How Many Methyl Groups in [{(η 5 ‐C 5 Me  n  H 5− n  ) 2 Zr} 2 (μ 2 ,η 2 ,η 2 ‐N 2 )] Are Needed for Dinitrogen Hydrogenation? A Theoretical Study | Zendy

BobadovaParvanova Petia | Zendy; Wang Qinfang | Zendy; Morokuma Keiji | Zendy; Musaev Djamaladdin G. | Zendy

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How Many Methyl Groups in [{(η 5 ‐C 5 Me n H 5− n ) 2 Zr} 2 (μ 2 ,η 2 ,η 2 ‐N 2 )] Are Needed for Dinitrogen Hydrogenation? A Theoretical Study

Author(s) -

BobadovaParvanova Petia,

Wang Qinfang,

Morokuma Keiji,

Musaev Djamaladdin G.

Publication year - 2005

Publication title -

angewandte chemie international edition

Language(s) - English

Resource type - Journals

SCImago Journal Rank - 5.831

H-Index - 550

eISSN - 1521-3773

pISSN - 1433-7851

DOI - 10.1002/anie.200501371

Subject(s) - reactivity (psychology) , steric effects , content (measure theory) , chemistry , group (periodic table) , crystallography , physics , stereochemistry , organic chemistry , mathematics , pathology , medicine , mathematical analysis , alternative medicine

Five are too many : It is known experimentally that [(Cp′) 2 Zr(N 2 )Zr(Cp′) 2 ] (Cp′=C 5 H 5− n Me n ) with n =4 activates N 2 for hydrogenation, but the complex with n =5 does not. This difference in reactivity arises from the N 2 coordination mode (see scheme): For n =0–4, N 2 is side‐on coordinated—the most suitable mode for hydrogenation—whereas for n =5, N 2 is end‐on coordinated owing to steric repulsion between five Me groups.

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