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Alkenes are unsaturated hydrocarbons containing at least one carbon-carbon double bond (C=C). They have the general formula CₙH₂ₙ (for non-cyclic alkenes with one double bond) and are far more reactive than alkanes. The C=C double bond is the defining feature and the site where almost all alkene reactions take place.
The C=C double bond consists of two components:
Each carbon in the C=C is sp² hybridised, giving a trigonal planar geometry with bond angles of approximately 120°. The three sp² orbitals lie in a plane and the remaining p orbital is perpendicular to this plane.
The pi bond has important consequences:
Like alkanes, alkenes are non-polar and have only van der Waals forces between molecules. Their boiling points follow the same trends:
Alkenes are insoluble in water and less dense than water.
To distinguish an alkene from an alkane, add bromine water (an aqueous solution of Br₂, which is orange-brown):
This is a rapid, reliable test for the presence of a C=C double bond.
The characteristic reactions of alkenes are addition reactions. In an addition reaction, atoms or groups add across the C=C double bond, breaking the pi bond and forming two new sigma bonds. The product is a single molecule with no atoms "left over" — there is no by-product.
This reaction map shows the key addition reactions an alkene can undergo:
graph TD
A["Alkene (C=C)"] -->|"+ H₂, Ni catalyst, 150 °C"| B["Alkane"]
A -->|"+ HBr, room temp"| C["Halogenoalkane"]
A -->|"+ H₂O(g), H₃PO₄, 300 °C"| D["Alcohol"]
A -->|"+ Br₂, room temp"| E["Dihalogenoalkane"]
A -->|"+ H₂SO₄, room temp"| F["Alkyl hydrogen sulfate"]
A -->|"Polymerisation"| G["Addition Polymer"]
F -->|"+ H₂O, hydrolysis"| D
CH₂=CH₂ + H₂ → CH₃CH₃
CH₂=CH₂ + HBr → CH₃CH₂Br
For unsymmetrical alkenes (e.g., propene), two products are possible. Markovnikov's rule predicts the major product (see below).
CH₂=CH₂ + H₂O → CH₃CH₂OH
CH₂=CH₂ + Br₂ → CH₂BrCH₂Br
CH₂=CH₂ + H₂SO₄ → CH₃CH₂OSO₃H (ethyl hydrogen sulfate)
The ethyl hydrogen sulfate can then be hydrolysed with water to produce ethanol:
CH₃CH₂OSO₃H + H₂O → CH₃CH₂OH + H₂SO₄
This is an indirect route to ethanol.
When HBr adds to an unsymmetrical alkene such as propene (CH₃CH=CH₂), two products are possible:
Markovnikov's rule states: The hydrogen atom adds to the carbon of the double bond that already has the most hydrogen atoms.
The major product is 2-bromopropane. The reason for this is the stability of the carbocation intermediate (explained in detail in the next lesson on electrophilic addition mechanisms).
Consider the reaction of 2-methylpropene with HCl:
The alkene is (CH₃)₂C=CH₂. The two C=C carbons are:
Markovnikov's rule: H adds to the carbon with more H atoms (the right carbon, CH₂). Cl adds to the left carbon. Product: 2-chloro-2-methylpropane, (CH₃)₃CCl.
The intermediate is a tertiary carbocation (CH₃)₃C⁺, which is very stable — confirming why this is the major product.
Alkenes can undergo addition polymerisation, in which many small alkene molecules (monomers) join together to form a long-chain molecule (polymer) with no by-product.
n CH₂=CH₂ → -(CH₂-CH₂)ₙ-
The repeating unit is drawn in brackets with bonds extending from either side to show the chain continues.
| Monomer | Polymer | Uses |
|---|---|---|
| Ethene (CH₂=CH₂) | Poly(ethene) / polyethylene | Plastic bags, bottles, packaging |
| Propene (CH₃CH=CH₂) | Poly(propene) / polypropylene | Ropes, carpets, containers |
| Chloroethene (CH₂=CHCl) | Poly(chloroethene) / PVC | Pipes, window frames, insulation |
| Tetrafluoroethene (CF₂=CF₂) | PTFE (Teflon) | Non-stick coatings |
During polymerisation, the pi bond in each monomer breaks and new sigma bonds form between monomers. The polymer is saturated — it contains no C=C double bonds.
