Edexcel International A Level Chemistry: Alcohols (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.

These questions cover Topic 10C: Alcohols of the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry specification (YCH11), Issue 1, September 2017: outcomes 10.15 to 10.20, including Core Practical 7. This is International AS (IAS) content in Unit 2: Energetics, Group Chemistry, Halogenoalkanes and Alcohols.

Before you start, revise from the alcohols study guide and the alcohols revision notes. Course pages: course hub and printable checklist.

Questions

1. Name each alcohol and state whether it is primary, secondary or tertiary.

(a) CH₃CH(OH)CH₂CH₂CH₂CH₃ [2] (b) (CH₃CH₂)₃COH [2] (c) HOCH₂C(CH₃)₂CH₂CH₂CH₃ [2]

2. Pentan-1-ol, C₅H₁₁OH, burns completely in oxygen.

(a) Write a balanced equation for the complete combustion. [1] (b) Calculate the mass of oxygen needed to burn 2.64 g of pentan-1-ol completely. (Mr: C₅H₁₁OH 88.0, O₂ 32.0) [3]

3. Phosphorus(V) chloride is added to a dry sample of butan-2-ol.

(a) State what you would see, and give a test that confirms the gas. [2] (b) Write an equation for the reaction. [1] (c) Explain why a positive result does not prove that the sample is an alcohol. [1]

4. 1-iodobutane is made from butan-1-ol.

(a) Give the reagents and condition. [1] (b) Write two equations to show how the iodoalkane forms. [2] (c) 5.55 g of butan-1-ol gives 9.20 g of 1-iodobutane. Calculate the percentage yield. (Mr: C₄H₉OH 74.0, C₄H₉I 183.9) [3]

5. 2-bromopropane is made by warming propan-2-ol with potassium bromide and 50% concentrated sulfuric acid.

(a) Write two equations for the reactions in the flask. [2] (b) Explain why the sulfuric acid is diluted to 50%. [2] (c) The overall equation is C₃H₇OH + KBr + H₂SO₄ → C₃H₇Br + KHSO₄ + H₂O. Calculate the atom economy by mass for 2-bromopropane. (Mr: C₃H₇Br 122.9, KHSO₄ 136.2, H₂O 18.0) [2]

6. Alcohols are dehydrated by heating with concentrated phosphoric acid.

(a) Name the two alkenes formed from 2-methylhexan-2-ol, (CH₃)₂C(OH)CH₂CH₂CH₂CH₃. [2] (b) Explain why neither alkene shows E-Z isomerism. [2] (c) 1.50 g of propan-2-ol (Mr 60.0) gives 444 cm³ of propene, measured at room temperature and pressure. Calculate the percentage yield. Use a molar gas volume of 24.0 dm³ mol⁻¹. [3]

7. Classify each reaction as substitution, elimination or oxidation.

(a) Ethanol with PCl₅ [1] (b) Propan-1-ol heated with concentrated phosphoric acid [1] (c) Butan-2-ol heated with acidified potassium dichromate(VI) [1]

8. Three alcohols, A, B and C, are isomers of C₅H₁₂O. Each is warmed with potassium dichromate(VI) in dilute sulfuric acid.

• A: the mixture stays orange.
• B (unbranched chain): the mixture turns green; the product, distilled as it forms, gives a red precipitate with Benedict’s solution.
• C (unbranched chain): the mixture turns green; the product does not react with Benedict’s solution or with sodium hydrogencarbonate. C has a symmetrical structure.

(a) Identify A and explain your answer. [2] (b) Identify B. Write equations, using [O], for its oxidation with distillation and under reflux. [3] (c) Identify C and name its oxidation product. [2] (d) Describe a test, with the result, that shows the reflux product of B is a carboxylic acid. [2]

9. A student makes crude 1-bromopentane by heating pentan-1-ol with potassium bromide and 50% concentrated sulfuric acid, then purifies it. 1-bromopentane is denser than water and does not mix with it.

