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This species is a derivative of benzene, with a single fluorine atom attached. Its melting point is -44 °C, which is lower than that of benzene, indicative of the remarkable effect of fluorination on the intermolecular interactions as seen throughout organofluorine chemistry. In contrast, the boiling points of PhF and benzene differ by only 4 °C.

Why is this true? Wikipedia doesn't elaborate on why.

From what I understand the melting point is the point at which the solid and liquid forms are in equilibrium.

Boiling point is the point at which the liquid and gas phases are in equilibrium.

I know that intermolecular forces play an integral role in melting and boiling point. However, why does it seem that the strength of the intermolecular forces in the solid phase is stronger than that in the liquid phase for fluorobenzene?

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The answer has everything to do with fluorine. Dunitz, in his early studies on fluorine interactions found that intermolecular $\ce{C-H..F}$ interactions can be solid-state structure (e.g. crystal packing) directing. In the case of fluorobenzene, the molecule adopts a solid-state structure that sacrifices long-range order so that these short-range $\ce{C-H..F}$ interactions can be preserved (also see here, and here where it is stated that, "Its [fluorobenzene] melting point is -44 °C, which is lower than that of benzene, indicative of the remarkable effect of fluorination on the intermolecular interactions as seen throughout organofluorine chemistry"). The loss of long-range order in the solid-state, as compared to benzene, produces a lower melting point in fluorobenzene (-44$\ce{^{o}}$C) than in benzene (5.5$\ce{^{o}}$C). Note too that the "size" of fluorine in organic molecules is very similar to that of hydrogen (that is why it has found such widespread use in pharmaceuticals) - it is often referred to as a hydrogen "mimic" (see here). So steric (or symmetry) arguments cannot be used to explain the packing disruption with fluorobenzene.

Once the molecule is in the liquid state the intermolecular $\ce{C-H..F}$ short-range interactions are also diminished due to increaseded molecular motion; as a result, fluorbenzene and benzene boil relatively close to one another (84$\ce{^{o}}$C and 80$\ce{^{o}}$C respectively).

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  • $\begingroup$ Then why does chlorobenzene have almost exactly the same melting point, -45C? and bromobenzene and iodobenzene also quite low? $\endgroup$ – DavePhD Dec 22 '14 at 23:39
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    $\begingroup$ @DavePhD Fluorine plays a determining role in answering the OP's question, take a look at the links. As to chlorobenzene, perhaps the increased MW (raise mp) roughly balances the lost C-H-F interaction (lowers mp). Perhaps it is the same reason that the bp of benzene = mp of naphthalene and the bp of naphthalene ~ mp of anthracene - coincidence. The OP's question deals specifically with fluorobenzene and in that case it is known that the C-H-F interaction determines the solid state structure. $\endgroup$ – ron Dec 22 '14 at 23:56
  • $\begingroup$ @ron What interactions are lost in napthalene which are balanced by the increased molecular weight of napthalene vs benzene? $\endgroup$ – Dissenter Dec 23 '14 at 5:35
  • $\begingroup$ It's not about fluorine as many of the specific examples show. Toluene also has a much lower melting point than benzene and it is all to do with crystal packing not the specific substituent added to the ring. Most non-substituted benzenes have much lower melting points regardless of the substituent because they all disrupt the crystal packing possible in a neat flat ring. $\endgroup$ – matt_black Dec 23 '14 at 14:08
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    $\begingroup$ @ron Steric arguments aren't going to explain disruption of packing, but charge arguments may. Benzene has + on the Hs and - charge in the ring and the + and - favorably interact, perfluorbenzene has - on the Fs and + in the ring, and the + and - favorably interact. Substituting one F for an H would decrease such favorable charge interaction. The hydrogen bonding effect may compensate for some of the lost interaction. Also, there are the residual entropy considerations that make the more symmetric structure higher melting point pubs.rsc.org/en/content/articlehtml/2004/ob/b407105k $\endgroup$ – DavePhD Dec 23 '14 at 15:50
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The answer really doesn't have much to do with fluorine. Beacall's 1928 The melting-points of benzene derivatives explains on the first page that it has long been known "that the more symmetrical the constitution of a benzene derivative, the greater the resistance it offers to the passage from the solid to the liquid state".

In the more modern Effect of molecular symmetry on melting temperature and solubility:

As a general rule, symmetrical molecules in crystalline form have higher melting temperatures and exhibit lower solubilities compared with molecules of similar structure but with lower symmetry. Symmetry in a molecule imparts a positive amount of residual entropy in the solid phase (i.e., more possible arrangements leading to the same structure). This means that the entropy of a crystal of symmetric molecules is greater than the entropy of crystal of a similar, but non-symmetric molecule.

Benzene and hexafluorbenzene have almost the same melting point, while monofluoro, monochloro, monobromo benzene all have much lower melting points.

Both benzene and hexafluoro-benezne have a very stable "herringbone" crystal structure, where the edge of one ring approaches the center of another ring in a perpendicular manner.

enter image description here

So realizing that the cyrstal structure of benzene is unusually stablized, it is understandable that the melting point of benzene is higher than monofluorobenzene, but the boiling point is not.

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  • $\begingroup$ what's stabilizing about the herringbone structure $\endgroup$ – Dissenter Dec 22 '14 at 23:40
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    $\begingroup$ What is the crystal structure of fluorobenzene? Is it more or less stable than that of benzene, and why? Without answers to those questions how can you draw a conclusion? $\endgroup$ – ron Dec 23 '14 at 0:19
  • $\begingroup$ @Dissenter Some describe it as the postive electrostatic potential of the ring edge interacting with the negative electrostatic potential of the ring face, while others describe it as a hydrogen bond interaction. Even the gas phase dimer has the edge-face T-shape geometry as lowest energy by 2 kcal/mol. $\endgroup$ – DavePhD Dec 23 '14 at 3:46
  • $\begingroup$ @ron, you are right, it would be a much better answer if I had references for the crystal structures of the various benzene derivatives in the answer $\endgroup$ – DavePhD Dec 23 '14 at 5:23

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