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bromine."},{"text":"hydroxide ions","correct":false,"explanation":"Hydroxide ions are reagents that can substitute the halogen in some reactions, but they are not the atoms replacing hydrogen in a halogenoalkane."},{"text":"alkyl groups","correct":false,"explanation":"Alkyl groups are hydrocarbon fragments formed by removing hydrogen from an alkane; they are not the substituted atoms that define a halogenoalkane."},{"text":"oxygen atoms","correct":false,"explanation":"Oxygen atoms occur in alcohols and other oxygen-containing compounds, but replacing alkane hydrogen with oxygen does not produce a halogenoalkane."}]},{"id":"22fe025a-69dd-4a89-b3b3-9332d3b67bb1","cardType":"cloze","points":null,"front":"In a C-X bond, carbon is [ ] and the halogen is [ ].","back":"In a C-X bond, carbon is δ+ and the halogen is δ−.","mcqOptions":[{"text":"In a C-X bond, carbon is δ+ and the halogen is δ−.","correct":true,"explanation":"The halogen is more electronegative than carbon, so it attracts the bonding electrons and becomes slightly negative, leaving carbon slightly positive."},{"text":"In a C-X bond, carbon is δ− and the halogen is δ+.","correct":false,"explanation":"Reversing the partial charges ignores the halogen’s greater electronegativity. The correct arrangement is carbon δ+ and the halogen δ−."},{"text":"In a C-X bond, carbon is δ0 and the halogen is δ0.","correct":false,"explanation":"A C-X bond is polar rather than non-polar because the halogen attracts bonding electrons more strongly. Carbon is δ+ and the halogen is δ−."},{"text":"In a C-X bond, carbon is δ+ and the halogen is δ+.","correct":false,"explanation":"Both atoms cannot be slightly positive: the shared electrons are pulled towards the more electronegative halogen. Carbon is δ+ while the halogen is δ−."}]},{"id":"e66402fd-35b4-4846-8b6f-4c74520fce92","cardType":"basic","points":2,"front":"Why is the C-X carbon the key reactive site?","back":"The δ+ carbon attracts electron-pair donors.","mcqOptions":[{"text":"The δ+ carbon attracts electron-pair donors.","correct":true,"explanation":"The polar C-X bond leaves carbon δ+, so it attracts electron-pair donors such as nucleophiles, making that carbon the key reactive site."},{"text":"The δ− carbon attracts electron-pair donors.","correct":false,"explanation":"The carbon is not δ−: the halogen withdraws bonding electrons, leaving carbon δ+. Therefore it attracts, rather than repels, electron-pair donors."},{"text":"The δ+ halogen attracts electron-pair donors.","correct":false,"explanation":"Electron-pair donors attack the electron-deficient carbon, not the electron-rich halogen. The δ+ carbon is the reactive site in the C-X bond."},{"text":"The δ+ carbon repels electron-pair donors.","correct":false,"explanation":"A δ+ carbon attracts electron-pair donors rather than repelling them. The partial positive charge explains why nucleophiles attack this site."}]},{"id":"cdae1182-4037-4677-aff2-971b9cd96515","cardType":"basic","points":2,"front":"What conditions favour nucleophilic substitution by hydroxide?","back":"Aqueous NaOH, heated under reflux.","mcqOptions":[{"text":"Aqueous NaOH, heated under reflux.","correct":true,"explanation":"Aqueous hydroxide ions act as nucleophiles, and reflux provides sustained heating so substitution proceeds without losing volatile reactants."},{"text":"Ethanolic NaOH, heated under reflux.","correct":false,"explanation":"Ethanolic hydroxide favours elimination by removing a hydrogen atom from a neighbouring carbon. Aqueous NaOH under reflux favours substitution."},{"text":"Aqueous NaOH, kept at room temperature.","correct":false,"explanation":"Aqueous hydroxide is appropriate, but room-temperature conditions do not provide the usual sustained heating needed for efficient substitution. Reflux is required."