Isomers help needed

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I need help on 4ci. I can't understand at all how you understand that question nd I don't even get hoe the given diagram of neural show the Z isomer. Please help me.

Thanks so much!

For cis/trans isomerism, you should have two different substituents on one of the alkene carbon (usually) so that you can distinguish whether the two groups of substituents on either side of the carbons of alkenes are positioned relative to each other. read up on nomenclature(naming system) of alkenes if still unsure

Z refers to cis (same side), E refers to trans (opposite side)

so there are two alkene double bonds for neral but the terminal alkene has two methyl groups which are the same on the same side of that alkene carbon hence E/Z isomerism is not existent at this alkene

so the only E/Z isomer possible is with the other alkene functional group.

sketch out the two other geometrical (E/Z) isomers with neral to see for yourself clearly

By this do you mean the alkenes in metal on the left hand side or the one on the right?
Thanks

well it is fairly obvious. there is only one alkene functional group where there are two methyl groups attached to the same alkene carbon on the same side. it should not be too difficult for you to spot it

By this do you mean the alkenes in metal on the left hand side or the one on the right?
Thanks

Look at the alkene C=C on the left of neral. Right side carbon has 2 CH3 groups bonded to it while the other carbon on the left has a hydrogen atom bonded to it as well as an aldehyde group. How does that work?

I thought if there's a C=C bond with say the left carbon having a hydrogen and methyl group bonded to it, then the right carbon has methyl group and a hydrogen bonded to it, so the 2 stuff bonded to one C of the C=C is the same as the 2 stuff bonded to the other C involved in the C=C. This does not follow for neral. So I'm confused with this.

Perhaps a little reading would freshen up your mind.


To get geometric isomers you must have:

a) restricted rotation (often involving a carbon-carbon double bond for introductory purposes);

b) two different groups on the left-hand end of the bond and two different groups on the right-hand end. It doesn't matter whether the left-hand groups are the same as the right-hand ones or not.

These are from chemguide page, you should read further (it got illustrations to help depict the situation better).
http://www.chemguide.co.uk/basicorg/isomerism/geometric.html