How can I make o-bromoaniline from aniline (as a major product)?

I know how to make p-bromoaniline from aniline, by acetylation followed by treatment with $\ce{Br2}$/$\ce{CH3COOH}$ and hydrolysis. But in this process the ortho isomer (if formed) will be the minor product.

  • $\begingroup$ Did you consider different sequences? Remember that aromatic nitro compounds can be reduced to the corresponding amines ;-) $\endgroup$ Mar 8, 2016 at 8:08
  • $\begingroup$ @Klaus Warzecha But even if I consider nitro benzene, the ortho isomer will be the minor product. Would the yield be satisfactory ? $\endgroup$ Mar 8, 2016 at 8:18
  • 3
    $\begingroup$ Remember how the nitro group directs. If you start with nitrobenzene, chances are good that the bromination will almost exclusively yield the meta-substituted product. But what if you try to nitrate bromobenzene? $\endgroup$ Mar 8, 2016 at 8:29

3 Answers 3


Didn't you consider the way through sulfanilic acid, the para-position will be protected. In this case both substituents will direct the electrophilic substitution in ortho- position with respect to amino group. Then remove sulfonic acid group by desulfonation and you will obtain your ortho-bromoaniline.

  • $\begingroup$ Wont it exist in zwitter ion form and loose its directing influence? $\endgroup$ Feb 17, 2020 at 8:46

As outlined in a comment direct bromination of nitrobenzene with $\ce{Br2}$ and $\ce{FeBr3}$ is likely to give the meta-substituted product.

If you want to avoid time-consuming and expensive procedures, your options are:

  • Change the order of events. Nitrate bromobenzene under mild(er) conditions, or
  • Find conditions, under which bromination of nitrobenzene furnishes the ortho-product

For the latter, have a look at Chem. Commun., 1996, 2679-2680 (DOI). The authors state:

We have observed that monosubstituted benzene derivatives react with a mixture of dimethyl sulfoxide and aqueous hydrobromic acid […] to yield regiospecifically only the monobromo derivatives. It is interesting to note that electron- withdrawing groups such as $\ce{NO2}$, $\ce{CHO}$ yield ortho-bromo derivatives, while electron-donating groups such as $\ce{NH2}$, $\ce{OH}$ or $\ce{Me}$ give only para-bromo derivatives.


I think that you probably cannot avoid separating ortho from para isomers (and possibly disubstitution products) after using a method like the one you suggest or a method of blocking the para- position.

Another method that is probably not taught to undergraduates (?) is the Directed ortho metallation (DoM) or Snieckus chemistry. The image below has been copied from that link. enter image description here


Your Answer

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

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