An organic compound (A), \( C_9H_{10}O \), gives a 2,4-DNP reaction, negative Tollens' test and iodoform test. On drastic oxidation with chromic acid, it forms B, having molecular formulae, \( C_7H_6O_2 \). Identify A and B.
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → A: Propiophenone, B: Benzoic acid
Each test removes some possibilities:
2,4-DNP test positive: A has a carbonyl group (aldehyde or ketone).
Tollens’ test negative: A is not an aldehyde, so it is a ketone.
Iodoform test negative: A is not a methyl ketone; it has no \( \ce{CH3CO-} \) group.
Drastic oxidation gives \( \ce{C7H6O2} \), benzoic acid: A has a benzene ring with one side chain, attached to the ring through carbon.
\( \ce{C9H10O} \) has \( \frac{2(9) + 2 - 10}{2} = 5 \) degrees of unsaturation: four for the benzene ring and one for C=O. With \( \ce{C6H5-} \) and C=O taken away, two carbon atoms remain. The ketones possible are \( \ce{C6H5COCH2CH3} \) and \( \ce{C6H5CH2COCH3} \); the second is a methyl ketone and would give iodoform. So A is propiophenone (1-phenylpropan-1-one).
On drastic oxidation with chromic acid, the C–C bond next to the carbonyl group breaks and the carbon attached to the ring ends up as –COOH:
\( \displaystyle \ce{C6H5COCH2CH3} \xrightarrow[\text{drastic}]{\ce{H2CrO4}} \ce{C6H5COOH + CH3COOH} \)
So B is benzoic acid, \( \ce{C7H6O2} \).
Why the other options are incorrect
Option (1) A: 3-Phenylpropanal, B: Benzoic acid — Incorrect. 3-Phenylpropanal, \( \ce{C6H5CH2CH2CHO} \), is an aldehyde and would give a positive Tollens’ test.
Option (3) A: 1-Phenylacetophenone, B: Benzoic acid — Incorrect. Acetophenone carrying another phenyl group has fourteen carbon atoms, so it cannot have the formula \( \ce{C9H10O} \).
Option (4) A: 3-Phenylpropanal, B: Phenylacetic acid — Incorrect. The aldehyde would give Tollens’ test, and phenylacetic acid is \( \ce{C8H8O2} \), not \( \ce{C7H6O2} \).