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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.

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  • $\begingroup$ No time to go through your Q but it can be related to this Q and my A at it. chemistry.stackexchange.com/q/89689/48509 $\endgroup$
    – Alchimista
    Commented Mar 12, 2018 at 9:50
  • $\begingroup$ Neopentane is completely symmetrical, whereas neohexane loses this symmetry. Neohexane thus has to be "flipped around a certain way" to fit into the crystal compared to neopentane where all the methyls are equivalent. $\endgroup$ Commented Oct 25, 2021 at 15:03

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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.

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  • $\begingroup$ 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. $\endgroup$ Commented Sep 9, 2018 at 22:20
  • $\begingroup$ Some call it "Order-disorder phase transitions", and yes, they are pretty much a thing. $\endgroup$ Commented May 8, 2019 at 8:29

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