SolutionsThe catalytic cycle: oxidative addition, transmetalation and reductive elimination
Exercise 1
Give the oxidation state of palladium and the number of valence electrons in: (a) Pd(PPh3)4; (b) Pd(PPh3)2; (c) trans-PhPdBr(PPh3)2. Which of them is the species that carries out oxidative addition?
(a) Pd(0), d10: 10 + 4 × 2 = 18 electrons, coordinatively saturated. (b) Pd(0): 10 + 2 × 2 = 14 electrons. (c) Pd(II), d8: 8 + 2 (Ph−) + 2 (Br−) + 4 (two phosphines) = 16 electrons, square planar. Oxidative addition is carried out by the unsaturated species (b), or even PdL, formed by dissociation of phosphines from (a); its product is (c).
Exercise 2
Why does reductive elimination require the two groups to be cis? Why are alkylpalladium species with hydrogens on the β carbon often a problem in couplings with alkyl groups?
In reductive elimination a bond forms between the two ligands as their bonds to the metal break, through a three-centre transition state: the two groups must be adjacent, cis. trans complexes must isomerize first. An alkyl group with β hydrogens can undergo β-hydride elimination, giving an alkene and a palladium hydride instead of the coupling product; this is why ligands that accelerate reductive elimination, nickel, or zinc reagents (Negishi) are used.