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24:[["$","$L33",null,{"feature":"flashcards"}],"$L34",["$","$L35",null,{"href":"/a-level/chemistry/ocr/a/carbonyl-compounds/flashcards","label":"Carbonyl compounds","subjectCode":"a_level_chemistry__ocr_a","surface":"flashcards"}],["$","$L36",null,{"runner":["$","div",null,{"className":"min-h-0 flex-1 overflow-hidden bg-background","children":["$","$L37",null,{"decks":[{"id":"0d2b5b43-909d-4586-a040-b3b3997162c4","title":"Carbonyl compounds Flashcards","topicId":"806e0e70-fddd-5988-9a32-b84a6b3c9be5","topicSlug":"carbonyl-compounds","cards":[{"id":"25561de2-01f2-40b3-8317-b5a479c602d4","cardType":"basic","points":1,"front":"A molecule contains a terminal –CHO group. What carbonyl compound type is it?","back":"Aldehyde.","mcqOptions":[{"text":"Aldehyde.","correct":true,"explanation":"A terminal –CHO group contains a carbonyl carbon bonded to hydrogen. Organic compounds with this functional group at the end of a chain are classified as aldehydes."},{"text":"Ketone.","correct":false,"explanation":"A terminal carbonyl group is not a ketone because one of its attachments is hydrogen rather than a second carbon group. The correct classification is an aldehyde."},{"text":"Carboxylic acid.","correct":false,"explanation":"Carboxylic acids contain a –COOH group, including an –OH attached to the carbonyl carbon. A terminal –CHO group instead identifies an aldehyde."},{"text":"Ester.","correct":false,"explanation":"Esters contain a carbonyl group bonded to an oxygen-containing carbon group, not hydrogen at the end of the chain. A terminal –CHO group is an aldehyde."}]},{"id":"7ea4fe53-2637-4780-a3f5-913f30cadce8","cardType":"basic","points":1,"front":"A carbonyl group is within a chain and bonded to two carbon groups. What type is it?","back":"Ketone.","mcqOptions":[{"text":"Ketone.","correct":true,"explanation":"A ketone contains a carbonyl group within a carbon chain, with the carbonyl carbon bonded to two carbon groups. Neither attachment is a hydrogen atom."},{"text":"Aldehyde.","correct":false,"explanation":"An aldehyde has a carbonyl carbon bonded to at least one hydrogen, usually at the end of a chain. Bonding to two carbon groups makes the compound a ketone."},{"text":"Ester.","correct":false,"explanation":"An ester has an oxygen atom between the carbonyl group and one carbon group, giving a –COO– linkage. A carbonyl bonded directly to two carbon groups is a ketone."},{"text":"Amide.","correct":false,"explanation":"An amide has a nitrogen-containing group attached to the carbonyl carbon. Two direct carbon attachments instead define a ketone."}]},{"id":"a5b6b5eb-6b70-432b-bcac-733a4e7f895a","cardType":"cloze","points":null,"front":"In a C=O bond, oxygen is [ ] and the carbonyl carbon is [ ].","back":"In a C=O bond, oxygen is δ− and the carbonyl carbon is δ+.","mcqOptions":[{"text":"In a C=O bond, oxygen is δ− and the carbonyl carbon is δ+.","correct":true,"explanation":"Oxygen attracts the shared electrons more strongly than carbon, so it gains a partial negative charge while the carbonyl carbon becomes partially positive."},{"text":"In a C=O bond, oxygen is δ+ and the carbonyl carbon is δ−.","correct":false,"explanation":"Reversing the partial charges ignores oxygen’s greater electronegativity. Oxygen should be δ− and the carbonyl carbon should be δ+ in the correct structure."},{"text":"In a C=O bond, oxygen is neutral and the carbonyl carbon is δ+.","correct":false,"explanation":"The carbonyl carbon is partially positive because oxygen withdraws electron density from the bond. Oxygen therefore cannot be neutral; it is δ− while carbon is δ+."},{"text":"In a C=O bond, oxygen is δ− and the carbonyl carbon is neutral.","correct":false,"explanation":"Although oxygen is correctly assigned a partial negative charge, the carbonyl carbon is not neutral. Electron withdrawal by oxygen makes it δ+."