Which of the following alkyl halides reacts fastest with trimethylamine in an $S_{\mathrm{N\; 2}}$ reaction?

• A. ${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{Br}$ (ethyl bromide)
• B. ${\mathrm{CH}}_3\mathrm{Br}$ (methyl bromide)
• C. ${(}^{\mathrm{CH}}{C}_3\mathrm{Br}$ (tert-butyl bromide)
• D. ${(}^{\mathrm{CH}}{\mathrm{CH}}_2\mathrm{Br}$ (isopropyl bromide)

Answer: (B)

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 an SN2 reaction involves a backside nucleophilic attack on the carbon attached to the leaving group, and the rate depends on steric hindrance around this carbon.
3. Understand that steric hindrance increases from methyl < primary < secondary < tertiary alkyl halides, making nucleophilic attack more difficult as the alkyl group becomes more substituted.
4. Since trimethylamine is a nucleophile, the alkyl halide with the least steric hindrance (methyl bromide) will react fastest in an SN2 reaction.
5. Conclude that methyl bromide reacts fastest because it offers the least steric hindrance for the nucleophile to attack, facilitating the SN2 mechanism.