From C7 Alkenes you know alkenes have a C=C double bond, decolourise bromine water, and make polymers. A-Level explains why: the double bond is a region of high electron density that reaches out and attacks electrophiles, through a mechanism (electrophilic addition) whose carbocation intermediate decides which product forms in the greater amount. Then it sharpens the polymer work into repeat units and structure–property links.
Structure, bonding & reactivity
Alkenes are unsaturated hydrocarbons: they contain a carbon–carbon double bond, C=C. That double bond is made of two shared pairs of electrons: one lies directly between the two carbon nuclei, and the second — from two p orbitals overlapping sideways — spreads its electron density above and below the plane of the molecule, where it is exposed and easy to reach. That exposed, electron-rich region is the key to everything alkenes do.
So alkenes are attacked by electrophiles (electron-pair acceptors), which are drawn to that electron-rich double bond — making alkenes far more reactive than the alkanes. Each carbon of the C=C is trigonal planar, with bond angles of about 120°.
Where the electron density sits
One bonding pair sits between the nuclei; the second — from p orbitals overlapping sideways — sits above and below the plane, exposed.
The shape at each carbon
Three bonding pairs around each C=C carbon: trigonal planar, about 120°.
- Reactivity comes from the C=C’s high electron density — the exposed electrons above and below the plane — that is why alkenes are attacked by electrophiles and alkanes are not.
- You only need “a region of high electron density” — AQA does not ask you to name σ and π bonds, so describe the double bond, not the orbital labels.
- Trigonal planar, 120° around each C=C carbon.
- The functional-group isomer of an alkene is a cycloalkane (same CnH2n) — recap the isomerism types from the introduction.
Recall E–Z isomerism about the C=C, and the curly-arrow rules.
🧪 Exam-style questions
In which reaction does the inorganic reagent act initially as an electrophile? Tick (✓) one box.
Which alkene shows E–Z isomerism? Tick (✓) one box.
Source: AQA A-Level Chemistry past papers.
The three electrophiles: HBr, H2SO4, Br2
Electrophile — an electron-pair acceptor. It is a species (often carrying a positive charge or a δ+ atom) that is attracted to a region of high electron density and accepts a pair of electrons to form a new covalent bond.
Because the C=C is a centre of high electron density, alkenes are attacked by electrophiles. Three are examined at A-level — meet them first, then we’ll draw the single three-arrow mechanism they all share.
| Electrophile | Where the δ+ comes from | Product with an alkene |
|---|---|---|
| HBr | already polar — the H is δ+ (Br is more electronegative) | a bromoalkane |
| conc H2SO4 | polar — the H is δ+ (drawn as H–OSO3H) | an alkyl hydrogensulfate, then an alcohol on adding water |
| Br2 | non-polar — but the alkene induces a dipole, making the nearer Br δ+ | a dibromoalkane — the basis of the bromine test |
The one to watch is Br2: it has no dipole of its own until the electron-rich C=C induces one — that is what turns a non-polar molecule into an electrophile. Whichever atom carries the δ+ is the one the C=C attacks first; from there, the mechanism is identical for all three.
Electrophilic addition — the mechanism
This is the reaction alkenes are known for, and the centrepiece of the topic. The idea is always the same: the electron-rich C=C reaches out and attacks an electrophile, and the two ends of the reagent end up added across the double bond. One product, nothing lost — addition.
Every electrophilic addition runs in the same three moves — get these and you can do all three reagents:
- Arrow 1 — from the C=C double bond to the δ+ atom of the electrophile. (The electron-rich double bond attacks.)
- Arrow 2 — from the electrophile’s bond to its more electronegative atom, which breaks heterolytically and leaves as a negative ion.
- Arrow 3 — from a lone pair on that negative ion to the positively charged carbon (the carbocation), forming the second new bond.
Adding hydrogen bromide (HBr)
H–Br is already polar: bromine is more electronegative, so the hydrogen is Hδ+. The C=C attacks that Hδ+, the H–Br bond breaks to give Br−, a carbocation forms, and the Br− then bonds to the positive carbon. Ethene + HBr → bromoethane.
Adding concentrated sulfuric acid (a route to alcohols)
Concentrated sulfuric acid provides the Hδ+ electrophile. Written as H–OSO3H, it adds across the C=C by exactly the same three arrows as HBr — the C=C attacks the Hδ+, the H–O bond breaks to give the hydrogensulfate ion, and a lone pair on its oxygen bonds to the carbocation. The product is an alkyl hydrogensulfate (for ethene: CH3CH2OSO3H). Warm that with water and it is hydrolysed to the alcohol, releasing the sulfuric acid again — so the H2SO4 acts as a catalyst overall.
The same alcohol is made industrially by direct hydration (steam + H3PO4 catalyst) — you’ll compare the two routes in Alcohols (3.3.5).
