HRID573957

反应详情

EQUATION

反应方程式

REACTANTS

反应物

HCID180

Acetone

C3H6O

HCID516892

Sodium Bicarbonate

CHNaO3

$结构：Cc1ccc(C(C)C)c(N2C(=O)CS/C2=N\C(=O)NCCc2ccc(-c3ncn(-c4ccc(OC(F)(F)F)cc4)n3)cc2)c1$

HCID78319090

未命名化合物

PRODUCTS

生成物

$结构：Cc1ccc(C(C)C)c(N2C(=O)C(C(C)(C)O)S/C2=N\C(=O)NCCc2ccc(-c3ncn(-c4ccc(OC(F)(F)F)cc4)n3)cc2)c1$

HCID140653474

未命名化合物

收率 19%

PROCEDURE

实验过程

To (Z)-1-(3-(2-isopropyl-5-methylphenyl)-4-oxothiazolidin-2-ylidene)-3-(4-(1-(4-(trifluoromethoxy)phenyl)-1H-1,2,4-triazol-3-yl)phenethyl)urea (F5) (0.40 g, 0.64 mmol) in acetone (30 mL) was added saturated aqueous sodium bicarbonate (3 mL). The reaction was heated to reflux for 12 hours, cooled, and concentrated. The reaction mixture was partitioned between ethyl acetate and water. The organic layer was dried and concentrated. Purification by flash column chromatography using 100-40% A/B, where A=1:1 dichloromethane/hexanes and B=3:1 ethyl acetate/acetone, as eluent provided the title compound as a white solid (0.095 g, 19%).

WORKUP

后处理

temperatureThe reaction was heated
temperatureto reflux for 12 hours
temperaturecooled
concentrationconcentrated
customThe reaction mixture was partitioned between ethyl acetate and water
customThe organic layer was dried
concentrationconcentrated
customPurification by flash column chromatography

来源与证据

来源类型: Open Reaction Database
ORD 记录: ord-0321c041fde64cf1ad7c8a24e358c784
来源数据集: uspto-grants-2015_05
DOI: 10.6084/m9.figshare.5104873.v1
专利: US09029560B2
来源说明: uspto-grants-2015_05

反应 SMILES

[CH:1]([C:4]1[CH:9]=[CH:8][C:7]([CH3:10])=[CH:6][C:5]=1[N:11]1[C:15](=[O:16])[CH2:14][S:13]/[C:12]/1=[N:17]\[C:18]([NH:20][CH2:21][CH2:22][C:23]1[CH:28]=[CH:27][C:26]([C:29]2[N:33]=[CH:32][N:31]([C:34]3[CH:39]=[CH:38][C:37]([O:40][C:41]([F:44])([F:43])[F:42])=[CH:36][CH:35]=3)[N:30]=2)=[CH:25][CH:24]=1)=[O:19])([CH3:3])[CH3:2].C(=O)(O)[O-].[Na+].[CH3:50][C:51]([CH3:53])=[O:52]>>[OH:52][C:51]([CH:14]1[S:13]/[C:12](=[N:17]\[C:18]([NH:20][CH2:21][CH2:22][C:23]2[CH:24]=[CH:25][C:26]([C:29]3[N:33]=[CH:32][N:31]([C:34]4[CH:35]=[CH:36][C:37]([O:40][C:41]([F:44])([F:43])[F:42])=[CH:38][CH:39]=4)[N:30]=3)=[CH:27][CH:28]=2)=[O:19])/[N:11]([C:5]2[CH:6]=[C:7]([CH3:10])[CH:8]=[CH:9][C:4]=2[CH:1]([CH3:3])[CH3:2])[C:15]1=[O:16])([CH3:53])[CH3:50]