 Solution

Phosphine binding breaks the hydroxo-bridged dimer into mononuclear Rh-OH units. Hydroxide/boronic-acid association forms a Rh-O-B linkage; transfer of the aryl group from boron to rhodium, with Rh-O bond cleavage, gives the Rh-aryl initiator. This is transmetalation, with rhodium remaining formally Rh(I).

Writing $\mathrm{Ar}=p-{\mathrm{C}}_6{\mathrm{H}}_4\mathrm{Me}$ and $L=\mathrm{PP}{\mathrm{h}}_3$, the balanced net activation isThe stoichiometric byproduct is boric acid, $\mathrm{B}(\mathrm{OH}{)}_3$. Each monomeric Rh(I) product is a 16-electron complex: $d^8$ Rh(I), four electrons from COD, two from phosphine and two from the aryl X ligand. Temporary phosphine dissociation makes room for alkyne binding during propagation.