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The relative rate of acid catalysed dehydration of following alcohols would be: enter image description here

According to me, we have to arrange them in order of their carbocation stability.

Q will be the most stable as it is stabilized by Resonance and Hyperconjugation. P should come next. After that I think it should be S>R as in case of R there is no hyperconjugation and hyperconjugation is a “stronger” effect than the Inductive effect.

But in my Book R>S is given. Is Inductive effect stronger here? Or is there some other reason to it?

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    $\begingroup$ R is capable of rearrangement (methide shift) to give a stable tertiary carbocation. S needs two hydride shift to get there. $\endgroup$
    – Mathew Mahindaratne
    Commented Aug 3, 2020 at 8:14
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    $\begingroup$ @Mathew In the dehydration of alcohols,the formation of the first carbocation is the rate determining step(that is,the one with the biggest activation energy), right? Subsequent rearrangements may lead to lower energy intermediates,but shouldn't the stability of the first carbocation(and thereby,the energy of the first transition state) be the governing factor for rate,and not what the cation can potentially rearrange into? $\endgroup$
    – Yusuf Hasan
    Commented Aug 3, 2020 at 10:20
  • $\begingroup$ @Yusuf Hasan: True, but R and S have no choice but make primary carbocation. However, that would be assisted by concomitant migration of methide and hydride migration, respectively. Even though this assistance is significant, the activation energy of each is still higher than P and Q. The activation energy of R is comparatively lower than that of S, since methide migrarion is giving a tertiary carbocation. Meantime, S has to go through two activation barriers (ist one is higher). The first activation barrier decide the rate. $\endgroup$
    – Mathew Mahindaratne
    Commented Aug 3, 2020 at 10:43
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    $\begingroup$ @MathewMahindaratne That poses another issue in my mind: If the migration is concomitant, then why is the activation energy of R significantly greater than P and Q? If such a high difference exists, then it must be because due to the formation of the first primary cation as opposed to P and Q straight off the bat forming secondary and tertiary cations respectively. If this is the case, then why isn't such an initial difference being observed between the energies of primary cation-like transition states of R and S, which should exist if there has to be a big difference b/w R & S and P &Q? $\endgroup$
    – Yusuf Hasan
    Commented Aug 3, 2020 at 11:05
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    $\begingroup$ @Yusuf Hasan: Think this way: In transition state, $\ce{C-^+OH2}$ bond is partially broken. That need very high energy regardless getting assist or not. That bond needed to be partially broken to get this assistance at the first place. $\endgroup$
    – Mathew Mahindaratne
    Commented Aug 3, 2020 at 11:20

2 Answers 2

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Your basic assumptions are correct.

It can be observed as such:

  • In Q the carbocation is stablised by resonance, inductive and hyperconjugative effects so it is quite stable and will be formed fastest.
  • Comparing P and R: In P, a secondary carbocation is formed and there are 4+1 or 5 hyperconjugative structures possible. Moreover, the inductive effect is also greater in P than in R. Hence, it forms carbocation faster.
  • Both R and S form primary carbocations. Inductive effect is more prevelent in R than in S, while S has 3 hyperconjugating structures. But R offers capacity to rearrange and stablise the carbocation, by methyl migration to produce highly stable tertiary carbocation, also stablised by resonance over the adjoining phenyl ring. But, S also forms a carbocation which is primary, but offers no stablising effect other than 2 +1 or 3 hyperconjugating structures.

Similarly, this rearrangement may also be attributed to the comparison of P and R, where hydride shift and methyl shift occur. The migration amplitude of hydride, being lower, leads to rearrangement of positive charge quickly and hence is kinetically also favoured.

Thus finally the order may be:

Q>P>R>S

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  • $\begingroup$ Won't both R and S rearrange to form the same carbocation? $\endgroup$
    – Robin Singh
    Commented Aug 3, 2020 at 6:48
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    $\begingroup$ R has to have only one rearrangement to form tertiary resonance stablised carbocation. But, carbocation in S is quite isolated and primary. It can be argued that in S the first hydride shift will form secondary carbocation comparatively more stable than the initial. But then a second rearrangement has to occur to get resonance stablised carbocation, similar to Q. Then also inductive effect and hyperconjugation will be higher in R than S in the final carbocation formed after multiple rearrangement steps jn S. $\endgroup$
    – Solid - NMR
    Commented Aug 3, 2020 at 7:01
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    $\begingroup$ Then the whole process of multiple rearrangement will be somewhat tough, and although Hammond Postulate provides for the carbocation rearrangement to occur i.e. from primary, secondary to tertiary and transition states are effectively stablised, the process will be long, tedious and the whole process will make the rate determining step slow. $\endgroup$
    – Solid - NMR
    Commented Aug 3, 2020 at 7:05
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An acid catalysed dehydration reaction of alcohols will always proceed with the removal of -OH as H2O molecule and therefore, will be dependent on the formed carbocation's stability because the rate deternining step for this process is formation of carbocation. Carbocation Carbocation can be stabilized by inductive effect, field effect, mesomeric effect (or) resonance and hyperconjugation effect.

Q should be the fastest because it's corresponding carbocation is the most stable because it is favoured by all the effects. P is the next fastest as it's carbocation gains maximum stability through a hydride shift as the secondary carbocation becomes tertiary carbocation which is stabilised by both resonance with the phenyl group and the hyperconjgation of H in the attached methyl and ethyl group. R would be faster than S because it's carbocation becomes stable after it undergoes a methyl shift which is less favourable than hydride shift but results into similar stability. S shall be the slowest because it has the most unstable carbocation after its formation.

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