Horner-Wadsworth-Emmons reaction tends to give predominantly which stereochemical outcome in alkenes?

• A. Z-alkenes
• B. Mixture of E and Z
• C. E-alkenes
• D. None

Answer: (C)

The Horner–Wadsworth–Emmons reaction is built from a stabilized phosphonate carbanion that attacks the carbonyl to form a betaine-like intermediate. The elimination that follows proceeds in an anti fashion, placing the leaving phosphonate oxide away from the developing σ-bond as the double bond forms. This anti elimination aligns the larger substituents on opposite sides of the new C=C bond, so the predominantly produced product is the E (trans) alkene. The stabilization of the phosphonate carbanion and the steric demands of the transition state bias the reaction toward E geometry, more so than toward Z. Under typical conditions with stabilized phosphonates, Z or mixtures are much less common, whereas E dominates. If you used unstabilized ylides (as in some Wittig scenarios), you’d see more Z or mixtures, but that’s a different regime from the Horner–Wadsworth–Emmons approach.