Edexcel International A Level Chemistry: Alkenes (YCH11)

This guide teaches Topic 5: Alkenes of the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry specification (YCH11), Issue 1, September 2017. It covers learning outcomes 5.1 to 5.8. It is International AS content, examined in the IAS unit called Unit 1: Structure, Bonding and Introduction to Organic Chemistry. The specification notes that related topics in Units 2, 4 and 5 assume this material.

Course links: the course hub for Edexcel International A Level Chemistry and the printable Edexcel Chemistry checklist. When you have read this guide, condense it with the alkenes revision notes and test yourself with the alkenes practice questions. For the wider Unit 1 picture, see the Unit 1 structure, bonding and organic chemistry guide. Not sure where your gaps are? Try one of the free diagnostics.

Topic 5 at a glance

Outcome	Skill
5.1	Give the general formula of alkenes; explain why alkenes and cycloalkenes are unsaturated; describe the C=C as one σ bond plus one π bond
5.2	Explain geometric isomerism using restricted rotation and the groups on each carbon of the C=C
5.3	Name geometric isomers with the E–Z system and explain when cis–trans naming breaks down
5.4	Describe addition of hydrogen, halogens, hydrogen halides and steam, and oxidation by acidified potassium manganate(VII) to a diol
5.5	Use bromine or bromine water to test for C=C
5.6	Draw electrophilic addition mechanisms for bromine and hydrogen bromide with ethene, and hydrogen bromide with propene; use carbocation stability; give evidence
5.7	Describe addition polymerisation; draw a repeat unit from a monomer and the monomer from a repeat unit
5.8	Explain how biodegradable polymers and the removal of toxic incineration gases limit polymer disposal problems

5.1 Structure of alkenes

An alkene is a hydrocarbon containing a carbon–carbon double bond. Open-chain alkenes with one C=C have the general formula CₙH₂ₙ. Ethene is C₂H₄, propene C₃H₆, butene C₄H₈. A cycloalkene with one C=C in a single ring has two fewer hydrogen atoms again: CₙH₂ₙ₋₂, so cyclohexene is C₆H₁₀.

Both are unsaturated: they contain a C=C, so atoms can be added across it.

Sigma and pi bonds

• The σ bond forms by head-on (end-to-end) overlap of orbitals along the line joining the two carbon nuclei.
• The π bond forms by sideways overlap of one p orbital on each carbon. Its electron density sits above and below the plane of the molecule.

Each carbon of the C=C is trigonal planar (about 120°). The π electrons are more exposed than the σ electrons, so the π bond is weaker and attracts electrophiles (electron-pair acceptors).

Worked example 1: how strong is the π bond?

Use these values: mean bond enthalpy of C=C = 612 kJ mol⁻¹; C–C = 347 kJ mol⁻¹. Estimate the π bond enthalpy.

1. The C=C is one σ bond plus one π bond. A C–C single bond is one σ bond.
2. π bond ≈ 612 − 347 = 265 kJ mol⁻¹.
3. The π bond is weaker than the σ bond, so addition reactions break only the π bond.

5.2 Geometric isomerism

The π bond only exists while the two p orbitals stay parallel. Twisting one end of the C=C would destroy that overlap, so there is restricted rotation about a C=C bond. Groups fixed on the same side stay on the same side.

Geometric isomers exist only when both of these are true:

1. there is a C=C double bond (restricted rotation), and
2. both C=C carbons have two unlike atoms or groups attached.

Pent-2-ene, CH₃CH=CHCH₂CH₃, qualifies: one carbon has H and CH₃, the other H and C₂H₅. Pent-1-ene does not, because carbon 1 carries two hydrogen atoms.

5.3 The E–Z naming system

Cis–trans naming works when the two carbons share a common group and there are only two different groups in total. It breaks down when three or four of the groups are different, because it is no longer clear which pair “cis” describes. E–Z always works, because it ranks groups by a rule.

Method

1. Look at one carbon of the C=C. Rank its two groups by the atomic number of the atom joined directly to that carbon. Higher atomic number = higher priority.
2. If those atoms tie, move outwards and compare the atoms attached to them at the first point of difference.
3. Repeat for the other carbon.
4. Top-ranked groups together on one side: Z (zusammen, together). Top-ranked groups across from each other: E (entgegen, opposite).