}]},{"id":"eed954f7-ffb2-41f9-95ad-34b12d40986e","cardType":"basic","points":2,"front":"Why do nucleophiles attack the carbonyl carbon?","back":"The carbonyl carbon is δ+ due to the polar C=O bond.","mcqOptions":[{"text":"The carbonyl carbon is δ+ due to the polar C=O bond.","correct":true,"explanation":"The C=O bond is polar because oxygen is more electronegative, leaving the carbonyl carbon partially positive. A nucleophile, which supplies electron density, is attracted to this site."},{"text":"The carbonyl carbon is δ− due to the polar C=O bond.","correct":false,"explanation":"Nucleophiles are electron-pair donors and are attracted to positive centres, not because the carbonyl carbon is electron-rich. The correct reason is its δ+ character from the polar bond."},{"text":"The carbonyl carbon is neutral due to the polar C=O bond.","correct":false,"explanation":"The carbonyl carbon is electrophilic rather than neutral because oxygen withdraws electron density. Its partial positive charge attracts nucleophiles."},{"text":"The carbonyl carbon is δ− due to the non-polar C=O bond.","correct":false,"explanation":"The carbonyl oxygen carries the partial negative charge, while the carbonyl carbon carries the partial positive charge. Nucleophiles therefore attack carbon, not because it is δ−."}]},{"id":"0a7c35d5-bff9-4125-91b7-047344d473b8","cardType":"cloze","points":null,"front":"Aldehydes and ketones commonly undergo [ ], where the C=O [ ].","back":"Aldehydes and ketones commonly undergo nucleophilic addition, where the C=O π bond opens.","mcqOptions":[{"text":"Aldehydes and ketones commonly undergo nucleophilic addition, where the C=O π bond opens.","correct":true,"explanation":"Nucleophilic addition involves attack on the electrophilic carbonyl carbon, followed by movement of the π-electron pair onto oxygen. The carbonyl π bond therefore opens."},{"text":"Aldehydes and ketones commonly undergo electrophilic substitution, where the C=O π bond opens.","correct":false,"explanation":"Aldehydes and ketones usually undergo addition at the polar carbonyl group, not electrophilic substitution. The π bond opens during nucleophilic addition."},{"text":"Aldehydes and ketones commonly undergo nucleophilic substitution, where the C=O σ bond breaks.","correct":false,"explanation":"The carbonyl π bond does not remain intact during the reaction, and the reaction is not nucleophilic substitution. It is nucleophilic addition involving opening of the π bond."},{"text":"Aldehydes and ketones commonly undergo nucleophilic addition, where the C=O σ bond breaks.","correct":false,"explanation":"The carbonyl π bond opens when nucleophiles add, rather than the σ bond breaking. Aldehydes and ketones therefore undergo nucleophilic addition."}]},{"id":"c075486a-6715-49c0-b3e0-91a8343887f6","cardType":"basic","points":1,"front":"In nucleophilic addition to C=O, where do the C=O π electrons move?","back":"To the oxygen atom.","mcqOptions":[{"text":"To the oxygen atom.","correct":true,"explanation":"When a nucleophile attacks the carbonyl carbon, the C=O π-electron pair moves onto oxygen. This accommodates the extra bonding at carbon and forms an alkoxide intermediate."},{"text":"To the carbonyl carbon.","correct":false,"explanation":"The π electrons cannot move to the nucleophile because they originate in the carbonyl bond. They move onto the oxygen atom, which becomes negatively charged temporarily."},{"text":"To the nucleophile.","correct":false,"explanation":"The incoming nucleophile forms a new bond to carbon; it does not receive the original π-electron pair. Those electrons move to oxygen."},{"text":"To the hydrogen atom.","correct":false,"explanation":"The carbonyl carbon is the site attacked by the nucleophile, but the π electrons move away from it onto oxygen. Confusing attack position with electron movement causes this error."