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22:[["$","$L31",null,{"feature":"flashcards"}],"$L32",["$","$L33",null,{"href":"/a-level/chemistry/eduqas/amines/flashcards","label":"Amines","subjectCode":"a_level_chemistry__eduqas","surface":"flashcards"}],["$","$L34",null,{"runner":["$","div",null,{"className":"min-h-0 flex-1 overflow-hidden bg-background","children":["$","$L35",null,{"decks":[{"id":"5c1c50eb-48ab-4f78-ae28-abdaef22ae52","title":"Amines Flashcards","topicId":"5253ec23-63ca-5588-b8aa-4915c99a51af","topicSlug":"amines","cards":[{"id":"bd3ab583-4f7d-48f6-903c-875cde726626","cardType":"cloze","points":null,"front":"Amines contain a nitrogen [ ], making them [ ].","back":"Amines contain a nitrogen lone pair, making them bases and nucleophiles.","mcqOptions":[{"text":"Amines contain a nitrogen lone pair, making them bases and nucleophiles.","correct":true,"explanation":"The non-bonding electron pair on nitrogen can accept a proton and attack electron-deficient species, so amines show both basic and nucleophilic behaviour."},{"text":"Amines contain a nitrogen positive charge, making them acids and electrophiles.","correct":false,"explanation":"A positive charge would make nitrogen electron-poor, but amines are electron-rich because nitrogen has a non-bonding pair. Therefore they behave as bases and nucleophiles, not acids and electrophiles."},{"text":"Amines contain a nitrogen unpaired electron, making them reducing agents and electrophiles.","correct":false,"explanation":"An unpaired electron describes a radical, not a typical amine. The nitrogen lone pair accounts for proton acceptance and attack on electron-deficient centres, giving basic and nucleophilic properties."},{"text":"Amines contain a nitrogen bonding pair, making them oxidising agents and electrophiles.","correct":false,"explanation":"The relevant electrons are non-bonding rather than shared in a bond. They make amines electron-pair donors, so the correct properties are basicity and nucleophilicity rather than oxidation and electrophilic behaviour."}]},{"id":"879cf688-bc2f-4f41-92be-06fa3388b551","cardType":"cloze","points":null,"front":"In an aliphatic amine, nitrogen is attached to an [ ]; in an aromatic amine, nitrogen is attached directly to a [ ].","back":"In an aliphatic amine, nitrogen is attached to an alkyl chain; in an aromatic amine, nitrogen is attached directly to a benzene ring.","mcqOptions":[{"text":"In an aliphatic amine, nitrogen is attached to an alkyl chain; in an aromatic amine, nitrogen is attached directly to a benzene ring.","correct":true,"explanation":"Aliphatic amines have nitrogen attached to an alkyl group or chain, whereas aromatic amines have nitrogen directly bonded to an aromatic benzene ring."},{"text":"In an aliphatic amine, nitrogen is attached to an alkene chain; in an aromatic amine, nitrogen is attached directly to a carbonyl group.","correct":false,"explanation":"An alkene chain is defined by a carbon-carbon double bond, and a carbonyl group contains C=O. The distinction here is alkyl attachment versus direct attachment to a benzene ring."},{"text":"In an aliphatic amine, nitrogen is attached to a benzyl group; in an aromatic amine, nitrogen is attached directly to a cyclohexane ring.","correct":false,"explanation":"A benzyl group is not the general defining feature of an aliphatic amine, while cyclohexane is non-aromatic. Aromatic amines require direct attachment to a benzene ring."},{"text":"In an aliphatic amine, nitrogen is attached to an alkyl chain; in an aromatic amine, nitrogen is attached directly to a phenol group.","correct":false,"explanation":"Although an alkyl chain correctly describes the aliphatic case, a phenol group contains an -OH substituent and is not the relevant structure. The aromatic attachment is directly to a benzene ring."}]},{"id":"2824deac-e959-49a5-974a-27067d323c7f","cardType":"basic","points":1,"front":"Which general formula identifies a primary amine?","back":"RNH₂","mcqOptions":[{"text":"RNH₂","correct":true,"explanation":"A primary amine has nitrogen bonded to one carbon-containing group, represented by R, and two hydrogen atoms, giving the general formula RNH₂."