I am trying to design a proper synthesis route for 2‐amino‐3‐(chlorosulfonyl)benzoic acid, though methyl 2‐amino‐3‐(chlorosulfonyl)benzoate—using an ester group in lieu of a carboxylic acid—would also be acceptable.

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I am thinking a good starting material would be toluene or benzene. Here is my proposed route.

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I determined the ortho-para ratio for toluene nitration from p. 816 of "Organic Chemistry, 6e" by Loudon & Parise.

Typically, para substitution predominates over ortho substitution, but not always. For example, nitration of toluene gives twice as much o-nitrotoluene as p-nitrotoluene. This result occurs because the nitration of toluene at either the ortho or para position is so fast that it occurs on every encounter of the reagents; that is, the energy barrier for the reaction is insignificant. Hence, the product distribution corresponds simply to the relative probability of the reactions. Because the ratio of ortho and para positions is 2 : 1, the product distribution is 2 : 1.

Also, according to the same page,

The boiling points of o- and p-nitrotoluene, 220 °C and 238 °C, respectively, are sufficiently different that these isomers can be separated by careful fractional distillation.

Is there a better method? Please identify any mistakes I may have made in my synthesis and suggest improvements. Note that I install t-butyl as a blocking group in order to direct the sulfonyl chloride group ortho to the amino group.

  • $\begingroup$ I would be concerned about how you are going to isolate the final product from the Aluminium residues $\endgroup$
    – Waylander
    Commented Jan 6, 2021 at 9:16
  • 2
    $\begingroup$ There's no way to isolate such compound. What magic do you have to prevent reaction between sulfonyl chloride and amine groups? $\endgroup$
    – Mithoron
    Commented Jan 6, 2021 at 13:01
  • 2
    $\begingroup$ I suspect that with 2 strong electron-withdrawing groups on either side the amine will not be very nucleophilic. That said the material looks horrendously difficult to handle. $\endgroup$
    – Waylander
    Commented Jan 6, 2021 at 14:09
  • 2
    $\begingroup$ This is a cmpd that is likely to self-condense with an amino and a sulfonyl chloride group. How do you avoid protonation of the amino group during sulfonation? $\endgroup$
    – user55119
    Commented Jan 6, 2021 at 16:24
  • 1
    $\begingroup$ The sulfonyl chloride group is very reactive and water sensitive so solvent/aqueous extractions are out. $\endgroup$
    – Waylander
    Commented Jan 8, 2021 at 16:39

1 Answer 1


Here is a proposed route starting from aniline. First protect the -NH2 with t-BOC

Directed lithiation of t-BOC-aniline (sBuLi/Et2O, -75C) and quench with MeSSMe to install -SMe in the 2 position.

Directed lithiation under the same conditions quenching with Ethyl Chloroformate to install -CO2Et in the 6 position.

Oxidise the -SMe group to sulfoxide (MCPBA, DCM)

Pummerer rearrangement of the sulfoxide (Ac2O, aq. base work up) gives the free thiol.

Oxidation of the thiol to the sulfonic acid, one example of possible conditions hydrogen peroxide with Rh catalysis

Form the sulfonyl chloride (conditions using trichlorotriazene here. Depending on your choice of conditions you can remove the -BOC group or preserve it. I think preserving it will give a flexible late stage intermediate that is reasonably easily handled. It is possible to directly form the sulfonyl chloride from the -SMe with chlorine in aq acetic acid conditions here but the conditions are harsh and risk removing the -BOC group

  • $\begingroup$ Would it be acceptable to convert the thiol-containing intermediate directly to the sulfonyl chloride using H2O2 and SOCl2 in acetonitrile at room temperature? $\endgroup$ Commented Jan 12, 2021 at 17:49
  • $\begingroup$ I think that would remove the tBOC group from the aniline as HCl is generated. Is this a paper exercise or are you actually doing it? $\endgroup$
    – Waylander
    Commented Jan 12, 2021 at 17:57
  • $\begingroup$ I am working it out on paper before I actually implement/"do" it. $\endgroup$ Commented Jan 12, 2021 at 18:18
  • $\begingroup$ What is your next step with this product? $\endgroup$
    – Waylander
    Commented Jan 12, 2021 at 19:14
  • $\begingroup$ Wikipedia (en.wikipedia.org/wiki/Pummerer_rearrangement) illustrates an extra carbon, with two H-atoms attached, in-between the two 'R' groups. If I treat the 'Me' (attached to the S-atom) as one group, and the benzene ring with its other substituents as the other 'R' group, that does not leave me any space for a CH2 in-between. I have seen a couple of examples on the internet with a CH in-between, but none with no H's in-between. I am curious if you have a source, given that the Pummerer rearrangement seems to usually involve an alpha-carbon. $\endgroup$ Commented Jan 12, 2021 at 22:53

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