}]},{"id":"0ad1aac6-e1bc-40f0-a342-964fe2c7259b","cardType":"basic","points":1,"front":"Complete the organic oxidation: RCHO + [O] → ?","back":"RCOOH.","mcqOptions":[{"text":"RCOOH.","correct":true,"explanation":"Oxidation of an aldehyde adds an oxygen atom overall, converting the –CHO group into the carboxylic acid group –COOH. The balanced organic equation is therefore RCHO + [O] → RCOOH."},{"text":"RCH₂OH.","correct":false,"explanation":"This product represents reduction of the aldehyde to a primary alcohol, involving addition of hydrogen rather than oxidation. Aldehyde oxidation gives the carboxylic acid RCOOH."},{"text":"RCOOR.","correct":false,"explanation":"An ester is not formed by simple oxidation of an aldehyde with one equivalent of [O]. The aldehyde functional group is oxidised to RCOOH."},{"text":"RCONH₂.","correct":false,"explanation":"An amide requires replacement or reaction involving nitrogen and is not the product of aldehyde oxidation. The carbonyl group becomes a carboxylic acid, RCOOH."}]},{"id":"1945e0ca-f2b0-42f5-a0a3-95b4b54fddce","cardType":"basic","points":2,"front":"What reagent and conditions oxidise aldehydes to carboxylic acids?","back":"Acidified K₂Cr₂O₇/dilute H₂SO₄, warmed under reflux.","mcqOptions":[{"text":"Acidified K₂Cr₂O₇/dilute H₂SO₄, warmed under reflux.","correct":true,"explanation":"Acidified potassium dichromate(VI) supplies the oxidising agent, while dilute sulfuric acid provides the acidic conditions. Reflux allows complete oxidation of the aldehyde to the carboxylic acid."},{"text":"NaBH₄/aqueous ethanol, warmed under reflux.","correct":false,"explanation":"Sodium borohydride is a reducing agent, so it converts aldehydes into primary alcohols rather than carboxylic acids. Acidified dichromate under reflux is required for complete oxidation."},{"text":"LiAlH₄/dry ether, warmed under reflux.","correct":false,"explanation":"Lithium aluminium hydride is a powerful reducing agent and would produce a primary alcohol. It cannot provide the oxidation needed to form a carboxylic acid."},{"text":"Acidified K₂Cr₂O₇/dilute H₂SO₄, distilled.","correct":false,"explanation":"Distillation removes the aldehyde as it forms and is used when preventing further oxidation is desired. Reflux with acidified dichromate instead promotes complete oxidation to the acid."}]},{"id":"81cdebb2-9a66-4184-8630-4365c985e529","cardType":"cloze","points":null,"front":"When aldehydes reduce dichromate(VI), the colour changes from [ ] to [ ].","back":"When aldehydes reduce dichromate(VI), the colour changes from orange to green.","mcqOptions":[{"text":"When aldehydes reduce dichromate(VI), the colour changes from orange to green.","correct":true,"explanation":"Acidified dichromate(VI) is orange and is reduced to chromium(III), which is green, when an aldehyde is oxidised to a carboxylic acid."},{"text":"When aldehydes reduce dichromate(VI), the colour changes from green to orange.","correct":false,"explanation":"This reverses the redox colour change. Dichromate(VI) starts orange, then reduction produces green chromium(III), so the correct change is from orange to green."},{"text":"When aldehydes reduce dichromate(VI), the colour changes from purple to green.","correct":false,"explanation":"Purple is associated with permanganate(VII), not dichromate(VI). Acidified dichromate(VI) begins orange and becomes green as chromium(VI) is reduced to chromium(III)."},{"text":"When aldehydes reduce dichromate(VI), the colour changes from orange to purple.","correct":false,"explanation":"The starting colour is correctly identified, but the product colour is not. Reduction of orange chromium(VI) gives green chromium(III), rather than purple manganese species."}]},{"id":"5883ce31-4210-43a6-831e-d76f14e457aa","cardType":"basic","points":2,"front":"What happens to ketones with acidified dichromate(VI), and why?","back":"

