Question

Explain Hoffmann’s exhaustive alkylation with suitable reactions.

Answer 1

Hofmann’s exhaustive alkylation of amines:

1. When a primary amine is heated with excess of primary alkyl halide it gives a mixture of secondary amine, tertiary amine along with tetraalkylammonium halide. This can be given as,
   \[\ce{\underset{\text{1° Amine}}{R - NH2} ->[R - X][-HX] \underset{\text{2° Amine}}{R2NH} ->[R - X][-HX] \underset{\text{3° Amine}}{R3N} ->[R - X][-HX] \underset{\text{Tetraalkyl ammonium halide}}{R4N+X-}}\]
2. If the excess alkyl halide is used tetraalkylammonium halide is obtained as a major product and the reaction is known as exhaustive alkylation of amines.
3. Tetraalkylammonium halides or quaternary ammonium salts are the derivatives of ammonium salts in which all the four hydrogen atoms attached to nitrogen in N⁺H₄ are replaced by four alkyl groups (same or different).
4. Tetraalkylammonium halides are crystalline solids.
5. Primary, secondary and tertiary amines consume three, two and one moles of alkyl halide respectively to get converted into quaternary ammonium salt.
6. The reaction is carried out in presence of mild base NaHCO₃, to neutralize the large quantity of HX formed.
7. If the alkyl halide is methyl iodide, the reaction is called exhaustive methylation of amines.
   e.g. When methylamine is heated with excess methyl iodide, it gives tetramethyl ammonium iodide.
   \[\ce{\underset{\text{Methylamine}}{CH3 - NH2} + \underset{\text{Methyl iodide}}{CH3 - I} ->[\Delta] \underset{\text{Dimethylamine}}{(CH3)2NH} + HI}\]
   \[\ce{(CH3)2 - NH + CH3 - I ->[\Delta] \underset{\text{Trimethyl amine}}{(CH3)3N} + HI}\]
   \[\ce{(CH3)3N + CH3 - I ->[\Delta] \underset{\text{Tetramethyl ammonium iodide}}{(CH3)4N+I-}}\]