A compound (X) with a molecualr formula `C_(5)H_(10)O` gives a positive 2, 4-DNP test but a negative Tollens' test. On oxidation it gives a carboxylic acid (Y) with a molecular formula `C_(3)H_(6)O_(2)`. Potassium salt of (Y) undergoes Kolbe's reaction and gives a hydrocarbon (Z). (X), (Y) and (Z) respectively are

Options

A. pentan - 3 - one, propanoic acid, butane
B. pentanal, pentanoic acid, octane
C. 2 - methylbutanone, butanoic acid, hexane
D. 2, 2 - demethylpropanone, propanoic acid, hexane.

Explanation

To solve the problem step by step, we will analyze the information provided and deduce the identities of compounds X, Y, and Z.

### Step 1: Analyze Compound X
- **Given**: Molecular formula of X is C₅H₁₀O.
- **Tests**: Positive 2,4-DNP test indicates that X is a carbonyl compound (either an aldehyde or a ketone). Negative Tollens' test indicates that X cannot be an aldehyde (as aldehydes give a positive Tollens' test).
- **Conclusion**: Therefore, compound X must be a ketone.

### Step 2: Determine the Structure of Compound X
- **Molecular Formula**: C₅H₁₀O suggests a ketone with a straight-chain structure.
- **Possible Structure**: One possible structure is 3-pentanone (CH₃CH₂COCH₂CH₃).

### Step 3: Oxidation of Compound X
- **Oxidation Reaction**: The ketone (X) undergoes oxidation to form a carboxylic acid (Y).
- **Molecular Formula of Y**: Given as C₃H₆O₂.
- **Conclusion**: The oxidation of 3-pentanone (C₅H₁₀O) can yield propanoic acid (C₃H₆O₂).

### Step 4: Identify Compound Y
- **Structure of Y**: The carboxylic acid formed from the oxidation of 3-pentanone is propanoic acid (CH₃CH₂COOH).

### Step 5: Potassium Salt of Compound Y
- **Formation of Salt**: The potassium salt of propanoic acid (Y) would be CH₃CH₂COOK.

### Step 6: Kolbe's Electrolysis Reaction
- **Kolbe's Reaction**: The potassium salt of Y undergoes Kolbe's electrolysis, which typically leads to the formation of hydrocarbons.
- **Expected Hydrocarbon**: The reaction of the carboxylate ion (CH₃CH₂COO⁻) will lead to the formation of butane (C₄H₁₀).

### Step 7: Identify Compound Z
- **Structure of Z**: The hydrocarbon formed is butane (C₄H₁₀).

### Final Summary
- **Compound X**: 3-Pentanone (C₅H₁₀O)
- **Compound Y**: Propanoic acid (C₃H₆O₂)
- **Compound Z**: Butane (C₄H₁₀)

### Conclusion
The identities of the compounds are:
- X: 3-Pentanone
- Y: Propanoic acid
- Z: Butane