Cis-Trans & E/Z Isomers | DP IB Chemistry: HL Revision Notes 2016 (2024)

• Configurational isomerism can be seen in unsaturated compounds, cyclic structures or compounds that contain at least one asymmetric carbon (sometimes called a chiral centre)
  • These structures have the same molecular formula and order of atoms (the atoms are connected similarly to each other) but different shapes
• As previously discussed, these can be grouped into further types of isomers:
  • Cis / trans
  • E / Z
  • Optical

Cis / trans isomers

• In saturated compounds, the atoms / functional groups attached to the single, σ-bonded carbons are not fixed in their position due to the free rotation about the C-C σ-bond
  • This causes conformational isomers, as previously discussed
• In unsaturated compounds, the groups attached to the C=C carbons remain fixed in their position
  • This is because free rotation of the bonds about the C=C bond is not possible due to the presence of a π bond
• Cis / transnomenclature can be used to distinguish between the isomers
  • Cis isomers have two functional groups on the same side of the double bond / carbon ring, i.e. both above the C=C bond or both below the C=C bond
  • Trans isomers have two functional groups on opposite sides of the double bond / carbon ring, i.e. one above and one below the C=C bond

The presence of a π bond in unsaturated compounds restricts rotation about the C=C bond forcing the groups to remain fixed in their position and giving rise to the formation of certain configurational isomers

Naming cis / trans isomers

• For cis / trans isomers to exist, we need two different atoms or groups of atoms on either side of the C=C bond
  • This means that 2-methylpropene cannot have cis / trans isomers as the methyl groups are both on the same side of the C=C bond:

2-methylpropene molecules do not have cis / trans isomers

• However, moving one of the methyl groups to the other side of the C=C bond causes cis / trans isomerism:

But-2-ene does have cis / trans isomers

• The atoms or groups of atoms on either side of the C=C bond do not have to be the same for cis / trans isomers:

1-chloroprop-1-ene also shows cis / trans isomerism

• However, the cis / trans naming system starts to fail once we have more than one atom or group of atoms on either side of the C=C bond
  • The cis / trans naming system can still be used with three atoms / groups of atoms but only if:
    • Two of the three atoms or groups of atoms are the same
    • These two atoms or groups of atoms are on opposite sides of the double bond
    

1,2-dichloropropene can be named using cis / trans

• • The cis / trans naming system cannot be used with three atoms / groups of atoms when they are all different
    • This requires the use of theE /Z naming system

1-bromo-2-chloropropene cannot be named using cis / trans

Cyclic cis / trans isomers

• Cis / trans isomerism can also occur in cyclic structures
  • Even though cyclic alkanes contain single carbon-carbon bonds, the rigid structure of the ring system does not allow for free rotation
    • Therefore, cis isomers can occur when the atoms (or groups of atoms) are on the same side of the ring, i.e. both above or both below
    • While trans isomers can occur when the atoms (or groups of atoms) are on the opposite side of the ring, i.e. one above and one below
    

Cis / trans isomerism in cyclic compounds

E / Z isomers

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

The general alkene, C₂R₄

• When the groups R₁, R₂, R₃and R₄ are all different (i.e. R₁ ≠ R₂ ≠ R₃ ≠ R₄), we have to use theE /Z naming system
  • This is based on Cahn-Ingold-Prelog (CIP) priority rules
• To do this, we look at theatomic 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
    • TheE 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
    • TheZ 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 isE-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 exhibitsE /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 anE orZ 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 CH₃group

• 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