So I learned that hyperconjugation is electron delocalization from a $\ce{C-H}$ $\unicode[Times]{x3C3}$ bond into an empty p orbital (on a carbocation for example) or an antibonding orbital. I get how it's stabilizing in the case of an empty p orbital, but how is it stabilizing in terms of delocalization into an antibonding orbital, since by definition electron density in antibonding orbitals is destabilizing?


3 Answers 3


This is nothing specific to antibonding or nonbonding orbitals. There is a very general principle behind it:

Mix an occupied and an empty orbital with each other, and the occupied orbital will be stabilised.

You’re not actually ‘transferring electron density into an antibonding orbital’. You are simply linearcombining two orbitals: one occupied one with a rather low energy and one unoccupied one with a rather high energy. With this linear combination, you receive two new orbitals, where the unoccupied one is further destabilised while the occupied one is further stabilised.

Your electron density is still in the (now more stable) occupied orbital and the antibonding/nonbonding/otherwise empty orbital (now less stable) is still empty. Only their shapes have changed a bit, because we mixed them.

  • $\begingroup$ What you just said contradicts the concept of hyperconjugation, because the occupied orbital is not associated with the atom that has the antibonding orbital and is being stabilized. You're basically saying you are gonna destabilize the antibonding orbital even more $\endgroup$
    – Michael
    Oct 8, 2015 at 15:02
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    $\begingroup$ @Michael No it doesn’t. It is the concept of hyperconjugation. $\endgroup$
    – Jan
    Oct 8, 2015 at 15:27
  • $\begingroup$ if that were true, then the more alkyl groups that are substituted in an alkene, the less stable the alkene would be because what you are saying that the double bond is more unstable. The opposite is true though the alkyl groups make the enthalpy of the reaction of breaking a double bond in an alkene more positive, not negative $\endgroup$
    – Michael
    Oct 9, 2015 at 0:32
  • $\begingroup$ @Michael No, you just got the argument the wrong way around. The (stable and occupied) $\pi$-orbital of the alkene mixes with the (instable and unoccupied) $\sigma^*$-orbital of the $\ce{C-H}$ bond. This interaction stabilises the $\pi$ bond and destabilises the $\sigma^*$ $\endgroup$
    – Jan
    Oct 9, 2015 at 7:23

The electron density increasing in the antibonding orbital only means that the original bond is reduced in strength. Consider propene, the hyperconjugative structure adds electron density to the $\pi^*$ orbital of carbon. Let us say that the double bond was between C1 and C2.Now the electron density from the alpha C-H bond at C3 moves into the antibonding orbital, thus weakening the C2 -C1 pi bond, and begins to create a double bond between C3 and C2. Thus one bond is weakened and another is strengthened. The stability is only on account of the delocalization, similar to resonance stability.


In organic chemistry, hyperconjugation is the interaction of the electrons in a sigma bond (usually C–H or C–C) with an adjacent empty (or partially filled) non-bonding p-orbital, antibonding σ or π orbital, or filled π orbital, to give an extended molecular orbital that increases the stability of the statement.

No of canonical structure= no of hyperconjugable $H$ atom +1

Higher is the no of canonical structures formed Higher is the stability.

It effects are by following ways:

Bond length: Hyperconjugation is suggested as a key factor in shortening of sigma bonds (σ bonds).

The heat of formation of molecules with hyperconjugation are greater than sum of their bond energies and the heats of hydrogenation per double bond are less than the heat of hydrogenation of ethylene

Stability of carbocations:

$\ce{(CH3)3C+ > (CH3)2CH+ > (CH3)CH2+ > CH3+}$

The C–C σ bond adjacent to the cation is free to rotate, and, as it does so, the three C–H σ bonds of the methyl group in turn undergoes the stabilization interaction. The more adjacent C-H bonds there are the larger hyperconjugation stabilization is.

In chemical bonding theory, an antibonding orbital is a type of molecular orbital that, if occupied by electrons, weakens the bond between two atoms and helps to raise the energy of the molecule relative to the separated atoms. Such an orbital has one or more nodes in the bonding region between the nuclei.

Antibonding molecular orbitals (MOs) are normally higher in energy than bonding molecular orbitals. Bonding and antibonding orbitals form when atoms combine into molecules.

as the spacing between the two atoms becomes smaller, the electron wave functions begin to overlap. Each energy level of the isolated atoms slits into two molecular orbitals belonging to the pair, one lower in energy than the original atomic level and one higher. The orbital which is lower than the orbitals of the separate atoms is the bonding orbital, which is more stable and promotes the bending.

A molecular orbital becomes antibonding when there is less electron density between the two nuclei than there would be if there were no bonding interaction at all. When a molecular orbital changes sign (from positive to negative) at a nodal plane between two atoms, it is said to be antibonding with respect to those atoms.

For example, the He2 molecule, in which both the 1sσ and 1sσ* orbitals are filled. Since the antibonding orbital is more antibonding than the bonding orbital is bonding, the molecule has a higher energy than two separated helium atoms, and it is therefore unstable.

negative hyperconjugation is the donation of electron density from a filled π- or p-orbital to a neighboring σ*-orbital.This phenomenon, a type of resonance, can stabilize the molecule or transition state. It also causes an elongation of the σ-bond by adding electron density to its antibonding orbital.And hence helping in delocalization of orbitals.

Delocalizationelectrons are electrons in a molecule, ion or solid metal that are not associated with a single atom or a covalent bond . its also the Distribution of electron density beyond a fixed place such as a single atom, lone pair, or covalent bond via resonance or inductive effects.

For example , 1.The oxygen lone pairs of ethoxide ion are not delocalized.

2.The oxygen lone pairs of trifluoroethoxide ion are delocalized via the trifluoromethyl group's electron-withdrawing inductive effect.

Negative hyperconjugationis most commonly observed when the σ*-orbital is located on certain C–F or C–O bonds,and does not occur to an appreciable extent with normal C–H bonds.

In negative hyperconjugation, the electron density flows in the opposite direction (from π- or p-orbital to empty σ*-orbital) than it does in the more common hyperconjugation (from σ-orbital to empty p-orbital).

decolalization of electron is though conjujation of $\ce{CH σ-bond}$ with π - orbital or atomic orbital( π - orbital which does not participate in resonance) in hyperconjugation

Increase in delocalization, increases the stability, by decreasing the total (potential) energy.

I hope this will help.

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    $\begingroup$ Why does adding electron density to the antibonding orbitals increase stability though? That is what the question is about. Most of this answer is only vaguely relevant and it could also do with using better grammar. $\endgroup$
    – bon
    Oct 8, 2015 at 10:49

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