OxfordAQA A-Level Chemistry: Alcohols (9620) – 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 practises 3.3.5 Alcohols, from the OxfordAQA International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards, covering 3.3.5.1 Oxidation of alcohols, 3.3.5.2 Elimination and Required practical 4. All of it is International AS content. Unit 2 (Organic 1 and Physical 1) assesses it; Unit 5, the practical and synoptic paper, covers all content.

Stuck? Return to the 9620 alcohols guide or the 9620 alcohols notes. Other topics: the course page for 9620 chemistry, the printable checklist and a free diagnostic (about 10 minutes). Take Ar values as C = 12.0, H = 1.0 and O = 16.0.

Questions

1. Each alcohol below is warmed with acidified potassium dichromate(VI). For each one, state whether it is primary, secondary or tertiary and give the final colour of the mixture.

(a) cyclohexylmethanol, C₆H₁₁CH₂OH [1] (b) 3-methylpentan-2-ol, CH₃CH(OH)CH(CH₃)CH₂CH₃ [1] (c) 1-methylcyclopentanol (the –OH and a methyl group are on the same ring carbon) [1]

2. Write equations, using [O] to represent the oxidising agent, for:

(a) the oxidation of 2-methylbutan-1-ol, CH₃CH₂CH(CH₃)CH₂OH, to 2-methylbutanal [1] (b) the oxidation of 2-methylbutan-1-ol directly to 2-methylbutanoic acid [1] (c) the oxidation of 3-methylpentan-2-ol to a ketone. Name the ketone. [1]

3. A student wants to make 2-methylbutanal. She heats 2-methylbutan-1-ol with an excess of acidified potassium dichromate(VI) under reflux for 40 minutes, then distils the mixture.

(a) Name the main organic product she obtains and explain why it forms. [2] (b) Describe two changes to her method that would give 2-methylbutanal instead. [2]

4. Two unlabelled bottles contain 2,2-dimethylpropanal, (CH₃)₃CCHO, and cyclohexanone. Describe a test using Fehling’s solution that distinguishes them. Give the condition, what is seen with each compound, and an equation, using [O], for the compound that reacts. [4]

5. When 3-methylbutan-2-ol, CH₃CH(OH)CH(CH₃)₂, is warmed with a suitable acid, it loses water.

(a) Give the name of an acid that catalyses this reaction. [1] (b) Outline how 3-methylbut-1-ene forms, showing the curly arrows and intermediates in words. [3] (c) A second alkene also forms. Name it and explain why it forms. [2]

6. Hexan-3-ol, CH₃CH₂CH(OH)CH₂CH₂CH₃, is heated with concentrated sulfuric acid. Four alkenes are formed. Name all four and explain why there are four. [3]

7. Alcohol E has the molecular formula C₆H₁₄O.

• When E is warmed with acidified potassium dichromate(VI), the mixture turns green. The organic product does not change the colour of Fehling’s solution.
• When E is heated with concentrated phosphoric acid, only one alkene forms.

Deduce the structure and name of E. Explain your reasoning. [4]

8. The overall equation for the oxidation of ethanol to ethanoic acid by acidified dichromate(VI) is:

2Cr₂O₇²⁻ + 16H⁺ + 3CH₃CH₂OH → 4Cr³⁺ + 3CH₃COOH + 11H₂O

A 10.0 cm³ sample of an aqueous ethanol solution reacts exactly with 20.0 cm³ of 0.0450 mol dm⁻³ potassium dichromate(VI) in excess acid. Calculate the concentration of ethanol in the sample in g dm⁻³. [3]

9. In Required practical 4, a student makes 4-methylcyclohexene, C₇H₁₂, by heating 4-methylcyclohexanol, C₇H₁₄O, with concentrated phosphoric acid and distilling the alkene off as it forms.

