# What happens in the attempted 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?

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:
• 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. – Ben Norris Sep 13 '13 at 13:45
• @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? – Philipp Sep 13 '13 at 16:51