Edexcel International A Level Chemistry: Carbonyl compounds (YCH11) – 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 from your board or school.

This set covers Topic 15B: Carbonyl compounds, outcomes 15.6 to 15.8, as set out in Issue 1 (September 2017) of Pearson Edexcel’s International Advanced Subsidiary/Advanced Level in Chemistry specification, YCH11. Unit 4 (Rates, Equilibria and Further Organic Chemistry) is where this IA2 topic is examined. If you have not yet covered it, start with the study guide and revision notes. For every other topic, visit the hub page for this course or print the checklist; the free diagnostic quizzes point out weak areas.

Use these relative atomic masses where needed: C 12.0, H 1.0, O 16.0, Ag 107.9, I 126.9.

Questions

1. Name or give formulae as instructed.

(a) Name (CH₃)₂CHCH(CH₃)CHO. [1] (b) Name CH₃CH₂COCH₂CH₂CH₂CH₃. [1] (c) Give the structural formula of 3,3-dimethylbutan-2-one. [1]

2. Four aldehydes share the molecular formula C₅H₁₀O.

(a) Give the structural formula and name of each. [4] (b) Identify the one aldehyde whose molecules contain a chiral carbon. [1] (c) Explain why none of the four gives a pale yellow precipitate when warmed with iodine and sodium hydroxide solution. [1]

3. Use these boiling temperatures: pentane (Mr 72) 36 °C; butanal (Mr 72) 75 °C; butan-1-ol (Mr 74) 118 °C.

(a) Account for the order of these boiling temperatures by comparing the forces between molecules in each liquid. [4] (b) Butanal is partly soluble in water but pentane is almost insoluble. Explain this difference. [2]

4. Describe two chemical tests that, together, identify each of propanal, butanone and pentan-3-one. Give the reagents, conditions and the result for each compound. [4]

5. 0.731 g of 3-methylbutanal, (CH₃)₂CHCH₂CHO, is warmed with excess Tollens’ reagent.

(a) Write an equation for the oxidation of 3-methylbutanal, using [O]. [1] (b) Write the half-equation showing how silver forms. [1] (c) Calculate the maximum mass of silver that can form. [3]

6. Lithium tetrahydridoaluminate(III) reduces carbonyl compounds.

(a) Name the organic product when 3-methylpentan-2-one is reduced, and classify it as a primary, secondary or tertiary alcohol. [2] (b) Give the reason the solvent must be ether with no water present. [1] (c) Write an equation for the reduction of cyclohexanone, C₆H₁₀O, to cyclohexanol, using [H]. [1] (d) 4.90 g of cyclohexanone gave 3.85 g of cyclohexanol. Calculate the percentage yield. [2]

7. Butanal reacts with hydrogen cyanide in the presence of potassium cyanide.

(a) State the role of the potassium cyanide. [1] (b) Describe the mechanism of this reaction. Your answer should refer to dipoles, lone pairs, curly arrows and the intermediate. [4] (c) Name the organic product. [1] (d) A polarimeter shows zero overall rotation for the product mixture, even though each product molecule is chiral. Explain this observation. [2]

8. A carbonyl compound Z has the molecular formula C₅H₁₀O. A student adds Z to 2,4-DNPH, filters off the solid derivative and measures its melting temperature: 136–141 °C. After recrystallisation the derivative melts at 143–144 °C. Use these derivative melting temperatures: pentanal 107 °C; 3-methylbutan-2-one 120 °C; 3-methylbutanal 123 °C; pentan-2-one 144 °C; pentan-3-one 156 °C.

(a) State what the student sees when Z is added to 2,4-DNPH. [1] (b) Explain why the crude derivative melted lower and over a range. [1] (c) Outline how the derivative is recrystallised. [2] (d) Identify Z. [1] (e) Give one further test, with its expected result, that supports your answer to (d) and rules out pentan-3-one. [2]

9. 1.80 g of butanone is warmed with excess iodine and sodium hydroxide solution.