},{"text":"R₂NH","correct":false,"explanation":"This formula has nitrogen bonded to two carbon-containing groups and one hydrogen, so it represents a secondary amine. A primary amine has only one R group: RNH₂."},{"text":"R₃N","correct":false,"explanation":"Three carbon-containing groups attached to nitrogen identify a tertiary amine. Primary amines contain one R group and two hydrogens, represented by RNH₂."},{"text":"NH₃","correct":false,"explanation":"This is ammonia, with no carbon-containing group attached to nitrogen. A primary amine must contain one organic group bonded to nitrogen, giving the form RNH₂."}]},{"id":"cda2d94f-3435-4359-8693-62a41da984fe","cardType":"cloze","points":null,"front":"Halogenoalkanes form primary amines using [ ] and heat in a [ ].","back":"Halogenoalkanes form primary amines using excess ethanolic ammonia and heat in a sealed tube.","mcqOptions":[{"text":"Halogenoalkanes form primary amines using excess ethanolic ammonia and heat in a sealed tube.","correct":true,"explanation":"Ethanolic ammonia provides ammonia as the nucleophile, while excess ammonia limits further substitution. Heating in a sealed tube keeps the volatile ammonia present at a high concentration."},{"text":"Halogenoalkanes form primary amines using excess aqueous ammonia and heat in a sealed tube.","correct":false,"explanation":"Aqueous ammonia is not the specified solvent for this preparation; ethanolic ammonia is used. The sealed tube and excess ammonia help promote substitution while limiting further reactions."},{"text":"Halogenoalkanes form primary amines using limited ethanolic ammonia and heat in a sealed tube.","correct":false,"explanation":"Using limited ammonia leaves amine product available to attack more halogenoalkane, increasing secondary and tertiary amine formation. Excess ethanolic ammonia favours the primary amine instead."},{"text":"Halogenoalkanes form primary amines using excess ethanolic ammonia and heat in a open vessel.","correct":false,"explanation":"An open vessel allows volatile ammonia to escape, reducing its concentration and making further substitution more likely. The reaction is heated in a sealed tube to retain ammonia."}]},{"id":"b02ab01f-1fc8-41b3-8e66-9a6b24b356eb","cardType":"basic","points":2,"front":"Why does excess ammonia favour formation of a primary amine from a halogenoalkane?","back":"Ammonia reacts first, reducing further substitution by the amine product.","mcqOptions":[{"text":"Ammonia reacts first, reducing further substitution by the amine product.","correct":true,"explanation":"Excess ammonia competes successfully with the primary amine product for the halogenoalkane, so the product is less likely to undergo further substitution to secondary or tertiary amines."},{"text":"The amine product reacts first, reducing further substitution by ammonia.","correct":false,"explanation":"The amine product is the species that can cause further substitution, so it must not be allowed to dominate. Excess ammonia reacts preferentially and suppresses additional alkylation."},{"text":"Ammonia reacts first, increasing further substitution by the amine product.","correct":false,"explanation":"Excess ammonia reduces the chance of the amine product attacking another halogenoalkane. Increasing the product’s opportunity for substitution would favour secondary and tertiary amines instead."},{"text":"The halogenoalkane reacts first, preventing substitution by ammonia.","correct":false,"explanation":"The halogenoalkane is the electrophile, but the key competition is between ammonia and the amine product as nucleophiles. Excess ammonia favours primary amine formation by limiting product alkylation."}]},{"id":"a506d840-d9dc-4e83-b5c4-9118efae565a","cardType":"cloze","points":null,"front":"Reduction of a nitrile: RCN + 4[H] → [ ].","back":"Reduction of a nitrile: RCN + 4[H] → RCH₂NH₂.","mcqOptions":[{"text":"RCH₂NH₂","correct":true,"explanation":"Reduction adds four hydrogen atoms to the nitrile group, converting its carbon and nitrogen into a primary amine while retaining the original R group."