TpPtMe(H) ((2):为什么甲基组有H/D杂乱,但没有甲损失?
Mark A Iron1, H Christine Lo, Jan M L Martin
1Department of Organic Chemistry, Weizmann Institute of Science, 76100 Rehovot, Israel.
Journal of the American Chemical Society
|June 13, 2002
概括
酸复合物TPPtMe (H) (2) 由于其刚性连接体结构,可以抵抗甲的损失. 然而,它很容易经历可逆的H/D与脱甲醇混杂.
科学领域:
- 有机金属化学 有机金属化学
- 计算化学的计算化学
- 反应机制 解 反应机制
背景情况:
- 复合物的与化和甲基联结物的反应性对于催化是至关重要的.
- 了解连接体对复杂稳定性和反应通路的影响是关键.
- TpPtMe (H) (2) 呈现了一个独特的稳定性案例,可以抵御甲的损失.
研究的目的:
- 调查TpPtMe的反应性,特别是其对甲损失的抵抗力.
- 阐明使用化溶剂观察到的H/D杂乱的机制.
- 确定刚性Tp连接体在观察到的反应性中的作用.
主要方法:
- 用密度函数理论 (DFT) 的计算来研究反应.
- 计算是在mPW1k/LANL2DZ+P//mPW1k/LANL2DZ理论层面进行的.
- 一个模型系统,[(NH(3)) ((3) PtMe(H) ((2)) ](+),用于比较分析.
主要成果:
- Tp联体的刚性阻止了消除甲所需的变体几何,使其在能量方面不利.
- 甲基和化物联体的可逆H/D杂乱很容易通过一个Eta2-CH-CH4中间体发生.
- 没有观察到甲损失,证实了Tp联体的稳定作用.
结论:
- Tp联体的刚性是抑制TPPtMe中的甲损失的主要因素.
- H/D编码是一个简单的过程,通过一个明显的中间路径进行.
- 对类似系统进行精确的能量计算需要特定的DFT级别,例如mPW1k/SDB-cc-pVDZ.
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