Primary amines

1. Overview

Amines are organic relatives of ammonia. Replace one hydrogen atom of NH₃ with an alkyl group and you have a primary amine, such as ethylamine, CH₃CH₂NH₂. At AS Level you need one way to make them: heat a halogenoalkane with ammonia dissolved in ethanol, in a sealed tube. This page explains each condition, the equations, and why the ammonia must be in large excess.

Key Definitions

• Amine: an organic compound related to ammonia, NH₃, in which one or more of its hydrogen atoms have been replaced by carbon-containing groups such as alkyl groups.
• Primary amine: an amine whose nitrogen atom is bonded to one carbon atom and two hydrogen atoms, RNH₂; for example ethylamine, CH₃CH₂NH₂.
• Nucleophile: an electron-pair donor; a species that uses a lone pair to form a new covalent bond with an electron-deficient atom, such as the δ+ carbon of a C-X bond. Ammonia is one because of the lone pair on its nitrogen.
• Nucleophilic substitution: a reaction in which a nucleophile attacks an electron-deficient carbon atom and replaces an atom or group bonded to it, such as the halogen in a halogenoalkane.
• Ammonia in ethanol (ethanolic ammonia): ammonia dissolved in ethanol, the reagent used to make amines from halogenoalkanes. Ethanolic means "in ethanol"; do not confuse it with ethanoic, which means the acid.
• Heating in a sealed tube (under pressure): heating a mixture in a closed container so that a volatile reactant, here ammonia, cannot escape; the pressure inside rises as it is heated.
• Alkylammonium salt: an ionic compound such as ethylammonium bromide, CH₃CH₂NH₃⁺ Br⁻, formed when ammonia replaces the halogen in a halogenoalkane; a strong alkali such as NaOH(aq) releases the free amine from it.

Content

What a primary amine is

In a primary amine, the nitrogen atom is bonded to one alkyl group, written R, and two hydrogen atoms: RNH₂. The -NH₂ group is the amine functional group. As in ammonia, the nitrogen keeps one lone pair of electrons, which lets amines act as nucleophiles.

To name a straight-chain amine with -NH₂ on the end carbon, add "amine" to the alkyl group's name (propylamine, not propamine):

Structural formula	Name
CH₃CH₂NH₂	ethylamine
CH₃CH₂CH₂NH₂	propylamine
CH₃(CH₂)₅NH₂	hexylamine

You may also see names such as propan-1-amine. When -NH₂ is not on an end carbon, a number shows where it is: CH₃CH(NH₂)CH₃ is 2-aminopropane.

Here "primary" means the nitrogen is bonded to one carbon atom, so 2-aminopropane is a primary amine too. (This differs from the halogenoalkane meaning in 15.1; classifying amines is not needed at AS Level.)

Making an amine from a halogenoalkane

• Reagent: ammonia, NH₃, in large excess.
• Conditions: dissolved in ethanol, heated in a sealed tube (under pressure).

For bromoethane, the bromine is replaced by -NH₂:

CH₃CH₂Br + NH₃ → CH₃CH₂NH₂ + HBr

In fact the HBr reacts with more ammonia to form ammonium bromide. So, with ammonia in excess, the equation can also be written:

CH₃CH₂Br + 2NH₃ → CH₃CH₂NH₂ + NH₄Br

Both forms are balanced; the second shows where the HBr ends up. In general, with X = Cl, Br or I: RX + 2NH₃ → RNH₂ + NH₄X. These equations are written without state symbols. Never add (aq): the solvent is ethanol, not water.

Chloro-, bromo- and iodoalkanes all react. For the same carbon chain, iodoalkanes react fastest, because the C-I bond is the weakest of the three carbon-halogen bonds (15.1).

Why each condition is needed

• Ethanol as the solvent. Halogenoalkanes barely dissolve in water, so with aqueous ammonia the reactants would form two layers and react only where they meet. Ethanol dissolves both. With water present, some halogenoalkane could also be hydrolysed to an alcohol.
• Heat. The reaction is slow at room temperature.
• A sealed tube. Ammonia is a gas at room temperature, so heating drives it out of an open flask. Reflux does not help: ammonia boils far below 0 °C, so a water-cooled condenser cannot liquefy it and much of it escapes. In a sealed tube the ammonia stays in, and the pressure rises as the tube is heated.
• A large excess of ammonia. This makes the primary amine the main product (see below).

