Question

Match 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	IV. Phenol + formaldehyde

Choose the correct answer from the options given below:

(1) A-II, B-IV, C-I, D-III

(2) A-III, B-I, C-IV, D-II

(3) A-IV, B-I, C-III, D-II

(4) A-IV, B-II, C-III, D-I

Answer 1

(2) A-III, B-I, C-IV, D-II

• Hexamethylenediamine on reaction with adipic acid forms Nylon 6, 6 which shows H-bonding due to presence of amide group.
• AlEt₃ + TiCl₄ is Ziegler-Natta catalyst used to prepare high density polyethylene..
• 2-chloro-1, 3-butadiene (chloroprene) is monomer of neoprene which is a rubber (an elastomer)
• Phenol formaldehyde forms Bakelite which is heavily branched (cross-linked) polymer