Which of these carbonyl compounds on reduction with Zn-Hg/HCl will give the same compound as product

Options

A.
B.
C.
D.

Explanation

To solve the problem of which carbonyl compounds will yield the same product upon reduction with Zn-Hg/HCl, we can follow these steps:

### Step 1: Identify the Carbonyl Compounds
We have four carbonyl compounds to analyze. Let's denote them as:
- **Compound A**: HC=O (an aldehyde)
- **Compound B**: C6H5C(=O)CH(CH3)2 (a ketone)
- **Compound C**: C6H5C(=O)H (another aldehyde)
- **Compound D**: C6H5C(=O)C(CH3)3 (a different ketone)

### Step 2: Understand the Reaction
The reduction of carbonyl compounds using Zn-Hg in the presence of HCl is known as Clemenson reduction. This reaction converts carbonyl groups (C=O) into methylene groups (CH2).

### Step 3: Perform the Reduction for Each Compound
1. **For Compound A (HC=O)**:
- Reduction: HC=O → CH2
- Product: CH2

2. **For Compound B (C6H5C(=O)CH(CH3)2)**:
- Reduction: C6H5C(=O) → C6H5CH(CH3)2
- Product: C6H5CH(CH3)2 (which can be simplified to CH2)

3. **For Compound C (C6H5C(=O)H)**:
- Reduction: C6H5C(=O) → C6H5CH2
- Product: C6H5CH2 (which can also be simplified to CH2)

4. **For Compound D (C6H5C(=O)C(CH3)3)**:
- Reduction: C6H5C(=O) → C6H5C(CH3)3
- Product: C6H5C(CH3)3 (which does not simplify to CH2)

### Step 4: Compare the Products
- **Products from A, B, and C**: All yield a structure that can be represented as CH2 (or similar structures that can be simplified to CH2).
- **Product from D**: Yields a different structure that does not match the others.

### Conclusion
The carbonyl compounds that yield the same product upon reduction with Zn-Hg/HCl are **A, B, and C**. Compound D does not yield the same product.

### Final Answer
**A, B, and C will give the same product upon reduction, while D will not.**