OCR A Level Chemistry A: Organic synthesis, 6.2.5 (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.

Organic synthesis, section 6.2.5 in Module 6 of OCR A Level Chemistry A (H432), is the target of every question below, following the May 2026 specification (Version 3.1) for courses first examined in 2017. Every outcome, 6.2.5(a) to (c), is tested, including the PAG6 techniques for an organic solid. H432 is the full A level. Questions 6, 7 and 10 are longer multi-stage problems that draw on reactions from across Module 6.

Revise with the 6.2.5 study guide and the 6.2.5 revision notes first. The hub for H432 chemistry holds the other topics, every outcome appears on the outcome list to print, and a diagnostic quiz points you to weak areas.

Questions supply any data you need. Take O = 16.0, Br = 79.9, H = 1.0 and C = 12.0.

Questions

1. A student has a crude sample of an organic solid, D, that contains small amounts of soluble and insoluble impurities. D is very soluble in hot ethanol and only slightly soluble in cold ethanol. Describe how the student should purify D by recrystallisation. [4]

2. Crystals are collected by filtration under reduced pressure.

(a) State two advantages of filtration under reduced pressure over gravity filtration. [2] (b) Explain why the crystals are washed with ice-cold solvent rather than warm solvent. [1]

3. Take the melting point of a pure compound, G, as 128 °C. Sample P of G melts from 120 °C to 125 °C. Sample R melts from 127 °C to 128 °C.

(a) State which sample is purer and explain your answer. [2] (b) Sample P was recrystallised correctly but tested straight after filtration. Suggest why its melting point was still low. [1]

4. Give the reagent(s) for each conversion.

(a) benzene to chlorobenzene [1] (b) propanone to 2-hydroxy-2-methylpropanenitrile [1] (c) 1-bromobutane to butylamine [1] (d) ethanoic acid to ethanoyl chloride [1] (e) nitrobenzene to phenylamine [1]

5. Compound M is methyl 4-(hydroxymethyl)benzoate, HOCH₂C₆H₄COOCH₃.

(a) Name the two functional groups in M, other than the benzene ring. [2] (b) Give the structure of the organic product when M reacts with: (i) excess acidified potassium dichromate(VI), heated under reflux (assume the ester group is unchanged); (ii) hot aqueous sodium hydroxide; (iii) ethanoyl chloride. [3] (c) State whether M reacts with sodium carbonate solution, and explain why. [1]

6. Devise a three-stage synthesis of 1-hydroxycyclohexane-1-carboxylic acid from cyclohexanol. For each stage, give the reagents and conditions and the structure of the product. [6]

7. N-phenylpentanamide, CH₃CH₂CH₂CH₂CONHC₆H₅, is to be made from benzene and pentanoic acid.

(a) Devise a route. Give reagents for each step and name each intermediate. [4] (b) Write an equation for the final step. [1] (c) State whether the product is a primary or a secondary amide. [1]

8. A student adds excess bromine water to 1.50 g of phenol. After recrystallisation, 4.12 g of 2,4,6-tribromophenol is obtained.

(a) Construct the equation for this bromination. [1] (b) Determine the percentage yield of 2,4,6-tribromophenol. [3] (c) The product melts from 89 °C to 93 °C. Take the melting point of pure 2,4,6-tribromophenol as 95 °C. Comment on its purity. [1]

9. A student plans to make phenyl 2-methylpropanoate from 2-methylpropan-1-ol and phenol.

Step 1: Heat 2-methylpropan-1-ol with acidified potassium dichromate(VI) and distil off the product as it forms. Step 2: Heat the product of step 1 with phenol and a few drops of concentrated sulfuric acid.

Identify the problem with each step and explain how to correct it. [4]

10. N-propylbutanamide, CH₃CH₂CH₂CONHCH₂CH₂CH₃, is to be made using 1-bromopropane as the only organic starting material.

(a) Devise a synthesis. Give the reagents and conditions for each step and the structure of each intermediate. [7] (b) Explain why one step uses an excess of a reagent. [1] (c) Classify the product as a primary or a secondary amide. [1]

Answers

1. Dissolve D in the minimum volume of hot ethanol, heating in a water bath [1]. Filter the hot solution to remove insoluble impurities [1]. Cool the filtrate, then stand it in ice so crystals of D form while soluble impurities stay dissolved [1]. Collect by suction filtration, rinse with a few drops of ice-cold ethanol, then dry [1]. [4] Examiner insight: “Minimum” and “hot” are two separate ideas; a description that says only “dissolve in ethanol” does not show why the product crystallises on cooling.

2. (a) Any two: faster [1]; leaves the solid drier [1]; easier to wash the solid on the filter paper. [2] (b) Ice-cold solvent rinses off the impure solution without dissolving much of the product [1]. [1] Examiner insight: In part (b) refer to the product’s solubility; “to cool the crystals” does not answer the question.

3. (a) Sample R [1]. It melts over a narrow range (1 °C) at the value for pure G; P melts lower and over a 5 °C range [1]. [2] (b) The crystals were still damp with solvent, which acts as an impurity [1]. [1] Examiner insight: Mention both features of an impure sample, a lower value and a wider range, when comparing two samples.

