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

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.