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24:[["$","$L33",null,{"feature":"flashcards"}],"$L34",["$","$L35",null,{"href":"/a-level/chemistry/eduqas/hydrocarbons/flashcards","label":"Hydrocarbons","subjectCode":"a_level_chemistry__eduqas","surface":"flashcards"}],["$","$L36",null,{"runner":["$","div",null,{"className":"min-h-0 flex-1 overflow-hidden bg-background","children":["$","$L37",null,{"decks":[{"id":"027112ac-ee83-4c74-b4cd-88bb926f02d3","title":"Hydrocarbons Flashcards","topicId":"2ab2d674-db0e-5811-ae54-4fc9c36348c9","topicSlug":"hydrocarbons","cards":[{"id":"cb0b3c3b-672c-47d0-98be-0a7b68d636c7","cardType":"cloze","points":null,"front":"Open-chain alkanes have the general formula [ ].","back":"Open-chain alkanes have the general formula C_nH_2n+2.","mcqOptions":[{"text":"C_nH_2n+2","correct":true,"explanation":"An open-chain alkane contains only single bonds, giving the saturated hydrocarbon formula C_nH_2n+2. Each additional carbon therefore adds two hydrogen atoms overall."},{"text":"C_nH_2n","correct":false,"explanation":"This is the general formula for an alkene with one double bond, not a saturated open-chain alkane. Alkanes have two more hydrogens, so the correct formula is C_nH_2n+2."},{"text":"C_n+1H_2n+2","correct":false,"explanation":"The expression has one too few carbon atoms in relation to hydrogen and does not represent the standard alkane series. Open-chain alkanes follow C_nH_2n+2."},{"text":"C_nH_2n+4","correct":false,"explanation":"This gives an excess of hydrogen compared with the saturated alkane formula. For an open-chain alkane containing n carbon atoms, the correct hydrogen count is 2n+2."}]},{"id":"70eb403b-e9f6-463f-822e-860645b77cf3","cardType":"cloze","points":null,"front":"An alkene with one C=C double bond has the general formula [ ].","back":"An alkene with one C=C double bond has the general formula C_nH_2n.","mcqOptions":[{"text":"C_nH_2n","correct":true,"explanation":"A single C=C bond introduces one degree of unsaturation compared with an alkane, reducing the hydrogen count by two. The general formula is therefore C_nH_2n."},{"text":"C_nH_2n+2","correct":false,"explanation":"This is the formula for an open-chain alkane, which contains only single bonds. One C=C bond removes two hydrogens, giving the correct alkene formula C_nH_2n."},{"text":"C_nH_2n-2","correct":false,"explanation":"This formula has two too few hydrogens for an alkene with only one double bond and instead suggests an additional degree of unsaturation. The correct formula is C_nH_2n."},{"text":"C_nH_2n+4","correct":false,"explanation":"The added four hydrogens would make the molecule more saturated than an alkene can be. A hydrocarbon with one C=C double bond has the formula C_nH_2n."}]},{"id":"fc633850-9b05-4cc0-beeb-fe3141f898c5","cardType":"basic","points":1,"front":"What products form when an alkane burns in plentiful oxygen?","back":"Carbon dioxide and water.","mcqOptions":[{"text":"Carbon dioxide and water.","correct":true,"explanation":"Plentiful oxygen allows complete combustion of the alkane. All carbon atoms are oxidised to carbon dioxide and the hydrogen atoms form water."},{"text":"Carbon monoxide and water.","correct":false,"explanation":"Carbon monoxide forms when oxygen is limited, so this describes incomplete rather than complete combustion. With plentiful oxygen, the products are carbon dioxide and water."},{"text":"Carbon dioxide and hydrogen.","correct":false,"explanation":"Hydrogen is not normally released as a combustion product because it reacts with oxygen to form water. Complete alkane combustion produces carbon dioxide and water."},{"text":"Sulfur dioxide and water.","correct":false,"explanation":"Sulfur dioxide requires sulfur in the fuel, but alkanes contain only carbon and hydrogen. In plentiful oxygen, their complete combustion gives carbon dioxide and water."}]},{"id":"0c127514-165b-4737-a2ff-0925bfde5d81","cardType":"basic","points":2,"front":"Why does large-scale fossil fuel use contribute to climate change?","back":"CO₂ is a greenhouse gas.","mcqOptions":[{"text":"CO₂ is a greenhouse gas.","correct":true,"explanation":"Burning fossil fuels transfers carbon into the atmosphere as carbon dioxide. CO₂ absorbs outgoing infrared radiation, so it is a greenhouse gas that contributes to warming."