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23:[["$","$L32",null,{"feature":"flashcards"}],"$L33",["$","$L34",null,{"href":"/a-level/chemistry/edexcel/arenes/flashcards","label":"Arenes","subjectCode":"a_level_chemistry","surface":"flashcards"}],["$","$L35",null,{"runner":["$","div",null,{"className":"min-h-0 flex-1 overflow-hidden bg-background","children":["$","$L36",null,{"decks":[{"id":"ddd96d47-7f07-4f00-b0b1-4b9c9f95c0dd","title":"Arenes Flashcards","topicId":"3332e219-cbb8-589c-9979-9a79486a3ca6","topicSlug":"arenes","cards":[{"id":"7edb9026-c494-4c06-a6f0-d7891b82d547","cardType":"basic","points":1,"front":"What is the molecular formula of benzene?","back":"C₆H₆","mcqOptions":[{"text":"C₆H₆","correct":true,"explanation":"Benzene contains six carbon atoms and six hydrogen atoms, giving the molecular formula C₆H₆. Its four degrees of unsaturation arise from one ring and three π bonds."},{"text":"C₆H₁₂","correct":false,"explanation":"This formula represents cyclohexane, where all carbon-carbon bonds are single. Benzene has three additional degrees of unsaturation from its delocalised π system, so the correct formula is C₆H₆."},{"text":"C₆H₅","correct":false,"explanation":"This is the formula for the phenyl group, formed when one hydrogen is removed from benzene. The complete neutral benzene molecule contains six hydrogens, giving C₆H₆."},{"text":"C₆H₁₄","correct":false,"explanation":"This formula belongs to hexane, a saturated acyclic alkane. Benzene has a ring and delocalised π bonding, so it contains fewer hydrogens: C₆H₆."}]},{"id":"e6d1d638-7789-4205-915c-55528d218575","cardType":"cloze","points":null,"front":"An arene contains at least one [ ] and only carbon and hydrogen.","back":"An arene contains at least one benzene ring and only carbon and hydrogen.","mcqOptions":[{"text":"benzene ring","correct":true,"explanation":"An arene is an aromatic hydrocarbon containing at least one benzene ring and consisting only of carbon and hydrogen. This structural feature distinguishes arenes from other hydrocarbons."},{"text":"carbonyl group","correct":false,"explanation":"A carbonyl group contains carbon and oxygen and is therefore not present in a hydrocarbon-only definition of an arene. The required structural feature is a benzene ring."},{"text":"cyclohexane ring","correct":false,"explanation":"A cyclohexane ring is a saturated six-membered ring and lacks benzene’s aromatic delocalisation. Arenes must contain at least one benzene ring, not merely any six-membered carbon ring."},{"text":"carboxyl group","correct":false,"explanation":"A carboxyl group contains carbon, oxygen and hydrogen, so a compound containing it is not composed only of carbon and hydrogen. Arenes contain at least one benzene ring."}]},{"id":"c38ebbde-b30e-4adf-bbc5-6254fe4faade","cardType":"basic","points":1,"front":"What is the group C₆H₅- called?","back":"The phenyl group.","mcqOptions":[{"text":"The phenyl group.","correct":true,"explanation":"Removing one hydrogen atom from benzene produces C₆H₅-, called the phenyl group. It is commonly written as Ph when used as a substituent in organic structures."},{"text":"The benzyl group.","correct":false,"explanation":"The benzyl group is C₆H₅CH₂-, so it contains an additional CH₂ unit between the ring and the point of attachment. C₆H₅- itself is the phenyl group."},{"text":"The phenol group.","correct":false,"explanation":"Phenol is a molecule containing a hydroxyl group attached directly to a benzene ring. It is not the name of the hydrocarbon substituent C₆H₅-; that substituent is phenyl."},{"text":"The benzene group.","correct":false,"explanation":"Benzene is the complete molecule C₆H₆, not the substituent formed by removing one hydrogen. The group C₆H₅- is specifically named phenyl."}]},{"id":"c9eff71b-82fc-4fa7-95ad-347639659d6e","cardType":"basic","points":1,"front":"What is the group C₆H₅CH₂- called?","back":"The benzyl group.","mcqOptions":[{"text":"The benzyl group.","correct":true,"explanation":"The group C₆H₅CH₂- is called benzyl. It consists of a phenyl ring attached to a CH₂ group, with the free bond extending from that CH₂ carbon."},{"text":"The phenyl group.","correct":false,"explanation":"Phenyl is C₆H₅-, with the free bond directly attached to the ring. The extra CH₂ unit in C₆H₅CH₂- makes this group benzyl instead."},{"text":"The benzoyl group.","correct":false,"explanation":"Benzoyl is C₆H₅CO-, containing a carbonyl group. The given group has CH₂ rather than CO, so its correct name is benzyl."