During an electrophilic aromatic substitution, the aromatic pi cloud interacts with an electrophile to form an arenium ion (AKA sigma-complex). The following image (from my organic chemistry class) suggests that aromaticity is always restored and that a nucleophile can't interact with the resonanced carbocation of the arenium ion.

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If I were to replace the base in the above image with a nucleophile that is an extremely poor base (such as iodine), could the nonaromatic cyclodiene be formed? Or would the reaction just halt instead?

Follow up question: If electrophilic aromatic subsitution doesn't work, how would one go about making the cyclodiene (with Nu and E substituents) from an arene? Is it possible?


1 Answer 1


The intermediate is non-aromatic arenium cation which is stabilized by resonance. Deprotonation of sigma complex is a fast process as it leads to much thermodynamically stable i.e., aromatic ring. Aromatization or deprotonation step is fast as Kinetic Isotopic effect is not observed. Hence even a weak base will extract the proton from the substituted site which is sp3 hybridized.

As for another nucleophile to attack the $\sigma$ complex, the positive charge is delocalized and deprotonation step is much faster. I think that it would be easier for the base to extract proton than to attack the Delocalized positive charge. The proton already has a $\delta$+ charge due to Hyperconjugation effect. I think that even if the base if a strong acid, it can dissociate to give H+ but the non-aromatic intermediate will not remain the solution as it is.

  • $\begingroup$ In the case of iodine, formation of HI will take place. $\endgroup$ Aug 3, 2016 at 3:15
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    $\begingroup$ this answer would be better if it was explained in greater detail $\endgroup$
    – Technetium
    Aug 3, 2016 at 4:16
  • $\begingroup$ Welcome to Chemistry.SE! Take the tour to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. For more information in general have a look at the help center. $\endgroup$ Aug 3, 2016 at 4:47

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