Aromatisation of a ring which includes nitrogen

Question

Complete the following reaction sequence:

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.

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

This would be the mechanism I propose for the reaction:

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

• 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. – orthocresol Sep 3 at 17:08
• Thanks for pointing it out @orthocresol! I shall make the changes.. – sai-kartik Sep 3 at 18:04
• @orthocresol would it be correct now? – sai-kartik Sep 3 at 18:27
• Yup, that’s pretty much what I’d have done, I think. – orthocresol Sep 3 at 18:57
• Also the last step should be removing of $\ce{H3O+}$ and not $\ce{H2O}$ but I think that was understood.. – sai-kartik Sep 3 at 18:59