Fischer indole synthesis: practice and mechanism
Phenylhydrazine and the ketone form a hydrazone, which rearranges and cyclises to the indole with loss of NH₃. The carbonyl carbon becomes C2 of the indole, the α-carbon C3.
Example
Phenylhydrazine forms a phenylhydrazone with the ketone. It tautomerises to the ene-hydrazine, which undergoes a [3,3]-sigmatropic rearrangement (new C–C bond to the ring). After ring closure and loss of NH₃ the aromatic indole forms; the carbonyl carbon becomes C2.
Mechanism step by step
- 1
Phenylhydrazine and the ketone form a phenylhydrazone, which tautomerises to the ene-hydrazine.
- 2
After protonation the ene-hydrazine undergoes a [3,3]-sigmatropic rearrangement: the N–N bond breaks and a C–C bond forms between the ring and the α-carbon.
- 3
Rearomatisation, ring closure of the aniline N onto the imine carbon and loss of NH₃ give the aromatic indole.
Practice tasks
Which product forms, which reagent is needed? After each answer you see the explanation.
Which main product forms?
Typical mistakes
- The phenylhydrazone is only the first intermediate. With acid and heat the [3,3]-rearrangement and ring closure to the indole follow.
- The groups are swapped: the former carbonyl carbon becomes C2 of the indole (next to NH) and carries the group; the α-carbon becomes C3.
Often confused with: Wolff–Kishner reduction · Imine formation · Oxime formation
Frequently asked questions
Which ketones work?
Ketones (and aldehydes other than acetaldehyde) with at least one α-CH₂ or α-CH₃. Unsymmetrical ketones with two different α-sides give mixtures.
Where do I meet the indole skeleton?
In tryptophan, serotonin, melatonin and many alkaloids and drugs (e.g. triptans for migraine).