The product formed by the treatment of ethanol and ethane nitrile in the presence of sulphuric acid is:

Options

A. Ethyl acetate
B. Diethyl ether
C. Ethyl methyl ketone
D. Butanal

Explanation

To solve the problem of determining the product formed by the treatment of ethanol (C2H5OH) and ethane nitrile (C2H5CN) in the presence of sulfuric acid (H2SO4), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
- Ethanol (C2H5OH)
- Ethane nitrile (C2H5CN)
- Sulfuric acid (H2SO4)

2. **Understand the Reaction**:
- Ethane nitrile (C2H5CN) can undergo hydrolysis in the presence of sulfuric acid. Hydrolysis involves the addition of water to break the nitrile group (C≡N) into a carboxylic acid.

3. **Hydrolysis of Ethane Nitrile**:
- The hydrolysis of ethane nitrile (C2H5CN) will yield ethanoic acid (C2H5COOH) and ammonia (NH3).
- The reaction can be represented as:
\[ C2H5CN + H2O \xrightarrow{H2SO4} C2H5COOH + NH3 \]

4. **Reaction with Ethanol**:
- The ethanoic acid (C2H5COOH) produced from the hydrolysis will then react with ethanol (C2H5OH) in a condensation reaction (esterification).
- The reaction can be represented as:
\[ C2H5COOH + C2H5OH \rightarrow C2H5COOC2H5 + H2O \]
- This produces ethyl acetate (C2H5COOC2H5) and water.

5. **Final Product**:
- The final product formed from the reaction of ethanol and ethane nitrile in the presence of sulfuric acid is **ethyl acetate (C2H5COOC2H5)**.

### Conclusion:
The correct answer is **Option A: Ethyl acetate**.