Abstract
Reaction of 3-bromobenzofuran with trifluoromethyl iodide and copper powder in pyridine gave 2-(trifluoromethyl)-, 2- and 3-(pentafluoroethyl)-, and 2,3-bis(trifluoromethyl)-benzofuran, as well as the expected product, 3-(trifluoromethyl)benzofuran. Bromoanisole also gave (trifluoromethyl)anisole and (pentafluoroethyl)anisole, but introduction of the perfluoroalkyl group occurred at the position originally occupied by the bromine. Formation of pentafluoroethyl compounds is explained by decomposition of trifluoromethylcopper to cuprous fluoride and difluorocarbene, which can then react with a further molecule of trifluoromethylcopper to form pentafluoroethylcopper. This then reacts with aryl halide to give pentafluoroethyl compounds. Perfluoroalkylcopper is thermally cleaved to perfluoroalkyl radical, which then reacts with pyridine to give perfluoroalkylpyridines. This mechanism must be involved in the formation of 2,3-bis(trifluoromethyl)benzofuran. Introduction of a perfluoroalkyl group to the position originally unoccupied with halogen might be due to the rather localized double bond in benzofuran.
Published Version
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