For example, take the reaction:

MeCOCl + H₂O -> MeCOOH + HCl

Bonds broken: C-Cl + H-O -> 330 + 460 = 790 kJ/mol
Bonds formed: C-O + H-Cl -> 350 + 432 = 782 kJ/mol

I realize that BDE is only an estimate and does not take into account the fact that HCl (and MeCOOH to some extent) will undergo heterolysis and donate a proton to water (or whatever other species is available to accept it), forming a stable anion. I assume it is this process that makes the reaction favorable, correct?

If so, then would this reaction in the gas phase favor the MeCOCl + H₂O side?

I will admit I don't immediately recognize the products as having more favorable solvent interactions when considered as neutral species:

  • H-Cl makes weaker hydrogen bonds than H-O (bad)
  • O is more electron-withdrawing and leaves C more electron-deficient (bad)
  • but creates a stronger dipole (good)
  • and less diffuse electron density because it is smaller than Cl (good)
  • but HCl has a weaker dipole than H2O (bad)
  • and less concentrated electron density (bad)

This is why I ask how important the ability of MeCOOH and HCl to donate their proton is for hydrolysis to be favorable.

How would using an aprotic solvent affect the equilibrium?

  • $\begingroup$ BDE's are for homolysis, not heterolysis. $\endgroup$
    – user55119
    Commented Apr 25, 2022 at 18:21
  • $\begingroup$ Yes, but it should still be a valid indicator for the relative stability of each species, shouldn't it? $\endgroup$ Commented Apr 25, 2022 at 18:24
  • 3
    $\begingroup$ "it should still be a valid indicator for the relative stability of each species, shouldn't it?" Absolutely not. You're completely ignoring all of the components that depend on charge, and that's particularly important, particularly in a protic or polar solvent. $\endgroup$
    – Zhe
    Commented Apr 25, 2022 at 19:16
  • $\begingroup$ I'm aware that these effects exist, that's why I was asking about the reaction in the gas-phase, where they become irrelevant. $\endgroup$ Commented Apr 25, 2022 at 19:39
  • 1
    $\begingroup$ @user123795 ...? $\endgroup$ Commented May 5, 2022 at 14:14

1 Answer 1


The data more or less explain why acid chlorides are made by reaction of acids with PCl5, SOCl2 and similar compounds rather than anhydrous HCl. In water or similar solvents [see Schotten-Baumann method] removal of the H by a base definitely promotes the reaction. I have often wondered Why what seem to be very reactive compounds when one is exposed to them are not so active in anhydrous reactions and why the need for the base. The final driving force is the reaction with the strong acid HCl or, in the case of sulfonyl chlorides the sulfonic acid. The same thoughts explain the lower reactivity of acid anhydrides.


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