Reaktion #1279

Reaktionsgleichung

oleum

hexafluoropropene

hexafluoropropene

→

HFP sultone

Ausbeute 62.0%

Reaktionsbedingungen

Aufarbeitung

1. 1
   SonstigeThere was consumed 42.0 g (0.28 mole) HFP
2. 2
   workup.DISTILLATIONwas distilled

Vorschrift

Using essentially the procedure of Example 1, the reactor was charged with 33.8 g of 67 wt. % oleum (0.28 mole SO3) and 58.0 g hexafluoropropene. After the initial exotherm (max. pressure 896 kPa at 39° C.) the mixture was heated for 13 hours at 43° C. The reactor was vented of excess hexafluoropropene (16.0 g recovered including HFP collected during followed distillation). There was consumed 42.0 g (0.28 mole) HFP. The two phase reaction solution (70.1 g) was distilled using a concentric tube column to yield HFP sultone (39.8 g, 62% yield based on HFP consumed, b.p. 47° C.). The distillation residue was then distilled using a mini-lab apparatus to give 5.4 g of an intermediate fraction with a b.p. of 56°-155° C., 20 g of a fraction with a b.p. of 156°-167° C., and 1.3 g ofundistilled residue. The fraction with b.p. 156°-167° C. was analyzed by 19F and 1H NMR to reveal a complex mixture of primarily fluorosulfonic acid and several acid fluorides, sulfonyl fluorides, and at least three tetrafluoropropionic acid derivatives of the structure CF3CHFCOX, where X=F, OSO2F and possibly OCOCHFCF3. The molar ratio of HSO3F to all CHF compounds was 88:35.

Quelle