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5.16: E and Z Alkene Isomers

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    191240

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    Double bonds can exhibit stereoisomerism if there is more than one way to arrange two groups at opposite ends of a double bond: either on the same side of the bond axis (commonly cis) or opposite sides of the double bond axis (commonly trans). There is a different way of denoting these two stereochemical configurations that is always used if the compound has a more complicated structure. Compounds that have more than one non-hydrogen group at either end of the double bond are not normally referred to as cis or trans isomers, but are instead called E or Z isomers.

    • In Z isomers, the highest-priority groups at each end of an alkene are on the same side of the double bond axis. The double bond axis is a straight line drawn through the two atoms that are double bonded to each other.
    • Z is from the German zusammen (together).
    • In E isomers, the highest-priority groups at each end of an alkene are on opposite sides of the double bond axis.
    • E is from the German entgegen (against).

    Highest-priority is decided based on atomic number.

    Bond-line structure of (1E)-1-bromo-2-chloro-2-fluoroethene. On the left is the bromine group; on the right are the chlorine and fluorine groups. On the left end of the double bond, bromine has priority due to higher atomic number. On the right end of the double bond, chlorine has priority over fluorine.
    Figure \(\PageIndex{1}\)a: An example of an E alkene.
    Bond-line structure of (1Z)-1-bromo-1-chloroprop-1-ene. On the left are the bromine and chlorine groups; on the right are a methyl group and a hydrogen. On the left end of the double bond, bromine has priority due to higher atomic number. On the right end of the double bond, carbon has priority over hydrogen.
    Figure \(\PageIndex{1}\)b: An example of a Z alkene.

    If the first atoms attached to one end of the double bond are the same, go one bond further on each chain and compare all the atoms that come next. If one of the chains at this point contains an atom with a higher atomic number than any of the atoms on the other chain, it is given higher priority.

    Bond-line structure of an alkene. On the left end of the double bond is an ethyl group with a fluorine on the first carbon and another ethyl group with a bromine on the terminal carbon. On the right side of the double bond is a hydrogen and a methyl group. On the left, both carbons have equal priority; however, the top branch with fluorine takes priority, as bromine is further away. Comparison stops when the first difference is found along the chain from the double bond. On the right side of the double bond, carbon has priority.
    Figure \(\PageIndex{1}\)c: A more complicated example of an E alkene.

    Sometimes, going one bond further leads to the same set of atomic numbers along both branches, but there are more of the higher atomic numbered atom on one branch than the other.

    Bond-line structure of a Z-alkene. On the left of the double bond is a methyl group and a hydrogen. On the right of the double bond is an ethyl group on top and an isopropyl group on bottom. On the left side of the double bond, carbon has priority. On the right side of the double bond, the isopropyl group takes priority as there are two carbons attached to the initial carbon instead of just one terminal methyl carbon on the ethyl group.
    Figure \(\PageIndex{1}\)d: Another complicated example, this time of a Z alkene.

    Remember, this is an arbitrary set of rules that have been adopted to name these compounds so that people can refer to one isomer or the other. They may not be the rules that you would have come up with on your own.

    Exercise \(\PageIndex{1}\)

    Determine whether the following molecules are E or Z.

    Exercise 5.16.1, a through h, showing several bond-line structure of alkenes.
    Answer A.:

    Z

    Answer B.:

    E

    Answer C.:

    E

    Answer D.:

    Z

    Answer E.:

    E

    Answer F.:

    E

    Answer G.:

    Z

    Answer H.:

    E

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