},{"text":"Aqueous KCN, heated under reflux.","correct":false,"explanation":"Aqueous KCN supplies cyanide ions, which form nitriles by substitution, not hydroxide substitution to produce an alcohol. The reagent must be aqueous NaOH."}]},{"id":"0308e489-5ad3-41e0-860f-aa069f5ce902","cardType":"basic","points":1,"front":"What organic product forms from 1-bromobutane with aqueous NaOH?","back":"Butan-1-ol.","mcqOptions":[{"text":"Butan-1-ol.","correct":true,"explanation":"Hydroxide ions replace bromine in 1-bromobutane by nucleophilic substitution, producing the primary alcohol butan-1-ol."},{"text":"Butan-2-ol.","correct":false,"explanation":"The carbon skeleton remains four carbons, but substitution at carbon 1 gives butan-1-ol. Butan-2-ol would require a different starting halogenoalkane."},{"text":"But-1-ene.","correct":false,"explanation":"But-1-ene forms by elimination with ethanolic hydroxide and heat, not by the aqueous hydroxide substitution conditions specified here."},{"text":"1-Bromobutane.","correct":false,"explanation":"The starting halogenoalkane is converted by replacing bromine with hydroxyl. Unreacted 1-bromobutane is therefore not the organic product of substitution."}]},{"id":"dd0aa15a-f95b-4a4c-8a73-a03d0b796fa4","cardType":"basic","points":2,"front":"In substitution, where does the curly arrow from OH^- go?","back":"From the O lone pair to the δ+ carbon.","mcqOptions":[{"text":"From the O lone pair to the δ+ carbon.","correct":true,"explanation":"The oxygen lone pair supplies the electron pair for the new C-O bond, so the curly arrow points from oxygen to the electron-deficient δ+ carbon."},{"text":"From the O lone pair to Br.","correct":false,"explanation":"The nucleophile attacks carbon rather than bromine. The arrow should start at the oxygen lone pair and point to the δ+ carbon."},{"text":"From the C-Br bond to the δ+ carbon.","correct":false,"explanation":"The C-Br bond contains the electrons that leave with bromide; it does not provide the attacking electron pair. The O lone pair attacks δ+ carbon."},{"text":"From the OH^- to the δ+ carbon.","correct":false,"explanation":"The arrow must begin at the oxygen lone pair, which forms the new bond. Bromide is the leaving group, while δ+ carbon is the electrophilic target."}]},{"id":"4ee2e25e-e256-4b3c-8739-6d1cf572d6a9","cardType":"basic","points":2,"front":"In substitution, where does the curly arrow from the C-Br bond go?","back":"From the C-Br bond to Br.","mcqOptions":[{"text":"From the C-Br bond to Br.","correct":true,"explanation":"Breaking the C-Br bond heterolytically sends its bonding electron pair onto bromine, forming a bromide ion as the leaving group."},{"text":"From the C-Br bond to C.","correct":false,"explanation":"The bonding pair leaves with bromine, not carbon. The arrow must point from the C-Br bond to Br to show heterolytic bond fission."},{"text":"From the O lone pair to Br.","correct":false,"explanation":"The oxygen lone pair forms the new C-O bond and is shown in a separate arrow. It does not represent cleavage of the C-Br bond."},{"text":"From Br to the C-Br bond.","correct":false,"explanation":"Bromine receives the bonding pair when the C-Br bond breaks; an arrow from bromine to carbon would show the wrong electron movement and reaction mechanism."}]},{"id":"6a1e2170-1dda-46c6-a04f-484017dfcdc0","cardType":"basic","points":2,"front":"What conditions favour elimination of a halogenoalkane?","back":"Ethanolic NaOH/KOH with heat.","mcqOptions":[{"text":"Ethanolic NaOH/KOH with heat.","correct":true,"explanation":"Ethanolic hydroxide acts as a strong base, while heat promotes removal of hydrogen and the halogen to form an alkene by elimination."},{"text":"Aqueous NaOH/KOH with heat.","correct":false,"explanation":"Aqueous hydroxide favours nucleophilic substitution to form an alcohol. Ethanolic hydroxide with heat is used to favour elimination instead."},{"text":"Ethanolic NaOH/KOH without heat.","correct":false,"explanation":"Ethanolic hydroxide is the correct reagent medium, but heating is needed to favour elimination effectively. Without heat, substitution is less clearly displaced."