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My attempt

In the first case:
$\ce{H+}$ adds to the $\ce{OH}$ group, giving us a carbocation. The carbocation thus formed is exceptionally stable due to back bonding. I wonder why would it go under ring expansion even though the strain is not a factor here as the ring strain in a cyclobutane ring is ~$26.3\ \mathrm{kcal/mol}$, and that in a cyclopropane ring is ~$27.5\ \mathrm{kcal/mol}$.

In the second case:
Again the $\ce{H+}$ adds to the $\ce{OH}$ group, giving us a tertiary carbocation with seven hyper-conjugating structures. Why would it go under ring expansion to give secondary carbocation with just two hyper-conjugating structures? I believe is based on ring strain in this case, as the ring strain in a five-membered ring is ~$6.2\ \mathrm{kcal/mol}$, while the ring strain in a six-membered ring is ~$0.1\ \mathrm{kcal/mol}$.


Source: Advanced Problems In Organic Chemistry, MS Chouhan, 11th edition; Chapter - Hydrocarbons (Alkenes); Question 180 in latest edition

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You overlooked the strain from the sp2 carbon in a small ring. The ring strain you have listed is based on sp3 carbon. sp2 supposed to have a planar triangular geometry (120°, compared to 109° for sp3) and will be more strained in a 3 or 4 membered ring. 6 membered ring will be best followed by 5 membered ring.

Also, you have discussed a lot about the stability of carboncations. Acid catalyzed elimination/rearrangement reaction is a thermodynamic condition. The stability of product is more important than the stability of intermediate.

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  • $\begingroup$ Then kindly explain why the following contraction takes place?here $\endgroup$ – yasir Jun 13 '15 at 14:11
  • $\begingroup$ I will expect the carboncation after ring contraction reaction to be trapped by some nucleophile instead of elimination product. With no $sp^{2}$ carbon center involved in the product, the ring strain data will make sense in this case. Thus, resonance structure stabilization will overcome the ring strain. This reaction should be a reaction somewhat under kinetic condition with careful control, and it may not happen under general thermodynamic condition(just heat up without any nucleophile), where I will expect ring opening to happen. $\endgroup$ – Ian Fang Jun 13 '15 at 14:33
  • $\begingroup$ i agree with you for the first case. But can you please elaborate on the second compound.As you said the stability of the product is of major concern and not the intermediate. So in case of thermodynamic product, whenever we have the chance to expand the ring from 5 membered to 6, the process will be favoured ? $\endgroup$ – yasir Jun 14 '15 at 12:20

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