DMSO and acetone have almost identical structures, except for us replacing the carbonyl carbon with sulfur to obtain DMSO. DMSO's melting point is $19~\mathrm{^\circ C}$, whereas acetone's is $-95\ \mathrm{^\circ C}$.

Is the dominant factor DMSO's larger molecular weight? Probably not. While acetone is $58\ \mathrm{g/mol}$ vs. DMSO's $78\ \mathrm{g/mol}$, 2-pentanone is $86\ \mathrm{g/mol}$, but still has a very low melting temperature of $-77\ \mathrm{^\circ C}$.

The only protons on all of these species are aliphatic, so hydrogen bonding is probably not the dominant contributor to this effect either.

What's going on?

  • 6
    $\begingroup$ Melting points are largerly ruled by crystal packing and thus are often contrintuitive. For example, toluene has lower melting temperature than benzene. DMSO and acetone have different molecular geometry (DMSO is trigonal pyramidal and acetone is trigonal planar) so they cannot be compared directly. DMSO also has larger dipole moment. $\endgroup$
    – permeakra
    Dec 5, 2016 at 16:14
  • $\begingroup$ I seeee, okay. I totally failed to notice that DMSO's sulfur has a lone pair. That makes a more sense. Thanks! $\endgroup$
    – user49404
    Dec 5, 2016 at 16:52

1 Answer 1


The first thing to remember is that boiling and melting points, while seemingly simple, are incredibly hard to predict. Yes, many teachers and professors give their pupils and students a number of simple prediction rules, but there is a lot of intrinsic balance and experience associated with applying them that they may often seem unintuitive.

Comparing DMSO and acetone, the first striking difference is the replacement of carbon with sulfur. This may seem like a small change but results in radical differences. For example, compare propane with dimethyl sulfide; or the latter with dimethyl ether. $\ce{Me2S}$ is a liquid with a boiling point of around $40~\mathrm{^\circ C}$, the others are gases at room temperature and standard pressure.

There are two factors that explain the difference between $\ce{Me2S}$ and propane/methyl ether — one does better in explaining the difference to propane the other does much better in explaining the difference to dimethyl ether.

  1. Polarity. Sulfur is more electronegative than carbon, thus the $\ce{Me2S}$ molecule is slightly polar while propane is essentially nonpolar. Since all of these molecules exhibit (approximate) $C_\mathrm{2v}$ geometry, the polarity of individual bonds does not cancel out. However, one would expect $\ce{Me2O}$ to be more polar and thus have an even higher boiling/melting point.

  2. London interactions. These are often simplified as ‘rising with molecular weight’, but in fact, the key factor is not molecular weight (which is a nuclear property) but polarisablility of electrons. Sulfur, being larger, having an additional shell and thus being much more diffuse than both carbon and oxygen is much more polarisable. Thus, the significant difference in the boiling and melting points of $\ce{Me2S}$ and $\ce{Me2O}$ can be attributed mainly to the stronger London interactions, made possible by adding eight (core) electrons and thereby moving the valence electrons further away from the core.

Both factors play a role in the difference of melting points of DMSO and acetone, although there is even more. While acetone has a trigonal planar structure featuring a $\ce{C=O}$ double bond, DMSO is trigonal pyramidal with an additional lone pair and formal positive charge on sulfur, to give a $\ce{S^+-O^-}$ single bond. While this feature reduces the London interactions somewhat — sulfur’s higher oxidation state means that the electrons are more stable and the orbitals appear smaller at the same cutoff value — it increases polarity significantly.

Indeed, it is better to compare similar molecules; isopropanol’s structure is much closer to DMSO’s than acetone’s is.


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