Diphenyliodonium hexafluorophosphate was prepared under acidic conditions using potassium iodate, benzene, and potassium hexachlorophosphate as starting materials. The reaction formula is shown in the figure below:
Figure 1: Reaction formula for the synthesis of Diphenyliodonium hexafluorophosphate
Method 1:
In a 250 mL three-necked flask, add 21.4 g (0.10 mol) potassium iodate, 18.5 mL (0.20 mol) benzene, 100 mL acetic acid, and 35 mL acetic anhydride. Slowly add 18 mL of concentrated sulfuric acid at 2–5 °C. After the addition is complete, stir the reaction at room temperature for 24 hours.
After the reaction system temperature was lowered to 10℃, 18.4 g (0.10 mol) of potassium hexachlorophosphate was slowly added. The resulting precipitate was filtered, washed, and recrystallized to obtain 41.2 g of white crystalline Diphenyliodonium hexafluorophosphate, with a yield of 41.6%. Method Two: 30.6 g (0.227 mol) of benzene, 100 mL of acetic acid, and 17 mL of concentrated sulfuric acid were added to a reaction flask. While controlling the reaction temperature below 20℃, 22 g (0.103 mol) of potassium iodate was added. The reaction was continuously stirred at 25℃ for 24 hours. After filtration, 100 mL of an aqueous solution of 18.952 g (0.103 mol) of potassium hexachlorophosphate was added to the filtrate. Immediately, a flocculent precipitate formed. After filtration and washing until almost colorless, recrystallization yielded 26.53 g (47.7%) of Diphenyliodonium hexafluorophosphate.
