The product A is4-methylcyclohexyl bromide + | vedclass

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(B) $S_N2$ reactions proceed with complete inversion of configuration at the chiral center.In the given reaction,the nucleophile $OH^-$ attacks from t

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(B) $S_N2$ reactions proceed with complete inversion of configuration at the chiral center.In the given reaction,the nucleophile $OH^-$ attacks from the side opposite to the leaving group $Br^-$.Therefore,the product $A$ will have the $HO$ group in the position opposite to the original $Br$ atom.This corresponds to the structure shown in option $(b)$.

The product $A$ is
$4-methylcyclohexyl bromide + OH^- \xrightarrow{S_N2} A$

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The product $A$ is$4-methylcyclohexyl bromide + OH^- \xrightarrow{S_N2} A$