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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

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  • $\begingroup$ Wouldnt the mechanism depend upon the concentration of strong base taken? $\endgroup$ – Del Pate Feb 17 '15 at 4:13
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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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  • $\begingroup$ Can I ask what book you are using? $\endgroup$ – jerepierre Dec 5 '14 at 16:33
  • $\begingroup$ Also if I'm not mistaken I've seen the first mechanism in Organic Chemistry by Clayden. $\endgroup$ – Dissenter Dec 5 '14 at 16:37
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    $\begingroup$ March's Advanced Organic Chemistry shows eight different mechanisms for ester hydrolysis under acidic and basic condtions. Some of these I would consider very unlikely, but the top mechanism that you've drawn is not included. $\endgroup$ – jerepierre Dec 5 '14 at 16:41
  • $\begingroup$ Neither Loudon or Maitland Jones show the top mechanism, only the bottom mechanism. Are your notes a direct copy of something your prof provided to you (on the board or otherwise)? $\endgroup$ – jerepierre Dec 5 '14 at 16:47
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    $\begingroup$ Another argument favoring @jerepierre 's position is: Why should the mechanism for basic hydrolysis of an ester be different than the mechanism for transesterification under basic conditions? In this latter case, there is no proton to remove from the tetrahedral intermediate, so the dioxa anion cannot form. The 2 reactions are very similar ($\ce{OH^{-} vs. OR^{-}}$), why would you invoke different mechanisms for 2 reactions so seemingly similar? $\endgroup$ – ron Dec 5 '14 at 16:51
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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 enter image description here

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  • $\begingroup$ Acid base reactions are always unidirectional? $\endgroup$ – Dissenter Dec 12 '14 at 8:23
  • $\begingroup$ yes where weaker conjugate acids and weaker conjugate bases are formed. $\endgroup$ – RE60K Dec 12 '14 at 8:28
  • $\begingroup$ Determining the rate law often can distinguish between mechanisms, but in this case, both proposed mechanisms could follow that rate law. $\endgroup$ – jerepierre Dec 12 '14 at 15:06
  • $\begingroup$ @jerepierre oh jerrie perry, This is the actual mechanism (surely), I just gave the related info to prove that this is not wrong.It's not the other way where I'm trying to prove this mechanism with the help of rate law $\endgroup$ – RE60K Dec 12 '14 at 16:39
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    $\begingroup$ @ADG It wasn't my question, so it's not up to me to accept your answer. This answer repeats the mechanism that is given in the question and gives some rates that show that mechanism is reasonable. However, a full answer also has to say why it is more reasonable than the alternative. That is what the questioner wants to know. As I read it, this answer argues, "That's just the way it is" without considering the alternative mechanism. Keep in mind that I agree with you that the second mechanism is more reasonable. $\endgroup$ – jerepierre Dec 12 '14 at 18:01

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