OCR A Level Chemistry A: Amines (H432)

This study guide teaches section 6.2.1, Amines, of the OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017. The section opens 6.2, Nitrogen compounds, polymers and synthesis, inside Module 6, Organic chemistry and analysis. It has two lettered learning outcomes, 6.2.1(a) and (b), and (b) has two parts; every one is covered below with worked examples. H432 is the full A level. The specification states that Module 6 content is assessed in Synthesis and analytical techniques (02) and Unified chemistry (03). No PAG is listed against this section.

Every H432 topic is linked from the OCR chemistry hub. Tick off each outcome on the printable checklist once you can do it without notes. The free diagnostics show where your gaps are across the course.

Outcomes in 6.2.1

Outcome	What you must be able to do	Additional guidance in the specification
6.2.1(a)	Explain the basicity of amines as proton acceptance by the nitrogen lone pair; write equations for amines reacting with dilute acids such as HCl(aq) to form salts	Comparing the basicity of different amines is not required; acids are restricted to inorganic acids
6.2.1(b)(i)	Prepare aliphatic amines by substituting haloalkanes with excess ethanolic ammonia, and with amines	Includes primary amines from ammonia and secondary or tertiary amines from amines; see also reduction of nitriles, 6.2.4(c)
6.2.1(b)(ii)	Prepare aromatic amines by reducing nitroarenes with tin and concentrated hydrochloric acid	None listed

What an amine is

An amine is derived from ammonia, NH₃, by replacing one or more hydrogen atoms with carbon groups. The number of carbon groups on the nitrogen sets the class.

Class	General formula	Example	Name
Primary	RNH₂	CH₃CH₂CH(NH₂)CH₃	butan-2-amine
Secondary	R₂NH	(CH₃)₂CHCH₂NHCH₂CH₃	N-ethyl-2-methylpropan-1-amine
Tertiary	R₃N	(CH₃CH₂)₃N	triethylamine

In an aliphatic amine the nitrogen is joined to alkyl carbon; in an aromatic amine it is joined directly to a benzene ring, as in phenylamine, C₆H₅NH₂. Primary amines are named with “-amine” on the alkane stem (pentan-2-amine) or as alkylamines (ethylamine); groups on nitrogen take the prefix N-. In every amine, nitrogen forms three bonds and keeps one lone pair, and that lone pair drives all of 6.2.1.

6.2.1(a): Amines as bases

Proton acceptance

Amines are Brønsted–Lowry bases: they accept protons. The lone pair on the nitrogen atom forms a dative covalent bond to an H⁺ ion. The amine becomes a substituted ammonium ion, just as ammonia becomes NH₄⁺.

RNH₂ + H⁺ → RNH₃⁺

In water, a small proportion of amine molecules take a proton from water molecules, so the solution contains OH⁻ ions and is alkaline:

RNH₂ + H₂O ⇌ RNH₃⁺ + OH⁻

Comparing the strength of different amines as bases is not required.

Salts with dilute inorganic acids

With a dilute acid, the proton transfer goes to completion and a salt forms. The salt is ionic: an alkylammonium cation and the acid’s anion. Name it by changing “-amine” to “-ammonium” and adding the anion.

• With hydrochloric acid: CH₃CH₂NH₂ + HCl → CH₃CH₂NH₃⁺Cl⁻ (ethylammonium chloride)
• With nitric acid: C₆H₅NH₂ + HNO₃ → C₆H₅NH₃⁺NO₃⁻ (phenylammonium nitrate)
• With sulfuric acid, two amine molecules accept the two protons: 2CH₃NH₂ + H₂SO₄ → (CH₃NH₃⁺)₂SO₄²⁻ (methylammonium sulfate)

Secondary and tertiary amines react the same way: (CH₃)₂NH + HCl → (CH₃)₂NH₂⁺Cl⁻. A strong alkali frees the amine from its salt:

RNH₃⁺ + OH⁻ → RNH₂ + H₂O

Worked example 1: salt from a primary amine

Question. 3.48 g of pentan-2-amine, CH₃CH₂CH₂CH(NH₂)CH₃, reacts completely with dilute hydrochloric acid. Write the equation, name the salt and calculate the mass of salt formed. Then find the mass of salt if dilute sulfuric acid is used instead. (A_r: C 12.0, H 1.0, N 14.0, Cl 35.5, O 16.0, S 32.1)

Step 1: equation. CH₃CH₂CH₂CH(NH₂)CH₃ + HCl → CH₃CH₂CH₂CH(NH₃⁺)CH₃ Cl⁻. The salt is pentan-2-ylammonium chloride.