},{"text":"RCH₂OH","correct":false,"explanation":"This confuses nitrile reduction with reduction of a carboxylic acid derivative to an alcohol. Nitriles reduce to primary amines, giving RCH₂NH₂ rather than an alcohol."},{"text":"RCHNH₂","correct":false,"explanation":"The missing hydrogen results from treating the nitrile carbon as though it gained only one hydrogen. Four reducing hydrogens produce the methylene group in RCH₂NH₂."},{"text":"RCH₂NHCH₃","correct":false,"explanation":"Adding a methyl group incorrectly suggests formation of a secondary amine. Nitrile reduction introduces no extra carbon, so the product is the primary amine RCH₂NH₂."}]},{"id":"ae5f2fac-ea10-48a2-92ee-37e6420a1c97","cardType":"basic","points":2,"front":"Which reagents reduce a nitrile to a primary amine?","back":"LiAlH₄ in dry ether, then dilute acid work-up.","mcqOptions":[{"text":"LiAlH₄ in dry ether, then dilute acid work-up.","correct":true,"explanation":"Lithium aluminium hydride supplies hydride for reducing the nitrile, and dry ether is required because the reagent reacts with water. Dilute acid then protonates the amine product."},{"text":"NaBH₄ in aqueous ethanol, then alkali work-up.","correct":false,"explanation":"Sodium borohydride is a milder reducing agent and is not the standard reagent for reducing nitriles to primary amines. The required sequence is LiAlH₄ in dry ether followed by dilute acid."},{"text":"LiAlH₄ in dry ether, then concentrated acid work-up.","correct":false,"explanation":"The reducing agent and solvent are appropriate, but concentrated acid is not the specified work-up and can create unnecessarily harsh conditions. The standard work-up uses dilute acid."},{"text":"H₂ with a nickel catalyst, then dilute acid work-up.","correct":false,"explanation":"Catalytic hydrogenation with nickel is not the standard A-level reagent combination for this nitrile reduction. LiAlH₄ in dry ether, followed by dilute acid, gives the primary amine."}]},{"id":"1ec67b6b-73a4-418e-976b-6f19342561b6","cardType":"basic","points":1,"front":"What happens to carbon chain length in the CN⁻ then reduction route?","back":"It becomes one carbon longer.","mcqOptions":[{"text":"It becomes one carbon longer.","correct":true,"explanation":"Cyanide adds one carbon because the carbon in CN⁻ becomes part of the nitrile chain. Reduction then converts the nitrile group into a primary amine without removing that carbon."},{"text":"It becomes one carbon shorter.","correct":false,"explanation":"The cyanide ion contributes a carbon atom to the product, so the chain length increases rather than decreases. Reduction changes the functional group but does not remove that carbon."},{"text":"It remains the same length.","correct":false,"explanation":"The reduction step alone preserves the nitrile carbon, but the preceding cyanide substitution adds one carbon to the original halogenoalkane chain. Therefore the final chain is one carbon longer."},{"text":"It becomes two carbons longer.","correct":false,"explanation":"Only one carbon is introduced by CN⁻; the nitrogen does not contribute a carbon. Reduction changes -C≡N into -CH₂NH₂, so the chain increases by exactly one carbon."}]},{"id":"5b9d5a99-8694-477a-9adc-c10127f44f33","cardType":"basic","points":2,"front":"Which reducing conditions convert nitrobenzene to phenylamine?","back":"Tin and concentrated HCl, heated under reflux.","mcqOptions":[{"text":"Tin and concentrated HCl, heated under reflux.","correct":true,"explanation":"Tin provides the reducing agent and concentrated hydrochloric acid supplies the acidic conditions; heating under reflux ensures nitrobenzene is reduced to phenylamine without losing volatile substances."},{"text":"Zinc and dilute HCl, heated under distillation.","correct":false,"explanation":"Zinc and dilute hydrochloric acid are a tempting alternative reducing system, but the specified preparation uses tin and concentrated HCl under reflux. Distillation would risk losing volatile materials."