
Journal of Fluorine Chemistry p. 87 - 112 (1989)
Update date:2022-08-17
Topics:
Santacesaria, E.
Serio, M. Di
Basile, G.
Carra, S.
It is shown that SbCl4F can be considered as the main fluorinating agent when HF and SbCl5 are used in fluorination with an HF/SbCl5 ratio of 1:1.A very simple way for preparing SbCl4F is described in the paper.Kinetic runs, performed in batches using both SbCl4F and HF+SbCl5 as fluorinating agents, allowed identification of the reaction pattern for the chloroform fluorination.Fluorination proceeds through four consecutive steps in which the first and the second are fast and reversible reactions, corresponding to the fluorine-chlorine exchange between HF/SbCl5 and SbCl4F/ChCl3 respectively.The main product, CHClF2 is formed in the third reaction.The last step to CHF3 is very slow.Assuming the formation of CHClF2 as reference, the relative reaction rates for the steps are: 150/7/1/0.03. Kinetic equations and parameters obtained by fitting batch runs have also been verified by interpreting experiments performed under steady state conditions.The kinetic model, employed for elaborating both batch and continuous runs, takes into account also the vapour-liquid equilibria for reactant and products.Suggestions are made on the reaction mechanisms.
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