F[C:2]1[CH:7]=[CH:6][CH:5]=[CH:4][C:3]=1[N+:8]([O-:10])=[O:9].[Br:11][C:12]1[CH:18]=[CH:17][C:15]([NH2:16])=[CH:14][CH:13]=1.C([O-])(C)(C)C.[K+]>CS(C)=O>[Br:11][C:12]1[CH:18]=[CH:17][C:15]([NH:16][C:2]2[CH:7]=[CH:6][CH:5]=[CH:4][C:3]=2[N+:8]([O-:10])=[O:9])=[CH:14][CH:13]=1The product (6.49 g) is obtained according to the method of stage 1 of Example 4, by using 3.73 mL of 2-fluoro-nitrobenzene and 7.31 g of 4-bromoaniline in the presence of 6.36 g of potassium tert-butanolate in 120 mL of DMSO.
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