# Which side of the C-o-C bond breaks in esters?

When an ester undergoes hydrolysis which side of the 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 -OH an alcohol also forms).

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

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

In the formation of an ester, wherin you react an alcohol with an acid in presence of conc.$\ce{H2SO4}$
$\ce{RCOOH + R'OH -> RCOOR' + H2O}$
$\ce{RCOO'H + R''OH -> RCOOR'' + H2O'}$
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.