Question

Which halogen compound in the following pair will react faster in \( \text{S}_\text{N}2 \) reaction and why?

Hint:

In \( \text{S}_\text{N}2 \) reactions, primary halides react faster than secondary or tertiary halides due to less steric hindrance. The nucleophile attacks the electrophilic carbon directly, leading to an inversion of configuration. Less bulky groups around the carbon make this attack easier and faster.

Answer 1

(a) The compound \( \text{CH}_3\text{CH}_2\text{Cl} \) (ethyl chloride) will react faster in an \( \text{S}_\text{N}2 \) reaction than \( \text{C}_6\text{H}_5\text{CH}_2\text{Cl} \) (benzyl chloride). The main reasons are:
1. Steric Hindrance:
- \( \text{CH}_3\text{CH}_2\text{Cl} \) is a primary halide with less steric hindrance around the carbon bonded to the halogen. This allows the nucleophile to attack the electrophilic carbon more easily.
- \( \text{C}_6\text{H}_5\text{CH}_2\text{Cl} \) (benzyl chloride) is a benzyl halide. Although benzyl halides are generally reactive in \( \text{S}_\text{N}2 \), they still experience more steric hindrance than a simple alkyl halide like ethyl chloride. 2. Nucleophilic Substitution Mechanism:
- \( \text{S}_\text{N}2 \) reactions involve a backside attack of the nucleophile on the carbon bonded to the halogen, resulting in the inversion of configuration. Less steric hindrance on the carbon results in a faster reaction. Therefore, \( \text{CH}_3\text{CH}_2\text{Cl} \) reacts faster in an \( \text{S}_\text{N}2 \) reaction than \( \text{C}_6\text{H}_5\text{CH}_2\text{Cl} \).