Cambridge AS & A Level9701

Primary amines

Chemistry 9701 Chapter Notes

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Primary aminesNitriles and hydroxynitriles
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1. What are Primary Amines?

Primary amines are organic derivatives of ammonia (NH₃) where one hydrogen atom has been replaced by an alkyl group (like ethyl, -C₂H₅) or an aryl group (like phenyl, -C₆H₅). This gives them the general formula R-NH₂. The key feature is the amino group (-NH₂), which contains a nitrogen atom with a lone pair of electrons. This lone pair is the source of their characteristic chemical properties, particularly their basicity and nucleophilicity. For simple amines, we name them by adding '-amine' to the alkyl group name, e.g., ethylamine. For more complex molecules, the -NH₂ group is treated as a substituent and named 'amino', with its position indicated by a number, e.g., 2-aminopentane.

General formula: R-NH₂ (where R is an alkyl or aryl group)

Key term

Primary Amine: An organic compound containing an amino group (-NH₂) bonded to a single alkyl or aryl group.

Common pitfall

Confusing primary amines (one R group on the N) with primary alcohols (OH on a carbon bonded to one other carbon). The classification depends on what's attached to the nitrogen atom, not the carbon atom.

Fun fact

The smell of decaying fish is due to amines like trimethylamine, but primary amines like putrescine and cadaverine, formed during protein breakdown, are responsible for the foul smell of rotting flesh.

Worked example 12 marks

Propan-1-amine and phenylamine are both primary amines. Draw the displayed formula for each, showing the lone pair on the nitrogen atom.

  1. 1

    Step 1: Identify the alkyl/aryl group. For propan-1-amine, it's a propyl group (CH₃CH₂CH₂-). For phenylamine, it's a phenyl group (C₆H₅-).

  2. 2

    Step 2: Attach the amino group (-NH₂) to the carbon skeleton. For propan-1-amine, it's on the first carbon. For phenylamine, it's attached to the benzene ring.

  3. 3

    Step 3: Draw the displayed formula, showing all atoms and bonds. Crucially, add two dots (or a line) on the nitrogen atom to represent the lone pair of electrons.

  4. 4

    Propan-1-amine: H H H | | | H-C-C-C-N-H | | | | H H H H

  5. 5

    Phenylamine: A hexagon with a circle inside for the benzene ring, with -N(H)-H attached to one vertex. Add a lone pair on the N atom.

Recap

  • Primary amines have the general structure R-NH₂.
  • The nitrogen atom in a primary amine has a lone pair of electrons.
  • Aliphatic amines have an alkyl group attached to the -NH₂ group (e.g., ethylamine).
  • Aromatic amines have an aryl group attached to the -NH₂ group (e.g., phenylamine).
  • The lone pair of electrons makes amines basic and nucleophilic.

Quick check

  1. Name the compound CH₃CH(NH₂)CH₂CH₃.1 mark

2. Making Aliphatic Primary Amines

There are several key routes to synthesise aliphatic primary amines, which are common exam questions. The choice of method depends on the starting material and whether the carbon chain needs to be extended.

  1. From Halogenoalkanes: Heating a halogenoalkane with an excess of hot, ethanolic ammonia causes nucleophilic substitution. The ammonia acts as the nucleophile. Using excess ammonia is vital to minimise the formation of secondary and tertiary amines, as the primary amine product is also a nucleophile.
  2. Reduction of Nitriles: This is a two-step process that extends the carbon chain by one carbon. First, a halogenoalkane is reacted with potassium cyanide (KCN) in ethanol to form a nitrile (R-CN). The nitrile is then reduced to a primary amine using a strong reducing agent like LiAlH₄ in dry ether, or by catalytic hydrogenation (H₂/Ni).
  3. Reduction of Amides: Primary amides can be reduced directly to primary amines using a powerful reducing agent like lithium tetrahydridoaluminate (LiAlH₄) in dry ether.

