[FutureDoc]

Is it because octane, being unbranched in a single plane, would have less points of contact than cubane, so hence, fewer intermolecular forces? Or am I looking at this completely wrong, because I thought as you got more intermolecular forces, the boiling point increased, but the melting point decreased?

Any help is appreciated greatly.

[FutureDoc]

I looked it over, and I think I got my reasoning backwards.

Octane, being unbranched, would have more surface area, and therefore more points of intermolecular attraction and more intermolecular forces. And as the intermolecular forces increase, the melting point decreases. So cubane, having less surface area, would have fewer intermolecular forces, and hence, a higher melting point?

Is this the correct way to state that?

[Yggdrasil]

Melting point should increase as the strength of intermolecular forces increase.  Unfortunately, I'm not sure why cubane would have a higher melting point.  Perhaps it could do with the efficiency of packing in the crystal (if cubane molecules can pack closer together in the crystal than octane molecules, then the cubanes' intermolecular forces will be stronger than those the octane solid).

[nj_bartel]

Could it possibly have something to do with antibonding orbital interactions?  Cyclobutane has a higher melting/boiling point than n-butane as well, so it's leading me to believe there's something special about the 90^o bond angles.

Then again it's the same deal with hexane/cyclohexane... =/