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Aldehydes: at the end of the carbon chain
Ketones: in the middle of the carbon chain

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

Aldehydes: at the end of the carbon chain
Ketones: in the middle of the carbon chain

","correct":true,"explanation":"An aldehyde’s carbonyl carbon is bonded to at least one hydrogen, placing the group at a chain end. A ketone’s carbonyl carbon is bonded to two carbon groups."},{"text":"

Aldehydes: in the middle of the carbon chain
Ketones: at the end of the carbon chain

","correct":false,"explanation":"This reverses the functional-group positions. Aldehydes have a terminal carbonyl group, while ketones have a carbonyl carbon bonded to two other carbons within the chain."},{"text":"

Aldehydes: at the end of the carbon chain
Ketones: at the end of the carbon chain

","correct":false,"explanation":"A terminal carbonyl group characterises an aldehyde, not a ketone. In a ketone, the carbonyl carbon is bonded to two carbon groups and therefore lies within the chain."},{"text":"

Aldehydes: in the middle of the carbon chain
Ketones: in the middle of the carbon chain

","correct":false,"explanation":"This incorrectly treats both functional groups as internal. Ketones have carbonyl groups within carbon chains, whereas aldehydes have carbonyl groups at chain ends."}]},{"id":"0e3f93a8-e65d-426c-ac9e-e52fdf0dcb89","cardType":"cloze","points":null,"front":"In the $C=O$ bond, the oxygen is more electronegative, giving the carbon a $\\text{[...]}$ charge and the oxygen a $\\delta^-$ charge.","back":"In the $C=O$ bond, the oxygen is more electronegative, giving the carbon a $\\boxed{\\delta^+}$ charge and the oxygen a $\\delta^-$ charge.","mcqOptions":[{"text":"$$\\delta^+$","correct":true,"explanation":"Oxygen attracts the bonding electrons more strongly than carbon, so electron density moves towards oxygen. The electron-deficient carbon consequently carries a partial positive charge, $\\delta^+$."},{"text":"$$\\delta^-$","correct":false,"explanation":"This reverses the polarity of the carbonyl bond. Oxygen is more electronegative and becomes $\\delta^-$, leaving the carbon electron-deficient and therefore $\\delta^+$."},{"text":"$$0$","correct":false,"explanation":"This assumes the bonding electrons are shared equally. Oxygen is more electronegative than carbon, so the bond is polar and carbon carries a $\\delta^+$ charge."},{"text":"$$\\delta^{2+}$","correct":false,"explanation":"This mistakes a partial charge for a full ionic charge. The carbonyl bond is polar covalent, so carbon has the partial positive charge $\\delta^+$."}]},{"id":"7b690871-ca4a-497f-ba17-961773984d8c","cardType":"cloze","points":null,"front":"In the $C=O$ bond, the oxygen is more electronegative, giving the carbon a $\\delta^+$ charge and the oxygen a $\\text{[...]}$ charge.","back":"In the $C=O$ bond, the oxygen is more electronegative, giving the carbon a $\\delta^+$ charge and the oxygen a $\\boxed{\\delta^-}$ charge.","mcqOptions":[{"text":"$$\\delta^-$","correct":true,"explanation":"Oxygen attracts the shared electrons in the carbonyl bond more strongly than carbon. Its increased electron density gives it a partial negative charge, represented by $\\delta^-$."},{"text":"$$\\delta^+$","correct":false,"explanation":"This reverses the carbonyl bond’s polarity. Oxygen attracts electron density because it is more electronegative, so oxygen is $\\delta^-$ and carbon is $\\delta^+$."},{"text":"$$0$","correct":false,"explanation":"This assumes no electronegativity difference between carbon and oxygen. Oxygen attracts the bonding electrons more strongly, so it carries the partial negative charge $\\delta^-$."},{"text":"$$\\delta^{2-}$","correct":false,"explanation":"This treats oxygen as if it had gained two electrons to form an oxide ion. In a polar covalent carbonyl bond, oxygen instead carries $\\delta^-$."}]},{"id":"6b82b6a4-64c7-4d84-8ace-3a20dceea025","cardType":"basic","points":1,"front":"What is the general mechanism type for reactions of aldehydes and ketones with $H^-$ or $CN^-$?","back":"Nucleophilic addition","mcqOptions":[{"text":"Nucleophilic addition","correct":true,"explanation":"The negatively charged nucleophile attacks the electron-deficient carbonyl carbon, and the $C=O$ double bond opens as its $\\pi$ electrons move onto oxygen."},{"text":"Electrophilic addition","correct":false,"explanation":"Electrophilic addition is associated with electron-rich multiple bonds such as $C=C$. Carbonyl carbon is electron-deficient, so $H^-$ or $CN^-$ attacks as a nucleophile in nucleophilic addition."},{"text":"Nucleophilic substitution","correct":false,"explanation":"Substitution requires a leaving group to be replaced, but aldehydes and ketones have no suitable leaving group on the carbonyl carbon. The reaction is nucleophilic addition."},{"text":"Electrophilic substitution","correct":false,"explanation":"Electrophilic substitution commonly occurs in aromatic systems and involves an electrophile replacing another group. Here a nucleophile attacks the polar carbonyl group, giving nucleophilic addition."