OCR A Level Chemistry A: Organic synthesis (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 4.2.3, Organic synthesis, of the OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017. The questions test every learning outcome, 4.2.3(a) to (c), from Module 4, Core organic chemistry, including the PAG5 techniques. H432 is the full A level. Questions 9 to 11 are longer problems that combine outcomes with earlier Module 4 chemistry.

Revise first with the Organic synthesis study guide and the Organic synthesis revision notes. Other topics are on the OCR chemistry hub, and the printable checklist lists every outcome. For a wider check, try the free diagnostics.

Any data you need, such as densities or boiling points, appears in the question itself.

Questions

1. A student sets up Quickfit apparatus to heat a mixture under reflux.

(a) State why Quickfit glassware can be assembled without corks or bungs. [1] (b) State which end of the condenser the cooling water should enter, and why. [1] (c) Explain why the top of the condenser must not be stoppered. [1]

2. 2-methylpropan-1-ol, (CH₃)₂CHCH₂OH, is oxidised with acidified potassium dichromate(VI).

(a) Name the apparatus set-up used to obtain 2-methylpropanal, and explain why it gives the aldehyde. [2] (b) State two changes to the method needed to obtain 2-methylpropanoic acid instead. [2]

3. A crude organic liquid, density 0.89 g cm⁻³, contains traces of hydrochloric acid and water. Describe how a separating funnel and a drying agent are used to give a dry organic liquid ready for redistillation. [4]

4. Compound B has the structural formula (CH₃)₂C(OH)CH₂CH=CHCH₂CH₂OH. Identify the three functional groups in B, classifying any alcohol groups as primary, secondary or tertiary. [3]

5. Predict the organic product, as a structural formula, when compound B from question 4 reacts with each reagent. Include an observation for (b).

(a) Bromine [1] (b) Excess acidified potassium dichromate(VI), heated under reflux [3] (c) Hydrogen with a nickel catalyst [1]

6. Compound B (Mr 144.0) and decane (Mr 142.0) have similar Mr values. Predict which has the higher boiling point and which is more soluble in water, and explain both predictions. [3]

7. Devise a two-stage synthesis of pentan-3-one from 3-bromopentane. Give the reagents and conditions for each stage and the structure of the intermediate. [4]

8. A student plans to make butan-1-ol from but-1-ene in two stages: add HBr, then heat with aqueous sodium hydroxide.

(a) Give the structure of the major organic product of stage 1 and explain why it forms. [3] (b) State the alcohol the student will mainly obtain from stage 2. [1]

9. A student makes 1-bromohexane by heating 10.0 cm³ of hexan-1-ol (density 0.820 g cm⁻³) under reflux with excess sodium bromide and sulfuric acid. After purification, 8.45 g of 1-bromohexane is collected. Take Br as 79.9, O as 16.0, C as 12.0 and H as 1.0.

(a) Write an equation for the reaction, using sodium bromide and sulfuric acid as reactants. [1] (b) Calculate the percentage yield. [4] (c) The crude product is shaken with sodium hydrogencarbonate solution. State the purpose of this step and why the tap must be opened often. [2] (d) After drying, the liquid is redistilled. State how the drying agent is removed first and which fraction is collected. Data: 1-bromohexane boils at about 155 °C. [2]

10. Use this information, which is not in Module 4: a haloalkane heated with potassium hydroxide dissolved in ethanol loses HX to form an alkene.

(a) Devise a two-stage route from 1-bromo-2-methylpropane, (CH₃)₂CHCH₂Br, to 2-bromo-2-methylpropane, (CH₃)₃CBr. Give reagents, conditions and the intermediate. [4] (b) Explain why 2-bromo-2-methylpropane is the major product of stage 2, and name the minor product. [3]

11. 2-methylbutan-2-ol is (CH₃)₂C(OH)CH₂CH₃.

(a) A student claims it can be converted into a ketone by heating with acidified potassium dichromate(VI). Explain why this is wrong. [2] (b) Devise a two-stage route from 2-methylbutan-2-ol to 2-methylbutane. [3] (c) Give the reagents for a one-stage conversion of 2-methylbutan-2-ol into 2-bromo-2-methylbutane. [1] (d) The alkene mixture from your first stage in (b) is collected by distillation. Suggest why this does not affect the final product. [1]

Answers

1. (a) The pieces have standard ground-glass joints that fit each other [1] (b) At the bottom (out at the top), so the jacket stays full of cold water [1] (c) Heating a sealed system builds pressure, so the apparatus could shatter [1] Examiner insight: In (b) the reason earns the credit along with the direction; “at the bottom” alone does not explain why the condenser works better that way.

2. (a) Distillation [1]; its boiling point is below that of the alcohol it came from, so it leaves the flask as soon as it forms and avoids further oxidation [1] (b) Heat under reflux instead of distilling [1]; use excess oxidising agent [1] Examiner insight: “Heat more strongly” is not a change of method; name reflux and excess dichromate explicitly.

3. Shake with water in the separating funnel, opening the tap often to release pressure [1]. The organic liquid (0.89 g cm⁻³) is less dense, so it is the upper layer; run off and discard the lower aqueous layer [1]. Wash with sodium hydrogencarbonate solution to remove the acid, venting the CO₂, and run off the aqueous layer again [1]. Transfer the organic layer to a dry conical flask, then swirl it with anhydrous MgSO₄ or CaCl₂ until it turns clear [1] Examiner insight: Use the density value given; an answer that says the organic layer is on top “because organic liquids float” does not show the reasoning the data was there to test.

