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Phenyl carbonation is said to be unstable because the + charge on it cannot be stabilized by Resonance or any other electronic effects, so my question is that if the resonance occurs in the direction shown in the below picture the + charge can be delocalized to some extent, so will it adapt that ?

Phenyl Carbocation

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    $\begingroup$ see the description of the phenyl carbocation in this earlier answer $\endgroup$ – ron Dec 27 '16 at 20:21
  • $\begingroup$ The carbocation is orthogonal to the pi system $\endgroup$ – RobChem Dec 28 '16 at 1:12
  • $\begingroup$ Those are valid resonance structures that you've drawn. They correspond to the charge-separated resonance structures one can draw for benzene. Since they involve charge separation, they contribute little to the description of the aromatic pi system. Further, the pi system is orthogonal to the empty $\ce{sp^2}$ orbital, consequently such resonance structures provide little, if any, stabilization of the cationic center. $\endgroup$ – ron Dec 28 '16 at 2:16
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Your resonance structure doesn't do anything to stabilize the cation. You've created a singlet carbene at the top most carbon, so it has both the cation and the anion on the same carbon. These do not cancel out. As drawn, you have two hydrogens on the top carbon. You need to replace it with a single lone pair and an empty orbital.

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  • $\begingroup$ Except for the fact that this is an inverse singlet carbene with the hybrid orbital being empty and the p orbital being populated … very interesting to think about the compound! =D $\endgroup$ – Jan Dec 27 '16 at 21:27
  • $\begingroup$ By the way, why would anybody downvote this (correct) answer? $\endgroup$ – Jan Dec 27 '16 at 21:27
  • $\begingroup$ @Jan Yeah, filled p orbital doesn't exist since it's really the resonance structure, but you're right that this is unusual. $\endgroup$ – Zhe Dec 27 '16 at 22:39
  • $\begingroup$ I don't see it as a "singlet carbene at the top most carbon". It is a series of charge-separated resonance structures of benzene which don't really contribute much to the overall description of the aromatic system. Further, as I pointed out in my earlier answer, the aromatic pi system and the unoccupied $\ce{sp^2}$ orbital are orthogonal. Consequently there is not much, if any, interaction between them. $\endgroup$ – ron Dec 28 '16 at 2:10
  • $\begingroup$ @Zhe thanks for clearing my doubt. You are right , even if the bond breaks in the direction shown carbon is left with an unfilled octet. $\endgroup$ – jyoti proy Dec 28 '16 at 8:37

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