OxfordAQA A-Level Chemistry: Alkenes (9620)

Here you will learn section 3.3.4 Alkenes from the OxfordAQA International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards. Its four parts come in order: structure, bonding and reactivity (3.3.4.1), addition reactions (3.3.4.2), addition polymers (3.3.4.3) and epoxyethane (3.3.4.4). Every point is International AS, with no International A-level only material. Unit 2 (Organic 1 and Physical 1) is where this section is examined, and because Unit 5 is synoptic, alkenes can turn up there as well.

Afterwards, condense it with the alkenes revision notes and try the alkenes practice questions. The OxfordAQA A-Level Chemistry hub lists every topic, the printable checklist tracks each specification point, and the free 10-minute diagnostics show your gaps. Polarity and intermolecular forces are taught in shapes, polarity and intermolecular forces, and mole and atom economy calculations in amount of substance.

Coverage

Spec point	You should be able to
3.3.4.1	Say why alkenes count as unsaturated; link the electron-rich C=C to attack by electrophiles
3.3.4.2	Draw curly-arrow mechanisms for adding HBr, H₂SO₄ and Br₂; describe the bromine water test
3.3.4.2	Predict which product dominates from an unsymmetrical alkene, using the carbocation stability order
3.3.4.3	Move between monomer, chain section and repeating unit; apply IUPAC names; account for low reactivity and the forces between chains
3.3.4.3	PVC uses and plasticisers; how polymer understanding has grown
3.3.4.4	Describe making epoxyethane and its hazards; explain its reactivity; equations and mechanisms with water and alcohols; importance of the products

Section 3.3.4 lists no required practical. The bromine test is part of Required practical 5 (tests for alcohols, aldehydes, alkenes and carboxylic acids), listed under 3.3.6.1.

3.3.4.1 Why alkenes react the way they do

Alkenes are unsaturated hydrocarbons: they have at least one C=C double bond. The double bond packs four electrons into the space between two carbon atoms, making it a centre of high electron density.

An electrophile is an electron pair acceptor, either positive or carrying a δ+ atom. Electrophiles are drawn to the double bond and take a pair of electrons from it. One carbon–carbon bond breaks and two new single bonds form: electrophilic addition. Alkanes have no such electron-rich region.

3.3.4.2 Electrophilic addition mechanisms

A curly arrow that makes a bond starts from a lone pair or a bond; one that breaks a bond starts from that bond. Each mechanism has the same shape:

1. Arrow from the C=C bond to the δ+ atom of the electrophile; a second arrow from the bond in the electrophile to its δ− atom. A carbocation (positive carbon) and a negative ion form.
2. Arrow from one lone pair of the anion to C⁺.

Hydrogen bromide

Bromine is more electronegative than hydrogen, so H–Br is polar, Hδ+–Brδ−. Cyclopentene is symmetrical, so it gives one product.

Step 1   cyclopentene  +  Hδ+–Brδ−   →   C₅H₉⁺ (cyclopentyl cation)  +  Br⁻
         arrow a: from the C=C bond to the H
         arrow b: from the H–Br bond to the Br
Step 2   C₅H₉⁺  +  :Br⁻   →   C₅H₉Br   (bromocyclopentane)
         arrow c: from a lone pair on Br⁻ to C⁺

Bromine, and the test for unsaturation

Br₂ is non-polar. Near the double bond, the high electron density repels the electrons in Br–Br, so the nearer Br becomes δ+ (an induced dipole) and acts as the electrophile.

Worked example 1. Outline the reaction of hept-1-ene with bromine.

Step 1   CH₃(CH₂)₄CH=CH₂ + Brδ+–Brδ−  →  CH₃(CH₂)₄C⁺H–CH₂Br + Br⁻
         arrows: C=C bond → δ+ Br;  Br–Br bond → δ− Br
Step 2   CH₃(CH₂)₄C⁺H–CH₂Br + :Br⁻    →  CH₃(CH₂)₄CHBrCH₂Br
         arrow: lone pair on Br⁻ → C⁺

Product: 1,2-dibromoheptane. Both added atoms are Br, so there is no major/minor pair with bromine.

Test for unsaturation: add bromine water and shake. A compound with C=C turns it from orange to colourless. A saturated compound leaves it orange.

