Enolate chemistry — elimination (E1cb)
Aldol condensation: dehydration of a β-hydroxy aldehyde
How the Aldol condensation works, in 3 steps. Each step shows the structures and the curved arrows that move the electrons.
In short. A 3-step mechanism. Typical reagents: NaOH or KOH (cat.). Conditions: H₂O/EtOH, heat 50–100 °C. Bonds formed: C=C π bond (conjugated); H₂O; bonds broken: α C–H; C–OH. Stereochemistry: Mostly E-enal (thermodynamic). Rate and selectivity: E1cB; loss of HO⁻ from the enolate is rate-determining.
At a glance
Step 1: Base removes the α-hydrogen
Base removes the acidic alpha-hydrogen adjacent to the carbonyl of the aldol addition product, forming an enolate conjugated with the C=O.
Step 2: E1cB: hydroxide expelled
The beta-hydroxyl is a poor leaving group alone, but with negative charge already delocalized in the enolate, it departs as hydroxide in a stepwise E1cb elimination, forming a new C=C.
Step 3: Conjugated enal
The product is an α,β-unsaturated aldehyde (an enal) — conjugation between the new double bond and the carbonyl is the thermodynamic driving force for this dehydration.
Abbreviations and terms
- EtOH
- Ethanol. Polar protic solvent.
- NaOH
- Sodium hydroxide. Strong aqueous base and nucleophile.
- KOH
- Potassium hydroxide. Strong aqueous/alcoholic base.
- E1cB
- Elimination, unimolecular conjugate base. Deprotonation first to a carbanion, then loss of the leaving group.
- pKa
- Acid-dissociation constant (–log). Lower pKa = stronger acid; used to judge which proton a base removes.
How it compares with related reactions
| Aldol condensation | Base-catalyzed aldol addition of acetaldehyde | Claisen condensation of ethyl acetate | |
|---|---|---|---|
| Reagents | NaOH or KOH (cat.) | cat. NaOH or KOH | Ethyl acetate (2 equiv), NaOEt (≥ 1 equiv); then H₃O⁺ |
| Conditions | H₂O/EtOH, heat 50–100 °C | H₂O / EtOH, 0–25 °C | EtOH, reflux; same alkoxide as ester OR group |
| Stereochemistry | Mostly E-enal (thermodynamic) | New stereocentre — racemic | None (achiral product) |
| Rate / selectivity | E1cB; loss of HO⁻ from the enolate is rate-determining | Heating promotes dehydration to the enal (aldol condensation) | Equilibrium pulled forward by deprotonating the β-keto ester (pKa ≈ 11) |
Test yourself
Try to answer each question before opening it.
Which reagents and conditions are typical for the Aldol condensation?
NaOH or KOH (cat.). Conditions: H₂O/EtOH, heat 50–100 °C.
Which bonds are formed and which are broken?
Formed: C=C π bond (conjugated); H₂O. Broken: α C–H; C–OH.
What is the stereochemical outcome?
Mostly E-enal (thermodynamic)
What controls the rate or the selectivity?
E1cB; loss of HO⁻ from the enolate is rate-determining
Study guides that help with this reaction
- Carbonyl chemistry: addition, acyl substitution and enolates
- SN1, SN2, E1 or E2? How to decide which mechanism wins
- Nucleophiles, electrophiles and leaving groups: what makes a reaction go