},{"text":"The ethylbenzene group.","correct":false,"explanation":"Ethylbenzene is a complete molecule containing a C₂H₅ substituent on benzene, not the name of C₆H₅CH₂-. A benzene ring attached through CH₂ forms benzyl."}]},{"id":"db228acd-cdb0-45b8-bdee-961c37a802d6","cardType":"basic","points":1,"front":"How many σ bonds does each carbon form in benzene?","back":"Three σ bonds.","mcqOptions":[{"text":"Three σ bonds.","correct":true,"explanation":"Each carbon in benzene forms two σ bonds to neighbouring carbon atoms and one σ bond to hydrogen. Its remaining p orbital contributes to the delocalised π system."},{"text":"Two σ bonds.","correct":false,"explanation":"Counting only the two carbon-carbon σ bonds misses the carbon-hydrogen σ bond. Each benzene carbon forms three σ bonds in total, while its p orbital forms part of the π system."},{"text":"Four σ bonds.","correct":false,"explanation":"Four σ bonds would give each carbon a typical tetrahedral arrangement and leave no unhybridised p orbital. Benzene carbons are sp² hybridised and form three σ bonds each."},{"text":"Six σ bonds.","correct":false,"explanation":"The six carbon atoms share six carbon-carbon σ bonds around the ring, but that does not mean each carbon forms six σ bonds. Each individual carbon forms three."}]},{"id":"5fc8b687-3f0a-4dc7-958e-ddec51ce2157","cardType":"cloze","points":null,"front":"Each carbon in benzene has one unhybridised [ ].","back":"Each carbon in benzene has one unhybridised p orbital.","mcqOptions":[{"text":"p orbital","correct":true,"explanation":"Each benzene carbon is sp² hybridised, leaving one unhybridised p orbital. These p orbitals overlap sideways around the ring to create the delocalised π system."},{"text":"s orbital","correct":false,"explanation":"The s orbital is involved in forming hybrid orbitals and is not the orbital left unhybridised in an sp² carbon. Benzene retains an unhybridised p orbital for π bonding."},{"text":"sp² orbital","correct":false,"explanation":"sp² describes hybrid orbitals formed by mixing one s and two p orbitals; it is not an unhybridised orbital. The remaining unhybridised orbital is a p orbital."},{"text":"d orbital","correct":false,"explanation":"Carbon’s valence bonding in benzene does not require d orbitals. The π system forms through sideways overlap of the unhybridised p orbitals on the six carbon atoms."}]},{"id":"12ed5810-beba-4a2c-a79a-6ed1cf57a662","cardType":"basic","points":1,"front":"What forms the continuous π system in benzene?","back":"Sideways overlap of six p orbitals.","mcqOptions":[{"text":"Sideways overlap of six p orbitals.","correct":true,"explanation":"Each of benzene’s six sp²-hybridised carbon atoms contributes one unhybridised p orbital. Sideways overlap around the ring produces one continuous delocalised π system."},{"text":"End-on overlap of six p orbitals.","correct":false,"explanation":"End-on overlap forms a σ bond along the internuclear axis, not the delocalised π system. Benzene’s p orbitals overlap sideways above and below the ring."},{"text":"Sideways overlap of six s orbitals.","correct":false,"explanation":"s orbitals are spherical and do not form the continuous π system by sideways overlap. The π electrons come from the six unhybridised p orbitals."},{"text":"Sideways overlap of three p orbitals.","correct":false,"explanation":"All six carbon atoms contribute one p orbital each, so three orbitals would omit half the ring. The continuous system arises from sideways overlap of six p orbitals."}]},{"id":"de4318d2-ac07-478e-81c8-72d05b68a720","cardType":"basic","points":1,"front":"Where is benzene’s delocalised π system located?","back":"Above and below the plane of the ring.","mcqOptions":[{"text":"Above and below the plane of the ring.","correct":true,"explanation":"The parallel p orbitals overlap sideways, placing the delocalised π electron density in two continuous regions above and below the planar carbon ring."},{"text":"In the plane of the ring.","correct":false,"explanation":"Electron density in the ring’s plane is associated mainly with σ bonds. The π system forms from p orbitals perpendicular to the plane, so it lies above and below it."},{"text":"Above the plane of the ring only.","correct":false,"explanation":"A p orbital has two lobes on opposite sides of the ring, so π electron density occurs in regions both above and below. One-sided localisation would not describe benzene’s π system."},{"text":"Below the plane of the ring only.","correct":false,"explanation":"The p orbitals extend on both sides of the planar ring, producing two π electron clouds. Therefore the delocalised system is above and below, not below alone."