonds made and broken are pretty much the same in the breakdown of ATP; loosely, we just trade in one P-O bond for another. This case is more complicated, but the simplest explanation is that ATP cleavage relieves repulsion between the multiple negative charges in the ATP molecule. Energy decreases in the resulting molecules, and the rest of the energy that used to be in the reactants is released.
In the reverse, when ADP is phosphorylated to make ATP, the system goes up in energy (the system just means everything in the reaction; it is everything on one side of the arrow or the other). That energy, however, is not really stored in any chemical bonds. It is distributed throughout the system, for example, in the motions of all of those atoms. The bonds may stretch, getting longer and shorter, but in addition the groups on the ends of the bonds can spin, and the molecules can tumble and zip around through space. There are lots of ways to distribute that energy throughout that entire collection of atoms; it isn't forced to sit in that one bond that was newly formed between two atoms.
So, although the idea of energy being stored in chemical bonds may be very useful in the biology classroom, it is only going to get in your way in the chemistry classroom. You need to be able to take off your biologist's hat and put on your chemist's lab coat when you need it.
Exercise $1$
Our economy is driven largely by the consumption of fossil fuels, such as heptane. Given the following reaction for the breakdown of heptane:
CH3CH2CH2CH2CH2CH2CH3 + 11 O2 → 7 CO2 + 8 H2O
Use the table of bond strengths to determine how much energy is released when a mol of heptane is consumed.
Bond O=O C-C C-H C=O O-H
Average Bond Strength (kcal/mol) 120 80 100 190 110
1. Start by determining the energy needed to break bonds.
2. Determine the energy released when new bonds are made.
3. Determine the overall energy change.
Answer
Bonds Broken:
C-C 6 x 80 kcal/mol = 480 kcal/mol
C-H 16 x 100 kcal/mol = 1,600 kcal/mol
O=O 7 x 120 kcal/mol = 840 kcal/mol
Total: 2,920 kcal/mol
Bonds Made:
C=O 14 x (- 190 kcal/mol) = - 2,660 kcal/mol
O-H 16 x (- 110 kcal/mol) = -1,760 kcal/mol
Total: -4,420 kcal/mol
Overall: 1,240 - 4,420 kcal/mol = -1,500 kcal/mol
Exercise $2$
Use the table of bond strengths to determine how much energy is released when a mol of octane is consumed.
CH3CH2CH2CH2CH2CH2CH2CH3 + 12.5 O2 → 8 CO2 + 9 H2O
Answer
Bonds Broken:
C-C 7 x 80 kcal/mol = 560 kcal/mol
C-H 18 x 100 kcal/mol = 1,800 kcal/mol
O=O 12.5 x 120 kcal/mol = 1,500 kcal/mol
Total: 3,860 kcal/mol
Bonds Made:
C=O 16 x (- 190 kcal/mol) = - 3,040 kcal/mol
O-H 18 x (- 110 kcal/mol) = -1,980 kcal/mol
Total: -5,020 kcal/mol
Overall: 3,860 - 5,020 kcal/mol = -1,160 kcal/mol
Exercise $3$
Given an approximate C-O bond strength of 85 kcal/mol, use the table of bond strengths to determine how much energy is released when a mol of glucose is consumed.
Answer
Bonds Broken:
C-C 6 x 80 kcal/mol = 480 kcal/mol
C-H 7 x 100 kcal/mol = 700 kcal/mol
C-O 7 x 85 kcal/mol = 595 kcal/mol
O-H 5 x 110 kcal/mol = 550 kcal/mol
O=O 6 x 120 kcal/mol = 840 kcal/mol
Total: 3,165 kcal/mol
Bonds Made:
C=O 12 x (- 190 kcal/mol) = - 2,280 kcal/mol
O-H 12 x (- 110 kcal/mol) = -1,320 kcal/mol
Total: -3,600 kcal/mol
Overall: 3,165 - 3,600 kcal/mol = -435 kcal/mol
Exercise $4$
Provide a mechanism for the hydrolysis of ATP to ADP.
Answer
Exercise $5$
Suggest a possible role for magnesium ion in the hydrolysis of ATP.
Answer
In the mechanism for hydrolysis, water acts as a nucleophile and ATP acts as an electrophile. That's a problem because ATP is negatively charged. It will not attract electrons very easily. By binding to magnesium ion (Mg2+), the charge on the ATP will be lowered, accelerating the reaction with water. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/III%3A_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_1/08%3A_Mechanisms_of_Glycolysis/8.01%3A_Introduct.txt |
In the cell, carbohydrates are used as a source of energy. The carbohydrates might have been recently ingested, or they might be released from long-term storage. Carbohydrates can be stored in the form of glycogen, in animals, or starch, in plants. Glycogen and starch are biomacromolecules. They are composed of large collections of glucose molecules bonded together (or enchained) into one, much bigger molecule. They have different structures: starch is composed of very long chains of glucose molecules, whereas glycogen is a highly branched structure, more like a coral or a tree. Specific enzymes can be used to release glucose molecules, one by one, from these structures.
Glycolysis is a biochemical pathway in which glucose is consumed and ATP is produced. ATP is like the spring that powers all of the windup toys of the cell. To compress that spring, energy has to be expended, so glucose is sent through a series of reactions that eventually release some energy that can be used for this purpose. Once the spring is charged, it can release its energy rapidly.
The consumption of glucose is associated with the catabolic process of respiration. In respiration, glucose is combined with oxygen, the reactants are converted to carbon dioxide and water, and energy is released. We could write that reaction as follows:
\[\ce{C6H12O6(s) + 6O2(g) -> 6CO2(g) + 6H2O(l) + energy}\]
However, that reaction is really the sum total of three different processes. In the first process, glycolysis, glucose is broken down partway, forming pyruvate. Some energy is released by the process. In the second process, the tricarboxylic acid (TCA) cycle or citric acid cycle, pyruvate is broken down further to release carbon dioxide. Again, some energy is produced by this process. Both of these pathways produce ATP. Both of these pathways also produce NADH. In the third process, oxidative phosphorylation, NADH is used to power an "electron transport chain", releasing additional energy that is harnessed in order to make more ATP. The TCA cycle will be covered in a later chapter in this section of the book. Oxidative phosphorylation is not discussed until the Reactivity III section of the book, which is concerned with single electron processes.
When looking at biochemical pathways, it is helpful to have a map of the process to get an overview of how all the steps fit together. Glycolysis is sometimes presented in two parts, so two maps are shown below. The first part, Phase One, actually consumes energy; this part is the initial investment needed for a later return.
The map of phase one of glycolysis starts with glucose and leads eventually to the formation of two G3P molecules. Glucose is the initial input, and G3P is the final output; everything else along the way is just an intermediate that is consumed soon after it is made. Along the way, additional inputs to the reaction are shown in red, and outputs are shown in blue. Enzymes and other catalytic factors, which are not consumed by the reaction, are shown in green.
Exercise \(1\)
Provide a net equation for phase one of glycolysis that shows all of the inputs on one side of the reaction arrow and all of the outputs on the other side of the reaction arrow.
Answer
glucose + 2 ATP → 2 G3P + 2 ADP
Note that the first phase of glycolysis actually consumes ATP rather than producing it. This initial investment of energy leads to a later return. The second part of glycolysis, Phase Two, occurs after the glucose has been cleaved in half; this phase leads to the release of energy.
The map of phase two of glycolysis starts with G3P and leads eventually to the formation of pyruvate. G3P is the initial input, and pyruvate is the final output; everything else along the way is just an intermediate that is consumed soon after it is made. Along the way, additional inputs to the reaction are shown in red, and outputs are shown in blue. Enzymes and other catalytic cofactors are shown in green.
Exercise \(2\)
Provide a net equation for phase one of glycolysis that shows all of the inputs on one side of the reaction arrow and all of the outputs on the other side of the reaction arrow.
Answer
G3P + NAD+ + PO43- + 2 ADP → pyr + NADH + 2 ATP + H2O
Exercise \(3\)
Provide a net equation for glycolysis (both phases combined) that shows all of the inputs on one side of the reaction arrow and all of the outputs on the other side of the reaction arrow.
Answer
First we need to realise that one glucose gives rise to two molecules of G3P, so the second phase occurs twice for every glucose molecule consumed.
2 G3P + 2 NAD+ + 2 PO43- + 4 ADP → 2 pyr + 2 NADH + 4 ATP + 2 H2O
Adding the equations for the two phases together gives:
glucose + 2 ATP + 2 G3P + 2 NAD+ + 2 PO43- + 4 ADP → 2 G3P + 2 ADP + 2 pyr + 2 NADH + 4 ATP + 2 H2O
That equation can be simplified, because some things appear on both the left and the right. It's just like algebra.
glucose + 2 NAD+ + 2 PO43- + 2 ADP → 2 pyr + 2 NADH + 2 ATP + 2 H2O
It helps to be able to think about sugars in both cyclic and open chain forms. Phase One can be presented in an alternative way that depicts the carbohydrates as open chains rather than rings.
These two different ways of depicting carbohydrates takes into account that most carbohydrates are dynamic structures. Rather than having one form, they have different forms that can interconvert between each other, and all of these forms are present at equilibrium. One form dominates, and that's usually a ring. For glucose, the six-membered ring or "pyranose" form makes up about 99% of molecules in solution. In water, the beta-form, in which the rightmost OH group in the Haworth projection is cis- to the CH2OH group, makes up about two thirds of the pyranose form. In nonpolar environments the ratio is reversed, with the alpha-form dominating, in which the rightmost OH group in the Haworth projection is trans- to the CH2OH group.
The open chain form makes up less than one percent of glucose molecules in solution. Nevertheless, sometimes it is the open-chain form that actually undergoes reaction, so it is useful to think of carbohydrates both ways.
Exercise \(4\)
Provide a mechanism for the formation of glucopyranose from the chain form of glucose. Assume biological conditions, such as the presence of a lysine residue to assist in proton transfer.
Answer
Exercise \(5\)
In the map of glycolysis, the α-D-glucopyranose form is shown. Draw β-D-glucopyranose.
Answer
Exercise \(6\)
Show the hemiacetals that would form in the following cases.
Exercise \(1\)
Show the pyran rings (six-membered rings with oxygen) that would form in the following cases.
Exercise \(8\)
Show the furan rings (five-membered rings with oxygen) that would form in the following cases. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/III%3A_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_1/08%3A_Mechanisms_of_Glycolysis/8.02%3A_Overview_.txt |
In the next couple of sections, we will look at the changes occurring in reactions of glycolysis and try to understand how those changes may be happening. In some cases, we will develop a formal understanding of the reaction, and return in a later section to see how that reaction is actually catalysed by an enzyme.
The first phase of glycolysis is all about taking the initial carbohydrate, glucose, and getting it into the right form for the energy-releasing, ATP-forming reactions of the second phase. All of the reactions of glycolysis are catalysed by enzymes. If you know anything about enzymes, you may know that there are many controls in place so that their reactions can be turned on or off.
One common element of enzyme control is phosphorylation. In phosphorylation, a phosphate group is added. The phosphate group could be added to the enzyme, or it could be added to the substrate -- the compound on which the enzyme carries out a reaction. If you look again at the map of Phase One, you will see a couple of phosphorylations taking place. Glucose is phosphorylated to start things off, and later on, a fructose phosphate is again phosphorylated to make a fructose bisphosphate.
Phosphorylation typically modifies the interaction of a substrate with its surroundings. It could be that phosphorylation, and the negative charge that results on the glucose, helps the glucose to bind more tightly with the next enzyme in the pathway. It is also believed that the anionic glucose phosphate is less likely to leave the cell, because it can't be taken up at the nonpolar membrane and transported across the membrane to the outside of the cell. A cell that needs energy would find it advantageous to prevent its glucose from escaping.
Exercise \(1\)
Provide a mechanism for the phosphorylation of glucose.
Answer
The second step of glycolysis is an isomerisation. The carbonyl of glucose, on the first carbon in the chain form, migrates to the second carbon in the chain.
The key to understanding this reaction is noting that, when the second carbon in the chain becomes a carbonyl, it loses a hydrogen.
The reaction actually occurs via an enol intermediate. Remember, an enol is just a tautomer of the original compound; that means it is an isomer in which the major difference is the position of one proton. In an enol tautomer, one proton has been moved from the alpha position, next to the carbonyl, over to the carbonyl oxygen.
Exercise \(2\)
Keto-enol tautomerism involves eventual transfer of a proton from one site to another within the same molecule. The reaction is subject to general catalysis, meaning it can be carried out either by acid or by base.
Provide a mechanism for the conversion of 2-propanone into its enol form in the presence of:
1. aqueous hydrochloric acid
2. aqueous sodium hydroxide
Exercise \(3\)
Provide a mechanism for the keto-enol tautomerism of glucose-6-phosphate. Assume acidic, biological conditions (for example, assisted by the acidic form of a lysine side chain).
Answer
Another enol-keto equilibrium is needed in order to complete the overall reaction. This time, it is the enol form going back into a keto form. That means the proton is being transferred from an OH group along the C=C bond (the enol position) back to the alpha position. But take a close look at this molecule. It actually has two different enol OH groups. Either one could lose its proton and turn back into a carbonyl. One of those events leads right back where we started from. The other one leads forward to fructose-6-phosphate.
Why did this step have to take place? Remember, the eventual end product of phase one of glycolysis is a three-carbon sugar, G3P. Right now we have a six carbon sugar (fructose, not glucose, but still a six-carbon sugar). The migration of the carbonyl from the first carbon to the second is what will allow the glucose molecule to break in half, between the third and the fourth carbon. It is going to break in an alpha position, next to a carbonyl. We have just moved the carbonyl to put the alpha position in the right place.
Exercise \(4\)
Provide a mechanism for the keto-enol tautomerism that forms fructose-6-phosphate. Assume basic, biological conditions (for example, assisted by the basic form of a lysine side chain).
Answer
Exercise \(5\)
Frequently, formation of an enol from a ketone could result in different products. Show the enol products (there are at east two in each case) that could form from each of the following compounds.
Exercise \(6\)
Each of the following enols could form two different keto tautomers via keto-enol tautomerism. Show both products in each case.
The subsequent step in the pathway is that second phosphorylation, forming fructose-1,6-bisphosphate. The compound changes from an anion to a dianion, and that change has consequences in how the molecule interacts with its environment. Once again, the molecule becomes even less likely to move into the nonpolar cell membrane. Furthermore, it can now interact with different enzymes in a way it couldn't when it was just a monoanion.
This phosphorylation requires the consumption of another molecule of ATP. Remember, this phase of glycolysis is not the energy-producing part. We are still getting ready for that part. It will happen in phase two, but we still have more to do in phase one on the next page.
Exercise \(7\)
Provide a mechanism for the formation of FBP from F6P. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/III%3A_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_1/08%3A_Mechanisms_of_Glycolysis/8.03%3A_Mechanism.txt |
The first few steps of glycolysis involve relatively minor changes. The sugar is phosphorylated, at two different stages, and a couple of keto-enol tautomerisations are employed to move the carbonyl position one carbon down the chain. The carbonyl migration allows the fructose molecule to undergo a scission or cleavage, in which the molecule is cut in half. That gives a couple of three-carbon sugars. Three-carbon sugars like these are important in phase two of glycolysis, the energy-producing phase.
To understand how the molecule splits in two, we're going to imagine a complementary case, because it might seem more familiar. The reaction shown below is called an aldol reaction. That's the one in which an enolate ion, with a negatively charged carbon next to a carbonyl, donates to a regular carbonyl. The enolate ion acts as the nucleophile, donating electrons to the neutral carbonyl, which acts as the electrophile. The electrophilic carbonyl pops open to make an alkoxide ion, which then picks up a proton to make the alcohol part of the aldol product.
Exercise \(1\)
Show why the enolate ion is very stable, even though other carbon anions typically are not easy to form.
The products of aldol reactions bear a pattern that is easy to spot: there is always a carbonyl-carbon-alcohol sequence, O=C-C-C-OH. The middle carbon was the nucleophilic carbon; the carbon attached to the OH is the former electrophilic carbonyl carbon. The thing is, aldol reactions are reversible. They can go backward again, re-forming their starting materials. So if you look at an aldol product, with that carbonyl-carbon-alcohol sequence, you can imagine the molecule splitting apart, with a break between the middle carbon and the alcohol carbon.
The open arrow in the picture above is a specific symbol in chemistry that means "can be made from". The aldol product can be made from the enolate and the carbonyl. But if an aldol reaction is reversible, then the reverse is also true: an enolate and a carbonyl can be formed from an aldol product, characterised by its O=C-C-C-OH pattern. This reverse reaction is called a retro-aldol reaction.
Exercise \(2\)
Predict the products of the following aldol reactions.
Exercise \(3\)
Predict the precursors that could be used to form these compounds through aldol reactions.
What makes retro-aldol reactions possible? The same thing that made the forward aldol reaction possible: the stability of an enolate anion. Carbon-carbon bonds are actually pretty difficult to break, because the pair of electrons forming the bond aren't likely to shift one way or the other. The enolate provides a release mechanism. A pair of electrons can shift to the carbon next to the carbonyl, or the alpha position, because they will form a stable anion there.
The enolate ion that would result from a retro-aldol reaction of FBP would be stable because the negative charge is delocalised. That charge delocalisation of an enolate anion is part of what makes retro-aldol reactions possible in the first place. You can't just have a C-C bond break in a random place and have one of the carbons form an anion; that would never happen. If the anion is somehow stabilised, however, then that changes everything.
Exercise \(4\)
Provide a mechanism for the retro-aldol reaction of fructose-1,6-bisphosphate.
Answer
That's not the whole story, though. In reality, the retro-aldol reaction of FBP undergoes iminium catalysis. That means the reaction is accelerated through the action of a lysine residue in the enzyme, fructose bisphosphate aldolase. The lysine, which contains a nucleophilic amine side chain, transforms the carbonyl of fructose-1,6-bisphosphate into an iminium ion. Biochemists often call this kind of structure a Schiff base. Its structure is very similar to the original, but it contains a C=N group instead of a C=O group. At common biological pH, chances are that the nitrogen in the Schiff base is protonated.
How does that help things? Well, charge stability is always a key factor during reactions. It costs energy to stabilise charges. If we can avoid generating the negative charge of an enolate anion, even though it's relatively stable, that might make things easier. The Schiff base is already positively charged because of its basicity, so when it accepts a pair of electrons it becomes neutral. It has no charge to need stabilising. That step ought to be less difficult than a similar step that results in an anion.
So, by adding a couple of steps to the reaction -- adding the lysine onto the sugar and then taking it off again -- the reaction is accelerated. It can seem confusing that we add steps to the reaction, making the overall reaction longer, but the reaction gets faster. That's because, even though there are additional steps, each individual step is a lot easier than it was before. It's like the difference between jumping across a river and stepping across on a series of stones. Taking the extra steps may be the faster and surest way to get there.
The products of FBP cleavage are glyceraldehyde-3-phosphate (G3P) and dihydroxyacetone phosphate (DHAP). The dihydroxyacetone phosphate subsequently undergoes isomerisation to give another molecule of glyceraldehyde-3-phosphate. This reaction occurs via initial formation of an enol tautomer. It's similar to the isomerisation of glucose-6-phosphate to fructose-6-phosphate that we saw earlier.
So, when the fructose-1,6-bisphosphate forms, it makes one molecule of glyceraldehyde-3-phosphate and one molecule of dihydroxyacetone phosphate, but that dihydroxyacetone phosphate molecule is then converted into another molecule of glyceraldehyde-3-phosphate. Thus, the overall outcome of this part of the reaction is to transform the six-carbon sugar (first glucose and later fructose) into glyceraldehyde-3-phosphate. This is where phase two of glycolysis begins.
Exercise \(5\)
Provide a mechanism for the isomerisation reaction of dihydroxyacetone phosphate to glyceraldehyde-3-phosphate.
Answer | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/III%3A_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_1/08%3A_Mechanisms_of_Glycolysis/8.04%3A_Mechanism.txt |
Reactions in biochemistry are usually catalysed by enzymes. In a catalysed reaction, an alternative pathway is available that makes it easier to get from reactants to products. That doesn't mean that there are fewer steps. In fact, normally there are more steps in a catalysed reaction than there are in an uncatalysed one. It does mean that the overall energy needed to traverse the catalysed barrier is lower than the energy needed to surpass the uncatalysed barrier. It's like taking the stairs up to the second floor rather than taking a running leap at the window: more steps, but overall it will save time.
So far, the reactions we have seen in glycolysis are just the overall reactions. By looking at the overall reactions, we get a pretty good sense about what is happening at each stage of the pathway. We even get some sense of how those reactions might happen, because we can identify familar nucleophiles and electrophiles that appear to be involved. Here, we will take a more detailed look at the catalytic pathways taken during the first phase of glycolysis.
Essentially every step of glycolysis involves catalysis, and so the reactions entail cofactors and detailed steps that we have glossed over until now. The first step, phosphorylation of glucose to afford glucose-6-phosphate, requires the consumption of ATP. During that step, the terminal hydroxy group of glucose takes up phosphate from ATP, leaving ADP.
It seems like that first step should be pretty straightforward, because we think of ATP as this high-energy power source for the cell, so it must be really reactive. ATP is not quite as reactive as you might think, though. That's a good thing. If it reacted too readily, it couldn't travel around the cell at all; it would get hydrolysed the first time it encountered a water molecule, and there really are an awful lot of those in a typical cell. In order to react, the ATP needs to be activated.
Part of the catalysis of the phosphorylation of glucose simply involves binding ATP to a magnesium ion. Once bound to the magnesium ion, the ATP becomes more electrophilic, because of that positive charge on the magnesium ion.
Although a nucleophile, such as water, is unlikely to donate to ATP -- partly because of the negative charge on the ATP -- it is more likely to donate to ATP once coordination takes place, because the magnesium ion has leveled out that negative charge.
Another aspect of catalysis in the phosphorylation of glucose involves the removal of a proton. A hydroxyl group is converted to a phosphate, and a proton is lost. Acid-base catalysis is quite common in biochemistry. There are only a handful of amino acids that commonly participate in deprotonation steps: aspartate, glutamate, lysine, and histidine. All of these residues have two structures in equilibrium: a protonated one and a non-protonated one. The non-protonated form is ready to remove a proton when needed.
Similarly, acid-base catalysis is carried out by nearby amino acid residues in the active site of the enzyme that carries out the isomerisation of glucose-6-phosphate to fructose-6-phosphate.
Phosphoglucoisomerase accomplishes this task by removing a proton from an alpha position, and also from an O-H group, as well as donation of protons to a different alpha position and a different oxygen.
An enol intermediate is the halfway point between the two isomers. If the proton were to be removed from that end hydroxy group, and a proton put back in the alpha position again, the structure would return to the original G6P. If, instead, a proton is removed from the second hydroxyl group along the chain, a different structure results.
Of course, the same sort of catalytic requirements arise again during the conversion of F6P to FBP. ATP must be activated by magnesium, and proton transfers must be carried out by acidic and basic amino acid residues.
The magnesium will bind to the ATP in order to reduce the amount of negative charge. That way, the nucleophilic alcohol can donate electrons more easily. The proton from the alcohol group can be removed by an amino acid side chain, such as the negatively charged carboxylate of aspartate or glutamate, or a neutral histidine.
A completely different kind of catalysis occurs during the scission step of phase one, when the six-carbon sugar is cleaved into a pair of three-carbon sugars. You may recall that this cleavage is accomplished via a retro-aldol reaction: an aldol reaction goes into reverse, spitting out an enolate or enol and a carbonyl.
That retro-aldol step is accomplished via iminium ion catalysis. Very often in biochemical reactions, a lysine residue binds with a carbonyl to form either an iminium ion, containing an electrophilic C=N bond, or an enamine, with a nucleophilic N-C=C unit. The process begins with donation of the lysine, in its non-protonated form, to the carbonyl of FBP.
Frequently, enzymatic reactions are presented in a more condensed form. Multiple steps are shown at once; but not just any steps. In the example below, the FBP reacts with both the lysine and the aspartate at the same time. We never draw three molecules coming together at once, because the probability of three molecules colliding at the same time is just about nil. In this case, that's not a problem; the aspartate and lysine are both part of the same molecule. Furthermore, by this point, the FBP has already bound to the enzyme, so the whole thing is one big assembly.
Subsequently, the carbinolamine undergoes displacement of water to yield the imine, C=N.
Imines and enamines are roughly equivalent to carbonyls and enolates, respectively. Enamines are very good nucleophiles, just like enolates. Enolates are a little better, because of the negative charge. However, even though they are neutral nucleophiles, enamines are almost as good, because the reaction is driven by a less electronegative nitrogen atom; which is more willing to donate its electrons. Furthermore, in the environment of the cell, enamines form much more easily than enolates. That's because there isn't a whole lot of LDA or even NaOH floating around inside your cells. As a result, enamines are often employed as nucleophile in cases where you might think of using an enolate.
Exercise \(1\)
Provide mechanisms for the following aldol-like reactions.
The imine unit isn't an inherently better electrophile than a carbonyl; after all, it contains a less polar C=N bond instead of a C=O bond. However, the nitrogen in an imine is much more basic than the oxygen in a carbonyl. It can be protonated quite easily under biological conditions. The resulting iminium ion, containing the C=N-H+ unit, is an activated electrophile. Of course it reacts much more quickly than a regular carbonyl.
Because carbonyls can easily form either imines or enamines, they will often be converted into those compounds in order to do reactions. This process is called enamine and imine catalysis.
Aldol reactions are actually reversible. A retro-aldol reaction is just the aldol reaction going in reverse. In that case, an enolate or an enol or an enamine might come out as a leaving group, rather than acting as the initial nucleophile.
Exercise \(2\)
Provide mechanisms for the following retro-aldol reactions.
In the context of a retro-aldol reaction, we need to think about the catalysis backwards. Instead of an iminium ion acting as an activated intermediate to receive a nucleophile, it is accepting electrons to form a leaving group. Instead of having an enolate leaving group in the retro-aldol reaction, we have an enamine leaving group.
The lysine group, which just came on board to catalyse the reaction, is liberated by addition of a water molecule.
There is one last reaction in phase one, the fifth overall in glycolysis. It's the conversion of DHAP to G3P; but that's just another keto-enol tautomerism. The catalytic mechanisms will be very much like those seen in the conversion of G6P to F6P.
Basic amino acid sites such as neutral histidine and anionic aspartate can readily remove a proton, whereas the corresponding conjugates can supply a proton. Tautomerism simply requires removal of a proton from an alpha position. When a proton is returned, it goes to the other end of the double bond in the enol. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/III%3A_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_1/08%3A_Mechanisms_of_Glycolysis/8.05%3A_Catalysis.txt |
Phase one of glycolysis is setting the stage. An initial investment of ATP is needed to get the system ready to produce more ATP; it's like a farmer who sows sunflower seeds in the spring to get a bigger crop of sunflower seeds in the fall.
