Nitrogen Compounds (A Level)

0. What Examiners Want

This guide covers Topic 34 of Cambridge International AS & A Level Chemistry 9701 (2025–27) and follows the learning outcomes below, word for word. The right-hand column shows where each one is taught. It builds on Topic 19 (primary amines and nitriles at AS), Topic 32 (phenol and azo coupling with a phenoxide) and Topic 33 (acyl chlorides, the route to amides). Amino acids lead straight into Topic 35, where the same amide link makes polyamides.

1. Amines: Classes and Preparation

An amine is a derivative of ammonia in which one or more H atoms are replaced by alkyl or aryl groups. The class depends on the number of carbon groups on the nitrogen atom. Every amine has a lone pair on N, and this lone pair controls nearly all of their chemistry.

Primary: RNH₂. Secondary: R₂NH. Tertiary: R₃N. Phenylamine is a primary arylamine.

There are four ways to make alkylamines in the syllabus, and one of them leads to a secondary amine.

Amber: how ethylamine is made. Green: what it does. The last two green boxes use the reactions you meet again for amides.

Halogenoalkane and ammonia. This is nucleophilic substitution: the lone pair on NH₃ attacks the δ+ carbon (Topic 31). Excess ammonia is used for two reasons. It makes the chance of a second halogenoalkane molecule meeting the product small (the amine product is also a nucleophile and would give a secondary and then a tertiary amine), and it takes up the HBr as NH₄Br. The reaction is carried out under pressure because ammonia is a gas and the mixture is heated.

Secondary amines. Heating a halogenoalkane with a primary amine in ethanol in a sealed tube gives the secondary amine. Ethylamine and bromoethane give diethylamine.

Reduction routes. The nitrile route adds a carbon atom: to make ethylamine start from bromomethane, which gives CH₃CN with KCN in ethanol (Topic 19). Reducing an N-substituted amide gives a secondary amine: CH₃CONHCH₃ → CH₃CH₂NHCH₃.

Because amines are bases, they dissolve in dilute acid to give salts, for example ethylammonium chloride, CH₃CH₂NH₃⁺Cl⁻. Adding NaOH(aq) to the salt releases the free amine.

Worked example. Give the reagents and conditions to make propylamine from (i) 1-bromopropane and (ii) 1-bromoethane, using two steps. [4]

(i) A large excess of ammonia in ethanol, heated under pressure [1]; CH₃CH₂CH₂Br + 2NH₃ → CH₃CH₂CH₂NH₂ + NH₄Br [1]. (ii) KCN in ethanol, heat under reflux, to give CH₃CH₂CN [1]; then H₂ with Ni catalyst (or LiAlH₄ in dry ether) [1].

2. Basicity of Amines

An amine is a Brønsted–Lowry base: it accepts H⁺ by donating the lone pair on nitrogen to form a dative (co-ordinate) bond. In water it sets up an equilibrium that produces OH⁻:

Strongest base first: diethylamine > ethylamine > ammonia > phenylamine > ethanamide. Values of pK_b are approximate.

• Ethylamine > ammonia. The ethyl group is electron-donating (positive inductive effect). It pushes electron density on to N, so the lone pair is more available and the RNH₃⁺ ion formed is stabilised.
• Diethylamine > ethylamine. Two electron-donating ethyl groups push more density on to N.
• Ammonia > phenylamine. In phenylamine the lone pair on N overlaps with the π system of the benzene ring and is delocalised into the ring. It is less available to bond with H⁺.
• Ethanamide is neutral (Section 5): the lone pair is delocalised on to the C=O group.

The same delocalisation makes phenylamine a weaker base and makes the ring more reactive towards electrophiles.

Worked example. Place CH₃CH₂NH₂, NH₃ and C₆H₅NH₂ in order of basicity, strongest first, and explain. [4]

CH₃CH₂NH₂ > NH₃ > C₆H₅NH₂ [1]. Basicity depends on the availability of the lone pair on N to accept H⁺ [1]. The ethyl group is electron-donating, so it increases the electron density on N [1]. In phenylamine the lone pair is delocalised into the benzene ring, so it is less available [1].

