Alkenes and Electrophilic Addition

Alkenes are unsaturated hydrocarbons of general formula CₙH₂ₙ that contain at least one carbon-to-carbon double bond. Unlike the alkanes you met in the previous lesson, where the only reaction of practical importance was free-radical halogenation, alkenes are reactive across a whole library of synthetic transformations because the C=C double bond is built from one strong σ component and one much weaker π component, and the π electron density is exposed above and below the molecular plane. In this lesson we examine the bonding, develop the general electrophilic-addition mechanism, work through the reactions with Br₂, HBr, H₂SO₄, water, and hydrogen, derive Markovnikov's rule (first formulated in 1869) from carbocation stability, and signpost the industrial significance of polymerisation that the next lesson develops in detail.

Spec mapping (AQA 7405): This lesson maps to §3.3.4 (alkenes), specifically §3.3.4.1 (structure, bonding and reactivity), §3.3.4.2 (addition reactions of alkenes) and §3.3.4.3 (addition polymers, signposted forward). It connects backwards to lesson 3 (alkanes — saturated comparison and free-radical contrast), forwards to lesson 5 (addition polymers — extends the mechanism to chain growth), lesson 7 (alcohols — hydration of ethene to ethanol is the industrial anchor) and lesson 8 (mechanisms master class — consolidates electrophilic addition alongside SN1/SN2/elimination/electrophilic aromatic). It also feeds into §3.3.6 (organic analysis), where Br₂(aq) decolourisation is the standard wet test for unsaturation. Refer to the official AQA 7405 specification for the exact wording.

Assessment objectives: AO1 covers definitions — what is an alkene, an electrophile, an electrophilic addition, Markovnikov's rule, and a carbocation. AO2 dominates the marks and demands fluent mechanism writing for Br₂, HBr, H₂O/H⁺ (H₂SO₄ or H₃PO₄), and the test reactions with bromine water and acidified potassium manganate(VII). AO3 reasoning tasks include predicting the major and minor products of HBr addition to unsymmetrical alkenes using carbocation stability, justifying anti-addition stereochemistry from the bromonium-ion intermediate, and applying the same logic to industrial decisions (e.g. why direct hydration of ethene to ethanol uses a phosphoric-acid catalyst at high pressure).

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Structure and Bonding in Alkenes

The two carbons of a C=C double bond are sp² hybridised. Each contributes three sp² hybrid orbitals (which form three σ bonds in a trigonal-planar arrangement at ~120°) and one unhybridised 2p orbital perpendicular to that plane. The σ bond between the two carbons arises from head-on overlap of one sp² orbital from each. The π bond arises from sideways overlap of the two unhybridised p orbitals — this places electron density in two lobes, one above and one below the plane of the σ framework.

The π bond has three consequences that drive everything else in this lesson:

• Restricted rotation. Rotating one CH₂ end of ethene relative to the other would require breaking the sideways p-orbital overlap. This costs roughly 260 kJ mol⁻¹ — utterly unavailable at room temperature. The molecule is therefore locked planar, which is why E/Z geometric isomerism arises whenever the two substituents on each sp² carbon are different (lesson 2 of this course).
• Planar geometry. Every atom directly bonded to the two sp² carbons lies in the same plane as the C=C. For ethene that is all six atoms; for propene the methyl-carbon nucleus is also in-plane, even though the CH₃ hydrogens sit above and below.
• Exposed electron density. The π electrons stick out above and below the plane, well away from the heavier σ framework. That is geometrically convenient for an approaching electrophile: a positively-charged or δ+ atom can reach the π cloud without first having to displace any σ-bonded substituent.

The π bond is also weaker than the σ bond — typical bond-energy decompositions give the σ component of C=C at around 350 kJ mol⁻¹ and the π component at only ~265 kJ mol⁻¹. Hence the total bond dissociation enthalpy of a C=C (~614 kJ mol⁻¹) is less than twice that of a C–C single bond (347 kJ mol⁻¹). In a reaction that breaks the π bond and forms two new σ bonds in its place, the overall energy balance is almost always strongly exothermic, which is the thermodynamic reason addition reactions of alkenes are so general.

