When ethanamide is treated with `EtMgBr`, followed by hydrolysis, the product is :

Options

A.
B.
C.
D.

Explanation

To solve the problem of what product is formed when ethanamide is treated with ethyl magnesium bromide (EtMgBr) followed by hydrolysis, we can follow these steps:

### Step 1: Identify the Reactants
The reactants in this reaction are:
- **Ethanamide (CH3CONH2)**: This is a two-carbon amide.
- **Ethyl magnesium bromide (EtMgBr)**: This is a Grignard reagent, which contains an ethyl group (C2H5) bonded to magnesium bromide.

### Step 2: Understand the Grignard Reaction
Grignard reagents react with carbonyl compounds (like amides) to form an intermediate. In this case, the carbonyl carbon of ethanamide will react with the ethyl group from the Grignard reagent.

### Step 3: Reaction Mechanism
1. The carbonyl carbon in ethanamide (C=O) is electrophilic and will react with the nucleophilic ethyl group from EtMgBr.
2. The nucleophilic attack leads to the formation of a tetrahedral intermediate. The carbonyl oxygen becomes negatively charged (alkoxide ion), and the nitrogen remains attached to the carbon chain.

The reaction can be represented as:
\[ \text{CH}_3\text{C(O)NH}_2 + \text{EtMgBr} \rightarrow \text{CH}_3\text{C(OH)(C_2H_5)NH}_2 \]

### Step 4: Hydrolysis of the Intermediate
After the formation of the intermediate, hydrolysis is performed. Hydrolysis involves adding water, which will protonate the alkoxide oxygen and convert the nitrogen into an amine.

The hydrolysis of the intermediate leads to:
\[ \text{CH}_3\text{C(OH)(C_2H_5)NH}_2 + \text{H}_2\text{O} \rightarrow \text{CH}_3\text{C(OH)(C_2H_5)} + \text{NH}_2\text{H} \]

### Step 5: Identify the Final Product
The final product after hydrolysis is 2-butanol (CH3-CH(OH)-C2H5), which is a four-carbon alcohol.

### Final Answer
The product formed when ethanamide is treated with ethyl magnesium bromide followed by hydrolysis is **2-butanol**.

---