The following is a question from Advanced problems in organic chemistry by MS Chouhan:
Product $A$ is rather straightforward:
and on acidic hydrolysis, the following reaction takes place:
Here the book says that this product rearranges to (b), and they termed it the "3-carbon system tautomerism"
I found no reference for this interesting mechanism. How exactly does this rearrangement take place?
Lost Anisole: Before fulfilling your request for a diagram of the mechanism for the formation of cyclohex-2-en-1-one (9), a few remarks about the problem as presented are in order. Your product A is correct. Unfortunately, anisole (1) will not reduce with lithium in liquid ammonia in the absence of an alcohol such as ethanol. As to the loss of the methyl group in the mechanism presented, one cannot simultaneously have both protons and hydroxide available in acid medium. The only nucleophilic species in aqueous acid apart from the counterion is water.
Enol ether 2 is the Birch reduction product of anisole (1). Protonation of the more electron-rich double bond affords species 3, which suffers attack by water at carbonyl carbon not at the methyl group, giving rise to the hemiketal 4. This species is a typical intermediate in the formation of acetals and ketals in acid medium when alcohol is the solvent. Here, with water in excess, the equilibrium is driven toward the unconjugated cyclohexenone 6 via protonated species 5.
Acid-catalyzed enolization of compound 6 will ultimately lead to the thermodynamically more stable dienol 8 as illustrated in 7. While dienol 8 may protonate at either the $\alpha$ or $\gamma$-positions, protonation at the $\alpha$-position leads back directly to cyclohexenone 6. This material will recycle with acid until $\gamma$-protonation occurs and the thermodynamic, conjugated cyclohexenone 9 is formed.
