Aldol condensation is a popular and classical method to prepare $$\alpha,\beta$$-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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• Option A: Cleaving the double bond at the ring junction yields a single open-chain diketone molecule. Therefore, it is an intramolecular aldol product.
  
  
• Option B: Cleaving the double bond in the 5-membered ring yields a single aromatic dialdehyde/diketone precursor (2-formylbenzaldehyde derivative). Therefore, it is an intramolecular aldol product.
  
  
• Option C: Cleaving the double bond in the cyclohexenone ring yields a single open-chain dicarbonyl precursor. Therefore, it is an intramolecular aldol product.
  
  
• Option D: This compound features a terminal exocyclic double bond ($$=\text{CH}_2$$) attached to the $\alpha$-carbon. Cleaving this bond breaks the molecule into two separate components: a cyclic ketone and formaldehyde ($$\text{HCHO}$$). Because it requires two independent starting molecules, it cannot be formed via an intramolecular pathway.
  

Correct Answer

Option D (The compound with the terminal exocyclic double bond, $$=\text{CH}_2$$)