When acetamide is treated with `Br_(2)` and caustic soda, then we get

Options

A. Bromoacetic acid
B. Acetic acid
C. Methylamine
D. Ethane

Answer

To solve the question of what product is formed when acetamide is treated with bromine (Br2) and caustic soda (KOH), we can follow these steps:

### Step 1: Identify the Reactants
The reactants in this reaction are:
- Acetamide (CH3CONH2)
- Bromine (Br2)
- Caustic soda (KOH)

### Step 2: Understand the Reaction Type
This reaction is an example of the Hofmann degradation reaction, which involves the conversion of amides to amines. In this case, acetamide will be converted to methyl amine.

### Step 3: Write the Reaction
The reaction can be written as follows:
\[ \text{CH}_3\text{CONH}_2 + \text{Br}_2 + 4 \text{KOH} \rightarrow \text{CH}_3\text{NH}_2 + 2 \text{KBr} + \text{K}_2\text{CO}_3 + 2 \text{H}_2\text{O} \]

### Step 4: Identify the Products
From the reaction, we can see that the products formed are:
- Methyl amine (CH3NH2)
- Potassium bromide (KBr)
- Potassium carbonate (K2CO3)
- Water (H2O)

### Step 5: Compare with Given Options
The options provided are:
1. Bromoacetic acid
2. Acetic acid
3. Methyl amine
4. Ethane

From our reaction, we see that the main organic product is methyl amine (CH3NH2).

### Step 6: Conclusion
Thus, when acetamide is treated with Br2 and caustic soda, the product formed is methyl amine, which corresponds to option C.

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