I took second-semester organic last summer, and we blazed through the more complicated reactions on the organic frontier. So I thought, why not begin a series this summer in which I write about some of these more complicated reactions? Today’s baby is the Mannich reaction, a process that has actually been receiving a lot of attention lately.
The Mannich reaction, in words, is an aminoalkylation at the site of an alpha-carbonyl (acidic) proton. The lone pair on the amino nitrogen first attacks the carbon on formaldehyde (or any aldehyde), forming a dipolar molecule neutralized when a hydrogen migrates from the nitrogen to the anionic aldehyde oxygen. This neutral molecule loses hydroxide to form an iminium ion, ironically similar to the process I described in the post on my organic lab project.
The alpha-acidic ketone loses a proton and then attacks the iminium carbon, and voila, aminoalkylation! Really, any acidic compound can be used as a nucleophile; ketones and carboxylic acids are of particular interest because the Mannich reaction can be employed in amino acid synthesis.
As you can see above, the M.R. can generate two adjacent chiral centers on the carbons between the carbonyl and amino functionality. In 2006, Barbas et al. published a beautiful method in JACS of controlling the stereochemistry of Mannich products using asymmetric catalysts. Using one catalyst they obtained only syn products, and using another only anti products. Their catalyst for the syn-Mannich reaction, (S)-proline, is even a naturally occurring amino acid! And if that didn’t make you soil yourself outright, the Barbas group also found that these syntheses were enantioselective, producing only (S,R) and (S,S) products!
How does it work? The first step in the syn reaction is enamine formation between the aldehyde and proline. Sterics dictate that proline’s COOH group and the enamine double bond face away from each other. The enamine can attack the imine to start the Mannich reaction, but it must do so such that the Si faces of the enamine and imine line up (because of H-bonding between proline’s COOH group and the imine N). This forces the enamine and imine hydrogens back, leading to syn aminoalkylation. The picture explains it better than I can:
In the cyclic transition state, you can see that the hydrogens are forced syn to each other. Conrotation from the transition state leads to intermediate 2, at which point the reaction is essentially done (reverse enamine-ization occurs to give the product 3c).
The anti process is similar, but the structure of the catalyst forces COOH and the enamine double bond to face each other, which leads to anti products.