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22:[["$","$L31",null,{"feature":"flashcards"}],"$L32",["$","$L33",null,{"href":"/a-level/chemistry/eduqas/aromaticity/flashcards","label":"Aromaticity","subjectCode":"a_level_chemistry__eduqas","surface":"flashcards"}],["$","$L34",null,{"runner":["$","div",null,{"className":"min-h-0 flex-1 overflow-hidden bg-background","children":["$","$L35",null,{"decks":[{"id":"18e0543d-9a80-4e43-bea7-7ce12896d790","title":"Aromaticity Flashcards","topicId":"6e64edf8-89ba-5e55-a386-2fd1f912cc2a","topicSlug":"aromaticity","cards":[{"id":"a0dd5e23-f7d7-4636-88f7-d194d2f71310","cardType":"cloze","points":null,"front":"Aromaticity is extra [ ] in a cyclic, planar, fully conjugated [ ].","back":"Aromaticity is extra stability in a cyclic, planar, fully conjugated π system.","mcqOptions":[{"text":"Aromaticity is extra stability in a cyclic, planar, fully conjugated π system.","correct":true,"explanation":"Continuous overlap of p orbitals delocalises π electrons around the ring, lowering the molecule’s energy and providing additional stability compared with a comparable non-aromatic structure."},{"text":"Aromaticity is extra reactivity in a cyclic, planar, fully conjugated π system.","correct":false,"explanation":"Aromatic delocalisation generally reduces reactivity rather than increasing it. Aromaticity is the extra stability produced by electron delocalisation in a cyclic, planar, fully conjugated π system."},{"text":"Aromaticity is extra stability in a cyclic, planar, fully conjugated σ system.","correct":false,"explanation":"Confusing σ and π bonding overlooks the unhybridised p orbitals responsible for delocalisation. Aromaticity is extra stability in a cyclic, planar, fully conjugated π system."},{"text":"Aromaticity is extra strain in a cyclic, planar, fully conjugated σ system.","correct":false,"explanation":"Aromaticity is not ring strain and does not arise from σ-electron delocalisation. It is extra stability caused by delocalisation within a cyclic, planar, fully conjugated π system."}]},{"id":"01747c12-e2c8-435e-9709-d9d200f554e3","cardType":"basic","points":1,"front":"What allows π electrons to delocalise in a conjugated system?","back":"Adjacent p orbitals overlap continuously.","mcqOptions":[{"text":"Adjacent p orbitals overlap continuously.","correct":true,"explanation":"Continuous sideways overlap between adjacent p orbitals creates an uninterrupted conjugated π system, allowing π electrons to spread across several atoms rather than remaining localised."},{"text":"Adjacent s orbitals overlap continuously.","correct":false,"explanation":"Overlapping s orbitals contribute to σ bonding, not the delocalised π system. π electrons delocalise because adjacent p orbitals overlap continuously along the conjugated structure."},{"text":"Adjacent p orbitals remain perpendicular.","correct":false,"explanation":"This confuses the orientation of p orbitals relative to the molecular plane with their orientation to one another. Adjacent p orbitals must be parallel enough to overlap continuously."},{"text":"Adjacent σ bonds rotate continuously.","correct":false,"explanation":"Free rotation would disrupt the alignment needed for conjugation rather than cause delocalisation. π electrons delocalise when adjacent p orbitals overlap continuously."}]},{"id":"152d4249-a12a-4acc-9bd4-ad0e7f0149f3","cardType":"cloze","points":null,"front":"Each benzene carbon is [ ] hybridised, giving a [ ] arrangement.","back":"Each benzene carbon is sp² hybridised, giving a trigonal planar arrangement.","mcqOptions":[{"text":"Each benzene carbon is sp² hybridised, giving a trigonal planar arrangement.","correct":true,"explanation":"Each carbon uses three sp² orbitals to form three σ bonds in a trigonal planar arrangement, leaving one unhybridised p orbital for π delocalisation."},{"text":"Each benzene carbon is sp³ hybridised, giving a tetrahedral arrangement.","correct":false,"explanation":"Sp³ hybridisation would leave no unhybridised p orbital for the delocalised π system and would make the ring non-planar. Benzene carbons are sp² hybridised and trigonal planar."