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(C) The reaction of benzyl bromides with ethanol proceeds via an $S_N1$ mechanism,which involves the formation of a carbocation intermediate.The stabi

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$p-$methoxybenzyl bromide

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(C) The reaction of benzyl bromides with ethanol proceeds via an $S_N1$ mechanism,which involves the formation of a carbocation intermediate.The stability of the carbocation determines the reactivity. Electron-donating groups $(EDG)$ stabilize the carbocation,while electron-withdrawing groups $(EWG)$ destabilize it.$1$. $p-$methoxybenzyl bromide: The methoxy group $(-OCH_3)$ is a strong electron-donating group by resonance ($+M$ effect),which significantly stabilizes the carbocation.$2$. $p-$methylbenzyl bromide: The methyl group $(-CH_3)$ is a weak electron-donating group by hyperconjugation ($+H$ effect).$3$. $p-$chlorobenzyl bromide: The chloro group $(-Cl)$ is electron-withdrawing by induction ($-I$ effect).$4$. $p-$nitrobenzyl bromide: The nitro group $(-NO_2)$ is a strong electron-withdrawing group by both resonance $(-M)$ and induction $(-I)$ effects.Therefore,$p-$methoxybenzyl bromide forms the most stable carbocation and reacts most readily with ethanol.

Among the following,the one which reacts most readily with ethanol is

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