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Nucleophilic catalysis — Michael/aldol tandem

Baylis-Hillman reaction

How the Baylis-Hillman reaction works, in 3 steps. Each step shows the structures and the curved arrows that move the electrons.

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In short. A 3-step mechanism. Typical reagents: Activated alkene (methyl acrylate), aldehyde, DABCO (10–100 mol %). Conditions: Neat or MeOH/CH₃CN, 25 °C, days (slow). Bonds formed: C–C (α-carbon to carbinol); C=C (regenerated); bonds broken: C=O π; α C–H; C–N. Stereochemistry: Racemic unless chiral amine/Brønsted acid used. Rate and selectivity: Proton transfer/elimination is rate-limiting; accelerated by protic additives.

At a glance

ReagentsActivated alkene (methyl acrylate), aldehyde, DABCO (10–100 mol %)
Solvent / conditionsNeat or MeOH/CH₃CN, 25 °C, days (slow)
Bonds formedC–C (α-carbon to carbinol); C=C (regenerated)
Bonds brokenC=O π; α C–H; C–N
StereochemistryRacemic unless chiral amine/Brønsted acid used
Rate / selectivityProton transfer/elimination is rate-limiting; accelerated by protic additives

Step 1: Michael addition of the amine

+NNOONOON1 · MICHAEL ADDITION OF THE AMINEDABCONUCLEOPHILEmethyl acrylateC₄H₆O₂ELECTROPHILEzwitterionic enolateC₁₀H₁₈N₂O₂INTERMEDIATETertiary amine adds to the β-carbon, leaving an ammonium enolate.

A tertiary amine (like DABCO) adds in a Michael fashion to an activated alkene (like an acrylate), generating a stabilized enolate with a positively charged ammonium tail.

Step 2: Aldol addition to the aldehyde

+NOONODABCOOOO2 · ALDOL ADDITION TO THE ALDEHYDEammonium enolateC₁₀H₁₈N₂O₂NUCLEOPHILEbenzaldehydeC₇H₆OELECTROPHILEalkoxide zwitterionINTERMEDIATEThe enolate α-carbon forms the new C–C bond.

That enolate then attacks an aldehyde's carbonyl carbon in an ordinary aldol addition, forming a new C–C bond and a new alkoxide.

Step 3: Proton transfer, then elimination of DABCO

proton transferE1cB+DABCOHOOODABCOOOOHDABCOOOOHOOOHNN3 · PROTON TRANSFER, THEN ELIMINATION OF DABCOalkoxide zwitterionINTERMEDIATEammonium enolateINTERMEDIATEammonium enolateINTERMEDIATEBaylis–Hillman adductC₁₁H₁₂O₃PRODUCTDABCOCATALYSTTop: alkoxide takes the α-H (formally via a 6-membered TS). Bottom: enolate expels DABCO to restore the alkene.

An intramolecular proton transfer, followed by elimination of the amine catalyst (an E1cb-type step), regenerates the double bond and releases the catalyst — the product combines both a new C–C bond and a retained alkene, all from one catalytic cycle.

Abbreviations and terms

MeCN
Acetonitrile. Polar aprotic solvent.
MeOH
Methanol. Polar protic solvent (also a nucleophile in solvolysis).
DABCO
1,4-Diazabicyclo[2.2.2]octane. Unhindered tertiary amine, a good nucleophilic catalyst (Baylis–Hillman).
TS
Transition state. The highest-energy point on the pathway between two minima; it cannot be isolated.
E1cB
Elimination, unimolecular conjugate base. Deprotonation first to a carbanion, then loss of the leaving group.

How it compares with related reactions

Baylis-Hillman reactionRobinson annulationAsymmetric phase-transfer catalysis
ReagentsActivated alkene (methyl acrylate), aldehyde, DABCO (10–100 mol %)Ketone + methyl vinyl ketone (MVK); NaOEt or KOHChiral cinchona-derived Q*⁺Br⁻ (1–10 mol%), 50% aq. KOH, glycine imine tBu ester, BnBr
ConditionsNeat or MeOH/CH₃CN, 25 °C, days (slow)EtOH, 25–80 °C; heat for dehydrationToluene/CH₂Cl₂, 0 °C, vigorous stirring
StereochemistryRacemic unless chiral amine/Brønsted acid usedQuaternary stereocentre; often racemic (proline gives ee)High ee: chiral ion pair shields one enolate face
Rate / selectivityProton transfer/elimination is rate-limiting; accelerated by protic additivesMichael first, then intramolecular aldol; E1cB dehydration lastAlkylation of the ion pair is rate-limiting

Test yourself

Try to answer each question before opening it.

Which reagents and conditions are typical for the Baylis-Hillman reaction?

Activated alkene (methyl acrylate), aldehyde, DABCO (10–100 mol %). Conditions: Neat or MeOH/CH₃CN, 25 °C, days (slow).

Which bonds are formed and which are broken?

Formed: C–C (α-carbon to carbinol); C=C (regenerated). Broken: C=O π; α C–H; C–N.

What is the stereochemical outcome?

Racemic unless chiral amine/Brønsted acid used

What controls the rate or the selectivity?

Proton transfer/elimination is rate-limiting; accelerated by protic additives

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