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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(b). Tertiary alcohol with two identical alkyl groups; Lucas test

  • 10 marks
  • Answered
Question

Write a method of preparation of a tertiary alcohol containing two identical alkyl groups. Write one method for distinguishing primary, secondary and tertiary alcohols.

Answer in brief

Ester + two equivalents of a Grignard reagent (e.g. ethyl acetate + 2 C₂H₅MgBr) gives a tertiary alcohol with two identical groups, 3-methylpentan-3-ol. The Lucas test separates 3° (immediate turbidity), 2° (about 5 min) and 1° (no turbidity) alcohols.

Key points

  • CH₃COOC₂H₅ + C₂H₅MgBr → tetrahedral alkoxide → butan-2-one + EtOMgBr.
  • Butan-2-one + C₂H₅MgBr → magnesium alkoxide; H₃O⁺ gives 3-methylpentan-3-ol, CH₃C(OH)(C₂H₅)₂: a tertiary alcohol with two identical ethyl groups.
  • RCOOEt + 2 R'MgX → R'₂C(OH)R. A ketone R₂CO with R'MgX also gives R₂C(OH)R'.

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. A Grignard reagent adds twice to an ester: the first addition gives a ketone, which is more reactive than the ester and takes the second equivalent at once. The two groups delivered are therefore identical.

Scheme: Ester + 2 Grignard: the two new alkyl groups come from the Grignard reagent.

  1. CH₃COOC₂H₅ + C₂H₅MgBr → tetrahedral alkoxide → butan-2-one + EtOMgBr.
  2. Butan-2-one + C₂H₅MgBr → magnesium alkoxide; H₃O⁺ gives 3-methylpentan-3-ol, CH₃C(OH)(C₂H₅)₂: a tertiary alcohol with two identical ethyl groups.
  3. General: RCOOEt + 2 R'MgX → R'₂C(OH)R. A ketone R₂CO with R'MgX also gives R₂C(OH)R'.

Lucas test (primary, secondary, tertiary)

Lucas reagent is conc. HCl saturated with anhydrous ZnCl₂. The alcohol dissolves; the alkyl chloride formed is insoluble and makes the solution turbid. The rate follows carbocation stability (SN1, ZnCl₂ helps OH leave): 3° > 2° > 1°.

AlcoholObservation at room temperatureReason
tertiaryturbidity at oncestable 3° carbocation forms fast
secondaryturbidity in about 5 minutes2° carbocation forms slowly
primaryno turbidity (only on heating)1° cation not formed; slow SN2

Conclusion. Ester + two equivalents of a Grignard reagent (e.g. ethyl acetate + 2 C₂H₅MgBr) gives a tertiary alcohol with two identical groups, 3-methylpentan-3-ol. The Lucas test separates 3° (immediate turbidity), 2° (about 5 min) and 1° (no turbidity) alcohols.

Application. Ester + excess Grignard reagent is used to make triphenylmethanol (from methyl benzoate) and tertiary terpene alcohols such as linalool analogues.

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

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