The major product of the above reaction is A. | vedclass

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(A) Step $1$: $4$-Methoxybenzyl bromide reacts with $NaCN$ to form $4$-methoxybenzyl cyanide $(CH_3O-C_6H_4-CH_2CN)$.Step $2$: In the presence of $OH^

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(A) Step $1$: $4$-Methoxybenzyl bromide reacts with $NaCN$ to form $4$-methoxybenzyl cyanide $(CH_3O-C_6H_4-CH_2CN)$.Step $2$: In the presence of $OH^-$,the alpha-hydrogen of the nitrile is abstracted to form a carbanion,which then performs a nucleophilic attack on the carbonyl carbon of cyclohexanone.Step $3$: This results in the formation of a cyanohydrin derivative: $CH_3O-C_6H_4-CH(CN)-C(OH)(C_6H_{10})$.Step $4$: Finally,catalytic hydrogenation using $H_2, Ni$ reduces the nitrile group $(-CN)$ to a primary amine group $(-CH_2NH_2)$.The final product is $1-(4-methoxyphenyl)-2-(cyclohexyl)-2-hydroxyethanamine$ derivative,which matches the structure in option $A$.

The major product of the above reaction is $A$.

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