Sulfuric acid

Cold, concentrated H₂SO₄ adds across C=C. An H of H₂SO₄ is the electrophile. The hydrogensulfate ion then bonds to the carbocation through its negatively charged oxygen.

Worked example 2. Show the product when 2,3-dimethylbut-2-ene reacts with cold, concentrated sulfuric acid.

Step 1   (CH₃)₂C=C(CH₃)₂ + H–OSO₂OH  →  (CH₃)₂C⁺–CH(CH₃)₂ + ⁻OSO₂OH
         arrows: C=C bond → H;  O–H bond → O
Step 2   (CH₃)₂C⁺–CH(CH₃)₂ + ⁻OSO₂OH  →  (CH₃)₂C(OSO₂OH)CH(CH₃)₂

The alkene is symmetrical, so both possible carbocations are the same tertiary ion: one product.

Major and minor products

In an unsymmetrical alkene the two carbons of C=C carry different groups. HBr or H₂SO₄ can then add in two ways, through two different carbocations, giving two products.

Carbocation	Alkyl groups on C⁺	Relative stability
primary	1	lowest
secondary	2	intermediate
tertiary	3	highest

Alkyl groups push electron density towards the positive carbon, spreading the charge, so more alkyl groups mean a more stable carbocation. The more stable carbocation forms more readily, giving the major product; the other route gives the minor product.

Worked example 3. 1-methylcyclohexene reacts with HBr. The ring carbon carrying the methyl group is C1; its partner in the double bond is C2. Predict and explain the major product.

• H adds to C2, leaving the positive charge on C1. C1 is bonded to the methyl group and to two ring carbons, so this is a tertiary carbocation. Br⁻ joins C1: 1-bromo-1-methylcyclohexane.
• H adds to C1, leaving the positive charge on C2. C2 is bonded to two carbons only, so this is a secondary carbocation. Br⁻ joins C2: 1-bromo-2-methylcyclohexane.
• The tertiary carbocation has three electron-releasing alkyl groups, so it is more stable. Major: 1-bromo-1-methylcyclohexane. Minor: 1-bromo-2-methylcyclohexane.

3.3.4.3 Addition polymers

Alkenes and substituted alkenes (an H replaced by another atom or group) join in long chains as each double bond opens. The polymer is the only product.

Repeating unit and monomer

The repeating unit contains the two carbons that came from one C=C, with every substituent kept on its original carbon, a single C–C bond, extension bonds at both ends, brackets and n.

Worked example 4. Draw the repeating unit of the polymer made from propenenitrile, CH₂=CHCN, and name the polymer.

  H   H                     H   H
  |   |                     |   |
  C = C        →       —(— C — C —)—n
  |   |                     |   |
  H   CN                    H   CN

IUPAC rule: poly + the monomer name in brackets, so poly(propenenitrile).

Worked example 5. Part of a polymer chain is –CF₂–CFCl–CF₂–CFCl–CF₂–CFCl–. Find the monomer.

• The two-carbon pattern that repeats is –CF₂–CFCl–. That is the repeating unit.
• Restore the double bond between those carbons: CF₂=CFCl, chlorotrifluoroethene.
• Polymer: poly(chlorotrifluoroethene).

Worked example 6. A poly(propenenitrile) sample has an average Mr of 2.12 × 10⁵. Find the average number of monomer units per chain. (Ar values: N = 14.0, C = 12.0, H = 1.0)

• Mr(C₃H₃N) = 3(12.0) + 3(1.0) + 14.0 = 53.0. The repeating unit has the monomer’s formula.
• Units = 2.12 × 10⁵ ÷ 53.0 = 4000.

Why addition polymers are unreactive

No C=C remains, so electrophiles have nothing to attack, and the backbone consists of strong C–C and C–H bonds with little or no polarity.

Forces between polyalkene chains

Poly(ethene) and poly(propene) chains are non-polar, so the forces between them are induced dipole–dipole (van der Waals) forces. Each chain has an enormous number of electrons and touches its neighbours along a great length, so the total attraction is large and the polymers are solids. Where a substituent makes a polar bond (C–Cl in PVC, C≡N in poly(propenenitrile)), permanent dipole–dipole forces act between chains as well.

