Potassium Iodide → Methyl 4-(2-iodoethyl)benzoate
Reaction equation
Cl[CH2:2][CH2:3][C:4]1[CH:13]=[CH:12][C:7]([C:8]([O:10][CH3:11])=[O:9])=[CH:6][CH:5]=1.[I-:14].[K+]>C(#N)C>[I:14][CH2:2][CH2:3][C:4]1[CH:13]=[CH:12][C:7]([C:8]([O:10][CH3:11])=[O:9])=[CH:6][CH:5]=1Participants
4
Reactant 2
Product 1
Reagents, catalysts & solvents 1
Procedure
70 g (352.4 mmol) of methyl 4-(2-chloroethyl)benzoate [CAS Reg. No. 65787-72-6] and 146.2 g (880.9 mmol) of potassium iodide are suspended in 800 ml of acetonitrile and stirred under reflux for three days. After the reaction has gone to completion, the reaction solution is cooled and filtered and the filtrate is evaporated to dryness under reduced pressure. The resulting residue is purified by flash chromatography on silica gel (mobile phase:petroleum ether/ethyl acetate 30:1→20:1). This gives 101 g (348.3 mmol, 98.8% of theory) of a yellowish oil.