The common route for aspirin synthesis (described here) employs acetic anhydride and salicylic acid, in acidic conditions.

The first step of such a reaction mechanism is the protonation of the carbonyl oxygen of acetic anhydride.

Why, instead, protonation of carbonyl oxygen in salycilic acid should not occur ? Of course, such a protonation is not desired, since it won't lead to aspirin: however, can it be neglected and why ?

Thanks for any hints.


1 Answer 1


Mechanisms are explanative, and not predictive.

The first step of the reaction mechanism is protonation of the anhydride because that is the first step of the reaction. That just is. If the anhydride is not protonated, step 2 simply won't happen.

This does not mean that the salicyclic acid is not protonated. It just means that the protonation of salicyclic acid is not necessary to explain how acetylsalicyclate forms. Proton exchanges are at equilibrium. It is likely that the salycilic acid is protonated during this process. However, that protonated salicyclic acid may then protonate acetic anhydride and proceed as indicated in your link. For the purposes of describing the mechanism, it is not necessary to distinguish where the proton that protonates the acetic anhydride comes from.

So your question really is self-answered, where you say, roughly, 'Such a protonation does not lead to product directly'.

  • $\begingroup$ I'm curious - what is the product of protonation of salicyclic acid and why is it insignificant? I know that reactions are always at equilibrium, but if the equilibrium constant is very big shouldn't it be almost irreversible? What is the product and why is it unstable? Thanks, $\endgroup$
    – studen
    May 2, 2015 at 17:28
  • $\begingroup$ I don't have easy access to a chemical sketching program, so I'll try to explain. Draw Salicyclic acid. Add a bond from the C=O (double bond) oxygen to a new hydrogen. Add a positive charge to to the same oxygen. This compound is unstable because it is highly acidic. Any positively charged oxygen will be highly electron withdrawing, and will pull electrons away from and bonded hydrogens. Related ... being at equilibrium, we cannot describe it as irreversible, even if the proportion of one of the species is vanishingly small. $\endgroup$
    – Lighthart
    May 4, 2015 at 7:11

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