[Cl:1][C:2]1[CH:7]=[CH:6][C:5]([C:8]2[N:12]([CH2:13][CH:14]([OH:19])[C:15]([F:18])([F:17])[F:16])[C:11](=[O:20])[N:10]([CH2:21]C3C=CC(C(OC)=O)=CC=3)[N:9]=2)=[CH:4][CH:3]=1.ClC1C=CC(C2N(CC(O)C(F)(F)F)C(=O)NN=2)=CC=1.BrC[C:54]1[CH:55]=[C:56]([CH:61]=[CH:62][CH:63]=1)[C:57]([O:59][CH3:60])=[O:58]>>[Cl:1][C:2]1[CH:3]=[CH:4][C:5]([C:8]2[N:12]([CH2:13][CH:14]([OH:19])[C:15]([F:18])([F:16])[F:17])[C:11](=[O:20])[N:10]([CH2:21][C:54]3[CH:55]=[C:56]([CH:61]=[CH:62][CH:63]=3)[C:57]([O:59][CH3:60])=[O:58])[N:9]=2)=[CH:6][CH:7]=1In the same way as for the compound from Example 24A, 267 mg (0.87 mmol) of the compound from Example 19A were reacted with 199 mg (0.87 mmol) of methyl 3-(bromomethyl)benzoate. This gave 302 mg (76% of theory) of the target compound.
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