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(D) The solvolysis reaction in formic acid proceeds via an $S_N1$ mechanism,which involves the formation of a carbocation intermediate.The rate of rea

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$III > IV > II > I$

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(D) The solvolysis reaction in formic acid proceeds via an $S_N1$ mechanism,which involves the formation of a carbocation intermediate.The rate of reaction depends on the stability of the carbocation formed after the loss of the chloride ion $(Cl^-)$.$(I)$ forms a primary allylic carbocation: $CH_2=CH-CH_2^+$.$(II)$ forms a secondary allylic carbocation: $CH_2=CH-CH^+(CH_3)$,which is more stable than $(I)$ due to the $+I$ effect of the methyl group.$(III)$ forms a primary allylic carbocation: $CH_2=C(CH_3)CH_2^+$,which is more stable than $(I)$ due to the $+I$ effect of the methyl group,but less stable than $(II)$ because it is primary.$(IV)$ forms a primary allylic carbocation: $CH_3CH=CH-CH_2^+$,which is more stable than $(I)$ due to the $+I$ effect of the methyl group,but less stable than $(II)$ and $(III)$ due to the position of the methyl group.Comparing the stability: $(II) > (III) > (IV) > (I)$.Wait,re-evaluating: $(II)$ is secondary allylic (most stable). $(III)$ is primary allylic with a methyl group at the $2$-position (more stable than $(IV)$). $(IV)$ is primary allylic with a methyl group at the $3$-position. $(I)$ is primary allylic (least stable).Therefore,the order is $(II) > (III) > (IV) > (I)$.

What is the correct decreasing order of the relative rate of solvolysis for the following alkyl halides in formic acid containing a small amount of water?
$(I) \ CH_2=CH-CH_2Cl$
$(II) \ CH_2=CH-CH(CH_3)Cl$
$(III) \ CH_2=C(CH_3)CH_2Cl$
$(IV) \ CH_3CH=CH-CH_2Cl$

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What is the correct decreasing order of the relative rate of solvolysis for the following alkyl halides in formic acid containing a small amount of water?$(I) \ CH_2=CH-CH_2Cl$$(II) \ CH_2=CH-CH(CH_3)Cl$$(III) \ CH_2=C(CH_3)CH_2Cl$$(IV) \ CH_3CH=CH-CH_2Cl$