A New Family of Carbenerhodium(I) Complexes: Ligand Variation as The Key to Success

Abstract

AbstractA synthetic methodology to obtain square‐planar carbenerhodium(I) complexes of the general composition trans‐[RhCl(CRR′)(L)2] where L is a tertiary phosphane, arsane, or stibane has been developed. The starting material trans‐[RhCl(C2H4)(SbiPr3)2] (3) reacts with diazoalkanes RR′CN2 [RR′ = Ph2, Ph(C6H4X), (C6H4X)2, Ph(CF3), C12H8] under mild conditions to give the compounds trans‐[RhCl(CRR′)(SbiPr3)2] (4–11) almost quantitatively. On treatment of 3 with EtO2CCHN2 and PhC(N2)C(O)R, the olefinrhodium and diazoalkanerhodium compounds trans‐[RhCl{(E)‐C2H2(CO2Et)2}(SbiPr3)2] (12) and trans‐[RhCl{N2C(R)C(O)Ph}(SbiPr3)2] (13, 14) are obtained instead of carbene complexes. Displacement of the SbiPr3 ligands in 4 (R = R′ = Ph) by PiPr3, PiPr2Ph, PiPrPh2, PPh3, PPh2Me, AsiPr3, and SbEt3 leads to the corresponding carbene complexes trans‐[RhCl‐(CPh2)(L)2] (15–21) in high yields. The results of the X‐ray crystal structure analyses of 4 and 15 (L = PiPr3) illustrate that the different donor—acceptor properties of SbiPr3 and PiPr3 have little influence on the RhC bond length. The reactions of 4 and 15 with CO and CNtBu afford, by metal‐assisted CC coupling, diphenylketene Ph2CCO (23) and the corresponding imine Ph2CCNtBu (26). On treatment of 4 and 15 with ethene, however, two different olefinic products, 3,3‐diphenyl‐1‐propene (31) and 1,1‐diphenyl‐1‐propene (32), are formed. Compound 15 reacts with KBr, NaOPh, and NaC5H5 by substitution of the chloride to give trans‐[RhBr(CPh2)‐(PiPr3)2] (33), trans‐[Rh(OPh)(CPh2)‐(PiPr3)2] (34) and [C5H5Rh(CPh2)‐(PiPr3)] (35), and with HCl by oxidative addition to yield [RhCl2(CHPh2)‐(PiPr3)2](36).