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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 · 2021 · Paper II

Q1(e). Natural rubber and polyethylene

Question

Give the monomer of natural rubber and distinguish natural rubber from polythene by IR spectroscopy.

Answer in brief

Natural rubber is cis-1,4-polyisoprene, derived from isoprene (2-methylbuta-1,3-diene); its residual double bonds distinguish it from saturated polyethylene.

Key points

  • Isoprene is CH₂=C(CH₃)CH=CH₂; the rubber repeat is [–CH₂C(CH₃)=CHCH₂–]ₙ with cis chain geometry.
  • Rubber shows C=C stretching near 1660 cm⁻¹, olefinic C–H near 3030 cm⁻¹ and a trisubstituted-alkene out-of-plane band near 830–840 cm⁻¹.
  • Polyethylene, [–CH₂CH₂–]ₙ, lacks substantial alkene bands; it shows CH₂ stretches near 2920 and 2850, bending near 1460 and rocking near 720–730 cm⁻¹.
  • Use the whole pattern; weak unsaturation from chain ends or defects does not make polyethylene a polyisoprene.

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(e), 2021 Paper II: solution pages 5 and 6
Solution pages 5 and 6 of 40

Coversnatural rubbercis-1,4-polyisopreneisoprenepolyethyleneIR spectroscopy