Explain why but-1-ene reacts with bromine water to give predominantly 2-bromobutanol rather than 1-bromobutanol as the major organic product.

IB DP Chemistry Higher Level (2023 syllabus) — R3.3 Electron sharing reactions · Explain · 4 marks · View as Markdown

Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).

Bromine water (Br₂(aq)) reacts with but-1-ene (CH₂=CHCH₂CH₃) via an electrophilic addition mechanism. The reaction proceeds through a bromonium ion intermediate rather than a simple carbocation.

Model answer (4 marks)

Br₂ approaches the π bond and the more electron‑rich bromine acts as an electrophile, forming a cyclic bromonium ion intermediate across the double bond.

The bromonium ion is unsymmetrical because the C‑2 atom can stabilise the developing positive charge better than C‑1; consequently the C–Br bond at C‑2 is weaker and the C‑2 centre carries a greater partial positive charge.

Water, as the nucleophile, attacks preferentially at C‑2, the more positively charged carbon, giving a 2‑bromobutyl oxonium ion.

Deprotonation of this intermediate yields 2‑bromobutanol; the attack occurs anti to the leaving Br, giving the anti‑addition product.

Examiner tips

  • Use the word ‘electrophile’ and ‘nucleophile’ to show understanding of the mechanism.
  • Show that the bromonium ion is unsymmetrical and explain why C‑2 is more positive.
  • Mention anti‑addition to justify the stereochemistry.
  • Keep the answer concise – 4 points, 1 sentence per point.

Common mistakes

  • Confusing the roles of Br₂ and water (calling Br₂ the nucleophile).
  • Failing to explain why the bromonium ion is unsymmetrical and why C‑2 is more positive.
  • Omitting the anti‑addition aspect of the attack.

Mark scheme (4 marks)

  1. Br₂ approaches the π bond and the more electron-rich bromine acts as an electrophile, forming a cyclic bromonium ion intermediate across the double bond
  2. The bromonium ion is unsymmetrical because C-2 can better stabilise partial positive charge than C-1, so the C–Br bond at C-2 is weaker / longer
  3. Water (the nucleophile) attacks preferentially at C-2 because it carries the greater partial positive charge
  4. Attack of water at C-2 followed by loss of a proton gives 2-bromobutanol; attack is anti (from the opposite face to Br) giving the anti-addition product

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