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 determine the increasing order of reactivity for the given alkyl halides in an SN₂ reaction, we need to consider the factors that influence SN₂ reaction rates:

• SN₂ reactions occur via a bimolecular mechanism where the nucleophile attacks the electrophilic carbon from the opposite side of the leaving group, resulting in inversion of configuration.
• The steric hindrance around the electrophilic carbon is a major factor affecting SN₂ reactions. Less sterically hindered primary alkyl halides react faster than more hindered secondary or tertiary alkyl halides.

Let's evaluate each compound:

1. (A) CH₃CH₂CH₂CH₂Br: This is a primary alkyl halide with minimal steric hindrance, making it quite reactive in SN₂.
2. (B) CH₃CH₂CH(Br)CH₃: This is a secondary alkyl halide, slightly more hindered than a primary alkyl halide.
3. (C) (CH₃)₃CBr: This is a tertiary alkyl halide with significant steric hindrance, making it the least reactive in SN₂.
4. (D) (CH₃)₂CHCH₂Br: This is also a secondary alkyl halide, less hindered than (B) due to the primary carbon bonded to bromine, thus more reactive than (B).

Thus, arranging them in the order of increasing reactivity towards SN₂ reactions, we get: (C) < (B) < (D) < (A)

Answer 2

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)