Rearrangement of a bicyclic dienone under acidic conditions

Question

Find the major product of the following reaction.

Attempt:

First step would be the formation of oxonium ion. After that, there are two possibilities. Either the upper double bond or the lower bond will shift and create a carbocation (A and B)

In product A there are two possible migrations. The better migrating group is the carbon which is a part of the second ring. But it can't migrate because of formation of a bridged carbocation. So methyl group migrates.

If B carbocation is formed, then the better migrating group can attack the carbocation.

My teacher told that the product A is wrong because a weaker migrating group is acting while in B, the stronger migrating group is acting. Is this correct? Because B involves ring contraction and more steps.

Answer 1

Your analysis of the possible products is excellent. I would consider both of the products to be major. Waring and coworkers studied this problem with a slightly different substrate (an extra methyl group at the 1-position, next to the carbonyl on the 'bottom' of the molecule). Under aqueous acidic conditions (sulfuric acid), the corresponding products were obtained in ~25% (mechanism A) and ~40% yield (product B). Only when run with sulfuric acid in acetic anhydride did mechanism B truly dominate (~90% vs 5% mechanism A).

There is actually a third pathway, where the methyl group shifts from ring juncture to ring juncture, and then goes through a mechanism similar to B. For the dienone in the question, the product is indistinguishable from mechanism B, but for the dienone studied in the publication, this gave another product in ~30% yield. I suspect that for the dienone in the question, the total yield of product B would be ~70%. That makes it appear as if mechanism B is dominant, but it is really two similar pathways operating that happen to give the same product.