The carbonyl compound producing an optically active product by reaction with `LiAlH_(4)` is

Options

A. propanone
B. butanone
C. 3-pentanone
D. benzophenone.

Answer

To solve the problem of identifying the carbonyl compound that produces an optically active product when reacted with lithium aluminum hydride (LiAlH₄), we need to analyze each option provided.

### Step-by-Step Solution:

1. **Understanding Optical Activity**:
- An optically active compound is one that can rotate plane-polarized light. For a compound to be optically active, it must be chiral, meaning it must have at least one carbon atom bonded to four different groups.

2. **Analyzing Each Option**:
- **Option 1: Propanol (CH₃CH₂CHO)**:
- When propanal reacts with LiAlH₄, it gets reduced to propanol (CH₃CH₂CH₂OH).
- The resulting compound has two identical groups (the two CH₃ groups), making it achiral and thus not optically active.
- **Conclusion**: Not optically active.

- **Option 2: Butanone (CH₃COCH₂CH₃)**:
- When butanone reacts with LiAlH₄, it is reduced to butanol (CH₃CH(OH)CH₂CH₃).
- In this product, the carbon atom bonded to the hydroxyl group (OH) has four different groups attached to it (CH₃, CH₂CH₃, OH, and H), making it chiral.
- **Conclusion**: Optically active.

- **Option 3: 3-Pentanone (CH₃COCH₂CH₂CH₃)**:
- When 3-pentanone reacts with LiAlH₄, it is reduced to 3-pentanol (CH₃CH(OH)CH₂CH₂CH₃).
- The carbon atom bonded to the hydroxyl group in 3-pentanol has two identical groups (the two CH₂CH₃ groups), making it achiral.
- **Conclusion**: Not optically active.

- **Option 4: Benzophenone (C₆H₅COC₆H₅)**:
- When benzophenone reacts with LiAlH₄, it is reduced to diphenylmethanol (C₆H₅C(OH)C₆H₅).
- The carbon atom bonded to the hydroxyl group has two identical phenyl groups (C₆H₅), making it achiral.
- **Conclusion**: Not optically active.

3. **Final Answer**:
- The only compound that produces an optically active product upon reaction with LiAlH₄ is **Butanone**.

### Summary:
The carbonyl compound that produces an optically active product by reaction with LiAlH₄ is **Butanone**.