I have checked the internet and read quite a few books, but I still am not able to understand why vinylic and arylic carbocations are highly unstable.

What I found while surfing the internet is: For arylic carbocation, the carbon bearing postive charge cannot participate in resonance with adjacent carbons because it's p orbital is perpendicular to the other orbitals. But, I didn't really understand what this means.


3 Answers 3



S-orbitals are lower in energy than p-orbitals, therefore the more s-character in an orbital, the lower its energy and the more stable any electrons occupying the orbital. In other words, electrons will be more stable (lower in energy) in an $\ce{sp}$ orbital than an $\ce{sp^2}$ orbital, and electrons in an $\ce{sp^2}$ orbital will be lower in energy than those in an $\ce{sp^3}$ orbital.

This is the basis of the inductive effect. For example, it explains why the electrons in an $\ce{sp^3}$ methyl group will be drawn towards a connected $\ce{sp^2}$ carbon in an aromatic ring - the methyl group is inductively electron donating to the aromatic ring as its $\ce{sp^3}$ electrons drift towards the lower energy $\ce{sp^2}$ aromatic carbon.

What we're really saying is that an $\ce{sp}$ hybridized carbon atom is more electronegative (it's lower in energy and the electrons prefer to move towards it) than an $\ce{sp^2}$ hybridized carbon atom, which in turn is more electronegative than an $\ce{sp^3}$ hybridized carbon atom.


Look at the figure below, notice that in the alkyl carbocation on the left the cationic center is attached to an $\ce{sp^3}$ carbon, whereas in the vinylic cation in the middle, the cationic center is attached to a more electronegative $\ce{sp^2}$ carbon. For a positive center to be attached to a more electronegative group is destabilizing. Hence the vinylic cation is less stable than a typical alkyl cation.

Things are even worse with the aryl carbocation on the right. Here the positive carbon is attached to 2 $\ce{sp^2}$ carbons. Destabilizing the aryl cation even further is its geometry. A vinyl cation prefers to be linear, but due to geometrical constraints imposed by the aromatic ring the aryl cation must be bent and the empty orbital is forced to be $\ce{sp^2}$ rather than $\ce{p}$. These three factors combine to make the aryl carbocation even higher in energy than the vinyl cation.

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  • $\begingroup$ I was taught that electrons from an sp3 carbon move to an sp2 carbon due to hyperconjugation, that is: the shift of a bonded electron into the unhybridised p orbital of the sp2 carbon. Is the shift of electrons due to a combination of both this and inductive effect? $\endgroup$ Commented Feb 20, 2016 at 1:55
  • 1
    $\begingroup$ Yes, that's how I see it, a mix of resonance and inductive effects. $\endgroup$
    – ron
    Commented Feb 20, 2016 at 3:58
  • $\begingroup$ @ron for hyperconjugation to occur, must the orbitals hyperconjugating be in the same plane? $\endgroup$
    – Muhammad
    Commented Feb 9, 2019 at 13:44
  • $\begingroup$ Yes, for hyperconjugation orbitals have to be in the same plane. $\endgroup$ Commented Aug 1, 2020 at 13:58

The instability of a vinyl cation cannot be blamed on its sp2 orbital being very electronegative as the positive charge resides in a p orbital, not an sp2 orbital. The instability derives from the inability of that p orbital to overlap with the the sp2 orbitals of the carbon on the other end of the double bond. The bond angles of that carbon are too large (120*) and their highly electronegative nature prevent stabilization of the cationic center. The only stabilization available to the vinyl cation comes from the saturated sp3 carbon on the other side and its stability is therefore similar to a primary cation.


We need to know one fact i.e. more the s-character in the hybridisation of a compound more is its electronegativity. This due to the fact that s-orbital is nearer to the nucleus and is thus attracted more towards the nucleus leading to an increased electronegativity.

As the carbon of vinyl carbocation is sp- hybridised (50% s-character, 50% p-character) it is highly electronegative. And positive charge on electronegative atom is highly unstable


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