I found an organic chemistry question which asked the major product of the following reaction -

enter image description here

The mechanism for the same in the answer was -

enter image description here

My questions regarding this are as follows -

  1. The molecule has two oxygen atoms, one as an epoxide and the other as ketone. How do we decide which oxygen atom accepts the proton? Seeing the mechanism provided in the answer, I reasoned that the epoxide oxygen is sp3 hybridized, hence less electronegative than the ketone oxygen, which is sp2 hybridized. Hence as the less electronegative atom it donates its lone pair to the proton. Am I correct? If not, what is the real reason?

  2. Acid-catalyzed opening of epoxide follows SN1 mechanism, which implies more stable carbocation must be formed. In the given mechanism, however, the carbocation is formed on the right side, where the carbonyl group exerts -I effect and destabilizes the carbocation. I could not make sense of this.

Are there any specific rules as to which site is attacked by the proton if multiple nucleophilic sites are present in the molecule? Also, is my reasoning in (2) correct?

I am also open to the fact that the given mechanism in the image may be wrong, in which case, I would also like to know the correct mechanism and product.

  • 1
    $\begingroup$ The second point is incorrect. The cation gians stability due to +R effect of the conjugated carbonyl. $\endgroup$ Apr 28 '21 at 10:44
  • 1
    $\begingroup$ @NisargBhavsar Thank you for your comment. You mean the oxygen atom on the carbonyl group acquires a positive charge by +R effect, while stabilizing the cation with a double bond between the carbonyl carbon and the cation? Positive charge on an oxygen atom seems a bit counterintuitive... $\endgroup$
    – TRC
    Apr 28 '21 at 10:49
  • 1
    $\begingroup$ Yes the +ve charge on oxygen is not the best but the delocalisation it provides is definitely better than nothing. $\endgroup$ Apr 28 '21 at 10:52
  • $\begingroup$ @NisargBhavsar Thank you so much for the clarification. It would be helpful if you could please let me know if my reasoning in (1) is correct or not $\endgroup$
    – TRC
    Apr 28 '21 at 10:55
  • 2
    $\begingroup$ I think for your point 1, you are overcomplicating stuff. If you protonate the ketone, then you can't draw any mechanism from that point, but if you protonate the epoxide, you can open the ring. Minor differences in electronegativities of oxygens is probably a secondary factor. $\endgroup$
    – S R Maiti
    Apr 28 '21 at 19:01

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