Lecture on Mixed or Crossed Aldol Condensation

Introduction

• Aldol Condensations: Previously covered simple aldol condensations with the same molecules.
• Mixed/Crossed Aldol Condensations: Involves different aldehydes; example: benzaldehyde and propanal.

Formation of Enolate Anion

• Propanal:
  • Identify the alpha carbon next to the carbonyl.
  • Contains two alpha protons that can form an enolate anion.
• Benzaldehyde:
  • No alpha protons due to carbon having four bonds.
  • Cannot form an enolate anion.

Mechanism of Reaction

• Step 1: Formation of Nucleophile
  • Sodium hydroxide deprotonates alpha carbon of propanal.
  • Forms a carbanion enolate anion with a negative charge.
• Step 2: Nucleophilic Attack
  • Enolate anion (nucleophile) attacks the electrophilic carbon of benzaldehyde.
  • Forms an alkoxide intermediate.

Product Formation

• Alkoxide Intermediate
  • Forms a new carbon-carbon bond.
  • Protonate the alkoxide to form aldol intermediate.
• Final Product
  • With heat, a conjugated product is formed (enal).

Example Reaction

• Reaction Setup: Cyclohexanone and a diketone.
• Alpha Carbon Identification:
  • Cyclohexanone has two potential alpha carbons.
  • Diketone has alpha carbons between two carbonyls (most acidic protons).

Reaction Steps

• Deprotonation
  • Sodium ethoxide deprotonates the more acidic protons in diketone.
  • Forms an enolate anion.
• Nucleophilic Attack
  • Enolate anion attacks the ketone's carbonyl group.
  • Forms an alkoxide intermediate.

Final Product Analysis

• Formation of Aldol
  • Protonate to form aldol.
  • Heat facilitates formation of conjugated product via complete aldol condensation.
• End Product: Stable conjugated structure with a new double bond.

Conclusion

• Goal: Understanding and predicting product formation in mixed aldol condensations.
• Key Considerations: Identifying alpha carbons, understanding enolate formation, and predicting nucleophilic attacks.
• Practical Application: Useful in exams to quickly determine reaction products.