Do more double bonds cause a bigger number of stereoisomers in cycloalkenes?

If *trans* double bonds cause cycloalkenes to be chiral, does a second *trans* or maybe *cis* double bond induce another chiral plane and hence four stereoisomers are possible for the given cyclodiene?


The question is based upon these previous questions:
[Are all higher cycloalkenes chiral?](https://chemistry.stackexchange.com/questions/35493/are-all-higher-cycloalkenes-chiral)
[Why is trans-cyclooctene chiral?](https://chemistry.stackexchange.com/questions/35470/why-is-trans-cyclooctene-chiral)





> 
> If trans double bonds cause cycloalkenes to be chiral
> 
> 
> 


Keep in mind that a *trans* double bond in a cycloalkene does not necessarily produce a chiral cycloalkene. Only when the chain connecting the ends of the double bond is short enough so that the *trans* double bond is twisted will a chiral conformation exist. 


So in large *trans*-cycloalkenes the double bond is not twisted and cannot produce chirality. 


Remember too, that even if the *trans* double bond is twisted and a chiral conformation results, the molecule may still be optically inactive if the chiral conformation rapidly interconverts with its enantiomeric conformation. This latter point was discussed in one of your [earlier questions](https://chemistry.stackexchange.com/questions/35493/are-all-higher-cycloalkenes-chiral) where it was pointed out that



> 
> At room temperature, *trans*-cyclooctene is the only *trans*-cycloalkene
>  where the two enantiomeric conformations are stable enough (*e.g.* the
>  barrier to reach the planar conformation is high enough) to be
>  isolated. If we lower the temperature to 0°C, then we can isolate the
>  enatiomers of trans-cyclononene. 
> 
> 
> 


Back to your question.



> 
> does a second trans or maybe cis double bond induce another chiral
>  plane
> 
> 
> 


The key here is that, as mentioned above, if the second double bond is also twisted, then there will be the potential for an increase in the number of chiral stereoisomers (of course inserting a double bond will always increase the number of stereoisomers because we have additional *cis* and *trans* double bond possibilities, but I think your question is focused on chiral stereoisomers, not geometric stereoisomers).


[![enter image description here](https://i.stack.imgur.com/aJyD5.png)](https://i.stack.imgur.com/aJyD5.png)


Look at the above figure on the left, of a *trans* double bond in a small ring. We are sighting down a *trans* double bond. To alleviate ring strain, the double bond if forced to **twist**. It is this twisting out of plane that infers chirality upon the double bond. Its insertion into the ring will produce an additional chiral plane and the R or S chirality of the new double bond will need to be specified in the name of the compound.


Note the figure on the right of a *cis* double bond in a small ring. To alleviate strain here, the $\ce{C\_{ring}-C=C}$ angle is pinched down; there is no out of plane twisting motion, the plane of the double bond still exists. Therefore, in the *cis* double bond case, there is no chirality associated with the double bond; no additional chiral isomers will be produced.



> 
> and hence four stereoisomer are possible for the given cyclodiene?
> 
> 
> 


Not always. Just as in more common compounds with 2 chiral centers, *meso* stereoisomers are possible. As an example, if the known E,E-cycloocta-1,5-diene 
adopted a conformation as pictured below, then it has a plane of symmetry and would be an achiral *meso* compound where one *trans* double bond has the R configuration and the other double bond the S configuration.


[![enter image description here](https://i.stack.imgur.com/cx9pm.png)](https://i.stack.imgur.com/cx9pm.png)


([image source](http://wtt-pro.nist.gov/wtt-pro/index.html?cmp=~e.e~-1.5-cyclooctadiene))


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