OCR A Level Chemistry A: Organic synthesis (H432)

This study guide teaches 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 section sits in Module 4, Core organic chemistry, and its learning outcomes are 4.2.3(a) to (c), all covered below with worked examples. H432 is the full A level. The specification states that Module 4 content is assessed in Synthesis and analytical techniques (02) and Unified chemistry (03). Outcome (a) is linked to PAG5, Synthesis of an organic liquid.

Every H432 topic is on the OCR chemistry hub; mark your progress with the printable checklist.

Outcomes in 4.2.3

Statement	Skill required	Specification guidance and links
4.2.3(a)(i)	Use Quickfit apparatus, including for distillation and heating under reflux	PAG5; HSW4
4.2.3(a)(ii)	Prepare and purify an organic liquid: separating funnel to remove an organic layer from an aqueous layer; drying with an anhydrous salt (e.g. MgSO₄, CaCl₂); redistillation	PAG5; HSW4
4.2.3(b)(i)	Identify the individual functional groups in a molecule with several of them	Functional groups met in Module 4; HSW3
4.2.3(b)(ii)	Predict the properties and reactions of such a molecule	HSW3
4.2.3(c)	Devise two-stage synthetic routes	Transformations between all functional groups met so far; extra information may be given on exam papers; HSW3

The PAG table in the specification lists heating under reflux, purification using a separating funnel, distillation and risk assessment for PAG5, with synthesis of a haloalkane as one suitable activity. It notes that reflux may be covered in PAG5 or PAG6. Section 6.2.5 in Module 6 later builds on this section.

4.2.3(a)(i): Quickfit apparatus

Quickfit glassware has standard ground-glass joints, so a flask, condenser, still head and adapter fit together without corks or bungs. Clamp the flask and condenser to a stand. Use a round-bottomed or pear-shaped flask no more than about half full, with anti-bumping granules so the liquid boils smoothly. Most organic liquids are flammable, so heat with an electric heating mantle or a water bath, not a Bunsen flame.

Heating under reflux

The condenser stands vertically on top of the flask. Cooling water enters at the bottom and leaves at the top, so the jacket stays full. Vapour rises, condenses and drips back into the flask. The top of the condenser stays open to the air.

Reflux lets you heat a slow reaction for a long time without losing volatile reactants or products.

Distillation

The flask is fitted with a still head. The thermometer bulb sits level with the side arm, so it reads the temperature of the vapour leaving for the condenser. The condenser slopes down to a receiver, again with water in at the bottom. You collect the liquid that distils over close to the boiling point of the substance you want.

Distillation removes a volatile product as it forms, or separates liquids with different boiling points.

Worked example 1: choosing the set-up

A primary alcohol is oxidised with acidified potassium dichromate(VI). Choose the set-up for each target product.

1. Aldehyde wanted. Use distillation. With no O–H group, the aldehyde lacks the hydrogen bonding of the alcohol, has the lower boiling point, and escapes from the flask before more oxidation can happen.
2. Carboxylic acid wanted. Use heating under reflux with excess oxidising agent. Any aldehyde that vaporises condenses and falls back, so it stays in contact with the dichromate until it is oxidised to the acid.
3. Either way, the colour turns from orange to green.

Safety

Never heat a closed system: a stoppered reflux condenser builds pressure and can shatter. A PAG5 risk assessment also covers corrosive acids, flammable liquids and eye protection.

4.2.3(a)(ii): Preparing and purifying an organic liquid

A haloalkane can be made from an alcohol using sodium bromide and sulfuric acid (4.2.1(e)). The crude product contains unreacted alcohol, water, acid and by-products. Purify it in this order.

1. React. Heat the alcohol, sodium bromide and sulfuric acid under reflux.
2. Isolate the crude product. Distil the haloalkane out of the reaction mixture.
3. Separate. Pour the distillate into a separating funnel with water. Stopper, invert and shake, opening the tap often to release pressure. Let the layers settle. The less dense liquid floats on top.
4. Wash. Remove leftover acid by shaking with sodium hydrogencarbonate solution. Carbon dioxide forms, so vent often. Run off the aqueous layer each time.
5. Dry. Run the organic layer into a dry conical flask. Add an anhydrous salt such as anhydrous magnesium sulfate or anhydrous calcium chloride and swirl. The liquid is dry when it changes from cloudy to clear and some drying agent moves freely instead of clumping.
6. Redistil. Decant or filter off the drying agent, then redistil. Collect only the fraction that boils within a narrow range around the product’s boiling point.

