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

These notes condense section 6.2.5, Organic synthesis, from Module 6 of H432, OCR’s A Level Chemistry A, using Version 3.1 of its specification (May 2026; first examined 2017). All of outcomes 6.2.5(a) to (c) are here. Components 02 and 03 examine Module 6, and H432 is one full A level. Full explanations and longer examples are in the 6.2.5 study guide. Then test yourself with the 6.2.5 practice questions.

Every other topic is linked from the central H432 course page. Tick outcomes on the paper checklist for H432, and run a diagnostic quiz to see what needs work. Organic liquids and two-stage routes sit in the Module 4 notes (4.2.3).

6.2.5(a): Making and purifying an organic solid

This outcome links to PAG6, Synthesis of an organic solid, in the specification.

Technique	Purpose	Detail that earns credit
Quickfit apparatus	Leak-free assembly	Ground-glass joints; clamp flask and condenser
Heating under reflux	Heat a slow reaction for a long time	Vertical condenser, open top; water in at the bottom
Distillation	Remove a volatile liquid	Thermometer bulb level with the side arm
Filtration under reduced pressure	Collect crystals quickly and fairly dry	Büchner funnel, side-arm flask, vacuum source
Recrystallisation	Remove impurities from a solid	Minimum hot solvent; cool; filter; wash cold; dry
Melting point	Check purity and identity	Capillary tube; heat slowly near the expected value

Method in steps: recrystallisation

1. Pick a solvent: product very soluble hot, slightly soluble cold.
2. Add hot solvent until the crude solid just dissolves (the minimum volume).
3. Filter while hot if insoluble solid remains.
4. Cool slowly, then in ice.
5. Filter under reduced pressure.
6. Wash with a little ice-cold solvent.
7. Dry, then weigh and take the melting point.

Why it works: the product saturates the cold solution and crystallises. Soluble impurities are too dilute to saturate it, so they stay dissolved.

Melting point rules

• Pure: sharp range (about 1 °C) at the known value.
• Impure: lower and wider range.
• Damp: melts low as well, so dry first.

Worked reminder: mass recovered

3.40 g of crude solid gives 2.55 g after recrystallisation. Recovery = 2.55 ÷ 3.40 × 100 = 75.0%. Some product always stays in the cold solvent, so recovery is never 100%.

Must-know distinctions

• Reflux vs distillation. Reflux returns vapour to the flask; distillation sends it to a receiver.
• Reduced pressure vs gravity filtration. Suction is faster and leaves a drier solid.
• Hot solvent vs cold wash. Use the minimum of hot solvent to dissolve, and only a little ice-cold solvent to rinse.
• PAG5 vs PAG6. PAG5 is the organic liquid (separating funnel, drying, redistillation); PAG6 is the organic solid.

6.2.5(b): Several functional groups in one molecule

First name every group, then go through likely reagents one at a time.

Reagent	Group that responds	What you see or get
Na₂CO₃(aq)	Carboxylic acid only	Effervescence (CO₂)
NaOH(aq)	Carboxylic acid and phenol	Salt forms; no gas
Br₂(aq)	Alkene (addition), phenol (substitution)	Orange colour removed; phenol also gives a white precipitate
2,4-DNP	Aldehyde or ketone	Orange precipitate
Tollens’ reagent, warm	Aldehyde only	Silver mirror
K₂Cr₂O₇/H₂SO₄	Primary or secondary alcohol, aldehyde	Orange to green
NaBH₄	Aldehyde or ketone	Alcohol; C=C untouched
Dilute HCl	Amine	Ammonium salt
Hot aqueous acid or alkali	Ester, amide, nitrile (acid hydrolysis)	Hydrolysis products

O–H and N–H groups hydrogen bond. They raise melting and boiling points and increase solubility in water.

Worked reminder: a hydroxy ester

Compound H is CH₃CH(OH)CH₂COOCH₃, methyl 3-hydroxybutanoate.

1. Groups. A secondary alcohol (OH on carbon 3) and a methyl ester.
2. Acidified dichromate, heat under reflux. Oxidation turns the secondary alcohol into a ketone, CH₃COCH₂COOCH₃. The colour changes from orange to green.
3. Hot aqueous NaOH. The ester is hydrolysed to sodium 3-hydroxybutanoate and methanol.
4. Sodium carbonate. No effervescence: there is no COOH group.
5. Chirality. Carbon 3 carries H, OH, CH₃ and CH₂COOCH₃, so it is a chiral centre.

