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 question, we need to identify which compound, when reduced with zinc and dilute hydrochloric acid (HCl), yields propane (C3H8).

### Step-by-Step Solution:

1. **Understanding the Reaction**:
The reduction of alkyl halides (R-X) with zinc and dilute HCl typically results in the formation of the corresponding alkane (R-H) and hydrogen halide (HX). The general reaction can be represented as:
\[ R-X + Zn + HCl \rightarrow R-H + ZnX + H_2 \]
where R is the alkyl group and X is the halogen.

2. **Identifying the Compounds**:
We have four options:
- a) Propanol (C3H8O)
- b) Propene (C3H6)
- c) Propyne (C3H4)
- d) 1-Chloropropane (C3H7Cl)

3. **Analyzing Each Compound**:
- **Propanol**: This is an alcohol and does not contain a halogen. It cannot be reduced to propane using zinc and dilute HCl.
- **Propene**: This is an alkene and also does not contain a halogen. It cannot be reduced to propane using zinc and dilute HCl.
- **Propyne**: This is an alkyne and does not contain a halogen. It cannot be reduced to propane using zinc and dilute HCl.
- **1-Chloropropane**: This is an alkyl halide. When reduced with zinc and dilute HCl, it will yield propane.

4. **Performing the Reduction**:
For 1-chloropropane (C3H7Cl):
\[ C3H7Cl + Zn + HCl \rightarrow C3H8 + ZnCl2 \]
This reaction shows that 1-chloropropane will yield propane (C3H8) upon reduction.

5. **Conclusion**:
The compound that gives propane on reduction with zinc and dilute hydrochloric acid is **1-chloropropane**.

### Final Answer:
The correct option is **d) 1-Chloropropane**.