Worked example 2: which layer is which?

The haloalkane in a preparation has density 1.21 g cm⁻³ (value supplied). After shaking with water, a student is unsure which layer to keep.

1. Water has density about 1.00 g cm⁻³.
2. The haloalkane is denser, so it forms the lower layer.
3. Run the lower layer into a clean flask and keep it. Discard the aqueous upper layer.
4. If still unsure, add a few drops of water: the layer that grows is the aqueous one.

Worked example 3: percentage yield of a preparation

This links to 2.1.3 in Module 2. A student heats 11.0 cm³ of 3-methylbutan-1-ol (density 0.810 g cm⁻³) under reflux with excess sodium bromide and sulfuric acid. After purification, 9.80 g of 1-bromo-3-methylbutane is collected. Take Br as 79.9, O as 16.0, C as 12.0 and H as 1.0.

The equation is: (CH₃)₂CHCH₂CH₂OH + NaBr + H₂SO₄ → (CH₃)₂CHCH₂CH₂Br + NaHSO₄ + H₂O

1. Mass of alcohol = 11.0 × 0.810 = 8.91 g.
2. M(C₅H₁₂O) = 88.0 g mol⁻¹, so n(alcohol) = 8.91 ÷ 88.0 = 0.10125 mol.
3. One mole of alcohol gives one mole of bromoalkane, so at most 0.10125 mol can form.
4. M(C₅H₁₁Br) = 150.9 g mol⁻¹, so the theoretical mass = 0.10125 × 150.9 = 15.28 g.
5. Percentage yield = 9.80 ÷ 15.28 × 100 = 64.1%.

Losses at each transfer, in aqueous layers and on the drying agent make yields well below 100% normal.

4.2.3(b): Molecules with several functional groups

You must recognise the functional groups met in Module 4 and predict how each behaves. Treat each group on its own, then check whether a reagent would attack more than one group.

Functional group	How to spot it	Typical reactions in Module 4
Alkene	C=C	Adds H₂ (Ni catalyst), halogens, hydrogen halides, steam (H₃PO₄ catalyst); decolourises bromine; addition polymerisation
Primary alcohol	–CH₂OH	Oxidised to aldehyde (distil) or carboxylic acid (reflux); dehydration; substitution to haloalkane
Secondary alcohol	>CHOH	Oxidised to ketone; dehydration; substitution
Tertiary alcohol	OH on a carbon bonded to three carbons	Resists oxidation; dehydration; substitution
Haloalkane	C–X	Hydrolysis by aqueous alkali to an alcohol; hydrolysis by water with AgNO₃ and ethanol
Aldehyde	–CHO	Further oxidation to carboxylic acid
Ketone	C=O between two carbons	No reaction with acidified dichromate
Carboxylic acid	–COOH	Acid reactions (2.1.4)
Alkane chain	C–C and C–H only	Combustion; radical substitution with Cl₂ or Br₂ in UV

An O–H group allows hydrogen bonding, raising the boiling point and giving some water solubility. A C=C bond is reactive because the π-bond has a relatively low bond enthalpy. A polar C–X bond makes the carbon a target for nucleophiles.

Worked example 4: predicting reactions

Compound A is CH₂=CHCH₂CH₂CH(OH)CH₂Cl, 1-chlorohex-5-en-2-ol.

(b)(i) Identify the groups. A has an alkene (C=C at the end of the chain), a secondary alcohol (OH on carbon 2, which is bonded to two other carbons) and a chloroalkane (C–Cl on carbon 1).

(b)(ii) Predict reactions.