(a) Give a balanced equation for this reaction, using NaOH. [1] (b) Name the organic product other than triiodomethane. [1] (c) 6.40 g of triiodomethane, CHI₃, is collected. Calculate the percentage yield. [3]

10. Compounds A and B are unbranched isomers with the formula C₆H₁₂O. Both give a yellow-orange precipitate with 2,4-DNPH. A gives a red precipitate when warmed with Benedict’s solution; B does not. B gives a pale yellow precipitate with iodine in alkali.

(a) Identify A and B, explaining each step of your reasoning. [4] (b) A is warmed with potassium dichromate(VI) and dilute sulfuric acid. State the colour change, write the equation using [O], and name the organic product. [3] (c) Name the product when B is reduced by LiAlH₄ in dry ether, and classify it. [2]

11. Cyclohexanone and hexan-3-one each react with HCN in the presence of KCN.

(a) Name the product from hexan-3-one. [1] (b) State whether each product contains a chiral carbon. [2] (c) Neither product sample rotates plane-polarised light. Explain why, for each one. [2]

Answers

1. (a) 2,3-dimethylbutanal [1] (b) Heptan-3-one [1] (c) CH₃COC(CH₃)₃ [1] Examiner insight: For a ketone, check the locant from both ends of the chain and keep the lower number.

2. (a) CH₃CH₂CH₂CH₂CHO, pentanal [1]; CH₃CH₂CH(CH₃)CHO, 2-methylbutanal [1]; (CH₃)₂CHCH₂CHO, 3-methylbutanal [1]; (CH₃)₃CCHO, 2,2-dimethylpropanal [1] (b) 2-methylbutanal: its C2 carries H, CH₃, C₂H₅ and CHO [1]. (c) None has a CH₃CO– group: in each, the C=O carbon carries H, not CH₃ [1]. Examiner insight: Before writing a fourth isomer, check it is not one you already have, drawn or numbered from the other end; a repeat is not a new isomer.

3. (a) The three molecules have about the same number of electrons, so their London forces are similar [1]. Butanal has, in addition, permanent dipole–dipole forces arising from its polar C=O bond [1]. Butanal has no H bonded to O, so it cannot form hydrogen bonds between its molecules [1]. Butan-1-ol forms hydrogen bonds through O–H; these are the strongest forces here and need the most energy to overcome [1]. (b) The δ− carbonyl oxygen of butanal uses a lone pair to accept a hydrogen bond from a water H [1]; pentane has no δ− atom with a lone pair, so it cannot hydrogen bond to water [1]. Examiner insight: Talk about forces between molecules; any mention of covalent bonds breaking on boiling contradicts the answer.

4. Warm each with Tollens’ reagent in a water bath [1]; only propanal gives a silver mirror [1]. Warm the other two with iodine and sodium hydroxide solution [1]; butanone gives a pale yellow precipitate, pentan-3-one gives none [1]. Examiner insight: Give the reagent, the condition and a result for every compound, including the negative ones.

5. (a) (CH₃)₂CHCH₂CHO + [O] → (CH₃)₂CHCH₂COOH [1] (b) Ag⁺ + e⁻ → Ag [1] (c) n(aldehyde) = 0.731 ÷ 86.0 = 0.00850 mol [1]; n(Ag) = 2 × 0.00850 = 0.0170 mol [1]; mass = 0.0170 × 107.9 = 1.83 g [1] Examiner insight: The 1 : 2 ratio comes from the two electrons each aldehyde loses; a 1 : 1 ratio halves the answer.

6. (a) 3-methylpentan-2-ol [1]; secondary [1] (b) LiAlH₄ reacts violently with water, so water must be absent [1]. (c) C₆H₁₀O + 2[H] → C₆H₁₂O [1] (d) n = 4.90 ÷ 98.0 = 0.0500 mol, so maximum mass = 0.0500 × 100.0 = 5.00 g [1]; yield = 3.85 ÷ 5.00 × 100 = 77.0% [1] Examiner insight: Reduction of a carbonyl needs 2[H]; a single [H] leaves the equation unbalanced.

