Stereoisomerism (AQA A Level Chemistry)

• To discuss E / Z isomers, we will use an alkene of the general formula C₂R₄:

The general alkene, C₂R₄

• When the groups R₁, R₂, R₃ and R₄ are all different (i.e. R₁ ≠ R₂ ≠ R₃ ≠ R₄), we have to use the E / Z naming system
  • This is based on Cahn-Ingold-Prelog (CIP) priority rules

• To do this, we look at the atomic number of the first atom attached to the carbon in question
  • The higher the atomic number; the higher the priority

• For example, 2-bromo-1-propen-1-ol has four different atoms or groups of atoms attached to the C=C bond
  • This means that it can have two different displayed formulae:

2-Bromo-1-propen-1-ol (compounds A and B)

Compound A

• Step 1: Apply the CIP priority rules
  • Look at R₁ and R₃:
    • Bromine has a higher atomic number than hydrogen so bromine has priority

  • Look at R₂ and R₄:
    • Oxygen has a higher atomic number than carbon so oxygen has priority

• Step 2: Deduce E or Z
  • E isomers have the highest priority groups on opposite sides of the C=C bond, i.e. one above and one below
    • The E comes from the German word "entgegen" meaning opposite

  • Z isomers have the highest priority groups on the same side of the C=C bond, i.e. both above or both below
    • The Z comes from the German word "zusammen" meaning together

  • In compound A, the two highest priority groups are on opposite sides (above and below) the C=C bond
    • Therefore, compound A is E-2-bromo-1-propen-1-ol

Compound B

• Step 1: Apply the CIP priority rules
  • Look at R₁ and R₃:
    • Bromine has a higher atomic number than hydrogen so bromine has priority

  • Look at R₂ and R₄:
    • Oxygen has a higher atomic number than carbon so oxygen has priority

• Step 2: Deduce E or Z
  • In compound B, the two highest priority groups are on the same side (both below) the C=C bond
    • Therefore, compound B is Z-2-bromo-1-propen-1-ol

More complicated E / Z isomers

• Compound X exhibits E / Z isomerism:

Compound X

• Step 1: Apply the CIP priority rules
  • Look at R₁ and R₃:
    • Carbon is the first atom attached to the C=C bond, on the left hand side

  • Look at R₂ and R₄:
    • Carbon is the first atom attached to the C=C bond, on the right hand side

  • This means that we cannot deduce if compound X is an E or Z isomer by applying the CIP priority rules to the first atom attached to the C=C bond
    • Therefore, we now have to look at the second atoms attached

  • Look again at R₁ and R₃:
    • The second atoms attached to R₁ are hydrogens and another carbon
    • The second atoms attached to R₃ are hydrogens and bromine
    • We can ignore the hydrogens as both R groups have hydrogens
    • Bromine has a higher atomic number than carbon, so bromine is the higher priority
      • Therefore, the CH₂Br group has priority over the CH₃CH₂ group

  • Look again at R₂ and R₄:
    • The second atoms attached to R₂ are hydrogens
    • The second atoms attached to R₃ are hydrogens and an oxygen
    • Oxygen has a higher atomic number than hydrogen, so oxygen is the higher priority
      • Therefore, the CH₂OH group has priority over the CH₃ group

• Step 2: Deduce E or Z
  • In compound X, the two highest priority groups are on the same side (both below) the C=C bond
    • Therefore, compound X is the Z isomer