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UPSC Civil Services (Mains)

Chemistry optional

  • 2 papers
  • 250 marks each
  • 3 h per paper
  • 5 of 8 questions, Q1 and Q5 compulsory

Paper II covers organic chemistry and spectroscopy.

UPSC Civil Services (Mains), Chemistry optional · 2019 · Paper II

Q3(c). Fischer indole synthesis with butan-2-one: two indoles

  • 20 marks
  • Answered
Question

In Fischer indole synthesis, phenylhydrazine upon reaction with 2-butanone in presence of PPA, gives a mixture of two indoles in equimolar ratio. Predict their structures and explain their formation.

Answer in brief

Phenylhydrazine and butan-2-one form one hydrazone but two ene-hydrazines. Each undergoes the [3,3]-sigmatropic shift, ring closure and loss of NH₃: the internal ene-hydrazine gives 2,3-dimethylindole, the terminal one 2-ethylindole, in about equal amounts.

Key points

  • Protonation of the ene-hydrazine by the acid (PPA) activates the N–N bond.
  • [3,3]-Sigmatropic shift (rate-determining, through a six-membered cyclic transition state): the weak N–N bond breaks and a new C–C bond forms between the ortho carbon of the ring and the enamine β-carbon. A di-imine…
  • Tautomerisation re-aromatises the ring and gives an o-aminophenyl imine.
  • The aniline NH₂ attacks the iminium carbon (5-exo ring closure): a 2-aminoindoline (aminal).

The full model answer

Written out in full, as it should go in the answer booklet. The solution page with its diagrams follows it.

Introduction. In the Fischer indole synthesis an aryl hydrazone of an enolisable ketone is heated with an acid (PPA, ZnCl₂, H₂SO₄). The key step is a [3,3]-sigmatropic rearrangement of the ene-hydrazine tautomer. Butan-2-one is an unsymmetrical ketone with two different α-carbons (CH₃ and CH₂), so two ene-hydrazines, and therefore two indoles, are possible.

Step 1: hydrazone and the two ene-hydrazines

Scheme: Condensation at the C=O.

Scheme: Imine–enamine tautomerism can use either α-carbon.

Step 2: [3,3]-shift, cyclisation and loss of ammonia (same for both)

  1. Protonation of the ene-hydrazine by the acid (PPA) activates the N–N bond.
  2. [3,3]-Sigmatropic shift (rate-determining, through a six-membered cyclic transition state): the weak N–N bond breaks and a new C–C bond forms between the ortho carbon of the ring and the enamine β-carbon. A di-imine (cyclohexadienone-imine) results.
  3. Tautomerisation re-aromatises the ring and gives an o-aminophenyl imine.
  4. The aniline NH₂ attacks the iminium carbon (5-exo ring closure): a 2-aminoindoline (aminal).
  5. Acid-catalysed loss of NH₃ and a proton gives the aromatic indole.

Figure: Sequence common to both ene-hydrazines.

The two indoles

Scheme: From ene-hydrazine (a) and (b); both C₁₀H₁₁N.

  • From ene-hydrazine (a), formed towards the CH₂ group (CH₃-C(NHNHPh)=CH-CH₃): that CH carbon becomes C-3 and carries its CH₃, and the former C=O carbon becomes C-2 with its CH₃: 2,3-dimethylindole.
  • From ene-hydrazine (b), formed towards the CH₃ group (CH₂=C(NHNHPh)-CH₂CH₃): C-3 is unsubstituted and the ethyl group stays on C-2: 2-ethylindole.

Ratio. The product ratio is set by the competition between the two ene-hydrazines. The internal ene-hydrazine is the more substituted, more stable enamine and gives 2,3-dimethylindole; the terminal one is less stable but is formed and rearranges readily in strongly acidic media. With butan-2-one in PPA the two routes are comparable, so a near 1: 1 mixture results. The ratio depends on the acid strength and temperature, which is why the choice of acid is used to steer the reaction with unsymmetrical ketones.

Conclusion. Phenylhydrazine and butan-2-one form one hydrazone but two ene-hydrazines. Each undergoes the [3,3]-sigmatropic shift, ring closure and loss of NH₃: the internal ene-hydrazine gives 2,3-dimethylindole, the terminal one 2-ethylindole, in about equal amounts.

Application. The Fischer synthesis makes the indole cores of the triptan anti-migraine drugs and of indomethacin; the regiochemistry problem with unsymmetrical ketones is controlled by the choice of acid.

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Model answer to Q3(c), 2019 Paper II: solution page 13
Solution page 13 of 23

All parts of Question 3