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Methanol → 2-(4-{[(4-Chloro-5-ethyl-1H-imidazol-2-yl)carbonyl]amino}-3-methylphenyl)-5-methyl-1,3-oxazole-4-carboxylic acid

HRID 1425112
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인쇄 미리보기

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HRID 1425112 반응식
[Cl:1][C:2]1[N:3]=[C:4]([C:9]([NH:11][C:12]2[CH:17]=[CH:16][C:15]([C:18]3[O:19][CH:20]=[C:21]([C:23]([O:25]C)=[O:24])[N:22]=3)=[CH:14][C:13]=2[CH3:27])=[O:10])[NH:5][C:6]=1[CH2:7][CH3:8].[OH-].[Li+].[CH3:30]O>O1CCCC1>[Cl:1][C:2]1[N:3]=[C:4]([C:9]([NH:11][C:12]2[CH:17]=[CH:16][C:15]([C:18]3[O:19][C:20]([CH3:30])=[C:21]([C:23]([OH:25])=[O:24])[N:22]=3)=[CH:14][C:13]=2[CH3:27])=[O:10])[NH:5][C:6]=1[CH2:7][CH3:8]
참가 물질실험 절차출처 및 근거반응 노트

참가 물질

5

반응물 3

HCID 887MethanolHCID 3939Lithium HydroxideHCID 67549102Methyl 2-(4-{[(4-Chloro-5-ethyl-1H-imidazol-2-yl)carbonyl]amino}-3-methylphenyl)-1,3-oxazole-4-carboxylate

생성물 1

HCID 596154292-(4-{[(4-Chloro-5-ethyl-1H-imidazol-2-yl)carbonyl]amino}-3-methylphenyl)-5-methyl-1,3-oxazole-4-carboxylic acid수율 80%

시약·촉매·용매 1

HCID 8028
Tetrahydrofuran
용매

실험 절차

The same operation as in Example (91d) was performed using methyl 2-(4-{[(4-chloro-5-ethyl-1H-imidazol-2-yl)carbonyl]amino}-3-methylphenyl)-1,3-oxazole-4-carboxylate obtained in Example (109f) (0.13 g, 0.33 mmol), 2 N lithium hydroxide (3 mL, 6 mmol), methanol (3 mL) and tetrahydrofuran (5 mL), to obtain 0.10 g of the title compound as a colorless solid (80%).

출처 및 근거

ORD 기록
ord-efac98b4c00a44468260251af6e56ae1
출처 데이터셋
uspto-grants-2015_01
DOI
10.6084/m9.figshare.5104873.v1
특허 번호
US08927588B2

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