An organic compound [X]. `C_(5)H_(8)O` reacts with hydroxylamine to form [Y]. In the presence of conc. `H_(2)SO_(4)` gives `delta`-lactam. [X] neither give Benedicts test nor it respond positively towards haloform test. The compound [X] is

Options

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B.
C.
D.

Answer

To solve the problem, we need to identify the organic compound [X] with the molecular formula C5H8O that reacts with hydroxylamine and forms a delta-lactam in the presence of concentrated sulfuric acid. Let's break down the information step by step.

### Step-by-Step Solution:

1. **Determine the Degree of Unsaturation (DU)**:
- The formula for calculating the degree of unsaturation (DU) is:
\[ DU = \frac{C + 1 - H}{2} \]
- For [X], we have:
- C = 5
- H = 8
- Therefore,
\[ DU = \frac{5 + 1 - 8}{2} = \frac{-2}{2} = 2 \]
- This indicates that there are two degrees of unsaturation, which could correspond to either two double bonds, one triple bond, or a ring structure.

**Hint**: Remember that each degree of unsaturation corresponds to a ring or a double bond.

2. **Analyze the Reactivity**:
- The compound [X] reacts with hydroxylamine (NH2OH) to form [Y]. This suggests that [X] is likely a ketone or an aldehyde since they typically react with hydroxylamine to form oximes.
- The fact that [X] does not give a Benedict's test indicates that it is not an aldehyde (as aldehydes are typically oxidized by Benedict's reagent).
- Therefore, [X] is likely a ketone.

**Hint**: Aldehydes give positive results in Benedict's test, while ketones do not.

3. **Consider the Haloform Test**:
- The haloform test is positive for methyl ketones (ketones with a CH3 group adjacent to the carbonyl). Since [X] does not respond positively to this test, it cannot be a methyl ketone.
- This further supports that [X] is a ketone without a methyl group adjacent to the carbonyl.

**Hint**: The haloform test is specific for methyl ketones; if the test is negative, the structure must be different.

4. **Formation of Delta-Lactam**:
- When [X] is treated with concentrated sulfuric acid, it forms a delta-lactam. This suggests that [X] must have a structure that can undergo a rearrangement to form a cyclic amide.
- The presence of a delta-lactam indicates that the starting compound [X] likely has a cyclic structure.

**Hint**: Delta-lactams are cyclic amides formed from amino acids or cyclic ketones.

5. **Identify the Compound**:
- Given that [X] is a ketone with 5 carbons, does not give a haloform test, and can form a delta-lactam, the most likely candidate is cyclopentanone.
- Cyclopentanone has the formula C5H8O, fits the criteria of not being a methyl ketone, and can undergo the necessary reactions to form a delta-lactam.

**Hint**: Look for cyclic structures with the correct number of carbons and functional groups that match the given reactions.

### Conclusion:
The compound [X] is **cyclopentanone**.