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23:[["$","$L32",null,{"feature":"flashcards"}],"$L33",["$","$L34",null,{"href":"/a-level/chemistry/edexcel/alkenes/flashcards","label":"Alkenes","subjectCode":"a_level_chemistry","surface":"flashcards"}],["$","$L35",null,{"runner":["$","div",null,{"className":"min-h-0 flex-1 overflow-hidden bg-background","children":["$","$L36",null,{"decks":[{"id":"d672e201-b8dd-48d9-915e-3230469df686","title":"Alkenes Flashcards","topicId":"82fcc437-7f50-5b51-8004-b3ff6e564aeb","topicSlug":"alkenes","cards":[{"id":"18855453-9e22-443e-bae1-37e71d055055","cardType":"basic","points":1,"front":"Which functional group identifies an alkene?","back":"The C=C double bond.","mcqOptions":[{"text":"The C=C double bond.","correct":true,"explanation":"Alkenes are hydrocarbons containing a carbon-carbon double bond. This C=C functional group determines their characteristic structure and reactions, including electrophilic addition."},{"text":"The C-C single bond.","correct":false,"explanation":"Carbon-carbon single bonds occur in alkanes and many other organic compounds, so they do not identify alkenes. The defining functional group is the C=C double bond."},{"text":"The C≡C triple bond.","correct":false,"explanation":"A carbon-carbon triple bond is the functional group of an alkyne, not an alkene. Alkenes are identified by the presence of a C=C double bond."},{"text":"The C-OH hydroxyl group.","correct":false,"explanation":"The hydroxyl group identifies an alcohol rather than an alkene. The functional group characteristic of an alkene is a carbon-carbon double bond, C=C."}]},{"id":"07695640-da71-4f9c-b2af-bb5b4b544925","cardType":"cloze","points":null,"front":"Acyclic alkenes with one C=C have the formula [ ].","back":"Acyclic alkenes with one C=C have the formula C_nH_2n.","mcqOptions":[{"text":"C_nH_2n","correct":true,"explanation":"An acyclic alkene with one C=C bond has two fewer hydrogen atoms than the corresponding alkane, giving the general formula C_nH_2n."},{"text":"C_nH_2n+2","correct":false,"explanation":"This is the general formula for an acyclic alkane. Introducing one C=C bond removes two hydrogen atoms, so the alkene formula is C_nH_2n."},{"text":"C_nH_2n-2","correct":false,"explanation":"This formula represents two degrees of unsaturation, such as an alkyne or a compound with two rings or double bonds. One acyclic C=C gives C_nH_2n."},{"text":"C_nH_n","correct":false,"explanation":"This incorrectly halves the expected hydrogen count and is not the general formula for simple alkenes. An acyclic alkene with one C=C has formula C_nH_2n."}]},{"id":"70cee4c9-f9e7-48d1-976a-49f0d598caa4","cardType":"basic","points":2,"front":"Why is an alkene unsaturated?","back":"Its C=C bond can open up to add atoms.","mcqOptions":[{"text":"Its C=C bond can open up to add atoms.","correct":true,"explanation":"The pi component of the C=C bond can break during an addition reaction, allowing new atoms or groups to bond to both carbon atoms."},{"text":"Its C-C bond can split to remove atoms.","correct":false,"explanation":"Unsaturation concerns the capacity to add atoms across a multiple bond, not remove them by splitting a single bond. The C=C bond can open to permit addition."},{"text":"Its C=C bond can only substitute hydrogen atoms.","correct":false,"explanation":"Alkenes characteristically undergo addition rather than only substitution. Their C=C bond can open as the pi bond breaks, allowing atoms to be added across it."},{"text":"Its C-H bonds can open up to add atoms.","correct":false,"explanation":"The relevant reactive feature is not a C-H bond. An alkene is unsaturated because its C=C bond can open up, allowing atoms to add across the double bond."}]},{"id":"5c0a2acb-c3a3-46fb-925d-b29be1809bb5","cardType":"cloze","points":null,"front":"A C=C double bond contains one [ ] and one [ ].","back":"A C=C double bond contains one sigma bond and one pi bond.","mcqOptions":[{"text":"A C=C double bond contains one sigma bond and one pi bond.","correct":true,"explanation":"Head-on orbital overlap forms the sigma bond, while sideways overlap of parallel p orbitals forms the pi bond. Together, these make the C=C double bond."