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 problem of what n-propyl chloride reacts with sodium metal in dry ether to produce, we can follow these steps:

### Step 1: Identify the Reactants
The reactant is n-propyl chloride, which has the chemical formula CH3CH2CH2Cl. This is a primary alkyl halide.

### Step 2: Understand the Reaction Conditions
The reaction is conducted in the presence of sodium metal (Na) and dry ether. This setup is characteristic of the Wurtz reaction, which is a coupling reaction that forms alkanes.

### Step 3: Write the Reaction Mechanism
In the Wurtz reaction, sodium metal reacts with the alkyl halide to form free radicals. The reaction can be summarized as follows:
1. Sodium donates an electron to the alkyl halide, generating a radical and sodium halide.
2. Two alkyl radicals can then couple together to form a new alkane.

### Step 4: Determine the Product
For n-propyl chloride, the reaction can be represented as:
\[ 2 \text{CH}_3\text{CH}_2\text{CH}_2\text{Cl} + 2 \text{Na} \rightarrow \text{C}_6\text{H}_{14} + 2 \text{NaCl} \]
This indicates that two molecules of n-propyl chloride will couple to form hexane (C6H14).

### Step 5: Conclusion
Thus, the product of the reaction of n-propyl chloride with sodium metal in dry ether is hexane.

### Final Answer
The reaction of n-propyl chloride with sodium metal in dry ether gives hexane (C6H14).

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