},{"text":"Aqueous HCl with heat.","correct":false,"explanation":"Acidic conditions do not provide the hydroxide base needed to remove a hydrogen atom. Elimination is favoured by ethanolic NaOH or KOH with heat."}]},{"id":"f7830238-a45e-4d14-8ce1-9576af12430d","cardType":"basic","points":1,"front":"What organic product forms when 1-bromopropane undergoes elimination?","back":"Propene.","mcqOptions":[{"text":"Propene.","correct":true,"explanation":"Elimination removes HBr from adjacent carbon atoms in 1-bromopropane, forming the alkene propene as the organic product."},{"text":"Propan-1-ol.","correct":false,"explanation":"This product would form by substitution with hydroxide ions, not elimination. Elimination removes HBr and forms the alkene propene."},{"text":"Propan-2-ol.","correct":false,"explanation":"This is an alcohol formed by substitution and would also require a different reaction pathway. Eliminating HBr from 1-bromopropane produces propene."},{"text":"Propane.","correct":false,"explanation":"Elimination forms a carbon-carbon double bond rather than replacing bromine with hydrogen. The organic product is the alkene propene."}]},{"id":"714f56c0-409d-4112-858f-6fc760a4d9f4","cardType":"basic","points":1,"front":"In halogenoalkane elimination, which hydrogen is removed?","back":"A hydrogen on the carbon next to the C-X carbon.","mcqOptions":[{"text":"A hydrogen on the carbon next to the C-X carbon.","correct":true,"explanation":"Elimination requires a hydrogen on the adjacent carbon so that its C-H bond can form a C=C bond as the halogen leaves."},{"text":"A hydrogen on the C-X carbon.","correct":false,"explanation":"Removing hydrogen from the carbon bonded to X does not create the required adjacent carbon-carbon double bond. The hydrogen must be on the carbon next to the C-X carbon."},{"text":"A hydrogen on the carbon two positions away from the C-X carbon.","correct":false,"explanation":"That position is too far away to form a double bond directly with the C-X carbon. The hydrogen must be on the carbon next to the C-X carbon."},{"text":"A hydrogen on the halogen atom.","correct":false,"explanation":"Halogens in halogenoalkanes do not carry the hydrogen removed during elimination. The hydrogen must be attached to the carbon next to the C-X carbon."}]},{"id":"5e2c8e8e-7894-49f0-88e3-3e1a43c96932","cardType":"basic","points":1,"front":"In elimination, what does the arrow from the C-H bond to the C-C bond form?","back":"A C=C double bond.","mcqOptions":[{"text":"A C=C double bond.","correct":true,"explanation":"The electron pair in the C-H bond moves between the two carbon atoms, creating a second shared pair and therefore a C=C double bond."},{"text":"A C-C single bond.","correct":false,"explanation":"A C-C single bond already exists before elimination, so this arrow cannot represent its formation. The electron pair creates a C=C double bond."},{"text":"A C-H double bond.","correct":false,"explanation":"The hydrogen is removed during elimination, so it cannot remain double-bonded to carbon. The electron pair forms a C=C double bond between adjacent carbons."},{"text":"A C-X double bond.","correct":false,"explanation":"The C-X bond breaks as the halogen leaves; it does not become a double bond. The C-H electron pair forms a C=C double bond."}]},{"id":"f6e2adfb-c535-42ca-92ae-2b4857a25b39","cardType":"cloze","points":null,"front":"For halogenoalkane hydrolysis rates, [ ] usually matters more than C-X bond polarity.","back":"For halogenoalkane hydrolysis rates, bond enthalpy usually matters more than C-X bond polarity.","mcqOptions":[{"text":"bond enthalpy","correct":true,"explanation":"Hydrolysis involves breaking the C-X bond, so its bond enthalpy strongly affects how readily the bond breaks. Polarity alone is less important for comparing these rates."