}]},{"id":"23a14933-3647-4f76-89bb-4b8cc68c2f78","cardType":"cloze","points":null,"front":"Benzene is best represented as a hexagon with a [ ] inside.","back":"Benzene is best represented as a hexagon with a circle inside.","mcqOptions":[{"text":"circle","correct":true,"explanation":"The circle represents the six π electrons delocalised across all six carbon atoms, rather than three fixed double bonds between particular pairs of carbons."},{"text":"dot","correct":false,"explanation":"A dot would usually represent a single electron, not the delocalised π system. Benzene is represented by a hexagon containing a circle."},{"text":"cross","correct":false,"explanation":"A cross has no accepted meaning in the displayed formula of benzene. A circle is used to show that its π electrons are delocalised."},{"text":"triangle","correct":false,"explanation":"A triangle would not represent the continuous delocalisation of π electrons around the ring. The accepted symbol inside the benzene hexagon is a circle."}]},{"id":"565861d7-1744-4a32-b2c4-d3b840845d24","cardType":"basic","points":2,"front":"Why are all C-C bonds identical in benzene?","back":"The six π electrons are delocalised over all six carbons.","mcqOptions":[{"text":"The six π electrons are delocalised over all six carbons.","correct":true,"explanation":"Delocalisation spreads the six π electrons evenly around the ring, making every carbon-carbon bond equivalent and intermediate in character between a single and double bond."},{"text":"The six π electrons are localised over three C=C bonds.","correct":false,"explanation":"This is the outdated alternating-bond model, which predicts two bond lengths. In benzene, the six π electrons are delocalised over all six carbons."},{"text":"The six σ electrons are delocalised over all six carbons.","correct":false,"explanation":"This confuses σ and π bonding. The σ bonds form the ring framework, while the six π electrons are delocalised over all six carbon atoms."},{"text":"The six π electrons are shared between alternate carbons.","correct":false,"explanation":"Restricting the electrons to alternate carbons would produce distinct single and double bonds. Instead, all six π electrons are delocalised over the whole ring."}]},{"id":"6c8e629c-2cf0-4695-b294-a139f721ff69","cardType":"basic","points":2,"front":"What bond length evidence supports benzene’s delocalised model?","back":"All C-C bonds are equal, about 0.139 nm.","mcqOptions":[{"text":"All C-C bonds are equal, about 0.139 nm.","correct":true,"explanation":"Every carbon-carbon bond has the same intermediate length, between typical single and double bond lengths. This supports a delocalised π system rather than fixed alternating bonds."},{"text":"All C-C bonds are equal, about 0.154 nm.","correct":false,"explanation":"The value 0.154 nm is typical of a carbon-carbon single bond. Benzene’s equal bonds are shorter, at about 0.139 nm, because of delocalised π bonding."},{"text":"All C-C bonds are equal, about 0.134 nm.","correct":false,"explanation":"The value 0.134 nm is closer to a localised carbon-carbon double bond. Benzene instead has equal intermediate carbon-carbon bonds of about 0.139 nm."},{"text":"The C-C bonds are alternating, about 0.134 and 0.154 nm.","correct":false,"explanation":"Alternating lengths would support fixed single and double bonds, but diffraction evidence shows that all benzene carbon-carbon bonds are equal at about 0.139 nm."}]},{"id":"519ef679-a2a0-4cd2-82ad-d03181235856","cardType":"basic","points":1,"front":"If benzene had alternating single and double bonds, what bond length pattern would be expected?","back":"Two different C-C bond lengths.","mcqOptions":[{"text":"Two different C-C bond lengths.","correct":true,"explanation":"Fixed alternating single and double bonds would have different bond orders and therefore different lengths, producing one longer single-bond length and one shorter double-bond length."},{"text":"One identical C-C bond length.","correct":false,"explanation":"One identical bond length is the observed evidence for delocalisation. A structure containing fixed alternating single and double bonds would predict two different carbon-carbon bond lengths."},{"text":"Three different C-C bond lengths.","correct":false,"explanation":"Alternation involves only two bond types: carbon-carbon single and carbon-carbon double bonds. The hypothetical model would therefore predict two different bond lengths, not three."