I am teaching organic chemistry to a high school student. While discussing Teflon, we referred to the corresponding section in the 10$^\text{th}$ edition of "Organic Chemistry" by Solomons and Fryhle, where it is mentioned that "fluorocarbons have a lower boiling point when compared to hydrocarbons of the same molecular weight:"

fluorocarbons have lower boiling points than hydrocarbons with the same molecular weight

Now, we know that carbon—fluorine bonds are more polar and higher polarity should result in a higher melting/boiling point due to higher interatomic forces. Moreover, higher molecular weights usually translate to higher boiling points. How do I explain it to my student.

Furthermore, the example taken here does not validate the statement made in the book because the molecular weights of $C_5F_{12}$ and $C_5H_{12}$ are about 288 and 72 g/mol, respectively, and not the same. I collected the data from here and here. On combining the two, the said relation does seems noticeable. What is the author trying to say here? Do we compare the molecular weights or the molecular formula?

data table for molecular weights and boiling points of fluorocarbons and hydrocarbons

I am also having trouble understanding the underlying principle as well, and if someone could elucidate it more, it would be really helpful.


For easy comparisons, I have plotted the boiling points of hydro- and fluoro-carbons for up to 8 carbon atoms:

plot for boiling points of hydrocarbons and fluorocarbons?

Except for the case of 1 carbon (methane and tetrafluoromethane), the trend seems to hold true.

  • 1
    $\begingroup$ You accidentally copied the boiling point of ethane instead of propane in your table. Fixing it grealy reduces the apparent discrepancy between propane and octafluoropropane. $\endgroup$ Commented Apr 16, 2023 at 15:35
  • $\begingroup$ chemistry.stackexchange.com/questions/124744/… $\endgroup$
    – Mithoron
    Commented Apr 16, 2023 at 15:38
  • $\begingroup$ @NicolauSakerNeto thank you for noticing the discrepancy, I have corrected the same. $\endgroup$
    – ananta
    Commented Apr 16, 2023 at 15:41
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    $\begingroup$ Molecules of halogenated hydrocarbons (check also CCl4 ) of the same/similar molar mass as respective hydrocarbons are much more compact with much weaker intermolecular van der Waals forces. $\endgroup$
    – Poutnik
    Commented Apr 16, 2023 at 15:44
  • 1
    $\begingroup$ Eventual slight partial charges of F atoms would lead rather to decrease of the overall attractive intermolecular forces by mutual repulsion. $\endgroup$
    – Poutnik
    Commented Apr 17, 2023 at 5:24

1 Answer 1


The following explanation is given by the Ref.1 for the vast difference in physical properties between perfluoro-alkanes and their hydrocarbon counterparts:

The physical properties of fluorinated organic compounds are in many cases very different to those observed for nonfluorinated molecules. In a very simple way, the majority of these changes can be rationalized by the atomic properties of the fluorine atom. For example, it has a high ionization potential and a low polarizability, which consequently leads to very weak interactions between perfluorinated moieties. Due to the electronegativity of fluorine, $\ce{C-F}$ bonds are always strongly polarized; nevertheless, perfluorocarbons belong to the most nonpolar compounds known. This is explained by the fact, that the dipole moments in a perfluorocarbon molecule cancel each other. Consequently, partly fluorinated compounds can have a significant polar character. Overall, the physical properties of hydrofluorocarbons are often very different from those of both, perfluorocarbons and hydrocarbons. Finally, the properties of fluoroorganic compounds are also determined by the excellent match of the corresponding $\mathrm{2s}$ and $\mathrm{2p}$ orbitals of carbon and fluorine.

Since fluorine is a larger substituent than hydrogen, the structural behavior of perfluorocarbons differs from that of their nonfluorinated analogues. Typically, linear hydrocarbons adapt a linear zig-zag conformation, which is not the case for the corresponding perfluorocarbons. The steric repulsion of fluorine substituents in relative 1,3-positions renders a zig-zag conformation very unfavorable, so that nonbranched perfluorocarbons have a helical structure. Expectedly, helical chains are less flexible than zig-zag conformations, so that perfluorocarbon molecules occur as rigid rods.

I'm not sure this would answer completely your question, nevertheless, it is worth mentioning. Interesting point of view (also note that perfluoro-alkane compounds have significantly lower dielectric constants than expected, representing their low polarizability).

If you are interested in learning boiling point comparisons of perfluoro-ethers and -amines to that of hydrocarbons, it is worth reading Reference 2.


  1. Axel Haupt, "Chapter 11: Fluoroorganic compounds – unusual properties and versatile applications," In Organic and Inorganic Fluorine Chemistry: Methods and Applications; Walter de Gruyter GmbH: Berlin, Germany, 2021, pp. 283-300 (DOI: https://doi.org/10.1515/9783110659337-014)(ISBN: 978-3-11-065933-7).
  2. A. A. Woolf, "Relative boiling points of fluoro-ethers, fluoroamines and other fluorocarbon derivatives to fluorocarbons," Journal of Fluorine Chemistry 1999, 94(1), 47-50 (DOI: https://doi.org/10.1016/S0022-1139(98)00336-4).

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