首页> 外文OA文献 >Reactivity of a (bis-NHC)tricarbonylruthenium(0) complex with methyl triflate and methyl iodide. Formation of methyl- and acetylruthenium(II) derivatives: Experimental results and mechanistic DFT calculations
【2h】

Reactivity of a (bis-NHC)tricarbonylruthenium(0) complex with methyl triflate and methyl iodide. Formation of methyl- and acetylruthenium(II) derivatives: Experimental results and mechanistic DFT calculations

机译:(双-NHC)三羰基钌(0)配合物与三氟甲磺酸甲酯和甲基碘的反应性。甲基和乙酰基钌(II)衍生物的形成:实验结果和力学DFT计算

摘要

The ruthenium(0) complex [Ru{κ2C2-MeIm(CH 2)3ImMe}(CO)3] (1), MeIm(CH2) 3ImMe = 1,3-bis(3-methylimidazol-2-yliden-1-yl)propane, which contains a chelating bis(N-heterocyclic carbene) ligand, reacts with MeOTf at room temperature to give the ionic ruthenium(II) methyl derivative [RuMe{κ2C2-MeIm(CH2)3ImMe}(CO) 3]OTf ([2]OTf), whereas an analogous reaction of 1 with MeI renders the neutral ruthenium(II) acetyl derivative [RuI{C(O)Me} {κ2C2-MeIm(CH2)3ImMe}(CO) 2] (3), in which the iodide and acetyl ligands occupy mutually trans coordination sites. The fact that [2]OTf reacts with [Et4N]I to give 3 evidences the participation of the cationic species 2+ in the synthesis of 3. The mechanisms of these reactions in THF solution have been modeled by DFT calculations. They have shown that 2+ can be made from complex 1 and either MeOTf or MeI. In both cases, two plausible reaction pathways have been identified. They start with a rate-determining SN2 substitution process in which the metal atom of 1 attacks the C atom of MeI or MeOTf, displacing the corresponding anion to give, depending on how compound 1 approaches MeI or MeOTf, 2+ or a less stable isomeric species 2′+ that is easily transformed into 2+. A subsequent CO migratory insertion in 2+ leads to an unsaturated (pentacoordinated) acetyl intermediate that easily adds the iodide anion to end in 3. DFT calculations have also shown that the reaction of 1 with MeOTf to give [2]OTf is thermodynamically more favorable than that of 1 with MeI to give [2]I due to the resonance stabilization and greater solvation energy of the triflate anion. These two effects are also responsible for the fact that the incorporation of the triflate anion with 2+ to give a putative triflate complex analogous to 3 is thermodynamically disfavored, whereas the incorporation of the iodide anion with 2+ to give 3 is thermodynamically favored. © 2013 American Chemical Society.
机译:钌(0)络合物[Ru {κ2C2-MeIm(CH 2)3ImMe}(CO)3](1),MeIm(CH2)3ImMe = 1,3-双(3-甲基咪唑-2-yliden-1-yl )含有螯合双(N-杂环卡宾)配体的丙烷在室温下与MeOTf反应,得到离子钌(II)甲基衍生物[RuMe {κ2C2-MeIm(CH2)3ImMe}(CO)3] OTf( [2] OTf),而1与MeI的类似反应会生成中性钌(II)乙酰基衍生物[RuI {C(O)Me} {κ2C2-MeIm(CH2)3ImMe}(CO)2](3),其中碘化物和乙酰基配体占据相互反式配位点。 [2] OTf与[Et4N] I反应生成3的事实证明阳离子物种2+参与了3的合成。这些反应在THF溶液中的机理已通过DFT计算进行了建模。他们表明2 +可以由复合物1和MeOTf或MeI制成。在这两种情况下,已经确定了两个合理的反应途径。他们从确定速率的SN2取代过程开始,其中1的金属原子攻击MeI或MeOTf的C原子,取代相应的阴离子以生成,取决于化合物1如何接近MeI或MeOTf,2 +或不稳定的异构体物种2'+很容易转化为2+。随后的CO迁移插入2+会导致不饱和的(五配位的)乙酰基中间体,该中间体容易将碘阴离子添加到3中。DFT计算还表明1与MeOTf的反应生成[2] OTf在热力学上更有利由于三氟甲磺酸根阴离子的共振稳定和更大的溶剂化能,因此比用MeI的1产生[2] I。这两个效应也是造成以下事实的原因:在热力学上不利的是将三氟甲磺酸根阴离子与2+结合以产生类似于3的推定的三氟甲磺酸盐络合物,而在热力学上不利于将碘化物阴离子与2+结合以产生3。 ©2013美国化学学会。

著录项

相似文献

  • 外文文献
  • 中文文献
  • 专利

客服邮箱:kefu@zhangqiaokeyan.com

京公网安备:11010802029741号 ICP备案号:京ICP备15016152号-6 六维联合信息科技 (北京) 有限公司©版权所有
  • 客服微信

  • 服务号