Heck reaction mechanism: bromobenzene and ethyl acrylate

Question

Bromobenzene reacts with ethyl acrylate using a Pd(0) catalyst and a base. Draw the catalytic cycle for the Heck reaction, name each step and give the product.

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Heck reaction mechanism: bromobenzene and ethyl acrylate — question structures

Expert answer

Catalyst activation: a Pd(II) pre-catalyst such as Pd(OAc)₂ is reduced in situ to the active Pd(0)L₂ (proposed: a phosphine ligand is oxidised to the phosphine oxide). he product is ethyl (E)-cinnamate, PhCH=CHCO₂Et. The by-product is Et₃NH⁺ Br⁻. Catalyst activation: a Pd(II) pre-catalyst such as Pd(OAc)₂ is reduced in situ to the active Pd(0)L₂, where L is PPh₃ (proposed: one phosphine is oxidised to the phosphine oxide). Catalytic cycle Oxidative addition (→): Pd(0)L₂ inserts into the C–Br bond of bromobenzene to give PhPd(II)Br(L)₂. This is slower for Ar–Br than for Ar–I, so heat is needed. Alkene coordination (⇌): a ligand leaves, and ethyl acrylate binds to Pd. Syn migratory insertion (→): the phenyl group migrates to the terminal CH₂ of the acrylate (the less hindered carbon), and Pd bonds to the carbon bearing CO₂Et. C–C rotation: this puts a β-hydrogen syn to Pd. Syn β-hydride elimination (⇌): this forms the C=C bond and a Pd–H species. The less hindered conformer, with Ph and CO₂Et trans, gives the E alkene. The product then dissociates from Pd. Base-assisted regeneration (→): Et₃N removes HBr from H–Pd–Br(L)₂, regenerating Pd(0)L₂.