= Solution
Let $\mathrm{Ar}=2,6$-dimethylphenyl and $\mathrm{Ar'}=2$-methoxyphenyl. After conversion of the Pd(OAc)$_2$ precatalyst into an active phosphine–Pd(0) species, <oxidative addition> of $\mathrm{Ar'Cl}$ gives $\mathrm{L_nPd^{II}(Ar')Cl}$. Carbonate-assisted replacement of chloride by the <alcohol>'s <alkoxide> produces $\mathrm{L_nPd^{II}(Ar')[OC(Me)_2Ar]}$.
In the key <beta-carbon elimination>, the <alkoxide> <oxygen> forms a C=O bond while the bond between its carbinol <carbon> and $\mathrm{Ar}$ cleaves, transferring $\mathrm{Ar}$ to <palladium>. This gives \b[<acetone> and a diarylpalladium(II) complex], $\mathrm{L_nPd^{II}(Ar')(Ar)}$. <reductive elimination> creates the new biaryl bond and regenerates Pd(0). Thus the <tertiary alcohol> functions as an aryl donor by <carbon>–<carbon> bond cleavage, rather than being incorporated as a carbinol substituent. The overall organic products are \b[the cross-coupled biaryl and <acetone>]. The carbinol <carbon> has no <hydrogen>, so ordinary <beta-hydride elimination> from this <alkoxide> cannot explain the observed aryl transfer. The bulky electron-rich phosphine helps access the aryl-chloride <oxidative addition> and the sterically congested biaryl-forming elimination.
Back to article page
