What I'm missing is how it goes from: +N2-chain-COOH to a ring?
You are vey close to the answer. The problem is your carbanion on C6. You do lose N2, but this doesn't leave an anion. Draw the curly arrow from the N-C bond in +N2-CH2.. to show where the electrons in that bond go when you lose nitrogen gas.
Do the electrons go to C, or the +N2 group?
Given the movement of electrons, what charge would you expect on C6?
Which part of this molecule could now attack C6?
I'VE GOT IT NOW!!!
Okay, so first off the amine group undergoes diazonization and after an elongated process winds up as a terminal +N2. Then the N2 breaks off (TAKING ITS ELECTRONS WITH IT) and making C6 a
carbocation! Now, since the oxygen is deprotonated in aqueous solution, the negative from the oxygen goes after the positive terminal carbocation and forms the ring!
Heh. I've been toiling over this for days.
Someone verify my answer?
EDIT: I also noticed when going over diazonization that the reaction has to take place at 0 degrees? What happens to the rxn if it isn't cooled? Why does it have to be cooled in the first place?
My best guess would be that the rxn proceeds too quickly if left to its own devices and the NO2 gas becomes a hazard.