The Student Room Group

Optical and geometrical isomers

Which one of the following can exhibit both geometrical and optical isomerism?
A (CH3)2C=CHCH(CH3)CH2CH3
B CH3CH2CH=CHCH(CH3)CH2CH3
C (CH3)2C=C(CH2CH3)2
D CH3CH2CH(CH3)CH(CH3)C=CH2

I’m really confused and also struggling to draw it out too Ik that optical isomerism is when a carbon has 4 diff groups attached but not sure about geometrical
Would appreciate any help
Reply 1
Geometrical isomerism's referring to E and Z isomers - it's to do with the arrangements of groups around the C=C bond. This website explains it well. https://www.chemguide.co.uk/basicorg/isomerism/ez.html
Reply 2
Original post by H_K_05
Geometrical isomerism's referring to E and Z isomers - it's to do with the arrangements of groups around the C=C bond. This website explains it well. https://www.chemguide.co.uk/basicorg/isomerism/ez.html

Thanks
Reply 3
Original post by zara ijaz
Thanks

I looked at the website and I understand what geometrical isomers are but I still don’t get the question
Could you please explain it
Thanks
Reply 4
So the first step would be to draw them all out. It may help to work outwards from the double bond bc there's some branching involved, and it will also allow you to spot geometric isomers a bit faster. After you've done that, you kinda have to do it by process of elimination.

So first you want to eliminate the ones that physically can't have geometrical isomers, i.e. when you've got two identical groups on the same side of the C=C bond. For example ethene-1-ol (if such a thing exists) couldn't have geometrical isomers, because there's two hydrogens on the same side of the C=C bond, so if you flip the groups you end up with the same chemical you had previously.

After that, you want to check if the chemical(s) which display geometrical isomerism also have a chiral centre (ie there's a carbon with 4 different groups coming off of it like you said). That should leave you with one option

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