# 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?

• The scheme does not illustrate your question — and the 2-phenylethanolate will immediately deprotonate the carboxylic acid to give 2-phenylethanol and the carboxylate (your left-hand products). – Jan Mar 18 '17 at 16:09

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.

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.