},{"text":"A C=C double bond contains one pi bond and one sigma bond.","correct":false,"explanation":"This reverses the terms relative to the stated order. The first bond formed between the carbon nuclei is the sigma bond, and the additional bond is the pi bond."},{"text":"A C=C double bond contains one sigma bond and one sigma bond.","correct":false,"explanation":"A double bond does not contain two sigma bonds because only one bond can form directly along the internuclear axis. It consists of one sigma bond and one pi bond."},{"text":"A C=C double bond contains one pi bond and one pi bond.","correct":false,"explanation":"A carbon-carbon double bond cannot consist of two pi bonds. One head-on overlap forms a sigma bond, while one sideways p-orbital overlap forms the pi bond."}]},{"id":"92237d92-aa3e-4e69-aa89-3ffccd08ca3b","cardType":"basic","points":1,"front":"How is the pi bond in an alkene formed?","back":"By sideways overlap of p orbitals.","mcqOptions":[{"text":"By sideways overlap of p orbitals.","correct":true,"explanation":"Parallel unhybridised p orbitals on the two carbon atoms overlap sideways above and below the molecular plane, creating the electron density of the pi bond."},{"text":"By head-on overlap of p orbitals.","correct":false,"explanation":"Head-on overlap places electron density along the internuclear axis and forms a sigma bond. The alkene pi bond instead forms through sideways overlap of parallel p orbitals."},{"text":"By sideways overlap of s orbitals.","correct":false,"explanation":"S orbitals are spherical and cannot produce the two-lobed sideways overlap required for a pi bond. The pi bond forms through sideways overlap of p orbitals."},{"text":"By head-on overlap of sp² orbitals.","correct":false,"explanation":"Head-on overlap of sp² orbitals contributes to the carbon-carbon sigma bond. The pi bond is formed separately by sideways overlap of the unhybridised p orbitals."}]},{"id":"2a97fb16-c212-40e7-9f42-c9ace8ec8a1d","cardType":"basic","points":2,"front":"Why are alkenes more reactive than alkanes?","back":"The exposed, weaker pi bond is electron-rich and attracts electrophiles.","mcqOptions":[{"text":"The exposed, weaker pi bond is electron-rich and attracts electrophiles.","correct":true,"explanation":"The pi electrons lie above and below the bond axis, making them relatively exposed. Their high electron density attracts electrophiles, and the weaker pi bond breaks readily."},{"text":"The stronger carbon-carbon sigma bond is electron-rich and attracts electrophiles.","correct":false,"explanation":"The carbon-carbon sigma bond is strong and its electrons are held directly between the nuclei, so it is less accessible. Alkene reactivity comes from the exposed, weaker pi bond."},{"text":"The exposed, weaker pi bond is electron-poor and attracts nucleophiles.","correct":false,"explanation":"This reverses the electron-density argument. A pi bond is electron-rich, so it attracts electron-deficient electrophiles rather than electron-rich nucleophiles during electrophilic addition."},{"text":"The exposed carbon-hydrogen bond is electron-rich and attracts electrophiles.","correct":false,"explanation":"Carbon-hydrogen bonds are not the feature responsible for the greater reactivity of alkenes. The exposed, weaker, electron-rich pi bond attracts electrophiles and breaks during addition."}]},{"id":"5278da92-3042-4914-842d-02a95562c8f2","cardType":"cloze","points":null,"front":"Each carbon in a C=C bond is [ ] with bond angles of about [ ].","back":"Each carbon in a C=C bond is trigonal planar with bond angles of about 120°.","mcqOptions":[{"text":"Each carbon in a C=C bond is trigonal planar with bond angles of about 120°.","correct":true,"explanation":"Each alkene carbon has three regions of electron density and is sp² hybridised. These arrange in a trigonal planar geometry with bond angles close to 120°."},{"text":"Each carbon in a C=C bond is tetrahedral with bond angles of about 109.5°.","correct":false,"explanation":"Tetrahedral geometry and 109.5° angles apply to carbon with four sigma bonds and sp³ hybridisation. Each C=C carbon is sp² hybridised, trigonal planar and approximately 120°."},{"text":"Each carbon in a C=C bond is trigonal planar with bond angles of about 180°.","correct":false,"explanation":"Trigonal planar geometry has three bonding directions separated by approximately 120°, not 180°. An angle of 180° is associated with a linear arrangement around an sp-hybridised atom."