Which one is not prepared by Wurtz reaction ?

Options

A. `C_(2)H_(6)`
B. `n-C_(4)H_(10)`
C. `CH_(4)`
D. `CH_(3)-overset(CH_(3))overset(|)CH-overset(CH_(3))overset(|)CH-CH_(3)`

Answer

To determine which compound cannot be prepared by the Wurtz reaction, we first need to understand the Wurtz reaction itself.

### Step-by-Step Solution:

1. **Understanding the Wurtz Reaction**:
The Wurtz reaction involves the coupling of alkyl halides in the presence of sodium metal and an ether solvent. The general reaction can be represented as:
\[ 2 \text{R-X} + 2 \text{Na} \xrightarrow{\text{ether}} \text{R-R} + 2 \text{NaX} \]
where R is an alkyl group and X is a halogen (like Cl, Br, or I).

2. **Mechanism of the Reaction**:
- Sodium donates an electron to the halogen atom in the alkyl halide, resulting in the formation of a free radical.
- Two alkyl radicals then combine to form an alkane.

3. **Odd vs. Even Number of Carbon Atoms**:
- The Wurtz reaction typically produces alkanes with an even number of carbon atoms. This is because two alkyl halide molecules are combined.
- If we start with alkyl halides that have an odd number of carbon atoms, the resulting product will still have an even number of carbon atoms.

4. **Identifying the Compound**:
- If we are given four compounds, we need to identify which one has an odd number of carbon atoms.
- For example, if the compounds are:
1. C2H6 (Ethane) - 2 carbon atoms (even)
2. C3H8 (Propane) - 3 carbon atoms (odd)
3. C4H10 (Butane) - 4 carbon atoms (even)
4. C6H14 (Hexane) - 6 carbon atoms (even)

5. **Conclusion**:
- From the example compounds, propane (C3H8) cannot be prepared by the Wurtz reaction because it has an odd number of carbon atoms. Therefore, it is the correct answer to the question.

### Final Answer:
The compound that cannot be prepared by the Wurtz reaction is **C3H8 (Propane)**.