Lithium iodide → 2-Chloro-3-ethoxyiminomethyl-4-methylsulfonylbenzoic acid
Reaction equation
[Cl:1][C:2]1[C:11]([CH:12]=[N:13][O:14][CH2:15][CH3:16])=[C:10]([S:17]([CH3:20])(=[O:19])=[O:18])[CH:9]=[CH:8][C:3]=1[C:4]([O:6]C)=[O:5].[I-].[Li+]>N1C=CC=CC=1>[Cl:1][C:2]1[C:11]([CH:12]=[N:13][O:14][CH2:15][CH3:16])=[C:10]([S:17]([CH3:20])(=[O:19])=[O:18])[CH:9]=[CH:8][C:3]=1[C:4]([OH:6])=[O:5]Participants
4
Reactant 2
Product 1
Reagents, catalysts & solvents 1
Procedure
A solution of 4.37 g (0.0137 mol) of methyl 2-chloro-3-ethoxyiminomethyl-4-methylsulfonylbenzoate was slowly added dropwise to 7.29 g (0.055 mol) of lithium iodide in 50 ml of dry pyridine. After stirring for 2 hours at reflux, the reaction mixture was cooled and the solvent was removed under reduced pressure. The residue was taken up in water and adjusted to pH=1-2 with dilute hydrochloric acid. The aqueous phase was extracted with ethyl acetate and the combined organic phases were washed with water, dried and concentrated under reduced pressure. 3.70 g of 2-chloro-3-ethoxyiminomethyl-4-methylsulfonylbenzoic acid were obtained. (Yield 89%).