10
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enter image description here

There are two possibilities:

  • Oxonium ion formation
  • Cleaving of epoxide

Firstly, I don't know if epoxide is more reactive than carbonyl group. If epoxide reacts with the proton, enter image description here

But the given answer is:

enter image description here

In the above mechanism, an unstable carbocation is formed because its near a carbonyl group which is electron withdrawing:

enter image description here

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    $\begingroup$ The carbocation formation don't occurs. It's a concerted step. $\endgroup$ – Koba Dec 20 '15 at 15:28
  • $\begingroup$ Why? How do you know it has to be concerted? $\endgroup$ – Aditya Dev Dec 20 '15 at 15:29
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    $\begingroup$ You are correct in your last structure. This carbocation is very unstable, because this the mechanism is concerted. See the two products, whats more stable? $\endgroup$ – Koba Dec 20 '15 at 15:40
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    $\begingroup$ Both are equally stable? Are you referring steric repulsion? But the product in my answer is stabilized by conjugation which is not there in the answer given. $\endgroup$ – Aditya Dev Dec 20 '15 at 15:43
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    $\begingroup$ In your answer the second carbocation seems unstable and the final product in your answer is more unstable in relationship the correct answer. I think that the product that you propound are a subproduct. $\endgroup$ – Koba Dec 20 '15 at 15:58
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I would go with your approach but your final product is unstable due to high repulsion's by $\pi$ electrons.

enter image description here

Hence your final product will react further to form a more thermodynamically feasible product.

enter image description here

NOTES:
In second structure two anti-periplanar migrations are possible and the one which leads to the more stable carbonation is favored (or consider the better migrating group based on electronegativity). Third structure rotates about spiral carbon due to repulsion's.

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  • $\begingroup$ Is any kind of solvent required for such a rearrangement? $\endgroup$ – Aaron John Sabu May 6 '17 at 14:33

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