OCR A Level Chemistry A: Alcohols (H432) – 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.

Here you practise section 4.2.1, Alcohols, of the OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017. Every learning outcome, 4.2.1(a)(i) to (e), is tested here, along with the PAG7 dichromate test. H432 is the full A level. Questions 4 and 9 also use enthalpy and yield methods; questions 10 and 11 draw several outcomes together.

Before you start, work through the study guide and the revision notes. Other topics are on the OCR chemistry hub; log what you have done on the printable checklist. The free diagnostics test the rest of the course.

Wherever a calculation needs them, take Aᵣ as H 1.0, C 12.0 and O 16.0.

Questions

1. State whether each alcohol is primary, secondary or tertiary.

(a) CH₃CH₂CH(CH₃)CH₂OH [1] (b) CH₃CH(OH)CH₂CH₂CH₂CH₃ [1] (c) CH₃CH₂C(CH₃)₂OH [1]

2. Use these values: propan-1-ol, Mᵣ 60.0, boiling point 97 °C; butane, Mᵣ 58.0, boiling point −1 °C.

(a) Explain the difference in boiling points. [3] (b) Propan-1-ol mixes with water in all proportions, but heptan-1-ol is only slightly soluble. Explain this difference. [2]

3. Butan-1-ol burns completely in oxygen.

(a) Write the balanced equation. [1] (b) A sample of 1.48 g of butan-1-ol is burned. Calculate the mass of carbon dioxide formed and the mass of oxygen used. [3]

4. A spirit burner of propan-1-ol is used to heat 150 g of water. Burning 0.528 g of the alcohol raises the water temperature by 21.0 K. The specific heat capacity of water is 4.18 J g⁻¹ K⁻¹.

(a) Calculate the enthalpy change of combustion of propan-1-ol, in kJ mol⁻¹, to 3 significant figures. [3] (b) Suggest one reason why this value is less exothermic than the accepted value. [1]

5. Hexan-1-ol is oxidised by acidified potassium dichromate.

(a) Name the two organic products that can form. [2] (b) Describe and explain the conditions you would use to obtain the first product rather than the second. [3] (c) Write an equation, using [O], for the oxidation of hexan-1-ol to the second product. [1]

6. Samples of butan-2-ol, 2-methylpropan-2-ol and butanone are each warmed with acidified potassium dichromate.

(a) Predict what is observed for each compound. [2] (b) Explain the result for 2-methylpropan-2-ol. [2] (c) Explain why this test cannot tell butanone apart from 2-methylpropan-2-ol. [1] (d) Write an equation, using [O], for the reaction of butan-2-ol. [1]

7. 5-methylhexan-2-ol, CH₃CH(OH)CH₂CH₂CH(CH₃)₂, is converted into alkenes.

(a) State the reagent and the condition needed. [2] (b) State the type of reaction. [1] (c) Name the two structural isomers of the alkene that can form. [2] (d) One of these alkenes exists as E and Z isomers. Identify it and explain why it shows E/Z isomerism. [2]

8. 2-bromopropane is made from propan-2-ol.

(a) State the reagents and the condition. [2] (b) Write two equations: one for forming the hydrogen halide and one for the reaction with propan-2-ol. [2] (c) Explain why this is a substitution reaction. [1]

9. A student oxidises 5.22 g of heptan-1-ol, C₇H₁₅OH, with excess acidified potassium dichromate and collects 4.10 g of heptanoic acid, C₆H₁₃COOH.

(a) State the colour change and give the reason for using reflux. [2] (b) Write an equation, using [O], for the reaction. [1] (c) Calculate the percentage yield. [3]

10. Alcohol P belongs to the homologous series CₙH₂ₙ₊₁OH and has Mᵣ = 88.0. When P is warmed with acidified potassium dichromate, the solution stays orange.

(a) Show that P has the molecular formula C₅H₁₁OH. [2] (b) Deduce the structure of P. Explain your reasoning. [3] (c) P is heated with concentrated phosphoric acid. Name the two alkenes that form. [2] (d) Name the organic product when P is heated with sodium bromide and sulfuric acid. [1] (e) Q is an isomer of P. Q is a secondary alcohol with a branched carbon chain. Identify Q and name the product of its oxidation by acidified dichromate. [2]

11. 3-methylbutan-1-ol, (CH₃)₂CHCH₂CH₂OH, is used to make three different compounds.

