The compound which gives propane on reduction with Zn and dilute hydrochloric acid is

Options

A. `CH_3CH_2CH_2OH`
B. `CH_3CH=CH_2`
C. `CH - C -= CH`
D. `CH_3CH_2CH_2 - Cl`

Answer

To solve the problem, we need to identify the compound that, when reduced with zinc (Zn) and dilute hydrochloric acid (HCl), yields propane (C3H8).

### Step-by-Step Solution:

1. **Understanding the Reaction**:
The reaction involves the reduction of an alkyl halide using zinc and dilute hydrochloric acid. The general reaction can be represented as:
\[ \text{R-X} + \text{Zn} + \text{HCl} \rightarrow \text{R-H} + \text{ZnX}_2 \]
where R is an alkyl group and X is a halogen (like Cl, Br, etc.).

2. **Identifying the Product**:
We want the product to be propane (C3H8). This means that the alkyl group (R) must correspond to a three-carbon alkyl group.

3. **Choosing the Alkyl Halide**:
The simplest alkyl halide that can yield propane upon reduction is 1-chloropropane (C3H7Cl). When 1-chloropropane is reduced, it will yield propane:
\[ \text{C3H7Cl} + \text{Zn} + \text{HCl} \rightarrow \text{C3H8} + \text{ZnCl}_2 \]

4. **Conclusion**:
Therefore, the compound that gives propane on reduction with zinc and dilute hydrochloric acid is **1-chloropropane**.

### Final Answer:
The compound which gives propane on reduction with Zn and dilute hydrochloric acid is **1-chloropropane**.