VKC ScienceChemistry for written exams Telegram

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 · 2018 · Paper II

Q6(b). Acetolysis of threo- and erythro-3-phenyl-2-butyl tosylates

  • 15 marks
  • Answered
Question

Write the product(s) with proper stereochemical outcome of acetolysis of each of threo- and erythro-3-phenyl-2-butyltosylate giving the reaction pathway in each case.

Answer in brief

Threo tosylate → achiral phenonium ion → racemic threo-3-phenyl-2-butyl acetate. Erythro tosylate → chiral phenonium ion → optically active erythro acetate (retention). Both result from double inversion through the phenyl-bridged ion.

Key points

  • In the conformation where Ph and OTs are anti, the two methyls of threo-tosylate end up on the same side of the phenonium ring: the bridged ion has a mirror plane (it is meso-like, achiral).
  • AcOH attacks C-2 or C-3 with equal probability (the two carbons are mirror images).
  • Attack at C-2 gives threo acetate with retention; attack at C-3 gives its enantiomer. An optically active threo tosylate therefore gives racemic threo acetate (Cram: about 96% threo, racemic; only about 4% erythro).
  • Here the methyls are on opposite sides of the phenonium ring; the ion has a C₂ axis and no mirror plane: it is chiral.

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 phenyl group on C-3 is anti-periplanar to the leaving OTs on C-2 and acts as a neighbouring group: its π electrons push out TsO⁻ from the back (anchimeric assistance) and form a bridged phenonium ion. Acetic acid then opens the three-membered ring from the back at C-2 or at C-3. Two inversions mean overall retention at the carbon attacked (Cram).

Scheme: Threo: the two opening paths give the two enantiomers, 50: 50: racemic threo acetate.

Threo isomer

  1. In the conformation where Ph and OTs are anti, the two methyls of threo-tosylate end up on the same side of the phenonium ring: the bridged ion has a mirror plane (it is meso-like, achiral).
  2. AcOH attacks C-2 or C-3 with equal probability (the two carbons are mirror images).
  3. Attack at C-2 gives threo acetate with retention; attack at C-3 gives its enantiomer. An optically active threo tosylate therefore gives racemic threo acetate (Cram: about 96% threo, racemic; only about 4% erythro).

Erythro isomer

Scheme: Erythro: both opening paths give the same molecule: optically active erythro acetate, retention.

  1. Here the methyls are on opposite sides of the phenonium ring; the ion has a C₂ axis and no mirror plane: it is chiral.
  2. Attack at C-2 or at C-3 gives the same compound (the C₂ axis makes the two carbons equivalent).
  3. Product: erythro acetate, optically active, with retention of configuration.

Evidence for the pathway

Without phenyl participation (open carbocation or SN2) the configuration at C-2 would be lost or inverted and erythro/threo mixtures would form. The retained erythro/threo identity, racemisation of threo but not of erythro, are explained only by the bridged phenonium ion. Some alkenes (2-phenylbut-2-ene) form by elimination as side products.

Conclusion. Threo tosylate → achiral phenonium ion → racemic threo-3-phenyl-2-butyl acetate. Erythro tosylate → chiral phenonium ion → optically active erythro acetate (retention). Both result from double inversion through the phenyl-bridged ion.

Application. Phenonium and other bridged ions explain retention and scrambling in solvolysis of β-aryl systems and in biosynthetic aryl migrations (NIH shift).

Practising this answer? Write it in a Mains test and get it checked by VKC. Civil Services Mains test series · Telegram

Model answer to Q6(b), 2018 Paper II: book edition page 26
Book edition, page 26
Model answer to Q6(b), 2018 Paper II: book edition page 27
Book edition, page 27

All parts of Question 6