Reaction of propanamide with ethanolic sodium hydroxide and bromine will give

Options

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

Answer

To solve the problem of what happens when propanamide reacts with ethanolic sodium hydroxide and bromine, we can break down the reaction into several steps.

### Step-by-step Solution:

1. **Identify the Reactants**:
- Propanamide has the structure CH3-CH2-CO-NH2.
- Sodium hydroxide (NaOH) is a strong base and will be used in an ethanolic solution.
- Bromine (Br2) is a halogen that will participate in the reaction.

2. **Role of Sodium Hydroxide**:
- Sodium hydroxide acts as a base in this reaction. It will deprotonate the amide nitrogen, making it a better nucleophile.

3. **Deprotonation of Propanamide**:
- The base (NaOH) will extract a hydrogen ion (H+) from the amide nitrogen, resulting in a negatively charged nitrogen (N-).
- The reaction can be represented as:
\[ CH3-CH2-CO-NH2 + NaOH \rightarrow CH3-CH2-CO-NH + H2O \]

4. **Nucleophilic Attack on Bromine**:
- The negatively charged nitrogen (N-) will act as a nucleophile and attack the bromine molecule (Br2).
- This results in the formation of an N-bromo derivative:
\[ CH3-CH2-CO-NH + Br2 \rightarrow CH3-CH2-CO-NHBr \]

5. **Formation of an Imine**:
- The N-bromo derivative can undergo further reaction. The nitrogen can lose a bromide ion (Br-) and form a double bond with the carbonyl carbon, leading to the formation of an imine:
\[ CH3-CH2-CO-NHBr \rightarrow CH3-CH2-C(=N)-Br \]

6. **Hydrolysis of the Imine**:
- The imine can then be hydrolyzed by water (from the solution), leading to the formation of an amine:
\[ CH3-CH2-C(=N)-Br + H2O \rightarrow CH3-CH2-C(=O)-NH2 + HBr \]

7. **Final Product**:
- The final product after hydrolysis is ethylamine (CH3-CH2-NH2) and carbon dioxide (CO2) as a byproduct.

### Final Answer:
The reaction of propanamide with ethanolic sodium hydroxide and bromine will yield ethylamine (C2H5NH2).