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(a) Why would cyclohexane expand into cycloheptane? Cyclohexane is stable.

(b) Why would cyclopentane not ring expand into cyclohexane? In the process a tertiary carbocation would be formed just like with the methyl shift, but you would also get rid of ring strain.

(c) Why is there only one product? Wouldn't the double bond have hydrogen addition on both carbon sides leading to a potential cyclobutane to cyclohexane expansion?

(e) Once again like in (c), why is there only one product shown?

(f) Wouldn't there be two products once more?

POTENTIAL EXPLANATION: By examining more mechanisms in class, I have noted that potentially the methyl/alkyl shifts could be occurring in concert with the hydrobromination so perhaps that is the reason why.

  • $\begingroup$ Maybe ask one question at a time? For (b): there is no ring strain for cyclopentane, so your premise is incorrect. To combine a couple of the questions, you could ask what determines whether a carbocation undergoes hydride shift, alkyl shift or phenyl shift, and what happens when more then one of these is possible. $\endgroup$
    – Karsten
    Nov 10, 2019 at 21:45

1 Answer 1


In example A, you are right that the cyclohexane ring is more stable than the cycloheptane ring. There are two choices for rearrangement in this case:

1) Ring expansion to form the cycloheptane ring

2) 1,2-hydride shift to yield the tertiary alkyl bromide

The second product will be more stable and thus be the major product

In example B, you are right, a ring expansion is possible, so there should be the ring expansion product too. You will have a mixture of products in this case (1,2-alkyl shift, ring expansion, and no carbocation rearrangement)

In examples C, E and F, it is possible for the H-A to add across the double bond either way, so again, a mixture of products would be obtained

A lot of times, only one product is shown for addition reactions such as hydrohalogenation; however, it is important to remember that there are often multiple products possible.

Chemists very rarely utilize reactions that involve a carbocation rearrangement because a mixture of products is obtained, which results in very poor yields

  • $\begingroup$ At least in examples (F), it is not possible for $\ce{HX}$ to add across the double bond either way, because protonation followed by hydride transfer gives very stable carbocation ($\ce{Ph2C^.R}$), which followed to give the given product. $\endgroup$ Nov 3, 2019 at 4:28
  • $\begingroup$ @MathewMahindaratne But the initial addition across the double bond should have rather similar transition states for both cases and not be regioselective, right? The hydride transfer does give a very stable carbocation, but that is in subsequent steps and wouldn't be rate-determining. As the initial carbocations formed are both secondary and because the initial process is endothermic they would have similar transition states energies and thus rates. $\endgroup$
    – Eliot Behr
    Nov 3, 2019 at 5:10
  • $\begingroup$ @ebehr: Do not forget the rate determining step is the addition of halo ion in the final step. Until then, all steps are in equilibrium, so as soon as most stable carbo cation formed, it would be attack by halide ion and reaction is done. $\endgroup$ Nov 3, 2019 at 5:19
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    $\begingroup$ @MathewMahindaratne link according to this it isn't though $\endgroup$
    – Eliot Behr
    Nov 3, 2019 at 6:04
  • $\begingroup$ @ebehr: Theoretically it is possible but practically would never happens. Please pay attention to relevant activation energies in the given site (msu.edu). If you pay attention, one is for formation of primary carbo cation. Practically it is impossible. $\endgroup$ Nov 3, 2019 at 6:17

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