Two molecules of glyceraldehyde-3-phosphate are produced in phase one of glycolysis. Both of those molecules enter into phase two. The first step in this phase is an oxidation reaction. In organic chemistry, the term oxidation suggests that a carbon atom is getting fewer bonds to hydrogen, or more bonds to oxygen. (The complementary term, reduction, suggests new carbon-hydrogen bonds are forming, or carbon-oxygen bonds are disappearing.) Notice that the carbonyl carbon, C=O, is becoming a carboxylic carbon, O-C=O.
That isn't all that's going on. There is also a phosphorylation step here, but this time ATP is not required. The source of the phosphate group is a simple phosphate ion.
Furthermore, a molecule of NADH is produced. You might remember that NAD+ can pick up a hydride ion (that's right, H- instead of H+) to become NADH. In this case, the hydride ion is coming from the aldehyde that is converted into a carboxyloid, the phosphoric anhydride group in BPG.
This first step has consequences for energy-packaging pathways further downstream. NADH is the starting material for oxidative phosphorylation, an elegant process in which electrons are passed from one metal ion to another within membrane-bound proteins; as the electrons move across the membrane, they draw oppositely-charged protons along with them. A proton gradient builds up, with protons on one side of the membrane outnumbering those on the other; this osmotic pressure is relieved when the protons find a channel to pour back through the membrane, but as they do so they turn a molecular millwheel that drives the production of more ATP. Remember, oxidative phosphorylation, along with glycolysis and the citric acid cycle, is one of the three pathways that together make up the process of respiration.
There is also a more immediate energy-packaging result. That first step produces 1,3-bisphosphoglycerate, which is primed to deliver a phosphate to a molecule of ADP. In addition to ATP, a molecule of 3-phosphoglycerate is left behind.
A slight modification of the 3-phosphoglycerate ensues. In this step, the phosphate group migrates from the 3-position to the 2-position, resulting in 2-phosphoglycerate.
Subsequently, the 2-phosphoglycerate undergoes a dehydration, the loss of water. You might recall that dehydrations sometimes occur after aldol reactions: the O=C-CH-C-OH loses a proton at the alpha position and a hydroxide at the beta position to give the enone group, O=C-C=C, and water, HOH. That step is often driven by the conjugated system that results. In this case, the conjugated product is phosphoenolpyruvate.
The final step in glycolysis is the loss of the phosphate group from phosphoenolpyruvate. This phosphate group is transferred to another molecule of ADP, forming ATP. The ATP can then be used to power processes elsewhere in the cell. Note that this is the second molecule of ATP produced during phase two. Since each molecule of glucose produces two three-carbon sugars that enter into phase two, a total of four molecules of ATP are produced per glucose. Remember, phase one required the consumption of two molecules of glucose, so this yield represents a doubling of the initial investment of ATP. It's like putting two dollars in the bank and getting four dollars back out.
That's the end of glycolysis, but that's not the end of the story. So far, glucose has only been broken down to a three-carbon sugar. When we think of respiration, we think of glucose breaking down all the way to carbon dioxide. That part continues in the tricarboxylic acid cycle, or TCA cycle. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/III%3A_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_1/08%3A_Mechanisms_of_Glycolysis/8.06%3A_Mechanism.txt |
In phase two of glycolysis, glyceraldehyde-3-phosphate proceeds through an oxidation reaction, in which an aldehyde hydrogen is replaced by a phosphate group. If you are scratching your head, wondering how that reaction might happen, then that's a good thing. The hydrogen is not a good leaving group, and you haven't seen much precedent for this kind of reaction.
This step looks like a very good place for catalysis. The reaction doesn't look like it will work very well, and so we need an alternative, lower-energy pathway. At the very least, this is a good situation for the enzymatic strategy of approximation, in which two substrates are held near each other by their respective placements in the active site. Approximation is needed because of that poor hydride leaving group, H-. It really can't be displaced from the molecule to become a hydride ion in free solution, because it isn't stable enough. It needs that NAD+ acceptor to be waiting for it as it comes off the molecule of glyceraldehyde-3-phosphate.
But the hydride isn't displaced by the phosphate directly; there is an invisible step in between. A cysteine residue from the enzyme first donates to the carbonyl. Sulfur is a very good nucleophile for carbon electrophiles; sulfur is a little less electronegative and more polarizable than oxygen, so it can donate electrons relatively easily.
That donation results in the formation of a hemithioacetal. In addition, it tethers the glyceraldehyde molecule to the enzyme. This feature is an example of the group transfer strategy, in which the substrate becomes temporarily attached to the enzyme that is working on it. This group transfer might be explained as gaining an entropic edge to help get through an enthalpically difficult step. This next part may be a little bit tricky, so let's just tie the substrate down for a moment.
So, now we are ready to pi-donate from the tetrahedral intermediate, pushing the hydride onto the waiting NAD+, like a fast pitch straight into the catcher's mitt.
Of course, group transfer always relies on the substrate being cut from the enzyme again. Otherwise, the transfer would be an example of irreversible inhibition; the substrate molecule would be stuck in the enzyme, forever blocking its active site. The thing about sulfur, though, is that its polarizability makes it a pretty good leaving group. Remember, too, that thioesters are more reactive than regular esters. Thioesters sit higher on the ski hill illustrating the relative reactivity of carboxyloids. That means that the use of a cysteine residue as the tethering group, rather than a lysine's nitrogen or a serine's oxygen, is particularly advantageous for the cleavage step.
The cleavage step involves donation from an inorganic phosphate ion. The carbonyl opens up to give a tetrahedral intermediate, and when that intermediate collapses again, the cysteine is released.
The second step of phase two looks a little simpler. It's just a displacement of a carboxylate leaving group from a phosphoanhydride.
Remember, like the thioester it was formed from, this anhydride was high on the ski hill, because the carboxylate anion is a good leaving group. The phosphate donates to the phosphoryl group, resulting in a five-coordinate phosphorus. This intermediate collapses, ejecting the carboxylate leaving group.
The third step also appears straightforward. It looks like the phosphate is just moving from one oxygen to another.
In practice, things are slightly more complicated. In animals, the phosphate that comes from the 3-position isn't the same as the phosphate that ends up at the 2-position. In other words, rather than just moving the phosphate from one place to another, a new phosphate is added before the old one is removed.
The new phosphate is delivered from a modified histidine residue in the anzyme. The same histidine can then pick up the old phosphate from the 3-position, which it will then hold onto, waiting to deliver it to the next substrate molecule that arrives in the enzyme.
The fourth step is, in principle, a simple dehydration.
Once again, this step requires a cation to activate it. In some cases, two magnesium ions appear to be involved: one to bind the phosphate and one to bind the carboxylate. Other cases employ a magesium ion and a potassium ion.
Either way, coordination to a metal ion probably lowers the pKa of the alpha hydrogen. Instead of deprotonating to produce a tri-anion or tetra-anion (depending whether the phosphate is already singly deprotonated or doubly deprotonated), the metal-bound substrate will produce an anion with overall lower negative charge that we would have otherwise. The deprotonation seems to be carried out by a lysine in some enzymes and a histidine in others.
After that, the beta- hydroxy group is lost. It needs to pick up a proton to become water. That proton may be supplied by a nearby glutamate, which shuttles the proton from elsewhere.
The final step in glycolysis is another phosphate transfer.
Once again, ADP is the nucelophile that displaces the leaving group from the phosphate electrophile. This time, the leaving group is an enolate anion. The reaction requires magnesium ions to hold the ADP in place. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/III%3A_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_1/08%3A_Mechanisms_of_Glycolysis/8.07%3A_Catalysis.txt |
Glycolysis is intimately linked to the release of energy in biological systems, and harnessing that energy to do work. That's what the field of thermodynamics is all about. In this section, we will take a very brief look at some of the energetic considerations of this pathway.
We have seen that glycolysis is a sequence of reactions leading from one intermediate compound in the pathway to the next. (To see that pathway again, click here.) Inevitably, there are energy changes associated with each of those reactions. Some of the reactions may be endothermic, others may be exothermic; some may be essentially irreversible, whereas others may occur in equilibrium. If we map these energy changes out from start to finish, we get a picture like the one below. It's a roller coaster, with lots of energetic drops but just as many hills, and it becomes difficult to think of glycolysis as a process that releases energy, except for the dramatic drop in the last couple of steps.
Where does a picture like this one come from? Well, it depicts a series of reactions, and the energy change associated with each reaction. We can determine the energy change associated with a specific reaction using calorimetry. A calorimeter is a well-insulated device in which we can perform a reaction. A thermometer tells us the temperature change as a result of the reaction. We can calibrate the device by releasing known amounts of heat and seeing how much its temperature rises. Consequently, we can also use that correlation backwards: given the temperature rise, we can deduce how much energy was released during a reaction.
Now, if enough people have studied this sort of thing for long enough, we can begin to compile a lot of data. Given enough data, you actually might not need to perform calorimetry to determine how much energy is involved in a reaction.
To illustrate why, consider one of the most common kinds of thermodynamic data you can find: heats of formation. The heat of formation of a compound is the energy involved when the compound is formed from the elements. So, for example, the heat of formation of methane would be the energy involved when hydrogen cas combines with carbon to form methane:
$\ce{2H2 + C -> CH4 \: \: \Delta H = ??} \nonumber$
It would be difficult to perform calorimetry in this case. First of all, there are just too many things that could happen if you managed to get hydrogen and carbon to combine; there are many other compounds made from hydrogen and carbon, so who knows what reaction would really occur?
But we find that heat of formation indirectly, using other data. We can burn methane:
$\ce{2O2 + CH4 -> CO2 + 2H2O \: \: \Delta H = -802 \frac{kJ}{mol}} \nonumber$
We can burn hydrogen to get water:
$\ce{H2 + 0.5 O2 -> H2O \: \: \Delta H = -285_{.}8 \frac{kJ}{mol}} \nonumber$
We can burn carbon to get carbon dioxide:
$\ce{O2 + C -> CO2 \: \: \Delta H = -393_{.}5 \frac{kJ}{mol}} \nonumber$
Well, that just seems like a series of random facts, but equations of reaction are quite a bit like algebraic equations, and those reaction arrows are quite a bit like equals signs. If we keep that in mind, we can manipulate these equations to get useful information. For example, what would happen if we took the middle reaction and multiplied it by two?
$\ce{2H2 + O2 -> 2H2O \: \: \Delta H = -571_{.}6 \frac{kJ}{mol}} \nonumber$
Just as in algebra, if we multiply every term in an equation by the same factor, we end up with an equivalent equation. It's a perfectly legal operation. Note that if we multiple the equation by two, we also multiply the energy by two; it's part of the equation.
Now, you probably already know what happens if we consider one of these equations in reverse:
$\ce{2CO2 + 2H2O -> 2O2 + CH4 \: \: \Delta H = +802 \frac{kJ}{mol}} \nonumber$
If the reaction is exothermic in one direction, then it must be endothermic in the other. One way is downhill, so the other way is uphill.
Look what happens if we add these three reactions together in their current forms:
$\ce{CO2 + 2H2O -> 2O2 + CH4 \: \: \Delta H = + 890_{.}3 \frac{kJ}{mol}} \nonumber$
$\ce{2H2 + O2 -> 2H2O \: \: \: \: \: \: \: \Delta H =-571_{.}6 \frac{kJ}{mol}} \nonumber$
$\ce{O2 + C -> CO2 \: \: \: \: \: \: \: \Delta H = -393_{.}5 \frac{kJ}{mol}} \nonumber$
$\ce{CO2 + 2H2O + 2O2 +2H2 + C -> 2O2 + CH4 + 2H2O + CO2 \: \: \Delta H = +802 \frac{kJ}{mol}} \nonumber$
Several things cancel on the left and right, leaving:
$\ce{2H2 + C -> CH4 \: \: \Delta H = -74_{.}8 \frac{kJ}{mol}} \nonumber$
What that means is that, if we have energetic information about some reactions, and we can combine the equations for those reactions to get a new equation of reaction, then we automatically get the energy associated with that new reaction.
Essentially, if we want to know about the energetics of producing methane from carbon and hydrogen, then it doesn't matter how we get from the carbon and the hydrogen to the methane. We can first take the carbon and combine it with oxygen, not hydrogen, and make carbon dioxide. Then, we can take hydrogen and combine it with oxygen, not carbon, to make water. If, finally, we combine the water and the carbon dioxide we have made and produce methane, then the energy of that whole, roundabout process is the same as if we converted the carbon and the hydrogen directly into methane.
This idea illustrates something called Hess' Law. The overall energy required to get from one set of reactants to another set of products is always the same, regardless of the path taken. Hess' Law is true because energy is a "state function". If we know the state that something is currently in - for example, methane in the gas phase at a certain temperature and pressure - then we know its energy. It doesn't matter what has happened to it before, or how it got to its current state.
Exercise $1$
Problem GL9.1.
In the next section, we are going to see a slightly different picture of the energetic terrain of glycolysis. Rather than the roller coaster ride we saw before, we will find that glycolysis exists mostly on an energetic plain, with just a couple of steep drops. The reason for that has to do with the relative concentrations of the different species under cellular conditions.
8.12: Solutions
Bonds Broken:
$C-C \: 6 \times 80 \frac{kcal}{mol} = 480 \frac{kcal}{mol} \nonumber$
$C-H \: 16 \times 100 \frac{kcal}{mol} = 1600 \frac{kcal}{mol} \nonumber$
$O=O \: 7 \times 120 \frac{kcal}{mol} = 840 \frac{kcal}{mol} \nonumber$
Total: 2,920 kcal/mol
Bonds Made:
$C=O \: 14 \times (-190 \frac{kcal}{mol}) = -2660 \frac{kcal}{mol} \nonumber$
$O-H \: 16 \times (-110 \frac{kcal}{mol}) = -1760 \frac{kcal}{mol} \nonumber$
Total: -4,420 kcal/mol
Overall: $1240- 4420 \frac{kcal}{mol} = -1500 \frac{kcal}{mol}$
Bonds Broken:
$C-C \: 7 \times 80 \frac{kcal}{mol} =560 \frac{kcal}{mol} \nonumber$
$C-H \: 18 \times 100 \frac{kcal}{mol} = 1800 \frac{kcal}{mol} \nonumber$
$O=O \: 12.5 \times 120 \frac{kcal}{mol} = 1500 \frac{kcal}{mol} \nonumber$
Total: 3,860 kcal/mol
Bonds Made:
$C=O \: 16 \times (-190 \frac{kcal}{mol}) = -3040 \frac{kcal}{mol} \nonumber$
$O-H \: 18 \times (-110 \frac{kcal}{mol}) = -1980 \frac{kcal}{mol} \nonumber$
Total: -5,020 kcal/mol
Overall: $3860 - 5020 \frac{kcal}{mol}= -1160 \frac{kcal}{mol}$
Bonds Broken:
$C-C \: 6 \times 80 \frac{kcal}{mol} = 480 \frac{kcal}{mol} \nonumber$
$C-H \: 7 \times 100 \frac{kcal}{mol} = 700 \frac{kcal}{mol} \nonumber$
$C-O \: 7 \times 85 \frac{kcal}{mol} = 595 \frac{kcal}{mol} \nonumber$
$O-H \: 5 \times 110 \frac{kcal}{mol} = 550 \frac{kcal}{mol} \nonumber$
$O=O \: 6 \times 120 \frac{kcal}{mol} = 840 \frac{kcal}{mol} \nonumber$
Total: 3,165 kcal/mol
Bonds Made:
$C=O \: 12 \times (-190 \frac{kcal}{mol}) = -2280 \frac{kcal}{mol} \nonumber$
$O-H \: 12 \times (-110 \frac{kcal}{mol} = -1320 \frac{kcal}{mol} \nonumber$
Total: -3,600 kcal/mol
Overall: $3165 - 3600 \frac{kcal}{mol} = -435 \frac{kcal}{mol}$
In the mechanism for hydrolysis, water acts as a nucleophile and ATP acts as an electrophile. That's a problem because ATP is negatively charged. It will not attract electrons very easily. By binding to magnesium ion (Mg2+), the charge on the ATP will be lowered, accelerating the reaction with water.
$glucose + 2ATP \rightarrow 2G3P + 2ADP \nonumber$
$G3O + NAD^{+} + PO_{4}^{3-} + 2ADP \rightarrow pyr + NADH + 2ATP + H_{2}O \nonumber$
First we need to realise that one glucose gives rise to two molecules of G3P, so the second phase occurs twice for every glucose molecule consumed.
$2G3P + 2NAD^{+} + 2PO_{4}^{3-} + 4ADP \rightarrow 2pyr + 2NADH + 4ATP + 2H_{2}O \nonumber$
Adding the equations for the two phases together gives:
$glucose + 2ATP + 2G3P + 2NAD^{+} + 2PO_{4}^{3-} + 4ADP \rightarrow 2G3P + 2ADP + 2pyr + 2NADH + 4ATP + 2H_{2}O \nonumber$
That equation can be simplified, because some things appear on both the left and the right. It's just like algebra.
$glucose + 2NAD^{+} + 2PO_{4}^{3-} + 2ADP \rightarrow 2pyr + 2NADH + 2ATP + 2H_{2}O \nonumber$ | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/III%3A_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_1/08%3A_Mechanisms_of_Glycolysis/8.08%3A_Thermodyn.txt |
Fatty acids are long-chain carboxylic acids. They are metabolically important because they can be used as high-energy fuel sources in the cell. Like gasoline, they contain long hydrocarbon chains, holding plenty of feedstocks for the formation of strong O-H and C=O bonds.
Fatty acids play a number of different roles, beyond serving as a fuel source. They can also be stored for the long term as triglycerides; we frequently see them modified in an ester form. These ester forms may be stored as fats, to be released when other energy stores are getting low, or they may be used to build cell membranes. For other purposes, they are modified into a thioester form. However, these modifications occur via carboxyloid substitutions that can be reversible. Hence, we can think of those other forms as pools of fatty acids because they can easily be converted into fatty acids if needed.
Where do fatty acids come from? Apart from a couple of them that we need to take in from certain foods, we can make them ourselves, if we have to, in our cells. So, we are going to take a look at the essential synthetic transformations that build up a fatty acid molecule from smaller molecules available in the cell.
The pathway for fatty acid biosynthesis is shown below. Inputs into the system are shown in red, outputs are shown in blue, and the enzymatic domains are shown in green. Interestingly, plants and bacteria use a separate enzyme for each of these steps, whereas fungi and animals employ a megasynthase, a giant enzyme that can perform all of the steps in sequence, like one big fatty acid factory.
The two units coming into the cycle are malonyl coenzyme A and acetyl coenzyme A. They join together to make a longer chain. The subscript n in the drawing is any even integer including zero. After the addition of the malonyl and acetyl thioesters at the top, we have a four carbon chain (n = 0). However, the product of the first cycle gets recycled and anothe rpair of carbons gets added, making a six-carbon chain, and so on.
In the following pages we'll develop some of the ideas about how all of this works from the molecular point of view.
10.02: Transformati
When we looked earlier at the citric acid cycle, we saw that acetyl coenzyme A, or AcCoA, was a key link between that process and glycolysis. This two-carbon remnant of glucose metabolism undergoes additional metabolic steps that harness energy in the form of ATP. But that's not the only purpose of AcCoA; it also serves as a building block for other compounds that are necessary for the survival of the cell or the organism.
Take myristic acid as an example, and think about how simple building blocks like AcCoA might be used to make it. AcCoA is a thioester, much like an ester but with a sulfur in place of one of the oxygen atoms. If we need to build myristic acid up from smaller pieces, we are going to need carbon-carbon bond-forming steps or homologation steps. You may not realise it, but there aren't an awful lot of those; their number just doesn't compare to the number of reactions that will simply switch out one heteroatom for another, such as an oxygen for a nitrogen. So, there may be a couple of choices, but not many.
The obvious carbon-carbon bond-forming reaction to perform with an ester is a Claisen condensation. In a Claisen condensation, the leaving group attached to the carbonyl is replaced by an enolate nucleophile. If the enolate nucleophile comes from AcCoA, then two more carbons will be added to the structure of the electrophile. So, if we were going to add two carbons to something in order to get myristic acid, it should be something with twelve carbons, like lauric acid.
Remember, the open arrow (below) means "comes from", so this picture means the target compound, myristic acid thioester, could come from the two synthons on the right.
This isn't exactly how it happens in the cell, but there are a couple of complications in the actual biosynthesis and we will work our way up to them.
Retrosynthetically, we could think of myristic acid (or the corresponding thioester, easily hydrolysed to give the free fatty acid) as potentially being made from two simpler synthetic precursors: lauric acid (or its thioester) and the enolate ion of AcCoA. The AcCoA would add another two carbons to the chain of the lauric acid, making a fourteen-carbon chain.
Exercise \(1\)
Show the product of a Claisen condensation between the enolate anion of acetyl CoA and another molecule of acetyl CoA.
Answer
Where would the lauric acid come from? We will need to look further into that question, but you may already have thought of another problem that we need to solve, first: Although a Claisen condensation of lauric acid thioester and AcCoA enolate would result in a two-carbon chain extension, it would also bring an unwanted oxygen into the structure. There is a missing link between this structure and myristic acid. That oxygen needs to be replaced by a pair of hydrogens.
We can certainly get started by replacing one of the C=O bonds with a C-H bond. That should bring to mind a familiar carbonyl reaction. In the lab, we could use a hydride reagent, such as sodium borohydride (although we would have to worry about whether that thioester might react, too). In the cell, the equivalent reagent would be another hydride donor such as NADH.
That gets us part of the way there. How do we get rid of the other bond to oxygen? How do we add a second hydrogen? The first question might be answered by thinking about something else that might be familar: an aldol addition. The reason that might be relevant is that the compound now contains a beta-hydroxy carbonyl. We see something similar in an aldol addition. Aldol additions frequently result in beta-hydroxy carbonyl units. However, they sometimes proceed though a subsequent elimination reaction, the loss of water to give an enone ("een-own"). That variation is sometimes called an aldol condensation to distinguish the two outcomes, although the two terms are used loosely.
Exercise \(2\)
What is the driving force for the dehydration reaction (loss of water) after the initial aldol addition?
Answer
The product becomes conjugated. In general, the more conjugation there is in the product of an aldol addition, the more likely is a subsequent condensation (elimination or dehydration). However, other conditions can lead to the loss of water.
In laboratory chemistry, that condensation often results in the presence of strong base and high heat -- conditions that might be a little hard on a cell. We'll find that the reaction conditions are much milder in the cell.
Exercise \(3\)
What thermodynamic factor leads increased heat to promote the dehydration reaction?
Answer
Entropy. The dehydration or elimination takes one molecule (the beta-hydroxy thioester) and converts it into two molecules (the water and the alpha,beta-unsaturated thioester. That change represents an increase in internal entropy. Because the entropy term in free energy is weighted by temperature (ΔG = ΔH - TΔS), it predominates as the temperature rises.
Now the problem of replacing that extra oxygen with two hydrogens is almost solved. The first hydrogen was added as a hydride, and that would work for the second one, too. This time, we need a 1,4-addition rather than a regular 1,2-addition. In the lab, that might be helped by adding a Lewis acid under the right conditions. In the cell, it might be aided by positioning of the hydride reducing agent in the right place in the enzyme, so that it delivers the hydride to the right place.
If we put all of those steps together, we can get from lauric acid thioester to myristic acid thioester in just a few steps.
Hold onto that thought, because we can use it again to make lauric acid from one that is two carbons shorter. In fact, we can keep repeating this cycle over and over until we work our way back to acetyl coenzyme A. Two molecules of AcCoA, one acting as the electrophile and the other as the enolate, could be used to make a four-carbon thioester. Adding another AcCoA would make a six-carbon thioester, and so on.
This dependence on a two-carbon building block is the reason for the vast preponderance of even-numbered carbon chains in natural fatty acids.
Next, we will take a look at how the cell actually shepherds these molecules through these kinds of reactions. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/III%3A_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_1/10%3A_Fatty_Acid_Synthesis/10.01%3A_Overview_of_.txt |
After the last page, we would have you believe that fatty acids are constructed from acetyl coenzyme A units, each acting as consecutive nucleophiles, building the structure two carbons at a time.
There are a number of aspects of the biosynthesis that we didn't consider in that synthetic overview, but we are going to add some extra levels of understanding now. The most obvious difference is that the biosynthesis does not even involve acetyl coenzyme A, although it does involve a related thioester. Prior to the reaction, the initial acetyl coenzyme A electrophile is attached to an acyl carrier protein (ACP) via a cysteine residue on the protein.
You can think of this as a regulatory step. The attachment of the acetyl group to the acyl carrier protein allows it to interact with the series of enzyme domains responsible for fatty acid biosynthesis.
This protein-attaching step is catalysed by malonyl-CoA-acetyl-CoA-ACP transcylase (everyone just calls it MAT). MAT uses a group transfer catalytic strategy, securing the acyl (or malonyl) group in place with a serine residue. The serine hydroxy group displaces the CoAS- thiolate from the acetyl, temporarily making it a regular ester. The serine is then displaced by a cysteine residue on the acyl carrier protein, or ACP. After that, the same sequence occurs with the malonyl group.
As in many such cases, the nucleophilicity of the key serine residue in MAT is enhanced through a hydrogen bond with a neighbouring histidine. This situation is pretty common in group transfer catalysis, especially in proteases. Frequently in these situations, a nearby aspartate or helps to in turn deprotonate the histidine (that's the so-called "catalytic triad" of ser-his-asp), but that part doesn't happen this time.
There is another big difference in the real biosynthetic route. In the Claisen condensation steps, the nucleophile does not really come from acetyl coenzyme A. Instead, it comes from malonyl coenzyme A. Acetyl coenzyme A and malonyl coenzyme A come together in a reaction catalysed by beta-ketoacyl-ACP synthase.
Notice that the malonyl group is also first attached to an acyl carrier protein, just like the acetyl group. That step is also carried out by MAT, and it happens in the same way.
What is the point of using malonate thioester instead of acetate thioester for one of the components? It doesn't look very efficient; a molecule of carbon dioxide is lost, rather than just the four-carbon product. There are a couple of possibilities to consider here. First, the malonyl group is more acidic than the acetyl group, so it should be easier to convert into an enolate ion via removal of a proton.
Exercise \(1\)
Show why the malonyl ester would be deprotonated so much more easily than the acetyl ester.
Answer
In the case of the malonyl enolate, there is additional delocalisation as demonstrated by resonance. This anion has extra stability.
Exercise \(2\)
If the reaction did happen this way, there would still be an extra CO2 in the structure after th Claisen reaction. Propose a mechanism for the Claisen reaction, and the loss of CO2 from the product.
Answer
There are plenty of cases where this "malonate ester strategy" make reactions involving enolates much easier. Most biochemists think there is something different going on here, though. Rather than forming the enolate via proton removal, it is believed that the enolate forms via a decarboxylation reaction.
Decarboxylation reactions happen pretty easily if the CO2 group is beta to another carbonyl. That's because of the relative stability of that enolate anion. This decarboxylation reaction, therefore, accomplishes two things at once: it gets rid of the extra carbon in the malonyl group, and it generates exactly the enolate needed for the Claisen condensation.
There is an additional reason why the pathway employs the malonyl group. The formation of carbon dioxide is exothermic because of the formation of a strong C=O bond, as well as being entropically favored. Those factors help make the decarboxylation a driving force for the reaction.