}]},{"id":"f9323eb8-bbea-429c-ada2-74db11bdc9ea","cardType":"basic","points":1,"front":"What is observed when Tollens' reagent is warmed with an aldehyde?","back":"A silver mirror forms on the inside of the test tube.","mcqOptions":[{"text":"A silver mirror forms on the inside of the test tube.","correct":true,"explanation":"The aldehyde reduces the silver(I) ions in Tollens’ reagent to metallic silver while being oxidised to a carboxylate ion, producing the characteristic silver mirror."},{"text":"A brick-red precipitate forms on the inside of the test tube.","correct":false,"explanation":"A brick-red precipitate is the positive result with Fehling’s solution, caused by copper(I) oxide. Tollens’ reagent is reduced to metallic silver, forming a silver mirror."},{"text":"A white precipitate forms on the inside of the test tube.","correct":false,"explanation":"A white precipitate is not the positive aldehyde result with Tollens’ reagent. The aldehyde reduces silver(I) ions to metallic silver, which coats the tube as a silver mirror."},{"text":"A purple solution forms on the inside of the test tube.","correct":false,"explanation":"The purple colour is associated with manganate(VII) oxidising agents, not Tollens’ reagent. A positive Tollens’ test deposits metallic silver as a mirror inside the tube."}]},{"id":"fbe46481-6c43-46a0-96ab-2aae0314b53c","cardType":"basic","points":1,"front":"What is observed when Fehling's solution is warmed with an aldehyde?","back":"The blue solution forms a brick-red precipitate.","mcqOptions":[{"text":"The blue solution forms a brick-red precipitate.","correct":true,"explanation":"The aldehyde reduces blue copper(II) ions in Fehling’s solution to copper(I) oxide, $Cu_2O$, which appears as the characteristic brick-red precipitate."},{"text":"The blue solution forms a silver mirror.","correct":false,"explanation":"A silver mirror is produced by Tollens’ reagent, not Fehling’s solution. Fehling’s copper(II) ions are reduced to brick-red copper(I) oxide by an aldehyde."},{"text":"The blue solution forms a white precipitate.","correct":false,"explanation":"A white precipitate is not formed in a positive Fehling’s test. The aldehyde reduces copper(II) ions to copper(I) oxide, producing a brick-red precipitate."},{"text":"The blue solution forms a green precipitate.","correct":false,"explanation":"Green may appear during incomplete colour change, but it is not the final positive observation. Sufficient warming with an aldehyde produces a brick-red copper(I) oxide precipitate."}]},{"id":"6885f57a-011c-4c81-bfdf-7187a27317a3","cardType":"basic","points":2,"front":"Why can aldehydes be distinguished from ketones using mild oxidising agents?","back":"Aldehydes have a hydrogen attached to the carbonyl carbon and can be oxidised; ketones cannot.","mcqOptions":[{"text":"Aldehydes have a hydrogen attached to the carbonyl carbon and can be oxidised; ketones cannot.","correct":true,"explanation":"The hydrogen bonded to an aldehyde’s carbonyl carbon allows mild oxidation to a carboxylic acid or carboxylate. Ketones lack this hydrogen and resist oxidation under these conditions."},{"text":"Ketones have a hydrogen attached to the carbonyl carbon and can be oxidised; aldehydes cannot.","correct":false,"explanation":"This reverses the structures and reactivities. Aldehydes have the hydrogen on the carbonyl carbon and undergo mild oxidation, whereas ketones have two carbon groups and do not."},{"text":"Aldehydes have a hydrogen attached to the carbonyl carbon and can be reduced; ketones cannot.","correct":false,"explanation":"Both aldehydes and ketones can be reduced to alcohols, so reduction does not distinguish them. Aldehydes can instead be oxidised by mild oxidising agents, while ketones cannot."},{"text":"Aldehydes have a hydroxyl group attached to the carbonyl carbon and can be hydrolysed; ketones cannot.","correct":false,"explanation":"A hydroxyl group attached to a carbonyl carbon describes a carboxylic acid, not an aldehyde. Aldehydes possess a hydrogen there and can be oxidised under mild conditions."}]},{"id":"944c92b4-2fb1-4e6c-87b9-2c88491cd22d","cardType":"basic","points":2,"front":"Identify the reagent and solvent used to reduce carbonyls to alcohols.","back":"$NaBH_4$ (sodium tetrahydridoborate(III)) in aqueous solution.","mcqOptions":[{"text":"$NaBH_4$ (sodium tetrahydridoborate(III)) in aqueous solution.","correct":true,"explanation":"Sodium tetrahydridoborate(III) supplies hydride ions that attack the carbonyl carbon. It can be used in aqueous solution to reduce aldehydes and ketones to alcohols."},{"text":"$LiAlH_4$ (lithium tetrahydridoaluminate(III)) in aqueous solution.","correct":false,"explanation":"Although lithium tetrahydridoaluminate(III) is a reducing agent, it reacts violently with water and cannot be used in aqueous solution. The required combination is $NaBH_4$ in aqueous solution."},{"text":"$K_2Cr_2O_7$ (potassium dichromate(VI)) in aqueous solution.","correct":false,"explanation":"Potassium dichromate(VI) is an oxidising agent, so it does not reduce carbonyl compounds to alcohols. The appropriate reducing agent is $NaBH_4$ used in aqueous solution."},{"text":"$HCN$ (hydrogen cyanide) in ethanolic solution.","correct":false,"explanation":"Hydrogen cyanide undergoes nucleophilic addition to form a hydroxynitrile rather than reducing the carbonyl to an alcohol. Carbonyl reduction uses $NaBH_4$ in aqueous solution."}]},{"id":"a1dfbfd1-e36b-497b-90bb-efd00d5c9e40","cardType":"basic","points":2,"front":"State the classification of the alcohol produced when reducing aldehydes vs. ketones.","back":"

