The compound listed below that would be produced in greatest yield is : Methyl iodide and n-propyl iodide reacts with sodium metal to produce organic compounds

Options

A. hexyl iodide
B. sodium propane
C. n-hexane
D. n-butane

Explanation

To solve the problem, we need to analyze the reaction between methyl iodide (CH3I) and n-propyl iodide (C3H7I) with sodium metal, which is a classic example of the Wurtz reaction. The Wurtz reaction allows for the coupling of alkyl halides to form alkanes.

### Step-by-Step Solution:

1. **Identify the Reactants**:
The reactants are methyl iodide (CH3I) and n-propyl iodide (C3H7I).

2. **Understand the Wurtz Reaction**:
In the Wurtz reaction, two alkyl halides react with sodium metal to form a higher alkane and sodium halide. The general reaction can be represented as:
\[ 2 \text{R-X} + 2 \text{Na} \rightarrow \text{R-R} + 2 \text{NaX} \]
where R is an alkyl group and X is a halogen.

3. **Possible Products**:
When methyl iodide (CH3I) and n-propyl iodide (C3H7I) react, the following products can be formed:
- **n-Butane (C4H10)**: This is formed when one molecule of methyl iodide reacts with one molecule of n-propyl iodide.
- **Ethane (C2H6)**: This is formed when two molecules of methyl iodide react with each other.
- **Propane (C3H8)**: This is formed when two molecules of n-propyl iodide react with each other.

4. **Determine the Major Product**:
In this reaction, the major product will be the one formed from the combination of the two different alkyl halides, which is n-butane (C4H10). The other products (ethane and propane) are formed from the reactions of the same type of alkyl halides and will be in lesser amounts.

5. **Conclusion**:
The compound produced in the greatest yield is n-butane (C4H10).

### Final Answer:
The compound produced in greatest yield is **n-butane (C4H10)**.

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