Why is the ester bond stable under Reformatsky reaction conditions?

The following reaction is a Reformatsky reaction.

Why is the ester bond not attacked in the last step? Couldn’t $\ce{OR3}$ be replaced by $\ce{OH}$ or $\ce{O(molecule)}$?

• You ask why the ester doesn't react with the H3O to form an actic acid and release an alcohol? – Avishai Barnoy Jan 12 '17 at 6:20
• @AvishaiBarnoy yes , I want to know that – Icandoahandstand99 Jan 12 '17 at 7:53
• It probably occurs as it's an equilibrium step.maybe the base adding is under cold condition and prevents further reaction. – Avishai Barnoy Jan 12 '17 at 11:15

Aside from the oxyanion here being rather closely associated to the zinc cation which has a larger positive charge than e.g. sodium and thus exercises a greater attractive force, there is also the fact that the alcohol in question is tertiary and tertiary alcohols are bad nucleophiles. In most of the cases, $\ce{R3}$ will be primary or at most secondary, thus the transesterification reverse reaction will be more rapid than the forward reaction.
If, however, $\ce{R3}$ is tertiary, too, you would be dealing with a huge steric bulk on that carbonyl carbon, attempting to have two tertiary $\ce{OR}$ groups on the tetrahedric intermediate. That is unlikely to happen for steric reasons.