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

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

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}&amp;\textbf{Boiling point / °C} &amp;\textbf{Compound}&amp;\textbf{Boiling point / °C} \\ \hline
\text{propane} &amp; -42 &amp; \text{methoxymethane} &amp; -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?
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.