Why is the stability of (Z)-cyclodecene 2 greater than that of (E)-cyclodecene 1?

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I'd think that since 1 contains a trans double bond instead of a cis double bond, it should experience less steric repulsion and hence be more stable.

However, I've read that the cis isomer 2 is more stable than the trans isomer 1 if the ring size is 11 or smaller. For ring sizes larger than 11, the opposite is true.

Why is this so? Also, what would happen if there were more double bonds?


1 Answer 1


Usually trans-olefins are more stable than their cis isomers for steric reasons, like you suggested. However in small and medium size rings this is not the case; here the cis-cycloalkene is more stable than the corresponding trans isomer.

trans-Cyclooctene is the smallest trans-cycloalkene that is stable at room temperature (trans-cyclohexene and trans-cycloheptene have been detected as short-lived intermediates). In your drawing up above of trans-cyclodecene (a) draw in the two hydrogens on the trans double bond. One of them is "inside" the ring and causes steric problems with the carbons on the other side of the ring as pictured below. Because of this destabilizing transannular interaction trans-cycloalkenes with up to around 11 carbons do not exist as planar molecules. They exist in a conformation as pictured below for trans-cyclooctene.

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(image source)

If you can, build a model of trans-cyclooctene. Notice how you have to pull and stretch the methylene chain in order to attach that last methylene back onto the other end of the double bond - there is a lot of strain in this molecule. Now build a model of cis-cyclooctene; it is easy to loop the methylene chain from one end of the double bond to the other end.

Not until you get to around trans-cyclododecene (12 carbons) do you have enough carbons to easily connect from one end of the trans double bond back to the other end. trans-Cycloalkenes with less than 12 carbons have additional strain because of this difficulty in spanning the double bond. Once you get to around 12 carbons there is no difficulty spanning the double bond and so once again trans-cycloalkenes become more stable than their cis counterparts.

Since double bond carbons are $\mathrm{sp^2}$ hybridized with 120° angles, if we add more double bonds, the 120° angles make it more difficult to loop the more flexible $\mathrm{sp^3}$ methylene (with their 109.5° angles) chain around the double bonds.

  • 2
    $\begingroup$ Brilliant answer, thank you, but I was also wondering if there was a way to understand this without making a model (although it would certainly make it easier to understand)? $\endgroup$
    – tkhanna42
    Commented May 6, 2015 at 15:19
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    $\begingroup$ Just try to imagine how hard it would be to loop the methylene chain from one end of a trans double bond to the other, when the chain is relatively short. $\endgroup$
    – ron
    Commented May 6, 2015 at 15:34
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    $\begingroup$ In figure b of the question, wouldn't the 2 sp3 hydrogens from both sides contribute to hindrance? $\endgroup$
    – Mixcels
    Commented May 6, 2015 at 17:07
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    $\begingroup$ @Mixcels Yes, probably some, but the big effect is from having to "stretch" 6 carbons from one end of the double bond to the other. $\endgroup$
    – ron
    Commented May 6, 2015 at 18:18
  • $\begingroup$ The problem with figure b here is that it's wrong. The lowest energy conformer is a crown conformation, it's hard to draw but this one isn't one. you really need to look at it in 3D like here comporgchem.com/blog/?p=1709 . The strain itself isn't a big problem, those are quite stable too. $\endgroup$
    – DSVA
    Commented Nov 29, 2016 at 19:56

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