The beta-ketoacyl-ACP synthase (or KS) also uses a group transfer strategy to hold the acetyl group in place. This time, a cysteine residue on KS displaces the thiolate of the acyl carrier protein.
Once the acetyl group is tied down in the active site of KS, the malonyl-ACP enters the active site and undergoes decarboxylation, followed by Claisen condensation. The final part of the Claisen condensation releases the enzyme, completing the group transfer catalysis.
The subsequent steps aim to replace the beta-keto group with hydrogens. The first hydride is delivered with the help of beta-ketoacyl-ACP reductase, or KR.
It's a pretty standard mechanism. Frequently in enzymatic reactions, the delivery of a proton happens almost at the same time as the delivery of a nucleophile to a carbonyl. In this case, the nucleophile is a hydride ion delivered by NADPH. The proton comes from some amino acid residue in a protonated state; lysine has been used in the illustration as an example.
Elimination of water is promoted by beta-hydroxyacyl-ACP dehydratase, or DH.
This looks like a tricky reaction. It depends once again on the ease of removing a proton from that alpha position. That's not terribly difficult, because a lone pair in the alpha position is stabilised by resonance. The other problem is pushing off the hydroxy group. That isn't a very good leaving group, so it poses some problem. However, it is thought that the histidine that deprotonates the alpha position then delivers the proton to the hydroxyl group. As a result, the hydroxyl group can leave as a water molecule.
This pathway for an elimination reaction (the loss of a proton from one carbon and a leaving group from the next) is called an E1CB reaction. The CB in this acronym stands for the conjugate base that is formed after the deprotonation.
The final portion of the cycle is another hydride addition, this time carried out by beta-enoyl-ACP reductase (ER).
The mechanism is pretty similar to the first hydride addition. Once again, the exact nature of the proton donor is a little unclear, but the hydride donor is another NADPH molecule. This time, the NADPH must be positioned so that it delivers the hydride to the 1,4-position rather than to the carbonyl.
The final product of this cycle could be released or it could be sent back for another turn of the cycle, extending the fatty acid chain by two more carbons.
10.04: Solutions to
Exercise 10.2.1:
Exercise 10.2.2
The product becomes conjugated. In general, the more conjugation there is in the product of an aldol addition, the more likely is a subsequent condensation (elimination or dehydration). However, other conditions can lead to the loss of water.
Exercise 10.2.3:
Entropy. The dehydration or elimination takes one molecule (the beta-hydroxy thioester) and converts it into two molecules (the water and the alpha,beta-unsaturated thioester. That change represents an increase in internal entropy. Because the entropy term in free energy is weighted by temperature (ΔG = ΔH - TΔS), it predominates as the temperature rises.
Exercise 10.3.1:
In the case of the malonyl enolate, there is additional delocalisation as demonstrated by resonance. This anion has extra stability.
Exercise 10.3.2: | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/III%3A_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_1/10%3A_Fatty_Acid_Synthesis/10.03%3A_Catalysis_in.txt |
• 1.1: Introduction- Reaction Kinetics
• 1.2: Reaction Rates
• 1.3: Activation Barriers
There are a number of factors that influence reaction rates, but the first one that we will look at is the activation barrier. An activation barrier is a sort of energetic hurdle that a reaction must get over. Some reactions have higher hurdles and some have lower hurdles. It's much easier to get over lower hurdles, so reactions with low activation barriers can proceed more quickly than ones with higher activation barriers.
• 1.4: Collisions and Phase
• 1.5: Collisions and Concentration
We know that in order for two molecules to react with each other, they must first contact each other. We think of that contact as a "collision". The more mobile the molecules are, the more likely they are to collide. Also, the closer the molecules are together, the more likely they are to collide.
• 1.6: Rate Laws
• 1.7: Elementary Reactions
The mechanism of a reaction is a series of steps leading from the starting materials to the products. After each step, an intermediate is formed. The intermediate is short-lived, because it quickly undergoes another step to form the next intermediate. These simple steps are called elementary reactions. Because an overall reaction is composed of a series of elementary reaction, the overall rate of the reaction is dependent on the rates of those smaller reactions.
• 1.8: Catalysis
The mechanism of a reaction is a sequence of elementary steps leading from the starting materials to a series of intermediates and eventually to the products. Each step involves an activation barrier. Each intermediate has some measure of stability. We can keep track of energy changes along this reaction pathway by using a reaction progress diagram.
• 1.9: Solutions to Selected Problems
01: Reaction Kinetics
How can you confirm through experiment that a reaction is happening in a particular way? What is the mechanism of the reaction? What intermediates are occurring, and in what order do the bond-making and bond-breaking steps take place?
There are lots of experiments people perform to work out how reactions happen. One of the methods used is chemical kinetics, in which the rate of a reaction is measured. By making changes in the reaction conditions and measuring the effect of the changes on the rate of reaction, we can infer what is going on at the molecular level.
• Chemical kinetics is the measurement of how quickly reactions occur.
• If changes in conditions affect the speed of reaction, we can learn something about how the reaction happens.
Kinetic studies are important in understanding reactions, and they have practical implications, too. For example, in industry, reactions are conducted in reactors in which compounds are mixed together, possibly heated and stirred for a while, and then moved to the next phase of the process. It is important to know how long to hold the reaction at one stage before moving on, to make sure that reaction has finished before starting the next one.
By understanding how a reaction takes place, many processes can be improved. For example, if we know that a particular intermediate is involved in a reaction, we might avoid the use of conditions (such as certain solvents) that are incompatible with that intermediate. We might also be able to think of reagents to add that would make certain steps in the reaction happen more easily.
Not only are kinetic studies important in industry, but they are also used to understand biological processes, especially enzyme-catalyzed reactions. They also play a role in environmental and atmospheric chemistry, as part of an effort to understand a variety of issues ranging from the fate of prescription pharmaceuticals in wastewater to the cascade of reactions involved in the ozone cycle. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/IV%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_2/01%3A_Reaction_Kinetics/1.01%3A_Introduction-_Re.txt |
To get started, let's think about what is meant by "rate". The rate of a reaction is just its speed. Just as your speed when driving down the highway can be described in terms of your progress over time (in miles or kilometers per hour), a reaction can be described in terms of the progress of the reaction over time.
• The reaction rate is the progress of the reaction over time.
Let's look at a very simple comparison of rates. We'll look at what might happen when we roll a ball along a sidewalk. We'll roll two balls, side by side: a red one and a blue one. We'll give the red ball a slightly harder push, and give the blue one a softer touch. Then we'll snap some pictures of the two balls as they roll along.
The following drawings are a sequence of these pictures. Let's assume we take a picture every second. You will notice that some kid has marked off a line every meter using sidewalk chalk.
What do we see? The red ball is going a little faster than the blue one. It is making more progress over time.
Sometimes, it's useful to make a graph of our observations.
A graph lets us see the relationships among the data we have observed. In this case, the data is just the distance the ball has moved (in meters) and the time (in seconds). This particular relationship is described as a "linear relationship"; that just means that, when we graph the distance the ball has moved over time, we see a straight line. This is true for both the blue ball and the red ball. In this case, it tells us that both balls are moving at a steady speed.
• Plotting data on a graph lets us see relationships very easily.
However, the red ball is moving more quickly than the blue ball. We can easily use the graph to measure exactly how much more quickly it is moving (you may also be able to do it just by looking at our pictures of the ball rolling along the sidewalk. We can use the slope of the line to determine the rate of change of distance with changing time.
• For a straight line, slope = "rise"/"run"
• "rise"/"run" = distance / time
• distance / time = speed
The red ball moves 8 m in 4 s. Its speed is 2 m/s. The blue ball moves 4 m in 4 s. Its speed is 1 m/s. The red ball is moving twice as quickly as the blue ball.
That sort of comparison of rates of change is similar to what is done in chemical kinetics. Furthermore, we might try to explain why there is a difference between the two rates (in this case, we pushed the red ball harder than the blue ball).
• Sometimes we can quantify comparisons using graphical analysis.
Let's look at another race between the red ball and the blue ball. This time, who knows? Maybe the blue one will win.
Clearly, the red ball is faster than the blue one again. But let's see what else is made clear by graphing the data.
The blue ball is still moving at a steady rate. However, this time the red ball is getting faster and faster. The slope of the line between the last two data points is much higher than it is between the first two data points. In this case the red ball is accelerating; its speed is increasing over time.
The red ball doesn't obey a linear relationship between distance and time. This is a nonlinear relationship, instead. It may fit a polynomial expression, or some other nonlinear function.
• Sometimes, rates are non-linear. The speed changes over time.
Again, the rate could be interpreted physically. The red ball may be accelerating because of some factor that is making it go faster and faster. Maybe the sidewalk is sloped on one side, so that the red ball is actually rolling slightly downhill.
Let's look at one more race between the red ball and the blue ball. This time, the balls have already been rolling for some time when we start taking pictures.
It looks like the blue ball starts out ahead, but it is soon passed by the red ball. When the progress is graphed, we can see a different kind of nonlinear relationship.
This time the red ball is decelerating. It is getting slower and slower. You can imagine that, if the blue ball keeps rolling steadily along like it has so far, it may eventually pass the red ball again. This deceleration of the red ball is a different type of non-linear relationship.
Why is the red ball slowing down, but the blue one is not? Maybe the sidewalk is uneven again, but this time the red ball is going uphill. You might be able to come up with other reasons, too.
Reaction Rates
Let's take a look at molecules and reactions. This time, instead of tracking distance over time, we will look at the number of molecules present over time. Maybe we are looking at two reactions. One reaction produces red molecules. The other produces blue molecules.
This is more like what we are talking about when we look at chemical kinetics. We are looking at changes in the amount of a compound over time. In this case, the picture makes it look like the red molecules are produced more quickly than the blue molecules. The red molecules and blue molecules are produced at two different rates.
• Reaction kinetics looks at changes in the amount of compounds present over time.
This example may look a little unrealistic. The molecules are appearing from out of nowhere, and of course they can't do that. Something can't be made from nothing. We'll see other examples in which the origin of the new molecules is more clear.
Exercise \(1\)
a) Draw a graph showing the rate of production of red molecules in the picture above (graph the number of red molecules vs. time).
b) Draw a graph showing the rate of production of blue molecules.
c) Compare the rates of production of the two kinds of molecules.
Exercise \(2\)
a) Make a drawing of molecules changing over time. This time, instead of more molecules appearing over time, make molecules disappear over time, at a constant rate.
b) Draw a graph corresponding with your series of pictures.
Exercise \(3\)
The following rows of flasks show two different scenarios.
a) Describe what is happening in the first row.
b) Graph what is happeing in the first row.
c) Describe what is happening in the second row.
d) Graph what is happening in the second row.
e) Compare what is happening in the first row to what is happening in the second row.
Exercise \(4\)
The following graphs show changes in the number of molecules over time. Describe what is happening to the number of molecules (increasing? decreasing?) and to the rate at which changes are taking place (staying the same? getting faster? getting slower?).
a)
b)
c)
d)
e)
f)
g)
Exercise \(5\)
For the following cases in problem RK4, make a qualitative graph showing what you think is happening to the rate of the reaction over time (e.g. Do you think the rate is increasing over time? How would you show that on a graph?)
a) cerulean molecules
b) orange molecules
c) green molecules
d) crimson molecules
e) yellow molecules
f) black molecules
g) pink molecules
Exercise \(6\)
How would you describe the similarities between the changes in the green molecules and the pink molecules in the previous questions? What about the differences?
Exercise \(7\)
Frequently, instead of just plotting changes in one compound, we might graph changes in two things at once. That way, we can look for relationships between them.
In the following graphs, pink molecules are reacting to form another type of molecule.
Are the pink molecules turning into black molecules or indigo molecules? How can you tell?
Exercise \(8\)
The following graphs also show changes in two different compounds at once. Describe what you think is happening in each case. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/IV%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_2/01%3A_Reaction_Kinetics/1.02%3A_Reaction_Rates.txt |
Why do reactions take place at different rates? Why do some happen quickly, and others proceed very slowly? Why might the same reaction proceed at different rates under different conditions? There are a number of factors that influence reaction rates, but the first one that we will look at is the activation barrier.
An activation barrier is a sort of energetic hurdle that a reaction must get over. Some reactions have higher hurdles and some have lower hurdles. It's much easier to get over lower hurdles, so reactions with low activation barriers can proceed more quickly than ones with higher activation barriers.
• A low activation barrier allows a reaction to happen quickly.
• A high activation barrier makes a reaction go more slowly.
A reaction can be exergonic overall (it can give off energy), but it will generally still have an activation barrier at the beginning. Even if the compounds go down in energy by the end of the reaction, they will generally go up in energy before that happens.
• Even if a reaction gives off energy overall, some energy must be put in at first to get the reaction started.
This situation is a little like investing in a business. A business generally requires some sort of financial investment to get started. If the business is successful, it will eventually make products and pay money back to the investors. If the business is unable to make back its initial investment, it may fail.
Reactions require some initial investment of energy. This energy may come from surrounding molecules or the environment in general. If the reaction is successful, it will proceed to make products and it will give energy back to its surroundings.
• It always "costs" a molecule energy to enter into a reaction; it "borrows" that energy from its environment.
• That initial investment of energy may be "paid back" as the reaction proceeds.
All reactions must overcome activation barriers in order to occur. The activation barrier is the sum of the energy that must be expended to get the reaction going. An activation barrier is often thought of, cartoonishly, as a hill the molecule has to climb over during the reaction. Once, there, it can just slide down the other side of the hill to become products. At the top of the hill, the molecule exists in what is called the "transition state". At the transition state, the structure is somewhere between its original form and the structure of the products.
The type of diagram shown above is sometimes called a "reaction progress diagram". It shows energy changes in the system as a reaction proceeds. One or more activation barriers may occur along the reaction pathways, as various elementary steps occur in the reaction.
In order to see more concretely what terms like "reaction progress" and "transition state" mean, let's look at a real reaction. Suppose a nucleophile, such as an acetylide ion, donates its electrons to an electrophilic carbonyl. The π bond breaks and an alkoxide ion is formed.
The reaction progess simply refers to how far the reaction has proceeded. Is it just starting out, is it almost finished, is it just halfway there? The transition state refers specifically to the highest energy point on the pathway from reactants to products. It refers to the structure at that point, and the energy associated with that structure.
In the following diagram, the term "reaction progress" has been replaced by an illustration that shows how far the reaction has proceed by that point in the energy curve that is above the reaction drawing. The structure in the square brackets is the transition state, coresponding to the highest point on the curve. The "double dagger" symbol (a little bit like a Patriarchal or Russian Orthodox cross, with two crosspieces on a vertical post) is the symbol that tells you that you are looking at a transition state structure.
The transition state doesn't refer to a regular chemical structure. It doesn't necessarily obey the rules of Lewis structures, because some new bonds have started to form and some old bonds have started to break; you can't really draw partial bonds in a Lewis structure.
Physically, the transition state structure is not something that can be isolated and stored in a bottle. Because it sits at the top of an energy curve, the transition state is motivated to turn into something else. No matter what direction it goes to change its structure, it will go to lower energy. Remember, things always proceed to lowest energy if possible. As soon as the transition state forms, it will either slide back into the original starting materials or slip forward into the final products.
• The transition state is inherently a high-energy, unstable structure, with a very short lifetime. As soon as it comes into existence, it disappears again.
Exercise $1$
Draw what you think the transition state might look like for the following elementary reactions.
More commonly, reaction progress diagrams aren't drawn like the one above. Instead, structures of reactants, transition states and products are simply shown along the potential energy curve, as shown below.
Reactions don't always happen in one step. Sometimes there is an intermediate, or more than one. An intermediate differs from a transition state in that it has finite lifetime. Although it is not as stable as the reactants or the products, it is stable enough that it does not immediately decay. Going either forward to products or back to reactants is energetically uphill.
Exercise $2$
Draw reaction progress diagrams for the following reactions. Note that the reactions may be composed of more than one elementary step.
Rate Constant
There is a measurable parameter that can be used to get an idea about the activation barrier of a reaction. It is called the rate constant. The rate constant for a reaction is related to how quickly the reaction proceeds. A large rate constant corresponds to a very fast reaction. A very small rate constant corresponds to a slow one.
• The rate constant is an index of the speed of the reaction.
Rate constants have different units depending on how the reaction proceeds, but just to give you a feel for how they vary, a reaction with a "first order" rate constant of 0.001 s-1 (or 10-3 s-1; you'll learn what "first order" means later) would be over in about an hour. A reaction with a first order rate constant of 10-6 s-1 might take a couple of weeks.
The rate constant gives direct insight into what is happening at the transition state, because it is based on the energy difference between the reactants and the transition state. Based on that information, we get some ideas of what is happening on the way to the transition state.
The rate constant can be broken down into pieces. Mathematically, it is often expressed as
$k = \frac{RT}{Nh} e^{- \Delta \frac{G}{RT}} \nonumber$
In which R = the ideal gas constant, T = temperature, N = Avogadro's number, h = Planck's constant and D G = the free energy of activation.
The ideal gas constant, Planck's constant and Avogadro's number are all typical constants used in modeling the behaviour of molecules or large groups of molecules. The free energy of activation is essentially the energy requirement to get a molecule (or a mole of them) to undergo the reaction.
Exercise $3$
For each of the following pairs, use < or > to indicate which quantity is larger.
a) e 2 or e 10
b) e 1/4 or e 1/2
c) e -3 or e -4
d) e -1/2 or e -1/3
Answer a
e 2 < e 10
Answer b
e 1/4 < e 1/2
Answer c
e -3 > e -4
Answer d
e -1/2 < e -1/3
Note that k really depends on just two variables:
• rate constant depends on ΔG or the energy required for the reactionli>
• rate constant depends on T or the temperature of the surroundings, which is an index of how much energy is available
The ratio of activation free energy to temperature compares the energy needed to the energy available. The more energy available compared to the energy needed, the lower this ratio becomes. As a result, the exponential part of the function becomes larger (since the power has a minus sign). That makes the rate constant bigger, and the reaction becomes faster.
Large groups of molecules behave like populations of anything else. They have averages, as well as outliers on the high and low end. However, the higher the temperature, the more energy a group of molecules will have on average.
In the following drawing, the blue curve represents the energy content in a population of molecules at low temperature. The peak of the curve is near the average energy for this collection of molecules. Some of the molecules have more energy than average (they are further to the right on the blue curve) and some have less (further to the left).
The black slice through this curve indicates how much energy is needed to get over the activation barrier for a particular reaction. Notice that, at low temperature, not that many molecules have enough energy to get over the barrier at any one time. The reaction will proceed very slowly. Nevertheless, more energy is probably available from the surroundings, and so after some time most of the molecules will have obtained enough energy so that they can eventually hop over the barrier.
The yellow curve represents molecules at a higher temperature, and the red curve is a population at a higher temperature still. As the temperature is increased, larger and larger fractions of the molecules have enough energy to get over the activation barrier, and so the reaction proceeds more quickly.
• the rate constant compares energy needed to energy available
• based on that comparison, a specific fraction of the population will be able to react at a time
Free Energy of Activation
There is really more to the activation energy than we have seen so far. The activation free energy is constant for a given reaction at a given temperature. But at different temperatures, ΔG changes. Just as in thermodynamics, it can be broken down in turn to:
$\Delta G = \Delta H - T \Delta S \nonumber$
in which ΔH = activation enthalpy and ΔS = activation entropy.
The activation enthalpy is the part that corresponds most closely with the energy required for the reaction, the way we have been describing the activation barrier so far.
The activation entropy deals with how the energy within the molecule must be redistributed for the reaction to occur. One of the major factors influencing energy distribution over the course of the reaction is molecular geometry.
For example, suppose two molecules need to come together for a reaction to take place. They need to collide with each other. However, a reaction might not happen each time the molecules collide. Sometimes, the molecules may be pointing the wrong way when they bump into each other, so that the reaction can't occur. Often, atoms need to be lined up in the proper place where they will be forming a new bond.
When molecules are restricted to only certain orientations or geometries, they have fewer degrees of freedom. With fewer degrees of freedom, energy can be stored in fewer ways. As a result, there is often an entropy cost in initiating a reaction.
On the other hand, a reaction might start in a very different way, with one molecule breaking a bond and dividing into two pieces. Because each individual piece can move independently from the other, the degrees of freedom increase. Energy can be stored in more ways than they could be before the reaction started. As a result, although this reaction would still have an activation barrier, the entropy component may actually lower that barrier a little bit.
Exercise $4$
In the following drawings, one orientation of reactants is more likely to lead to the product shown. Select which one will be most successful in each set and explain what is wrong with each of the others.
Exercise $5$
Because the activation barrier depends partly on the energy needed to break bonds as the molecule heads into the transition state, comparative bond strength can be a useful factor in getting a qualitative feel for relative activation barriers.
The metal-carbonyl (M-CO) bond strengths of the coordination complexes M(CO)6 have been estimated via photoacoustic calorimetry and are listed below, by metal.
Cr: 27 kcal/mol Mo: 32 kcal/mol W: 33 kcal/mol
a) Based on that information, sketch qualitative activation barriers for the loss of a CO ligand from Cr(CO)6, Mo(CO)6 and W(CO)6.
b) Predict the relative rates for these three reactions (fastest? slowest?).
Answer a
We would expect the lowest barrier for breaking the Cr-CO bond. The barrier to break the Mo-CO bond would be just slightly lower than to break the W-CO bond.
Answer b
Based on this information alone, we might expect the Cr-CO cleavage to occur most rapidly. Mo-CO cleavage would be slightly faster than W-CO cleavage.
Exercise $6$
Comparing the strengths of bonds that will be broken in a reaction is often a good way to get a first estimate of relative activation barriers.
a) Use the following bond strengths to estimate the barriers to addition of a nucleophile (such as NaBH4) to the following double bonds: C=O (180 kcal/mol); C=N (147 kcal/mol); C=C (145 kcal/mol). Make a sketch of the three reaction progress diagrams.
b) In general, C=O bonds are the most reactive of these three groups toward electrophiles, followed by C=N bonds. Are these relative barriers consistent with this observation?
c) What other factor(s) might be important in determining the barrier of the reaction?
d) Modify your reaction progress diagram to illustrate these other factors.
Answer a
On this basis alone, we would expect the lowest barrier for C=C cleavage, followed by C=N and then C=O with the highest barrier.
Answer b
This trend is exactly the opposite of what we just predicted based on bond strengths.
Answer c
There may be a few different reasons for these differences. For example, the electrophilicity of the carbon may be a factor. Based on electronegativity differences, the C=O carbon should be most positive, the C=N carbon less so and the C=C carbon not at all. That electrophilicity may raise the reactant a little in energy.
Alternatively, there may be charge stabilization factors in the first-formed intermediate, which may be reflected in the transition state on the way there. These three differing atoms (O, N, C) are all found in a row of the periodic table, so electronegativity differences should dominate charge stability. The alkoxide ion would be most stable, the amide ion of medium stability and the alkyl anion least stable of all. That trend would lower the barrier to alkoxide formation and raise the barrier to formation of a carbon-based alkyl anion. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/IV%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_2/01%3A_Reaction_Kinetics/1.03%3A_Activation_Barri.txt |
Reaction rates depend on the energy required (the activation barrier) and the energy available. They also may involve collisions between molecules.
If two molecules need to collide in order for a reaction to take place, then factors that influence the ease of collisions will be important.
The more energy there is available to the molecules, the faster they will zip around. The faster they zip around, the more likely they are to bump into each other. So higher temperatures ought to lead to more collisions and a greater frequency of reactions between molecules. In the drawing below, the cold, sluggish molecules on the left don't appear to be in danger of colliding with anything, but the hot, zippy molecules on the right look like they are due for a crash at any time.
We already knew that higher temperatures increased reaction rates. This new observation is just an additional reason why temperature is important in kinetics.
• Temperature affects molecule mobility
• The higher the temperature, the more mobile the molecules will be, and the more likely they are to collide and react.
Phase also has a pronounced effect on the mobility of molecules. Molecules in the gas phase are quite free to move around, and they do so pretty quickly. On the other hand, they are pretty well spread out. Nevertheless, collisions in the gas phase happen pretty easily, which might help gas-phase reactions happen more readily.
At the opposite extreme, molecules in the solid phase are not very mobile at all. (Reactions may involve atoms or ions, rather than moleules, but the same arguments apply.) Not many collisions happen. As a result, reactions often happen extremely slowly in the solid state. Reactions are mostly limited to the grain boundaries: the surfaces of the grains, where they are in contact with each other. Nothing happens in the middle of a lump of solid, which remains unreacted.
If you heat a solid up, the molecules can move around a little more. They may even leave their crystal lattice (if the solid is a crystalline one) and diffuse very slowly through the solid. Many solid state reactions are run at elevated temperature.
There are also many solid state reactions that are conducted in combination with gas-phase reactants. The solid reactants are often heated in a furnace while gas-phase reactants flow over them.
Many reactions are performed in the liquid state, either because the reactants are already liquid of because the solid reactants are heated past their melting point. In the liquid phase, molecules are much more mobile and collisions are much more frequent than in the solid phase.
It is also very common to run reactions in solution. In solution, a compound that is meant to undergo reaction is dissolved in a solvent. The solvent needs to be a liquid at the temperature at which the reaction will be run, so that molecules will be very mobile, but will still be close together, so collisions are favored.
There are many advantages to running reactions in solution. The reactant molecules are very mobile and pretty close together, so that collisions are facilitated. If the reaction is exothermic and gives of a lot of heat, the excess heat can be absorbed by the solvent molecules and carried away. That can be important in controlling reactions and avoiding decomposition. Also, we will see that the rate of collisions can be controlled by adding more solvent or less, in order to slow the reaction down or speed it up. In this way, the reaction rate can be controlled to some extent.
There are many liquids that are commonly used as solvents. Dichloromethane, toluene, dimethylformamide, tetrahydrofuran and acetonitrile are some common "organic" solvents, so called because they are based on carbon, which forms the basis of molecules in organisms. These different solvents offer a range of polarities, so different ones can dissolve different reactants.
Water may be the most common solvent on the planet, and it is non-toxic, so it is very appealing for use in large-scale, industrial reactions. However, it is not very good at dissolving non-polar reactants.
The reactant compound could be a liquid or a solid. It just has to have strong enough intermolecular attractions with the solvent molecules so that it can become dissolved. Individual molecules of the reactant become lost among the solvent molecules and swim around with ease.
The disadvantage of using a solvent is that the solvent must be removed at the end of the reaction, so that the desired product can be isolated and used. That means the solvent may eventually be thrown away as waste. That practice is less efficient and less environmentally friendly, although the solvent could possibly be recycled.
One method of dissolving reactants that is potentially greener is the use of supercritical fluids. In this approach, gases such as carbon dioxide are pressurized until they turn into liquids. In this form, carbon dioxide is a pretty good solvent, and reactions can be run when reactants are dissolved in it. At the end of the reaction, a valve is opened, releasing the pressure, and the carbon dioxide turns back into a gas. It can be stored and re-pressurized for another reaction.
Exercise \(1\)
Diffusion in the solid state is greatly enhanced by defects in the crystal lattice. Show why with the following drawing.
