Reaction of ethyl chloride with sodium in presence of dry ether leads to

Options

A. Ethane
B. Propane
C. n-butane
D. n-pentane

Explanation

To solve the question regarding the reaction of ethyl chloride with sodium in the presence of dry ether, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants:**
- Ethyl chloride (C2H5Cl) is an alkyl halide.
- Sodium (Na) is a metal that will participate in the reaction.
- Dry ether is used as a solvent to prevent any unwanted reactions with sodium.

2. **Understand the Reaction Type:**
- The reaction is a type of coupling reaction known as the Wurtz reaction (or Wood's reaction), which typically involves the coupling of two alkyl halides in the presence of sodium to form a higher alkane.

3. **Write the Reaction:**
- The general reaction can be represented as:
\[ 2 \text{C}_2\text{H}_5\text{Cl} + 2 \text{Na} \xrightarrow{\text{dry ether}} \text{C}_4\text{H}_{10} + 2 \text{NaCl} \]
- Here, two moles of ethyl chloride react with two moles of sodium to produce n-butane (C4H10) and sodium chloride (NaCl).

4. **Identify the Product:**
- The product formed is n-butane (C4H10), which is an alkane with four carbon atoms.

5. **Conclusion:**
- Therefore, the reaction of ethyl chloride with sodium in the presence of dry ether leads to the formation of n-butane and sodium chloride.

### Final Answer:
The reaction of ethyl chloride with sodium in the presence of dry ether leads to the formation of **n-butane (C4H10)** and sodium chloride (NaCl).

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