},{"text":"Each benzene carbon is sp hybridised, giving a trigonal planar arrangement.","correct":false,"explanation":"Sp hybridisation produces a linear arrangement, not the three σ-bond geometry found in benzene. Each carbon is sp² hybridised, producing a trigonal planar arrangement."},{"text":"Each benzene carbon is sp² hybridised, giving a tetrahedral arrangement.","correct":false,"explanation":"The hybridisation is correct, but sp² hybridisation gives three bonding orbitals separated by approximately 120°, so the arrangement is trigonal planar rather than tetrahedral."}]},{"id":"d89c8101-39a2-42e4-9232-fda1154743e9","cardType":"basic","points":1,"front":"How is each unhybridised p orbital positioned in benzene?","back":"Perpendicular to the plane of the ring.","mcqOptions":[{"text":"Perpendicular to the plane of the ring.","correct":true,"explanation":"Each unhybridised p orbital projects above and below the ring plane. This perpendicular orientation allows sideways overlap with adjacent p orbitals around the entire benzene ring."},{"text":"Parallel to the plane of the ring.","correct":false,"explanation":"Placing the p orbitals parallel to the ring plane would prevent the required sideways overlap above and below it. Each p orbital is perpendicular to the plane."},{"text":"Tangential to the plane of the ring.","correct":false,"explanation":"A tangential orientation would not form the continuous π cloud above and below benzene. Each unhybridised p orbital stands perpendicular to the ring plane."},{"text":"Diagonal to the plane of the ring.","correct":false,"explanation":"Tilted p orbitals would overlap less effectively and would not describe benzene’s planar conjugated structure. Each unhybridised p orbital is perpendicular to the ring plane."}]},{"id":"568dd4c4-32e2-40b7-bfc4-21496a2bb7fd","cardType":"cloze","points":null,"front":"Benzene’s delocalised π cloud lies [ ] the ring and contains [ ] π electrons.","back":"Benzene’s delocalised π cloud lies above and below the ring and contains six π electrons.","mcqOptions":[{"text":"Benzene’s delocalised π cloud lies above and below the ring and contains six π electrons.","correct":true,"explanation":"The six perpendicular p orbitals overlap to form a delocalised π cloud on both sides of the ring. Each carbon contributes one electron, giving six π electrons."},{"text":"Benzene’s delocalised π cloud lies above and below the ring and contains three π electrons.","correct":false,"explanation":"Counting one electron pair per drawn double bond as one electron gives this error. Each of benzene’s six carbons contributes one electron, so the cloud contains six π electrons."},{"text":"Benzene’s delocalised π cloud lies within the plane of the ring and contains six π electrons.","correct":false,"explanation":"The σ-bond framework lies in the ring plane, which can cause this confusion. The π cloud forms from perpendicular p orbitals and therefore lies above and below the ring."},{"text":"Benzene’s delocalised π cloud lies above and below the ring and contains twelve π electrons.","correct":false,"explanation":"Counting both lobes of each occupied p orbital as separate electrons doubles the total incorrectly. Benzene has one π electron from each carbon, giving six π electrons."}]},{"id":"9bf286c8-206e-4fcd-95ea-17c06946e946","cardType":"basic","points":2,"front":"Why are all six C-C bonds identical in benzene?","back":"π electrons are delocalised over the whole ring.","mcqOptions":[{"text":"π electrons are delocalised over the whole ring.","correct":true,"explanation":"The π electrons are shared across all six carbon atoms rather than confined to three bonds. Consequently, every C-C bond has the same intermediate bond order and length."},{"text":"π electrons are localised in three double bonds.","correct":false,"explanation":"A fixed Kekulé structure would contain alternating single and double bonds with different lengths. Benzene’s bonds are identical because its π electrons are delocalised over the whole ring."},{"text":"σ electrons are delocalised over the whole ring.","correct":false,"explanation":"The σ framework consists of localised bonds between neighbouring atoms. It is the π electrons that are delocalised across the whole ring, making all six C-C bonds identical."