How Many Methyl Groups in [{(η5-C5MenH5−n)2Zr}2(μ2,η2,η2-N2)] Are Needed for Dinitrogen Hydrogenation? A Theoretical Study

Five are too many: It is known experimentally that [(Cp′)2Zr(N2)Zr(Cp′)2] (Cp′=C5H5−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 suitabl...

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Bibliographic Details
Published in:Angewandte Chemie International Edition 2005-11, Vol.44 (43), p.7101-7103
Main Authors: Bobadova-Parvanova, Petia, Wang, Qinfang, Morokuma, Keiji, Musaev, Djamaladdin G.
Format: Article
Language:English
Subjects:
computer chemistry
density functional calculations
homogeneous catalysis
Hydrogenation
ligand effects
Models, Chemical
Molecular Structure
Nitrogen - chemistry
nitrogen fixation
Organometallic Compounds - chemical synthesis
Organometallic Compounds - chemistry
Stereoisomerism
Zirconium - chemistry
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Summary:Five are too many: It is known experimentally that [(Cp′)2Zr(N2)Zr(Cp′)2] (Cp′=C5H5−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.
ISSN:1433-7851
1521-3773
DOI:10.1002/anie.200501371