},{"text":"Each carbon in a C=C bond is linear with bond angles of about 120°.","correct":false,"explanation":"Linear geometry would require two regions of electron density and bond angles of 180°. Each alkene carbon has three regions, giving trigonal planar geometry and angles near 120°."}]},{"id":"134b89a1-53c7-4eca-bb01-5faa9713c04e","cardType":"basic","points":2,"front":"Why is rotation about a C=C bond restricted?","back":"Rotation would break the pi bond.","mcqOptions":[{"text":"Rotation would break the pi bond.","correct":true,"explanation":"Rotation would move the parallel p orbitals out of alignment, eliminating their sideways overlap. This requires the pi bond to break, so free rotation is prevented."},{"text":"Rotation would break the sigma bond.","correct":false,"explanation":"The sigma bond lies along the internuclear axis and can remain intact during rotation. The restriction arises because rotation destroys sideways p-orbital overlap and breaks the pi bond."},{"text":"Rotation would form another pi bond.","correct":false,"explanation":"Rotation misaligns the p orbitals rather than creating additional overlap. It would destroy the existing sideways overlap and therefore require the pi bond to break."},{"text":"Rotation would make both carbons tetrahedral.","correct":false,"explanation":"Rotation alone does not change the number of electron-density regions or rehybridise the carbon atoms. It is restricted because misaligning the p orbitals would break the pi bond."}]},{"id":"78eeec76-03a3-41d4-8667-c20062c13c14","cardType":"basic","points":1,"front":"When naming an alkene, what must get the lowest possible number?","back":"The C=C double bond.","mcqOptions":[{"text":"The C=C double bond.","correct":true,"explanation":"The carbon chain is numbered from the end that gives the C=C double bond the lowest possible locant, because the double bond takes priority over substituents."},{"text":"The first alkyl substituent.","correct":false,"explanation":"Prioritising an alkyl substituent is a common numbering error. The C=C double bond must receive the lowest possible number, even if this gives a substituent a higher number."},{"text":"The longest carbon branch.","correct":false,"explanation":"Branches do not determine the numbering direction in this way. The parent chain must contain the C=C bond, which is then assigned the lowest possible locant."},{"text":"The first carbon atom.","correct":false,"explanation":"Carbon 1 simply begins whichever numbering direction is chosen; it is not itself the feature being prioritised. The C=C double bond must get the lowest possible number."}]},{"id":"1bc65e6d-d538-4d7e-9e12-661bb4fc8511","cardType":"basic","points":1,"front":"Name CH₃CH₂CH=CHCH₃.","back":"pent-2-ene.","mcqOptions":[{"text":"pent-2-ene.","correct":true,"explanation":"The longest chain contains five carbons, giving pent-, and numbering from the nearest end places the C=C bond between carbons 2 and 3."},{"text":"pent-3-ene.","correct":false,"explanation":"Numbering from the left gives 3, but alkene chains must be numbered from the end nearest the C=C bond. This gives the correct name pent-2-ene."},{"text":"but-2-ene.","correct":false,"explanation":"This incorrectly counts only four carbons in the parent chain. The longest chain containing the C=C bond has five carbons, so the correct name is pent-2-ene."},{"text":"pent-1-ene.","correct":false,"explanation":"The five-carbon parent chain is correct, but the double bond does not begin at carbon 1. Numbering from the nearest end places it at carbon 2, giving pent-2-ene."}]},{"id":"29cdcf9d-d088-4740-ab01-a903a1b95b78","cardType":"basic","points":2,"front":"Why are but-1-ene and but-2-ene positional isomers?","back":"The C=C bond is in a different position.","mcqOptions":[{"text":"The C=C bond is in a different position.","correct":true,"explanation":"Both molecules have the same carbon skeleton and molecular formula, but the C=C bond begins at carbon 1 in but-1-ene and carbon 2 in but-2-ene."},{"text":"The carbon chain has a different length.","correct":false,"explanation":"Both compounds have the same four-carbon chain, so chain length is unchanged. They are positional isomers because the C=C bond occupies a different position."