},{"text":"Six different C-C bond lengths.","correct":false,"explanation":"Although benzene contains six carbon-carbon bonds, an alternating model assigns them to only two categories. It would predict two different lengths, repeated three times each."}]},{"id":"c4900c7c-88d1-4f2d-b764-555d9454de6a","cardType":"basic","points":2,"front":"What is the predicted hydrogenation enthalpy for three separate C=C bonds if one C=C gives -120 kJ mol^-1?","back":"-360 kJ mol^-1","mcqOptions":[{"text":"-360 kJ mol^-1","correct":true,"explanation":"Three independent carbon-carbon double bonds would each release 120 kJ mol^-1, so multiplication by three gives a predicted enthalpy of -360 kJ mol^-1."},{"text":"-120 kJ mol^-1","correct":false,"explanation":"This uses the hydrogenation enthalpy for only one double bond. Three separate C=C bonds would release three times as much energy, giving -360 kJ mol^-1."},{"text":"-240 kJ mol^-1","correct":false,"explanation":"This multiplies by two rather than three, effectively omitting one C=C bond. Three bonds at -120 kJ mol^-1 give -360 kJ mol^-1."},{"text":"+360 kJ mol^-1","correct":false,"explanation":"The magnitude is obtained correctly, but the sign is reversed. Hydrogenation is exothermic, so three separate double bonds have a predicted enthalpy of -360 kJ mol^-1."}]},{"id":"337690ba-530d-4ffa-894b-a986c854a207","cardType":"basic","points":2,"front":"What does benzene’s less exothermic hydrogenation show?","back":"Benzene is more stable than the hypothetical triene model.","mcqOptions":[{"text":"Benzene is more stable than the hypothetical triene model.","correct":true,"explanation":"Benzene releases less energy on hydrogenation because it begins at a lower energy than the hypothetical triene. Delocalisation therefore gives benzene additional stability."},{"text":"Benzene is less stable than the hypothetical triene model.","correct":false,"explanation":"This reverses the energy comparison. Benzene’s less exothermic hydrogenation means it starts at a lower energy and is more stable than the hypothetical triene."},{"text":"Benzene is equally stable as the hypothetical triene model.","correct":false,"explanation":"Equal stability would give the same hydrogenation enthalpy as three separate double bonds. The less exothermic observed value shows that benzene is more stable."},{"text":"Benzene is more reactive than the hypothetical triene model.","correct":false,"explanation":"The enthalpy evidence concerns thermodynamic stability, not simply reaction rate. Benzene’s smaller energy release shows that it is more stable than the hypothetical triene."}]},{"id":"0b45f139-9cb4-459b-872f-b3891b21bf5c","cardType":"basic","points":2,"front":"Using -360 and -208 kJ mol^-1, what is benzene’s stabilisation energy?","back":"152 kJ mol^-1","mcqOptions":[{"text":"152 kJ mol^-1","correct":true,"explanation":"The stabilisation energy is the difference between the predicted and actual hydrogenation enthalpy magnitudes: 360 - 208 = 152 kJ mol^-1."},{"text":"568 kJ mol^-1","correct":false,"explanation":"This adds the magnitudes of the two hydrogenation enthalpies instead of finding their difference. The stabilisation energy is 360 - 208 = 152 kJ mol^-1."},{"text":"-152 kJ mol^-1","correct":false,"explanation":"This keeps the enthalpy sign when reporting an energy difference. Stabilisation energy is conventionally given as a positive magnitude, so the value is 152 kJ mol^-1."},{"text":"52 kJ mol^-1","correct":false,"explanation":"This results from an arithmetic place-value error when subtracting 208 from 360. The correct difference, and therefore benzene’s stabilisation energy, is 152 kJ mol^-1."}]},{"id":"2f2f0de7-50a4-49d5-8028-de8cfd5654c8","cardType":"basic","points":2,"front":"Why does benzene not readily decolourise bromine water at room temperature?","back":"Addition would destroy the stable delocalised π system.","mcqOptions":[{"text":"Addition would destroy the stable delocalised π system.","correct":true,"explanation":"An addition reaction would break the continuous delocalised π system and lose benzene’s stabilisation energy. Benzene therefore resists bromine addition under these conditions."},{"text":"Addition would preserve the stable delocalised π system.","correct":false,"explanation":"This reverses the effect of addition. Adding bromine would interrupt π-electron delocalisation, so benzene resists the reaction because its stable delocalised system would be destroyed."},{"text":"Bromine cannot react with the localised C=C bonds.","correct":false,"explanation":"Bromine readily adds to localised C=C bonds in alkenes. Benzene behaves differently because it has a stable delocalised π system rather than three localised double bonds."