Alkenes — OCR A-Level Chemistry

This row is a study-guidance heading, not an assessed answer target. Use it to organise your reading of the alkenes section: learn the C=C double bond as a region of high electron density, which makes alkenes reactive towards electrophiles, and link that to the characteristic addition reactions (with hydrogen, halogens, hydrogen halides and water) and to addition polymerisation. Read the specification statements under this heading to see exactly which properties and reactions are required, then check each against your notes. Build a summary table of reaction, reagent, conditions and product, and practise drawing displayed and skeletal formulae. Treat any percentages, timings or table layouts in your textbook as presentation, not examinable content.

Alkenes are unsaturated hydrocarbons: their molecules contain only carbon and hydrogen and include at least one C=C double bond, so they hold fewer hydrogen atoms than the corresponding alkane. The C=C is not two identical bonds. One σ-bond forms by direct end-on overlap of orbitals along the line between the two carbon nuclei. The π-bond forms by sideways overlap of adjacent p-orbitals, giving electron density above and below the C–C axis. Because the π-electron cloud must stay continuous for overlap to persist, rotation about the C=C is restricted, which is why alkene stereoisomerism such as E/Z exists.

Each carbon of a C=C is bonded to three atoms or groups: the other carbon and two substituents. Around each carbon there are three regions of electron density, and electron pairs repel one another to positions of maximum separation. Three regions therefore adopt a trigonal planar arrangement with bond angles of about 120°. The two carbons and their attached atoms lie in the same plane, and the π-electron density sits above and below that plane. The double bond counts as one region of electron density at each carbon, not two, so the shape is trigonal planar rather than tetrahedral.

Alkene stereoisomerism arises because rotation about the C=C double bond is restricted, so groups attached to each carbon stay on fixed sides of the plane. If each doubly bonded carbon carries two different groups, two stereoisomers exist: the E isomer has the higher-priority groups on opposite sides, the Z isomer has them on the same side. Priority follows Cahn–Ingold–Prelog rules: compare atoms directly attached by atomic number, and if tied, work outward along the chain. A molecule such as but-2-ene, CH₃CH=CHCH₃, shows this: each carbon holds H and CH₃, so E and Z forms occur. Where one carbon carries two identical groups, for example in propene, no stereoisomerism is possible. Distinguish this geometric (E/Z) isomerism from optical isomerism, which needs a chiral centre.

This row is course guidance rather than an assessed answer target: it signals that learners must be able to explain defined terms within the alkenes topic. Read the specification statements around it to identify which terms are required — typically structural isomerism, stereoisomerism, E/Z isomerism and the meaning of a chiral centre — and build a precise definition for each. A good method is to write each term, a one-sentence definition and a worked example, such as but-2-ene for E/Z isomerism. Check each definition against the specification wording and against a textbook, and note where a term is defined by a rule (priority order) rather than a single sentence. Use the definitions to answer explanation-style questions accurately.

Stereoisomers are compounds that share the same structural formula yet differ in how their atoms are arranged in space. The structural formula fixes which atoms are joined to which, so stereoisomers have identical connectivity and identical molecular formulae; only the three-dimensional placement of groups differs. In this alkenes section the key example is E/Z isomerism, which arises because rotation about the C=C double bond is restricted, locking the groups into different spatial arrangements. A useful test is to draw the full displayed structure: if two molecules have the same atoms joined in the same order but the groups sit differently around a double bond, they are stereoisomers. Contrast this with structural isomers, which differ in connectivity, and with conformers, which interconvert by rotation about single bonds.

