Química Orgánica 2 · Sustitución de acilo, aminas y sales de diazonio
Pent-1-ene by Hofmann elimination of pentan-2-amine
Condiciones: CH3I (excess), NaOH; then Ag2O, H2O; Δ
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Siguiente reproduce las flechas curvas de cada paso y las explica. Los pasos escritos están abajo.
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Third methylation gives the quaternary salt
Two SN2 methylations of pentan-2-amine, each followed by removal of the new N–H proton, give the tertiary amine. Its nitrogen lone pair displaces iodide from a third iodomethane, giving the quaternary ammonium iodide, which has no N–H left.
Las flechas curvas
- The nitrogen lone pair forms the third N–CH3 bond.
- The C–I pair leaves as iodide.
Hofmann E2
Ag2O in water supplies hydroxide as the counter-ion (the iodide ends up as insoluble AgI; it is drawn as a spectator). On heating, hydroxide removes a hydrogen from the methyl group (C1) anti to the C–N bond while trimethylamine leaves in one concerted step. The bulky NMe3⁺ group and the early, carbanion-like transition state favour the least hindered β-hydrogen.
Las flechas curvas
- Hydroxide takes the β-hydrogen on the terminal methyl.
- The C–H pair forms the new C=C π bond.
- The C–N pair leaves with neutral trimethylamine.
Why the terminal alkene
With a trimethylammonium leaving group the less substituted alkene dominates (Hofmann rule).
Pent-1-ene
Pent-1-ene is the major product, with trimethylamine and water. Anti-Zaitsev (Hofmann) regiochemistry is the hallmark of quaternary ammonium eliminations.
Qué vas a practicar
Excess CH3I with NaOH first converts pentan-2-amine into N,N-dimethylpentan-2-amine. Draw the third SN2 methylation to the quaternary ammonium salt and the Hofmann E2 of the quaternary ammonium hydroxide that gives pent-1-ene, the less substituted alkene.
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