Amines

In a nutshell

Amines are organic compounds made from ammonia by replacing one or more of its hydrogen atoms with carbon groups. Like ammonia, they keep a lone pair on the nitrogen, and that one lone pair explains almost everything in this subtopic.

The lone pair lets an amine accept a proton, so amines are bases that react with acids to make salts. The same lone pair makes an amine a nucleophile, which is how amines are made from haloalkanes. Aromatic amines such as phenylamine are made a different way: by reducing a nitroarene with tin and concentrated hydrochloric acid.

Assumed knowledge: Acids, Bonding and structure, Haloalkanes, Aromatic compounds.

Core content

Basicity and preparation of amines: what an amine is

An amine contains nitrogen bonded to at least one carbon group (alkyl or aryl). Amines are classified by how many carbon groups are bonded to the nitrogen:

• Primary amine: one carbon group on N, e.g. ethylamine, CH₃CH₂NH₂.
• Secondary amine: two carbon groups on N, e.g. diethylamine, (CH₃CH₂)₂NH.
• Tertiary amine: three carbon groups on N, e.g. triethylamine, (CH₃CH₂)₃N.

All three keep one lone pair on the nitrogen, just like ammonia.

Phenylamine, C₆H₅NH₂, is the aromatic amine you need: an NH₂ group bonded directly to a benzene ring.

Naming amines

• Questions usually use names built from the alkyl group: ethylamine, propylamine, diethylamine.
• The fully systematic style names it like an alcohol: CH₃CH₂CH₂NH₂ is propan-1-amine (same compound as propylamine).
• A compound with two NH₂ groups is a diamine, e.g. hexane-1,6-diamine, H₂N(CH₂)₆NH₂.

Basicity and preparation of amines: amines as bases

The lone pair on the nitrogen accepts a proton (H⁺), so amines act as bases.

A Brønsted-Lowry base is a species that accepts a proton. When the amine accepts H⁺, the nitrogen's lone pair forms a dative covalent bond to the H⁺.

CH₃CH₂NH₂ + H⁺ → CH₃CH₂NH₃⁺

The nitrogen now has four bonds and a + charge. The product is an alkylammonium ion (here, ethylammonium), named like the ammonium ion, NH₄⁺.

Aliphatic amines that dissolve in water take a proton from water, so their solutions are alkaline:

CH₃CH₂NH₂(aq) + H₂O(l) ⇌ CH₃CH₂NH₃⁺(aq) + OH^-(aq)

Depth limit (OCR guidance): comparing the base strength of different amines (for example ammonia vs ethylamine vs phenylamine) is not required. Learn the proton-acceptance explanation, not a ranking.

Still don't get it? · why one lone pair makes an amine both a base and a nucleophile

Think of the lone pair as a spare hand that the nitrogen is not using. It can grab whatever is positive and on offer.

Step by step:

1. Nitrogen has five outer electrons. In an amine, three are used in bonds, so two are left over as a lone pair.
2. A proton, H⁺, has no electrons at all. It is desperate for a pair.
3. The nitrogen gives its lone pair to the H⁺. Both electrons in the new bond came from N, so it is a dative covalent bond.
4. Nitrogen now shares four pairs and has "given away" part of its electron share, so it carries a + charge.
5. If, instead of H⁺, the spare hand reaches for a slightly positive (δ+) carbon in a haloalkane, the same lone pair forms a bond to carbon. That is the amine acting as a nucleophile.

Exam wording: an amine is a base because the lone pair on the nitrogen accepts a proton. An amine is a nucleophile because it is an electron pair donor. Same lone pair, two jobs.

Basicity and preparation of amines: reactions with dilute acids to form salts

An amine reacts with a dilute acid to form a salt. The amine accepts H⁺ from the acid, and the acid's anion becomes the counter-ion.

• With hydrochloric acid: CH₃CH₂NH₂(aq) + HCl(aq) → CH₃CH₂NH₃⁺Cl^-(aq), ethylammonium chloride
• With sulfuric acid (H₂SO₄ donates two protons, so two amines react): 2CH₃CH₂NH₂(aq) + H₂SO₄(aq) → (CH₃CH₂NH₃⁺)₂SO₄^2-(aq), ethylammonium sulfate
• With nitric acid: CH₃CH₂NH₂(aq) + HNO₃(aq) → CH₃CH₂NH₃⁺NO₃^-(aq), ethylammonium nitrate
• Phenylamine reacts in the same way: C₆H₅NH₂ + HCl → C₆H₅NH₃⁺Cl^-, phenylammonium chloride

The pattern for the formula: the nitrogen gains one H and a + charge. A secondary amine with one H on N ends up with two H on N⁺.

