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So I was practicing for exams, and this is the specific question.enter image description here

I get that structure 2 is possible through polymerization and 3 doesn't fit to be the polymer of cyclohexene. I'm kinda consfused about 1 tho, I see that the double bonds have broken and joined with another cyclohexene, but I'm not sure about how that works. Like, is it similar to the addition of electrophiles..if so why isn't this seen with non-cyclic alkenes?

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    $\begingroup$ You're going to need an initiator of some kind. Probably a radical generator. That radical will begin a chain reaction that propagates by adding into the double bond to produce a new secondary radical. $\endgroup$ – Zhe Oct 31 '16 at 20:40
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    $\begingroup$ 1 is essentially a cyclic form of 2 with $n = 1.5$. So theoreticly, it could form. $\endgroup$ – Jan Nov 2 '16 at 1:42
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    $\begingroup$ $n = 0.75$ of course. Not sure where my brain was in November 2016 … $\endgroup$ – Jan Dec 26 '17 at 15:29
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I'm fairly sure this question is merely testing one on their ability to count atoms. If the structure has a chemical formula of $\ce{(C6H10)_m}$ where $m \in \mathbb{Z}^+$, then it is – at least in principle – a possible polymerisation product of cyclohexene $\ce{C6H10}$. Whether it is formed in real life is a different question, but my belief is that that is not part of the question, so you're probably overthinking this if you're wondering about how exactly 1 is formed.

1 ($\ce{C18H30} \equiv \ce{(C6H10)3}$) and 2 ($\ce{(C24H40)_n} \equiv \ce{(C6H10)_{4n}}$) fulfill the above criterion, but 3 ($\ce{(C21H36)_n}$) does not.

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  • $\begingroup$ I think OP went further from his answer - this isn't homework or overthinking, but curiosity about actual trimerisation reactions. $\endgroup$ – Mithoron Dec 30 '17 at 1:03
  • $\begingroup$ It's overthinking for the purpose of answering that question, so I decided not to talk about it - however, if we don't restrict ourselves to that question, it's certainl ynot a bad thing to be curious. $\endgroup$ – orthocresol Dec 30 '17 at 1:16
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This is a less mathematical, and thus slightly simpler, alternative to the user orthocresol's answer.

There's absolutely no need to get confused when coming to addition polymerisation questions. When I learnt about this reaction in my chemistry class this year, my chemistry taught us that a very simple way of handling the question is using this converting to "H form" strategy: Break the $ \pi $ bond and draw the substituents and the two carbons in the shape of the letter H, as shown in the image below. Notice that each of the two carbons have only formed three bonds and each can make another single bond to another atom.

enter image description here

Image source: https://getrevising.co.uk/https_proxy/747

enter image description here

For your question, the 'H form' of cyclohexene is shown above. Now, it is very easy to see that 1 is in fact possible because you are connecting three of these 'H units' in a circular fashion.

Actually, for this particular question, this conversion to the "H form" strategy may seem trivial but for alkenes where the substituents may be "ugly"-looking carbon chains, this strategy can be prove to be very useful.

I hope you find my answer insightful.

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I think the main reason behind the rare form of the first polymer might be that it is very less probable for three cyclohexene molecules to be aligned the way they should be in order the first polymer forms that are somewhat I thought would look like this - enter image description here

There might be other reasons, or this might not be a reason at all but this is a very logical reason.

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    $\begingroup$ There’s no stipulation that it has to react in a concerted manner. $\endgroup$ – orthocresol Dec 26 '17 at 13:55
  • $\begingroup$ Yes orthocresol I wrote this might not be the reason, please read the answer before commenting. $\endgroup$ – Sri Krishna Sahoo Dec 26 '17 at 14:02
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    $\begingroup$ Of course I read it before commenting. Even if you've already said you might be wrong, I'm well within my rights to point out why you might be wrong. $\endgroup$ – orthocresol Dec 26 '17 at 14:52

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