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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

Q2(b). 1,2-Distyrylbenzene from phthalic acid by Wittig reaction

  • 15 marks
  • Answered
Question

Outline the synthesis of the following compound using Wittig reaction starting with phthalic acid:
[structure]

Answer in brief

Phthalic acid → dimethyl phthalate (MeOH/H⁺) → 1,2-benzenedimethanol (LiAlH₄) → phthalaldehyde (PCC) 2 PhCH=PPh₃ 1,2-distyrylbenzene + 2 Ph₃P=O; E/Z mixtures are equilibrated to E,E with I₂.

Key points

  • phthalic acid + methanol (H₂SO₄) gives dimethyl phthalate. (The acid itself can also be reduced by LiAlH₄, but the ester reacts more cleanly.)
  • LiAlH₄ reduces both ester groups to CH₂OH: 1,2-benzenedimethanol.
  • PCC (or Swern oxidation) stops at the aldehyde: phthalaldehyde (benzene-1,2-dicarbaldehyde).
  • benzyl bromide + PPh₃ (SN2) gives benzyltriphenylphosphonium bromide; BuLi (or NaOEt) removes the benzylic proton: PhCH=PPh₃ (a semi-stabilised ylide).

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 target is 1,2-bis(2-phenylethenyl)benzene. Disconnecting both C=C bonds (Wittig disconnection) gives either phthalaldehyde + 2 PhCH=PPh3 or the bis-ylide of o-xylene + 2 PhCHO. Phthalic acid provides the ortho-disubstituted ring; its two CO₂H groups are first brought to the aldehyde (or CH₂Br) level.

Route: phthalaldehyde and benzylidenetriphenylphosphorane

Scheme: Steps 1–3: the diacid is taken to the dialdehyde.

  1. Esterification: phthalic acid + methanol (H₂SO₄) gives dimethyl phthalate. (The acid itself can also be reduced by LiAlH₄, but the ester reacts more cleanly.)
  2. Reduction: LiAlH₄ reduces both ester groups to CH₂OH: 1,2-benzenedimethanol.
  3. Mild oxidation: PCC (or Swern oxidation) stops at the aldehyde: phthalaldehyde (benzene-1,2-dicarbaldehyde).
  4. Ylide: benzyl bromide + PPh₃ (SN2) gives benzyltriphenylphosphonium bromide; BuLi (or NaOEt) removes the benzylic proton: PhCH=PPh₃ (a semi-stabilised ylide).
  5. Double Wittig reaction: two equivalents of the ylide add to the two CHO groups. Each addition forms a betaine/oxaphosphetane, which collapses to C=C and Ph₃P=O (strong P=O bond, the driving force).

Scheme: Two Wittig olefinations on one molecule.

Alternative route through the bis-ylide

1,2-Benzenedimethanol + PBr₃ gives α,α'-dibromo-o-xylene; with two PPh₃ it forms the bis(phosphonium) salt, C₆H₄(CH₂PPh₃⁺)₂ 2Br⁻. Two equivalents of base give the bis-ylide, which reacts with 2 PhCHO to give the same product. This route avoids handling the easily oxidised dialdehyde.

Stereochemistry. Benzylic (semi-stabilised) ylides give mixtures of E and Z alkenes. The mixture is isomerised to the more stable (E,E) isomer by a trace of iodine (light) or heating.

Conclusion. Phthalic acid → dimethyl phthalate (MeOH/H⁺) → 1,2-benzenedimethanol (LiAlH₄) → phthalaldehyde (PCC) 2 PhCH=PPh₃ 1,2-distyrylbenzene + 2 Ph₃P=O; E/Z mixtures are equilibrated to E,E with I₂.

Application. Double Wittig reactions on aromatic dialdehydes make stilbene-type optical brighteners and the precursors of helicenes and phenanthrenes (photocyclisation of stilbenes).

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Model answer to Q2(b), 2019 Paper II: solution page 8
Solution page 8 of 23

All parts of Question 2