},{"text":"π electrons are shared between alternate carbons.","correct":false,"explanation":"Restricting electron sharing to alternate carbons would produce distinct single and double bonds. In benzene, π electrons are delocalised over every carbon in the whole ring."}]},{"id":"e7e229d3-b9d5-450b-818e-37e7c6482bf9","cardType":"basic","points":1,"front":"What is the C-C bond length in benzene?","back":"About 0.139 nm.","mcqOptions":[{"text":"About 0.139 nm.","correct":true,"explanation":"Benzene’s C-C bonds have an intermediate bond order because of π delocalisation, so their length is about 0.139 nm, between typical single and double C-C bonds."},{"text":"About 0.154 nm.","correct":false,"explanation":"This is approximately the length of a localised C-C single bond. Benzene’s delocalised π bonding shortens every ring bond to about 0.139 nm."},{"text":"About 0.134 nm.","correct":false,"explanation":"This is approximately the length of a localised C=C double bond. Benzene’s bonds have an intermediate bond order, so each is slightly longer at about 0.139 nm."},{"text":"About 0.120 nm.","correct":false,"explanation":"This is closer to a carbon-carbon triple bond length and implies excessive bond order. Delocalisation gives benzene’s C-C bonds the intermediate length of about 0.139 nm."}]},{"id":"927e5ccc-445e-4b4b-ad8b-0f49e155e21f","cardType":"basic","points":1,"front":"What hydrogenation enthalpy is expected for three isolated C=C bonds if one is -120 kJ mol⁻¹?","back":"-360 kJ mol⁻¹.","mcqOptions":[{"text":"-360 kJ mol⁻¹.","correct":true,"explanation":"Three independent C=C bonds would each release 120 kJ mol⁻¹ on hydrogenation, so their expected total enthalpy change is $3\\times-120=-360$ kJ mol⁻¹."},{"text":"-120 kJ mol⁻¹.","correct":false,"explanation":"This includes the hydrogenation of only one C=C bond and omits the other two. Three isolated bonds would give $3\\times-120=-360$ kJ mol⁻¹."},{"text":"-240 kJ mol⁻¹.","correct":false,"explanation":"This multiplies the enthalpy by two rather than three, effectively counting only two C=C bonds. Three isolated bonds are expected to release -360 kJ mol⁻¹."},{"text":"+360 kJ mol⁻¹.","correct":false,"explanation":"The magnitude correctly accounts for three bonds, but the sign is reversed. Hydrogenation is exothermic, so the expected enthalpy change is -360 kJ mol⁻¹."}]},{"id":"ed520185-655f-48ff-ae37-86fccee83523","cardType":"basic","points":1,"front":"What is benzene’s actual hydrogenation enthalpy to cyclohexane?","back":"About -208 kJ mol⁻¹.","mcqOptions":[{"text":"About -208 kJ mol⁻¹.","correct":true,"explanation":"Experimental hydrogenation of benzene to cyclohexane releases about 208 kJ mol⁻¹. This is substantially less exothermic than hydrogenating three isolated carbon-carbon double bonds."},{"text":"About -360 kJ mol⁻¹.","correct":false,"explanation":"The -360 kJ mol⁻¹ value is the predicted enthalpy for three independent double bonds, not benzene’s measured value. Benzene’s actual hydrogenation enthalpy is about -208 kJ mol⁻¹."},{"text":"About -152 kJ mol⁻¹.","correct":false,"explanation":"This confuses the aromatic stabilisation energy with the hydrogenation enthalpy. The experimental hydrogenation of benzene to cyclohexane releases about 208 kJ mol⁻¹, giving -208 kJ mol⁻¹."},{"text":"About -120 kJ mol⁻¹.","correct":false,"explanation":"This treats benzene as if it contained only one ordinary alkene bond. Complete hydrogenation involves the whole ring and has an actual enthalpy change of about -208 kJ mol⁻¹."}]},{"id":"fc32598d-942b-4812-8d44-8f51fe9e5882","cardType":"basic","points":2,"front":"What does benzene’s less exothermic hydrogenation show?","back":"Benzene starts at a lower enthalpy, so is extra stable.","mcqOptions":[{"text":"Benzene starts at a lower enthalpy, so is extra stable.","correct":true,"explanation":"Because both comparisons produce cyclohexane, the smaller energy release means benzene begins at lower enthalpy than the hypothetical non-delocalised structure. This demonstrates its additional aromatic stability."},{"text":"Benzene starts at a higher enthalpy, so is less stable.","correct":false,"explanation":"A higher starting enthalpy would produce a more exothermic hydrogenation to the same product. Benzene’s less exothermic value shows that it starts at lower enthalpy and is extra stable."