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Halfsandwich‐Type Triisopropylstibane Ruthenium Complexes with 16‐, 17‐, and 18‐Electron Configurations and an Unsymmetrical Binuclear 18/16‐Electron Ru 2 (Sb i Pr 3 ) Species
Author(s) -
Braun Thomas,
Laubender Matthias,
Gevert Olaf,
Werner Helmut
Publication year - 1997
Publication title -
chemische berichte
Language(s) - English
Resource type - Journals
SCImago Journal Rank - 0.667
H-Index - 136
eISSN - 1099-0682
pISSN - 0009-2940
DOI - 10.1002/cber.19971300505
Subject(s) - chemistry , ruthenium , metallacycle , molar ratio , yield (engineering) , crystallography , medicinal chemistry , electron paramagnetic resonance , stereochemistry , x ray crystallography , catalysis , organic chemistry , physics , materials science , nuclear magnetic resonance , diffraction , optics , metallurgy
The paramagnetic ruthenium complex [(η 5 ‐C 5 Me 5 )RuCl 2 ‐{Sb i Pr 3 }] ( 2 ), which is prepared from [(η 5 ‐C 5 Me 5 )RuCl 2 ] n ( 1 ) and Sb i Pr 3 , reacts with Mg/Hg in THF to give the coordinatively unsaturated 16‐electron species (η 5 ‐C 5 Me 5 )RuCl)(Sb i Pr 3 )] ( 3 ). Compound 3 is obtained in much better yield from [(η 5 ‐C 5 Me 5 )Ru(μ 3 ‐Cl)] 4 ( 4 ) and Sb i Pr 3 (molar ration 1:4). While 3 reacts with CO to give the 18‐electron complex [(η 5 ‐C 5 Me 5 )RuCl(CO)(Sb i Pr 3 )] ( 5 ), on treatment of 3 with HCˇO 2 Me the metallacycle [(η 5 ‐C 5 Me 5 )RuCl{k 2 (C,C)‐C(CO 2 Me}=CHC(CO 2 Me)=CH}(Sb i Pr 3 )] ( 6 ) is obtained. The reaction of 4 with Sb i Pr 3 in the molar ratio of 1:2 affords the binuclear compound [(η 5 ‐C 5 Me 5 )(Sb i Pr 3 )Ru(μ‐Cl) 2 Ru(‐C 5 Me 5 )] ( 7 ). The molecular structures of 2 and 7 have been determined by X‐ray crystallography.