400608-25-5Relevant academic research and scientific papers
Protonation and hydrogenation experiments with Iridium(0) and Iridium(-1) tropp complexes: Formation of hydrides
Boehler, Carsten,Avarvari, Narcis,Schoenbcrg, Harfmut,Woerle, Michael,Rueegger, Heinz,Gruetzmacher, Hansjoerg
, p. 3127 - 3147 (2007/10/03)
The reactions of three different tetracoordinated Ir complexes, [Ir(troppph)2]n (n = + 1, 0, - 1), which differ in the formal oxidation state of the metal from + 1 to - 1, with proton sources and dihydrogen were investigated (tropp = 5-(diphenylphosphanyl)dibenzo[a,d] cycloheptene). It was found that the cationic 16-electron complex [Ir(troppph)2]+ (2) cannot be protonated but reacts with NaBH4 to the very stable 18-electron IrI hydride [IrH(troppph)2] (5), which is further protonated with medium strong acids to give the 18-electron IrIII dihydride [IrH2(troppph)2]+ (6; pKs in CH2Cl2/THF/H2O 1:1:2 ca. 2.2). Both, the neutral 17-electron Ir0 complex [Ir(troppph)2] (3) and the anionic 18-electron complex [Ir(troppph)2]- (4) react rapidly with H2O to give the monohydride 5. In reactions of 3 with H2O, the terminal IrI hydroxide [Ir(OH)(troppph)2] (8) is formed in equal amounts. All these complexes, apart from 5, which is inert, do react rapidly with dihydrogen. The complex 2 gives the dihydride 6 in an oxidative addition reaction, while 3, 4, and 8 give the monohydride 5. Interestingly, a salt-type hydride (i.e., LiH) is formed as further product in the unexpected reaction with [Li(thf)x]+-[Ir(troppph)2] - (4). Because 3 undergoes disproportionation into 2 and 4 according to 2 3?2 + 4 (Kdisp = 2.7 · 10-5), it is likely that actually the diamagnetic species and not the odd-electron complex 3 is involved in the reactions studied here, and possible mechanisms for these are discussed.
