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The following is a question from Advanced problems in organic chemistry by MS Chouhan:

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Product $A$ is rather straightforward:

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and on acidic hydrolysis, the following reaction takes place:

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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?

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    $\begingroup$ Not sure I've ever seen this being given a fancy name, but you tautomerise to form the conjugated enol, and then re-tautomerise back to the ketone, but protonating the gamma carbon instead of the alpha carbon... $\endgroup$ Apr 15, 2022 at 11:09
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    $\begingroup$ This mechanism is wrong at the second step. Water attacks the cation at the carbonyl carbon to form the hemiacetal that collapses to the deconjugated ketone. Enolization of this ketone to form the conjugated dienol and gamma protonation leads to (b), the conjugated enone. $\endgroup$
    – user55119
    Apr 15, 2022 at 22:42
  • $\begingroup$ @user55119 it would really help if you could maybe sketch it out! $\endgroup$ Apr 16, 2022 at 7:17

1 Answer 1

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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.

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  • $\begingroup$ "As to the loss of the methyl group in ... the counterion is water." Can you please elaborate a bit on this? And yes, thanks for this excellent analysis! $\endgroup$ Apr 16, 2022 at 18:14
  • $\begingroup$ If hydrochloric acid is the source of hydronium ion, then chloride is the counterion. If you have pH 1, then the hydroxide concentration is $\pu{[10^-15]}$. Not much hydroxide. $\endgroup$
    – user55119
    Apr 16, 2022 at 19:13

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