`C_2H_5Ioverset(Na)underset("ether")rarrX.` Here X is :

Options

A. Methane
B. Methane iodide
C. Ethane
D. Butane

Explanation

To solve the question `C_2H_5I + Na (in ether) → X`, we need to identify the product X formed when ethyl iodide (C2H5I) reacts with sodium in the presence of ether. This reaction is known as the Wurtz reaction.

### Step-by-Step Solution:

1. **Identify the Reactants**:
- The reactant is ethyl iodide (C2H5I).
- Sodium (Na) is used as a reducing agent.
- Ether is used as a solvent.

2. **Understand the Wurtz Reaction**:
- The Wurtz reaction involves the coupling of alkyl halides in the presence of sodium to form higher alkanes.
- The general reaction can be represented as:
\[ 2 R-X + 2 Na \rightarrow R-R + 2 NaX \]
- Here, R is the alkyl group and X is the halogen.

3. **Apply the Wurtz Reaction to C2H5I**:
- In this case, R is C2H5 (ethyl group) and X is I (iodine).
- When ethyl iodide reacts with sodium, two ethyl groups will couple together:
\[ 2 C_2H_5I + 2 Na \rightarrow C_4H_{10} + 2 NaI \]
- This means that the product X will be butane (C4H10).

4. **Write the Final Answer**:
- The product X formed from the reaction of ethyl iodide with sodium in ether is **butane (C4H10)**.

### Final Answer:
X = C4H10 (butane)