(a) Classify 3-methylbutan-1-ol. [1] (b) For each conversion, state the reagent(s) and conditions, and give the structural formula of the organic product.

(i) to a carboxylic acid [2] (ii) to an alkene [2] (iii) to a bromoalkane [2]

(c) Explain why only one structural isomer of the alkene forms in (b)(ii). [1]

Answers

1. (a) Primary: one carbon is attached to the carbon carrying the OH [1] (b) Secondary: the C–OH carbon is C2, bonded to C1 and C3 [1] (c) Tertiary: three carbons are attached to the carbon carrying the OH [1] Examiner insight: In (c), drawing out the condensed formula shows the C–OH carbon has no hydrogen and three carbon neighbours.

2. (a) Both have similar numbers of electrons, so similar London forces [1]. Propan-1-ol also has hydrogen bonds between –OH groups of neighbouring molecules [1]. Hydrogen bonds are the stronger force, so separating the molecules takes more energy, giving the higher boiling point [1] (b) Propan-1-ol’s –OH forms hydrogen bonds with water molecules [1]. Heptan-1-ol has a longer non-polar chain that cannot hydrogen-bond, so it disrupts the hydrogen bonding in water and solubility is low [1] Examiner insight: “Alcohols have hydrogen bonds” earns little alone; place the hydrogen bonds between molecules and compare their strength with London forces.

3. (a) C₄H₉OH + 6O₂ → 4CO₂ + 5H₂O [1] (b) n(C₄H₉OH) = 1.48 ÷ 74.0 = 0.0200 mol [1]; n(CO₂) = 4 × 0.0200 = 0.0800 mol, mass = 0.0800 × 44.0 = 3.52 g [1]; n(O₂) = 6 × 0.0200 = 0.120 mol, mass = 0.120 × 32.0 = 3.84 g [1] Examiner insight: Doubling to 2C₄H₉OH + 12O₂ is equally correct, but forgetting the alcohol’s own oxygen gives 6.5O₂ per molecule and makes every later mole ratio wrong.

4. (a) q = 150 × 4.18 × 21.0 = 13 167 J = 13.2 kJ [1]; n(propan-1-ol) = 0.528 ÷ 60.0 = 0.00880 mol [1]; ΔcH = −13.167 ÷ 0.00880 = −1.50 × 10³ kJ mol⁻¹ [1] (b) Heat is lost to the surroundings instead of reaching the water (or combustion is incomplete) [1] Examiner insight: In q = mcΔT, m is the water’s mass, not the alcohol’s; give the negative sign; a positive value describes an endothermic change, which combustion is not.

5. (a) Hexanal [1] and hexanoic acid [1] (b) Excess hexan-1-ol (limited oxidising agent), gentle heat [1]. Distil the hexanal off as it forms [1]. Hexanal molecules do not hydrogen-bond to each other, so it boils lower than hexan-1-ol and leaves before further oxidation [1] (c) CH₃(CH₂)₄CH₂OH + 2[O] → CH₃(CH₂)₄COOH + H₂O [1] Examiner insight: Link distilling the aldehyde off as it forms to stopping further oxidation; “distil” alone explains nothing.

6. (a) 2-methylpropan-2-ol and butanone: stay orange [1]; butan-2-ol: orange to green [1] (b) It is a tertiary alcohol [1]; there is no hydrogen atom on the carbon bonded to the –OH, so it cannot be oxidised [1] (c) Neither compound is oxidised by acidified dichromate, so both give the same result (no colour change) [1] (d) CH₃CH(OH)CH₂CH₃ + [O] → CH₃COCH₂CH₃ + H₂O [1] Examiner insight: Describe the colour change from start to finish; “turns green” alone is incomplete.