},{"text":"The molecular formula is different in each.","correct":false,"explanation":"Isomers must share the same molecular formula, and both compounds are C₄H₈. Their difference is the position of the C=C bond along the chain."},{"text":"The carbon skeleton is arranged differently.","correct":false,"explanation":"A different carbon skeleton would make them chain isomers. But-1-ene and but-2-ene retain the same straight-chain skeleton and differ only in the C=C bond’s position."}]},{"id":"796ff61f-aec6-4a84-9d74-2d153bf45ad3","cardType":"cloze","points":null,"front":"E/Z stereoisomerism requires each C=C carbon to have [ ] attached.","back":"E/Z stereoisomerism requires each C=C carbon to have two different groups attached.","mcqOptions":[{"text":"two different groups","correct":true,"explanation":"Each carbon of the C=C bond must have two different attached groups; otherwise, swapping groups does not produce distinguishable E and Z arrangements."},{"text":"two identical groups","correct":false,"explanation":"Identical groups on either C=C carbon make the two apparent arrangements indistinguishable. E/Z stereoisomerism requires each double-bonded carbon to have two different groups attached."},{"text":"two hydrogen atoms","correct":false,"explanation":"Attaching two hydrogens to either C=C carbon prevents E/Z stereoisomerism because that carbon has identical groups. Each double-bonded carbon needs two different groups."},{"text":"four different groups","correct":false,"explanation":"It is unnecessary for all four groups around the C=C bond to differ. The requirement is that each individual double-bonded carbon has two different groups attached."}]},{"id":"6113fcd6-6ffd-4417-9cb4-3650ba524689","cardType":"basic","points":1,"front":"In E/Z priority rules, which attached atom has higher priority?","back":"The atom with the higher atomic number.","mcqOptions":[{"text":"The atom with the higher atomic number.","correct":true,"explanation":"Cahn-Ingold-Prelog priority is determined first by the atomic numbers of the atoms directly attached to each C=C carbon; the higher atomic number receives higher priority."},{"text":"The atom with the higher relative mass.","correct":false,"explanation":"Priority is generally based on atomic number, not relative atomic mass. Comparing mass is relevant only when distinguishing isotopes of the same element."},{"text":"The atom with the lower atomic number.","correct":false,"explanation":"This reverses the Cahn-Ingold-Prelog rule. Of the atoms directly attached to a C=C carbon, the one with the higher atomic number has higher priority."},{"text":"The atom with the greater bond polarity.","correct":false,"explanation":"Bond polarity does not set E/Z priority. The attached atoms are compared by atomic number, with the atom of higher atomic number receiving higher priority."}]},{"id":"c0327127-0354-49c0-9a6c-547143c0ac2a","cardType":"cloze","points":null,"front":"Higher-priority groups on the same side give [ ]; on opposite sides give [ ].","back":"Higher-priority groups on the same side give Z; on opposite sides give E.","mcqOptions":[{"text":"Higher-priority groups on the same side give Z; on opposite sides give E.","correct":true,"explanation":"Z comes from the German zusammen, meaning together, so the higher-priority groups are on the same side. E corresponds to them being on opposite sides."},{"text":"Higher-priority groups on the same side give E; on opposite sides give Z.","correct":false,"explanation":"This reverses the E/Z assignments. Higher-priority groups together on the same side give Z, while higher-priority groups on opposite sides give E."},{"text":"Higher-priority groups on the same side give Z; on opposite sides give Z.","correct":false,"explanation":"Z applies only when the higher-priority groups are on the same side. When those groups are on opposite sides of the C=C bond, the configuration is E."},{"text":"Higher-priority groups on the same side give E; on opposite sides give E.","correct":false,"explanation":"E applies only when the higher-priority groups are on opposite sides. Groups on the same side give Z, from zusammen, meaning together."