Addition polymers such as polyethylene are non-biodegradable because the C-C backbone is unreactive and resistant to microbial attack. This leads to environmental persistence. Strategies to address this include recycling, incineration for energy recovery, and development of biodegradable alternatives.
| Property | Alkanes | Alkenes |
|---|---|---|
| General formula | CₙH₂ₙ₊₂ | CₙH₂ₙ |
| Saturation | Saturated | Unsaturated |
| Bonding | C-C single bonds (σ only) | C=C (σ + π) |
| Shape at C=C | Tetrahedral (109.5°) | Trigonal planar (120°) |
| Reactivity | Low | High |
| Typical reaction | Free radical substitution | Electrophilic addition |
| Bromine water test | No reaction (stays orange) | Decolourised (orange → colourless) |
| Intermolecular forces | Van der Waals only | Van der Waals only |
The key message: alkenes are far more reactive than alkanes because the pi bond provides an accessible region of high electron density that electrophiles can attack. The pi bond is also weaker than the sigma bond, so it breaks preferentially during addition reactions.
Edexcel 9CH0 specification Topic 6 — Organic Chemistry I, sub-topic 6.4 covers the alkene homologous series: structure (sp2-hybridised C atoms in the C=C, planar geometry around the double bond, restricted rotation), the σ-π bonding in the C=C, geometric (E/Z) isomerism, and characteristic addition reactions with H2/Ni catalyst, halogens (X2), hydrogen halides (HX), water (H2O with H2SO4 catalyst), and the test for unsaturation by decolourisation of bromine water (refer to the official specification document for exact wording). Sub-topic 6.4 is examined directly on Paper 2 (mechanism, products, observations) and Paper 3 (synoptic with NMR/IR identification, polymerisation, industrial chemistry). The data booklet does not list specific bond enthalpies for π bonds; candidates are expected to know that C=C (612) is less than 2 × C-C (2 × 348 = 696), reflecting that the π bond is weaker than the σ.
Question (8 marks):
(a) Draw the full electrophilic addition mechanism for the reaction of HBr with propene (CH3−CH=CH2). Use curly arrows to show electron movement, and identify both the major and minor products. (5)
(b) State Markovnikov's rule and use carbocation stability to explain which product is the major one. (3)
Solution with mark scheme:
(a) Step 1 — initial polarisation and π-electron attack. The H-Br bond is polar (Hδ+−Brδ−). The π electrons of the C=C attack the partially-positive H, breaking the H-Br bond heterolytically.
The π electrons form a new C-H bond on the terminal C; the secondary carbocation forms on C2 (the more substituted carbon).
M1 — curly arrow from C=C π bond to the H of HBr, showing π electron movement.
M1 — curly arrow from H-Br bond to Br, showing heterolytic fission.
M1 — secondary carbocation CH3−CH+−CH3 correctly drawn.
Step 2 — bromide attack. The bromide ion attacks the carbocation, forming the C-Br bond.
M1 — curly arrow from Br⁻ lone pair to C+; product = 2-bromopropane (major).
A1 — minor product is 1-bromopropane, formed via the primary carbocation CH3−CH2−CH2+.
(b) Step 1 — Markovnikov's rule (statement).
M1 — Markovnikov's rule states that in the addition of HX to an asymmetric alkene, the H atom adds to the carbon of the C=C that already carries more H atoms, and the X adds to the carbon with fewer H atoms.
Step 2 — carbocation stability.
M1 — the rule is rationalised by carbocation stability: tertiary > secondary > primary > methyl. Tertiary carbocations are stabilised by the inductive (electron-donating) effect of three alkyl groups, while methyl carbocations have no such stabilisation.
Step 3 — apply to propene.
A1 — addition of HBr to propene generates either a secondary carbocation (CH3−CH+−CH3, which becomes 2-bromopropane) or a primary carbocation (CH3−CH2−CH2+, which becomes 1-bromopropane). The secondary cation is more stable, so 2-bromopropane is the major product.
Total: 8 marks (M5 A3).
Question (6 marks): A student tests three samples — hex-1-ene, hexane, and benzene — by adding bromine water to each.
(a) Predict the observation for each sample, justifying your answer in terms of bonding. (3)
(b) Write the equation and structural formula for the product of hex-1-ene + Br2 (in non-aqueous conditions). (2)
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