How to determine what organic molecule has the highest boiling point?

Compare and answer which one has the highest boiling point:

• $$\ce{CH3CH2CH2CH3}$$ (butane) ... [$$\pu{−1}$$ to $$\pu{1^\circ C};\ 30$$ to $$\pu{34^\circ F};\ 272$$ to $$274\ \pu K$$]
• $$\ce{CH3NH2}$$ (methylamine) ... [$$−6.6$$ to $$−6.0\ ^\circ\pu C;\ 20.0$$ to $$21.1\ ^\circ\pu F;\ 266.5$$ to $$267.1\ \pu K$$]
• $$\ce{CH3OH}$$ (methanol) ... [$$64.7\ ^\circ\pu C\ (148.5\ ^\circ\pu F;\ 337.8 \pu K)$$]
• $$\ce{CH2F2}$$ (difluoromethane) ... [$$−52\ ^\circ\pu C\ (−62\ ^\circ\pu F;\ 221\ \pu K)$$]

My effort: At first, without referring to factual data (but the factual data has been provided in parenthesis, see above-options, Source: wikipedia.org), I reached the following conclusion:

• Butane appears to have the lowest boiling point because there is only Van-der-Waals forces, which are weaker than dipole-dipole interactions, present in the rest of molecules.

• Next I would compare between methylamine and methanol because they both have dipole-dipole interactions and most importantly that these molecules associate themselves with hydrogen bonds, but as oxygen is more electronegative than nitrogen, hydrogen bonds formed within methanol molecules would be stronger than those formed between methylamine molecules. So methylamine takes the 3rd position.

• Lastly, it becomes easy for me to differentiate between methanol and difluoromethane, because according to my knowledge the latter has got fluorine which is more electronegative than oxygen, present in former. Moreover difluoromethane has got 2 fluorine, which add weight to my above argument in the way that there will be even more extensive hydrogen bonding within the molecules.

Inferences and comparison with the source's answer

I am full of doubts as to why difluoromethane is not the answer. Revisiting my arguments again and again, I always ended up with the same conclusion i.e., mine answer would always contrast with the source's answer.

Afterthoughts & Deductive Conclusions
A few days after extensive thinking and arguing myself, I ended with the following reason, which seemed plausible to my brain. It may be possible that the because of: the presence of two fluorine atoms, small and compact nature of carbon-fluorine bond (according to bond length - bond strength - bond dissociation enthalpy relation/ theory), and the overall small and virtually spherically shape (But, in fact, it is tetrahedral geometry) of difluoromethane; the two fluorine atoms might be finding themselves in a position, in which they cannot simultaneously form hydrogen bonds with the hydrogens of other molecule. My above explanation can also imply that they might be hindering into each other's hydrogen bonds (reason can be same as mentioned in list in the previous sentence).
This reasons and explanation thus explain why the boiling point of difluoromethane falls below that of methanol. Remember that in methanol, there is less compact nature of carbon-oxygen (similar size of orbitals of carbon and oxygen as well as bigger size of oxygen atom as compared to that of fluorine), a single carbon-oxygen bond is weaker than a single carbon-fluorine bond (based on electronegativity difference and concept of dipole-development & interactions), and the last that is a molecule of methanol is less symmetrical (3-H and 1-O) as compared to difluoromethane, which is more symmetrical (2-H and 2-F).

• Solution1: $$\ce{C4H10 < CH3NH2 < CH3OH < CH2F2}$$ (in increasing order from left to right)
• Solution2: $$\ce{C4H10 < CH3NH2 < CH2F2 < CH3OH}$$ (in increasing order from left to right)
• Solution3: None (it's should be totally based on your own conclusions and knowledge, backed up with info. from appreciable sources)
• It's simpler than that. Hydrogens that sit on carbon just don't form hydrogen bonds. – Ivan Neretin Apr 16 '16 at 13:14
• @Ivan That is a crude approximation, that is not true in many cases. It is, however, known that such hydrogen bonds are weaker. – Martin - マーチン Apr 16 '16 at 13:31
• Yes, but a crude approximation is just what we need here. – Ivan Neretin Apr 16 '16 at 13:33
• So, should we consider the 'crude approximation' of another Hydrogen atom's interaction in our analysis. – Shivanshu Gupta Apr 17 '16 at 10:16

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.