Organic synthesis and applications
In aldehydes, carbonyl group is bonded to carbon and hydrogen while in ketones, it is bonded to two carbon atoms. The carbonyl carbon atom is sp2-hybridised and a trigonal coplanar structure. The bond angles are approximately 120°. Methanal is gas at room temperature. Ethanal is a volatile liquid. Other aldehydes and ketones are liquid or solid at room temperature. The boiling points of aldehydes and ketones are higher than hydrocarbons and ether of comparable molecular masses. It is due to weak molecular association in aldehydes and ketones arising out of the dipole-dipole interactions. Their boiling points are lower than those of alcohols of similar molecular masses due to absence of intermolecular hydrogen bonding. The lower members of aldehydes and ketones are miscible with water because they form hydrogen bonding with water. However, the solubility of aldehydes and ketones decreases rapidly on increasing the length of alkyl chain.
Aldehydes are generally more reactive than ketones in the nucleophilic addition reactions due to steric and electronic reasons. An organic compound (A) with molecular formula C8H8O forms an orange-red precipitate with 2,4-DNP reagent and gives yellow precipitate on heating with iodine in the presence of sodium hydroxide. It neither reduces Tollens' or Fehling's reagent, nor does it decolourise bromine water or Baeyer's reagent. On drastic oxidation with chromic acid, it gives a carboxylic acid (B) having molecular formula C7H6O2.
Identify the compounds (A), and (B), in the above reaction.
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Acetophenone; Benzoic acid.
An unknown carbonyl compound is identified by reading each test result as a clue:
Molecular formula C8H8O: degree of unsaturation = (2 × 8 + 2 − 8)/2 = 5, which fits a benzene ring (4) plus one C=O (1).
Orange-red precipitate with 2,4-DNP: a C=O group (aldehyde or ketone) is present.
Yellow precipitate with I2/NaOH (iodoform test): a CH3CO– group is present.
No reaction with Tollens’ or Fehling’s reagent: it is a ketone, not an aldehyde.
Does not decolourise bromine water or Baeyer’s reagent: there is no C=C outside the aromatic ring.
The only structure that fits is \( \ce{C6H5COCH3} \), acetophenone (A). It gives the iodoform test:
\( \displaystyle \ce{C6H5COCH3 + 3I2 + 4NaOH -> C6H5COONa + CHI3 v + 3NaI + 3H2O} \)
On drastic oxidation the side chain is oxidised and the carbon attached to the ring becomes –COOH, giving \( \ce{C7H6O2} \), benzoic acid (B):
\( \displaystyle \ce{C6H5COCH3} \xrightarrow{\text{chromic acid}} \ce{C6H5COOH} \)
So A is acetophenone and B is benzoic acid.
Why the other options are incorrect
Option (1) Benzylic acid; Benzaldehyde — Incorrect. There is no compound called benzylic acid, and A must be the \( \ce{C8H8O} \) carbonyl compound, not an acid; benzaldehyde (\( \ce{C7H6O} \)) is an aldehyde that would reduce Tollens’ reagent.
Option (3) Benzoic acid; Acetophenone — Incorrect. The pair is reversed; benzoic acid does not give the 2,4-DNP or iodoform test.
Option (4) Phenyl ethyl ketone; Benzoic acid — Incorrect. \( \ce{C6H5COC2H5} \) is \( \ce{C9H10O} \), and it has no \( \ce{CH3CO} \)– group, so it does not give the iodoform test.