Given below are two statements :
Statement I : Ethene at 333 to 343 K and 6-7 atm pressure in the presence of AlEt₃ and TiCl₄ undergoes addition polymerization to give LDP.
Statement II : Caprolactam at 533-543 K in H₂ O through step growth polymerizes to give Nylon 6.
In the light of the above statements, choose the correct answer from the options given below:

Solution

Chemistry ma’am · 3 steps

Ziegler–Natta catalyst with ethene gives straight chains, so the product is HDPE, not LDP.

1. Statement I. AlEt₃ with TiCl₄ is the Ziegler–Natta catalyst. At 333 to 343 K and 6 to 7 atm it adds ethene units into straight, unbranched chains that pack closely. That polymer is high density polythene (HDP), so Statement I is false.

   Polythene	Conditions	Chain shape
   LDP	350 to 570 K, 1000 to 2000 atm, traces of O₂ or a peroxide (free radical)	highly branched
   HDP	333 to 343 K, 6 to 7 atm, AlEt₃ + TiCl₄ (Ziegler–Natta)	linear

2. Statement II. Caprolactam heated with water at 533 to 543 K opens its ring and joins up through amide links, one step at a time. This is step growth (condensation) polymerisation, and the product is Nylon 6. Statement II is true.
3. So. Statement I is false and Statement II is true, which is option (c).

The idea to keep. For polythene, match the conditions. High pressure with a free radical initiator gives branched LDP. Low pressure with the Ziegler–Natta catalyst gives linear HDP.

Answer from the NTA final answer key. Solution by PrepWiser.