},{"text":"CO is a greenhouse gas.","correct":false,"explanation":"Carbon monoxide is toxic, but it is not the main greenhouse gas responsible for this fossil-fuel effect. The relevant explanation is that CO₂ is a greenhouse gas."},{"text":"SO₂ is a greenhouse gas.","correct":false,"explanation":"Sulfur dioxide contributes mainly to acid rain and aerosol effects rather than being the key greenhouse gas released by fossil-fuel combustion. The correct reason involves CO₂."},{"text":"CO₂ destroys greenhouse gases.","correct":false,"explanation":"Carbon dioxide does not remove infrared radiation from the atmosphere; it absorbs outgoing infrared radiation and traps heat. Thus, CO₂ is a greenhouse gas."}]},{"id":"0df12fff-a42e-4ce6-a644-10ad0164b226","cardType":"basic","points":2,"front":"Why is carbon monoxide from incomplete combustion dangerous?","back":"CO binds strongly to haemoglobin in the blood.","mcqOptions":[{"text":"CO binds strongly to haemoglobin in the blood.","correct":true,"explanation":"Carbon monoxide binds strongly to haemoglobin, forming carboxyhaemoglobin. This reduces haemoglobin’s ability to carry oxygen, potentially causing unconsciousness or death."},{"text":"CO destroys red blood cells in the blood.","correct":false,"explanation":"Carbon monoxide does not mainly destroy red blood cells; its danger comes from binding strongly to haemoglobin and preventing oxygen transport through the blood."},{"text":"CO makes haemoglobin bind oxygen more strongly in the blood.","correct":false,"explanation":"Carbon monoxide is dangerous because it reduces oxygen transport, not because it makes haemoglobin bind oxygen more strongly in the lungs. It binds strongly to haemoglobin itself."},{"text":"CO reacts harmlessly with plasma in the blood.","correct":false,"explanation":"Carbon monoxide does not react with plasma to form a harmless substance. It binds strongly to haemoglobin, reducing the blood’s capacity to transport oxygen."}]},{"id":"202c327b-b78d-4f62-aec2-55f8b7341f4d","cardType":"basic","points":2,"front":"Why do alkanes usually not react with acids, bases or electrophiles?","back":"Their C-C and C-H σ-bonds are strong and not very polar.","mcqOptions":[{"text":"Their C-C and C-H σ-bonds are strong and not very polar.","correct":true,"explanation":"Alkanes contain strong, relatively non-polar C-C and C-H sigma bonds. Acids, bases and electrophiles therefore have no especially reactive polar bond or electron-rich site to attack."},{"text":"Their C-C and C-H π-bonds are strong and not very polar.","correct":false,"explanation":"The C-C and C-H bonds in alkanes are sigma bonds, not pi bonds. Their strong, low-polarity sigma bonds explain the usual lack of reaction."},{"text":"Their C-C and C-H σ-bonds are weak and very polar.","correct":false,"explanation":"Alkane bonds are not highly polar, and their lack of reactivity is not caused by weak bonds. Strong, relatively non-polar C-C and C-H sigma bonds are responsible."},{"text":"Their C-C and C-H ionic bonds are strong and not very polar.","correct":false,"explanation":"Alkanes do not contain ionic bonds. Their covalent C-C and C-H sigma bonds are strong and not very polar, making attack by these reagents difficult."}]},{"id":"9e57e39f-c0ac-47e0-8314-3fe70bcb1bf6","cardType":"cloze","points":null,"front":"In [ ], each atom takes one electron from the bond, forming radicals.","back":"In homolytic fission, each atom takes one electron from the bond, forming radicals.","mcqOptions":[{"text":"homolytic fission","correct":true,"explanation":"Homolytic fission splits the bonding pair equally, giving one electron to each atom. Each product therefore has an unpaired electron and is a radical."},{"text":"heterolytic fission","correct":false,"explanation":"Heterolytic fission gives both bonding electrons to one atom, producing ions rather than two radicals. Equal one-electron splitting is called homolytic fission."