From Halogenoalkane: CH₃CH₂Br + NH₃ (excess) → CH₃CH₂NH₂ + HBr

From Nitrile: CH₃CN + 4[H] → CH₃CH₂NH₂ (Reagents: LiAlH₄ or H₂/Ni)

From Amide: CH₃CONH₂ + 4[H] → CH₃CH₂NH₂ + H₂O (Reagent: LiAlH₄)

Key term

Nucleophile: A species that is rich in electrons and donates a lone pair to form a new covalent bond with an electron-deficient centre (a nucleus).

Examiner insight

Examiners frequently test synthesis pathways. Marks are awarded for knowing the specific reagents and conditions, and for correctly identifying when a carbon atom is added to the chain (the nitrile route).

Worked example 14 marks

Devise a two-step synthesis to prepare propylamine (CH₃CH₂CH₂NH₂) starting from bromoethane (CH₃CH₂Br). State the reagents and conditions for each step.

  1. 1

    Step 1: Analyse the starting material and product. Bromoethane has 2 carbons; propylamine has 3. We need to add a carbon atom. This points to the nitrile route.

  2. 2

    Step 2: Form the nitrile. React bromoethane with potassium cyanide in ethanol and heat under reflux. This is a nucleophilic substitution. CH₃CH₂Br + KCN → CH₃CH₂CN + KBr. The product is propanenitrile.

  3. 3

    Step 3: Reduce the nitrile. Reduce propanenitrile to propylamine using a suitable reducing agent. For example, use LiAlH₄ in dry ether, or H₂ gas with a nickel catalyst at high temperature and pressure. CH₃CH₂CN + 4[H] → CH₃CH₂CH₂NH₂.

Recap

  • To make a primary amine from a halogenoalkane without changing carbon chain length, use excess hot ethanolic ammonia.
  • To make a primary amine and add one carbon atom, use the nitrile route (KCN, then reduction).
  • Nitriles can be reduced to amines using LiAlH₄ or H₂/Ni.
  • Amides can be reduced to amines using LiAlH₄.
  • Using excess ammonia prevents further reaction to form secondary and tertiary amines.

Quick check

  1. Why is an excess of ammonia used when preparing ethylamine from bromoethane?1 mark
  2. What reducing agent can be used to convert propanamide to propylamine?1 mark

3. Preparing Phenylamine (Aniline)

Phenylamine (also known as aniline) is the simplest aromatic amine. It is a vital starting material for producing dyes, pharmaceuticals, and polymers. The standard laboratory and industrial preparation involves the reduction of nitrobenzene. This is achieved by heating nitrobenzene with tin (Sn) and concentrated hydrochloric acid (HCl). The tin and acid react to produce hydrogen in situ, which reduces the nitro group (-NO₂) to an amino group (-NH₂). Because the reaction is carried out in acidic conditions, the phenylamine product is protonated to form the phenylammonium ion (C₆H₅NH₃⁺). To obtain the free phenylamine, a strong base such as sodium hydroxide (NaOH) is added at the end of the reaction to deprotonate the salt.

Overall reduction: C₆H₅NO₂ + 6[H] → C₆H₅NH₂ + 2H₂O

Reagents for reduction: Sn and concentrated HCl, heat

Formation of salt: C₆H₅NH₂ + HCl → C₆H₅NH₃⁺Cl⁻

Liberation of free amine: C₆H₅NH₃⁺Cl⁻ + NaOH → C₆H₅NH₂ + NaCl + H₂O

Key term

Reduction: A chemical reaction that involves the gaining of electrons, often characterized by the removal of oxygen or the addition of hydrogen.

Examiner insight

This is a very common recall question. You must state both 'tin' and 'concentrated hydrochloric acid' for the reagent mark. Simply writing 'acid' is not sufficient.

Fun fact

The first commercially successful synthetic dye, mauveine, was discovered by William Henry Perkin in 1856 when he was trying to synthesise quinine from a derivative of aniline.

Worked example 13 marks

Write a balanced equation for the reduction of nitrobenzene to form phenylamine. State the reagents and conditions required.

