Aldehydes, Ketones and Carboxylic Acids · Question 29

JEE Main 2025 (Online) 4th April Morning ShiftMCQMedium📈 Topic in 32% papers+4−1

Aldol condensation is a popular and classical method to prepare , -unsaturated carbonyl compounds. This reaction can be both intermolecular and intramolecular. Predict which one of the following is not a product of intramolecular aldol condensation?

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Answer

(2) [figure]

Solution

In aldol condensation the -carbon of one carbonyl group adds to another C=O group, and dehydration gives the C=C between the - and -carbons. Breaking this C=C shows the starting compound:

(A) Breaking the ring-fusion C=C gives cyclodecane-1,6-dione; its transannular (intramolecular) aldol condensation gives this bicyclic enone.

(D) Inden-1-one ← 2-acetylbenzaldehyde, CH3CO-C6H4-CHO (intramolecular).

(B) Breaking the exocyclic C=CH2 bond gives -tetralone + HCHO, i.e. two different molecules. So this enone is the product of an intermolecular (crossed) aldol condensation, not an intramolecular one.

Answer: (2)

JEE Main 2025 (Online) 4th April Morning Shift

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