The coupling between`C_2 H_5 MgBr` and `MeBr` gives propane in the presence of :

Options

A. `MeOTs`
B. `EtOTs`
C. `AgBr`
D. `CuCl_2`

Answer

To solve the problem, we need to understand the coupling reaction between ethyl magnesium bromide (C2H5MgBr) and methyl bromide (MeBr) that produces propane (C3H8). The question asks for the conditions under which this reaction occurs.

### Step-by-Step Solution:

1. **Understanding Grignard Reagents**:
- Grignard reagents, such as C2H5MgBr, are organomagnesium compounds that can react with alkyl halides to form new carbon-carbon bonds.

2. **Identifying the Reaction**:
- The coupling reaction can be represented as:
\[ \text{C}_2\text{H}_5\text{MgBr} + \text{MeBr} \rightarrow \text{C}_3\text{H}_8 + \text{MgBr}_2 \]
- Here, the ethyl group (C2H5) from C2H5MgBr couples with the methyl group (CH3) from MeBr to form propane (C3H8).

3. **Role of the Medium**:
- The reaction requires a suitable medium that allows the coupling to occur. Commonly used mediums include copper(I) halides, such as CuCl or CuBr, which facilitate the coupling reaction.

4. **Exclusion of Certain Reagents**:
- The question mentions several options, and we need to determine which one does not allow the coupling to occur.
- It is noted that AgBr (silver bromide) does not facilitate this reaction because it forms a yellow precipitate and inhibits the coupling process.

5. **Conclusion**:
- Therefore, the coupling between C2H5MgBr and MeBr gives propane in the presence of copper(I) halides or other suitable conditions, but not in the presence of AgBr.

### Final Answer:
The coupling between C2H5MgBr and MeBr gives propane in the presence of copper(I) halides (like CuCl or CuBr) but not in the presence of AgBr.

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