Interaction of Nitric Oxide with Paramagnetic and Diamagnetic Alkyls of Titanium, Zirconium, Vanadium, Niobium, and Tantalum

Article abstract of DOI:10.1039/DT9800001888

An earlier observation that there is a profound difference in the way paramagnetic and diamagnetic alkyl compounds of transition metals react with nitric oxide is extended.The reaction of NO with bis(η⁵-cyclopentadienyl)-dimethylniobium(IV), d¹, can be followed by nuclear magnetic resonance spectra and intermediates detected.The final product of the reaction is bis(η⁵-cyclopentadienyl)methyloxoniobium(V), which with excess of NO, gives the N-methyl-N-nitrosohydroxylaminato-complex Nb(η-C5H5)2(O).An intermediate niobioaziridine collapses with methylnitrene transfer to give azomethane, and in the presence of styrene, an aziridine.Tetrakis(trimethylsilylmethyl)vanadium(IV) gives with NO the previously known VO(CH2SiMe3)3 while the titanium(III), d¹, compounds Ti(η-C5H5)2R (R = Ph or CH2Ph) lose both cyclopentadienyl and aryl or benzyl groups to give trimeric species Ti3(η-C5H5)3O4(NO) that has oxo-bridges and a bent Ti-NO group.The diamagnetic alkyls of titanium and zirconium, TiR4, TiR3Cl, TiR2Cl2, ZrR4, and ZrR2Cl2 (R = CH2SiMe3 or CH2CMe3) and NO give N-alkyl-N-nitrosohydroxylaminato-complexes.The titanium compounds are six-co-ordinate with unidentate and bidentate ON(R)NO groups, while the zirconium compounds are eight-co-ordinate with chelate groups only.The pentamethyls MMe5 (M = Nb or Ta) and NO give NbMe23 and 2>2 respectively.

Full text of DOI:10.1039/DT9800001888

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