What this chapter covers19.1

This outcome asks for one reaction: a halogenoalkane heated with ammonia gives an amine. The reaction is short to state, but each of its conditions is there for a reason, and the marks go to students who give all of them. Ammonia has a lone pair, so it attacks the δ+ carbon of a halogenoalkane. It is a gas, so the reaction is done in a sealed tube. It is a base, so a second molecule is needed to take a proton. The amine it produces also has a lone pair, so the ammonia has to be in excess. This page works through each of those four points in turn.

The verb is recall. What is being tested is the reagent and its conditions: ammonia, in ethanol, heated under pressure. The same reaction also appears in the halogenoalkane unit as outcome 15.1.3(c), where it is one of four nucleophilic substitutions. That unit is where the mechanism and the reactivity order are examined, so this page uses them to explain the conditions rather than as separate outcomes. The statement does not ask for an equation, but if you are asked to write one, the easy mistake is to use one ammonia molecule where the balanced equation needs two.

The page is in three parts. The first explains what a primary amine is and then goes through the reaction: the equation, each condition and the reason for it, the mechanism, and why the ammonia is used in excess. The second part applies the route: choosing the right halogenoalkane for a target amine, how the halogen affects the rate, and how this reaction compares with the other things a halogenoalkane can become. The third part is the self-test, the definitions and the reference data.

What a primary amine is13.1 · 19.1

An amine is a compound in which one or more of the hydrogen atoms of ammonia has been replaced by an alkyl group. In a primary amine exactly one has been replaced, so the nitrogen is bonded to one carbon atom and two hydrogen atoms. The functional group is –NH₂, the amino group. The 13.1 functional-group table lists it as "amine (primary only)", and primary amines are the only amines this part of the course deals with.

The lone pair is the key feature of both molecules. It is what makes ammonia a nucleophile in the reaction that makes an amine, and it is also why the amine can go on reacting, which is the problem section 7 deals with.

Names

You will see three styles of name for the same compound, and it is worth recognising all of them. Cambridge papers most often use the alkylamine name: methylamine, ethylamine, propylamine. The systematic name puts -amine where an alcohol has -ol, with a locant for the carbon carrying the –NH₂ group: ethanamine, propan-1-amine, propan-2-amine. An older style uses amino- as a prefix: aminoethane. Where the –NH₂ group is not on the end of an unbranched chain, the systematic name is the clearer one, because "propylamine" does not say which carbon carries the group.

The two small amines are gases, or nearly so, at room temperature, and all of them smell strongly of fish. As with ammonia, the N–H bonds let them hydrogen-bond, so each amine boils well above the alkane of similar size but below the alcohol with the same skeleton: ethylamine at 17 °C, ethanol at 78 °C. An N–H bond is less polar than an O–H bond, because nitrogen is less electronegative than oxygen.

The reaction and its equation19.1.1(a)

The reaction happens in two stages, and the overall equation is easier to write correctly once you know what they are. In the first stage an ammonia molecule attacks the carbon and pushes the halide ion off. The nitrogen is then bonded to four atoms, so it carries a positive charge. The product at this point is not the amine but its salt, ethylammonium bromide:

CH₃CH₂Br + NH₃ → CH₃CH₂NH₃⁺ + Br⁻

In the second stage another ammonia molecule, acting as a base, removes a proton from the –NH₃⁺ group. This releases the free amine and forms an ammonium ion. This stage is an equilibrium, and the large excess of ammonia pushes it to the right:

CH₃CH₂NH₃⁺ + NH₃ ⇌ CH₃CH₂NH₂ + NH₄⁺

Adding the two stages and pairing the ammonium ion with the bromide ion gives the overall equation in the box above. The inorganic product is ammonium bromide, not hydrogen bromide. Any HBr formed would meet a large excess of a base and be neutralised straight away, and the second ammonia molecule in the equation is that base.

Every halogenoalkane in the syllabus reacts in the same way, and the model below shows it. Pick a starting compound and the model replaces the halogen with nitrogen on the same carbon, fills in the hydrogens, names the amine, and writes all three equations from the structure, checking each one for atoms and charge. Try bromoethane and iodoethane: they give the same amine. Only the by-product changes, from ammonium bromide to ammonium iodide.

Each condition, and why it is there19.1.1(a)

A mark scheme for this reaction usually needs the reagent and the conditions. The quickest way to remember the conditions is to know what problem each one solves.

Use the trainer below to test the set. It includes a few conditions from other syllabus reactions that are commonly written here by mistake, such as heating under reflux, adding an acid catalyst or using ethanolic NaOH. Each line explains itself once you check.

Why the tube is sealed19.1.1(a)

Heating under reflux works for most organic reactions because the reactants are liquids that boil above the temperature of the cooling water. The condenser turns their vapour back into liquid and it runs back into the flask. The model below shows why that approach fails here. It plots the boiling point of each substance involved. Move the heating temperature and every substance that boils below it turns red. In an open vessel, those are the substances you would lose.

Two things stand out. Ammonia boils far below any temperature you would heat the mixture to, so it is a gas even in a cold flask, and nothing short of a closed vessel keeps it in the reaction. Several of the reactants and products are also volatile. Bromoethane boils at 38 °C and ethylamine at 17 °C, so heating the mixture anywhere near the boiling point of ethanol would drive them off too. In a sealed tube nothing can escape. As the tube is heated, the pressure inside rises, and the higher pressure keeps more of the ammonia dissolved in the ethanol.

