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

The SN2 reaction is a bimolecular nucleophilic substitution reaction. Its rate depends on the concentration of both the substrate (alkyl halide) and the nucleophile. Steric hindrance plays a crucial role in SN2 reactions.

Here's how steric hindrance affects SN2 reactivity:

Primary alkyl halides are the most reactive due to minimal steric hindrance.

Secondary alkyl halides are less reactive due to increased steric hindrance.

Tertiary alkyl halides are the least reactive (or essentially unreactive) due to significant steric hindrance.

Let's analyze the given alkyl halides:

(A) CH₃CH₂CH₂CH₂Br (1-bromobutane): Primary alkyl halide.

(B) CH₃CH₂CH(Br)CH₃ (2-bromobutane): Secondary alkyl halide.

(C) (CH₃)₃CBr (2-bromo-2-methylpropane): Tertiary alkyl halide.

(D) (CH₃)₂CHCH₂Br (1-bromo-2-methylpropane): Primary alkyl halide.

Comparing (A) and (D), both are primary alkyl halides. However, (D) has branching closer to the reaction site, which adds slightly more steric hindrance than (A).

Therefore, the increasing order of reactivity towards SN2 reactions is:

(C) < (B) < (D) < (A)

The correct answer is:

Option 4: (C) < (B) < (D) < (A)