Cambridge AS & A Level9701

Primary and secondary amines

Chemistry 9701 Chapter Notes

What this chapter covers

Primary and secondary aminesPhenylamine and azo compoundsAmidesAmino acids
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In this chapter

  1. 1Classifying Amines: Primary, Secondary, and Tertiary
  2. 2The Basicity of Amines
  3. 3Comparing Basicity: Ethylamine, Ammonia, and Phenylamine
  4. 4Preparation of Aliphatic Amines
  5. 5Preparation of Phenylamine

1. Classifying Amines: Primary, Secondary, and Tertiary

Amines are organic compounds derived from ammonia (NH₃) where one or more hydrogen atoms have been replaced by an alkyl or aryl group. They are classified based on how many of these groups are attached to the nitrogen atom. A primary (1°) amine has one alkyl/aryl group (e.g., ethylamine, C₂H₅NH₂). A secondary (2°) amine has two groups (e.g., dimethylamine, (CH₃)₂NH). A tertiary (3°) amine has three groups (e.g., trimethylamine, (CH₃)₃N). This classification is crucial for understanding their reactions and properties.

Primary Amine: R-NH₂

Secondary Amine: R₂NH (or R-NH-R')

Tertiary Amine: R₃N (or R-N(R')R'')

Key term

Primary Amine: An amine in which the nitrogen atom is bonded to one alkyl or aryl group and two hydrogen atoms.

Examiner insight

Examiners expect you to be able to identify the class of an amine from its structural, displayed, or skeletal formula.

Common pitfall

Confusing the classification of amines with alcohols. Amine class is determined by bonds to the nitrogen atom, while alcohol class is determined by bonds to the carbon atom attached to the -OH group.

Worked example 13 marks

Classify the following amines as primary, secondary, or tertiary:(a) CH₃CH(NH₂)CH₃(b) (CH₃CH₂)₂NH(c) N(CH₃)₃.

  1. 1

    Step 1: Examine the nitrogen atom in compound (a), CH₃CH(NH₂)CH₃. The nitrogen atom is bonded to one carbon chain (the propyl group) and two hydrogen atoms. Therefore, it is a primary amine.

  2. 2

    Step 2: Examine the nitrogen atom in compound (b), (CH₃CH₂)₂NH. The nitrogen atom is bonded to two separate alkyl groups (two ethyl groups) and one hydrogen atom. Therefore, it is a secondary amine.

  3. 3

    Step 3: Examine the nitrogen atom in compound (c), N(CH₃)₃. The nitrogen atom is bonded to three separate alkyl groups (three methyl groups) and no hydrogen atoms. Therefore, it is a tertiary amine.

Recap

  • Amines are organic derivatives of ammonia, NH₃.
  • Primary (1°) amines have the general formula RNH₂.
  • Secondary (2°) amines have the general formula R₂NH.
  • Tertiary (3°) amines have the general formula R₃N.
  • The classification depends on the number of alkyl/aryl groups attached directly to the nitrogen atom.

Quick check

  1. What class of amine is phenylamine (C₆H₅NH₂)?1 mark
  2. Draw the displayed formula for diethylamine.1 mark

2. The Basicity of Amines

Like ammonia, amines act as weak bases. This is due to the lone pair of electrons on the nitrogen atom. According to the Brønsted-Lowry theory, a base is a proton (H⁺) acceptor. The nitrogen's lone pair is available to form a dative covalent bond with a proton from an acid. This reaction forms an ammonium salt. For example, ethylamine reacts with hydrochloric acid to form ethylammonium chloride.

General reaction with H⁺: RNH₂ + H⁺ → RNH₃⁺

Reaction with acid: CH₃CH₂NH₂ + HCl → CH₃CH₂NH₃⁺Cl⁻

Key term

Brønsted-Lowry Base: A substance that acts as a proton (H⁺ ion) acceptor.

Fun fact

The characteristic smell of fish is caused by amines like trimethylamine. Squeezing lemon juice (citric acid) over fish neutralizes these amines by converting them into non-volatile salts, which eliminates the odour.

Worked example 13 marks

Write a balanced equation for the reaction between methylamine (CH₃NH₂) and sulfuric acid (H₂SO₄). Name the product.

  1. 1

    Step 1: Identify the reactants as a base (methylamine) and a strong acid (sulfuric acid). The amine will accept a proton.

  2. 2

    Step 2: Recognize that sulfuric acid is diprotic (can donate two H⁺ ions). Therefore, two molecules of methylamine are needed to react with one molecule of sulfuric acid.

  3. 3

    Step 3: Write the protonation of two methylamine molecules: 2CH₃NH₂ → 2CH₃NH₃⁺.

  4. 4

    Step 4: Write the full balanced equation: 2CH₃NH₂ + H₂SO₄ → (CH₃NH₃)₂SO₄.

  5. 5

    Step 5: Name the salt product. The cation is methylammonium (from methylamine) and the anion is sulfate. The product is methylammonium sulfate.

Recap

  • Amines are weak bases due to the lone pair of electrons on the nitrogen atom.
  • The nitrogen lone pair can accept a proton (H⁺), forming a dative covalent bond.
  • Amines react with acids in a neutralization reaction to form ammonium salts.
  • This basic character is a key chemical property of all amines.

Quick check

  1. What specific structural feature is responsible for the basicity of amines?1 mark

3. Comparing Basicity: Ethylamine, Ammonia, and Phenylamine

Not all amines are equally basic. Their strength depends on the availability of the nitrogen's lone pair of electrons to accept a proton. We can compare ethylamine (an aliphatic amine), ammonia, and phenylamine (an aromatic amine). The order of basicity is: Ethylamine > Ammonia > Phenylamine.

  1. Ethylamine is a stronger base than ammonia because the ethyl group is electron-donating (has a positive inductive effect). It 'pushes' electron density towards the nitrogen atom, making the lone pair more available and better at attracting a proton.
  2. Phenylamine is a weaker base than ammonia because the lone pair on the nitrogen atom is delocalised into the pi-electron system of the benzene ring. This spreads the electron density away from the nitrogen, making the lone pair less available to accept a proton.

Key term

Inductive Effect: The effect where electron density in a sigma bond is shifted towards a more electronegative atom, or pushed away by an electron-donating group.

Examiner insight

A full explanation of relative basicity must compare the availability of the lone pair on the nitrogen atom and correctly link this to the electronic effects of the attached groups (inductive effect for alkyl groups, delocalisation for aryl groups).

Fun fact

Many powerful medicines, including morphine and codeine, are alkaloids, which are naturally occurring amines. Their basicity is key to how they are extracted from plants and how they function in the body.

Worked example 14 marks

Explain why phenylamine is a much weaker base than ethylamine.

  1. 1

    Step 1: State that basicity depends on the availability of the lone pair of electrons on the nitrogen atom.

  2. 2

    Step 2: Explain the effect of the ethyl group in ethylamine. The ethyl group is an electron-donating alkyl group. It has a positive inductive effect, which increases the electron density on the nitrogen atom. This makes the lone pair more available to accept a proton.

  3. 3

    Step 3: Explain the effect of the benzene ring in phenylamine. The lone pair on the nitrogen atom is delocalised into the pi system of the benzene ring. This decreases the electron density on the nitrogen atom.

  4. 4

    Step 4: Conclude by linking electron availability to basicity. Because the lone pair in phenylamine is less available than in ethylamine, phenylamine is a weaker base.

Recap

  • The strength of a base depends on the availability of the nitrogen lone pair.
  • Alkyl groups are electron-donating and increase basicity.
  • Aromatic rings are electron-withdrawing (by delocalisation) and decrease basicity.
  • The order of strength is: aliphatic amine > ammonia > aromatic amine.
  • Ethylamine is a stronger base than ammonia.
  • Phenylamine is a weaker base than ammonia.

Quick check

  1. Arrange the following in order of increasing basicity: ammonia, phenylamine, diethylamine.2 marks

4. Preparation of Aliphatic Amines

Aliphatic amines can be synthesised in the lab via two main routes.

  1. From Halogenoalkanes: A primary amine can be formed by heating a halogenoalkane with an excess of ammonia dissolved in ethanol in a sealed tube. This is a nucleophilic substitution reaction where ammonia acts as the nucleophile. Using excess ammonia is critical to minimise the formation of secondary and tertiary amines, as the primary amine product is also a nucleophile.
  2. From Nitriles: This is a two-step process that increases the carbon chain length by one. First, a halogenoalkane is reacted with potassium cyanide (KCN) in ethanol to form a nitrile. Second, the nitrile is reduced to a primary amine using a strong reducing agent like LiAlH₄ in dry ether, or by catalytic hydrogenation with H₂ over a nickel catalyst.

