Textbooks say removing proton from thiamine's C atom will produce a resonance stabilized carbanion which is hence bioactive. That's OK, but how comes that 1) we abstract the acidic proton with the OH base whose pK is lower that for thiamine, and 2) the remaining anion acts - in bio context - in aqua medium (?), hence will immediately deprotonate water and lose its negative charge.

deprotonation of thiamine's thiazolium ring


In the ionized structure shown, which looks like an ylid, the carbanion is formed on an $\mathrm{sp^2}$ orbital and no resonance stabilization of this charge is possible.

Regarding your question about the ionization, remember that biochemical reactions don't always happen in water, they may happen in the active site of enzymes, which have a completely different environment: the dielectric constant is different, the $\mathrm{p}K_\mathrm{a}$ of substances change and the proximity of certain groups stabilize reactive intermediates that would not be formed in water.

  • $\begingroup$ Your second point is good, but first is incorrect - formal charge on carbon is only in mesomeric structure with dipolar bond, there's another one - carbenic. $\endgroup$ – Mithoron Aug 15 '20 at 20:46
  • $\begingroup$ I see that a carbene can be formed, but he specifically said "will produce a resonance stabilized carbanion", which does not seem to be the case $\endgroup$ – Rafael L Aug 18 '20 at 22:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.