Last week, I tried synthesizing acetylsalicylic acid - the reaction is shown below - using $\ce{H2SO4}$ as a catalyst. However, as the title suggests the synthesis failed as I used too much $\ce{H2SO4}$ - approximately four times more than the prescribed volume. Needless to say, I had to redo the synthesis.

Synthesis of aspirin

But now I try to understand why the synthesis failed - I know that catalysts are used in small amounts, but I thought that much of it would only lead to equilibrium state being reached faster.

I think that maybe the acidity of the reaction environment could've somehow explained why the synthesis didn't take place. But I don't understand why - shouldn't more protons in the environment lead to faster protonation of the carbonyl oxygen of acetic anhydride? Could you please explain it to me?

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    $\begingroup$ You want to protonate the carbonyl group on the anhydride without protonating the phenolic hydroxy group of salicyclic acid, because a protonated hydroxy group is a terrible nucleophile. Adding a large excess of aqueous acid would also hydrolyze the anhydride... $\endgroup$ – J. LS May 1 '15 at 18:08
  • $\begingroup$ @J.LS Thanks for answer. I'm not sure if I understand correctly - there are two competitive processes - protonation of the phenolic OH group and of the carbonyl group of acetic anhydride. The concentration of protons affect the equilibrium between the two (leaning on the salicyclic acid side when concentrations are high). And that's what happened? What do you mean by hydrolization of anhydrate? That the protons and OH group of acetic anhydride formed water and the reaction do not occur environment with water? $\endgroup$ – studen May 1 '15 at 18:45
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    $\begingroup$ @Studen - You will just protonate both. The anhydrate will undergo an acid catalyzed reaction with water to give two equivalents of acetic acid. $\endgroup$ – J. LS May 1 '15 at 18:59
  • $\begingroup$ @J.LS Because I added too much H2SO4 most of the anhydrate reacted with water forming acetic acid. It may be a stupid question, but where did the extra oxygen come from (to make two acectic acids from one anhydrate I need another oxygen - I get it from water, but the oxygen in water comes from? air?). Is protonation of the two compounds independent of the concentration of H2SO4? Because I thought that they are both protonated always. $\endgroup$ – studen May 1 '15 at 19:12
  • $\begingroup$ Wait, wait, how do you know you have to redo the synthesis? It's not that obvious. $\endgroup$ – Mithoron May 1 '15 at 20:02

But I don't understand why - shouldn't more protons in the environment lead to faster protonation of the carbonyl oxygen of acetic anhydride? Could you please explain it to me?

The short answer is "side reactions". Without more information about your exact experimental conditions, or even better, data about the obtained yields of the desired product and of byproducts, it isn't possible to explain conclusively what went wrong, but here are some ideas.

  1. Acetyl sulfate formation. In the comments you you implied that you used anhydrous sulfuric acid. Under those conditions, acetic anhydride can undergo solvolysis to yield acetyl sulfate and water. $$\ce{CH3COOOCCH3 + 2H2SO4 <=> 2 CH3COOSO3H + H2O} $$ Acetyl sulfate might be a less active acetyl transfer reagent than acetic anhydride, or the water formed may hydrolyze some of the acetic anhydride or even the formed acetyl sulfate.

  2. Dehydration/sulfation of the starting material instead of acetyl transfer. If you really did use anhydrous sulfuric acid, it is a powerful dehydrating reagent. It could be possible that you are sulfating the alcohol or even the acid group (or both?) of the substrate, forming either o-sulfoxybenzoic acid (sulfation at $\ce{-OH}$ group) or sulfosalicylic acid (sulfation at $\ce{-COOH}$ group).

The heat of mixing of concentrated sulfuric acid with water is about $-74~\mathrm{kJ~mol^{-1}}$, judging from Figure 34b of a NIST treatise on the subject. The heat of reaction for hydrolysis of acetic anhydride is only $-41~\mathrm{kJ~mol^{-1}}$. Thus under anhydrous conditions, I think sulfated products will be more thermodynamically favored than acetylated products. Sulfated products will be far more soluble in water or sulfuric acid than would be acetylated products.

I would think that using more water in the reaction would help. The reason is that the heat of mixing of sulfuric acid is a very strong function of water concentration, but the heat of hydrolysis of acetic anhydride is not. So raising the concentration of water a little bit will disfavor sulfation significantly by lowering the heat of mixing of sulfuric acid, but will leave the heat of hydrolysis of acetic anhydride the same. It's worth noting that this argument is purely thermodynamic, not kinetic. But I don't think sulfation reactions are kinetically hindered under ambient conditions, so although it isn't always true, I think thermodynamic arguments are valid in this case.


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