How to determine what organic molecule has the highest boiling point?
There are commonly three types of intermolecular forces quoted when it comes to melting and boiling points: dipole-dipole interactions, London interactions and hydrogen bonding. Hydrogen bonding is by far the strongest force of them all. In all cases except for very rare ones, for a hydrogen to take part in hydrogen bonding, it must be bonded to $\ce{O, N, F}$ — one of the three very electronegative atoms.

In difluoromethane, there are no hydrogens bound to $\ce{O, N, F}$ — they are both bound to carbon. It is mainly carbon that experiences fluorine’s electronegativity and by the time the inductive effect reaches the hydrogens it has been substantially lowered. On the other hand, methanol and methyl amine both have hydrogens bonded to the electronegative atom — nitrogen or oxygen — so these compounds actually take part in hydrogen bonding. Since we have two compounds that participate in hydrogen bond networks it is safe to assume that these will be the top runners.

As you mentioned, methanol’s hydrogen bonds should be slightly stronger since oxygen has a higher electronegativity. I think your reasoning as to why methanol is stronger than methyl amine is sound enough.

Difluoromethane’s principal intermolecular forces will be dipole-dipole interactions. However, you should also note that the presence of fluorine atoms strongly reduces the underlying London interactions, since $\ce{C-F}$ bonds are more localised and fluorine’s lone pairs are much less disperse than most other carbon-heteroatom bonds — this is due to fluorine’s very high electronegativity. Thus, while we have significant dipole-dipole interactions, we are somewhat lacking London interactions in spite of the rather large molecular weight of $52$.

This is important when comparing it to both methyl amine $M_\mathrm{r} = 31$ and butane $M_\mathrm{r} = 58$. Butane, given its larger electron count ($26$ bonding electrons versus $14$ bonding and lone pair electrons in methyl amine) and its very non-spherical shape can give rise to much stronger London interactions, especially since all its bonds are almost or entirely unpolar $\ce{C-H}$ or $\ce{C-C}$ bonds. A general rule of thumb is the heavier a molecule, the stronger the London interactions and they quickly catch up with what dipoles or hydrogen bonds can do.

So putting it all together:


Difluoromethane has very weak London interactions but some dipole-dipole interactions. It is expected to have a rather low boiling point.
Methyl amine can form hydrogen bonds but is also rather small and thus lacking London interactions. It is less polar than methanol giving rise to lesser dipole-dipole interactions. Put it somewhere in the middle.
Butane is significantly larger in volume than all the others and also has much more electrons that can take part in London interactions. Thus, London interactions are predicted to be rather strong but nothing else is there.
Finally, methanol of course forms strong hydrogen bonds, is a good dipole and can take part in London interactions well. A clear winner.


I wouldn’t know how exactly to rank butane and methyl amine but I would put difluoromethane at lowest and methanol at clearly highest boiling point. This is consistent with the experimental data.