Match List - I with List - II.
| List - I | List - II |
| (A) Nylon | (I) Terylene |
| (B) Polyester | (II) Caprolactam |
| (C) Bakelite | (III) Isoprene |
| (D) Rubber | (IV) Novolac |
Choose the correct answer from the options given below :
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(II), (B)-(I), (C)-(IV), (D)-(III)
Each polymer is matched with its monomer or with a named member/intermediate of its class:
(A) Nylon — This matches (II) Caprolactam. Nylon 6 is obtained by heating caprolactam with water at a high temperature (ring-opening polymerisation); it is a polyamide.
\( \displaystyle n\,\text{Caprolactam} \xrightarrow[533\text{–}543\ \mathrm{K}]{\ce{H2O}} \ce{-[NH-(CH2)5-CO]_{n}-}\ \ (\text{Nylon 6}) \)
(B) Polyester — This matches (I) Terylene. Terylene (dacron) is the best-known polyester, made by condensing ethylene glycol with terephthalic acid; water is lost at each ester link.
\( \displaystyle n\,\ce{HOCH2CH2OH} + n\,\ce{HOOC-C6H4-COOH} \xrightarrow{\text{420-460 K}} \ce{-[OCH2CH2O-CO-C6H4-CO]_{n}-} + 2n\,\ce{H2O} \)
(C) Bakelite — This matches (IV) Novolac. Phenol and formaldehyde first give a linear polymer, novolac; heating novolac with more formaldehyde cross-links it into bakelite.
\( \displaystyle \ce{C6H5OH + HCHO} \xrightarrow{\ce{H+}\text{ or }\ce{OH-}} \text{Novolac (linear)} \xrightarrow[\Delta]{\ce{HCHO}} \text{Bakelite (cross-linked)} \)
(D) Rubber — This matches (III) Isoprene. Natural rubber is cis-1,4-polyisoprene, a linear polymer of isoprene (2-methyl-1,3-butadiene).
\( \displaystyle n\,\ce{CH2=C(CH3)-CH=CH2} \longrightarrow \ce{-[CH2-C(CH3)=CH-CH2]_{n}-}\ \ (\text{cis-1,4-polyisoprene}) \)
Shortcut: The pair (A)–(II) appears in Options (3) and (4), and (B)–(I) then leaves only Option (3).
Note: This question is from the pre-rationalised NCERT book (2022-23). The topic is not there in the latest NCERT book (2026-27), and it is also out of the CUET 2026 syllabus.