Step 2: moles of amine. M(C₅H₁₃N) = 5(12.0) + 13(1.0) + 14.0 = 87.0 g mol⁻¹. n = 3.48 ÷ 87.0 = 0.0400 mol.

Step 3: mass of chloride salt. The ratio is 1 : 1. M(C₅H₁₄NCl) = 87.0 + 36.5 = 123.5 g mol⁻¹. Mass = 0.0400 × 123.5 = 4.94 g.

Step 4: with sulfuric acid. 2C₅H₁₃N + H₂SO₄ → (C₅H₁₄N⁺)₂SO₄²⁻. Each mole of salt uses two moles of amine, so n(salt) = 0.0200 mol. M = 2(88.0) + 96.1 = 272.1 g mol⁻¹. Mass = 0.0200 × 272.1 = 5.44 g.

6.2.1(b)(i): Aliphatic amines from haloalkanes

Primary amines from ammonia

A haloalkane reacts with ammonia to give a primary amine. The ammonia molecule uses its lone pair as a nucleophile: it attacks the δ+ carbon bonded to the halogen, and the halide ion leaves. This gives an alkylammonium ion, which then loses H⁺ to a second ammonia molecule. Overall, two ammonia molecules are used for each haloalkane:

RX + 2NH₃ → RNH₂ + NH₄X

Conditions: excess ammonia, dissolved in ethanol, heated. Because ammonia is volatile, the mixture is usually heated in a sealed container.

Why each condition matters:

• Ethanol as solvent. In aqueous ammonia, hydroxide ions and water are present and can substitute the halogen instead, giving an alcohol as a by-product.
• Excess ammonia. The primary amine product also has a lone pair, so it is a nucleophile too. It can attack another haloalkane molecule to give a secondary amine, which can react again to give a tertiary amine. With ammonia in large excess, a haloalkane molecule is much more likely to meet ammonia than the product, so the primary amine is the main product. The excess also removes the HX formed, as the ammonium salt.

With only one ammonia in the equation, the product is the salt RNH₃⁺X⁻; the second ammonia releases the free amine.

Secondary and tertiary amines from amines

Replacing ammonia with an amine extends the same reaction. A primary amine and a haloalkane give a secondary amine; a secondary amine and a haloalkane give a tertiary amine. Again, a second molecule of the amine takes the H⁺:

RNH₂ + R′X → RNHR′ + HX, then HX + RNH₂ → RNH₃⁺X⁻

Overall: 2RNH₂ + R′X → RNHR′ + RNH₃⁺X⁻

R₂NH + R′X → R₂NR′ + HX, with the HX taken up by more R₂NH

Some further substitution always happens, so the product is a mixture that must be separated.

Worked example 2: yield of a primary amine

Question. A chemist heats 8.245 g of 1-bromo-3,3-dimethylbutane, (CH₃)₃CCH₂CH₂Br, with excess ethanolic ammonia in a sealed tube and isolates 3.64 g of 3,3-dimethylbutan-1-amine. Write an equation and calculate the percentage yield. (A_r: Br 79.9)

Step 1: equation. (CH₃)₃CCH₂CH₂Br + 2NH₃ → (CH₃)₃CCH₂CH₂NH₂ + NH₄Br

Step 2: moles of haloalkane. M(C₆H₁₃Br) = 72.0 + 13.0 + 79.9 = 164.9 g mol⁻¹. n = 8.245 ÷ 164.9 = 0.0500 mol.

Step 3: moles of amine obtained. M(C₆H₁₅N) = 72.0 + 15.0 + 14.0 = 101.0 g mol⁻¹. n = 3.64 ÷ 101.0 = 0.03604 mol.

Step 4: yield. The ratio is 1 : 1, so yield = 0.03604 ÷ 0.0500 × 100 = 72.1%.

Worked example 3: building a secondary amine

Question. Give the reagent needed to convert 2-methylpropan-1-amine into N-ethyl-2-methylpropan-1-amine, and write an overall equation that shows why two moles of the starting amine are used.

Answer. React the amine with a haloalkane that supplies the ethyl group, such as iodoethane, CH₃CH₂I, in ethanol.

