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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

Q4(c)(ii). Reimer–Tiemann reaction of p-cresol: normal and abnormal products

  • 10 marks
  • Answered
Question

p-methylphenol on reaction with OH⁻/CHCl₃ gives two products. Write the structure of both and propose a suitable mechanism.

Answer in brief

The normal Reimer–Tiemann product is 2-hydroxy-5-methylbenzaldehyde (ortho formylation via:CCl₂); the abnormal product is 4-(dichloromethyl)-4-methylcyclohexa-2,5-dienone, formed by attack at the methyl-bearing para carbon.

Key points

  • Ortho attack (normal product). The phenoxide attacks:CCl₂ through its ortho carbon, giving a -CCl₂⁻ carbanion on a cyclohexadienone. Proton transfer restores the aromatic ring: an o-(dichloromethyl)phenoxide.
  • The benzal chloride is hydrolysed by OH⁻ (two SN steps via a quinone methide) to the aldehyde; acidification gives 2-hydroxy-5-methylbenzaldehyde (major).
  • Para attack (abnormal product). Attack at C-4, which carries CH₃, gives a dienone with a quaternary carbon. There is no H on C-4 to lose, so aromaticity cannot be restored, and the CHCl₂ group, attached to an sp3 carbon,…

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. CHCl₃ and OH⁻ generate dichlorocarbene (:CCl₂), an electrophile that attacks the electron-rich ring of the phenoxide (Reimer–Tiemann reaction). With p-cresol the para position is blocked by CH₃, so attack at the ortho carbon gives the normal aldehyde, while attack at the substituted para carbon gives a dienone that cannot re-aromatise.

CHCl₃ + OH⁻ ⇌ CCl₃⁻ + H₂O CCl₃⁻:CCl₂ + Cl⁻ (α-elimination)

Scheme: Normal (ortho-formyl) product and abnormal (4-dichloromethyl dienone) product.

  1. Ortho attack (normal product). The phenoxide attacks:CCl₂ through its ortho carbon, giving a -CCl₂⁻ carbanion on a cyclohexadienone. Proton transfer restores the aromatic ring: an o-(dichloromethyl)phenoxide.
  2. The benzal chloride is hydrolysed by OH⁻ (two SN steps via a quinone methide) to the aldehyde; acidification gives 2-hydroxy-5-methylbenzaldehyde (major).
  3. Para attack (abnormal product). Attack at C-4, which carries CH₃, gives a dienone with a quaternary carbon. There is no H on C-4 to lose, so aromaticity cannot be restored, and the CHCl₂ group, attached to an sp3 carbon, resists hydrolysis: 4-(dichloromethyl)-4-methylcyclohexa-2,5-dien-1-one (minor).

Conclusion. The normal Reimer–Tiemann product is 2-hydroxy-5-methylbenzaldehyde (ortho formylation via:CCl₂); the abnormal product is 4-(dichloromethyl)-4-methylcyclohexa-2,5-dienone, formed by attack at the methyl-bearing para carbon.

Application. The Reimer–Tiemann reaction makes salicylaldehyde (for coumarin) and vanillin-type aldehydes; the abnormal dienones are the classical evidence for the carbene mechanism.

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Model answer to Q4(c)(ii), 2019 Paper II: book edition page 18
Book edition, page 18
Model answer to Q4(c)(ii), 2019 Paper II: book edition page 19
Book edition, page 19

All parts of Question 4