 Solution

Let $\mathrm{Ar}=2,6$-dimethylphenyl and $\mathrm{A}{\mathrm{r}}^{\prime }=2$-methoxyphenyl. After conversion of the Pd(OAc)${}_2$ precatalyst into an active phosphine–Pd(0) species, oxidative addition of $\mathrm{A}{\mathrm{r}}^{\prime }\mathrm{Cl}$ gives ${\mathrm{L}}_{\mathrm{n}}\mathrm{P}{\mathrm{d}}^{\mathrm{II}}(\mathrm{A}{\mathrm{r}}^{\prime })\mathrm{Cl}$. Carbonate-assisted replacement of chloride by the alcohol's alkoxide produces ${\mathrm{L}}_{\mathrm{n}}\mathrm{P}{\mathrm{d}}^{\mathrm{II}}(\mathrm{A}{\mathrm{r}}^{\prime })[\mathrm{OC}(\mathrm{Me}{)}_2\mathrm{Ar}]$.

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 acetone and a diarylpalladium(II) complex, ${\mathrm{L}}_{\mathrm{n}}\mathrm{P}{\mathrm{d}}^{\mathrm{II}}(\mathrm{A}{\mathrm{r}}^{\prime })(\mathrm{Ar})$. 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 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.