Worked example. Calculate the pH of 0.10 mol dm⁻³ ethylamine, K_b = 5.0 × 10⁻⁴ mol dm⁻³ (K_w = 1.00 × 10⁻¹⁴). [3]

[OH⁻] = √(K_b × c) = √(5.0 × 10⁻⁴ × 0.10) = 7.07 × 10⁻³ mol dm⁻³ [1]. pOH = 2.15 [1]. pH = 14.00 − 2.15 = 11.85 [1].

3. Phenylamine

Preparation. Phenylamine, C₆H₅NH₂, cannot be made from chlorobenzene and ammonia (the C–Cl bond is too strong, Topic 31). It is made from benzene in two steps: nitration to nitrobenzene (Topic 29: conc. HNO₃ + conc. H₂SO₄, 25–60 °C), then reduction of the nitro group by heating with tin and concentrated hydrochloric acid under reflux. Finally NaOH(aq) is added.

The acid gives the phenylammonium salt. NaOH(aq) liberates the free amine, which is separated by steam distillation.

Reactions of phenylamine. The delocalised lone pair makes the ring electron-rich, so the –NH₂ group is 2-, 4- and 6-directing and strongly activating (compare phenol, Topic 32). Phenylamine also gives the diazonium salt and the other reactions of a primary amine.

Br₂(aq) gives a white precipitate at room temperature; the diazonium salt is made below 10 °C, warmed to give phenol, or coupled with phenol in NaOH(aq).

With bromine water. At room temperature, with no catalyst, Br₂(aq) is decolourised and a white precipitate of 2,4,6-tribromophenylamine forms. All three activated positions are substituted:

C₆H₅NH₂ + 3Br₂ → C₆H₂Br₃NH₂ + 3HBr

Diazotisation. Nitrous acid, HNO₂ (nitric(III) acid), is unstable, so it is made in the mixture from NaNO₂ and dilute HCl: NaNO₂ + HCl → HNO₂ + NaCl. The temperature must be kept below 10 °C (ice bath), because the benzenediazonium ion decomposes above this, giving off N₂.

C₆H₅NH₂ + HNO₂ + HCl → C₆H₅N₂⁺Cl⁻ + 2H₂O

Warming with water. If the diazonium salt solution is deliberately warmed, it is hydrolysed to phenol and nitrogen gas is evolved (effervescence):

C₆H₅N₂⁺ + H₂O → C₆H₅OH + N₂ + H⁺

Worked example. Phenylamine is converted into phenol in two steps. Give the reagents and conditions. [4]

Step 1: NaNO₂ and dilute HCl (or HNO₂) [1] below 10 °C [1]. Step 2: warm the solution of benzenediazonium chloride with water [1]; N₂ is given off [1].

4. Diazonium Salts and Azo Dyes

A benzenediazonium ion is a weak electrophile. It attacks the electron-rich ring of a phenol (as the phenoxide ion) in a coupling reaction, giving an azo compound. The azo group, –N=N–, is a bridge between the two rings.

The azo group joins two aromatic rings. Other aryl amines and phenols give other dyes.

• The diazonium ion substitutes at position 4, opposite the –OH group (position 2 is possible if 4 is blocked).
• The reaction is carried out in NaOH(aq): phenol is converted to the phenoxide ion, a much more reactive (more strongly activated) ring than phenol itself.
• Azo compounds are coloured because the delocalised π system extends over both rings and through the –N=N– bridge. Extensive delocalisation also makes them stable. They are widely used as dyes.
• Other azo dyes are made by the same route: change the amine (so that the diazonium ion differs) and/or the coupling component (a different phenol, naphthol or aromatic amine such as C₆H₅N(CH₃)₂, which gives a yellow dye).

Worked example. State the reagents and conditions for each step in making an azo dye from phenylamine and phenol. [5]

Step 1: NaNO₂ + dilute HCl [1], below 10 °C [1], gives C₆H₅N₂⁺Cl⁻ [1]. Step 2: add the diazonium solution to phenol dissolved in NaOH(aq) [1], below 10 °C, to give the orange azo dye with the –N=N– group [1].

5. Amides

An amide contains the group –CONH–. In names, the italic N shows which group is on the nitrogen: CH₃CONHCH₃ is N-methylethanamide, and CH₃CON(CH₃)₂ is N,N-dimethylethanamide.

Preparation. Acyl chlorides react with ammonia or a primary amine at room temperature (Topic 33; this is a condensation: HCl is lost). The HCl formed is taken up by a second molecule of the amine, so two moles are used:

CH₃COCl + 2NH₃ → CH₃CONH₂ + NH₄Cl

CH₃COCl + 2CH₃NH₂ → CH₃CONHCH₃ + CH₃NH₃⁺Cl⁻

Hydrolysis breaks the amide link; LiAlH₄ reduces the C=O group to CH₂.