6.2.5(c): Multi-stage routes

Route map by goal

Goal	Steps
Add one carbon to a chain	Haloalkane + CN⁻ in ethanol, or aldehyde/ketone + NaCN(aq)/H⁺(aq)
Add a carbon chain to a ring	Friedel–Crafts: haloalkane or acyl chloride + halogen carrier
Amine, same carbon count	Haloalkane + excess ethanolic NH₃
Amine, one more carbon	Haloalkane → nitrile (CN⁻) → amine (H₂/Ni)
Aromatic amine	Arene → nitroarene (conc. HNO₃/H₂SO₄) → amine (Sn/conc. HCl)
Carboxylic acid	Oxidise primary alcohol or aldehyde; hydrolyse nitrile or ester
Ester of an alcohol	Acid + alcohol (conc. H₂SO₄); or acid anhydride or acyl chloride + alcohol
Ester of a phenol	Acid → acyl chloride (SOCl₂) → + phenol
Primary or secondary amide	Acyl chloride + NH₃, or + primary amine
Alcohol from carbonyl	NaBH₄

Method in steps

1. List the groups in the start and target, and count carbons.
2. Spot the carbon-adding step, if any.
3. Work backwards one step at a time.
4. On a ring, order substitutions by directing effects: OH and NH₂ direct to 2 and 4; NO₂ directs to 3.
5. Make reactive groups such as NH₂ late in the route.
6. Give reagents, conditions and every intermediate.
7. Use any extra reaction the paper supplies exactly as given.

More must-know distinctions

• Ethanolic NH₃ vs CN⁻ then H₂/Ni. Both give a primary amine from a haloalkane. Only the nitrile route adds a carbon.
• NaBH₄ vs H₂/Ni. NaBH₄ reduces only C=O in aldehydes and ketones. H₂/Ni reduces C=C and C≡N.
• Carboxylic acid vs acyl chloride. Both make esters of alcohols. Only the acyl chloride readily esterifies a phenol, and it reacts without an acid catalyst.
• Benzene vs phenol nitration. Benzene needs concentrated HNO₃ and concentrated H₂SO₄. Phenol reacts with dilute HNO₃ and gives a mixture of 2- and 4-nitrophenol that must be separated.

Small worked reminders

• Propanal to 2-hydroxybutanoic acid. NaCN(aq)/H⁺(aq) gives 2-hydroxybutanenitrile; heating under reflux with dilute acid hydrolyses CN to COOH.
• Benzene to 1-phenylbutan-1-ol. Butanoyl chloride with AlCl₃ (Friedel–Crafts acylation) gives the ketone C₆H₅COCH₂CH₂CH₃; NaBH₄ then reduces it to the secondary alcohol C₆H₅CH(OH)CH₂CH₂CH₃, which has a chiral centre.
• Two-stage yield. Stages of 80% and 65% give 0.80 × 0.65 × 100 = 52.0% overall.

Quick self-test

1. Why is the minimum volume of hot solvent used in recrystallisation?
2. Why are the collected crystals washed with ice-cold solvent?
3. A sample melts from 131 °C to 135 °C. Take the pure compound’s melting point as 137 °C. Is the sample pure?
4. Crude product 3.40 g; recrystallised product 2.55 g. What percentage was recovered?
5. Which reagent shows that a molecule with a phenol group also has a COOH group?
6. Which test tells an aldehyde from a ketone?
7. Give the two stages to turn 1-chlorobutane into pentanoic acid.
8. Phenyl butanoate is needed from phenol. What should react with the phenol, and why not butanoic acid?
9. Name the reaction and reagents that put an ethyl group onto a benzene ring.
10. Which reagent reduces a C=O in a molecule but leaves its C=C unchanged?
11. A route has two stages with yields of 80% and 65%. What is the overall yield?

Answers

1. So the solution is saturated when cold and most product crystallises; extra solvent keeps product dissolved.
2. To rinse off solution holding impurities without dissolving much product.
3. No. It melts below 137 °C and over a 4 °C range.
4. 2.55 ÷ 3.40 × 100 = 75.0%.
5. Sodium carbonate solution: effervescence of CO₂. Phenol alone gives none.
6. Tollens’ reagent. Aldehydes give a silver mirror; ketones leave it unchanged.
7. KCN in ethanol, heat under reflux, gives pentanenitrile; then dilute acid, heat under reflux, gives pentanoic acid.
8. Butanoyl chloride; phenol is not readily esterified by a carboxylic acid.
9. Friedel–Crafts alkylation: chloroethane with a halogen carrier such as AlCl₃.
10. NaBH₄ (H₂/Ni would also reduce the C=C).
11. 52.0%.

Where marks are usually lost

• Writing “dissolve in hot solvent” without “minimum volume”.
• Washing crystals with room-temperature solvent.
• Saying an impure sample has a higher melting point, or omitting the wider range.
• Describing gravity filtration when filtration under reduced pressure is asked for.
• Giving aqueous KCN for the nitrile step; the specification pairs CN⁻ with ethanol.
• Using ammonia when the target amine has one more carbon than the haloalkane.
• Esterifying a phenol with a carboxylic acid.
• Ordering ring substitutions against the directing effects, or making NH₂ before an acyl chloride step.
• Using NaBH₄ to reduce a carboxylic acid or a C=C bond.
• Missing an intermediate’s structure in a multi-stage route.

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.