Cleavage of F–C(sp2) bonds by MHR(CO)(PtBu2Me)2 (M = Os and Ru; R = H, CH3 or Aryl): Product dependence on M and R

Both MH(Ph)(CO)L2 (L = PtBu2Me; M = Ru and Os) react with vinyl fluoride to form M–F bonds; however, Ru eliminates benzene, while Os eliminates ethylene. In contrast, Ru(H)2(CO)L2 and Os(H)2(CO)(1-butene)L2 both react with vinyl fluoride to give ethylene and MHF(CO)L2. Ethylene production from both dihydrides is attributed to β-F migration to M from an MCH2CH2F transient, while the unique behavior of RuH(Ph)(CO)L2 (giving the C–F oxidative addition product Ru(η1-vinyl)F(CO)L2) is attributed to the difficulty of achieving RuIV, and the ability of the strongly π-acidic vinyl fluoride to rapidly trigger reductive elimination of benzene. The products of reaction of RuH(Ar)(CO)L2 with vinyl fluoride are redirected more towards ethylene formation when Ar carries fluorine substituents. The reaction products of OsH(R)(CO)L2 with vinyl fluoride revert to R-H elimination when R is methyl. Finally, the more π-acidic H2CCF2 triggers very rapid CH4 elimination from OsH(CH3)(CO)L2; cleavage of the second C–F bond yields the vinylidene OsF2(CCH2)(CO)L2. All selectivity is rationalized via the fate of the adduct MH(R)(C2H4 − nFn)(CO)L2.

Graphical abstractThe very different products of vinyl C–F bonds reacting with unsaturated Ru vs. Os Hydride compounds are explained in terms of the lifetimes of the adducts of metal complex with fluoro-olefin.Figure optionsDownload full-size imageDownload as PowerPoint slide