OCR A Level Chemistry A: Organic synthesis, 6.2.5 (H432)

This guide teaches section 6.2.5, Organic synthesis, from Module 6 of OCR A Level Chemistry A, H432. It follows Version 3.1 of the OCR specification, dated May 2026 and first assessed in 2017, and covers every learning outcome, 6.2.5(a) to (c). Each part has a worked example. The specification treats H432 as a single full A level and examines Module 6 in components 02 and 03. Outcome (a) links to PAG6, Synthesis of an organic solid, whose example activity in the specification is the synthesis of aspirin.

Other H432 sections are reached from the course hub for H432 chemistry. The outcome checklist (printable) lets you log progress, and you can use our diagnostic quizzes to locate weak spots.

Outcomes in 6.2.5

Statement	What you must be able to do	Specification links
6.2.5(a)(i)	Organic preparation: Quickfit apparatus, distillation, heating under reflux	PAG6; HSW4; see also 4.2.3(a)
6.2.5(a)(ii)	Purify an organic solid: filtration under reduced pressure, recrystallisation, melting points	PAG6; HSW4
6.2.5(b)(i)	Identify each functional group in a molecule that has several	Module 6 functional groups; 4.2.3(b); HSW3
6.2.5(b)(ii)	Predict the properties and reactions of that molecule	HSW3
6.2.5(c)	Devise multi-stage synthetic routes	All functional groups studied; extra information may be given on papers; HSW3

This section extends the Module 4 synthesis guide (4.2.3), which deals with organic liquids and two-stage routes.

6.2.5(a)(i): Organic preparation

• Heating under reflux keeps a slow reaction hot for a long time. The vertical condenser is open at the top, and vapour condenses back into the flask.
• Distillation removes a volatile liquid. The thermometer bulb sits beside the side arm of the still head.

Water enters the condenser at its lower end. Use an electric heater and anti-bumping granules.

When the product is a solid, the preparation usually ends by making it leave solution: cool in ice, or change the pH so an insoluble form precipitates. You then collect and purify it.

6.2.5(a)(ii): Purifying an organic solid

Filtration under reduced pressure

A Büchner funnel holds a flat filter paper and sits in a side-arm flask joined to a vacuum source such as a water pump. Wet the paper with cold solvent so it seals, then pour in the mixture. Suction pulls the liquid through quickly and leaves the crystals drier than gravity filtration. Rinse them with a few drops of ice-cold solvent to remove clinging solution.

Recrystallisation

Choose a solvent in which the product is very soluble when hot but only slightly soluble when cold.

1. Add hot solvent bit by bit until the crude solid just dissolves: use the minimum volume.
2. Filter off any insoluble bits while hot.
3. Cool slowly, then in ice, so crystals form.
4. Filter under reduced pressure; rinse with a few drops of ice-cold solvent.
5. Dry the crystals, e.g. in a desiccator.

Soluble impurities are present in small amounts, so they never reach saturation and stay in solution. Excess solvent lowers the yield.

Measuring a melting point

Pack dry, powdered solid into a capillary tube sealed at one end. Heat it in a melting point apparatus, slowly near the expected value, and record the range from first liquid to complete melting.

• A pure substance melts sharply, over about 1 °C, close to its known value.
• Impurities lower the melting point and spread melting over several degrees.

Worked example 1: benzoic acid from methyl benzoate

A student heats 5.00 cm³ of methyl benzoate (density 1.09 g cm⁻³, supplied) under reflux with excess aqueous sodium hydroxide. After cooling, she adds dilute hydrochloric acid until the mixture is acidic, and benzoic acid precipitates. After recrystallisation from hot water and drying, 3.62 g of benzoic acid is obtained. Use H = 1.0, O = 16.0 and C = 12.0.

C₆H₅COOCH₃ + NaOH → C₆H₅COONa + CH₃OH C₆H₅COONa + HCl → C₆H₅COOH + NaCl

1. Reflux is used because alkaline hydrolysis (6.1.3(d)(ii)) is slow and volatile substances must not escape.
2. Sodium benzoate is ionic and stays dissolved; benzoic acid is much less soluble in cold water, so acidifying precipitates it.
3. Mass of ester = 5.00 × 1.09 = 5.45 g. M(C₈H₈O₂) = 136.0 g mol⁻¹, so n = 5.45 ÷ 136.0 = 0.04007 mol.
4. One mole of ester gives one mole of acid. M(C₇H₆O₂) = 122.0 g mol⁻¹, so the theoretical mass = 0.04007 × 122.0 = 4.889 g.
5. Percentage yield = 3.62 ÷ 4.889 × 100 = 74.0%.