},{"text":"Sodium borohydride and aqueous NaOH, heated under reflux.","correct":false,"explanation":"Sodium borohydride is not the standard reagent for reducing nitrobenzene in this preparation, and alkaline conditions are unsuitable. The required conditions are tin, concentrated HCl and reflux."},{"text":"Iron and concentrated HNO₃, heated under reflux.","correct":false,"explanation":"Iron can act as a reducing metal in some reactions, but nitric acid is an oxidising acid rather than the required acidic medium. Nitrobenzene is reduced using tin and concentrated HCl."}]},{"id":"5af0fe63-d074-4fc7-8a37-1dcb46f48664","cardType":"basic","points":2,"front":"Why is NaOH(aq) added after reducing nitrobenzene in acid?","back":"It releases free phenylamine from the phenylammonium salt.","mcqOptions":[{"text":"It releases free phenylamine from the phenylammonium salt.","correct":true,"explanation":"NaOH reacts with the phenylammonium salt, removing a proton and converting the ionic salt into uncharged, free phenylamine, which can then be separated."},{"text":"It converts free phenylamine into the phenylammonium salt.","correct":false,"explanation":"Adding NaOH does the opposite: hydroxide removes a proton from the phenylammonium ion. The correct purpose is to release free phenylamine from the phenylammonium salt."},{"text":"It oxidises phenylamine into nitrobenzene.","correct":false,"explanation":"Aqueous NaOH is not being used as an oxidising agent here. It neutralises the acid associated with the phenylammonium salt and releases free phenylamine."},{"text":"It removes phenylamine by reacting it with hydroxide ions.","correct":false,"explanation":"Hydroxide ions do not remove phenylamine by forming a new product; they deprotonate phenylammonium ions. This releases the neutral free phenylamine from its salt."}]},{"id":"77f2a347-05ee-416d-b772-ce2342d405ba","cardType":"basic","points":2,"front":"Why are aliphatic amines usually stronger bases than ammonia?","back":"Alkyl groups push electron density towards nitrogen, increasing lone-pair availability.","mcqOptions":[{"text":"Alkyl groups push electron density towards nitrogen, increasing lone-pair availability.","correct":true,"explanation":"Alkyl groups have a positive inductive effect, pushing electron density towards nitrogen. Its lone pair is therefore more available to accept a proton than ammonia’s lone pair."},{"text":"Alkyl groups withdraw electron density from nitrogen, increasing lone-pair availability.","correct":false,"explanation":"Alkyl groups donate rather than withdraw electron density by the inductive effect. This donation increases lone-pair availability, which is why aliphatic amines are usually stronger bases than ammonia."},{"text":"Alkyl groups push electron density towards nitrogen, decreasing lone-pair availability.","correct":false,"explanation":"The direction of the inductive effect is correct, but increased electron density makes the nitrogen lone pair more available, not less. Reduced availability would weaken basicity."},{"text":"Alkyl groups delocalise the nitrogen lone pair into a benzene ring, increasing lone-pair availability.","correct":false,"explanation":"Alkyl groups do not provide a benzene ring for lone-pair delocalisation. Their relevant effect is inductive donation of electron density towards nitrogen, increasing basicity."}]},{"id":"c15187b1-de6b-4c33-a19e-185d5d748fc7","cardType":"basic","points":2,"front":"Why is phenylamine weaker as a base than ammonia?","back":"The nitrogen lone pair is delocalised into the benzene ring.","mcqOptions":[{"text":"The nitrogen lone pair is delocalised into the benzene ring.","correct":true,"explanation":"In phenylamine, the nitrogen lone pair overlaps with the benzene ring’s pi system and becomes delocalised. It is therefore less available to accept a proton than ammonia’s lone pair."},{"text":"The nitrogen lone pair is localised on the benzene ring.","correct":false,"explanation":"The lone pair is not simply localised on the ring; it is delocalised across the nitrogen and benzene pi system. This delocalisation reduces its availability for proton acceptance."