Why do you need molecules like $\ce{Br2, I2}$ as one of the reagents in oxidation of thiols? I am saying this because when you look at the mechanism of reaction, it seems that you can have same product of disulfide by simply joining two protonated thiols (by strong base like $\ce{OH^{-}}$) together. Thanks.

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    $\begingroup$ Probably so the electrons have somewhere to go. $\endgroup$ – ringo Feb 9 '17 at 16:32

Try drawing a mechanism without halogens present, starting with your base DE-protonating one thiol. See how far you get. The problem is that you'd have to somehow kick out H- (hydride), which is a terrible leaving group (enzymes can accomplish this, but standard lab reactions cannot do this easily). The fact is, combing two thiols to give the disulfide is an oxidation reaction, and there is an electron pair being lost from the reactants. It has to go somewhere, i.e. to reduce a diatomic halogen (playing the role of an oxidizing agent).

The thiol-disulfide redox equilibrium you are asking about has great biological significance; see disulfide bridges (for protein structure) and glutathione (an antioxidant).


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