},{"text":"Cyclohexane ends at a higher enthalpy, so is less stable.","correct":false,"explanation":"The comparison concerns reactants that form the same cyclohexane product, so the product’s enthalpy does not explain the difference. Benzene starts at lower enthalpy because aromatic delocalisation stabilises it."},{"text":"Benzene starts at the same enthalpy, so is equally stable.","correct":false,"explanation":"Equal starting enthalpies would give equal hydrogenation enthalpies when the final product is identical. Benzene instead starts lower in enthalpy because its delocalised ring is extra stable."}]},{"id":"be512ad5-d4d0-49fa-a4d2-77aa3a0598d3","cardType":"basic","points":1,"front":"Using -360 and -208 kJ mol⁻¹, what is benzene’s aromatic stabilisation energy?","back":"About +152 kJ mol⁻¹.","mcqOptions":[{"text":"About +152 kJ mol⁻¹.","correct":true,"explanation":"The difference between the predicted release of 360 kJ mol⁻¹ and the measured release of 208 kJ mol⁻¹ is 152 kJ mol⁻¹. This represents benzene’s aromatic stabilisation."},{"text":"About -152 kJ mol⁻¹.","correct":false,"explanation":"The magnitude is obtained correctly, but the sign is reversed. Aromatic stabilisation is conventionally stated as the positive energy required to remove that stability, about +152 kJ mol⁻¹."},{"text":"About +568 kJ mol⁻¹.","correct":false,"explanation":"This adds the magnitudes, 360 and 208, rather than finding their difference. Aromatic stabilisation is calculated as 360 - 208, giving about +152 kJ mol⁻¹."},{"text":"About +76 kJ mol⁻¹.","correct":false,"explanation":"This results from an incorrect subtraction or inappropriate division of the enthalpy values. The stabilisation is the 152 kJ mol⁻¹ difference between -360 and -208 kJ mol⁻¹."}]},{"id":"5760c491-2193-4c96-bd9d-9fe73bef12d3","cardType":"basic","points":2,"front":"Why does benzene resist addition reactions?","back":"Addition would destroy aromaticity and lose delocalisation stability.","mcqOptions":[{"text":"Addition would destroy aromaticity and lose delocalisation stability.","correct":true,"explanation":"Addition would convert ring carbons from planar, conjugated bonding and interrupt the delocalised pi system. Losing this aromatic stabilisation makes addition energetically unfavourable."},{"text":"Addition would preserve aromaticity and retain delocalisation stability.","correct":false,"explanation":"Addition does not preserve benzene’s complete delocalised pi system; it interrupts conjugation at ring carbons. Benzene resists addition because this destroys aromaticity and sacrifices delocalisation stability."},{"text":"Addition would repel electrophiles and prevent electron-pair attraction.","correct":false,"explanation":"Benzene does contain delocalised pi electrons that can interact with electrophiles. Addition is resisted because the resulting product loses aromaticity, not because electrophiles cannot approach the ring."},{"text":"Addition would break sigma bonds and open the carbon ring.","correct":false,"explanation":"Addition does not primarily fail because carbon-carbon sigma bonds must be broken. It is unfavourable because disrupting the delocalised pi system destroys the ring’s aromatic stabilisation."}]},{"id":"ac70b758-bdaf-431d-9239-92bd7de51c36","cardType":"cloze","points":null,"front":"At room temperature, benzene does [ ] bromine water.","back":"At room temperature, benzene does not rapidly decolourise bromine water.","mcqOptions":[{"text":"not rapidly decolourise","correct":true,"explanation":"Benzene’s aromatic delocalisation makes ordinary electrophilic addition unfavourable, so bromine water is not rapidly decolourised at room temperature without a suitable catalyst."},{"text":"rapidly decolourise","correct":false,"explanation":"Rapid decolourisation is characteristic of alkenes, whose localised double bonds readily add bromine. Benzene is aromatically stabilised and does not rapidly decolourise bromine water at room temperature."},{"text":"immediately decolourise","correct":false,"explanation":"Immediate decolourisation incorrectly applies the alkene bromine-water test to benzene. At room temperature, benzene does not rapidly decolourise bromine water because addition would destroy aromaticity."