Match the List I with List II List I List II (A) Weak intermolecular forces of attraction I Hexamethylenedia mine + adipic acid (B) Hydrogen bonding II AlEt3 + TiCl4 (C) Heavily branched polymer III 2-chloro-1, 3-butadiene (D) High density polymer III phenol+ formaldehyde
choose the correct answer from the options given below:
| List I | List II | ||
| (A) | Weak intermolecular forces of attraction | I | Hexamethylenedia mine + adipic acid |
| (B) | Hydrogen bonding | II | AlEt3 + TiCl4 |
| (C) | Heavily branched polymer | III | 2-chloro-1, 3-butadiene |
| (D) | High density polymer | III | phenol+ formaldehyde |
choose the correct answer from the options given below:
- A-II, B-III, C-I, D-IV
- A-IV, B-I, C-III, D-II
- A-III, B-I, C-IV, D-I
- A-IV, B-II, C-III, D-I
The Correct Option is C
Solution and Explanation
1. A. Weak intermolecular forces of attraction: 2–chloro–1,3–butadiene (chloroprene) is the monomer of neoprene, which has weak intermolecular forces of attraction, being an elastomer. Correct match: III. 2–chloro–1,3–butadiene.
2. B. Hydrogen bonding: Hexamethylenediamine reacts with adipic acid to form Nylon-6,6. The presence of the amide group allows hydrogen bonding between polymer chains. Correct match: I. Hexamethylenediamine + adipic acid.
3. C. Heavily branched polymer: - Phenol reacts with formaldehyde to form Bakelite, which is a heavily branched (cross-linked) polymer. Correct match: IV. Phenol + formaldehyde.
4. D. High density polymer: - High-density polyethylene is prepared using the Ziegler-Natta catalyst (\( \text{AlEt}_3 + \text{TiCl}_4 \)). Correct match: II. \( \text{AlEt}_3 + \text{TiCl}_4 \). ### Final Matching: \[ \text{A–III, B–I, C–IV, D–II.} \]
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