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Questions on aldehydes and ketones: nucleophilic addition and the tests that tell them apart
The carbonyl group and its polarity, oxidation of aldehydes by Tollens' reagent and Fehling's solution with the half-equations, reduction of aldehydes and ketones by NaBH₄ to primary and secondary alcohols, the nucleophilic addition mechanisms for hydride and for cyanide, hydroxynitriles from KCN and acid, the hazard of HCN, racemic products, 2,4-DNP with derivative melting points, the triiodomethane test, and LiAlH₄ in dry ether.
18 original questions · 59 marks · the aldehydes and ketones: nucleophilic addition and the tests that tell them apart notes · Carbonyls and carboxylic acids · Flashcards for this unit
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Describe how Tollens' reagent and Fehling's solution can each be used to distinguish propanal from propanone. Give the observations and the relevant equations.
Mark scheme
Indicative content
- each compound is warmed in a water bath with freshly made Tollens' reagent, which contains the [Ag(NH₃)₂]⁺ ion
- propanal gives a silver mirror on the inside of the tube, while propanone gives no change and the solution stays colourless
- the silver complex is reduced: [Ag(NH₃)₂]⁺ + e⁻ → Ag + 2NH₃; allow Ag⁺ + e⁻ → Ag
- warmed with Fehling's solution, propanal gives a brick-red precipitate while propanone leaves the solution blue; allow red or orange-red precipitate
- the copper(II) is reduced, Cu²⁺ + e⁻ → Cu⁺, and copper(I) oxide, Cu₂O, precipitates in the alkaline solution
- propanal is oxidised, CH₃CH₂CHO + H₂O → CH₃CH₂COOH + 2H⁺ + 2e⁻ or with [O]; propanone has no H on its carbonyl carbon and is not oxidised
Levels
- Level 3 (5–6 marks) both tests are described with their observations for both compounds, the half-equations for the reduction of silver and of copper(II) are given, the oxidation of propanal is written, and the answer explains why propanone does not react
- Level 2 (3–4 marks) both tests give the silver mirror and the brick-red precipitate for propanal and no change for propanone; only one equation is right, or the answer does not say why propanone resists oxidation
- Level 1 (1–2 marks) one test is described with its positive result, often with the observations of the two tests swapped; no equations and no reason propanone does not react
Outline the mechanism for the reaction of propanone with KCN followed by dilute acid, and name the organic product.
Mark scheme
- B1 curly arrow from the lone pair on the carbon of CN⁻ to the carbonyl carbon
- B1 curly arrow from the C=O double bond to the oxygen atom
- B1 intermediate (CH₃)₂C(CN)O⁻ with the negative charge on oxygen
- B1 curly arrow from a lone pair on O⁻ to H⁺
- B1 the product is 2-hydroxy-2-methylpropanenitrile
Outline the mechanism for the reduction of propanal by NaBH₄. Show the hydride ion as the nucleophile and water as the source of the second hydrogen.
Mark scheme
- B1 curly arrow from the lone pair on H⁻ to the carbonyl carbon
- B1 curly arrow from the C=O double bond to the oxygen atom
- B1 intermediate CH₃CH₂CH₂O⁻ drawn with the negative charge on oxygen
- B1 curly arrow from a lone pair on O⁻ to an H of H₂O, with an arrow from the O–H bond to the water's oxygen, giving propan-1-ol
Compare the reaction of HBr with propene and the reaction of HCN with propanal, in terms of the attacking species and the mechanism.
Mark scheme
- B1 propene reacts by electrophilic addition, propanal by nucleophilic addition
- B1 in propene the electron-rich C=C bond attacks the δ+ hydrogen of HBr, the electrophile
- B1 in propanal the CN⁻ nucleophile attacks the δ+ carbonyl carbon, using the lone pair on its carbon
- B1 the alkene gives a carbocation intermediate, whereas the carbonyl compound gives an alkoxide ion with a negative charge on oxygen
Describe how 2,4-dinitrophenylhydrazine is used to show that an unknown liquid is a carbonyl compound and then to identify it.
Mark scheme
- B1 a few drops of the liquid are added to 2,4-DNP solution, and a yellow or orange precipitate shows an aldehyde or ketone
- B1 the precipitate is filtered off and recrystallised from the minimum volume of hot ethanol
- B1 the dried crystals' melting point is measured
- B1 the melting point is compared with a data table of the melting points of 2,4-DNP derivatives to identify the compound
Explain why the carbon atom of a carbonyl group is attacked by nucleophiles while the carbon atoms of a C=C bond are not.
Mark scheme
- B1 oxygen is more electronegative than carbon, so the C=O bond is polar
- B1 the carbonyl carbon carries a δ+ charge, which attracts the lone pair of a nucleophile
- B1 the C=C bond is non-polar and a region of high electron density, so it repels nucleophiles and attracts electrophiles instead
Explain why KCN followed by acid is used to make a hydroxynitrile rather than hydrogen cyanide, and state the hazard of the method.
