Select the single best answer. Which $S{\mathrm{N}}_2$ reaction in the following pair is faster? (A) $\mathrm{CH}{3}_2\mathrm{CHBr}$ with $\mathrm{OH}-$ or (B) $\mathrm{CH}{3}_3\mathrm{CBr}$ with $\mathrm{OH}-$

• A. Neither reaction will proceed via the $S{\mathrm{N}}_2$ mechanism
• B. Both reactions proceed at the same rate
• C. Reaction (A): $\mathrm{CH}{3}_2\mathrm{CHBr}$ with $\mathrm{OH}-$ is faster
• D. Reaction (B): $\mathrm{CH}{3}_3\mathrm{CBr}$ with $\mathrm{OH}-$ is faster

Answer: (C)

Verified step by step guidance

1. Identify the substrates involved in the SN2 reactions: (A) is 1-bromopropane (a primary alkyl bromide) and (B) is 2-bromopropane (a secondary alkyl bromide).
2. Recall that SN2 reactions proceed via a backside attack mechanism, which is sterically hindered by bulky groups near the electrophilic carbon.
3. Understand that primary alkyl halides (like in reaction A) have less steric hindrance compared to secondary alkyl halides (like in reaction B), making nucleophilic attack easier and faster.
4. Consider the nucleophile, hydroxide ion (OH⁻), which is a strong nucleophile and will favor SN2 mechanisms when steric hindrance is low.
5. Conclude that reaction (A) with the primary alkyl bromide will proceed faster via the SN2 mechanism than reaction (B) with the secondary alkyl bromide due to lower steric hindrance.