After searching about the preparation of sulfonyl chlorides, I couldn't find much. I found that for preparation of tosyl chloride, chlorosulfonic acid can be used, as in:

Toluene chlorosulfonation

Can chlorosulfonic acid be used to prepare sulfonyl chlorides in general? What are some other ways to prepare sulfonyl chlorides? If there is any.

The other part of the question, I'd like to know how the mechanism of this reaction would work. Having the equilibrium of the reaction:

$$\ce{3 ClSO2(OH) <=> SO2Cl+ + 2 SO3Cl- + H3O+}$$

How exactly is the $\ce{SO2Cl}$ formed? I didn't quite understand how the $\ce{Cl}$ became a cation. Apparently the benzene attacks that cation in order to form the desired product.

  • 1
    $\begingroup$ As for other ways of making sulfonyl chlorides, that is too broad to cover in a single post (I suggest removing it from this question, in general we don't encourage having two questions in one), but organic-chemistry.org/synthesis/C1S/sulfonylchlorides.shtm has a good summary. $\endgroup$ May 29, 2019 at 23:26
  • $\begingroup$ Related: What is the mechanism of chlorosulfonation of benzene? $\endgroup$
    – andselisk
    May 30, 2019 at 4:50
  • $\begingroup$ Not so far mentioned are other ways to prepare sulfonyl chlorides. The most obvious of these is oxidation of thiols, hydrogen peroxide or Oxone are among the reagents that will do this, then chlorination with e.g SOCl2 $\endgroup$
    – Waylander
    May 30, 2019 at 6:52

1 Answer 1


The way to understand this is to think of the autoprotolysis of water:

$$\ce{H2O + H2O <=> H3O+ + OH-} \tag{1}$$

In general autoprotolysis can be represented in the following form (you can recover eq. (1) by setting $\ce{A} = \ce{OH}$):

$$\ce{HA + HA <=> H2A+ + A-} \tag{2}$$

Now set $\ce{HA}$ as $\ce{ClSO3H}$ (or $\ce{(HO)SO2Cl}$, for clarity) and you have

$$\ce{(HO)SO2Cl + (HO)SO2Cl <=> (H2O+)SO2Cl + ^-OSO2Cl} \tag{3}$$

The cation produced in eq. (3) fragments to give water and the reactive chlorosulfonyl cation:

$$\ce{(H2O+)SO2Cl -> ^+SO2Cl + H2O} \tag{4}$$

The water in eq. (4) reacts with a third molecule of acid in an ordinary acid-base reaction:

$$\ce{H2O + (HO)SO2Cl -> H3O+ + ^-OSO2Cl} \tag{5}$$

The combined equation you have is simply the sum of eqs. (3), (4), and (5).

  • $\begingroup$ What about the reaction with benzene? I think the $\ce{SO2Cl}$ reacts with the double ring. But how is the double ring formed again to get the desired product? Maybe the hydrogen of the carbon which got attached to the S is abstracted by the O of $\ce{OSO2Cl}$? $\endgroup$ May 30, 2019 at 0:08
  • 2
    $\begingroup$ @DennerEvaristo, your image seems to be taken from Clayden, which is a good textbook. Please consider reading the chapter on electrophilic aromatic substitution, it will answer your question. Also, it's a double bond, not a double ring. $\endgroup$ May 30, 2019 at 0:18

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.