Química Orgânica 2 · Síntese e transformação de aminas
Fischer indole synthesis: follow the two nitrogens
Condições: Acetophenone + phenylhydrazine, Brønsted acid (HA; for example acetic acid with catalytic H₂SO₄), heat. H₃O⁺/water explicitly represents the acid/solvent proton shuttle; the counterion and bulk solvent are omitted. The selected reactive branch uses ene-hydrazine protonated at the aryl-bound nitrogen for the [3,3] rearrangement, not every protonation equilibrium. Hydroxyl protonation/water loss and the aminal solvent relay group successive events, not elementary kinetics. Acetophenone has only one enolizable α carbon; the two arene ortho sites are equivalent. Ph abbreviates only the unchanged acetophenone phenyl group. Hydrazone E/Z equilibration, transient aminal stereochemistry and competing reactions are not assessed. Ammonia is trapped as NH₄⁺, consuming one acid equivalent overall. The linked heavy-atom 3D panel uses seven milestone shapes and displays the nearest prior milestone, not every intermediate.
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Choose the attacking nitrogen
Track both nitrogens throughout: the nitrogen that first bonds to the carbonyl carbon is not the nitrogen retained in the indole ring.
Activate acetophenone with acid
Brønsted acid increases carbonyl electrophilicity. This selected reactive encounter uses unprotonated phenylhydrazine, not the full distribution of acid–base species in solution.
As setas curvas
- A carbonyl oxygen lone pair accepts the hydronium proton.
- The hydronium O–H pair stays on oxygen, giving water.
Add the terminal nitrogen
The carbinolamine adduct has a neutral OH and a four-bonded, positively charged terminal nitrogen. The N–N bond remains intact.
As setas curvas
- The terminal nitrogen lone pair bonds to the carbonyl carbon.
- The C=O π pair moves onto oxygen.
Neutralize the carbinolamine nitrogen
A solvent-mediated proton transfer gives the neutral carbinolamine and regenerates a hydronium ion. Every transferred hydrogen retains its atom identity.
As setas curvas
- Water accepts one proton from the terminal ammonium nitrogen.
- The N–H pair restores the terminal nitrogen lone pair.
Protonate hydroxyl and release water
Hydroxyl protonation and water departure share a bookkeeping scene, as two successive acid-catalyzed events rather than a claim of a single elementary step. The original ketone oxygen is now in the water ledger.
As setas curvas
- The hydroxyl oxygen accepts the hydronium proton.
- The hydronium O–H pair returns to oxygen.
- The terminal nitrogen lone pair forms the C=N π bond.
- The C–O pair leaves with the protonated hydroxyl group as water.
Reveal the neutral hydrazone
The acetophenone phenylhydrazone is Ph–C(CH₃)=N–NH–Ph. Condensation has eliminated one water; the acid catalyst remains available for tautomerization.
As setas curvas
- Water accepts the remaining proton on the positively charged imine nitrogen.
- The N–H pair stays on the imine nitrogen, giving the neutral hydrazone.
Which carbon becomes indole’s 3-position?
The original carbonyl carbon becomes indole’s 2-position and retains its phenyl substituent; the methyl carbon becomes the 3-position. The two ortho positions of the unsubstituted phenylhydrazine ring are equivalent.
Protonate the hydrazone nitrogen
Protonating the hydrazone activates it for acid-catalyzed tautomerization. The next carbon deprotonation makes an ene-hydrazine, not a free carbanion.
As setas curvas
- The imine nitrogen lone pair accepts a hydronium proton.
- The hydronium O–H pair stays on oxygen.
Form the ene-hydrazine
The ene-hydrazine has Ph–NH–NH–C(Ph)=CH₂ connectivity. One hydrogen on the α carbon remains implicit and one is drawn explicitly for the final aromatization.
As setas curvas
- Water accepts an α proton from the methyl carbon.
- The α C–H pair forms the C=C bond.
- The C=N π pair returns to the terminal nitrogen.
Activate the aryl-bound nitrogen
This pedagogical branch uses the ene-hydrazine protonated at the aryl-bound nitrogen. The six-atom rearrangement array is ortho carbon–ipso carbon–aryl-bound nitrogen–terminal nitrogen–carbonyl-derived carbon–α carbon; the drawing places those atoms around an open six-membered loop.
As setas curvas
- The aryl-bound nitrogen lone pair accepts the hydronium proton.
- The hydronium O–H pair stays on water oxygen.
