Past exam of the chemistry B course of the University of Cambridge / 2022 / Paper 2 / B8 / c

Step 1 is oxidative addition of Si–H to Co(I), giving Co(III)–H and Co(III)–SiR${}_3$. It cleaves the Si–H bond and forms two metal–ligand bonds, increasing the formal metal oxidation state by two. The phosphine dissociation shown before this numbered step makes room for the reaction; alkene binding and phosphine exchange are ligand-substitution events rather than oxidative additions.

Step 2 is migratory insertion of the coordinated alkene into Co–H, equivalently intramolecular hydride migration to the alkene. One alkene carbon gains H and the other gains the Co–C bond. Co remains formally Co(III); a two-electron alkene ligand plus a hydride is replaced by one alkyl ligand, reducing the electron count by two before phosphine reassociation.

Step 3 is C–Si reductive elimination between the alkyl and silyl ligands. It releases the hydrosilylated alkane and regenerates Co(I). The electron count for the drawn sequence is consistent: the one-PMe${}_3$ alkene adduct is 16-electron; oxidative addition gives 18; insertion gives 16; PMe${}_3$ association gives 18; reductive elimination returns the 16-electron catalyst. Because insertion occurs into Co–H, this is the conventional Chalk-Harrod mechanism.

oxidative addition of an intact Si–H bond initially places H and SiR${}_3$ cis in the metal coordination sphere; this is the local arrangement generated by addition of a sigma bond. hydride and alkene must also attain adjacent sites for intramolecular insertion, and the eventual alkyl and silyl groups must be cis for direct reductive elimination. Subsequent rearrangements should not be confused with the stereochemistry of the bond-addition event itself.

For an alkyl-substituted terminal alkene, insertion puts Co on the terminal carbon and H on the substituted carbon, producing $\mathrm{Co}-\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{R}$. It avoids the more crowded secondary metal–alkyl arrangement. Replacement of that Co–C bond by C–Si therefore gives the linear product $\mathrm{RC}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{Si}{\mathrm{R}}_3$, an Anti-Markovnikov addition outcome. Hydrometallation is syn at the coordinated alkene, although this simple terminal substrate generally does not retain two substituents whose relative stereochemistry could diagnose the complete sequence.

The opposite insertion orientation gives a benzylic metal–carbon bond, $\mathrm{Co}-\mathrm{CH}(\mathrm{Ph})\mathrm{C}{\mathrm{H}}_3$. C–Si reductive elimination then gives the branched productA plausible explanation is stabilization of the developing and final benzylic Co–C interaction by the adjacent aromatic pi system, which can outweigh the steric preference for a primary alkyl. That makes an electronically different choice from an ordinary alkyl substituent. This is a mechanistic rationale for the proposed cycle, not proof of its rate-determining step. In the absence of a chiral catalyst or another source of asymmetric induction, the new stereogenic carbon would give a racemic mixture.