Question

Write the complete mechanism for the diazotization reaction of primary amine.

Answer 1

Write the complete mechanism for the diazotization reaction of primary amine.

Answer:

The complete diazotization reaction equation is shown below:

$$\mathrm{RNH_2} + \mathrm{NaNO_2} + 2 \mathrm{HX} \longrightarrow \mathrm{RN^+} = \mathrm{NX^-} + \mathrm{NaX} + 2 \mathrm{H_2O}$$

The R substituent is in most cases the aryl radical due to the low stability of the alkyl diazonium salts.

The complete mechanism involves a stage of a partial protonation of nitrous acid in a solution of a strong acid (HCl, H₂SO₄, HBr, etc.):

I) $\mathrm{HNO_2} + \mathrm{H_3O^+} \rightleftharpoons \mathrm{H} - \mathrm{O} - \mathrm{N} = \mathrm{O} + \mathrm{H_2O}$

On the second stage, protonated nitrous acid reacts with an inorganic anion, present in solution:

II a) $\mathrm{H_2NO_2^+} + \mathrm{NO_2^-} \rightleftharpoons \mathrm{O} = \mathrm{N} - \mathrm{O} - \mathrm{N} = \mathrm{O} + \mathrm{H_2O}$ (reaction in a solution of sulphuric, phosphoric or perchloric acid)

II b) $\mathrm{H_2NO_2^+} + \mathrm{Cl^-} \rightleftharpoons \mathrm{Cl} - \mathrm{N} = \mathrm{O} + \mathrm{H_2O}$ (reaction in a solution of hydrochloric acid)

II c) $\mathrm{H_2NO_2^+} + \mathrm{Br^-} \rightleftharpoons \mathrm{Br} - \mathrm{N} = \mathrm{O} + \mathrm{H_2O}$ (reaction in a solution of hydrobromic acid)

After that, there are a number of fast, consequent reactions, which lead, to the main product of reaction - diazonium salt:

III) $\mathrm{RNH_2} + \mathrm{X} - \mathrm{N} = \mathrm{O} \rightleftharpoons \mathrm{slow} \mathrm{R} - \mathrm{N} - \mathrm{O} + \mathrm{X^-}$

$$\mathrm{X = Br, Cl, ONO.}$$

IV) $\mathrm{R} - \mathrm{N} - \mathrm{N} = \mathrm{O} + \mathrm{H_2O} \rightleftharpoons \mathrm{fast} \mathrm{R} - \mathrm{N} - \mathrm{O} + \mathrm{H_3O^+}$

V) $\mathrm{R} - \mathrm{N} - \mathrm{N} = \mathrm{O} + \mathrm{H_3O^+} \rightleftharpoons \mathrm{fast} \mathrm{R} - \mathrm{N} - \mathrm{N} = \mathrm{O} + \mathrm{H_2O}$

VI) $\mathrm{R} - \mathrm{N} - \mathrm{N} = \mathrm{O} + \mathrm{H_2O} \rightleftharpoons \mathrm{fast} \mathrm{R} - \mathrm{N} - \mathrm{N} = \mathrm{O} + \mathrm{H_3O^+}$

VII) $\mathrm{R} - \mathrm{N} - \mathrm{OH} + \mathrm{H_3O^+} \rightleftharpoons \mathrm{fast} \mathrm{R} - \mathrm{N} - \mathrm{N} = \mathrm{O} + \mathrm{H_2O}$

VIII) $\mathrm{R} - \mathrm{N} - \mathrm{N} - \mathrm{H} \longrightarrow \mathrm{fast} \mathrm{R} - \mathrm{N} = \mathrm{N} + \mathrm{H_2O}$

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