Question

For SN₂ reaction, the increasing order of the reactivity of the following alkyl halides is:
(A) CH₃CH₂CH₂CH₂Br (B) CH₃CH₂CH(Br)CH₃   (C) (CH₃)₃CBr   (D) (CH₃)₂CHCH₂Br

• (A) $<$ (B) $<$ (C) $<$ (D)
• (A) $<$ (C) $<$ (B) $<$ (D)
• (B) $<$ (A) $<$ (D) $<$ (C)
• (C) < (B) < (D) < (A)

Answer 1

The Correct Option is D

To establish the increasing order of reactivity for the given alkyl halides in an SN₂ reaction, the factors influencing SN₂ reaction rates must be considered:

• SN₂ reactions involve a bimolecular mechanism. The nucleophile attacks the electrophilic carbon from the side opposite the leaving group, leading to inversion of configuration.
• Steric hindrance around the electrophilic carbon significantly impacts SN₂ reaction rates. Primary alkyl halides, with less steric hindrance, react faster than more hindered secondary or tertiary alkyl halides.

Evaluation of each compound:

1. (A) CH₃CH₂CH₂CH₂Br: This primary alkyl halide exhibits minimal steric hindrance, thus demonstrating high SN₂ reactivity.
2. (B) CH₃CH₂CH(Br)CH₃: This secondary alkyl halide possesses slightly more steric hindrance compared to a primary alkyl halide.
3. (C) (CH₃)₃CBr: This tertiary alkyl halide is characterized by substantial steric hindrance, rendering it the least reactive in SN₂ reactions.
4. (D) (CH₃)₂CHCH₂Br: This is also a secondary alkyl halide. However, it is less sterically hindered than (B) due to the bromine being attached to a primary carbon, resulting in greater reactivity than (B).

Therefore, the order of increasing reactivity towards SN₂ reactions is: (C) < (B) < (D) < (A)