Whenever you can lower the energy of a system, the system is more stable.
This is common in resonance. Allylic carbocations are stablized through resonance. The electrons in the pi-bond can delocalize into the empty p-orbital and this delocalization of electrons lowers the energy of the system. Allylic carbocations are stabilized through the delocalization of electrons through resonance.
The effect is also true for other carbocations. When you form a carbocation you have a full positive charge on carbon, and an empty p orbital (fig. 1). When it's a methyl carbocation, there are three hydrogen atoms attached to the carbocation, and that's it (fig 2). we know that to be very unstable.
As soon as you have one methyl group on the carbocation, things change. Now, there is a carbon-hydrogen sigma bond that (when the rotation is right) is almost parallel to the empty p-orbital on the carbocation (fig 3). Why is that important? Because the orbital of the carbon-hydrogen sigma bond (the one with two electrons) is very close in proximity to the empty p-orbital. Because of the close proximity of an orbital high in electron density (the filled sigma bond) and an orbital low in electron density (the empty p-orbital), those orbitals can interact. I'm not going to say they bond, or they overlap, but they interact. The electron density in the filled sigma bond interacts with the empty p-orbital (fig 4)
This doesn't form a new bond, and doesn't mean there isn't a carbocation, and doesn't mean the carbon now is free of its full positive charge... but the electrons in the C-H sigma bond do interact with the empty p-orbital. The filled molecular orbital of the C-H sigma bond (with its pair of electrons) and the empty p-orbital interact in a stabilizing manner. The electron density is now spread out a bit and to some extent encompasses part of the empty p-orbital. The electrons are somewhat delocalized over 2 atoms (even though there is still a full sigma bond between C-H), and the delocalization of electrons - along with the relief of some of the lack of electron density on carbon - lowers the energy of those electrons. I've drawn a pseudo energy diagram in fig 5, but I made the electrons dashed lines. I don't want you to think that a double bond is forming between the two carbon atoms (although you could draw that resonance structure if you wanted... you'd also have a free proton). The interaction between the electrons in the C-H sigma bond and the empty p-orbital provides a net stabilizing interaction.
Same thing with hyperconjugation. In this case, the electrons in one sigma bonding orbital delocalize and interact with the sigma star antibonding orbital of the neighboring group (fig 6). The same energy diagram applies, except the C-H sigma bonding electrons are interacting with a C-H sigma* antibonding orbital - not a p orbital. The delocalization of electrons lowers the energy of the electrons and provides a net stabilization to the system.
After re-reading the op, you say "the hyperconjugated electron goes into an antibonding orbital." This is key. The electron doesn't leave the sigma bonding orbital and relocate into the antibonding orbital. If it did, that would be very destabilizing, as you'd remove electrons from a lower, more stable orbital and relocate the to a higher, much less stable orbital. I used to think of this the same way for quite a while, but it's not how it works. The key is the delocalization of electrons from a lower-energy filled orbital with a higher-energy unfilled orbital. This delocalization provides a net stabilization to the pair of electrons, and provides a net stabilization to the molecule as a whole.
We mentioned the anomeric effect, so I'll talk about it here for a bit. Although, it's the same analysis. For a 6-membered ring with an endocyclic oxygen and an electronegative element as a substituent on the carbon atom next to the endocyclic oxygen atom (fig 7), that exocyclic electronegative substituent prefers to reside in an axial orientation. Sterics would tell us the large substituent should be equatorial, and in the absence of the neighboring endocyclic oxygen atom that would be true. But this particular orientation - the endocyclic oxygen atom next to the exocyclic electronegative atom - the substituent prefers an axial orientation.
Why? In one explanation, the same orbital overlap scenario applies. There are two lone pairs on the endocyclic oxygen atom. As with all substituents in a 6-membered ring, one lone pair is axial, one lone pair is equatorial. Here's the key - when the electronegative subtituent is axial, the axially-oriented lone pair of electrons is synperiplanar with the C-X sigma* antibonding orbital (fig 8 ). In this case, the lone pair of electrons can delocalize with the antibonding orbital just like the energy diagram in fig 5. This lowers the energy of the electrons and leads to a net stabilization of the system. This is the principle behind the anomeric effect. The delocalization of electrons leads to a net stabilization of the molecule.
This is much more dramatic for anomerically-stabilized systems than for, say, ethane. Why? The lone pair is higher in energy and the C-X antibonding orbital is lower in energy, and the two energy levels are closer together than a C-H sigma bond and a C-H sigma* antibonding orbital. With the energy levels closer together, the overlap is more favorable. There is a whole series of good electron donors (lone pairs are among the best) and electron acceptors (empty p-orbitals are among the best).
Hope this helps.
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Topic: Question regarding hyperconjugation and FMO (Read 17366 times)