Química Orgánica 2 · Sustitución de acilo y elección del derivado
Phthalimide from phthalic anhydride and ammonia
Condiciones: NH3, Δ
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El mecanismo, paso a paso
Siguiente reproduce las flechas curvas de cada paso y las explica. Los pasos escritos están abajo.
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Ammonia adds to a carbonyl
Ammonia attacks one carbonyl carbon of phthalic anhydride; the C=O π pair moves onto oxygen.
Las flechas curvas
- The ammonia lone pair forms the new N–C bond.
- The C=O π electrons move onto oxygen.
Ring opens to the carboxylate
The O⁻ re-forms C=O and the C–O bond to the ring oxygen breaks, opening the anhydride. One carbon is now an amide (still protonated), the other a carboxylate.
Las flechas curvas
- An oxygen lone pair re-forms C=O.
- The ring C–O bond breaks; electrons go to the carboxylate oxygen.
Ammonia deprotonates the amide nitrogen
A second ammonia removes the N–H proton, giving the neutral amide and ammonium phthalamate.
Las flechas curvas
- The ammonia lone pair bonds to the N–H proton.
- The N–H pair returns to nitrogen.
Ammonium protonates the carboxylate
On heating the ammonium salt hands its proton to the carboxylate, giving phthalamic acid and free ammonia.
Las flechas curvas
- A carboxylate lone pair bonds to the ammonium proton.
- The N–H pair returns to nitrogen, releasing ammonia.
Amide nitrogen closes the ring
With heat, the amide nitrogen attacks the neighbouring acid carbonyl carbon. The five-membered ring forms easily because the two groups are held side by side on the benzene ring.
Las flechas curvas
- The amide nitrogen lone pair forms the new N–C bond, closing the ring.
- The C=O π electrons move onto oxygen.
Ammonia removes the N–H proton
Ammonia deprotonates the positively charged ring nitrogen of the tetrahedral intermediate.
Las flechas curvas
- The ammonia lone pair bonds to the N–H proton.
- The N–H pair returns to nitrogen.
Ammonium protonates the OH
The ammonium ion protonates the OH on the tetrahedral carbon, turning it into water, a good leaving group.
Las flechas curvas
- An OH lone pair bonds to the ammonium proton.
- The N–H pair returns to nitrogen.
Water leaves as C=O re-forms
The O⁻ lone pair re-forms the carbonyl and expels water, completing the cyclodehydration to the imide.
Las flechas curvas
- An oxygen lone pair re-forms C=O.
- The C–O bond breaks; water leaves.
Why the ring closes
Ring closure forms a five-membered imide whose N–H sits between two carbonyls.
Phthalimide
Phthalimide forms with water, and ammonia is regenerated. The imide N–H is acidic (pKa about 8.3) because the anion is delocalised onto two carbonyls, which is what the Gabriel synthesis exploits.
Qué vas a practicar
Turn phthalic anhydride into phthalimide: ammonia opens the anhydride to the amide-acid, then heating closes the imide ring with loss of water.
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