The compound [Ru3(CO)12] reacts at room temperature in tetrahydrofuran with tri(2-pyridyl)phosphine, P(2-C5H4N)3′ when catalysed by [N(PPh3)2]Cl, to give a product derived by cleavage of a phosphorus-2-pyridyl bond, [Ru3(µ-2-C5H4N){µ3-P(2-C5H4N)2}(CO)9]1. No simple substitution compounds could be isolated. The likely intermediates [Ru3(CO)n{P(2-C5H4N)3}](n= 10 and/or 11) spontaneously decarbonylate while a 2-pyridyl group migrates from phosphorus to ruthenium atoms. A similar reaction using phenyldi(2-pyridyl) phosphine, PPh(2-C5H4N)2′ leads to the competitive transfer of 2-pyridyl and pheyl groups to give almost equal quantities of two products, [Ru3(µ-2-C5H4N){µ3-PPh(2-C5H4N)}-(CO)9] and [Ru3(µ-PhCO){µ3-P(2-C5H4N)2}(CO)9]. The crystal structure of 1 shows that the 2-pyridyl group that has migrated bridges two ruthenium atoms with no Ru-Ru bond between them and that the di(2-pyridyl) phosphido ligand is triply bridging with one nitrogen co-ordinated and one free. As in the reported reaction of [Ru3(CO)12] with PPh2(2-C5H4N) to give [Ru3(µ-PhCO){µ3-PPh(2-C5H4N)}(CO)9], the reason why P-C bond cleavage occurs under remarkably mild conditions and the mechanism of the processes have still to be determined.
No takes yet. Share an insight, caveat, or question.
Deeming et al. (1993) studied this question.
Synapse has enriched 5 closely related papers on similar clinical questions. Consider them for comparative context: