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The energetics of the mechanism of saponification were previously under debate, but I will admit, I misrepresented a point.

Here is the revised mechanism of saponification proposed by someone other than myself. It is the one on top.

The previously suggested mechanism is in the middle.

The textbook mechanism is the bottom one.

Now, it does seem that if the mechanism were concerted then proton abstraction would be much more plausible from a thermodynamic standpoint. If the second mechanism were akin to scaling Mt. Everest, then the first one looks to me to be more like climbing the rock wall at the gym.

We start with the same reactant - an ester - but in the second mechanism we are going through a dianion intermediate. In the first mechanism however we are simply going through a concerted transition state en route to the stable carboxylate.

So, again, I ask - which is (kinetically) correct?

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    $\begingroup$ The reason I prefer the textbook mechanism is that #1 requires bringing 2 anions together, and #2 requires forming a dianionic species - both energetically unfavorable. $\endgroup$
    – iad22agp
    Dec 7, 2014 at 0:28
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    $\begingroup$ Another way of thinking about these three mechanisms is that the top is an E2 mechanism, the middle is an E1-conjugate base mechanism, and the bottom is an E1 mechanism. The middle and bottom represent ends of a mechanistic spectrum and the top represents the mid-point of those two mechanisms. $\endgroup$
    – jerepierre
    Dec 9, 2014 at 17:02

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The question borders on a semantic argument. In the third mechanism (textbook), we would assume that the deprotonation event is very fast. The first mechanism (concerted) is basically saying that the deprotonation is so fast that it cannot be distinguished from the alkoxide leaving.

We know that acid-base reactions are fast. We also know that an anionic oxygen 'kicking out' an alkoxide is fast, at least relative to nucleophilic attack on a carbonyl (the rate determining step for any of these reactions AND the related trans-esterification is nucleophilic attack on a carbonyl to give the tetrahedral intermediate). What we're trying to answer is how fast is fast?

Here is my argument for the third mechanism. While acid-base reaction is fast, the top mechanism is essentially an E2 reaction, which does have a stereoelectronic preference for the hydrogen to be anti-periplanar to the leaving group. I have no doubt that conformation is readily accessible (the C-O bond will be rotating freely). The step is also bimolecular. In the third mechanism, an alkoxide kicking out a leaving group has no stereoelectronic preference. There are no alternate conformations to even consider. The C-O bond may be rotating, but it doesn't matter: there is always an electron pair arranged properly to kick out the alkoxide. Furthermore, the step is unimolecular.

The above is speculation, because I have never seen an alternative to the standard textbook saponification mechanism before. To be honest I never considered the concerted mechanism, and although I think it is a reasonable mechanism, I still prefer the textbook mechanism.

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    $\begingroup$ I'm also not sure how we would even design an experiment to address the issue. In both cases, the step in question is after the rate determining step. That makes it difficult to study empirically. It's not necessarily impossible, but definitely beyond my expertise in physical organic chemistry. $\endgroup$
    – jerepierre
    Dec 9, 2014 at 16:54
  • $\begingroup$ good points, and I like the point about the periplanar requirement. Unfortunately we don't cover stereoelectronics at all as the goal of the instructor is for us to do organic "by feel." I also think that while the first mech. is reasonable like you said, the proposer's reasoning behind it is still crappy - there's nothing wrong with a -COOH being formed in basic solution through electron rearrangement. $\endgroup$
    – Dissenter
    Dec 9, 2014 at 17:00
  • $\begingroup$ @Dissenter I feel like if the base approaches from the appropriate trajectory and the proton is lined up properly, the top mechanism will occur. Otherwise, the bottom mechanism will occur. It might also be somewhere in between... If the bond to the leaving group is mostly broken, then the anti-periplanar preference will be diminished. We could call that a concerted, but asynchronous pathway. $\endgroup$
    – jerepierre
    Dec 9, 2014 at 17:06
  • $\begingroup$ you do agree that it's okay to make the -COOH through electron rearrangement though, correct? But not through protonation in basic solution, right? $\endgroup$
    – Dissenter
    Dec 9, 2014 at 17:08
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    $\begingroup$ @Dissenter yes, like we discussed on your previous question, I think it's possible to make the acid through the bottom mechanism. That type of things happens all the time, making an intermediate that's reactive under the conditions. E.g., reaction of a Grignard with an ester, an aldehyde is formed as an intermediate that reacts immediately with another molecule of the Grignard reagent. $\endgroup$
    – jerepierre
    Dec 9, 2014 at 17:11

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