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(A) The rate of $S_{N}1$ reaction depends on the stability of the carbocation intermediate formed after the departure of the leaving group $(Br^-)$.In

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$IV > III > II > I$

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(A) The rate of $S_{N}1$ reaction depends on the stability of the carbocation intermediate formed after the departure of the leaving group $(Br^-)$.In these compounds,the carbocation formed is a substituted benzyl carbocation: $Ar-CH_2^+$.The stability of the benzyl carbocation is increased by electron-donating groups $(EDG)$ at the para position due to the $+I$ effect and hyperconjugation.The electron-donating power of the alkyl groups follows the order: isopropyl $(-(CH_3)_2CH)$ > ethyl $(-CH_2CH_3)$ > methyl $(-CH_3)$ > hydrogen $(-H)$.Therefore,the stability of the carbocations follows the order: $(IV) > (III) > (II) > (I)$.Consequently,the rate of $S_{N}1$ hydrolysis follows the same order: $(IV) > (III) > (II) > (I)$.

Arrange the following compounds in order of decreasing rate of hydrolysis for $S_{N}1$ reaction:
$(I) \ C_6H_5-CH_2-Br$
$(II) \ p-CH_3-C_6H_4-CH_2-Br$
$(III) \ p-CH_3CH_2-C_6H_4-CH_2-Br$
$(IV) \ p-(CH_3)_2CH-C_6H_4-CH_2-Br$

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$tert$-Butylbromide reacts with $aq.$ $NaOH$ by $S_N1$ mechanism while $n$-butylbromide reacts by $S_N2$ mechanism. Why?

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Which compound in each of the following pairs will react faster in $S_N2$ reaction with $OH^{-}$?
$(i)$ $CH_3-Br$ or $CH_3-I$
$(ii)$ $CH_3-C(CH_3)_2-Cl$ or $CH_3-Cl$

The correct order of $S_{N}1$ reaction rate for the given compounds is:

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Arrange the following compounds in order of decreasing rate of hydrolysis for $S_{N}1$ reaction:$(I) \ C_6H_5-CH_2-Br$$(II) \ p-CH_3-C_6H_4-CH_2-Br$$(III) \ p-CH_3CH_2-C_6H_4-CH_2-Br$$(IV) \ p-(CH_3)_2CH-C_6H_4-CH_2-Br$