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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

Q4(a). Stobbe condensation; Fries rearrangement

  • 15 marks
  • Answered
Question

(i) Stobbe condensation of acetophenone with succinic acid gives two products. Write their structure and stereochemistry.
(ii) Phenyl acetate reacts with AlCl₃ in an inert solvent to give two isomers. Write their structures and indicate how you will separate them.

Answer in brief

(i) The Stobbe products are the (E)- and (Z)-3-(ethoxycarbonyl)-4-phenylpent-3-enoic acids, formed through a γ-lactone. (ii) Fries rearrangement gives 2- and 4-hydroxyacetophenone; the ortho isomer distils with steam.

Key points

  • EtO⁻ removes an α-H of diethyl succinate; the enolate adds to the C=O of acetophenone (aldol addition).
  • The new alkoxide attacks the other ester carbonyl (five-membered ring, favourable) and expels EtO⁻: a γ-lactone (a paraconic ester). Intermediate lactones of this kind have been isolated.
  • The base removes the acidic H next to the remaining ester; the lactone C–O bond breaks (E1cb), giving a carboxylate. This irreversible step (a stable carboxylate) drives the reaction.
  • Acidification gives 3-(ethoxycarbonyl)-4-phenylpent-3-enoic acid, C₁₄H₁₆O₄, as two geometric isomers: itemize

The full model answer

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

Reading of the question: In the Stobbe condensation the succinic acid is used as its diethyl ester (diethyl succinate) with a base such as NaOEt or KOC(CH₃)₃.

(i) Stobbe condensation

Introduction. A ketone condenses with diethyl succinate under base. The special feature of the Stobbe condensation is that one ester group is used up inside the molecule: a γ-lactone forms and opens, so the product is the half-ester (one COOH, one COOEt) of an alkylidenesuccinic acid. With an unsymmetrical ketone the new C=C can have two geometries.

Scheme: The alkoxide formed in the aldol step attacks the distant ester group and expels EtO⁻.

Scheme: Elimination opens the lactone and puts the C=C conjugated with the remaining ester.

  1. EtO⁻ removes an α-H of diethyl succinate; the enolate adds to the C=O of acetophenone (aldol addition).
  2. The new alkoxide attacks the other ester carbonyl (five-membered ring, favourable) and expels EtO⁻: a γ-lactone (a paraconic ester). Intermediate lactones of this kind have been isolated.
  3. The base removes the acidic H next to the remaining ester; the lactone C–O bond breaks (E1cb), giving a carboxylate. This irreversible step (a stable carboxylate) drives the reaction.
  4. Acidification gives 3-(ethoxycarbonyl)-4-phenylpent-3-enoic acid, C₁₄H₁₆O₄, as two geometric isomers: itemize
  5. On C-3, COOEt (O,O,O) outranks CH₂COOH (C,H,H); on C-4, phenyl outranks CH₃.
  6. (E)-isomer: phenyl trans to COOEt. (Z)-isomer: phenyl cis to COOEt. itemize
  7. Both isomers form because the elimination step can proceed from either conformation of the lactone; they are separated by crystallisation. Hydrolysis of either gives the corresponding alkylidenesuccinic acid.

(ii) Fries rearrangement of phenyl acetate

Scheme: The acyl group moves from O to the ring, ortho or para.

  1. AlCl₃ complexes the ester oxygen; the acylium ion CH₃C#O⁺ is released and attacks the activated ring (Fries rearrangement, an intramolecular/intermolecular Friedel–Crafts acylation). Low temperature favours the para isomer, high temperature the ortho isomer (stabilised as an Al chelate).
  2. Separation by steam distillation: 2-hydroxyacetophenone has an intramolecular hydrogen bond (O–H…O=C, six-membered ring), so it is less associated and steam-volatile. 4-Hydroxyacetophenone forms intermolecular hydrogen bonds with itself and with water and stays behind in the flask.

Conclusion. (i) The Stobbe products are the (E)- and (Z)-3-(ethoxycarbonyl)-4-phenylpent-3-enoic acids, formed through a γ-lactone. (ii) Fries rearrangement gives 2- and 4-hydroxyacetophenone; the ortho isomer distils with steam.

Application. Stobbe half-esters are cyclised to naphthols and tetralones in syntheses of steroids and lignans; the Fries rearrangement is used to make hydroxyaryl ketones such as the analgesic intermediate 4-hydroxyacetophenone (paracetamol route).

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Model answer to Q4(a), 2016 Paper II: solution page 10
Solution page 10 of 25