Reaction of propanamide with ethanolic sodium hydroxide and bromine will give

Options

A. Ethylamine
B. Methylamine
C. Propylamine
D. Aniline

Answer

To solve the question about the reaction of propanamide with ethanolic sodium hydroxide and bromine, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
- The reactants are propanamide (CH3CH2CONH2), ethanolic sodium hydroxide (NaOH in ethanol), and bromine (Br2).

2. **Understand the Reaction Type**:
- This reaction is known as the Hofmann bromamide degradation reaction. It involves the conversion of an amide to a primary amine.

3. **Write the Structure of Propanamide**:
- Propanamide can be represented as:
```
CH3-CH2-C(=O)-NH2
```

4. **Reaction with Sodium Hydroxide and Bromine**:
- When propanamide reacts with sodium hydroxide and bromine, the carbonyl carbon (C=O) is attacked, leading to the degradation of the amide.

5. **Formation of the Primary Amine**:
- The carbon chain will lose one carbon atom, resulting in the formation of a primary amine. In this case, the product will be ethanamine (CH3CH2NH2).

6. **Additional Products**:
- Along with the primary amine, other products formed include sodium bromide (NaBr) and carbon dioxide (CO2). The reaction may also produce bromoform (CHBr3) and water (H2O).

7. **Final Products**:
- The major product is ethanamine (CH3CH2NH2), and the other products are NaBr and possibly CHBr3.

### Final Answer:
The reaction of propanamide with ethanolic sodium hydroxide and bromine will give **ethanamine (CH3CH2NH2)** as the primary product.