Nitrogen compounds

Making a primary amine: a halogenoalkane with ammonia

A primary amine contains the -NH2 group joined to a carbon atom, so ethylamine is CH3CH2NH2 and propylamine is CH3CH2CH2NH2. The carbon-nitrogen bond in the middle of the molecule is made in the laboratory by heating a halogenoalkane with ammonia. Ammonia is the nucleophile: the nitrogen atom carries a lone pair, and that lone pair attacks the delta-positive carbon of the C-Br bond, because bromine is more electronegative than carbon and has pulled electron density away from it. The bromide ion leaves with the bonding pair, so the mechanism is nucleophilic substitution.

The conditions matter as much as the reagents. The halogenoalkane and the ammonia are heated together in ethanol in a sealed tube, so the reaction runs under pressure. Ethanol is the solvent because it dissolves both the halogenoalkane and the ammonia and brings the two reactants into the same phase; water is not used, because the halogenoalkane would also be hydrolysed to an alcohol and ammonia is much less soluble in water. The tube is sealed because the mixture has to be heated above the boiling point of ammonia (-33 °C) to react at a useful rate, and in an open flask the ammonia would escape as a gas.

An excess of ammonia is used, and there is a real reason for it. Once the primary amine has been formed it has a lone pair of its own, so it can act as a nucleophile and attack a second molecule of the halogenoalkane, giving a secondary amine, then a tertiary amine, and finally a quaternary ammonium salt. Keeping the concentration of ammonia far higher than the concentration of the amine makes reaction with ammonia by far the most likely step, so the primary amine is the main organic product. The overall yield is still only moderate and the further products are present as impurities. You are not asked to classify amines at AS Level, but you are expected to explain why the excess is used.

The equation for the reaction in excess ammonia is CH3CH2Br + 2NH3 -> CH3CH2NH2 + NH4Br: one molecule of ammonia substitutes the bromine and a second molecule removes the hydrogen bromide that is released. The mixture at the end contains the amine as its salt, CH3CH2NH3+Br-, because the amine is a stronger base than ammonia and takes the proton. Adding sodium hydroxide solution liberates the free amine, which can then be distilled off.

Nitriles: adding one carbon atom to a halogenoalkane

A nitrile contains the -CN group. The point most candidates drop is that the carbon of the CN group counts as part of the chain: CH3CH2CN has three carbon atoms and is propanenitrile, not ethanenitrile. In the laboratory a nitrile is made by heating a halogenoalkane with potassium cyanide dissolved in ethanol, under reflux. KCN supplies the cyanide ion, CN-, and the mechanism is again nucleophilic substitution.

In the mechanism a lone pair on the carbon atom of the cyanide ion attacks the delta-positive carbon of the halogenoalkane, because the halogen has drawn electron density towards itself. At the same time the pair of electrons in the C-Br bond moves onto the bromine atom, which leaves as a bromide ion. The first arrow, from the CN- lone pair to the carbon, shows the new carbon-carbon bond being formed; the second arrow, from the C-Br bond to the bromine, shows the old bond breaking and the pair going with the leaving group.

This reaction is the standard way to lengthen a carbon chain by one carbon atom, and that is why it appears in almost every synthesis route at this level. Bromoethane (2 C) gives propanenitrile (3 C), and propanenitrile hydrolyses to propanoic acid (3 C), so the two reactions together convert a two-carbon halogenoalkane into a three-carbon carboxylic acid. Nothing else in the AS course adds a single carbon in one step.

The conditions are ethanol as the solvent and heat under reflux, because the reaction is slow at room temperature and ethanol dissolves both KCN and the halogenoalkane. Reflux means boiling the mixture in a flask fitted with a vertical condenser, so that vapour condenses and runs back into the flask; the mixture can then be heated for as long as the reaction needs without the solvent boiling away. Potassium cyanide is very poisonous and releases hydrogen cyanide with any acid, so it is weighed out and used in a fume cupboard, and the mixture is never acidified while unreacted cyanide is still present.

• Reagents: KCN, the source of the cyanide ion nucleophile.
• Solvent: ethanol, which dissolves both KCN and the halogenoalkane.
• Conditions: heat under reflux.
• Chain length: the CN carbon is counted in the chain, so 1-bromopropane (3 C) gives butanenitrile (4 C).

Hydroxynitriles: nucleophilic addition to an aldehyde or ketone

Aldehydes and ketones react with hydrogen cyanide to give hydroxynitriles, compounds that carry an -OH group and a -CN group on the same carbon atom. Ethanal gives 2-hydroxypropanenitrile, CH3CH(OH)CN, and propanone gives 2-hydroxy-2-methylpropanenitrile, CH3C(OH)(CH3)CN. The mechanism is nucleophilic addition, and it is the first reaction in the course in which a nucleophile attacks a carbonyl group instead of a halogenoalkane.

