Questions: Which compound is most likely to follow second-order kinetics in a substitution reaction? (A) CH3Br (B) (CH3)3CCH2Br (C) CH3CH2Br (D) (CH3)2CHBr

Solution Steps

Second-order kinetics in substitution reactions typically involve bimolecular nucleophilic substitution (S\(_\text{N}2\)) reactions. In S\(_\text{N}2\) reactions, the rate of reaction depends on the concentration of both the substrate and the nucleophile. These reactions are favored by primary alkyl halides because they have less steric hindrance, allowing the nucleophile to attack the electrophilic carbon more easily.

• (A) \(\mathrm{CH}_{3} \mathrm{Br}\): This is a primary alkyl halide with minimal steric hindrance, making it a good candidate for S\(_\text{N}2\) reactions.
• (B) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCH}_{2} \mathrm{Br}\): This is a tertiary alkyl halide, which is more likely to undergo S\(_\text{N}1\) reactions due to steric hindrance.
• (C) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Br}\): This is a primary alkyl halide, similar to (A), and is also a good candidate for S\(_\text{N}2\) reactions.
• (D) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHBr}\): This is a secondary alkyl halide, which can undergo both S\(_\text{N}1\) and S\(_\text{N}2\) reactions, but is less favorable for S\(_\text{N}2\) compared to primary alkyl halides.

Given the analysis, compounds (A) \(\mathrm{CH}_{3} \mathrm{Br}\) and (C) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Br}\) are both primary alkyl halides and are likely to follow second-order kinetics. However, \(\mathrm{CH}_{3} \mathrm{Br}\) is the simplest primary alkyl halide with the least steric hindrance, making it the most likely to follow S\(_\text{N}2\) kinetics.

Final Answer