Examine the two $S{N}_2$ reactions below: (A) ${\mathrm{CH}}_3\mathrm{Br}$ with $\mathrm{OH}-$ and (B) ${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{Br}$ with $\mathrm{OH}-$. Which reaction will proceed at a faster rate?

• A. Reaction A, because methyl bromide is less sterically hindered than ethyl bromide.
• B. Reaction B, because ethyl bromide is more reactive in $S{N}_2$ reactions.
• C. Neither reaction will proceed because bromide is a poor leaving group.
• D. Both reactions proceed at the same rate because they have similar leaving groups.

Answer: (A)

Verified step by step guidance

1. Identify the type of reaction: Both reactions are SN2 reactions, which proceed via a backside attack by the nucleophile and involve a single transition state.
2. Consider the structure of the alkyl halides: Reaction A involves methyl bromide (CH\_3Br), and Reaction B involves ethyl bromide (CH\_3CH\_2Br).
3. Recall that SN2 reaction rates are highly sensitive to steric hindrance around the electrophilic carbon. Less steric hindrance means easier nucleophilic attack and faster reaction.
4. Compare steric hindrance: Methyl bromide has no alkyl groups attached to the carbon bearing the leaving group, while ethyl bromide has one alkyl group, making it more hindered.
5. Conclude that the reaction with methyl bromide (Reaction A) will proceed faster because the nucleophile can more easily attack the less hindered carbon, despite both having the same leaving group (bromide).