}]},{"id":"97f16a70-ea6a-4682-a8a8-f0cc0e211280","cardType":"cloze","points":null,"front":"In alkene addition, the C=C becomes a [ ] as the [ ] breaks.","back":"In alkene addition, the C=C becomes a C-C single bond as the pi bond breaks.","mcqOptions":[{"text":"In alkene addition, the C=C becomes a C-C single bond as the pi bond breaks.","correct":true,"explanation":"The weaker pi component of the C=C bond breaks during addition, allowing two new sigma bonds to form while the original C-C sigma bond remains."},{"text":"In alkene addition, the C=C becomes a C-C single bond as the sigma bond breaks.","correct":false,"explanation":"The existing C-C sigma bond remains intact during addition. It is the weaker pi bond that breaks, leaving a C-C single bond between the carbon atoms."},{"text":"In alkene addition, the C=C becomes a C-C triple bond as the pi bond breaks.","correct":false,"explanation":"Breaking the pi bond reduces the bond order rather than increasing it. The C=C double bond therefore becomes a C-C single bond, not a triple bond."},{"text":"In alkene addition, the C=C becomes a C=C double bond as the sigma bond breaks.","correct":false,"explanation":"Addition removes the alkene’s pi bond rather than preserving the double bond or breaking its sigma component. The product contains a C-C single bond."}]},{"id":"ee2666c2-6776-42a4-9990-3b08768ca184","cardType":"basic","points":2,"front":"Why is alkene addition usually electrophilic?","back":"The pi bond donates an electron pair to an electrophile.","mcqOptions":[{"text":"The pi bond donates an electron pair to an electrophile.","correct":true,"explanation":"The pi bond has a region of high electron density and acts as an electron-pair donor. It attacks an electron-deficient electrophile to begin the addition mechanism."},{"text":"The pi bond accepts an electron pair from an electrophile.","correct":false,"explanation":"This reverses the electron movement: electrophiles accept electron pairs rather than donate them. The electron-rich pi bond donates an electron pair to the electrophile."},{"text":"The sigma bond donates an electron pair to an electrophile.","correct":false,"explanation":"The C-C sigma bond is stronger and less exposed than the pi bond, so it does not initiate the reaction. The pi bond donates the electron pair."},{"text":"The pi bond donates a proton to an electrophile.","correct":false,"explanation":"A pi bond supplies an electron pair, not a proton. Its electron density attacks the electrophile, which accepts that pair and initiates electrophilic addition."}]},{"id":"6aded7ab-f950-4122-8881-a9050d4760d9","cardType":"cloze","points":null,"front":"A carbocation contains a [ ].","back":"A carbocation contains a positively charged carbon atom.","mcqOptions":[{"text":"positively charged carbon atom","correct":true,"explanation":"A carbocation is an ion in which the carbon atom carries a positive charge, giving it only six electrons in its outer shell and making it electron-deficient."},{"text":"negatively charged carbon atom","correct":false,"explanation":"A negatively charged carbon atom is a carbanion, not a carbocation. The correct definition is a positively charged carbon atom, which is electron-deficient and highly reactive."},{"text":"positively charged hydrogen atom","correct":false,"explanation":"A positively charged hydrogen atom is a proton, not a carbocation. Carbocations specifically contain a positively charged carbon atom within an organic molecule."},{"text":"uncharged carbon atom","correct":false,"explanation":"An uncharged carbon atom does not describe a carbocation. The defining feature is a positive charge on carbon, so the correct term is positively charged carbon atom."}]},{"id":"f38273b5-4337-4ad2-af5f-272ce39bf203","cardType":"basic","points":1,"front":"What is the A-Level carbocation stability order?","back":"tertiary > secondary > primary.","mcqOptions":[{"text":"tertiary > secondary > primary.","correct":true,"explanation":"Tertiary carbocations are most stable because three alkyl groups donate electron density and provide greater stabilisation, followed by secondary and then primary carbocations."},{"text":"primary > secondary > tertiary.","correct":false,"explanation":"This reverses the effect of alkyl groups. Alkyl groups stabilise the positive charge through electron donation, so the correct order is tertiary > secondary > primary."