Insight into the mechanism of decarbonylation of methanol by ruthenium complexes; a deuterium labelling study

In the reaction of [RuHClP3] (P = PPh3) with NaOMe in methanol, the product is [RuH2(CO)P3]. Short reaction times show that the final product is formed through [RuH4P3] as the major intermediate. Using NaOCD3 in CD3OD, the first formed product is [RuH4P′3] (P′ is PPh3 partially deuterated in the ortho positions of the aromatic rings). Further reaction leads to a mixture of [RuHnD2−n(CO)P3] (n = 0, 22%; n = 1, 2 isomers each 28%; n = 2, 22%). Mechanistic aspects of both steps of the reaction are explored and, together with previously published calculations, they provide definitive mechanisms for both dehydrogenation and decarbonylation in these interesting systems.