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:


[![ester saponification mechanism](https://i.stack.imgur.com/x7x9o.jpg)](https://i.stack.imgur.com/x7x9o.jpg)


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.$\ce{H2SO4}$


$\ce{RCOOH + R'OH -> RCOOR' + H2O}$


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


$\ce{RCOO'H + R''OH -> RCOOR'' + H2O'}$


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.***


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I agree with SubZero's answer to some extend. To explain the hydrolysis, use of formation of ester is a good idea but there is a flow in that explanation.


The formation of an ester, when an alcohol reacts with an acid in the presence of catalytic amount of concentrated $\ce{H2SO4}$ can be given as:


$$\ce{RCOOH + R'OH <=>[cat. H2SO4] RCOOR' + H2O} \tag1$$


However, when the $\ce{O}$ of $\ce{-OH}$ group in $\ce{-C(=O)OH}$ part is replaced with an isotope of oxygen (say $\ce{O^{18}}$), SubZero's explanation of the isotope oxygen loosing as $\ce{H2O^{18}}$ is incorrect as given here:


$$\ce{RC(=O)O^{18}-H + R'OH -> RCOOR' + H2O^{18}} \tag2$$


In reality, the product mixture should contains both $\ce{H2O^{18}}$ and $\ce{H2O^{16}}$ (normal water) as shown in following equations:


$\ce{RC(=O)O^{18}-H + R'OH <=> RC(-OH)(-H\overset{+}{O}-R')O^{18}H <=>[H+ transfer] \\ RC(-\overset{+}{O}H2)(-O-R')O^{18}H \text{ or } RC(-OH)(-OR')\overset{+}{O}^{18}H2} \tag3$


Thus, there would be two different water eliminations:


1. From $\ce{RC(-\overset{+}{O}H2)(-O-R')O^{18}H}$, the products are $\ce{RC(=O^{18})-OR' + H2O}$; and
2. From $\ce{RC(-OH)(-OR')\overset{+}{O}^{18}H2}$, the products are $\ce{RC(=O)-OR' + H2O^{18}}$.


What this mechanism reveals is that the acid loses either oxygen as $\ce{H2O^\*}$ $(\ce{O^\* = O^{16} \text{ and } O^{18}})$ and the alcohol does not lose its oxygen, but supply its $\ce{H+}$ to make the water molecule (as a consequence, you get labelled ester and non-labelled water, and vice versa).


However, in base catalyzed hydrolysis of an ester (saponification), this scrimmage would not happen. Suppose you want to hydrolyze $\ce{RC(=O)O^{18}-R'}$ in basic medium. The only products you would get are as in the following equation:


$$\ce{RC(=O)O^{18}-R' + ^-OH -> RCOO^- + R'-O^{18}H} \tag4$$


You will not get $\ce{RC(=O)^{18}O^-}$ and $\ce{R'-OH}$ in the mixture.


**Mechanism:**


$$\ce{RC(=O)O^{18}-R' + ^-OH <=> RC(-O^-)(OH)-O^{18}-R' <=> \\ RC(=O)-OH + ^-O^{18}-R' -> RC(=O)-O^- + HO^{18}-R'} \tag5$$


Similarly, if you used the ester, $\ce{RC(=O^{18})O-R'}$, to hydrolyze this would happen:


$$\ce{RC(=O^{18})O-R' + ^-OH -> RC(=O^{18})O^- + R'-OH} \tag6$$


You will not get $\ce{RC(=O)O^-}$ and $\ce{R'-^{18}OH}$ in the mixture.


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