Key Definition: An electrophile is an electron-pair acceptor — any species attracted to and able to accept a pair of electrons from a region of high electron density. Examples relevant to A-Level: H⁺, the δ+ end of a polar H–X or X–X bond, NO₂⁺, carbocations themselves, and SO₃ in electrophilic aromatic sulfonation.

Key Point: "Electrophilic addition" means the rate-determining step is attack on an electrophile by an electron-rich π system, and the overall outcome is the conversion of one π bond and one X–Y bond into two new σ bonds. Both halves of that definition are examinable.

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The General Electrophilic Addition Mechanism

Every alkene addition reaction at A-Level fits the same two-step electrophilic template:

1. Step 1 (slow, rate-determining): The π electrons of the C=C attack the electrophile. The π bond breaks; one new σ bond forms from one of the alkene carbons to the electrophile. The other alkene carbon — which has lost its share of the π electrons — becomes a carbocation (a tricoordinate, sp²-hybridised carbon bearing a formal +1 charge in an empty p orbital).
2. Step 2 (fast): A nucleophile (often the leaving anion from step 1) donates a lone pair to the carbocation, forming the second new σ bond.

The curly-arrow convention is unambiguous:

• A double-headed arrow shows the movement of a pair of electrons.
• The tail of the arrow sits on the source of the electron pair (a bond or a lone pair).
• The head of the arrow points to where that pair ends up (an atom, the centre of a new bond, or onto a leaving atom).

Exam Tip: Markers reward arrows that start from the correct electron source. From a C=C, the arrow must start on the line of the double bond, not on one of the atoms. When heterolytic fission of a bond places both electrons on one atom, the arrow goes from the centre of the bond to the atom that keeps the pair.

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Reaction 1: Addition of Bromine — The Test for Unsaturation

Bromine reacts rapidly with alkenes, and the visual change is the most famous wet test in school chemistry:

• Add a few drops of bromine water (an orange/brown aqueous solution of Br₂) to the unknown.
• If a C=C is present, the orange colour is decolourised almost instantly to a clear, colourless solution.
• If no C=C is present (e.g. a saturated alkane), the orange colour persists.

The overall equation for ethene and propene is straightforward:

CH₂=CH₂ + Br₂ → CH₂BrCH₂Br (1,2-dibromoethane)

CH₃CH=CH₂ + Br₂ → CH₃CHBrCH₂Br (1,2-dibromopropane)

The mechanism, however, deserves care because the intermediate is not a simple open carbocation but a cyclic bromonium ion.

Mechanism — Br₂ addition to ethene

1. As the Br₂ molecule approaches the electron-rich π cloud, the electric field of the π electrons polarises Br₂: the closer bromine atom becomes δ+ and the farther bromine δ−. This is an induced dipole — bromine is non-polar in isolation, but is easily polarised because its outer electrons are loosely held (large atomic radius, low electronegativity gradient within the molecule).
2. The π electrons attack the δ+ bromine. A curly arrow runs from the C=C centre to that bromine. Simultaneously, a curly arrow runs from the Br–Br bond to the δ− bromine, depositing the bond pair as Br⁻.
3. Crucially, the δ+ bromine retains a lone pair and forms a three-membered cyclic bromonium ion: the bromine bridges both former alkene carbons, holding a positive formal charge.
4. The released Br⁻ attacks the bromonium ion from the opposite face to the bridging Br (because that face is now blocked by the bromonium ring). This back-side attack opens the three-membered ring and places the second Br on the opposite face from the first.

The stereochemical signature is anti-addition: the two new C–Br bonds end up on opposite faces of the former C=C plane. With cyclohexene the product is trans-1,2-dibromocyclohexane exclusively — cis product is not formed. AQA accepts either the cyclic bromonium-ion mechanism or the simpler open-carbocation drawing, but the more sophisticated drawing wins the marks for stereochemical reasoning at A*.