Adding bromine — and the test for a C=C
Bromine, Br2, is non-polar — so how can it be an electrophile? As it approaches the electron-rich double bond, the C=C repels the electrons in Br–Br, pushing them onto the far bromine. That induces a dipole, Brδ+–Brδ−, and the nearer Brδ+ is now the electrophile. From there it is the same mechanism: the first Br adds to one carbon, a carbocation forms on the other, and the Br− then bonds to it. Addition across the C=C gives a dibromoalkane (for ethene: 1,2-dibromoethane).
This is the basis of the test for unsaturation: add bromine water (an orange solution) to the sample. An alkene decolourises it — orange solution → colourless solution — because the Br2 is used up adding across the double bond. An alkane leaves it an orange solution.
For the bromine test always give both the reagent and the colour change: “add bromine water; it is decolourised (orange solution to colourless solution)”. “Goes clear” scores nothing — clear means see-through, not colourless. And for Br2 you must say the dipole is induced; do not call Br2 polar.
All three reagents run the same three arrows — only the electrophile changes. Try placing each arrow yourself, for whichever reagent you like:
Interactive — place the three curly arrows
The arrow starts from…
…and ends at…
🧪 Exam-style questions
Propene reacts with hydrogen bromide by electrophilic addition. Outline the mechanism. Your answer should include curly arrows, the structure of the intermediate, and relevant charges.
Show answer
Arrow from the C=C double bond to the H of H–Br, and arrow from the H–Br bond to the Br. 1 mark
Correct carbocation intermediate drawn: the secondary carbocation CH3–C+H–CH3, with the + charge shown on the middle carbon. 1 mark
Arrow from a lone pair on Br− to the positive carbon; product 2-bromopropane. 1 mark
Describe a simple chemical test you could use to distinguish between separate samples of propane and propene. Give the reagent and the result for each.
Show answer
Add bromine water to each. 1 mark
Propene decolourises it (orange solution → colourless solution); propane leaves it an orange solution / no change. 1 mark
Source: AQA A-Level Chemistry past papers.
Major & minor products — carbocation stability
When the alkene is unsymmetrical (the two ends of the C=C are different), the electrophile’s H+ can add to either carbon — and that gives two possible carbocations, so two possible products. One forms in greater amount: the major product. The other is the minor product.
Which wins? The one that goes through the more stable carbocation. So the whole skill is: draw both carbocations, decide which is more stable, and that tells you where the H (and therefore the Br/OH) ends up.
tertiary (3°) > secondary (2°) > primary (1°)
Why: alkyl groups are electron-releasing (they push electron density onto the positive carbon through the inductive effect). The more alkyl groups attached to the C+, the more the positive charge is spread out and stabilised. A tertiary carbocation has three; a primary has only one.
It is worth seeing the three types side by side. Count the carbon groups directly attached to the positive carbon — that is what makes it 1°, 2° or 3°, and the more there are, the more stable it is:
Interactive — two carbocations, one major product
- Justify the major product by carbocation stability — not by quoting “Markovnikov’s rule”. AQA wants the reason (the more stable carbocation), which the rule only summarises.
- Classify the carbocation by the carbon carrying the + charge: count how many carbon groups are directly attached to it (1° = one, 2° = two, 3° = three).
- Both products do form — “major” means the greater amount, not the only one.
🧪 Exam-style questions
3-Methylbut-1-ene reacts with hydrogen bromide. What is the major product? Tick (✓) one box.
2-Ethylbut-1-ene reacts with hydrogen bromide. What is the major product? Tick (✓) one box.
Source: AQA A-Level Chemistry past papers.
Addition polymers
Because the C=C can open up and form new bonds, many alkene molecules (monomers) can join end-to-end into a very long chain — an addition polymer. The double bond becomes a single bond as each monomer links to the next, and nothing else is produced (that is what makes it addition, not condensation).
- Take the monomer and open the C=C to a single bond.
- Draw the two carbons with their groups, and extend a bond out through each side of the brackets (this shows the chain continues).
- Put “n” outside the bracket. For example, ethene → poly(ethene): [–CH2–CH2–]n; chloroethene → PVC: [–CH2–CHCl–]n.
AQA asks for this drawing in three directions: from a monomer, open the C=C (as above); from a section of the polymer chain, bracket any two adjacent backbone carbons with their groups — that pair is the repeat unit; and from a repeat unit back to the monomer, put the C=C back between the two carbons. Nothing else moves — every side group stays exactly where it was, because addition polymerisation loses no atoms.
Naming follows one rule: poly(name of the monomer), brackets included — propene → poly(propene), chloroethene → poly(chloroethene) (better known as PVC), phenylethene → poly(phenylethene).
