I was taught that the following reaction happens:

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It is called the "ipso-substitution". My professor had told me that it occurs in the case of $\ce{-SO3H}$ and $\ce{-COOH}$. But he had ended the topic there. My textbook doesn't detail any more either.

My searches landed me up at the fact that ipso-substitution is also employed in the preparation of picric acid from nitrobenzene. But I couldn't find answers all my other questions, which are:

  1. Does the ipso-substitution occur only for these two groups, or are there more? I guess there is a "threshold" value for some parameter, above (or below) which ipso substitution cannot occur, but I am unable to come up with what that parameter is. I guess it's the high stability of the leaving group $\ce{SO3-}$, but I am not sure.
  2. Is ipso-substitution affected by the choice of the electrophile and the solvent? What if, instead of bromine water, I chose bromine in acetic acid? Or something even simpler as $\ce{FeCl3/Cl2}$. Would ipso substitution occur then?
  3. I know that no organic reaction produces 100% of only one compound, as its other isomers are always also formed. Ipso substitution should be no exception. But how do we decide whether the ipso product is a major product or a minor product? I know how to decide major/minor products between ortho/meta/para substitution (electron density/steric hindrance/H-bonding), but I haven't been taught any logic to decide the same for ipso substitution.
  • $\begingroup$ I can't say that there is a set of rules for ipso substitution. For the loss of a more complex group by an ipso protonation, see chemistry.stackexchange.com/questions/50866/… $\endgroup$
    – user55119
    Apr 13, 2018 at 20:45
  • 4
    $\begingroup$ All "ipso" substitution means is that it occurs where the leaving group was. That's all. Essentially all of the $\mathrm{S}_{N}\mathrm{Ar}$ reactions you learn are ipso substitutions. $\endgroup$
    – Zhe
    May 22, 2018 at 20:51
  • 3
    $\begingroup$ Trick to remember. All o/p positions wrt strongest activating group becomes br on addition on bromine water(irrespective of the group's there) $\endgroup$
    – user600016
    Jan 4, 2019 at 15:16

1 Answer 1


The shorthand way that we draw structures in organic chemistry, with implicit hydrogens, leads us to often forget that the hydrogens are there, or to neglect considering them in our analyses. In electrophilic aromatic substitution (EAS) reactions, the generally accepted mechanism involves attack of the aromatic ring on the electrophile to give a Wheland intermediate, which then loses a proton from the $\mathrm{sp^3}$-hybridized ring carbon to restore aromaticity. Ipso substitution, then, is really nothing special - we simply substitute a position on the aromatic ring that bears a non-hydrogen (e.g., $\ce{X}$) group, presumably via an analogous mechanism. For this to be viable, both steps of the mechanism need to be favorable:

  1. $\ce{X}$ needs to enhance, or at least not greatly reduce, the nucleophilicity of the carbon it is attached to. For example, inductive effects would great disfavor ipso attack at groups like $\ce{F}$ or $\ce{CF3}$.
  2. $\ce{X+}$ should be a reasonable species to lose in the second step, to regain aromaticity, or there should be some mechanism by which $\ce{X}$ can be lost, with electrons flowing back towards the aromatic ring.

Sulfonic acids and carboxylic acids are rather electron-withdrawing, but if you drill down into the details of these ipso-substitution reactions, many are actually conducted on the conjugate base (i.e. sulfonate or carboxylate), which also provides a pathway for loss of a neutral group in the second step, i.e. $\ce{SO3}$ or $\ce{CO2}$. Alternatively, harsh conditions are used, e.g., sulfonation/desulfonation.

It turns out that ipso attack is more common than it first appears, because the product of the initial ipso attack can re-arrange. So, for example, nitration of p-cresol involves 40% attack at the ipso carbon bearing the methyl group, followed by acid-catalyzed rearrangement to the expected product.[1] There is more discussion here.

Closer to the original thrust of your question - there are some groups that are even better at being substituted in an EAS reaction than a proton. Trialkylsilyl and trialkylstannyl groups are often used. Both $\ce{Si}$ and $\ce{Sn}$ are less electronegative than carbon and highly polarizable, making the ipso carbon a good nucleophile. These groups also help to stabilize the Wheland intermediate through hyperconjugation. Finally, they both have energetically-favorable pathways to leave with electron density flowing back towards the aromatic ring. For these reasons, ipso substitution at $\ce{SiR3}$ or $\ce{SnR3}$ substituted carbons can often be achieved rapidly, under mild conditions, with high regioselectivity. This is often helpful in specialist applications, like radiolabelling.

I haven't specifically addressed each of the original three numbered questions, but they can be tackled using the principles above. Ipso substitution is possible for any group - it's a competition, $\ce{H}$ vs $\ce{X}$. Is ipso substitution affected by reaction conditions? I'm sure it could be - factors like protonation states, hard vs soft electrophiles, reversibility, etc. would have an influence.


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