What will be the product formed when cyclohexanone undergoes Aldol co…

The correct answer is Option (4) → 2-cyclohexylidenecyclohexan-1-one

Aldehydes and ketones having at least one α-hydrogen react in the presence of dilute alkali to form β-hydroxy aldehydes (aldol) or β-hydroxy ketones (ketol), which readily lose water on heating to give α,β-unsaturated carbonyl compounds. The whole reaction is called aldol condensation.

Cyclohexanone has four α-hydrogens. The base removes one of them to give a carbanion (enolate), which attacks the carbonyl carbon of a second cyclohexanone molecule to give the ketol, 2-(1-hydroxycyclohexyl)cyclohexan-1-one:

    \( \displaystyle \ce{2C6H10O} \xrightarrow{\text{dil. }\ce{NaOH}} \ce{C12H20O2} \)

On heating, the ketol loses water between the OH group and the α-hydrogen, forming a C=C bond between the two rings, conjugated with the C=O group:

    \( \displaystyle \ce{C12H20O2} \xrightarrow[-\ce{H2O}]{\Delta} \underset{\text{2-cyclohexylidenecyclohexan-1-one}}{\ce{C12H18O}} \)

So the product is the one in Option (4): a cyclohexanone ring carrying a cyclohexylidene group (=\( \ce{C6H10} \)) on the carbon next to C=O.

Why the other options are incorrect

• Option (1) cyclohexanol — Incorrect. This is the reduction product of a single cyclohexanone molecule; aldol condensation joins two molecules and keeps the C=O group.

• Option (2) cyclohex-2-en-1-one — Incorrect. This has only six carbons, so it is not a product of two molecules; it is the dehydrogenated form of cyclohexanone.

• Option (3) 4-cyclohexylcyclohexan-1-one — Incorrect. The rings are joined by a single bond at C-4, far from the carbonyl group, and there is no C=C; aldol condensation forms the new bond at the α-carbon (C-2) and gives an α,β-unsaturated ketone.

NCERT (2026-27 edition):
Aldol Condensation

Note: The same/similar question has been asked in different years by NTA as per the details given below: