Why do alkanes have higher boiling point than their ether counterparts?

Question

Based on my understanding of inter-molecular forces, I expect dipole-dipole interactions to be significantly stronger than van der Waal's 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, infact 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 Waal's interactions). Alkanes on, the other hand, have van der Waal's 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? Am I missing some important concept here? Or is there any other factor that I am not considering?

Answer 1

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.

Answer 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.