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.

Coversnatural rubbercis-1,4-polyisopreneisoprenepolyethyleneIR spectroscopy
All parts of Question 1
- 1(a) Aromaticity classifications5 marks
- 1(b) Photochemical hydrogen migration in a diene5 marks
- 1(c) Isotope-labelled acetate pyrolysis5 marks
- 1(d) Kinetic enolate aldol sequence5 marks
- 1(e) Natural rubber and polyethylene5 marks
- 1(f) Phenol to a reduced aryl ether5 marks
- 1(g) Photochemical cleavage and alkene isomerization5 marks
- 1(h) Hyperfine lines of an isotopic radical5 marks