`SN^(1)` reaction of optically active alkyl halides leads to ________

Options

A. Inversion of configuration
B. Retention of configuration
C. Racemisation
D. None of these

Explanation

To solve the question regarding the outcome of an SN1 reaction of optically active alkyl halides, let's break it down step by step.

### Step-by-Step Solution:

1. **Understanding SN1 Mechanism**:
- The SN1 reaction involves two main steps: formation of a carbocation intermediate and nucleophilic attack.

2. **Starting Material**:
- We start with an optically active alkyl halide, which has a chiral center. This means it can exist in two enantiomeric forms (R and S configurations).

3. **Formation of Carbocation**:
- In the first step of the SN1 reaction, the alkyl halide loses the halide ion (X⁻) to form a carbocation. This step is the rate-determining step and results in the loss of optical activity because the carbocation is planar and can be attacked from either side.

4. **Nucleophilic Attack**:
- In the second step, a nucleophile (for example, methanol) attacks the carbocation. Since the carbocation is planar, the nucleophile can attack from either the front or the back side of the carbocation.

5. **Formation of Products**:
- This leads to the formation of two different products (enantiomers) because the nucleophile can approach the carbocation from two different orientations. As a result, we get a racemic mixture (equal amounts of both enantiomers).

6. **Conclusion**:
- Therefore, the SN1 reaction of optically active alkyl halides leads to racemization, meaning we obtain a mixture of both enantiomers.

### Final Answer:
The SN1 reaction of optically active alkyl halides leads to **racemization**.