Hydrolysis. The amide link is broken by water in the presence of acid or alkali, by heating under reflux:

• with dilute acid (HCl): the products are the carboxylic acid and the ammonium salt of the amine: CH₃CONHCH₃ + H₂O + HCl → CH₃COOH + CH₃NH₃⁺Cl⁻;
• with aqueous alkali (NaOH): the products are the sodium salt of the acid and the free amine: CH₃CONHCH₃ + NaOH → CH₃COO⁻Na⁺ + CH₃NH₂.

An unsubstituted amide, RCONH₂, gives ammonia (or NH₄⁺ in acid) in place of the amine.

Reduction. LiAlH₄ in dry ether reduces the C=O group of an amide to CH₂, giving an amine: CH₃CONH₂ → CH₃CH₂NH₂; CH₃CONHCH₃ → CH₃CH₂NHCH₃.

Why amides are much weaker bases than amines. In an amide the lone pair on N is delocalised over the N–C=O system, partly on to the electronegative oxygen. The C=O group is electron-withdrawing. The lone pair is therefore not available to accept H⁺, and amides are neutral in water.

The partial double bond between C and N makes the amide group planar.

Worked example. Compare the products of heating N-methylethanamide under reflux with (i) dilute HCl and (ii) NaOH(aq). [4]

(i) Ethanoic acid [1] and methylammonium chloride, CH₃NH₃⁺Cl⁻ [1]. (ii) Sodium ethanoate [1] and methylamine [1].

Worked example. Explain why ethanamide is a much weaker base than ethylamine. [3]

In ethylamine the ethyl group is electron-donating and the lone pair on N is available to accept H⁺ [1]. In ethanamide the lone pair on N is delocalised into the C=O group [1], so it is not available to bond with H⁺ [1].

6. Amino Acids and Zwitterions

An amino acid has an amino group (–NH₂, basic) and a carboxylic acid group (–COOH, acidic). The 2-aminocarboxylic acids, RCH(NH₂)COOH, are the building blocks of proteins. The group R may be neutral (H in glycine, CH₃ in alanine), acidic (an extra –COOH, as in glutamic acid) or basic (an extra –NH₂, as in lysine).

The carbon next to the –COOH group carries the –NH₂ group. Every amino acid except glycine has a chiral carbon.

All the 2-amino acids except glycine have a chiral carbon (four different groups on the 2-carbon: H, NH₂, COOH and R), so they exist as two optical isomers (enantiomers). Glycine has two H atoms on that carbon, so it has none.

Zwitterions. The acidic –COOH group donates H⁺ to the basic –NH₂ group within the same molecule, giving a zwitterion, RCH(NH₃⁺)COO⁻. It carries a positive and a negative charge, and no net charge. Because of this ionic character amino acids are crystalline solids with relatively high melting points and are soluble in water (strong ionic attractions between the ions, not the weak forces of a small molecule).

Adding acid gives the cation, adding alkali gives the anion. At the isoelectric point the species has no net charge.

Acid and base properties. An amino acid is amphoteric: it reacts with acids and with alkalis.

• In acid (low pH): the –COO⁻ group accepts H⁺, giving RCH(NH₃⁺)COOH, a cation.
• In alkali (high pH): the –NH₃⁺ group loses H⁺, giving RCH(NH₂)COO⁻, an anion.

Because the zwitterion can remove added H⁺ or OH⁻, a solution of an amino acid is a buffer.

The value of the pI depends on the R group. Acidic amino acids have a low pI (glutamic acid 3.2) because extra acid must be present to remove the charge on the side chain; basic amino acids have a high pI (lysine 9.7); neutral ones are near 6 (glycine 6.0, alanine 6.0). For a neutral amino acid the pI is halfway between the two pK_a values: for glycine ½(2.3 + 9.6) = 6.0 (5.95).

Worked example. Write two equations to show how an aqueous solution of alanine acts as a buffer. [2]

CH₃CH(NH₃⁺)COO⁻ + H⁺ → CH₃CH(NH₃⁺)COOH [1]; CH₃CH(NH₃⁺)COO⁻ + OH⁻ → CH₃CH(NH₂)COO⁻ + H₂O [1].

