Formation of New η 5 -Rhodium(III) Complexes from η 5 -Rh(I) Rhodacarborane-Containing Charge-Compensated Ligands
2004; American Chemical Society; Volume: 23; Issue: 10 Linguagem: Inglês
10.1021/om030635h
ISSN1520-6041
AutoresRosario Núñez, Oscar Tutusaus, Francesç Teixidor, Clara Viñas, R. Sillanpää, Raikko Kivekäs,
Tópico(s)Radiopharmaceutical Chemistry and Applications
ResumoA series of new Rh(I) half-sandwich complexes of formula [3,3-(PPh3)2-8-L-closo-3,1,2-RhC2B9H10)] (L = SMe2 (2a), SEt2 (2b), S(CH2)4 (2c), SEtPh (2d)) and [1-Me-3,3-(PPh3)2-8-L-closo-3,1,2-RhC2B9H9)] (L = SMe2 (2e), SEt2 (2f)) have been prepared by reaction of the respective monoanionic charge-compensated ligands [10-L-nido-7,8-C2B9H10]- and [7-Me-10-L-7,8-C2B9H9]- with [RhCl(PPh3)3]. Complex [3,3-(cod)-8-SMe2-closo-3,1,2-RhC2B9H10] (3) has also been prepared by reaction of K[10-SMe2-nido-7,8-C2B9H10] with [RhCl(cod)]2. Rh(I) complexes 2a−d may be easily oxidized to the corresponding Rh(III) complexes 4a−d under N2 atmosphere in some halogenated solvents such as CCl4 and CHCl3. The complexes have been fully characterized by IR and NMR spectroscopy, and the crystal structures of 2a, 3, and 4a have been elucidated by single-crystal X-ray diffraction analysis. An EPR spectrum analysis clearly evidences the formation of free radicals as intermediates in the evolution of 2a−d to 4a−d complexes. The capacity to stabilize both Rh(I) and Rh(III) oxidation states by the [10-SMe2-nido-7,8-C2B9H10]- system may be attributed to its donor capacity together with the presence of labile ligands in the molecule.
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