Hydrochloric Acid → HM-30181 hydrochloride
반응식
[CH3:1][O:2][C:3]1[CH:4]=[C:5]2[C:10](=[CH:11][C:12]=1[O:13][CH3:14])[CH2:9][N:8]([CH2:15][CH2:16][C:17]1[CH:22]=[CH:21][C:20]([N:23]3[N:27]=[N:26][C:25]([C:28]4[CH:33]=[C:32]([O:34][CH3:35])[C:31]([O:36][CH3:37])=[CH:30][C:29]=4[NH:38][C:39]([C:41]4[O:42][C:43]5[C:48]([C:49](=[O:51])[CH:50]=4)=[CH:47][CH:46]=[CH:45][CH:44]=5)=[O:40])=[N:24]3)=[CH:19][CH:18]=1)[CH2:7][CH2:6]2.[ClH:52]>>[ClH:52].[CH3:1][O:2][C:3]1[CH:4]=[C:5]2[C:10](=[CH:11][C:12]=1[O:13][CH3:14])[CH2:9][N:8]([CH2:15][CH2:16][C:17]1[CH:22]=[CH:21][C:20]([N:23]3[N:27]=[N:26][C:25]([C:28]4[CH:33]=[C:32]([O:34][CH3:35])[C:31]([O:36][CH3:37])=[CH:30][C:29]=4[NH:38][C:39]([C:41]4[O:42][C:43]5[C:48]([C:49](=[O:51])[CH:50]=4)=[CH:47][CH:46]=[CH:45][CH:44]=5)=[O:40])=[N:24]3)=[CH:19][CH:18]=1)[CH2:7][CH2:6]2참가 물질
3
실험 절차
The procedure of Example 33 was repeated except for using 1.5 g of the compound obtained in Example 21 and 0.1 ml of hydrochloric acid as starting materials to obtain 1.4 g of the title compound (yield 89%).
출처 및 근거
- ORD 기록
- ord-cbd9426eabe2455e88bca7f6c5b4b625
- 출처 데이터셋
- uspto-grants-2009_12