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Heterocycle synthesis — sigmatropic

Fischer indole synthesis

How the Fischer indole synthesis works, in 3 steps. Each step shows the structures and the curved arrows that move the electrons.

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In short. A 3-step mechanism. Typical reagents: Phenylhydrazine, ketone (acetone), acid (ZnCl₂, PPA, H₂SO₄, AcOH). Conditions: 80–180 °C, Brønsted or Lewis acid. Bonds formed: C–C (aryl–CH₂); C–N (ring); aromatic indole; bonds broken: N–N; C=O; C–NH₂. Stereochemistry: None in product; regiochemistry from ene-hydrazine formation. Rate and selectivity: [3,3]-shift is rate-limiting and irreversible.

At a glance

ReagentsPhenylhydrazine, ketone (acetone), acid (ZnCl₂, PPA, H₂SO₄, AcOH)
Solvent / conditions80–180 °C, Brønsted or Lewis acid
Bonds formedC–C (aryl–CH₂); C–N (ring); aromatic indole
Bonds brokenN–N; C=O; C–NH₂
StereochemistryNone in product; regiochemistry from ene-hydrazine formation
Rate / selectivity[3,3]-shift is rate-limiting and irreversible

Step 1: Hydrazone, then ene-hydrazine

+H⁺−H₂OH⁺, tautomeriseNH₂NHONNHNNHNHNH1 · HYDRAZONE, THEN ENE-HYDRAZINEphenylhydrazineC₆H₈N₂NUCLEOPHILEacetoneC₃H₆OELECTROPHILEphenylhydrazoneC₉H₁₂N₂INTERMEDIATEphenylhydrazoneC₉H₁₂N₂INTERMEDIATEene-hydrazineC₉H₁₂N₂INTERMEDIATEImine ⇌ enamine tautomerism puts C=C next to N.

A phenylhydrazine condenses with a ketone, losing water to form a phenylhydrazone; acid then tautomerizes it to the more reactive ene-hydrazine form.

Step 2: [3,3]-sigmatropic shift

[3,3]NHNHNHNH2 · [3,3]-SIGMATROPIC SHIFTene-hydrazine (chair-like arrangement)INTERMEDIATEbis-imineC₉H₁₂N₂INTERMEDIATEConcerted, like the Claisen/Cope: C–C forms to the ortho carbon while N–N breaks.

This ene-hydrazine undergoes a [3,3]-sigmatropic rearrangement — the same concerted, closed-loop arrow-pushing pattern as the Claisen and Cope rearrangements you already know — forming a new C–C bond directly to the aromatic ring.

Step 3: Rearomatise, cyclise, lose NH₃

H⁺, tautomerise(ring aromatic)5-exo attackH⁺, −NH₃+NHNHNH₂NHNH₂NHNH₂NHNHNH₃3 · REAROMATISE, CYCLISE, LOSE NH₃bis-imineC₉H₁₂N₂INTERMEDIATEo-aminophenyl imineC₉H₁₂N₂INTERMEDIATEcyclic aminalC₉H₁₂N₂INTERMEDIATEcyclic aminalC₉H₁₂N₂INTERMEDIATE2-methylindoleC₉H₉NPRODUCTammoniaH₃NBYPRODUCTElimination of NH₃ from the aminal creates the aromatic pyrrole ring of the indole.

Loss of ammonia and ring closure (an intramolecular condensation) aromatizes the five-membered ring, giving an indole — a core scaffold found throughout natural products and pharmaceuticals.

Abbreviations and terms

AcOH
Acetic acid. Weak acid and polar protic solvent.
H2
Hydrogen gas. Reductant for catalytic hydrogenation.
Lewis acid
Lewis acid. Electron-pair acceptor (AlCl₃, BF₃, TiCl₄…) that activates carbonyls and halides.
H2SO4
Sulfuric acid. Strong acid and dehydrating agent.

How it compares with related reactions

Fischer indole synthesisBischler indole synthesisBischler-Napieralski reaction
ReagentsPhenylhydrazine, ketone (acetone), acid (ZnCl₂, PPA, H₂SO₄, AcOH)Aniline (excess), α-haloketone (PhCOCH₂Br)β-Arylethylamide, POCl₃ (or P₂O₅, Tf₂O)
Conditions80–180 °C, Brønsted or Lewis acidStep 1: 25–80 °C; step 2: acid (or aniline·HBr), 150–200 °CToluene or MeCN, reflux
StereochemistryNone in product; regiochemistry from ene-hydrazine formationNoneProduct is a 3,4-dihydroisoquinoline; none created
Rate / selectivity[3,3]-shift is rate-limiting and irreversibleCyclisation (high T) is slow; SN2 alkylation is fastNitrilium formation then SEAr; electron-rich arenes react best

Test yourself

Try to answer each question before opening it.

Which reagents and conditions are typical for the Fischer indole synthesis?

Phenylhydrazine, ketone (acetone), acid (ZnCl₂, PPA, H₂SO₄, AcOH). Conditions: 80–180 °C, Brønsted or Lewis acid.

Which bonds are formed and which are broken?

Formed: C–C (aryl–CH₂); C–N (ring); aromatic indole. Broken: N–N; C=O; C–NH₂.

What is the stereochemical outcome?

None in product; regiochemistry from ene-hydrazine formation

What controls the rate or the selectivity?

[3,3]-shift is rate-limiting and irreversible

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