“…OEt 2 gives a mixture of the monofluorides 86a,b ± 88a,b. 221 The reaction of XeF 2 with 1-R-2,3,4,5-tetrafluorobenzene (R = H, F, Br, NO 2 ) or 1-R-2,3,4,6-tetrafluorobenzene (R = H, CF 3 ) in HF or CH 2 Cl 2 ± BF 3 . OEt 2 results in the substitution of F for the hydrogen atom in the aromatic ring.…”