What are the major products?

I assumed that their is only one that is possible, which would be replacing the chloro group because the resonance structure supports that one. The bromo group doesn't seem to be able to be replaced. Would adding the ethoxide group, allow for the bromo group to replaced?


The reaction over here is a $\text{S}_{\text{N}}(\text{Ar})$ reaction. Normally, nucleophilic substitutions do not occur, but in presence of strong nucleophiles and electron withdrawing groups, it is possible.

Now, coming to your reaction, let's see what really happens:

Consider the resonation structures of 4-chloro-3-bromonitrobenzene: enter image description here You can see that the resonance structures produce a carbocation at the chlorine attached carbon, but not at the bromine attached one. Nucleophiles always tend to attack electrophilic centers, so it is the chlorine that will get replaced.

Continuing with the reaction, we will get: enter image description here

So, the product would be 3-bromo-4-methoxynitrobenzene.

  • 3
    $\begingroup$ Formal charges on the nitro group are missing. $\endgroup$ – orthocresol Apr 19 '17 at 16:29
  • $\begingroup$ @orthocresol I am aware of that, I just made it simpler. $\endgroup$ – Pritt says Reinstate Monica Apr 19 '17 at 16:32
  • 2
    $\begingroup$ @PrittBalagopal If it's incorrect, it's not simpler. $\endgroup$ – jerepierre Apr 25 '17 at 16:50
  • 1
    $\begingroup$ @jerepierre I have added formal charges :) $\endgroup$ – Pritt says Reinstate Monica Apr 28 '17 at 14:43
  • $\begingroup$ @orthocresol I have added formal charges :) $\endgroup$ – Pritt says Reinstate Monica Apr 28 '17 at 14:43

This reaction is difficult to occur because it is a nucleophilic substitution to a benzene ring but in the presence of a strong electron withdrawing group this reaction moves ahead. The point here to be noted is that -OEt is a strong nucleophile and attacks that carbon position where there is higher positive charge and thereby replacing the halogen out there.the third resonance structure is formed more due to stability factor

And thereby we obtain the product.... Hope this was helpful


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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