Describe the mechanism in three moves. First, a lone pair on the carbon atom of the cyanide ion attacks the delta-positive carbon of the C=O group, which is delta-positive because oxygen is more electronegative than carbon. Second, at the same moment, the pair of electrons in the C=O pi bond moves onto the oxygen atom, giving a negatively charged intermediate in which the carbon-oxygen bond is a single bond and the oxygen carries the negative charge. Third, the lone pair on that negative oxygen picks up a proton from HCN or from water in the mixture, and the neutral hydroxynitrile is left behind.

KCN is a catalyst, and understanding why is worth marks. HCN is a weak acid, so only a very small fraction of it is ionised and the concentration of CN- in the mixture is far too low for a fast reaction. Adding a little KCN supplies a high concentration of CN- directly, so the nucleophile attacks far more often and the reaction speeds up; the CN- is regenerated at the end of the mechanism, so only a catalytic amount is needed. Because hydrogen cyanide is an extremely poisonous gas, in practice it is generated in the flask by adding dilute sulfuric acid to potassium cyanide, and the work is done in a fume cupboard with the apparatus sealed as far as possible.

Naming follows the rules you already know, with the -CN carbon counted in the chain and the -OH given the lowest possible number: ethanal (2 C) gives 2-hydroxypropanenitrile (3 C, because the CN carbon is part of the chain) and propanone (3 C) gives 2-hydroxy-2-methylpropanenitrile (4 C). The older name for a hydroxynitrile is a cyanohydrin, and you may still meet it in textbooks.

Hydrolysing a nitrile to a carboxylic acid

A nitrile is hydrolysed by heating it under reflux with dilute acid, such as dilute hydrochloric or dilute sulfuric acid. The -CN group is converted into a -COOH group and the nitrogen finishes up as ammonium ions. For propanenitrile the equation is CH3CH2CN + H2O + HCl -> CH3CH2COOH + NH4Cl. With dilute sulfuric acid two moles of nitrile are needed for each mole of acid: 2CH3CH2CN + 4H2O + H2SO4 -> 2CH3CH2COOH + (NH4)2SO4.

The second route is heating with dilute alkali, such as aqueous sodium hydroxide, and then acidifying the mixture. Alkaline hydrolysis gives the carboxylate salt and ammonia: CH3CH2CN + NaOH + H2O -> CH3CH2COONa + NH3. Acidifying that salt liberates the carboxylic acid: CH3CH2COONa + HCl -> CH3CH2COOH + NaCl. The acidification step is easy to forget and it earns a mark, because no free carboxylic acid is present while the mixture is still alkaline - the hydroxide keeps it deprotonated as the carboxylate ion.

What happens in the flask is a two-stage hydrolysis. A water molecule attacks the carbon of the -CN group and the group is converted first into an amide, -CONH2, and the amide is then hydrolysed to the carboxylic acid. Acid catalyses both stages, which is why dilute acid is used rather than water alone, and why the mixture is heated under reflux for a long time: neither stage is fast at room temperature.

Hydroxynitriles are hydrolysed in exactly the same way, but the -OH group survives the hydrolysis, so a hydroxynitrile gives a 2-hydroxycarboxylic acid. 2-Hydroxypropanenitrile, made from ethanal and HCN, gives 2-hydroxypropanoic acid, CH3CH(OH)COOH, the acid of sour milk. This is what makes the hydroxynitrile step worth doing: it puts an -OH group and a -COOH group on adjacent carbon atoms, starting from an aldehyde, in two steps.

• Dilute acid, reflux: carboxylic acid plus ammonium ions, for example CH3CH2CN + H2O + HCl -> CH3CH2COOH + NH4Cl.
• Dilute alkali, reflux, then acidify: carboxylate salt plus ammonia first, then the free carboxylic acid after acidification.
• Hydroxynitrile hydrolysis keeps the -OH group, giving a 2-hydroxycarboxylic acid such as 2-hydroxypropanoic acid.

Curly arrows, and the two mechanisms to keep apart

A curly arrow shows the movement of a pair of electrons. It starts from where the pair is - a lone pair on an atom, or the middle of a bond - and points to where the pair goes: onto an atom to become a lone pair, or between two atoms to become a new covalent bond. When a bond breaks, the arrow goes from the bond to the atom that keeps the pair, normally the more electronegative one. Electronegativity is why these reactions happen at all: the carbon of a C-Br bond is delta-positive because bromine is more electronegative, and the carbon of a C=O bond is delta-positive because oxygen is more electronegative, so both are open to attack by a nucleophile.

