Acetoisonitrile on reaction with `C_2 H_5 MgBr` followed by hydrolysis, gives compound `(A)`, which one further hydrolysis gives `(B)` and `( C)`.
`(B)` and `( C)` are :

Options

A. `MeNH_2` and `EtCHO`
B. `EtNH_2` and `MeCHO`
C. `MeNH_2` and `EtCOOH`
D. `EtNH_2` and `MeCOOH`

Answer

To solve the problem, we will follow the steps of the reaction of acetoisonitrile with ethyl magnesium bromide (C2H5MgBr) and the subsequent hydrolysis.

### Step-by-Step Solution:

1. **Identify the Reactants**:
Acetoisonitrile is CH3C≡N (where the isonitrile group is attached to the carbon of the carbonyl). The Grignard reagent is C2H5MgBr.

2. **Formation of the Intermediate**:
When acetoisonitrile reacts with C2H5MgBr, the nucleophilic ethyl group (C2H5-) attacks the carbon atom of the isonitrile group. This results in the formation of an intermediate compound:
\[ CH3C≡N + C2H5MgBr \rightarrow CH3C(C2H5)N^(-)MgBr^(+) \]
Here, the nitrogen atom carries a negative charge.

3. **Hydrolysis of the Intermediate**:
Upon hydrolysis (in acidic medium), the magnesium bromide (MgBr) will react with water, leading to the protonation of the nitrogen atom and the formation of a carbonyl compound:
\[ CH3C(C2H5)N^(-)MgBr^(+)\ + H2O \rightarrow CH3C(C2H5)OH + NH3 \]
The product formed here is an alcohol (compound A):
\[ A: CH3C(C2H5)OH \]

4. **Further Hydrolysis**:
The next step involves further hydrolysis of compound A. The alcohol group can undergo dehydration (loss of water) to form a carbonyl compound:
\[ CH3C(C2H5)OH \rightarrow CH3C(C2H5)O + H2O \rightarrow CH3C(C2H5)O \]
This leads to the formation of compound B (an amide) and compound C (an aldehyde):
\[ B: CH3C(C2H5)O \quad (Methanamide) \]
\[ C: CH3C(C2H5)O \quad (Ethyl aldehyde) \]

5. **Final Products**:
After the complete hydrolysis, the final products are:
- Compound B: Methanamide (CH3NH2)
- Compound C: Propanal (C2H5CHO)

### Conclusion:
Thus, the final compounds formed after the reactions are:
- Compound B: Methanamide (CH3NH2)
- Compound C: Propanal (C2H5CHO)