How the reaction happens

Ammonia acts as a nucleophile. The C-X bond is polar: the halogen is δ− and its carbon is δ+.

1. The lone pair on the nitrogen of NH₃ forms a new bond to the δ+ carbon.
2. The C-X bond breaks heterolytically and the halogen leaves as X⁻. For a primary halogenoalkane, steps 1 and 2 happen together (the SN2 mechanism in 15.1).
3. The nitrogen now has four bonds and a positive charge: the product is an alkylammonium salt.

CH₃CH₂Br + NH₃ → CH₃CH₂NH₃⁺ Br⁻

4. A second ammonia molecule takes an H⁺ ion from the ethylammonium ion, freeing the amine:

CH₃CH₂NH₃⁺ + NH₃ ⇌ CH₃CH₂NH₂ + NH₄⁺

This step is reversible; a large excess of ammonia pushes it to the right. Adding the two equations gives the overall equation with 2NH₃. Any amine left as its salt can be released with NaOH(aq), just as alkali displaces ammonia from ammonium salts (12.1).

Why ammonia must be in excess

The amine product still has a lone pair on its nitrogen, so it is a nucleophile too. An ethylamine molecule that meets a bromoethane molecule can replace the bromine, giving (CH₃CH₂)₂NH, with two ethyl groups on the nitrogen. That product can react again, and so on until four alkyl groups are bonded to one nitrogen.

With equal amounts of reactants, the result is a mixture of all these products. With a large excess of ammonia, each halogenoalkane molecule is far more likely to meet ammonia than an amine, so the primary amine is the main product. (Making amines with two alkyl groups on purpose is A Level work, in 34.1.)

Choosing the right reagent

Each reagent gives a different product:

Reagent and conditions	Product
NaOH(aq), heat	alcohol
NaOH in ethanol, heat	alkene
KCN in ethanol, heat	nitrile
NH₃ in ethanol, heat in a sealed tube	amine

Three use ethanol as the solvent, so always name the reagent too.

Worked example 1: making hexylamine

Question: Hexylamine, CH₃(CH₂)₅NH₂, can be made from 1-bromohexane, CH₃(CH₂)₅Br. (a) Give the reagent and conditions. (b) Name the type of reaction and the role of the reagent. (c) Write an overall equation with the reagent in excess.

Step 1: spot the change. The carbon chain is unchanged and Br has become NH₂, so the reagent is ammonia.

(a) Excess NH₃ dissolved in ethanol, heated in a sealed tube (under pressure).

(b) Nucleophilic substitution. Ammonia is the nucleophile: the lone pair on its nitrogen bonds to the δ+ carbon carrying the bromine, and Br⁻ leaves.

(c) A second NH₃ takes the hydrogen bromide as ammonium bromide:

CH₃(CH₂)₅Br + 2NH₃ → CH₃(CH₂)₅NH₂ + NH₄Br

Check: each side has 6 C, 19 H, 2 N and 1 Br.

Answer: (a) excess NH₃ in ethanol, heated in a sealed tube; (b) nucleophilic substitution, ammonia is the nucleophile; (c) as above.

Worked example 2: percentage yield of hexylamine

Question: A student heats 8.25 g of 1-bromohexane, C₆H₁₃Br, with a large excess of ammonia in ethanol in a sealed tube, and obtains 3.62 g of pure hexylamine, C₆H₁₅N. (a) Calculate the percentage yield. (b) Suggest two reasons, linked to this reaction, why the yield is below 100%. ($A_{\text{r}}$: H 1.0, C 12.0, N 14.0, Br 79.9)

Step 1: molar masses.

