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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.