If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

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.

Want to join the conversation?

  • orange juice squid orange style avatar for user MatthewRMcCann
    What happens if you have a complete tie on one side? I'm guessing it just doesn't get an assignment.
    (9 votes)
    Default Khan Academy avatar avatar for user
    • leafers tree style avatar for user justjacjab
      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.
      (13 votes)
  • purple pi pink style avatar for user Michelle Verstraaten
    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.

    Thanks in advance!
    (4 votes)
    Default Khan Academy avatar avatar for user
    • mr pants purple style avatar for user Ryan W
      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.
      (7 votes)
  • leaf green style avatar for user robinnem
    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?
    (6 votes)
    Default Khan Academy avatar avatar for user
  • piceratops ultimate style avatar for user Robert Turtle
    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.
    (3 votes)
    Default Khan Academy avatar avatar for user
  • duskpin seedling style avatar for user mthayagan
    What happens if the two groups wth higher priority are on the same carbon? Would that have a Z configuration?
    (2 votes)
    Default Khan Academy avatar avatar for user
  • leafers seedling style avatar for user buingocmyan
    Between a branch (eg. isopropyl) and a long chain (>5C), which one has higher priority?
    (2 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user flyingkart.up
    Sir incase of tie what we do
    (2 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user Aneesh Mamidi
    At , why are there single bonds between the carbon and 2 oxygens? Isn't there only one oxygen?
    (1 vote)
    Default Khan Academy avatar avatar for user
  • leaf green style avatar for user William@11
    If E-Z and cis-trans can both apply, which one do you use?
    (1 vote)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user Sea
    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)
    Default Khan Academy avatar avatar for user
    • leaf red style avatar for user Richard
      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)

