Heterocyclic
Intermediate
Fischer indole synthesis of (3-ethylphenyl)hydrazine and 4-methyl-2-pentanone
Question
(3-Ethylphenyl)hydrazine is heated with 4-methyl-2-pentanone in ethanol containing conc. HCl at reflux. Draw the Fischer indole mechanism and give the major product. Explain which ortho position of the aryl ring forms the new C–C bond.
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Expert answer
The major product is 6-ethyl-3-isopropyl-2-methyl-1H-indole.
Hydrazone formation (⇌): the hydrazine adds to the protonated ketone, and loss of water gives the hydrazone.
Tautomerisation (⇌): the hydrazone tautomerises to the ene-hydrazine, with H₂O or ethanol as the base. The more substituted ene-hydrazine forms towards the CH₂ in strong acid, which gives the 2-methyl-3-isopropyl pattern.
N-protonation (⇌).
[3,3]-sigmatropic rearrangement (→): the N–N bond breaks as a new C–C bond forms between the ene-hydrazine carbon and an ortho carbon of the aryl ring. This is the key step.
Rearomatisation (→): loss of H⁺ restores the aromatic ring.
Cyclisation (⇌): the NH₂ attacks the imine carbon to form a cyclic aminal.
Loss of NH₃ (→): ammonia leaves as NH₄⁺, and the aromatic indole forms. This step is irreversible.
Regiochemistry: the ethyl group is meta to the hydrazine, so there are two ortho positions. One sits between the N and the ethyl group, and the other is on the far side, away from the ethyl. The [3,3] rearrangement favours the less hindered ortho position (the one away from the ethyl group), because the chair-like transition state is less crowded there. This puts the ethyl at C6 of the indole. Attack at the hindered position would give the 4-ethyl isomer as a minor product.