The mechanism15.1.3(c) · 15.1.5

This is nucleophilic substitution, the same reaction type as hydrolysis by hydroxide ions or substitution by cyanide ions. With a primary halogenoalkane such as bromoethane it goes by the S_N2 mechanism of 15.1.5: the nucleophile attacks as the leaving group departs, in a single step. With ammonia there is one step after that which hydroxide and cyanide do not need. The nucleophile is neutral, so the nitrogen ends up with a positive charge and a spare proton, and the proton has to be removed.

The arrows follow the rule in 13.2.2: each curly arrow starts at a lone pair or at a bond, and ends where that pair of electrons goes. In the substitution step, the arrow from nitrogen starts at its lone pair and not at the letter N, and the second arrow starts at the C–Br bond, not at the carbon. In the proton-transfer step, the second ammonia's lone pair goes to the hydrogen, and the N–H bonding pair moves back onto the nitrogen.

A secondary halogenoalkane such as 2-bromopropane reacts by a mixture of S_N1 and S_N2 (15.1.6), but the product is the same whichever mechanism operates, because the –NH₂ group ends up on the carbon the bromine left. Tertiary halogenoalkanes are left out of the models on this page. Ammonia is a base as well as a nucleophile, and with a tertiary halogenoalkane, elimination to an alkene competes strongly with substitution.

Why the ammonia is in excess19.1.1(a)

The last stage of the mechanism gives an amine with a lone pair on its nitrogen, which means the amine is a nucleophile just as ammonia is. If it meets a halogenoalkane molecule, it attacks it in exactly the same way, and the product has two alkyl groups on the nitrogen. That product can react again, and so can the next one, until the nitrogen carries four alkyl groups and a positive charge:

RNH₂ → R₂NH → R₃N → R₄N⁺X⁻

Left to itself, the reaction therefore gives a mixture, and the primary amine is only part of it. The fix is statistical. If there is far more ammonia than amine in the tube, a halogenoalkane molecule is far more likely to collide with ammonia than with an amine molecule. The model below makes that argument quantitative. It puts the four competing substitutions side by side, integrates them until the halogenoalkane is used up, and reports how much of the halogenoalkane ends up in each product.

At the model's starting setting, where the amine reacts twice as fast as ammonia, only about a third of the halogenoalkane becomes primary amine at 2 : 1, the exact ratio the equation needs. At 10 : 1 most of it does, and at 50 : 1 almost all of it does. The curve rises steeply at first and then levels off. Each extra mole of ammonia helps less than the one before, which is why an excess is used but not an unlimited one. Set the second slider to zero, so the amine does not react at all, and the problem disappears at every ratio. The mixture forms only because the product is a nucleophile too.

Choosing the starting halogenoalkane19.1.1(a)

Questions often run the reaction backwards: here is an amine, so what would you make it from? Two rules settle almost every case, and both follow from the reaction being a substitution.

• The chain length does not change. Ammonia brings a nitrogen and some hydrogens, but no carbon. A four-carbon amine comes from a four-carbon halogenoalkane.
• The position does not change. The –NH₂ group goes onto the carbon the halogen left. Propan-2-amine comes from 2-bromopropane, and propylamine from 1-bromopropane.

Any halogen will do. A chloroalkane, a bromoalkane and an iodoalkane with the same skeleton all give the same amine, and they differ only in how fast they react (section 9). Bromoalkanes are the usual examples in questions.

The wrong options in the model are built from the target, and each one matches a real mistake. The option one carbon short is what you get by confusing this route with the cyanide route of 19.2, which does add a carbon. The option with the halogen on the wrong carbon is a positional isomer, and it gives a different amine. The alcohol is a plausible guess, because alcohols also come from halogenoalkanes, but an –OH group is not displaced by ammonia under these conditions.

Which halogen reacts fastest15.1.7

The carbon–halogen bond breaks in the rate-determining step, so the rate depends mainly on how strong that bond is. The C–I bond is the weakest and the C–Cl bond the strongest, so with the same carbon skeleton an iodoalkane reacts fastest with ammonia and a chloroalkane slowest. This is the same order as for every other nucleophile in 15.1.3, and it is the order 15.1.7 asks you to explain.

The class of the halogenoalkane affects which mechanism operates (15.1.6): S_N2 for primary, S_N1 for tertiary, and a mixture for secondary. It does not change what the –NH₂ group replaces. It does matter when the substrate is tertiary: ammonia is a base, and with a tertiary halogenoalkane much of the product is an alkene formed by elimination. Syllabus questions on this route use primary and secondary halogenoalkanes.

One halogenoalkane, four products15.1.3 · 15.1.4 · 19.1.1(a)

Much of the difficulty in recalling this reaction is keeping it separate from the halogenoalkane's other reactions. Hydroxide ions, cyanide ions and ammonia all attack the same δ+ carbon, and ethanolic hydroxide removes HX instead. The reagent decides what replaces the halogen, but the solvent and the way the mixture is heated also matter: NaOH gives an alcohol in water and an alkene in ethanol. The model below accepts any combination and gives a product only when the combination matches a syllabus statement.

Writing it in an exam19.1.1(a)

This outcome is tested mainly by recall: a reaction scheme with a gap where a reagent and conditions belong, or a question asking how to turn a named halogenoalkane into an amine. A complete answer has three parts, and it is common to lose a mark by leaving one of them out.

Self-test19.1

Sixteen questions covering the reagent and conditions, the equation, the reason for each condition, choosing a starting material, and how this route differs from the halogenoalkane's other reactions. Each answer comes with a short explanation.

Definitions to learn19.1

Data used on this page19.1

These are the reference values used by the models and tables on this page. They are typical literature values, not the official data booklet, and sources can differ by a degree or two for boiling points and by a few kJ mol⁻¹ for bond enthalpies. Where an examination question gives data, use the data it gives.