Why does neopentane have a higher melting point than n-pentane?

I know that we can say that, due to stacking, it has a higher melting point as it freezes easily. But then, why doesn't this happen in the case of n-hexane too?

n-Hexane has a melting point of −95 °C whereas 2,2-dimethylbutane has a melting point of −98 °C. Instead, if we go by stacking, 2,2-dimethylbutane should have a higher melting point.

enter image description here enter image description here

image source


Many symmetrical molecules undergo a phase change before they melt. This is usually interpreted as the ability to rotate without disrupting the lattice structure. This has a much greater effect on the entropy of the solid than on its energy. This reduces the entropy of fusion more than it reduces the enthalpy of fusion, thereby raising the melting point which equals delta-H/delta-S. This is very dramatic in the comparison of melting points of octane, cyclo-octane, and bicyclo-octane.

|improve this answer|||||
  • $\begingroup$ Is there a name for this phase change? Or can you give a source describing it? It sounds reasonable, but I've never heard of it. $\endgroup$ – SendersReagent Sep 9 '18 at 22:20
  • $\begingroup$ Some call it "Order-disorder phase transitions", and yes, they are pretty much a thing. $\endgroup$ – Ivan Neretin May 8 '19 at 8:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.