Enolate chemistry — ester condensation
Claisen condensation of ethyl acetate
How the Claisen condensation of ethyl acetate 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: Ethyl acetate (2 equiv), NaOEt (≥ 1 equiv); then H₃O⁺. Conditions: EtOH, reflux; same alkoxide as ester OR group. Bonds formed: C–C σ; β-keto ester; bonds broken: α C–H; C–OEt (leaving group). Stereochemistry: None (achiral product). Rate and selectivity: Equilibrium pulled forward by deprotonating the β-keto ester (pKa ≈ 11).
At a glance
Step 1: Ester enolate formation
Ethoxide deprotonates the alpha carbon of one ethyl acetate molecule, forming an ester enolate.
Step 2: Enolate attacks a second ester
This enolate attacks the carbonyl carbon of a second ethyl acetate molecule — nucleophilic acyl substitution again — forming a tetrahedral alkoxide intermediate.
Step 3: Loss of EtO⁻; deprotonation drives it
Ethoxide is expelled as a leaving group, and a final, essentially irreversible deprotonation of the very acidic β-ketoester product pulls the whole equilibrium forward, giving ethyl acetoacetate.
Abbreviations and terms
- OEt
- Ethoxy. –OCH₂CH₃.
- EtOH
- Ethanol. Polar protic solvent.
- NaOEt
- Sodium ethoxide. Alkoxide base matched to ethyl esters (avoids transesterification).
- pKa
- Acid-dissociation constant (–log). Lower pKa = stronger acid; used to judge which proton a base removes.
How it compares with related reactions
| Claisen condensation of ethyl acetate | Aldol condensation | Base-catalyzed aldol addition of acetaldehyde | |
|---|---|---|---|
| Reagents | Ethyl acetate (2 equiv), NaOEt (≥ 1 equiv); then H₃O⁺ | NaOH or KOH (cat.) | cat. NaOH or KOH |
| Conditions | EtOH, reflux; same alkoxide as ester OR group | H₂O/EtOH, heat 50–100 °C | H₂O / EtOH, 0–25 °C |
| Stereochemistry | None (achiral product) | Mostly E-enal (thermodynamic) | New stereocentre — racemic |
| Rate / selectivity | Equilibrium pulled forward by deprotonating the β-keto ester (pKa ≈ 11) | E1cB; loss of HO⁻ from the enolate is rate-determining | Heating promotes dehydration to the enal (aldol condensation) |
Test yourself
Try to answer each question before opening it.
Which reagents and conditions are typical for the Claisen condensation of ethyl acetate?
Ethyl acetate (2 equiv), NaOEt (≥ 1 equiv); then H₃O⁺. Conditions: EtOH, reflux; same alkoxide as ester OR group.
Which bonds are formed and which are broken?
Formed: C–C σ; β-keto ester. Broken: α C–H; C–OEt (leaving group).
What is the stereochemical outcome?
None (achiral product)
What controls the rate or the selectivity?
Equilibrium pulled forward by deprotonating the β-keto ester (pKa ≈ 11)
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