Aldehydes: primary ($1^o$) alcohols
Ketones: secondary ($2^o$) alcohols

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

Aldehydes: primary ($1^o$) alcohols
Ketones: secondary ($2^o$) alcohols

","correct":true,"explanation":"Reduction converts the carbonyl group into an alcohol group. An aldehyde forms a primary alcohol, whereas a ketone forms a secondary alcohol because its carbonyl carbon is bonded to two carbon groups."},{"text":"

Aldehydes: secondary ($2^o$) alcohols
Ketones: primary ($1^o$) alcohols

","correct":false,"explanation":"This reverses the classifications. Aldehydes have one carbon group attached to the carbonyl carbon and form primary alcohols, while ketones have two carbon groups and form secondary alcohols."},{"text":"

Aldehydes: primary ($1^o$) alcohols
Ketones: tertiary ($3^o$) alcohols

","correct":false,"explanation":"Reduction does not add another carbon group to a ketone, so a tertiary alcohol cannot form. Ketones retain their two carbon groups and therefore produce secondary alcohols."},{"text":"

Aldehydes: secondary ($2^o$) alcohols
Ketones: tertiary ($3^o$) alcohols

","correct":false,"explanation":"This incorrectly raises each alcohol classification by one. Reduction adds hydrogen across the carbonyl group without adding carbon groups, giving primary alcohols from aldehydes and secondary alcohols from ketones."}]},{"id":"5da27a22-0cb6-447a-bc0a-d8e02107d36f","cardType":"basic","points":2,"front":"Write the balanced equation for the reduction of propanone using $[H]$.","back":"$$$CH_3COCH_3 + 2[H] \\to CH_3CH(OH)CH_3$$","mcqOptions":[{"text":"$$$CH_3COCH_3 + 2[H] \\to CH_3CH(OH)CH_3$$","correct":true,"explanation":"Reduction adds two hydrogen atoms across the carbonyl group: one bonds to the carbonyl carbon and the other converts the carbonyl oxygen into an $-OH$ group."},{"text":"$$$CH_3COCH_3 + [H] \\to CH_3CH(OH)CH_3$$","correct":false,"explanation":"Only one hydrogen equivalent has been included, so the hydrogen atoms are not balanced. Carbonyl reduction requires $2[H]$ to form the secondary alcohol propan-2-ol."},{"text":"$$$CH_3COCH_3 + 2[H] \\to CH_3CH_2CH_2OH$$","correct":false,"explanation":"This treats propanone as though reduction could rearrange its carbon skeleton. The carbonyl carbon remains bonded to both methyl groups, so the product is $CH_3CH(OH)CH_3$."},{"text":"$$$CH_3COCH_3 + 2[H] \\to CH_3CH(OH)CH_3 + H_2O$$","correct":false,"explanation":"Adding water as a product makes the equation unbalanced and confuses reduction with a condensation reaction. Both the original oxygen and the two added hydrogens are contained in propan-2-ol."}]},{"id":"90b9d801-89c3-4162-98a5-7491e4cc86d2","cardType":"cloze","points":null,"front":"In the reduction of carbonyls, $NaBH_4$ acts as a source of $\\text{[...]}$ (hydride) ions.","back":"In the reduction of carbonyls, $NaBH_4$ acts as a source of $\\boxed{H^-}$ (hydride) ions.","mcqOptions":[{"text":"$$H^-$","correct":true,"explanation":"Sodium borohydride supplies hydride ions, $H^-$, which act as nucleophiles. A hydride ion attacks the electron-deficient carbon atom of the polar carbonyl group."},{"text":"$$H^+$","correct":false,"explanation":"This confuses hydride ions with protons. $NaBH_4$ supplies nucleophilic $H^-$ ions, which attack the partially positive carbonyl carbon during reduction."},{"text":"$$OH^-$","correct":false,"explanation":"Hydroxide ions are not the reducing species supplied by sodium borohydride. $NaBH_4$ provides $H^-$ ions, which add to the carbonyl carbon."},{"text":"$$CN^-$","correct":false,"explanation":"Cyanide ions are nucleophiles used to form hydroxynitriles, not to reduce carbonyl compounds. Sodium borohydride acts as a source of hydride ions, $H^-$."