(a) Explain why the alkene distils from the flask while most of the alcohol stays behind. [2] (b) State where the thermometer bulb is placed, and why. [1] (c) State why anti-bumping granules are added. [1] (d) 6.84 g of 4-methylcyclohexanol gives 4.32 g of 4-methylcyclohexene. Calculate the percentage yield. [3]

10. A company plans to make poly(propene) from propan-2-ol obtained from plant material.

(a) State the reagent and condition used to convert propan-2-ol to propene, and write an equation. [2] (b) Draw the repeating unit of poly(propene). [1] (c) Give one advantage of this route over making propene from crude oil. [1]

11. 2-Methylpentan-1-ol, CH₃CH₂CH₂CH(CH₃)CH₂OH, is oxidised by acidified potassium dichromate(VI) in two experiments.

• Experiment A: 2-methylpentan-1-ol in excess, gentle warming, product distilled off as it forms.
• Experiment B: excess oxidising agent, heated under reflux, then distilled.

(a) Name the organic product of each experiment and write an equation, using [O], for each. [4] (b) Explain why the product of Experiment A has a lower boiling point than 2-methylpentan-1-ol. [2] (c) Each product is warmed with Tollens’ reagent. State what is seen in each case. [2] (d) In Experiment A, 5.61 g of 2-methylpentan-1-ol gives 3.50 g of product. Calculate the percentage yield. [2]

Answers

1. (a) Primary; turns green [1] (b) Secondary; turns green [1] (c) Tertiary; stays orange [1] Examiner insight: Each mark needs both the class and the colour; a correct class with the wrong colour does not score.

2. (a) CH₃CH₂CH(CH₃)CH₂OH + [O] → CH₃CH₂CH(CH₃)CHO + H₂O [1] (b) CH₃CH₂CH(CH₃)CH₂OH + 2[O] → CH₃CH₂CH(CH₃)COOH + H₂O [1] (c) CH₃CH(OH)CH(CH₃)CH₂CH₃ + [O] → CH₃COCH(CH₃)CH₂CH₃ + H₂O; 3-methylpentan-2-one [1] Examiner insight: Unbalanced [O] equations are not credited; count the oxygens, and remember that 2[O] is needed for the one-step route to the acid.

3. (a) 2-methylbutanoic acid [1]; under reflux the aldehyde formed first condenses and returns to the flask, where the excess oxidising agent oxidises it further [1] (b) Use the alcohol in excess (limited oxidising agent) [1]; set the apparatus for distillation and remove the aldehyde as soon as it forms, instead of refluxing [1] Examiner insight: “Use distillation” alone gives only one mark in (b); the second needs the change in amounts of reactants.

4. Warm each with Fehling’s solution in a water bath [1]; 2,2-dimethylpropanal: blue solution gives a red precipitate [1]; cyclohexanone: stays blue [1]; (CH₃)₃CCHO + [O] → (CH₃)₃CCOOH [1] [4] Examiner insight: A test needs an observation for both compounds, so leaving out “stays blue” for the ketone costs a mark even if everything else is right.

5. (a) Concentrated H₃PO₄ (or concentrated H₂SO₄) [1] (b) Arrow from an oxygen lone pair (on –OH) to H⁺, giving the protonated alcohol R–OH₂⁺ [1]; arrow from the C–O bond to O, water leaves, giving the secondary carbocation CH₃C⁺HCH(CH₃)₂ [1]; arrow from a C–H bond on C1 to the C1–C2 bond, forming C=C and releasing H⁺ [1] (c) 2-methylbut-2-ene, (CH₃)₂C=CHCH₃ [1]; H⁺ can also be lost from C3, the other carbon next to the positive carbon, and C3 carries a hydrogen [1] Examiner insight: Arrows must begin at a lone pair or a bond; an arrow drawn from H⁺ or from a positive charge is not credited.

6. (E)-hex-2-ene and (Z)-hex-2-ene [1]; (E)-hex-3-ene and (Z)-hex-3-ene [1]; H⁺ can be lost from C2 or C4 (which are not equivalent), giving two positions of C=C, and each alkene has two different groups on each double-bond carbon, so each shows E/Z isomerism [1] Examiner insight: “Hexene” without a position number and an E/Z label is not a full name, so it cannot gain a naming mark.

