A compound (A) `C_(5)H_(10) Cl_(2)` on hydrolysis gives `C_(5)H_(10)O` which reacts with `NH_(2)OH`, forms iodoform but does not give Fehling test (A) is :

Options

A. `H_(3)C-overset(CI)overset(|)underset(CI)underset(|)C-CH_(2)CH_(2)CH_(3)`
B. `CH_(3)CH_(2)- overset(CI) overset(|) underset(CI)underset(|)C- CH_(2)CH_(3) `
C. `overset(CI)overset(|) underset(CI)underset(|) CHCH_(2) CH_(2)CH_(2)CH_(3)`
D. `CH_(3)CH_(2) overset(CI)overset(|)CH- overset(CI)overset(|)CHCH_(3)`

Answer

To solve the problem, we need to identify the compound (A) with the molecular formula \( C_5H_{10}Cl_2 \) that, upon hydrolysis, yields \( C_5H_{10}O \). This compound also reacts with hydroxylamine (\( NH_2OH \)) to form iodoform but does not give a positive Fehling's test.

### Step-by-Step Solution:

1. **Identify the Compound (A)**:
The molecular formula of compound (A) is \( C_5H_{10}Cl_2 \). This indicates that it is a dichloro compound with five carbon atoms.

2. **Hydrolysis Reaction**:
When compound (A) undergoes hydrolysis, it produces \( C_5H_{10}O \). This suggests that the hydrolysis replaces the chlorine atoms with hydroxyl groups, forming an alcohol or a ketone.

3. **Determine the Structure of \( C_5H_{10}O \)**:
The product \( C_5H_{10}O \) must contain a carbonyl group (C=O) for it to react with hydroxylamine to form an iodoform. The structure must also not be an aldehyde or a compound that would give a positive Fehling's test.

4. **Check the Options**:
We need to analyze the given options to see which one fits the criteria:
- **Option A**: \( CH_3CCl_2CH_2CH_2CH_3 \)
- **Option B**: \( CH_3CH_2CCl_2CH_2CH_3 \)
- **Option C**: \( CCl_2HCH_2CH_2CH_3 \)
- **Option D**: \( CH_3CH_2CHClCHClCH_3 \)

5. **Analyze Each Option**:
- **Option A**: Hydrolysis gives \( CH_3C(OH)CH_2CH_2CH_3 \) which can form \( CH_3C(=O)CH_2CH_2CH_3 \) (a ketone) and can react with \( NH_2OH \) to form iodoform. It does not give a Fehling's test.
- **Option B**: Hydrolysis gives a product that cannot form iodoform and does not fit the criteria.
- **Option C**: Hydrolysis gives a product that would yield a positive Fehling's test (indicating an aldehyde).
- **Option D**: Hydrolysis gives a product that does not fit the criteria.

6. **Conclusion**:
The only option that meets all the criteria (produces \( C_5H_{10}O \), reacts with \( NH_2OH \) to form iodoform, and does not give a Fehling's test) is **Option A**.

### Final Answer:
**A is the correct answer: \( CH_3CCl_2CH_2CH_2CH_3 \)**