Question

In an \( S_{\text{N}2} \) substitution reaction of the type: \[ R - Br + Cl^- \longrightarrow R - Cl + Br^- \]
Which one of the following has the highest relative rate?

• A. CH₃-CH-CH₂Br (with a CH₃ group attached to the second carbon)
• B. CH₃-CH(CBr)-CH₃ (with two CH₃ groups attached to the second carbon)
• C. CH₃CH₂Br
• D. CH₃CH₂CH₂Br

Hint:

\( S_{\text{N}2} \) reactions favor primary halides over secondary and tertiary ones due to lower steric hindrance.

Answer 1

The Correct Option is C

Step 1: \( S_{\text{N}2} \) Mechanism Analysis
The \( S_{\text{N}2} \) reaction occurs through a backside attack. Steric hindrance is a critical factor influencing the reaction rate. A decrease in steric hindrance adjacent to the leaving group results in an accelerated reaction.
Step 2: Steric Hindrance Evaluation
Options (A) and (B) involve secondary and tertiary carbons, respectively, which are sterically hindered, thus impeding \( S_{\text{N}2} \) reactions.
Option (D) presents a primary halide, but it possesses an extended alkyl chain.
Option (C) features a primary halide with minimal steric hindrance, leading to the most rapid \( S_{\text{N}2} \) reaction.
Final Answer: \( CH_3CH_2Br \) exhibits the highest relative reaction rate.