0
$\begingroup$

I am having trouble counting the stereo-isomers in this structure:

enter image description here

There are no chiral carbons. The double bonds means there maybe cis-trans geometrical isomerism but the groups are different. So that approach failed.

Answer given is: 4

$\endgroup$
1
  • 1
    $\begingroup$ The only thing that is different when the groups are different is that cis/trans becomes known as Z/E. Other than that, it is very much still there. $\endgroup$ Feb 21, 2018 at 10:17

1 Answer 1

1
$\begingroup$

Though all the four groups around double bond are different, the geometrical isomerism is still present. If the groups are not same then the highest priority group on one side of the carbon is determined through CIP sequence rules.

Now if two highest priority groups are on the same side of double bond (syn position), the isomer is called Z-isomer. If two highest priority groups are on the opposite side of double bond (anti-position), that isomer is called E-isomer.

Here there will be four combinations, i.e (2Z,6Z); (2Z,6E); (2E,6Z); (2E, 6E). Thus, it gives four possible geometrical isomers.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.