钌二氢配合物对 P(CF) 的 C-F 键活化：“缺失”的 Ru(PPh)H(卤化物)配合物，Ru(PPh)HF 的分离和反应性。

C-F Bond Activation of P(CF) by Ruthenium Dihydride Complexes: Isolation and Reactivity of the "Missing" Ru(PPh)H(halide) Complex, Ru(PPh)HF.

机构信息

Department of Chemistry , University of Bath , Claverton Down , Bath BA2 7AY , U.K.

出版信息

Inorg Chem. 2018 Nov 5;57(21):13749-13760. doi: 10.1021/acs.inorgchem.8b02286. Epub 2018 Oct 10.

DOI:10.1021/acs.inorgchem.8b02286

Abstract

The major product of the reaction between Ru(IMe)(PPh)H (1; IMe = 1,3,4,5-tetramethylimidazol-2-ylidene) and P(CF) (PCF) is the five-coordinate complex Ru(IMe)(PF{CF})(CF)H (2), which is formed via a complex series of C-F/P-C bond cleavage and P-F bond formation steps. In contrast, hydrodefluorination of all six ortho C-F bonds in PCF occurs with Ru(PPh)H to afford Ru(PPh)HF (3). NaBAr abstracted the fluoride ligand in 3 to give [Ru({η-CH}PPh)(PPh)H][BAr], while Bpin reacted with 3 in CD to yield a mixture of [Ru({η-CD)(PPh)H] and Ru(PPh)H. The treatment of 3 with HBpin (5 equiv) and HSiR (R = Et, Ph; 2 equiv) afforded Ru(PPh)(σ-HBpin)H and Ru(PPh)(SiR)H, respectively. No stable substitution products were generated when 3 was reacted with MeSiX (X = CF, CF).

摘要

反应产物 Ru(IMe)(PPh)H（1；IMe=1,3,4,5-四甲基咪唑-2-亚基）与 P(CF)（PCF）之间的主要产物是五配位配合物 Ru(IMe)(PF{CF})(CF)H（2），它是通过一系列复杂的 C-F/P-C 键断裂和 P-F 键形成步骤形成的。相比之下，Ru(PPh)H 对 PCF 中所有六个邻位 C-F 键的氢氟去氟化作用生成 Ru(PPh)HF（3）。NaBAr 从 3 中抽走氟化物配体，得到[Ru({η-CH}PPh)(PPh)H][BAr]，而 Bpin 在 CD 中与 3 反应生成[Ru({η-CD)(PPh)H]和 Ru(PPh)H 的混合物。3 与 HBpin（5 当量）和 HSiR（R=Et，Ph；2 当量）反应分别得到 Ru(PPh)(σ-HBpin)H 和 Ru(PPh)(SiR)H。当 3 与 MeSiX（X=CF，CF）反应时，没有稳定的取代产物生成。