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Indian Forest Service (Mains)

Chemistry optional

  • 2 papers
  • 200 marks each
  • 3 h per paper

Paper II covers organic chemistry and spectroscopy.

Indian Forest Service (Mains), Chemistry optional · 2025 · Paper II

Q1(d). Elimination next to an existing alkene

Question

Give major and minor products of the drawn bromide with methoxide. [structure of the bromide, CH₂=CHCH₂CH(Br)CH(CH₃)₂, given in the paper]

Answer in brief

Major: CH₂=CHCH=CHCH(CH₃)₂; minor: CH₂=CHCH₂CH=C(CH₃)₂.

Key points

  • Read the substrate as CH₂=CHCH₂CH(Br)CH(CH₃)₂; methoxide can remove a β H on either side of the brominated carbon.
  • Removal from the CH₂ next to the original alkene creates a conjugated 1,3-diene; conjugation favours this route despite the competing alkene being more substituted.
  • Removal from the isopropyl methine gives the isolated, more substituted double bond: a non-conjugated 1,4-diene.
  • E2 elimination requires the C–H and C–Br bonds to be anti-periplanar; acyclic bond rotation permits both productive conformations.

The full model answer

As written for the answer booklet: the diagram carries the working, the lines stay short. Read here only; it is not offered as a download.

Model answer to Q1(d), 2025 Paper II: solution pages 5 and 6
Solution pages 5 and 6 of 30
Read the whole 2025 solution2025 question paper

Topic notes: elimination reactions (E1, E2, E1cB, Saytzeff and Hofmann, anti-periplanar, Chugaev, Cope), with every Civil Services, IFoS and Geo-Scientist PYQ →

CoversE2 eliminationconjugated dieneSaytzeff vs Hofmannanti-periplanardehydrohalogenation

All parts of Question 1