[SirRoderick]

Right, so I posted this a while back I think in relation to a project I'm doing at university. I'm still stumped on the mechanism of the reaction displayed here (top one)



So it is a rearrangement in sulphuric acid.
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.

Now as for the second reaction, it is done in presence of NaEtO, 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.

[orgopete]

I answered before, but I guess you thought I was wrong. This is what I think the problem should have been, semi-mechanism included. I cannot think of how the posted reaction could take place. The second reaction requires some new bond cleavage reactions. Could there be an error? Since the second reaction really, really looked like an error, I was skeptical of the first as well. Check my post. So this is how I thought the reaction might plausibly take place, assuming some missing information.

[bolo]

Surely there is an OH group missing. I suppose there is at first an nitroso-enol formed which undergoes an attack from the benzene ring.