Two useful facts about these salts:

• They are ionic, so they are soluble in water (phenylamine itself is only slightly soluble, but phenylammonium chloride dissolves).
• Adding a strong base such as NaOH(aq) removes the H⁺ again and releases the free amine: C₆H₅NH₃⁺Cl^- + NaOH → C₆H₅NH₂ + NaCl + H₂O

Depth limit (OCR guidance): the acids are restricted to inorganic acids (HCl, H₂SO₄, HNO₃).

Basicity and preparation of amines: aliphatic amines from haloalkanes

Reagent and conditions: a haloalkane with excess ammonia dissolved in ethanol (excess ethanolic ammonia). The mixture is heated in a sealed container so the volatile ammonia cannot escape.

Type of reaction: nucleophilic substitution. The ammonia's lone pair attacks the δ+ carbon, and the halide ion leaves.

Overall equation, making a primary amine:

CH₃CH₂Br + 2NH₃ → CH₃CH₂NH₂ + NH₄Br

Why two NH₃? The first ammonia substitutes for Br. The amine formed is a base, so it is protonated. A second ammonia takes that H⁺ back off, freeing the amine and forming NH₄⁺Br^-.

Why ethanol as the solvent, not water? In water, the haloalkane would also be hydrolysed (substituted by water or OH^-), giving an alcohol as a by-product.

The mechanism (applying 4.2.2). This is the same nucleophilic substitution you met with OH^-, but the nucleophile is a neutral NH₃ molecule. That is why a second step is needed to remove H⁺.

The OCR spec names the reagent and the reaction type here, not this mechanism. It is shown because you can be asked to apply the nucleophilic substitution mechanism to a new nucleophile.

Basicity and preparation of amines: secondary and tertiary amines from amines

The primary amine product still has a lone pair on N, so it is a nucleophile too. It can attack another haloalkane molecule:

• Primary amine + haloalkane → secondary amine 2CH₃CH₂NH₂ + CH₃CH₂Br → (CH₃CH₂)₂NH + CH₃CH₂NH₃⁺Br^-
• Secondary amine + haloalkane → tertiary amine 2(CH₃CH₂)₂NH + CH₃CH₂Br → (CH₃CH₂)₃N + (CH₃CH₂)₂NH₂⁺Br^-

In each case one amine molecule substitutes and a second amine molecule takes the H⁺.

Why excess ammonia? With a large excess, a haloalkane molecule is far more likely to meet an NH₃ molecule than a molecule of amine product. This maximises the primary amine and limits further substitution.

Making secondary and tertiary amines on purpose: start from the amine instead of ammonia. A primary amine with a haloalkane gives a secondary amine; a secondary amine gives a tertiary amine.

Still don't get it? · why you get a mixture of amines

Imagine a party where every guest who arrives is handed a ticket to a raffle. The catch: once someone wins, they get handed a new ticket and can win again.

Step by step:

1. The haloalkane molecules are the prizes. The nucleophiles (anything with a lone pair on N) are the people trying to win them.
2. At the start, the only nucleophile is ammonia, so the first prizes go to ammonia, making primary amine.
3. But the primary amine still has a lone pair. It is now also in the raffle, competing with ammonia for the remaining haloalkane.
4. When the primary amine wins, it becomes a secondary amine, which still has a lone pair, and so on to a tertiary amine.
5. If there are vastly more ammonia molecules than amine molecules, almost every prize goes to an ammonia. That is the point of the excess.

Exam wording: the amine product is also a nucleophile (it has a lone pair on N), so it can react with more haloalkane to form secondary and tertiary amines. Excess ammonia makes formation of the primary amine more likely.

Basicity and preparation of amines: aromatic amines from nitroarenes

Aromatic amines are made by reducing a nitroarene.

