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Hydrocarbons

E–Z isomerism in alkenes

  • 3 things to know
  • AS Unit 2

Overview

Fifteen recall questions on E–Z isomerism in alkenes.

What you need to know

Answer each one before you reveal it.

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Key term

Why does the double bond prevent rotation and so allow E–Z isomers to exist?

Key term

But-2-ene exists as two isomers but but-1-ene does not. Explain.

Key term

Does 3-methylpent-2-ene show E–Z isomerism? Explain which group takes priority on each carbon.

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    What two conditions must be met for an alkene to show E–Z isomerism?

    Model answer

    There must be restricted rotation, which the carbon–carbon double bond provides, and each of the two doubly bonded carbon atoms must carry two different groups; if either carbon has two identical groups the two possible arrangements are the same molecule.

  2. 2

    Why does the double bond prevent rotation and so allow E–Z isomers to exist?

    Model answer

    Rotating about the double bond would require the sideways p orbital overlap of the π-bond to be broken, which needs far more energy than is available at ordinary temperatures, so the groups are locked on their own side of the double bond and the two arrangements cannot interconvert.

  3. 3

    What do the labels E and Z actually mean?

    Model answer

    The higher priority group on each of the two doubly bonded carbons is identified, and the isomer is labelled Z if those two groups lie on the same side of the double bond and E if they lie on opposite sides.

  4. 4

    How is the priority of the groups on a doubly bonded carbon decided?

    Model answer

    Priority is decided by the atomic number of the atom directly attached to the doubly bonded carbon, the higher atomic number taking priority; if the two attached atoms are the same, you move outwards to the next atoms along each branch until a difference is found.

  5. 5

    But-2-ene exists as two isomers but but-1-ene does not. Explain.

    Model answer

    In but-2-ene each doubly bonded carbon carries a hydrogen atom and a methyl group, which are different, so the methyl groups can be on the same side or on opposite sides; in but-1-ene the end carbon carries two hydrogen atoms, so swapping them produces the identical molecule.

  6. 6

    Identify the E and Z isomers of but-2-ene.

    Model answer

    On each doubly bonded carbon the methyl group outranks the hydrogen atom, so the isomer with the two methyl groups on the same side of the double bond is Z-but-2-ene and the one with them on opposite sides is E-but-2-ene.

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