Why e z isomerism

When multiple bond is part of the group, the multiple bond is treated as if it was singly bonded to multiple of those atoms. So, if we compare the order of these three groups, it is:. All Rights Reserved. Skip to content Geometric Isomers of Alkenes In the discussions about 1,2-dimethylcyclohexane in Chapter 4 , we have learned that there are two geometric isomers possible for that compound, that are cis and trans. Figure 5. With a bit of practice, it takes a few seconds to work out in any but the most complex cases.

Before you read on, have a go at working out the relative priorities of the two groups on the left-hand end of the double bond, and the two on the right-hand end.

There's another bit of rule that I haven't specifically told you yet, but it isn't hard to guess what it might be when you start to look at the problem. If you can work this out, then you won't have any difficulty with any problem you are likely to come across at this level. In both the top and bottom groups, you have a CH 2 group attached directly to the carbon-carbon double bond, and the carbon in that CH 2 group is also attached to another carbon atom.

In each case, the list will read C H H. There is no difference between the priorities of those groups, so what are you going to do about it? The answer is to move out along the chain to the next group. And if necessary, continue to do this until you have found a difference. Next along the chain at the top left of the molecule is another CH 2 group attached to a further carbon atom.

The list for this group is again C H H. But the next group along the chain at the bottom left is a CH group attached to two more carbon atoms. Its list is therefore C C H. Comparing these lists atom by atom, leads you to the fact that the bottom group has the higher priority.

The chlorine has a higher atomic number than carbon, and so the bottom right group has the higher priority of these two groups. The extra point I am trying to make with this bit of the example is that you must just focus on one bit of a chain at a time. We never get around to considering the bromine at the extreme top right of the molecule. We don't need to go out that far along the chain - you work out one link at a time until you find a difference.

Anything beyond that is irrelevant. For the record, this molecule is a Z - isomer because the higher priority groups at each end are on the same side of the double bond.

This is clearly a cis- isomer. It has two CH 3 groups on the same side of the double bond. But work out the priorities on the right-hand end of the double bond.

The two directly attached atoms are carbon and bromine. Bromine has the higher atomic number and so has the higher priority on that end. At the other end, the CH 3 group has the higher priority. That means that the two higher priority groups are on opposite sides of the double bond, and so this is an E - isomer - NOT a Z -.

Never assume that you can convert directly from one of these systems into the other. The only safe thing to do is to start from scratch in each case. Does it matter that the two systems will sometimes give different results? The purpose of both systems is to enable you to decode a name and write a correct formula.

Properly used, both systems will do this for you - although the cis-trans system will only work for very straightforward molecules. If this is the first set of questions you have done, please read the introductory page before you start. The E-Z system The problem with the cis-trans system for naming geometric isomers Consider a simple case of geometric isomerism which we've already discussed on the previous page.

How the E-Z system works We'll use the last two compounds as an example to explain how the system works. So the two isomers are: Summary E - : the higher priority groups are on opposite sides of the double bond. There may seem to be a simple correspondence, but it is not a rule.

Be sure to determine cis,trans or E,Z separately, as needed. If the compound contains more than one double bond, then each one is analyzed and declared to be E or Z. The configuration at the left hand double bond is E; at the right hand double bond it is Z. Thus this compound is 1E,4Z -1,5-dichloro-1,4-hexadiene. This is 1-chloroethyl-1,3-butadiene -- ignoring, for the moment, the geometric isomerism.

There is no geometric isomerism at the second double bond, at , because it has 2 H at its far end. What about the first double bond, at ? On the left hand end, there is H and Cl; Cl is higher priority by atomic number. Both of these groups have C as the first atom, so we have a tie so far and must look further. What is attached to this first C? For the ethyl group, the first C is attached to C, H, and H. For the ethenyl group, the first C is attached to a C twice, so we count it twice; therefore that C is attached to C, C, H.

Since the priority groups, Cl and ethenyl, are on the same side of the double bond, this is the Z-isomer; the compound is Z chloroethyl-1,3-butadiene. The first C has one atom of high priority but also two atoms of low priority. How do these "balance out"? Answering this requires a clear understanding of how the ranking is done.

The simple answer is that the first point of difference is what matters; the O wins. To illustrate this, consider the molecule at the left.

Is the double bond here E or Z? One cis and one trans. Since there is more than one pi bond, you have to specify which pi bond is cis and which is trans. In a trans alkene, the substituents are facing away from each other. When the groups try to move away from each other, they cause strain on the molecule.

All of this leads to an unhappy and higher energy cis conformation. Remember, substituents will be cis and trans if they are locked in place. Pi bonds are one way to lock them in place.

Rings are another matter. For example, in 1,2-dimethylcyclohexane, I can show both substituents going into the page or both going out of the page. If I show one going into the page and one going out of the page. They are trans to each other. Even though the carbons are sp3 and sigma bound to each other, the molecule itself cannot rotate because of the ring structure.

The only way to turn cis-1,2-dimethylcyclohexane into trans-1,2-dimethylcyclohexane, is to break open the ring, rotate, and reform the ring. What happens if we have a pi bond with 2 different atoms or groups on the sp2 carbon?

You can draw this molecule in 2 different ways. But will you compare the red methyl or red ethyl to the green methyl when choosing cis or trans?

The E Z system requires ranking the groups on either side of the pi bond. We must determine if the higher priority groups are next to each other, Z think cis , or away from each other, E think trans.

The video below is from my chirality series, but teaches this concept in detail. Start watching from Not mass of the group, not size of the group. The higher the atomic number of the atom directly attached, the higher the priority. This applies to molecules that have more than just 1 carbon on either side of double bond.