# Why does sulfonation of alcohols even work?

I am now near the end of organic chemistry 2, and I'm trying to understand the patterns of reactions so that I can predict them rather than brute force memorizing all of them. Looking at the sulfonation of alcohols using mesyl chloride, I don't think I would ever predict that. I don't understand why the oxygen should attack the sulfur in mesyl chloride. To me it looks like a nucleophilic substitution reaction, but alcohols aren't even powerful nucleophiles. Not to mention that the end of this reaction creates HCl a very powerful acid. I thought when reactions occur they like to create relatively weaker acids and bases? What is driving the reaction, and what patterns should I be noticing here for future predictions?

There are always two sides of a reaction. In the case of an alcohol attacking a sulphonic acid chloride, we can look closely at the alcohol and say: ‘Oh my, this is a bad nucleophile and we cannot deprotonate it with the conditions we have. Probably, this reaction won’t work’. But we can also look at the sulphonic acid chloride and say: ‘Oh my, this is a very strong electrophile and it has a very good leaving group. This reaction will work like a charm!’

So it is important, that you compare the reactants extensively. In this case, the electrophilicity of the sulphur(VI) is high enough to even allow rather bad nucleophiles to work. On the other hand, oxygen atoms aren’t even that bad a nucleophile. Give them enough time and alcohols will attack via $\mathrm{S_N2}$ — it might just take forever.

We can also take electrostatics into consideration. Oxygen is rather negatively polarised since it has a very high electronegativity. The sulphur atom, on the other hand, is strongly positively polarisied, since it is connected to two formal oxide anions and a chlorine atom — all three are more electronegative than sulphur by a noticeable margin. A rather negative particle such as the alcohol oxygen, will happily attack a rather positive particle such as the sulphonic acid chloride sulphur.

Mechanistically, this attack forms an intermediate pentavalent sulphur which will then displace the best leaving group: The chloride atom. Therefore, even the hydrochloric acid that is produced isn’t as bad as it may look. We loose the chloride first and the proton much later. We aren’t immediately removing a strong acid but rather a good leaving group and the proton just has to deal with it.

However, I will allow the objection that a significant concentration of $\ce{HCl}$ would favour the reverse reaction. Thus, when employed synthetically, a base is usually added that serves no purpose other than to capture the proton and keep it somewhere where everybody can be happy.

• Driving force is the liberation of the chloride and turning a weak $\ce{S-Cl}$ bond into a strong $\ce{S-O}$ one.
• @AbidRizvi No, that was not what I was trying to say. You really can’t compare the two. However, the pentavalent sulphur is only a transition state: It has an electron dectet on sulphur which wants to abide by the octet rule as do all main group elements. You can probably better think of the pentavalent sulphur as a halted transition state of an $\mathrm{S_N2}$ reaction. That said, $\ce{HCl}$ is ‘good’ in this context not because it is a good acid or whatever. What we are initally loosing is a chloride $\ce{Cl-}$. Chloride is very stable by itself — see all the chloride salts. (tbc) – Jan Dec 3 '15 at 16:49
• (cont’d) As I tried to say, the proton only leaves later and it doesn’t have to end up as $\ce{HCl}$ although it could. And thermodynamically, forming $\ce{HCl}$ molecules isn’t that bad either; hydrogen and chlorine gas react strongly exothermicly to form it. – Jan Dec 3 '15 at 16:51