},{"text":"The nitrogen lone pair is delocalised into the alkyl group.","correct":false,"explanation":"Phenylamine contains a benzene ring, not an alkyl group attached to nitrogen. Delocalisation into the benzene ring makes the lone pair less available and lowers basicity."},{"text":"The nitrogen lone pair is more available because of the benzene ring.","correct":false,"explanation":"Conjugation with the benzene ring withdraws lone-pair availability from nitrogen rather than increasing it. Consequently, phenylamine is a weaker base than ammonia."}]},{"id":"183fcfcd-b480-4631-918b-3b22c7c5afd6","cardType":"cloze","points":null,"front":"Typical basicity order: [ ] > [ ] > phenylamine.","back":"Typical basicity order: ethylamine > ammonia > phenylamine.","mcqOptions":[{"text":"Typical basicity order: ethylamine > ammonia > phenylamine.","correct":true,"explanation":"Ethylamine is more basic than ammonia because the ethyl group donates electron density to nitrogen. Phenylamine is weakest because its lone pair is delocalised into the benzene ring."},{"text":"Typical basicity order: ammonia > ethylamine > phenylamine.","correct":false,"explanation":"This reverses the first two positions. Ethylamine is usually more basic than ammonia because its alkyl group pushes electron density towards nitrogen and increases lone-pair availability."},{"text":"Typical basicity order: ethylamine > phenylamine > ammonia.","correct":false,"explanation":"Phenylamine is not stronger than ammonia: its nitrogen lone pair is delocalised into the benzene ring. The typical order is ethylamine, then ammonia, then phenylamine."},{"text":"Typical basicity order: phenylamine > ethylamine > ammonia.","correct":false,"explanation":"This incorrectly treats phenylamine as the strongest base. Delocalisation into the benzene ring weakens phenylamine, while ethylamine is strengthened by the electron-donating ethyl group."}]},{"id":"9960207c-f5df-48ca-ac30-8c093c249e90","cardType":"cloze","points":null,"front":"Ethanoylation of a primary amine: RNH₂ + CH₃COCl → [ ] + [ ].","back":"Ethanoylation of a primary amine: RNH₂ + CH₃COCl → CH₃CONHR + HCl.","mcqOptions":[{"text":"Ethanoylation of a primary amine: RNH₂ + CH₃COCl → CH₃CONHR + HCl.","correct":true,"explanation":"The amine nitrogen attacks ethanoyl chloride and replaces the chlorine, forming the N-substituted amide CH₃CONHR. The removed hydrogen combines with chloride to form HCl."},{"text":"Ethanoylation of a primary amine: RNH₂ + CH₃COCl → CH₃COOR + HCl.","correct":false,"explanation":"CH₃COOR is an ester, which would result from an alcohol attacking ethanoyl chloride. A primary amine forms the amide CH₃CONHR, with HCl produced."},{"text":"Ethanoylation of a primary amine: RNH₂ + CH₃COCl → CH₃CONHR + NaCl.","correct":false,"explanation":"No sodium-containing reactant is present, so sodium chloride cannot be the by-product. The amide forms alongside HCl when the amine reacts directly with ethanoyl chloride."},{"text":"Ethanoylation of a primary amine: RNH₂ + CH₃COCl → CH₃COOH + RNHCl.","correct":false,"explanation":"This incorrectly represents hydrolysis and an implausible amine chloride product. The amine substitutes for chlorine at the acyl carbon, producing CH₃CONHR and HCl."}]},{"id":"990c64c1-ca8a-4e9c-a70f-8eda6c72c85d","cardType":"basic","points":1,"front":"What mechanism type occurs when a primary amine reacts with ethanoyl chloride?","back":"Nucleophilic addition-elimination","mcqOptions":[{"text":"Nucleophilic addition-elimination","correct":true,"explanation":"The amine’s lone pair attacks the electrophilic carbonyl carbon, giving an addition intermediate. Chloride then leaves and the carbonyl group reforms, so the mechanism is nucleophilic addition-elimination."},{"text":"Nucleophilic substitution","correct":false,"explanation":"Although chloride is ultimately replaced, the carbonyl group first undergoes nucleophilic addition before elimination. Therefore the acyl chloride reaction is classified as nucleophilic addition-elimination."