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: Understanding the \( S_{\text{N}2} \) Mechanism
The \( S_{\text{N}2} \) reaction proceeds via a backside attack, where steric hindrance significantly affects the rate of reaction.
The lower the steric hindrance around the leaving group, the faster the reaction.
Step 2: Evaluating Steric Hindrance
Option (A) and (B): Both have sterically hindered secondary and tertiary carbons, making \( S_{\text{N}2} \) difficult.
Option (D): A primary halide but still has a longer chain.
Option (C): A primary halide with the least steric hindrance, making it the fastest in \( S_{\text{N}2} \).
Final Answer: \( CH_3CH_2Br \) has the highest relative rate.