The product of the chemical reaction,
`CH_(3)-overset(O)overset(||)C-NH_(2)+Br_(2)+4NaOH to` ? Will be

Options

A. `CH_(4)`
B. `CH_(3)NBr`
C. `CH_(3)Obr`
D. `CH_(3)NH_(2)`

Explanation

To solve the problem, we need to analyze the reaction step by step. The reaction given is:

\[ \text{CH}_3\text{C(=O)NH}_2 + \text{Br}_2 + 4\text{NaOH} \]

### Step 1: Identify the Reactants
The reactants are:
- Acetamide (\( \text{CH}_3\text{C(=O)NH}_2 \))
- Bromine (\( \text{Br}_2 \))
- Sodium Hydroxide (\( \text{NaOH} \))

### Step 2: Mechanism of Reaction
1. **Deprotonation**: The hydroxide ion (\( \text{OH}^- \)) from sodium hydroxide acts as a base and abstracts a proton from the amide group of acetamide. This generates a negatively charged nitrogen species.

\[ \text{CH}_3\text{C(=O)NH}_2 + \text{OH}^- \rightarrow \text{CH}_3\text{C(=O)N}^- + \text{H}_2\text{O} \]

2. **Bromination**: The negatively charged nitrogen then attacks one of the bromine atoms in \( \text{Br}_2 \), leading to the formation of a brominated intermediate.

\[ \text{CH}_3\text{C(=O)N}^- + \text{Br}_2 \rightarrow \text{CH}_3\text{C(=O)NBr} + \text{Br}^- \]

3. **Further Deprotonation**: The base (sodium hydroxide) abstracts another proton from the nitrogen, resulting in a more stable intermediate.

\[ \text{CH}_3\text{C(=O)NBr} + \text{OH}^- \rightarrow \text{CH}_3\text{C(=O)N}^- + \text{Br}^- + \text{H}_2\text{O} \]

4. **Formation of Product**: The negatively charged nitrogen can now undergo a rearrangement where a methyl group from the carbon chain migrates to the nitrogen, forming a double bond between carbon and nitrogen.

\[ \text{CH}_3\text{C(=O)N}^- \rightarrow \text{CH}_3\text{N}H + \text{CO}_2 \]

### Step 3: Final Products
The final products of the reaction are:
- Methylamine (\( \text{CH}_3\text{NH}_2 \))
- Sodium bromide (\( \text{NaBr} \))
- Water (\( \text{H}_2\text{O} \))
- Carbon dioxide (\( \text{CO}_2 \))

### Step 4: Summary of the Reaction
The overall reaction can be summarized as:

\[ \text{CH}_3\text{C(=O)NH}_2 + \text{Br}_2 + 4\text{NaOH} \rightarrow \text{CH}_3\text{NH}_2 + 2\text{NaBr} + \text{Na}_2\text{CO}_3 + 2\text{H}_2\text{O} \]

### Final Answer
The main product of the reaction is **Methylamine (\( \text{CH}_3\text{NH}_2 \))**.

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