We learn that polar compounds have higher melting and boiling points than nonpolar compounds due to strong intermolecular forces such as hydrogen bonds and dipole-dipole interactions.

Naphthalene is an extremely simple and nonpolar compound, yet it has a significantly higher melting point than diphenylamine (a similar enough structure that is not only heavier, but has a dipole-inducing amine) and water (famously super polar).

What's going on here?


1 Answer 1


In comparisons such as this water needs to be excluded as it is exceptional in almost every respect due to its highly directional and strong hydrogen bonds.

As a broad generalisation the melting of solids is more dependant on the nature of the repulsive forces between molecules than on their attractive ones. Experimentally the enthalpy of melting ( Latent Heat or just Heat of Fusion) is approximately ten times less than the corresponding heat of vaporisation which indicates that repulsive forces play a greater role in melting than in vaporisation: in a solid molecules tend to be slightly closer together than in a liquid, thus may experience larger repulsive forces. Van-der-Waals forces tend to have weak orientational dependence, but repulsive forces do not as they are short range and depend on the asymmetric shape of the molecules and can have a large effect.

Thus melting points are rather dependent on the geometry of the molecules, if they can pack well into a lattice then a high melting point is expected. If the molecule has a dipole this can cause both attractive and repulsive interactions depending on the relative positioning of molecules; head to tail can be attractive if molecules are face to face, but repulsive if in line one behind the other and vice versa if their relative orientation is changed.

In the case of naphthalene and the amine, clearly the naphthalene can stack nicely and so increase its attractive interaction relative to repulsive ones, whereas the amine is twisted and cannot so easily pack meaning that repulsion can be stronger and attractive interactions weaker (even with dipole) and melting points lower. As an example consider the $\ce{C6H14}$ isomers, n-hexane mp -95 C, 2-methyl-pentane mp -154 C but boiling points are 69 & 60 C respectively and $\ce{C8H18}$ isomers, n-octane mp -57, 2-me-heptane mp -120 and bp 126 & 119 C respectively.

( Attractive forces largely determine the enthalpy of vaporisation, and by Trouton's rule their boiling points. The enthalpy of vaporisation is closely related to the (attractive) cohesive energy in a liquid.)


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