Fischer indole synthesis: push the electrons yourself

The Fischer indole synthesis mechanism, step by step

The key [3,3]-sigmatropic rearrangement breaks the N–N bond and forms a C–C bond; this exercise uses the ene-hydrazine protonated at the aryl-bound nitrogen. Acetophenone has only one enolizable α carbon: its methyl carbon becomes indole C3, while its carbonyl carbon becomes C2 and keeps the phenyl group. Allen and Wilson’s 1943 ¹⁵N-labelling work established that the aryl-bound nitrogen is retained and the terminal nitrogen is eliminated as ammonia. Loss of a proton at C3, not removal of the ring N–H, makes the five-membered ring aromatic.

The 20 steps of the demo, in the exercise's own words. Try drawing each one above before you read it.

1. 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.

2. 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.

3. 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.

4. 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.

5. 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.

6. 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.

7. 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.

8. 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.

9. 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.

10. 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.

11. 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.

12. 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.

13. 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.

14. 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.

15. 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.

16. 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.

17. 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.

18. 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.

19. 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.

20. 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.

How the mechanism is drawn

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.