No reaction under normal A-Level conditions.
C-C bonds would need breaking.

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

No reaction under normal A-Level conditions.
C-C bonds would need breaking.

","correct":true,"explanation":"Ketones resist oxidation by acidified dichromate(VI) because oxidation would require breaking a carbon-carbon bond, unlike aldehyde oxidation to a carboxylic acid."},{"text":"

Reaction under normal A-Level conditions.
C-H bonds would need breaking.

","correct":false,"explanation":"Ketones do not normally react because their oxidation would require carbon-carbon bond cleavage, not merely removal of a hydrogen from a carbonyl carbon."},{"text":"

No reaction under normal A-Level conditions.
C=O bonds would need breaking.

","correct":false,"explanation":"The carbonyl bond is not the key obstacle in ketone oxidation. Further oxidation requires breaking a C-C bond, so acidified dichromate(VI) gives no reaction under normal conditions."},{"text":"

Reaction under normal A-Level conditions.
C-C bonds would need forming.

","correct":false,"explanation":"Oxidising a ketone would require carbon-carbon bond cleavage, not formation, and ketones therefore show no reaction with acidified dichromate(VI) under normal A-Level conditions."}]},{"id":"6e2f91bb-154e-4b7c-b101-dabbb799c395","cardType":"basic","points":1,"front":"What does NaBH₄ reduction of an aldehyde produce?","back":"A primary alcohol.","mcqOptions":[{"text":"A primary alcohol.","correct":true,"explanation":"An aldehyde contains one carbon group attached to the carbonyl carbon. Reduction adds hydrogen across the C=O bond, giving a primary alcohol."},{"text":"A secondary alcohol.","correct":false,"explanation":"A secondary alcohol is formed by reducing a ketone, whose carbonyl carbon is attached to two carbon groups. Aldehyde reduction instead gives a primary alcohol."},{"text":"A tertiary alcohol.","correct":false,"explanation":"Tertiary alcohols have three carbon groups attached to the carbon bearing the hydroxyl group. NaBH₄ reduction of an aldehyde produces a primary alcohol, not a tertiary one."},{"text":"A carboxylic acid.","correct":false,"explanation":"Conversion of an aldehyde to a carboxylic acid is oxidation, typically using acidified dichromate(VI). NaBH₄ is a reducing agent and produces a primary alcohol."}]},{"id":"e5793073-89cd-4d23-96a3-26b7a237bea0","cardType":"basic","points":1,"front":"What does NaBH₄ reduction of a ketone produce?","back":"A secondary alcohol.","mcqOptions":[{"text":"A secondary alcohol.","correct":true,"explanation":"A ketone has two carbon groups attached to its carbonyl carbon. NaBH₄ adds hydrogen across the C=O bond, producing an alcohol with two carbon substituents: a secondary alcohol."},{"text":"A primary alcohol.","correct":false,"explanation":"Primary alcohols result from reducing aldehydes, where the carbonyl carbon has one carbon substituent. A ketone has two, so its reduction gives a secondary alcohol."},{"text":"A tertiary alcohol.","correct":false,"explanation":"A tertiary alcohol requires three carbon groups attached to the carbon bearing the hydroxyl group. Ketone reduction creates only two such groups, giving a secondary alcohol."},{"text":"A carboxylic acid.","correct":false,"explanation":"Carboxylic acid formation is an oxidation pathway for aldehydes, not the reduction of ketones. NaBH₄ reduces the ketone carbonyl to a secondary alcohol."