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# E–Z system

EZ notation can be used to describe the configuration of a double bond with two or more substituents. To do so, each substituent is assigned a priority using the Cahn–Ingold–Prelog rules. If the two higher priority substituents are on same side of the double bond, the configuration of the bond is Z. If the two higher priority substituents are on opposite sides of the double bond, the configuration of the bond is E.

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• What happens if you have a complete tie on one side? I'm guessing it just doesn't get an assignment.
• Well, you're right to say that it doesn't get an assignment because then there wouldn't be 2 different configurations of that molecule. If both of the groups on one side of the double bond are the same then both drawings of that molecule represent the same configuration. E and Z are only necessary to differentiate the isomers of a molecule. All you'd have to do is flip the entire diagram over and you'd get the other one. An example of this was shown in the previous video.
• When working on the molecule around , the speaker says "This is our tiebreaker". But what if we had written HOO for the top carbon atom and OHH for the lower carbon atom? Do we have to write these atoms directly bonded to the carbons also in order of priority? Because obviously, the order in which they are written will affect what is the "first point of difference".

Another question: does it mean an ethyl group will always have a higher priority than a methyl group, and a propyl group than an ethyl group, etc.? Because this will be useful to know in assigning configurations to double bonds quickly.

• Yes you put those atoms directly bonded to that carbon in order of highest atomic weight to lowest, so it's OOH vs OHH as he has. The first point of difference is the second letter, O vs H. Obviously O wins here.

For your second bit, yes. Ethyl has a higher priority than methyl, propyl higher than ethyl, because when you compare them one will eventually have CHH where the other has HHH. Watch out for isopropyl etc. though, that will be higher priority than butyl.
• Can't the Hydrogen be on either side of the carbon? Can't it also be bonded downward from the carbon, making the configuration of the alkene Z?
• If H and CH₃ are both "down" on the left hand molecule, the configuration is Z.
The molecule is then identical with the molecule on the right, which is also Z.
• Am I correct in saying that the first examples are not strictly alkenes but enols of 2-bromopropanal and they wouldn't really have an independent existance from it i.e. they would only be present in small equilibrium concentrations in samples of 2-bromopropanal and would likely undergo rapid interconversion via tautomerisation.
• Good point! Those are enols and so would be in equilibrium with each other by way of 2-bromopropanal.
• What happens if the two groups wth higher priority are on the same carbon? Would that have a Z configuration?
• That can't be the case. With E/Z we are looking for where the highest priority groups on either side of the double bond are.

If you have a molecule like 2-methylpropene (isobutylene) it does not have E/Z isomerism.
• Between a branch (eg. isopropyl) and a long chain (>5C), which one has higher priority?
• If E-Z and cis-trans can both apply, which one do you use?
(1 vote)
• Use E/Z because most of the time you won't have a symmetric alkene so cis/trans becomes unclear.
• At , why are there single bonds between the carbon and 2 oxygens? Isn't there only one oxygen?
(1 vote)
• The determination of E/Z configuration across the double bond at in the video does not fully make sense because the initially determined longest carbon chain would be 3 carbons long. Apparently, the double bond takes priority in being the lowest number, so starting from the leftmost carbon, it is Carbon 1, 2, and 3 including the carbon in what is originally deemed as a COOH functional group bonded to the two double bonded carbons C=C. However, would this carbon not be included in the original parent chain of carbons, thus making the functional groups Br, Cl, and CH2OH with Br and Cl being the highest priority groups bonded to the C=C double bond and thus an E configuration?
(1 vote)
• E/Z configuration only applies to carbon/carbon double bonds. So when determining which configuration we are only concerned with the groups directly bonded to those two carbons which constitute the double bond. In which case we're deciding if the highest priority groups are next to each other (Z) or across from each other (E).

The other carbons belonging to the hydroxylmethyl group (-CH2OH) and the aldehyde (-CH=O) do not use E/Z configuration simply because they are not carbon/carbon double bonds. They would be involved if we were naming the molecule as a whole, but for just the E/Z part we only need to know their priority.

Also, we have an aldehyde functional group; a COOH functional group is a carboxylic acid.

Hope that helps.
(1 vote)
• At -> (onward), could we not just identify the longer substituent chain coming off of carbon 4, for brevities sake. Conversely, though, if we had any different groups we would have to go thru this process that Jay does?
(1 vote)