For two compounds that are symmetrical, why does the more compact one have a higher melting point? In this post, I refer to two molecules B (Di-tert-butyl ether) and C (Dibutyl ether). 



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




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?
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.