Complete the following reaction sequence:

1‐methylidene‐2,3‐dihydroindene‐2‐carboxylic acid reaction scheme



My solution

I was able to complete all the steps except for the last one. This is what I ended up with:


What I'm not getting is how the $\ce{-OH}$ groups are removed and the ring is aromatised. If anyone could provide me with a mechanism for the same, it would be really helpful.

  • 1
    $\begingroup$ Hint: An aldimine forms which reacts with the ketone to form the desired compound. $\endgroup$ – Safdar Sep 3 '20 at 6:30
  • $\begingroup$ Does the aldimine form first and then attack the ketone( which means my final product may be incorrect) $\endgroup$ – sai-kartik Sep 3 '20 at 6:37
  • $\begingroup$ That's would be rather easy part - simple elimination. Perhaps NH3 misled you about pH, as it's acid catalysed? $\endgroup$ – Mithoron Sep 3 '20 at 14:26
  • $\begingroup$ @Mithoron yes, i would think that was the problem.. I got it now, though! $\endgroup$ – sai-kartik Sep 3 '20 at 16:10

This would be the mechanism I propose for the reaction:

Proposed mechanism

I have taken the $\ce{pH}$ of the medium to be $4.5$. That should explain the protonation of the $\ce{-OH}$ groups.

  • $\begingroup$ Your final product is missing a methyl group. And I wouldn't draw both at the same time (but that's nitpicking). By the way, you should use the N lone pair to kick out the water next to it, then deprotonate. $\endgroup$ – orthocresol Sep 3 '20 at 17:08
  • $\begingroup$ Thanks for pointing it out @orthocresol! I shall make the changes.. $\endgroup$ – sai-kartik Sep 3 '20 at 18:04
  • $\begingroup$ @orthocresol would it be correct now? $\endgroup$ – sai-kartik Sep 3 '20 at 18:27
  • $\begingroup$ Yup, that’s pretty much what I’d have done, I think. $\endgroup$ – orthocresol Sep 3 '20 at 18:57
  • $\begingroup$ Also the last step should be removing of $\ce{H3O+}$ and not $\ce{H2O}$ but I think that was understood.. $\endgroup$ – sai-kartik Sep 3 '20 at 18:59

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