OCR A Level Chemistry A: Amines (H432) – Revision Notes

These revision notes condense section 6.2.1, Amines, of the OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017. They cover every learning outcome, 6.2.1(a), (b)(i) and (b)(ii), in Module 6, Organic chemistry and analysis. H432 is the full A level, and the specification states that Module 6 is assessed in components 02 and 03. The Amines study guide has the full explanations and longer worked examples.

Once the self-test below feels easy, go on to the Amines practice questions. Every H432 topic is linked from the OCR chemistry hub, and you can track outcomes on the printable checklist. For a quick check of the whole course, use the free diagnostics.

Key words

Term	Meaning
Amine	Ammonia with one or more H atoms replaced by carbon groups
Primary amine	One carbon group on N: RNH₂
Secondary amine	Two carbon groups on N: R₂NH
Tertiary amine	Three carbon groups on N: R₃N
Aliphatic amine	N bonded to alkyl carbon, e.g. pentan-3-amine
Aromatic amine	N bonded directly to a benzene ring, e.g. phenylamine
Nitroarene	Arene carrying an NO₂ group, e.g. nitrobenzene
Base (Brønsted–Lowry)	Proton acceptor

Naming reminders: primary amines take “-amine” on the stem (hexan-2-amine) or the alkylamine form (propylamine). Groups on nitrogen in secondary and tertiary amines use N-, as in N-methylethanamine or N,N-dimethylpropan-1-amine.

6.2.1(a): Basicity and salts

The explanation in one line: the lone pair on the nitrogen atom accepts a proton (H⁺), forming a dative covalent bond, so the amine is a base.

• RNH₂ + H⁺ → RNH₃⁺ (an alkylammonium ion)
• In water: RNH₂ + H₂O ⇌ RNH₃⁺ + OH⁻, so the solution is alkaline
• Comparison of basicity between amines is not required by the specification
• Reactions with acids are restricted to inorganic acids

Salts: method in steps

1. Put an extra H on the nitrogen and a + charge on it.
2. Pair it with the acid’s anion: Cl⁻, NO₃⁻ or SO₄²⁻.
3. For H₂SO₄, use two amine molecules per acid.
4. Name it: “-amine” becomes “-ammonium”, then the anion name.

Amine + acid	Salt formula	Salt name
CH₃CH₂CH(NH₂)CH₂CH₃ + HCl	(CH₃CH₂)₂CHNH₃⁺Cl⁻	pentan-3-ylammonium chloride
C₆H₅NH₂ + HCl	C₆H₅NH₃⁺Cl⁻	phenylammonium chloride
(CH₃)₃N + HNO₃	(CH₃)₃NH⁺NO₃⁻	trimethylammonium nitrate
2CH₃CH₂CH₂NH₂ + H₂SO₄	(CH₃CH₂CH₂NH₃⁺)₂SO₄²⁻	propylammonium sulfate

Reversing it: RNH₃⁺ + OH⁻ → RNH₂ + H₂O. Warming a salt with NaOH(aq) gives back the free amine.

6.2.1(b)(i): Aliphatic amines by substitution

Making	From	Reagent	Conditions
Primary amine	Haloalkane	NH₃	Excess, in ethanol, heat in a sealed tube
Secondary amine	Haloalkane	Primary amine	In ethanol, heat
Tertiary amine	Haloalkane	Secondary amine	In ethanol, heat

Equations:

• Primary: RX + 2NH₃ → RNH₂ + NH₄X
• Secondary: 2RNH₂ + R′X → RNHR′ + RNH₃⁺X⁻
• Tertiary: 2R₂NH + R′X → R₂NR′ + R₂NH₂⁺X⁻

Reasons for each condition:

• Ammonia (or amine) as nucleophile: its N lone pair attacks the δ+ carbon of the C–X bond; X⁻ leaves.
• Second ammonia or amine molecule: removes H⁺ from the N–H⁺ formed, so it ends up as the ammonium salt.
• Excess ammonia: because the product amine keeps a lone pair, it can substitute again; with NH₃ in large excess, collisions of haloalkane with NH₃ are far more likely, so the primary amine dominates.
• Ethanol, not water: OH⁻ or water would substitute the halogen and form an alcohol.
• Sealed tube: stops volatile ammonia escaping on heating.

Cross-link: a nitrile reduced with H₂/Ni (6.2.4(c)) also gives a primary amine, RCN + 2H₂ → RCH₂NH₂. Starting from a haloalkane via a nitrile adds one carbon, and the amine never meets unreacted haloalkane, so there is no further substitution. More in Carbon–carbon bond formation.

Worked reminder

(Bromomethyl)cyclohexane, C₆H₁₁CH₂Br, with excess ethanolic ammonia:

C₆H₁₁CH₂Br + 2NH₃ → C₆H₁₁CH₂NH₂ + NH₄Br

Product: cyclohexylmethanamine, a primary amine with the same carbon skeleton as the haloalkane.

6.2.1(b)(ii): Aromatic amines by reduction

Reagents: tin (Sn) and concentrated hydrochloric acid, heat under reflux. Then: excess NaOH(aq).

Nitroarenes come from nitration of arenes: see Aromatic compounds.

Equation with [H], as the specification’s guidance for organic redox asks:

ArNO₂ + 6[H] → ArNH₂ + 2H₂O

Nitroarene	[H] needed	Product
One NO₂ group	6[H]	One NH₂, plus 2H₂O
Two NO₂ groups	12[H]	Two NH₂, plus 4H₂O
Three NO₂ groups	18[H]	Three NH₂, plus 6H₂O

Why add NaOH? In concentrated HCl the amine forms as ArNH₃⁺. OH⁻ removes the extra proton: ArNH₃⁺ + OH⁻ → ArNH₂ + H₂O.

