How many stereoisomers are possible for hepta-1,6-diene?

I am asked to identify the number of stereoisomers that are possible for hepta-1,6-diene $(\ce{H2C=CHCH2CH2CH2CH=CH2})$.


I drew the bond-line structure as follows, and I think that there can be *cis* and *trans* isomers, but my answer is wrong. The correct answer states that since neither of the double bonds exhibit stereoisomerism, this compound does not have any stereoisomers. 


Originally, I thought that single bonds can be considered too. But I think I my error is that by looking at the single bonds, we are considering the constitutional isomers of this compound, rather than the stereoisomers (which are determined by the 'fixed' double bonds). Since double bonds lack the free rotation that single bonds have, for any compound that may exhibit stereoisomerism, we can only consider the double bonds. Also, are we considering only one double bond at a time? 


Is this reasoning correct? Can someone clarify and expound more on this please?


![enter image description here](https://i.stack.imgur.com/Nysn2.png)




**In the case of your original drawing with the terminal double bonds, there are no *E* or *Z* (*trans* or *cis*) isomers!**


![cis and trans-substituted double bonds](https://i.stack.imgur.com/PpmJI.png)


*E* and *Z* isomers are only possible if both ends of the $\ce{C=C}$ double bond each carry a substituent other than a hydrogen atom.




---


As far as the other structure is concerned, *E* and *Z* isomers look as follows:


![ethyl-isopropyl-substituted double bond](https://i.stack.imgur.com/96Q1n.png)


 5

There are no stereoisomers (cis/trans) of this compound because both the double bonds have three identical substituents and one different substituent. In order to have stereoisomerism across a double bond you need to have two different groups on each side of the bond.


For example but-2-ene has one methyl group and one hydrogen on each side of the double bond and so has two stereoisomers: *trans*-but-2-ene and *cis*-but-2-ene: 


![enter image description here](https://i.stack.imgur.com/2lw0h.jpg)


For double bonds with at least three different substituents the [E/Z notation](http://en.wikipedia.org/wiki/E-Z_notation) system is used. Z denotes substituents which are on the same side of a double bond and E denotes those on the opposite side. The four substituents are assigned a priority in accordance with the [Cahn-Ingold-Prelog priority rules](http://en.wikipedia.org/wiki/Cahn%E2%80%93Ingold%E2%80%93Prelog_priority_rules) and then the stereochemistry is assigned based on the relative position of the two highest priority substituents. In this example, bromine has a higher priority than chlorine so the label is assigned based on the position of the bromine and the chlorine across the double bond, making this the Z isomer.


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


 2

There is no Cis/Trans isomerism across the double bonds because the molecule you have there is a terminal alkene which is an alkene with double bonds at the first 2 and last 2 carbons or just at the first 2 carbons if there are no higher priority groups or just at the last 2 carbons if there are higher priority groups closer to the end with all single bonds. 


If you had other substituents you could have R and S stereoisomers and even combinations of R and S at different chiral carbons.


When you look at the differences in bonding than yes you are looking at structural or constitutional isomers.


But you can use cis/trans for multiple double bonds. For example oct-2,6-diene(or 2,6-octadiene) can be:


oct-(2 cis, 6 cis)-diene


oct-(2 cis, 6 trans)-diene


oct-(2 trans, 6 trans)-diene


or


oct-(2 trans, 6 cis)-diene


 2