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You don’t see carbocations on double bonds very much, and here's a good reason: compared to sp3, there is more s character in the orbitals, so the empty orbital is held more closely to the nucleus. This is destabilizing if we’re talking about positive charge.

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This doesn't really make sense. Positive charge arises from the lack of electrons. How does holding an empty orbital (regardless of whether that makes any physical sense) closer affect your stability if the empty orbital is ... empty?

So, what is the exact reason sp2 (aren't most carbocations sp2 though?) and sp carbocations are generally less stable than their sp3 counterparts?

Also, how can we have an sp3 carbocation? Huh?!

enter image description here

The only reason I can think of is that it's energetically expensive to make an sp carbocation ... pulling electrons which are held more tightly to the nucleus away is probably harder than pulling away electrons which are not as close to the nucleus.

I can't think of any reason why the end products would be any less/more stable ... if anything an sp carbocation should be more stable than an sp2 carbocation in that in the sp carbocation the electrons are held closer to the nucleus and therefore better stabilizing the nucleus.

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Why are sp hybridized carbocations high energy?

Here are links to 2 earlier answers where this question has been addressed for vinylic and aryl carbocations.

link 1

link 2

When vinylic carbocations are generated, the empty orbital (the orbital holding no electrons) is intitially $\ce{sp^2}$ hybridized. This carbonium ion orbital being $\ce{sp^2}$ hybridized is using some s character, the remaining orbitals around the carbonium ion carbon will therefore have less s character available for use in their bonds. The more s character in a bond, the lower the energy of the electrons in that bond because s orbitals are lower in energy than p orbitals.

If the vinylic carbocation rehybridizes such that the empty orbital becomes a p orbital, then the remaining orbitals that contain electrons will become lower in energy as they rehybridize and incorporate the s character that was initially used in the empty $\ce{sp^2}$ carbocation orbital. Once rehybridized the carbocations overall energy has been lowered and the carbocation is more stable.

The high energy of the initially generated $\ce{sp^2}$ hybridized vinylic carbocation is why it is so difficult to generate vinylic carbocations.

Note that some vinylic carbocations cannot rehybridize to an $\ce{sp^2}$ carbocation carbon with a p orbital due to geometrical constraints, and they are even harder to generate. Examples would be the phenyl carbocation and vinylic carbocations in small rings, like the cyclohexenyl carbocation (see figure below).

Also, how can we have an sp3 carbocation?

They exist, but again, realize that since the empty orbital is not p hybridized they will be higher in energy and harder to generate then carbocations in which the empty orbital is p hybridized. Examples of carbocations in which the empty orbital is roughly $\ce{sp^3}$ hybdridized would include bridgehead carbocations.