},{"text":"Benzene contains only a stable network of σ bonds.","correct":false,"explanation":"Benzene contains both a σ-bond framework and a delocalised π system. It resists bromine addition because that addition would destroy the stable delocalised π system."}]},{"id":"53780697-62f1-434c-80ef-1117a47e186a","cardType":"cloze","points":null,"front":"Alkenes usually undergo electrophilic [ ], but benzene mainly undergoes electrophilic [ ].","back":"Alkenes usually undergo electrophilic addition, but benzene mainly undergoes electrophilic substitution.","mcqOptions":[{"text":"Alkenes usually undergo electrophilic addition, but benzene mainly undergoes electrophilic substitution.","correct":true,"explanation":"Alkenes usually add electrophiles across their C=C bond. Benzene instead replaces a hydrogen atom while restoring its delocalised aromatic π system, so electrophilic substitution is favoured."},{"text":"Alkenes usually undergo electrophilic substitution, but benzene mainly undergoes electrophilic addition.","correct":false,"explanation":"The reactions have been reversed: alkene π bonds normally undergo electrophilic addition, whereas benzene mainly undergoes electrophilic substitution because addition would permanently destroy aromatic stability."},{"text":"Alkenes usually undergo electrophilic elimination, but benzene mainly undergoes electrophilic addition.","correct":false,"explanation":"Elimination is not the usual reaction of an alkene, and benzene avoids addition because it would lose aromaticity. The correct pair is electrophilic addition for alkenes and substitution for benzene."},{"text":"Alkenes usually undergo electrophilic oxidation, but benzene mainly undergoes electrophilic elimination.","correct":false,"explanation":"Oxidation and elimination are not the characteristic electrophilic reactions being compared. Alkenes generally undergo electrophilic addition, while benzene undergoes substitution to regain its aromatic π system."}]},{"id":"eabf08d3-ca1c-4f13-a208-efba802711db","cardType":"basic","points":1,"front":"What intermediate forms when benzene’s π system attacks an electrophile?","back":"An arenium ion or σ complex.","mcqOptions":[{"text":"An arenium ion or σ complex.","correct":true,"explanation":"Attack by benzene’s π system temporarily removes aromaticity and produces a positively charged intermediate. This species is called an arenium ion or σ complex."},{"text":"A carbanion or π complex.","correct":false,"explanation":"The intermediate is positively charged, not negatively charged, and is classified as an arenium ion or σ complex. Confusing the charge gives the wrong intermediate."},{"text":"A free radical or σ complex.","correct":false,"explanation":"Electrophilic substitution proceeds through a carbocation-like intermediate rather than a species with an unpaired electron. The correct intermediate is an arenium ion, also called a σ complex."},{"text":"A nitronium ion or π complex.","correct":false,"explanation":"The nitronium ion is an electrophile used specifically in nitration, not the intermediate formed after benzene attacks. The resulting positively charged intermediate is an arenium ion or σ complex."}]},{"id":"6b722111-2e8d-4029-9ee8-d336a201b408","cardType":"basic","points":2,"front":"Why does deprotonation make electrophilic substitution favourable in benzene?","back":"It restores the delocalised aromatic π system.","mcqOptions":[{"text":"It restores the delocalised aromatic π system.","correct":true,"explanation":"Deprotonation removes the hydrogen from the carbon bonded to the electrophile, allowing the π electrons to reform the delocalised aromatic system and regain benzene’s stability."},{"text":"It creates a localised aromatic π system.","correct":false,"explanation":"A localised π system would not provide benzene’s characteristic aromatic stabilisation. Deprotonation is favourable because it restores the delocalised aromatic π system."},{"text":"It forms a stable saturated ring system.","correct":false,"explanation":"The ring does not become saturated during substitution; aromatic π bonding is recovered. The driving force is restoration of the delocalised aromatic π system."},{"text":"It removes the electrophile from the ring.","correct":false,"explanation":"Deprotonation removes a proton, not the electrophile, which remains attached as the substituent. The resulting electron movement restores the delocalised aromatic π system."