When an ester undergoes hydrolysis which side of the C-O-C breaks for instance in the following example:

enter image description here

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 -OH an alcohol also forms).

Basically which carbon does the original O of the C-O-C stays with?

  • $\begingroup$ could you explain the difference b/w the 2 products other than the complimentary ion,to stabilize the -ve charge? $\endgroup$ – Supernova Mar 18 '17 at 10:03
  • $\begingroup$ 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.

| improve this answer | |

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.