In Wurtz reaction , if we take `CH_(3)Cl and C_(2)H_(5)Cl` then product , will be

Options

A. propane + ethane
B. propane
C. propane + ethane + butane
D. propane + butane

Explanation

To solve the question regarding the products formed in the Wurtz reaction using `CH₃Cl` (methyl chloride) and `C₂H₅Cl` (ethyl chloride), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reaction Conditions**:
The Wurtz reaction involves the reaction of alkyl halides with sodium in the presence of dry ether. Here, we are using `CH₃Cl` and `C₂H₅Cl`.

2. **Understand the Mechanism**:
In the presence of sodium, the alkyl halides undergo homolytic cleavage, leading to the formation of free radicals. Sodium donates an electron to the halide, resulting in the formation of alkyl radicals:
- From `CH₃Cl`, we get the methyl radical `CH₃·`.
- From `C₂H₅Cl`, we get the ethyl radical `C₂H₅·`.

3. **Formation of Radicals**:
The radicals formed are:
- `CH₃·` (methyl radical)
- `C₂H₅·` (ethyl radical)

4. **Radical Combination**:
The radicals can combine in different ways:
- Two methyl radicals can combine to form ethane (`C₂H₆`):
\[ CH₃· + CH₃· \rightarrow C₂H₆ \]
- Two ethyl radicals can combine to form butane (`C₄H₁₀`):
\[ C₂H₅· + C₂H₅· \rightarrow C₄H₁₀ \]
- A methyl radical and an ethyl radical can combine to form propane (`C₃H₈`):
\[ CH₃· + C₂H₅· \rightarrow C₃H₈ \]

5. **Conclusion**:
Therefore, the products formed from the Wurtz reaction of `CH₃Cl` and `C₂H₅Cl` in the presence of sodium and dry ether will be:
- Ethane (`C₂H₆`)
- Propane (`C₃H₈`)
- Butane (`C₄H₁₀`)

### Final Answer:
The products will be ethane, propane, and butane.