Nucleophilic substitution is the pattern for a halogenoalkane: one arrow from the lone pair of the nucleophile to the delta-positive carbon, and one from the C-Br bond to the bromine. The carbon keeps four bonds throughout, because one bond is made as another breaks. Ammonia and the cyanide ion both follow this pattern, which is why the same style of conditions - the nucleophile dissolved in ethanol with heat - covers both reactions, the only difference being whether the nucleophile is NH3 or CN-.

Nucleophilic addition is the pattern for a carbonyl compound: the nucleophile attacks the delta-positive carbon of the C=O group, the pi bond of the C=O breaks and its pair goes onto the oxygen, and the oxygen then takes a proton from the reagent. Nothing leaves the molecule, so the two parts of the reagent add across the double bond. If you are unsure which mechanism a question wants, look at the substrate: a halogenoalkane, which has a good leaving group, gives substitution, and a C=O double bond gives addition.

Amines as bases: the lone pair on the nitrogen atom

The same lone pair that makes ammonia and amines nucleophiles also makes them bases. A base accepts a proton, and the lone pair on the nitrogen forms a dative covalent bond to the H+ ion. Ethylamine therefore reacts with dilute hydrochloric acid to give ethylammonium chloride: CH3CH2NH2 + HCl -> CH3CH2NH3+Cl-. This is why amines dissolve readily in dilute acid, and why the amine made from a halogenoalkane and ammonia comes out of the flask as a salt rather than as the free amine.

Amines are stronger bases than ammonia. An alkyl group releases electron density towards the nitrogen atom by the inductive effect, and the additional electron density makes the lone pair better able to accept a proton, so the position of RNH2 + H2O <-> RNH3+ + OH- lies further to the right than for NH3 in the same equilibrium. A solution of methylamine therefore has a higher pH than a solution of ammonia of the same concentration. The more alkyl groups there are on the nitrogen, the more electron density is released, so the base strength increases in that order.

That difference has a practical use when you free the amine from its salt. Adding sodium hydroxide pushes the equilibrium back towards the free amine: CH3CH2NH3+Cl- + NaOH -> CH3CH2NH2 + NaCl + H2O. The free amine is volatile and can be distilled off and collected. In a preparation you would heat the halogenoalkane with ethanolic ammonia in a sealed tube, add dilute acid to convert the amine into its salt so that it stays in the aqueous layer, and then add alkali and distil the free amine over.

Keep the two roles of the lone pair apart when you answer questions. Asked why an amine is a nucleophile, you say the lone pair attacks a delta-positive carbon and forms a new carbon-nitrogen bond. Asked why an amine is a base, you say the lone pair accepts a proton and forms a dative covalent bond. Both answers begin with the same lone pair, but they are separate marks.

Conditions, hazards, and how the four reactions fit together

Each reaction in this topic has its own set of conditions, and examiners award the reagent, the solvent and the temperature or pressure as separate points, so learn them as a package rather than as a list of chemicals. The table below collects them; the detail that distinguishes the reactions is whether the nucleophile attacks a halogenoalkane or a carbonyl group, and whether the nucleophile is ammonia or the cyanide ion.

All of these reactions are carried out hot and for a long time, and reflux is the technique that makes that possible: boiling the mixture in a flask fitted with a vertical condenser, so that the vapour condenses and returns to the flask. Heating in an open flask would simply boil the solvent away and leave a dry residue. The halogenoalkane and ammonia reaction is the exception, because it needs a temperature above the boiling point of ammonia: it goes in a sealed tube, so that the pressure rises instead of the ammonia escaping.

The hazards are specific and worth learning. Hydrogen cyanide is a gas that is lethal at low concentrations in air, so it is never handled directly: it is generated in the flask from potassium cyanide and dilute acid, in a fume cupboard. Potassium cyanide is itself very poisonous and gives off HCN with any acid, so it is never left in contact with acid outside a fume cupboard. Ammonia in a sealed tube is under pressure, so the tube is heated in a metal block or an oven and cooled before it is opened, and ethanol is flammable, so these mixtures are heated with an electric heating mantle or a water bath rather than over a naked flame.

See the whole set as one toolkit for building molecules. Ammonia turns a halogenoalkane into an amine. The cyanide ion turns a halogenoalkane into a nitrile, which is one carbon longer and which hydrolyses to a carboxylic acid. The cyanide ion also adds across the C=O group of an aldehyde or ketone to give a hydroxynitrile, which hydrolyses to a 2-hydroxy acid. Learn those four conversions with their conditions and you can answer any synthesis question set on this topic.