CARBONYL COMPOUNDS
In aldehydes, the carbonyl group is bonded to a carbon and hydrogen while in the ketones, it is bonded to two carbon atoms. Aldehydes and ketones undergo nucleophilic addition reactions. A nucleophile attacks the electrophilic carbon atom of the polar carbonyl group from a direction approximately perpendicular to the plane of sp2 hybridized orbitals of carbonyl carbon. The hybridization of carbon changes from sp2 to sp3 in the nucleophilic addition. Aldehydes give Fehling’s test positive while aromatic aldehydes do not respond to this test .
An organic compound ‘A’ with molecular formula C3H5N, undergoes Stephen reaction to produce another compound ‘B’. Compound ‘B’ gives orange precipitate with 2,4-DNP but it does not undergo Haloform reaction. However, the reaction of compound B with dilute alkali produces a compound ‘C’ which is β-hydroxy compound with increased number of carbon atoms.
Identify the name reaction involved in formation of compound ‘C’ from ‘B’ is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Aldol reaction.
Aldehydes and ketones that have at least one α-hydrogen react in the presence of dilute alkali to give β-hydroxy aldehydes (aldols) or β-hydroxy ketones (ketols). This is the aldol reaction. Two molecules join, so the product has more carbon atoms than the starting compound.
Working out the compounds of the passage:
A (\( \ce{C3H5N} \)) undergoes the Stephen reaction, which is a reaction of nitriles, so A is propanenitrile, \( \ce{CH3CH2CN} \).
B is the aldehyde from Stephen reduction, propanal, \( \ce{CH3CH2CHO} \). It gives an orange precipitate with 2,4-DNP (a carbonyl compound) but no haloform reaction (it has no \( \ce{CH3CO} \)– group).
C is formed from B with dilute alkali and is a β-hydroxy compound with more carbon atoms: the aldol.
\( \displaystyle \ce{CH3CH2CN} \xrightarrow[\text{(ii) }\ce{H3O^+}]{\text{(i) }\ce{SnCl2 + HCl}} \ce{CH3CH2CHO} \)
\( \displaystyle \ce{2CH3CH2CHO} \xrightarrow{\text{dil. NaOH}} \ce{CH3CH2CH(OH)CH(CH3)CHO} \)
So C is formed from B by the aldol reaction.
Why the other options are incorrect
Option (1) Wolff Kishner reduction — Incorrect. It reduces C=O to \( \ce{CH2} \) using hydrazine and KOH in ethylene glycol; no β-hydroxy compound is formed.
\( \displaystyle \ce{R2C=O} \xrightarrow{\ce{NH2NH2}} \ce{R2C=NNH2} \xrightarrow[\text{heat}]{\ce{KOH}\text{/ethylene glycol}} \ce{R2CH2 + N2} \)
Option (2) Hell-Volhard-Zelinsky reaction — Incorrect. It halogenates the α-carbon of a carboxylic acid using \( \ce{X2} \) and red phosphorus.
\( \displaystyle \ce{R-CH2-COOH} \xrightarrow[\text{(ii) }\ce{H2O}]{\text{(i) }\ce{X2}\text{/red P}} \ce{R-CH(X)-COOH} \)
Option (3) Etard reaction — Incorrect. It oxidises toluene to benzaldehyde using chromyl chloride.
\( \displaystyle \ce{C6H5CH3} \xrightarrow{\ce{CrO2Cl2}\text{, }\ce{CS2}} \ce{C6H5CH(OCrOHCl2)2} \xrightarrow{\ce{H3O+}} \ce{C6H5CHO} \)
NCERT (2026-27 edition):
Aldol Condensation
