Lecture 14 ovember 3, 2011 OK I want to continue briefly with the topic of proline catalysis that we discussed last time. In particular, the idea of using secondary amines to catalyze carbonyl chemistry reaction is OT 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: The problem is that your product still has a-protons, and can keep on forming enolates to get more methyl groups added: 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 unsymmetric ketone there are two possible enolates that you can form: You can theoretically control which one you form (kinetic or thermodynamic) by careful control of the reaction conditions; however, in real life you will always 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: 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. Once 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. Once 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:
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: That concludes our discussion of Stork enamine alkylation. Let s go back and do a few examples using proline catalysis before we move on to our next topic. The key take-home message that I would like you all to get is that 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). Also note that you can only form an enamine if you have an a-proton. That should help you to figure out where to form the enamine in your reaction. 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). ow if I tell you that the reaction is catalyzed by proline, you should be able to draw the structure of the key enamine intermediate: CO 2 H R 1 R 2 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: CO 2 H O O H 3 C CH 3 2. HC catalysis: So far we have been talking only about using secondary amines (proline and imidazolidinone) as chiral organocatalysts. You can actually use a variety of small organic molecules to catalyze reactions. One other example is -heterocyclic carbenes (HCs). Let s do a little background about carbenes before we discuss how they can be used as enantioselective organic catalysts. Most of the time when we talk about carbon, we consider cases where the carbon has formed four bonds (either sigma and/or pi bonds) with its four valence electrons: A carbene is a neutral molecule where the carbon atom has formed two bonds and has two unshared electrons. An unsubstituted carbene is highly unstable, but it is possible to generate more stable carbene analogues with a variety of substitution patterns. One class of carbenes is called -heterocyclic carbenes (HCs). Some examples of HCs are shown below: S imidazole-derived triazole-derived thiazole-derived
These carbenes are substantially more stable than the simple carbenes we talked about last time, and the basis for their stability can be rationalized either by electronics or sterics. Think about the electronic stabilization of these carbenes, and we will talk about it briefly next time. One of the first isolated carbenes was formed in 1992, using strong base (sodium hydride) as well as catalytic potassium tert butoxide: This was more successful than the first attempt to isolate a carbene in 1961, which led to dimerization of the product: So that s it for how you make HCs. ow I want to spend the rest of class talking about reactions that are catalyzed by HCs. 1. Benzoin condensation. This reaction is catalyzed by a thiazolium-derived HC. It was discovered because researchers knew that thiamine (vitamin B12) catalyzed the benzoin condensation, and they realized that the active part of it was the thiazolium. The structure of thiamine is shown below: The overall benzoin condensation is shown below, and involves two molecules of benzaldehyde reacting to form benzoin: 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 (C-) can act to convert one molecule from a nucleophile to an electrophile. And the mechanism of thiazolium catalysis is shown below:
Basically, the thiazolium 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. The key intermediate here is molecule 156 that has a nucleophilic carbon (carbanion) which is the same carbon that used to be the electrophilic carbonyl carbon. You have effectively transformed the nucleophile to the electrophile. This intermediate also has another resonance form where it can be neutral The neutral form is called the Breslow intermediate named after my PhD advisor Ron Breslow. There are two points I want to emphasize here. First is that when you see this substitution pattern of an alcohol on a carbon right next to a carbonyl, you should think retrosynthetically that this could have come from a benzoin-type condensation. This is different from recognizing an aldol product, which has a 1,3 relationship between the carbonyl and the alcohol. Both of these structural motifs are shown below: Second point that you should keep in mind is that the role of the HC is to take something that is a strong electrophile, an aldehyde, and convert it into a nucleophile that is then able to attack another molecule of the aldehyde. It does that through the so-called Breslow intermediate, which has the carbene in a stable neutral enol form. This reversal of polarity from electrophile to nucleophile is called umpolung which is a German word that basically means switching. You can also have a cross-benzoin reaction, where a ketone and aldehyde react. Usually this works only in an intramolecular setting. One example of a cross benzoin is shown here:
and the newly formed bond is highlighted in red. Here are some intramolecular benzoin products: The reason we don t form any of the seven-membered ring product is that nobody likes a sevenmembered ring (ETROPY). You should make sure that for each of them you know how to figure out the starting material (retrosynthesis!). This concludes our discussion of chirality. ext time we are going to start discussing organic polymers.