OCR A Level Chemistry A: Alkenes (H432)

This guide teaches section 4.1.3, Alkenes, from the OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017. The section belongs to Module 4, Core organic chemistry, and its lettered learning outcomes run from 4.1.3(a) to (l); every one is taught below. H432 is the full A level. The specification states that Module 4 is assessed in Synthesis and analytical techniques (02) and Unified chemistry (03). PAG7, qualitative analysis of organic functional groups, is linked to outcome (f).

All H432 topics are listed on the OCR chemistry hub. Tick off outcomes as you go on the printable checklist, and use the free diagnostics to find weak spots.

Outcomes in 4.1.3

Outcome	What you must be able to do	Notes from the specification
4.1.3(a)	C=C as one σ-bond plus one π-bond; restricted rotation	Hybridisation not required; HSW1
4.1.3(b)	Trigonal planar shape and 120° angle at each C=C carbon	M4.1, M4.2
4.1.3(c)(i)	Stereoisomers, E/Z and cis–trans isomerism	E = trans, Z = cis only reliable with an H on each C=C carbon
4.1.3(c)(ii)	CIP rules to identify E and Z isomers	No priorities for multiple bonds inside R groups
4.1.3(d)	Possible E/Z or cis–trans isomers from a structural formula	M4.2, M4.3
4.1.3(e)	Reactivity from the low π-bond enthalpy
4.1.3(f)	Addition of hydrogen, halogens, hydrogen halides and steam	Bromine test for unsaturation; PAG7
4.1.3(g)	Define and use the term electrophile
4.1.3(h)	Draw the electrophilic addition mechanism	Carbocation or halonium intermediate accepted for halogens; HSW1, 2, 8
4.1.3(i)	Markownikoff’s rule for H–X and unsymmetrical alkenes	Carbocation stability order only; no explanation needed
4.1.3(j)	Addition polymers: repeat unit from monomer, monomer from polymer
4.1.3(k)	Benefits of processing waste polymers	HSW9, 10
4.1.3(l)	Environmental benefits of biodegradable and photodegradable polymers	HSW9, 10

4.1.3(a) and (b): The double bond and its shape

An alkene is an unsaturated hydrocarbon: it contains at least one C=C double bond, so more atoms can be added to it. The double bond is made of two different kinds of covalent bond.

• σ-bond: formed by orbitals overlapping directly between the two bonding carbon atoms, on the line joining the nuclei.
• π-bond: formed by the sideways overlap of a p-orbital on each carbon. Its electron density lies above and below the two carbons rather than on the line between them.

Because the π-bond depends on the two p-orbitals staying parallel, the carbons cannot twist about the C=C axis without breaking it. This is restricted rotation. A single C–C bond is a σ-bond only, so free rotation is possible there.

Each carbon in C=C has three regions of electron density: two single bonds and the double bond, which counts as one region. These repel as far apart as possible, so the arrangement around each carbon is trigonal planar with bond angles of about 120°. The two carbons and the four atoms attached to them all lie in one plane.

4.1.3(c) and (d): Stereoisomerism in alkenes

Stereoisomers share one structural formula, yet their atoms are arranged differently in space.

E/Z isomerism is stereoisomerism caused by restricted rotation about a C=C bond. It needs two things:

1. a C=C bond (so the groups cannot swap sides by rotation), and
2. each C=C carbon must carry two different groups.

If either carbon carries two identical groups, such as =CH₂ or =C(CH₃)₂, no E/Z isomers exist.

Cis–trans isomerism is a special case of E/Z isomerism in which two of the substituent groups on each C=C carbon are the same. Cis means those identical groups are on the same side; trans means opposite sides.

CIP priority rules

To name an isomer E or Z, rank the two groups on each carbon separately.

1. Look at the atom joined directly to the C=C carbon. Higher atomic number means higher priority. So Br > Cl > O > C > H.
2. If those atoms are the same, move out one bond and compare the sets of atoms attached there, highest first. Ethyl (C attached to C, H, H) beats methyl (C attached to H, H, H).
3. Z (zusammen): the higher-priority groups are on the same side. E (entgegen): they are on opposite sides.

