Hello I was thinking about the formation mechanism of the double condensation to form nile blue. However, I am stuck, as I can't explain why the condensation of the hydroxyl moiety happens.
In this example, I suppose the nitroso group gets protonated, and the electrophilic nitrogen atom is attacked, and condensation happens, as well as the reformation of the aromatic ring. But why does the hydroxy group also get condensated and thus forming a fused ring?
You are correct that the first step of the mechanism involves protonation of the nitroso group. Be aware that protons are being exchanged on the amines during the course of the condensation. There are a number of subtleties in the mechanistic pathway that may be invoked but I have selected what, in my experience, is a likely route. Be forewarned, there is a surprise in store.
Condensation of aryl nitroso compound 1 and β-naphthol 2 affords intermediate 3, which rearomatizes to oxime 4. Dehydration of the oxime may well be stepwise but, in order to save space, I have shown it as a concerted process to form quinoid structure 5. The dehydration of 4 could lead to the double bond stereoisomer of 5 but that product would be sterically hindered. Addition of the phenolic oxygen to the quinoid moiety provides 6 that rearomatizes to tetracycle 7. Here is the surprise! The leuco-form of the dye is formed, not the dye itself. continued
Let's presume that the reaction is not conducted under an inert atmosphere. Then oxygen is a likely suspect. The ground state of oxygen is a triplet, which can abstract an electron from the amino group of 7 to form radical cation 8 and the radical anion of oxygen 9. The latter species is protonated by acid to afford the hydroperoxy radical that abstracts a hydrogen atom from radical cation 8 to produce the tautomeric forms, 10 and 11, of the dye.
