Reduction by `LiAIH_(4)` of hydorlysed product of an ester gives:

Options

A. Two alcohols
B. Two aldehyde
C. One acid and one alcohol
D. Two acids

Answer

To solve the question regarding the reduction by Lithium Aluminum Hydride (LiAlH4) of the hydrolyzed product of an ester, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Ester Hydrolysis**:
- An ester (RCOOR') can be hydrolyzed in the presence of water (H2O) to form a carboxylic acid (RCOOH) and an alcohol (R'OH).
- The reaction can be represented as:
\[ RCOOR' + H2O \rightarrow RCOOH + R'OH \]

2. **Identifying the Hydrolyzed Products**:
- After hydrolysis, we have two products: a carboxylic acid (RCOOH) and an alcohol (R'OH).

3. **Reduction by LiAlH4**:
- Lithium Aluminum Hydride (LiAlH4) is a strong reducing agent that can reduce carboxylic acids and aldehydes to primary alcohols.
- The reaction of the carboxylic acid with LiAlH4 can be represented as:
\[ RCOOH + LiAlH4 \rightarrow RCH2OH + Al(OH)3 \]
- Here, the carboxylic acid (RCOOH) is reduced to a primary alcohol (RCH2OH).

4. **Final Products**:
- The final products of the reduction are:
- The alcohol formed from the hydrolysis (R'OH).
- The alcohol formed from the reduction of the carboxylic acid (RCH2OH).
- Therefore, the overall result of the reduction of the hydrolyzed product of an ester by LiAlH4 is two alcohols.

### Conclusion:
The reduction by LiAlH4 of the hydrolyzed product of an ester gives two alcohols: one from the original hydrolysis and one from the reduction of the carboxylic acid.