In the wikipedia article "Step-growth Polymerization/Kinetics", it is stated that:

The simple esterification is an acid-catalyzed process in which the protonation of the acid is followed by interaction with the alcohol to produce an ester and water

My question is: How are acids even protonated? Aren't they the ones supposed to be giving out protons? How are they the ones accepting the protons?

  • 1
    $\begingroup$ Probably this could help you : organic-chemistry.org/namedreactions/… $\endgroup$
    – Raghav
    Commented Jan 26, 2017 at 5:32
  • 6
    $\begingroup$ Everything can be protonated, given enough effort. $\endgroup$ Commented Jan 26, 2017 at 6:53
  • 1
    $\begingroup$ @IvanNeretin Although I would love to see the stuff that protonates superacids — on paper only, that is! $\endgroup$
    – Jan
    Commented Jan 26, 2017 at 21:50
  • 1
    $\begingroup$ @Jan Naked proton would protonate any neutral molecule, I guess. $\endgroup$ Commented Jan 26, 2017 at 22:22
  • $\begingroup$ Acidic and basic is all relative--we just happen to have arbitrarily used water as the point where we switch from calling it an acid to a base. If you instead think of things as "more acidic" and "less acidic" (or "more basic" and "less basic"), then all that's happened is that you have something in solution more acidic than COOH. $\endgroup$
    – chipbuster
    Commented Jan 26, 2017 at 23:27

1 Answer 1


A general acid-base reaction is one of the kind $(1)$.

$$\ce{\underset{\text{acid}}{HA} + \underset{\text{base}}{B} <=> \underset{\text{conj. acid}}{HB+} + \underset{\text{conj. base}}{A-}}\tag{1}$$

All that the acid needs to react as an acid is some hydrogen atom bonded to something, and all that the base needs to react as a base (although it is not shown here) is a lone pair which can accept the acid’s proton. Depending on the relative strength of the acid and the conjugate acid, this equilibrium will be shifted. If the acid is stronger, it will be shifted towards the product; if the conjugate acid is stronger, towards the reactant. If the acidity are within reasonable range of each other, you can always expect a catalytic amount of the stronger acid to be protonated.

The above was the general introduction; now how does this apply to your question?

The ‘acid’ you are talking about is a carboxylic acid. These have the functional group $\ce{-COOH}$ and each of the oxygens in there has two lone pairs. Thus technically, a carboxylic acid could be protonated up to four times.[1] We only need one further protonation, so we can define the carboxylic acid to be compound $\ce{B}$ above — the base in the acid-base reaction.

In fact, acid-catalysed esterification is typically catalysed by the addition of a small drop of sulfuric acid. An average carboxylic acid has a $\mathrm{p}K_\mathrm{a}$ value of $\approx 5$, sulfuric acid $-3$. The hydronium ion $\ce{H3O+}$, that forms if water takes the place of the base in $(1)$, has a $\mathrm{p}K_\mathrm{a}$ of $-1.7$ or $0$, depending on who you ask. Other alcoholic oxonium ions have similar $\mathrm{p}K_\mathrm{a}$ values. The lower the $\mathrm{p}K_\mathrm{a}$, the stronger the acid. Thus, we can assume that most carboxylic acid molecules remain as $\ce{-COOH}$, while the catalytic amounts of sulfuric acid (practically completely deprotonated once) protonated a few solvent molecules at equilibrium. The stronger acid — sulfuric acid — practically governs that none of the weaker acid — the carboxylic acid — can lose its proton.[2]

Even though it is equilibrium, a few of the acid molecules — the ‘next-weakest links in the chain’ — will also be protonated at a given point in time, even though we can expect the $\mathrm{p}K_\mathrm{a}$ value of a protonated carboxylic acid to be rather low. This small proportion is enough to get the reaction going.


[1]: However, each successive protonation introduces a positive charge which always makes further protonation less favourable. A positively charged compound is, generally, also more inclined to transfer hydrogen to something else, i.e. is more acidic.

[2]: This can actually be extended. There are a number of superacids with even lower $\mathrm{p}K_\mathrm{a}$ values — in fact, those values are so low that they can no longer accurately be calculated and instead a different value, the Hammett acidity is used. Some of the strongest superacids are even able to protonate methane which is unsuspiscious of providing a lone pair an acidic proton could interact with. Hence Ivan’s comment of:

Everything can be protonated, given enough effort.

  • $\begingroup$ Except the strongest superacid known? $\endgroup$
    – hBy2Py
    Commented Jan 26, 2017 at 22:28
  • $\begingroup$ @hBy2Py Until we find a stronger one. $\endgroup$
    – Jan
    Commented Jan 26, 2017 at 22:39
  • $\begingroup$ I got lost in this. I think the OP is looking for the mechanism for Fischer Esterification. organic-chemistry.org/namedreactions/… $\endgroup$
    – MaxW
    Commented Jan 26, 2017 at 22:55
  • $\begingroup$ @MaxW I think, OP gets the idea of a Fischer esterification but wants to know why the thing that they learnt gives protons (‘acid’) is suddenly acting like the thing that they learnt accepts protons (‘base’). Tbh, I feel like I wrote better answers, too. (I got lost in my thoughts writing it …) $\endgroup$
    – Jan
    Commented Jan 26, 2017 at 23:00
  • $\begingroup$ The point is that the mechanism clearly shows the carboxylic acid accepting another proton to form $\ce{R-C-(OH)2^+}$. So it is, as you stated, acting as a base. I'd also point out that this reaction happens in "relatively" anhydrous solutions. $\endgroup$
    – MaxW
    Commented Jan 26, 2017 at 23:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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