}]},{"id":"9950811f-41f0-4d4f-9040-94b2f7e2067b","cardType":"cloze","points":null,"front":"NaBH₄ is treated as a source of [ ], which act as the [ ].","back":"NaBH₄ is treated as a source of hydride ions, H⁻, which act as the nucleophile.","mcqOptions":[{"text":"NaBH₄ is treated as a source of hydride ions, H⁻, which act as the nucleophile.","correct":true,"explanation":"NaBH₄ supplies hydride ions, H−. These electron-rich ions attack the electron-deficient carbonyl carbon, so they act as nucleophiles in the addition mechanism."},{"text":"NaBH₄ is treated as a source of hydroxide ions, OH⁻, which act as the nucleophile.","correct":false,"explanation":"Hydroxide ions are not the reducing species supplied by NaBH₄. Hydride ions, H−, attack the carbonyl carbon as nucleophiles and lead to alcohol formation."},{"text":"NaBH₄ is treated as a source of hydride ions, H⁻, which act as the electrophile.","correct":false,"explanation":"Hydride ions have a lone pair and negative charge, making them electron-pair donors. They attack the carbonyl carbon as nucleophiles, while the carbonyl carbon is the electrophilic centre."},{"text":"NaBH₄ is treated as a source of hydrogen atoms, H, which act as the free radical.","correct":false,"explanation":"The mechanism is not a free-radical reaction involving neutral hydrogen atoms. NaBH₄ is treated as supplying hydride ions, H−, which act as nucleophiles."}]},{"id":"560141a7-40b1-4136-9f36-a3ce6e7f173d","cardType":"basic","points":2,"front":"What typical conditions are used for NaBH₄ reduction?","back":"NaBH₄ in water or aqueous ethanol at room temperature.","mcqOptions":[{"text":"NaBH₄ in water or aqueous ethanol at room temperature.","correct":true,"explanation":"NaBH₄ is sufficiently reactive to reduce aldehydes and ketones in water or aqueous ethanol at room temperature, providing a mild and practical reduction method."},{"text":"NaBH₄ in dry ether under reflux.","correct":false,"explanation":"Dry ether under reflux is not the typical A-Level condition for NaBH₄ reduction. The reagent is normally used in water or aqueous ethanol at room temperature."},{"text":"NaBH₄ in concentrated sulfuric acid at room temperature.","correct":false,"explanation":"Concentrated sulfuric acid is not the usual solvent and can react dangerously with borohydride. Standard conditions use water or aqueous ethanol at room temperature."},{"text":"NaBH₄ in aqueous ethanol under ultraviolet light.","correct":false,"explanation":"Ultraviolet light is unnecessary because NaBH₄ reduces the carbonyl directly under mild conditions. The usual setup is aqueous ethanol or water at room temperature."}]},{"id":"88f3deea-b1ff-424f-a7c8-663e05fc97ae","cardType":"basic","points":1,"front":"After H⁻ attacks a carbonyl and the π bond opens, what intermediate forms?","back":"An alkoxide ion.","mcqOptions":[{"text":"An alkoxide ion.","correct":true,"explanation":"Hydride attack pushes the carbonyl π-electrons onto oxygen, giving oxygen a negative charge. The resulting intermediate is an alkoxide ion before protonation forms the alcohol."},{"text":"A carbocation.","correct":false,"explanation":"The π-electrons move onto oxygen rather than leaving the carbon electron-deficient. This creates an alkoxide ion, not a positively charged carbocation."},{"text":"A carboxylate ion.","correct":false,"explanation":"A carboxylate ion contains two oxygen atoms and is associated with carboxylic acids. Carbonyl reduction initially gives a one-oxygen alkoxide ion before protonation."},{"text":"A protonated carbonyl.","correct":false,"explanation":"Protonation is not the immediate result of hydride attack. The carbonyl π bond opens onto oxygen, forming an alkoxide ion that is protonated in a later step."