Which hydrocarbon is formed by the action of sodium on ethyl iodide?

Options

A. Methane
B. Ethane
C. Ethylene
D. n-Butane

Answer

To determine which hydrocarbon is formed by the action of sodium on ethyl iodide, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**: The reactants in this reaction are ethyl iodide (C2H5I) and sodium (Na). Ethyl iodide is an alkyl halide.

2. **Understand the Reaction Type**: The reaction between an alkyl halide and sodium is known as the Wurtz reaction. This reaction typically involves the coupling of two alkyl halides in the presence of sodium metal to form a higher alkane.

3. **Write the Reaction**: In the Wurtz reaction, two equivalents of ethyl iodide react with two equivalents of sodium:
\[ 2 \text{C}_2\text{H}_5\text{I} + 2 \text{Na} \rightarrow \text{C}_4\text{H}_{10} + 2 \text{NaI} \]
Here, C4H10 is the hydrocarbon formed, and NaI is the byproduct.

4. **Determine the Product**: The product of this reaction is n-butane (C4H10). The structure of n-butane can be represented as:
\[ \text{CH}_3-\text{CH}_2-\text{CH}_2-\text{CH}_3 \]

5. **Conclusion**: Therefore, the hydrocarbon formed by the action of sodium on ethyl iodide is n-butane.

### Final Answer:
The hydrocarbon formed is **n-butane (C4H10)**.

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