}]},{"id":"251e08f0-1dbb-4373-8689-33515409447e","cardType":"basic","points":1,"front":"What electrophile reacts with benzene during nitration?","back":"The nitronium ion, NO₂⁺.","mcqOptions":[{"text":"The nitronium ion, NO₂⁺.","correct":true,"explanation":"The nitronium ion, NO₂⁺, is the electrophile that attacks benzene during nitration. It is generated from concentrated nitric acid and sulfuric acid."},{"text":"The nitrate ion, NO₃^-.","correct":false,"explanation":"The nitrate ion is negatively charged and is not the electrophile in nitration. Benzene is attacked by the positively charged nitronium ion, NO₂⁺."},{"text":"The nitrite ion, NO₂^-.","correct":false,"explanation":"The nitrite ion has the wrong charge and is not sufficiently electrophilic for this reaction. Nitration uses the positively charged nitronium ion, NO₂⁺."},{"text":"The nitric acid molecule, HNO₃.","correct":false,"explanation":"Nitric acid supplies nitrogen and oxygen but is converted into the active electrophile in the acid mixture. The species that attacks benzene is NO₂⁺, the nitronium ion."}]},{"id":"412968a5-48e5-4882-b9e9-697e8b6e0f25","cardType":"cloze","points":null,"front":"Nitration of benzene uses concentrated [ ] and concentrated [ ].","back":"Nitration of benzene uses concentrated HNO₃ and concentrated H₂SO₄.","mcqOptions":[{"text":"Nitration of benzene uses concentrated HNO₃ and concentrated H₂SO₄.","correct":true,"explanation":"Concentrated sulfuric acid protonates nitric acid, enabling formation of the nitronium ion. The nitronium ion is the electrophile that attacks benzene in nitration."},{"text":"Nitration of benzene uses concentrated HCl and concentrated H₂SO₄.","correct":false,"explanation":"Hydrochloric acid does not provide the nitrogen-containing species needed to generate NO₂⁺. The mixture requires concentrated HNO₃ and concentrated H₂SO₄."},{"text":"Nitration of benzene uses concentrated HNO₃ and concentrated H₃PO₄.","correct":false,"explanation":"Although phosphoric acid is acidic, it is not the standard reagent used to generate the nitronium ion in this reaction. The required acid mixture contains concentrated HNO₃ and H₂SO₄."},{"text":"Nitration of benzene uses concentrated H₂SO₄ and concentrated H₂O.","correct":false,"explanation":"Water dilutes sulfuric acid and cannot supply the nitrogen needed for nitration. Concentrated HNO₃ provides the nitrogen, while concentrated H₂SO₄ helps form the electrophile."}]},{"id":"f959198a-9057-44e9-ad8e-d468692c927e","cardType":"basic","points":1,"front":"What temperature is used for mononitration of benzene?","back":"About 50-55 °C.","mcqOptions":[{"text":"About 50-55 °C.","correct":true,"explanation":"Mononitration is carried out at about 50-55 °C, which provides a suitable reaction rate while limiting further substitution of nitrobenzene."},{"text":"About 25-30 °C.","correct":false,"explanation":"This lower temperature can make nitration too slow for the intended preparation. The usual controlled temperature for mononitration is about 50-55 °C."},{"text":"About 80-85 °C.","correct":false,"explanation":"A temperature this high increases the likelihood of further nitration and formation of dinitrobenzene. Mononitration is normally controlled at about 50-55 °C."},{"text":"About 100-105 °C.","correct":false,"explanation":"Strong heating makes multiple substitution more likely and reduces control of the product mixture. The standard temperature for mononitration is about 50-55 °C."}]},{"id":"cba8361c-67ae-47cf-a098-80a899603d6a","cardType":"basic","points":2,"front":"Why must nitration temperature be controlled?","back":"Strong heating can cause further nitration to dinitrobenzene.","mcqOptions":[{"text":"Strong heating can cause further nitration to dinitrobenzene.","correct":true,"explanation":"Nitrobenzene can undergo a second nitration under harsher conditions. Controlling the temperature helps favour mononitration and limits formation of dinitrobenzene."},{"text":"Strong heating can cause hydrogenation to cyclohexane.","correct":false,"explanation":"Hydrogenation requires hydrogen and a suitable catalyst, not simply stronger nitration conditions. The relevant risk from excessive heating is further nitration to dinitrobenzene."},{"text":"Strong heating can cause hydrolysis to phenol.","correct":false,"explanation":"Hydrolysis of benzene to phenol is not the competing process in the nitrating mixture. Excessive heating instead promotes further nitration, producing dinitrobenzene."},{"text":"Strong heating can cause reduction to phenylamine.","correct":false,"explanation":"Reduction of the nitro group to an amine requires reducing conditions, such as tin and hydrochloric acid. In nitration, excessive heating promotes further nitration to dinitrobenzene."