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My book (see bottom) implies that hydrohalogenation of 4-methyl-cyclohexene follows this mechanism:

mechanism

Why not undergo consecutive hydride shifts to furnish a tertiary carbocation?

I would guess that the second step, nucleophilic attack by bromine, is much faster and prevents this from happening. But when I try to confirm this by looking up some examples I see this one at masterorganicchemistry.com:

hydride shift example

This implies the following very similar mechanism:

hydride shift example mechanism

The carbocation in the example is a smaller ring, meaning it should be less stable and even more reactive (less able to perform hydride shifting.) But then again the charge is closer to the tertiary carbon and easier accessible.

Could someone explain what is happening?

Vollhardt, P.; Schore, N. Molecules: Structure and Function, 6th edition; W. H. Freeman: New York, 2014​, p491 exercise 12-6:

Exercise 12-6

Predict the outcome of the addition of HBr to (a) 1-hexene; (b) trans-2-pentene; (c) 2-methyl-2-butene; (d) 4-methylcyclohexene. How many isomers can be formed in each case?

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  • $\begingroup$ It's an exercise and the book only gives the starting material as "4-methyl-cyclohexene" and in the back the two structures which I drew in my pictures (1-bromo-[4/3]-methyl-cyclohexane.) $\endgroup$
    – ttdijkstra
    Jan 21, 2017 at 13:20
  • $\begingroup$ Maybe the textbook is just wrong? $\endgroup$
    – DHMO
    Jan 21, 2017 at 14:12
  • $\begingroup$ I'll settle for that for now. $\endgroup$
    – ttdijkstra
    Jan 21, 2017 at 15:31

1 Answer 1

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The reason for not having consecutive hydride shifts is that generally, a new carbocation is formed through shift only when it is more stable than the immediately previous one.

To form a tertiary carbocation, 3 consecutive hydride shifts would have to occur consecutively (when + is at c-4 from the alkyl group, your first product) in which more than one intermediates have same stability (same number of alpha-H). Hydride shifts will not occur for your second product to form tertiary carbocation as the next intermediate is infact less stable (3 alpha-H as compared to 4 alpha-H previously).

enter image description here

Since the ring already has 6 members (least strain) so ring expansion/contraction will not occur either.

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  • $\begingroup$ On second thought, there will also be (+)ve inductive effect on the c+ in my first product, increasing it's stability but I think that HyperConjugation has greater effect on stability of carbocation than inductive effect so loss of alpha-H should have a greater effect. $\endgroup$
    – Tushar Pandey
    Jan 21, 2017 at 17:02
  • $\begingroup$ Please check out the mechanism of squalene cyclisation which includes four consecutive methyl and hydride Wagner-Meerwein rearrangements, neither of which is truly downhill. $\endgroup$
    – Jan
    Jan 21, 2017 at 19:08
  • $\begingroup$ For rearrangement of carbocation there should be some driving force like increase in stability of carbocation. $\endgroup$
    – narendra kumar
    Jan 22, 2017 at 14:17

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