(a) Explain why the mixture is heated under reflux. [2] (b) Next, the student washes the crude liquid with aqueous sodium hydrogencarbonate. State what this wash removes, and why gas pressure builds up during it. [2] (c) State which layer contains the 1-bromopentane. [1] (d) Name a drying agent and state how you know the liquid is dry. [2] (e) Use this value: 1-bromopentane boils at 130 °C. The student’s final product boils between 124 °C and 133 °C. Comment on this. [1]

10. Core Practical 7. A student makes propanal from 4.50 cm³ of propan-1-ol (density 0.80 g cm⁻³) using acidified potassium dichromate(VI). Use these boiling temperatures: propanal 49 °C, propan-1-ol 97 °C, propanoic acid 141 °C.

(a) Explain why the oxidising agent is added slowly to an excess of the alcohol. [2] (b) Explain why the receiving flask is cooled in ice. [1] (c) Explain, in terms of intermolecular forces, why propanal distils over first. [3] (d) The student collects 1.74 g of propanal. Calculate the percentage yield. (Mr: propan-1-ol 60.0, propanal 58.0) [3] (e) Give a test, with the result, to show the distillate contains an aldehyde. [1]

11. You are given four unlabelled liquids: propan-1-ol, propan-2-ol, 2-methylpropan-2-ol and propanoic acid. Describe a sequence of tests to identify each one. Give reagents and observations. [6]

Answers

1. (a) hexan-2-ol [1]; secondary [1] (b) 3-ethylpentan-3-ol [1]; tertiary, the C–OH carbon is bonded to three carbons [1] (c) 2,2-dimethylpentan-1-ol [1]; primary [1] Examiner insight: In (c) the longest chain has five carbons, not three; a name built on propan- gains no credit.

2. (a) 2C₅H₁₁OH + 15O₂ → 10CO₂ + 12H₂O [1] (b) n(pentan-1-ol) = 2.64 ÷ 88.0 = 0.0300 mol [1]; n(O₂) = 7.5 × 0.0300 = 0.225 mol [1]; mass = 0.225 × 32.0 = 7.20 g [1] Examiner insight: The 15 : 2 ratio from the equation must be used; an answer that uses 1 : 1 cannot gain the ratio mark.

3. (a) Steamy fumes [1]; they turn damp blue litmus red, or form white smoke with ammonia [1] (b) CH₃CH(OH)CH₂CH₃ + PCl₅ → CH₃CHClCH₂CH₃ + POCl₃ + HCl [1] (c) Any compound with an –OH group, such as a carboxylic acid (or water), also gives HCl [1] Examiner insight: “A gas is given off” is too vague for (a); say “steamy fumes” and give a test for HCl.

4. (a) Red phosphorus and iodine, warmed [1] (b) 2P + 3I₂ → 2PI₃ [1]; 3C₄H₉OH + PI₃ → 3C₄H₉I + H₃PO₃ [1] (c) n(butan-1-ol) = 5.55 ÷ 74.0 = 0.0750 mol [1]; theoretical mass = 0.0750 × 183.9 = 13.8 g [1]; yield = 9.20 ÷ 13.8 × 100 = 66.7% [1] Examiner insight: Dividing 9.20 by 5.55 compares masses of different substances and earns no credit; work through moles.

5. (a) KBr + H₂SO₄ → KHSO₄ + HBr [1]; CH₃CH(OH)CH₃ + HBr → CH₃CHBrCH₃ + H₂O [1] (b) Fully concentrated sulfuric acid oxidises HBr to bromine [1]; so less HBr is available and the yield of bromoalkane falls [1] (c) 122.9 ÷ (122.9 + 136.2 + 18.0) × 100 [1] = 122.9 ÷ 277.1 × 100 = 44.4% [1] Examiner insight: Atom economy uses the molar masses of all products in the equation; leaving out KHSO₄ gives 87.2%, which is wrong for this overall equation.

6. (a) 2-methylhex-1-ene [1]; 2-methylhex-2-ene [1] (b) In 2-methylhex-1-ene, C1 carries two hydrogen atoms [1]; in 2-methylhex-2-ene, C2 carries two methyl groups; E-Z isomerism needs two different groups on each C=C carbon [1] (c) n(propan-2-ol) = 1.50 ÷ 60.0 = 0.0250 mol [1]; theoretical volume = 0.0250 × 24.0 = 0.600 dm³ = 600 cm³ [1]; yield = 444 ÷ 600 × 100 = 74.0% [1] Examiner insight: In (c), convert to the same unit before dividing: 444 cm³ against 600 cm³.

