Which of the following compound forms two stereoisomers when treated with `NH_(2)OH` in weakly acidic medium?

Options

A. `CH_(3)COCH_(3)`
B. `HCHO`
C. `CH_(3)COCH_(2)H_(5)`
D. `C_(2)H_(5)COCH_(2)CH_(3)`

Explanation

To determine which compound forms two stereoisomers when treated with hydroxylamine (NH2OH) in a weakly acidic medium, we need to analyze the reaction of carbonyl compounds (aldehydes or ketones) with hydroxylamine. The key to forming stereoisomers lies in the presence of different substituents on the carbonyl carbon.

### Step-by-Step Solution:

1. **Understand the Reaction**:
- When a carbonyl compound (C=O) reacts with hydroxylamine, it undergoes a condensation reaction to form an oxime. The general reaction is:
\[ R_1C(=O)R_2 + NH_2OH \rightarrow R_1C(=N-OH)R_2 + H_2O \]
- Here, \(R_1\) and \(R_2\) are the substituents on the carbonyl carbon.

2. **Identify the Conditions for Stereoisomer Formation**:
- For stereoisomers to form, the carbon attached to the nitrogen must have two different groups. This is because stereoisomers arise from different spatial arrangements of groups around a chiral center.

3. **Analyze the Options**:
- **Option A**: Two identical groups (e.g., CH3-CO-CH3)
- This will yield only one product, as both substituents are the same.
- **Option B**: Two identical groups (e.g., H-CHO)
- Similar to Option A, this will also yield only one product.
- **Option C**: Two different groups (e.g., CH3-CO-CH2CH3)
- This will yield two stereoisomers because the carbonyl carbon has two different substituents (CH3 and CH2CH3).
- **Option D**: Two identical groups (e.g., C2H5-CO-C2H5)
- This will yield only one product, as both substituents are the same.

4. **Conclusion**:
- The only option that forms two stereoisomers is **Option C** (with CH3 and CH2CH3 as substituents). This is because it has a chiral center after the reaction with hydroxylamine, allowing for the formation of geometrical isomers (cis and trans).

### Final Answer:
The compound that forms two stereoisomers when treated with NH2OH in a weakly acidic medium is **Option C**.