Hydrochloric Acid → 5-Chloroisoquinoline-4-carboxylic acid
Reaction equation
[Cl:1][C:2]1[CH:11]=[CH:10][CH:9]=[C:8]2[C:3]=1[C:4]([C:12]([O:14]C)=[O:13])=[CH:5][N:6]=[CH:7]2.[OH-].[K+].Cl>CO.O>[Cl:1][C:2]1[CH:11]=[CH:10][CH:9]=[C:8]2[C:3]=1[C:4]([C:12]([OH:14])=[O:13])=[CH:5][N:6]=[CH:7]2Participants
6
Reactant 3
Product 1
Reagents, catalysts & solvents 2
Procedure
To a stirred solution of methyl 5-chloroisoquinolin-4-carboxylate (Intermediate-42) (3.8 g, 17.1 mmol) in MeOH (30 mL) was added, KOH (9.62 g, 171 mmol) solution in H2O (20 mL) and the reaction mixture was heated to 70° C. for 3 h. After completion (TLC), it was concentrated under vacuum to give crude product (3.5 g) that was diluted with water, acidified with 1N HCl (pH=3) and the aqueous phase was extracted with EtOAc, dried and concentrated under vacuum to afford the desired acid derivative (3 g, 84%) as an off white solid.
Workup
- concentrationAfter completion (TLC), it was concentrated under vacuum
- customto give crude product (3.5 g) that
- extractionthe aqueous phase was extracted with EtOAc
- customdried
- concentrationconcentrated under vacuum
Sources & evidence
- ORD record
- ord-1cecb7aa22be4b98b26767766448b1a9
- Source dataset
- uspto-grants-2015_04
- Patent number
- US09006442B2
Reaction notes
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