Question

An organic compound ‘A’ having molecular formula C\(_8\)H\(_8\)O, gives orange-red precipitate with 2,4-DNP (2,4-Dinitrophenylhydrazine) reagent. ‘A’ gives yellow precipitate on heating with iodine in presence of NaOH. ‘A’ neither reduces Tollen’s reagent, Fehling’s solution nor decolourises bromine water. It forms a carboxylic acid ‘B’ having molecular formula C\(_7\)H\(_6\)O\(_2\), on strong oxidation with chromic acid. Identify compounds ‘A’ and ‘B’ and explain the main reactions.

Hint:

Iodoform test always confirms CH\(_3\)–CO group; oxidation of side chain in aromatic ketones gives benzoic acid.

Answer 1

Step 1: 2,4-DNP test.
Compound A gives orange-red precipitate → confirms presence of a carbonyl group (aldehyde or ketone).
Step 2: Iodoform test.
Yellow precipitate with iodine + NaOH → confirms presence of CH\(_3\)–CO group. Hence A must be a methyl ketone.
Step 3: Negative tests.
- No Tollen’s/Fehling’s reduction → not an aldehyde.
- No bromine water decolourisation → not an alkene.
Step 4: Oxidation product.
On oxidation with chromic acid, A gives a monocarboxylic acid B (C\(_7\)H\(_6\)O\(_2\)) = benzoic acid. \[ C_6H_5COCH_3 \xrightarrow[\text{}]{[O]} C_6H_5COOH \] Step 5: Conclusion.
- A = Acetophenone (C\(_6\)H\(_5\)COCH\(_3\))
- B = Benzoic acid (C\(_6\)H\(_5\)COOH)