Química Orgánica 2 · Síntesis y transformación de aminas
2,5-Dimethylpyrrole by the Paal–Knorr synthesis
Condiciones: NH4OAc, AcOH, Δ
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Arrastra desde los electrones que se mueven hasta donde van, o toca uno y luego el otro. Cada flecha se revisa; Pista te ayuda cuando te atascas. No necesitas cuenta.
El mecanismo, paso a paso
Siguiente reproduce las flechas curvas de cada paso y las explica. Los pasos escritos están abajo.
Leer los 14 pasos escritos
Ammonia adds to a ketone
Ammonium acetate supplies ammonia and acetic acid. Ammonia attacks one ketone carbon and the C=O π pair moves onto oxygen.
Las flechas curvas
- The ammonia lone pair forms the C–N bond.
- The C=O π pair moves onto oxygen.
Acetic acid protonates the alkoxide
Acetic acid, the acid present, protonates the alkoxide.
Las flechas curvas
- The alkoxide takes the proton.
- The O–H pair stays on oxygen as acetate.
Acetate removes an N–H proton
Acetate deprotonates the ammonium nitrogen, giving a neutral amino alcohol whose nitrogen is free to attack again.
Las flechas curvas
- Acetate takes the N–H proton.
- The N–H pair returns to nitrogen.
Nitrogen closes the ring
The nitrogen attacks the second ketone carbon, forming a five-membered ring, the favoured ring size.
Las flechas curvas
- The nitrogen lone pair forms the ring C–N bond.
- The C=O π pair moves onto oxygen.
Protonate the new alkoxide
Acetic acid protonates the alkoxide.
Las flechas curvas
- The alkoxide takes the proton.
- The O–H pair stays on oxygen.
Acetate removes the N–H proton
Acetate deprotonates the ring nitrogen, giving the neutral 2,5-dihydroxypyrrolidine.
Las flechas curvas
- Acetate takes the N–H proton.
- The N–H pair returns to nitrogen.
Protonate the first OH
Acetic acid protonates one OH so it can leave as water.
Las flechas curvas
- The OH oxygen takes the proton.
- The O–H pair stays on oxygen.
Lose water: first iminium
The ring nitrogen lone pair pushes out water, giving an iminium ion.
Las flechas curvas
- The nitrogen lone pair forms the C=N π bond.
- The C–O pair leaves with water.
Deprotonate α to the iminium
Acetate removes a ring C–H next to the iminium carbon, giving an enamine (the first ring C=C).
Las flechas curvas
- Acetate takes the C–H proton.
- The C–H pair forms the C=C π bond.
- The C=N π pair moves onto nitrogen.
Protonate the second OH
Acetic acid protonates the remaining OH.
Las flechas curvas
- The OH oxygen takes the proton.
- The O–H pair stays on oxygen.
Lose water: second iminium
The nitrogen lone pair pushes out the second water.
Las flechas curvas
- The nitrogen lone pair forms the C=N π bond.
- The C–O pair leaves with water.
Driving force
The last step creates something special.
Deprotonate: aromatic pyrrole
Acetate removes the last ring C–H next to the iminium carbon; the new C=C completes the aromatic ring.
Las flechas curvas
- Acetate takes the C–H proton.
- The C–H pair forms the C=C π bond.
- The C=N π pair moves onto nitrogen.
2,5-Dimethyl-1H-pyrrole
2,5-Dimethyl-1H-pyrrole forms with two waters; acetic acid is regenerated.
Qué vas a practicar
Draw ammonia adding to one ketone, ring closure of the nitrogen onto the second ketone, and the two dehydrations and deprotonations that give 2,5-dimethyl-1H-pyrrole.
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