Question

The increasing order of reactivity of the following compounds in \( S_N1 \) reaction is:
A. \( (CH_3)_3CBr \)
B. \( CH_3CH_2CH_2CH_2Br \)
C. \( (CH_3)_2CHCH_2Br \)
D. \( CH_3CH(Br)CH_2CH_3 \)

• A. A < B < C < D
• B. B < C < D < A
• C. D < C < B < A
• D. B < D < A < C

Hint:

\( S_N1 \) rate follows carbocation stability \( 3^{\circ} > 2^{\circ} > 1^{\circ} \); watch for possible rearrangement in the isobutyl halide.

Answer 1

The Correct Option is B

Step 1: Concept.
\( S_N1 \) reactions go through a carbocation intermediate. The rate depends on how stable that carbocation is: more stable carbocation means faster \( S_N1 \). Stability order of carbocations is \( 3^{\circ} > 2^{\circ} > 1^{\circ} \).

Step 2: Classify each halide.
A \((CH_3)_3CBr\) is tertiary (gives a stable \(3^{\circ}\) carbocation) - most reactive.
D \(CH_3CH(Br)CH_2CH_3\) is secondary (gives \(2^{\circ}\) carbocation).
B \(CH_3CH_2CH_2CH_2Br\) is a straight-chain primary halide.
C \((CH_3)_2CHCH_2Br\) is primary too, but its cation can rearrange (hydride shift) to a stable \(3^{\circ}\) cation, so it is slightly more reactive than the plain primary B.

Step 3: Arrange in increasing order.
Least to most reactive: B < C < D < A.

Conclusion: Option (ii).
\[\boxed{B < C < D < A}\]