When an ester undergoes hydrolysis which side of the $\ce{C-O-C}$ breaks for instance in the following example:

ester saponification mechanism

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?

  • $\begingroup$ 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). $\endgroup$ – 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.


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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.