Exercise \(2\)
The following drawing represents a reaction between molecules in the gas phase and molecules in the solid phase.
1. Which material is in the gas phase and which one is in the solid phase?
2. Both pictures contain the same mass of each material. One of these materials appears to be distributed differently in each picture, however. Explain how that difference may affect the reaction rate.
Answer a
The small, red dots are in the gas phase, so they are distributed throughout the container. The larger, gray dots are the solid, which lies along the bottom of the container.
Answer b
The solid on the left is divided into finer particles, with much more surface area. If the gas reacts on the surface of the solid, reaction will be much faster on the left.
Exercise \(3\)
Three common solvents are shown below. Compare and contrast these three solvents in terms of how they interact with other molecules.
Exercise \(4\)
Indicate how well you think each of the three solvents in the above question (THF, water, DMF) could dissolve each of the following compounds. Justify your answers in terms of interactions between molecules. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/IV%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_2/01%3A_Reaction_Kinetics/1.04%3A_Collisions_and_P.txt |
We know that in order for two molecules to react with each other, they must first contact each other. We think of that contact as a "collision". The more mobile the molecules are, the more likely they are to collide. Also, the closer the molecules are together, the more likely they are to collide.
In the following drawings, the molecules are closer together in the picture on the right than they are in the picture on the left. The molecules are more likely to collide and react in the picture on the right.
We might describe the two drawing above in terms of population density. Both drawings appear to offer the same amount of space, but they have different amounts of molecules in them.
The difference is a lot like the difference between human population densities in various locations around the world. Some places, such as Mexico City or Tokyo, are very crowded; they have high population densities. Some places, such as the Australian Outback or the Canadian Arctic, have low population densities.
Exercise \(1\)
In which location do you think you are likely to bump into another person: the Upper East Side of New York City or 75 degrees north, 45 degrees west, Greenland?
Answer
Upper East Side. Lots more people per square foot.
Exercise \(2\)
Rank the following places in terms of population density (the number of people per square kilometer).
1. Russia: pop. 143 million; area 17 million km2
2. Bahrain: pop. 1.2 million; area 750 km2
3. Argentina: pop. 41 million; area 2.7 million km2
4. China: pop. 1.3 billion; area 9.6 million km2
5. Malawi: pop. 15 million; area 118 thousand km2
6. Vatican City: pop. 850; area 0.44 km2
7. Jamaica: pop. 2.7 million; area 10,990 km2
Answer a
8.41 people km-2
Answer b
1,600 people km-2
Answer c
15.2 people km-2
Answer d
13.5 people km-2
Answer e
12.7 people km-2
Answer f
1,930 people km-2
Answer g
24.6 people km-2
Answer
Rank: Bahrain > Vatican > Jamaica > Argentina > China > Malawi > Russia
Sometimes, a lot of people are living in a small area, and the population density is high. Sometimes, the population is large, but the area is, too. Population density depends on two different factors: the number of people and the area in which they are spread out.
Concentration is the term we use to describe the population density of molecules (and other chemical entities such as atoms or ions). It describes the number of molecules there are, but also how much room, or volume, they have to move around. Thus, whereas a human population density may be described in terms of people per square kilometer, the concentration of a solution may be described in terms of molecules per liter.
Exercise \(3\)
Describe what is happening to the concentrations of the three solutions as you go from left to right.
a)
b)
Exercise \(4\)
In the cases above, describe what would happen to rates of collisions between molecules as you go from right to left in the drawings.
Answer
In both cases, the molecules are becoming more densely packed. Collisions would become more frequent as we go from left to right.
Exercise \(5\)
a) In the following drawings, state what is happeing to the concentration of molecules of each type as you go from left to right.
b) Explain what would happen to the rate of collisions between red molecules and blue molecules as you move from left to right.
c) How would your answer about rates change if the situation here were reversed: if the number of blue molecules stayed the same and the number of red molecules increased?
d) How would your answer about rates change if the numbers of both the red and the blue molecules were increasing at the same time?
Answer a
The number of blue molecules is increasing, but the number of red molecules is staying the same.
Answer b
Collisions between red and blue molecules would become more frequent as we go from left to right. The higher density of blue molecules makes collisions more likely.
Answer c
The answer would stay the same.
Answer d
The answer would stay the same qualitatively, but would differ quantitatively. The number of collisions would increase more sharply if the concentrations of both red and blue molecules increased, rather than just one concentration increasing.
We usually don't count individual molecules in a solution. We deal with groups of molecules because it's more convenient. Individual molecules are just too small to work with. In dealing with molecules in bulk, we usually use a unit called a mole, often abbreviated to mol. It's kind of like dealing with eggs by the dozen. Molecules are easier to keep track of by the mole, rather than individually.
Exercise \(6\)
1. In the following solutions, how many dozen blue molecules are there in each case?
2. What is happening to the concentration of the beaker as you go from one beaker to the next? Quantify your answer.
Answer a
A half dozen, a dozen, two dozen.
Answer b
The concentration is doubling as we go left to right from one beaker to the next.
In reality, we don't count molecules or even moles of molecules. When we want to work with a compound, we just weigh it out on the balance. We can then use the known weight of the compound to figure out how many moles we have.
Of course, if we are going to be measuring out molecules by weight, we'll need to know how much each molecule weighs. For example, if we need an equal number of red molecules and blue molecules, and red molecules weigh three times as much as blue molecules, we'll need to weigh out three times as much of the red stuff as the blue stuff.
Exercise \(7\)
You are developing a new "extreme sport" amusement park ride. Each ride (for one person) is powered by one mouse and one elephant. You have plenty of elephants to get started, but will need to go and buy some mice.
1. If an elephant weighs 6,800 kg, how many grams does it weigh?
2. If you have 47,600 kg of elephants, how many rides can you set up?
3. If a mouse weighs 25 g, how many g of mice will you need to buy?
4. Suppose you decide to get three extra mice (sometimes accidents happen around elephants). What is the total weight of mice you would need, including these extras?
Answer a
The elephant weighs 6,800 kg x 1,000 g/kg = 6,800,000 g = 6.8 x 106 g.
Answer b
We have 47,600 kg x 1 elephant/6,800 kg = 7 elephants. Very lucky.
Answer c
7 mice x 25 g/mouse = 175 g.
Answer d
10 mice x 25 g/mouse = 250 g.
Exercise \(8\)
Suppose each of the following beakers contains an equal weight of each kind of molecule. The molecular weight of a blue molecule is 60 g/mol. What is the molecular weight of a red, an orange and a grey molecule?
The weight of a molecule can be determined by adding up the weights of all its atoms. For example, a carbon dioxide molecule has a molecular weight of 44 amu (carbon is 12 amu plus two oxygens at 16 amu apiece). The weight of a mole of carbon dioxide is the same as the molecular weight, but in grams instead of amu. A mole of carbon dioxide is 44 g. In other words, the molecular weight (MW) of carbon is 44 g/mol.
Answer
The weight of 7 blue molecules = the weight of 2 red molecules. 1 red molecule = 7/2 x the weight of a blue molecule. MWred = 3.5 x MWblue = 3.5 x 60 g mol-1 = 210 g mol-1.
The weight of 6 blue molecules = the weight of 6 orange molecules. MWorange = MWblue = 60 g mol-1.
The weight of 6 blue molecules = the weight of 1 grey molecule. MWgrey = 6 x MWblue = 6 x 60 g mol-1 = 360 g mol-1.
Exercise \(9\)
How much does a mole of each of the following molecules weigh?
1. nitric oxide, NO2
2. glucose, C6H12O6
3. benzaldehyde, C7H6O
4. phosphorus pentoxide, P2O5
Answer a
A mole of material corrsponds to the numerical equivalent of the sum of atomic masses in an molecule, in grams.
atomic masses in NO2: 14 amu (N) + 32 amu (2 x O) = 46 amu. 1 mol of NO2 is 46 g of NO2.
Answer b
180 g
Answer c
106 g
Answer d
142 g
Exercise \(10\)
How many moles of each of the following compounds are there in the given weights?
1. 3 grams of glucose
2. 10 grams of benzaldehyde
3. 30 grams of phosphorus pentoxide
Answer a
3 g x 1 mol/180 g = 0.017 mol
Answer b
10 g x 1 mol/106 g = 0.094 mol
Answer c
30 g x 142 mol/g = 0.21 mol
Exercise \(11\)
What is the concentration of each of the following solutions (in moles per liter)?
1. 5 g of glucose in 50 mL of water
2. 11 g of benzaldehyde in 25 mL of THF
3. 9 g of menthol (MW 156 g/mol) in 60 mL of DMF
Answer a
5 g x 1 mol/180 g = 0.028 mol; 0.028 mol / 50 mL = 5.6 x 10-4 mol/mL x 1000 mL/L = 0.56 mol L-1.
Answer b
11 g x 1 mol/106 g = 0.104 mol; 0.104 mol / 25 mL = 4.16 x 10-3 mol/mL x 1000 mL/L = 4.2 mol L-1.
Answer c
9 g x 1 mol/156 g = 0.058 mol; 0.058 mol / 60 mL = 9.61 x 10-4 mol/mL x 1000 mL/L = 0.96 mol L-1. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/IV%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_2/01%3A_Reaction_Kinetics/1.05%3A_Collisions_and_C.txt |
So far, we have talked about changes in the number of molecules over time as a reaction progresses. The number of reactant molecules decreases as the number of product molecules increases. Practically, the easiest way to measure the speed of a reaction is to measure the concentration over time. We can measure either the concentration of the reactants or the products. Remember, concentration refers to how densely populated a solution is with a particular compound.
The concentration of black dots is higher in the beaker on the right than in the beaker on the left.
Reactions are often monitored by some sort of spectroscopy. In spectroscopy, "light" or some other frequency of electromagnetic radiation shines through a sample in which a reaction is taking place. The light can interact with the molecules in the sample. The molecules absorb particular frequencies of light, so if the light encounters the molecules on its way through the sample, a little of the light at those frequencies is absorbed. Less light makes it all the way through the sample; the amount that does make it through is measured by a detector on the other side.
If the concentration of the sample is different, a different amount of light from the spectrometer will be absorbed. For instance, suppose the sample is more concentrated. The more molecules there are, the more light is absorbed. And because the beam of light travels through the sample in a straight line, the more concentrated the solution, the more molecules it will encounter.
It is pretty simple to calibrate the instrument to be able to determine concentration from the amount of light absorbed. In addition, the light may interact with the reactant molecules and product molecules in different ways. That means you can monitor the absorption of a frequency that you know is absorbed by reactant molecules, but not by product molecules, and you can detect changes in reactant concentration. You could also do the same thing to detect changes in product concentration.
We sometimes write the rate of the reaction as:
$Rate = \frac{d[product]}{dt} \nonumber$
Meaning, the rate is the change in concentration of product with change in time.
Concentration could be measured in any units. Frequently, we are dealing with a solution, and we use units such as grams per liter or, much more commonly, moles per liter. The change in time is most often measured in seconds.
We could also write the rate of the reaction as:
$Rate = \frac{-d[reactant]}{dt} \nonumber$
Meaning, the rate is the change in concentration of reactant with change in time. The minus sign just means that the reaction is getting consumed over time as it turns into product, so its concentration is decreasing.
Kinetic studies are important in understanding reactions. Not only are they important in industry, but they are also used to understand biological processes, especially enzyme-catalyzed reactions. They also play a role in environmental and atmospheric chemistry, as part of an effort to understand a variety of issues ranging from the fate of prescription pharmaceuticals in wastewater to the cascade of reactions involved in the ozone cycle.
Exercise $1$
Suppose the rate of the reaction between the black circles and the white circles depends only on the concentration of the black circles. That is, rate = k [black circle]. Compare the rate in each case to the rate of the reaction that would occur in the original beaker.
Exercise $2$
Suppose the rate of the reaction between the black circles and the white circles depends on both the concentrations of the black circles and the white circles. That is, rate = k [black circle][white circle]. Compare the rate in each case to the rate of the reaction that would occur in the original beaker.
Exercise $3$
Suppose the rate of a reaction in the beaker depends on the surface area of the solid at the bottom of the beaker. That is, rate = k x (surface area of white circles). Compare the rate in each case to the rate of the reaction that would occur in the original beaker.
Exercise $4$
Often in studying reaction kinetics, the changing concentration of a reactant or a product is plotted against time. In one method, many data points are collected very early in a reaction (when fewer than 5% of the material has reacted), and the slope of the resulting line is used to determine the "initial rate". Explain why this method might not work if the data points are plotted all the way until the reaction is finished.
Answer
At the very start of the reaction, the concentrations of reactants have not changed very much. That means the rate of the reaction remains roughly constant as the first few percent of reactants are consumed. Plotting [product] vs. time gives a straight line with the slope = rate. However, over the course of the reaction, the concentration of reactants goes down as the reactants are consumed. That means the rate of product formation slows down and a plot of [product] vs. time becomes curved. We will be unable to measure the slope in a simple way.
Exercise $5$
Suppose the following plots were obtained before 5% conversion for the reaction:
$A + B \rightarrow C \nonumber$
What do you know about the rate law for the reaction?
Answer
The slope of the first curve, with [A]o = 1 mol L-1, can be estimated by observing that [C] increases from zero to 25 mmol L-1 in 50 seconds. The slope is about 25/50 = 0.5 mmol L-1 s-1. The slope of the next curve, with [A]0 = 2 mol L-1, is 1.0 mmol L-1 s-1 (50 mmol L-1 / 50 s). The initial concentration is doubled, and the rate doubles. The slope of the final curve, with [A]0 = 4 mol L-1, is 2.0 mmol L-1 s-1 (100 mmol L-1 / 50 s). The initial concentration is doubled, and the rate doubles. The reaction is linearly dependent on the concentration of A. Whatever happens to [A] also happens to the rate.
In terms of rate laws, there is a mathematical approach to demonstrating this relationship.
Suppose $Rate = k [A]^{x}$; x is the power of the mathematical relationship.
The ratio of rates in two experiments is $\frac{Rate_{1}}{Rate_{2}} = \frac{k[A_{1}]^{x}}{k[A_{2}]^{x}}$.
If we take the logarithm of both sides: $ln \frac{Rate_{1}}{Rate_{2}} = ln (\frac{[A_{1}]}{[A_{2}]})^{x} = x ln \frac{A_{1}}{A_{2}}$
Then $x = ln \frac{Rate_{1}}{Rate_{2}}/ ln \frac{[A_{1}]}{[A_{2}]}$
Exercise $6$
Suppose the following plots were obtained before 5% conversion for the reaction:
$A + B \rightarrow C \nonumber$
a) What is a possible rate law for the reaction?
b) Two different rate laws could explain this data. What is the second possible rate law?
c) Propose an experiment to distinguish between these two possible rate laws.
Answer a
Each time the concentrations double (for example, from 2 to 4 mol L-1), the rate quadruples (for example, from 50/50 = 1 mmol L-1 s-1 to 200/50 = 4 mmol L-1s-1). One explanation is $Rate = k[A][B]$
Answer b
Another explanation is $Rate = k[A]^{2}$ or $Rate = k[B]^{2}$.
Answer c
We could run a series of experiments in which [A] is changed while holding [B] constant (or vice versa).
Exercise $7$
Suppose the following data were obtained by monitoring the following reaction to completion:
$A + B \rightarrow C \nonumber$
a) How long does it take until the reaction is essentially finished, if the starting concentration of A is:
i) 1 mol/L?
ii) 2 mol/L?
iii) 4 mol/L?
b) What do you know about the rate law for the reaction? Explain.
Exercise $8$
Suppose the following data were obtained by monitoring the following reaction to completion:
$A + B \rightarrow C \nonumber$
Compare this graph to the one in the previous problem. What differences can you detect in the curves? Do you think this reaction has the same rate law as the previous one? | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/IV%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_2/01%3A_Reaction_Kinetics/1.06%3A_Rate_Laws.txt |
The mechanism of a reaction is a series of steps leading from the starting materials to the products. After each step, an intermediate is formed. The intermediate is short-lived, because it quickly undergoes another step to form the next intermediate. These simple steps are called elementary reactions.
Because an overall reaction is composed of a series of elementary reaction, the overall rate of the reaction is somehow dependent on the rates of those smaller reactions. But how are the two related?
Let's look at two cases. We'll keep it simple and both cases will be two-step reactions.
The first case begins with a reaction having a pretty low activation barrier. Maybe there is a lot of bond-making involved at the beginning, so this reaction gets an easy start. However, after that, things get harder. The second step has a higher activation barrier.
Both of these reactions occur at particular rates. The first step has a low barrier, so it occurs quickly. The second step has a high barrier, so it happens only slowly.
In that case, the intermediate will be formed quickly and it will sit around for a while before the second step has a chance to happen. The second step is like a bottleneck or a traffic jam. No matter how quickly things were going in the first step, the reaction has to wait to get through the second step. The rate of the overall reaction really depends on the second step. We call this step the "rate-determining step".
Each step has its own rate constant associated with it. We'll call these constants k1 and k2. The lower the barrier, the faster the rate, and the larger the rate constant.
We can predict rate laws for elementary reactions (although the same thing isn't true for overall reactions). The rate law for an elementary reaction is simply its rate constant, times the concentrations of any species involved in that step.
The rate of formation of product will be determined by the rate of that second step, the rate-determining step. The rate of that second step is the rate constant, k2, times the concentration of the intermediate.
$\frac{d[Product]}{dt} = k_{2}[Intermediate]$
But what is the concentration of the intermediate? The intermediate is produced in the first step, so its concentration depends on how quickly it is produced from the reactants. However, if the activation barrier is low, that first step may be reversible; it may really be an equilibrium reaction. So there is a third step in this reaction, and it goes backwards from the intermediate to the reactant. We'll call the rate constant for this step k-1.
In terms of kinetics, an equilibrium is really just a ratio of forward and reverse steps. If the forward step is much faster, this ratio is bigger than one, and products are favored. If the reverse step is much faster, the ratio is smaller than one, and reactants are favored.
$K_{eq} = \frac{k_{1}}{K_{-1}} = \frac{[Intermediate]}{[Reactant]}$
Or, rearranging,
$\frac{k_{1}}{k_{-1}} [Reactant]=[Intermediate]$
Now we know the concentration of the intermediate. That means the rate of the overall reaction is
$\frac{d[Product]}{dt}= \frac{k_{2}k_{1}}{k_{-1}}[Reactant]$
The take-home lesson is this: the second step is the rate determining step, so that step tells us how quickly the product forms. However, that step depends on an intermediate formed in an earlier step, so that earlier step also influences how quickly the product forms.
• The rate-determining step controls the rate of the reaction.
• The steps prior to the rate determining step influence the rate of the reaction by supplying intermediates needed for the rate determining step.
Now let's think about the opposite case. Suppose the first step has a very high barrier and the second step has a lower one. We have to wait and wait and wait for the intermediate to be produced, but once it's there, it reacts pretty quickly to give products.
In this case, the first step is the rate-determining step. That's the step that controls the rate of the reaction. As soon as that step proceeds, the rest of the reaction can occur quickly.
In this case,
$\frac{d[Product]}{dt} = k_{1}[Reactant]$
• Any steps after the rate-determining step don't influence the rate of the reaction.
In some cases, we can use our judgement to try to predict which transition states are highest in energy in a multi-step mechanism. However, much of the knowledge about what is happening between the reactant and the product would really come from computational chemistry. In computational chemistry, the energies of the intermediates and the transition states can be calculated using quantum mechanics. This task is not necessarily easy.
Exercise $1$
Consider the borohydride reduction of a ketone in methanol.
1. Draw a mechanism for this reaction. (If you know about the competing reaction between borohydride and methanol, ignore it. If you didn't know anything about that, forget I said anything.)
2. Assign rate constants (k1, k2, k3...) for each elementary step in the reaction.
3. One of the steps along this reaction is probably reversible. Which one? Why?
4. What do you think is the rate-determining step in this reaction?
5. Draw a reaction progress diagram for this reaction.
6. Predict a rate law for this reaction.
Answer a
Answer b
The first step proceeds at k1; the second at k2.
Answer c
The second step is probably reversible. It is just the exchange of a proton from one OH group to another.
Answer d
Answer e
$Rate = k_{1}[ketone][NaBH_{4}] \nonumber$
Exercise $2$
Consider the ammonolysis of acetyl chloride.
1. Draw a mechanism for this reaction.
2. Assign rate constants (k1, k2, k3...) for each elementary step in the reaction.
3. Draw alternative reaction progress diagrams for this reaction. In each case, assume a different rate determining step.
4. Predict the rate law corresponding to each of your possible reaction progress diagrams. Can the rate law be used to distinguish between all of the possibilities?
5. Consider your alternative reaction progress diagrams and decide which one is most likely. Justify your choices for eliminating the other one(s).
6. Predict a rate law for this reaction.
Answer a
Answer b
The first step proceeds at k1; the second at k2; the third at k3.
Answer c
There are three different steps, each of which might be rate determining.
The first:
The second:
The third:
Answer d
First one: $Rate = k_{1}[CH_{3}COCl][NH_{3}]$
Second one: $Rate = k_{1}k_{2}[CH_{3}COCl][NH_{3}] Third one: \(Rate = k_{1}k_{2}k_{3}[CH_{3}COCl][NH_{3}]^{2}$
The rate law could be used to distinguish the third from the first two. However, it would be impossible to tell the difference between the first two using the rate law alone.
Answer e
We can probably rule out the third possibility right away. Proton transfers tend to happen pretty quickly, especially in the presence of a reasonable base such as ammonia. Measuring the rate law would quickly confirm this assumption.
Scenarios 1 and 2 are much harder to distinguish. In both cases, a bond is being broken as another bond is being made. We could make an educated guess that the formation of the good chloride leaving group is pretty easy; that would make step 1 the rate determining step.
However, we are always left with some ambiguity in subtle cases like this one. In order to get a better idea about which transition states are the highest, we would have to perform computational chemistry.
Exercise $3$
There are two other possible mechanisms for ammonolysis of acetyl chloride. You already know they aren't the right mechanism, but let's look at the reason we know that.
1. The first possibility is that the chloride ion leaves in the first step. The ammonia comes in afterwards.
1. Draw this mechanism.
2. Provide a reaction progress diagram.
3. Identify the rate determining step and the rate law.
2. The second possibility is that the chloride gets pushed out as the ammonia comes in (instead of the C=O π bond breaking). Draw this mechanism.
1. Draw this mechanism.
2. Provide a reaction progress diagram.
3. Identify the rate determining step and the rate law.
3. Given the experimental rate law, $\text{Rate} = k [\ce{CH3COCl}][\ce{NH3}]$, can either of these possibilities be ruled out?
Answer a i
Answer a ii
Answer a iii
$Rate = k_{1}[CH_{3}COCl] \nonumber$
Answer b i
Answer b ii
Answer b iii
$Rate = k_{1}[CH_{3}COCl][NH_{3}] \nonumber$
Answer c
On the basis of the rate law, possibility (a) can be ruled out, but not possibility (b)
Another commonly used experiment in kinetics is a kinetic isotope effect experiment. Although two isotopes of an atom have almost identical properties, the difference in mass between the two isotopes can cause subtle differences in reaction rates. For rather complicated reasons, two isotopes display different sensitivities to geometry changes that occur up to and including the rate determining step.
For example, if the carbonyl carbon in acetyl chloride is replaced by a 13C isotope (it is normally 12C), it will react at a different rate tthan the normal compound, but only if that carbon undergoes a geometry change at some point before it finishes the rate determining step. If the two compounds react at the same rate, no such geometry change has occurred.
Exercise $4$
The ratio of reaction rates of CH313COCl : CH312COCl is about 0.9.
1. What does this observation tell you about the mechanism?
2. Does it rule out any possibilities in Exercise $3$?
Answer a
There is a geometry change prior to or during the rate determining step.
Answer b
Possibility (b) can be ruled out.
Exercise $5$
a) Draw the product of the following reaction.
b) Provide a mechanism for the reaction.
c) Briefly describe what is meant by the term "thermodynamics" in the context of reactivity.
d) Briefly define what is meant by the term "kinetics" in this context.
e) In which step of the mechanism is ΔS positive?
f) Write a rate law for the reaction based on your mechanism.
g) What happens to reaction rate as temperature increases?
h) What does collision theory tell us about rates of reactions?
i) The following rate constants were observed for the reaction involving different substituents X.
-X k mol L-1 s-1
-OCH3 0.17
-SCN 1.76
Exercise $6$: the Bayliss-Hillman reaction
The Bayliss-Hillman reaction is used extensively in syntheses of pharmaceuticals. However there are two possible mechanisms, shown below (Price, Broadwater, Walker and McQuade, J. Org. Chem., 2005, 70, 3980-3987).
1. Provide arrows to show electron flow for each mechanism.
2. Provide a rate law for each mechanistic proposal, based on concentrations of compounds 1, 2, and 3.
3. Indicate the order in each compound based on the graphs below.
d) Which mechanism is consistent with the data?
Answer a
Answer b
Mechanism 1: k [1][2][3]
Mechanism 2: k [1][2][3]2
Answer c
First order in [1]; first order in [2]; second order in [3]
Answer d
Mechanism 2. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/IV%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_2/01%3A_Reaction_Kinetics/1.07%3A_Elementary_React.txt |
The mechanism of a reaction is a sequence of elementary steps leading from the starting materials to a series of intermediates and eventually to the products. Each step involves an activation barrier. Each intermediate has some measure of stability. We can keep track of energy changes along this reaction pathway by using a reaction progress diagram.
We can change the speed of a reaction by flooding the system with reactant or by adding more energy to help it get over the barrier, but the reaction still follows this same energy pathway.
However, if we add a catalyst, that's no longer true. A catalyst introduces a completely different pathway that was not there before.
Some reactions simply can't proceed without a catalyst. There is too high a barrier to proceed. That may be true even if the overall reaction is exergonic and should be favored to proceed.
If molecules approach this barrier from the left, they encounter a very large barrier. They can't proceed, even though it would be energetically favorable for them to get to their destination. The molecules get stuck.
The same is true if molecules approach from the right. They can't get over the barrier, even though thermodynamically their destination is just a short hop. These molecules are stuck.
Everything would be fine if we could just get rid of that darned barrier. The molecules could freely move back and forth, and settle out where they are supposed to be.
A catalyst doesn't remove the barrier, but it offers a new pathway with a lower barrier. The reaction is able to proceed back and forth.
How can there suddenly be a new pathway? A reaction progress diagram plots what happens to the energy along one particular coordinate of interest to the molecule. We may be following a bond as it lengthens and breaks through the course of a reaction. But there are always other energetic possibilities that remain unseen in this diagram. What we are looking at is merely a slice through a potential energy surface. A potential energy surface is like a landscape, a mountain range in which elevation corresponds to energy. The reaction progress diagram depicts a single pathway from one energetic valley to another. In one of these valleys is the reactant; the product is in the other. The path from one valley to another leads uphill, over a mountain pass, and down into the other valley again.
Suppose you are going to visit a friend. You live in one valley and the friend lives in another. Every day you walk the same path to your friend's house. You aren't a fool, so you take the easiest route, over the lowest mountain pass.
