Reaction of p-amino phenol with benzene diazonium salt

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:

My answer- o-amino m-hydroxy azobenzene

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:

Actual answer- m-amino o-hydroxy azobenzene

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.


1 Answer 1


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.

Example one

Example two

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.

  • $\begingroup$ @orthocresol Thank you very much. Yes I do know that chemistry is mostly an experimental science, and that most theories were proposed only to reason the observations. But I was worried because the actual product clashed with my principles, and since I found no literature about the speciality of this reaction, I doubted the answer. So do you suggest doing the same reaction under acidic conditions(most of which are), it could yield a product with ortho coupling w.r.t. -NH2 $\endgroup$ Dec 25, 2019 at 17:22
  • $\begingroup$ @SirArthur7 Are most azo couplings done under acidic conditions...? I'm not aware of that - as far as I know, they are base-catalysed or at least performed in neutral pH. If you acidify too much you will protonate the aniline which kills its nucleophilicity anyway. On top of that, there is another question of whether the diazonium ion reacts ortho to NH2, or with the NH2 group itself to form a triazene (see chemistry.stackexchange.com/q/107344). So I'm afraid I don't quite know how this would turn out, and I don't want to make any suggestions as to what the outcome would be. $\endgroup$ Dec 26, 2019 at 3:40
  • $\begingroup$ Well I just said what I saw on this very site, that aniline undergoes azo coupling under acidic conditions,(chemistry.stackexchange.com/questions/29251/…) And yes I am aware about the possibility of formation of triazene which rearranges to form more stable amino azobenzene(mentioned in the question itself). I asked the question only to ascertain myself, actually the problem is here at school we don't get apparatus or chemicals to run a reaction, we need to make use of pen paper and basic principles. $\endgroup$ Dec 26, 2019 at 4:10
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    $\begingroup$ @SirArthur7 Ah, ok. I'm not very convinced by Klaus's answer there: if you protonate the aniline, then there is no more lone pair in conjugation with the ring, and the ortho position will no longer be nucleophilic. But I looked it up and you are right that azo couplings do take place at acidic pH's as well. I think it is sensible to guess that under (very mildly) acidic conditions, the preferred product comes about via reaction ortho to NH2. But again, I do not want to definitively claim that to be the case without some sort of evidence. $\endgroup$ Dec 26, 2019 at 4:23
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    $\begingroup$ "most people avoid answering, but couldn't understand why" - because we don't fully understand it. The truth is that most things don't have a nice, one-line answer which we can rattle off, and if you are taught otherwise, you should probably take it with a healthy pinch of salt. $\endgroup$ Dec 26, 2019 at 4:27

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