On the mechanism of the hydrogen transfer from H2O-CO to gamma-keto-alfa-hydroxy carboxylic acid to yield gamma-keto acids catalyzed by a PdCl2(PPh3)2 precursor in combination with hydrochloric acid

The catalytic system PdCl2(PPh3)2-HCl is highly active and selective in the hydrogen transfer reaction from H2O-CO to PhCOCH2CHOHCOOH which yields the corresponding gamma-keto acid PhCOCH2CH2COOH, with concomitant evolution of CO2. An increase of temperature, pressure of carbon monoxide and catalyst concentration have a beneficial effect on the reaction rate, which appears to be of the first order in the substrate and passes through a maximum when varying the concentration of HCl. It is proposed that one important function of HCl is to give rise to chloride PhCOCH2CHClCOOH which interacts with a palladium hydride that takes origin from the decarboxylation of a species having a Pd-COOH moiety, which in turn results from the interaction of H2O and CO on the metal center. The yield passes through a maximum on increasing the concentration of H2O. This trend is attributed to the fact that. on one hand, H2O favors the formation of the Pd-COOH species, while, on the other hand, it may compete with other reacting molecules for coordination to the metal center. Moreover. H2O does not favor the formation of the chloride. When employed in relatively high concentration, the catalyst precursor has been recovered as a complex of palladium(0), Pd3(CO)3(PPh3)3 or Pd(CO)(PPh3)3, the latter in the presence of PPh3. The reduction to palladium(O) takes place only in the presence of H2O and is likely to occur via the intermediacy of a Pd-COOH species, which after CO2 evolution gives the reduced complex probably via reductive elimination of HCl from the hydride intermediate trans-PdHCl(PPh3)2. Moreover, PhCOCH=CHCOOH in combination with HCl (equivalent to PhCOCH2CHClCOOH) reacts with Pd(CO)(PPh3)3 to give the hydrogenated product PhCOCH2CH2COOH and PdCl2((PPh3)2. On the basis of these results, and knowing that HCl reacts with Pd(CO)(Ph3)3 to give the hydride PdHCl(PPh3)2, it is proposed that the catalytic cycle proceeds through the following steps: (i) H2O and CO interact with the metal center of the precursor yielding a Pd-COOH species, (ii) this gives off CO2 with formation of a hydride. (iii) this interacts with chloride PhCOCH2CHClCOOH to yield the product PhCOCH2CH2COOH and the palladium(I1) precursor back to the catalytic cycle.