• Reagents: tin (Sn) and concentrated hydrochloric acid
• Conditions: heat (under reflux)
• Type of reaction: reduction

Using [H] for the reducing agent, nitrobenzene to phenylamine:

C₆H₅NO₂ + 6[H] → C₆H₅NH₂ + 2H₂O

Balancing tip: both O atoms leave as 2H₂O (that uses 4[H]), and the N gains 2 H (another 2[H]). Total 6[H].

Because the reaction mixture is strongly acidic, the amine is protonated as it forms: you actually get phenylammonium chloride, C₆H₅NH₃⁺Cl^-. Adding NaOH(aq) afterwards releases the free phenylamine.

The same method works on any nitroarene. Every NO₂ group needs 6[H] and gives 2H₂O.

Basicity and preparation of amines: preparation routes compared

The last row is the nitrile route from Carbon-carbon bond formation (6.2.4). The haloalkane + ammonia route keeps the same number of carbons; the nitrile route adds one, because the CN carbon becomes part of the chain:

CH₃CH₂CN + 2H₂ → CH₃CH₂CH₂NH₂

Worked examples

No calculation is listed for 6.2.1, so these examples model the equations, names and routes the marks are for.

Example 1: salts of a diamine

Write an equation for hexane-1,6-diamine with excess dilute hydrochloric acid, and give the formula of the salt.

1. Count the basic sites: there are two NH₂ groups, each with a lone pair.
2. Each NH₂ accepts one H⁺, so 2HCl react.
3. Each nitrogen becomes NH₃⁺, so the cation carries 2+, balanced by 2Cl^-.

H₂N(CH₂)₆NH₂ + 2HCl → [H₃N(CH₂)₆NH₃]²⁺ 2Cl^-

Check: C₆H₁₆N₂ + 2HCl gives C₆H₁₈N₂Cl₂. Atoms balance, and the two + charges cancel the two Cl^-.

Where marks go: using 1 HCl, or writing the salt without charges (e.g. "H₂N(CH₂)₆NH₂Cl₂").

Example 2: predicting substitution products

1-bromobutane is heated with excess ethanolic ammonia. Give the organic product and an equation. Then explain how a small amount of dibutylamine could also form.

1. Ammonia's lone pair substitutes for Br, so the organic product is butylamine (butan-1-amine), CH₃CH₂CH₂CH₂NH₂. Same four carbons as the haloalkane.
2. Equation, with a second NH₃ taking the H⁺: CH₃CH₂CH₂CH₂Br + 2NH₃ → CH₃CH₂CH₂CH₂NH₂ + NH₄Br
3. Butylamine still has a lone pair on N, so it is also a nucleophile.
4. It can attack another 1-bromobutane molecule, giving the secondary amine dibutylamine, (CH₃CH₂CH₂CH₂)₂NH.
5. The excess ammonia keeps this to a small amount.

Example 3: model mechanism checklist

Outline the mechanism for ammonia with 1-bromopropane (Figure 3). Each line is a point a marker looks for:

1. δ+ on the carbon bonded to Br, δ- on Br.
2. Curly arrow from the lone pair on N of NH₃ to the δ+ carbon.
3. Curly arrow from the C-Br bond to Br, with Br^- shown as a product.
4. Intermediate CH₃CH₂CH₂NH₃⁺, with the + on N.
5. A second NH₃ uses its lone pair to remove H⁺ from N⁺, and the N-H bonding pair returns to N.
6. Products: propylamine and NH₄⁺ (with Br^-, i.e. NH₄Br).

Example 4: a two-step route to an aromatic amine

Give a two-step synthesis of phenylamine from benzene, with reagents, conditions and an equation for each step.

1. Nitration: benzene with concentrated HNO₃ and concentrated H₂SO₄, warm to about 50 °C. C₆H₆ + HNO₃ → C₆H₅NO₂ + H₂O
2. Reduction: nitrobenzene with tin and concentrated HCl, heat under reflux, then add NaOH(aq) to release the amine. C₆H₅NO₂ + 6[H] → C₆H₅NH₂ + 2H₂O

Where marks go: writing "dilute HCl", or "HCl" with no tin, or 3[H] or 4[H] in the reduction equation.