  1. 1

    Step 1: Write the formula for the reactant (nitrobenzene, C₆H₅NO₂) and the product (phenylamine, C₆H₅NH₂).

  2. 2

    Step 2: Show the reduction using [H] to represent the reducing agent. The nitro group (-NO₂) becomes an amino group (-NH₂), which requires 6 hydrogen atoms. The two oxygen atoms are removed as water.

  3. 3

    Step 3: Balance the equation: C₆H₅NO₂ + 6[H] → C₆H₅NH₂ + 2H₂O.

  4. 4

    Step 4: State the reagents: Tin (Sn) and concentrated hydrochloric acid (HCl).

  5. 5

    Step 5: State the condition: Heat.

Recap

  • Phenylamine is prepared by the reduction of nitrobenzene.
  • The reagents used are tin (Sn) and concentrated hydrochloric acid (HCl).
  • The condition required is heating.
  • The initial product is the phenylammonium salt due to the acidic conditions.
  • Sodium hydroxide is added to liberate the free phenylamine base.

Quick check

  1. What is the name of the starting compound for the preparation of phenylamine?1 mark
  2. Why is sodium hydroxide added after the reduction of nitrobenzene is complete?1 mark

4. The Basicity of Amines

Amines are weak bases. Their basicity stems from the lone pair of electrons on the nitrogen atom, which can accept a proton (H⁺) from an acid, forming a dative covalent bond. The strength of an amine as a base depends on the availability of this lone pair. The more available the lone pair (i.e., the higher its electron density), the more readily it can accept a proton, and the stronger the base. We can compare the basicity of ethylamine (an aliphatic amine), ammonia, and phenylamine (an aromatic amine):

  • Ethylamine is a stronger base than ammonia. The ethyl group is electron-donating (it has a positive inductive effect). It 'pushes' electron density onto the nitrogen atom, making the lone pair more available and attractive to a proton.
  • Phenylamine is a weaker base than ammonia. The lone pair on the nitrogen atom overlaps with the delocalised pi-electron system of the benzene ring. This delocalisation spreads the lone pair's electron density into the ring, making it less available to accept a proton.

General base reaction: RNH₂ + H₂O ⇌ RNH₃⁺ + OH⁻

Reaction with acid: CH₃CH₂NH₂ + HCl → CH₃CH₂NH₃⁺Cl⁻ (ethylammonium chloride)

Order of basicity: Ethylamine > Ammonia > Phenylamine

Key term

Positive Inductive Effect: The effect where an electron-donating group, such as an alkyl group, pushes electron density along a sigma bond, increasing electron density elsewhere in the molecule.

Examiner insight

For questions on relative basicity, examiners look for clear explanations using correct terminology. You must mention the 'positive inductive effect' for alkylamines and the 'delocalisation of the lone pair into the benzene ring' for phenylamine to gain full marks.

Worked example 14 marks

Explain why ethylamine is a stronger base than ammonia, but phenylamine is a weaker base than ammonia.

  1. 1

    Step 1: State the fundamental principle. Basicity in amines depends on the availability of the lone pair of electrons on the nitrogen atom to accept a proton.

  2. 2

    Step 2: Explain ethylamine vs. ammonia. In ethylamine, the ethyl group is an electron-donating group (positive inductive effect). It pushes electron density onto the nitrogen atom, increasing the electron density of the lone pair and making it more available to accept a proton compared to ammonia.

  3. 3

    Step 3: Explain phenylamine vs. ammonia. In phenylamine, the lone pair on the nitrogen atom is delocalised into the pi-system of the benzene ring. This reduces the electron density on the nitrogen atom, making the lone pair less available to accept a proton compared to ammonia.

Recap

  • Amines act as bases because the nitrogen lone pair can accept a proton.
  • The strength of the base depends on the availability of the lone pair.
  • Alkyl groups are electron-donating and increase basicity.
  • Aryl groups are electron-withdrawing (by delocalisation) and decrease basicity.
  • The order of strength is: alkylamines > ammonia > arylamines.