E/Z isomerism is a form of stereoisomerism shown by alkenes. Rotation about the C=C double bond is restricted because the π bond locks the two carbon atoms in one plane, so groups cannot swap sides by rotation. Each carbon atom of the C=C group must carry two different groups for E/Z isomers to exist; if either carbon carries two identical groups, the two arrangements are the same molecule. Assign priority at each carbon using Cahn–Ingold–Prelog rules: higher atomic number wins. The Z isomer has the higher-priority groups on the same side of the double bond; the E isomer has them on opposite sides. For example, but-2-ene has a methyl and a hydrogen on each carbon, so E and Z forms exist.

cis–trans isomerism is a restricted form of E/Z isomerism. In E/Z isomerism each carbon of the C=C group carries two different substituent groups, so the two ends can be ranked by priority. cis–trans is the special case where, on each carbon, one of the two attached groups is the same group as on the other carbon — for example two hydrogen atoms, one on each carbon. With that shared group present, the molecule can be drawn with the two identical groups on the same side (cis) or on opposite sides (trans) of the C=C group. Restricted rotation about the C=C group holds the two arrangements in place, so they are separate stereoisomers. If a carbon carried two identical groups, it would have no E/Z isomerism at all, so cis–trans cannot apply there.

CIP priority rules rank the two groups on each carbon of the C=C group. Compare the atoms directly attached to the alkene carbon and give higher priority to the atom of greater atomic number. If those atoms are identical, move outward along the chain, comparing the sets of atoms attached at the first point of difference, and treat a double bond as two single bonds to the same atom. Once each carbon has a higher-priority and a lower-priority group, assign Z when the two higher-priority groups lie on the same side of the double bond and E when they lie on opposite sides. For example, in 1-bromo-1-chloropropene the carbon bearing Br and Cl gives Br priority (atomic number 35 over 17), and the other carbon bearing CH₃ and H gives CH₃ priority (carbon over hydrogen).

Stereoisomers arise when the same structural formula allows different spatial arrangements that cannot be interconverted by rotation about single bonds. For an alkene, restricted rotation about the C=C bond fixes the groups on each carbon in a plane. To decide whether E/Z isomers are possible, first check that each carbon of the double bond carries two different groups; if either carbon carries two identical groups, no E/Z isomerism exists. Where both carbons carry two different groups, assign priorities using the Cahn–Ingold–Prelog rule: the group with the higher atomic number directly attached to the double-bond carbon takes priority. If the two higher-priority groups lie on opposite sides, the isomer is E; if on the same side, it is Z. For simple cases the cis–trans label may be used, where cis places the main carbon chains on the same side and trans places them on opposite sides.

Alkenes contain a C=C double bond made of a sigma bond and a pi bond. The pi electrons are exposed above and below the molecular plane, so alkenes are nucleophiles and undergo electrophilic addition. In each addition reaction the pi bond breaks and two new sigma bonds form, converting the unsaturated alkene into a saturated product. Typical reagents include hydrogen with a nickel catalyst to give an alkane, steam with a phosphoric acid catalyst to give an alcohol, hydrogen halides to give a haloalkane, and bromine to give a dibromoalkane. Bromine water decolourises from orange to colourless, which is the standard test for a C=C bond. With an unsymmetrical alkene such as propene, the hydrogen halide adds to give the more stable carbocation intermediate, so the major product follows Markovnikov's rule. Write the balanced equation and name the product.

The C=C group contains a σ-bond and a π-bond. The π-bond has a relatively low bond enthalpy, so less energy is needed to break it than to break the σ-bond. In a reaction such as electrophilic addition, the π-bond breaks and two new σ-bonds form to the two carbons. Forming those new σ-bonds releases energy, and the overall enthalpy change is exothermic because the energy released in bond formation exceeds the energy needed to break the π-bond. The low bond enthalpy of the π-bond therefore explains why alkenes are more reactive than comparable alkanes, whose σ-bonds are stronger. Breaking bonds always requires energy input; the reactivity comes from the favourable balance of breaking the weaker π-bond and forming stronger σ-bonds.

