# Competition of H2O and NH3 leaving groups in nucleophilic acyl substitutions

I've learned in several textbooks, that in the nucleophilic acyl substitution $\ce{H2O}$ is a better leaving group than $\ce{NH3}$. This can be explained by the "thumb rule" of the $\mathrm{p}K_\mathrm{a}$. $\ce{NH3}$ is more basic $(\mathrm{p}K_\mathrm{b}(\ce{NH3})= 4.3)$ than $\ce{H2O}$ ($\mathrm{p}K_\mathrm{b}(\ce{H2O})=15.4)$ and therefore a worser leaving group.

But what is the reason for the fact, that the following common reaction happen (and the worser leaving group has left the molecule)?

My idea: Is it more a kind of "equilibrium discussion"? The reaction can happen but the equilibrium lies not really on the "product side"? Compared to another nucleophilic acyl substitution (e.g. water with a acyl chloride) this equilibrium lies far more on the educt side. But nevertheless, this reaction happens and can be exploited for example in the Strecker Synthesis through several methods (Le Chatelier, etc.)?

Additional idea: The excess of water can push the equillibrium to the product.

• This reaction only occurs under forcing conditions (conc HCl, heat) so it is not correct to think of it as comparable to acyl chloride + water – Waylander Jul 5 '18 at 11:45
• @Waylander With my comparision I wanted to outline the difference reactivity. You have to "force" the equibrillium with harsh condition (as you state) to react and undergo this rather "unhappy" Substitution. – Nilsfrank99 Jul 5 '18 at 12:40