Sigma Percentile
JEE Main 2019
LEVELJEE Advanced

Animated Solution for Chemistry - Aldehydes and Ketones: The major product obtained in the following reaction is

Select Answer:

Visualized Solution

Identifying the Acidic Proton

  • The reactant is a cyclopentanone derivative with an ester group and a ketone side chain.
  • The base will abstract the most acidic proton.
  • The proton at is flanked by two electron-withdrawing groups (ketone and ester), making it an active methylene group.

Enolate Formation

  • Base abstracts the acidic proton to form a resonance-stabilized enolate anion.

Intramolecular Nucleophilic Attack

  • The enolate anion acts as a nucleophile.
  • It attacks the electrophilic carbonyl carbon of the side chain.
  • This intramolecular aldol addition forms a new 5-membered ring.

Dehydration to Final Product

  • The resulting alkoxide is protonated to form a -hydroxy ketone.
  • Heating () promotes base-catalyzed dehydration.
  • Water is eliminated to form a stable, conjugated -unsaturated system.

Conclusion

  • The final product is a 5,5-fused bicyclic system (pentalene derivative).
  • This matches option (b).

The Sigma Insight: Aldol Condensation

Solution Diagram

Analyzing the Setup

When faced with a complex organic molecule and a strong base like sodium ethoxide (), the first step is always to hunt for the most acidic proton. In our reactant, we have a cyclopentanone ring adorned with an ester group () and a ketone side chain.
Look closely at the carbon atom situated directly between the ring ketone and the ester group. This is an active methylene carbon. The protons here are highly acidic because the resulting negative charge can be delocalized into both adjacent carbonyl oxygen atoms via resonance.

The Master Equation

Enolate Formation and Attack
The base abstracts this acidic proton, generating a stable enolate anion. Now, this molecule is primed for action. The enolate carbon is a strong nucleophile, and it's looking for an electrophilic target.
Conveniently, the molecule has a built-in electrophile: the carbonyl carbon of the side chain ketone. The enolate swings around and attacks this carbonyl carbon. This is a classic intramolecular aldol addition.
Why does it attack this specific carbon? It's all about ring strain. The attack forms a new five-membered ring, which is kinetically and thermodynamically very favorable. If it had attacked elsewhere, it might have formed a strained four-membered ring or a less favorable bridged system.

Final Calculation

Dehydration
The initial product of this attack is an alkoxide, which quickly picks up a proton to become a -hydroxy ketone. However, the reaction conditions specify heat ().
Heating a -hydroxy ketone in the presence of a base inevitably leads to dehydration (loss of water). The molecule eliminates the hydroxyl group and an adjacent proton to form a double bond.
Nature always favors stability, so the double bond forms in a position that maximizes conjugation with the adjacent carbonyl group. The result is a beautiful, stable -unsaturated bicyclic system—specifically, a 5,5-fused pentalene derivative. This perfectly matches the structure shown in option (b).

Similar Questions