Synthesis and Reactivity of Ir(I) and Ir(III) Complexes with MeNH<sub>2</sub>, Me<sub>2</sub>C═NR (R = H, Me), <i>C,N-</i>C<sub>6</sub>H<sub>4</sub>{C(Me)═N(Me)}-2, and <i>N,N′</i>-RN═C(Me)CH<sub>2</sub>C(Me<sub>2</sub>)NHR (R = H, Me) Ligands
作者:José Vicente、María Teresa Chicote、Inmaculada Vicente-Hernández、Delia Bautista
DOI:10.1021/ic801184v
日期:2008.10.20
Complexes [Ir(Cp*)Cl(n)(NH2Me)(3-n)]X(m) (n = 2, m = 0 (1), n = 1, m = 1, X = Cl (2a), n = 0, m = 2, X = OTf (3)) are obtained by reacting [Ir(Cp*)Cl(mu-Cl)]2 with MeNH2 (1:2 or 1:8) or with [Ag(NH2Me)2]OTf (1:4), respectively. Complex 2b (n = 1, m = 1, X = ClO 4) is obtained from 2a and NaClO4 x H2O. The reaction of 3 with MeC(O)Ph at 80 degrees C gives [Ir(Cp*)C,N-C6H4C(Me)=N(Me)}-2}(NH2Me)]OTf (4), which
配合物[Ir(Cp *)Cl(n)(NH2Me)(3-n)] X(m)(n = 2,m = 0(1),n = 1,m = 1,X = Cl(2a) ,n = 0,m = 2,X = OTf(3))是通过使[Ir(Cp *)Cl(mu-Cl)] 2与MeNH2(1:2或1:8)或与[Ag( NH2Me)2] OTf(1:4)。由2a和NaClO4 x H2O获得络合物2b(n = 1,m = 1,X = ClO 4)。3与MeC(O)Ph在80℃下反应得到[Ir(Cp *)C,N-C6H4 C(Me)= N(Me)}-2}(NH2Me)] OTf(4),依次与RNC反应得到[Ir(Cp *)C,N-C6H4 C(Me)= N(Me)}-2}(CNR)] OTf(R =(t)Bu(5), XY(6))。[Ir(mu-Cl)(COD)] 2与[Ag N(R)= CMe2} 2] X(1:2)反应生成[Ir