}]},{"id":"936504ae-c20a-457f-a8e2-22a5288bfaa3","cardType":"cloze","points":null,"front":"In HCN addition, [ ] produces the attacking [ ] ions.","back":"In HCN addition, NaCN(aq)/H⁺(aq) produces the attacking CN⁻ ions.","mcqOptions":[{"text":"In HCN addition, NaCN(aq)/H⁺(aq) produces the attacking CN⁻ ions.","correct":true,"explanation":"Aqueous sodium cyanide provides cyanide ions, while acid supplies HCN and maintains the appropriate equilibrium. Cyanide ions attack the electrophilic carbonyl carbon."},{"text":"In HCN addition, NaOH(aq)/H⁺(aq) produces the attacking OH⁻ ions.","correct":false,"explanation":"This describes hydroxide chemistry rather than HCN addition. The attacking species is CN−, produced using aqueous NaCN with acid present, and it forms a hydroxynitrile."},{"text":"In HCN addition, NaCN(aq)/H⁺(aq) produces the attacking H⁺ ions.","correct":false,"explanation":"Acid provides H+ but H+ does not attack the carbonyl as the nucleophile. NaCN supplies CN−, whose lone pair attacks the electrophilic carbonyl carbon."},{"text":"In HCN addition, KBr(aq)/H⁺(aq) produces the attacking Br⁻ ions.","correct":false,"explanation":"Bromide ions are not the nucleophile used in hydroxynitrile formation. Aqueous NaCN with acid provides CN−, which adds to the carbonyl carbon before protonation."}]},{"id":"283f5a89-6574-44d2-9dc9-9143beab3871","cardType":"basic","points":1,"front":"What compound type forms when HCN adds to an aldehyde or ketone?","back":"A hydroxynitrile.","mcqOptions":[{"text":"A hydroxynitrile.","correct":true,"explanation":"HCN adds across the carbonyl group, converting the C=O group into a hydroxyl group while the CN group remains attached to the same carbon, forming a hydroxynitrile."},{"text":"A hydroxycarboxylic acid.","correct":false,"explanation":"HCN addition does not form a hydroxycarboxylic acid; the nitrile group remains in the product. The correct compound type is a hydroxynitrile, containing both hydroxyl and nitrile groups."},{"text":"A nitrile.","correct":false,"explanation":"A cyanohydrin is another name sometimes used for this product, but it is not the classification expected here. The required compound type is a hydroxynitrile."},{"text":"An amino alcohol.","correct":false,"explanation":"HCN does not produce an amino alcohol because no nitrogen-hydrogen bond is introduced. The CN group stays as a nitrile, so the product is a hydroxynitrile."}]},{"id":"58588186-89e2-4b9d-9e67-3b99bf964c82","cardType":"basic","points":1,"front":"Through which atom does CN⁻ attack the carbonyl carbon?","back":"Through its carbon atom.","mcqOptions":[{"text":"Through its carbon atom.","correct":true,"explanation":"The carbon atom in CN⁻ has the lone pair used to form the new bond to the electrophilic carbonyl carbon, so cyanide attacks through carbon."},{"text":"Through its nitrogen atom.","correct":false,"explanation":"The nitrogen atom is not the attacking end in this reaction; using it would give a different connectivity. Cyanide forms the new carbon-carbon bond through its carbon atom."},{"text":"Through the oxygen atom.","correct":false,"explanation":"The carbonyl oxygen is part of the electrophilic molecule, not an atom in CN⁻ that attacks. Cyanide attacks through its own carbon atom."},{"text":"Through its hydrogen atom.","correct":false,"explanation":"The hydrogen atom belongs to HCN rather than CN⁻, and it does not form the new bond to the carbonyl carbon. Attack occurs through cyanide’s carbon atom."}]},{"id":"e452028b-19fe-48a2-8a89-4faf25040b8d","cardType":"basic","points":2,"front":"Why does HCN addition increase the carbon chain by one?","back":"The –CN group contains carbon.","mcqOptions":[{"text":"The –CN group contains carbon.","correct":true,"explanation":"The carbonyl compound gains a CN group during addition, and that group contributes one carbon atom to the molecule. Therefore, the carbon chain increases by one."