The effect of hyperconjugation on the stability of alkenes with MO theory

Hyperconjugation stabilizes carbocations and that makes sense because electrons are given to the empty p orbital.


But how does it stabilize alkenes? Can molecular orbital theory be used to explain it?




Since you are familiar with how carbocations are stabilized via hyperconjugation I will keep this brief: With alkenes, it really is the same principle at work. Instead of the empty p orbital, you have to consider the empty $\pi^{\*}$ orbitals of the $\ce{C=C}$ double bond. They interact with the filled high-lying neighboring $\ce{C-H}$ or $\ce{C-C}$ $\ce{\sigma}$ bonds in $\pi$-bond-like manner


![](https://i.stack.imgur.com/F8juF.png)


and so from those two fragment orbitals you will get two molecular orbitals: one from the in-phase combination of $\pi^{\*} (\ce{C=C})$ with $\sigma (\ce{C-X})$, which will be lower in energy then the fragment orbitals, and one from the out-of-phase combination of $\pi^{\*} (\ce{C=C})$ with $\sigma (\ce{C-X})$, which will be higher in energy then the fragment orbitals. Since in this interaction there are only two electrons involved, i.e. the ones in $\sigma (\ce{C-X})$, you get a stabilization from the interaction since those two electrons will now occupy the in-phase MO and have thus lowered their energy. The out-of-phase MO is empty and doesn't contribute to the energy of the molecule.


![enter image description here](https://i.stack.imgur.com/Kcdem.png)


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When the sp2 hybridized Carbon atom directly attached to sp3 hybridized Carbon atom having at least 1 hydrogen atom(called α hydrogen), this condition is known as hyperconjugation. 
Stability of any hyperconjugative hydrocarbon depends upon no. of α hydrogen present in compound.. 


More the no. of α hydrogen, more is the stability of compound 


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