},{"text":"secondary > primary > tertiary.","correct":false,"explanation":"Although secondary carbocations are more stable than primary ones, tertiary carbocations are most stable because they have three electron-donating alkyl groups. The correct order is tertiary > secondary > primary."},{"text":"tertiary > primary > secondary.","correct":false,"explanation":"The first position is correct, but primary and secondary are reversed. Secondary carbocations have two stabilising alkyl groups, so the correct order is tertiary > secondary > primary."}]},{"id":"d1ac5cbc-9305-4b43-bc3e-802942eb39b3","cardType":"basic","points":1,"front":"For an unsymmetrical alkene, which addition pathway gives the major product?","back":"The one forming the more stable carbocation.","mcqOptions":[{"text":"The one forming the more stable carbocation.","correct":true,"explanation":"The major pathway is the one that forms the more stable carbocation intermediate, because this lower-energy pathway proceeds more readily and produces more of the major product."},{"text":"The one forming the less stable carbocation.","correct":false,"explanation":"The less stable carbocation forms through a higher-energy pathway and is therefore less favoured. The major pathway is the one forming the more stable carbocation."},{"text":"The one forming the more stable carbanion.","correct":false,"explanation":"This reaction pathway involves carbocation intermediates, not carbanions. The major product comes from the pathway forming the more stable carbocation."},{"text":"The one forming the least substituted carbocation.","correct":false,"explanation":"Greater substitution generally stabilises a carbocation through alkyl-group electron donation. Therefore, the major pathway forms the more stable carbocation rather than the least substituted one."}]},{"id":"f2dc2d25-e125-4075-ba87-8f16cbb9c846","cardType":"basic","points":1,"front":"What is the major product when propene reacts with HBr?","back":"2-bromopropane.","mcqOptions":[{"text":"2-bromopropane.","correct":true,"explanation":"HBr adds across propene so that hydrogen forms the more stable secondary carbocation. Bromide then bonds to the positively charged carbon, producing 2-bromopropane as the major product."},{"text":"1-bromopropane.","correct":false,"explanation":"This product would result from forming a less stable primary carbocation during addition. The major product is 2-bromopropane because the reaction proceeds through a more stable secondary carbocation."},{"text":"2-bromopropene.","correct":false,"explanation":"Addition of HBr removes the C=C bond and adds hydrogen and bromine across it, so the product is saturated. The correct major product is 2-bromopropane."},{"text":"1-bromopropene.","correct":false,"explanation":"This retains a double bond and does not represent addition of HBr across propene. The major saturated product is 2-bromopropane, formed through a secondary carbocation."}]},{"id":"43c4d996-b660-4b55-890c-878a17ad8fd7","cardType":"basic","points":1,"front":"What is the bromine water observation for a C=C bond?","back":"Orange/brown to colourless.","mcqOptions":[{"text":"Orange/brown to colourless.","correct":true,"explanation":"The alkene C=C bond reacts with bromine by addition, consuming the coloured bromine. Consequently, bromine water changes from orange/brown to colourless."},{"text":"Colourless to orange/brown.","correct":false,"explanation":"The colour change is in the wrong direction. Bromine water starts orange/brown and loses its colour when bromine reacts with the alkene C=C bond."},{"text":"Remains orange/brown.","correct":false,"explanation":"An unchanged orange/brown colour suggests no reaction, as with a saturated compound under these conditions. An alkene decolourises bromine water, giving orange/brown to colourless."},{"text":"Blue to colourless.","correct":false,"explanation":"Blue is associated with tests such as iodine-starch chemistry, not bromine water. The alkene test starts orange/brown and changes to colourless when the C=C bond reacts."}]},{"id":"40106c46-75a8-4ac1-97bc-2bfd69df5b7a","cardType":"basic","points":3,"front":"What conditions are typical for alkene hydrogenation?","back":"

