Trifluoromethyl iodide reacts with [AuMeL] to give [AuMe2(CF3)L] and [AuIL](L = PMe3 or PMe2Ph), or [Au(CF3)L] and Mel (L = PPh3), or a mixture of these products (L = PMePh2). In some cases reaction of [AuMe(PMe3)] with CF3I gives [AuMe(CF3)I(PMe3)]. Evidence is presented that the reactions proceed, at least in part, by a free-radical chain mechanism.
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Johnson et al. (1976) studied this question.
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