The product formed by the reaction of acetamide with `Br_(2)` in presence of NaOH is:

Options

A. `CH_(3)CN`
B. `CH_(3)CHO`
C. `CH_(3)CH_(2)OH`
D. `CH_(3)NH_(2)`

Answer

To solve the question regarding the product formed by the reaction of acetamide with bromine in the presence of NaOH, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
The reactants are acetamide (CH₃CONH₂), bromine (Br₂), and sodium hydroxide (NaOH).

2. **Recognize the Reaction Type**:
The reaction between acetamide and bromine in the presence of NaOH is known as the Hofmann degradation reaction. This reaction typically leads to the formation of a primary amine.

3. **Write the Reaction**:
The reaction can be represented as follows:
\[ \text{CH}_3\text{CONH}_2 + \text{Br}_2 + 4 \text{NaOH} \rightarrow \text{CH}_3\text{NH}_2 + 2 \text{NaBr} + 2 \text{H}_2\text{O} + \text{Na}_2\text{CO}_3 \]
Here, acetamide reacts with bromine and sodium hydroxide to produce methyl amine (CH₃NH₂), sodium bromide (NaBr), water (H₂O), and sodium carbonate (Na₂CO₃).

4. **Identify the Main Product**:
The main product of the reaction is methyl amine (CH₃NH₂), which is a primary amine.

5. **Conclusion**:
Therefore, the product formed by the reaction of acetamide with Br₂ in the presence of NaOH is methyl amine (CH₃NH₂).

### Final Answer:
The product formed is **methyl amine (CH₃NH₂)**.

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