The problem is that your product still has a-protons, and can keep on forming enolates to get more methyl groups added:

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1 Lecture 13 Notes November 7, 2012 We are going to start with some examples of proline-catalyzed reactions. In each case, you can draw a mechanism that would involve proline as a chiral catalyst, and one which has no proline at all and just relies on simple deprotonation of the alpha carbon to generate an enolate. Both enolates and enamines have the overall effect of making the alpha carbon nucleophilic. Example 1: The new bond is shown here in red. This is a simple aldol reaction with an enolizable ketone (nucleophile) and an aldehyde without any a-protons (electrophile). Now if I tell you that the reaction is catalyzed by proline, you should be able to draw the structure of the key enamine intermediate: Example 2: This is also an aldol reaction, although with a slightly more complicated substrate. Another complicating factor here is that the aldehyde has an enolizable proton, however, in this case the ketone s protons are more likely to react. The key enamine intermediate here (generated via reaction with the proline catalyst) is: The idea of using secondary amines to catalyze carbonyl chemistry reaction is NT new. It was developed by Gilbert Stork in a process called enamine alkylation. Stork was trying to solve the following problem: Let s say you want to add a methyl group to the alpha carbon on cyclohexanone. We all know (I hope) that you can form the enolate, which will act as a nucleophile to kick off a leaving group and add the desired methyl group:

2 The problem is that your product still has a-protons, and can keep on forming enolates to get more methyl groups added: B: - H H 3 C I R H 3 C R H 3 C The resulting product mixture that you get will be very difficult to purify. So people wanted a way to control these reactions to only get mono-addition of an alkyl group. The second problem that Stork was trying to solve was a regiochemistry one say you start with an unsymmetrical ketone there are two possible enolates that you can form: The kinetic enolate will form faster because that proton is easiest to pull off; however, the thermodynamic enolate is more stable because it is more substituted. You can theoretically control which one you form (kinetic or thermodynamic) by careful control of the reaction conditions. You will get a kinetic enolate under the following conditions with a hindered base (which will be too bulky to attack the substituted alpha carbon), with low temperatures (which will not provide enough energy to force things to react at the substituted carbons). You will get a thermodynamic enolate under the following conditions a small base that can pull off the alpha proton, and elevated temperatures (which allow reactions to equilibrate). In real life you will often get a mixture of the two. Each of these enolates is going to give you a different product if you treat them with methyl iodide: And again, if you form a mixture of products they are going to be pretty difficult to separate and purify.

3 So to solve these two problems, Stork developed the idea that you can first treat the ketone with a secondary amine to form an enamine. This enamine will react very selectively to give mono-addition of an alkyl group, and also will help set the regiochemistry of alkylation. nce the alkylation step is over, you can hydrolyze off the amine to regenerate the ketone. Usually this is done using a cyclic secondary amine like pyrrolidine. So again, there are three steps: (1) Form enamine. (2) React enamine. (3) Get rid of enamine. We can go through each step in turn. 1. Form enamine: The first step in forming an enamine is to form the iminium: This iminium can then lose a proton to generate two potential enamines: You could make an argument in favor of each of these enamines. It turns out that you form only the less substituted enamine- the one that results from removing the red proton. The reason for this is that the other enamine has a lot of steric hindrance between the methyl group and the protons on the pyrrolidine ring: And the steric hindrance is made worse when these are forced to be coplanar b/c of the double bond on that side of the ring. This explains why you only form one of two potential enamines, which solves the regiochemistry problem, because now you will only get substitution on the less hindered side of the cyclohexanone ring. 2. React the enamine. nce you form the enamine, it will react as a nucleophile and attack through the a-carbon, much the same way like an enolate would react:

4 You form an iminium as your product from this reaction. The iminium does not react further it just sits around and waits for water to hydrolyze it. 3. Get rid of the amine. You can accomplish this through iminium hydrolysis, which is basically just the reverse of putting the nitrogen on: This solves the regiochemistry problem, because only one enamine can form, and it solves the problem of overaddition, because the iminium (which is the initial product) does not react further until it gets hydrolyzed by water. Never miss an opportunity to learn some nomenclature: Summary of proline catalysis: You can draw the product of these carbonyl reactions by basically ignoring the proline and figuring out how the two molecules would react on their own. The only real difference when you have proline is that the mechanism goes through an enamine/iminium, and that you can get chirality in your product (as a result of having the chiral enamine). I want to switch gears to talk about a different reaction, which is also often catalyzed by organic molecules, called the benzoin reaction: This reaction can be catalyzed by organic carbenes, and also by the cyanide ion. I am going to show the mechanism of cyanide catalysis for simplicity s sake. The overall reaction is shown below, and involves two molecules of benzaldehyde reacting to form benzoin:

5 Basically you are forming a bond directly between the carbonyl carbon s on two different molecules. When you see this you should be very confused both carbons are electrophilic! How can they form a bond between them? That s where the role of the catalyst comes in. Either a thiazolium carbene, or simpler catalysts like cyanide (CN-) can act to convert one molecule from a nucleophile to an electrophile. And the mechanism of cyanide catalysis is shown below: Basically, the cyanide acts as a nucleophile and attack the carbonyl first. The resulting intermediate acts as a nucleophile to attack a second carbonyl and form the key new carbon-carbon bond. We are going to do some practice problems using a benzoin reaction tomorrow, and then move on to a general review for the test that will be next Tuesday.