7. (a) substitution [1] (b) elimination [1] (c) oxidation [1] Examiner insight: Answer (b) with the specification’s term, elimination, rather than only “dehydration”.

8. (a) 2-methylbutan-2-ol [1]; it is the only tertiary isomer, and tertiary alcohols are not oxidised, so the dichromate stays orange [1] (b) Pentan-1-ol [1]; CH₃(CH₂)₃CH₂OH + [O] → CH₃(CH₂)₃CHO + H₂O [1]; CH₃(CH₂)₃CH₂OH + 2[O] → CH₃(CH₂)₃COOH + H₂O [1] (c) Pentan-3-ol [1]; product pentan-3-one [1] (d) Add sodium hydrogencarbonate (or sodium carbonate) [1]; effervescence, and the gas turns limewater milky [1] Examiner insight: In (b), the reflux equation needs 2[O]; writing [O] with a carboxylic acid product leaves the equation unbalanced.

9. (a) The reaction is slow, so it needs heating for a long time [1]; the condenser returns volatile reactants and product to the flask so none is lost [1] (b) It neutralises and removes acid left in the crude product [1]; carbon dioxide forms, so pressure must be released [1] (c) The lower layer [1] (d) Anhydrous magnesium sulfate (or calcium chloride or sodium sulfate) [1]; the liquid turns from cloudy to clear, with some solid still loose [1] (e) It boils over a wide range, so it is impure [1] Examiner insight: For (a), give both ideas: prolonged heating, and no loss of volatile substances.

10. (a) With the alcohol in excess, there is too little oxidising agent to take propanal further [1]; slow addition lets propanal distil off as it forms, before it is oxidised to propanoic acid [1] (b) Propanal is very volatile, so cooling stops it evaporating from the receiver [1] (c) Propan-1-ol and propanoic acid molecules have O–H groups and form hydrogen bonds with each other [1]; propanal has no O–H, so only weaker London forces and permanent dipole forces act between its molecules [1]; less energy is needed to separate propanal molecules, so it has the lowest boiling temperature [1] (d) mass = 4.50 × 0.80 = 3.60 g; n = 3.60 ÷ 60.0 = 0.0600 mol [1]; theoretical mass of propanal = 0.0600 × 58.0 = 3.48 g [1]; yield = 1.74 ÷ 3.48 × 100 = 50.0% [1] (e) Warm with Benedict’s or Fehling’s solution: blue to red precipitate [1] Examiner insight: In (c), “propanal has weaker bonds” is unlikely to score; name hydrogen bonding between alcohol molecules and its absence in propanal.

11. Add sodium hydrogencarbonate to each [1]; only propanoic acid fizzes (CO₂) [1]; warm the other three with acidified potassium dichromate(VI) [1]; 2-methylpropan-2-ol stays orange, the other two turn green [1]; distil the products and warm each with Benedict’s solution [1]; the product from propan-1-ol (propanal) gives a red precipitate, the product from propan-2-ol (propanone) does not [1] [6] Examiner insight: Each test needs a reagent and an observation for each liquid; naming a reagent alone identifies nothing.

Where marks are usually lost

• Classifying by branching elsewhere in the molecule instead of by the C–OH carbon.
• Writing a colour change for a tertiary alcohol with dichromate(VI).
• Using [O] instead of 2[O] for the carboxylic acid.
• Leaving POCl₃ or HCl out of the PCl₅ equation.
• Naming one alkene when two non-equivalent neighbouring carbons can each lose a hydrogen.
• Skipping the density step when a liquid volume is given.
• Using the wrong test: Benedict’s is for aldehydes, sodium hydrogencarbonate for acids.

Next steps

• Review with the alcohols revision notes
• Re-read the alcohols study guide
• Edexcel International A Level Chemistry course hub
• Printable Edexcel Chemistry checklist
• Find your gaps with all free 10-minute diagnostics
• Book a free trial class

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry (XCH11/YCH11), Specification, Issue 1, September 2017, Pearson Education Limited. Topic 10: Organic Chemistry: Halogenoalkanes, Alcohols and Spectra, part 10C: Alcohols (outcomes 10.15 to 10.20, Core Practical 7).