# Why is the halogenation of an alcohol a multistep process?

Refer to the mechanism posted on this site, if you will: http://www.chem.ucalgary.ca/courses/351/Carey5th/Ch08/ch8-10-3.html

When the alcohol attacks the sulfur center, the pi electrons in the oxygen sulfur bond are taken by the oxygen. Then pyridine sucks up the proton. Afterwards the oxygen donates those pi electrons back and the chlorine anion leaves.

Why doesn't the chlorine anion just leave? What's up with the oxygen shuffling electrons around?

• The mechanism in the referred website should not be considered a $\ce{S_{N}2}$ mechanism, as it has a tetrahedral intermediate and therefore is more likely to be an addition-elimination reaction. A $\ce{S_{N}2}$ mechanism is concerted, where nucleophile attaches, while simultaneously the leaving group detaches. Dec 9 '14 at 10:53

## 2 Answers

If the chlorine left straight away I.e after step 1 in the mechanism you linked to then you would have three charges on the structure - charges on two oxygens and a charge on sulfur. Doesn't seem too stable to me.

Instead what happens after step 1 is a proton transfer. Proton transfers are generally rapid. This proton transfer is also thermodynamically favored as it relieves the oxygen of its positive formal charge.

• If would still be 1 charge wouldn't it? The alcohol oxygen is +1, and the chloride is -1. The O double bonded to S is neutral.
– Jeff
Dec 9 '14 at 6:46

As a general rule $S_{N}^{2}$ reaction do not happen at at an '$sp^{2}$' center (check here for help on hybridization: wikipedia). Unlike in an '$sp^{3}$' carbon, there is no likely transition state to accommodate the 'nucleophile' and 'nucleofuge'. Instead of this 'concerted' reaction, an addition-elimination takes place. Like @Dissenter already stated, the proton transfer reaction is favourable and fast, driving the reaction to completion.