Quick check

  1. Which is the stronger base, methylamine or phenylamine? Explain why in one sentence.2 marks

5. Reaction of Phenylamine with Bromine

Phenylamine reacts readily with aqueous bromine (bromine water) at room temperature. Unlike benzene, which requires a halogen carrier catalyst (like FeBr₃) for bromination, phenylamine is much more reactive. This is because the -NH₂ group is a powerful activating group. The lone pair of electrons on the nitrogen atom is delocalised into the benzene ring, significantly increasing the electron density of the ring. This makes the ring highly susceptible to attack by electrophiles like the Br⁺ in Br₂(aq). The reaction is rapid and extensive, resulting in the substitution of three hydrogen atoms on the ring at positions 2, 4, and 6. The product, 2,4,6-tribromophenylamine, is a white solid that precipitates out of the solution. This reaction is often used as a test for phenylamine.

C₆H₅NH₂(aq) + 3Br₂(aq) → C₆H₂Br₃NH₂(s) + 3HBr(aq)

Key term

Activating Group: A substituent on a benzene ring that increases the ring's reactivity towards electrophilic substitution and directs incoming electrophiles to the 2- (ortho) and 4- (para) positions.

Examiner insight

Examiners often ask you to contrast the reaction of phenylamine with bromine against the reaction of benzene with bromine. The key differences to mention are the conditions (room temp vs. catalyst) and the extent of substitution (trisubstitution vs. monosubstitution).

Fun fact

Phenol (C₆H₅OH) behaves very similarly, also having a powerful activating group (-OH) and forming a white precipitate of 2,4,6-tribromophenol with bromine water.

Worked example 13 marks

Describe the observations when aqueous bromine is added to a solution of phenylamine. Draw the structure of the organic product formed.

  1. 1

    Step 1: Recall the reaction. Phenylamine reacts with bromine water.

  2. 2

    Step 2: State the initial and final appearance. The reddish-brown colour of the aqueous bromine disappears (it is decolourised).

  3. 3

    Step 3: State the other observation. A white precipitate is formed.

  4. 4

    Step 4: Identify and draw the product. The product is 2,4,6-tribromophenylamine. Draw a benzene ring with an -NH₂ group at position 1, and Br atoms at positions 2, 4, and 6.

Recap

  • Phenylamine reacts with aqueous bromine at room temperature without a catalyst.
  • The -NH₂ group activates the benzene ring towards electrophilic attack.
  • Three bromine atoms substitute onto the ring at positions 2, 4, and 6.
  • The product is a white precipitate of 2,4,6-tribromophenylamine.
  • During the reaction, the bromine water is decolourised.

Quick check

  1. Why does phenylamine react with bromine more readily than benzene does?1 mark

6. Diazotisation and Azo Dyes

This is a two-stage process that is unique to primary aromatic amines like phenylamine and is used to create brightly coloured azo dyes.

Stage 1: Diazotisation Phenylamine is reacted with nitrous acid (HNO₂) to form a benzenediazonium salt. A critical condition is that the temperature must be kept below 10°C (ideally 0-5°C). This is because the diazonium salt is unstable and will decompose at higher temperatures. Since nitrous acid is also unstable, it is generated in situ (in the reaction mixture) by reacting sodium nitrite (NaNO₂) with a strong acid, usually cold, dilute hydrochloric acid.

Stage 2: Coupling Reaction The cold solution of the benzenediazonium salt is then immediately reacted with a coupling agent. A common coupling agent is phenol, dissolved in sodium hydroxide solution (to form the more reactive phenoxide ion). The diazonium ion (C₆H₅N₂⁺) acts as an electrophile and attacks the electron-rich phenol ring, forming an azo compound. This reaction is called a coupling reaction. The product contains an azo group (-N=N-) which links the two benzene rings. This creates a large, stable, delocalised electron system which absorbs visible light, causing the compound to be coloured. The product from coupling with phenol is an orange precipitate.

