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Starting with an example, n-butane has boiling point of $\pu{55^\circ C}$, whereas that of iso-butane is $\pu{-10.2^\circ C}$.

Why there is so much difference in the boiling points?

*EDIT : On the contrary, branched alkanes are more stable. For example, 2-methylpropane is more stable than n-butane. Why?*

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  • $\begingroup$ @Abdul Ahad. Do not equate low boiling point and stability because the two things are not related, at least not in general. $\endgroup$ – Alchimista Jul 24 '17 at 17:48
  • $\begingroup$ @Alchimista No! Obviously not.What I'm saying is that how come a compound (n-alkane) having a higher boiling point is less stable ? $\endgroup$ – Abdul Ahad Jul 24 '17 at 17:50
  • $\begingroup$ Is the same. They, stability and high or low boiling point, are two different things and in general they are unrelated! $\endgroup$ – Alchimista Jul 24 '17 at 17:53
  • $\begingroup$ Well! Thanks for clearing the fact that they are unrelated.If you could provide me with an explanation on the second part of the post,it'll be much better.So the question can be isolated as "Why n-alkanes are less stable than iso- ?" $\endgroup$ – Abdul Ahad Jul 24 '17 at 17:57
  • $\begingroup$ The point is that myself I cannot recall an easy answer I started to constructing one. I proceed by considering linear and branched carbocations and or radicals. But it did not seemed to me a proper way to go. With my surprise I was somehow on the track. I just googled your question and found Protobranching. Still I have to read. It is indeed an electronic effect getting stronger as the number of Cs in germinal position - respect a given C - is higher. I forgot what I learned / I was told about this years ago. Probably some attempt to rationalise the situation via steric effects, now dismissed $\endgroup$ – Alchimista Jul 24 '17 at 19:33

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