Lithium Hydroxide → HCID 58978927, compound

HRID 1180561Notes

Reaction equation

HRID 1180561 reaction equation
[C:1]([N:9]1[C:15]2[CH:16]=[CH:17][CH:18]=[CH:19][C:14]=2[CH2:13][N:12]([S:20]([C:23]2[CH:28]=[CH:27][C:26]([O:29][CH2:30][CH:31]=[C:32]=[CH:33]C)=[CH:25][CH:24]=2)(=[O:22])=[O:21])[CH:11]([C:35]([O:37]C)=[O:36])[CH2:10]1)(=[O:8])[C:2]1[CH:7]=[CH:6][CH:5]=[CH:4][CH:3]=1.[OH-].[Li+]>O.CO.O1CCCC1>[C:1]([N:9]1[C:15]2[CH:16]=[CH:17][CH:18]=[CH:19][C:14]=2[CH2:13][N:12]([S:20]([C:23]2[CH:24]=[CH:25][C:26]([O:29][CH2:30][CH:31]=[C:32]=[CH2:33])=[CH:27][CH:28]=2)(=[O:22])=[O:21])[CH:11]([C:35]([OH:37])=[O:36])[CH2:10]1)(=[O:8])[C:2]1[CH:3]=[CH:4][CH:5]=[CH:6][CH:7]=1

Participants

6

Reactant 2

HCID 3939Lithium HydroxideHCID 58978916HCID 58978916, compound

Product 1

HCID 58978927HCID 58978927, compoundYield 99%

Reagents, catalysts & solvents 3

HCID 887MethanolSolventHCID 962WaterSolventHCID 8028TetrahydrofuranSolvent

Procedure

The procedure of Example 42 was followed using the product from Example 40 (270 mg, 0.521 mmol), lithium hydroxide (55 mg, 1.3 mmol) in water (0.4 ml), methanol (1 ml) and tetrahydrofuran (1 ml) to give 260 mg (99%) of the product.

Sources & evidence

ORD record
ord-4a43af45d9fb48c0997a2ed437148506
Source dataset
uspto-grants-2007_10
Patent number
US07282496B2

Reaction notes

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