Propanamide on reaction with bromine in aqueous NaOH gives:

Options

A. Propanamine
B. Etanamine
C. N-Methyl ethanamine
D. Propanenitrile

Answer

To solve the question, "What does propanamide yield when it reacts with bromine in aqueous NaOH?", we can follow these steps:

### Step 1: Identify the Reactants
The reactant is propanamide, which has the structural formula CH3-CH2-CO-NH2.

### Step 2: Understand the Role of Sodium Hydroxide
Sodium hydroxide (NaOH) acts as a base in this reaction. Its role is to deprotonate the amide nitrogen, making it more nucleophilic.

### Step 3: Deprotonation of Propanamide
When NaOH is added to propanamide, the hydroxide ion (OH-) will extract a proton (H+) from the nitrogen atom of the amide, resulting in a negatively charged nitrogen:
\[ \text{CH}_3\text{CH}_2\text{C(=O)NH}_2 + \text{OH}^- \rightarrow \text{CH}_3\text{CH}_2\text{C(=O)N}^- + \text{H}_2\text{O} \]

### Step 4: Nucleophilic Attack on Bromine
The negatively charged nitrogen can now act as a nucleophile and attack the bromine (Br2). This leads to the formation of N-bromo-propanamide:
\[ \text{CH}_3\text{CH}_2\text{C(=O)N}^- + \text{Br}_2 \rightarrow \text{CH}_3\text{CH}_2\text{C(=O)NBr} + \text{Br}^- \]

### Step 5: Elimination of Carbon Dioxide
In the presence of NaOH, the N-bromo-propanamide can undergo further reactions, leading to the elimination of carbon dioxide (CO2) and the formation of an amine. This is a part of the Hofmann degradation reaction:
\[ \text{CH}_3\text{CH}_2\text{C(=O)NBr} \rightarrow \text{CH}_3\text{CH}_2\text{NH}_2 + \text{CO}_2 + \text{HBr} \]

### Step 6: Final Product
The final product of the reaction is ethylamine (CH3-CH2-NH2).

### Conclusion
Thus, the reaction of propanamide with bromine in aqueous NaOH yields ethylamine.

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