}]},{"id":"42672274-fc4c-4855-baca-6b5e50b62078","cardType":"cloze","points":null,"front":"The reaction of a carbonyl with $KCN$ followed by dilute acid produces a [ ], which contains both an $-OH$ group and a $-CN$ group.","back":"The reaction of a carbonyl with $KCN$ followed by dilute acid produces a hydroxynitrile, which contains both an $-OH$ group and a $-CN$ group.","mcqOptions":[{"text":"hydroxynitrile","correct":true,"explanation":"Cyanide adds to the carbonyl carbon and subsequent protonation forms an $-OH$ group. The product contains both hydroxyl and nitrile groups, so it is a hydroxynitrile."},{"text":"hydroxyketone","correct":false,"explanation":"This overlooks the nitrile group introduced by the $CN^-$ nucleophile. The product contains both $-OH$ and $-CN$ groups and is therefore called a hydroxynitrile."},{"text":"carboxylic acid","correct":false,"explanation":"Carboxylic acids form through oxidation of suitable carbonyl compounds, not cyanide addition. Reaction with $KCN$ followed by dilute acid produces a hydroxynitrile containing $-OH$ and $-CN$."},{"text":"halogenoalkane","correct":false,"explanation":"No halogen atom is introduced in this reaction. Cyanide addition followed by protonation gives a hydroxynitrile, identified by the presence of both $-OH$ and $-CN$ groups."}]},{"id":"955d0592-ee95-434a-ad60-3e753cc22152","cardType":"cloze","points":null,"front":"The reaction of a carbonyl with $KCN$ followed by dilute acid produces a hydroxynitrile, which contains both an $\\text{[...]}$ group and a $-CN$ group.","back":"The reaction of a carbonyl with $KCN$ followed by dilute acid produces a hydroxynitrile, which contains both an $\\boxed{-OH}$ group and a $-CN$ group.","mcqOptions":[{"text":"$$-OH$","correct":true,"explanation":"After $CN^-$ attacks the carbonyl carbon, the negatively charged oxygen is protonated by the acid. This converts the original carbonyl oxygen into an $-OH$ group."},{"text":"$$-COOH$","correct":false,"explanation":"This confuses cyanide addition with oxidation to a carboxylic acid. Protonation of the carbonyl oxygen produces an $-OH$ group, while the attacking nucleophile supplies the $-CN$ group."},{"text":"$$-CHO$","correct":false,"explanation":"The carbonyl group is consumed during nucleophilic addition, so an aldehyde group is not retained. The product instead contains an $-OH$ group alongside the newly added $-CN$ group."},{"text":"$$-NH_2$","correct":false,"explanation":"No nitrogen-containing amino group is formed because the nitrogen remains within the nitrile group. Protonation produces an $-OH$ group on the former carbonyl carbon."}]},{"id":"293dac1d-53c4-4470-8e30-b355c8b57348","cardType":"cloze","points":null,"front":"The reaction of a carbonyl with $KCN$ followed by dilute acid produces a hydroxynitrile, which contains both an $-OH$ group and a $\\text{[...]}$ group.","back":"The reaction of a carbonyl with $KCN$ followed by dilute acid produces a hydroxynitrile, which contains both an $-OH$ group and a $\\boxed{-CN}$ group.","mcqOptions":[{"text":"$$-CN$","correct":true,"explanation":"The cyanide ion attacks through its carbon atom and becomes attached to the former carbonyl carbon. The product therefore contains a nitrile, $-CN$, group alongside the $-OH$ group."},{"text":"$$-COOH$","correct":false,"explanation":"A carboxyl group would arise from oxidation or hydrolysis under different conditions. Cyanide addition introduces a nitrile group, $-CN$, while protonation creates the $-OH$ group."},{"text":"$$-NH_2$","correct":false,"explanation":"The nitrogen supplied by cyanide remains triple-bonded within the nitrile group rather than becoming an amino group. The second functional group in the product is $-CN$."},{"text":"$$-CHO$","correct":false,"explanation":"The original carbonyl group reacts and is not retained as an aldehyde group. Cyanide addition introduces $-CN$, producing a molecule containing both $-OH$ and $-CN$ groups."}]},{"id":"1e104eff-9001-4103-9eb4-db65d9fdbbf7","cardType":"basic","points":2,"front":"Why is a mixture of $KCN$ and dilute acid preferred over $HCN$ gas?","back":"