7. (a) It supplies cyanide ions, the nucleophile, because HCN is a weak acid and barely ionises [1]. (b) C=O is polar, δ+ on C and δ− on O; CN⁻ has a lone pair on carbon [1]. A curly arrow starts at the carbon lone pair of CN⁻ and ends on the carbonyl carbon [1]. A curly arrow from the C=O π bond to O gives an intermediate with O⁻ and a new C–CN bond [1]. A lone pair on O⁻ forms a bond to the H of HCN (arrow from the H–C bond to C), giving the product and regenerating CN⁻ [1]. (c) 2-hydroxypentanenitrile [1] (d) Attack on the planar C=O happens from above and below equally often [1]; so the two enantiomers form 50 : 50, a racemic mixture, and their opposite rotations cancel [1]. Examiner insight: Start the first arrow at the lone pair on carbon, not at the negative charge on N.

8. (a) A yellow-orange precipitate [1]. (b) Impurities lower the melting temperature and make it melt over a range [1]. (c) Add just enough hot solvent to dissolve the solid, then let it cool so crystals form [1]; filter, wash with a little cold solvent and dry [1]. (d) Pentan-2-one [1] (e) Warm with iodine and sodium hydroxide solution: a pale yellow precipitate forms [1]; pentan-2-one has a CH₃CO– group but pentan-3-one does not [1]. Examiner insight: Match the purified value, not the crude one; the crude range points nowhere useful.

9. (a) CH₃COCH₂CH₃ + 3I₂ + 4NaOH → CHI₃ + CH₃CH₂COONa + 3NaI + 3H₂O [1] (b) Sodium propanoate [1] (c) n(butanone) = 1.80 ÷ 72.0 = 0.0250 mol [1]; maximum mass CHI₃ = 0.0250 × 393.7 = 9.843 g [1]; yield = 6.40 ÷ 9.843 × 100 = 65.0% [1] Examiner insight: Check the by-product has one fewer carbon than the ketone; butanone gives a three-carbon carboxylate.

10. (a) Both give a precipitate with 2,4-DNPH, so both have C=O [1]. A reduces Benedict’s solution, so A is an aldehyde: unbranched, so hexanal [1]. B does not, so B is a ketone [1]. B has a CH₃CO– group (iodoform), so C=O is at C2: hexan-2-one [1]. (b) Orange to green [1]; CH₃(CH₂)₄CHO + [O] → CH₃(CH₂)₄COOH [1]; hexanoic acid [1] (c) Hexan-2-ol [1]; a secondary alcohol [1] Examiner insight: Link each observation to the structural feature it shows; a bare name does not show the reasoning.

11. (a) 2-ethyl-2-hydroxypentanenitrile [1] (b) Cyclohexanone product: no chiral carbon, as the two ring branches from that carbon are identical [1]. Hexan-3-one product: chiral, as the carbon carries OH, CN, ethyl and propyl [1]. (c) The cyclohexanone product has no chiral centre, so it has no optical isomers [1]. The hexan-3-one product forms as a racemic mixture because CN⁻ attacks both faces of the planar C=O equally [1]. Examiner insight: Two inactive samples need two different reasons: achiral for one, racemic for the other.

Where marks are usually lost

• Leaving the locant out of a ketone name, or numbering from the wrong end.
• Saying aldehydes and ketones hydrogen bond to each other.
• Explaining solubility with “polar” alone, without the hydrogen bond to water.
• Forgetting to warm the Tollens’, Fehling’s or Benedict’s tubes, or giving the wrong colours.
• Using a 1 : 1 ratio for silver in Tollens’ calculations.
• Writing [H] instead of 2[H] in reduction equations.
• Curly arrows from the wrong atom, or no intermediate shown.
• Calling a racemic product “optically active”.
• Using 2,4-DNPH to separate aldehydes from ketones.

Next steps

• All free 10-minute diagnostics, to see which topics need work
• Carbonyl compounds revision notes for a final recap
• Printable chemistry checklist to tick off 15.6 to 15.8
• Carbonyl compounds study guide to relearn anything you missed
• Want live teaching? Book a free trial class
• Hub page for this course, listing the other units

Official syllabus

Source: Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry, specification XCH11/YCH11, Issue 1, September 2017 (Pearson Education Limited). This set covers Topic 15B: Carbonyl compounds, outcomes 15.6 to 15.8, in Unit 4.