One day, instead of visiting our friend on foot, we are going to take the train. The train takes a completely different pathway than the one we are used to. It doesn't even go over the same mountain pass; it may take another route that was inaccessible by foot, or it may simply tunnel through the mountain. Furthermore, when the train reaches its destination, it picks up more passengers and makes the same journey again, over and over.
In molecular turns, it is the reactant that takes the train, a low-energy pathway, on its way to the product. That train is a catalyst, and it has some very important features.
• A catalyst takes a reaction pathway that is lower in energy than the usual one.
• A catalyst returns again and again to take more molecules through the reaction.
That recycling of the catalyst is sometimes referred to as "turnover". The turnover number of a catalyst is the number of times the catalyst is able to return and carry out the reaction again. (Eventually something may go wrong and the catalyst may stop working.) The reactant in a catalytic reaction is often called the substrate, particularly in cases in which the catalyst is an enzyme of a transition metal catalyst. The speed at which the catalyst is able to carry out the reaction on new substrates is called the turnover frequency. These are important parameters in describing the efficiency of a catalyst.
Exercise \(1\)
Polymerization catalysts take small molecules called monomers and connect them together using the same reaction over and over again to make a long polymer chain. What was the minimum turnover number of the catalyst used to make each of the following polymers? (Mn = number average molecular weight, a statistical estimate of the average size of polymer chain.)
1. polylactide, Mn = 3,000, from lactide, C6H8O4.
2. polystyrene, Mn = 250 thousand, from styrene, C6H5C2H3.
3. high-modulus polyethylene (HMPE), Mn = 6 million, from ethene, C2H4.
Exercise \(2\)
Catalysis often begins with a binding step, followed by one or more subsequent steps needed to carry out the reaction. For the following catalysed reactions, draw the binding step using curved arrows. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/IV%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_2/01%3A_Reaction_Kinetics/1.08%3A_Catalysis.txt |
Exercise 1.2.7:
a) Scarlet molecules are disappearing over time. Ultramarine molecules are appearing over time. However, the number of scarlet molecules seems to stabilize around half its original number; the number of ultramarine molecules reaches about the same level. It is possible that the scarlet molecules are converted into the ultramarine ones, but reach an equilibrium. At this equilibrium, there happen to be about equal amounts of the two compounds.
b) Because the rate of growth of the tan molecules seems to track the rate of loss of the brown ones, it seems likely that the brown ones are being converted to tan molecules. At some point, the change levels out, so an equilibrium is reached. At this equilibrium, tan molecules outnumber brown ones.
c) Viridian molecules seem to be converting into gold ones but the system reaches equilibrium, at which point there are still more viridian molecules than gold ones.
Exercise 1.3.3:
a) e 2 < e 10
b) e 1/4 < e 1/2
c) e -3 > e -4
d) e -1/2 < e -1/3
Exercise 1.3.4:
a) Orientation (i) looks most likely to result in connection of the oxygen to carbon with displacement of bromide. In orientation (iii), the oxygen appears ready to collide with the bromine atom. and in oreientation (ii), it may collide with a hydrogen atom.
In addition, you will see later that bringing the the oxygen in along the C-Br axis (but away from the bromine) is also more likely to break the C-Br bond, for reasons involving molecular orbital overlap.
b) Orientation (ii) looks like the carbon anion in the acetylide ion is most likely to bond with the carbonyl carbon. In option (i), the carbon is going to collide with the carbonyl oxygen. In option (iii), it may collide with the alpha carbon, next to the carbonyl. Having the nucleophile approach from outside the plane of the carbonyl, as in option (ii), lowers the chance of collision with atoms other than the carbonyl carbon.
Again, there are also molecular orbital reasons that make this approach the preferred one.
Exercise 1.3.5:
a) We would expect the lowest barrier for breaking the Cr-CO bond. The barrier to break the Mo-CO bond would be just slightly lower than to break the W-CO bond.
b) Based on this information alone, we might expect the Cr-CO cleavage to occur most rapidly. Mo-CO cleavage would be slightly faster than W-CO cleavage.
Exercise 1.3.6:
1. On this basis alone, we would expect the lowest barrier for C=C cleavage, followed by C=N and then C=O with the highest barrier.
2. This trend is exactly the opposite of what we just predicted based on bond strengths.
3. There may be a few different reasons for these differences. For example, the electrophilicity of the carbon may be a factor. Based on electronegativity differences, the C=O carbon should be most positive, the C=N carbon less so and the C=C carbon not at all. That electrophilicity may raise the reactant a little in energy.
Alternatively, there may be charge stabilization factors in the first-formed intermediate, which may be reflected in the transition state on the way there. These three differing atoms (O, N, C) are all found in a row of the periodic table, so electronegativity differences should dominate charge stability. The alkoxide ion would be most stable, the amide ion of medium stability and the alkyl anion least stable of all. That trend would lower the barrier to alkoxide formation and raise the barrier to formation of a carbon-based alkyl anion.
Exercise 1.4.1:
The vacancies in the crystal lattics give other atoms places to move into, sothey act as a path through which atoms can move. Diffusion through the solid is greatly accelerated.
Exercise 1.4.2:
a) The small, red dots are in the gas phase, so they are distributed throughout the container. The larger, gray dots are the solid, which lies along the bottom of the container.
b) The solid on the left is divided into finer particles, with much more surface area. If the gas reacts on the surface of the solid, reaction will be much faster on the left.
Exercise 1.4.3:
All three of these solvents have dipole moments. Also, all three molecules have oxygen lone pairs, so they are able to accept hydrogen bonds from hydrogen bond donors.
However, some of the solvents are much more polar than others. Water is capable of donating hydrogen bonds, because of its partially positive hydrogen attached to oxygen. It is the most polar of these solvents.
DMF is also very polar, because it has a polar C=O bond. This particular carbonyl is more like +N=C-O- because of lone pair donation from the nitrogen, so it is quite polar and will interact strongly with other species via dipole-dipole forces (or ion-dipole forces, if the other molecule is a salt).
THF only has a moderate dipole compared to the others. Although it will still interact via dipole-dipole (or ion-dipole) forces, it does so less effectively than water or DMF.
Exercise 1.4.4:
Water would be a very good solvent for (a) and (d), because both of those molecules would be very good at hydrogen bonding. Although water may be able to dissolve small amounts of the others, their solubility would be limited by the need for water molecules to release hydrogen bonds to each other in order to make room for the non-polar portions of these molecules.
THF would be able to dissolve the other molecules pretty well: (b), (c), (e) and (f). All of those molecules contain polar bonds, like THF, and could interact via dipole-dipole forces. THF would be able to dissolve small amounts of (a) and (d) but may not be polar enough to overcome the stronger intermolecular forces between these molecules.
DMF may be able to dissolve all of these molecules to a moderate extent. Although it is not a protic solvent, its dipole is enough to help overcome hydrogen bonding among (a) and (d).
Exercise 1.5.1:
Upper East Side. Lots more people per square foot.
Exercise 1.5.2:
a) 8.41 people km-2
b) 1,600 people km-2
c) 15.2 people km-2
d) 13.5 people km-2
e) 12.7 people km-2
f) 1,930 people km-2
g) 24.6 people km-2
Rank: Bahrain > Vatican > Jamaica > Argentina > China > Malawi > Russia
Exercise 1.5.3:
a) The volume is constant, but the number of molecules is increasing. The concentration is increasing.
b) The number of molecules is constant, but the volume is decreasing. The concentration is increasing.
Exercise 1.5.4:
In both cases, the molecules are becoming more densely packed. Collisions would become more frequent as we go from left to right.
Exercise 1.5.5:
a) The number of blue molecules is increasing, but the number of red molecules is staying the same.
b) Collisions between red and blue molecules would become more frequent as we go from left to right. The higher density of blue molecules makes collisions more likely.
c) The answer would stay the same.
d) The answer would stay the same qualitatively, but would differ quantitatively. The number of collisions would increase more sharply if the concentrations of both red and blue molecules increased, rather than just one concentration increasing.
Exercise 1.5.6:
a) A half dozen, a dozen, two dozen.
b) The concentration is doubling as we go left to right from one beaker to the next.
Exercise 1.5.7:
a) The elephant weighs $6800kg \times 1000 \frac{g}{kg} = 6800000g = 6.8 \times 10^{6} g$
b) We have $47600 kg \times \frac{1 elephant}{6800 kg} = 7 elephants$. Very lucky.
c) $7 mice \times 25 \frac{g}{mouse} = 175g$
d) $10 mice \times 25 \frac{g}{mouse} = 250g$
Exercise 1.5.8:
The weight of 7 blue molecules = the weight of 2 red molecules. 1 red molecule = 7/2 x the weight of a blue molecule. $MW_{red} = 3.5 \times MW_{blue} = 3.5 \times 60 \frac{g}{mol} = 210 \frac{g}{mol}$
The weight of 6 blue molecules = the weight of 6 orange molecules. $MW_{orange} = MW_{blue} 60 \frac{g}{mol}$
The weight of 6 blue molecules = the weight of 1 grey molecule. $MW_{grey} = 6 \times MW_{blue} = 6 \times 60 \frac{g}{mol} = 360 \frac{g}{mol}$
Exercise 1.5.9:
A mole of material corrsponds to the numerical equivalent of the sum of atomic masses in an molecule, in grams.
a) atomic masses in NO2: 14 amu (N) + 32 amu (2 x O) = 46 amu. 1 mol of NO2 is 46 g of NO2.
b) 180 g
c) 106 g
d) 142 g.
Exercise 1.5.10:
a) $3g \times \frac{1mol}{180g} = 0.017 mol$
b) $10g \times \frac{1mol}{106 g} = 0.094mol$
c) $30g \times \frac{142mol}{g} =0.21 mol$
Exercise 1.5.11:
a) $5g \times \frac{1mol}{180g} = 0.028mol; \frac{0.028 mol}{50mL} = 5.6 \times 10^{-4} \frac{mol}{mL} \times 1000 \frac{mL}{L} = 0.56 \frac{mol}{L}$
b) $11g \times \frac{1mol}{106g} = 0.104mol; \frac{0.104 mol}{25mL} = 4.16 \times 10^{-3} \frac{mol}{mL} \times 1000 \frac {mL}{L} = 4.2 \frac{mol}{L}$
c) $9g \times \frac{1mol}{156g} = 0.058 mol; \frac{0.058mol}{60mL} = 9.61 \times 10^{-4} \frac{mol}{mL} \times 1000 \frac{mL}{L} 0.96 \frac{mol}{L}$
Exercise 1.6.1:
a) Rate of (a) = 1/2 x Rate of original
b) Rate of (b) = 1/2 x Rate of original
c) Rate of (c) = 3/4 x Rate of original
d) Rate of (d) = 1/2 x Rate of original; although the number of molecules is the same as the original, the volume is doubled. As a result, the concentration is cut in half.
e) Rate of (e) = 2 x Rate of original; although the number of molecules is the same as the original, the volume is halved. As a result, the concentration is doubled.
f) Rate of (f) = Rate of original; although the number of molecules is doubled, the volume is also doubled, leaving the concentration unchanged.
Exercise 1.6.2:
a) Rate of (a) = 1/2 x Rate of original
b) Rate of (b) = 1/2 x 1/2 = 1/4 x Rate of original
c) Rate of (c) = 2 x Rate of original
d) Rate of (d) = 1/2 x 1/2 = 1/4 x Rate of original; although the number of molecules is the same as the original, the volume is doubled. As a result, the concentration is cut in half.
e) Rate of (e) = 2 x 2 = 4 x Rate of original; although the number of molecules is the same as the original, the volume is halved. As a result, the concentration of each reactant is doubled.
f) Rate of (f) = Rate of original; although the number of molecules is doubled, the volume is also doubled, leaving the concentration unchanged.
Exercise 1.6.3:
a) Rate of (a) = Rate of original; the surface area of the white solid appears to be the same.
b) Rate of (b) = 1/2 x Rate of original; the surface area of the white solid appears to be cut in half.
c) Rate of (c) = 2 x Rate of original; the surface area of the white solid appears to be doubled.
d) The smaller sizes of the particles in (d) makes it harder to answer this question. Let's assume these white solids are spherical and that the radius of a sphere in (d) is half that of a sphere in the original. The surface area of a sphere is $A = 4 \pi r^{2}$. The ratio of surface areas of one sphere to another is $\frac{A_{1}}{A_{2}} = \frac{4 \pi r_{1}^{2}}{4 \pi r_{2}^{2}} = (\frac{r_{1}}{r_{2}})^{2}$. The ratio of surface areas of an original sphere to a sphere in (d) is $\frac{A_{d}}{A_{0}} = (\frac{1}{2})^{2} = \frac{1}{4}$ However, there are 12 spheres in (d) and only 4 spheres in the original. Thus, the ratio of total surface areas $\frac{A_{dT}}{A_{0T}} = (\frac{12}{4}) \times (\frac{1}{4}) = \frac{3}{4}$. The estimated rate of (d) = 3/4 x the Rate of original.
e) Let's assume these white solids are spherical and that the radius of a sphere in (e) is one quarter that of a sphere in the original. The ratio of surface areas of an original sphere to a sphere in (e) is $\frac{A_{e}}{A_{0}} = (\frac{1}{4})^{2} = \frac{1}{16}$. However, there are 40 spheres in (e) and only 4 spheres in the original. Thus, the ratio of total surface areas $\frac{A_{eT}}{A_{0T}} = (\frac{40}{4}) \times (\frac{1}{16}) = \frac{5}{8}$ The estimated rate of (e) = 5/8 x the Rate of original.
f) Rate of (f) = Rate of original; the surface area of the white solid appears to be the sam
Exercise 1.6.4:
At the very start of the reaction, the concentrations of reactants have not changed very much. That means the rate of the reaction remains roughly constant as the first few percent of reactants are consumed. Plotting [product] vs. time gives a straight line with the slope = rate. However, over the course of the reaction, the concentration of reactants goes down as the reactants are consumed. That means the rate of product formation slows down and a plot of [product] vs. time becomes curved. We will be unable to measure the slope in a simple way.
Exercise 1.6.5:
The slope of the first curve, with [A]o = 1 mol L-1, can be estimated by observing that [C] increases from zero to 25 mmol L-1 in 50 seconds. The slope is about $\frac{25}{50} = 0.5 \frac{mmol}{Ls}$. The slope of the next curve, with $[A]_{0} = 2 \frac{mol}{L}$, is $1.0 \frac{mmol}{Ls} (\frac{50 \frac{mmol}{L}}{50s})$. The initial concentration is doubled, and the rate doubles. The slope of the final curve, with $[A]_{0} = 4 \frac{mol}{L}$, is $2.0 \frac{mol}{Ls} (\frac{100 \frac{mmol}{L}}{50s})$. The initial concentration is doubled, and the rate doubles. The reaction is linearly dependent on the concentration of A. Whatever happens to [A] also happens to the rate.
In terms of rate laws, there is a mathematical approach to demonstrating this relationship.
Suppose $Rate = k[A]^{x}$; x is the power of the mathematical relationship.
The ratio of rates in two experiments is $\frac{Rate_{1}}{Rate_{2}} = (\frac{k[A_{1}]^{x}}{k[A_{2}]^{x}}) = (\frac{[A_{1}]}{[A_{2}]})^{x}$
If we take the logarithm of both sides: $\ln (\frac{Rate_{1}}{Rate_{2}}) = \ln ((\frac{[A_{1}]}{[A_{2}]})^{x}) = x \ln (\frac{[A_{1}]}{[A_{2}]})$
Then $x = \frac{ln(\frac{Rate_{1}}{Rate_{2}})}{ln(\frac{[A_{1}]}{[A_{2}]})}$
Exercise 1.6.6:
1. Each time the concentrations double (for example, from 2 to 4 mol L-1), the rate quadruples (for example, from 50/50 = 1 mmol L-1 s-1 to 200/50 = 4 mmol L-1s-1). One explanation is Rate = k[A][B].
2. Another explanation is $Rate = k[A]^{2} \: or \: Rate = k[B]^{2}$
3. We could run a series of experiments in which [A] is changed while holding [B] constant (or vice versa).
Exercise 1.7.1:
a)
b) The first step proceeds at k1; the second at k2.
c) The second step is probably reversible. It is just the exchange of a proton from one OH group to another.
d)
e) $Rate = k_{1} [ketone][NaBH_{4}]$
Exercise 1.7.2:
a)
b) The first step proceeds at k1; the second at k2; the third at k3.
c) There are three different steps, each of which might be rate determining.
The first:
The second:
The third:
d) First one: $Rate = k_{1}[CH_{3}COCl][NH_{3}]$
Second one: $Rate = k_{1}k_{2}[CH_{3}COCl][NH_{3}]$
Third one: $Rate = k_{1}k_{2}k_{3}[CH_{3}COCl][NH_{3}]^{2}$
The rate law could be used to distinguish the third from the first two. However, it would be impossible to tell the difference between the first two using the rate law alone.
e) We can probably rule out the third possibility right away. Proton transfers tend to happen pretty quickly, especially in the presence of a reasonable base such as ammonia. Measuring the rate law would quickly confirm this assumption.
Scenarios 1 and 2 are much harder to distinguish. In both cases, a bond is being broken as another bond is being made. We could make an educated guess that the formation of the good chloride leaving group is pretty easy; that would make step 1 the rate determining step.
However, we are always left with some ambiguity in subtle cases like this one. In order to get a better idea about which transition states are the highest, we would have to perform computational chemistry.
Exercise 1.7.3:
a) i)
ii)
iii) $Rate = k_{1}[CH_{3}COCl]$
b) i)
ii)
iii) $Rate = k_{1}[CH_{3}COCl][NH_{3}]$
c) On the basis of the rate law, possibility (a) can be ruled out, but not possibility (b)
Exercise 1.7.4:
1. There is a geometry change prior to or during the rate determining step.
2. Possibility (b) can be ruled out.
Exercise 1.7.6
a)
b) Mechanism 1: $k[1][2][3]$
Mechanism 2: $k[1][2][3]^{2}$
c) First order in [1]; first order in [2]; second order in [3]
d) Mechanism 2.
Exercise 1.8.2: | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/IV%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_2/01%3A_Reaction_Kinetics/1.09%3A_Solutions_to_Sel.txt |
Intuitively, you know that a reaction goes faster as the temperature is raised, as more reactant molecules have the energy needed to overcome the activation barrier to the reaction. The Arrhenius equation relates reaction rate constants (k) and temperature. One of the forms of the Arrhenius equation is:
$ln k = \frac{-Ea}{RT} + ln A \nonumber$
where Ea is the activation energy for the reaction, T is the absolute temperature (in Kelvin) at which a corresponding k is determined, R is the gas constant, and A is a pre-exponential factor. The activation energy may then be extracted from a plot of ln k vs. 1/T, which should be linear. This plot is called an "Arrhenius plot".
Exercise $1$
Recall that y = mx + b.
1. In a so-called “Arrhenius plot” plot, what is the slope?
2. What is the intercept?
Exercise $2$
Using the following data, construct an Arrhenius plot and determine the activation energy (in both kcal/mol and kJ/mol) and the pre-exponential factor.
1/T (K-1)
ln k (unitless)
0.00152 3.7
0.00157 3.2
0.00160 2.9
0.00165 2.2
0.00170 1.6
Exercise $3$
Using the following data, construct an Arrhenius plot and determine the activation energy (in both kcal/mol and kJ/mol) and the pre-exponential factor.
T (°C) k (mol L-1 s-1)
40 1.3 x 10-4
50 2.2 x 10-4
60 4.0 x 10-4
70 7.5 x 10-4
80 1.4 x 10-3
In practice, activation energies are not often cited in the current literature. Instead, a similar but more useful equation called the Eyring equation is used. The Eyring equation is:
$ln (\frac{k}{T}) = \frac{- \Delta H^{\ddagger}}{RT} + ln (\frac{k_{B}}{h}) + \frac{\Delta S^{\ddagger}}{R} \nonumber$
where k, T and R are the same as in the Arrhenius equation, kB is Boltzmann’s constant, h is Planck’s constant and ΔH and ΔS are the enthalpy and entropy of activation, respectively.
Exercise $4$
1. What should be plotted to make an Eyring plot?
2. What is equal to the slope?
3. What is equal to the intercept?
Note that the activation parameters (ΔH and ΔS ) are not the same as the entropy and enthalpy of the reaction, which can usually be calculated from tables of values. Since they depend on how the reaction proceeds, not just the initial and final states of the reaction, they must be determined experimentally. Once that has been done, interpretation of the numerical values provides insight into the mechanism of the reaction.
4.01%3
Aliphatic nucleophilic substitution is a mouthful, but each piece tells you something important about this kind of reaction.
In substitution reactions, one piece of a molecule is replaced by another. For example, ligands can be replaced in transition metal complexes. Oxygen atoms in organic carbonyl compounds can be replaced by nitrogen atoms or sulfur atoms, in a particular variation of carbonyl addition reactions.
These reactions all involve the addition of a nucleophile to an electrophilic atom or ion. They are all nucleophilic substitution reactions.
Aliphatic systems involve chains of saturated hydrocarbons, in which carbons are attached to each other only through single bonds. Aliphatic nucleophilic substitution is the substitution of a nucleophile at a tetrahedral or sp3 carbon.
Aliphatic nucleophilic substitutions do not play a glamourous, central role in the world of chemistry. They don't happen in every important process, the way carbonyl additions and carboxyloid substitutions appear to in biochemistry. Instead, they are ubiquitous little reactions that play important, small roles in all kinds of places.
For example, polyethylene gloycol (PEG) is a commonly used polymer in lots of biomedical applications. PEG frequently has hydroxyl groups at each end of the polymer. Capping the ends of the polymer through reaction with another group can lead to very different physical properties.
For another example, many biochemical processes require prenylation of proteins. That would involve a nucleophilic substitution in which a sulfur in a cysteine residue adds to a tetrahedral carbon in a prenyl group, replacing a phosphate group.
In order to be an electrophile, that tetrahedral carbon should have at least some partial positive charge on it. In the simplest cases, this electrophilic carbon is attached to a halogen: chlorine, bromine or iodine. These compounds are called alkyl halides (or alkyl chlorides, alkyl bromides and alkyl iodides).
Exercise \(1\)
Draw structures of the following alkyl halides.
a) 2-bromopentane b) 2-methyl-2-chlorobutane c) benzyl iodide d) allyl chloride
Answer
Lots of things can be nucleophiles in these reactions. Sometimes, the nucleophile is a neutral compound with a lone pair, such as ammonia or water (or, by extension, an amine or an alcohol).
Exercise \(2\)
Sometimes, addition of a mild base is helpful in reactions of neutral nucleophiles. Show, with mechanistic arrows, how sodium carbonate (K2CO3) would play a role in the reaction.
Answer
The third row analogs of these nucleophiles, in which the nucleophlic atom is a phosphorus or a sulfur, are also good nucleophiles in these reactions.
Sometimes, the nucleophile is an anion. Cyanide anion is a good nucleophile, as are the structurally similar acetylides.
Enols, enolates and enamines are also very good nucleophiles in this type of reaction.
Semi-anionic nucleophiles such as Grignard (or organomagnesium) reagents and alkyl lithium reagents can sometimes act as nucleophiles in this reactions, but they are not very reliable. Complications often lead to other reactions instead. Gilman (or organocopper) reagents, in which a carbon atom is attached to a copper atom, can usually react with alkyl halides. However, they probably act via a different mechanism from the ones described in this chapter.
Exercise \(3\)
Put a box around the nucleophilic portion of the following molecules. Note: lone pairs have not been drawn in. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/IV%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_2/02%3A_Mathematical_Tools_in_Reaction_Kinetics/2..txt |
Aliphatic nucleophilic substitution clearly involves the donation of a lone pair from the nucleophile to the tetrahedral, electrophilic carbon bonded to a halogen. We might expect this carbon to be electrophilic because of the halogen attached to it. For that reason, it attracts to nucleophile. However, the mechanism of the reaction might happen in a couple of different ways.
Exercise \(1\)
Compare the electronegativity of carbon to that of fluorine, chlorine, bromine and iodine.
1. On this basis alone, explain why the carbon attached to the halogen would be electrophilic.
2. Which compound should be most electrophilic based on electronegativity: fluoromethane, chloromethane, bromomethane or iodomethane?
3. Use the following bond strengths to estimate the qualitative trend in activation barriers for nucleophilic substitution in the four compounds in part (b): C-F 115 kcal/mol; C-Cl 84 kcal/mol; C-Br 72 kcal/mol; C-I 58 kcal/mol.
4. Fluorocarbons are quite stable towards aliphatic nucleophilic substitution; in general, they do not undergo this reaction. Explain why.
Answer a
The electronegativity of carbon (2.55 on Pauling scale) is less than that of fluorine (3.98), chlorine (3.16), bromine (2.96) or iodine (2.66).
On that basis, the carbon attached to a halogen is electrophilic because it has a partial positive charge resulting from the polar carbon-halogen bond.
Answer b
We would expect an alkyl fluoride to be the most electrophilic of these compounds, based on electronegativity.
Answer c
Assuming the energy required for breaking the carbon-halogen bond plays a major role in the activation barrier (not guaranteed), we would expect the activation barrier to be lowest with the alkyl iodide, then the alkyl bromide, then the alkyl chloride and finally the alkyl fluoride. This prediction contrasts with what we might expect based on electronegativity.
Answer d
The stability of alkyl fluorides towards this reactions suggests that there is, in fact, a prominent role played by bond strengths, at least in that case. The carbon-fluoride bond is strong enough to hinder nucleophilic substitution in this compound.
In considering possible mechanisms for this reaction, we ought to think about overall bond-making and bond-breaking steps. In the addition of sodium cyanide to alkyl chloride to make an alkyl nitrile, there is one bond-making step (the C-C bond) and one bond-breaking step (the C-Cl bond). The simplest reaction mechanism would involve some combination of these steps.
Two possibilities immediately present themselves:
Mechanism A
The C-C bond forms and then the C-Cl bond breaks.
Mechanism B
The C-Cl bond breaks and then the C-C bond forms.
However, some familiarity with bonding in the second row of the periodic table may suggest to you that mechanism A is not very likely. That mechanism would require forming five bonds to carbon before the C-Cl bond eventually breaks. We can safely ignore this possibility.
Instead, there may be a third possibility to consider.
Mechanism C
The C-Cl bond breaks and the C-C bond forms at the same time.
Mechanism C is a concerted mechanism; two bond-making and -breaking events happen at once. However, no octet rules are violated.
Reaction progress diagrams for these two reactions would look like the illustrations below.
Mechanism B, ionization and then addition of nucleophile:
Mechanism C, direct displacement of leaving group by nucelophile:
Exercise \(2\)
Compare mechanism B and C in terms of your expectations of the following parameters:
1. Activation enthalpy.
2. Activation entropy.
Answer a
In mechanism B, the dissociative one, we would expect a higher activation enthalpy. The first step, which appears to be rate determining, is a bond-breaking step, which will cost energy. In mechanism C, the bond-breaking is compensated by some bond-making; overall, this probably costs less energy.
