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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

Q1(d). Baeyer–Villiger oxidation; Bischler–Napieralski synthesis

  • 10 marks
  • Answered
Question

Identify the product (X) in the following reactions and write the name of the reaction involved:
(i) [structure] → (mCPBA) (X)(ii) [structure] → ((1) CH₃COCl(2) P₂O₅, (3) Pd/C100°C) (X)

Answer in brief

(i) Baeyer–Villiger oxidation: the ring-fusion carbon migrates to O, giving the seven-membered lactone X. (ii) Bischler–Napieralski reaction followed by Pd/C dehydrogenation: X = 1-methylisoquinoline.

Key points

  • The peracid adds to the C=O to give the tetrahedral Criegee intermediate, C(OH)(O-O-COAr).
  • In the rate-determining step one α-carbon migrates from C to O as ArCO₂⁻ leaves. Migratory aptitude: tertiary > secondary > primary. At C-1 the ring-fusion CH (C-8a, secondary) migrates rather than the CH₂ (C-2, primary), and…
  • The C-6 ketone has two CH₂ neighbours, gives no regiochemical preference, and is the slower one.
  • N-(2-phenylethyl)acetamide.

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) A peracid converts a ketone into an ester (a cyclic ketone into a lactone) by the Baeyer–Villiger oxidation; the more substituted carbon migrates. (ii) A β-arylethylamide closes to a dihydroisoquinoline by the Bischler–Napieralski reaction; Pd/C then dehydrogenates it to the aromatic isoquinoline.

(i) Baeyer–Villiger oxidation

Reading of the question: the 1,6-dione is read with one equivalent of mCPBA, which attacks the C-1 ketone; excess peracid would also convert the C-6 ketone into a lactone.

Scheme: O is inserted between C-1 and the ring-fusion carbon, which migrates with retention.

  1. The peracid adds to the C=O to give the tetrahedral Criegee intermediate, C(OH)(O-O-COAr).
  2. In the rate-determining step one α-carbon migrates from C to O as ArCO₂⁻ leaves. Migratory aptitude: tertiary > secondary > primary. At C-1 the ring-fusion CH (C-8a, secondary) migrates rather than the CH₂ (C-2, primary), and it moves with retention of configuration.
  3. The C-6 ketone has two CH₂ neighbours, gives no regiochemical preference, and is the slower one.

X is the ε-lactone: the original ring A becomes a seven-membered ring containing -O-CO⁻, fused to the cyclohexanone ring (an octahydro-1-benzoxepine-2,7-dione, the second C=O kept).

(ii) Bischler–Napieralski reaction, then dehydrogenation

Scheme: Acylation, cyclodehydration (SEAr on a nitrilium ion) and aromatisation.

  1. Acetyl chloride acylates the amine: N-(2-phenylethyl)acetamide.
  2. P₂O₅ activates the amide oxygen; loss of water gives an electrophilic nitrilium ion (or imidoyl phosphate), which attacks the ortho position of the ring (intramolecular electrophilic substitution, six-membered ring closure). Loss of H⁺ gives 1-methyl-3,4-dihydroisoquinoline.
  3. Pd/C at 100°C removes two H from C-3 and C-4: the ring becomes aromatic.

X = 1-methylisoquinoline.

Conclusion. (i) Baeyer–Villiger oxidation: the ring-fusion carbon migrates to O, giving the seven-membered lactone X. (ii) Bischler–Napieralski reaction followed by Pd/C dehydrogenation: X = 1-methylisoquinoline.

Application. Baeyer–Villiger oxidation makes caprolactone (for polycaprolactone) and steroid lactones; the Bischler–Napieralski route is the standard way into isoquinoline alkaloids such as papaverine.

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Model answer to Q1(d), 2019 Paper II: solution page 4
Solution page 4 of 23

All parts of Question 1