Hexafluorocyclotriphosphazene is prepared by the reaction of potassium fluoride and 2,2,4,4,6,6-hexachloro-1,3,5-triaza-2,4,6-triphosphorine. The steps are as follows: Put 104.3g of hexachlorocyclotriphosphazene, 113.3g of potassium fluoride, 0.011g of ionic liquid catalyst [Nbmm]OH, and 217.6g of anhydrous acetonitrile in an electric stirrer, thermometer, In the flask with reflux condenser, the reaction was carried out at 30°C, and the reaction was closed after 2h, filtered, The filtrate was rectified to obtain hexafluorocyclotriphosphazene with a yield of 98.7%.
