Stereoisomerism in Alkenes (OCR AS Chemistry A)

Types of Stereoisomerism

• One of the features of alkenes is their stereoisomerism, which can be classed as E / Z or cis / trans isomerism

  • Cis / trans isomerism occurs where two of the atoms or groups of atoms attached to each carbon atom of the C=C bond are the same

  • E / Z isomerism is an example of stereoisomerism where different atoms or groups of atoms are attached to each carbon atom of the C=C bond 

Cis / trans isomers

• In saturated compounds, the atoms / 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

• 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 / trans nomenclature can be used to distinguish between the isomers

  • Cis isomers have two 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 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

Deducing E / Z Isomers

• Cis / trans naming is relatively straightforward when you only have two different groups on either side of the double bond

• However, with three or four different groups the cis / trans system doesn't work, so it has been superseded by E / Z isomerism, which covers all possibilities when 2, 3 or 4 different groups are involved.

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 the E / Z naming system

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

E / Z isomers

• 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, 1-bromo-1-propen-2-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:

1-bromo-1-propen-2-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-1-bromo-1-propen-2-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-1-bromo-1-propen-2-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