Reaction between 1,7‐dichloroheptan‐4‐one and alcoholic KOH

The given answer is 1,1-dicyclopropylmethanone:


Step 1 is abstraction of acidic H by the strong base. Then, there's surely going to be anchimeric assistance. But there can be two places where the first carbanion can attack: either at the nearer carbon to form the cyclopropyl group or at the far end one to form a five-membered ring. If it does the former, then this process will be repeated and another cyclopropyl group would be made.

Why not make a more stable five-membered ring in the first step, and then again perform E2 to get 2-ethenylcyclopentan-1-one?


  • $\begingroup$ Entropy requirements for the transition states maybe? 5-Chloro-2-pentanone also forms the cyclopropane under these conditions: orgsyn.org/demo.aspx?prep=cv4p0597 $\endgroup$ – Dennis Cao Jun 26 '18 at 0:38
  • $\begingroup$ Carbanions are very unstable species I think the product will be kinetically controlled. $\endgroup$ – Avnish Kabaj Jun 26 '18 at 6:29
  • 2
    $\begingroup$ @Mithoron the answer is not wrong, the dicyclopropylketone can be isolated in ~55% yields: orgsyn.org/demo.aspx?prep=CV4P0278 $\endgroup$ – Dennis Cao Jun 26 '18 at 15:55
  • $\begingroup$ @DennisCao Is it possible that the carbonyl carbon here conjugates with the cyclopropyl rings and that conjugation leads to stability which makes it the thermodynamic product? $\endgroup$ – Archer Jul 14 '18 at 18:23
  • $\begingroup$ @Abcd hmm I doubt that would be significant enough, and we should be thinking about the transition state stability not the product stability $\endgroup$ – Dennis Cao Jul 14 '18 at 18:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.