When ethane nitrile is treated with `C_2H_5 MgBr`, followed by hydrolysis, the product is :

Options

A.
B.
C.
D.

Explanation

To solve the problem of what product is formed when ethane nitrile is treated with C2H5 MgBr followed by hydrolysis, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
- Ethane nitrile can be represented as CH3C≡N (where the C≡N indicates a nitrile group).
- C2H5 MgBr is an organomagnesium compound, specifically an ethyl magnesium bromide (Grignard reagent).

2. **Understand the Reaction**:
- Grignard reagents react with nitriles to form an intermediate imine, which upon hydrolysis gives a ketone.
- The general reaction of a nitrile with a Grignard reagent can be summarized as:
\[ R-C≡N + R'MgX \rightarrow R-C(=NR')-R' + H2O \text{ (upon hydrolysis)} \]

3. **Write the Reaction**:
- For our specific case:
\[ CH3C≡N + C2H5MgBr \rightarrow CH3C(=N-C2H5) \text{ (intermediate)} \]

4. **Hydrolysis of the Intermediate**:
- Upon hydrolysis (adding water and acid), the imine formed will convert to a ketone:
\[ CH3C(=N-C2H5) + H2O \rightarrow CH3C(=O)C2H5 + NH3 \]
- The product here is an ethyl ketone, specifically:
\[ CH3C(=O)C2H5 \text{ (which is ethyl methyl ketone)} \]

5. **Final Product**:
- The final product after hydrolysis is:
\[ CH3C(=O)C2H5 \]

### Conclusion:
The product formed when ethane nitrile is treated with C2H5 MgBr followed by hydrolysis is **ethyl methyl ketone** (also known as 3-pentanone).