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(D) The reactivity of compounds in an $S_N1$ reaction depends on the stability of the carbocation intermediate formed after the departure of the leavi

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$(IV)$ > $(II)$ > $(III)$ > $(I)$

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(D) The reactivity of compounds in an $S_N1$ reaction depends on the stability of the carbocation intermediate formed after the departure of the leaving group $(Br^-)$.The stability order of the carbocations is as follows:$(I)$ $C_6H_5-CH_2^+$ (Benzyl carbocation,stabilized by one phenyl ring).$(II)$ $(C_6H_5)_2CH^+$ (Diphenylmethyl carbocation,stabilized by two phenyl rings).$(III)$ $C_6H_5-CH^+(CH_3)$ ($1$-Phenylethyl carbocation,stabilized by one phenyl ring and one methyl group).$(IV)$ $(C_6H_5)_2C^+(CH_3)$ ($1$,$1$-Diphenylethyl carbocation,stabilized by two phenyl rings and one methyl group).Comparing the stability: $(IV) > (II) > (III) > (I)$.Since the rate of $S_N1$ reaction is directly proportional to the stability of the carbocation,the order of reactivity is $(IV) > (II) > (III) > (I)$.

Predict the order of reactivity of the following compounds in $S_N1$ reaction.
$(I)$ $C_6H_5-CH_2Br$
$(II)$ $C_6H_5-CH(C_6H_5)Br$
$(III)$ $C_6H_5-CH(CH_3)Br$
$(IV)$ $C_6H_5-C(CH_3)(C_6H_5)Br$

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Predict the order of reactivity of the following compounds in $S_N1$ reaction.$(I)$ $C_6H_5-CH_2Br$$(II)$ $C_6H_5-CH(C_6H_5)Br$$(III)$ $C_6H_5-CH(CH_3)Br$$(IV)$ $C_6H_5-C(CH_3)(C_6H_5)Br$