作者:Juan M. Coronas、Guillermo Muller、Mercedes Rocamora
DOI:10.1016/0022-328x(86)80013-4
日期:1986.2
R-R′, RBr and R′Br. The formation of the products R-R′ is increasingly favoured the greater the electronegativity of the organic ligands. The reductive elimination giving RBr takes place more readily the greater the electronegativity of the organic ligand R. The product of the reductive elimination reaction is “NiIBr”, “NiIR”, or “NiIR′”, which in the presence of bromine give Ni2+, [NiBr(RBr)L2], phosphonium
已经对化合物t- [NiRR'L 2 ](L = PMe 2 Ph和PEt 3; R =芳基或乙烯基)和[Ni(mes)(邻甲苯基)bipy]的分解进行了研究( mes = mesityl)通过电化学方法或溴氧化诱导。镍的无有机金属化合物III,分离在上述反应中,但Ni的五配位中间体III推测。Ni III中间体形成后很容易发生分解。如果分解是电化学诱导的,则中间体分解,仅给出偶联产物RR'。当溴用作氧化剂时,Ni III仅在配体的体积允许与中心原子配位的情况下形成中间体。因此,[Ni(C 2 Cl 3)(mes)(PMe 2 Ph)2完全不分解,仅获得[Ni(C 2 Cl 3)(mesBr 2)(PMe 2 Ph)2 ]。中间体“ Ni III RR'BrL 2进行还原性消除,得到RR',RBr和R'Br。有机配体的电负性越大,越来越倾向于形成产物RR'。有机配体R的电负性越大,给出RB