},{"text":"electrophilic substitution","correct":false,"explanation":"Electrophilic substitution describes a reaction mechanism, not the equal splitting of a covalent bond. The process forming two radicals is homolytic fission."},{"text":"nucleophilic addition","correct":false,"explanation":"Nucleophilic addition involves a nucleophile adding to a multiple bond and does not describe one electron going to each atom. That bond cleavage is homolytic fission."}]},{"id":"61b5ef1a-74f1-4cee-b015-149d7126e189","cardType":"basic","points":1,"front":"What condition is needed for photochlorination of an alkane?","back":"Ultraviolet light.","mcqOptions":[{"text":"Ultraviolet light.","correct":true,"explanation":"Ultraviolet light supplies the energy needed for homolytic cleavage of the chlorine molecule, generating radicals that initiate the photochlorination chain reaction."},{"text":"Infrared light.","correct":false,"explanation":"Visible light generally does not provide the required energy for chlorine homolysis in this reaction. Photochlorination requires ultraviolet light to initiate radicals."},{"text":"A nickel catalyst.","correct":false,"explanation":"A catalyst may alter reaction rate but is not the photochemical condition that generates chlorine radicals. The required condition is ultraviolet light."},{"text":"High pressure.","correct":false,"explanation":"Heating can promote some reactions, but the standard condition for alkane photochlorination is irradiation that causes homolytic chlorine fission: ultraviolet light."}]},{"id":"352a7977-ba53-4c29-8396-44e49e50016c","cardType":"basic","points":1,"front":"Write the initiation step in methane photochlorination.","back":"Cl₂ → 2Cl•.","mcqOptions":[{"text":"Cl₂ → 2Cl•.","correct":true,"explanation":"Ultraviolet light causes homolytic fission of the chlorine molecule, with each chlorine atom receiving one electron and forming two chlorine radicals."},{"text":"Cl₂ → 2Cl^-.","correct":false,"explanation":"This shows ionic fission, which would produce chloride ions rather than radicals. Photochlorination begins when UV light splits the covalent Cl-Cl bond homolytically."},{"text":"CH₄ → CH₃• + H•.","correct":false,"explanation":"Methane is not split during initiation. The chlorine molecule absorbs UV light and undergoes homolytic fission, generating the chlorine radicals that start the chain."},{"text":"Cl• + CH₄ → HCl + •CH₃.","correct":false,"explanation":"This is the first propagation step, not initiation. Initiation produces chlorine radicals from chlorine molecules before those radicals react with methane."}]},{"id":"579f07c1-f9fc-43e2-93b5-4b584b214472","cardType":"basic","points":2,"front":"What is the first propagation step in methane photochlorination?","back":"Cl• + CH₄ → HCl + •CH₃.","mcqOptions":[{"text":"Cl• + CH₄ → HCl + •CH₃.","correct":true,"explanation":"A chlorine radical abstracts a hydrogen atom from methane, forming hydrogen chloride and a methyl radical that continues the chain reaction."},{"text":"Cl• + CH₄ → CH₃Cl + H•.","correct":false,"explanation":"The first propagation step removes hydrogen as HCl, not as a hydrogen radical. Chloromethane forms in the following propagation step when the methyl radical reacts with Cl₂."},{"text":"•CH₃ + Cl₂ → CH₃Cl + Cl•.","correct":false,"explanation":"This is the second propagation step because it uses the methyl radical formed first. The initial propagation step must produce that methyl radical from methane."},{"text":"Cl₂ + CH₄ → CH₃Cl + HCl.","correct":false,"explanation":"This is the overall substitution equation, not an individual propagation step. The chain mechanism separates it into two radical reactions, with one radical involved in each step."}]},{"id":"9091a789-94f7-4db7-a9d9-308bed79461f","cardType":"basic","points":2,"front":"What is the second propagation step in methane photochlorination?","back":"•CH₃ + Cl₂ → CH₃Cl + Cl•.","mcqOptions":[{"text":"•CH₃ + Cl₂ → CH₃Cl + Cl•.","correct":true,"explanation":"The methyl radical reacts with chlorine, forming chloromethane and regenerating a chlorine radical, so the chain can continue through further propagation steps."