1. Bromine. The C=C adds Br₂, so bromine is decolourised. Product: CH₂BrCHBrCH₂CH₂CH(OH)CH₂Cl.
2. H₂ with a nickel catalyst. Only the C=C reacts. Product: CH₃CH₂CH₂CH₂CH(OH)CH₂Cl.
3. Acidified potassium dichromate(VI), heated under reflux. The secondary alcohol is oxidised to a ketone; the colour turns from orange to green. Product: CH₂=CHCH₂CH₂COCH₂Cl.
4. Aqueous sodium hydroxide, heated. The C–Cl bond is hydrolysed. Product: CH₂=CHCH₂CH₂CH(OH)CH₂OH, a diol.
5. Properties. The OH group hydrogen bonds, so A is less volatile than an alkene of similar Mr and partly soluble in water.

4.2.3(c): Two-stage synthetic routes

The specification says you should devise two-stage routes using transformations between all the functional groups met up to this point. Exam papers may also supply extra reactions to extend your toolkit. Use them as given.

Module 4 reaction toolkit

From	To	Reagents and conditions
Alkane	Haloalkane	Cl₂ or Br₂, UV radiation (gives a mixture of products)
Alkene	Alkane	H₂, Ni catalyst
Alkene	Dihaloalkane	Br₂ or Cl₂
Alkene	Haloalkane	HBr or HCl (Markownikoff’s rule for unsymmetrical alkenes)
Alkene	Alcohol	Steam, H₃PO₄ catalyst
Alcohol	Alkene	H₃PO₄ or H₂SO₄ catalyst, heat
Alcohol	Haloalkane	NaBr and H₂SO₄ (halide ions with acid)
Haloalkane	Alcohol	NaOH(aq), heat under reflux
Primary alcohol	Aldehyde	K₂Cr₂O₇ / H₂SO₄, distil the aldehyde off as it forms
Primary alcohol	Carboxylic acid	Excess K₂Cr₂O₇ / H₂SO₄, reflux
Secondary alcohol	Ketone	K₂Cr₂O₇ / H₂SO₄, reflux

Method

1. Compare the starting material and the target. Which groups change? No Module 4 reaction adds carbon atoms, so the skeleton must match.
2. Work backwards: what single step makes the target, and from what?
3. Check that the intermediate can be made from the starting material in one step.
4. Write each step with reagents, conditions and the structure of the intermediate.

Worked example 5: 1-chloropentane to pentanoic acid

1. The target is a carboxylic acid with the same five-carbon chain, which comes from a primary alcohol.
2. Step 1: NaOH(aq), heat under reflux. 1-chloropentane → pentan-1-ol, CH₃(CH₂)₃CH₂OH.
3. Step 2: excess K₂Cr₂O₇ with H₂SO₄, heat under reflux. Pentan-1-ol → pentanoic acid, CH₃(CH₂)₃COOH.

Worked example 6: pentan-2-ol to pentane

1. Step 1: heat with an acid catalyst such as H₃PO₄. Pentan-2-ol loses water to give pent-1-ene and pent-2-ene (E and Z).
2. Step 2: H₂ with a Ni catalyst. Every one of these alkenes gives pentane, so the mixture from step 1 does not matter.

Worked example 7: cyclopentane to cyclopentanol

1. An alkane has no functional group, so start with radical substitution.
2. Step 1: Br₂ in UV radiation gives bromocyclopentane. All ten hydrogens are equivalent, so only one monobromo product forms, though further substitution still occurs (4.1.2(g)).
3. Step 2: NaOH(aq), heat under reflux, gives cyclopentanol.

Common errors

• Stoppering a reflux condenser, or putting the distillation thermometer bulb in the liquid.
• Giving “heat” with no mention of reflux or distillation for an alcohol oxidation.
• Trying to oxidise a tertiary alcohol in a route.
• Using a reagent that also attacks a group you wanted to keep.

Next steps

• Condensed tables and a quick self-test: Organic synthesis revision notes
• Original questions with marked answers: Organic synthesis practice questions
• The wider practical techniques: Practical endorsement study guide
• Yield calculations: Amount of substance
• Related: Alcohols, Haloalkanes, Module 6 organic synthesis
• Find your gaps first: free diagnostics

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.