While preparing diazonium salts, what prevents the freshly formed diazonium salt from reacting with the still unreacted amines left in the solution and perform diazo coupling?

If this happened then there won't be an appreciable yield of the diazo salt.

So what's the reason?

  • 2
    $\begingroup$ Reaction do not have 100% yield in reality.. $\endgroup$ Aug 30, 2020 at 7:54
  • $\begingroup$ Yes but if coupling happens then percentage of diazonium salt left would be practically negligible right? $\endgroup$ Aug 30, 2020 at 7:59
  • $\begingroup$ Why do you say so? Have you attempted to extract a diazo compound from a solution and faced this problem? (I assume this is a query for something you think should happen, I haven't done this either so I cannot answer for methods of extraction) But, a goal of synthesis is to maximise yield and so I would assume that a simple way would be that you add enough of the reagent to minimise the amount of amines present in the final solution, maximising yield of the diazo compound. $\endgroup$ Aug 30, 2020 at 8:04
  • 1
    $\begingroup$ The NO+ species is more reactive than the diazonium salt so the amine reacts with that if it is present. $\endgroup$
    – Waylander
    Aug 30, 2020 at 8:51
  • 2
    $\begingroup$ The excess of nitrous acid and the order of addition are indeed important. But I am surprised nobody mentioned pH. It needs to be kept around 5-6 to allow enough unprotonated amine to be present and react with nitronium, while preventing side reactions, e.g. addition of water to the newly formed diazonium cation, or indeed addition of free amine to it. That, as far as I can recall from my organic chemistry days, is the most important parameter to consider. $\endgroup$ Aug 30, 2020 at 14:31

1 Answer 1


During diazonium ion preparation, first, sodium nitrite is mixed with hydrochloric acid to produce nitrous acid. The nitrous acid can be protonated under acidic conditions, which undergoes the loss of water and produce the nitrosonium ion ($\ce{O=N+}$; this reaction is usually performed at or around $\pu{0 ^\circ C}$):

  • An aromatic amine can attack the electrophilic nitrosonium ion to form a nitrosoamine $(\ce{Ar-NH-N=O})$.
  • The nitrosoamine can undergo a proton transfer (tautomerism) to give corresponding imminol $(\ce{Ar-N=N-OH})$.
  • Under acidic conditions, this $\ce{N=N-OH}$ group can be protonated and left as a water molecule to give the diazonium salt, which is resonance stabilized $(\ce{(Ar-N#N)+Cl-})$.

The key word here is all of these happen under acidic conditions. Therefore, the aromatic nucleus of aniline derivative is not active because of protonation. Also, during the synthesis, the starting aniline derivative is used as the limiting reagent. In addition, many of diazonium salts are unstable and hence, they are always prepared as needed under acidic conditions with good stirring, kept at or near $\pu{0 ^\circ C}$ to minimize the side reactions such as the reaction with water to produce a corresponding phenol until used in the coupling reaction (usually immediately).

The diazonium salt reacts as an electrophile with an electron-rich coupling component (a phenol or an aniline derivative), through an electrophilic aromatic substitution mechanism under basic conditions (specially when coupling reagent is a phenol derivative). To complete the synthesis of some azo dyes, it is also required heating.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.