},{"text":"Cl• + CH₄ → HCl + •CH₃.","correct":false,"explanation":"This is the first propagation step, where a chlorine radical abstracts hydrogen from methane. The second step instead consumes the methyl radical and chlorine molecule."},{"text":"•CH₃ + HCl → CH₄ + Cl•.","correct":false,"explanation":"Hydrogen chloride is a product of the first propagation step, not the reactant in the second. The methyl radical reacts with Cl₂ to make chloromethane."},{"text":"2Cl• → Cl₂.","correct":false,"explanation":"This is a termination reaction because two chlorine radicals combine and the radical chain stops. The second propagation step must form chloromethane and regenerate a radical."}]},{"id":"62be67e7-3ba2-48b4-98ff-7642ff7b118d","cardType":"basic","points":2,"front":"What happens to radicals in a termination step?","back":"Radicals combine to form a molecule, removing radicals.","mcqOptions":[{"text":"Radicals combine to form a molecule, removing radicals.","correct":true,"explanation":"In termination, two radicals react together to form a stable molecule. Because the products contain no unpaired electrons, radicals are removed and the chain ends."},{"text":"Radicals split to form more radicals, removing radicals.","correct":false,"explanation":"Splitting radicals would increase the number of reactive species rather than remove them. Termination occurs when radicals combine to form stable molecules."},{"text":"Radicals react with oxygen to form ions, removing radicals.","correct":false,"explanation":"Termination does not require oxygen or ionic products. The defining event is combination of radicals into a molecule, which removes the unpaired electrons."},{"text":"Radicals transfer electrons to form ions, removing radicals.","correct":false,"explanation":"Electron transfer to form ions is not the general radical-chain termination process. Radicals terminate when they combine and produce a stable molecule without radicals."}]},{"id":"e104e6ad-f15e-43fd-bc77-1b3fa5607abf","cardType":"cloze","points":null,"front":"A C=C double bond contains one [ ] and one [ ].","back":"A C=C double bond contains one σ-bond and one π-bond.","mcqOptions":[{"text":"A C=C double bond contains one σ-bond and one π-bond.","correct":true,"explanation":"A carbon-carbon double bond consists of one head-on overlap forming a sigma bond and one sideways overlap forming a pi bond."},{"text":"A C=C double bond contains two σ-bonds and no π-bond.","correct":false,"explanation":"A double bond contains only one sigma bond; the second bond is a pi bond formed by sideways p-orbital overlap. Two sigma bonds would not describe a C=C bond."},{"text":"A C=C double bond contains one π-bond and one δ-bond.","correct":false,"explanation":"A carbon-carbon double bond contains sigma and pi bonding. Delta bonds are associated with more unusual orbital interactions and are not the second component here."},{"text":"A C=C double bond contains two π-bonds and no σ-bond.","correct":false,"explanation":"The first bond between two carbon atoms must be a sigma bond from direct orbital overlap. Only the additional bond in a C=C is a pi bond."}]},{"id":"30a0569c-b43f-45b2-a2d0-55b80bceb617","cardType":"basic","points":2,"front":"Why are alkenes more reactive than alkanes?","back":"The exposed, electron-rich π-bond attracts electrophiles.","mcqOptions":[{"text":"The exposed, electron-rich π-bond attracts electrophiles.","correct":true,"explanation":"The pi electrons are exposed above and below the carbon-carbon framework, making them available to attract electrophiles and causing alkenes to undergo addition reactions readily."},{"text":"The exposed, electron-rich σ-bond attracts nucleophiles.","correct":false,"explanation":"The reactive region of an alkene is its exposed pi bond, which attracts electrophiles. Sigma bonds are stronger and less exposed, so they are not the main cause."},{"text":"The exposed, electron-rich π-bond attracts nucleophiles.","correct":false,"explanation":"Nucleophiles are electron-pair donors and are repelled by the electron-rich pi bond. Electrophiles, which accept electrons, are attracted to it instead."