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In this post, I refer to two molecules B (Di-tert-butyl ether) and C (Dibutyl ether).

Molecule B, Di-tert-butyl ether

Wikipedia: Di-tert-butyl ether (B)
b.p.  107.2 C
m.p. - 61   C

Molecule C, Dibutyl ether

ChemSpider: Dibutyl ether (C)
b.p.  142 C
m.p. - 96 C

I understand that Compound C has a higher boiling point than compound B, because of its greater surface area and therefore stronger intermolecular VDWs.

I don't understand, however, why Compound B has a higher melting point than compound C. A lecturer told me that this is due to compound B being more compact and symmetrical. I think that I understand the concept that some structures pack into a crystalline lattice better than others, such that the structure of the packing influences the strength of intermolecular forces within the solid, therefore affecting the melting point.

However, it seems to me that both B and C above are equally symmetrical in at least one conformation. Beyond that, C has conformations of greater surface area than B, which should further strengthen its intermolecular bonds in a solid state. I would expect the melting point of C to be higher than B; yet this is not the case.

I feel like the answer may lie in the fact that C has many more conformers than B. Even so, I can't seem to draw a link anywhere. Shouldn't the solid formed take the most stable crystalline structure, much like how water tends to just one of many crystal structures?

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  • $\begingroup$ Welcome to Chemistry! Take the tour to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. If you receive useful answers, consider accepting one. $\endgroup$ – pentavalentcarbon Jan 29 '18 at 13:59
  • $\begingroup$ I comment because I can't give you solid answer. The teacher refer to a kind of compactness and if you let me say so "sphericity". This reduces surface forces but has a positive effect on paking. Most important is the conformational spread of the linear chain. They bring an entropic term that favours melting and solubility. In other words in your solid there is potential for the molecule to be not only in the long confirmation sketched. Even if they adopted it, to get there they have already got energy at the expense of the wdw forces. Ergo the latter are less effective in binding the solid $\endgroup$ – Alchimista Jan 29 '18 at 14:16
  • $\begingroup$ Melting is a pretty tricky thing, unlike boiling. You can't really compare it like that. True, one would expect higher melting point in a molecule having central symmetry, but your molecules don't have it, and other kinds of symmetry are less relevant. As for the number of conformers, it hardly matters much, since the crystal is typically built of only one conformer. $\endgroup$ – Ivan Neretin Jan 29 '18 at 14:21
  • $\begingroup$ You basically see it in DG = DH - TDS, if DS is big enough. $\endgroup$ – Alchimista Jan 29 '18 at 14:39
  • $\begingroup$ @Alchimista Thank you, your comment has been very helpful! I can at least kind of grasp the intuition behind it now. $\endgroup$ – ning Jan 29 '18 at 16:16
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The terms used by your teacher are quite in line with at least a qualitative explanation.

I really suspect that there are various ways to express compactness and sort of "sphericity" of molecules. I am not a specialist so forgive me for being imprecise on this.

However, if we forget a rigorous classification in symmetry group, a potato is more spherical symmetrical than a banana.

In such a sense the ramified isomer is definitely more symmetric and "round" than its linear counterpart.

This is even more evident if all conformers are taken into account: in the linear molecule a chain end can even turn toward the center while the opposite being extended.

Now this does not necessarily explain the observed behaviour as for the more spherical molecule can pack easily but at expense of VdW forces that depend on molecular surface.

However, it relates to the fact that the linear isomer has a number of conformers much higher than the ramified one (rotation about 4 bonds versus 1 for each side of the ether)*** and results in the first having a bigger conformational entropic term.

Although the crystals will generally contain one isomer, the crystal formation take place at expense of the intramolecular forces. As such, when keeping entalpy terms comparable, bigger is the entropy term then less stronger is the binding.

$\Delta G = \Delta H - T\Delta S$,

For comparable enthalpies and very different conformational entropy terms, than the melting occurs at lower T for the entropy rich molecule, i.e. that one for wich more conformations become accessible.

This obviously is not the case for the liquid to gas transition, as the conformational S term is similar in both phases and negligible faced to the gas entropic content.

***Conformers due to the rotation around the bonds in the terbutyl groups do not really alter the shape of the molecules (connection with the compactness mentioned above). More rigorously at each side we should have (calling x the number of conformers around a bond) that the number of conformations is circa $x^4$ and about $3x^2$ for the linear and the branched molecules, respectively.

Additional note: conformational entropic terms are crucial to many phenomena in polymers and proteins (solubility, crystallization, protein folding/unfolding, etc). A treatment of this requires statistics that I have forgotten. So take my numerical estimated about conformers with a grain of salt. I am satisfied if I did convey the message.

Add. Note 2. Another useful example is, for instance, a stiff and insoluble chain such as polythiophene.

It is insoluble as for electron delocalisation makes it flat and rigid. Once side substituted with alkyl chains, the entropic term brought in by the flexible chain is the main responsible for the acquired solubility.

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