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.
What is the name of the orange-red solid?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 2,4-Dinitrophenylhydrazone
Aldehydes and ketones react with derivatives of ammonia, H2N–Z. The nitrogen adds to the carbonyl carbon and water is lost, giving a compound with a C=N bond:
\( \displaystyle \ce{R2C=O + H2N-Z -> R2C=N-Z + H2O} \)
The compound in the passage is acetophenone, so with 2,4-DNP it forms acetophenone 2,4-dinitrophenylhydrazone (Ar = 2,4-dinitrophenyl):
\( \displaystyle \ce{C6H5(CH3)C=O + H2N-NH-Ar -> C6H5(CH3)C=N-NH-Ar + H2O} \)
So the orange-red solid is a 2,4-dinitrophenylhydrazone.
Why the other options are incorrect
Option (1) Phenylhydrazone — Incorrect. It forms with phenylhydrazine, which has no nitro groups.
Option (3) Thiosemicarbazone — Incorrect. It forms with thiosemicarbazide (\( \ce{NH2NHCSNH2} \)).
Option (4) Hydrazone — Incorrect. It forms with hydrazine itself.
NCERT (2026-27 edition):
Chemical Reactions – …
