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(D) In all three reactions,the treatment with $alc. KOH$ leads to dehydrohalogenation (elimination of $HBr$).Reaction $(1)$: The starting material is

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$A = B = C$ (all products are identical)

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(D) In all three reactions,the treatment with $alc. KOH$ leads to dehydrohalogenation (elimination of $HBr$).Reaction $(1)$: The starting material is $2$-bromocyclohex-$2$-en-$1$-one. Elimination of $HBr$ results in cyclohexa-$1,3$-dien-$1$-ol (enol form),which tautomerizes to form phenol.Reaction $(2)$: The starting material is $4$-bromocyclohex-$2$-en-$1$-one. Elimination of $HBr$ results in cyclohexa-$1,3$-dien-$1$-ol (enol form),which tautomerizes to form phenol.Reaction $(3)$: The starting material is $4$-bromocyclohex-$2$-en-$1$-one (isomer). Elimination of $HBr$ results in cyclohexa-$1,3$-dien-$1$-ol (enol form),which tautomerizes to form phenol.Since all three reactions yield phenol as the final stable product,$A = B = C$.

The products obtained in the above reactions $(1)$,$(2)$,and $(3)$ are:

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