Alkyl halides react with metallic sodium in dry ether producing

Options

A. alkanes with same number of carbon atoms
B. alkanes with double the number of carbon atoms
C. alkenes with triple the number of carbon atoms
D. alkenes with same number of carbon atoms.

Explanation

To solve the question regarding the reaction of alkyl halides with metallic sodium in dry ether, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reaction Type**:
- The reaction of alkyl halides with metallic sodium in dry ether is known as the Wurtz reaction.

2. **Write the General Reaction**:
- The general form of the Wurtz reaction can be represented as:
\[ 2 \, R-X + 2 \, Na \xrightarrow{\text{dry ether}} R-R + 2 \, NaX \]
- Here, \( R-X \) represents the alkyl halide, \( Na \) is the metallic sodium, and \( NaX \) is the sodium halide by-product.

3. **Analyze the Products**:
- In this reaction, two molecules of alkyl halide combine to form a new alkane. The number of carbon atoms in the product is double that of the original alkyl halides.

4. **Determine the Correct Answer**:
- Since the Wurtz reaction produces alkanes with double the number of carbon atoms from the original alkyl halides, the correct answer to the question is:
- **Alkanes with double the number of carbon atoms**.

### Conclusion:
Thus, alkyl halides react with metallic sodium in dry ether to produce **alkanes with double the number of carbon atoms**.

---