Worked example 2: where cis–trans fails

2-chlorobut-2-ene is CH₃CCl=CHCH₃. Carbon 2 carries Cl and CH₃; carbon 3 carries CH₃ and H.

• “Cis” is ambiguous: it could refer to the two CH₃ groups or to another pair.
• Carbon 2: Cl (atomic number 17) beats C (6), so Cl has priority.
• Carbon 3: C (6) beats H (1), so CH₃ has priority.
• Isomer with Cl and the carbon-3 CH₃ on the same side: (Z)-2-chlorobut-2-ene. With them on opposite sides: (E)-2-chlorobut-2-ene.

Notice that the Z isomer has its two methyl groups on opposite sides, so calling it “cis” would mislead.

Worked example 3: first point of difference

3-methylhex-3-ene is CH₃CH₂C(CH₃)=CHCH₂CH₃.

• Carbon 3 carries CH₃ and CH₂CH₃. Both join through carbon, so look one atom further out. The methyl carbon bears three hydrogens; the ethyl CH₂ bears one carbon and two hydrogens. Carbon outranks hydrogen there, so ethyl wins.
• Carbon 4 carries H and CH₂CH₃: ethyl wins.
• Both ethyl groups on the same side gives (Z)-3-methylhex-3-ene; on opposite sides, (E)-3-methylhex-3-ene.

5.4 Reactions of alkenes

Reaction	Reagent and conditions	Example product from ethene
Hydrogenation	H₂, nickel catalyst, heat (about 150 °C)	ethane
Halogen addition	Br₂ or Cl₂, room temperature	1,2-dibromoethane (a di-substituted halogenoalkane)
Hydrogen halide addition	HBr, HCl or HI gas, room temperature	bromoethane (mono-substituted)
Hydration	steam, acid catalyst (phosphoric acid), high temperature and pressure	ethanol
Oxidation	acidified potassium manganate(VII), room temperature	ethane-1,2-diol

Equations:

• CH₂=CH₂ + H₂ → CH₃CH₃
• CH₂=CH₂ + Br₂ → CH₂BrCH₂Br
• CH₂=CH₂ + HBr → CH₃CH₂Br
• CH₂=CH₂ + H₂O → CH₃CH₂OH
• CH₂=CH₂ + [O] + H₂O → HOCH₂CH₂OH

[O] is oxygen from the oxidising agent; the purple solution turns colourless and the diol has an –OH on each carbon of the old C=C. With an unsymmetrical alkene, HX and steam give two products; the major one comes from the more stable carbocation (see 5.6), so propene and steam give mainly propan-2-ol.

5.5 Testing for C=C

Shake the sample with bromine water (orange-brown). An alkene decolourises it quickly at room temperature. An alkane leaves it orange-brown, because without UV light it does not react. A solution of bromine in an organic solvent works the same way.

With bromine water the main product is a bromo-alcohol (2-bromoethanol from ethene), because water competes with bromide ions in the second step. The colour change is the same.

5.6 Electrophilic addition mechanisms

An electrophile accepts an electron pair. Each curly arrow shows an electron pair moving, and the specification expects it to start from a bond or a lone pair.

Ethene and bromine

As non-polar Br₂ nears the π bond, its electrons are repelled: the near bromine becomes δ+, the far one δ−.

Step 1  arrow from the C=C (pi bond) to the Br(delta+) atom
        arrow from the Br-Br bond onto the Br(delta-) atom
        H2C=CH2 + Br-Br  ->  H2C(+)-CH2Br  +  Br(-)
                             carbocation

Step 2  third arrow: Br(-) lone pair -> C(+)
        H2C(+)-CH2Br + Br(-)  ->  BrCH2-CH2Br

Ethene and hydrogen bromide

HBr is already polar, Hδ+–Brδ−. One arrow runs from the π bond to the Hδ+; a second from the H–Br bond onto Br. The intermediate is CH₃CH₂⁺. A lone pair on Br⁻ then attacks C⁺ to give bromoethane.