}]},{"id":"b0e4ace4-e996-4f90-9336-6e7407804bbf","cardType":"basic","points":1,"front":"What organic product forms in nitration of benzene?","back":"Nitrobenzene, C₆H₅NO₂.","mcqOptions":[{"text":"Nitrobenzene, C₆H₅NO₂.","correct":true,"explanation":"Electrophilic substitution replaces one hydrogen atom on benzene with a nitro group, producing nitrobenzene with the molecular formula C₆H₅NO₂."},{"text":"Phenylamine, C₆H₅NH₂.","correct":false,"explanation":"Phenylamine contains an amino group and is formed by reducing a nitro group, not by nitrating benzene. Direct nitration produces nitrobenzene, C₆H₅NO₂."},{"text":"Benzenesulfonic acid, C₆H₅SO₃H.","correct":false,"explanation":"Benzenesulfonic acid forms during electrophilic substitution with sulfur trioxide or fuming sulfuric acid. Nitration introduces a nitro group and produces nitrobenzene, C₆H₅NO₂."},{"text":"Cyclohexane, C₆H₁₂.","correct":false,"explanation":"Cyclohexane forms by hydrogenation, which adds hydrogen across all of benzene’s π bonds. Nitration substitutes one hydrogen with NO₂, giving nitrobenzene, C₆H₅NO₂."}]},{"id":"abb51a7a-a33c-4358-ac50-b9d5a52eecac","cardType":"basic","points":1,"front":"What type of catalyst is needed for benzene halogenation?","back":"A halogen carrier, e.g. AlBr₃ or FeBr₃.","mcqOptions":[{"text":"A halogen carrier, e.g. AlBr₃ or FeBr₃.","correct":true,"explanation":"Benzene halogenation requires a halogen carrier such as AlBr₃ or FeBr₃. This Lewis acid polarises the halogen molecule and generates the electrophile needed for substitution."},{"text":"A metal catalyst, e.g. Ni or Pt.","correct":false,"explanation":"Nickel and platinum are hydrogenation catalysts, not halogen carriers. Benzene halogenation instead requires a Lewis acid such as AlBr₃ or FeBr₃ to generate a stronger electrophile."},{"text":"An acid catalyst, e.g. H₂SO₄ or H₃PO₄.","correct":false,"explanation":"Concentrated acids catalyse reactions such as nitration, but they are not the usual catalysts for benzene bromination. A halogen carrier such as AlBr₃ or FeBr₃ is required."},{"text":"A radical initiator, e.g. UV light or peroxide.","correct":false,"explanation":"Ultraviolet light promotes free-radical halogenation rather than electrophilic substitution of benzene. Benzene bromination uses a halogen carrier such as AlBr₃ or FeBr₃."}]},{"id":"48f3f566-bc81-402c-8528-4a80a02d3f33","cardType":"basic","points":2,"front":"Why is a halogen carrier needed for bromination of benzene?","back":"It generates a stronger electrophile, often Br⁺.","mcqOptions":[{"text":"It generates a stronger electrophile, often Br⁺.","correct":true,"explanation":"Bromine alone is not electrophilic enough to attack the stable benzene ring. The halogen carrier polarises Br₂, producing a much stronger electrophile often represented as Br⁺."},{"text":"It generates a reactive free radical, often Br•.","correct":false,"explanation":"Free bromine radicals are associated with radical substitution under ultraviolet light, not benzene’s electrophilic substitution mechanism. The carrier instead generates a stronger electrophile, often represented as Br⁺."},{"text":"It generates a stronger nucleophile, often Br^-.","correct":false,"explanation":"The halogen carrier does not convert bromine into a nucleophile because an electron-rich benzene ring is attacked by electrophiles. It generates a stronger electrophile, often represented as Br⁺."},{"text":"It generates a stronger acid, often H⁺.","correct":false,"explanation":"Producing hydrogen ions is not the halogen carrier’s essential role in bromination. It polarises Br₂ and generates a stronger electrophile, often represented as Br⁺."}]},{"id":"cca6115a-8572-462d-ada2-eefab62d893f","cardType":"cloze","points":null,"front":"Bromination of benzene: C₆H₆ + Br₂ → [ ] + [ ].","back":"Bromination of benzene: C₆H₆ + Br₂ → C₆H₅Br + HBr.","mcqOptions":[{"text":"Bromination of benzene: C₆H₆ + Br₂ → C₆H₅Br + HBr.","correct":true,"explanation":"Electrophilic substitution replaces one hydrogen atom on benzene with bromine, forming bromobenzene, C₆H₅Br. The displaced hydrogen combines with the other bromine atom to form HBr."},{"text":"Bromination of benzene: C₆H₆ + Br₂ → C₆H₆Br₂ + no other product.","correct":false,"explanation":"Benzene bromination is substitution, not addition, because the aromatic ring regains its delocalised structure. The correct products are C₆H₅Br and HBr, not a dibromo addition product."