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.

enter image description here

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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.