},{"text":"readily decolourise","correct":false,"explanation":"Readily decolourising bromine water would imply easy bromine addition, as occurs with an alkene. Benzene resists this reaction because retaining its aromatic delocalisation is energetically favourable."}]},{"id":"f86a0df1-2267-4fcd-b5a2-ae1ba401e62b","cardType":"basic","points":1,"front":"Which reaction type is characteristic of arenes because aromaticity is restored?","back":"Electrophilic substitution.","mcqOptions":[{"text":"Electrophilic substitution.","correct":true,"explanation":"Arenes characteristically undergo electrophilic substitution: an electrophile temporarily disrupts aromaticity, then loss of H⁺ restores the delocalised ring while replacing a hydrogen atom."},{"text":"Electrophilic addition.","correct":false,"explanation":"Electrophilic addition is typical of ordinary alkenes and would leave benzene’s aromatic system permanently disrupted. Arenes instead favour electrophilic substitution because aromaticity is restored."},{"text":"Nucleophilic substitution.","correct":false,"explanation":"Nucleophilic substitution is not the characteristic reaction of an electron-rich benzene ring under standard arene conditions. Benzene typically attracts electrophiles and undergoes electrophilic substitution."},{"text":"Free-radical substitution.","correct":false,"explanation":"Free-radical substitution is characteristic of reactions such as alkane halogenation, not the standard reactions of benzene’s pi system. Arenes characteristically undergo electrophilic substitution."}]},{"id":"9a5f804f-ec2f-458c-a7f5-f50ee8a97cb2","cardType":"basic","points":1,"front":"What intermediate forms when benzene first attacks an electrophile?","back":"An arenium ion / sigma complex.","mcqOptions":[{"text":"An arenium ion / sigma complex.","correct":true,"explanation":"Attack on the electrophile forms a positively charged, non-aromatic cyclohexadienyl intermediate called an arenium ion or sigma complex. Subsequent deprotonation restores the aromatic ring."},{"text":"A carbanion / sigma anion.","correct":false,"explanation":"A carbanion would carry negative charge, whereas benzene’s attack on an electrophile produces a positively charged intermediate. The correct intermediate is an arenium ion, also called a sigma complex."},{"text":"A free radical / radical complex.","correct":false,"explanation":"A radical intermediate requires homolytic bond formation and unpaired electrons, which are not involved in electrophilic aromatic substitution. Benzene instead forms an arenium ion or sigma complex."},{"text":"A benzyne intermediate / aryne complex.","correct":false,"explanation":"Benzyne is associated with certain elimination-addition nucleophilic substitution mechanisms, not initial electrophilic attack on benzene. Electrophilic substitution forms an arenium ion or sigma complex."}]},{"id":"40cc4fb4-a1bc-473c-86a1-fa807b7ddcff","cardType":"basic","points":2,"front":"How is aromaticity restored in electrophilic substitution?","back":"A base removes H⁺; C-H electrons return to the ring.","mcqOptions":[{"text":"A base removes H⁺; C-H electrons return to the ring.","correct":true,"explanation":"A base removes the proton from the carbon bonded to the electrophile. The electrons from that C-H bond re-form the ring’s pi system, restoring aromaticity."},{"text":"A base removes H⁻; C-H electrons return to the ring.","correct":false,"explanation":"Hydride loss would remove both the hydrogen nucleus and its electrons rather than regenerate the pi bond correctly. A base removes H⁺, and the C-H electrons return to the ring."},{"text":"A base removes H⁺; C-H electrons move to the electrophile.","correct":false,"explanation":"Sending the C-H electrons to the electrophile would not re-form the missing pi bond. A base removes H⁺ while those electrons return to the ring and restore aromaticity."},{"text":"An acid adds H⁺; C-H electrons return to the ring.","correct":false,"explanation":"Adding H⁺ would not eliminate the hydrogen attached during formation of the sigma complex and would not restore conjugation. A base must remove H⁺ so C-H electrons re-form the ring."}]},{"id":"facd2973-3c91-4afd-995f-fa46461759ae","cardType":"basic","points":3,"front":"What reagent package gives mononitration of benzene?","back":"