(CH₃)₂CHCH₂NH₂ + CH₃CH₂I → (CH₃)₂CHCH₂NHCH₂CH₃ + HI

The HI is a strong acid. It protonates a second molecule of the basic starting amine:

2(CH₃)₂CHCH₂NH₂ + CH₃CH₂I → (CH₃)₂CHCH₂NHCH₂CH₃ + (CH₃)₂CHCH₂NH₃⁺I⁻

A further reaction with iodoethane gives the tertiary amine (CH₃)₂CHCH₂N(CH₂CH₃)₂.

A second route: nitriles

The specification points you to 6.2.4(c), where nitriles are reduced to primary amines, for example with H₂ and a nickel catalyst: RC≡N + 2H₂ → RCH₂NH₂. Making the nitrile from a haloalkane and cyanide adds one carbon, so 1-bromopropane leads to butan-1-amine by this route, whereas ammonia substitution keeps the chain length. The nitrile route also avoids further substitution, because the amine only forms once the haloalkane has been used up.

6.2.1(b)(ii): Aromatic amines from nitroarenes

Aromatic amines are made by reducing a nitroarene (an arene carrying an NO₂ group, made by nitration in 6.1.1).

Reagents and conditions: tin and concentrated hydrochloric acid, heated under reflux. Then add excess aqueous sodium hydroxide.

Why the second step is needed: the reduction takes place in strong acid, so the basic amine product is protonated as soon as it forms. The mixture contains the arylammonium ion, not the free amine. Adding excess NaOH removes the proton and releases the amine:

C₆H₅NH₃⁺ + OH⁻ → C₆H₅NH₂ + H₂O

For organic redox equations, the specification’s guidance (at 6.1.2(a)) says to use [H] for the reducing agent. Each NO₂ group needs 6[H]: four to remove the two oxygen atoms as water, and two to become the NH₂ group.

C₆H₅NO₂ + 6[H] → C₆H₅NH₂ + 2H₂O

Check: 11 H and 2 O on each side.

Worked example 4: reducing a substituted nitroarene

Question. A student reduces 8.076 g of 1-bromo-3-nitrobenzene, BrC₆H₄NO₂, and obtains 4.81 g of 3-bromophenylamine, BrC₆H₄NH₂, after the alkali step. Give the reagents, write the equation using [H], and calculate the percentage yield.

Step 1: reagents. Tin and concentrated HCl, heated under reflux; then excess NaOH(aq).

Step 2: equation. BrC₆H₄NO₂ + 6[H] → BrC₆H₄NH₂ + 2H₂O. The bromine on the ring is unchanged.

Step 3: moles. M(C₆H₄BrNO₂) = 72.0 + 4.0 + 79.9 + 14.0 + 32.0 = 201.9 g mol⁻¹, so n = 8.076 ÷ 201.9 = 0.0400 mol. M(C₆H₆BrN) = 72.0 + 6.0 + 79.9 + 14.0 = 171.9 g mol⁻¹, so n = 4.81 ÷ 171.9 = 0.0280 mol.

Step 4: yield. 0.0280 ÷ 0.0400 × 100 = 70.0%.

For a dinitro compound, double the reducing agent: C₆H₄(NO₂)₂ + 12[H] → C₆H₄(NH₂)₂ + 4H₂O.

Choosing the right preparation

Target	Starting material	Reagent and conditions
Primary aliphatic amine, same chain length	Haloalkane	Excess NH₃ in ethanol, heat (sealed)
Primary aliphatic amine, one more carbon	Haloalkane, via nitrile (6.2.4)	CN⁻ in ethanol, then H₂/Ni
Secondary or tertiary aliphatic amine	Primary or secondary amine + haloalkane	Heat in ethanol
Aromatic amine	Nitroarene	Sn/conc. HCl, reflux; then excess NaOH(aq)

To put NH₂ on a benzene ring, use the nitroarene route, not ammonia substitution.

Common errors

• Explaining basicity as “amines contain OH⁻” or “amines donate electrons” without naming the lone pair on nitrogen and the proton it accepts.
• Using a 1 : 1 ratio with sulfuric acid.
• Leaving out “ethanolic” or “excess” when giving conditions for ammonia substitution.
• Writing RX + NH₃ → RNH₂ + HX and forgetting that HX reacts with a second base molecule.
• Giving “Sn/HCl” without “concentrated”, or omitting the NaOH step.
• Balancing a nitroarene reduction with 4[H] or 2[H] instead of 6[H] per NO₂ group.

Next steps

• Condensed tables and a quick self-test: Amines revision notes
• Original questions with mark-by-mark answers: Amines practice questions
• Brønsted–Lowry acids and bases: Acids
• Moles and percentage yield: Amount of substance
• Find your gaps: free diagnostics

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.