Azo coupling with arenediazonium salts - why isn't nitrogen gas lost?

Question

Arenediazonium compounds contain a very good leaving group - nitrogen gas - and thus undergo nucleophilic substitution reactions rapidly. With water, for instance, they give phenols.

In the formation of azo dyes from arenediazonium salts, however, the arenediazonium cation acts as an electrophile to carry out a substitution reaction on activated benzene rings like those of phenols and anilines. Why doesn't displacement of $\ce{N2}$ happen here as well? Normally the carbon attached to $\ce{-N2+}$ acts as the electrophilic centre, so phenols and anilines should react at that $\ce{C}$ rather than at the $\ce{-N2+}$ group. Moreover, phenols and anilines have more nucleophilic centres at their $\ce{O}$ and $\ce{N}$ than at the carbon atoms of their benzene ring - so why aren't the final products be diphenyl ether with phenol, and N-phenylaniline with aniline?

Answer 1

A simple explanation would be that the displacement of a diazonium group by oxygen to form a phenol or phenyl ether linkage typically requires heating, indicating that it is a slower process than the very fast room temperature coupling reaction at aromatic carbon to form an azo dye. Why? One explanation is that the diazonium cation and the phenol or amine being coupled are a "soft" acid and base respectively, whereas water is a "hard" base. But a substituted phenol or aniline that could not form an azo dye might well form a phenyl ether with a diazonium salt, as you propose.

Answer 2

Another possibility is that the aryl nucleophiles that form azo dyes are bulky and attack on the aromatic ring is stetrically hindered. The exposed nitrogen in the diazonium function offers an alternative that is reasonably electrophilic ($sp$ hybridization + nitrogen instead of carbon --> more electronegative) and has more room.