H₂ gas
Ni catalyst
Heated, about 150 °C

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

H₂ gas
Ni catalyst
Heated, about 150 °C

","correct":true,"explanation":"Hydrogenation adds H₂ across the alkene C=C bond. Industrially, a nickel catalyst and heating to about 150 °C provide suitable conditions for the reaction."},{"text":"

Steam
H₃PO₄ catalyst
High temperature and pressure: about 300 °C, 60 atm

","correct":false,"explanation":"These are hydration conditions, which add water to an alkene and form an alcohol. Hydrogenation instead requires H₂ gas, a nickel catalyst and heating to about 150 °C."},{"text":"

Br₂ liquid
No catalyst
Room temperature, in the dark

","correct":false,"explanation":"Bromine adds across a C=C bond without a nickel catalyst, forming a dibromoalkane. Hydrogenation requires H₂, nickel and heating to about 150 °C."},{"text":"

O₂ gas
Ni catalyst
Heated, about 150 °C

","correct":false,"explanation":"Oxygen would not hydrogenate the alkene; it could promote oxidation or combustion instead. The required reagent is H₂ gas with nickel catalyst and heating."}]},{"id":"dd5370bb-e866-47c1-85f1-5a319955255c","cardType":"basic","points":3,"front":"What typical industrial conditions hydrate alkenes to alcohols?","back":"

Steam
H₃PO₄ catalyst
High temperature and pressure: about 300 °C, 60 atm

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

Steam
H₃PO₄ catalyst
High temperature and pressure: about 300 °C, 60 atm

","correct":true,"explanation":"Industrial hydration adds steam across ethene’s C=C bond to form ethanol. Phosphoric acid catalyses the reaction, while about 300 °C and 60 atm provide suitable conditions."},{"text":"

H₂ gas
Ni catalyst
Heated, about 150 °C

","correct":false,"explanation":"These conditions hydrogenate an alkene to form an alkane, rather than hydrate it to an alcohol. Hydration requires steam, H₃PO₄ and about 300 °C at 60 atm."},{"text":"

Br₂ liquid
No catalyst
Room temperature and pressure

","correct":false,"explanation":"Bromine adds across the C=C bond to form a dibromoalkane and does not produce an alcohol. Industrial hydration uses steam and an H₃PO₄ catalyst."},{"text":"

