How many methyl groups in [{(η5-C5MenH5-n)2Zr}2(μ2,η2,η2-N2)] are needed for dinitrogen hydrogenation?: A theoretical study

被引：27

作者：

Bobadova-Parvanova, P

Wang, QF

Morokuma, K ^{[1
]}

Musaev, DG

机构：

[1] Emory Univ, Cherry L Emerson Ctr Sci Computat, Atlanta, GA 30322 USA

[2] Emory Univ, Dept Chem, Atlanta, GA 30322 USA

来源：

ANGEWANDTE CHEMIE-INTERNATIONAL EDITION | 2005年 / 44卷 / 43期

关键词：

computer chemistry; density functional calculations; homogeneous catalysis; ligand effects; nitrogen fixation;

D O I：

10.1002/anie.200501371

中图分类号：

O6 [化学];

学科分类号：

0703 ;

摘要：

Five are too many: It is known experimentally that [(Cp') 2Zr(N2)Zr(Cp')2] (Cp' = C5H 5-nMen) with n = 4 activates N2 for hydrogenation, but the complex with n = 5 does not. This difference in reactivity arises from the N2 coordination mode (see scheme): For n = 0-4, N2 is side-on coordinated - the most suitable mode for hydrogenation - whereas for n = 5, N2 is end-on coordinated owing to steric repulsion between five Me groups. © 2005 Wiley-VCH Verlag GmbH & Co. KGaA.

引用

页码：7101 / 7103

页数：3

共 14 条

[1] Call the binuclear dinitrogen complex [P2N2]Zr(μ-η2-N2)Zr[P2N2] activate more than one hydrogen molecule?: A theoretical study [J].