},{"text":"The exposed, electron-poor σ-bond attracts electrophiles.","correct":false,"explanation":"The pi bond is electron-rich, not electron-poor, and it is this exposed pi system that attracts electrophiles. The sigma bond is not responsible for alkene reactivity."}]},{"id":"85aa0d8d-b82a-4b74-9108-f17bad3c2afd","cardType":"basic","points":2,"front":"What structural condition is required for E-Z isomerism around C=C?","back":"Each C=C carbon has two different groups attached.","mcqOptions":[{"text":"Each C=C carbon has two different groups attached.","correct":true,"explanation":"E-Z isomerism requires two different groups attached to each carbon of the double bond. This prevents identical arrangements from making the two geometries indistinguishable."},{"text":"Each C=C carbon has two identical groups attached.","correct":false,"explanation":"Identical groups on either alkene carbon make the possible arrangements equivalent, so distinct E-Z isomers cannot exist. Each carbon must instead have two different groups."},{"text":"Only one C=C carbon has two different groups attached.","correct":false,"explanation":"Both carbon atoms must have two different attached groups. If one carbon has identical groups, priority-based arrangements cannot produce a pair of E-Z isomers."},{"text":"Each C=C carbon has three different groups attached.","correct":false,"explanation":"A carbon in a C=C double bond has only two groups attached besides its double-bond connection. The required condition is two different groups on each carbon."}]},{"id":"a921a458-aa9e-45df-b872-66594a919a51","cardType":"cloze","points":null,"front":"In E-Z isomerism, [ ] has higher-priority groups on the same side; [ ] has them on opposite sides.","back":"In E-Z isomerism, Z has higher-priority groups on the same side; E has them on opposite sides.","mcqOptions":[{"text":"In E-Z isomerism, Z has higher-priority groups on the same side; E has them on opposite sides.","correct":true,"explanation":"Z comes from the German zusammen, meaning together, so higher-priority groups are on the same side. E means entgegen, or opposite, so they are across the double bond."},{"text":"In E-Z isomerism, E has higher-priority groups on the same side; Z has them on opposite sides.","correct":false,"explanation":"The labels have been reversed. Z means the higher-priority groups are together on the same side, whereas E means they are opposite each other."},{"text":"In E-Z isomerism, Z has higher-priority groups on opposite sides; E has them on the same side.","correct":false,"explanation":"This also reverses the meanings of the labels. The mnemonic zusammen identifies Z as same-side, while E describes higher-priority groups on opposite sides."},{"text":"In E-Z isomerism, E has identical groups on the same side; Z has identical groups on opposite sides.","correct":false,"explanation":"E-Z notation compares the higher-priority group on each carbon, not identical groups. Z places those priority groups together, while E places them opposite."}]},{"id":"fe2abc9e-cd20-44b8-bd39-1b247ef6d4ff","cardType":"cloze","points":null,"front":"E-Z priority is usually decided by [ ].","back":"E-Z priority is usually decided by atomic number.","mcqOptions":[{"text":"atomic number","correct":true,"explanation":"Atomic number determines the priority because the atom with the greater nuclear charge is ranked higher directly at the point of attachment."},{"text":"electronegativity","correct":false,"explanation":"Electronegativity describes attraction for bonding electrons, not the priority rule normally used to compare atoms in E-Z notation. Atomic number determines the priority instead."},{"text":"mass number","correct":false,"explanation":"Mass number includes the number of neutrons, which does not normally determine E-Z priority. The attached atoms are ranked using atomic number instead."},{"text":"number of electron shells","correct":false,"explanation":"The number of electron shells is not the standard priority criterion and can give the wrong order for different elements. Atomic number is used instead."