The Reformatsky reaction uses zinc metal and an acid work-up to convert an α-haloester and an aldehyde or ketone to a β-hydroxyester. Zinc metal is first inserted into the carbon-halogen bond of the α-haloester, followed by the formation of an O-zinc enolate, which undergoes an aldol reaction with the carbonyl compound, passing through a six-membered chair-like transition state. An acid work-up cleaves the zinc-oxygen bond, producing a zinc (II) salt and a β-hydroxyester.

  • Reagents: (Zn, Mg, Cd, etc.) or Metal Salt (CrCl2, TiCl2, etc.) Catalyst, Inert Solvent (Et2O, THF, 1,4-dioxane, etc.), Acid Work-Up
  • Reactant: α-Haloester, Aldehyde or Ketone
  • Product: β-Hydroxyester
  • Type of Reaction: Zinc Enolate (Aldol) Condensation
  • Bond Formation: C-C and C-OH

Lab Tips

Several techniques have been used to activate the zinc metal and improve yields. For example, pretreatment of zinc dust with a solution of copper acetate provides a more reactive zinc-copper couple and exposure to trimethylsilyl chloride can also activate the zinc. Note also that this reaction can be conducted with highly hindered ketones and can be adapted for intramolecular aldol reactions.


Top Citations

Original Paper

Related Reactions

  • Blaise Reaction
  • Grignard Reaction
  • Weinreb Ketone Synthesis

Related Compounds

  • Zinc Catalyst
  • Catalytic Acid
  • Aldehyde
  • Ketone
By shuhan yang


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