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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 · 2017 · Paper II

Q3(c). Tosylation, S_N₂ by acetate and hydrolysis: a Walden cycle

  • 10 marks
  • Answered
Question

Write the stereo structures of L to N and comment on the specific rotation of N. Specific rotation of the starting alcohol is +x.
[[α]_D²⁵=+x] → (TsCl, Py) L → (KOAc, DMF, Δ) M → (⁻OH) N

Answer in brief

L = (R)-tosylate, M = (S)-acetate, N = (S)-1-phenylpropan-2-ol, the enantiomer of the starting alcohol. So [α]_D²⁵(N) = -x (equal magnitude, opposite sign, if the SN2 step is clean and no racemisation occurs).

Key points

  • L. The alcohol oxygen attacks sulfur of TsCl; pyridine takes HCl. The C–O bond is not broken, so configuration is retained: (R)-1-phenylpropan-2-yl tosylate. OTs is an excellent leaving group.
  • M. Acetate in the polar aprotic solvent DMF attacks C-2 from the back side (SN2); TsO⁻ leaves: Walden inversion. M = (S)-1-phenylpropan-2-yl acetate.
  • N. Hydroxide attacks the acyl carbon of the ester (BAC2, acyl–oxygen cleavage); the alkyl C–O bond is untouched, so configuration is retained: N = (S)-1-phenylpropan-2-ol.

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 printed alcohol is (R)-1-phenylpropan-2-ol. Only one of the three steps breaks a bond to the stereocentre; that step decides the outcome.

Scheme: Inversion happens only in the SN2 step at carbon.

  1. L. The alcohol oxygen attacks sulfur of TsCl; pyridine takes HCl. The C–O bond is not broken, so configuration is retained: (R)-1-phenylpropan-2-yl tosylate. OTs is an excellent leaving group.
  2. M. Acetate in the polar aprotic solvent DMF attacks C-2 from the back side (SN2); TsO⁻ leaves: Walden inversion. M = (S)-1-phenylpropan-2-yl acetate.
  3. N. Hydroxide attacks the acyl carbon of the ester (BAC2, acyl–oxygen cleavage); the alkyl C–O bond is untouched, so configuration is retained: N = (S)-1-phenylpropan-2-ol.

Conclusion. L = (R)-tosylate, M = (S)-acetate, N = (S)-1-phenylpropan-2-ol, the enantiomer of the starting alcohol. So [α]_D²⁵(N) = -x (equal magnitude, opposite sign, if the SN2 step is clean and no racemisation occurs).

Application. This tosylate–acetate cycle (Phillips and Kenyon, with 1-phenyl-2-propanol) proved that SN2 displacement inverts configuration; it is used to invert secondary alcohol centres in synthesis (cf. the Mitsunobu reaction).

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Model answer to Q3(c), 2017 Paper II: solution page 11
Solution page 11 of 23
Read the whole 2017 solution2017 question paper

All parts of Question 3