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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-

enter image description here

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.

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  • $\begingroup$ One reaction is esterification and the other is acid catalysed hydrolysis.. does this answer your question? $\endgroup$ Apr 29 at 6:09
  • $\begingroup$ @SafdarFaisal I'm not sure I get what you mean $\endgroup$ Apr 29 at 6:17
  • $\begingroup$ But then how can we be sure that formation of carbocation is RDS. Determination of RDS is experimental, right? $\endgroup$ Apr 29 at 6:31
  • $\begingroup$ Yes, What is the RDS in acid catalysed hydrolysis of esters? $\endgroup$ Apr 29 at 6:32
  • $\begingroup$ It's as given in the above picture. Second step, right? I want to know what will be RDS in esterification. $\endgroup$ Apr 29 at 6:34
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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

enter image description here

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

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    $\begingroup$ 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. chemistry.stackexchange.com/questions/49548/… $\endgroup$
    – S R Maiti
    Apr 29 at 8:16
  • $\begingroup$ 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. $\endgroup$ Apr 29 at 9:40
  • $\begingroup$ 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. $\endgroup$
    – S R Maiti
    Apr 29 at 9:54
  • $\begingroup$ 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 $\endgroup$ Apr 29 at 10:02
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    $\begingroup$ @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. $\endgroup$
    – Zhe
    Apr 29 at 14:22

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