CBSE Class 12 Chemistry: Marking Scheme Solution

Question Analysis:

• Molecular formula $\text{C}_5\text{H}_{10}\text{O}$ (Degree of unsaturation = 1, likely a ketone or aldehyde).
• Forms 2,4-DNP derivative: Confirms presence of a carbonyl group ($>\text{C}=\text{O}$).
• Does not reduce Tollens' reagent: Confirms it is a ketone, not an aldehyde.
• Positive iodoform test: Confirms presence of a methyl ketone group ($\text{CH}_3\text{CO}-$).
• Vigorous oxidation gives ethanoic acid ($\text{CH}_3\text{COOH}$) and propanoic acid ($\text{CH}_3\text{CH}_2\text{COOH}$): Confirms the structure is Pentan-2-one.

Step 1 (Deduction of Structure): [1 Mark]

Based on the chemical tests, the compound must be a methyl ketone with 5 carbons. The cleavage of pentan-2-one during vigorous oxidation occurs at the carbonyl group:

Structure of 'X': Pentan-2-one

$$\text{CH}_3 - \text{CO} - \text{CH}_2 - \text{CH}_2 - \text{CH}_3$$

Step 2 (Reaction with 2,4-DNP): [1 Mark]

The reaction involves the nucleophilic addition-elimination of 2,4-dinitrophenylhydrazine with the carbonyl group of pentan-2-one to form a hydrazone derivative.

Structure of the product:

$$\text{CH}_3 - \text{C}(\text{NNH-C}_6\text{H}_3(\text{NO}_2)_2) - \text{CH}_2 - \text{CH}_2 - \text{CH}_3$$

Step 3 (Iodoform Test Products): [1 Mark]

The reaction of a methyl ketone with $\text{I}_2/\text{NaOH}$ yields a yellow precipitate of iodoform ($\text{CHI}_3$) and the sodium salt of the corresponding carboxylic acid (sodium butanoate).

Products:

• Iodoform: $\text{CHI}_3$ (Yellow ppt)
• Sodium Butanoate: $\text{CH}_3\text{CH}_2\text{CH}_2\text{COONa}$

💡 EL10 Concept Insight:

Think of the iodoform test like a "methyl-ketone detector"—if a molecule has a $\text{CH}_3$ group attached directly to a $\text{C}=\text{O}$, the iodine "chops" that methyl group off to make a bright yellow powder ($\text{CHI}_3$), leaving the rest of the chain behind as a salt.