5
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enter image description here
The compound given above is subjected to the following reaction (in that order):

  1. $\ce{t-BuOK}$;
  2. $\ce{CHCl3}$ and $\ce{CH3ONa}$;
  3. $\ce{H3O+}$ and heat.

Find out the final product.

Source

Attempt:

The question involves exocyclic double bond which is less stable than endocyclic double bond, and so I get that the compound should rearrange to five-membered ring containing a double bond.

In the first reaction tertiary butoxide is a bully base, so it should bring about elimination. Hence $\ce{MeOH}$ will be removed and a double bond will be formed:

enter image description here

However, I could not get how the reaction proceeds. Any ideas?

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  • $\begingroup$ Disregard previous comment. Carbene formation requires loss of bromide group... $\endgroup$ – Zhe Oct 31 '17 at 18:45
  • $\begingroup$ Are you sure you drew that bromide in the right place in the product? $\endgroup$ – Zhe Oct 31 '17 at 18:47
  • $\begingroup$ @Zhe I thought that since endocyclic double bond is more stable than exocuclic double bond the ring should expand $\endgroup$ – user471651 Nov 1 '17 at 1:57
  • $\begingroup$ You still haven't answer my question. Are you sure that the bromide is in the 3-position and not the 2-position? $\endgroup$ – Zhe Nov 1 '17 at 13:11
  • $\begingroup$ @Zhe I am not sure and I am not able to think of any mechanism of conversion of exocyclic double bond to endocyclic double bond. $\endgroup$ – user471651 Nov 1 '17 at 13:50

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