Answer b
In mechanism B, the dissociative case, we expect a more positive entropy of activation. As the bond to the halide begins to break, the halide and carbocation fragments begin to move independently of each other, gaining degrees of freedom and increasing in entropy. In mechanism C, the incoming nucleophile appears to coordinate its motion with that of the departing halide; as a result, there are fewer degrees of freedom in this case.
There isn't necessarily a reason to believe that mechanism B is the correct mechanism and mechanism C is the wrong one, or vice versa. Either one may be possible. You may need to do some work in order to figure out which one really happens. Some experiments may help to highlight what is going on.
Exercise \(3\)
If charged intermediates are suspected along a reaction pathway, insight can sometimes be gained by running a reaction in a more polar solvent and comparing its rate to that of the reaction in a less polar solvent.
1. Are charged intermediates present, either in mechanism B or C?
2. Explain how each of these mechanisms might behave in a more polar solvent.
Answer a
Charged intermediates are present in the dissociative mechanism (B).
Answer b
It seems like a more polar solvent would favor both mechanisms, because both involve the interaction of an anionic nucleophile with an electrophile and loss of an anionic leaving group. However, the dissociative case (B) involves a build-up of charge in the intermediate. It is possible that a more epolar solvent could reduce the barrier to that buildup of charge separation, accelerating this mechanism.
Exercise \(4\)
Sometimes, a distinction between two possible mechanisms can be gained by comparing rate laws expected from each mechanism.
1. What do you think is the likely rate-determining step in mechanism B?
2. What do you expect will be the rate law for mechanism B?
3. What do you think is the likely rate-determining step in mechanism C?
4. What do you expect will be the rate law for mechanism C?
Answer a
The rate-determining step is probably the bond-breaking one (the first one).
Answer b
Because the nucleophile has not yet participated at that point, \(Rate = k[R-X]\), if R-X = the alkyl halide.
Answer c
There is only one step; it is the rate-determining step, by default.
Answer d
\(Rate = k[R-X][Nu]\). | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/IV%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_2/04%3A_Aliphatic_Nucleophilic_Substitution/4.02%3.txt |
In aliphatic nucleophilic substitution, a nucleophile (abbreviated Nu) replaces a halogen "leaving group" (abbreviated LGp) from a tetrahedral carbon. Aliphatic nucleophilic substitution may take place through two different mechanisms:
• C-LGp bond breaking, followed by C-Nu bond formation.
or
• C-Nu bond formation at the same time as C-LGp bond breaking.
A look at the reaction progress diagrams for these two reactions illustrates some big differences. We will look at cyanide anion, a nucleophile, substituting for chloride in 2-chloropropane.
In the first case, some energy must be added in order to break the carbon-chlorine bond. The chlorine forms an anion, leaving a cation on the carbon. This ion pair is an intermediate along the reaction pathway. The cyanide ion then connects with this cation to form the nitrile product. Thus, there are two elementary steps in this mechanism.
Most likely, the first step is the rate-determining step. Breaking bonds costs energy, whereas making bonds releases energy. It is hard to imagine that there could be a significant barrier to the second step; the anion and cation should come together almost automatically.
The rate law for this stepwise mechanism is:
$Rate = k[PrCl] \nonumber$
that is, the rate depends on the first elementary step, but not on the second one. The second step happens pretty much automatically as soon as the first one has finally gotten around to happening.
In the second case, the nucleophile displaces the chloride directly in one step. There is only one elementary step in this reaction, and it requires both compounds to come together at once.
The rate law for this concerted mechanism is:
$Rate = k[PrCl][^{-}CN] \nonumber$
These two rate laws are very different, and offer an additional way for us to tell how this reaction is taking place. In principle, if we try the reaction with different concentrations of cyanide (but keep the 2-chloropropane concentration constant), we can see whether that has an effect on how quickly the product appears. If it has the predictable effect, maybe the reaction happens in one step. If not, maybe it is a two-step reaction.
Because the rate laws for these two mechanisms are so different, there has arisen a catchy shorthand for describing these reactions based on their rate laws, coined by C.K. Ingold. The rate of the stepwise reaction depends only on one concentration and is referred to as a "unimolecular reaction"; Ingold's shorthand for this kind of nucleophilic substitution was "SN1".
The rate of the concerted reaction depends on two different concentrations and is referred to as a bimolecular reaction; Ingold's shorthand for this reaction was "SN2".
Exercise $1$
Suppose you run this reaction with three different concentrations of cyanide: 0.1 mol/L, 0.2 mol/L and 0.3 mol/L. You keep the 2-chloropropane concentration constant at 0.05 mol/L.
1. The reaction turns out to be proceeding via a SN1 mechanism. Plot a graph of rate vs.[-CN].
2. The reaction turns out to be proceeding via a SN2 mechanism. Plot a graph of rate vs.[-CN].
Now you switch things up and run this reaction with three different concentrations of 2-chloropropane: 0.1 mol/L, 0.2 mol/L and 0.3 mol/L. You keep the 2-chloropropane concentration constant at 0.05 mol/L.
1. The reaction turns out to be proceeding via a SN1 mechanism. Plot a graph of rate vs.[iPrCl].
2. The reaction turns out to be proceeding via a SN2 mechanism. Plot a graph of rate vs.[iPrCl].
Why would the mechanism proceed in one way and not the other? Molecular choices between pathways like this are often described on the basis of "steric and electronic effects"; in other words, it's either something to do with charge or something to do with crowdedness. We will see soon how these effects can influence the course of the reaction, and how the mechanism can itself have consequences in the formation of different products.
Exercise $2$
How might crowdedness or steric effects influence the pathway taken by the reaction between cyanide and 2-chloropropane?
Exercise $3$
How might charge stability influence the pathway taken by the reaction between cyanide and 2-chloropropane? | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/IV%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_2/04%3A_Aliphatic_Nucleophilic_Substitution/4.03%3.txt |
Apart from measuring rates of reaction and deducing the rate law, there are other lines of evidence that can suggest how the reaction is occurring. For aliphatic nucleophilic substitution, stereochemistry of the products provides some additional evidence.
Suppose you carry out a nucleophilic substitution reaction using a chiral starting material. You decide to convert (S)-3-chlorooctane into the corresponding azide. Azides are pretty widely used reagents (but slightly dangerous and potentially explosive). They are employed in a class of reactions called "click chemistry"; you've just heard about these reactions and you want to try one out for yourself.
You have complete control over the mechanism of the reaction (not so easy in reality, but in this thought experiment you can set the dial on your stir plate to the desired mechanism). You choose to make the reaction occur through an SN2 pathway.
You know the product will be chiral, so you plan to check its optical rotation. The trouble is, once you have finished the reaction, the optical rotation is exactly the opposite of what you were expecting, based on the values of other compounds like this one. You did the reaction successfully but got the unexpected enantiomer.
You're not worried. You've been taking this nifty chemistry class and you have an idea of something else to try. This time you select an SN1 pathway.
You finish the reaction and get the right product, but it shows no optical rotation whatsoever. This time you got a racemic mixture.
This is just a thought experiment, but what would it all mean? Why might changing mechanism influence the stereochemistry?
This is just a thought experiment, but the results are generally true: in an SN2 reaction, the chiral center undergoes an inversion. The three-dimensional arrangement of groups around the chiral center is the opposite of how it started. In an an SN1 reaction, the chiral center undergoes racemization. There is a 50:50 mixture of enantiomers.
Exercise \(1\)
1. Propose reasons why the stereochemistry would flip in an SN2 reaction. A drawing will help.
2. Propose reasons why an equal mixture of stereoisomers would result from an SN1 reaction. A drawing will help.
Answer
Exercise \(2\)
You may have noticed that two different solvents were used in the two reactions above. Propose a reason why this change in solvent may lead to a change in mechanism.
Answer
In an SN2 reaction, the nucleophile donates electrons to the electrophilic carbon, displacing the leaving group from the other side. The nucleophile donates its electrons to the lowest unoccupied molecular orbital, which displays a lage lobe on the side of the carbon opposite the leaving group.
As a result, the nucleophile always approaches from the opposite side of the electrophilic carbon as the location of the leaving group, and ends up on the opposite side from where the leaving group group was.
• SN2 reactions at chiral centers lead to reversal or "inversion" of the chiral center.
On the other hand, in an SN1 reaction, the nucleophile enters only after the leaving group has left. At that point, the electrophilic carbon is a cation, so it is trigonal planar because it only has three groups attached to it. The nucleophile could easily approach from either side.
SN1 reactions at chiral centers lead to racemic mixtures.
Exercise \(3\)
Show the products of the following reactions, and indicate stereochemical configuration of bothe the starting material and the product in each case.
Exercise \(4\)
The lesson here can be restated as follows: the mechanism affects the stereochemical outcome of the reaction. Explain why that fact is important in the context of making a chiral drug for the pharmaceutical market. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/IV%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_2/04%3A_Aliphatic_Nucleophilic_Substitution/4.04%3.txt |
Regiochemistry is the term for where changes take place in a reaction. It can be another indication of how the reaction occurred.
In aliphatic nucleophilic substitution, the answer seems pretty obvious: the reaction takes place at the electrophilic carbon, the one attached to the electronegative halogen. That's where the leaving group is. When the leaving group is replaced, that's where the nucleophile will be. But this isn't always true.
• In an SN2 reaction, the nucleophile is always found on the carbon where the leaving group used to be.
• In an SN1 reaction, the nucleophile is usually found on the carbon where the leaving group used to be. Sometimes it moves.
Under some circumstances, unexpected changes occur. The following two reactions are examples of such surprises. These reactions happen to take place via an SN1 mechanism.
So, the regiochemistry of this reaction may be more complicated than we thought.
What is happening in these two reactions? In one of them, the bromine is just hopping from one place to another along the molecule. Some of the original compound remains, too, so there is a mixture. If you look carefully, though, the bromine has switched places with a hydrogen atom. It doesn't seem like that hydrogen atom could come off very easily.
In the other reaction, something very similar is happening. Bromide and chloride both have lone pairs, so they can both be nucleophiles as well as leaving groups, and one can replace the other. There is lots of chloride around, so it beats any bromide to the electrophilic carbon. Once again, though, some of the chloride seems to end up in the wrong place.
This sort of behaviour is characteristic of carbocations. It is called a rearrangement, in which part of the molecule unexpectedly switches places.
Again, one of the products forms in a simple enough way.
The formation of the other product involves a "1,2-hydride shift". In this event, a hydrogen anion hops from one carbon to the next, leaving a cation where it used to be.
Exercise \(1\)
Draw the reaction above with all the hydrogens drawn in the structures, to confirm the formal charges and the positions of the hydrogens.
Answer
Exercise \(2\)
Explain the observed product ratios in the above reaction.
Exercise \(3\)
Predict the products of the following reactions.
The hydride can hop one carbon away because of the proximity to the empty p orbital with which it can overlap and form a new bond.
The barrier for a hydride shift is not very high, provided a carbocation is available on the very next carbon. As a result, an equilibrium between cations is established pretty quickly. Below, there is an equilibrium between two secondary cations on the 2-methylpentyl skeleton.
However, that particular structure has another possible cation that is more stable. Once a tertiary cation forms, the hydride isn't likely to hop back.
On the other hand, there is also a primary position. A hydride shift could give a primary cation, but that isn't likely to happen, because it would be too far uphill.
Altogether, there is an energy surface linking several different possible cations. In this case, however, one tertiary cation would quickly dominate.
It isn't just hydride ions that are able to undergo 1,2-migrations. Alkyl anions (such as methyl, CH3-) and aryl anions (such as phenyl, C6H5-) can also undergo 1,2-shifts, rearranging to give stable cations.
For example, the following reaction is apparently just a substitution of a bromo group for a hydroxy group. The regiochemistry indicates a cation was formed, however, because the new group is found at the site of the most stable carbocation.
The mechanism here involves protonation of the hydroxy group; the reaction takes place in strong, concentrated acid. The resulting cation is able to undergo a 1,2-methyl shift leading to a new carbocation. The bromide ion connects at that new position. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/IV%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_2/04%3A_Aliphatic_Nucleophilic_Substitution/4.05%3.txt |
If two possible mechanisms can occur, there may be some factors that have an influence on the course of the reaction, tipping it one way or the other. One of the most important factors determining the mechanism of aliphatic nucleophilic substitution is the structure of the alkyl halide.
The SN1 mechanism involves formation of a carbocation. Ion stability is often a very important factor influencing how easily a reaction occurs. It stands to reason that, the more stable the cation that forms, the more easily an SN1 mechanism can occur.
There is a simple trend that more substituted carbocations are more stable than less-substituted ones. By more-substituted, we mean carbocations in which the carbon bearing the positive charge is attached to more carbons and fewer hydrogens. A tertiary carbocation is more stable than a secondary, a secondary is more stable than a primary carbocation, and a primary is more stable than a methyl cation.
This trend is usually explained by hyperconjugation in the more substituted cation. In hyperconjugation, neighbouring σ-bonding orbitals overlap with the empty π-orbital that is the center of the carbocation.
That electronic donation from the occupied σ-bonding orbitals helps delocalize the positive charge, lowering the positive charge on the central carbon and placing a little of it on the surrounding ones. Energetically, the interaction is favorable because the electrons in the σ-bonding orbitals are lowered stabilized by delocalization.
Overall, SN1 reactions occur much more easily when the halide is attached to a more substituted carbon. The resulting carbocation forms more readily in what is otherwise the hardest step in the reaction. SN1 reactions occur most easily at tertiary carbons, moderately well at secondary ones, and very sluggishly at primary ones, if at all.
Exercise \(1\)
Although they may be considered primary alkyl halides, compounds like benzyl chloride and allyl bromide are capable of undegoing SN1 reactions. Show why.
Answer
SN2 reactions do not undergo highly charged intermediates, so ion stability is less important in that pathway. On the other hand, there is a stark contrast between SN1 reactions and SN2 reactions in terms of steric effects. SN1 reactions occur via a trigonal planar intermediate, which is less sterically crowded than the starting material. SN2 reactions do not occur via an intermediate, but the transition state through which they proceed is actually five-coordinate; the electrophilic carbon gets more crowded as the reaction proceeds. Steric effects are much more important in this concerted pathway.
Comparing the approach to the transition state in a series of alkyl halides of different substitution, we can see some steric differences. The pictures shows a snapshot very early in the reaction, when the nucleophile is just approaching the electrophile but the reaction is not really committed yet. Comparing these pictures, it seems most likely that the nucleophile will keep going forward in the top case, with the methyl. In the bottom case, with the tertiary halide, chances are that the nucleophile will just bounce off the methyl groups before it can connect with the electrophilic carbon.
As a result, SN2 reactions are more likelyto occur with less substituted alkyl halides. They occur very easily with methyl halides and primary alkyl halides. They occur moderately well at secondary alkyl halides, but only with difficulty at tertiary alkyl halides.
Exercise \(2\)
In each of the following cases, is nucleophilic substitution likely to proceed via an SN1 or an SN2 reaction? | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/IV%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_2/04%3A_Aliphatic_Nucleophilic_Substitution/4.06%3.txt |
There are often several factors that can influence the course of a reaction. Probably the most important is the structure of the alkyl halide, but the solvent can also play a role.
The crucial difference between SN1 and SN2 reactions is the ionization step in the SN1 pathway. Factors that stabilize ions, and assist in ionization, promote this pathway.
In general, more polar solvents are often helpful in nucleophilic substitutions; the nucleophile may be an ionic compound itself, and a more polar solvent will help it to dissolve. However, especially polar solvents may provide additional stability to ions.
The most polar solvents tend to be those that are capable of hydrogen bonding, such as water and alcohols. These are sometimes called "polar, protic solvents". The "protic" part refers to hydrogen bonding; in hydrogen bonding, the hydrogen atom attached to a very electronegative oxygen or nitrogen develops a significant positive charge, like a proton.
Polar, protic solvents can stabilize ions through very strong intermolecular attractions. The protic hydrogen can strongly interact with anions, whereas the lone pair on the oxygen atom can stabilize cations.
These stabilizing interactions can strongly stabilize the intermediates in SN1 reactions. In the same way, the transition state leading into the intermediate is also significantly stabilized. The barrier for this reaction is lowered and the reaction can occur more quickly.
In addition, polar, protic solvents may play an additional role in stabilizing the nucleophile. If the nucleophile is stabilized, it is less likely to react until a sufficiently strong electrophile becomes available. As a result, polar, protic solvents may also depress the rate of SN2 reactions. Once the alkyl halide ionizes and a more attractive electrophile becomes available, the nucleophile can spring into action.
Exercise \(1\)
The following are a baker's dozen of potential solvents.
1. Identify the protic solvents in this group.
2. Identify the completely non-polar solvents in this group.
3. Identify the polar, aprotic solvents.
4. Rank the polar solvents from most polar to least polar.
5. Identify two basic compounds, frequently used to facilitate proton transfers in reactions.
Answer a
ethanol, isopropanol, trifluoroacetic acid
Answer b
hexane, toluene
Answer c
THF, acetonitrile, DMF, dichloromethane, ether, DMSO, triethylamine, pyridine
Answer d
DMSO > DMF > ACN > pyridine > DCM > THF > ether > TEA, based on dielectric constants. In general, the ones with multiple bonds between two different atoms are the most polar.
Answer e
pyridine and triethylamine. The lone pair on the nitrogen atom is basic toward protons. The trend in basicity is triethylamine > pyridine >> acetonitrile; as the percent s character in the lone pair increases, the electrons are lower in energy and less available for donation.
Exercise \(2\)
Describe the SN1 reaction as slow, medium or fast in the following cases.
1. NaCN and PhCH2Cl in acetonitrile, CH3CN.
2. NaSH and 2-bromo-2-methylheptane in methanol.
3. KI and 1-chloropentane in 2-propanol.
Exercise \(3\)
Describe the SN2 reaction as slow, medium or fast in the following cases.
1. LiCCCH3 and 2-bromopentane in THF.
2. NaN3 and 3-chloro-3-methyloctane in DMF.
3. LiOPh and 1-bromohexane in methanol.
4.08%3
The nucleophile can sometimes play a pronounced role in nucleophilic substitutions. The following relative rates have been observed when these nucleophiles reacted with methyl bromide in methanol:
note: Ph = phenyl, C6H5; Ac = acetyl, CH3C=O; Et = ethyl, CH3CH2.
Presumably, some of the species react much more quickly with methyl bromide because they are better nucleophiles than others.
Exercise \(1\)
Sometimes we can draw general conclusions about kinetic factors by looking at sub-groups among the data. Determine how the following factors influence nucleophilicity (the ability of a species to act as a nucleophile). Support your ideas with groups of examples from the data (preferably more than just a pair of entries).
1. charge on the nuclophile
2. size of the atom bearing the charge
3. electronegativity of the atom bearing the charge
4. delocalization of charge
Exercise \(2\)
Nucleophilicity plays a strong role in the rate of one type of substitution mechanism, but not the other.
1. In which mechanism is it important? Support your idea.
2. Is the reaction of methyl bromide likely to proceed via this mechanism? Why or why not?
Exercise \(3\)
A trend very similar to the data above is found in substitution reactions of py2PtCl2 (py = pyridine) in methanol. Draw a mechanism for this substitution and explain why nucleophilicity plays an important role.
Answer
Exercise \(4\)
Very fast nucleophiles are sometimes more likely to undergo SN2 reactions than SN1 reactions. Explain why. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/IV%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_2/04%3A_Aliphatic_Nucleophilic_Substitution/4.07%3.txt |
Enolates and related nucleophiles deserve a closer look because they are very common and because they have their own issues of regiochemistry.
Remember, an enolate is just the conjugate base of an enol. An enolate can also be thought of as the conjugate base of a related carbonyl. Because the enolate is a delocalized anion, it can be protonated in two different places to get two different conjugates.
Enols typically are not seen because of a rapid equilibrium with that related carbonyl compound. As soon as an enol forms, if there is any way for it to transfer a proton to get to the carbonyl, it will do so. This kind of equilibrium is called "tautomerism", involving the transfer of one proton from one place to another within the molecule. The enol and its related carbonyl are referred to as "tautomers". "Tautomers" describes the relationship between these two molecules.
Enamines are very similar to enolates, but with a nitrogen atom in place of the oxygen. Hence, they are amines instead of alcohols.
Enamine, enolates and enols are all turbo-charged nucleophiles. The nucleophilic atom is the alpha carbon. Although that carbon can be thought of as a double bonded carbon, with no lone pair, that position is motivated to donate electrons because of pi donation from the oxygen (or nitrogen).
One of the issues with these nucleophiles has to do with asymmetry about the carbonyl (or the would-be carbonyl). If one alpha position next to the carbonyl isn't the same as the other one, two possible enolates could result from removal of a proton.
That means we potentially have two different nucleophiles from the same starting compound. Sometimes, mixtures of products result from enolate reactions.
Nevertheless, enolates and enamines are very broadly used in the synthesis of important things like pharmaceuticals, precisely because they can be controlled so well. If you want the enolate on one side of the carbonyl -- we'll call it the more-substituted side -- then you can have it. If you want the enolate on the other side of the carbonyl -- the less-substituted side -- you can have that, instead.
To get the proton off and turn a carbonyl compound into an enolate requires a base. Some control over which proton is removed might come from the choice of base. Let's think about what is different about those two sides of the carbonyl. One side is more substituted. It has more stuff on it. It's more crowded. Maybe to get the proton off the more crowded position, you need a smaller base.
Conversely, to get the proton exclusively from the least crowded position, and have very little chance of getting it from the more crowded spot, you could use a really big base.
So let's go all the way through a reaction, looking at two possible nucleophilic substitution products. Suppose we take 2-methylpentan-3-one, deprotonate it to make an enolate anion, and add chloroethane to complete the reaction. The overall reaction is called an α-alkylation, because the alkyl electrophile that is added goes to the alpha position, next to the carbonyl.
You can see the two different products of the reactions: one of them is 4,4-dimethylhexan-3-one. The other one is 2,4-dimethylhexan-3-one. From what we have said so far, if we use a big base, maybe it won't fit as well on the more hindered side (the top pathway above), so it will go to the least-hindered side (the bottom path).
We can make the base bulkier by adding alkyl substituents. If it just has little hydrogens attached to the basic atom, it won't be very crowded. Also, in this case we used amide bases because they are very strong bases. (Amide is the inorganic chemistry term for the conjugate base of ammonia, or NH2-.) There is an equilibrium between a ketone and an enolate. A stronger base pushes the equilibrium all the way to the enolate side, with no leftover ketone.
• Bulky bases can promote deprotonation on the least substituted side (the least crowded side) of the carbonyl.
So if we need to take the lower pathway, we have found a way to do it. LDA (lithium diisopropylamide) will take the least hindered proton and alkylation will occur on that side. The trouble is, the much smaller lithium amide can fit on either side. It has no preference, so it leads to a mixture of products.
How can we force things through the upper pathway?
There's something else about enolates that is apparent only when you look at the ions in one resonance form. Enolate ions can be thought of as alkenes, of course. Depending on which proton we remove, we get two different alkenes. There may be factors that make one of these two alkenes more stable. If so, there may be ways to form that one instead of the other.
In general, more-substituted alkenes are more stable than less-substituted ones. The more substituted alkene is formed via loss of the proton at the more crowded position. Enolate stability should influence things, and it does, but there are some subtleties involved. To understand the subtleties, we are going to need to look at the potential energy diagram for enolate formation.
The barrier might be a little higher going from the middle of the picture to the left than going from middle to right (it's more crowded, after all). However, once the reaction gets over that barrier, it settles out in a more stable place (the enolate is more stable). An initially greater investment, in this case, yields a greater return in the end.
• The thermodynamic product refers to the one that is more stable. In this case, it also happens to be harder to get to, becuas eof a higher barrier.
• The kinetic product refers to the one that is formed more easily, because of a lower barrier. in this case, it also happens to be less stable.
An alternative view of that surface places the reactant on the left and both products on the right. The two possible pathways are then superimposed on each other. The conclusion we draw is the same: the barrier is higher along the red path than the blue path because of steric hindrance. Eventually, the red path leads to the lower energy product.
Just looking at the diagram, we can infer that it takes more energy to get over the barrier leading to the more stable enolate. If we supply more energy to start with, we might have a better chance of reaching the more stable enolate. Of course, the easiest way to do that is to heat the reaction up. If we don't want to form the more stable enolate, we could cool the reaction down. That way, there wouldn't be enough energy to get to the more stable enolate.
• Warmer temperatures can lead to formation of the thermodynamic product.
• Colder temperatures will encourage formation of the kinetic product.
Forming a product based on its relative stability means relying on thermodynamics. Getting to the thermodynamic product means you have to give the system a chance to find its way there. One way to do that is to allow the deprotonation to happen reversibly. Given multiple chances, the more stable enolate will form eventually.
• Time is a key factor in forming the thermodynamic product. The longer the reaction time, the more likely the system will eventually reach the most stable position.
An obvious way to allow reversibility is to conduct the reaction in a protic solvent, such as ethanol. If the enolate takes a proton from ethanol, it becomes a ketone again. It has a second chance to get deprotonated again. Eventually, it should form the most stable enolate.
• Protic solvents allow reversibility in enolate formation. Conducting enolate formation in a protic solvent should allow eventual formation of the more stable enolate.
On the other hand, if you intend to take the proton off the least substituted position, you don't want any reversibility. Given the chance, the wrong enolate will eventually form. You want to just take what you get the first time, which is more likely to be the blue pathway because of its lower barrier. Earlier, we used a bulky amide base to form the least stable enolate because we didn't want to allow the possibility of second chances.
This is a case in which we need kinetic control to get one product: we want the least-substituted enolate, and we depend on it forming more quickly than the ther enolate. After all, if the other enolate forms, it is more stable; it isn't likely to come back.
• Base strength can be important in formation of a thermodynamic enolate. A strong base, such as hydroxide or alkoxide, forms a protic conjugate acid that could easily reprotonate the enolate.
• A very strong base, such as amide, forms a protic conjugate acid that is much less likely to reprotonate the enolate.
There are some even more subtle factors that are used to help things along just a little bit more. For example, the counterions for the bases can have a modest effect one way or the other. The effect has to do with the degree of covalency between the counterion and the enolate. For example, lithium is a smaller, harder alkali ion that forms strong bonds with oxygen. Potassium is a larger, softer alkali ion that forms weaker bonds with oxygen. We are just using these terms in a relative way; lithium and potassium ion are both hard, but lithium ion is even harder than potassium ion.
If the enolate is bound to its counterion, it is less likely to undergo reversal to the ketone. Lithium counterions are sometimes used to promote formation of the kinetic enolate. Potassium counterions are sometimes used to promote formation of the thermodynamic enolate.
• Smaller counterions like lithium promote formation of the kinetic enolate; larger counterions promote the formation of thermodynamic enolates.
There is one final factor to talk about, which comes up in the case of sodium hydride. The use of sodium hydride generally leads to the thermodynamic enolate. That may be a surprise. After all, sodium hydride is a very strong base. We would not expect the counterion, H2, to be acidic. That means the use of sodium hydride allows only one chance to get to the thermodynamic product. So how does NaH manage to do that?