Video transcript

- [Voiceover] If we look at the molecule on the left and try to use our cis-trans terminology, we quickly realize that we can't use it. To use cis or trans, we would need to have two identical groups to compare. And here we have four different groups attached to our double bond, so we need to use a different system to find the configuration of our double bond. We're going to use the E-Z system. So to use the E-Z system, you need to think about atomic number to assign priority to the groups attached to your double bond. So let's start with the carbon on the right side of our double bond. We look at the atoms directly bonded to that carbon. There's a hydrogen directly bonded to that carbon, and there's an oxygen directly bonded to that carbon. Next, you need to compare those atoms in terms of atomic number. Hydrogen has an atomic number of one, and oxygen has an atomic number of eight. The higher the atomic number, the higher the priority, so the group that contains oxygen is the higher priority group. So this is the higher priority group, and the hydrogen would be number two here. Now let's look at the left side of our double bond, so let's look at this carbon, and we look at the atoms directly bonded to that carbon. There's a bromine directly bonded to that carbon, and there's a carbon. We go over here, and we see that carbon has atomic number of six, and bromine has an atomic number of 35. The higher the atomic number, the higher the priority, so the bromine gets higher priority, so the bromine would get a number one, and we give a number two to the methyl group. Next, I like to draw a line to think about sides of our double bond. If our two higher priority groups are on opposite sides of our double bond, that is the E configuration, and the E comes from the German word for opposite. So if the two higher priority groups are on opposite sides of the double bond, it's the E configuration. Let's look at the example on the right. So it's very similar. If we start on the right side of the double bond, this carbon is still bonded to an H and an OH, and the OH group gets a higher priority because the oxygen has the higher atomic number, so this gets a one, and the hydrogen gets a two. So that's the same as the previous example. When we go to the left side of the double bond and we look at this carbon, now I've switched the methyl group and the bromine. We know the bromine gets higher priority because the bromine has the higher atomic number, and so the methyl group gets a number two. So when I draw a line here, it's easier to see that the two higher priority groups are on the same sides. The OH and the bromine are on the same side of our double bond. This is the Z configuration. Z comes from the German word meaning together, so the two higher priority groups are together. They're on the same side. A good way to remember this is to think about the two higher priority groups being on ze zame zide, and that's one way to remember that the two higher priority groups being on the same side is Z. So the E-Z system is more inclusive than the cis-trans terminology, so E-Z is often a better way to come up with the configuration of a double bond. Let's assign a configuration to this double bond, and let's start with the carbon on the left side. We look at the atoms directly bonded to that carbon. There's a bromine, and there's a chlorine. Since bromine has the higher atomic number, bromine gets the higher priority, so bromine gets a one, and chlorine gets a two. If we go to the right side of our double bond and we look at the atoms directly bonded to this carbon, we know that here's a carbon, and we know that here's a carbon, so we have a tie because obviously carbon has the same atomic number. So to break a tie, we need to keep going. So what I'm going to do is re-draw this molecule, and I'm going to do that over here on the right. So we have a carbon bonded to a bromine and a chlorine, and the carbon on the right is bonded to another carbon. And this carbon, let me change colors here, this carbon is this one, which is double bonded to an oxygen. So for the purpose of assigning priority, we're going to pretend like this carbon is bonded to two oxygens. Obviously, it has a double bond to only one oxygen, but this will help us to assign priority. And then we also have this carbon bonded to a hydrogen. What about this carbon down here? Well, this carbon is bonded to two hydrogens and directly bonded to an oxygen, and the oxygen is bonded to a hydrogen. So let's go back to thinking about priority. We started with the carbon on the right side, and we got to these two carbons, and then we had a tie. So, next, we need to think about what atoms are directly bonded to those carbons. So we'll start with this carbon up here. This carbon is directly bonded to an oxygen, an oxygen, and a hydrogen, so we write down oxygen, oxygen, hydrogen. Down here, this carbon is directly bonded to an oxygen and two hydrogens, so oxygen, hydrogen, hydrogen. To assign priority, we look for the first point of difference. So we start with the two oxygens, and we have a tie, so we go to the next atom, and we have an oxygen versus a hydrogen. Obviously, oxygen has a higher atomic number than hydrogen, so this is our tiebreaker. Right here is our first point of difference, and so this group gets higher priority, so this group gets a number one for priority, and this group gets a number two. Next, we draw a line here for our double bond, and we look at our two higher priority groups. Our two higher priority groups are on the same side, so ze zame zide. This is the Z configuration for our double bond. So how do we incorporate the E-Z system into IUPAC nomenclature? Well, let's say our goal is to name this alkene. Well, we'd find the longest carbon chain that includes our double bond, and we'd give our lowest number possible to our substituents, so that means starting from the right side here. So this would be carbon one, carbon two, carbon three, four, five, six, and seven. So what do we call a seven carbon alkene? That would be heptene. So over here I will write heptene. And our double bond starts at carbon three, so let me write 3-heptene. Next, we think about the substituents coming off of our carbon chain. We have a methyl group coming off of carbon two and an ethyl group coming off of carbon four. Now we need to put those in alphabetical order. So 4-ethyl would come first, so let me go ahead and put that right here, so 4-ethyl, and then we have 2-methyl 3-heptene. So that's the older way of naming it. You could've also have written here 4-ethyl 2-methyl hept-3-ene, so either one is accepted. Now we have to think about our double bonds. Is it an E configuration or is it a Z configuration? So let's start with the carbon on the right side of our double bond here. We know there's a hydrogen coming up this way, and so we are comparing the atoms directly bonded to that carbon in red. So we have a hydrogen here, and then we have a carbon right here. Carbon has the higher atomic number, so this group gets higher priority, so we get a number one for this group, and hydrogen would get a number two. For the left side, so we're thinking about this carbon, we think about the atoms directly bonded to that carbon. Well, there's a carbon here and a carbon here, so there's a tie, so I'm actually going to re-draw part of the molecule here to help us figure out the higher priority group to help us break that tie. So here's our double bond. Here's our hydrogen. Here's our carbon. That's the right side of our double bond. Again, only drawing in part of the molecule. On the left side of our double bond, we would have a carbon down here bonded to two hydrogens, and then this carbon has three hydrogens, so that's our ethyl group. And then up here, we would have a carbon bonded to two hydrogens bonded to another carbon with two hydrogens, and finally bonded to a carbon with three hydrogens. Now we have to figure out which of those two groups on the left side is the higher priority group. So, again, let's start with the carbon on the left side of our double bond. We look directly at the atom bonded to that, and it's a carbon, so we need to continue on. So this carbon is directly bonded to a carbon and then a hydrogen and then a hydrogen, so we have carbon, hydrogen, hydrogen. What about this carbon? That carbon's bonded to a carbon and two hydrogens, so we have carbon, hydrogen, hydrogen. We look for first point of difference. We have carbon versus carbon, hydrogen versus hydrogen, and hydrogen versus hydrogen, so we don't have anything. We still don't know which is the higher priority group, so we need yet another tiebreaker. So let's go to our next carbon. So that's this carbon right here versus this carbon down here. We'll go back up to this top carbon. This carbon is bonded to a carbon and then hydrogen, hydrogen, so we have carbon, hydrogen, hydrogen. Down here, let's look at this carbon. We have hydrogen, hydrogen, hydrogen, so hydrogen, hydrogen, hydrogen. Finally, we have our first point of difference because we're comparing a carbon to a hydrogen, so carbon has the higher atomic number, so this group gets higher priority. So this group gets the higher priority. Let's go back up to this drawing. We're talking about this group, higher priority, and our ethyl group gets a number two here. So, finally, we go ahead and draw in our line, so we can see which side our higher priority groups or which sides, I should say, our higher priority groups are on. So our higher priority groups are on opposite sides of our double bond, so we know that is E. So the configuration of the double bond is E, so we put that in our names. We put our E here, and we put it in parentheses. So the final name for our compound is (E)-4-ethyl-2-methyl-3-heptene, or you could say (E)-4-ethyl-2-methylhept-3-ene.