JEE Chemistry 2025 Question with Solution

Given: The substrate undergoes intramolecular aldol condensation.

Find: The major cyclic product formed.

The solution explains that the given compound contains two carbonyl groups and forms an enolate at an $\alpha$-position. In an intramolecular aldol reaction, the enolate attacks the other carbonyl group within the same molecule, so cyclization occurs.

A $5$-membered ring is favored because intramolecular aldol condensations usually prefer formation of $5$- or $6$-membered rings, and here the $5$-membered cyclic product is stated to be the major product.

After cyclization, the intermediate $\beta$-hydroxy carbonyl compound undergoes dehydration to give an $\alpha,\beta$-unsaturated carbonyl compound.

Therefore, the major product is the cyclic $\alpha,\beta$-unsaturated carbonyl compound with a $5$-membered ring. The solution identifies this as option C.

The correct option is C.

Given: An intramolecular aldol condensation of a dicarbonyl compound.

Find: Which option represents the major product.

The extracted solution states the following sequence:

1. Identify the enolizable hydrogen adjacent to a carbonyl group.
2. Form the enolate ion.
3. Carry out intramolecular nucleophilic attack on the other carbonyl carbon.
4. Form a cyclic intermediate.
5. Remove water by dehydration.

Because ring closure that gives a $5$-membered ring is more stable, that pathway dominates over alternatives. The final product is therefore a conjugated cyclic enone/enal-type product formed after dehydration.

The source solution explicitly concludes: the correct answer is the third option, and the page marks The Correct Option is C.