In the base catalysed hydrolysis the hydroxide ion adds to the ester to form a tetrahedral intermediate, after which the alkoxide ions leave and forms carboxylate ion and alcohol.
Also we find the rate of reaction to be quite high.
So my questions are:

  1. Most importantly, alkoxide is a very strong base. So why is it acting as a leaving group here (apparently, a very good leaving group due to the high rate of reaction)?
  2. Why is the tertrahedral intermediate not as stable as the corresponding ones of carbonyl compounds?
  3. Why is this reaction essentially irreversible, considering the fact that alkoxide is a stronger base the hydroxide ion? (Therefore it should be able to displace the hydroxide in the hypothetical backward reaction.)
  • $\begingroup$ It is irreversible because the alkoxide deprotonates the carboxylic acid wihich is the initial product of the hydrolysis. $\endgroup$
    – Waylander
    Feb 12 '19 at 16:14
  • $\begingroup$ @Waylander So,why can't the carboxylate anion once again take up a proton either from the aqueous phase, or from the alcohol, because we have seen earlier in the mechanism scheme of this reaction only the alkoxide can be removed as a leaving group? $\endgroup$ Feb 12 '19 at 16:39
  • $\begingroup$ For (2), tetrahedral intermediates with a leaving group are not stable. Not sure what corresponding ones you're talking about, but those are probably not stable. $\endgroup$
    – Zhe
    Feb 12 '19 at 16:42
  • $\begingroup$ @YUSUFHASAN It can, but the proton on it is the most acidic (its an acid after all) so it is the most likely to be removed in a system where protons are being exchanged between all the species present. $\endgroup$
    – Waylander
    Feb 12 '19 at 17:03
  • $\begingroup$ @Waylander But,what I really want to know is related to (1), why is the alkoxide even leaving in the first place? Because,if such a bad leaving group is being removed,then it seems to me that after a few unsuccessful collisions, a protonated carboxylic acid can be formed, and the alkoxide can initiate a reverse reaction which may soon become favourable under thermodynamic control. $\endgroup$ Feb 12 '19 at 17:09

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