Horner–Wadsworth–Emmons reaction: practice and mechanism
The stabilised phosphonate anion turns the aldehyde into a C=C bond with an ester group at the other end. Almost only the E isomer forms.
Example
NaH deprotonates the phosphonate to the stabilised carbanion, which attacks the aldehyde. Via an oxaphosphetane, phosphate is eliminated and a C=C bond forms. Stabilised phosphonates give almost only the E alkene (α,β-unsaturated ester).
Mechanism step by step
- 1
NaH deprotonates the CH₂ group between phosphonate and ester; the carbanion is stabilised by both groups.
- 2
The carbanion attacks the aldehyde; the alkoxide closes a four-membered ring with the phosphorus (oxaphosphetane).
- 3
The four-membered ring falls apart into the E alkene and a water-soluble phosphate.
Practice tasks
Which product forms, which reagent is needed? After each answer you see the explanation.
Which main product forms?
Typical mistakes
- Stabilised phosphonates give the thermodynamically more stable E alkene because the intermediates can equilibrate. Z selectivity needs special variants (Still–Gennari).
- The addition does not stop at the β-hydroxy ester. The phosphate is eliminated and the C=C double bond forms.
Often confused with: Wittig reaction · Aldol condensation
Frequently asked questions
What is the advantage over the Wittig reaction?
The by-product (dialkyl phosphate) is water-soluble and easy to remove, unlike triphenylphosphine oxide. The anions are also more nucleophilic and E-selective.
When E, when Z?
Stabilised ylides/phosphonates (with ester or ketone) → E. Non-stabilised Wittig ylides (alkyl only) → mainly Z.