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

Choose the correct answer from the options given below:

The correct answer is (C) < (B) < (D) < (A).

Step-by-Step Explanation:

1. Understanding the S_N2 Reaction Mechanism:
In a bimolecular nucleophilic substitution (S_N2) reaction, the nucleophile attacks the electrophilic carbon atom from the back side, opposite to the leaving group (bromide ion, in this case), in a single concerted step. Because this transition state involves both the nucleophile and the alkyl halide, the reaction is highly sensitive to steric hindrance. The less crowded the carbon atom undergoing attack, the faster the reaction rate.

2. Analyzing the Structure of the Alkyl Halides:
Let us classify the given alkyl halides based on the degree of substitution of the carbon atom bonded to the halogen (Br):

(A) CH₃CH₂CH₂CH₂Br (1-bromobutane):
This is a primary (1°) alkyl halide with a straight, unbranched chain. The carbon attached to the bromine is bonded to only one other carbon, offering minimal steric hindrance.

(B) CH₃CH₂CH(Br)CH₃ (2-bromobutane):
This is a secondary (2°) alkyl halide. The carbon attached to the bromine is bonded to two other carbons. This increases the steric hindrance around the electrophilic carbon compared to primary alkyl halides.

(C) (CH₃)₃CBr (2-bromo-2-methylpropane / tert-butyl bromide):
This is a tertiary (3°) alkyl halide. The carbon attached to the bromine is bonded to three methyl groups. The steric hindrance is extremely high, making it virtually unreactive towards S_N2 substitution.

(D) (CH₃)₂CHCH₂Br (1-bromo-2-methylpropane / isobutyl bromide):
This is a primary (1°) alkyl halide, but it features a methyl branch at the β-carbon (the carbon adjacent to the one carrying the bromine). This β-branching introduces steric hindrance, making it slower to react than the unbranched primary halide (A), but it remains more reactive than the secondary halide (B).

3. Determining the Reactivity Order:
S_N2 reactivity generally follows the order:
Methyl halide > Primary (1°) > Secondary (2°) > Tertiary (3°)

Comparing the given structures:

• (C) is tertiary (3°), so it has the lowest reactivity (least reactive).
• (B) is secondary (2°), which is more reactive than tertiary but less reactive than primary.
• (D) and (A) are both primary (1°). However, (D) has steric hindrance on the adjacent (β) carbon due to the methyl branch, which slows down the nucleophilic attack compared to the unbranched chain in (A). Therefore, (D) is less reactive than (A).

Combining these observations, the increasing order of reactivity towards S_N2 reaction is:
(C) < (B) < (D) < (A)