Worked example. Explain why glycine is a solid at room temperature with a high melting point, whereas ethylamine is a gas at room temperature. [2]

Glycine exists as a zwitterion, H₃N⁺CH₂COO⁻, so there are strong ionic attractions between oppositely charged ions in the lattice [1]. Ethylamine has only weak hydrogen bonding and van der Waals’ forces between its molecules [1].

7. Peptides

The –COOH group of one amino acid reacts with the –NH₂ group of another, and a molecule of water is eliminated. The new –CONH– link is an amide bond, called a peptide bond. Two amino acids give a dipeptide, three give a tripeptide, and many give a polypeptide or protein. This is a condensation reaction.

The N-terminus (free –NH₂) is written first. Two different amino acids can form two different dipeptides.

• A dipeptide still has a free –NH₂ at one end (the N-terminus) and a free –COOH at the other (the C-terminus), so the chain can grow.
• Two different amino acids, X and Y, can form two dipeptides, X–Y and Y–X. Three different amino acids can be joined in 3! = 6 different tripeptide sequences.
• Hydrolysis of a peptide: heating under reflux with dilute HCl (or NaOH(aq)) breaks every peptide bond and gives the amino acids again (as their ammonium salts in acid).

Worked example. Draw the dipeptide ala–glu (alanine at the N-terminus), formed from alanine and glutamic acid. Show the peptide bond. [2]

CH₃CH(NH₂)–CO–NH–CH(CH₂CH₂COOH)COOH [1 for the amide link, displayed as C(=O)–N–H], with the side chain of glutamic acid intact [1].

Worked example. The tripeptide Gly–Ala–Ser is refluxed with dilute hydrochloric acid. State the bond broken and give the products. [3]

The peptide (amide) bonds, C–N [1]. Glycine, alanine and serine are formed [1], as their ammonium chlorides because the acid is in excess [1].

8. Electrophoresis

Electrophoresis separates charged species in an electric field. A drop of the mixture is placed on the centre line (the origin) of a strip of paper or a gel soaked in a buffer of known pH, and a direct voltage is applied. Positively charged ions move towards the cathode (−), negatively charged ions towards the anode (+). Afterwards a locating agent (such as ninhydrin) shows the spots. You do not need to describe the apparatus in detail, but you must interpret results.

Compare the pH of the buffer with the pI of each species to predict the direction.

Rule. For each amino acid or peptide, compare the pH of the buffer with its pI:

• pH < pI: the species is positive and moves towards the cathode;
• pH = pI: no net charge, it does not move;
• pH > pI: negative, moves towards the anode.

The rate depends on the charge and on the size: the greater the charge and the smaller the species, the faster it moves. Two species with the same charge are therefore separated by size, the larger moving more slowly.

Worked example. A mixture of alanine (pI 6.0), glutamic acid (pI 3.2) and the dipeptide ala–glu is analysed by electrophoresis in a buffer at pH 6.0. Predict the positions of the three species. [3]

Alanine stays at the origin because pH = pI, so it has no net charge [1]. Glutamic acid (pH > pI) is negatively charged and moves towards the anode [1]. Ala–glu also moves to the anode but more slowly than glutamic acid, because it is larger [1].

Worked example. Lysine (pI 9.7) and glycine (pI 6.0) are separated at pH 8.0. Predict the direction of each. [2]

Lysine: pH 8.0 < pI 9.7, so it is positive and moves to the cathode [1]. Glycine: pH 8.0 > pI 6.0, so it is negative and moves to the anode [1].

9. Quick Sheet and Checklist

• [ ] I can state the reagents and conditions for each preparation of a primary and secondary amine.
• [ ] I can rank ammonia, ethylamine, diethylamine, phenylamine and ethanamide by basicity and explain each.
• [ ] I can describe the preparation of phenylamine and its reaction with bromine water.
• [ ] I can describe diazotisation, the hydrolysis to phenol and the coupling to an azo dye, with conditions.
• [ ] I can describe how amides are made, hydrolysed and reduced, and explain why they are not basic.
• [ ] I can draw a zwitterion and the ions at low and high pH, and define the isoelectric point.
• [ ] I can draw a dipeptide with the peptide bond displayed and describe its hydrolysis.
• [ ] I can predict the result of electrophoresis from the pH and the pI values.