},{"text":"Electrophilic addition","correct":false,"explanation":"The amine acts as a nucleophile, not an electrophile, and attacks the carbonyl carbon. Addition followed by loss of chloride makes this nucleophilic addition-elimination."},{"text":"Free-radical substitution","correct":false,"explanation":"No radicals or radical initiation are involved. The nitrogen lone pair attacks the carbonyl carbon, followed by chloride elimination, giving a nucleophilic addition-elimination mechanism."}]},{"id":"f578ef79-e59c-4f84-8a96-7316d1a3e486","cardType":"basic","points":2,"front":"How is nitric(III) acid generated in the reaction mixture?","back":"Sodium nitrite reacts with hydrochloric acid.","mcqOptions":[{"text":"Sodium nitrite reacts with hydrochloric acid.","correct":true,"explanation":"Aqueous sodium nitrite reacts with hydrochloric acid in the mixture to generate nitric(III) acid, HNO₂, which is needed for the diazotisation reaction."},{"text":"Sodium nitrate reacts with hydrochloric acid.","correct":false,"explanation":"Sodium nitrate contains nitrate, not nitrite, and does not generate nitric(III) acid under these conditions. Sodium nitrite reacts with hydrochloric acid to form HNO₂."},{"text":"Sodium nitrite reacts with sulfuric acid.","correct":false,"explanation":"Sulfuric acid can provide acidic conditions in other reactions, but the stated preparation uses hydrochloric acid. Nitric(III) acid is generated from sodium nitrite and hydrochloric acid."},{"text":"Ammonia reacts with hydrochloric acid.","correct":false,"explanation":"Ammonia and hydrochloric acid form ammonium chloride, not nitric(III) acid. The required nitrite source is sodium nitrite, which reacts with hydrochloric acid to produce HNO₂."}]},{"id":"9abb42a9-48d9-4f9f-a655-f5f22ef03d90","cardType":"cloze","points":null,"front":"Primary aliphatic amines with cold HNO₂: RNH₂ + HNO₂ → [ ] + [ ] + H₂O.","back":"Primary aliphatic amines with cold HNO₂: RNH₂ + HNO₂ → ROH + N₂ + H₂O.","mcqOptions":[{"text":"Primary aliphatic amines with cold HNO₂: RNH₂ + HNO₂ → ROH + N₂ + H₂O.","correct":true,"explanation":"Cold nitrous acid converts a primary aliphatic amine into the corresponding alcohol, while the unstable intermediate breaks down and releases nitrogen gas."},{"text":"Primary aliphatic amines with cold HNO₂: RNH₂ + HNO₂ → RCl + N₂ + H₂O.","correct":false,"explanation":"Confusing this reaction with nucleophilic substitution leads to RCl. Cold nitrous acid instead replaces the amino group with OH, producing ROH and releasing N₂."},{"text":"Primary aliphatic amines with cold HNO₂: RNH₂ + HNO₂ → ROH + NO + H₂O.","correct":false,"explanation":"Nitrogen monoxide is associated with other nitrogen redox chemistry, but the diazonium intermediate here loses stable N₂ gas and forms the corresponding alcohol, ROH."},{"text":"Primary aliphatic amines with cold HNO₂: RNH₂ + HNO₂ → RNO₂ + N₂ + H₂O.","correct":false,"explanation":"Nitrous acid does not simply introduce a nitro group into a primary aliphatic amine. The unstable aliphatic diazonium intermediate forms ROH and releases N₂."}]},{"id":"20053d07-9c7b-4032-809c-a89547aba130","cardType":"basic","points":1,"front":"What does phenylamine form with HNO₂/HCl at 0-10 °C?","back":"Benzenediazonium chloride","mcqOptions":[{"text":"Benzenediazonium chloride","correct":true,"explanation":"At 0-10 °C, phenylamine undergoes diazotisation with nitrous acid generated in hydrochloric acid, producing benzenediazonium chloride while limiting its thermal decomposition."},{"text":"Phenylammonium chloride","correct":false,"explanation":"Phenylammonium chloride forms when phenylamine is merely protonated by hydrochloric acid. In the presence of nitrous acid at 0-10 °C, diazotisation produces benzenediazonium chloride."},{"text":"Nitrobenzene","correct":false,"explanation":"Nitrobenzene is not formed by diazotising phenylamine. Cold HNO₂/HCl converts the amino group into a diazonium group, giving benzenediazonium chloride."},{"text":"Benzenediazonium nitrate","correct":false,"explanation":"The counter-ion comes from the strong acid used during diazotisation. Because HCl is present, the product is benzenediazonium chloride rather than benzenediazonium nitrate."