$HCN$ is a highly toxic gas
$KCN$ provides a higher concentration of $CN^-$ nucleophiles

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

$HCN$ is a highly toxic gas
$KCN$ provides a higher concentration of $CN^-$ nucleophiles

","correct":true,"explanation":"Using $HCN$ gas directly creates a severe inhalation hazard, and weakly ionised $HCN$ supplies relatively few cyanide ions. $KCN$ provides a higher concentration of the required $CN^-$ nucleophile."},{"text":"

$HCN$ is a highly corrosive gas
$KCN$ provides a lower concentration of $CN^-$ nucleophiles

","correct":false,"explanation":"This mistakes the principal hazard and reverses the concentration comparison. $HCN$ is avoided because it is highly toxic, while $KCN$ supplies a higher concentration of $CN^-$ nucleophiles."},{"text":"

$HCN$ is a highly flammable gas
$KCN$ provides a higher concentration of $H^+$ nucleophiles

","correct":false,"explanation":"The key hazard is $HCN$'s extreme toxicity, not flammability, and $H^+$ is an electrophile rather than a nucleophile. $KCN$ supplies a higher concentration of $CN^-$ nucleophiles."},{"text":"

$HCN$ is a chemically unreactive gas
$KCN$ provides a lower activation energy for the reaction

