AS Chemistry: Aldehydes and Ketones — Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Related: Aldehydes and Ketones revision notes

---

Questions

1. Explain why the carbonyl carbon is susceptible to nucleophilic attack. [2]

2. Ethanal, CH₃CHO, reacts with HCN in the presence of a small amount of KCN.

(a) Name the mechanism. [1] (b) Name the product. [1] (c) Describe the mechanism, stating where each curly arrow starts. [3] (d) Explain why KCN is required and not HCN alone. [2]

3. The product in question 2 contains a chiral centre, and the reaction gives equal amounts of its two optical isomers.

(a) Explain what is meant by a chiral centre, and identify the chiral centre in the product. [2] (b) Explain, in terms of the shape of the carbonyl compound, why the two optical isomers are formed in equal amounts. [3] (c) Propanone reacts with HCN in the same way. Explain why its product has no optical isomers. [1]

4. Describe two chemical tests that distinguish an aldehyde from a ketone, stating the reagent and both observations in each case. [6]

5. Explain how you would distinguish propanal from propanone using infrared spectroscopy, and why this is difficult. [3]

6. State the reagent for reducing a ketone to a secondary alcohol. [1]

7. A student adds 2,4-dinitrophenylhydrazine (2,4-DNPH) to an unknown liquid and obtains an orange precipitate.

(a) What does this observation confirm about the unknown liquid? [1] (b) Explain why this test alone cannot tell the student whether the liquid is an aldehyde or a ketone. [1] (c) Describe how the identity of the specific carbonyl compound could then be confirmed. [2]

8. Explain how you would prepare a sample of propanal from propan-1-ol in the laboratory, and explain why distillation is critical to this particular method — even though propanone (made from propan-2-ol by the same oxidising agent) would not be over-oxidised even if the mixture were heated under reflux instead. [4]

---

Answers

1. Oxygen is more electronegative than carbon [1], so the C=O bond is polar and the carbon carries a partial positive charge (δ+) that attracts electron-rich nucleophiles [1].

2. (a) Nucleophilic addition [1]. (b) 2-hydroxypropanenitrile, CH₃CH(OH)CN [1]. (c) An arrow from the lone pair on CN⁻ to the δ+ carbonyl carbon [1]; an arrow from the C=O π bond to the oxygen [1]; the resulting alkoxide is then protonated by HCN or water [1]. (d) The nucleophile is the cyanide ion CN⁻, not HCN [1]; KCN provides a sufficient concentration of CN⁻, since HCN is a weak acid and dissociates only slightly [1].

3. (a) A carbon atom bonded to four different groups [1]; in 2-hydroxypropanenitrile it is C2, the carbon bonded to H, CH₃, OH and CN [1]. (b) The carbonyl carbon in ethanal, the reactant, is planar (trigonal, sp²) [1], so CN⁻ attacks it with equal probability from either face [1]; attack from the two faces gives the two different optical isomers, so they form in equal amounts [1]. (The alkoxide intermediate formed after attack is tetrahedral, not planar — it is the shape of the starting carbonyl compound that this reveals. The name racemic mixture for such a product is A Level, 29.4.) (c) Its product, 2-hydroxy-2-methylpropanenitrile, has two identical methyl groups on the carbon bonded to OH and CN, so that carbon is not bonded to four different groups — there is no chiral centre, and attack from either face gives the same molecule [1].

4. Tollens’ reagent (ammoniacal silver nitrate), warmed [1] — the aldehyde gives a silver mirror [1]; the ketone gives no change [1]. Fehling’s or Benedict’s solution, warmed [1] — the aldehyde gives a brick-red precipitate [1]; the ketone gives no change [1]. Both work because aldehydes are readily oxidised to carboxylic acids and ketones are not.

5. Both show a strong C=O absorption near 1700 cm⁻¹ [1], so the carbonyl peak alone does not distinguish them [1]. They can only be told apart by comparing the whole fingerprint region with reference spectra [1].

6. NaBH₄ (sodium tetrahydridoborate) [1].

7. (a) That the liquid contains a carbonyl group — it is an aldehyde or a ketone [1]. (b) Both aldehydes and ketones give the same orange precipitate with 2,4-DNPH, so the test confirms a carbonyl is present but does not distinguish which type [1]. (c) Recrystallise the orange precipitate (the 2,4-DNPH derivative) [1], then measure its melting point and compare it against a data table of known derivatives to identify the specific compound [1].

8. Oxidise propan-1-ol with acidified potassium dichromate(VI) [1], and distil off the propanal as it forms rather than heating under reflux [1]. This is necessary because propanal, an aldehyde, would otherwise be oxidised further to propanoic acid if it remained in contact with the oxidising mixture [1]. Propan-2-ol is oxidised the same way and, by the syllabus method, also distilled to give propanone — but here distillation isn’t preventing over-oxidation, since a ketone cannot be oxidised further under these conditions regardless of whether the mixture is distilled or refluxed; distillation matters for a genuinely different reason in each case [1].

---

Where marks are usually lost

• Saying HCN is the nucleophile rather than CN⁻.
• Stating a racemate forms without explaining the planar carbonyl reactant (the shape being revealed is the starting carbonyl compound’s, not the tetrahedral alkoxide intermediate formed after attack).
• Claiming two optical isomers from a symmetrical ketone such as propanone — its hydroxynitrile has two identical methyl groups on the new carbon and no chiral centre; only an aldehyde other than methanal or an unsymmetrical ketone gives a chiral product.
• Giving only one observation in a distinguishing test — both are needed.
• Claiming IR alone easily distinguishes an aldehyde from a ketone.
• Claiming that 2,4-DNPH distinguishes an aldehyde from a ketone — it only confirms a carbonyl group is present; Tollens’ or Fehling’s is needed to tell them apart.
• Using reflux instead of distillation when preparing an aldehyde from a primary alcohol, which allows over-oxidation to the carboxylic acid.
• Forgetting that a hydroxynitrile has one more carbon atom than the carbonyl compound it was made from.

Preparing and testing carbonyl compounds — the full picture

Starting material	Method	Product
Primary alcohol	Oxidise with acidified K₂Cr₂O₇, distil immediately	Aldehyde
Primary alcohol	Oxidise with acidified K₂Cr₂O₇, reflux (excess oxidant)	Carboxylic acid
Secondary alcohol	Oxidise with acidified K₂Cr₂O₇, distil	Ketone
Aldehyde or ketone	Reduce with NaBH₄	Primary or secondary alcohol
Aldehyde or ketone	React with HCN/KCN	Hydroxynitrile

The recurring theme across this table and the distinguishing tests is that an aldehyde sits at an intermediate oxidation state — it can be pushed further to a carboxylic acid or pulled back to a primary alcohol — while a ketone is a dead end for oxidation, which is exactly why Tollens’ and Fehling’s give a positive result only with the aldehyde.

For the full reaction schemes and curly-arrow mechanism, see the Carbonyl Compounds study guide; for condensed recall notes covering the same preparation and test tables, see the Aldehydes and Ketones revision notes.