}]},{"id":"b7666c3b-e768-4c84-9370-38865124f7de","cardType":"basic","points":1,"front":"What observation shows bromine water has reacted with an alkene?","back":"It changes from orange/brown to colourless.","mcqOptions":[{"text":"It changes from orange/brown to colourless.","correct":true,"explanation":"Bromine water is orange/brown because of dissolved bromine. An alkene reacts with it by addition, removing the bromine colour and producing a colourless mixture."},{"text":"It changes from colourless to orange/brown.","correct":false,"explanation":"The colour change is reversed here: bromine water starts orange/brown and becomes colourless when bromine reacts with the alkene. It does not begin colourless."},{"text":"It changes from orange/brown to green.","correct":false,"explanation":"Green is not the expected product colour in the bromine water test. Reaction with an alkene removes the orange/brown bromine colour, giving a colourless solution."},{"text":"It remains orange/brown.","correct":false,"explanation":"An unchanged orange/brown colour indicates that bromine has not been consumed. A reacting alkene decolourises bromine water, changing it to colourless."}]},{"id":"18dedabc-7cc8-4fa7-9767-01cca60feb80","cardType":"basic","points":2,"front":"Why does the nearer bromine atom in Br₂ become Brδ+ during alkene addition?","back":"The C=C π-bond induces a dipole in Br₂.","mcqOptions":[{"text":"The C=C π-bond induces a dipole in Br₂.","correct":true,"explanation":"The electron-rich C=C π-bond repels electrons in Br2, inducing a dipole so the nearer bromine becomes Brδ+ and the farther bromine becomes Brδ−."},{"text":"The C=C σ-bond induces a dipole in Br₂.","correct":false,"explanation":"The σ-bond framework is not responsible for polarising Br2 in this reaction. The C=C π-bond induces the dipole that makes the nearer bromine Brδ+."},{"text":"Br₂ induces a dipole in the C=C.","correct":false,"explanation":"Br2 does not induce the dipole in the alkene to explain the bromine charge. The alkene’s C=C π-bond induces a dipole in Br2 instead."},{"text":"The C=C π-bond attracts electrons from Br₂.","correct":false,"explanation":"The π-bond is electron-rich and repels electrons in Br2; it does not attract them. This induced dipole, rather than electron attraction, creates Brδ+."}]},{"id":"76e7df67-b867-4da5-9a8e-6593b2703468","cardType":"basic","points":1,"front":"What is the major product when HBr adds normally to propene?","back":"2-bromopropane, CH₃CHBrCH₃.","mcqOptions":[{"text":"2-bromopropane, CH₃CHBrCH₃.","correct":true,"explanation":"Normal addition follows Markovnikov orientation: protonation forms the more stable secondary carbocation, then bromide attacks it to give 2-bromopropane."},{"text":"1-bromopropane, CH₃CH₂CH₂Br.","correct":false,"explanation":"1-Bromopropane would result from forming a primary carbocation, which is less stable than the secondary carbocation formed in the normal addition pathway."},{"text":"2-bromopropene, CH₃CBr=CH₂.","correct":false,"explanation":"Addition of HBr adds one hydrogen and one bromine across the double bond without creating another double bond. The normal product is 2-bromopropane."},{"text":"1,2-dibromopropane, CH₃CHBrCH₂Br.","correct":false,"explanation":"HBr addition adds only one bromine atom across the C=C bond, not two. A dibromide would require bromine, Br2, rather than hydrogen bromide."}]},{"id":"a15aae7e-769a-4be2-b3e6-ead46471a765","cardType":"cloze","points":null,"front":"In HBr addition to propene, the secondary carbocation is favoured because alkyl groups [ ] towards the positive carbon.","back":"In HBr addition to propene, the secondary carbocation is favoured because alkyl groups donate electron density towards the positive carbon.","mcqOptions":[{"text":"donate electron density","correct":true,"explanation":"Alkyl groups donate electron density through the σ-bond framework, stabilising a nearby positive carbon. Therefore, the secondary carbocation is favoured over the primary alternative."},{"text":"withdraw electron density","correct":false,"explanation":"Alkyl groups do not withdraw electron density from the positive carbon in this context. Their electron donation stabilises the carbocation and favours the secondary intermediate."},{"text":"donate protons","correct":false,"explanation":"Alkyl groups do not donate protons during carbocation stabilisation. Their relevant effect is donation of electron density towards the positively charged carbon."},{"text":"increase electron deficiency","correct":false,"explanation":"Alkyl groups reduce the electron deficiency of a carbocation rather than increase it. Donation of electron density makes the secondary carbocation more stable."}]},{"id":"9b934759-a38d-4508-97c4-8fc6b6ab98e7","cardType":"basic","points":3,"front":"What conditions are used for catalytic hydrogenation of an alkene?","back":"

