According to Clayden's Organic Chemistry (2e) (p.485)

... and the acid chloride (tosyl chloride, $\ce{TsCl}$) needed to make tosylates can be made from the acid in the usual way (p. 215) with $\ce{PCl5}$ . It can also be made directly from toluene by sulfonation with chloro-sulfonic acid $\ce{ClSO2OH}$. This reaction favours the ortho sulfonyl chloride, which is isolated by distillation. Chlorosulfonation of toluene

My question is, why is the ortho product favoured? If anything there should be steric hindrance due to the methyl group and the $\ce{sp}^{3}$ hybridized sulphur atom, destabilizing the ortho-product.

Edit 1: I've found the same question here which does provide an answer based on kinetic and thermodynamic control, however, I share the same doubt as the author at the end of the answer.

Edit 2: Some more data from the source quoted in the linked answer (see Edit 1) enter image description here So, I guess, it's just not detailed enough in Clayden? Please share your thoughts.

  • $\begingroup$ According to my teacher, steric hindrance is the reason for the reaction favouring para-toluenesulfonic acid during sulfonation of toluene. $\endgroup$
    – zxen
    Jan 26 at 17:13
  • $\begingroup$ What more than you found would you know? Steric isn't the issue. If it was t-butyl, or methyls on both sides, the situation would be different. $\endgroup$
    – Mithoron
    Jan 26 at 20:32

1 Answer 1


The fact that raising the temperature favors the para product is a clue. Higher temperatures, which allow a wider variety of choices kinetically and more reversibility of these choices, tend to favor thetmodynamically more stable products; lower temperatures favor products that form kinetically through a lower-energy intermediate.

Here, steric hindrance in the product certainly makes the para isomer more favorable thetmodynamically, so if we raise the temperature sufficiently we are sure to get that isomer as expected. But at lower temperatures where kinetics and activation energy are more important, the ortho intermediate has the advantage of being stabilized by hydrogen bonding.

Hydrogen bonding to hydrogen atoms attached to carbon? Well, the methyl group on the intermediate formed when the electrophile attacks toluene does not quite have your ordinary carbon-hydrogen bonds. With the intermediate becoming positively charged and electron-deficient, the methyl group will hyperconjugate with the ring: its bonding orbitals will overlap with the formally electron-deficient ring, producing an extension of the pi conjugation in exchange for weakening the carbon-hydrogen sigma bonding in the methyl group. We may represent this interaction as a contributing structure in which a carbon-hydrogen bond is exchanged for this extended pi interaction:

enter image description here

Imparting part of the electron deficiency and positive charge then primes the methyl hydrogen atoms for hydrogen bonding in several ways related to both the electrostatic and the molecular-orbital renderings of hydrogen bonds:

  • The methyl hydrogens gain positive charge and thus attract electrons from a potential acceptor atom in the electrophile.

  • When the bonding is shifted away from the methyl group towards the ring, the unoccupied antibonding orbitals shift in the reverse direction. Therefore there is more probability density for the electrons from the acceptor atom to overlap.

  • Withdrawing electron density from the bonding orbitals in the methyl group means there is less repulsion to oppose overlap with the antibonding orbitals which occurs in hydrogen bonding.

So when we specify the electrophilic function $E$ as the chlorosulfonyl group, the oxygen atoms can act as hydrogen-bond acceptors and they will, creating a third contributing structure in which the hydrogen bonding in effect replaces the lost sigma bonding from the hyperconjugated methyl group:

enter image description here

This extra delocalization can occur, however, only with the ortho intermediate as otherwise the methyl group and oxygen atoms are too far apart. The hydrogen bonding interaction will eventually disappear when the intermediate is deprotonated and becomes neutral again, but the intermediate has already done the deed of directing the reaction towards the ortho product. Unless, of course, the temperature is high enough to allow this choice to be reversed.


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