This specification point introduces the addition reactions of alkenes, in which the carbon–carbon double bond opens and two new groups add across it. The published statement is a heading that lists the reagents to be studied, so read it alongside the full specification to see exactly which reagents are named. Typical examples include reaction with hydrogen in the presence of a nickel catalyst to form an alkane, with a halogen such as bromine to give a dihalogenoalkane, with a hydrogen halide such as hydrogen bromide to give a halogenoalkane, and with steam in the presence of an acid catalyst to give an alcohol. Learn the reagents, conditions and the structural change in each case, and be ready to draw the product from a given alkene. Treat the exact list of reagents as fixed by the specification rather than adding reactions from memory.

An alkene reacts with hydrogen gas over a suitable catalyst, for example nickel, to form the corresponding alkane. The C=C double bond becomes a C–C single bond as each carbon gains a hydrogen atom, so the product has the same carbon skeleton but is saturated. This is a surface-catalysed addition: hydrogen and the alkene are adsorbed onto the nickel surface, where the reaction occurs, so it is not an electrophilic addition and no carbocation intermediate forms. For example, ethene with hydrogen over nickel gives ethane. In multiple-choice questions, check the reagent, the catalyst, the product class and the type of addition before choosing an option.

This row covers the addition of a halogen across the C=C bond to form a dihaloalkane, and the use of bromine as a test for unsaturation. When bromine, Br₂, adds to an alkene, one bromine atom bonds to each carbon of the former double bond, giving a compound with two halogen atoms on adjacent carbons, a dihaloalkane. For example, ethene with bromine gives 1,2-dibromoethane, C₂H₄ + Br₂ → C₂H₄Br₂. In the test, bromine water is added to the sample: an alkene decolourises the orange-brown bromine water because the bromine reacts with the C=C bond, whereas an alkane leaves the colour unchanged. Decolourisation therefore indicates unsaturation.

Alkenes undergo electrophilic addition with hydrogen halides (HX, where X is Cl, Br or I) to give haloalkanes. The π electrons of the C=C bond attack the δ⁺ hydrogen of the polar H–X molecule, forming a carbocation and a halide ion; the halide ion then bonds to the carbocation. With an unsymmetrical alkene such as propene, two products are possible, and the major product forms via the more stable carbocation (the Markovnikov outcome). For example, propene with HBr gives mainly 2-bromopropane, with 1-bromopropane as the minor product. The reaction is carried out by passing the gaseous hydrogen halide into the alkene, often in an inert solvent, at room temperature. Learners should be able to name the product haloalkane and explain why one product predominates.

Alkenes react with steam (water vapour) in the presence of an acid catalyst such as phosphoric acid, H₃PO₄, to form alcohols. This is the industrial hydration of ethene to make ethanol, carried out at high temperature (about 300 °C) and high pressure (about 60–70 atm) over a solid H₃PO₄ catalyst. The reaction is an electrophilic addition: the alkene is protonated to give a carbocation, which is then attacked by water, and loss of a proton gives the alcohol. With an unsymmetrical alkene such as propene, the major product follows carbocation stability, giving mainly propan-2-ol. Learners should write the equation, name the alcohol product and explain the role of the acid catalyst.

An electrophile is a species that accepts an electron pair, so it is attracted to regions of high electron density such as the C=C double bond in an alkene. The π bond is exposed above and below the plane of the molecule, so an approaching electrophile draws electron density towards itself. A bromine molecule is non-polar on its own, but as it nears the alkene the electron-rich π bond induces a dipole: the nearer bromine atom becomes δ⁺ and accepts an electron pair, while the other becomes δ⁻. The same logic explains why HBr and other polar or polarisable species act as electrophiles. Recognise the term by its defining feature, electron pair acceptance, not by charge alone, since neutral molecules can be electrophiles once polarised.

