I was told to propose a mechanism for the following conversion:


One obvious step is the conversion of cyclic ketal to ketone because of acidic conditions. But I am unable to understand how the three membered ring opened to give an alkene? (However, I can understand that the three membered ring is unstable)

I feel that somehow, a carbocation is being generated and E1 type elimination is taking place.

What is happening exactly? And does this type of thing take place with any three membered ring under given conditions?

  • $\begingroup$ Break the three mbered ring to form an enol. Rearrange the carbocation and heat it to form an alkene in the mean while your enol tautomerizes to a ketone in acidic medium. Havent the solutions been released along with the paper though? $\endgroup$ Feb 4, 2019 at 18:22
  • $\begingroup$ @AvnishKabaj No solution was given with the paper. $\endgroup$ Feb 5, 2019 at 1:15
  • $\begingroup$ Since you get hydrolysis of the ketal, you should list the hydronium ion over the arrow. With just a proton as a catalyst, the ring cleavage can occur with the retention of the ketal. Temporary opening of the ketal to the enol ether and protonation of the ketone causes fragmentation. The ketal can then reform. $\endgroup$
    – user55119
    Feb 5, 2019 at 2:26

1 Answer 1


Possible mechanism for cyclopropyl ring opening of given compound in acid medium:

Possible mechanism for cyclopropyl ring opening

  • $\begingroup$ Is there any name to the step in which the alkene attacks to open the cyclopropyl ring? Is it related to Neighbouring Group Participation? $\endgroup$ Feb 5, 2019 at 1:26
  • $\begingroup$ Not really to my understanding. But, you see when the first enol formed, it was conjugated all the way to the other ketone (you can write few resonance to see how stabilizing it is). $\endgroup$ Feb 5, 2019 at 2:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.