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I have been looking everywhere for an explanation to a reaction mechanism concerning the oxidation of DTT (or just thiols to disulfides in general) using oxygen as the oxidant. I'm aware that oxidizing DTT with oxygen ($\ce{O2}$) will give me the oxidized form of DTT and $\ce{H2O2}$. However, I'm running my reaction in $\ce{MeOH}$ and $10\% \; \ce{KOH}\; (w/w)$ in $\ce{H2O}$ while bubbling it through with air. So, I get that the air is providing the oxygen source, but why run it in basic conditions? Why not just in water?

Below is how I think the oxidation might look in the presence of only oxygen as if DTT was left out in the open:

This is how I think it might look in the presence of only oxygen

I have a hard time seeing why the basic conditions are needed besides it keeping the thiols de-protonated? Is the high $\mathrm{pH}$ to prevent the formation of $\ce{H2O2}$?

Does the mechanism below make sense?

This is just a guess

I hope all this rambling made somewhat sense and I really REALLY hope that there is someone out there who can help me because I'm stuck.

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    $\begingroup$ This mech. shows what appears to be SET - not something I'd expect from thiol - it's quite possible that it rather abstracts hydrogen. pH is rater for tuning redox potential, so further oxidation won't happen. $\endgroup$ – Mithoron Oct 30 at 0:12
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    $\begingroup$ Welcome to ChemSE! Given the low pKa of thiols, the bis-thiol certainly would be present as the dianion. Your second equivalents of water in place of 2 protons. Balance the right side with 2 hydroxides. While your diradical mechanism is viable, formally one would have thiolate add to oxygen. The second thiolate can displace on sulfur to eliminate peroxide. These intermediates may be protonated at any stage of the process. Certainly, a thiolate is more nucleophilic than the neutral thiol in water. $\endgroup$ – user55119 Oct 30 at 0:21
  • $\begingroup$ Thank you so much for your answer! :) So I should replace the peroxide on the right side of the equation with 2 hydroxides? I'm not quite sure what you meant by 'your second equivalent of water in place of 2 protons'? $\endgroup$ – line112 Oct 30 at 11:38
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    $\begingroup$ Did you use any catalyst in this case? $\endgroup$ – Mathew Mahindaratne Oct 30 at 16:34
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    $\begingroup$ Sorry for the confusion in the second sentence. Because you are working with aqueous alkali, you have no H+ but only water as a proton source. Replace H+ with H2O and place HO- on the right side to balance the equation. BTW, while I don't have to ping you, you as OP have to ping the person (@user55119) making the comment. I only came across your response by chance. $\endgroup$ – user55119 Nov 1 at 19:47

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