},{"text":"C-X bond polarity","correct":false,"explanation":"Polarity can make the carbon electrophilic, but it does not determine bond-breaking ease as directly as bond enthalpy. The key factor is bond enthalpy."},{"text":"molecular mass","correct":false,"explanation":"A larger halogen atom may increase molecular mass, but mass itself does not control C-X bond cleavage. The important comparison is bond enthalpy."},{"text":"carbon chain length","correct":false,"explanation":"Changing the carbon chain can affect steric factors, but it is not the main reason different C-X bonds hydrolyse at different rates. The key factor is bond enthalpy."}]},{"id":"dbd085d0-a2a9-4903-aec8-386263f57a95","cardType":"basic","points":2,"front":"Why does 1-iodobutane hydrolyse faster than 1-chlorobutane?","back":"The C-I bond has lower bond enthalpy, so breaks more easily.","mcqOptions":[{"text":"The C-I bond has lower bond enthalpy, so breaks more easily.","correct":true,"explanation":"The C-I bond is weaker than the C-Cl bond because it has a lower bond enthalpy. It therefore breaks more readily during hydrolysis."},{"text":"The C-Br bond has lower bond enthalpy, so breaks more easily.","correct":false,"explanation":"This compares the wrong halogen and does not explain the iodide-versus-chloride rate. The C-I bond has lower bond enthalpy than C-Cl, so breaks more easily."},{"text":"The C-Cl bond has lower bond enthalpy, so breaks more easily.","correct":false,"explanation":"The bond enthalpy trend is reversed here: C-Cl is stronger than C-I. The C-I bond has lower bond enthalpy, making 1-iodobutane hydrolyse faster."},{"text":"The C-I bond has higher polarity, so breaks more easily.","correct":false,"explanation":"Greater polarity does not mean a bond is easier to break. The faster reaction results from the C-I bond having lower bond enthalpy than the C-Cl bond."}]},{"id":"6ef74adf-bbe3-421d-a03a-5b66dee48668","cardType":"basic","points":3,"front":"Rank halogenoalkanes by ease of substitution.","back":"Iodoalkanes > bromoalkanes > chloroalkanes >> fluoroalkanes.","mcqOptions":[{"text":"Iodoalkanes > bromoalkanes > chloroalkanes >> fluoroalkanes.","correct":true,"explanation":"Substitution becomes easier as the carbon-halogen bond becomes weaker. C-I is weakest, followed by C-Br and C-Cl, while the very strong C-F bond makes fluoroalkanes much less reactive."},{"text":"Fluoroalkanes > chloroalkanes > bromoalkanes > iodoalkanes.","correct":false,"explanation":"This reverses the effect of C-X bond strength. Stronger bonds resist substitution, so the correct order is iodoalkanes > bromoalkanes > chloroalkanes >> fluoroalkanes."},{"text":"Chloroalkanes > fluoroalkanes > iodoalkanes > bromoalkanes.","correct":false,"explanation":"The order incorrectly places fluoroalkanes above the more reactive heavier halogenoalkanes. The correct ranking follows decreasing bond strength: iodoalkanes > bromoalkanes > chloroalkanes >> fluoroalkanes."},{"text":"Bromoalkanes > iodoalkanes > fluoroalkanes > chloroalkanes.","correct":false,"explanation":"Iodoalkanes are more reactive than bromoalkanes because C-I bonds have lower bond enthalpy, while fluoroalkanes are least reactive. The correct order is iodoalkanes > bromoalkanes > chloroalkanes >> fluoroalkanes."}]},{"id":"d8d35216-8e7a-4363-bfc8-c6b60d20a279","cardType":"basic","points":2,"front":"Why are fluoroalkanes resistant to substitution despite C-F polarity?","back":"The C-F bond is very strong.","mcqOptions":[{"text":"The C-F bond is very strong.","correct":true,"explanation":"Although the C-F bond is highly polar, its high bond enthalpy means it requires considerable energy to break. Fluoroalkanes therefore resist substitution."},{"text":"The C-F bond is very weak.","correct":false,"explanation":"A weak C-F bond would make substitution easier, not harder. Fluoroalkanes resist substitution because the C-F bond is very strong."