Concentrated HNO₃
Concentrated H₂SO₄ acid catalyst
Warm to about 323 K / 50 °C

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

Concentrated HNO₃
Concentrated H₂SO₄ acid catalyst
Warm to about 323 K / 50 °C

","correct":true,"explanation":"Concentrated nitric acid provides the nitrating species, concentrated sulfuric acid acts as the acid catalyst, and gentle warming to about 323 K allows mononitration without promoting further substitution."},{"text":"

Dilute HNO₃
Concentrated HCl acid catalyst
Cool to about 273 K / 0 °C

","correct":false,"explanation":"Dilute nitric acid with hydrochloric acid does not generate the required nitronium ion efficiently. Mononitration uses concentrated HNO₃ and concentrated H₂SO₄, warmed to about 323 K."},{"text":"

Concentrated HNO₃
Concentrated H₃PO₄ acid catalyst
Heat to about 373 K / 100 °C

","correct":false,"explanation":"Phosphoric acid is not the standard catalyst for generating NO₂⁺, and stronger heating risks further reaction. The correct package uses concentrated H₂SO₄ and warming to about 323 K."},{"text":"

Concentrated H₂SO₄
Concentrated HNO₃ acid catalyst
Warm to about 423 K / 150 °C

","correct":false,"explanation":"The acids have been assigned the wrong roles and the temperature is excessive for controlled mononitration. Concentrated HNO₃ is the reagent, concentrated H₂SO₄ the catalyst, at about 323 K."}]},{"id":"5343312e-7de0-4db8-acb6-0694a95baf55","cardType":"cloze","points":null,"front":"In nitration: HNO₃ + H₂SO₄ → [ ] + [ ].","back":"In nitration: HNO₃ + H₂SO₄ → NO₂⁺ + HSO₄⁻ + H₂O.","mcqOptions":[{"text":"In nitration: HNO₃ + H₂SO₄ → NO₂⁺ + HSO₄⁻ + H₂O.","correct":true,"explanation":"Sulfuric acid protonates nitric acid, allowing water to leave and producing the electrophile NO₂⁺. The remaining species are HSO₄⁻ and H₂O, which complete the ionisation equation."},{"text":"In nitration: HNO₃ + H₂SO₄ → NO₃⁻ + H₃SO₄⁺.","correct":false,"explanation":"This treats nitric acid as producing nitrate rather than the electrophile needed for aromatic substitution. The correct products are NO₂⁺ and HSO₄⁻ + H₂O from acid-assisted dehydration."},{"text":"In nitration: HNO₃ + H₂SO₄ → NO₂⁻ + HSO₄⁺ + H₂O.","correct":false,"explanation":"The nitronium ion has a positive charge, not a negative one, and hydrogen sulfate is an anion. The correct products are NO₂⁺ and HSO₄⁻ + H₂O."},{"text":"In nitration: HNO₃ + H₂SO₄ → NO⁺ + HSO₄⁻ + H₂O₂.","correct":false,"explanation":"NO⁺ is not the electrophile generated in nitration, and the reaction forms water rather than hydrogen peroxide. Acid-assisted dehydration produces NO₂⁺ alongside HSO₄⁻ and H₂O."}]},{"id":"3481d695-d7ba-4db1-9259-d65b95b75361","cardType":"basic","points":2,"front":"What reagent package brominates benzene?","back":"

Br₂
FeBr₃ catalyst, dry conditions

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

Br₂
FeBr₃ catalyst, dry conditions

","correct":true,"explanation":"Bromine provides the brominating reagent, while dry FeBr₃ polarises Br₂ and generates the electrophile Br⁺. Anhydrous conditions prevent water from interfering with the Lewis acid catalyst."},{"text":"

Br₂
FeBr₃ catalyst, aqueous conditions

","correct":false,"explanation":"Water can react with or deactivate the Lewis acid catalyst, so aqueous conditions are unsuitable. Benzene bromination requires Br₂ with dry FeBr₃ under anhydrous conditions."},{"text":"

