n - propyl chloride reacts with sodium metal in dry ether to give

Options

A. `CH_(3)CH_(2)CH_(2)CH_(2)CH_(2)CH_(3)`
B. `CH_(3)CH_(2)CH_(3)`
C. `CH_(3)CH_(2)CH_(2)CH_(3)`
D. `CH_(3)CH_(2)CH_(2)CH_(2)CH_(2)CH_(3)`

Answer

To solve the question regarding the reaction of n-propyl chloride with sodium metal in dry ether, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
- The reactant is n-propyl chloride, which has the chemical formula CH₃CH₂CH₂Cl (or C₃H₇Cl).
- Sodium metal (Na) is the other reactant, and dry ether is the solvent used in this reaction.

2. **Understand the Reaction Type**:
- This reaction is known as the Wurtz reaction. It involves the coupling of alkyl halides in the presence of sodium metal to form alkanes.

3. **Write the Reaction Equation**:
- The reaction involves two moles of n-propyl chloride and two moles of sodium:
\[ 2 \text{CH}_3\text{CH}_2\text{CH}_2\text{Cl} + 2 \text{Na} \xrightarrow{\text{dry ether}} \text{C}_6\text{H}_{14} + 2 \text{NaCl} \]
- Here, sodium replaces the chlorine atoms in the alkyl halide, leading to the formation of an alkane.

4. **Determine the Product**:
- The product formed is hexane (C₆H₁₄). This is because two n-propyl groups (C₃H₇) combine to form a six-carbon alkane.

5. **Final Answer**:
- The final product of the reaction is hexane.