}]},{"id":"7aa9b7f5-4f9f-4c59-a1a5-eaa5a0b6849f","cardType":"basic","points":2,"front":"What happens if benzenediazonium salts are warmed?","back":"They decompose, releasing nitrogen gas and often forming phenol.","mcqOptions":[{"text":"They decompose, releasing nitrogen gas and often forming phenol.","correct":true,"explanation":"Benzenediazonium salts are thermally unstable. Warming causes loss of very stable nitrogen gas, and reaction with water commonly replaces the diazonium group with OH to form phenol."},{"text":"They become more stable, preventing the release of nitrogen gas.","correct":false,"explanation":"Benzenediazonium salts are kept cold precisely because warming destabilises them. They decompose, releasing nitrogen gas and commonly reacting with water to form phenol."},{"text":"They polymerise, releasing hydrogen gas and often forming benzene.","correct":false,"explanation":"Heating does not normally polymerise benzenediazonium salts or release hydrogen. They decompose with loss of nitrogen gas, often replacing the diazonium group with OH to form phenol."},{"text":"They decompose, releasing hydrogen gas and often forming chlorobenzene.","correct":false,"explanation":"Although warming causes decomposition, the gas released is N₂ rather than H₂. In aqueous conditions, OH commonly replaces the diazonium group, producing phenol rather than chlorobenzene."}]},{"id":"4d9a0b42-92fd-434b-a5b7-8486fa725fc1","cardType":"basic","points":1,"front":"What is the role of the diazonium ion in azo coupling?","back":"It acts as the electrophile.","mcqOptions":[{"text":"It acts as the electrophile.","correct":true,"explanation":"The positively charged diazonium ion accepts electron density from an activated aromatic ring during azo coupling, so it functions as the electrophile in electrophilic substitution."},{"text":"It acts as the nucleophile.","correct":false,"explanation":"The activated phenol or aromatic amine ring supplies electron density and therefore acts as the nucleophilic partner. The positively charged diazonium ion is the electrophile."},{"text":"It acts as the catalyst.","correct":false,"explanation":"The diazonium ion is consumed when the new carbon-nitrogen bond forms, so it is not a catalyst. Its positive charge makes it the electrophile."},{"text":"It acts as the leaving group.","correct":false,"explanation":"Nitrogen can leave from diazonium salts during decomposition reactions, but azo coupling retains both nitrogen atoms. In coupling, the diazonium ion acts as the electrophile."}]},{"id":"af1f5c14-a223-408a-a8ca-3c337741e239","cardType":"basic","points":2,"front":"Why are alkaline conditions used when coupling diazonium salts with phenol?","back":"They form phenoxide ions, which activate the ring more strongly.","mcqOptions":[{"text":"They form phenoxide ions, which activate the ring more strongly.","correct":true,"explanation":"Alkali deprotonates phenol to produce phenoxide ions. Their negative charge donates electron density into the aromatic ring, making it more reactive towards the diazonium electrophile."},{"text":"They form phenyl cations, which activate the ring more strongly.","correct":false,"explanation":"Alkaline conditions do not produce unstable phenyl cations. They deprotonate phenol to form phenoxide ions, which donate electron density and strongly activate the aromatic ring."},{"text":"They form phenoxide ions, which deactivate the ring more strongly.","correct":false,"explanation":"Phenoxide ions activate rather than deactivate the ring because their negative charge is delocalised into it. This increased electron density promotes attack on the diazonium electrophile."},{"text":"They form phenol molecules, which activate the ring more strongly.","correct":false,"explanation":"Phenol molecules are already present before alkali is added. The alkali converts them into phenoxide ions, which activate the aromatic ring more strongly than neutral phenol."