HBr
FeBr₃ catalyst, dry conditions

","correct":false,"explanation":"HBr supplies bromide and a proton but does not provide Br₂ for FeBr₃ to polarise into the electrophile. The correct brominating reagent is Br₂ under dry conditions."},{"text":"

Br₂
AlCl₃ catalyst, dry conditions

","correct":false,"explanation":"AlCl₃ is commonly associated with Friedel-Crafts reactions and is not the standard halogen carrier for bromination. Benzene bromination uses dry FeBr₃ to generate Br⁺."}]},{"id":"a0765a92-dfa9-4952-8b23-562d4058241d","cardType":"cloze","points":null,"front":"For benzene bromination: Br₂ + FeBr₃ → [ ] + [ ].","back":"For benzene bromination: Br₂ + FeBr₃ → Br⁺ + FeBr₄⁻.","mcqOptions":[{"text":"For benzene bromination: Br₂ + FeBr₃ → Br⁺ + FeBr₄⁻.","correct":true,"explanation":"FeBr₃ polarises the Br-Br bond, producing the electrophile Br⁺ and the counter-ion FeBr₄⁻. Br⁺ then attacks benzene in electrophilic aromatic substitution."},{"text":"For benzene bromination: Br₂ + FeBr₃ → Br⁻ + FeBr₄⁺.","correct":false,"explanation":"The charges are reversed: bromination requires an electron-poor Br⁺ electrophile, while the iron-containing species is the anion FeBr₄⁻. Reversing these charges prevents the correct mechanism."},{"text":"For benzene bromination: Br₂ + FeBr₃ → Br₂⁺ + FeBr₃⁻.","correct":false,"explanation":"FeBr₃ does not simply create Br₂⁺ and FeBr₃⁻. It polarises Br₂ so that Br⁺ acts as the electrophile and FeBr₄⁻ forms as its counter-ion."},{"text":"For benzene bromination: Br₂ + FeBr₃ → HBr + FeBr₄⁻.","correct":false,"explanation":"HBr is not formed during electrophile generation because no hydrogen source is involved at this stage. The Lewis acid produces Br⁺ and FeBr₄⁻ from Br₂ and FeBr₃."}]},{"id":"a221e0fa-d01a-4487-954b-f463a0c01483","cardType":"basic","points":2,"front":"What reagent package methylates benzene by Friedel-Crafts alkylation?","back":"

CH₃Cl
Anhydrous AlCl₃

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

CH₃Cl
Anhydrous AlCl₃

","correct":true,"explanation":"Chloromethane supplies the methyl group, while anhydrous AlCl₃ acts as a Lewis acid and generates the methyl electrophile for Friedel-Crafts alkylation."},{"text":"

CH₃OH
Anhydrous AlCl₃

","correct":false,"explanation":"Methanol does not provide the required methyl carbocation equivalent with AlCl₃ under standard conditions. The reagent package uses CH₃Cl and anhydrous AlCl₃."},{"text":"

CH₃Cl
Aqueous AlCl₃

","correct":false,"explanation":"Water hydrolyses and deactivates AlCl₃, so the Lewis acid must be anhydrous. Friedel-Crafts methylation uses CH₃Cl with anhydrous AlCl₃."},{"text":"

