HRID1175209

反应详情

EQUATION

反应方程式

REACTANTS

反应物

HCID1118

Sulfuric acid

H2O4S

HCID8900

Heptane

C7H16

结构：COC(=O)N[C@H](C(=O)N[C@@H](Cc1ccccc1)[C@@H](O)CN(Cc1ccc(-c2ccccn2)cc1)NC(=O)[C@@H](NC(=O)OC)C(C)(C)C)C(C)(C)C

HCID148192

Atazanavir

C38H52N6O7

结构：COC(=O)N[C@H](C(=O)N[C@@H](Cc1ccccc1)[C@@H](O)CN(Cc1ccc(-c2ccccn2)cc1)NC(=O)[C@@H](NC(=O)OC)C(C)(C)C)C(C)(C)C.O=S(=O)(O)O

HCID158550

Atazanavir sulfate

C38H54N6O11S

PRODUCTS

生成物

HCID86601659

未命名化合物

AUXILIARIES

试剂、催化剂与溶剂

溶剂HCID702

Ethanol

C2H6O

3 次

溶剂HCID1140

Toluene

C7H8

溶剂HCID8058

n-Hexane

C6H14

PROCEDURE

实验过程

The Form E3 is prepared by slurrying atazanavir free base in ethanol, treating the slurry with concentrated sulfuric acid employing a molar ratio of acid:free base with the range from about 1:1 to about 1.1:1, heating the resulting solution at from about 30 to about 40° C., seeding the solution with ethanol wet E3 crystals of atazanavir sulfate, treating the mixture with heptane (or other solvent such as hexane or toluene), filtering, and drying to yield atazanavir bisulfate Form E3 (triethanol solvate).

WORKUP

后处理

temperaturefree base with the range from about 1:1 to about 1.1:1, heating the resulting solution at from about 30 to about 40° C.
filtrationfiltering
customdrying

来源与证据

来源类型: Open Reaction Database
ORD 记录: ord-5e6057628aea4b25b7c78660272eaa1b
来源数据集: uspto-grants-2010_11
DOI: 10.6084/m9.figshare.5104873.v1
专利: US07838678B2
来源说明: uspto-grants-2010_11

反应 SMILES

[CH3:1][C:2]([C@H:5]([NH:47][C:48]([O:50][CH3:51])=[O:49])[C:6]([NH:8][C@H:9]([C@@H:17]([OH:46])[CH2:18][N:19]([NH:33][C:34]([C@@H:36]([NH:41][C:42]([O:44][CH3:45])=[O:43])[C:37]([CH3:40])([CH3:39])[CH3:38])=[O:35])[CH2:20][C:21]1[CH:22]=[CH:23][C:24]([C:27]2[CH:28]=[CH:29][CH:30]=[CH:31][N:32]=2)=[CH:25][CH:26]=1)[CH2:10][C:11]1[CH:12]=[CH:13][CH:14]=[CH:15][CH:16]=1)=[O:7])([CH3:4])[CH3:3].[S:52](=[O:56])(=[O:55])([OH:54])[OH:53].CC([C@H](NC(OC)=O)C(N[C@H]([C@@H](O)CN(NC([C@@H](NC(OC)=O)C(C)(C)C)=O)CC1C=CC(C2C=CC=CN=2)=CC=1)CC1C=CC=CC=1)=O)(C)C.OS(O)(=O)=O.CCCCCCC>C(O)C.C1(C)C=CC=CC=1.CCCCCC>[CH3:4][C:2]([C@H:5]([NH:47][C:48]([O:50][CH3:51])=[O:49])[C:6]([NH:8][C@H:9]([C@@H:17]([OH:46])[CH2:18][N:19]([NH:33][C:34]([C@@H:36]([NH:41][C:42]([O:44][CH3:45])=[O:43])[C:37]([CH3:38])([CH3:39])[CH3:40])=[O:35])[CH2:20][C:21]1[CH:22]=[CH:23][C:24]([C:27]2[CH:28]=[CH:29][CH:30]=[CH:31][N:32]=2)=[CH:25][CH:26]=1)[CH2:10][C:11]1[CH:12]=[CH:13][CH:14]=[CH:15][CH:16]=1)=[O:7])([CH3:1])[CH3:3].[S:52](=[O:54])(=[O:53])([OH:56])[O-:55]