Reactions of low-valent metal complexes with fluorocarbons. Part XIV. Hexafluoroisopropylideneamine-nickel and -platinum complexes
- 1 January 1970
- journal article
- research article
- Published by Royal Society of Chemistry (RSC) in J. Chem. Soc. A
- p. 3161-3165
- https://doi.org/10.1039/j19700003161
Abstract
Ethylenebis(triphenylphosphine)nickel reacts with the imine hexafluoroisopropylideneamine or the N-methylamine to form stable crystalline complexes [Ni(Ph3P)2{(CF3)2CNR}](R = H or Me). Reaction of bis(cycloocta-1,5-diene)nickel with N-methylhexafluoroisopropylideneamine affords [Ni(1,5-C8H12){(CF3)2CNMe}], which on treatment with triphenylphosphine, methyldiphenylphosphine, bipyridyl, or 1,2-bis(dimethylphosphino)-ethane, gives stable imine complexes, cyclo-octa-1,5-diene being displaced. In contrast, trimethyl phosphite displaces both the diene and the imine from [Ni(1,5-C8H12){(CF3)2CNMe}]. The analogous platinum complexes [Pt(Ph3P)2{(CF3)2CNR}](R = H or Me) are obtained by reaction of trans-stilbenebis(triphenylphosphine)platinum with the corresponding imine. Trifluoroacetic acid reacts with [Pt(Ph3P)2-{(CF3)2CNH}] to form the platinum(IV) complex [Pt(Ph3P)2{(CF3)2CNH}(CF3CO2)2], whereas a similar reaction with the N-methyl analogue gives [Pt(Ph3P)2(H){(CF3)2CNMe}(CF3CO2)]. Treatment of [Pt(Ph3P)2{(CF3)2CNH}] with hexafluoroacetone leads to a ring-expansion reaction and the formation of a five-membered heterocyclic ring compound. Hexafluoroacetone displaces N-methylhexafluoroisopropylideneamine from [Pt(Ph3P)2{(CF3)2-CNMe}] to give the known complex [Pt(Ph3P)2{(CF3)2CO}].Keywords
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