},{"text":"The C-F bond is non-polar.","correct":false,"explanation":"The C-F bond is strongly polar because fluorine is much more electronegative than carbon. Resistance to substitution is caused by its very high bond strength."},{"text":"The C-F bond has low bond enthalpy.","correct":false,"explanation":"Low bond enthalpy would indicate a bond that breaks readily. The C-F bond has high bond enthalpy, which makes fluoroalkanes resistant to substitution."}]},{"id":"4d429f2a-3701-43f9-88ba-c161bcd5e2d3","cardType":"basic","points":2,"front":"Why does AgNO₃ not immediately precipitate from an unhydrolysed halogenoalkane?","back":"There are no free halide ions until C-X hydrolyses.","mcqOptions":[{"text":"There are no free halide ions until C-X hydrolyses.","correct":true,"explanation":"Before hydrolysis, the halogen remains covalently bonded to carbon, so no free halide ions are present to react with silver ions and form a precipitate."},{"text":"There are free halide ions immediately before C-X hydrolyses.","correct":false,"explanation":"The halogen is still covalently attached before hydrolysis, so free halide ions are absent. They appear only after C-X hydrolysis and then precipitate with silver ions."},{"text":"There are no free silver ions until C-X hydrolyses.","correct":false,"explanation":"Silver ions are supplied by AgNO₃ before hydrolysis. It is free halide ions that are missing until the C-X bond hydrolyses."},{"text":"There are free halogen molecules until C-X hydrolyses.","correct":false,"explanation":"Hydrolysis does not initially produce neutral halogen molecules; it releases halide ions from the C-X bond. These ions then form the silver halide precipitate."}]},{"id":"ef86ecff-c4f0-41b6-b936-804407c7afca","cardType":"cloze","points":null,"front":"In the silver nitrate test, a [ ] precipitate identifies chloride ions.","back":"In the silver nitrate test, a white precipitate identifies chloride ions.","mcqOptions":[{"text":"white","correct":true,"explanation":"Silver chloride, AgCl, forms a white precipitate when chloride ions react with aqueous silver ions, distinguishing chloride from bromide and iodide."},{"text":"cream","correct":false,"explanation":"A cream precipitate indicates silver bromide, not silver chloride. Chloride ions react with silver ions to form a white precipitate of AgCl."},{"text":"yellow","correct":false,"explanation":"A yellow precipitate indicates silver iodide, not silver chloride. The precipitate produced by chloride ions is white silver chloride, AgCl."},{"text":"colourless","correct":false,"explanation":"Silver chloride is white rather than colourless because it forms a visible solid precipitate. Chloride ions therefore give a white precipitate with silver nitrate."}]},{"id":"03c60088-0c9e-4c90-88f2-b861258ec753","cardType":"cloze","points":null,"front":"In the silver nitrate test, a [ ] precipitate identifies bromide ions.","back":"In the silver nitrate test, a cream precipitate identifies bromide ions.","mcqOptions":[{"text":"cream","correct":true,"explanation":"Bromide ions react with aqueous silver ions to produce cream silver bromide, AgBr, whose colour lies between white AgCl and yellow AgI."},{"text":"white","correct":false,"explanation":"A white precipitate identifies silver chloride, not silver bromide. Bromide ions react with silver ions to produce a cream precipitate of AgBr."},{"text":"yellow","correct":false,"explanation":"A yellow precipitate identifies silver iodide, not silver bromide. Bromide ions give a cream precipitate when they react with aqueous silver ions."},{"text":"colourless","correct":false,"explanation":"Silver bromide forms a visible cream solid rather than a colourless mixture. Bromide ions are therefore identified by a cream precipitate with silver nitrate."