Worked example 2: reading melting point results

Take the melting point of pure benzoic acid as 122 °C. The crude solid melted from 117 °C to 120 °C. The recrystallised solid melted from 121 °C to 122 °C.

1. The crude sample melts below 122 °C and over 3 °C, so it is impure.
2. The recrystallised sample melts over 1 °C, close to 122 °C, so it is much purer.

6.2.5(b): Molecules with several functional groups

You need the Module 4 groups plus these. Treat each group separately, then check whether one reagent attacks two.

Group	How to spot it	Reactions to predict
Arene	Benzene ring	Electrophilic substitution: nitration, halogenation, Friedel–Crafts
Phenol	OH bonded directly to the ring	Reacts with NaOH, not carbonates; brominated without a halogen carrier
Aldehyde	–CHO	Oxidised to acid; reduced by NaBH₄; HCN addition; 2,4-DNP and Tollens’ positive
Ketone	C=O with a carbon on each side	Reduced by NaBH₄; HCN addition; 2,4-DNP positive, Tollens’ negative
Carboxylic acid	–COOH	Acid reactions; esterified; SOCl₂ gives acyl chloride
Ester	–COO– between two carbons	Hydrolysed by hot aqueous acid or alkali
Acyl chloride	–COCl	Gives esters, acids and amides
Amine	N bonded to carbon	Base: forms salts with acids
Amide	–CONH– or –CONH₂	Hydrolysed by acid or alkali
Nitrile	–C≡N	Reduced to amine; hydrolysed to acid

Groups that hydrogen bond (O–H, N–H, COOH) raise melting and boiling points and increase water solubility.

Worked example 3: a keto acid

Compound K is HOC₆H₄COCH₂CH₂COOH, with the OH and the side chain at positions 1 and 4 of the ring.

(b)(i) Groups. K contains a phenol (OH on the ring), a ketone (C=O between a ring carbon and a chain carbon) and a carboxylic acid.

(b)(ii) Predictions.

1. Sodium carbonate solution. Only the COOH reacts, giving effervescence (CO₂).
2. Excess sodium hydroxide. Both acidic groups react: NaOC₆H₄COCH₂CH₂COONa forms.
3. 2,4-DNP. An orange precipitate shows the carbonyl. Tollens’ reagent gives no silver mirror, because a ketone is not oxidised.
4. NaBH₄. The ketone is reduced to a secondary alcohol, HOC₆H₄CH(OH)CH₂CH₂COOH. The COOH is unchanged. The new CH(OH) carbon is a chiral centre (6.2.2(d)).
5. Ethanol with concentrated H₂SO₄, heated. The COOH forms an ethyl ester. The phenol OH is not readily esterified by a carboxylic acid.
6. Properties. Two O–H groups hydrogen bond, so K is a solid with a fairly high melting point.

6.2.5(c): Multi-stage synthetic routes

You may need to link any functional groups in the specification, in several steps. A paper may supply an unfamiliar reaction; use it as stated.

Module 6 reaction toolkit

Combine this with the Module 4 table in the 4.2.3 guide.

From	To	Reagents and conditions
Arene	Nitroarene	Concentrated HNO₃ and concentrated H₂SO₄
Arene	Haloarene	Cl₂ or Br₂ with a halogen carrier, e.g. AlCl₃ or FeBr₃
Arene	Alkylarene or aromatic ketone	Haloalkane or acyl chloride with a halogen carrier (Friedel–Crafts)
Phenol	2,4,6-tribromophenol; 2- and 4-nitrophenol	Br₂; dilute HNO₃
Nitroarene	Aromatic amine	Tin and concentrated HCl, then aqueous alkali
Haloalkane	Amine	Excess ethanolic NH₃ (an amine gives a secondary or tertiary amine)
Haloalkane	Nitrile (one more carbon)	CN⁻ (e.g. KCN) in ethanol
Aldehyde or ketone	Hydroxynitrile (one more carbon)	NaCN(aq) and H⁺(aq)
Aldehyde or ketone	Alcohol	NaBH₄
Aldehyde	Carboxylic acid	K₂Cr₂O₇ and H₂SO₄, heat under reflux
Nitrile	Amine	H₂ with Ni catalyst
Nitrile	Carboxylic acid	Dilute aqueous acid, heat under reflux
Carboxylic acid	Ester	Alcohol with concentrated H₂SO₄ catalyst
Acid anhydride	Ester	Alcohol
Ester	Acid + alcohol, or carboxylate + alcohol	Hot aqueous acid, or hot aqueous alkali
Carboxylic acid	Acyl chloride	SOCl₂
Acyl chloride	Ester, acid, primary or secondary amide	Alcohol or phenol; water; NH₃; primary amine