7. Green colour: E is oxidised, so it is primary or secondary [1]; the product does not react with Fehling’s, so it is a ketone and E is secondary [1]; only one alkene, so only one neighbouring carbon of the C–OH carbon carries H (the other neighbour has no H) [1]; E is 3,3-dimethylbutan-2-ol, CH₃CH(OH)C(CH₃)₃ [1] [4] Examiner insight: A bare name earns one mark at most; the reasoning marks need each observation linked to a deduction.

8. n(Cr₂O₇²⁻) = 20.0/1000 × 0.0450 = 9.00 × 10⁻⁴ mol [1]; n(ethanol) = 9.00 × 10⁻⁴ × 3/2 = 1.35 × 10⁻³ mol [1]; concentration = 1.35 × 10⁻³ ÷ 0.0100 = 0.135 mol dm⁻³, × 46.0 = 6.21 g dm⁻³ [1] [3] Examiner insight: Using the ratio the wrong way round (× 2/3) gives 2.76 g dm⁻³; show the ratio as a labelled step so your method is visible.

9. (a) The alkene has a lower boiling point than the alcohol [1]; the alcohol forms hydrogen bonds between its molecules but the alkene has only weaker van der Waals forces [1] (b) Level with the side arm to the condenser, so it measures the temperature of the vapour that distils [1] (c) To give smooth boiling (prevent violent bumping) [1] (d) n(alcohol) = 6.84 ÷ 114.0 = 0.0600 mol [1]; n(alkene) = 4.32 ÷ 96.0 = 0.0450 mol [1]; yield = 0.0450 ÷ 0.0600 × 100 = 75.0 % [1] Examiner insight: Dividing the masses (4.32 ÷ 6.84) gives 63.2 % and earns no credit for the yield; both quantities must be in moles first.

10. (a) Heat with concentrated phosphoric (or sulfuric) acid [1]; CH₃CH(OH)CH₃ → CH₃CH=CH₂ + H₂O [1] (b) –[CH₂–CH(CH₃)]– with extension bonds at both ends [1] (c) It does not use monomers derived from crude oil, a finite resource [1] Examiner insight: A repeating unit without the two extension bonds, or with a C=C left in it, is not credited.

11. (a) A: 2-methylpentanal [1]; CH₃CH₂CH₂CH(CH₃)CH₂OH + [O] → CH₃CH₂CH₂CH(CH₃)CHO + H₂O [1]; B: 2-methylpentanoic acid [1]; CH₃CH₂CH₂CH(CH₃)CH₂OH + 2[O] → CH₃CH₂CH₂CH(CH₃)COOH + H₂O [1] (b) 2-Methylpentan-1-ol molecules form hydrogen bonds with each other (O–H group) [1]; 2-methylpentanal has no O–H, so no hydrogen bonding between its molecules and less energy is needed to separate them [1] (c) 2-Methylpentanal (A): a silver mirror forms [1]; 2-methylpentanoic acid (B): no change [1] (d) n(2-methylpentan-1-ol) = 5.61 ÷ 102.0 = 0.0550 mol; n(2-methylpentanal) = 3.50 ÷ 100.0 = 0.0350 mol [1]; yield = 0.0350 ÷ 0.0550 × 100 = 63.6 % [1] Examiner insight: In (b), “the aldehyde has weaker bonds” scores nothing; the answer must name hydrogen bonding between molecules of the alcohol and its absence in the aldehyde.

Where marks are usually lost

• Giving a colour change for a tertiary alcohol; it stays orange.
• Leaving H₂O out of alcohol → aldehyde or ketone equations, or using one [O] for the one-step route to the acid.
• Describing distillation for an aldehyde without “as it forms” and the reason.
• No observation for the compound that does not react in a Fehling’s or Tollens’ test.
• Curly arrows starting at H⁺ or a positive charge, or no H⁺ released in the last step.

Next steps

• Recap the facts: 9620 alcohols notes
• Relearn a method: 9620 alcohols guide
• Practise mole steps: 9620 moles and equations guide
• Every 9620 topic: course page for 9620 chemistry
• Track progress: printable checklist
• Find gaps: all free 10-minute diagnostics
• Book a free trial class

Official syllabus

OxfordAQA, International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards: section 3.3.5 Alcohols (International AS), 3.3.5.1 Oxidation of alcohols and 3.3.5.2 Elimination, with Required practical 4.