I would like to understand the term stability in organic chemistry. We know empirically that isobutane (2-methylpropane) is more stable than n-butane. For example, their heats of formation (taken from the NIST WebBook) are $\pu{-134.2 kJ/mol}$ and $\pu{-125.6 kJ/mol}$ respectively, putting isobutane at ca. $\pu{10 kJ/mol}$ lower in energy.

But why is this so? Is there a physical or chemical explanation for this?


1 Answer 1


I would like to understand the term stability in organic chemistry.

We can speak of kinetic stability or thermodynamic stability when we discuss chemical compounds. If a compound is kinetically stable, that indicates that it has a high barrier (large activation energy) to cross and is therefore less reactive than a compound that has a low barrier (small activation energy) to reaction. In the following chart we can say that the reactants are kinetically less stable and the products are kinetically more stable since the barrier to the forward reaction is smaller than the barrier to the back reaction.

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(image source)

If one compound has a lower heat of formation than another compound, we can say that the former compound is thermodynamically more stable than the latter compound. Again using the above chart, we can say that the product is thermodynamically more stable than the reactant.

We know empirically that i-butane is more stable than n-butane. But why is it more stable?

Steric interactions do not provide the answer (reference).

We use arguments based on carbon-hydrogen hyperconjugation to explain many things.

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For example, why more highly substituted double bonds are more stable than less substituted double bonds. Carbon-carbon hyperconjugation can also be used to explain various chemical phenomena. In the case at hand, let's compare carbon-carbon hyperconjugated resonance structures for n-butane and 2-methylpropane.

enter image description here

The hyperconjugated resonance structure for butane involves a primary carbocation, while the hyperconjugated resonance structure for 2-methylpropane involves a more stable secondary carbocation. This more stable carbon-carbon hyperconjugated resonance structure explains the increased thermodynamic stability of 2-methylpropane compared to the isomeric n-butane.

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    $\begingroup$ I'm not fully convinced by the resonance structures; I feel like this is cherry picking the ionic resonance structure which gives the correct explanation. Why don't we consider, for example, $$\ce{CH3CH2CH2CH3 <-> CH3CH2CH^{-}CH3 + H+}$$ and $$\ce{CH3CH(CH3)2 <-> CH3C^{-}(CH3)2 + H+}$$ and say that on the basis of a secondary carbanion being more stable than a tertiary carbanion, this ionic resonance form of n-butane must contribute more than it does in isobutane, and hence n-butane must be more stable? $\endgroup$ Commented Apr 20, 2019 at 23:28

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