Sodium Iodide → 4-(3-Iodopropoxy)benzoic acid, methyl ester
Reaction equation
Cl[CH2:2][CH2:3][CH2:4][O:5][C:6]1[CH:15]=[CH:14][C:9]([C:10]([O:12][CH3:13])=[O:11])=[CH:8][CH:7]=1.[I-:16].[Na+]>CC(C)=O.C(OCC)C>[I:16][CH2:2][CH2:3][CH2:4][O:5][C:6]1[CH:15]=[CH:14][C:9]([C:10]([O:12][CH3:13])=[O:11])=[CH:8][CH:7]=1Participants
5
Reactant 2
Product 1
Reagents, catalysts & solvents 2
Procedure
Dissolve 4-(3-chloropropoxy)benzoic acid, methyl ester (6.97 g, 30.5 mmol) in anhydrous acetone (100 mL) and add powdered sodium iodide (16.0 g, 107 mmol). Heat at reflux under an argon atmosphere for 38 hours. Dilute with ethyl ether (100 mL) and filter through Celite® filter aid. Wash the filtrate with water and brine, then dry (MgSO4). Evaporate the solvent in vacuo to give 4-(3-iodopropoxy)benzoic acid, methyl ester as a yellow oil.