JEE Main Chemistry question

[A] ←NaBH₄– a cyclohexene-fused ketone with an exocyclic =CH₂ group –B₂H₆/H₂O₂, OH⁻→ [B]. [A] and [B] are

(A)A reduced allylic alcohol (ring C=C retained, C=O reduced to CH–OH) and the anti-Markovnikov hydroboration–oxidation alcohol (exocyclic =CH₂ converted to –CH₂OH)Correct
(B)Swapped identities of (a)
(C)Both the same hydroxymethyl-cyclohexanol product
(D)Both the same hydroxyl-cyclohexanone product

Answer: (A)

Solution

NaBH₄ selectively reduces the ketone carbonyl to a secondary alcohol while leaving the alkene (ring or exocyclic) untouched, giving [A], an allylic cyclohexenol. Separately, B₂H₆/H₂O₂,OH⁻ (hydroboration–oxidation) selectively adds across the exocyclic alkene (anti-Markovnikov) while leaving the ketone

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Question ID VK-149352