7. (a) Concentrated phosphoric acid (or concentrated sulfuric acid) as catalyst [1]; heat [1] (b) Elimination (dehydration) [1] (c) 5-methylhex-1-ene [1] and 5-methylhex-2-ene [1] (d) 5-methylhex-2-ene [1]. Both C=C carbons carry two different groups (C2: H and CH₃; C3: H and CH₂CH(CH₃)₂), and rotation about the C=C is restricted [1] Examiner insight: In (c), listing the E and Z forms as two structural isomers confuses structural isomerism with stereoisomerism.

8. (a) Sodium bromide [1] and sulfuric acid, heated [1] (b) NaBr + H₂SO₄ → NaHSO₄ + HBr [1]; CH₃CH(OH)CH₃ + HBr → CH₃CHBrCH₃ + H₂O [1] (c) The –OH group is replaced by a bromine atom [1] Examiner insight: Bromine is the reagent for alkenes, not for this conversion; naming Br₂ here gains no credit.

9. (a) Orange to green [1]. The condenser sends the volatile aldehyde intermediate back into the flask, and there it is oxidised to the acid [1] (b) CH₃(CH₂)₅CH₂OH + 2[O] → CH₃(CH₂)₅COOH + H₂O [1] (c) n(heptan-1-ol) = 5.22 ÷ 116.0 = 0.0450 mol [1]; theoretical mass of acid = 0.0450 × 130.0 = 5.85 g [1]; yield = 4.10 ÷ 5.85 × 100 = 70.1% [1] Examiner insight: Dividing 4.10 g by 5.22 g compares two different substances and gives a meaningless 78.5%; find the theoretical mass first.

10. (a) Mᵣ of CₙH₂ₙ₊₁OH = 12n + 2n + 1 + 16 + 1 = 14n + 18 [1]; 14n + 18 = 88.0, so n = 5 and P is C₅H₁₁OH [1] (b) Not oxidised, so P is tertiary [1]. The C–OH carbon carries three groups made from four carbons: CH₃, CH₃ and C₂H₅ [1]. P is 2-methylbutan-2-ol, CH₃CH₂C(CH₃)₂OH [1] (c) 2-methylbut-1-ene, CH₂=C(CH₃)CH₂CH₃ [1], and 2-methylbut-2-ene, (CH₃)₂C=CHCH₃ [1] (d) 2-bromo-2-methylbutane [1] (e) Q is 3-methylbutan-2-ol, CH₃CH(OH)CH(CH₃)₂ [1]; it is oxidised to 3-methylbutan-2-one, CH₃COCH(CH₃)₂ [1] Examiner insight: In (b), link the colour test to “tertiary, so not oxidised”; a correct name without that reasoning is only part of the answer.

11. (a) Primary [1] (b)(i) Excess acidified potassium dichromate, heat under reflux [1]; (CH₃)₂CHCH₂COOH [1] (ii) Concentrated H₃PO₄ (or H₂SO₄), heat [1]; (CH₃)₂CHCH=CH₂ [1] (iii) NaBr and H₂SO₄, heat [1]; (CH₃)₂CHCH₂CH₂Br [1] (c) The –OH is on C1, so the only neighbouring carbon is C2; the C=C can only form between C1 and C2 [1] Examiner insight: Be specific: “acid” alone is too vague for (b)(ii), and (b)(i) needs both “excess” and “reflux”.

Where marks are usually lost

• Comparing an alcohol with an alkane of very different Mᵣ, so the argument about London forces fails.
• Forgetting the oxygen in the alcohol when balancing combustion equations.
• Leaving out water, or writing [O] where 2[O] is needed, in oxidation equations.
• Using reflux for the aldehyde.
• Giving “no reaction” for a tertiary alcohol without the observation (stays orange).
• Missing the second alkene from an alcohol whose C–OH carbon has two unlike H-bearing neighbours.
• Naming bromine, not NaBr and H₂SO₄, for making a bromoalkane.

Next steps

• Recap the tables and equations: revision notes
• Worked examples and explanations: study guide
• Calorimetry practice: Enthalpy changes practice
• Yield and moles: Amount of substance practice
• Check the whole course: all free 10-minute diagnostics
• More OCR chemistry topics: course hub
• Printable checklist: OCR A Level Chemistry checklist
• Book a free trial class: /trial/

Official syllabus

OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017, section 4.2.1 Alcohols, learning outcomes (a) to (e). Published by OCR.