通过硫羰基还原简单生成二钌 μ-碳化配合物。

Simple generation of a dirhodium μ-carbido complex via thiocarbonyl reduction.

机构信息

Research School of Chemistry, Australian National University, Canberra, Australian Capital Territory, ACT 2601, Australia.

出版信息

Dalton Trans. 2018 Jul 24;47(29):9570-9574. doi: 10.1039/c8dt02524j.

DOI:10.1039/c8dt02524j

PMID:29969507

Abstract

The reaction of [RhCl(CS)(PPh3)2] with excess catecholborane affords the cumulenic carbido complex [Rh2(μ-C)Cl2(PPh3)4] which undergoes phosphine and halide substitution to afford a range of complexes in which the Rh[double bond, length as m-dash]C[double bond, length as m-dash]Rh spine remains intact. Amongst these, the reactions with K[L] (L = H2B(pz)2, H2B(pzMe2)2, HB(pz)3; pz = pyrazol-1-yl) afford [Rh2(μ-C)(PPh3)2(L)2] whilst with K[HB(pzMe2)3] the unsymmetrical complex [Rh2H(μ-C)(μ-C6H4PPh2-2){HB(pzMe2)3}2] is obtained in which the carbido ligand spans d6-Rh(iii) and d8-Rh(i) centres.

摘要

[RhCl(CS)(PPh3)2]与过量邻苯二硼烷反应生成累积型碳化物配合物[Rh2(μ-C)Cl2(PPh3)4]，该配合物可进行膦和卤化物取代，得到一系列 Rh[双键,长度 as m-dash]C[双键,长度 as m-dash]Rh 键保持完整的配合物。其中，与 K[L]（L = H2B(pz)2、H2B(pzMe2)2、HB(pz)3；pz = 吡唑-1-基）反应生成[Rh2(μ-C)(PPh3)2(L)2]，而与 K[HB(pzMe2)3]反应则得到不对称配合物[Rh2H(μ-C)(μ-C6H4PPh2-2){HB(pzMe2)3}2]，其中碳化物配体跨越 d6-Rh(iii)和 d8-Rh(i)中心。