ClC1C(CO)=CC=C(Cl)N=1.[Br:11][CH2:12][C:13]1[C:14]([Cl:21])=[N:15][C:16]([Cl:20])=[C:17](F)[CH:18]=1>>[Br:11][CH2:12][C:13]1[C:14]([Cl:21])=[N:15][C:16]([Cl:20])=[CH:17][CH:18]=1Compound 13B (0.88 g, 80%) was prepared from 13A (0.82 g, 4.6 mmol) using a procedure similar to the synthesis of 1B. HPLC Rt=2.44 min.
불러오는 중…