Mark scheme
- B1 HCN is a weak acid, so its solution contains few CN⁻ ions and the reaction is slow
- B1 KCN is fully ionised and supplies the CN⁻ nucleophile directly
- B1 cyanide with acid releases HCN, a highly toxic gas, so the reaction is done in a fume cupboard
Explain why the reaction of butanal with KCN followed by dilute acid gives a racemic mixture.
Mark scheme
- B1 the product, 2-hydroxypentanenitrile, has a chiral centre bonded to H, OH, CN and C₃H₇
- B1 the carbonyl group of butanal is planar, so the cyanide ion attacks from either side with equal probability
- B1 the two enantiomers are therefore formed in equal amounts
Name the organic products formed when butanal and butanone are each reduced by NaBH₄, and give the equation for the butanone reduction using [H].
Mark scheme
- B1 butanal gives butan-1-ol, a primary alcohol
- B1 butanone gives butan-2-ol, a secondary alcohol
- B1 CH₃COCH₂CH₃ + 2[H] → CH₃CH(OH)CH₂CH₃; allow molecular formulae if the structure is clear
Calculate the maximum mass of silver deposited when 0.220 g of ethanal reacts completely with Tollens' reagent. Mr of ethanal = 44.0; Ar of Ag = 107.9.
Mark scheme
- M1 amount of ethanal = 0.220 ÷ 44.0 = 5.00 × 10⁻³ mol
- M1 each ethanal releases two electrons, so the amount of silver = 2 × 5.00 × 10⁻³ = 1.00 × 10⁻² mol
- A1 mass of silver = 1.00 × 10⁻² × 107.9 = 1.08 g, to 3 significant figures
Calculate the mass of 2-hydroxy-2-methylpropanenitrile obtained from propanone, using the data given in the table.
Quantity Value mass of propanone 2.90 g Mr of propanone 58.0 Mr of 2-hydroxy-2-methylpropanenitrile 85.0 percentage yield 60.0% Mark scheme
- M1 amount of propanone = 2.90 ÷ 58.0 = 0.0500 mol
- M1 amount of product = 0.0500 × 0.600 = 0.0300 mol, since the reaction is 1 : 1
- A1 mass = 0.0300 × 85.0 = 2.55 g, to 3 significant figures
Suggest the product when propenal, CH₂=CHCHO, is reduced by NaBH₄, and the product when it is reduced by hydrogen with a nickel catalyst, and explain the difference.
Mark scheme
- B1 NaBH₄ gives prop-2-en-1-ol, CH₂=CHCH₂OH
- B1 hydrogen and nickel gives propan-1-ol, CH₃CH₂CH₂OH
- B1 the hydride ion is a nucleophile, which attacks the δ+ carbonyl carbon but is repelled by the electron-rich C=C bond, whereas H₂ with nickel adds across both double bonds
Suggest the identity of compound Y, C₄H₈O, which gives no silver mirror with Tollens' reagent and is reduced by NaBH₄ to an alcohol with a chiral centre. Explain each piece of evidence.
Mark scheme
- B1 no silver mirror means Y is not an aldehyde, and its reduction by NaBH₄ shows a C=O group, so Y is a ketone
- B1 the only ketone with formula C₄H₈O is butanone, CH₃COCH₂CH₃
- B1 its reduction product, butan-2-ol, has carbon 2 bonded to H, OH, CH₃ and C₂H₅, so it is chiral
Evaluate the conclusion of a student who warms an unknown liquid with Fehling's solution, sees the solution stay blue, and concludes that the liquid is a ketone.
Mark scheme
- B1 a negative Fehling's test shows only that the liquid is not an aldehyde
- B1 alcohols, esters, carboxylic acids and many other compounds also leave Fehling's solution blue, so the conclusion is not justified
- B1 a test showing a carbonyl group is needed as well, such as a C=O absorption near 1700 cm⁻¹ in the infrared spectrum, with no broad O–H absorption
Suggest which of butanal, butanone, butan-1-ol and butan-2-ol give a pale yellow precipitate when warmed with iodine and aqueous sodium hydroxide, and explain your choice.
Mark scheme
- B1 butanone and butan-2-ol give the precipitate of triiodomethane, CHI₃
- B1 butanone contains the CH₃CO– group, and butan-2-ol contains CH₃CH(OH)–, which alkaline iodine oxidises to CH₃CO–
- B1 butanal and butan-1-ol have neither group, so they give no precipitate
Explain why aldehydes are oxidised by Tollens' reagent but ketones are not.
Mark scheme
- B1 an aldehyde has a hydrogen atom on its carbonyl carbon, which is replaced by OH to give a carboxylic acid
- B1 a ketone has no such hydrogen, and oxidising it would need a C–C bond to break, which a mild oxidising agent cannot do
Explain why the 2,4-DNP derivative is recrystallised before its melting point is measured.
Mark scheme
- B1 impurities lower the melting point and make the solid melt over a range of temperatures
- B1 so only a pure derivative gives a sharp melting point that can be matched to the data table
Explain why LiAlH₄ is used in dry ether, while NaBH₄ can be used in water or ethanol.
Mark scheme
- B1 LiAlH₄ is a more powerful source of hydride ions and reacts violently with water, releasing hydrogen
- B1 so it must be kept dry until the reduction is complete, and dilute acid is added only afterwards