Electrophilic addition begins when the π electrons of the C=C bond attack an electrophile, forming a new bond and leaving a carbocation. The bond in the electrophile breaks heterolytically, meaning both electrons of the shared pair go to one atom, producing ions. For example, with HBr the δ⁺ hydrogen accepts the π pair, the H–Br bond breaks heterolytically so Br gains both electrons as Br⁻, and a carbocation intermediate forms. A lone pair from Br⁻ then attacks the positively charged carbon, giving the addition product. Draw the mechanism with curly arrows: one from the C=C π bond to the electrophilic atom, one from the bond being broken to the more electronegative atom, and one from the nucleophile to the carbocation. Show charges on intermediates and use heterolytic fission language throughout.

Markownikoff's rule predicts the major product when a hydrogen halide H–X adds to an unsymmetrical alkene such as propene. The rule works because the reaction proceeds through a carbocation intermediate, and the more stable carbocation forms faster. For propene, adding H⁺ to the terminal carbon gives a secondary carbocation on the middle carbon; adding H⁺ to the middle carbon gives a primary carbocation on the terminal carbon. The secondary carbocation is more stable because two alkyl groups donate electron density and disperse the positive charge, so the major product is 2-bromopropane, CH₃CHBrCH₃. The minor product is 1-bromopropane. Stability order is tertiary > secondary > primary. Predict the product by placing the halogen on the carbon that can best support the positive charge in the intermediate.

Alkenes undergo addition polymerisation because their C=C double bond opens to form new single bonds linking monomer units into a long chain. The repeat unit has the same atoms as the monomer but with the double bond replaced by two single bonds extending to neighbouring units. For ethene, the polymer is poly(ethene) with repeat unit –CH₂–CH₂–. For propene, the polymer is poly(propene) with the methyl group on alternate backbone carbons. Draw the repeat unit with the two free bonds shown through the brackets and the subscript n outside. Identify the monomer by reversing the process: take the repeat unit, restore the C=C between the backbone carbons and remove the side-group arrangement that came from the double bond. Addition polymers are saturated and generally unreactive, which raises environmental issues because they are not readily biodegradable.

Addition polymerisation joins many alkene monomers into one long saturated-chain polymer without eliminating any small molecule, so the empirical formula of the polymer matches that of the monomer. The C=C double bond of each alkene opens and forms two new single bonds, one to each neighbouring monomer unit, giving a backbone of carbon atoms with the original substituents hanging from it. For ethene the repeat unit is –CH₂–CH₂–; for chloroethene it is –CH₂–CHCl–; for propene it is –CH₂–CH(CH₃)–. Substituted alkenes behave the same way, but the substituent (for example Cl, CH₃ or C₆H₅) appears unchanged on every second backbone carbon. Conditions vary with the monomer: ethene is polymerised at high pressure with a catalyst, while other alkenes may need specific catalysts or conditions. The polymer is named by adding 'poly' to the monomer name, for example poly(chloroethene).

To deduce a repeat unit from a monomer, locate the C=C double bond and open it so each of those two carbons forms a single bond to the next repeat unit. The two carbons of the original double bond become the two-carbon backbone section enclosed in brackets, with the substituents carried over unchanged. For example, propene CH₂=CHCH₃ gives the repeat unit –CH₂–CH(CH₃)–, and chloroethene CH₂=CHCl gives –CH₂–CHCl–. Place the bracket lines through the single bonds that link one repeat unit to the next, and show the repeat unit with the correct number of hydrogen atoms and substituents. The empirical formula of the repeat unit must equal that of the monomer, which is a useful check. Naming follows by adding 'poly' to the monomer name.

Addition polymerisation joins alkene monomers through the C=C double bond, so the polymer backbone is a continuous chain of two-carbon units. To identify a monomer from a drawn polymer section, first locate the repeating unit: the shortest section bounded by the two single bonds formed from the original double bond. Then reverse the change by restoring the C=C between those two backbone carbons and removing the single bonds that link the unit to its neighbours. Any substituent groups attached to the backbone carbons stay attached and keep their positions. For example, a repeating unit written as –CH₂–CHCl– comes from chloroethene, CH₂=CHCl; a unit of –CH₂–C(CH₃)(COOCH₃)– comes from methyl methacrylate. Check the number of carbon atoms and the identity and placement of every substituent before selecting an option.

