Arrange the following alkyl halides in the in | vedclass

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(A) The reactivity towards nucleophilic substitution ($S_N1$ or $S_N2$) depends on the stability of the carbocation formed (for $S_N1$) or steric hind

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$II > III > I > IV$

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(A) The reactivity towards nucleophilic substitution ($S_N1$ or $S_N2$) depends on the stability of the carbocation formed (for $S_N1$) or steric hindrance (for $S_N2$).$(I)$ is a secondary alkyl chloride,which is reactive.$(II)$ is an allylic bromide $(CH_2=CH-CH(Br)-CH_3)$. Allylic halides are highly reactive due to resonance stabilization of the carbocation.$(III)$ is an allylic chloride $(CH_2=CH-CH(Cl)-CH_3)$. It is also reactive but less than the bromide $(II)$ because $Br^-$ is a better leaving group than $Cl^-$.$(IV)$ is a vinylic bromide $(CH_3-CH=C(Br)-CH_3)$. Vinylic halides are extremely unreactive towards nucleophilic substitution because the $C-X$ bond has partial double bond character due to resonance.Comparing the leaving groups,$Br^-$ is a better leaving group than $Cl^-$. Thus,the order of reactivity is $II > III > I > IV$.

Arrange the following alkyl halides in the increasing order of their reactivity towards nucleophilic substitution reaction:
$(I)$ $CH_3CH_2CH(Cl)CH_3$
$(II)$ $CH_2=CHCH(Br)CH_3$
$(III)$ $CH_2=CHCH(Cl)CH_3$
$(IV)$ $CH_3CH=C(Br)CH_3$

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Arrange the following alkyl halides in the increasing order of their reactivity towards nucleophilic substitution reaction:$(I)$ $CH_3CH_2CH(Cl)CH_3$$(II)$ $CH_2=CHCH(Br)CH_3$$(III)$ $CH_2=CHCH(Cl)CH_3$$(IV)$ $CH_3CH=C(Br)CH_3$