Worked example — propene + bromine

Propene plus liquid bromine in an inert solvent (CCl₄ or hexane) gives 1,2-dibromopropane, CH₃CHBrCH₂Br. Both new C–Br bonds end up on adjacent carbons (a 1,2-dihalide, never a 1,1- or 1,3-product), and they sit on opposite faces of the original molecular plane. The reaction is quantitative at room temperature and produces no rearrangement.

Practical-skills box: The bromine-water test is the standard classroom demonstration of unsaturation. Use a few cm³ of pale-orange bromine water in a test tube; add one drop of the alkene; shake briefly. Decolourisation within seconds is positive. In water the mechanism gives a mixed product — partly the dibromide and partly the bromohydrin (HOCH₂CH₂Br), because water is also a competent nucleophile against the bromonium ion. For the wet test it is enough that the orange colour disappears.

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Reaction 2: Addition of Hydrogen Halides

Hydrogen halides H–X (X = Cl, Br, I) add across the C=C to give haloalkanes:

CH₂=CH₂ + HBr → CH₃CH₂Br (bromoethane)

CH₃CH=CH₂ + HBr → CH₃CHBrCH₃ (2-bromopropane, major) + CH₃CH₂CH₂Br (1-bromopropane, minor)

Mechanism — HBr addition to propene

1. The H–Br bond is permanently polar: Br is more electronegative than H, so H is δ+ and Br is δ−. No induced dipole is needed.
2. The π electrons attack the δ+ hydrogen. A curly arrow from the C=C delivers an electron pair to H, forming a new C–H bond on one of the two alkene carbons. A second curly arrow takes the H–Br bond pair onto bromine, releasing Br⁻.
3. The other alkene carbon — the one that did not receive the hydrogen — becomes the carbocation.
4. The Br⁻ attacks the carbocation with its lone pair, forming the C–Br bond.

There are two ways the proton can add — to the terminal CH₂ giving a secondary carbocation on the middle carbon, or to the middle CH giving a primary carbocation on the terminal carbon. The two pathways are not equally probable, which is the point of Markovnikov's rule (next section).

Exam Tip: The most common AO2 mistake is to draw a curly arrow from the C=C to the Br of HBr. Always go to the δ+ atom. When the electrophile is a polar molecule, identify the δ+ atom first using electronegativity, then start the arrow from the π bond to that atom.

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Reaction 3: Hydration to Alcohols (Industrial Anchor)

Water is too weak a nucleophile to add directly to a C=C, but in the presence of a strong acid catalyst the alkene is first protonated to give a carbocation, and water then adds to that. Two industrial routes share the same overall outcome:

Route A — direct hydration (modern industrial method for ethanol):

CH₂=CH₂ + H₂O → CH₃CH₂OH

Conditions: gas-phase, 300 °C, 60 atm pressure, supported phosphoric acid (H₃PO₄ on a silica or alumina support) as catalyst. The mechanism is electrophilic addition: H⁺ from the acid catalyst protonates the alkene to give a carbocation, water attacks the carbocation with a lone pair, and the resulting oxonium ion loses a proton (regenerating the catalyst) to give the alcohol.

Route B — sulfuric-acid two-step (older route, included in AQA):

Step 1: CH₂=CH₂ + H₂SO₄ → CH₃CH₂OSO₃H (ethyl hydrogensulfate)

Step 2: CH₃CH₂OSO₃H + H₂O → CH₃CH₂OH + H₂SO₄

The first step is electrophilic addition of the polar H–OSO₃H bond across the C=C (concentrated sulfuric acid, room temperature). The second is a hydrolysis under warm dilute conditions. The H₂SO₄ is regenerated and is therefore a catalyst overall.

Both routes give ethanol, the industrial backbone product. Both obey Markovnikov regiochemistry when applied to an unsymmetrical alkene: propene + H₂O / H⁺ gives propan-2-ol (CH₃CH(OH)CH₃) as the major product, not propan-1-ol. The biofermentation route to ethanol (sugar + yeast → ethanol + CO₂) is a chemistry-and-society topic developed in lesson 7.

