Lecture Notes: Enolate Anions Formation and Base Selection

Introduction

• Previous Discussion: Formation of enolate anions in reactions.
• Current Focus: Detailed exploration of enolate anion formation and base selection.

Enolate Anion Formation

• Base Example: Ethoxide anion (Na+ OEt−) from sodium ethoxide.
• Reaction with Acetaldehyde:
  • Identify alpha carbon next to carbonyl carbon.
  • Alpha carbon has three hydrogens; each is an alpha proton.
  • Base (ethoxide anion) can remove one alpha proton, forming enolate anion.

Mechanism

• Carbanion Formation:

  • Removal of alpha proton leaves electrons on carbon (carbanion).
  • Draw resonance structure, electrons move to form a double bond, creating an oxyanion.

• Resonance Structures:

  • Carbanion form and oxyanion form of enolate anion.

Equilibrium Considerations

• Base and Product Formation: Ethoxide forms ethanol upon protonation.

• Equilibrium State:

  • Aldehyde (pKa ≈ 17) vs Ethanol (pKa ≈ 16).
  • Calculate pKeq = 17 - 16 = 1, Keq = 10^(-1) = 0.1 (favors reactants).

• Weaker Acid Principle:

  • Formation favors weaker acid (aldehyde).
  • Both aldehyde and enolate anion present at equilibrium.

Complete Enolate Anion Formation

• Alternative Base - Hydride Anion:

  • Source: Sodium hydride (Na+ H−) or potassium hydride (K+ H−).
  • Similar mechanism: Hydride removes alpha proton, forms carbanion.

• Reaction Completion:

  • Hydrogen gas formation drives reaction to completion.
  • Results in complete enolate anion formation.

Conclusion

• Upcoming Topic: Use of LDA for complete enolate anion formation.