Arguments based on intermolecular interactions are often used to explain
boiling point trends. But (as you've discovered from the dichlorobenzene trends) such explanations are sometimes incomplete.
Thermochemical dichlorobenzene data taken from the NIST Webbook.
Looking at the structures of the three compounds, you would expect that the highly asymmetric ortho isomer (with two big electronegative atoms on one side of the molecule) would be the most polar, and the para isomer (with a highly symmetrical shape) would be nonpolar. The meta isomer will be polar too, but (if the overall dipole moment is estimated by adding up the bond dipoles) less so than the ortho isomer.
|Isomer||mp, °C||bp, K|
The stronger the interactions, the more heat will be required to break them. So you would expect that the boiling and melting points would follow the trend ortho > meta > para. The melting points clearly don't follow this trend!
Here is a thermodynamic explanation. The difference in Gibbs free energy between the two phases at equilibrium is zero. At the melting point, we can write
G = 0 =
which shows that we have to look at both enthalpy AND entropy trends to do a good job of comparing melting (and boiling) points for related compounds. Intermolecular interactions relate to trends in enthalpies of transition. Changes in conformational freedom determine basic trends in entropies of transition.
Thermochemical dichlorobenzene data taken from the NIST Webbook
||Hfus° at mp||Sfus° at mp|
Enthalpy of fusion is really just the enthalpy of the liquid minus the enthalpy of the solid. The gap between the solid and liquid enthalpies is highest for the para isomer, and almost identical for the meta and ortho isomers. It seems that there is more breaking of intermolecular attractions for liquifying solid para isomer, as well as a larger increase in conformational freedom.
But why? Maybe the symmetry of the para isomers allows the molecules in the solid to pack much more closely than is possible for the meta or ortho isomers. The close-packed para molecules interact via London forces. That lowers the enthalpy of the solid and makes the enthalpy gap between solid and liquid larger for para.
The high entropy of fusion for para is consistent with this hypothesis. If the solid para isomer was packed in a much more orderly way than the ortho and meta isomers, you'd expect the entropy increase on melting to be highest for the para- provided that you assume that the absolute entropy of liquid dichlorobenzene is about the same whatever the isomer.
It's interesting that the meta and ortho enthalpies and entropies of fusion are so similar. Perhaps there isn't as large a difference between their polarities as a simple
vector addition of bond dipoles would lead you to predict.
Author: Fred Senese firstname.lastname@example.org