Rearrange through six atoms
The acid-promoted [3,3]-sigmatropic rearrangement breaks N–N and makes C–C in one six-electron reorganization. The ring is temporarily dearomatized: the ortho carbon still bears its hydrogen, and the aryl-bound nitrogen is now the cationic member of a diimine.
As setas curvas
- The arene π pair forms the new bond from the ortho carbon to the α carbon.
- The ene-hydrazine π pair forms the side-chain C=N bond.
- The N–N σ pair forms the ipso C=N bond as the N–N bond breaks.
Restore the benzene ring
Loss of the ortho proton restores the benzene ring. The aryl-bound nitrogen is now an aniline nitrogen with two hydrogens and a lone pair; the original terminal nitrogen is the side-chain imine nitrogen. Their N–N bond does not reform.
As setas curvas
- Water accepts the proton on the ortho carbon that acquired the new C–C bond.
- The ortho C–H pair restores the ring π bond.
- The ipso C=N π pair returns to nitrogen, giving an aniline NH₂ group.
Activate the imine for ring closure
Protonation makes the imine carbon more electrophilic. The neighboring aniline nitrogen can attack intramolecularly to close a five-membered ring.
As setas curvas
- The side-chain imine nitrogen accepts a hydronium proton.
- The hydronium O–H pair stays on oxygen.
Close the five-membered ring
The new bond from the aryl-bound nitrogen to the original carbonyl carbon completes the fused five-membered ring. This cyclic aminal is a protonated 2-amino-2-phenylindoline: the ring nitrogen is ammonium and the original terminal nitrogen is a neutral exocyclic NH₂ group.
As setas curvas
- The aniline nitrogen lone pair bonds to the original carbonyl carbon.
- The iminium π pair returns to the exocyclic nitrogen.
Relay a proton to the departing nitrogen
A water-mediated proton relay neutralizes the ring nitrogen and converts the exocyclic NH₂ group into an NH₃⁺ leaving group. The two solvent transfers are grouped for bookkeeping, not asserted to be a single elementary event.
As setas curvas
- Water accepts one proton from the ring ammonium nitrogen.
- The ring N–H pair restores the ring nitrogen lone pair.
- The exocyclic nitrogen accepts a proton from the solvent relay.
- The relay O–H pair returns to water oxygen.
Which nitrogen leaves as ammonia?
Allen and Wilson’s 1943 ¹⁵N-labeling work established this nitrogen provenance: the aryl-bound nitrogen is retained in indole, while the terminal nitrogen is eliminated as ammonia.
Expel ammonia
Ammonia leaves, not NH₂⁻. The immediate ring product is an indoleninium ion; the carbon derived from the methyl group still has two hydrogens, so the five-membered ring is not aromatic yet.
As setas curvas
- The ring nitrogen lone pair forms a C=N π bond.
- The exocyclic C–N pair leaves with the original terminal nitrogen as neutral ammonia.
Aromatize the five-membered ring
Deprotonation at the carbon that becomes indole’s 3-position gives aromatic 2-phenylindole and regenerates hydronium. The ring N–H bond remains intact; removing that proton is not the aromatization step.
As setas curvas
- Water accepts the remaining explicit proton on the carbon that becomes indole’s 3-position.
- The C–H pair forms the C=C π bond in the five-membered ring.
- The C=N π pair returns to nitrogen, whose lone pair participates in indole aromaticity.
Capture ammonia in acid
In acidic solution ammonia is trapped as NH₄⁺. The acid catalyzes the skeletal transformations, but one acid equivalent is consumed in ammonium formation; its counterion is not drawn.
As setas curvas
- The ammonia lone pair accepts a hydronium proton.
- The hydronium O–H pair stays on oxygen, giving water.
2-Phenylindole: trace the two nitrogens
The product is 2-phenylindole. Phenylhydrazine supplies the fused benzene ring and the indole N–H nitrogen. Acetophenone supplies the 2-position (its original carbonyl carbon, still bearing phenyl) and the 3-position (its original methyl carbon, now CH). The ketone oxygen left in water. The terminal phenylhydrazine nitrogen was lost as NH₃ and is shown captured by acid as NH₄⁺. All explicit hydrogens remain accounted for in the product and solvent ledger; indole’s pyrrolic N–H is retained.
O que você vai praticar
Turn acetophenone and phenylhydrazine into 2-phenylindole through hydrazone formation, ene-hydrazine tautomerization, a [3,3] rearrangement, cyclization and ammonia loss, tracking both nitrogen atoms and every explicit proton.
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