A mixture of ethyl bromide and methyl bromide is subjected to Wurtz reaction. The mixture of alkanes so formed, consists of

Options

A. propane and butane
B. ethane and propane
C. ethane, propane and butane
D. ethane and butane

Explanation

To solve the problem of determining the mixture of alkanes formed when a mixture of ethyl bromide and methyl bromide is subjected to the Wurtz reaction, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
- Ethyl bromide (C2H5Br) and methyl bromide (CH3Br) are the two alkyl halides involved in the Wurtz reaction.

2. **Understand the Wurtz Reaction**:
- The Wurtz reaction involves the coupling of two alkyl halides in the presence of sodium metal (Na) and dry ether. This results in the formation of alkanes.

3. **Possible Combinations**:
- When two different alkyl halides are used, we can form several combinations:
- Two ethyl bromides (C2H5Br + C2H5Br)
- One ethyl bromide and one methyl bromide (C2H5Br + CH3Br)
- Two methyl bromides (CH3Br + CH3Br)

4. **Calculate the Products**:
- **From two ethyl bromides**:
- C2H5 + C2H5 → C4H10 (Butane)
- **From one ethyl bromide and one methyl bromide**:
- C2H5 + CH3 → C3H8 (Propane)
- **From two methyl bromides**:
- CH3 + CH3 → C2H6 (Ethane)

5. **List the Products**:
- The resulting alkanes from the Wurtz reaction will be:
- Butane (C4H10)
- Propane (C3H8)
- Ethane (C2H6)

6. **Conclusion**:
- Therefore, the mixture of alkanes formed consists of ethane, propane, and butane.

### Final Answer:
The mixture of alkanes formed consists of ethane (C2H6), propane (C3H8), and butane (C4H10).

---