xrightarrow[Acetone]{NaI} underset{(Major)}{ | vedclass

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(C) The reaction proceeds in two steps,both involving $S_N2$ mechanisms.Step $1$: The starting material (assumed to be a chiral alkyl chloride) reacts

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(C) The reaction proceeds in two steps,both involving $S_N2$ mechanisms.Step $1$: The starting material (assumed to be a chiral alkyl chloride) reacts with $NaI$ in acetone (Finkelstein reaction) to form the alkyl iodide $X$. The $S_N2$ mechanism results in the inversion of configuration at the chiral center.Step $2$: The alkyl iodide $X$ reacts with $aq. KOH$ to form the alcohol $Y$. This is also an $S_N2$ reaction,which causes another inversion of configuration at the chiral center.Since two successive inversions occur,the final product $Y$ will have the same configuration as the original starting material.Comparing the structures,option $C$ shows the correct inversion for $X$ relative to the starting material (if we assume the starting material had $Cl$ on the right) and the subsequent inversion for $Y$ relative to $X$.

$ \xrightarrow[Acetone]{NaI} \underset{(Major)}{X} $ $ \xrightarrow[{S_N2}]{aq. KOH} \underset{(Major)}{Y} $
$X$ and $Y$ should be

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$ \xrightarrow[Acetone]{NaI} \underset{(Major)}{X} $ $ \xrightarrow[{S_N2}]{aq. KOH} \underset{(Major)}{Y} $$X$ and $Y$ should be