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$\ce{H2O}$ molecules can act as a nucleophile. So it should attack the carbonyl carbon. However, I think that that when the $\ce{O-}$ "kicks back" it will once again remove the water molecule keeping the $\ce{Cl}$ intact as water is definitely a better leaving group. But in my book they showed a reaction intermediate, which involved an acyl chloride and they used water molecule as a nucleophile to convert acyl chloride to carboxylic acid. I'm confused as to how that is possible? Isn't water a better leaving group than $\ce{Cl-}$ ?

  • $\begingroup$ Don't really remember, but in general proton transfers are very fast. For mechanistic organic chemistry Clayden is a good place to start. $\endgroup$ – orthocresol May 13 '17 at 16:25

You need to lose a proton, then chloride becomes a better leaving group than hydroxide. In general, proton transfers are incredibly fast; in this scenario there is tons of water around to accept the proton, and the tetrahedral intermediate with a positively charged oxygen is fairly acidic anyway.


Mechanism from Clayden 2nd ed. (p 199).

In the absence of an external base (e.g. pyridine), water will act as the base itself here. Another way of approaching it is to note that hydrolysis of an acyl chloride will release one equivalent of HCl:

$$\ce{RCO\color{red}{Cl} + \color{red}{H}OH -> RCOOH + \color{red}{HCl}}$$

so in the absence of a second base, you will necessarily have to produce $\ce{H3O+}$.


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