Arrange the following alkyl halides in decreasing order of the rate of ẞ-elimination reaction with alcoholic KOH:

Question

Accepted Answer

To determine the correct order of the rate of ẞ-elimination reaction with alcoholic KOH for the given alkyl halides, we need to consider the factors that influence the rate of ẞ-elimination. These factors include the structure of the alkyl halide, the stability of the resulting alkene, and the nature of the leaving group.

1. **Structure of the Alkyl Halide:**
   - ẞ-elimination reactions are generally favored in more substituted alkyl halides because they lead to more stable alkenes. This is due to the fact that more substituted alkenes are more stable due to hyperconjugation and the inductive effect.
   - In the given options:
     - [A] $CH_3-C(H)(CH_3)-CH_2Br$ is a secondary alkyl halide.
     - [B] $CH_{3}-CH_{2}-Br$ is a primary alkyl halide.
     - [C] $CH_{3}-CH_{2}-CH_{2}Br$ is also a primary alkyl halide.

2. **Stability of the Resulting Alkene:**
   - The more substituted the alkene, the more stable it is. This stability increases the rate of ẞ-elimination.
   - From [A], the resulting alkene is more substituted compared to those from [B] and [C].

3. **Nature of the Leaving Group:**
   - All the given compounds have bromine as the leaving group, which is a good leaving group. Therefore, the leaving group factor is constant across all options.

Based on these considerations, the order of reactivity for ẞ-elimination with alcoholic KOH is determined by the degree of substitution of the alkyl halide and the stability of the resulting alkene. Therefore, the correct order is:

- **[A] > [C] > [B]**

This order is consistent with Option 4: $A > C > B$.

**Explanation for Incorrect Options:**

- **Option 1 ($A > B > C$):** This option incorrectly suggests that [B] is more reactive than [C]. However, both [B] and [C] are primary alkyl halides, and the resulting alkene from [C] is more substituted than that from [B], making [C] more reactive than [B].

- **Option 2 ($C > B > A$):** This option incorrectly places [A] as the least reactive. [A] is a secondary alkyl halide and forms a more substituted alkene, making it more reactive than both [B] and [C].

- **Option 3 ($B > C > A$):** This option incorrectly suggests that [B] is the most reactive. As a primary alkyl halide, [B] is less reactive than [A], which is secondary and forms a more stable alkene.

Thus, the correct order of reactivity, supporting Option 4, is $A > C > B$.