2-Chloropropane with sodium and dry ether gives:

Options

A. 2,3-Dimethyl butane
B. n-Hexane
C. Propene
D. 1-Hexene

Answer

To solve the question regarding the reaction of 2-chloropropane with sodium in dry ether, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
The reactant is 2-chloropropane, which has the molecular formula C3H7Cl. It consists of three carbon atoms, with a chlorine atom attached to the second carbon.

2. **Understand the Reaction Conditions**:
The reaction is carried out in the presence of sodium (Na) and dry ether. This indicates that we are likely dealing with a Wurtz reaction, which is a coupling reaction of alkyl halides.

3. **Write the Reaction**:
The general reaction for the Wurtz reaction can be represented as:
\[ 2 \text{R-X} + 2 \text{Na} \rightarrow \text{R-R} + 2 \text{NaX} \]
Here, R represents the alkyl group (in this case, the propyl group), and X is the halogen (Cl in this case).

4. **Determine the Product**:
In the case of 2-chloropropane, when two molecules of 2-chloropropane react with sodium, the chlorine atoms are replaced, and a new carbon-carbon bond is formed between the two propyl groups. The product formed is 2,3-dimethylbutane.

5. **Write the Structural Formula of the Product**:
The structural formula of 2,3-dimethylbutane can be represented as:
\[ \text{CH}_3-\text{C}(\text{CH}_3)-\text{C}(\text{CH}_3)-\text{CH}_3 \]
This shows that there are four carbon atoms in total, with two methyl groups attached to the second and third carbon.

6. **IUPAC Naming**:
The IUPAC name for the product is 2,3-dimethylbutane, as it has a four-carbon chain (butane) with two methyl groups on the second and third carbons.

### Final Answer:
The reaction of 2-chloropropane with sodium in dry ether gives **2,3-dimethylbutane**.

---