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For the Tosylate ion, the negatively charged oxygen is stabilized via resonance with the oxygen atoms. But what are the orbitals of sulfur that are involved in the overlap? Is the sulfur $\ce{sp^2}$ hybridized and thus have a p-orbital for resonance?

If it is $\ce{sp^2}$, would that mean that it is possible for the electrons in oxygen to delocalise into the pi-electron cloud of benzene via the continuous overlap of p-orbitals in sulfur, carbon and oxygen atoms?

  • 5
    $\begingroup$ The sulphur has 4 $\sigma$ bonds and no lone pairs, so it is most likely $sp^3$ hybridized. Not written as an answer as I'm not completely sure. $\endgroup$
    – TRC
    Commented Jun 7, 2021 at 9:16
  • 1
    $\begingroup$ Yes, @TRC is right, the sulfur centre is tetrahedral, so the oxygen is not in the same plane as the ring. $\endgroup$
    – S R Maiti
    Commented Jun 12, 2021 at 14:23

2 Answers 2


Thanks to @Shoubhik R Maiti for confirming that my reasoning was correct

I hope you know the meanings of the terms $\sigma$ bond, $\pi$ bond and how to find hybridization of an atom. If not, you can get numerous good resources on Google.

To find the number of hybrid orbitals, we take the sum number of $\sigma$ bonds + number of lone pairs. Here, sulphur has six valence electrons, all bonded. So there are totally 4 $\sigma$ bonds and zero lone pairs. Resulting in hybridization $sp^3$, giving it a near-tetrahedral geometry (near because the aryl groups and oxygen atoms different, slightly distorting it from perfect tetrahedral). Since it is tetrahedral, the oxygen and phenyl group are not in the same plane, so the continuous overlap which you suggested is not possible.

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What you are asking is if the electrons of the oxygen atom can delocalize into the benzene ring. Firstly I need to mention that the negative charge is equally distributed on all three oxygen atoms, due to equivalent resonance. Now, from the canonical structure you have given in the question, there can be two cases of oxygen's electrons delocalizing into the benzene ring:

  1. The electrons of negatively charged oxygen bond with the sulphur atom. In that case the $\sigma$ bond between sulphur and benzene must completely break to accommodate the fresh electron pair from oxygen. You will be left with $\ce{SO3}$ molecule and a toluene with negative charge. This obviously can't happen.

  2. The electrons from one of the double bonded oxygen atoms delocalizes into the ring. Then you shall have an oxygen atom with a positive charge on it, singly bonded to the sulphur, and the benzene ring double bonded to the sulphur with a negative charge on one of the carbons. This is energetically unfavourable because it involves unlike charges' separation (which takes energy), and has + charge on electronegative oxygen and vice-versa, making it an extremely unstable resonance structure. Hence even if it is a contributing canonical structure, its contribution to the final resonance hybrid is very, very small, almost negligible.

Hence in no case is it possible to have electrons from oxygen atoms delocalizing into the benzene ring.

  • 3
    $\begingroup$ +1 I would also add that even though direct delocalisation cannot take place, hyperconjugation/ negative hyperconjugation can take place to some extent, but since oxygen is electronegative, the electrons from benzene ring are withdrawn. This is why SO3H is a meta-directing group. $\endgroup$
    – S R Maiti
    Commented Jun 13, 2021 at 7:41
  • 1
    $\begingroup$ The $\ce{-SO3-}$ group ist tetrahedral, that is the only reason why it can be described with sp³ hybrid orbitals. This is disregarding whether or not hybrid orbitals are indeed a good description in this case. Then there is the problem of double bonded oxygen, which is a stretch. If ever, this is a zero order approximation. I'm sorry, but this answer falls short. It is one of those 'good resources' that unfortunately feed into the myths that hybridisation could be a cause for anything. It is a mere mathematical interpretation, making that clear from the start would go a long way for science. $\endgroup$ Commented Jun 13, 2021 at 17:14
  • $\begingroup$ It doesn't matter whether you are aware or that or not; your answer uses reverse reasoning, it is wrong. Unfortunately it is employing those hand-wavy models which have been popular and continue to be so as these are fed by answers like this. I wholeheartedly disagree that this does not lead to wrong results. Apart from this, nobody who would need to read the correct answer will ever do so, and I do not need to post an answer for people who already know that. || Also, Newton's laws are a valid and good approximation; the comparison is moot and frankly quite insulting. $\endgroup$ Commented Jun 14, 2021 at 22:05
  • $\begingroup$ @Martin-マーチン The hybridization approach is something that is taught in high school in chemical bonding everywhere in India (I do not know of other countries' curriculum). I have found it in a number of introductory textbooks too, and since all of it is wrong, it is necessary first to amend those textbooks first so that people like me don't learn and use the wrong concept right from the start. The myth is largely fed and popularized by the people who include it in the curriculum, much more so than answers of Chem SE. I don't see any point in carrying ahead this discussion in any case. $\endgroup$
    – TRC
    Commented Jun 15, 2021 at 3:26

Simply put, the oxygens' lone pairs are NOT delocalised into the benzene ring, and the stabilisation of the sulfonate group must be allocated more to ionic interactions(or loss of such; see below) than to resonance effects(the latter of which does not really matter that much, due to the instability of the "toluenide anion"-containing resonance canonical).

As many people have pointed out in many different answers posted on this site, there are no double bonds in sulfate groups and the like- the sulfur has a +2 charge while the oxygens have -1 charges each. These charges attract each other, resulting in the "double-bond-like" sulfur-oxygen distances. If an electrophile attacks and bonds to the oxygen, then the oxygen would not be able to have a negative charge anymore, resulting in destabilisation from the loss of an electrostatic attraction interaction.

Now, both nitric and toluenesulfonic acids are strong acids. In the nitrate case, there are two factors (pi-resonance and (+1 charge nitrogen)-(-2/3 charge oxygen) electrostatic attractions at work here, while in the sulfonate case, there is only one factor at work here, namely the (+2 charge sulfur)-(-1 charge oxygen) electrostatic attractions. Note that the elctrostatic attractions in the sulfonate anion are much stronger than those in the nitrate anion.

The destabilisation of the loss of electrostatic attractions upon protonation of the former is nearly of the same order of magnitude of the destabilisation from the disturbing of the pi-system upon protonation of the latter(if any, since the proton certainly would not want to attack the pi-system and violate VSEPR theory), and hence both acids become nearly equally strong acids.

Note: all of the above is true because the "toluenide anion" is unstable. In situations like the trifluoromethanesulfonate anion, the three oxygens' sp³ lone pairs, in the same direction as the carbon-sulfur bond, CAN delocalise into the sulfur-carbon antibonding orbital(which is mainly of sulfur 3p character; remember that one side of a 3p orbital has one node, just like the whole of a 2sp³ orbital), since the "trifluoromethanide anion" is relatively stable due to the inductively electron withdrawing fluorines. In support of this argument, optimisation calculations done on Mopac using the PM7 level of theory indicate that, while there is a charge <-1 on each oxygen in the toluenesulfonate anion(i.e. no delocalisation of the oxygen lone pairs), while there is a charge slightly >-1 on each oxygen in the trifluoromethanesulfonate anion(i.e. some delocalisation of the oxygen lone pairs).


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