Aldol & Claisen Condensations
Crossed and intramolecular aldol reactions, Claisen and Dieckmann condensations, directed enolates with LDA
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🔗 Aldol & Claisen Condensations
Carbon–carbon bond formation between two carbonyl partners through the α-carbon enolate.
Aldol reaction (β-hydroxy carbonyl) vs. aldol condensation (α,β-unsaturated carbonyl) — heat or base drives E1cb dehydration after the aldol step.
Crossed aldol strategies
- Use a non-enolizable partner (benzaldehyde, formaldehyde, di-substituted ketone) as the electrophile.
- Pre-form the enolate with LDA at −78 °C to suppress equilibration (kinetic enolate of unsymmetrical ketones).
- Mukaiyama aldol — silyl enol ether + Lewis acid for full regiocontrol.
Intramolecular aldol — 5- and 6-membered rings dominate (entropic and ring-strain drivers).
Claisen condensation — two ester molecules combine to give a β-ketoester (Acetoacetic ester from ethyl acetate + NaOEt). Requires ≥ 2 α-Hs on the donor and an alkoxide base matched to the ester (no transesterification).
Dieckmann condensation — intramolecular Claisen of a 1,6- or 1,7-diester to give cyclic β-ketoesters.
📚 Practice Problems
1Problem 1medium
❓ Question:
Predict the product of the crossed aldol condensation (heat, NaOH) of acetone with benzaldehyde (1 equiv each). Justify selectivity and predict the final product after dehydration.
💡 Show Solution
Selectivity — Benzaldehyde has no α-H, so it cannot form an enolate. Acetone's enolate attacks the benzaldehyde carbonyl. With heat and NaOH, the β-hydroxyketone undergoes E1cb dehydration.
Aldol product — 4-hydroxy-4-phenylbutan-2-one. Final product (after dehydration) — (E)-4-phenylbut-3-en-2-one (benzalacetone, an α,β-unsaturated enone). The trans (E) geometry dominates because of conjugation with the phenyl group and minimization of steric clash.
2Problem 2hard
❓ Question:
(a) Write the Claisen condensation product of two ethyl propanoate (CH₃CH₂CO₂Et) molecules using NaOEt / EtOH. (b) Why must the base be NaOEt (not NaOMe)? (c) Why does ethyl 2-methylpropanoate (ethyl isobutyrate) fail under the same conditions?
💡 Show Solution
(a) Product — Ethyl 2-methyl-3-oxopentanoate (a β-ketoester with the structure CH₃CH₂–CO–C(CH₃)H–CO₂Et). Mechanism: NaOEt removes an α-H to form the enolate; enolate attacks the carbonyl of the second ester; loss of ethoxide gives the β-ketoester; the acidic α-H of the β-ketoester (pKa ≈ 11) is then deprotonated by NaOEt, providing the thermodynamic driving force for the otherwise unfavorable equilibrium.
(b) Matched base — NaOMe would transesterify the ester (CH₃CH₂CO₂Et → CH₃CH₂CO₂Me + EtOH), giving a messy mixture. Matching the alkoxide to the ester ester group prevents transesterification.
(c) Failure of ethyl isobutyrate — After the second deprotonation step (which drives the equilibrium), the β-ketoester would have to give up its α-H. Ethyl isobutyrate has only one α-H, and the β-ketoester product would have zero α-Hs (the α-carbon is quaternary). Without that final deprotonation, the equilibrium does not shift forward and only starting material is recovered.
⚠️ Common Mistakes: Aldol & Claisen Condensations
Avoid these 3 frequent errors
🌍 Real-World Applications: Aldol & Claisen Condensations
See how this math is used in the real world
📝 Worked Example: Stoichiometry — Limiting Reagent
mol of reacts with mol of . How many grams of water are produced? Which is the limiting reagent? ()
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