Diazotisation of 1,2-diaminobenzene

Usually, $\ce{NaNO2}$ and $\ce{HCl}$ are the reagents used to generate an aryldiazonium chloride from an aniline. However, in this compound there are two amine groups adjacent to each other, so it seems unlikely that a diazonium chloride would form. Is it possible for the two amine groups to react during the reaction and form some cyclic product?


1 Answer 1


Your suspicion that a ring forms is a good one. You can convert one of the amine groups to the diazonium chloride:

Diazotisation of one amine in 1,2-diaminobenzene

However, in addition to substitution reactions by loss of $\ce{N2}$, diazonium compounds will react with nucleophiles at the terminal nitrogen atom. The best known examples of this behavior are diazo couplings, where the nucleophile is an activated arene. However, other nucleophiles, especially intramolecular nucleophiles, can attack diazonium cations. The other amine is poised to do this:

Cyclisation to form benzotriazole

  • $\begingroup$ I have a doubt sir. Why does the nucleophilic attack happen on the diazonium chloride by the other amine group rather than conversion of both amine group to diazonium chloride? $\endgroup$ Sep 12, 2013 at 12:56
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    $\begingroup$ Certainly that could happen, but it is not likely. The formation of the first diazonium is probably much faster than the formation of the second. Consider that the process involves incremental increase in formal charge $\ce{0->+1->+2}$. Additionally, intramolecular reactions are generally faster (no diffusion or mass transport required). Formation of the second diazonium is an intermolecular nucleophilic attack of the second amine on a $\ce{NO2^+}$ ion. Formation of a ring is an intramolecular nuclephilic attack. $\endgroup$
    – Ben Norris
    Sep 13, 2013 at 13:45
  • $\begingroup$ @BenNorris But isn't the geometry a problem: The $\ce{N2^{+}}$ group and the $\ce{NH2}$ group are help away from one another by the ring without much freedom to bend. Wouldn't this hinder the intramolecular reaction or are the $\pi^{*}$ orbital of $\ce{N2^{+}}$ and the lone pair on $\ce{NH2}$ just pointing well enough towards each other that there is not much need for freedom of movement? $\endgroup$
    – Philipp
    Sep 13, 2013 at 16:51
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    $\begingroup$ @Philipp, I agree the geometry appears to be unhelpful, but the reaction works $\endgroup$
    – Ben Norris
    Sep 13, 2013 at 17:27
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    $\begingroup$ I did a recent Chem Abst search on the cyclization of the diazonium salt to the benzotriazole. No hits but that doesn't in itself prove anything. There were hits on conversion of the diamine to the benzotriazole by diazotization. The former diazotization used mineral acid while the latter used acetic acid or montmorillonite clay. Is there a change in mechanism? Might the cyclization occur earlier than formation of the diazonium salt, perhaps at the -NH-N=O stage? This would avoid the strain issue for cyclization. The formulation of the question is suspect IMHO. $\endgroup$
    – user55119
    Mar 10, 2019 at 23:25

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