4. Tertiary alcohol, (CH₃)₂C(OH)– [1]; alkene, C=C [1]; primary alcohol, –CH₂OH [1] Examiner insight: Writing “alcohol” twice without classifying each OH misses the point of the question, since the two OH groups behave differently with oxidising agents.

5. (a) (CH₃)₂C(OH)CH₂CHBrCHBrCH₂CH₂OH [1] (b) Orange to green [1]; the primary alcohol is oxidised to a carboxylic acid [1]; the tertiary alcohol is unchanged: (CH₃)₂C(OH)CH₂CH=CHCH₂COOH [1] (c) (CH₃)₂C(OH)CH₂CH₂CH₂CH₂CH₂OH [1] Examiner insight: In (b) the tertiary OH must still be shown in the product; oxidising both OH groups is a common way to lose the structure mark.

6. Compound B has the higher boiling point and is more soluble in water [1]. B has two O–H groups, so it forms hydrogen bonds between its molecules, which need more energy to break than the London forces between decane molecules [1]. B can also hydrogen bond with water molecules, but decane cannot [1] Examiner insight: Name the intermolecular forces for both compounds; “B has stronger bonds” suggests covalent bonds break on boiling.

7. Stage 1: NaOH(aq) [1], heat under reflux [1], giving the intermediate pentan-3-ol, CH₃CH₂CH(OH)CH₂CH₃ [1]. Stage 2: K₂Cr₂O₇ with H₂SO₄, heat under reflux, giving CH₃CH₂COCH₂CH₃ [1] Examiner insight: Each stage needs reagents and conditions; “NaOH” without “aqueous” can be read as a different reaction, so state the solvent.

8. (a) CH₃CH₂CHBrCH₃, 2-bromobutane [1]. The H adds to the CH₂ carbon, giving a secondary carbocation, CH₃CH₂C⁺HCH₃, rather than a primary one [1]; the secondary carbocation is more stable, so this product is major (Markownikoff’s rule) [1] (b) Butan-2-ol [1] Examiner insight: Refer to the relative stability of the carbocations; quoting Markownikoff’s rule as “H goes to the carbon with more H” without the carbocations is a weaker explanation.

9. (a) CH₃(CH₂)₄CH₂OH + NaBr + H₂SO₄ → CH₃(CH₂)₄CH₂Br + NaHSO₄ + H₂O [1] (b) Mass of hexan-1-ol = 10.0 × 0.820 = 8.20 g [1]; n = 8.20 ÷ 102.0 = 0.0804 mol [1]; theoretical mass of C₆H₁₃Br = 0.0804 × 164.9 = 13.26 g [1]; yield = 8.45 ÷ 13.26 × 100 = 63.7% [1] (c) To neutralise and remove acid left in the crude product [1]; CO₂ gas is produced, so pressure builds in the stoppered funnel [1] (d) Decant or filter the liquid away from the solid drying agent [1]; collect the fraction that boils close to 155 °C [1] Examiner insight: Keep unrounded values through (b) and round only at the end; rounding the amount to 0.08 mol early shifts the yield noticeably.

10. (a) Stage 1: KOH in ethanol, heat [1], giving 2-methylpropene, (CH₃)₂C=CH₂ [1]. Stage 2: HBr [1], adding across the C=C to give (CH₃)₃CBr [1] (b) H adds to the CH₂ carbon, forming the tertiary carbocation (CH₃)₃C⁺ [1]; a tertiary carbocation is more stable than the primary alternative, so (CH₃)₃CBr is major [1]. Minor product: 1-bromo-2-methylpropane [1] Examiner insight: When a question supplies a reaction, use its reagents and conditions as written; replacing ethanolic KOH with aqueous KOH changes the reaction to hydrolysis.

11. (a) It is a tertiary alcohol [1]; the C–OH carbon has no hydrogen atom to lose, so it resists oxidation and no ketone forms [1] (b) Stage 1: heat with an acid catalyst such as H₃PO₄ or H₂SO₄, giving an alkene such as 2-methylbut-2-ene [1]. Stage 2: H₂ [1] with a Ni catalyst, giving (CH₃)₂CHCH₂CH₃ [1] (c) NaBr and H₂SO₄ [1] (d) Every alkene in the mixture has the same carbon skeleton, so all give 2-methylbutane on hydrogenation [1] Examiner insight: In (b) the catalyst must be named; “add hydrogen” alone does not give the conditions for the stage.

Where marks are usually lost

• Describing a reflux set-up with a stoppered condenser or a thermometer that serves no purpose.
• Choosing reflux for an aldehyde, or distillation for a carboxylic acid.
• Ignoring density data and guessing which layer is organic.
• Describing drying as “removing impurities” rather than removing water.
• Leaving out redistillation, or not stating the fraction collected.
• Missing a functional group, or not classifying each alcohol group.
• Oxidising a tertiary alcohol in a predicted product or a route.
• Forgetting Markownikoff’s rule when HBr adds to an unsymmetrical alkene.
• Reagents without conditions, especially “NaOH” with no “aqueous” or “heat”.

Next steps

• Organic synthesis revision notes
• Organic synthesis study guide
• Practical endorsement revision notes
• OCR A Level Chemistry hub
• Printable checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017, section 4.2.3 Organic synthesis, learning outcomes (a) to (c). Published by OCR.