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(C) The $S_{N}1$ reaction proceeds via the formation of a carbocation intermediate.The stability of the carbocation determines the reactivity of the s

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$II > I > III$

Vedclass · Answer

(C) The $S_{N}1$ reaction proceeds via the formation of a carbocation intermediate.The stability of the carbocation determines the reactivity of the substrate.In the given compounds,the carbocations formed are benzyl carbocations substituted at the $p$-position.$(II)$ $p$-Methoxybenzyl chloride forms a $p$-methoxybenzyl carbocation,where the $-OCH_{3}$ group exerts a strong $+M$ effect,significantly stabilizing the carbocation.$(I)$ Benzyl chloride forms a benzyl carbocation with no substituent,providing moderate stability.$(III)$ $p$-Nitrobenzyl chloride forms a $p$-nitrobenzyl carbocation,where the $-NO_{2}$ group exerts a strong $-I$ and $-M$ effect,which destabilizes the carbocation.Therefore,the order of stability of the carbocations (and thus reactivity) is: $p$-Methoxybenzyl carbocation > Benzyl carbocation > $p$-Nitrobenzyl carbocation.The correct order is $II > I > III$.

The order of reactivity of the following compounds towards the $S_{N}1$ reaction will be:
$(I)$ Benzyl chloride
$(II)$ $p$-Methoxybenzyl chloride
$(III)$ $p$-Nitrobenzyl chloride

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The order of reactivity of the following compounds towards the $S_{N}1$ reaction will be:$(I)$ Benzyl chloride$(II)$ $p$-Methoxybenzyl chloride$(III)$ $p$-Nitrobenzyl chloride