","correct":false,"explanation":"The preference is not based on inertness or catalysis. $HCN$ is a highly toxic gas, whereas $KCN$ provides a higher concentration of cyanide nucleophiles for addition."}]},{"id":"f5eb5407-0008-453b-83ef-212a6ef640e3","cardType":"basic","points":1,"front":"How is carbon-1 identified when naming a hydroxynitrile?","back":"The carbon in the $-CN$ group is always carbon-1.","mcqOptions":[{"text":"The carbon in the $-CN$ group is always carbon-1.","correct":true,"explanation":"In nitrile nomenclature, the carbon atom of the $-C\\equiv N$ group is included in the parent chain. It receives the lowest possible number and is designated carbon-1."},{"text":"The carbon bonded to the $-CN$ group is always carbon-1.","correct":false,"explanation":"This begins numbering one carbon too late. The carbon within the $-CN$ group itself is included in the parent chain and is always designated carbon-1."},{"text":"The carbon bonded to the $-OH$ group is always carbon-1.","correct":false,"explanation":"This incorrectly gives numbering priority to the hydroxyl-bearing carbon. For a hydroxynitrile, the carbon within the nitrile group is carbon-1, and numbering continues from there."},{"text":"The carbon furthest from the $-CN$ group is always carbon-1.","correct":false,"explanation":"This reverses the required direction of numbering. The nitrile carbon is part of the parent chain and must be carbon-1, so numbering starts at the $-CN$ end."}]},{"id":"7dd70728-de38-43b7-8c22-6b8e27b7a889","cardType":"basic","points":3,"front":"Why does the nucleophilic addition of $CN^-$ to an aldehyde usually produce a racemic mixture?","back":"

The carbonyl group is planar
$CN^-$ attacks equally from above or below the plane
An equal amount of each enantiomer is formed

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

The carbonyl group is planar
$CN^-$ attacks equally from above or below the plane
An equal amount of each enantiomer is formed

","correct":true,"explanation":"The planar carbonyl group can be attacked with equal probability from either side. If a new chiral centre forms, the two possible attacks produce equal amounts of opposite enantiomers."},{"text":"

The carbonyl group is tetrahedral
$CN^-$ attacks equally from either carbon atom
An equal amount of each enantiomer is formed

","correct":false,"explanation":"This confuses the shape of the product with that of the reactant. The carbonyl group is planar, allowing attack from above or below its plane to form equal amounts of enantiomers."},{"text":"

The carbonyl group is planar
$CN^-$ attacks only from above the plane
An equal amount of each enantiomer is formed

","correct":false,"explanation":"Attack from only one side would favour one stereoisomer rather than produce a racemic mixture. In an achiral environment, cyanide attacks equally from above and below the planar carbonyl group."},{"text":"

The carbonyl group is planar
$CN^-$ attacks equally from above or below the plane
An unequal amount of each enantiomer is formed

