Here, the answer given is (iv), and I don't see the reason why methoxyethene is not forming an addition polymer with radical initiators. Clearly, the first option will form polystyrene and the second, polyacrylonitrile. But after that, I feel lost. After looking at the answer key, I felt that there would be some role of the OCH3 here as it may react with the peroxide initiator, but I'm not quite sure what. As far as I know, the radical is generated at the double bond which then keeps polymerizing independent of the substituent. Can somebody tell me what is the mechanism of polymerization in the (iv) part?
$\begingroup$
$\endgroup$
3
-
$\begingroup$ The second option would not yield PVC. $\endgroup$– user7951Commented Nov 3, 2018 at 9:55
-
$\begingroup$ @Loong Yep, corrected that. It should be polyvinyl cyanide, shouldn't it? $\endgroup$– Yusuf HasanCommented Nov 3, 2018 at 10:40
-
$\begingroup$ No, Polyacrylonitrile. $\endgroup$– MithoronCommented Nov 3, 2018 at 16:02
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
The compound IV is a great compound for cationic polymerization , the explanation is the oxygen which is giving electron with resonance : The form where the oxygen has a plus charge and the end carbon a minus is the reason why this compound will be able to start a radical polymerization but with a really low constant of polymerization (kinetic constant), and also the reason why the cationic polymerization will have high constant.
Compounds with strong (electron donor and tiny) resonance will not be good for radical mechanism, but if you put a phenyl group behind the ether function it will be better because the resonance effect will be "diluted".
