Acetic anhydride on readuction with `LiAlH"_(4)` in ether gives :

Options

A. acetaldehde
B. Ethyl alcohol
C. acetone
D. ethane

Answer

To solve the question regarding the reduction of acetic anhydride with lithium aluminum hydride (LiAlH₄) in ether, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Structure of Acetic Anhydride**:
Acetic anhydride has the formula (CH₃CO)₂O. It consists of two acyl groups (acetyl groups) linked by an oxygen atom.

2. **Understand the Role of Lithium Aluminum Hydride (LiAlH₄)**:
LiAlH₄ is a strong reducing agent that can reduce carbonyl compounds (like aldehydes, ketones, and anhydrides) to their corresponding alcohols.

3. **Reduction Mechanism**:
When acetic anhydride is treated with LiAlH₄, the carbonyl groups (C=O) in the anhydride will be reduced. Each carbonyl group will be converted to a hydroxyl group (-OH) resulting in the formation of alcohols.

4. **Products Formation**:
Since acetic anhydride has two identical acetyl groups, the reduction will yield two molecules of ethanol (ethyl alcohol). The reaction can be represented as:
\[ (CH₃CO)₂O + 2 \text{LiAlH₄} \rightarrow 2 \text{C₂H₅OH} \]
Thus, the product is ethanol.

5. **Final Answer**:
The reduction of acetic anhydride with LiAlH₄ in ether gives ethanol (C₂H₅OH).

### Summary of the Reaction:
- Acetic Anhydride + LiAlH₄ → Ethanol