","correct":false,"explanation":"Equal attack on the two faces of a planar carbonyl produces equal, not unequal, amounts of the two enantiomers. That 1:1 composition is what defines a racemic mixture."}]},{"id":"c7ac1d66-b33d-436d-9e40-9f39b15a1bbe","cardType":"basic","points":2,"front":"Why is the resulting mixture from the reaction of propanal and $KCN$ optically inactive?","back":"It is a racemic mixture where the effects of the two enantiomers on plane-polarised light cancel out.","mcqOptions":[{"text":"It is a racemic mixture where the effects of the two enantiomers on plane-polarised light cancel out.","correct":true,"explanation":"A racemic mixture contains equal amounts of two enantiomers. They rotate plane-polarised light by equal angles in opposite directions, so their optical effects cancel completely."},{"text":"It is a racemic mixture where the effects of the two enantiomers on plane-polarised light reinforce each other.","correct":false,"explanation":"Enantiomers rotate plane-polarised light in opposite directions, not the same direction. Equal quantities therefore cancel one another’s rotations, making the racemic mixture optically inactive."},{"text":"It is an achiral mixture where none of the molecules can rotate plane-polarised light at all.","correct":false,"explanation":"The individual products are chiral and each enantiomer is optically active. The mixture is inactive only because equal and opposite rotations from the two enantiomers cancel."},{"text":"It is an unequal mixture where the effects of the two enantiomers on plane-polarised light cancel out.","correct":false,"explanation":"Unequal amounts of enantiomers would leave a net optical rotation because one enantiomer’s effect would predominate. Complete cancellation requires an equal, racemic mixture."}]},{"id":"991c71d1-0eee-4037-ac00-21819217a5b9","cardType":"cloze","points":null,"front":"During the mechanism of reduction, the $C=O$ double bond breaks to form an intermediate [ ] ion with an $O^-$ charge.","back":"During the mechanism of reduction, the $C=O$ double bond breaks to form an intermediate alkoxide ion with an $O^-$ charge.","mcqOptions":[{"text":"alkoxide","correct":true,"explanation":"Hydride addition breaks the carbonyl π bond and transfers its electrons to oxygen. This produces an alkoxide ion containing a negatively charged oxygen atom before protonation."},{"text":"carbocation","correct":false,"explanation":"A carbocation has a positively charged carbon and is not the intermediate formed here. Carbonyl reduction produces an alkoxide ion because the π electrons move onto oxygen."},{"text":"carbanion","correct":false,"explanation":"This confuses the location of the negative charge. When the carbonyl π bond breaks during reduction, its electrons move to oxygen, producing an alkoxide ion rather than a carbanion."},{"text":"oxonium","correct":false,"explanation":"An oxonium ion has a positively charged oxygen, commonly after protonation. The immediate reduction intermediate has an $O^-$ charge and is therefore an alkoxide ion."}]},{"id":"b697ad54-a491-4d5d-a79e-702578a937fd","cardType":"cloze","points":null,"front":"During the mechanism of reduction, the $C=O$ double bond breaks to form an intermediate alkoxide ion with an $\\text{[...]}$ charge.","back":"During the mechanism of reduction, the $C=O$ double bond breaks to form an intermediate alkoxide ion with an $\\boxed{O^-}$ charge.","mcqOptions":[{"text":"$$O^-$","correct":true,"explanation":"When the carbonyl π bond breaks, its electron pair moves onto the electronegative oxygen atom. Oxygen therefore carries a negative charge in the alkoxide intermediate."},{"text":"$$O^+$","correct":false,"explanation":"This reverses the charge produced by movement of the carbonyl π electrons. Those electrons move onto oxygen, so the alkoxide intermediate contains $O^-$ rather than $O^+$."},{"text":"$$C^-$","correct":false,"explanation":"The negative charge is not left on carbon because the carbonyl π electrons move towards oxygen. The resulting alkoxide intermediate therefore contains an $O^-$ charge."},{"text":"$$C^+$","correct":false,"explanation":"A positively charged carbon is not formed during hydride addition to the carbonyl group. Breaking the π bond places its electrons on oxygen, giving an $O^-$ charge."}]},{"id":"cbc51f4e-464e-4b70-86e1-acea2836e83e","cardType":"basic","points":1,"front":"Name the complex ion present in Tollens' reagent.","back":"$[Ag(NH_3)_2]^+$ (diamminesilver(I) ion)","mcqOptions":[{"text":"$[Ag(NH_3)_2]^+$ (diamminesilver(I) ion)","correct":true,"explanation":"Tollens’ reagent contains the diamminesilver(I) complex, formed when aqueous ammonia coordinates to silver(I) ions. Aldehydes reduce its silver ions to metallic silver."