(How to manipulate JSmol structures)

London Dispersion Forces are the weakest intermolecular forces. They apply to both polar and non-polar molecules but are a smaller contributor in polar compounds. When non-polar molecules near one another the their electrons are temporarily polarized (think formation of a tiny bar magnets). They are attracted to one another. The more electrons available, the greater the attraction. While the two alkanes in question have the same formula, mass and number of electrons, the exposure of the electrons differs owing to the shapes of the two alkanes. The difference in behavior reduces to a question of surface area. The greater the surface area; the greater the attraction between molecules. Greater attractive forces favor the liquid state over the gaseous state. 2,2,3,3-Tetramethylbutane is more spherical than n-octane. Efficient surface area is more diffult than in the case of n-octane. Hence, its boiling point is lower than n-octane, which has a larger surface area. However, the spheroid shape of 2,2,3,3-tetramethylbutane favors the solid state over the liquid state. In the solid, crystalline state, efficient packing of molecules -- or atoms for that matter -- is essential (Think of the stacking of cannon balls in a pyramid). On the other hand, n-pentane has many more conformations available to it than 2,2,3,3-tetramethylbutane and, therefore, more difficult to organize into crystals. Consequently, n-octane has a much lower melting point (freezing point) than 2,2,3,3-tetramethylbutane. The more spherical of two isomers will tend to have a smaller gap between its melting and boiling point.