If one considers boiling points (in °C) of primary alcohols, one finds the following:

  • methanol: 65
  • ethanol: 79
  • 1-propanol: 97
  • 1-butanol: 117
  • 1-pentanol: 138

This trend is due to Van der Waals forces increasing with molecular weight.

Now if one focuses on melting points, I would expect the same trend, but these are the experimental values (in °C):

  • methanol: −98
  • ethanol: −117
  • 1-propanol: −127
  • 1-butanol: −90
  • 1-pentanol: −79

the lowest value being the melting point of 1-propanol.

The general idea stating that with all else being equal, more massive molecules have higher boiling and melting points, does not fully apply for this set of alcohols.

So my question: why melting and boiling points of primary alcohols do not follow the same trend?


1 Answer 1


Melting points are a bit trickier to compare than boiling points, especially when you're looking at the lightest examples of a group of molecules. This graph for the unbranched alkanes exemplifies nicely how the trends differ in smoothness.

As you increase the weight of molecules, the observed trend depends on how you increase the weight. For example, if you start with methanol and increase the weight by adding more $\ce{-HCOH -}$ units (methanol, ethylene glycol, glycerol, and so on) you would find a very clear increase in both melting and boiling points. In some sense this is a more legitimate trend to analyse, because the relative amounts of different types of intermolecular interactions stays approximately constant (in particular, all molecules have one hydroxyl per carbon atom, and hence they all have about the same amount of hydrogen bonding per atom in the molecule). In your sequence of alcohols, the strong hydrogen bonding allowed by the hydroxyl group becomes "diluted" as the molecule gains an ever larger alkyl chain, which can only sustain much weaker van der Waals interactions. In the limit of alcohols with very large alkyl chains, the alcohols' melting and boiling points do not differ tremendously from their parent saturated hydrocarbons. Compare the melting and boiling points of 1-hexadecanol (49°C and 344°C) and hexadecane (18°C and 287°C).

Also, the temperature at which a substance freezes is dependent not only on the strength of intermolecular interactions, but how the solid packs. Even substances with strong interactions may only freeze at very low temperatures if they pack very poorly when solidified. Extreme examples to which this effect contributes are ionic liquids. They tend to have a very low vapour pressure at reasonable temperatures (as you would expect of many salts), but some melt even below 0°C. The freezing of such bulky entities would entail lots of ordering and therefore a very large decrease in entropy, so it is disfavoured (or equivalently, their melting is highly entropically favoured, so the free energy of the liquid phase becomes lower than the free energy of the solid phase even at relatively low temperatures). The exact geometry of the molecules in the solid is also important, as some favourable interactions may be suppressed in the solid state due to steric hindrance which may be less present in the more freely-moving molecules of a liquid. Furthermore, molecules with a high amount of symmetry tend to have high melting points, because freezing is entropically favoured due to molecules more readily fitting in the solid lattice.

Yet another effect to consider is how large a surface area each molecule has available for intermolecular interactions. This is often brought up to explain differences between melting and boiling points of branched and unbranched organic compounds; when comparing structural isomers, branched compounds have a lower available area for interaction than unbranched compounds, so the latter tend to have higher melting and boiling points (branching may sometimes create a combination of opposing packing and surface area effects, though).

So why does your sequence of alcohols behave the way it does? The fact that the trend in boiling point is monotonic while the melting point is not suggests that even though there is a relative shift in importance of types of intermolecular interaction as the alkyl chain increase (from hydrogen bonding to van der Waals interactions), it is probably not the source of the melting point drop. Hence it likely has to do with geometric factors in the solid. The alcohol's alkyl chain probably disrupts the hydrogen bonding network of the solid as it grows from methyl to propyl in a very severe fashion, by making the solid pack less well or by partially hindering the number or strength of hydrogen bonds in the geometry of the solid. Perhaps looking at the crystalline structures of the solid alcohols may provide further insight.

Edit: I wrote a wrong answer initially. I've updated it, but now I realize I'm still not quite sure it's correct. I originally meant to write just a comment, and as I kept writing more I forgot that my very first thought was actually incorrect, so the whole text wasn't built as well as it should be. Anyone is free to pick apart or repurpose what I wrote!

Edit 2: User Uncle Al posted an interesting list of compounds and their melting/boiling points, showing the importance of molecular symmetry and solid packing.


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