Which side of the C-O-C bond breaks in esters during hydrolysis in basic medium? When an ester undergoes hydrolysis which side of the $\ce{C-O-C}$ breaks for instance in the following example:

I believe the first is correct but is it a rule that the salt of a carboxylic acid is formed (and then of course in the presence of $\ce{-OH}$ an alcohol also forms)?
Basically which carbon does the original O of the $\ce{C-O-C}$ stay with?
To answer this,Think about how an ester is formed.

In the formation of an ester, wherin you react an alcohol with an acid in presence of conc.$\ce{H2SO4}$

$\ce{RCOOH + R'OH -&gt; RCOOR' + H2O}$

Now what we have found by replacing the oxygen with an isotope of oxygen is that

$\ce{RCOO'H + R''OH -&gt; RCOOR'' + H2O'}$

What this reveals is that the acid loses an $\ce{OH- group}$ and the alcohol loses an $\ce{H+}$

So summing up, O atom stays with the carbon which is not attached to the $C=O$ group.