Carbenerhodium Complexes of the Half-Sandwich-Type: Synthesis, Substitution, and Addition Reactions

A series of carbenerhodium(I) complexes of the general composition [(η5-C5H5)Rh(=CRR′)(L)] (2 a–2 i) with R=R′=aryl and L=SbiPr3 or PR3 has been prepared from the square-planar precursors trans-[RhCl(=CRR′)(L)2] and NaC5H5 in excellent yields. Reaction of the triisopropylstibane derivative 2 a, which contains a rather labile Rh−Sb bond, with CO, PMe3, and CNR (R=Me, CH2Ph, tBu) leads to the displacement of the SbiPr3 ligand and affords the substitution products [(η5-C5H5)Rh(=CPh2)(L)] (3–7). In contrast, treatment of the triisopropylphosphane compound 2 c with CO and CNtBu leads to the cleavage of the Rh=CPh2 bond and gives besides [(η5-C5H5)Rh(PiPr3)(L)] (10, 12) by metal-assisted CC coupling diphenylketene Ph2C=C=O (11) or the corresponding imine Ph2C=C=NtBu (13). While the reaction of 2 a, c with C2H4 yields [(η5-C5H5)Rh(C2H4)(L)] (14, 15) and the trisubstituted olefin Ph2C=CHCH3 (16), treatment of 2 a, c with RN3 leads to the cleavage of both the Rh−EiPr3 and Rh=CPh2 bonds and gives the chelate complexes [(η5-C5H5)Rh(κ2-RNNNNR)] (19, 20). The substitution products 3 (L=CO) and 4 (L=PMe3) react with an equimolar amount of sulfur or selenium by addition of the chalcogen to the Rh=CPh2 bond to generate the complexes [(η5-C5H5)Rh(κ2-ECPh2)(L)] (21–24) with thio- or selenobenzophenone as ligand. Similarly, treatment of 3 with CuCl affords the unusual 1:2 adduct [(η5-C5H5)(CO)Rh(μ-CPh2)(CuCl)2] (25), which reacts with NaC5H5 to form [(η5-C5H5)(CO)Rh(μ-CPh2)Cu(η5-C5H5)] (26). The molecular structures of 3 and 22 have been determined by X-ray crystallography.