Formation of nitrous acid: NaNO₂(aq) + HCl(aq) → HNO₂(aq) + NaCl(aq)

Diazotisation: C₆H₅NH₂(aq) + HNO₂(aq) + HCl(aq) → C₆H₅N₂⁺Cl⁻(aq) + 2H₂O(l) (Temp < 10°C)

Coupling reaction: C₆H₅N₂⁺Cl⁻(aq) + C₆H₅OH(aq) → C₆H₅N=NC₆H₄OH(s) + HCl(aq) (in alkaline conditions)

Key term

Diazonium Salt: An organic compound with the general structure R-N₂⁺X⁻, where R is an aryl group and X⁻ is an anion, which is a key intermediate in the synthesis of azo dyes.

Examiner insight

Examiners expect you to know that nitrous acid is unstable and must be made in situ. Stating 'NaNO₂ and HCl' as the reagents for diazotisation is better than just writing 'HNO₂'.

Common pitfall

Forgetting the critical temperature condition (<10°C) for diazotisation. This is the most common reason for losing marks on this topic.

Worked example 15 marks

A student wants to prepare an orange azo dye starting from phenylamine. Outline the two main stages of the synthesis, giving all essential reagents and conditions.

  1. 1

    Stage 1: Diazotisation. Dissolve phenylamine in cold (below 10°C) dilute hydrochloric acid. Slowly add a cold aqueous solution of sodium nitrite (NaNO₂). The temperature must be kept below 10°C throughout the addition. This forms a solution of benzenediazonium chloride.

  2. 2

    Stage 2: Coupling. Prepare a separate solution of phenol dissolved in aqueous sodium hydroxide (this forms the sodium phenoxide ion). Keep this solution cold. Slowly add the cold diazonium salt solution from Stage 1 to the cold alkaline phenol solution. An orange precipitate (the azo dye) will form immediately.

Recap

  • Diazotisation is the reaction of phenylamine with nitrous acid below 10°C.
  • Nitrous acid is made in situ from sodium nitrite (NaNO₂) and hydrochloric acid (HCl).
  • The low temperature is essential to prevent the unstable diazonium salt from decomposing.
  • The diazonium salt then couples with an electron-rich compound like phenol in alkaline conditions.
  • The product is a coloured azo dye, containing the -N=N- azo linkage.

Quick check

  1. What is the name of the functional group responsible for the colour in azo dyes?1 mark
  2. What would happen if the diazotisation reaction was carried out at 25°C?1 mark

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

  1. Place the following compounds in order of increasing basicity (weakest first): ammonia, phenylamine, ethylamine. Explain your reasoning.4 marks
  2. Starting from nitrobenzene, describe how you would prepare a pure sample of phenylamine. Your answer should include reagents, conditions, and any necessary purification steps. Include relevant equations.5 marks
  3. Devise a three-step synthesis for butylamine starting from 1-bromopropane. For each step, state the reagents, conditions, and draw the structure of the organic product.6 marks
  4. Compare and contrast the reaction of benzene and phenylamine with aqueous bromine.4 marks
  5. Draw the displayed formula of 2-aminobutane. Is this a primary, secondary or tertiary amine? Explain your answer.3 marks
  6. Write the two equations that represent the formation of an orange dye starting from phenylamine and phenol. State all necessary reagents and conditions.5 marks
  7. A student attempts to prepare ethylamine by heating bromoethane with a limited amount of ammonia. Explain why this is a poor method for producing a high yield of ethylamine and name the other types of organic products that would be formed.3 marks
  8. Give two different methods for synthesising propylamine from a two-carbon starting material. One method should start with bromoethane and the other with ethanamide.4 marks
  9. The diazonium salt formed from phenylamine is unstable above 10°C and decomposes. Write an equation for this decomposition reaction and name the organic product.2 marks
  10. Explain, in terms of electronic structure, why the -NH₂ group directs incoming electrophiles to the 2- and 4- positions on the benzene ring in phenylamine.3 marks

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