Equal amount of an `RCl(C_4 H_9Cl)` is reacted at the same temperature with equal volume of `0.2 M` and `0.4 M` solution of `KOH`, respectively, in two separate experiments. The time taken for the reaction of `50 %` of `(C_4 H_9 Cl)` was found to be same, the alkyl halide is :

Options

A.
B.
C.
D.

Explanation

To solve the problem, we need to analyze the information given about the reaction of the alkyl halide \( C_4H_9Cl \) with two different concentrations of \( KOH \).

### Step-by-Step Solution:

1. **Understanding the Reaction**:
- We have an alkyl halide \( C_4H_9Cl \) reacting with \( KOH \).
- The concentrations of \( KOH \) used are \( 0.2 M \) and \( 0.4 M \).

2. **Observing the Reaction Rate**:
- The time taken for the reaction to reach 50% completion is the same for both concentrations of \( KOH \).
- This indicates that the rate of reaction is independent of the concentration of the nucleophile \( KOH \).

3. **Identifying the Reaction Mechanism**:
- Since the rate of reaction does not depend on the concentration of \( KOH \), we can conclude that the reaction follows an \( SN1 \) mechanism.
- In an \( SN1 \) mechanism, the rate-determining step involves the formation of a carbocation, which is dependent only on the concentration of the alkyl halide.

4. **Determining the Type of Alkyl Halide**:
- The \( SN1 \) mechanism is favored by tertiary (3°) alkyl halides, followed by secondary (2°) and then primary (1°) alkyl halides.
- Given that the reaction is independent of the nucleophile concentration, we can infer that \( C_4H_9Cl \) must be a tertiary alkyl halide.

5. **Conclusion**:
- Therefore, the alkyl halide \( C_4H_9Cl \) is a tertiary alkyl halide.

### Final Answer:
The alkyl halide is a tertiary alkyl halide (3° halide).

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