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(A) The reactivity of alkyl halides towards $S_N2$ reactions follows the order: $Primary > Secondary > Tertiary$. This is due to the steric hindrance,

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$n$-Butyl iodide

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(A) The reactivity of alkyl halides towards $S_N2$ reactions follows the order: $Primary > Secondary > Tertiary$. This is due to the steric hindrance,which increases from primary to tertiary alkyl halides,making it difficult for the nucleophile to attack the electrophilic carbon. Among the given options: $(A)$ $n$-Butyl iodide is a primary $(1^{\circ})$ alkyl halide. $(B)$ $sec$-Butyl iodide is a secondary $(2^{\circ})$ alkyl halide. $(C)$ Isobutyl iodide is a primary $(1^{\circ})$ alkyl halide,but it has steric hindrance at the $\beta$-carbon. $(D)$ $tert$-Butyl iodide is a tertiary $(3^{\circ})$ alkyl halide. Since $n$-Butyl iodide is a primary alkyl halide with the least steric hindrance,it shows the highest reactivity for $S_N2$ reactions.

Which of the following has the highest reactivity for $S_N2$ reactions?

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