Propene and hydrogen bromide

H⁺ can add to either carbon of the C=C:

• H to the end carbon gives CH₃C⁺HCH₃, a secondary carbocation (two alkyl groups on C⁺) → 2-bromopropane, major.
• H to the middle carbon gives CH₃CH₂CH₂⁺, a primary carbocation (one alkyl group) → 1-bromopropane, minor.

Stability order: tertiary > secondary > primary. Alkyl groups release electron density towards C⁺ (positive inductive effect). The more stable carbocation forms faster, so its product dominates.

Evidence for the mechanism

• Ethene in bromine water containing sodium chloride gives 1-bromo-2-chloroethane and 2-bromoethanol as well as 1,2-dibromoethane. Br adds first, then any nearby nucleophile (Br⁻, Cl⁻ or H₂O) attacks the carbocation. No 1,2-dichloroethane forms, so chloride cannot start the reaction.
• The major and minor products with propene match the order of carbocation stability.

5.7 Addition polymerisation

Alkene monomers join into one long saturated chain as each π bond opens; no other product forms.

Monomer → repeat unit: draw the two C=C carbons joined by a single bond, keep all their substituents, add a bond out of each side, put square brackets round it and write n after.

Repeat unit → monomer: take the two main-chain carbons, put a C=C between them and remove the extension bonds.

Worked example 4: Perspex

Perspex is made from methyl 2-methylpropenoate, CH₂=C(CH₃)COOCH₃.

1. Repeat unit: –[CH₂–C(CH₃)(COOCH₃)]ₙ–. The ester group hangs off the chain unchanged.
2. Mr of the monomer, C₅H₈O₂ = 5(12.0) + 8(1.0) + 2(16.0) = 100.0.
3. A chain with Mr 1.50 × 10⁶ contains 1.50 × 10⁶ ÷ 100.0 = 15 000 repeat units.

5.8 Polymer disposal

Poly(alkenes) have strong, non-polar C–C and C–H bonds, so they are unreactive and persist in landfill. The specification names two responses.

• Biodegradable polymers are designed to be broken down by microorganisms, for example polymers made from plant starch or lactic acid. Many need conditions such as industrial composting to degrade quickly.
• Removing toxic incineration gases. Burning polymers releases energy for electricity, but PVC releases hydrogen chloride. Exhaust gases pass through scrubbers where a base such as slaked lime (calcium hydroxide) neutralises it: Ca(OH)₂ + 2HCl → CaCl₂ + 2H₂O.

Further suggested practicals

The specification lists these as further suggested practicals (not Core Practicals):

• comparing alkanes and alkenes by combustion, bromine water and acidified potassium manganate(VII)
• preparing cyclohexene from cyclohexanol
• extracting limonene from orange peel by steam distillation
• preparing Perspex from methyl 2-methylpropenoate

Worked example 5: counting double bonds in limonene

Limonene, C₁₀H₁₆ (Mr 136.0), from orange peel. A 0.544 g sample reacts exactly with 32.0 cm³ of 0.250 mol dm⁻³ bromine solution. How many C=C bonds does one molecule contain?

1. n(limonene) = 0.544 ÷ 136.0 = 0.00400 mol.
2. n(Br₂) = 32.0/1000 × 0.250 = 0.00800 mol.
3. Ratio Br₂ : limonene = 0.00800 ÷ 0.00400 = 2. One Br₂ adds per C=C, so there are two C=C bonds.

Common errors

• Writing CₙH₂ₙ for a cycloalkene. One ring plus one C=C gives CₙH₂ₙ₋₂.
• Giving only one condition for geometric isomerism. You need the C=C and two different groups on each carbon.
• Ranking E–Z priority by group size instead of atomic number.
• Starting a curly arrow from a carbon atom or from Br⁻ itself rather than from the π bond or a lone pair.
• Explaining carbocation stability without electron release by alkyl groups.
• Keeping the C=C inside a repeat unit, or missing the extension bonds and n.

Where next

Summarise with the alkenes revision notes, then try the alkenes practice questions and read preparing for the Edexcel IAL Chemistry units.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry (XCH11/YCH11) specification, Issue 1, September 2017, Pearson Education Limited – Unit 1, Topic 5: Alkenes, outcomes 5.1 to 5.8.