Question

In an \( \text{S}_\text{N}2 \) substitution reaction of the type \( \text{R} - \text{Br} + \text{Cl}^- \xrightarrow{\text{DMF}} \text{R} - \text{Cl} + \text{Br}^- \), which one of the following has the highest relative rate?

• A. \( \text{CH}_3 \text{CH}_2 \text{CH}_2 \text{Br} \)
• B. \( \text{CH}_3 \text{CH}_2 \text{C} \text{H}_2 \text{Br} \)
• C. \( \text{CH}_3 \text{CH}_2 \text{C} \text{H}_2 \text{Br} \)
• D. \( \text{CH}_3 \text{CH}_2 \text{Br} \)

Hint:

In \( \text{S}_\text{N}2 \) reactions, less steric hindrance around the leaving group leads to a faster reaction rate.

Answer 1

The Correct Option is A

Step 1: Understand the \( \text{S}_\text{N}2 \) mechanism.
In \( \text{S}_\text{N}2 \) reactions, the leaving group is replaced by the nucleophile in a single step. The steric hindrance plays a role in determining the rate.
Step 2: Conclusion.
The least sterically hindered alkyl bromide will have the highest rate in an \( \text{S}_\text{N}2 \) reaction, so \( \text{CH}_3 \text{CH}_2 \text{CH}_2 \text{Br} \) will have the highest relative rate. Final Answer: \[ \boxed{\text{CH}_3 \text{CH}_2 \text{CH}_2 \text{Br}} \]