• $M(\text{C}_6\text{H}_{13}\text{Br}) = 6(12.0) + 13(1.0) + 79.9 = 164.9$ g mol⁻¹
• $M(\text{C}_6\text{H}_{15}\text{N}) = 6(12.0) + 15(1.0) + 14.0 = 101.0$ g mol⁻¹

Step 2: amount of 1-bromohexane. $$n = \frac{8.25}{164.9} = 0.05003\ \text{mol}$$

Step 3: theoretical mass. Ammonia is in excess, so 1-bromohexane is limiting, and 1 mol of it gives at most 1 mol of hexylamine. $$m = 0.05003 \times 101.0 = 5.053\ \text{g}$$

Step 4: percentage yield (actual mass ÷ theoretical mass × 100, from 2.4). $$\frac{3.62}{5.053} \times 100 = 71.6\%$$

Keep the extra figures until the end: rounding $n$ to 0.0500 mol first gives 71.7%.

(b) Two reasons linked to this reaction:

• Some hexylamine reacted further with 1-bromohexane, giving products with more than one hexyl group on the nitrogen.
• Some hexylamine stayed as its salt, hexylammonium bromide, and was not released.

Answer: (a) 71.6% (3 s.f.); (b) further substitution of the amine, and amine left as its salt.

Key Equations

Making a primary amine, ammonia in excess (R = alkyl group; X = Cl, Br or I; NH₃ in ethanol, heated in a sealed tube):

RX + 2NH₃ → RNH₂ + NH₄X

The same reaction, showing the hydrogen halide: RX + NH₃ → RNH₂ + HX

The two steps: RX + NH₃ → RNH₃⁺ X⁻ (the salt), then RNH₃⁺ + NH₃ ⇌ RNH₂ + NH₄⁺ (pushed right by excess NH₃).

Common Mistakes to Avoid

• Mistake: incomplete conditions, such as "NH₃, heat". Fix: give every part. For 1-bromohexane: "excess NH₃ in ethanol, heated in a sealed tube".
• Mistake: writing "heat under reflux". Fix: write "heat in a sealed tube" or "under pressure"; ammonia gas escapes through a reflux condenser.
• Mistake: giving aqueous ammonia, NH₃(aq). Fix: use ammonia in ethanol. 1-bromohexane barely dissolves in water, so in NH₃(aq) it would form a separate layer.
• Mistake: mixing up the halogenoalkane reagents. Fix: match reagent to product. Heating 1-bromohexane with NaOH in ethanol gives hex-1-ene, not hexylamine.
• Mistake: an unbalanced equation, such as CH₃CH₂Br + NH₃ → CH₃CH₂NH₂ + Br⁻. Fix: its right side has one H too few and a charge of −1. The bromine ends up in HBr, or in NH₄Br with excess ammonia: CH₃CH₂Br + 2NH₃ → CH₃CH₂NH₂ + NH₄Br.
• Mistake: explaining the excess only as "to speed it up". Fix: the amine is a nucleophile too, so without a large excess it reacts further: ethylamine attacks bromoethane to give (CH₃CH₂)₂NH.

Exam Tips

• State the reagent and conditions: ammonia (or NH₃), plus every condition: in ethanol, heat, sealed tube or under pressure. Adding "excess" is good practice. Write "ethanolic" clearly: "ethanoic" means the acid.
• Name the reaction or mechanism as nucleophilic substitution, with ammonia as the nucleophile because of the lone pair on its nitrogen.
• Write an equation with the structural formulas from the question. Either RX + NH₃ → RNH₂ + HX or RX + 2NH₃ → RNH₂ + NH₄X is balanced; if the question mentions excess ammonia or an ammonium halide, use the 2NH₃ form. Count the hydrogens on each side.
• Multi-step routes: the only amine-making reaction you must recall at AS starts from a halogenoalkane. From an alcohol or alkene, make the halogenoalkane first (16.1 or 14.2), then add ammonia, and give conditions for each step.
• Work backwards by putting a halogen on the carbon that carries -NH₂, keeping the same skeleton: CH₃CH₂CH(NH₂)CH₂CH₂CH₃ comes from CH₃CH₂CHBrCH₂CH₂CH₃.