[IrI(bpa–H)(cod)] (2). All attempts to prepare the rhodium analog of 2 failed and led to the spontaneous formation of 7. The thermodynamic differences are readily explained by a lower stability of the M–I oxidation state for iridium as compared to rhodium. The observed reductive double deprotonation leads to the formation of unusual structures and unexpected reactivity, which underlines the general importance
配合物[Rh I(
BPA)(cod)] +([ 4 ] +,
BPA = PyCH 2 NHCH 2 Py)与一摩尔当量碱的协同还原性双去质子化生成双
金属物质[(cod)Rh(
BPA– 2H)Rh(cod)](7),显示出大的Rh –I,Rh I对其电子结构的贡献。7中双去质子化的
配体在两个不同的区室中托管两个“ Rh(cod)”片段:一个由Py供体和一个中央氮供体组成的“方形平面区”和一个由另一个
吡啶和一个“
亚胺C═N”供体。在这个过程中“
亚胺供体”的形成是从
BPA-2H}大量电子转移的结果2-
配体与
铑中心之一,以形成中性
亚胺配体B
PI(B
PI = PyCH 2 N═CHPy) 。因此,[Rh I(
BPA)(cod)] +的去质子化代表还原过程,有效地导致
金属氧化态从Rh I还原为Rh –I。双核
铱对应物,复合物8也可以制备,但在1摩尔当量的游离
BPA配体存在时不稳定,导致定量形成中性酰胺基单核化合物[Ir