Groups such as CH₃ or Cl on the ring are left unchanged; only the NO₂ is reduced.

Worked reminder

2,4-dinitromethylbenzene, CH₃C₆H₃(NO₂)₂, is reduced:

CH₃C₆H₃(NO₂)₂ + 12[H] → CH₃C₆H₃(NH₂)₂ + 4H₂O

H count: 6 + 12 = 18 on the left; 10 + 8 = 18 on the right.

Calculations: method in steps

Amine questions often carry a mole calculation. The same four steps work for neutralisations and yields.

1. Write the balanced equation first, so the mole ratio is on the page (1 : 1 for HCl or HNO₃; 2 : 1 amine : H₂SO₄; 1 : 1 haloalkane : amine; 1 : 1 nitroarene : amine for one NO₂ group).
2. Work out M for each substance you need, using the A_r values given.
3. Convert the known mass, or concentration × volume, into moles.
4. Apply the ratio, then convert to the quantity asked for: mass, volume of acid, or percentage yield = actual moles ÷ theoretical moles × 100.

Carry unrounded values through to the last step, and give the answer to the precision of the data.

Must-know distinctions

• Aliphatic vs aromatic route. Haloalkane plus ammonia for amines on alkyl carbon; nitroarene plus Sn/HCl for amines on a ring.
• Ammonium salt vs amine. RNH₃⁺X⁻ is ionic; RNH₂ is the neutral base. Acid makes the salt; alkali frees the amine.
• Excess ammonia vs excess haloalkane. Excess NH₃ favours the primary amine. Without it, secondary and tertiary amines form too.
• Ethanolic vs aqueous ammonia. Ethanolic avoids the alcohol by-product.
• [H] vs [O]. Reduction of NO₂ uses [H]; [O] is for oxidations such as aldehyde to carboxylic acid.
• Same chain length vs one carbon more. Ammonia substitution keeps the chain; the nitrile route lengthens it by one carbon.

Quick self-test

1. Explain, in terms of structure, why ethylamine is a base.
2. Write the equation for phenylamine reacting with dilute nitric acid and name the salt.
3. Write the formula of the salt formed from methylamine and sulfuric acid.
4. Give the reagent and conditions for making butan-1-amine from 1-iodobutane.
5. Why is the ammonia used in excess in question 4?
6. Write an equation for the reaction of methylamine with chloroethane, using two moles of methylamine.
7. Give the reagents for converting nitrobenzene into phenylamine, including the final step.
8. Write the equation, using [H], for reducing 1,4-dinitrobenzene to benzene-1,4-diamine.
9. 1.46 g of 2-methylpropan-2-amine, (CH₃)₃CNH₂, is neutralised by 0.500 mol dm⁻³ HCl. Calculate the volume of acid needed. (A_r: C 12.0, H 1.0, N 14.0)
10. Suggest how a secondary amine could be turned into a tertiary amine.

Answers

1. The nitrogen lone pair can accept a proton (H⁺), forming a dative bond.
2. C₆H₅NH₂ + HNO₃ → C₆H₅NH₃⁺NO₃⁻, phenylammonium nitrate.
3. (CH₃NH₃⁺)₂SO₄²⁻, methylammonium sulfate (two amines per H₂SO₄).
4. Excess ammonia in ethanol, heated in a sealed tube. CH₃CH₂CH₂CH₂I + 2NH₃ → CH₃CH₂CH₂CH₂NH₂ + NH₄I.
5. The butan-1-amine also has a lone pair and would react with more 1-iodobutane to give secondary and tertiary amines; excess ammonia makes this much less likely.
6. 2CH₃NH₂ + CH₃CH₂Cl → CH₃NHCH₂CH₃ + CH₃NH₃⁺Cl⁻. The product is N-methylethanamine.
7. Tin and concentrated HCl, heat under reflux; then excess NaOH(aq) to release the amine from its salt.
8. C₆H₄(NO₂)₂ + 12[H] → C₆H₄(NH₂)₂ + 4H₂O.
9. M = 4(12.0) + 11(1.0) + 14.0 = 73.0 g mol⁻¹; n = 1.46 ÷ 73.0 = 0.0200 mol; ratio 1 : 1, so V = 0.0200 ÷ 0.500 = 0.0400 dm³ = 40.0 cm³.
10. Heat it with a haloalkane in ethanol; the N lone pair substitutes the halogen, giving R₂NR′.

Where marks are usually lost

• Saying “amines are bases because they donate electrons” without the proton and the lone pair on N.
• Writing the H⁺ on carbon instead of nitrogen in the salt.
• Using a 1 : 1 ratio for sulfuric acid salts.
• Dropping “excess” or “ethanolic” from the ammonia conditions, or writing “aqueous ammonia”.
• Giving RX + NH₃ → RNH₂ + HX without noting that HX reacts with more base, then using the wrong mole ratio.
• Writing “HCl” instead of “concentrated HCl” with tin, or leaving out the NaOH step.
• Balancing NO₂ → NH₂ with too few [H]; each nitro group needs 6[H].
• Proposing ammonia substitution to make phenylamine from a halobenzene instead of the nitroarene route.
• Forgetting that the nitrile route adds a carbon, so naming the wrong amine.

Official syllabus

OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017, section 6.2.1 Amines, learning outcomes (a) and (b)(i)–(ii). Published by OCR.