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(B) The reaction proceeds as follows:$1$. The starting material is $6$-chlorocyclohex-$2$-en-$1$-one.$2$. Addition of $HBr$ across the double bond fol

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phenol

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(B) The reaction proceeds as follows:$1$. The starting material is $6$-chlorocyclohex-$2$-en-$1$-one.$2$. Addition of $HBr$ across the double bond follows Markovnikov's rule,resulting in the formation of $6$-chloro-$3$-bromocyclohexan-$1$-one.$3$. Treatment with alcoholic $KOH$ $(alc. KOH)$ causes dehydrohalogenation (elimination of $HBr$ and $HCl$),leading to the formation of cyclohexa-$2,4$-dien-$1$-one.$4$. This intermediate undergoes tautomerization to form the more stable aromatic product,phenol $(C_6H_5OH)$.

The major product of the following reaction is

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