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Pathway 1: Rate determining step is the formation of carbocation. In this pathway, a conjugated tertiary carbocation is generated as an intermediate.

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Pathway 2: One of the resonance form of intermediate carbocation proceeds to produce a more stable product (due to benzene ring) through alkyl migration. Though, the resonating form of carbocation is conjugated secondary carbocation which is less stable than the carbocation formed through pathway 1.

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My questions are:

1- Is tert-butyl (don't know the correct term) migration possible in pathway- 2?

2- What are the major and minor products? Which factor will decide the major product: stability of final compound or the most stable intermediate?

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    $\begingroup$ The reactant/intermediate is clearly not aromatic, but the product is. What about the $\ce{\Delta H, \Delta G}$ of reaction? $\endgroup$
    – Rahul Verma
    Aug 8, 2020 at 10:40
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    $\begingroup$ This reaction is under thermodynamic control. All of steps are reversible but the last step to form aromatic compound. Hence, aromatic compound is the major. The product from first path is still susceptible to protonate under acidic conditions. $\endgroup$
    – Mathew Mahindaratne
    Aug 8, 2020 at 16:26
  • $\begingroup$ " resonating form of carbocation is conjugated secondary carbocation which is less stable than the carbocation formed through pathway 1" - that's wrong on many levels. $\endgroup$
    – Mithoron
    Aug 8, 2020 at 21:38
  • $\begingroup$ I have never seen a migration of a t-butyl group before. That seems sterically problematic. $\endgroup$
    – Zhe
    Aug 8, 2020 at 22:08
  • $\begingroup$ @Zhe Migratory aptitude might depend on the reaction. According to wikipedia for pinacol rearrangement the order is Hydride > $\ce{(CH3)3}$> Phenide > $\ce{C2H5}$ > $\ce{CH3}$ and per this site for Baeyer-Villiger rearrangement the order is 3º-alkyl > 2º-alkyl ~ benzyl ~ phenyl > 1º-alkyl > methyl (depends on electron density). $\endgroup$
    – Apurvium
    Aug 9, 2020 at 6:24

1 Answer 1

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Pathway 2 is unlikely. The tert-butyl group stays put, i.e., it remains attached to the same carbon throughout the reaction. It is the rest of the molecule that rearranges. The mechanism of this reaction was demonstrated 70 years ago. Compare this reaction with Possible nonclassical ion from a bicyclic system on ChemSE. There is information at the link on labeling to elaborate the mechanism.

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  • $\begingroup$ (a) I have gone through your linked answer (but could not access the references mentioned therein) and what I understood is that isotope labeling can decide the correct mechanism between the two discussed above. So, is it determined in the cited paper? (b) In your mechanism, is there any driving force between (1) and (2)? Or this is not important and final stable product decides it? $\endgroup$
    – Apurvium
    Aug 9, 2020 at 7:19
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    $\begingroup$ (a) In 1950 they didn't use isotopic labeling which is one way of testing the mechanism. They expected 4 but got 6 in the link. That result led to the spiro mechanism. (b) The cation 2 is resonance stabilized. The migratory aptitude of methyl is less than RCH2-. chemistry.stackexchange.com/questions/116051/… $\endgroup$
    – user55119
    Aug 9, 2020 at 11:56
  • $\begingroup$ Sorry, I should have asked for the driving force between (2) and (3). $\endgroup$
    – Apurvium
    Aug 9, 2020 at 12:04
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    $\begingroup$ There is not much difference (strain) between a 5- ans 6-membered ring. The delocalized cation is the same.in 2, 3 and 4. Aromatization drives the reaction. $\endgroup$
    – user55119
    Aug 12, 2020 at 15:14

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