Acctamide reacts with `NaOBr` in alkaline medium to form:

Options

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

Answer

To solve the problem of what acetamide reacts with NaOBr in an alkaline medium to form, we will follow these steps:

### Step 1: Identify the Reactants
The reactants in this reaction are acetamide (CH3C(=O)NH2) and sodium hypobromite (NaOBr) in an alkaline medium.

### Step 2: Understand the Reaction Type
This reaction is a type of degradation reaction known as the Hofmann degradation or Hofmann bromamide reaction. In this reaction, primary amides are converted into primary amines.

### Step 3: Write the Structure of Acetamide
Acetamide can be represented as:
\[ \text{CH}_3\text{C(=O)NH}_2 \]

### Step 4: Reaction with NaOBr
When acetamide reacts with NaOBr in an alkaline medium, the carbonyl group (C=O) is converted to a primary amine. The reaction can be summarized as follows:
\[ \text{CH}_3\text{C(=O)NH}_2 + \text{NaOBr} \rightarrow \text{CH}_3\text{NH}_2 + \text{by-products} \]

### Step 5: Identify the Product
The product formed from the reaction is methylamine (CH3NH2). The carbonyl group is removed, resulting in the formation of the amine.

### Conclusion
Thus, the final product of the reaction of acetamide with NaOBr in alkaline medium is methylamine (CH3NH2).

### Final Answer
The product formed is **methylamine (CH3NH2)**.

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