The decreasing order of the rate of S_{N^1} r | vedclass

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(A) The rate of $S_{N^1}$ reaction is directly proportional to the stability of the intermediate carbocation formed.$1$. For $I$: The carbocation is $

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$I > II > III > IV$

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(A) The rate of $S_{N^1}$ reaction is directly proportional to the stability of the intermediate carbocation formed.$1$. For $I$: The carbocation is $Ph-C^+(CH_3)-Ph$,which is stabilized by two phenyl groups and one methyl group (tertiary and resonance stabilized).$2$. For $II$: The carbocation is $Ph-CH^+-Ph$,which is stabilized by two phenyl groups (secondary and resonance stabilized).$3$. For $III$: The carbocation is $Ph-CH^+-CH_3$,which is stabilized by one phenyl group and one methyl group (secondary and resonance stabilized).$4$. For $IV$: The carbocation is $Ph-CH_2^+$,which is stabilized by only one phenyl group (primary and resonance stabilized).Comparing the stability: $I > II > III > IV$. Therefore,the order of reactivity is $I > II > III > IV$.

The decreasing order of the rate of $S_{N^1}$ reaction for the following compounds is:
$I$. $Ph-C(Br)(CH_3)-Ph$
$II$. $Ph-CH(Br)-Ph$
$III$. $Ph-CH(Br)-CH_3$
$IV$. $Ph-CH_2Br$

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The decreasing order of the rate of $S_{N^1}$ reaction for the following compounds is:$I$. $Ph-C(Br)(CH_3)-Ph$$II$. $Ph-CH(Br)-Ph$$III$. $Ph-CH(Br)-CH_3$$IV$. $Ph-CH_2Br$