Logically thinking, I thought that it would be high pressure and high temperature that would create high density poly(ethene)... However, that's not the case. Is it because of the catalyst? Why is the condition low pressure/temperature?

  • $\begingroup$ The actual density difference at high temperatures should be negligible, the name is due to the higher crystallinity and therefore density of cold HDPE. $\endgroup$
    – Karl
    Commented Nov 30, 2015 at 3:00

1 Answer 1


In fact, when polymerizing ethylene using a radical initiator at high temperature and high pressure, we favor the formation of low-density polyethylene. This is because the radical at the end of the growing chain can be transferred to any carbon atom of the chain by radical rearrangement (favored at high temperature) and due to close distances among molecules (high pressure) a branch chain can grow on the principal chain, and we get a branched polyethylene. The chains of a branched polymer cannot align together, and eventually the density of the resultant polymer is low. We have got LDPE.

On the other hand, using a specific catalyst, known as Ziegler-Natta catalyst and under mild conditions of temperature and pressure, we get linear chains of polyethylene that can be aligned together and the density of the resultant polymer is high. We have got HDPE.


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