On the mechanism of the hydrogen transfer from H 2OCO to γ-keto-α-hydroxy carboxylic acids to yield γ-keto acids catalyzed by a PdCl 2(PPh 3) 2 precursor in combination with hydrochloric acid

Abstract

The catalytic system PdCl 2(PPh 3) 2HCl is highly active and selective in the hydrogen transfer reaction from H 2OCO to PhCOCH 2CHOHCOOH which yields the corresponding γ-keto acid PhCOCH 2CH 2COOH, with concomitant evolution of CO 2. 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 PhCOCH 2CHClCOOH 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 H 2O and CO on the metal center. The yield passes through a maximum on increasing the concentration of H 2O. This trend is attributed to the fact that, on one hand, H 2O 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, H 2O 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), Pd 3(CO) 3(PPh 3) 3 or Pd(CO)(PPh 3) 3, the latter in the presence of PPh 3. The reduction to palladium(0) takes place only in the presence of H 2O and is likely to occur via the intermediacy of a PdCOOH species, which after CO 2 evolution gives the reduced complex probably via reductive elimination of HCl from the hydride intermediate trans-PdHCl(PPh 3) 2. Moreover, PhCOCHCHCOOH in combination with HCl (equivalent to PhCOCH 2CHClCOOH) reacts with Pd(CO)(PPh 3) 3 to give the hydrogenated product PhCOCH 2CH 2COOH and PdCl 2(PPh 3) 2. On the basis of these results, and knowing that HCl reacts with Pd(CO)(PPh 3) 3 to give the hydride PdHCl(PPh 3) 2, it is proposed that the catalytic cycle proceeds through the following steps: (i) H 2O and CO interact with the metal center of the precursor yielding a PdCOOH species, (ii) this gives off CO 2 with formation of a hydride, (iii) this interacts with chloride PhCOCH 2CHClCOOH to yield the product PhCOCH 2CH 2COOH and the palladium(II) precursor back to the catalytic cycle.