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The following mechanism is given in Peter Sykes when they are "talking about de-esterification" of an ester where the alcohol part is bulky-

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Here the second step is given as RDS. According to me, in the reversed reaction(esterification), this step will remain RDS and so the reaction is bimolecular. Someone argues that RDS will be the formation of carbocation then and so the reaction is unimolecular.

Robin Singh
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After discussion with OP in chat, the issue in logic can be rectified as follows. Simply put:

$\ce{A<=>B<=>[slow]C <=> D}$ doesn't imply that $\ce{B <=> C}$ is the RDS in both $\ce{A -> D}$ and $\ce{D -> A}$.

Rather than this, we need to consider $\ce{A->D}$ and $\ce{D-> A}$ as two separate reactions. Doing so for the two reactions given in the question (acid catalysed hydrolysis ($\ce{A ->D}$) and esterification ($\ce{D -> A}$)), we see the mechanisms are as follows:

Acid catalyzed hydrolysis

Mechanism of acid catalyzed hydrolysis

Here the RDS is the formation of the carbocation $\ce{^+CMe3}$.

Esterification of an acid and a bulky alcohol

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Here the RDS is the formation of carbocation but it happens at a different step and not the same step.

Therefore the two reactions have different RDS steps as they are different reactions

Safdar Faisal
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    I don't understand the logic of your answer. The forward and reverse reaction must have the same rate determining step, unless there are low energy intermediates in the reaction coordinate (and I think this doesn't have that). Forward and reverse reactions aren't different reactions. https://chemistry.stackexchange.com/questions/49548/is-the-rate-determining-step-the-step-with-the-largest-ea – S R Maiti Apr 29 '21 at 08:16
  • The product in the RDS remains the same - formation of the trimethyl carbocation. It just happens at a different step in the reaction. the trimethyl carbocation is the intermediate here. – Safdar Faisal Apr 29 '21 at 09:40
  • If they are the forward and backward version of the same reaction, it's not possible to get the RDS at a different step of the reaction, due to microscopic reversibilty. If you agree that the forward and reverse reaction has the same reaction coordinate, then they really must have the RDS at the same step. – S R Maiti Apr 29 '21 at 09:54
  • That's my point. They aren't the same reaction, both require slightly different conditions. tbh, the arrows are misguiding, the slow reaction in the first one becomes a fast reaction in the second. the formation of the carbocation leads to a valley between two peaks in the energetics graph. One step has the same rate forward and backward at equilibrium but in the whole reaction each step is distinct – Safdar Faisal Apr 29 '21 at 10:02
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    @ShoubhikRMaiti Sorry, that reasoning is not correct. Forward and reverse definitely have the same reaction coordinate, but that doesn't mean they have the same RDS. Consider A -> TS1 -> B -> TS2 -> C, with relative energies: A = 0, TS1 = 5, B = -20, TS2 = -10, C = -15. This means that the RDS in the forward direction is the one with TS2, but the RDS in the reverse direction is the one with TS1. – Zhe Apr 29 '21 at 14:22
  • @SafdarFaisal Your claim that $\ce{A <=> B <=> C}$ does not imply $\ce{A <=> C}$ is not correct due to microscopic reversibility and the definition of equilibrium. – Zhe Apr 29 '21 at 14:23
  • @Zhe I understand that. What I meant was what I gave as the logic in this answer.. – Safdar Faisal Apr 29 '21 at 14:44
  • @Zhe I already mentioned that caveat in my first comment, "unless there are low energy intermediates in the reaction coordinate". Your example has B as the low energy intermediate. Does this esterification reaction have a low energy intermediate? – S R Maiti Apr 29 '21 at 17:19
  • @ShoubhikRMaiti That's fair. I should've read your comment more carefully. I think you can however come up with tricky situations where the intermediate is intermediate in energy between the reactant and product, i.e., set C = -25. – Zhe Apr 30 '21 at 00:45
  • @Zhe Yes of course. If there is any intermediate that's lower than the reactant in energy, then the rds would be different for forward and reverse. – S R Maiti Apr 30 '21 at 07:38
  • @ShoubhikRMaiti Sorry to drag this out, but that's not necessarily true. It depends on the relative energies of the transition states even in that scenario. I think the point here is that the number of scenarios increases quite rapidly with the number of steps, and it's hard to make generalizations for all classes of reaction coordinate diagrams. – Zhe Apr 30 '21 at 17:53
  • @Zhe Ok, but my point is that for the usual organic reactions, starting from the reactant we have a series of high energy intermediates before the product. In those cases, the highest TS is definitely the rate determining TS, which means the same step has to be the rate determining step in both directions. Which is why I think this answer is probably wrong. Do you agree? – S R Maiti Apr 30 '21 at 18:01
  • @ShoubhikRMaiti The problem here is that the equilibrium, under practical conditions, lies on one end or the other. Under that treatment, the forwards and reverse reactions (run to generate predominantly either one or the other product) are different and therefore have different rate determining steps, which addresses the OP's actual question. But for the given equilibrium, the RDS ought to involve the same transition state. So, I think people are arguing about different things here without specifying the underlying assumptions. – Zhe Apr 30 '21 at 23:15