Beta Amino Acid Rearrangement

Here's the answer to the mechanism question I posed at the end of the last post. Some have suggested a 4-membered ring intermediate. While that cannot be ruled out, a mechanism that does not include the high strain of a bridged 4-membered ring seems more plausible.

Since this is an aminoacid, it will exist in it's zwitterionic form. Thus, the quaternary ammonium will not be acylated. The carboxylate is converted to a mixed anhydride. Then it undergoes a beta-elimination of the ammonium to open the 6-membered ring. This is followed by an acylation of the resulting amine to form the rearranged lactam.

This reaction was reported by Henry Rapoport (JACS 1970, 92, 5781). He cites an older paper by Ferles (Coll. Czech. Chem. Commun., 1964, 29, 2323.

Update: As liquidcarbon points out in the comments, the free amine of the ring-opened intermediate above would likely be acetylated in refluxing acetic anhydride. Another possible route to the product would involve an intramolecular acylation forming a bridging 4-membered ring, followed by beta elimination. Possible, but I'm not sure how well the bridgehead hydrogen sigma orbital would overlap with the sigma star orbital of the C-N bond.

Mechanism Challenge Answered

Tynchtyk, over at Chemist in a Transition State, posted a very interesting transformation and challenged us to propose a mechanism. Here is the reaction.

At first glance, this looks like some kind of reductive amination reaction. However, on closer inspection, you can see that there is one less carbon in the product than the starting material. Furthermore, there are no reducing agents present, only acid (and presumably water). Of course the obvious starting point is to react the secondary amine with the aldehyde to form a cyclic imminium structure. Once generated, this is nicely set up to undergo a [3,3]-sigmatropic rearrangement to transfer an allyl group to the imminium carbon. The resulting formaldehyde imminium product is then hydrolyzed in the presence of water to afford the product plus an equivalent of formaldehyde. The full mechanism is shown below. Notice I am a stickler for showing every proton transfer step! No shortcuts here.

Tynchtyk says this problem appeared in a science olypiad for High School Students in Moscow. I wish our high school education here in the states was up to this kind of challenge.

Thanks, Tynchtyk, nice problem! In the spirit of problem solving, let me pose a new challenge. This is one of my favorite transformations.