Need plausible reaction mechanism here. (Partial responses also VERY welcome)

[SirRoderick]

In relation to a project I'm doing at university. I'm stumped on the mechanism of the reactions displayed here.

 

 

So the first is a rearrangement in sulphuric acid. Afterwards we added crushed ice (well, we add it TO crushed ice of course)

Now I don't have a scanner right here so I can't upload my sketches. But it was suggested to me that it might be easier if I flipped over the starting molecule left-to-right. (so the NOH group is on the right) He then suggested that perhaps the N forms a bond with the ring structure now to the left. This would be joined by a protonation of the O group wich would split off as water. I'm not sure this is entirely plausible.

 

I'm quite simply stumped on the reaction mechanism here, which is required for my presentation, well I need to at least have a plausible thought lets say.

 

Now as for the second reaction, it is done in presence of NaEtO in ethanol (dry), which I assume acts as a base and attacks the proton on NOH. This would then through resonance allow for the addition reaction of a nitrosobenzene. My question there is, what the hell happens to the O in the Ph-N=O that is added?

 

Any help would be most appreciated. I really am out of practice with this sort of thing.

Edited May 29, 2011 by SirRoderick

[opsomath]

Something seems very odd here. The starting material is C₁₄H₁₁O₂N, and the "rearrangement" product is C₁₄H₁₁ON. In other words, this reaction represents the loss of O, and is therefore a reduction. You do not have any reducing agents listed, just sulfuric acid: did you leave anything out?

 

Also, I infer you add nitrosobenzene in the second step. This reaction is simple enough; ethoxide deprotonates the N-oxide, the double bond on the indole adds at the 3-position to N=O, and you get a bond. Proton shuffle to protonate the resulting N-O^- compound, then another equiv. of EtO- takes off the 3-indole proton and eliminates OH^- in an E2 reaction between the 3-indole carbon and the nitrosobenzene nitrogen, and you get a double bond between the two atoms. The net result is the formation of a double bond and water as a byproduct. Very similar to the condensation of an arylamine and a nitrosobenzene to form an azobenzene and water.

Edited June 30, 2011 by opsomath

[Horza2002]

Are you sure about the product of the first step? In your starting material, you don't have any carbons with hydrogens bound to then (you have the ketone and iminie carbons), but in the product you do....it is possible a rearrangement/migration goes on, but not completely sure.