The answer appears to be related to phase. NaH is an insoluble solid. As a result, the reaction between NaH in the solid phase and the ketone in the solution phase happens only at the surface of the NaH. This reaction is very slow. It happens over a long period of time. Remember, time was one of the factors that favored formation of the thermodynamic product.
Exercise \(1\)
Fill in the blanks in the following synthesis. Includes: aliphatic nucleophilic substitution, silyl ethers, carboxylic substitution, carbonyl addition (anionic nucleophiles, neutral nucleophiles, enolates)
3. Stereocontrol of the allylation step:
Is the base used a kinetic or a thermodynamic base? Why?
Explain the stereocontrol of the allylation step in the preparation of Fragment C (shown here). Be sure to include pictures.
Answer
3. Kinetic -- to fully deprotonate (not equilibrate or you lose stereocontrol) and for the chelation control.
Due to chelation effects shown above, the LDA preferentially removes one hydrogen to form only the Z-enolate that will then do the SN2 to allyl bromide on only one face. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/IV%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_2/04%3A_Aliphatic_Nucleophilic_Substitution/4.09%3.txt |
Aliphatic Nucelophilic Substitutions can be useful reactions. A minor drawback is the low natural occurrence of alkyl halides. Alcohols are much more common.
We could do nucleophilic substitutions on alcohols. The trouble is, oxygens are less polarizable than halides. A hydroxide ion is less stable, and harder to form than a halide ion. They don't make very good leaving groups, comparatively.
One way around that problem would be to protonate the oxygen. Attached to the carbon, it is a cation. Once it leaves, it becomes a neutral. The issue of ion stability is sidestepped. The trouble is, plunking a compound into concentrated acid is not always a reliable way to get things done.
For that reason, it is pretty helpful to be able to turn alcohols into alkyl halides, or otherwise turn hydroxyls into stable, anionic leaving groups.
One of the most common synthetic methods of converting alcohols into good candidates for nucleophilic substitution is to convert the hydroxyl into a halide through the use of a phosphorus reagent. Phosphorus tribromide is frequently used to make alkyl bromides from alcohols.
This reaction itself involves a sequence of nucleophilic substitution reactions. In the first, the oxygen atom in a hydroxyl group acts as the nucleophile and replaces a bromide on phosphorus. In the second, the displaced bromide ion rebounds to displace the oxygen atom from the tetrahedral carbon. This mechanism is aided by the strength of the strong phosphorus-oxygen bond that is formed. The phosphite that forms is a very good leaving group.
Another common method is to turn the hydroxyl into a sulfonate ester, such as a mesylate or tosylate. Again, the oxygen atom acts as a nucleophile, displacing a halide from the sulfur in a sulfonyl chloride. This is very similar to the bromination with phosphorus tribromide, but the sulfonate ester waits, poised to be displaced by a nucleophile. In fact, tosylates are generally even better leaving groups than halides.
Biologically, something very similar to both of these processes sometimes happens. The alcohol unit is converted into a phosphate. The alcohol can be phosphorylated by a molecule of ATP. Again, the phosphate portion of the molecule is a very good leaving group.
Exercise \(1\)
Provide a mechanism for the reaction of phosphorus tribromide with 2-pentanol, based on the description provided.
Answer
Exercise \(2\)
Provide a mechanism for the reaction of mesyl chloride with 2-pentanol, based on the description provided.
Answer
Exercise \(3\)
Sometimes in formation of sulfonate esters, halogenation occurs by accident, forming an alkyl chloride. Show how that might happen by continuing on from the mechanism of formation of a mesylate ester from mesyl chloride and 2-pentanol.
Answer
Exercise \(4\)
One way of preventing side reactions during synthesis of sulfonate esters, like the one in the previous question, is to perform a two-phase reaction. For example, the reaction might be performed in a mixture of water and dichloromethane, with a little added sodium hydroxide to act as a base. Show how this approach would limit the chlorination reaction.
Answer
Exercise \(5\)
Provide reagents (or series of reagents) to accomplish the following transformations.
Exercise \(6\)
Mitsunobu addition is a one-step method of replacing an alcohol's OH group with a nucleophile. In the following case, the original oxygen atom has been completely replaced. You may be able to guess where it goes based on other reactions you have seen here.
Propose a mechanism for this reaction, but don't worry about what DEAD does ( that is a more advanced topic).
4.11%3
Oxygen is a very common element in all kinds of compounds, whether they are biological molecules, minerals from the earth or petrochemicals. Exploiting oxygen's electronegativity and giving it a little help to become a leaving group is a common way to make connections and build new molecules in nature, the laboratory or the production facility.
Sometimes oxygen doesn't need much help to become a leaving group. Epoxides, or oxiranes, are three-membered ring ethers. They are good electrophiles, and a C-O bond breaks easily when a nucleophile donates electrons to the carbon.
Exercise \(1\)
Explain why the C-O bond in an epoxide breaks easily.
Answer
Exercise \(2\)
Use a potential energy diagram to show why epoxides are susceptible to react with nucleophiles, whereas other ethers are not.
Epoxides are very useful in the synthesis of important molecules. The Nu-C-C-O motif that is formed in nucleophilic addition to an epoxide is very valuable. Whereas other nucleophilic additions simply replace a halide or leaving group with a nucleophile, exchanging one reactive site with another, addition to an epoxide makes a product that has gone from having one reatcive site to two reactive sites. That can open the door to lots of useful strategies when trying to make a valuable commodity.
Exercise \(3\)
Show how you could carry out the following transformation. More than one step is involved.
Exercise \(4\)
One of the most widespread uses of epoxides is in making polymers. The polyethylene glycol produced in polymerization of an epoxide is frequently used in biomedical applications. Provide a mechanism with arrows for the following polymerization of ethylene oxide, in the presence of:
1. an acid catalyst
2. a basic catalyst.
Exercise \(5\)
Tetrahydrofuran can also be polymerized, forming polytetramethylene glycol.
1. Compare the rate of polymerization of THF with that of ethylene oxide.
2. Polymerization of THF generally requires an acid catalyst, rather than a basic one. Why? | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/IV%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_2/04%3A_Aliphatic_Nucleophilic_Substitution/4.10%3.txt |
Sometimes, elimination reactions occur instead of aliphatic nucleophilic substitutions. In an elimination reaction, instead of connecting to the electrophilic carbon, the nucleophile takes a proton from the next carbon away from it. The halide or other leaving group is still displaced. A double bond forms between the two carbons.
Thus, there are actually more than two competing mechanisms occurring at once here. In addition to unimolecular and bimolecular substitution, a reaction involving deprotonation is also possible.
Instead of acting as a nucleophile, the tert-butoxide anion acts as a base. It forms a bond to a proton, becoming tert-butanol. This proton must always come from the carbon next to the leaving group. The bromide still leaves, and the two adjacent carbons form a second bond together.
Exercise \(1\)
Draw a mechanism for the elimination reaction above. Assume the reaction is bimolecular and concerted, so that the C-H bond and the C-Br bond break at the same time, forming the C=C bond.
Answer
The mechanism of an elimination reaction is almost exactly the same as an aliphatic nucleophilic substitution, except that the nucelophile misses its mark. It hits a proton instead of a carbon and acts as a base instead of a nucleophile. This process can happen at the same time as the leaving group's departure or it can happen afterwards. These mechanisms are called E1 and E2.
Exercise \(2\)
Draw another mechanism for the elimination reaction above, but this time, suppose the reaction is unimolecular.
Answer
Exercise \(3\)
Given the mechanism in Exercise \(2\), other products would also be expected.
1. What are they? (Think about the reactive intermediate and what else could happen to it.)
2. What does their absence in the original reaction scheme above suggest about the most probable mechanism of the reaction?
Answer a
products of cation rearrangement via hydride shifts: 2-heptene instead of 3-heptene.
Answer b
The absence of rearrangement suggests the absence of cations. The mechanism for the reaction shown must be concerted rather than via the ionic intermediate.
Why might a reaction undergo elimination rather than substitution? The most important reason concerns the nature of the nucleophile. The more basic the nucleophile, the more likely it will induce elimination.
• Basic nucleophiles lead to elimination.
What makes something basic, rather than nucleophilic? As a very rough rule of thumb, we can beging by thinking of bases as less stable versions of nucleophiles. Nucleophiles are very often anions, and bases are generally less stable anions. So what factors make anions more stable? Those factors would make the anion more like a nucleophile and less like a base.
One of the most important factors here is polarizability. Remember, polarizable atoms are large atoms. In the main group of the periodic table, they include sulfur, phosphorus, chloride, bromide, iodide, etc. These anions are stable because the negative charge is spread out over a larger atom. Any time charge is spread out, it tends to result in greater stability. On the other hand, smaller, less polarizable atoms include oxygen, nitrogen and carbon. Anions of carbon, nitrogen and oxygen tend to be more basic. Anions of bromine, iodine, and sulfur are not basic. What, never? No, never. Well, hardly ever.
• Large, polarizable atoms such as Br, I, or S make stable anions.
• These stable anions are more likely to be nucleophiles than bases.
You might remember this factor from a discussion of anion stability in acid-base chemistry. Other considerations from acidity will be useful, too.
A second important factor that helped to stabilize anions was resonance. If a negative charge can be delocalised via resonance, the anion becomes much more stable. For example, an alkoxide ion, such as methoxide, might be very basic, because the negative charge is on an oxygen atom. Oxygen is not a large, polarizable atom. However, an adjacent carbonyl in an acetate anion makes all the difference. This anion is resonance stabilised.
• Resonance stabilised anions are relatively stable.
• These stable anions are more likely to be nucleophiles than bases.
Another factor sometimes plays a role in the case of carbon or nitrogen. It is the idea of hybridisation. Remember that the geometry of an atom determines which atomic orbitals are involved in bonding. Tetrahedral carbons are thought of as sp3 hybridised, meaning that the carbon uses an s electron and three p electrons in sigma bonding. A trigonal planar carbon is sp2 hybridised. That means it uses only two p orbitals and an s orbital in forming sigma bonds.
Different hybridisation leads to some subtle differences in properties. For example, the C-H bonds of sp2 carbons are a little stronger than those of sp3 carbons (maybe 105 to 110 kcal/mol for the former, and 95 to 100 kcal/mol for the latter). That's because the electrons in the sp2 C-H bonds are at slightly lower energy. That, in turn, is because an s orbital is a little lower in energy than a p orbital.
For similar reasons, an sp2 carbanion is more stable than an sp3 carbanion. The electrons on the sp2 carbon are lower in energy than the electrons on the sp3 carbon. An sp carbanion is more stable, still. As a result, although an ethyl anion (CH3CH2-) is extremely basic and a vinyl anion (H2C=CH-) is still highly basic, an acetylide or alkynyl anion (HC=C-), though basic, is much more nucleophilic than the other two.
• A sp hybridized carbon anion is much more stable than an sp2 or sp3 carbon anions.
• These relatively stable anions (remember, we are still talking about an anion on carbon), although pretty basic, are usually nucleophiles.
If any of these three factors apply (polarizability, resonance, sp hybridization in carbon), the anion is more likely to be a nucleophile than a base.
There is a fourth factor which is almost a non-sequitur. It goes without saying that a neutral compound -- in the sense that the compound that has no charge at all -- does not require charge stabilisation at all. Thus, if a nucleophile has no charge, it is relatively stable, and will often act as a nucleophile rather than a base.
• Neutral (uncharged) compounds are stable; they do not require charge stabilisation.
• Neutral compounds are usually nucelophiles rather than bases.
A fifth factor -- relative electronegativity within a row -- does play a minor role here. Remember, within a row of the periodic table, size changes are minor. All of the atoms are small. Differences in polarizability are not much of a factor. Instead, differences in electronegativity influence anion stability. The more electronegative the atom, the more stable the anion. As a result, an oxygen anion is more stable than a nitrogen anion or carbon anion. As a result, we sometimes make a distinction here, referring to hydroxide ion and alkoxide ions as strong bases, but amide anions and alkyl anions as very strong bases.
Exercise \(4\)
Classify the following anions as very strong base, strong base, or weak base.
a) NaNH2 b) NaOH c) CH3CO2Na d) NaH e) CH3OH
f) NaBr g) CH3COCH2Li h) CH3CH2CH2CH2Li i) KI j) H2O
k) CH3CCNa l) (CH3CH2)3N m) Na2CO3 n) LiCl o) CH3OK
p) CH3C6H4ONa q) NaSH r) [(CH3)2CH]2NLi s) (CH3)3P t) (CH3)3COK
Answer a
very strong
Answer b
strong
Answer c
weak (resonance)
Answer d
very strong
Answer e
weak (neutral)
Answer f
weak (polarizable)
Answer g
weak (resonance)
Answer h
very strong
Answer i
weak (polarizable)
Answer j
weak (neutral)
Answer k
medium-weak (C anion but sp)
Answer l
weak (neutral)
Answer m
weak (resonance)
Answer n
weak (polarizable)
Answer o
strong
Answer p
weak (O anion but delocalised)
Answer q
weak (polarizable)
Answer r
very strong
Answer s
weak (polarizable)
Answer t
strong
Very strong bases include carbon and nitrogen anions and semi-anions. Examples include butyllithium and sodium amide. Very strong bases are highly likely to engage in elimination, rather than substitution.
Strong bases include non-stabilized oxygen anions. Examples include sodium hydroxide as well as alkoxides such as potassium tert-butoxide or sodium ethoxide. Strong bases favor elimination, too. Nevertheless, they can sometimes undergo either elimination or substitution, depending on other factors (see below).
Weak bases include cyanide, stabilized oxygen anions such as carboxylates and aryloxides, sulfur anions, fluoride ion and neutral amines. Weak bases are much more likely to undergo substitution than elimination.
Very weak bases include heavy halides such as chloride, bromide or iodide, as well as neutral phosphorus or sulfur nucleophiles. Very weak bases undergo elimination only rarely.
Exercise \(5\)
Typically, strong bases and very strong bases are more likely to react via the E2 mechanism; they react so quickly that the deprotonation step triggers C-LGp ionization, rather than the other way around. However, E1 mechanisms also occur with these bases, especially at low concentrations. Explain why.
Exercise \(6\)
Why is it that an anion such as cyanide is a weak base, whereas CH3Li is a strong base?
Another factor is sterics. The more crowded the electrophile, the more likely the nucleophile will encounter a proton on its way to the electrophilic carbon.
As a nucleophile approaches tert-butyl bromide, coming from the side opposite the bromine in order to undergo nucleophilic substitution, it is pretty likely to collide with a proton on its way to the electrophilic carbon. The same thing has a good chance of happening with iso-propyl bromide. However, it is much less likely to happen with bromoethane. Finally, bromomethane doesn't even have a beta-hydrogen, so the chance of elimination in that case is zero.
• Crowding leads to elimination.
Note that the crowding could involve either the structure of the base or the structure of the electrophile. A large, bulky base may be more likely to deprotonate than find its way in to the electrophilic carbon atom.
• Bulkier nucleophiles can act as bases.
Exercise \(7\)
Given the following pairs of nucleophiles, which one is more likely to undergo elimination?
Exercise \(8\)
Although acetylides (such as sodium acetylide, Na CCH) are actually more basic than alkoxides (such as sodium isopropoxide, Na OCH(CH3)2), acetylides frequently undergo substitution rather than elimination. Propose a reason for this difference.
A third factor is temperature. An elimination reaction involves the cleavage of two bonds, whereas a substitution reaction requires only one bond to break. Thus, an elimination reaction is more energy-intensive, and it is more likely to occur at higher temperatures, when more energy is available.
• Higher temperatures lead to elimination.
Exercise \(9\)
An additional factor in the energy dependence of eliminations and substitutions is entropy.
1. Use simple rules about to determine which products are favored by entropy: Elimination or substitution?
2. Given the relationship ΔG = ΔH - T ΔS, which thermodynamic factor dominates free energy change at high temperature?
3. Therefore, which product is favored at high temperature? | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/IV%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_2/04%3A_Aliphatic_Nucleophilic_Substitution/4.12%3.txt |
Sometimes, an elimination reaction could lead to formation of a double bond in more than one place. If the halide is on one carbon and there are protons that could be removed on either side, then taking one proton or the other might lead to two different products. This reaction could have different regiochemical outcomes, meaning it could happen at two different places in the molecule.
What factors might influence which product forms? We might think about product stability, in case there are corresponding differences in barriers leading to those products. We already know about stereochemical effects in alkene stability, but what about other effects?
It is well-established that alkene stability is influenced by degree of substitution of the double bond. The greater the number of carbons attached to the double bond, the more stable it is.
That effect is related to hyperconjugation. Specifically, it's an interaction between bonding orbitals and antibonding orbitals on neighbouring carbons. The interaction allows the bonding electrons to drop a little lower in energy through delocalization. The interaction also pushes the antibonding orbitals a little higher in energy, but since they have no electrons, they don't contribute to the real energy of the molecule. Overall, the molecules goes down in energy.
• In most cases, the most-substituted alkene results from elimination reactions.
However, alkene stability isn't the only factor that plays a role in elimination. Steric hindrance can play a role, too.
In a case in which there are two different hydrogens from which to select, the one leading to the more-substituted double bond is sometimes a little bit crowded. That leaves the base with fewer viable pathways to approach the proton.
On the other hand, the removal of a proton leading to the less stable alkene is often less crowded, allowing the base to approach much more easily from a number of angles.
• In some cases, especially with very bulky bases, the least substituted alkene forms, even though it is less stable.
Exercise \(1\)
In the following reactions, more than one elimination product is possible. Draw the products. Circle the most stable product.
Exercise \(2\)
In the following reactions, more than one elimination product is possible. Draw the products. Circle the product formed via removal of the most accessible proton.
4.14%3
Sometimes, elimination reactions may lead to multiple stereoisomers; that is, they could lead to either the cis or the trans isomer, or in more complicated structures, either the Z or the E isomer.
Of course, if there were some inherent stability difference between these isomers, that could be a factor that plays a role in influencing the outcome. Elimination reactions aren't generally reversible, so products are not directly determined by alkene isomer stabilities. Nevertheless, sometimes the barrier leading to a more stable product is a little lower than the barrier leading to a less stable product.
We do know that in simple cis vs. trans cases, the trans isomer is generally lower in energy because of fewer steric interactions between the substituents on the double bond. In the absence of other information, we could take that as a starting point. Let's see whether elimination reactions generally lead to trans isomers.
It turns out that sometimes this is true: eliminations often lead to the more stable product. Sometimes it isn't true, though. The answer depends on the mechanism.
• E1 eliminations generally lead to the more stable stereochemistry.
• E2 eliminations may or may not lead to the more stable stereochemistry.
Instead, in an E2 reaction, stereochemistry of the double bond -- that is, whether the E or Z isomer results -- is dictated by the stereochemistry of the starting material, if it is diastereomeric. In other words, if the carbon with the hydrogen and the carbon with the halogen are both chiral, then one diastereomer will lead to one product, and the other diastereomer will lead to the other product.
The following reactions of potassium ethoxide with dibromostilbene (1,2-dibromo-1,2-diphenylethane) both occurred via an E2 mechanism. Two different diastereomers were used. Two different stereoisomers (E vs. Z) resulted.
Exercise \(1\)
Provide the stereochemical configurations of the following compounds from the above reactions:
1. 1,2-dibromo-1,2-diphenylethane (upper example)
2. 1-bromo-1,2-diphenylethene (upper example)
3. 1,2-dibromo-1,2-diphenylethane (lower example)
4. 1-bromo-1,2-diphenylethene (lower example)
In E2 eliminations, the spatial relationship between the proton and leaving group determines the product stereochemistry. That's because pi bond formation happens at the same time that the halide leaves and at the same time that the base removes the proton. All of these events have to be coordinated together. The central, tricky event is the pi bond formation. The leaving group can leave in any direction, and the base can approach from many directions, but unless the pi bond is ready to form, nothing else happens.
Let's slow the reaction down and imagine it takes place in slightly different stages.
As the leaving group leaves, it takes its electrons with it. It begins to leave a positive charge behind. That positive charge will be centered on the carbon from which the halide is departing. That carbocation, if it fully formed, would have only three neighbors to bond with. It would be trigonal planar. It would have an unoccupied, non-bonding p orbital.
As the base takes the proton, the hydrogen leaves behind the electrons from the C-H bond that held it in place. These electrons stay behind on the carbon atom. They are left in a non-bonding carbon valence orbital, a p orbital or something quite like it.
Now we have a filled p orbital next to an empty p orbital. They overlap to form a pi bond.
Of course, in an E2 reaction, things don't happen in stages. Everything happens at once. That means that, as the base removes the proton, the pi bond must already start forming. Because a pi bond requires parallel alignment of two p orbitals, and the p orbitals are forming from the C-H and C-LGp bonds, then those bonds must line up in order for the elimination to occur.
So let's look again at that dibromostilbene example.
In the first case, we need to spin the molecule so that we can see how the H on one carbon and the Br on the other are aligned and ready to eliminate via an E2 reaction. The substituents coming towards us in the reactant will still be coming towards us in the product. The substituents pointing away from us in the reactant will still be pointing away from us in the product.
So the relationships between the substituents on the nascent double bond are determined by their relationship once the reactant is aligned for the E2 reaction.
In the second case, we can spin the molecule but quickly realize the C-H and C-Br bonds are not lined up in this conformer. We need a bond rotation. Once we have made a conformational change, the C-H and C-Br bonds line up. It doesn't matter if this conformer is not favored; if there is going to be any E2 reaction at all, this is the conformer it will have to go through.
Again, the relationships between on the new double bond are determined by their relationship once the reactant is aligned for the E2 reaction.
Exercise \(2\)
Sometimes it is easier to see the relationships between substituents by using a Newman projection. Draw Newman projections showing how the two isomers above proceed to different products in an E2 reaction.
Exercise \(3\)
Predict the product of each of the following E2 reactions. Note that the compounds differ in the incorporation of a 2H isotope (deuterium, or D) in place of a regular 1H isotope (protium, or H).
Conformational analysis of cyclohexanes requires the use of diamond lattice projections ("chairs"). In a chair, a periplanar requirement would only be met when two neighbouring groups are both axial. One would be "axial up" whereas the other would be "axial down". In contrast, if two neighbouring groups are both equatorial, they are actually gauche to one another. The dihedral angle between them would be 60 degrees, and so a π bond would not readily form during an E2 elimination. If one group were axial and the other equatorial, the two groups would still be gauche to each other.
• An E2 elimination can only happen if a hydrogen and a neighbouring leaving group are anti to each other in a chair.
• These two groups must be trans to each other.
• These two groups must both be in axial positions.
Exercise \(4\)
In which of these two chair conformations is an E2 reaction ready to occur?
Exercise \(5\)
In the following cases, elimination might proceed very quickly, very slowly or not at all. Indicate the propensity to react through an E2 mechanism in each of the following compounds. If it could react via this mechanism, show the product.
Exercise \(6\)
Predict the E2 elimination products from the following compounds.
In contrast to E2 reactions, E1 reactions do not occur in one step. That means there is time for reorganization in the intermediate. Once the leaving group leaves, the cation can sort itself into the most stable conformer. When the proton is taken, generally the most stable stereoisomer results because it comes from the most stable conformer of the cation. Any steric interactions in the alkene would also have occurred in the cation, so this interaction would have been sorted out at that point.
Thus, in the case of the dibromostilbenes examined before, E1 elimination would result in the same product in either case.
• E1 reactions can, in principle, lead to either stereochemistry of alkene.
• Free rotation around bonds in the carbocation intermediate allows the cation to adopt either conformer prior to elimination.
• However, steric interactions will lead to a preponderance of one conformer.
• The more stable conformer will lead to the more stable alkene.
Exercise \(7\)
Predict the products of the following compounds under E1 conditions. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/IV%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_2/04%3A_Aliphatic_Nucleophilic_Substitution/4.13%3.txt |
The factors that influence whether an elimination reaction proceeds through an E1 or E2 reaction are almost exactly the same as the factors that influence the SN1/SN2 pathway. Cation stability, solvents and basicity play prominent roles. However, basicity may be the single most important of these factors.
Exercise \(1\)
By analogy with substitution reaction, in which elimination mechanism does cation stability play a strong role: E1 or E2?
Answer
Cation stability is important in an E1 reaction.
Exercise \(2\)
Draw an example of an alkyl halide that is likely to undergo an E1 elimination.
Answer
Any tertiary alkyl halide would be a good example. Benzylic alkyl halides would also be good examples if they are either secondary or tertiary.
Exercise \(3\)
By analogy with substitution reactions, what mechanism would be promoted by protic solvents: E1 or E2?
Answer
Protic solvents could promote E1 reactions.
Exercise \(4\)
Basicity refers to the strength of the base. Which mechanism is more likely to occur with strong bases: E1 or E2?
Answer
A strong base could promote an E2 reaction.
Because strong bases can be thought of as very reactive nucleophiles, it might not be surprising that they tilt the likely reaction towards E2. These compounds are so reactive that they are much more likely to abstract a proton than be guided to the electrophilic carbon. They are also sufficiently reactive that they can intercept the electrophile on a timescale faster than it takes for the electrophile to ionize. Even under conditions that might otherwise be expected to lead to an SN1 or E1 reaction, if the base is strong, E2 will frequently prevail.
Of course, that general rule is subject to limitations. If there is no beta-hydrogen to abstract, then there will be no elimination of any kind. The compound will be limited to substitution reactions. Furthermore, if the C-H bond cannot line up with the C-LGp bond in such a way as to allow them to break together, forming a pi bond at the same time, then an E2 mechanism would be impossible. However, the lack of possibility for an antiperiplanar relationship is not that common. If we are dealing with an alkyl chain, rotations around sigma bonds will be sufficient to bring the C-H and C-LGp bonds into alignment. Only in cyclic structures, in which rotation is much more restricted, would there be a possibility that an antiperiplanar relationship would be completely prevented.
So we need to keep E2 in mind when we are dealing with strong bases. Those compounds are generally unstabilized oxygen anions: hydroxide and alkoxides (such as CH3O- or CH3CH2O-). Very strong bases would also fall into this category. These bases are generally unstabilized carbon and nitrogen anions or semianions (such as amide, NH2-, or alkyllithiums such as CH3Li or CH3CH2CH2CH2Li). Bases that are stabilized in some way, such as resonance stabilized carboxylates (like CH3CO2-) or enolates (like CH3COCH2-) are much weaker, and are much more likely to act as nucleophiles than as bases.
E1 reactions, like SN1 reactions, are really only possible when cation stability allows them to occur. Remember that cation formation is the slow step in these mechanisms. If the cation is too unstable to form in the first place, then there is no going down this road. That means tertiary alkyl halides are good candidates for E1, but primary alkyl halides are not. Under the right conditions, such as in the presence of a protic solvent, secondary cations might also be coaxed into being.
Whether at that point the cation undergoes an SN1 or an E1 reaction depends less on the nucleophile than you might expect. At this point we are dealing with more or less weak bases; a strong base would have carried out an E2 while it had the chance. Instead, the two possibilities may compete, giving mixtures of products, but temperature is often the major factor determining which way the reaction will go. At higher temperature, E1 reactions are favored for entropic considerations. At lower temperature, SN1 may dominate.
