Methanol → HCID 59615429, compound

HRID 43636Notes

Reaction equation

HRID 43636 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][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]

Participants

5

Reactant 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

Product 1

HCID 59615429HCID 59615429, compoundYield 80%

Reagents, catalysts & solvents 1

HCID 8028TetrahydrofuranSolvent

Procedure

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%).

Sources & evidence

ORD record
ord-c6c2d670a1144a3eae870d4ca1764b3e
Source dataset
uspto-grants-2013_09
Patent number
US08536197B2

Reaction notes

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