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Alkene synthesis — stabilized ylide

Horner-Wadsworth-Emmons (HWE) reaction

How the Horner-Wadsworth-Emmons (HWE) 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: Phosphonate ester (EtO)₂P(O)CH₂CO₂Et, NaH or NaOMe, aldehyde. Conditions: THF or DME, 0–25 °C. Bonds formed: C=C (E-alkene), P–O (phosphate); bonds broken: α C–H, C=O, C–P. Stereochemistry: Strong E-selectivity (trans). Rate and selectivity: Aldehyde addition rate-limiting; water-soluble phosphate easily removed.

At a glance

ReagentsPhosphonate ester (EtO)₂P(O)CH₂CO₂Et, NaH or NaOMe, aldehyde
Solvent / conditionsTHF or DME, 0–25 °C
Bonds formedC=C (E-alkene), P–O (phosphate)
Bonds brokenα C–H, C=O, C–P
StereochemistryStrong E-selectivity (trans)
Rate / selectivityAldehyde addition rate-limiting; water-soluble phosphate easily removed

Step 1: Base forms the phosphonate carbanion

+OHPOOOOO1 · BASE FORMS THE PHOSPHONATE CARBANIONmethoxideCH₃O⁻BASEtrimethyl phosphonoacetateC₅H₁₁O₅PSUBSTRATEacidic C–H(P=O and C=O)The carbanion is delocalized onto the ester carbonyl and the P=O.

Base deprotonates a phosphonate ester alpha to a carbonyl, forming a resonance-stabilized carbanion — more stable, and more selective, than a classic Wittig ylide.

Step 2: Carbanion adds; oxaphosphetane forms

+addition,ring closurePOOOOOOδ⁺δ⁻OOOPOOO2 · CARBANION ADDS; OXAPHOSPHETANE FORMSphosphonate carbanionC₅H₁₀O₅P⁻NUCLEOPHILEbenzaldehydeC₇H₆OELECTROPHILEoxaphosphetaneC₁₂H₁₆O₆P⁻INTERMEDIATEReversible addition; the trans-oxaphosphetane is favoured.

This carbanion attacks an aldehyde's carbonyl carbon, forming a betaine/oxaphosphetane intermediate just like the Wittig.

Step 3: Collapse: (E)-alkene + phosphate

retro-[2+2]+OOOPOOOOOOPOOO3 · COLLAPSE: (E)-ALKENE + PHOSPHATEoxaphosphetaneC₁₂H₁₆O₆P⁻INTERMEDIATEmethyl (E)-cinnamateC₁₀H₁₀O₂PRODUCTdimethyl phosphateC₂H₆O₄P⁻BYPRODUCTWater-soluble phosphate is washed away, unlike Ph₃P=O.

Collapse of the ring ejects a soluble phosphate byproduct (easy to remove, unlike Wittig's phosphine oxide) and gives the alkene, strongly favoring the E isomer.

Abbreviations and terms

THF
Tetrahydrofuran. Cyclic ether solvent; its oxygen lone pairs solvate Li⁺/Mg²⁺ so organometallics stay dissolved and reactive.
NaOMe
Sodium methoxide. Alkoxide base matched to methyl esters.
NaH
Sodium hydride. Insoluble, irreversible base: removes H⁺ and bubbles off H₂.
TPPO
Triphenylphosphine oxide. The thermodynamic sink of Wittig/Mitsunobu/Appel reactions.

How it compares with related reactions

Horner-Wadsworth-Emmons (HWE) reactionWittig reactionJulia-Kocienski olefination
ReagentsPhosphonate ester (EtO)₂P(O)CH₂CO₂Et, NaH or NaOMe, aldehydePh₃P=CH₂ (ylide, from Ph₃P⁺CH₃ Br⁻ + n-BuLi); acetaldehyde1-phenyl-1H-tetrazol-5-yl (PT) sulfone, KHMDS (or NaHMDS, LiHMDS), aldehyde
ConditionsTHF or DME, 0–25 °CTHF, 0 °C → 25 °C; anhydrous, inert atmosphereTHF or DME, –78 °C → 25 °C
StereochemistryStrong E-selectivity (trans)Unstabilised ylides: Z-alkene; stabilised: E-alkeneHighly E-selective (anti-adduct, then anti elimination)
Rate / selectivityAldehyde addition rate-limiting; water-soluble phosphate easily removedDriven by the strong P=O bond of Ph₃P=OOne-pot; Smiles shift and elimination are fast after addition

Test yourself

Try to answer each question before opening it.

Which reagents and conditions are typical for the Horner-Wadsworth-Emmons (HWE) reaction?

Phosphonate ester (EtO)₂P(O)CH₂CO₂Et, NaH or NaOMe, aldehyde. Conditions: THF or DME, 0–25 °C.

Which bonds are formed and which are broken?

Formed: C=C (E-alkene), P–O (phosphate). Broken: α C–H, C=O, C–P.

What is the stereochemical outcome?

Strong E-selectivity (trans)

What controls the rate or the selectivity?

Aldehyde addition rate-limiting; water-soluble phosphate easily removed

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