CH₃Br
Anhydrous FeBr₃

","correct":false,"explanation":"Although methyl bromide is an alkyl halide, FeBr₃ is the halogen carrier used for bromination rather than the standard catalyst for this alkylation. The expected package is CH₃Cl and anhydrous AlCl₃."}]},{"id":"08ff1a57-b5f6-49ca-9ee1-1a0531c5d3ac","cardType":"cloze","points":null,"front":"For Friedel-Crafts methylation: CH₃Cl + AlCl₃ → [ ] + [ ].","back":"For Friedel-Crafts methylation: CH₃Cl + AlCl₃ → CH₃⁺ + AlCl₄⁻.","mcqOptions":[{"text":"For Friedel-Crafts methylation: CH₃Cl + AlCl₃ → CH₃⁺ + AlCl₄⁻.","correct":true,"explanation":"Anhydrous AlCl₃ accepts a chloride ion from CH₃Cl, generating the methyl electrophile CH₃⁺ and the counter-ion AlCl₄⁻ for attack on benzene."},{"text":"For Friedel-Crafts methylation: CH₃Cl + AlCl₃ → CH₃⁻ + AlCl₄⁺.","correct":false,"explanation":"The methyl species is electron-deficient and positively charged, not a carbanion. Chloride transfer gives CH₃⁺ and the negatively charged complex ion AlCl₄⁻."},{"text":"For Friedel-Crafts methylation: CH₃Cl + AlCl₃ → CH₃Cl⁺ + AlCl₃⁻.","correct":false,"explanation":"The reaction is represented as chloride abstraction, not as formation of intact CH₃Cl⁺. The correct ion pair is CH₃⁺ and AlCl₄⁻."},{"text":"For Friedel-Crafts methylation: CH₃Cl + AlCl₃ → CH₄ + AlCl₄⁻.","correct":false,"explanation":"No proton source is present to convert the methyl group into methane. AlCl₃ removes chloride from CH₃Cl, generating the electrophile CH₃⁺ and AlCl₄⁻."}]},{"id":"c98cb6e6-e73e-46bb-8a60-956219b7d504","cardType":"basic","points":2,"front":"Why is the C-Cl bond in chlorobenzene shorter and stronger than in chloroethane?","back":"Cl lone pair overlaps with benzene π system, giving partial double-bond character.","mcqOptions":[{"text":"Cl lone pair overlaps with benzene π system, giving partial double-bond character.","correct":true,"explanation":"A chlorine lone pair overlaps with the delocalised π system, allowing resonance and giving the C-Cl bond partial double-bond character. This makes it shorter and stronger than in chloroethane."},{"text":"Cl lone pair repels the benzene π system, giving partial single-bond character.","correct":false,"explanation":"The lone pair does not merely repel the π system; it overlaps with it and contributes to delocalisation. That resonance gives partial double-bond character and strengthens the C-Cl bond."},{"text":"The benzene π system overlaps with a chlorine σ* orbital, giving partial ionic character.","correct":false,"explanation":"The key interaction is donation from a chlorine lone pair into the benzene π system, not primarily into a chlorine σ* orbital. This produces partial double-bond character in C-Cl."},{"text":"Cl lone pair overlaps with the benzene σ system, giving partial triple-bond character.","correct":false,"explanation":"The relevant orbital is the benzene π system, and the resulting bond order is only partly double, not triple. Lone-pair delocalisation explains the bond's increased strength and shorter length."}]},{"id":"84b2866e-54bf-403b-ac34-a241bf7e92ce","cardType":"basic","points":1,"front":"What does partial double-bond character make chlorobenzene resist?","back":"Nucleophilic substitution / hydrolysis under mild conditions.","mcqOptions":[{"text":"Nucleophilic substitution / hydrolysis under mild conditions.","correct":true,"explanation":"Partial double-bond character strengthens the C-Cl bond and makes it less susceptible to attack by nucleophiles. Consequently, chlorobenzene resists nucleophilic substitution or hydrolysis under mild conditions."},{"text":"Electrophilic substitution / nitration under mild conditions.","correct":false,"explanation":"The aromatic ring remains able to undergo electrophilic substitution, such as nitration, under suitable conditions. The strengthened C-Cl bond specifically makes chlorobenzene resist nucleophilic substitution or hydrolysis."},{"text":"Free-radical substitution / chlorination under mild conditions.","correct":false,"explanation":"Free-radical chlorination involves different conditions and is not the reaction prevented by the strengthened aryl C-Cl bond. The relevant resistance is to nucleophilic substitution or hydrolysis."},{"text":"Electrophilic addition / hydrogenation under mild conditions.","correct":false,"explanation":"Aromatic rings resist addition because of aromatic stabilisation, not specifically because the C-Cl bond has partial double-bond character. That bond character makes nucleophilic substitution or hydrolysis difficult."}]}]}],"examCountdown":{"examDate":"2027-05-28","paperLabel":"Component 1"},"heading":"Aromaticity","title":"Aromaticity Flashcards","topbar":{"backHref":"/a-level/chemistry/eduqas/flashcards"},"topicSidebar":"$L36"}]}],"topicId":"6e64edf8-89ba-5e55-a386-2fd1f912cc2a","walled":"$L37"}],"$L38","$L39"]