Question

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:

• A. A-II, B-III, C-I, D-IV
• B. A-IV, B-I, C-III, D-II
• C. A-III, B-I, C-IV, D-I
• D. A-IV, B-II, C-III, D-I

Answer 1

The Correct Option is C

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: \[ \text{A–III, B–I, C–IV, D–II.} \]