},{"text":"Bromination of benzene: C₆H₆ + Br₂ → C₆H₄Br₂ + H₂.","correct":false,"explanation":"Replacing two ring hydrogens would require further bromination and additional Br₂; it is not the stated single-substitution equation. One Br replaces one H, producing C₆H₅Br and HBr."},{"text":"Bromination of benzene: C₆H₆ + Br₂ → C₆H₅Br + H₂.","correct":false,"explanation":"The displaced hydrogen forms hydrogen bromide, not hydrogen gas. Electrophilic substitution therefore produces bromobenzene, C₆H₅Br, together with HBr."}]},{"id":"118dcbc2-7a16-40b0-acd2-b49c636be7f2","cardType":"basic","points":1,"front":"Why must Friedel-Crafts reactions use anhydrous conditions?","back":"AlCl₃ reacts with water.","mcqOptions":[{"text":"AlCl₃ reacts with water.","correct":true,"explanation":"AlCl₃ reacts readily with water, destroying the Lewis acid catalyst and producing hydrolysis products. Friedel-Crafts reactions must therefore use dry reagents and apparatus."},{"text":"Benzene reacts with water.","correct":false,"explanation":"Benzene does not react appreciably with water under Friedel-Crafts conditions. Anhydrous conditions are required because the AlCl₃ catalyst reacts with water and loses its effectiveness."},{"text":"The product reacts with water.","correct":false,"explanation":"The acylated product is not the reason dry conditions are required during Friedel-Crafts reactions. Water instead reacts with AlCl₃, destroying the Lewis acid catalyst."},{"text":"HCl reacts with water.","correct":false,"explanation":"Hydrogen chloride may be formed during substitution, but its reaction with water is not the key issue. Anhydrous conditions protect AlCl₃ from hydrolysis."}]},{"id":"bd8a339a-0e20-4a54-a397-f7c1de841a95","cardType":"basic","points":2,"front":"Why is methylbenzene more reactive than benzene in electrophilic substitution?","back":"The methyl group donates electron density into the ring.","mcqOptions":[{"text":"The methyl group donates electron density into the ring.","correct":true,"explanation":"The methyl group releases electron density into the aromatic ring through its electron-donating effect. This makes the ring more attractive to electrophiles and stabilises substitution intermediates."},{"text":"The methyl group withdraws electron density from the ring.","correct":false,"explanation":"Withdrawing electron density would deactivate the aromatic ring rather than make it more reactive. The methyl group donates electron density into the ring, increasing its susceptibility to electrophilic attack."},{"text":"The methyl group donates an electrophile to the ring.","correct":false,"explanation":"The methyl group does not supply an electrophile; the attacking reagent does. It donates electron density into the benzene ring, making the ring more reactive towards electrophiles."},{"text":"The methyl group breaks electron delocalisation in the ring.","correct":false,"explanation":"Disrupting delocalisation would reduce aromatic stability rather than explain methylbenzene’s activation. The methyl group donates electron density into the intact delocalised ring."}]},{"id":"e33ed3bd-17a4-4bba-ae61-b8fbfd9f3b05","cardType":"basic","points":1,"front":"Which products mainly form in electrophilic substitution of methylbenzene?","back":"Mainly 2-substituted and 4-substituted products.","mcqOptions":[{"text":"Mainly 2-substituted and 4-substituted products.","correct":true,"explanation":"A methyl group activates the ring and directs incoming electrophiles mainly to the positions adjacent and opposite to itself. These give the 2-substituted and 4-substituted products."},{"text":"Mainly 3-substituted and 5-substituted products.","correct":false,"explanation":"The 3-position is meta to the methyl group and is not favoured; the 5-position is its equivalent by symmetry. Methylbenzene instead forms mainly 2-substituted and 4-substituted products."},{"text":"Mainly 2-substituted and 3-substituted products.","correct":false,"explanation":"Although substitution at position 2 is favoured, position 3 is not because it is meta to the methyl group. The major products are 2-substituted and 4-substituted."},{"text":"Mainly 3-substituted and 4-substituted products.","correct":false,"explanation":"Position 4 is favoured, but position 3 is not. The electron-donating methyl group directs electrophilic substitution mainly to the 2- and 4-positions."}]}]}],"examCountdown":{"examDate":"2027-05-28","paperLabel":"Paper 1"},"heading":"Arenes","title":"Arenes Flashcards","topbar":{"backHref":"/a-level/chemistry/edexcel/flashcards"},"topicSidebar":"$L37"}]}],"topicId":"3332e219-cbb8-589c-9979-9a79486a3ca6","walled":"$L38"}],"$L39","$L3a"]