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 need to analyze the reaction of a mixture of ethyl iodide (C2H5I) and n-propyl iodide (C3H7I) with sodium metal in the presence of ethoxyethane (an ether). This reaction follows the Wurtz reaction mechanism, which is used to couple alkyl halides to form higher alkanes.

### Step-by-Step Solution:

1. **Identify the Reactants**:
- Ethyl iodide (C2H5I)
- 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 ether.
- The general reaction can be represented as:
\[ 2 R-X + 2 Na \rightarrow R-R + 2 NaX \]
- Here, R represents the alkyl group and X represents the halogen.

3. **Possible Products from the Reaction**:
- When ethyl iodide reacts with sodium, it can form:
- Ethane (C2H6) from two ethyl iodide molecules.
- When n-propyl iodide reacts with sodium, it can form:
- Propane (C3H8) from two n-propyl iodide molecules.
- When ethyl iodide and n-propyl iodide react together, they can form:
- Butane (C4H10) from one ethyl iodide and one n-propyl iodide.
- Pentane (C5H12) from one ethyl iodide and one n-propyl iodide.
- Hexane (C6H14) from two n-propyl iodide molecules.

4. **List of Possible Hydrocarbons**:
- Ethane (C2H6)
- Propane (C3H8)
- Butane (C4H10)
- Pentane (C5H12)
- Hexane (C6H14)

5. **Identify the Hydrocarbon Not Formed**:
- From the list of possible products, we see that propane (C3H8) cannot be formed from the reaction of ethyl iodide and n-propyl iodide in the Wurtz reaction.
- The reason is that propane requires two identical n-propyl iodide molecules, which are not present in the mixture.

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