Hydrolysis under basic conditions

Which route actually occurs? **Is it really true that the carboxylic acid never forms under basic conditions?** 


Why not just form the -COOH and deprotonate it afterward?? Or is there a higher activation every barrier to the second route? 


I understand that carboxylic *acids* don't stick around long in basic solutions but that doesn't necessarily mean they don't exist in basic solutions *at all*. 


![enter image description here](https://i.stack.imgur.com/XlFfR.jpg)




I find the top mechanism unlikely for two reasons.


1) In the top mechanism, the proton removed in the first step should be way less acidic than a typical hydroxyl since the intermediate is already negatively charged. Going to a di-anion should be very far uphill, and therefore, slow. If that di-anion was ever formed, the next step would be exceedingly fast


2) In the bottom mechanism, the first step should be fast. It's intramolecular and entropically favored. The second step is instantaneous.


I can't imagine any text book includes the top mechanism. Check your book for saponification of esters.


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The reaction proceeds by a mechanism named $B\_{AC\_2}$ (Base-Acyl Bond-2 molecularity).


The rate is given by: (note that it is order 2 and molecularity 2) $$r=k\ce{[ester][OH- ]}$$


The $\bf rds$ is the attack of hydroxide ion. Since the last step is an acid-base reaction, it is unidirectional and reverse esterification is not possible under general conditions (I think you understand what I mean by this). Thus relative rates are also given which are well explained by this mechanism (kinda proof, considering $\bf rds$):


![enter image description here](https://i.stack.imgur.com/Cv2FP.png)
![enter image description here](https://i.stack.imgur.com/U44nr.png)


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