Optically active (S)-2-iodooctane with Nal in acetone gives a product which does not show optical activity. Explain.

Explanation

To explain why optically active (S)-2-iodooctane with sodium iodide (NaI) in acetone gives a product that does not show optical activity, we can break down the process into clear steps:

### Step-by-Step Solution:

1. **Understanding the Reactants**:
- We start with (S)-2-iodooctane, which is an optically active compound due to its chiral center at the second carbon atom. The presence of iodine (I) makes it a good leaving group.

**Hint**: Identify the structure of (S)-2-iodooctane and recognize the chiral center.
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