Why is 2-methylpropene less in energy than its alkene counterparts?

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?




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](https://i.stack.imgur.com/c09kw.jpg)


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](https://i.stack.imgur.com/hzIS2.jpg)


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](https://i.stack.imgur.com/E1cp2.jpg)


[image source](http://quimicaorganica.webcindario.com/Eliminacion1.pdf)


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