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(D) The reaction of an alkyl halide with $NaI$ in acetone is the Finkelstein reaction,which proceeds via an $S_N2$ mechanism.$1$. The $S_N2$ mechanism

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(D) The reaction of an alkyl halide with $NaI$ in acetone is the Finkelstein reaction,which proceeds via an $S_N2$ mechanism.$1$. The $S_N2$ mechanism involves a backside attack by the nucleophile $(I^-)$,leading to the inversion of configuration at the chiral center (Walden inversion).$2$. The starting material is a benzylic chloride. The $Cl$ atom attached to the chiral center is replaced by $I$ with inversion of configuration.$3$. The para-chloro group on the benzene ring is an aryl halide and does not undergo $S_N2$ substitution under these conditions.$4$. Therefore,the product is the one where the chiral center has undergone inversion,and the para-chloro group remains unchanged. This corresponds to option $D$.

Predict the product of the following reaction:
(Image: $A$ chiral carbon attached to a phenyl ring with a para-chloro group,a deuterium atom $(D)$,a hydrogen atom $(H)$,and a chlorine atom (Cl) in a specific stereochemical configuration. The reaction is with $NaI$ in acetone.)

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Predict the product of the following reaction:(Image: $A$ chiral carbon attached to a phenyl ring with a para-chloro group,a deuterium atom $(D)$,a hydrogen atom $(H)$,and a chlorine atom (Cl) in a specific stereochemical configuration. The reaction is with $NaI$ in acetone.)