For \(S_N2\) reaction, the increasing order of the reactivity of the following alkyl halide is:

(A) \(CH_3CH_2CH_2CH_2Br\)

(B) \(CH_3CH_2CH(Br)CH_3\)

(C) \((CH_3)_3CBr\)

(D) \((CH_3)_2CHCH_2Br\)

Choose the correct answer from the options given below:

C < B < D < A

The correct answer is option 4. C < B < D < A.

Let us evaluate the reactivity order of the alkyl halides based on their structures and the \(S_N2\) reaction mechanism.

C. \((CH_3)_3CBr\): It is a tertiary alkyl halide. Tertiary alkyl halides are the least reactive in \(S_N2\) reactions because the carbon bearing the leaving group is highly sterically hindered and the attack of the nucleophile from the backside will be most difficult. Therefore, \((CH_3)_3CBr\) should have the lowest reactivity.

B. \(CH_3CH_2CH(Br)CH_3\):It is a secondary alkyl halide. Secondary alkyl halides are more reactive than tertiary alkyl halides but less reactive than primary alkyl halides in \(S_N2\) reactions. Therefore, \(CH_3CH_2CH(Br)CH_3\) should have higher reactivity compared to compound C but lower reactivity compared to compound D and compound A.

D. \((CH_3)_2CHCH_2Br\): It is a primary alkyl halide.Therefore, \((CH_3)_2CHCH_2Br\) should have a higher reactivity tocompound C and D. But as the carbon carrying the leaving group is attached to carbon having branching so it is less reactive compared to the compound A. 

A. \(CH_3CH_2CH_2CH_2Br\): It is a primary alkyl halide. Primary alkyl halides are the most reactive in \(S_N2\) reactions because they are easily accessible for the back side attack of the nucleophile due to low steric hinderance. Therefore, \(CH_3CH_2CH_2CH_2Br\) should have the highest reactivity.

Based on the structural considerations and \(S_N2\) reaction mechanism, the correct order of increasing reactivity in \(S_N2\) reactions is: \(C < B < D < A\)

Therefore, the correct answer is 4. C < B < D < A