Question

An optically active compound \( X \) has molecular formula \( C_4H_8O_3 \), it evolves \( \text{CO}_2 \) with aqueous NaHCO\(_3\). \( X \) reacts with LiAlH\(_4\) to give an achiral compound. \( X \) is

• A. \( \text{CH}_3\text{COOH} \)
• B. \( \text{CH}_3 \text{CHO} \)
• C. \( \text{CH}_3 \text{COOH} \)
• D. \( \text{C}_6 \text{H}_5 \text{COOH} \)

Hint:

When acetic acid reacts with NaHCO\(_3\), carbon dioxide is released, and the reduction of acetic acid yields an achiral product.

Answer 1

The Correct Option is A

Step 1: Analyzing the reaction.
The given reactions and the molecular formula suggest that \( X \) is acetic acid. The reaction with NaHCO\(_3\) shows the presence of a carboxyl group, and the reduction to an achiral compound supports this.

Step 2: Conclusion.
The correct compound is acetic acid, corresponding to option (1).