Synthesis and Isomeric Analysis of Ru<sup>II</sup> Complexes Bearing Pentadentate Scaffolds
作者:Marcos Gil-Sepulcre、Jordan C. Axelson、Joan Aguiló、Lluís Solà-Hernández、Laia Francàs、Albert Poater、Lluís Blancafort、Jordi Benet-Buchholz、Gonzalo Guirado、Lluís Escriche、Antoni Llobet、Roger Bofill、Xavier Sala
DOI:10.1021/acs.inorgchem.6b01755
日期:2016.11.7
RuII-pentadentate polypyridyl complex [RuII(κ-N5-bpy2PYMe)Cl]+ (1+, bpy2PYMe = 1-(2-pyridyl)-1,1-bis(6–2,2′-bipyridyl)ethane) and its aqua derivative [RuII(κ-N5-bpy2PYMe)(H2O)]2+ (22+) were synthesized and characterized by experimental and computational methods. In MeOH, 1+ exists as two isomers in different proportions, cis (70%) and trans (30%), which are interconverted under thermal and photochemical conditions
甲钌II -pentadentate多吡啶配合物的[Ru II(κ-N 5 -bpy2PYMe)CL] +(1 +,bpy2PYMe = 1-(2-吡啶基)-1,1-双(6-2,2'联吡啶)乙烷)及其衍生物AQUA的[Ru II(κ-N 5 -bpy2PYMe)(H 2 O)] 2 +(2 2+被合成和表征通过实验和计算方法)。在MeOH中,1 +以两种不同比例的异构体形式存在,分别为顺式(70%)和反式(30%),它们在热和光化学条件下通过一系列过程相互转化:氯代去配位,吡啶基的去配位/配位和氯代再配位。下在二氯甲烷中氧化条件,反式- 1 2+产生的[Ru III(κ-N 4 -bpy2PYMe)氯2 ] +中间体的吡啶配位体的由氯的交换之后-抗衡离子,这也解释了反式/顺式异构化观察当系统被带回Ru(II)时。相反,独联体- 1 2+与直接平衡反式- 1 2+,与不存在κ-N的4 -双-