From Wade's "Organic Chemistry" [1, p. 1001]:

MECHANISM 21-3 Conversion of an Acid Chloride to an Ester
Mechanism 21-3

What I don't understand is the elimination of $\ce{Cl}$, even though a much better leaving group with a positive charge on it is present (which results in the backwards reaction).

Why doesn't the alcohol or acid chloride deprotonate the tetrahedral intermediate before step 2 and after step 2 $\ce{Cl-}$ can attack the protonated alcohol/acid chloride to turn into $\ce{HCl}$?


  1. Wade, L. G. Organic Chemistry; Pearson Education, 2012,. ISBN 978-0-321-81139-4.
  • 1
    $\begingroup$ I agree that would make a bit more sense. $\endgroup$ Commented Dec 31, 2017 at 12:23
  • $\begingroup$ Again, I added the better picture and full reference; however, page and mechanism numbers would be different. $\endgroup$
    – andselisk
    Commented Dec 31, 2017 at 13:01
  • $\begingroup$ Alcohol isn't "much better leaving group". Also depictions of mechanisms in books are usually not describing what exactly happens. More precisely they only show constructive pathway and not show surrounding molecules and their involvement. $\endgroup$
    – Mithoron
    Commented Dec 31, 2017 at 17:42

1 Answer 1


Indeed, there is no reason why the protonated alcohol should not be eliminated upon the lone pair rebound. It probably happens more often than not. However, it does not lead to an observable product: it just goes back to the reactants. Thus, we can write an equilibrium arrow on the first step.

$$\ce{R-C(=O)-Cl + R'-OH <=> R-C(-O^-)(-Cl)-OH+-R'}\tag{1}$$

Once the chloride has been kicked out, there is no way back as the ester is a much weaker electrophile than the acid chloride (and chloride a weaker nucleophile than the alcohol). Thus, the reason why we only care about the lone pair rebound eliminating chloride is because only that is productive.

$$\ce{R-C(-O^-)(-Cl)-OH+-R' -> R-C(=O)-OH+-R' + Cl-}\tag{2}$$

These arrows are also given in your source like that; if everything up to a single step is given as equilibria while that one step is given as non-reversible, that should be sufficient driving force to let a reaction happen.

At which point in time the proton is eliminated is somewhat up to debate, of little relevance for the overall reaction and dependent on what is in solution. If just a pure acid chloride and a pure alcohol are mixed the only potential bases that could accept the proton of the tetrahedral intermediate are other acid chloride molecules or other alcohols — or the tetrahedral intermediate’s $\ce{O-}$ atom. All of these are very weak bases and I see no preference for the proton to go anywhere. After we have eliminated chloride, that is added to the set of potential bases. It is also not a strong base; maybe even a weaker base than some of the other options. The big difference is that it is volatile and can diffuse out of the reaction mixture. Thus, with no other bases added, I would actually prefer this sequence of steps.

Once you add a base, I would also prefer deprotonation before chloride is eliminated.

  • $\begingroup$ Clayden's also writes the mechanism to have the deprotonation happen first. But what is the reason for saying that deprotonation happens before chloride leaves? $\endgroup$ Commented Jan 1, 2018 at 12:45
  • $\begingroup$ I believe the mechanism shown in the question just once to show the deprotonation after chloride leaves because it wants to show that the chloride is used as the base to deprotonate the intermediate. I wouldn't think it is accurate to show that the chloride ion is used as the base to deprotonate the intermediate as it is likely not the most basic species in solution capable of deprotonating the intermediate. Am I right? $\endgroup$ Commented Jan 1, 2018 at 12:48
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    $\begingroup$ @TanYongBoon A proton can only leave if it has somewhere to go. Thus, always ask yourself where (whither) the proton would go. It is not really happy on the alcoholic oxygen if that carries a formal positive charge. It is not really happy on the acyl oxygen, except for the short moment in time when that is in a tetrahedral intermediate. It is also not really happy on an $\ce{R-OH2+}$ group. And it’s not really happy on chloride but at least the molecule is neutral. Frankly, if there is something that can accept the proton, proton transfer will be the fastest reaction in the mix. $\endgroup$
    – Jan
    Commented Jan 1, 2018 at 12:56

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