}]},{"id":"0e93e228-8901-4656-8877-679c87dc0692","cardType":"basic","points":1,"front":"Which group is the azo chromophore?","back":"-N=N-","mcqOptions":[{"text":"-N=N-","correct":true,"explanation":"An azo group contains two nitrogen atoms joined by a double bond, -N=N-. When incorporated into a conjugated molecule, it contributes to visible-light absorption."},{"text":"-NH-NH-","correct":false,"explanation":"The -NH-NH- linkage is a hydrazo group containing a nitrogen-nitrogen single bond. The azo chromophore instead has the double-bonded structure -N=N-."},{"text":"-NO₂","correct":false,"explanation":"The -NO₂ unit is a nitro group, not an azo group. The characteristic azo chromophore consists of two nitrogen atoms joined as -N=N-."},{"text":"-C=O-","correct":false,"explanation":"The -C=O- unit is a carbonyl group and occurs in compounds such as aldehydes and ketones. The azo chromophore is the nitrogen linkage -N=N-."}]},{"id":"c71dbbe2-e87a-4dd9-8ab3-eea193e74871","cardType":"basic","points":2,"front":"Why are azo dyes coloured?","back":"Their extended delocalised π-electron system absorbs visible light.","mcqOptions":[{"text":"Their extended delocalised π-electron system absorbs visible light.","correct":true,"explanation":"Delocalisation across the azo group and aromatic rings creates closely spaced electronic energy levels. Visible photons can therefore promote electrons between them, removing particular wavelengths from white light."},{"text":"Their localised σ-bond system absorbs visible light.","correct":false,"explanation":"Localised σ bonds generally require high-energy ultraviolet radiation for electronic excitation. Azo dyes are coloured because an extended delocalised π-electron system absorbs wavelengths in the visible region."},{"text":"Their azo groups emit visible light spontaneously after thermal excitation.","correct":false,"explanation":"Azo dyes normally appear coloured through selective absorption, not spontaneous emission. Their extended conjugated π system absorbs certain visible wavelengths, and the remaining transmitted or reflected light is observed."},{"text":"Their ionic groups scatter particular wavelengths of visible light.","correct":false,"explanation":"Solubilising ionic groups can affect a dye’s properties but do not create its colour by scattering. Colour arises when the extended delocalised π system selectively absorbs visible light."}]},{"id":"d300d11b-bd67-486f-a89f-9c09d1fae3c0","cardType":"basic","points":2,"front":"What effect does increased conjugation have on absorption wavelength?","back":"It lowers the energy gap, so absorption occurs at a longer wavelength.","mcqOptions":[{"text":"It lowers the energy gap, so absorption occurs at a longer wavelength.","correct":true,"explanation":"Greater conjugation increases electron delocalisation and brings the relevant energy levels closer together. Since lower-energy photons have longer wavelengths, absorption shifts towards longer wavelengths."},{"text":"It raises the energy gap, so absorption occurs at a shorter wavelength.","correct":false,"explanation":"Increased conjugation brings electronic energy levels closer together rather than farther apart. The lower transition energy corresponds to absorption at a longer, not shorter, wavelength."},{"text":"It lowers the energy gap, so absorption occurs at a shorter wavelength.","correct":false,"explanation":"A lower energy gap cannot correspond to a shorter wavelength because photon energy is inversely proportional to wavelength. Increased conjugation therefore shifts absorption to a longer wavelength."},{"text":"It increases absorption intensity, but leaves the absorption wavelength unchanged.","correct":false,"explanation":"Conjugation primarily changes the spacing between electronic energy levels, so it shifts the absorption wavelength. Greater conjugation lowers the gap and moves absorption to longer wavelengths."}]}]}],"examCountdown":{"examDate":"2027-05-28","paperLabel":"Component 1"},"heading":"Amines","title":"Amines Flashcards","topbar":{"backHref":"/a-level/chemistry/eduqas/flashcards"},"topicSidebar":"$L36"}]}],"topicId":"5253ec23-63ca-5588-b8aa-4915c99a51af","walled":"$L37"}],"$L38","$L39"]