One mole of ethly acetate on treatment with an excess of `LiAlH_(4)` in dry ether and subsequent acidification produces

Options

A. 1 mol acetic acid + 1 mol ethyl alcohol
B. 1 mol ethyl alcohol + 1 mol methyl alcohol
C. 2 moles of ethyl alcohol
D. 1 mol of 2-butanol .

Explanation

To solve the question regarding the reaction of one mole of ethyl acetate with excess lithium aluminum hydride (LiAlH₄) in dry ether followed by acidification, we can break down the process step by step.

### Step-by-Step Solution:

1. **Identify the Reactants**:
- We start with ethyl acetate (C₂H₅COOCH₃), which is an ester.

2. **Reaction with LiAlH₄**:
- Lithium aluminum hydride (LiAlH₄) is a strong reducing agent. When ethyl acetate is treated with LiAlH₄, it undergoes reduction.
- The carbonyl group (C=O) in the ester is reduced to an alcohol.

3. **Reduction Mechanism**:
- The LiAlH₄ donates hydride ions (H⁻) to the carbonyl carbon of ethyl acetate, leading to the formation of an intermediate alkoxide.
- The reaction can be represented as:
\[ C_2H_5COOCH_3 + LiAlH_4 \rightarrow C_2H_5CH(OH)CH_3 + LiAlO_2 \]
- Here, the ester is converted to a primary alcohol, specifically ethanol (C₂H₅OH) and another alcohol.

4. **Acidification**:
- After the reduction, the reaction mixture is typically acidified (for example, using dilute hydrochloric acid).
- This step protonates the alkoxide formed during the reduction, yielding the corresponding alcohol.

5. **Final Product**:
- The final product after the acidification is ethanol (C₂H₅OH).

### Conclusion:
Thus, the treatment of one mole of ethyl acetate with excess LiAlH₄ followed by acidification produces ethanol.

### Final Answer:
**Ethanol (C₂H₅OH)**