Unravelling the reaction path of rhodium-MonoPhos-catalysed olefin hydrogenation

The mechanism of the asymmetric hydrogenation of methyl (Z)-2-acetamidocinnamate (mac) catalysed by [Rh(MonoPhos)(2)(nbd)]SbF(6) (MonoPhos: 3,5-dioxa-4-phosphacyclohepta[2,1-a:3,4-a']dinaphthalen-4-yl)dimethylamine) was elucidated by using (1)H, (31)P and (103)Rh NMR spectroscopy and ESI-MS. The use of nbd allows one to obtain in pure form the rhodium complex that contains two units of the ligand. In contrast to the analogous complexes that contain cis,cis-1,5-cyclooctadiene (cod), this complex shows well-resolved NMR spectroscopic signals. Hydrogenation of these catalyst precursors at 1 bar total pressure gave rise to the formation of a bimetallic complex of general formula [Rh(MonoPhos)(2)](2)(SbF(6))(2); no solvate complexes were detected. In the dimeric complex both rhodium atoms are ligated to two MonoPhos ligands but, in addition, each rhodium atom also binds to one of the binaphthyl rings of a ligand that is bound to the other rhodium metal. Upon addition of mac, a mixture of diastereomeric complexes [Rh(MonoPhos)(2)(mac)]SbF(6) is formed in which the substrate is bound in a chelate fashion to the metal. Upon hydrogenation, these adducts are converted into a new complex [Rh(MonoPhos)(2){mac(H)(2)}]SbF(6) in which the methyl phenylalaninate mac(H)(2) is bound through its aromatic ring to rhodium. Addition of mac to this complex leads to displacement of the product by the substrate. No hydride intermediates could be detected and no evidence was found for the involvement at any stage of the process of complexes with only one coordinated MonoPhos. The collected data suggest that the asymmetric hydrogenation follows a Halpern-like mechanism in which the less abundant substrate-catalyst adduct is preferentially hydrogenated to phenylalanine methyl ester.