enter image description here

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  • $\begingroup$ The bridgehead in the bicyclo[2.2.1]heptan-1-ylium cation is only about 11 degrees out of plane. The empty orbital will have negligible s character. $\endgroup$ – Martin - マーチン Feb 4 '15 at 5:20
  • $\begingroup$ @Martin Really, only 11° out of plane? Why then does the corresponding tosylate solvolyze 10^17 times more slowly than bicycle[3.3.3]undecyl-1-tosylate? In any case, just make the ring system smaller to illustrate my point - right? $\endgroup$ – ron Feb 4 '15 at 20:01
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    $\begingroup$ I was also very surprised. Since you raised the point I was curious how much it would be like sp3, I expected an out of plane between 30-40 degrees. I honestly did not think about this bridgehead carbons. Decreasing the ring size should work illustrating the problem, like bicyclo[1.1.1]pentane, tetrahedrane or cubane cations. If I have the time I run some calculations and add them to my answer. $\endgroup$ – Martin - マーチン Feb 5 '15 at 2:38
  • $\begingroup$ I checked the out of plane angle (opa) for methane as an ideal sp3 hybridised molecule and it is only 19.5 deg. So 11.7 deg for bicyclo[2.2.1]... is only little smaller and absolutely reasonable. However, NBO analysis predicts only 1% s character in the empty orbital. In bicyclo[1.1.1]... the opa is 25.5 deg, but NBO says only 7% s character in the empty orbital. In cubane the opa is 27.9 deg, but again NBO only wants 1% s character. Tetrahedrane is impossible to converge and fragments immediately. All this leaves me more confused than before. $\endgroup$ – Martin - マーチン Feb 9 '15 at 6:52
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    $\begingroup$ I guess it is safe to assume, that in any case, the s orbital will be used to the maximum possible extend, even if that means significantly distorting the geometry, or non-optimal overlap. But we are here on the verge of guessing anyway, since it is not really possible to observe empty orbitals. And then again, there is hyperconjugation and long range effects that play a much bigger role. $\endgroup$ – Martin - マーチン Feb 9 '15 at 6:55
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Let's just tackle the obvious part first: There is no sp3 carbocation. One reason for this is the same, as for why higher substituted carbocations are more stable than lower substituted. Electrons are more stable in orbitals with high s-character.
In $\ce{\overset{\oplus}{C}(CH3)3}$ the cationic carbon is sp2 hybridised while the empty orbital is 100% p character. All s character is used in occupied orbitals.
In $\ce{H2C=\overset{\oplus}{C}-CH3}$, one would assume from the common depiction, that the central cationic carbon is also (roughly) sp2 hybridised. My first draft of answering did also assume this. A quick DF-BP86/def2-SVP calculation revealed something different. The $\angle(\ce{CCC})$ bond angle comes out to be almost exactly $180^\circ$. The $\pi$ bond is obviously only of p character. The underlying $\sigma$ bonds can now contain the whole s character, making it sp hybridised, and as a result the p orbital of the cationic carbon atom is empty. There will be interactions with some carbon hydrogen bonds. (Hyperconjugation, see below.) This effect is most likely much less than in the prior case, causing it to be less stable. In addition to this, the molecule is much more constrained in its geometry. This linear arrangement is probably also only true for this very simple arrangement. Dependent on the moieties attached to it, a bent structure might be more stable and a different bonding angle will be observed. When that happens, there will be some s character in the empty orbital.
In the benzyl cation, due to symmetry restrictions, $D_\mathrm{2h}$, the empty orbital must be almost exactly sp2. Hence the underlying $\sigma$ bonds will roughly take away 67% s character. The empty orbital will therefore have a very high s character, about 33%, as a result. There is almost no bond widening possible to decrease this and successively increase s character in the $\sigma$ bonds. So the s character found in empty orbitals effectively destabilised the remaining occupied orbitals. (One could also think about it as some kind of ring strain.)
In $\ce{H3C-C#\overset{\oplus}{C}}$, symmetry is even higher, $C_\mathrm{\infty{}v}$ for the triple bond. The underlying $\sigma$ bond will have slightly more s character than in an ideal sp orbital. The empty orbital will, however, be binding a lot of s character. This is effectively destabilising the occupied orbitals again. Since the empty orbital contains the most s character in the orbital, it will also have the most destabilised $\sigma$ system of the series.

In summary, the more s character is in unoccupied orbitals, the less s character can be in occupied orbitals. This is destabilising these orbitals, causing the carbocation also to comparably less stable.
The quote expresses this feature in a quite confusing way.


Note
There are of course other reasons for higher substituted carbocations to be more stable than lower. One of the main effects is hyperconjugation of the $\sigma(\ce{C-H})$ bond with the empty p orbital.

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  • $\begingroup$ so we can apply Bent's rule even to empty orbitals? @Martin $\endgroup$ – Dissenter Feb 3 '15 at 5:45
  • $\begingroup$ @Dissenter I believe that Bent's rule serves a different purpose. There might be similarities, but as it is only an observed concept one should not stretch the boundaries of it too far. $\endgroup$ – Martin - マーチン Feb 4 '15 at 5:49

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