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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 · 2016 · Paper II

Q5(c). Pyrrolidine ring by the Hofmann–Löffler–Freytag reaction

  • 10 marks
  • Answered
Question

How will you carry out the following conversion? Write the steps involved in it:
[structure] → [structure]

Answer in brief

NO₂ → NH₂ (reduction) → N–Cl (NCS) → δ-chloroamine (H₂SO₄, hν; 1,5-H shift) → 2-ethyl-2-methyloctahydroindole (NaOH, cyclisation): the Hofmann–Löffler–Freytag reaction.

Key points

  • Reduce the nitro group to -NH₂ (H₂/Pd–C, Sn/HCl or LiAlH₄): 2-(2-methylbutyl)cyclohexan-1-amine.
  • N-Chlorinate with N-chlorosuccinimide (or NaOCl): R-NH-Cl.
  • Dissolve in conc. H₂SO₄ (or CF₃COOH) and irradiate or warm. The protonated N-chloroamine R-NH₂+-Cl cleaves homolytically to the aminium radical cation R-NH₂⁺.
  • The aminium radical abstracts the δ-hydrogen through a six-membered transition state (N, C-1, C-2, CH₂, C–H: a 1,5-H shift). The tertiary δ-C–H is weak and well placed, so it reacts selectively; a tertiary carbon radical forms.

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. The product is an octahydroindole: a new five-membered ring in which N is bonded to the carbon that carries the methyl and ethyl groups. In the starting material that carbon is an unactivated tertiary C–H four bonds away from N (the δ-carbon). Such a remote C–H is functionalised by the Hofmann–Löffler–Freytag reaction: an N-centred radical abstracts the δ-H by a 1,5-hydrogen shift.

Scheme: Steps 1–2: reduction of NO₂ and N-chlorination.

Scheme: Steps 3–4: radical 1,5-H abstraction, chlorine transfer, then intramolecular N-alkylation.

  1. Reduce the nitro group to -NH₂ (H₂/Pd–C, Sn/HCl or LiAlH₄): 2-(2-methylbutyl)cyclohexan-1-amine.
  2. N-Chlorinate with N-chlorosuccinimide (or NaOCl): R-NH-Cl.
  3. Dissolve in conc. H₂SO₄ (or CF₃COOH) and irradiate or warm. The protonated N-chloroamine R-NH₂+-Cl cleaves homolytically to the aminium radical cation R-NH₂⁺.
  4. The aminium radical abstracts the δ-hydrogen through a six-membered transition state (N, C-1, C-2, CH₂, C–H: a 1,5-H shift). The tertiary δ-C–H is weak and well placed, so it reacts selectively; a tertiary carbon radical forms.
  5. The carbon radical takes Cl from another N-chloroammonium ion (chain propagation), giving the δ-chloroamine and a new aminium radical.
  6. Base (NaOH) releases the amine, whose N displaces the tertiary chloride intramolecularly (five-membered ring closure): 2-ethyl-2-methyloctahydro-1H-indole. The ring fusion keeps the relative configuration of the two ring carbons of the starting material.

Conclusion. NO₂ → NH₂ (reduction) → N–Cl (NCS) → δ-chloroamine (H₂SO₄, hν; 1,5-H shift) → 2-ethyl-2-methyloctahydroindole (NaOH, cyclisation): the Hofmann–Löffler–Freytag reaction.

Application. The Hofmann–Löffler–Freytag reaction makes pyrrolidine and pyrrolizidine alkaloids (nicotine was made this way) and functionalises remote C–H bonds in steroids (the related Barton nitrite reaction gives aldosterone acetate).

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Model answer to Q5(c), 2016 Paper II: solution page 17
Solution page 17 of 25