Unit 2 · Spec 3.3.4

Alkenes: Structure, Electrophilic Addition & Polymers

Carbon-carbon double bond geometry, E/Z isomerism, electrophilic addition mechanisms, carbocation stability hierarchy, and addition polymerisation.

1. Structure, Bonding and Stereoisomerism

Alkenes are unsaturated hydrocarbons containing at least one carbon-carbon double bond (C=C):

  • σ (sigma) bond: Beyond the OxfordAQA specification: formed by the direct end-on overlap of hybridised orbitals along the internuclear axis. Strong and stable.
  • π (pi) bond: The π bond prevents rotation about the double bond and has high electron density, making alkenes open to attack by electrophiles. Beyond the OxfordAQA specification: it is formed by the sideways overlap of adjacent unhybridised \(p\) orbitals above and below the carbon-carbon plane.
  • Trigonal Planar Geometry: Each carbon in the double bond is surrounded by 3 bonding regions, producing a planar arrangement with bond angles of 120°.

E/Z Stereoisomerism

Restricted rotation around the C=C bond gives rise to \(E/Z\) isomerism when each carbon atom of the double bond is bonded to two different groups:

  • Cahn-Ingold-Prelog (CIP) Priority Rules: Assign priority to atoms directly attached to each double-bonded carbon based on their atomic number (\(Z\)). Higher atomic number = higher priority.
  • Z-isomer (zusammen / together): High priority groups are on the same side of the double bond plane.
  • E-isomer (entgegen / opposite): High priority groups are on opposite sides of the double bond plane.

For the full CIP rules (including the first point of difference) and drawn E and Z examples, see Introduction to Organic Chemistry.

2. Electrophilic Addition Mechanisms

Electrophile

An electron-pair acceptor. Attracted to regions of high electron density, such as the π cloud of a C=C double bond.

Electrophile Reagent & Conditions Intermediate Product
HBr / HCl Room temperature, gaseous or concentrated aqueous Carbocation Halogenoalkane
Bromine (Br₂) Bromine (pure, or dissolved in an organic solvent), room temperature Carbocation 1,2-dibromoalkane
Conc H₂SO₄ Cold concentrated sulfuric acid Carbocation Alkyl hydrogensulfate (hydrolysed by warm water to form alcohol)

HBr with propene

  1. The C=C electrons attack the δ+ H of H-Br. The H-Br bond pair moves onto Br. This gives a carbocation and Br⁻.
  2. Br⁻ uses a lone pair to attack the positive carbon.
Electrophilic addition: HBr with propene H₃C C H C H H H δ+ Br δ− H₃C C + H CH₃ secondary carbocation Br − CH₃CHBrCH₃ 2-bromopropane (major)

Br₂ with ethene

  • Br₂ is non-polar. As it approaches, the C=C electrons repel the electrons in the Br-Br bond. This induces a dipole: Brδ+-Brδ−.
  • The C=C attacks the Brδ+. The Br-Br bond pair moves onto the other Br. This gives a carbocation and Br⁻.
  • Br⁻ then attacks the carbocation. The product is 1,2-dibromoethane.
Electrophilic addition: Br₂ with ethene C C H H H H Br δ+ Br δ− induced dipole C + C H H H H Br Br − CH₂BrCH₂Br 1,2-dibromoethane

Concentrated H₂SO₄ with ethene

  • Addition of cold concentrated H₂SO₄ gives an alkyl hydrogensulfate. With ethene this is ethyl hydrogensulfate, CH₃CH₂OSO₃H.
  • The mechanism is like HBr: the C=C attacks a δ+ H, then the hydrogensulfate ion attacks the carbocation.
Electrophilic addition: H₂SO₄ with ethene C C H H H H H δ+ O δ− SO₃H C + C H H H H H O − SO₃H CH₃CH₂OSO₃H ethyl hydrogensulfate Cold concentrated H₂SO₄

Warming the alkyl hydrogensulfate with water hydrolyses it to the alcohol:

\[ \text{CH}_3\text{CH}_2\text{OSO}_3\text{H} + \text{H}_2\text{O} \rightarrow \text{CH}_3\text{CH}_2\text{OH} + \text{H}_2\text{SO}_4 \]

Bromine water test for C=C

Shake the compound with bromine water. An alkene turns it from orange to colourless.

3. Carbocation Stability and Markovnikov's Rule

When an unsymmetrical alkene (such as propene) reacts with an unsymmetrical electrophile (such as HBr), two different carbocation intermediates can form. This gives a major and a minor product.

  • Propene + HBr gives mainly 2-bromopropane, via the secondary carbocation.
  • Secondary carbocations are more stable than primary ones, because alkyl groups release electrons and stabilise the positive charge.

Carbocation Hierarchy

\[ \text{Tertiary } (3^\circ) > \text{Secondary } (2^\circ) > \text{Primary } (1^\circ) \]

Positive Inductive Effect: Alkyl groups are electron-releasing. They push electron density towards the positively charged carbon, dispersing the positive charge and stabilising the carbocation.

