Selectivity of diazo-coupling with p-amino phenol I came across this question and I've been thinking about it since, without being able to find a convincing major product. This question is related to one I found on this site, Diazo Coupling reaction with para-substituted phenol?, but it doesn't totally clear my doubt, and hence I specifically posted this reaction.

My thoughts were, since benzene diazonium is a weak electrophile, it would prefer to attack the most nucleophilic or electron rich site, and $\ce{-NH2}$ being better nucleophile than $\ce{-OH}$, I feel that would be more preferred site of coupling. Also I found out that aniline reacts with diazonium in basic medium to form diazo-aminobenzene which later rearranges to form more stable amino azobenzene; only to support my opinion. Thus my product was:



But the actual answer given was ortho coupling w.r.t. to $\ce{-OH}$, and it supports the product shown in the previously mentioned question, and @orthocresol commented that it occurs due to formation of phenoxide in basic medium: 



But still it wasn't very convincing for me, as no one posted a solid answer to the question, neither are their any suitable articles I found on the internet that compare the reactivities of different functional groups towards diazonium. 
I would be very obliged if someone could provide a deeper insight to the matter, and clear my misconceptions(if any), and a critical view about the barriers or aids in formation of the products.
I think the right way to approach this question isn't "we don't know the product, let's try to reason it out from first principles"; rather it should be more like "we found that this is the product when we ran the reaction in the lab, how can we rationalise this observation using a model of nucleophilicity that we know?"

So, in that previous comment which you quote, I did not mean to say "phenol will be deprotonated, hence this product will definitely be formed"; rather I meant something more like "if this is indeed the correct product, then a possible reason is because the phenol is deprotonated, making it (or the positions ortho to it) more nucleophilic than the aniline".

That said, I did just look it up, and there are indeed some references from the primary literature which support the claim that the product above is formed.





So, since we have experimental evidence (and insofar as this evidence is sound), you should not doubt what the product is; the only question left is why it is formed. And so I offer you the same explanation: that phenol is deprotonated under the basic conditions, and that phenoxide is more nucleophilic than aniline (at the ortho positions).

As for why phenoxide is more nucleophilic than aniline: I could offer you some wishy-washy explanations, like how a negative charge makes the oxygen atom more electron-donating towards the ring (and that outweighs the natural nucleophilicity ranking of N > O), but I don't think there is a simple one-line answer which is sufficiently rigorous for me to accept. I tried to look for some other evidence, but it is very difficult to compare C-reactivity of phenoxide vs aniline, since most EAS reactions take place in acidic media. There are some studies which conclusively show that phenoxide-O is more nucleophilic than aniline-N, but that isn't what we're looking for either.

So... all in all, it is better to use this experimental observation (that the azo coupling occurs ortho to oxygen) as proof that phenoxide-C is more nucleophilic than aniline-C, rather than to try to do it the other way round, i.e. "phenoxide-C is more nucleophilic than aniline-C, hence it must be true that this reaction will form this product". After all, chemistry is first and foremost an experimental science.

It's a subtle difference, but adopting the correct scientific approach will be better for you as a chemist in the future.