Planning method

1. Name the groups in the start and the target. Count the carbons. One extra carbon means a nitrile or hydroxynitrile step; a new carbon on a ring means Friedel–Crafts.
2. Work backwards from the target, one step at a time, until you reach something you can make from the start.
3. On a ring, place substituents in an order that the directing effects allow (6.1.1(k)–(l)). Introduce groups that would react with a later reagent, such as NH₂, as late as possible.
4. For each step give the reagent, the conditions and the intermediate’s structure.

Worked example 4: pent-1-ene to 2-methylpentan-1-amine

The target, CH₃CH₂CH₂CH(CH₃)CH₂NH₂, has six carbons; the start has five. A CH₂NH₂ group at the end of a branch comes from reducing a nitrile.

1. HBr adds to pent-1-ene. By Markownikoff’s rule the major product is 2-bromopentane, CH₃CH₂CH₂CHBrCH₃.
2. KCN in ethanol, heat under reflux. CN⁻ substitutes for Br⁻, giving 2-methylpentanenitrile, CH₃CH₂CH₂CH(CH₃)CN. This adds the sixth carbon.
3. H₂ with a nickel catalyst reduces C≡N to CH₂NH₂: 2-methylpentan-1-amine.

Ammonia on 2-bromopentane would give a five-carbon amine.

Worked example 5: benzene to 1-(3-aminophenyl)ethanone

Information supplied: the COCH₃ group is 3-directing. The target has NH₂ and COCH₃ at positions 1 and 3 of the ring.

1. Ethanoyl chloride, CH₃COCl, with AlCl₃ (Friedel–Crafts acylation) gives phenylethanone, C₆H₅COCH₃.
2. Concentrated HNO₃ and concentrated H₂SO₄. The COCH₃ group directs the nitro group to position 3: 1-(3-nitrophenyl)ethanone.
3. Tin and concentrated HCl, then aqueous NaOH, reduce NO₂ to NH₂: 1-(3-aminophenyl)ethanone.

The amine must come last. NH₂ is 2- and 4-directing, and it would react with ethanoyl chloride to form an amide.

Worked example 6: phenyl propanoate from propan-1-ol and phenol

1. Excess K₂Cr₂O₇ and H₂SO₄, heat under reflux, oxidises propan-1-ol to propanoic acid.
2. SOCl₂ converts the acid to propanoyl chloride, CH₃CH₂COCl.
3. Phenol reacts with propanoyl chloride: C₆H₅OH + CH₃CH₂COCl → CH₃CH₂COOC₆H₅ + HCl.

Step 2 is needed because phenol is not readily esterified by a carboxylic acid.

Common errors

• Using a large volume of hot solvent, so product stays dissolved.
• Washing crystals with solvent that is not ice-cold.
• Calling an impure sample’s melting point “higher” or “sharper”.
• Using ammonia when the target amine needs an extra carbon.
• Esterifying a phenol with a carboxylic acid.
• Making NH₂ before a later acyl chloride or substitution step.
• Leaving out conditions such as reflux, catalyst or solvent.

Next steps

• Condensed tables and a self-test: revision notes for 6.2.5
• Marked practice: questions on 6.2.5
• The reactions behind the toolkit: arenes and phenols (6.1.1), aldehydes and ketones (6.1.2), acids, esters and acyl chlorides (6.1.3), amines (6.2.1)
• Techniques across the course: PAG and endorsement techniques
• Check your gaps: diagnostic quizzes

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.