H₂ gas
Nickel catalyst
About 150 °C

","mcqOptions":[{"text":"

H₂ gas
Nickel catalyst
About 150 °C

","correct":true,"explanation":"Catalytic hydrogenation adds H2 across the C=C bond using nickel as the catalyst, with about 150 °C providing suitable industrial reaction conditions."},{"text":"

H₂ gas
Platinum catalyst
About 150 °C

","correct":false,"explanation":"Platinum can catalyse hydrogenation in some laboratory procedures, but these stated conditions use nickel. The expected conditions are H2, nickel and about 150 °C."},{"text":"

O₂ gas
Nickel catalyst
About 150 °C

","correct":false,"explanation":"Hydrogenation requires hydrogen gas, not oxygen, because hydrogen atoms add across the double bond. The catalyst is nickel at about 150 °C."},{"text":"

H₂ gas
Nickel catalyst
Room temperature

","correct":false,"explanation":"Room temperature is not the stated industrial condition for this catalytic hydrogenation. The required conditions are H2 gas, nickel catalyst and about 150 °C."}]},{"id":"051e3ee5-6bec-49a8-b2e4-b89a6eb9daf0","cardType":"basic","points":2,"front":"What happens to the C=C bond during addition polymerisation?","back":"It opens into C-C single bonds between monomers.","mcqOptions":[{"text":"It opens into C-C single bonds between monomers.","correct":true,"explanation":"During addition polymerisation, the alkene π-bond is broken and each carbon forms a new σ-bond to a neighbouring monomer, producing C-C single bonds in the chain."},{"text":"It remains as C=C double bonds between monomers.","correct":false,"explanation":"The double bond does not remain in the polymer backbone during addition polymerisation. It opens so neighbouring monomers become linked by C-C single bonds."},{"text":"It opens into C=C single bonds between monomers.","correct":false,"explanation":"A C=C bond cannot open into C=C single bonds; single bonds are C-C bonds. The alkene double bond opens to form C-C single links between monomers."},{"text":"It breaks into C-H bonds between monomers.","correct":false,"explanation":"Addition polymerisation changes the alkene π-bond into carbon-carbon links, not carbon-hydrogen bonds. Hydrogen is not added between the monomer units in this process."}]},{"id":"c9c0ba49-f447-43d4-97e9-e97c089b1d3b","cardType":"basic","points":2,"front":"What repeat unit forms from propene, CH₂=CHCH₃?","back":"[-CH₂-CH(CH₃)-]_n.","mcqOptions":[{"text":"[-CH₂-CH(CH₃)-]_n.","correct":true,"explanation":"The double bond opens and each propene monomer contributes a two-carbon backbone unit, while its CH3 group remains attached to the second carbon of the repeat unit."},{"text":"[-CH(CH₃)-CH₂-]_n.","correct":false,"explanation":"This places the methyl substituent on the wrong backbone carbon. Propene forms a repeat unit with CH3 attached to the second carbon: [-CH2-CH(CH3)-]n."},{"text":"[-CH₂-CH₂-]_n.","correct":false,"explanation":"The methyl group from propene is not lost during polymerisation; it remains as a side group on the carbon chain. The repeat unit is [-CH2-CH(CH3)-]n."},{"text":"[-CH₂-C(CH₃)=CH₂-]_n.","correct":false,"explanation":"Polymerisation removes the double bond from the backbone, so a C=C bond should not appear in the repeat unit. Propene gives saturated C-C links instead."}]}]}],"examCountdown":{"examDate":"2027-05-28","paperLabel":"Component 1"},"heading":"Hydrocarbons","title":"Hydrocarbons Flashcards","topbar":{"backHref":"/a-level/chemistry/eduqas/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"2ab2d674-db0e-5811-ae54-4fc9c36348c9","walled":"$L39"}],"$L3a","$L3b"]