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Amide Complexes of Zirconium, Rhodium, and Iridium:  Synthesis and Reactivity. X-ray Crystal Structures of (η5-C5H5)2Zr(NHC6H4-o-SMe)2 and [Rh(μ-SC6H4-o-NHMe)(COD)]2

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journal contribution
posted on 15.03.1998 by Rosa Fandos, Martín Martínez-Ripoll, Antonio Otero, María José Ruiz, Ana Rodríguez, Pilar Terreros
The reaction of Cp2ZrCl2 (Cp = η5-C5H5) with 2 equiv of the lithium amide derivative LiNHC6H4-o-SMe affords the new zirconium complex Cp2Zr(NHC6H4-o-SMe)2 (2). The structure of 2 has been determined by X-ray diffraction. When the reaction is carried out in an 1:1 ratio, the complex Cp2ZrCl(NHC6H4-o-SMe) (3) is generated as the major product. Reaction of “Cp*2Zr” (Cp* = η5-C5Me5) with 2-(methylmercapto)aniline yields a hydride−amide complex Cp*2ZrH(NHC6H4) (4). Reaction of complex 2 with [RhCl(COD)]2 generates complex 3 and the new rhodium amide complex Rh(NHC6H4-o-SMe)(COD), which has been also directly synthesized by reacting [RhCl(COD)]2 with LiNHC6H4-o-SMe. Thermolysis of complex 6, at 100 °C, produces to a new rhodium thiolate complex [Rh(μ-SC6H4-o-NHMe)(COD)]2 (5). Its structure has been determined by X-ray diffraction methods. Reaction of [IrCl(COD)]2 with LiNHC6H4-o-SMe gives the iridium(III) complex Ir(Me)(SC6H4NH)(COD) (7) by oxidative addition of the S−Me bond.