}]},{"id":"af355ee9-f441-4e66-9c97-290fa8acb17e","cardType":"cloze","points":null,"front":"In the silver nitrate test, a [ ] precipitate identifies iodide ions.","back":"In the silver nitrate test, a yellow precipitate identifies iodide ions.","mcqOptions":[{"text":"yellow","correct":true,"explanation":"Iodide ions react with aqueous silver ions to form yellow silver iodide, AgI, distinguishing iodide from chloride and bromide ions."},{"text":"white","correct":false,"explanation":"A white precipitate is silver chloride, not silver iodide. Iodide ions react with silver ions to form a yellow precipitate of AgI."},{"text":"cream","correct":false,"explanation":"A cream precipitate is silver bromide, not silver iodide. The silver nitrate test produces a yellow precipitate when iodide ions are present."},{"text":"colourless","correct":false,"explanation":"Silver iodide is a yellow solid rather than a colourless mixture. Its visible yellow precipitate is the positive silver nitrate test for iodide ions."}]},{"id":"5109db87-61b1-4c65-bd54-32bd13d58190","cardType":"basic","points":2,"front":"Why acidify with dilute nitric acid before adding AgNO₃ after NaOH hydrolysis?","back":"It removes excess OH^-, preventing misleading silver precipitates.","mcqOptions":[{"text":"It removes excess OH^-, preventing misleading silver precipitates.","correct":true,"explanation":"Hydrolysis leaves excess hydroxide ions, which could react with added silver ions and create misleading precipitates. Dilute nitric acid neutralises this hydroxide without introducing halide ions."},{"text":"It removes excess halide ions, preventing misleading silver precipitates.","correct":false,"explanation":"Nitric acid does not remove the halide ions being tested; doing so would prevent the desired result. It neutralises excess OH^- that could form misleading silver precipitates."},{"text":"It increases the AgNO₃ concentration, producing clearer silver precipitates.","correct":false,"explanation":"Acidification is not intended to increase the AgNO₃ concentration. Dilute nitric acid neutralises residual OH^-, stopping hydroxide-derived silver precipitates from confusing the halide test."},{"text":"It catalyses the hydrolysis reaction, ensuring complete silver precipitation.","correct":false,"explanation":"The mixture is acidified after hydrolysis, so nitric acid does not catalyse the hydrolysis reaction. Its purpose is to remove excess OH^- before silver nitrate is added."}]},{"id":"bbaeb43b-d1b3-45b0-93c1-cade9ee49775","cardType":"basic","points":2,"front":"What does reflux allow when heating volatile reactants?","back":"Heating for longer without losing reactants or products.","mcqOptions":[{"text":"Heating for longer without losing reactants or products.","correct":true,"explanation":"A reflux condenser cools vapours so they condense and return to the reaction flask, permitting prolonged heating without volatile reactants or products escaping."},{"text":"Heating for longer while separating reactants from products.","correct":false,"explanation":"Reflux does not separate a mixture into collected fractions; that describes distillation. It condenses vapours back into the flask, allowing extended heating without material loss."},{"text":"Heating for longer by increasing evaporation of products.","correct":false,"explanation":"A reflux apparatus returns condensed vapour to the flask rather than increasing evaporation. This prevents volatile reactants or products from being lost during prolonged heating."},{"text":"Heating for longer below the mixture’s boiling point.","correct":false,"explanation":"Reflux does not keep the reaction mixture below its boiling point. It allows boiling while condensing escaping vapours and returning reactants or products to the flask."}]},{"id":"5ca1c25a-731f-44e4-b5cc-f29d7b37663e","cardType":"basic","points":1,"front":"What is the cooling-water direction in a Liebig condenser?","back":"In at the bottom, out at the top.","mcqOptions":[{"text":"In at the bottom, out at the top.","correct":true,"explanation":"Water enters at the bottom so the condenser jacket fills completely and provides efficient counter-current cooling before leaving from the top."