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

A.
B.
C.
D.

Answer

To solve the problem, we need to identify the organic compound [X] with the molecular formula \( C_5H_8O \) based on the given reactions and tests. Let's break it down step by step.

### Step 1: Analyze the Molecular Formula
The compound [X] has the molecular formula \( C_5H_8O \). This indicates that it could be an unsaturated compound, possibly an aldehyde or ketone due to the presence of a carbonyl group (C=O).

**Hint:** The degree of unsaturation can be calculated using the formula: \( \text{Degree of Unsaturation} = \frac{(2C + 2 - H)}{2} \).

### Step 2: Reactivity with Hydroxylamine
The problem states that [X] reacts with hydroxylamine (NH2OH) to form [Y]. This indicates that [X] contains a carbonyl group (either an aldehyde or a ketone) since hydroxylamine forms oximes with carbonyl compounds.

**Hint:** Hydroxylamine reacts with carbonyl compounds to form oximes, which are characterized by the functional group -C=N-OH.

### Step 3: Formation of Delta-Lactam
In the presence of concentrated sulfuric acid (H2SO4), [X] gives a delta-lactam. A lactam is a cyclic amide, and a delta-lactam specifically refers to a lactam where the nitrogen atom is four carbons away from the carbonyl carbon.

**Hint:** A delta-lactam can be formed from a compound that has a suitable structure to allow for cyclization, typically involving a carbonyl group and an amine.

### Step 4: Tests for Aldehyde and Ketone
The compound [X] does not give a positive Benedict's test, which means it is not an aldehyde that can be oxidized to a carboxylic acid. Additionally, it does not respond positively to the haloform test, indicating that it is not a methyl ketone (which would have a structure like CH3CO-).

**Hint:** The absence of a positive haloform test suggests that there is no methyl ketone structure (CH3CO-) present in [X].

### Step 5: Identify Possible Structures
Given the molecular formula \( C_5H_8O \) and the information from the tests, we can deduce that [X] is likely a cyclic compound since it can form a lactam. The options provided in the question likely include cyclopentanone or cyclobutanone derivatives.

**Hint:** Consider the structure of cyclopentanone and cyclobutanone to see which one fits the criteria of not giving the tests mentioned.

### Step 6: Conclusion
After analyzing the possible structures, we find that cyclopentanone (which has a carbonyl group but does not fit the criteria for the tests mentioned) is the most plausible candidate for [X].

Thus, the compound [X] is **cyclopentanone**.

### Final Answer
The compound [X] is **cyclopentanone**.

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