Based on my understanding of intermolecular forces, I expect dipole-dipole interactions to be significantly stronger than van der Waals' interactions. Hence, I expect ethers (which obviously have dipole-dipole interactions) to have higher boiling points (as boiling point $\propto $ inter-molecular forces) than alkanes (assuming approximately the same molecular mass). But, the experimental data suggests otherwise.

$$ \begin{array}{|c|c|c|c|} \hline \textbf{Compound}&\textbf{Boiling point / °C} &\textbf{Compound}&\textbf{Boiling point / °C} \\ \hline \textit{n}\text{-pentane} & 36.1 & \text{ethoxyethane} & 34.6 \\ \hline \textit{n}\text{-heptane} & 98.42 & \text{1-propoxypropane} & 90 \\ \hline \textit{n}\text{-nonane} & 151 & \text{1-butoxybutane} & 141 \\ \hline \textit{n}\text{-undecane} & 196 & \text{1-pentoxypentane}& 184 \\ \hline \end{array} $$

As evident, experimental data suggests that alkanes are on par with ethers when it comes to inter-molecular forces, in fact the former are slightly higher.

I tried to reason this discrepancy as follows. Ethers have weak dipole, so their dipole-dipole interactions will also be weak (though stronger than their van der Waals' interactions). Alkanes on, the other hand, have van der Waals' forces only but I suspect that size is playing a dominant role here. Ethers have a v-shaped bent structure, and therefore, have reduced surface area as compared to alkanes. Therefore, I believe that alkanes lead over ether in this case and this compensates the weak dipole of ethers. Though this is only a hypothesis from my part.

But there is another problem:

\begin{array}{|c|c|c|c|} \hline \textbf{Compound}&\textbf{Boiling point / °C} &\textbf{Compound}&\textbf{Boiling point / °C} \\ \hline \text{propane} & -42 & \text{methoxymethane} & -24 \\ \hline \end{array}

Now at this point I am out of ideas. From $\ce{Et-O-Et}$ to $\ce{Pe-O-Pe}$ we had ethers leading over alkanes which felt odd but when we compare di methyl ether and propane, everything just seems fine with ether having the dominating lead (as expected initially). What is real cause?

  • $\begingroup$ I did not intentionally skip decane and octane, it's just that there is no symmetrical ether counterpart of these (I am avoiding asymmetrical ethers). $\endgroup$
    – sarthak-ag
    Commented Mar 11, 2018 at 15:59
  • 1
    $\begingroup$ Why did you avoid non-symmetrical ethers? That might give you important insight into why this trend is such. Also, polyethers. Someone actually studying this problem would probably have looked at different permutations. $\endgroup$
    – Zhe
    Commented May 16, 2018 at 18:09
  • $\begingroup$ I actually did look at non-symmetrical ethers, but the trend was way too irregular. I had to consider many possibilities like one carbon difference non-symmetrical ethers, 2 carbon difference, etc. It extensively messed up my data and no conclusion could be drawn. Their irregular dipole is also a problem both understanding and (i suppose) explaining. Under these consideration, I thought I would first keep my observations confined to the simplest case, once I get a strong theory I would expand it to other cases. $\endgroup$
    – sarthak-ag
    Commented May 19, 2018 at 2:43

2 Answers 2


I second that the Van der Waals forces will dominate, because the dipole of the symmetrical ethers is perpendicular to the long axis of the molecule. In this case the number of electrons over a large surface area is going to play a bigger role in holding the molecules together.

Oxygen has the same number of electrons as -CH₂-, but because oxygen is more electronegative, it’s holding those electrons more tightly, making it less polarisable, and reducing the magnitude of the induced dipoles.

I suspect that ᵀ he effect is less apparent for dimethyl ether because the polar C-O bonds play a bigger role because the partial-positive charges on the carbons is not attentispread down the carbon chain, muting it, but remains a concentrated, bare, bright spot of charge, which can interact with the lone pairs on other ether oxygens. Additionally, there is less steric hindrance from the floppy alkyl groups. To prevent this kind of dipole-dipole attraction.


I suspect London forces dominate in the longer hydrocarbons, whereas they are disrupted by the ether bond. For shorter hydrocarbons, the dipole of the O-C bond is greater than the London forces, but this is reversed as the # of carbons increases.

  • 2
    $\begingroup$ Isn't this more of a observation rather than a concrete theoretical reason? $\endgroup$
    – sarthak-ag
    Commented Mar 12, 2018 at 4:38
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    $\begingroup$ I don't recognize the distinction between observation and theory in your (?) question. The bp data you've tabulated is observational/empirical. The theory is that cooperativity of Van der Wals forced exceeded the energy of the dipole interaction for values on n, where n = # of carbons, with n being $\endgroup$
    – Adam Worth
    Commented Apr 14, 2018 at 23:32
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    $\begingroup$ @AdamWorth I believe that your comment is a potential answer, can you please elaborate further, explaining why the Van der Waals forces exceeded the dipole interactions? $\endgroup$
    – Abhigyan
    Commented May 12, 2018 at 7:23

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