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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 · 2020 · Paper II

Q4(c). Arndt–Eistert homologation; SN1 capture by pyridine

  • 20 marks
  • Answered
Question

(i) Identify the product and briefly discuss the steps involved in the following reaction:
[structure] → (1) SOCl₂, 2) CH₂N₂3) AgNO₃) ?
(ii) The rate constant for the reaction of diphenylchloromethane with pyridine and triethylamine is same. Write the structure of the major product in the following reaction. Justify your answer:
[structure] + [structure] + NEt₃ → Major product
(1: 1, large excess)

Answer in brief

(i) (3S,4R)-4-(2,5-dimethoxy-3-nitrophenyl)-3-methylpentanoic acid, the next homologue, formed by acid chloride → diazoketone → Wolff rearrangement (retention) → ketene → acid. (ii) N-(diphenylmethyl)pyridinium chloride: the SN1 benzhydryl cation is captured by the less hindered pyridine nitrogen.

Key points

  • SOCl₂ converts RCOOH into the acid chloride (SO₂ and HCl are lost); both stereocentres are untouched.
  • the nucleophilic carbon of diazomethane attacks the acid chloride; loss of Cl⁻ and of an α-proton gives the α-diazoketone RCOCHN₂. A second equivalent of CH₂N₂ mops up the HCl (as CH₃Cl + N₂), which would otherwise give the…
  • N₂ is lost and the alkyl group R moves from the carbonyl carbon to the neighbouring carbon (Wolff rearrangement), giving a ketene R-CH=C=O. The shift is intramolecular, so the migrating stereocentre keeps its configuration…
  • Water adds to the ketene, giving the homologous acid: (3S,4R)-4-(2,5-dimethoxy-3-nitrophenyl)-3-methylpentanoic acid. The starting acid is (2R,3R); the change of the descriptor at the migrating carbon from R to S is due only…

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. (i) The three reagents carry out the Arndt–Eistert homologation: the carboxylic acid is converted into its next higher homologue (one more CH₂), with the Wolff rearrangement as the key step. (ii) Equal rate constants with two different nucleophiles show that the nucleophile is not in the rate-determining step: the reaction is SN1.

(i) Arndt–Eistert homologation

Scheme: Acid chloride, then acylation of diazomethane: the second equivalent of CH₂N₂ removes HCl.

Scheme: The migrating carbon keeps its configuration; the ketene adds water to give the acid with one extra CH₂.

  1. SOCl₂ converts RCOOH into the acid chloride (SO₂ and HCl are lost); both stereocentres are untouched.
  2. CH₂N₂: the nucleophilic carbon of diazomethane attacks the acid chloride; loss of Cl⁻ and of an α-proton gives the α-diazoketone RCOCHN₂. A second equivalent of CH₂N₂ mops up the HCl (as CH₃Cl + N₂), which would otherwise give the chloromethyl ketone.
  3. AgNO₃ (Ag⁺ catalyst, aqueous solvent): N₂ is lost and the alkyl group R moves from the carbonyl carbon to the neighbouring carbon (Wolff rearrangement), giving a ketene R-CH=C=O. The shift is intramolecular, so the migrating stereocentre keeps its configuration (retention).
  4. Water adds to the ketene, giving the homologous acid: (3S,4R)-4-(2,5-dimethoxy-3-nitrophenyl)-3-methylpentanoic acid. The starting acid is (2R,3R); the change of the descriptor at the migrating carbon from R to S is due only to the new priority order (CH₂COOH replaces COOH), not to inversion.

(ii) Diphenylchloromethane, pyridine and triethylamine

Scheme: The cation, stabilised by two phenyl rings, is captured by the unhindered pyridine nitrogen.

  1. Mechanism. Because the rate is the same with pyridine as with triethylamine, the slow step does not involve the amine: it is ionisation of Ph₂CHCl to the benzhydryl cation (secondary, but delocalised over two rings). Rate =k[Ph₂CHCl].
  2. Product-determining step. The two amines then compete for the cation. With equal (large) amounts, the product is decided by how easily each nitrogen can reach the crowded cationic carbon, not by basicity.
  3. Pyridine wins. Its nitrogen lone pair is in an sp² orbital in the plane of a flat ring and is sterically unhindered. The nitrogen of NEt₃ is surrounded by three flexible ethyl groups; bonding it to a carbon that already carries two phenyl groups would create a very crowded quaternary ion, which forms slowly and readily dissociates again.
  4. NEt₃ is the stronger base, but the cation has no β-hydrogen that would allow E1 elimination; triethylamine mainly takes up any HCl.

Conclusion. (i) (3S,4R)-4-(2,5-dimethoxy-3-nitrophenyl)-3-methylpentanoic acid, the next homologue, formed by acid chloride → diazoketone → Wolff rearrangement (retention) → ketene → acid. (ii) N-(diphenylmethyl)pyridinium chloride: the SN1 benzhydryl cation is captured by the less hindered pyridine nitrogen.

Application. Arndt–Eistert homologation turns α-amino acids into β-amino acids without racemisation (used for β-peptides); the competition experiment in (ii) is the standard way to tell SN1 from SN2 (rate independent of the nucleophile).

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Model answer to Q4(c), 2020 Paper II: book edition page 16
Book edition, page 16
Model answer to Q4(c), 2020 Paper II: book edition page 17
Book edition, page 17

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