For the reaction

$$\ce{R-OH + R'-COOH ->[H+] R-COO-R' +H2O}$$

the reaction mechanism is given here.

What seems unusual to me is that the acid appears to be donating $\ce{OH-}$ rather than $\ce{H^+}$.

Can the mechanism be written as follows instead?

$$\ce{R'-COOH <=> R'-COO^- + H^+}$$

$$\ce{HA-> H^+ + A^-}$$

The inorganic acid pushes the equilibrium for the organic acid to the left and it remains undissociated.

Then, $\ce{H^+}$ from the inorganic acid first attacks the $\ce{O}$ atom in $\ce{R-OH}$ resulting in the formation of a water molecule and a $\ce{R^+}$ ion.

Next, as the equilibrium for organic acid is pushed to the right, the bisulfate ion removes the $\ce{H^+}$ from the acid creating a carboxylate ion. This carboxylate ion then reacts with $\ce{R^+}$ ion to form the ester.

In this mechanism, the carboxylic acid donates the $\ce{H^+}$.

I know one can't prove reaction mechanisms to be correct; however, which one the mechanisms given above is more likely to occur?


The accepted mechanism is far more likely than your proposed alternate mechanism for a number of reasons.

First, let’s consider an esterification reaction employing a chiral secondary alcohol such as butan-2-ol. Protonation of the alcoholic oxygen is no difficulty and will happen in equilibrium. To generate the corresponding ester from the conjugate acid of butan-2-ol, we would require some nucleophilic attack mechanism: the acid would either attack according to $\mathrm{S_N1}$ or $\mathrm{S_N2}$. If we assume an $\mathrm{S_N2}$-like mechanism, the asymmetric carbon atom would be inverted. If we assume an $\mathrm{S_N1}$ mechanism, it would racemise. Experimental evidence shows, however, that the alcoholic carbon atom strictly retains is stereochemistry — and this principle is used in many total synthesis in which asymmetric alcohols are first generated stereochemically and then transformed into esters of the same stereochemistry. To give you just one of many examples, consider the total synthesis of aetheramide by Gerstmann and Kalesse.[1]

Example esterification of a chiral butanol
Scheme 1: Example esterification of (R)-butan-2-ol

Stereoselective esterification performed by Gerstmann and Kalesse
Scheme 2: Esterification performed during the synthesis of aetheramide by Gerstmann and Kalesse.[1]

Second, and immediately tied to the scheme as referenced above, many alternate esterification methods exist in the literature. In principle, they all work by replacing the acidic particle ($\ce{H+}$) of a Fischer esterification with a less reactive and/or more specific compound. Practically all methods — Steglich or Yamaguchi to name just two — work by adding a compound that is known to interact with the carboxylic acid and not as much with alcohols. However, this is only a weak indication since different methods could follow different mechanisms — even though the quantity of the evidence does seem overwhelming.

Third and finally, the acidic hydrolysis of esters — the reverse reaction of the Fischer esterification — has been studied by $\ce{^18 O}$ marked water. Polanyi and Szabo could show that upon hydrolysis the heavy oxygen atom remains bound to the acid fragment and not to the alcohol.[2] This offers intriguing evidence for the mechanism of nucleophilic attacks on a carbonyl group. I had thought that a corresponding isotope-marking experiment had been performed for the Fischer esterification, too. Unfortunately, I was unable to find a reference. Do note, however, that using isotopically marked oxygen on either the acid or the alcohol would, by careful analysis of the product, rule out or confirm one mechanistic pathway.

ester hydrolysis mechanism as confirmed by Polanyi and Szabo using isotopic marking.
Scheme 3: Mechanism of hydrolysis as confirmed by Polanyi and Szabo using $\ce{^18O}$ marked water.[2]


[1]: L. Gerstmann, M. Kalesse, Chem. Eur. J. 2016, 22, 11210–11212. DOI: 10.1002/chem.201602682.

[2]: M. Polanyi, A. L. Szabo, Trans. Faraday Soc. 1934, 30, 508–12. DOI: 10.1039/tf9343000508.

  • $\begingroup$ "If we assume an SN1SN1 mechanism, it would racemise" Why? For alkyl shift happening aat the butane carbenium ion? $\endgroup$ Jun 13 '17 at 16:15
  • 3
    $\begingroup$ @Mockingbird360 because the $\mathrm{S_N1}$ mechanism requires a planar carbenium intermediate which has lost all stereochemical information and is therefore attacked from both sides equally $\endgroup$
    – Jan
    Jun 13 '17 at 16:20

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.