},{"text":"$[Cu(NH_3)_4]^{2+}$ (tetraamminecopper(II) ion)","correct":false,"explanation":"This is a copper-ammonia complex associated with copper(II) chemistry, not Tollens’ reagent. Tollens’ reagent contains $[Ag(NH_3)_2]^+$, the diamminesilver(I) ion."},{"text":"$[Ag(CN)_2]^-$ (dicyanoargentate(I) ion)","correct":false,"explanation":"This silver-cyanide complex contains cyanide ligands and is not Tollens’ reagent. Tollens’ reagent uses ammonia ligands and contains $[Ag(NH_3)_2]^+$, the diamminesilver(I) ion."},{"text":"$[AgCl_2]^-$ (dichloroargentate(I) ion)","correct":false,"explanation":"This formula represents a silver complex with chloride ligands, not the ammoniacal silver complex used in Tollens’ test. The required ion is $[Ag(NH_3)_2]^+$."}]},{"id":"03684829-2496-4ed6-b04d-bc013372635f","cardType":"cloze","points":null,"front":"In the Fehling's test, blue $Cu^{2+}$ ions are reduced to $\\text{[...]}$ (copper(I) oxide).","back":"In the Fehling's test, blue $Cu^{2+}$ ions are reduced to $\\boxed{Cu_2O}$ (copper(I) oxide).","mcqOptions":[{"text":"$$Cu_2O$","correct":true,"explanation":"In Fehling’s test, blue copper(II) ions are reduced to copper(I) oxide, $Cu_2O$. This forms the characteristic brick-red precipitate observed with an aldehyde."},{"text":"$$CuO$","correct":false,"explanation":"Copper(II) oxide contains copper in the +2 oxidation state, so it is not the reduced product. Fehling’s test produces copper(I) oxide, $Cu_2O$."},{"text":"$$Cu(OH)_2$","correct":false,"explanation":"Copper(II) hydroxide is a blue copper(II) compound and does not represent reduction to copper(I). The brick-red product of Fehling’s test is $Cu_2O$."},{"text":"$$Cu_2O_3$","correct":false,"explanation":"This formula would not represent copper(I) oxide and results from balancing the charges incorrectly. Copper(I) ions and oxide ions combine in a 2:1 ratio as $Cu_2O$."}]},{"id":"5e61f076-3a46-44c7-869d-3a219a3e4510","cardType":"basic","points":1,"front":"What is the specific biological hazard of cyanide ions ($CN^-$)?","back":"They block cellular respiration.","mcqOptions":[{"text":"They block cellular respiration.","correct":true,"explanation":"Cyanide inhibits enzymes involved in the mitochondrial electron transport chain, preventing aerobic respiration and ATP production. This rapidly deprives vital tissues of usable energy."},{"text":"They destroy red blood cells.","correct":false,"explanation":"This confuses cyanide poisoning with substances that cause haemolysis. Cyanide’s defining hazard is inhibition of the electron transport chain, which blocks cellular respiration."},{"text":"They cause uncontrolled cell division.","correct":false,"explanation":"Uncontrolled cell division is associated with carcinogenic effects, not acute cyanide toxicity. Cyanide blocks cellular respiration by inhibiting enzymes in the mitochondrial electron transport chain."},{"text":"They prevent blood clotting.","correct":false,"explanation":"Failure of blood clotting is linked to anticoagulants rather than cyanide. Cyanide blocks cellular respiration, preventing cells from producing sufficient ATP by aerobic pathways."}]},{"id":"48c1a9c1-4cf8-46d2-93c6-dc5c5649f750","cardType":"basic","points":1,"front":"Where must the curly arrow start when drawing the first step of a nucleophilic addition mechanism?","back":"From the lone pair (or negative charge) on the nucleophile.","mcqOptions":[{"text":"From the lone pair (or negative charge) on the nucleophile.","correct":true,"explanation":"A curly arrow shows the movement of an electron pair, so it begins at the nucleophile’s lone pair or negative charge and points towards the electron-deficient carbonyl carbon."},{"text":"From the carbonyl carbon (or partial positive charge) on the aldehyde.","correct":false,"explanation":"The carbonyl carbon accepts an electron pair and is therefore the arrow’s destination, not its starting point. The arrow begins at the nucleophile’s lone pair or negative charge."},{"text":"From the carbonyl oxygen (or partial negative charge) on the aldehyde.","correct":false,"explanation":"Starting at oxygen would imply that its electrons initiate the nucleophilic attack. The attacking electron pair comes from the nucleophile’s lone pair or negative charge."},{"text":"From the positive ion (or counter-ion) bonded to the nucleophile.","correct":false,"explanation":"A counter-ion such as $K^+$ does not supply the attacking electron pair. The curly arrow must begin at the nucleophile’s lone pair or negative charge."}]}]}],"examCountdown":{"examDate":"2027-05-28","paperLabel":"Paper 1"},"heading":"Aldehydes and ketones (A-level only)","title":"Aldehydes and ketones (A-level only) Flashcards","topbar":{"backHref":"/a-level/chemistry/aqa/flashcards"},"topicSidebar":"$L37"}]}],"topicId":"10220ad7-f7e2-54e1-8ed6-61ac3b12a4fb","walled":"$L38"}],"$L39","$L3a"]