Interactive — fix the repeat unit
Properties — why addition polymers are so unreactive
An addition polymer is saturated (no C=C left) and its backbone is made of strong, non-polar C–C and C–H bonds. There is nothing for common reagents to attack, so the polymers are chemically inert — which is useful, but also means they are non-biodegradable and persist in the environment.
Between the chains, the only forces are weak van der Waals forces. Because the chains are so long, though, there are a great many contact points, and these add up — enough to make the polymer a solid with a real melting point (far higher than the small monomer’s).
How polymers are made, used and modified is knowledge that has developed over time — and the clearest example of deliberately tuning a polymer’s properties is PVC.
PVC and plasticisers
Poly(chloroethene) — PVC — is naturally rigid: the polar C–Cl groups sit along chains that pack fairly closely, so it is hard and used for pipes and window frames. To make the flexible form (cable insulation, flooring), a small-molecule plasticiser is added: it slots between the chains and pushes them apart, weakening the van der Waals forces between them so the chains can slide — making the PVC soft and bendy.
Explain a plasticiser in terms of forces between chains: it gets between the polymer chains, pushing them further apart so the van der Waals forces between chains are weaker, letting them slide over each other. Do not say it “breaks bonds” — the covalent bonds are untouched.
🧪 Exam-style questions
An addition polymer has the repeat unit [–CH2–CH(CN)–]n. Which compound is the monomer? Tick (✓) one box.
Poly(ethene) has a much higher melting point than ethene. Which statement best explains why? Tick (✓) one box.
Draw the repeat unit of the addition polymer formed from chloroethene, CH2=CHCl.
Show answer
Repeat unit shows the two carbons as –CH2–CHCl– (C=C opened to a single bond, Cl on one carbon). 1 mark
Continuation bonds drawn passing through both brackets, with subscript n: [–CH2–CHCl–]n. 1 mark
Source: AQA A-Level Chemistry past papers.
Capstone: the mechanism-marks checklist
Examiner reports flag the same few lost marks on alkenes: fuzzy curly arrows, a wrongly drawn carbocation, and “major product” asserted without a reason. This last section is about writing the answer an examiner can tick.
- Arrow start points — from the C=C (not from a carbon atom), and from a lone pair on the anion (not from the minus sign).
- Arrow end points — to the δ+ atom, and to the positive carbon. A double-headed arrow means a pair of electrons moving.
- The carbocation — drawn correctly, with the + charge on the right carbon. This is where the major-product mark lives.
- The justification — name the more stable carbocation (3° > 2° > 1°) as the reason for the major product.
Interactive — build the mark-scheme answer
“Outline the mechanism for the reaction of but-1-ene with hydrogen bromide, and explain why 2-bromobutane is the major product.” [5 marks] Select every statement that earns a mark — and nothing that doesn’t. Order doesn’t matter: AQA credits each point on its own.
🧪 Capstone question
But-1-ene reacts with hydrogen bromide to give mainly 2-bromobutane. Outline the mechanism, and explain why 2-bromobutane is the major product.
Show answer
Mechanism named/shown as electrophilic addition; arrow from the C=C to the H of H–Br. 1 mark
Arrow from the H–Br bond to Br (heterolytic fission), forming Br−. 1 mark
Secondary carbocation CH3CH2–C+H–CH3 drawn with the + on the correct carbon. 1 mark
Arrow from a lone pair on Br− to the positive carbon → 2-bromobutane. 1 mark
This secondary carbocation is more stable than the primary alternative (alkyl groups release electron density and spread the charge), so 2-bromobutane is the major product. 1 mark
Source: AQA A-Level Chemistry past papers.
Next you meet the alcohols (3.3.5) these additions produce — how they are made, and their oxidation and elimination reactions.
- Alkenes are unsaturated (C=C). The C=C is a centre of high electron density, exposed above and below the plane, so alkenes are attacked by electrophiles. Trigonal planar, 120°.
- Electrophilic addition: the C=C electrons attack the electrophile (arrow 1), the electrophile’s bond breaks heterolytically (arrow 2), a carbocation forms, then a nucleophile’s lone pair adds to it (arrow 3). Reagents: HBr, conc H2SO4, Br2.
- Bromine test: an alkene decolourises bromine water (orange solution → colourless solution), because Br2 forms an induced dipole and adds across the C=C.
- Major product of an unsymmetrical alkene comes from the more stable carbocation: stability is tertiary > secondary > primary (alkyl groups release electrons and spread the + charge).
- Addition polymers: the C=C opens and monomers join with nothing lost. Draw the repeat unit in brackets with bonds passing through and subscript n (e.g. poly(ethene) [–CH2–CH2–]n); monomer ↔ repeat unit just moves the C=C, and the name is poly(monomer).
- Addition polymers are unreactive (saturated, non-polar bonds) so non-biodegradable; chains are held by van der Waals forces. PVC is rigid, but a plasticiser pushes the chains apart, weakening the forces and making it flexible.