Acetamide is treated separately with the following reagents. Which one of these would give methylamine ?

Options

A. `PCl_(5)`
B. `Br_(2) and NaOH`
C. `P_(4)O_(10)`
D. `LiAlH_(4)`

Answer

To determine which reagent will convert acetamide (CH3CO-NH2) to methylamine (CH3-NH2), we can analyze the reactions that acetamide undergoes with different reagents.

### Step-by-Step Solution:

1. **Identify the Structure of Acetamide**:
- Acetamide has the structure CH3CO-NH2. It contains a carbonyl group (C=O) and an amine group (NH2).

2. **Understand the Desired Product**:
- Methylamine has the structure CH3-NH2. This indicates that we need to remove the carbonyl group (C=O) and convert the acetamide into a primary amine.

3. **Recall the Hofmann Rearrangement**:
- The Hofmann rearrangement is a reaction where primary amides are converted to primary amines with the loss of one carbon atom. This reaction typically involves the use of bromine (Br2) and a strong base (like NaOH).

4. **Apply the Hofmann Rearrangement to Acetamide**:
- When acetamide (CH3CO-NH2) is treated with bromine in the presence of a strong base (NaOH), the carbonyl group is removed, and the product formed is methylamine (CH3-NH2).

5. **Evaluate the Given Reagents**:
- The question provides several reagents. We need to identify which one can facilitate the Hofmann rearrangement:
- **Option 1**: Bromine with a base (NaOH) - This will yield methylamine.
- **Option 2**: Lithium Aluminium Hydride (LiAlH4) - This is a reducing agent that would convert acetamide to ethylamine (CH3CH2-NH2), not methylamine.
- **Option 3**: Phosphorus pentoxide (P4O10) - This is a dehydrating agent and would not yield methylamine.
- **Option 4**: Phosphorus pentachloride (PCl5) - This would convert acetamide to an acyl chloride, not methylamine.

6. **Conclusion**:
- The reagent that will convert acetamide to methylamine is bromine in the presence of a strong base (NaOH), as per the Hofmann rearrangement.

### Final Answer:
The reagent that would give methylamine from acetamide is **Bromine (Br2) with a strong base (NaOH)**.