`CH_3Br` can be prepared by

Options

A. `CH_3COOAg+Br_2 overset Delta rarr`
B. `Ch_3COOH overset(P//Br_2) rarr overset (Sodallme.Delta) rarr`
C. Both (1) and (2)
D. None of the above

Answer

To prepare CH₃Br (methyl bromide), we can consider two main reactions that can yield this alkyl halide. Let's break down the steps for each reaction.

### Step-by-Step Solution:

1. **Identify the Compound**:
- CH₃Br is an alkyl halide, specifically methyl bromide, where CH₃ is the methyl group and Br is the bromine atom.

2. **Consider the First Reaction**:
- The first option is CH₃CoAg + Br₂. This is known as the Hans Dekker reaction.
- In this reaction, CH₃CoAg (methylcopper) reacts with bromine (Br₂).
- The reaction is a step-down reaction, meaning that the carbon count in the substrate decreases.

3. **Reaction Mechanism**:
- The reaction proceeds as follows:
\[ \text{CH}_3\text{CoAg} + \text{Br}_2 \rightarrow \text{CH}_3\text{Br} + \text{CO}_2 + \text{Ag} \]
- Here, the methyl group (CH₃) is retained, and bromine is added, resulting in the formation of CH₃Br, while carbon dioxide (CO₂) and silver (Ag) are released.

4. **Consider the Second Reaction**:
- The second option is CH₃CoH + PBr₂. This is known as the HVZ (Hunsdiecker-Van Duyne) reaction.
- In this reaction, CH₃CoH (methylcobalt) reacts with phosphorus tribromide (PBr₂).

5. **Reaction Mechanism**:
- The reaction proceeds as follows:
\[ \text{CH}_3\text{CoH} + \text{PBr}_2 \rightarrow \text{CH}_3\text{Br} + \text{COOH} + \text{PBr}_3 \]
- In this case, the methyl group (CH₃) is retained, and bromine is added, resulting in the formation of CH₃Br.

6. **Conclusion**:
- Both reactions lead to the formation of CH₃Br. Therefore, both options are valid methods to prepare CH₃Br.

### Final Answer:
CH₃Br can be prepared by both the Hans Dekker reaction (CH₃CoAg + Br₂) and the HVZ reaction (CH₃CoH + PBr₂).