# Recognising a rate-determining step in E1

If there is a hydride/methyl shift possible in E1, then is the carbocation formation the rate-determining step (RDS), or the shift, or is it considered a cumulative step and happens all at once?

Anindya Prithvi has said everything you needed to hear in two sentences. It is true that rate determining step for E1 mechanism is the carbocation formation. Thus, I present here the detail energy diagram for reaction progress of acid-catalyzed dehydration:

The first step is the protonation of starting compound (SM), which is shown in the right hand top corner in blue box. The next step is the rate determining step, which has the biggest activation energy and progress through the transition state 2. The resultant intermediate is the relatively higher energy 2°-carbocation, which is quickly rearrange to lower energy 3°-carbocation by 1,2-hydride shift through the transition state 3. It is worth noting that this 1,2-hydride shift does not involve net bond-breaking or bond-forming (it is believed to be undergoing concomitant bond breaking-making mechanism involving orbital overlapping). Thus its transition state (transition state 3) would not be higher in energy than that of transition state of RDS. The 3°-carbocation would finally give away the expected product through the transition state 4. It doesn't matter if all other steps happen at once or separately. All those steps have to wait to happen until the formation of the 2°-carbocation. That's why it is the rate determining.

• I think it is important to mention that the hydride shift occurs in an already-high-energy intermediate (i.e. the carbocation). Furthermore, the hydride shift does not involve net bond-breaking or bond-forming. Hence, the energetic requirement (i.e. the activation energy) for the hydride shift would not be higher than that of the carbocation formation. Jul 26, 2020 at 1:44
• @Tan Yong Boon: Good point. I thought my energy diagram itself would explain that. Nevertheless, I edited the text accordingly. Thanks for careful reading. Jul 26, 2020 at 2:09
• Why would the energy difference have a large effect on the kinetics of carbocation shifts, according to arhenius eqn, the k stands pretty close to an order of $10^{10}$
– user96208
Jul 28, 2020 at 18:53
• This is a catalysed reaction, at one point you have all species simultaneously. These are therefore not uncoupled reaction steps (the protonation will influence the whole mechanism). This is a rather simple reaction, but for more complex catalytic transformation, the rate determining step approximation is insufficient. But in a first order approximation, for a simple reaction like this, it works quite well; however, one should be aware that there are limitations. Jul 29, 2020 at 13:58

The RDS for E1 mechanism is the formation of carbocation. The rearrangement steps are fast and are hence not considered in the RDS.

• Didn't I said it in my first sentence? I didn't ask OP to accept my answer. I just give a overall view of the reaction to benefit other students in mind. It was OP's decision. OP can still change it back to where it was. However, keep in mind that answers should be descriptive. Should not look like a comment. This is for future perspectives. Jul 28, 2020 at 19:56
• The concept of rate determining step is a crude approximation and does not always lead to the right conclusions and often does not explain side products. A catalysed mechanism cannot directly be compared with an uncatalysed one. The energetics of all involved species determine the overall kinetics (including all possible steps and paths) of the reaction. While this comment answer might be correct (it better: sufficient) in a basic highschool textbook approach, it is by far too simple for most practical means. Jul 29, 2020 at 13:52
• There is a problem here. The concept of RDS is a crude approximation, like I mentioned. So the key point of the answer should be that the RDS Approximation ist insufficient. It's important to realise that for a catalytic mechanism all steps depend on each other, therefore you cannot separate them and therefore you cannot compare any rate constants, because they simply do not exist on their own. The catalytic cycle is described by one rate law, which is inturn characterised by the entire energetics of the cycle itself. So in conclusion: I say your answer is wrong. Jul 29, 2020 at 19:17
• The question is about a catalytic reaction, discussing it in any other terms is wrong. The kinetics of a reaction is entirely determine by energetics of the intermediates. The rds only works well for linear (consecutive) mechanisms with only elementary steps, and maybe not even there. Reaction mechanisms are usually not as simple as the textbook approaches suggest; discussing the energetics of the reaction is the only way to get close to understanding it. But I can't continue this discussion any further at this point. Jul 29, 2020 at 20:00