Ethane → 1,1,1,6,6,6-Hexafluoro-2,5-bis(trifluoromethyl)-2,5-hexanediol
Reaction equation
[CH3:1][CH3:2].[F:3][C:4]([F:12])([F:11])[C:5]([C:7]([F:10])([F:9])[F:8])=[O:6]>>[F:3][C:4]([F:12])([F:11])[C:5]([C:7]([F:10])([F:9])[F:8])([OH:6])[CH2:1][CH2:2][C:5]([C:7]([F:10])([F:9])[F:8])([C:4]([F:12])([F:11])[F:3])[OH:6]Participants
Procedure
By starting the experiment with a 1:1 molar ratio of ethane to hexafluoroacetone, 47.3 g 1,1-bis(trifluoromethyl)propanol-1, b.p. 75°-80° C/705 mm. and 6.0 g of 1,1,4,4-tetrakis(trifluoromethyl) butan-1,4-diol, m.p. 96°-98° C, were obtained. From these data it will become apparent that increasing the molar ratio of ethane to hexafluoroacetone will increase the yield of the propanol-1 with respect to the diol. The converse is also true. That is, the more hexafluoroacetone one uses with respect to ethane, the more diol one will achieve with respect to the monofunctional alcohol. When an excess of ethane is used, i.e., when it is desired to form primarily 1,1-bis(trifluoromethyl) propanol-1, the amount of excess used is immaterial. That is, ten or even a hundred moles of ethane can be used for every mole of hexafluoroacetone with excellent results. On the other hand, it has been found that it is not advisable to use more than about twelve moles of hexafluoroacetone per mole of ethane if one wishes to have the diol as the predominant product. If a larger excess than this of the hexafluoroacetone is used, by-products begin to predominate and it becomes difficult to separate the diol from them.
Sources & evidence
- ORD record
- ord-73445e5b60f942d2aa00b95c85ada11b
- Source dataset
- uspto-grants-1977_02
- Patent number
- US04008139
Reaction notes
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