Worked example 1: when cis is not Z

2-bromobut-2-ene is CH₃CBr=CHCH₃. Name the isomer in which the two methyl groups are on the same side.

1. C2 carries Br and CH₃; C3 carries CH₃ and H. Each carbon has two different groups, so E/Z isomers exist.
2. Both carbons carry CH₃, so cis–trans naming also applies: methyls on the same side means cis.
3. CIP at C2: Br (atomic number 35) beats C (6), so Br has priority.
4. CIP at C3: C beats H, so CH₃ has priority.
5. The priority groups are the Br on C2 and the CH₃ on C3. In this isomer the CH₃ on C2 is on the same side as the C3 methyl, so the Br is on the opposite side. The isomer is E.

So this molecule is cis and E at the same time. This is why the specification warns that E = trans and Z = cis only holds reliably when each C=C carbon carries a hydrogen.

Worked example 2: counting stereoisomers

How many stereoisomers does hexa-2,4-diene, CH₃CH=CH–CH=CHCH₃, have?

1. At C2=C3, C2 has CH₃ and H and C3 has H and CH=CHCH₃, so this bond can be E or Z. The same holds for C4=C5.
2. Combinations: (2E,4E), (2Z,4Z), (2E,4Z) and (2Z,4E).
3. The molecule is symmetrical, so (2E,4Z) turned round is (2Z,4E): one compound.
4. Answer: three stereoisomers.

4.1.3(e): Why alkenes are reactive

The π-bond has a relatively low bond enthalpy, so little energy is needed to break it, and its exposed electron density attracts attacking species. Alkenes therefore react far more readily than alkanes.

Worked example 3: estimating the π-bond enthalpy

Mean bond enthalpies (use these values): C=C 614 kJ mol⁻¹, C–C 348 kJ mol⁻¹. Estimate the bond enthalpy of the π-bond.

1. C=C is one σ-bond plus one π-bond. Take the σ-bond as similar to a C–C single bond.
2. π-bond ≈ 614 − 348 = 266 kJ mol⁻¹.
3. This is less than the 348 kJ mol⁻¹ needed for the σ-bond, so the π-bond breaks first when an alkene reacts.

4.1.3(f): Addition reactions

In each reaction the π-bond breaks and two atoms or groups add, one to each C=C carbon. The product is saturated.

Reagent	Conditions	Product type	Example
H₂	nickel catalyst, heat	alkane	hex-2-ene + H₂ → hexane
Br₂ or Cl₂	room temperature	dihaloalkane	pent-2-ene + Br₂ → 2,3-dibromopentane
HBr, HCl or HI	room temperature	haloalkane	hex-3-ene + HCl → 3-chlorohexane
H₂O (steam)	acid catalyst, e.g. H₃PO₄	alcohol	but-2-ene + H₂O → butan-2-ol

For example, but-2-ene with steam: CH₃CH=CHCH₃ + H₂O → CH₃CH(OH)CH₂CH₃.

The bromine test for unsaturation (PAG7)

Shake the sample with bromine water. If the sample contains C=C, the orange colour disappears and the mixture becomes colourless, because bromine adds across the double bond. A saturated compound leaves the orange colour unchanged. Record “orange to colourless”, not “clear”, which describes transparency rather than colour.

4.1.3(g) and (h): Electrophiles and the mechanism

An electrophile is an electron pair acceptor. It is attracted to the π-bond’s electron density. The δ+ hydrogen in H–Br and a δ+ bromine in polarised Br₂ both act as electrophiles.

Electrophilic addition involves heterolytic fission: a covalent bond breaks and both bonding electrons go to one atom, forming ions.

Worked example 4: hex-3-ene with hydrogen chloride

H–Cl is polar, with H δ+ and Cl δ−.