Addition polymers are chemically inert and non-biodegradable because their saturated carbon backbones resist attack, so waste polymers persist in landfill and can pollute land and water. Burning polymers releases carbon dioxide and can release toxic gases unless controlled. Common responses are to recycle polymers by sorting and reprocessing, to recover energy by controlled combustion, and to develop alternatives. Alternatives include biodegradable polymers such as poly(lactic acid) made from renewable plant material, and polymers designed to degrade. Feedstock recycling breaks polymers back into useful small molecules. When answering, link each problem to a specific property of the polymer and match each alternative to the problem it addresses, rather than listing options without justification.

Waste polymers are a resource as well as a disposal problem, and OCR frames the sustainability benefits around specific processing routes. Feedstock recycling breaks polymer chains back to monomers or useful small molecules, so the carbon can re-enter the chemical industry instead of being lost; this reduces demand for fresh crude-oil-derived feedstock. Combustion with energy recovery uses the high energy content of polymers to generate electricity or heat, displacing some fossil fuel that would otherwise be burned. Sorting and mechanical recycling keep material in use, though mixed and contaminated waste limits it. A strong answer names the route, states the resource or energy benefit and links it to reduced landfill or reduced use of virgin feedstock.

This statement sits within the section on waste polymers, where combustion is one route for processing waste plastics and recovering energy. Burning waste polymers releases heat that can generate electricity or heat buildings, so the material is used as a fuel rather than sent to landfill. The carbon and hydrogen in the polymer oxidise to carbon dioxide and water, releasing energy in an exothermic reaction. Evaluate the benefits, such as reducing landfill volume and recovering useful energy, against the drawbacks, including emission of carbon dioxide and potentially toxic gases unless the combustion is controlled and gases are scrubbed. Link the idea to sustainability: energy recovery is one option alongside recycling and feedstock reuse, and it does not remove the need to reduce plastic waste at source.

This clause sits within the alkenes topic and concerns using alkenes as an organic feedstock: the C=C bond is the reactive site that lets alkenes be converted into plastics and other organic chemicals. Ethene is the key example. Ethene molecules undergo addition polymerisation, joining through their double bonds to form poly(ethene), a plastic. Other organic chemicals are made by adding reagents across the C=C bond, for example steam in hydration to make ethanol, or hydrogen in hydrogenation. The feedstock is therefore valuable because the double bond can be opened and new single bonds formed, building larger or functionalised molecules. A multiple-choice question may ask why alkenes are useful feedstocks, which product comes from ethene, or what feature of the alkene enables these conversions.

When halogenated plastics such as PVC are disposed of by combustion, the halogen they contain is released as a toxic hydrogen halide; for PVC this is hydrogen chloride, HCl. Because HCl is harmful, waste gases must be treated to remove it before release. A common method is to pass the gases through an alkaline scrubber, where the HCl is neutralised, for example by reaction with calcium hydroxide or sodium hydroxide, forming a salt and water. The wider principle is that the halogen content of a plastic determines which toxic waste product forms on burning, and that this product must be captured rather than emitted. Recognising the halogen in the polymer lets you predict the hydrogen halide produced and choose a suitable removal method.

Addition polymerisation makes polymers such as poly(ethene) and poly(propene) from alkene monomers, and these polymers are generally unreactive and persist in landfill for a very long time. Biodegradable polymers are broken down by microorganisms, so they decompose in soil and reduce the volume of persistent plastic waste. Photodegradable polymers contain groups that absorb ultraviolet light, so sunlight breaks the polymer chains and the material fragments, again reducing long-lived litter. Both routes lessen the accumulation of waste and the harm to wildlife and habitats that persistent plastics cause. A good answer links each polymer type to its specific breakdown mechanism and to the environmental benefit, rather than claiming that all plastics simply disappear.