# Mechanism of acid-catalyzed Robinson annulation?

My professor posed my organic chemistry class a challenge problem for bonus points and I've been racking my brain over the last three days trying to work it out and I want to know if my solution is plausible. The problem asks us to draw the best curved arrow mechanism for the reaction in the bottom left corner of the photo included in this post (a substituted cyclobutane molecule to a substituted cycloalkene). At first, this seemed like a simple intramolecular Robinson annulation, however it appears to be catalyzed using the hydronium ion, rather than the hydroxide ion. No other reactants are stated, including light, heat, nor a double equivalence of the starting compound. Working through every mechanism that I could think of, I eventually ended up with the solution in the picture below. I assumed that reaction rates and percent yields were unimportant, as long as the mechanism is plausible. Does this solution break any rules of organic chemistry? Is there a better solution?

• Overall, a good solution. You don't need the enol in the first line. The allylic anion in the fourth structure is too basic. Protonate as you did in the first step but push electrons to fragment the ring. Enols do not sit around for several steps. They tautomerize to carbonyls. Formation of the cyclohexenone is fine except I would have opted for the enol of the ketone in the 3rd structure of the penultimate line to assist in the loss of water. This would lead directly to 3-methylcyclohexenone. Oct 28 '20 at 15:11

Addendum: The process is not a Robinson annulation, which involves an initial Michael addition followed by an aldol condensation. The latter reaction includes the transformation 3 $$\rightarrow$$ 8. The transformation 1 $$\rightarrow$$ 3 may described as a retro-aldol reaction.