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Reaction 4: Hydrogenation

Alkenes add hydrogen across the C=C to give the corresponding alkane:

CH₂=CH₂ + H₂ → CH₃CH₃

CH₃CH=CHCH₃ + H₂ → CH₃CH₂CH₂CH₃

Conditions: H₂ gas with a finely-divided nickel catalyst at around 150 °C and a few atmospheres of pressure (or with platinum/palladium at room temperature in research labs). This is not electrophilic addition — the mechanism is heterogeneous catalysis, with both reactants chemisorbed on the metal surface — but AQA does not require the surface mechanism. You do need the conditions, the equation, and one practical application: catalytic hydrogenation of unsaturated vegetable oils is the industrial route to margarine and to shortening, converting C=C-containing liquid oils into saturated solid fats.

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Reaction 5: Oxidation with Cold, Dilute, Acidified KMnO₄

When a cold, dilute, acidified solution of potassium manganate(VII) is added to an alkene, the purple MnO₄⁻ is reduced — typically to brown MnO₂ on warming, or pale Mn²⁺ in stronger acid — and the alkene is oxidised to a diol (two adjacent –OH groups across the former C=C):

CH₂=CH₂ + [O] + H₂O → HOCH₂CH₂OH (ethane-1,2-diol)

The colour change (purple → colourless or brown) is the second standard wet test for unsaturation alongside the bromine-water test. AQA does not require the mechanism, but it does require the observation (purple decolourised) and the product class (a diol).

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Markovnikov's Rule and Carbocation Stability

When HBr (or any unsymmetrical electrophile such as HCl, HI, H₂O/H⁺, H₂SO₄) adds to an unsymmetrical alkene, two products are possible — and they are formed in distinctly unequal amounts. The 1869 generalisation by the Russian chemist after whom the rule is named is empirical:

Markovnikov's Rule: When a hydrogen halide adds to an unsymmetrical alkene, the hydrogen attaches to the carbon of the C=C that already carries more hydrogen atoms. Equivalently, the halide attaches to the more substituted carbon.

For propene + HBr:

CH₃CH=CH₂ + HBr →

• Major product (Markovnikov): CH₃CHBrCH₃ — 2-bromopropane. H went to the CH₂ end (which already had two H atoms); Br went to the CH end (the more substituted carbon).
• Minor product (anti-Markovnikov): CH₃CH₂CH₂Br — 1-bromopropane. H went to the CH end; Br went to the terminal CH₂.

Mechanistic explanation — carbocation stability

The two pathways diverge in step 1 of the mechanism, where the proton is added and a carbocation is generated. The two possible carbocations are:

• H adds to CH₂ → secondary carbocation (CH₃)CH⁺(CH₃) sits on the central carbon, with two alkyl groups flanking it.
• H adds to CH → primary carbocation CH₃CH₂CH₂⁺ sits on the terminal carbon, with only one alkyl group flanking it.

The secondary carbocation is more stable than the primary one. The activation energy to reach the secondary intermediate is therefore lower, so by the Arrhenius relationship rate ∝ exp(−Eₐ/RT) the secondary pathway is faster, and the secondary pathway produces 2-bromopropane (Br⁻ attacks the secondary cation). The major product is the one whose carbocation intermediate is more stable.

The stability order is tertiary > secondary > primary > methyl, governed by two effects:

• Inductive effect: Alkyl groups are weakly σ-donating relative to hydrogen. Each alkyl group bonded to a positively-charged carbon pushes electron density toward the empty p orbital and reduces the formal positive charge density. More alkyl groups = more stabilisation.
• Hyperconjugation: Each C–H bond on a carbon adjacent (α) to a carbocation can donate its σ-bond pair into the empty p orbital of the cation. A tertiary cation has nine α-C–H bonds available for hyperconjugation; a methyl cation has none. A* candidates can cite this contribution by name.

Carbocation type	Example	Stability	α-C–H bonds available for hyperconjugation
Methyl	CH₃⁺	Least stable	0
Primary (1°)	CH₃CH₂⁺	Low	3
Secondary (2°)	(CH₃)₂CH⁺	Intermediate	6
Tertiary (3°)	(CH₃)₃C⁺	Most stable	9