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chemical reaction

For the carbocation (A) given in the reaction, first we can rearrange it by ring expansion (4 to 5) then we have two choices either H-shift or methyl-shift. My teacher told me that Hydrogen is the best migrator so according to me product (D) should form but in my textbook (C) is given as final product ?? Where am I wrong ? (or is my textbook wrong ?)

Also I have been taught that whenever there is a choice among H-shift, alkyl-shift or phenyl-shift we prefer H-shift since H is the best migrator. So why methyl-shift is happening here ? Please explain.

What's the concept I am missing ??

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  • $\begingroup$ C is tertiary carbocation whereas D is a secondary carbocation. $\endgroup$ Dec 29, 2021 at 14:51

4 Answers 4

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Your question is an example of Cationic Rearrangements. When a reaction is set to progress under correct condition for thermodynamic control, the final product is based on the thermodynamic stability of intermediates. The migratory aptitude matters if the intermediate stability increased by the migration (see the examples given in here).

For instance, in $\bf{A} \rightarrow \bf{B}$ rearrangement, there are two migratory choices: hydride shift or alkide shift. Although hydride is superior to alkide in migratory aptitude, why is alkide shift dominating here to give the ring expansion? Because it gives the most stable intermediate releasing ring strain energy (5-menbered ring versus 4-membered ring). Now, in intermediate $\bf{B}$ rearrangement to $\bf{C}$ (by methide shift) or $\bf{D}$ (by hydride shift), it has two choices as indicated. Again, although hydride is much superior to methide in migratory aptitude, it prefers methide shift because it would give more stable tertiary carbocation instead of energetically similar secondary carbocation by hydride shift.

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Yours textbook answer is correct. We can't just say that hydrogen is a best migrator. Shifting of groups depends on the stability of transition state and final product.
In this example, you can observe that carbocation in C is a tertiary carbon with 6 hyperconjugative structures whereas in D carbocation is a secondary carbon with only 4 hyperconjugative structures. Therefore, carbocation in C is more stable than in D. Hence compound C is the major intermediate.

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This is essentially a question of thermodynamic control versus kinetic control. The statement "my teacher told me that Hydrogen is the best migrator" has to do with the migration barrier, which is related to kinetic control. Since H atom is very light, its vibrational frequency can be high, which may facilitate the migration. Being a light atom, sometimes H can even display wave character and therefore can undergo quantum tunneling without needing to fully climb up the reaction barrier. However, the yield of final products is mainly determined by the stability of the products, which is related to thermodynamic control. As long as you let the reaction take place for sufficient length of time, you will get the more stable structure as the main product, unless the barrier connecting it with the reactant is prohibitively high at the given reaction temperature.

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Image is taken from http://www.chem.ucalgary.ca/courses/351/Carey5th/Ch10/ch10-3-3.html

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  • $\begingroup$ As long as C and D are not interconvertible under the reaction conditions, then both C and D are formed under kinetic control. If C and D are in equilibrium with each other, perhaps through reprotonation of the olefin related to D, then the reaction is under thermodynamic control. A thermodynamic product may be formed under kinetic control even though it is not being formed under thermodynamic conditions. Base elimination of 2-bromobutane to give the Zaitsev product (2-butene) is a case in point. 1-Butene and 2-butene are not in equilibrium under the reaction conditions. $\endgroup$
    – user55119
    Dec 30, 2021 at 22:07
  • $\begingroup$ Good point! But I noticed a difference b/w the base elimination of 2-bromobutane and the cationic rearrangements. Once HBr is eliminated, it is dissipated into the solvent, making the equilibrium between 1- and 2-buene unlikely. But cationic rearrangement is different in that the molecule does not lose any atoms. Not sure if a direct conversion between C and D by [1,3] H-shift is possible (probably too far). However, indirect conversion between C and D, e.g. through B, will lead to an equilibrium between C and D. Consequently, the reaction shown above most likely is thermodynamic control. $\endgroup$
    – Josiah_H
    Dec 30, 2021 at 22:54
  • $\begingroup$ Under basic conditions there is no HBr because it is neutralized. In the solvolysis (SN1) of 2-bromobutane, the alkenes formed are under thermodynamic control owing to the presence of HBr. See this. @Vishal Anand $\endgroup$
    – user55119
    Dec 30, 2021 at 23:24
  • $\begingroup$ That makes sense (under basic conditions there is no HBr). It may be the reason that the less stable 1-butene can be formed by kinetic control. Good to know that the presence of HBr can lead to thermodynamic control, despite the fact that it leaves the molecule during elimination. (It may be a different situation in case H2 is formed but pumped away by vacuum.) In any case, the cationic rearrangement shown above is most likely thermodynamic control. $\endgroup$
    – Josiah_H
    Dec 30, 2021 at 23:30
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Carbonium ion rearrangement results in a more stable carbonium ion and probably happens in the actual formation of the carbonium ion not in a time delayed subsequent reaction. If this is the case there is little sense in first looking at the migrating moiety but to first look for the most stable carbonium ion then second to look for the better migrating group to give that carbonium ion. In your reaction the ring opening is probably driven by loss of ring strain with an aid from the formation of the tertiary carbonium ion by methyl migration. If there is formation of the secondary carbonium ion there is no energetic reason for rearrangement to another secondary but there is for rearrangement to the tertiary. These are my impressions on these reactions based on some limited experience and some long almost forgotten book learning.
I found your reaction. It seems the secondary cation is formed blowing my one step proposal out of the water. https://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/rearrang.htm Time to check the original data.

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