# Why does ester hydrolysis occur in basic medium

Ester hydrolysis occurs in basic medium and this is the medium which is usually preferred for the same. But I am facing a very fundamental doubt and not able to understand the reason.

Ester Hydrolysis goes like this:

$$\ce{RCOOR' + OH- <=> RCOOH + OR'-}$$

Now according to me $$\ce{OR'-}$$ should be a better base because of +I inductive effect of alkyl group. If it is a better base then why does it not attack the acid back again !? In short, why does the equilibrium not reverse ?

• $\ce{OR'-}$ will take a proton from a water molecule as it is a better base and will form an alcohol, in turn creating more $\ce{OH-}$ which will further increase the hydroxide concentration and thus decreasing the probability of $\ce{OR'-}$ to attack the species. – Nisarg Bhavsar Apr 18 at 5:52
• The acid with be present as the carboxylate salt thus not an electrophile – Waylander Apr 18 at 7:28
• The presence of $\ce{OH-}$ ions destroys $\ce{RCOOH}$ which prevent the reverse reaction, which would produce the ester again. – Maurice Apr 18 at 8:29

In basic medium, there is a huge amount of $$\ce{OH-}$$ and they will deprotonate the carboxylic acid as soon as it forms.
$$\ce{RCOOR' + OH- <=> RCOOH + R'O-}$$ $$\ce{R'O- + H2O <=> R'OH + OH-}$$ $$\ce{RCOOH + OH- -> RCOO- + H2O}$$ The last deprotonation step is almost irreversible and consumes the base. This is why, an excess of base is usually added for ester hydrolysis.
The $$\ce{RCOO-}$$ is a negative ion, so it repels any nucleophile that comes close to it, so nucleophilic attack cannot happen, and the reverse reaction is prevented (as Maurice has mentioned in the comments). In essence it is this last step that drives the reaction to the product side.