A mixture of ethyl iodide and n - propyl iodide is treated with sodium metal in presence of ethoxyethane . The hydrocarbon which is not formed is

Options

A. butane
B. propane
C. pentane
D. hexane

Answer

To solve the problem, we will analyze the reaction of ethyl iodide and n-propyl iodide with sodium metal in the presence of ethoxyethane, which is a classic example of the Wurtz reaction.

### Step-by-Step Solution:

1. **Identify the Reactants**:
- The reactants are ethyl iodide (C2H5I) and n-propyl iodide (C3H7I).

2. **Understand the Wurtz Reaction**:
- The Wurtz reaction involves the coupling of two alkyl halides in the presence of sodium metal and an ether solvent. This can lead to the formation of various alkanes.

3. **Possible Coupling Products**:
- When two different alkyl halides are used, the possible products can be:
- R1-R1 (coupling of two molecules of the first alkyl halide)
- R2-R2 (coupling of two molecules of the second alkyl halide)
- R1-R2 (coupling of one molecule of each alkyl halide)

4. **Determine the Possible Products**:
- From ethyl iodide (C2H5I):
- Coupling with itself: C2H5-C2H5 → Butane (C4H10)
- From n-propyl iodide (C3H7I):
- Coupling with itself: C3H7-C3H7 → Hexane (C6H14)
- Coupling between ethyl and n-propyl iodide:
- C2H5-C3H7 → Pentane (C5H12)

5. **List the Products**:
- The possible products from the reaction are:
- Butane (C4H10)
- Hexane (C6H14)
- Pentane (C5H12)

6. **Identify the Hydrocarbon Not Formed**:
- Among the products listed, we need to identify which hydrocarbon is not formed:
- Butane: Formed
- Hexane: Formed
- Pentane: Formed
- Propane (C3H8): Not formed

### Conclusion:
The hydrocarbon that is not formed in this reaction is **Propane (C3H8)**.