Sometimes, E1 reactions can be carried out in the presence of a strong base, but the base is very dilute. As a result, the base is not as likely to encounter the alkyl halide until the alkyl halide has already had time to dissociate into anion and cation. At that point, the base just comes in and deprotonates the cation.
Exercise \(5\)
Predict the major products of the following reactions, and the mechanism type (SN1, SN2, E1, E2) by which the products are formed. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/IV%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_2/04%3A_Aliphatic_Nucleophilic_Substitution/4.15%3.txt |
Nucleophilic substitution reactions can be useful in organic synthesis. Mostly they are used for the interconversion of functional groups. For example, an alkyl halide might be transformed into an alcohol, or into an ether.
The trouble is, these seemingly simple steps can be very difficult. That's because the hydroxide needed to make an alcohol from an alkyl halide is really quite basic. So is the alkoxide that would be needed in order to turn an alkyl halide into an ether. Instead of substitution, you may get an elimination reaction.
Who cares? Well, you would care if you were working on the synthesis of an antimalarial drug that could save millions, or an anticancer drug, or anything of that nature. Maybe that ether forms the last crucial part of the pharmacophore that will bind the drug to its target. Without this reaction, it may be thousands of times less effective. So this little reaction could be very important.
Clearly, the best thing to do would be to make sure a substitution reaction happened, and not that elimination. We need the reagent to be a nucleophile, not a base.
In the case of the alcohol synthesis, we could use water as the nucleophile, rather than hydroxide. Water is certainly less basic than hydroxide ion. It is still nucelophilic, though, because it still has a lone pair. If we add it to an alkyl chloride, the water will displace the chloride, and then the extra proton will be plucked off the positive oxygen atom by the chloride, leaving us with the alcohol.
That last step would form hydrogen chloride, a corrosive acid, and that could cause problems. To counteract that possibility, we will want to add a weak base so that the HCl gets neutralized. Sodium carbonate (Na2CO3) or sodium bicarbonate (NaHCO3) may be good options, because they are mildly basic and they dissolve in water.
Alternatively, we could just use sodium hydroxide as the base. That gets us back to the original problem. However, in order to avoid elimination, we would use very dilute sodium hydroxide. We would keep its concentration low enough that the alkyl halide is much more likely to react with the water than with the hydroxide ion, for the simple reason that it is much more likely to run into a water molecule than a hydroxide ion. However, once that oxygen donor atom picked up a positive charge, it would be more attractive to the hydroxide ion, and the hydroxide would then come in for the proton.
In the same way, if we wanted to make an ether, we might use an alcohol as the nucleophile rather than the much more basic alkoxide ion. We would add a weak base to sponge up the extra proton and avoid formation of a strong acid.
• A neutral nucleophile is less basic than an anionic one, and may avoid elimination reactions
• A weak base can be used to scavenge protons from the reaction
Exercise \(1\)
Provide a mechanism for the following reaction.
There is another approach to limiting the amount of elimination during a substitution step to form an alcohol. It also involves the use of a more stable nucleophile than a hydroxide ion. However, it employs a more reactive anionic nucleophile, rather than the neutral water. If an acetate ion is used instead, very little elimination usually occurs. An ester is formed as a product. There isn't much elimination because the acetate ion is resonance stabilised. More stable nucleophiles often undergo substitution rather than elimination.
Of course, we didn't want an ester; we wanted an alcohol. No problem. Esters can be saponified relatively easily -- that is, broken down into an alcohol and a carboxylate. Just add a hydroxide and water. Now the stronger carbonyl electrophile is a better target for the hydroxide and the reaction is pretty well assured to get to the right place.
Overall, the reaction is actually a sequence of several events.
• Acetate esters are readily converted into alcohols under basic conditions
• By turning an alkyl halide into an ester first, and then into an alcohol, we can limit competition with elimination
We can't take the same approach in ether synthesis. An ether is an oxygen bridge between two tetrahedral or sp3 carbons; we can't have resonance stabilisation and still have those two sp3 carbons. Instead, another strategy is sometimes employed during addition of an anionic nucleophile to an alkyl halide. An alkoxide ion is still employed, but care is taken in how the alkoxide and the alkyl halide are chosen. Because the ether is symmetric -- it is two tetrahedral carbons attached to an oxygen -- either side could originate as the alkoxide and either side could originate as the alkyl halide.
If the alkyl halide is chosen so that steric crowding is minimized, there is a lower chance of an accidental collision between the alkoxide and a beta hydrogen on that alkyl halide. In some cases, we might even be able to choose the alkyl halide so that elimination is not possible at all. If possible, we can use an alkyl halide that doesn't have any beta hydrogens.
In general, the use of alkoxide ions as nucleophiles can be pretty successful if done carefully, and this approach to making ethers even has its own name. It's called the Williamson ether synthesis.
• In the Williamson ether synthesis, the less crowded half of the ether is formed from the alkyl halide
• In some cases, an alkyl halide may also be chosen because elimination is not physically possible with that structure
Exercise \(2\)
Propose Williamson ether syntheses of the following compounds.
Exercise \(3\)
Provide products of the following reactions. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/IV%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_2/04%3A_Aliphatic_Nucleophilic_Substitution/4.16%3.txt |
Amines can be synthesized through nucleophilic substitution. Using an alkyl halide and the proper nucleophile, the halide can be replaced by an amino group.
If an amide ion were used as the nucleophile, elimination would be a pretty sure thing. An amide ion is even more basic than a hydroxide ion. The nitrogen atom is less electronegative than the oxygen atom of hydroxide. This was the same problem with making alcohols and ethers with hydroxide or alkoxide ions, but now the problem is more severe.
In the case of alcohol and ether syntheses, one approach was to simply use a neutral form of the nucleophile rather than an anion. We could do that in this case. Ammonia still has a lone pair and it is a pretty good nucleophile. We don't need a negative charge on the nitrogen for it to displace a halogen from an alkyl halide.
Because nitrogen is a litle less electronegative than oxygen, ammonia is a better nucleophile than water. This substitution works even better than the substitution of water for a halide. However, there are other problems.
Ammonia is nucleophilic, but it is also basic. That can be helpful. In the synthesis of alcohols and ethers, addition of a neutral nucleophile had to be accompanied by a weak base, otherwise the buildup of acid in the reaction might cause unexpected side reactions, including the breaking of ether linkages that you were trying to make. However, it means that half of the ammonia that you put into the reaction would get used up in acid-base reactions. You would need to add twice as much ammonia as alkyl halide. One molecule of ammonia would act as the nucleophile in each reaction, and one would act as the base.
Remember, the order of steps does matter here. Sure, you could imagine some miniscule equilibrium in which one ammonia has plucked a proton off another, but a quick look at a pKa table tells you that's not very likely. The proton will only be removed after the first ammonia has donated its lone pair and the neutral nitrogen atom becomes part of a much more acidic ammonium ion.
There is another problem, though. Once that ammonium ion has been deprotonated, the nitrogen gets its lone pair back again. It becomes nucleophilic again. Because nitrogen is pretty nucleophilic, there is nothing stopping this newly-formed amine from reacting with another alkyl halide. It will do so, pretty reliably. That will lead to formation of some secondary amine, in addition to the primary amine that you may have been aiming for. A secondary amine has two alkyl groups attached to the nitrogen, rather than just one.
Of course, now that we have started down that road, there's no going back. As soon as that secondary ammonium ion is deprotonated, it gets a lone pair restored. Once it has a lone pair, it becomes pretty nucleophilic. It will donate to another alkyl halide and form a tertiary ammonium ion. That tertiary ammonium ions will get deprotonated almost immediately.
As soon as the tertiary amine forms, of course it is just going to do the same thing. It will donate its lone pair to an alkyl halide. It forms a quaternary ammonium ion. This time, however, there is no proton on the nitrogen. It can't easily be deprotonated. It remains as a quaternary ammonium salt. Quaternary ammonium salts are really quite stable; they are used in household consumer products all the time. Surfactants (like those used in cleaning sprays) and anti-static agents (which might show up in dryer sheets and shampoos) frequently use quaternary ammonium salts.
Now, let's think about our ratio again. Suppose we started with that 2:1 ratio of ammonia to alkyl halide, knowing that as soon as a first ammonia molecule bonded with an alkyl, a second ammonia would immediately take a proton. By forming a quaternary ammonium salt, we have actually used up four alkyl halides with one ammonia nucleophile. We have also used up three other ammonia molecules as bases. We have used the reactants up in a 1:1 ratio and we will have half of the ammonia left over.
So it sounds like a 2:1 ratio of ammonia to alkyl halide might get us a primary amine, but a 1:1 ratio might get us to a quaternary ammonium salt. What if we had stopped at a secondary amine? Then we would use up two alkyl halides as electrophiles, one ammonia as a nucleophile, and two ammonias as bases. That's a 3:2 ratio of ammonia to alkyl halide, or 1.5:1. Similar analysis would tell you that stopping at a teriary amine would require a 1.3:1 ratio of ammonia to alkyl halide.
The point is, formation of these different amines require ratios of ammonia to alkyl halide that are really very similar. Realistically, there is no way we could add these exact ratios of reagents to a reaction and expect things to stop at the right place. Remember, a reaction is taking place with millions of molecules at once, and probability says that all four pathways will be followed by significant fractions of the reactants. All of these ratios of ammonia to alkyl halide will lead to the same thing: a mixture of all possible products. There would invariably be some leftover amine or leftover alkyl halide, too, once the other reagent had run out.
Practically speaking, there are only two products that you can make from this reaction. You could make a primary amine if you used a vast excess of ammonia, so that you ensured that any alkyl halide was much more likely to run into an ammonia molecule long before it ran into an amine molecule.
In this case, you would have leftover ammonia. That should be pretty easy to remove because of its low boiling point.
Alternatively, you could make a quaternary ammonium salt if you used a large excess of alkyl halide.
In this scenario, you would have leftover alkyl halide. The difference in solubility between the alkyl halide and the quaternary ammonium salt could help separate these two materials.
You might recall that in the section on the synthesis of alcohols and ethers, an alternative strategy used a resonance-stabilised, anionic oxygen nucleophile to make alcohols. The same strategy is often used with the synthesis of amines. This approach is called a Gabriel amine synthesis.
The nucleophile in this case is a phthalimide ("FTAL-im-id" or "FTAL-im-eyed") ion. The phthalimide ion is easily formed by a strong base such as potassium hydroxide, becaus the anion obtained is pretty stable. Even though the negative charge is on a nitrogen atom, the two carbonyls serve to delocalise the charge and make this ion less reactive.
Notice the difference in nucleophilicity between the neutral phthalimide and the phthalimide ion. In the neutral compound, the single lone pair on the nitrogen is delocalised. It is not available to act as a nucleophile. This is generally true with amides and imides; the neighbouring carbonyl ties up the nitrogen lone pair so that it is neither basic nor nucleophilic.
However, the phthalimide ion has a spare lone pair. If one of the lone pairs is delocalised, then the other is still available to act as a nucleophile. Both lone pairs cannot be delocalized because they are orthogonal to each other -- that means they are forced to be in different areas of space in order to minimize electron repulsion.
The phthalimide anion is thus able to act as a nucleophile. It can donate to an alkyl halide and displace the halide anion.
The phthalate part of the molecule has now served its purpose. It was just there to deliver the nitrogen in a way that was stable, yet suitably reactive. We can get rid of it now through base-catalysed hydrolysis. Remember, that is a carboxyloid substitution reaction. The resulting phthalate salt is easily removed because of its very different solubility properties.
Exercise \(1\)
Provide a mechanism for the hydrolysis of the imide shown above.
Answer
Exercise \(2\)
Provide products of the following reactions.
Exercise \(3\)
For the reaction of nucleophile with iodomethane in acetone, rank the nucleophiles in order of reactivity (1 = fastest, 3 = slowest).
1. _ HO- _ HS- _ H2O
2. _ H2O _ H3O+ _ NH3
3. _ CH3CH2NH2 _ (CH3)2CHNH2 _ (CH3)3CNH2
Answer a
2 HO- 1 HS- 3 H2O; the anions are more nucleophilic than the neutral, but sulfur is more polarizable than oxygen
Answer b
2 H2O 3 H3O+ 1 NH3; the neutrals are more nucleophilic than the cation, but nitrogen is less electronegative than oxygen
Answer c
1 CH3CH2NH2 2 (CH3)2CHNH2 3 (CH3)3CNH2; steric effects
Exercise \(4\)
DNA bases (adenine, guanine, cytosine, and thymidine) contain nucleophilic nitrogen atoms, which is why many halogenated compounds are carcinogenic. Alkylated DNA can still function in its process of replication, though it will do so abnormally, resulting in mutations in the DNA and, ultimately, cancerous cells.
a) Propose a mechanism, with arrows, for the alkylation of guanine:
Damaging DNA via alkylation can also be used to treat cancer. The key is that cancer calls grow and divide more rapidly than normal calls, and thus are more susceptible to mechanisms that damage DNA and impair its functions. Mechlorethamine is one such drug that cross-links DNA.
The mechanism for the activation of mechlorethamine is shown below:
b) Using that step, provide a mechanism, with arrows, for the formation of two cross-linked guanine molecules.
c) What makes the aziridinium ring so electrophilic?
Answer a & b
Answer c
Ring strain promotes opening of the ring. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/IV%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_2/04%3A_Aliphatic_Nucleophilic_Substitution/4.17%3.txt |
Silicon is in the same group as carbon in the periodic table, so in some ways it might be expected to behave in some ways that are similar to carbon. It might not be surprising that silicon can undergo nucleophilic substitution reactions like carbon. However, there are differences in both the mechanism and the reactions that are likely to occur.
Probably the most common example of nucleophilic substitution at silicon occurs during the formation of silicone polymers. Silicon forms a pretty strong bond to oxygen, so a chloride leaving group is easily displaced from silicon to form a silanol. If a dichlorosilane is exposed to water, a silane diol is produced.
Just by adjusting the ratio of silicon that is reacting with water, we could imagine forming different products. A 2:1 ratio of H2O : Me2SiCl2 might lead to a silane diol, Me2Si(OH)2. However, a 3:2 ratio could end up forming a silicone dimer, Me2(HO)SiOSiMe2(OH). That's because the hydroxy group on the first silicon atom is still a nucleophile, and it could donate to a second silicon atom, forming a bridge between two silicon atoms.
If the water is introduced gradually to the dichlorosilane, polymerization results, because a chlorosilane is more likely to react with a neighbouring silanol than with scarce water molecules. Lots of Si-O-Si bridges form, leading to a polymeric material: silicone rubber.
• Silicones are polymers containing -(R2SiO)- repeating units.
• Silicones are formed via polymerization of dichlorosilane compounds with water.
The side product of this reaction is hydrogen chloride, HCl. When it contacts water, HCl dissociates to give hydrochloric acid. Hydrochloric acid is a strong acid and a severe health risk. Because of that, there are alternative formulations for making silicone that are a little safer. If the chlorine is replaced with an acetate group, polymerization is still possible upon exposure to water. However, the side product is the somewhat safer acetic acid. This route is used in commercially available silicones for household use. For example, silicone caulk can be used to waterproof around a bathtub or sink. The strong smell of vinegar when it is left to cure is the acetic acid being produced when the acetoylsilane reacts with moisture in the air.
Apart from its use in polymer materials, substitution at silicon plays an important role in organic synthesis. Silyl ethers can be made in ways similar to the formation of regular ethers. The usual approach is through addition of an alcohol to a chlorosilane. The presence of a weak base, such as a tertiary amine, prevents the buildup of corrosive hydrochloric acid. That's what would result if the chloride leaving group were the only base available to pick up the extra proton.
Exercise \(1\)
Draw structures for the following silyl ethers.
a) the trimethylsilyl ether of cyclohexanol
b) the triisoproylsilyl ether of (S)-2-heptanol
c) the tert-butyldiphenylsilyl
d) the triethylsilyl ether of pentanol
e) the tert-butyldimethylsilyl ether of (S)-5-methylheptan2-ol
Answer
The chief utility of silyl ethers is that they are just as easy to break down as they are to make in the first place. They can be converted back to alcohols in the presence of hydrochloric acid, although the process can be very slow. That's one of the reasons that HCl must be removed during siliyl ether formation. There is an equilibrium at work here, and at some point a buildup of HCl would drive the reaction backward, preventing further formation of the silyl ether.
Silyl ethers can be cleaved much more quickly in the presence of fluoride ion. The usual reagent for this transformation is tetrabutylammonium fluoride (TBAF). The tetrabutylammonium ion is pretty soluble in most organic solvents; that makes the reagent much easier to use.
The reason that reversibility is useful is because it offers a way to temporarily "cover up" and alcohol group. Alcohols can be problematic during many reactions. The slightly acidic OH group, like the slightly acidic OH group of a water molecule, can interfere with sensitive reagents. Compounds that are highly basic, such as alkylmetal compounds (e.g. BuLi, EtMgBr, etc) and metal amides (e.g. NaNH2, LDA, etc) will deprotonate that OH group instead og going on with their intended business with the reactant. For example, if a Grignard reagent is supposed to add to a carbonyl, but it encounters a hydroxy group, it will simply pull the proton from the hydroxy group. Then it will be neutralised. It won't be a nucleophile anymore, so nothing will happen after that.
Instead, if the hydroxy group is first protected, the reaction will proceed with no possibility of an accidental acid-base event. Later, the hydroxy group can easily be replaced.
• Silyl ethers can be used to protect mildly acidic alcohols
• Silyl ethers can be removed easily when protection is no longer needed
Common silyl ethers include trimethyl silyl (Me3Si, or TMS); triethylsilyl (Et3Si, or TES); tri-iso-propylsilyl (iPr3Si, or TIPS); tert-butyldimethylsilyl (tBuMe2Si, or TBS); and tert-butyldiphenylsilyl (tBuPh2Si, TBDPS). The fact that a variety of silyl ethers are commonly available allows chemists to choose from different ones. As a result, several different silyl ethers might be used to protect alcohols in different positions in a more complicated molecule. They can then choose which silyl group to remove, in which order. Selective removal of silyl ethers is possible because they are very sensitive to steric effects. The more crowded the silyl ether, the harder it is to remove.
• Less crowded silyl ethers are removed most easily
• More crowded silyl ethers are removed more slowly
• Less crowed silyl ethers can be removed, leaving more crowded ones intact
Exercise \(2\)
In each of the following cases, rank the order in which the silyl ethers could be removed.
Exercise \(3\)
Use a triethylsilyl ether to help complete the following transformations.
4.19%3
Exercise \(1\)
Fill in the missing items in the following synthesis. Includes: aliphatic nucleophilic substitution, carboxylic substitution, enolates, addition to carbonyl (anionic nucleophiles).
Exercise \(2\)
Fill in the missing items in the following synthesis. Includes: addition to carbonyls (anionic nucleophiles, neutral nucleophiles, aliphatic nucleophilic substitution, Mitsunobu reaction).
Exercise \(3\)
Fill in the missing items in the following synthesis. Includes: aliphatic nucleophilic substitution,sulfonate substitution, Mitsunobu reaction, silyl ether cleavage, nucleophilic substitution of epoxides, elimination / dehydration.
Exercise \(4\)
Fill in the missing items in the following synthesis. Includes: addition to carbonyls (ylides), nucleophilic substitution of epoxides, carboxylic substitution.
Exercise \(5\)
Fill in the missing items in the following synthesis. Includes: aliphatic nucleophilic substitution, carboxylic substitution, sulfonate substitution, addition to nitriles, addition to carbonyls (ylides).
Exercise \(6\)
Fill in the missing items in the following synthesis. Includes: aliphatic nucleophilic substitution, carbonyl addition (anionic nucleophiles, acetals, ylides), carboxylic substitution, silyl ethers.
Exercise \(7\)
Fill in the missing items in the following synthesis. Includes: aliphatic nucleophilic substitution, sulfonate substitution, elimination.
Exercise \(8\)
Predict products of the following reactions.
Exercise \(9\)
Predict products of the following reactions.
Exercise \(10\)
Predict products of the following reactions.
Exercise \(11\)
Predict products of the following reactions.
Exercise \(12\)
Predict products of the following reactions.
Exercise \(13\)
Predict products of the following reactions. | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/IV%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_2/04%3A_Aliphatic_Nucleophilic_Substitution/4.18%3.txt |
Exercise 4.1.1:
Exercise 4.1.2:
Exercise 4.1.3:
Exercise 4.2.1:
The electronegativity of carbon (2.55 on Pauling scale) is less than that of fluorine (3.98), chlorine (3.16), bromine (2.96) or iodine (2.66).
1. On that basis, the carbon attached to a halogen is electrophilic because it has a partial positive charge resulting from the polar carbon-halogen bond.
2. We would expect an alkyl fluoride to be the most electrophilic of these compounds, based on electronegativity.
3. Assuming the energy required for breaking the carbon-halogen bond plays a major role in the activation barrier (not guaranteed), we would expect the activation barrier to be lowest with the alkyl iodide, then the alkyl bromide, then the alkyl chloride and finally the alkyl fluoride. This prediction contrasts with what we might expect based on electronegativity.
4. The stability of alkyl fluorides towards this reactions suggests that there is, in fact, a prominent role played by bond strengths, at least in that case. The carbon-fluoride bond is strong enough to hinder nucleophilic substitution in this compound.
Exercise 4.2.2:
1. In mechanism B, the dissociative one, we would expect a higher activation enthalpy. The first step, which appears to be rate determining, is a bond-breaking step, which will cost energy. In mechanism C, the bond-breaking is compensated by some bond-making; overall, this probably costs less energy.
2. In mechanism B, the dissociative case, we expect a more positive entropy of activation. As the bond to the halide begins to break, the halide and carbocation fragments begin to move independently of each other, gaining degrees of freedom and increasing in entropy. In mechanism C, the incoming nucleophile appears to coordinate its motion with that of the departing halide; as a result, there are fewer degrees of freedom in this case.
Exercise 4.2.3:
1. Charged intermediates are present in the dissociative mechanism (B).
2. It seems like a more polar solvent would favor both mechanisms, because both involve the interaction of an anionic nucleophile with an electrophile and loss of an anionic leaving group. However, the dissociative case (B) involves a build-up of charge in the intermediate. It is possible that a more epolar solvent could reduce the barrier to that buildup of charge separation, accelerating this mechanism.
Exercise 4.2.4:
1. The rate-determining step is probably the bond-breaking one (the first one).
2. Because the nucleophile has not yet participated at that point, \(Rate = k [R-X]\), if R-X = the alkyl halide.
3. There is only one step; it is the rate-determining step, by default.
4. \(Rate = k [R-X][Nu]\).
Exercise 4.4.1:
Exercise 4.4.2:
Exercise 4.4.3:
Exercise 4.5.1:
Exercise 4.5.3:
Exercise 4.6.1:
Exercise 4.6.2:
Keep in mind that there are other factors that can influence the reaction pathway; what we have here are just the most likely mechanisms.
a) SN2 b) Both pathways are very possible c) Both pathways are very possible d) SN2
e) SN2 f) SN2 g) SN1 h) SN1 i) SN1 j) SN1
Exercise 4.7.1:
1. ethanol, isopropanol, trifluoroacetic acid
2. hexane, toluene
3. THF, acetonitrile, DMF, dichloromethane, ether, DMSO, triethylamine, pyridine
4. DMSO > DMF > ACN > pyridine > DCM > THF > ether > TEA, based on dielectric constants. In general, the ones with multiple bonds between two different atoms are the most polar.
5. pyridine and triethylamine. The lone pair on the nitrogen atom is basic toward protons. The trend in basicity is triethylamine > pyridine >> acetonitrile; as the percent s character in the lone pair increases, the electrons are lower in energy and less available for donation.
Exercise 4.8.3:
Exercise 4.9.1:
3. Kinetic -- to fully deprotonate (not equilibrate or you lose stereocontrol) and for the chelation control.
Due to chelation effects shown above, the LDA preferentially removes one hydrogen to form only the Z-enolate that will then do the SN2 to allyl bromide on only one face.
Exercise 4.10.1:
Exercise 4.10.2:
Exercise 4.10.3:
Exercise 4.10.4:
Exercise 4.10.5:
Exercise 4.10.6:
DEAD acts as an oxidizing agent to convert the phosphorus product to a stable side-product, triphenylphosphine oxide, Ph3P=O.
Exercise 4.11.1:
Exercise 4.12.1:
Exercise 4.12.2:
Exercise 4.12.3:
1. products of cation rearrangement via hydride shifts: 2-heptene instead of 3-heptene.
2. The absence of rearrangement suggests the absence of cations. The mechanism for the reaction shown must be concerted rather than via the ionic intermediate.
Exercise 4.12.4:
a) very strong b) strong c) weak (resonance) d) very strong e) weak (neutral)
f) weak (polarizable) g) weak (resonance) h) very strong i) weak (polarizable) j) weak (neutral)
k) medium-weak (C anion but sp) l) weak (neutral) m) weak (resonance) n) weak (polarizable) o) strong
p) weak (O anion but delocalised) q) weak (polarizable) r) very strong s) weak (polarizable) t) strong
Exercise 4.13.1:
Exercise 4.13.2:
Exercise 4.14.3:
Exercise 4.14.4:
Exercise 4.14.5:
Exercise 4.14.6:
Exercise 4.14.7:
Exercise 4.15.1:
Cation stability is important in an E1 reaction.
Exercise 4.15.2:
Any tertiary alkyl halide would be a good example. Benzylic alkyl halides would also be good examples if they are either secondary or tertiary.
Exercise 4.15.3:
Protic solvents could promote E1 reactions.
Exercise 4.15.4:
A strong base could promote an E2 reaction.
Exercise 4.15.5:
Exercise 4.16.1:
Exercise 4.16.2:
Exercise 4.16.3:
Exercise 4.17.1:
Exercise 4.17.2:
Exercise 4.17.3:
1. 2 HO- 1 HS- 3 H2O; the anions are more nucleophilic than the neutral, but sulfur is more polarizable than oxygen
2. 2 H2O 3 H3O+ 1 NH3; the neutrals are more nucleophilic than the cation, but nitrogen is less electronegative than oxygen
3. 1 CH3CH2NH2 2 (CH3)2CHNH2 3 (CH3)3CNH2; steric effects
Exercise 4.17.4:
a)
b)
c) Ring strain promotes opening of the ring.
Exercise 4.18.1:
Exercise 4.18.2:
Exercise 4.18.3:
Exercise 4.19.1:
Answers may assume aqueous workup after the reagents shown. Only one answer shown per box; similar answers may also work.
Answers may assume aqueous workup after the reagents shown. Only one answer shown per box; similar answers may also work.
Exercise 4.19.3:
Answers may assume aqueous workup after the reagents shown. Only one answer shown per box; similar answers may also work.
Exercise 4.19.4:
Exercise 4.19.5:
Exercise 4.19.6:
Exercise 4.19.7:
Exercise 4.19.8:
Exercise 4.19.9:
Exercise 4.19.10:
Exercise 4.19.11:
Exercise 4.19.12:
Exercise 4.19.13: | textbooks/chem/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/IV%3A__Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_2/04%3A_Aliphatic_Nucleophilic_Substitution/4.20%3.txt |
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