Consider the following four reactions: (I) C_ | vedclass

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(A) The given reaction is an $S_N1$ hydrolysis of a secondary benzylic halide. In $S_N1$ reactions,the leaving group $(Cl^-)$ initially forms an ion p

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$I > II > III > IV$

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(A) The given reaction is an $S_N1$ hydrolysis of a secondary benzylic halide. In $S_N1$ reactions,the leaving group $(Cl^-)$ initially forms an ion pair with the carbocation,which shields the front side and favors attack from the back side (inversion). In less polar solvents (higher acetone concentration),the ion pair is more stable and persists longer,leading to a higher proportion of inverted product. As the solvent polarity increases (more water),the ion pair dissociates more easily into a free carbocation,leading to more racemization and a lower proportion of inverted product. Therefore,the decreasing order of the proportion of inverted product is $I > II > III > IV$.

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

Among the given pairs,in which pair does the first compound react faster than the second compound in an $S_{N}1$ reaction?

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