Markovnikov's Rule

In electrophilic addition to an unsymmetrical alkene, the hydrogen atom attaches to the carbon that already has the greater number of hydrogen atoms.

The reaction proceeds via the more stable carbocation intermediate, yielding the major product.

Worked Example: Addition of HBr to Propene
Predict and explain the major product formed when propene reacts with hydrogen bromide.

Propene (CH₃CH=CH₂) can add H⁺ in two ways:

  • Route A: H⁺ adds to CH₂, forming a secondary carbocation (CH₃CH⁺CH₃) attached to two electron-releasing methyl groups.
  • Route B: H⁺ adds to CH, forming a primary carbocation (CH₃CH₂CH₂⁺) attached to only one alkyl group.

The secondary carbocation is more stable due to greater positive inductive charge dispersal. Attack by Br⁻ produces 2-bromopropane as the major product (1-bromopropane is the minor product).

4. Addition Polymers

Alkenes undergo addition polymerisation under high pressure and catalyst conditions. The π bond breaks, and thousands of monomer units join to form long saturated polymer chains.

Naming addition polymers

Write poly, then the monomer name in brackets. Propene gives poly(propene). Chloroethene gives poly(chloroethene), also called PVC.

From monomer to repeating unit

  • Draw the C=C as a C-C single bond.
  • Keep all the groups on the two carbons.
  • Draw a bond out of each end, through square brackets, and add n.
Monomer to repeating unit: poly(propene) and poly(chloroethene) n C C H H H CH₃ propene (monomer) C C H H H CH₃ n poly(propene): repeating unit n C C H H H Cl chloroethene (monomer) C C H H H Cl n poly(chloroethene), PVC: repeating unit
Worked Example: Monomer from a Chain Section
A polymer chain section is -CH₂-CHCl-CH₂-CHCl-CH₂-CHCl-. Give the repeating unit and the monomer.

Step 1: find the two-carbon unit that repeats: -CH₂-CHCl-.

Step 2: put a C=C between those two carbons: CH₂=CHCl. The monomer is chloroethene, and the polymer is poly(chloroethene).

Properties of polyalkenes

  • Unreactive: the chain is saturated, with strong non-polar C-C and C-H bonds.
  • Forces between chains: van der Waals forces in poly(ethene). PVC also has permanent dipole-dipole forces, because the C-Cl bond is polar.

PVC and plasticisers

  • Pure PVC is rigid. It is used for window frames and pipes.
  • A plasticiser's small molecules get between the chains. They push the chains apart and weaken the forces between them, so the chains can slide.
  • This makes flexible PVC, used for cable insulation and flooring.

Polymer science has developed over time; for example, poly(ethene) was discovered by accident in 1933.

Environmental Inertness

Addition polymers (such as poly(ethene) and poly(chloroethene)) have saturated chains held together by strong C − C and C − H bonds, which have little or no polarity. Poly(chloroethene) also has polar C − Cl bonds, but these are strong and unreactive. They are chemically inert and non-biodegradable, creating long-term landfill disposal challenges. Waste polymer management relies on sorting, mechanical recycling, feedstock recycling, and controlled incineration for energy recovery.

Go deeper

These UK AQA pages cover the same ideas in more detail. They may include content that is not in 9620.

5. Practice Questions

Practice Problem (3 Marks)
Draw the repeating unit of poly(propene) and explain why addition polymers are not biodegradable.
Show Answer & Mark Scheme

Repeating Unit: \(-[\text{CH}(\text{CH}_3)-\text{CH}_2]_n-\) (1 mark)

Explanation:

  • The polymer chain contains only strong, saturated C − C and C − H bonds. (1 mark)
  • The polymer chains have no C=C bond for electrophiles to attack, and their bonds are strong, so they cannot be broken down by biological organisms or chemical hydrolysis. (1 mark)
Practice Problem (4 Marks)
Outline the mechanism for the reaction of bromine with ethene. Explain how a non-polar Br₂ molecule can act as an electrophile.
Show Answer & Mark Scheme
  • The C=C electrons repel the Br-Br electrons, inducing a dipole: Brδ+-Brδ−. (1 mark)
  • Curly arrow from the C=C to the Brδ+, and from the Br-Br bond to the other Br. (1 mark)
  • Carbocation intermediate and Br⁻. (1 mark)
  • Curly arrow from a lone pair on Br⁻ to the C⁺, giving 1,2-dibromoethane. (1 mark)
Practice Problem (3 Marks)
Propene reacts with HBr to give two products. Name the major product and explain why it is the major product.
Show Answer & Mark Scheme
  • 2-bromopropane. (1 mark)
  • It forms via a secondary carbocation, which is more stable than the primary carbocation. (1 mark)
  • Alkyl groups release electrons and stabilise the positive charge. The secondary carbocation has two alkyl groups. (1 mark)
Practice Problem (2 Marks)
Explain how a plasticiser makes PVC flexible.
Show Answer & Mark Scheme
  • Plasticiser molecules get between the polymer chains and push them apart. (1 mark)
  • This weakens the forces between the chains, so the chains can slide over each other. (1 mark)