Given the following alkyl halides: 1. $\mathrm{CH}{3}_{\mathrm{Br}}$ (methyl bromide), 2. $\mathrm{CH}{3}_{\mathrm{CH}}{2}_{\mathrm{Br}}$ (ethyl bromide), 3. $\left(\mathrm{CH}{3}_2\right)\mathrm{CHBr}$ (isopropyl bromide), 4. $\left(\mathrm{CH}{3}_3\right)\mathrm{CBr}$ (tert-butyl bromide), rank them from most to least reactive in an $S{N}_2$ reaction.

• A. $1$ > $2$ > $3$ > $4$
• B. $3$ > $2$ > $1$ > $4$
• C. $2$ > $1$ > $3$ > $4$
• D. $4$ > $3$ > $2$ > $1$

Answer: (A)

Verified step by step guidance

1. Identify the type of alkyl halide for each compound: methyl bromide is a primary alkyl halide with no alkyl substituents on the carbon bearing the bromine; ethyl bromide is a primary alkyl halide with one alkyl substituent; isopropyl bromide is a secondary alkyl halide; tert-butyl bromide is a tertiary alkyl halide.
2. Recall that in an \(S_{N}2\) reaction, the nucleophile attacks the electrophilic carbon from the backside, and steric hindrance around this carbon greatly affects the reaction rate.
3. Understand that steric hindrance increases from methyl to tertiary alkyl halides, so the order of reactivity in \(S_{N}2\) reactions generally decreases as the substitution on the carbon increases: methyl > primary > secondary > tertiary.
4. Rank the compounds accordingly: methyl bromide (1) will be the most reactive, followed by ethyl bromide (2), then isopropyl bromide (3), and finally tert-butyl bromide (4) as the least reactive due to the highest steric hindrance.
5. Express the final order of reactivity for the \(S_{N}2\) reaction as: \$1 > 2 > 3 > 4$.