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C–F Bond Activation of P(C6F5)3 by Ruthenium Dihydride Complexes: Isolation and Reactivity of the “Missing” Ru(PPh3)3H(halide) Complex, Ru(PPh3)3HF
Inorganic Chemistry ( IF 4.3 ) Pub Date : 2018-10-10 00:00:00 , DOI: 10.1021/acs.inorgchem.8b02286
Mateusz K. Cybulski 1 , Caroline J. E. Davies 1 , John P. Lowe 1 , Mary F. Mahon 1 , Michael K. Whittlesey 1
Affiliation  

The major product of the reaction between Ru(IMe4)2(PPh3)2H2 (1; IMe4 = 1,3,4,5-tetramethylimidazol-2-ylidene) and P(C6F5)3 (PCF) is the five-coordinate complex Ru(IMe4)2(PF2{C6F5})(C6F5)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(PPh3)4H2 to afford Ru(PPh3)3HF (3). NaBArF4 abstracted the fluoride ligand in 3 to give [Ru({η6-C6H5}PPh2)(PPh3)2H][BArF4], while B2pin2 reacted with 3 in C6D6 to yield a mixture of [Ru({η6-C6D6)(PPh3)2H]+ and Ru(PPh3)4H2. The treatment of 3 with HBpin (5 equiv) and HSiR3 (R = Et, Ph; 2 equiv) afforded Ru(PPh3)3(σ-HBpin)H2 and Ru(PPh3)3(SiR3)3H3, respectively. No stable substitution products were generated when 3 was reacted with Me3SiX (X = CF3, C6F5).

中文翻译:

二氢化钌配合物对P(C 6 F 53的CF键活化:“缺失”的Ru(PPh 33 H(卤化物)配合物Ru(PPh 33 HF的分离和反应性

Ru(IMe 42(PPh 32 H 21 ; IMe 4 = 1,3,4,5-四甲基咪唑-2-亚烷基)与P(C 6 F 53( PCF)是五坐标复数Ru(IMe 42(PF 2 {C 6 F 5 })(C 6 F 5)H(2),由复数C–F / P–C形成键断裂和PF键形成步骤。相反,PCF中所有六个邻位CF键的加氢脱氟都与Ru(PPh34 H 2提供Ru(PPh 33 HF(3)。NaBAr ˚F 4抽象的氟化物配体3,得到的[Ru({η 6 -C 6 H ^ 5 } PPH 2)(PPH 32 H] [巴˚F 4 ],而B 22与反应3用C 6 d 6,得到的混合物的[Ru({η 6 -C 6 d 6)(PPH 32 H] +和Ru(PPh 34 H 2。治疗3与HBpin(5当量)和HSIR 3(R =的Et,Ph值; 2当量),得到的Ru(PPH 33(σ-HBpin)H 2和Ru(PPH 33(SIR 33 ħ 3,分别。当3与Me 3 SiX(X = CF 3,C 6 F 5)反应时,没有生成稳定的取代产物。
更新日期:2018-10-10
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