It's possible to select a perfluorinated hydrocarbon that is immiscible in another organic solvent, forming a distinct phase boundary. Such systems may become miscible at elevated temperature which is exploited in the design of some catalytic processes1.

Why is it that many fluorous solvents are immiscible with organic solvents? Are their polarities sufficiently different or is there another reason?

(1) Housecroft, C.E.; Sharpe, A.G.; Inorganic Chemistry 2e; Pearson Prentice Hall; 2005, pp. 798-799


Fluorous solvents have the odd property of being both hydrophobic and lipophobic (Ref) and thus are not miscible with either aqueous or many organic solvents as you've noted. Fluorine, as the most electronegative element, does odd things to a molecule (and is often used in pharmaceutical compounds just because of some of these odd properties).

In the perfluorinated hydrocarbons, the carbon-fluorine bonds are quite polar with the electron density higher toward the fluorine. The solvents themselves, however, are not polar due to free rotation about the C-C bonds, so they are not miscible with polar solvents and are hydrophobic.

Because the F is so electronegative, the electrons are quite tightly held and the molecules have a low polarizability and unusually weak London dispersion forces, so they aren't miscible with non-polar solvents either. These weak London forces are illustrated by the fact that the perfluorinated hydrocarbons have nearly the same boiling point as their corresponding normal hydrocarbon, despite having more mass. (Ref)

You mentioned heat as one way to render the organic and fluorous solvents miscible. Adding gaseous $\ce{CO2}$ to the biphasic system also makes the phases miscible in many cases due to the high solubility of $\ce{CO2}$ in both fluorous and organic solvents. (Ref)

This reference suggest that shape may also be a consideration when considering flourous solvent's miscibility with alkanes, also through poor/weak interaction of London dispersion forces in the two liquids.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.