},{"text":"The –OH group contains carbon.","correct":false,"explanation":"The hydroxyl group formed during the reaction contains oxygen and hydrogen, not carbon. The extra carbon comes from the CN group, which is added to the carbonyl carbon."},{"text":"The C=O group contains carbon.","correct":false,"explanation":"The carbonyl oxygen is retained as the hydroxyl oxygen, so it does not add a carbon atom. The chain length increases because the CN group contains carbon."},{"text":"The H atom contains carbon.","correct":false,"explanation":"The hydrogen from HCN may protonate the intermediate, but it cannot increase the carbon chain. The additional carbon is supplied by the CN group."}]},{"id":"8c27864d-dfa8-4c8f-965f-75bf108ec310","cardType":"cloze","points":null,"front":"With 2,4-DNPH, aldehydes and ketones give an [ ], showing a [ ] is present.","back":"With 2,4-DNPH, aldehydes and ketones give an orange or yellow precipitate, showing a carbonyl group is present.","mcqOptions":[{"text":"With 2,4-DNPH, aldehydes and ketones give an orange or yellow precipitate, showing a carbonyl group is present.","correct":true,"explanation":"2,4-DNPH reacts with aldehydes and ketones to form an orange or yellow precipitate. This positive test shows that a carbonyl group is present."},{"text":"With 2,4-DNPH, aldehydes and ketones give a blue solution, showing a carbon-carbon double bond is present.","correct":false,"explanation":"A blue solution is not the characteristic 2,4-DNPH result, and a carbon-carbon double bond is not what this test identifies. The correct result is an orange or yellow precipitate indicating a carbonyl group."},{"text":"With 2,4-DNPH, aldehydes and ketones give a silver mirror, showing a hydroxyl group is present.","correct":false,"explanation":"A silver mirror is produced by Tollens’ reagent with aldehydes, not by 2,4-DNPH. 2,4-DNPH gives an orange or yellow precipitate, showing a carbonyl group."},{"text":"With 2,4-DNPH, aldehydes and ketones give a white precipitate, showing a halogen atom is present.","correct":false,"explanation":"A white precipitate is associated with some halide tests, not the carbonyl test using 2,4-DNPH. The expected orange or yellow precipitate indicates a carbonyl group."}]},{"id":"5738e489-f95c-4048-a710-e28ccf63a116","cardType":"basic","points":1,"front":"What can 2,4-DNPH not distinguish after a positive result?","back":"Aldehydes from ketones.","mcqOptions":[{"text":"Aldehydes from ketones.","correct":true,"explanation":"A positive 2,4-DNPH test confirms a carbonyl group, but both aldehydes and ketones contain one. Therefore, the test cannot identify which of these two types is present."},{"text":"Carbonyl compounds from non-carbonyl compounds.","correct":false,"explanation":"2,4-DNPH can distinguish compounds with and without a carbonyl group because aldehydes and ketones form a precipitate. It cannot distinguish aldehydes from ketones after a positive result."},{"text":"Aromatic aldehydes from aliphatic aldehydes.","correct":false,"explanation":"The test does not distinguish aldehydes according to whether they are aromatic or aliphatic. Its limitation is that a positive result cannot separate aldehydes from ketones."},{"text":"Ketones from non-carbonyl compounds.","correct":false,"explanation":"A positive 2,4-DNPH result is useful for detecting a carbonyl group, so it does distinguish carbonyl compounds from many non-carbonyl compounds. The unresolved distinction is aldehydes versus ketones."}]},{"id":"a1289857-8e70-4adb-8fcb-2b688a38b9f3","cardType":"basic","points":3,"front":"How can the original carbonyl compound be identified from its 2,4-DNPH derivative?","back":"