O₂ gas
V₂O₅ catalyst
High temperature and pressure: about 400 °C, 10 atm

","correct":false,"explanation":"These conditions suggest oxidation chemistry, not addition of water to an alkene. Hydration uses steam, phosphoric acid and approximately 300 °C at 60 atm."}]},{"id":"fd9dd956-dd9d-4b0d-ba6d-9290618bc00f","cardType":"basic","points":1,"front":"Why should industrial ethene hydration be written with a reversible arrow?","back":"It is an equilibrium reaction.","mcqOptions":[{"text":"It is an equilibrium reaction.","correct":true,"explanation":"Ethene hydration is reversible because ethanol can dehydrate back to ethene and steam. Both forward and reverse reactions occur, establishing an equilibrium mixture."},{"text":"It is an irreversible reaction.","correct":false,"explanation":"Ethanol can dehydrate to ethene and steam under suitable conditions, so the reaction is reversible rather than irreversible. That is why an equilibrium arrow is used."},{"text":"It is a substitution reaction.","correct":false,"explanation":"Hydration adds steam across the C=C bond, making it an addition reaction. Its reversible nature means it is correctly described as an equilibrium reaction."},{"text":"It is a redox reaction.","correct":false,"explanation":"No overall oxidation-state change is the key reason for the arrow notation. Ethene hydration is a reversible addition reaction and therefore an equilibrium reaction."}]},{"id":"f0d061f2-0ccd-46e2-9364-15832e2be38c","cardType":"basic","points":2,"front":"Why is alkene polymerisation called addition polymerisation?","back":"Monomers join with no small molecule lost.","mcqOptions":[{"text":"Monomers join with no small molecule lost.","correct":true,"explanation":"The alkene C=C bonds open so that monomers link together without eliminating any atoms or molecules, meaning the polymer is the only product formed."},{"text":"Monomers join with a water molecule lost.","correct":false,"explanation":"Elimination of water occurs in condensation polymerisation, not alkene addition polymerisation. Addition polymerisation opens C=C bonds and joins monomers with no small molecule lost."},{"text":"Monomers join with a hydrogen molecule lost.","correct":false,"explanation":"Confusing addition with an elimination reaction makes hydrogen seem like a by-product. In addition polymerisation, C=C bonds open and monomers join with no small molecule lost."},{"text":"Monomers join with carbon dioxide lost.","correct":false,"explanation":"Carbon dioxide is not eliminated when alkene monomers polymerise. Their C=C bonds open to form links, with no small molecule lost from the monomers."}]},{"id":"a7b708cf-b83f-4a1d-bb04-d556b6e4dee9","cardType":"basic","points":1,"front":"What repeat unit forms from ethene in addition polymerisation?","back":"[-CH₂-CH₂-]_n.","mcqOptions":[{"text":"[-CH₂-CH₂-]_n.","correct":true,"explanation":"Ethene’s C=C bond opens during polymerisation, forming a saturated chain in which each monomer contributes the repeat unit -CH₂-CH₂-."},{"text":"[-CH₂-CH(CH₃)-]_n.","correct":false,"explanation":"The CH₃ side group comes from propene, so this is the repeat unit of poly(propene). Ethene instead forms [-CH₂-CH₂-]_n."},{"text":"[-CH₂-CHCl-]_n.","correct":false,"explanation":"The chlorine substituent comes from chloroethene, making this the repeat unit of poly(chloroethene). Ethene has no chlorine and forms [-CH₂-CH₂-]_n."},{"text":"[-CH=CH-]_n.","correct":false,"explanation":"This incorrectly leaves a C=C bond in each repeat unit. Ethene’s double bond opens during addition polymerisation, producing the saturated unit [-CH₂-CH₂-]_n."}]}]}],"examCountdown":{"examDate":"2027-05-28","paperLabel":"Paper 1"},"heading":"Alkenes","title":"Alkenes Flashcards","topbar":{"backHref":"/a-level/chemistry/edexcel/flashcards"},"topicSidebar":"$L37"}]}],"topicId":"82fcc437-7f50-5b51-8004-b3ff6e564aeb","walled":"$L38"}],"$L39","$L3a"]