},{"text":"In at the top, out at the bottom.","correct":false,"explanation":"Feeding water from the top can leave air pockets and an incompletely filled jacket. Water should enter at the bottom and leave at the top."},{"text":"In at the bottom, out at the bottom.","correct":false,"explanation":"Using only the bottom connection would not establish continuous cooling-water flow through the jacket. Water enters at the bottom and exits at the top."},{"text":"In at the top, out at the top.","correct":false,"explanation":"Using only the top connection would not keep the condenser jacket full with a steady flow. Water must enter at the bottom and leave at the top."}]},{"id":"d8375810-4d06-4293-87f4-38f114813f7a","cardType":"cloze","points":null,"front":"CFCs have no [ ], so they are unreactive in the lower atmosphere and can reach the [ ].","back":"CFCs have no C-H bonds, so they are unreactive in the lower atmosphere and can reach the stratosphere.","mcqOptions":[{"text":"CFCs have no C-H bonds, so they are unreactive in the lower atmosphere and can reach the stratosphere.","correct":true,"explanation":"Without C-H bonds, CFCs resist attack by radicals in the lower atmosphere. They survive long enough to reach the stratosphere, where ultraviolet radiation releases chlorine radicals."},{"text":"CFCs have no C-Cl bonds, so they are unreactive in the lower atmosphere and can reach the stratosphere.","correct":false,"explanation":"CFCs do contain C-Cl bonds; their absence of C-H bonds explains their lower-atmosphere stability. This allows them to reach the stratosphere before photodissociation occurs."},{"text":"CFCs have no C-H bonds, so they are unreactive in the lower atmosphere and can reach the troposphere.","correct":false,"explanation":"CFCs can already enter and pass through the troposphere, which is the lower atmosphere. Their lack of C-H bonds lets them survive and reach the stratosphere."},{"text":"CFCs have no C-F bonds, so they are unreactive in the lower atmosphere and can reach the mesosphere.","correct":false,"explanation":"CFCs contain strong C-F bonds, while their missing C-H bonds make them resistant to lower-atmosphere reactions. They consequently survive to reach the stratosphere, not the mesosphere."}]},{"id":"b0e2e1fe-ed30-4b21-a006-397ccef18bc9","cardType":"basic","points":2,"front":"Why can one chlorine radical destroy many ozone molecules?","back":"The chlorine radical is regenerated, so acts catalytically.","mcqOptions":[{"text":"The chlorine radical is regenerated, so acts catalytically.","correct":true,"explanation":"A chlorine radical reacts with ozone but is re-formed in a later propagation step. Because it is regenerated, it catalyses repeated ozone-destruction cycles."},{"text":"The chlorine radical is consumed, so acts stoichiometrically.","correct":false,"explanation":"Chlorine radicals are not permanently consumed in each ozone reaction. They are regenerated in the reaction cycle, allowing one radical to destroy many ozone molecules catalytically."},{"text":"The ozone molecule is regenerated, so acts catalytically.","correct":false,"explanation":"Ozone does not regenerate the chlorine radical by itself. The radical is re-formed during a later propagation reaction, enabling catalytic destruction of further ozone molecules."},{"text":"The chlorine radical remains unchanged, so acts catalytically.","correct":false,"explanation":"Chlorine radicals do not merely lower activation energy and remain unchanged in a single step. They react to form intermediates, then are regenerated in the overall catalytic cycle."}]}]}],"examCountdown":{"examDate":"2027-05-28","paperLabel":"Component 1"},"heading":"Halogenoalkanes","title":"Halogenoalkanes Flashcards","topbar":{"backHref":"/a-level/chemistry/eduqas/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"16a8acfe-2f2f-540e-8eef-b49d7e89be1a","walled":"$L39"}],"$L3a","$L3b"]