Question

Which among the following will be the major product of the given reaction?

• A.

  

• B.

  

• C.

  

• D.

  

Hint:

A 1,4-diketone undergoes intramolecular aldol condensation in dilute base to give a five-membered cyclopentenone; the more substituted (tetrasubstituted) conjugated enone is favoured.

Answer 1

The Correct Option is B

Step 1: The substrate is CH₃COCH₂CH₂COCH₂CH₂COOH (4,7-dioxooctanoic acid), a 1,4-diketone. Only five-membered rings can form (3-membered rings are not feasible; 6-membered would need a 1,5-dicarbonyl).
Step 2: Two enolates can close a five-membered ring: (i) the CH₃ enolate attacking the other ketone gives 3-(2-carboxyethyl)cyclopent-2-enone (trisubstituted C=C); (ii) the CH₂ enolate on the carbon next to the CH₂COOH chain attacking the methyl ketone gives a ring with CH₃ on the \(\beta\)-carbon and CH₂COOH on the \(\alpha\)-carbon (tetrasubstituted C=C).
Step 3: Under heating with 10% KOH, aldol is followed by dehydration under thermodynamic control; the more substituted, more stable tetrasubstituted conjugated enone is the major product: 2-(carboxymethyl)-3-methylcyclopent-2-en-1-one, i.e. (2-methyl-5-oxocyclopent-1-en-1-yl)acetic acid.
Step 4: Acidification (H₃O⁺) regenerates the free COOH.
Answer: (B)