Hydrogen Bromide → Carvedilol HBr
반응식
[CH3:1][O:2][C:3]1[CH:4]=[CH:5][CH:6]=[CH:7][C:8]=1[O:9][CH2:10][CH2:11][NH:12][CH2:13][CH:14]([OH:30])[CH2:15][O:16][C:17]1[CH:18]=[CH:19][CH:20]=[C:21]2[NH:29][C:28]3[CH:27]=[CH:26][CH:25]=[CH:24][C:23]=3[C:22]=12.[BrH:31]>C(O)C.C(O)(=O)C>[CH3:1][O:2][C:3]1[CH:4]=[CH:5][CH:6]=[CH:7][C:8]=1[O:9][CH2:10][CH2:11][NH:12][CH2:13][CH:14]([OH:30])[CH2:15][O:16][C:17]1[CH:18]=[CH:19][CH:20]=[C:21]2[NH:29][C:28]3[CH:27]=[CH:26][CH:25]=[CH:24][C:23]=3[C:22]=12.[BrH:31]참가 물질
5
실험 절차
Carvedilol free base is dissolved in ethanol, and anhydrous forms HBr is added (HBr in acetic acid). The solution is cooled, and crystallization ensues. The product is filtered, washed with process solvent, and dried.