1. Curly arrow from the C=C bond to the H; curly arrow from the H–Cl bond to the Cl. The H bonds to C3, leaving a carbocation with the positive charge on C4, plus Cl⁻.
2. Curly arrow from a lone pair on Cl⁻ to the positive carbon.
3. The product is 3-chlorohexane.

Show the dipole, the carbocation’s positive charge and the lone pair on Cl⁻.

With a halogen, the π-bond’s electron density induces a dipole in the approaching Br₂. The nearer Br becomes δ+ and is attacked; Br–Br breaks heterolytically, giving a carbocation (or bromonium ion; the specification accepts either) and Br⁻, which bonds to the positive carbon. Pent-2-ene gives 2,3-dibromopentane.

4.1.3(i): Markownikoff’s rule

When H–X adds to an unsymmetrical alkene, two carbocations are possible. The major product comes from the more stable carbocation. Stability order: tertiary > secondary > primary.

Worked example 5: 2-methylbut-1-ene with HBr

2-methylbut-1-ene is CH₂=C(CH₃)CH₂CH₃.

1. If H adds to C1, the positive charge sits on C2, bonded to three other carbons: a tertiary carbocation.
2. If H adds to C2, the positive charge sits on C1, bonded to one carbon: a primary carbocation.
3. The tertiary carbocation is more stable, so it forms more often.
4. Major product: 2-bromo-2-methylbutane. Minor product: 1-bromo-2-methylbutane.

4.1.3(j): Addition polymers

Many alkene monomers join through their C=C bonds to form a long saturated chain.

Worked example 6: repeat unit from a monomer

3,3-dimethylbut-1-ene is CH₂=CHC(CH₃)₃.

1. Draw the C=C with its four groups, then change C=C to C–C with a bond out of each end.
2. Repeat unit: –[CH₂–CH(C(CH₃)₃)]–, with extension bonds through the brackets and C(CH₃)₃ as a side branch.

Worked example 7: monomer from a polymer section

A section reads –CH(CH₃)–CH(CH₃)–CH(CH₃)–CH(CH₃)–.

1. Every main-chain carbon is the same, so the repeat unit is two adjacent ones: –CH(CH₃)–CH(CH₃)–.
2. Put C=C between them and remove the extension bonds: CH₃CH=CHCH₃.
3. Monomer: but-2-ene.

4.1.3(k) and (l): Waste polymers and alternatives

Cheap oil-derived polymers are unreactive, so they persist in landfill. Processing waste makes better use of resources.

• Combustion for energy: burning waste polymers generates heat for electricity, so less fuel is extracted.
• Organic feedstock: waste polymers are broken into small molecules used to make new plastics and other organic chemicals, reducing demand for crude oil.
• Removing toxic waste products: burning PVC and other halogenated plastics releases toxic, acidic hydrogen chloride. Passing the waste gases through a base neutralises the HCl.

Biodegradable polymers are broken down by microorganisms into small molecules such as water and carbon dioxide; many are made from plant material, reducing dependence on oil. Photodegradable polymers contain bonds that break when they absorb light, so the polymer breaks up. Both reduce persistent plastic waste.

Common errors

• Giving 109.5° at a C=C carbon instead of 120°.
• Claiming E/Z isomers for a molecule with =CH₂ at one end, such as but-1-ene.
• Ranking CIP groups by size or mass instead of by atomic number of the first atom.
• Assuming cis always means Z; check the priorities each time.
• Drawing the first curly arrow from a carbon atom rather than from the double bond, or omitting the halide ion’s lone pair.
• Writing the repeat unit with a C=C still in it, or putting a side group into the main chain.

Next steps

• Condensed tables and a quick self-test: Alkenes revision notes
• Original questions with mark-by-mark answers: Alkenes practice questions
• Shapes and bond angles: Bonding and structure
• Check your understanding: free diagnostics

Official syllabus

OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017, section 4.1.3 Alkenes, learning outcomes (a) to (l). Published by OCR.