Why is Wurtz reaction not preferred for the preparation of alkanes containing odd number of carbon atoms? Illustrate your answer by taking one example.

Explanation

### Step-by-Step Solution:

1. **Understanding Wurtz Reaction**:
The Wurtz reaction involves the coupling of two alkyl halides in the presence of sodium metal (Na) and dry ether to form alkanes. The general reaction can be represented as:
\[ 2 R-X + 2 Na \rightarrow R-R + 2 NaX \]
where \( R \) is the alkyl group and \( X \) is a halogen.

2. **Formation of Odd and Even Carbon Alkanes**:
When two different alkyl halides are used, the reaction can lead to the formation of multiple products, including alkanes with both even and odd numbers of carbon atoms. This is due to the possibility of different combinations of the alkyl groups.

3. **Example with Chloroethane and Chloropropane**:
Let's consider the reaction between 1-chloroethane (C2H5Cl) and 1-chloropropane (C3H7Cl):
\[ \text{1-chloroethane} + \text{1-chloropropane} + 2 \text{Na} \rightarrow \text{Products} \]
The possible products from this reaction include:
- Hexane (C6H14) from the coupling of two ethyl groups.
- Pentane (C5H12) from the coupling of one ethyl and one propyl group.
- Butane (C4H10) from the coupling of two propyl groups.

4. **Resulting Products**:
Thus, the products formed are:
- C6H14 (Hexane) - even number of carbons
- C5H12 (Pentane) - odd number of carbons
- C4H10 (Butane) - even number of carbons

5. **Conclusion**:
The Wurtz reaction is not preferred for the preparation of alkanes with an odd number of carbon atoms because it leads to a mixture of products, making it difficult to isolate the desired alkane. The presence of multiple products (both odd and even) complicates the reaction outcome.