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(C) The reactivity of alkyl halides towards $S_{N}2$ reactions is primarily governed by steric hindrance. The order of reactivity is $Primary > Second

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$I > III > II > IV$

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(C) The reactivity of alkyl halides towards $S_{N}2$ reactions is primarily governed by steric hindrance. The order of reactivity is $Primary > Secondary > Tertiary$.$(I)$ $CH_{3}CH_{2}CH_{2}CH_{2}Cl$ is a primary alkyl halide with minimal steric hindrance.$(III)$ $(CH_{3})_{2}CHCH_{2}Cl$ is a primary alkyl halide but has more steric hindrance than $(I)$ due to the branching at the $\beta$-carbon.$(II)$ $CH_{3}CH_{2}CH(Cl)CH_{3}$ is a secondary alkyl halide.$(IV)$ $(CH_{3})_{3}CCl$ is a tertiary alkyl halide,which is the least reactive towards $S_{N}2$ due to maximum steric hindrance.Therefore,the correct order is $I > III > II > IV$.

Predict the order of reactivity of the following four isomers towards $S_{N}2$ reaction.
$(I)$ $CH_{3}CH_{2}CH_{2}CH_{2}Cl$
$(II)$ $CH_{3}CH_{2}CH(Cl)CH_{3}$
$(III)$ $(CH_{3})_{2}CHCH_{2}Cl$
$(IV)$ $(CH_{3})_{3}CCl$

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Predict the order of reactivity of the following four isomers towards $S_{N}2$ reaction.$(I)$ $CH_{3}CH_{2}CH_{2}CH_{2}Cl$$(II)$ $CH_{3}CH_{2}CH(Cl)CH_{3}$$(III)$ $(CH_{3})_{2}CHCH_{2}Cl$$(IV)$ $(CH_{3})_{3}CCl$