4. (a) Cl₂ with a halogen carrier such as AlCl₃ [1] (b) NaCN(aq) and H⁺(aq), e.g. dilute H₂SO₄ [1] (c) Excess ammonia in ethanol [1] (d) SOCl₂ [1] (e) Tin and concentrated HCl [1] [5] Examiner insight: For part (c), state both “excess” and “ethanol”; ammonia alone does not show how further substitution to secondary and tertiary amines is limited.

5. (a) Primary alcohol [1]; ester [1]. [2] (b) (i) HOOCC₆H₄COOCH₃ [1]; (ii) HOCH₂C₆H₄COO⁻Na⁺ (and CH₃OH) [1]; (iii) CH₃COOCH₂C₆H₄COOCH₃ [1]. [3] (c) No: M has no carboxylic acid group, so no CO₂ is produced [1]. [1] Examiner insight: In (b)(i) reflux with excess oxidant takes the primary alcohol to the acid, not the aldehyde; match the product to the conditions given.

6. Stage 1: acidified potassium dichromate(VI), heat under reflux [1]; cyclohexanone [1]. Stage 2: NaCN(aq) and H⁺(aq) [1]; 1-hydroxycyclohexane-1-carbonitrile, with OH and CN on one ring carbon [1]. Stage 3: dilute HCl(aq), heat under reflux [1]; 1-hydroxycyclohexane-1-carboxylic acid, with OH and COOH on one ring carbon [1]. [6] Examiner insight: The CN step is what adds the carbon of the COOH group; check that your target has one more carbon than cyclohexanol before you start.

7. (a) Concentrated HNO₃ and concentrated H₂SO₄ give nitrobenzene [1]. Tin and concentrated HCl, then aqueous NaOH, give phenylamine [1]. SOCl₂ converts pentanoic acid into pentanoyl chloride [1]. Phenylamine reacts with pentanoyl chloride to give N-phenylpentanamide [1]. [4] (b) C₆H₅NH₂ + CH₃CH₂CH₂CH₂COCl → CH₃CH₂CH₂CH₂CONHC₆H₅ + HCl [1] [1] (c) Secondary amide [1] [1] Examiner insight: Name the intermediates clearly; “reduce it” without the reagents tin and concentrated HCl does not show which reduction you mean.

8. (a) C₆H₅OH + 3Br₂ → C₆H₂Br₃OH + 3HBr [1] [1] (b) n(phenol) = 1.50 ÷ 94.0 = 0.01596 mol [1]. M(C₆H₃Br₃O) = 330.7 g mol⁻¹, so theoretical mass = 0.01596 × 330.7 = 5.277 g [1]. Yield = 4.12 ÷ 5.277 × 100 = 78.1% [1] [3] (c) Impure: it melts below 95 °C and over a 4 °C range [1] [1] Examiner insight: One mole of phenol gives one mole of product, so the theoretical mass uses moles of phenol directly; the 3:1 ratio applies to bromine.

9. Step 1: distilling removes the aldehyde, 2-methylpropanal, before further oxidation [1]. Heat under reflux with excess dichromate to make 2-methylpropanoic acid [1]. Step 2: phenol is not readily esterified by a carboxylic acid [1]. Convert the acid into 2-methylpropanoyl chloride with SOCl₂, then react it with phenol [1]. [4] Examiner insight: For each step, name the fault and the fix; a correct reagent with no reason given shows only half the evaluation.

10. (a) Branch 1: excess ammonia in ethanol [1] gives propylamine, CH₃CH₂CH₂NH₂ [1]. Branch 2: KCN in ethanol, heat under reflux [1], gives butanenitrile, CH₃CH₂CH₂CN [1]. Dilute HCl(aq), heat under reflux, gives butanoic acid, CH₃CH₂CH₂COOH [1]. SOCl₂ gives butanoyl chloride, CH₃CH₂CH₂COCl [1]. Final step: propylamine reacts with butanoyl chloride to give N-propylbutanamide and HCl [1]. [7] (b) Excess ammonia makes it less likely that propylamine reacts further with 1-bromopropane to form secondary and tertiary amines [1]. [1] (c) Secondary amide [1]. [1] Examiner insight: The acyl part has four carbons but the starting material has three, so one branch must add a carbon through the nitrile; a route without this step cannot reach the target.

Where marks are usually lost

• Leaving “minimum” or “hot” out of a recrystallisation description.
• Washing crystals with warm solvent, or not washing them at all.
• Describing gravity filtration when filtration under reduced pressure is asked for.
• Saying an impure sample melts higher, or ignoring the range.
• Using a carboxylic acid with acid catalyst to esterify a phenol.
• Distilling when the carboxylic acid is wanted from a primary alcohol.
• Forgetting that the nitrile step adds a carbon, so the count is wrong.
• Giving “NH₃” for an amine with no mention of excess or ethanol.
• Missing the structure of an intermediate in a multi-stage route.

Next steps

• Recap the key facts: 6.2.5 revision notes
• Relearn a method: 6.2.5 study guide
• Module 4 questions on liquids and two-stage routes: 4.2.3 practice set
• Every H432 topic: chemistry hub
• Track your outcomes: printable outcome list
• Find weak areas: all free 10-minute diagnostics
• Prefer live help? Book a free trial class and work through routes with a tutor.

Official syllabus

Aligned to section 6.2.5 (Organic synthesis, outcomes (a) to (c)) of OCR’s H432 A Level Chemistry A specification, issued as Version 3.1 in May 2026 for the course first examined in 2017. Published by OCR.