The reduction of 4 - octyne with `H_(2)` in the presence of `Pd //CaCo_3` quinoline gives

Options

A. Trans -4- octene
B. cis -4- octene
C. A mixture of cis and trans -4 octene
D. A completely reduced productc `C_(8)H_(18)`

Explanation

To solve the question regarding the reduction of 4-octyne with hydrogen in the presence of Pd/CaCO3 and quinoline, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Compound**:
- The compound we are starting with is 4-octyne, which is an alkyne with the structure:
\[ CH_3-CH_2-C≡C-CH_2-CH_2-CH_2-CH_3 \]

2. **Understand the Reaction Conditions**:
- The reduction is carried out using hydrogen gas (H2) in the presence of palladium on calcium carbonate (Pd/CaCO3) and quinoline.
- Pd/CaCO3 is a catalyst that promotes the hydrogenation of alkynes.

3. **Determine the Type of Reduction**:
- The catalyst used (Pd/CaCO3) is known to selectively produce the cis isomer when reducing alkynes.
- This is because the catalyst allows for the addition of hydrogen to occur in a syn fashion, leading to the formation of the cis isomer.

4. **Perform the Reduction**:
- When 4-octyne undergoes hydrogenation, it will be converted to a corresponding alkene.
- Specifically, the reduction of 4-octyne will yield cis-4-octene:
\[ CH_3-CH_2-CH=CH-CH_2-CH_2-CH_2-CH_3 \quad \text{(cis-4-octene)} \]

5. **Final Product Identification**:
- The final product of this reaction is therefore cis-4-octene.

### Conclusion:
The reduction of 4-octyne with H2 in the presence of Pd/CaCO3 and quinoline gives **cis-4-octene**.