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 · 2016 · Paper II

Q3(a). Toluene to m-bromotoluene: an amino group as a temporary director

  • 10 marks
  • Answered
Question

How will you carry out the following conversion?
[structure] → [structure]

Answer in brief

Toluene → 4-nitrotoluene → p-toluidine → p-methylacetanilide → 2-bromo-4-methylacetanilide → 2-bromo-4-methylaniline → diazonium salt → 3-bromotoluene. The amino group places Br and is then removed.

Key points

  • Nitration of toluene gives 2- and 4-nitrotoluene; the para isomer is separated (fractional crystallisation/distillation).
  • Reduction (Sn/HCl or H₂/Ni) gives 4-methylaniline (p-toluidine).
  • Acetylation with acetic anhydride gives 4-methylacetanilide. -NHCOCH₃ is still a strong ortho/para director but is less activating than -NH₂, so bromination stops at monosubstitution.
  • the para position of NHAc is taken by CH₃, so Br enters ortho to NHAc, which is meta to CH₃. Product: 2-bromo-4-methylacetanilide.

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. CH₃ is an ortho/para director, so direct bromination of toluene gives 2- and 4-bromotoluene, never the 3-isomer. The bromine must be placed by a stronger ortho/para director that sits para to CH₃; that group (-NHCOCH₃) is afterwards removed by diazotisation and reduction (deamination).

Scheme: Steps 1–3: nitration, reduction, acetylation.

Scheme: Steps 4–5: bromination directed by NHAc, then removal of the acetyl group.

Scheme: Steps 6–7: diazotisation and replacement of N₂⁺ by H.

  1. Nitration of toluene gives 2- and 4-nitrotoluene; the para isomer is separated (fractional crystallisation/distillation).
  2. Reduction (Sn/HCl or H₂/Ni) gives 4-methylaniline (p-toluidine).
  3. Acetylation with acetic anhydride gives 4-methylacetanilide. -NHCOCH₃ is still a strong ortho/para director but is less activating than -NH₂, so bromination stops at monosubstitution.
  4. Bromination: the para position of NHAc is taken by CH₃, so Br enters ortho to NHAc, which is meta to CH₃. Product: 2-bromo-4-methylacetanilide.
  5. Hydrolysis (acid or base) releases the amine: 2-bromo-4-methylaniline.
  6. Diazotisation with NaNO₂/HCl at 0–5°C gives the arenediazonium chloride.
  7. Hypophosphorous acid (H₃PO₂) replaces -N₂⁺ by H (radical chain reduction), leaving 3-bromotoluene (m-bromotoluene).

Conclusion. Toluene → 4-nitrotoluene → p-toluidine → p-methylacetanilide → 2-bromo-4-methylacetanilide → 2-bromo-4-methylaniline → diazonium salt → 3-bromotoluene. The amino group places Br and is then removed.

Application. Temporary amino or acetamido directors removed by H₃PO₂ give meta-disubstituted benzenes that cannot be made directly, such as 1,3,5-tribromobenzene from aniline.

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

Model answer to Q3(a), 2016 Paper II: book edition page 10
Book edition, page 10
Model answer to Q3(a), 2016 Paper II: book edition page 11
Book edition, page 11

All parts of Question 3