Concentrated sulfuric acid is often used as a catalyst in Fischer esterification reactions. To my knowledge, the role it plays in the reaction is not as a reducing agent or dehydrating agent but only serves the role of creating an acidic environment, as seen in the mechanism. enter image description here

Thus, dilute sulfuric acid would be sufficient. In fact, it would be better since it ionises to give an even more acidic solution. However, chemists choose to use the concentrated one instead, suggesting that the role of sulfuric acid is more of a reducing agent. Is there a reason for this?

  • $\begingroup$ That is absolutly not true what you say about H2SO4 acidity vs concentration. Conc. H2SO4 can be milion times stronger then diluted. This seems to be common misconception. In reality addition of water only weakens any strong acid. $\endgroup$
    – Mithoron
    Jun 29, 2017 at 14:54
  • $\begingroup$ @Mithoron What do you mean by "strong". If you mean shows acidic properties well or to ionise completely in solution , then I don't think what you said is true. My teacher once mentioned the example of 20M nitric acid being much less acidic than 2M nitric acid because the acid fails to get ionised in the much more concentrated solution $\endgroup$ Jun 29, 2017 at 22:54
  • $\begingroup$ Well I guess you both didn't hear about Hammett acidity function. Nothing needs to disassociate to be acidic, water's levelling effect weakens every acid which is stronger by itself (like H2SO4 molecule) then H3O+ Also acids are often used in other solvents then water, much less basic. $\endgroup$
    – Mithoron
    Jun 29, 2017 at 23:11
  • $\begingroup$ This brings back old times, I was telling a new user about it more then 2 years ago, being relatively new here myself: chemistry.stackexchange.com/questions/27905 Also this is related: chemistry.stackexchange.com/questions/53580 $\endgroup$
    – Mithoron
    Jun 30, 2017 at 0:02

2 Answers 2


In fact, sulfuric acid is not a must. Many other H-acids, as well as Lewis acids and their combinations would be sufficient.

The trick here is not to get concentrated acid, but to reduce amount of water in the system. Consider the simpliest transformation, which is typically performed with a large excess of methanol and trace amounts of acid:

$$\ce{R-C(O)OH + CH3OH_{(excess)} <=>>[HCl] R-C(O)OCH3 + H2O}$$

This is an equilibrium reaction. The catalyst allows us significantly faster achieve the equilibrium state, but it does not shift it. Catalyst is equally promoting both direct and reverse reactions. A large excess of alcohol, in accordance with the law of mass action, shifts the equilibrium toward the formation of the ether and thus increases the degree of conversion of the acid into ether.

But if water is not eliminated from the system, or is initially present, hydrolysis occurs:

$$\ce{R-C(O)OCH3 + H2O_{(excess)} <=>>[HCl] R-C(O)OH + CH3OH }$$

Hydrolysis of ether will happen even faster, if a strong base is present, as in this case reaction equilibrium shifts further to the right due to the salt formation.

The bottom line is, in order for the reaction to go to the end, you either add an excess of acid (or alcohol), or remove water as it is produced. Starting with concentrated or water-free acid you are not only simplifying your life, but you are also getting better yield.

  • $\begingroup$ But isn't water concentration constant? Also why we need to add more alcohol (Le Chatelier principle) if also alcohol as a liquid have a constant concentration? Is because the volume of the products is reduced therefore and their concentration whereas for reactants only the volume of the acid is reduced (alcohol as solvent is assumed to have constant concentration?) ? $\endgroup$
    – ado sar
    May 26, 2020 at 16:25

In fact, it would be better since it ionises to give an even more acidic solution.

Yes, a dilute acid does dissociate more, but that doesn't mean it furnishes more $\ce{H3O+}$ ions to the solution.

As a simpler example, consider hydrochloric acid. It dissociates as:

$$\ce{HCl <=>[K] H+ + Cl-}$$

It's equilibrium equation would be:

$$K = \frac{\ce{[H+][Cl-]}}{\ce{[HCl]}}$$

Let's say the initial concentration of $\ce{HCl}$ was $x$, and the final concentration of $\ce{H+}$ is $y$. Plugging these values into the equilibrium equation, you get:

$$K = \frac{y^2}{x-y}$$

Rearranging the terms,

$$y^2 + Ky -Kx=0$$

Using the quadratic formula:

$$y = \frac{\sqrt{K^2+4Kx}-K}{2}$$

Plotting this on a graph, we get:

enter image description here

You can clearly see that the amount of $\ce{H+}$ you get increases as you increase the amount of $\ce{HCl}$ you add.

In short, while low concentrations increase the extent of dissociation, high concentrations increase the total amount of $\ce{H+}$ ions furnished.

  • 1
    $\begingroup$ I get your point. However, I was thinking along the lines of self-ionisation of sulfuric acid, which occurs to a large extent. In fact, according to Wikipedia, the self-ionisation constant of sulfuric acid is on the magnitude of 0.0001. Compare that to that of water to give you a better picture of the large extent of ionisation. $\endgroup$ Jun 29, 2017 at 10:48
  • $\begingroup$ So I was thinking that it was due to this large extent of self-dissociation which gives the highly acidic solution. $\endgroup$ Jun 29, 2017 at 10:49

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