Methanol → HCID 59615459, compound

HRID 43585Notes

Reaction equation

HRID 43585 reaction equation
[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]([CH2:28][CH3:29])=[C:21]([C:23]([O:25]CC)=[O:24])[N:22]=3)=[CH:14][CH:13]=2)=[O:10])[NH:5][C:6]=1[CH2:7][CH3:8].[OH-].[Li+].CO>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]([CH2:28][CH3:29])=[C:21]([C:23]([OH:25])=[O:24])[N:22]=3)=[CH:14][CH:13]=2)=[O:10])[NH:5][C:6]=1[CH2:7][CH3:8]

Participants

5

Reactant 3

HCID 887MethanolHCID 3939Lithium HydroxideHCID 67548256Ethyl 2-(4-{[(4-chloro-5-ethyl-1H-imidazol-2-yl)carbonyl]amino}phenyl)-5-ethyl-1,3-oxazole-4-carboxylate

Product 1

HCID 59615459HCID 59615459, compoundYield 87%

Reagents, catalysts & solvents 1

HCID 8028TetrahydrofuranSolvent

Procedure

The same operation as in Example (91d) was performed using ethyl 2-(4-{[(4-chloro-5-ethyl-1H-imidazol-2-yl)carbonyl]amino}phenyl)-5-ethyl-1,3-oxazole-4-carboxylate obtained in Example (99c) (0.12 g, 2.88 mmol), 2 N lithium hydroxide (2 mL, 4 mmol), methanol (2 mL) and tetrahydrofuran (5 mL), to obtain 97 mg of the title compound as a light pink solid (87%).

Sources & evidence

ORD record
ord-47c37a86afec48b4b740271b8f3b501b
Source dataset
uspto-grants-2013_09
Patent number
US08536197B2

Reaction notes

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