Filter/purify the solid derivative.
Measure its melting point.
Compare with data-book values.

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

Filter/purify the solid derivative.
Measure its melting point.
Compare with data-book values.

","correct":true,"explanation":"The solid 2,4-DNPH derivative is purified, its melting point is measured, and that value is compared with data-book values to identify the original carbonyl compound."},{"text":"

Distil/purify the original compound.
Measure its boiling point.
Compare with density values.

","correct":false,"explanation":"Boiling the derivative and measuring the original liquid’s density does not provide the characteristic identification method. The derivative should be purified, its melting point measured, and the value compared with data."},{"text":"

Filter/purify the solid derivative.
Measure its colour.
Compare with mass values.

","correct":false,"explanation":"Measuring colour and recording the derivative’s mass are not sufficiently distinctive for identification. The accepted method uses a purified derivative’s melting point compared with data-book values."},{"text":"

Dissolve/purify the solid derivative.
Measure its pH.
Compare with acidity values.

","correct":false,"explanation":"The derivative is not identified by dissolving it and measuring pH. Purification followed by melting-point measurement and comparison with data-book values gives the compound’s identity."}]},{"id":"9c80770a-5b97-444b-9290-58de53d25356","cardType":"cloze","points":null,"front":"Tollens’ reagent is [ ]; aldehydes give a [ ].","back":"Tollens’ reagent is ammoniacal silver nitrate; aldehydes give a silver mirror or grey silver precipitate.","mcqOptions":[{"text":"Tollens’ reagent is ammoniacal silver nitrate; aldehydes give a silver mirror or grey silver precipitate.","correct":true,"explanation":"Tollens’ reagent is ammoniacal silver nitrate. Aldehydes reduce silver ions to silver, producing a silver mirror or, under some conditions, a grey silver precipitate."},{"text":"Tollens’ reagent is alkaline copper(II) solution; aldehydes give a brick-red copper(I) oxide precipitate.","correct":false,"explanation":"Fehling’s solution is an alkaline copper(II) reagent and gives a brick-red precipitate with suitable aldehydes. Tollens’ reagent is ammoniacal silver nitrate and gives silver metal."},{"text":"Tollens’ reagent is acidified 2,4-DNPH solution; aldehydes give an orange or yellow precipitate.","correct":false,"explanation":"2,4-DNPH is used to detect carbonyl compounds and produces an orange or yellow precipitate. Tollens’ reagent is ammoniacal silver nitrate, giving silver with aldehydes."},{"text":"Tollens’ reagent is aqueous bromine; aldehydes give a colourless solution.","correct":false,"explanation":"Bromine water tests for certain unsaturated compounds and is not Tollens’ reagent. The correct reagent is ammoniacal silver nitrate, which gives a silver mirror or grey silver precipitate."}]},{"id":"143c13be-3e4f-4e67-a017-b463ab49ed1a","cardType":"basic","points":1,"front":"What result do ketones give with Tollens’ reagent?","back":"No reaction.","mcqOptions":[{"text":"No reaction.","correct":true,"explanation":"Ketones generally resist oxidation by Tollens’ reagent under the test conditions, so they do not reduce silver ions and produce no visible reaction."},{"text":"A silver mirror.","correct":false,"explanation":"A silver mirror is the positive Tollens’ test for an aldehyde, which reduces silver ions to silver. Ketones generally do not react under these conditions."},{"text":"A grey silver precipitate.","correct":false,"explanation":"Ketones do not usually reduce Tollens’ reagent to form a grey silver precipitate. This result is another possible positive observation for aldehydes, not ketones."},{"text":"An orange or yellow precipitate.","correct":false,"explanation":"An orange or yellow precipitate is associated with the 2,4-DNPH carbonyl test, not Tollens’ reagent. Ketones give no reaction with Tollens’ reagent."}]}]}],"examCountdown":{"examDate":"2027-05-28","paperLabel":"Paper 1"},"heading":"Carbonyl compounds","title":"Carbonyl compounds Flashcards","topbar":{"backHref":"/a-level/chemistry/ocr/a/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"806e0e70-fddd-5988-9a32-b84a6b3c9be5","walled":"$L39"}],"$L3a","$L3b"]