Aldehydes and Ketones, having at least one methyl group linked to the carbonyl carbon atom (methyl Ketones), are oxidised by sodium hypohalite to sodium salts of corresponding carboxylic acids having one carbon atom less than that of carbonyl compound is converted to haloform. This oxidation does not affect a carbon–carbon double bond, if present in a molecule.
Which of the following compound will give Aldol condensation?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (D)
Note 1: NTA dropped this question in the official Answer Key. However, to provide additional question practice to the students, the question has been slightly modified so that the question has one clear answer.
Aldol condensation needs at least one α-hydrogen (an H on the carbon next to C=O). Dilute alkali removes this hydrogen to give a carbanion (enolate), which adds to the C=O group of a second molecule; the β-hydroxy carbonyl compound then loses water on heating.
(A) Benzaldehyde, C6H5CHO: the carbon next to CHO is a ring carbon with no hydrogen: no α-H, no aldol (it gives the Cannizzaro reaction).
(B) 2,2-Dimethylpropanal, (CH3)3C–CHO: the α-carbon carries three \( \ce{CH3} \) groups and no H: no aldol (Cannizzaro instead).
(C) C6H5CH=C(CH3)–CHO: the α-carbon is the C=C carbon bearing \( \ce{CH3} \); it has no hydrogen, so no enolate forms at the α-position.
(D) Cyclohexanone: both carbons next to C=O are \( \ce{CH2} \) groups, so it has four α-hydrogens and undergoes aldol condensation:
\( \ce{2C6H10O} \xrightarrow{\text{dil. NaOH}} \ce{C12H20O2} \xrightarrow{\Delta} \ce{C12H18O + H2O} \)
Note 2: It seems that there is a mistake in the original NTA paper, which prints “… than that of carbonyl compound is converted to haloform” when it should read “… than that of carbonyl compound. The methyl group is converted to haloform”.
NCERT (2026-27 edition):
Aldol Condensation
