Other isomers include 1-butene, cis 2-butene, and trans 2-butene. Why would 2-methylpropene be less in energy if there is more steric hindrance?


1 Answer 1


There are two things we need to understand before we can answer the question.

1) More highly substituted double bonds are generally more stable than less substituted double bonds.

This is because the $\ce{sp^3}$ hybridized carbon in the alkyl group is electron donating towards the $\ce{sp^2}$ hybridized carbon in the double bond. Electron density likes to flow from low to high s-character orbitals since the more s-character in an orbital, the lower the energy of the electrons in that orbital. As the electrons flow from the $\ce{sp^3}$ carbon to the $\ce{sp^2}$ carbon the energy of the electrons (and therefore the energy of the molecule) is lowered.

A "shorthand" way to illustrate this electron flow is by drawing hyperconjugated resonance structures like the following.

enter image description here

Consequently 2-butene (where we can draw 6 hyperconjugated resonance structures) is more stable than 1-butene (where we can only draw 2 hyperconjugated resonance structures).

2) Carbanion stability decreases in the order primary > secondary > tertiary

As we've just shown above, alkyl groups are electron releasing. If we have a negative charge on a carbon (a carbanion), the last thing we want is to have more alkyl groups attached to that carbon releasing more electron density onto an already negative carbon atom.

Now we can return to your question and try to understand why 2-methylpropene is more stable than trans-2-butene. Carbon-hydrogen hyperconjugation is at the center of the answer. Let's compare some of the hyperconjugated resonance structures that can be drawn for these two molecules.

enter image description here

Remembering that a) more highly substituted double bonds are more stable and b) that primary carbanions are more stable than secondary carbanions, we can compare hyperconjugated resonance structures A and B. Resonance structure A contains a disubstituted double (more stable) and a primary carbanion (more stable), while resonance structure B contains a monosubstituted double bond (less stable) and a secondary carbanion (less stable). Therefore, resonance structure A should be more stable than resonance structure B. Since resonance structure A is more stable than resonance structure B, 2-methylpropene will be more stable than trans-2-butene.

Finally, we note that cis-2-butene is less stable that trans-2-butene for steric reasons. This allows us to place these $\ce{C_4}$ isomers in the following order of stability:

2-methylpropene is more stable than trans-2-butene which is more stable than cis-2-butene, which is more stable than 1-butene. If we examine the heats of combustion for these isomers, we see that our order of stability is confirmed.

enter image description here

image source

  • $\begingroup$ Why do you say that a primary carbanion is more stable than a secondary carbanion? $\endgroup$ Commented Jun 26, 2019 at 13:41
  • 1
    $\begingroup$ @Cyclopropane As Point 2 above notes, alkyl groups are electron releasing, hence attaching them to a carbanion (a carbon already bearing a negative charge) further destabilizes the ion; 3 alkyl groups more than 2, more than 1. $\endgroup$
    – ron
    Commented Jul 12, 2020 at 19:04

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