PVC and plasticisers

Poly(chloroethene) is commonly called PVC. Unplasticised PVC is rigid: water and sewage pipes, doors, bank cards. A plasticiser (most commonly a phthalate ester) is blended in as small molecules that lodge among the chains and hold them apart. With the chains spaced out, the forces between them weaken and they slide more easily. Plasticised PVC is flexible: cable insulation, inflatable products, flooring.

Knowledge that developed over time

PVC was first made in 1872 by the German chemist Eugen Baumann. In 1926 Waldo Semon at the B.F. Goodrich Company found a way to plasticise it by blending it with additives, which opened up its wide range of uses.

3.3.4.4 Epoxyethane

Production and hazards

Epoxyethane, C₂H₄O, is a three-membered ring of two CH₂ groups and an oxygen atom. It is made by the partial oxidation of ethene with oxygen over a silver catalyst at about 220–280 °C and 1–3 MPa:

2CH₂=CH₂ + O₂ → 2C₂H₄O

Every reactant atom ends up in the product, so the atom economy is 100%. The hazards:

• Complete combustion of ethene to CO₂ and water competes with it and is far more exothermic, so the temperature must be controlled.
• Epoxyethane is extremely flammable and its vapour forms explosive mixtures with air.
• It is toxic and classified as a human carcinogen, so leaks must be prevented.

Why epoxyethane is so reactive

The ring forces the bond angles to about 60°, far below 109.5°, so the ring is strained and its C–O bonds are relatively weak. Each C–O bond is also polar (Cδ+–Oδ−), so the ring carbons attract nucleophiles. Breaking one C–O bond opens the ring and releases the strain.

Reactions with water and alcohols

Water and alcohols have lone pairs on oxygen and add across the ring:

C₂H₄O + H₂O   → HOCH₂CH₂OH      (ethane-1,2-diol)
C₂H₄O + ROH   → ROCH₂CH₂OH      (e.g. R = CH₃: 2-methoxyethanol)

Mechanism, shown for an alcohol ROH (for water, R = H):

Step 1  RO(H): lone pair → ring carbon (Cδ+);  ring C–O bond → O
        gives  RO⁺(H)–CH₂–CH₂–O⁻
Step 2  proton transfer from the O⁺–H to the O⁻
        gives  RO–CH₂–CH₂–OH

In industry a little dilute acid speeds up the water reaction, and a large excess of water limits further reaction of the diol (which still has OH groups) with more epoxyethane.

Worked example 7. What mass of ethane-1,2-diol forms from 1.32 tonnes of epoxyethane, assuming full conversion? (Mr: C₂H₄O = 44.0, C₂H₆O₂ = 62.0)

• Ratio 1 : 1, so mass = 1.32 × 62.0 ÷ 44.0 = 1.86 tonnes.

Importance of the products

• Ethane-1,2-diol is used as antifreeze and engine coolant, and to make polyester such as PET. Most epoxyethane produced goes to make ethane-1,2-diol. It is toxic if swallowed, so spilled or dumped coolant is a hazard, although it breaks down in the environment within weeks.
• Glycol ethers (such as 2-methoxyethanol) are solvents in paints and cleaners.
• With a long-chain alcohol, several epoxyethane molecules add in turn: R–OH + nC₂H₄O → R–(OCH₂CH₂)ₙ–OH. The hydrocarbon tail is non-polar and the chain of ether groups ending in OH is polar, so these are non-ionic surfactants, made on a very large scale for detergents. In waterways they can harm aquatic life, so their fairly rapid breakdown matters.

Common errors

• First arrow drawn from a carbon atom instead of the C=C bond; no induced dipole on Br₂.
• Hydrogensulfate bonded to carbon through sulfur.
• The tertiary carbocation called “more reactive” instead of “more stable”.
• Leaving C=C in a repeating unit, or drawing a three-carbon repeating unit.
• Polymer unreactivity blamed on “strong intermolecular forces”.
• Epoxyethane ring opening drawn as electrophilic attack; water and alcohols attack as nucleophiles.

Official syllabus

OxfordAQA, International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards: section 3.3.4 Alkenes (International AS), points 3.3.4.1 to 3.3.4.4.