Common exam mistakes

• Explaining basicity without the lone pair or the proton. "Amines are basic because they contain nitrogen" scores nothing. The mark needs: the lone pair on the nitrogen accepts a proton (H⁺).
• Calling the proton a hydrogen atom. A base accepts H⁺ (a proton), not "a hydrogen".
• Wrong salt formula. Writing ethylammonium chloride as CH₃CH₂NH₂Cl or CH₃CH₂NH₄⁺Cl^-. The nitrogen gains one H: CH₃CH₂NH₃⁺Cl^-, with the + on the nitrogen.
• 1:1 ratio with sulfuric acid. H₂SO₄ has two protons, so two amine molecules react per H₂SO₄, giving (RNH₃⁺)₂SO₄^2-.
• Leaving out "excess" or "ethanolic". The reagent is excess ammonia in ethanol. "Aqueous ammonia" or "NH₃" alone loses the conditions mark.
• Thinking haloalkane + ammonia lengthens the chain. It does not: bromoethane gives ethylamine. To gain a carbon, go via the nitrile (CN^-, then H₂/Ni).
• Mechanism slips: a negative charge on NH₃, the first arrow drawn from N or an H rather than from the lone pair, the + on carbon instead of nitrogen in the intermediate, or using Br^- to remove the H⁺ (it is a second NH₃).
• Forgetting that the product can react again. Asked why a mixture forms, say the amine has a lone pair so is also a nucleophile, and it reacts with more haloalkane.
• Wrong reagents for the nitroarene reduction: "dilute HCl", "HCl" with no tin, or "Sn catalyst". Tin is a reactant (it is the reducing agent), and the acid is concentrated.
• Unbalanced [H] equation. Nitrobenzene to phenylamine needs 6[H] and gives 2H₂O. Students often write 2[H] or forget the water.
• Forgetting the NaOH step. In the acid, the product is the salt phenylammonium chloride; NaOH is needed to release phenylamine.
• Ranking base strengths. Comparing the basicity of different amines is not required by OCR, so time spent explaining inductive effects or delocalisation in phenylamine earns nothing here.

Key definitions

• Amine (as a base): the lone pair on the nitrogen accepts a proton (H⁺).
• Brønsted-Lowry base: a species that accepts a proton (H⁺).
• Nucleophile: an electron pair donor.
• Dative covalent (coordinate) bond: a covalent bond in which the shared pair of electrons is supplied by one of the bonding atoms only.
• Primary / secondary / tertiary amine: an amine with one / two / three carbon (alkyl or aryl) groups bonded to the nitrogen atom.
• Salt of an amine: the ionic compound formed when an amine accepts a proton from an acid, e.g. CH₃CH₂NH₃⁺Cl^- (ethylammonium chloride).
• Preparation of a primary aliphatic amine: substitution of a haloalkane with excess ethanolic ammonia.
• Preparation of an aromatic amine: reduction of a nitroarene using tin and concentrated hydrochloric acid.

Specification

• I can explain the basicity of amines in terms of the lone pair on nitrogen accepting a proton. (6.2.1 a)
• I can write equations for amines (including phenylamine and diamines) reacting with dilute inorganic acids such as HCl(aq), H₂SO₄(aq) and HNO₃(aq), and name the salts formed. (6.2.1 a)
• I can describe how to prepare a primary aliphatic amine by substitution of a haloalkane with excess ethanolic ammonia, and write the equation. (6.2.1 b i)
• I can explain why excess ammonia is used, and how secondary and tertiary amines form from a haloalkane and an amine. (6.2.1 b i)
• I can describe the preparation of an aromatic amine by reduction of a nitroarene using tin and concentrated hydrochloric acid, and write the equation using [H]. (6.2.1 b ii)
• I can recall that nitriles can also be reduced to amines (with H₂/Ni), giving a chain one carbon longer than the haloalkane. (6.2.1 b guidance, see 6.2.4 c)

Related notes

• Acids: acids, bases and neutralisation forming salts
• Bonding and structure: lone pairs and dative covalent bonds, as in NH₄⁺
• Haloalkanes: nucleophiles and the nucleophilic substitution mechanism
• Basic concepts of organic chemistry: naming and curly arrows
• Acids, bases and buffers: Brønsted-Lowry acids and bases
• Aromatic compounds: nitration of benzene, the step before making phenylamine
• Amino acids, amides and chirality: the amine group in amino acids reacting with acids, and amines making amides
• Polyesters and polyamides: diamines as monomers
• Carbon-carbon bond formation: making amines by reducing nitriles
• Organic synthesis: fitting amine preparations into multi-step routes