From Halogenoalkane: CH₃CH₂Br + NH₃(excess, ethanolic) → CH₃CH₂NH₂ + NH₄Br

Nitrile formation: CH₃Br + KCN → CH₃CN + KBr

Nitrile reduction: CH₃CN + 4[H] → CH₃CH₂NH₂ (Reagents: LiAlH₄ in dry ether, or H₂/Ni)

Key term

Nucleophile: A species that is rich in electrons and donates a lone pair to form a new covalent bond.

Common pitfall

Forgetting that the nitrile synthesis route (from a halogenoalkane) adds a carbon atom to the molecule's backbone.

Worked example 14 marks

Outline a two-step synthesis to prepare propylamine (CH₃CH₂CH₂NH₂) starting from bromoethane (CH₃CH₂Br).

  1. 1

    Step 1: Identify that the carbon chain needs to be extended by one carbon (from 2 carbons in bromoethane to 3 in propylamine). This indicates the nitrile route is required.

  2. 2

    Step 2: Write the first step: formation of the nitrile. React bromoethane with potassium cyanide in ethanol and heat under reflux. Equation: CH₃CH₂Br + KCN → CH₃CH₂CN + KBr. The product is propanenitrile.

  3. 3

    Step 3: Write the second step: reduction of the nitrile. Reduce propanenitrile to propylamine using a suitable reducing agent and conditions. Equation: CH₃CH₂CN + 4[H] → CH₃CH₂CH₂NH₂. State the reagents and conditions, for example, LiAlH₄ in dry ether, followed by hydrolysis.

  4. 4

    Step 4 (Alternative reduction): Alternatively, state catalytic hydrogenation as the reduction method. Equation: CH₃CH₂CN + 2H₂ → CH₃CH₂CH₂NH₂. State the reagents and conditions: H₂ gas with a nickel catalyst at high temperature and pressure.

Recap

  • Heat a halogenoalkane with excess, hot, ethanolic ammonia to form a primary amine.
  • Excess ammonia is used to prevent the formation of secondary and tertiary amines.
  • Amines can be made by reducing nitriles with H₂/Ni or LiAlH₄.
  • The nitrile route increases the length of the carbon chain by one.

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 propanenitrile to propylamine?1 mark

5. Preparation of Phenylamine

Phenylamine (also known as aniline) is the simplest aromatic amine. It is prepared by the reduction of nitrobenzene. This is a very important reaction in industrial chemistry, particularly in the manufacture of dyes. The reduction is typically carried out by heating nitrobenzene with tin (Sn) and concentrated hydrochloric acid (HCl). The acid provides the protons for the reaction, and the tin is the reducing agent. Because the reaction is carried out in acidic conditions, the initial product is the phenylammonium ion (C₆H₅NH₃⁺). A strong alkali, such as sodium hydroxide (NaOH), must be added at the end to deprotonate the salt and liberate the free phenylamine, which is an oily liquid.

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

Reaction with Acid: C₆H₅NO₂ + Sn + 6HCl → C₆H₅NH₃⁺Cl⁻ + SnCl₄ + 2H₂O (simplified)

Liberation of Amine: C₆H₅NH₃⁺ + OH⁻ → C₆H₅NH₂ + H₂O

Key term

Reduction: A reaction involving the gain of electrons, or in organic chemistry, often an increase in the hydrogen-to-carbon ratio or decrease in the oxygen-to-carbon ratio.

Examiner insight

For full marks on describing this preparation, candidates must name both tin and concentrated HCl as the reagents and mention the need for a final step with alkali to get the free amine.

Worked example 14 marks

Describe how you would prepare a sample of phenylamine starting from nitrobenzene. Your answer should include reagents, conditions, and the equation for the reduction.

  1. 1

    Step 1: State the reagents used for the reduction: Tin (Sn) and concentrated hydrochloric acid (HCl).

  2. 2

    Step 2: State the condition: Heat the mixture, for example, by heating under reflux.

  3. 3

    Step 3: Write a balanced equation for the reduction of nitrobenzene. The simplest form is using [H] to represent the reducing agent: C₆H₅NO₂ + 6[H] → C₆H₅NH₂ + 2H₂O.

  4. 4

    Step 4: Explain the final step to obtain the free amine. The reaction mixture is acidic, so the phenylammonium ion is formed. Add a strong alkali, such as aqueous sodium hydroxide, to the mixture to deprotonate the salt and release the phenylamine oil.

Recap

  • Phenylamine is prepared by the reduction of nitrobenzene.
  • The reagents are tin (Sn) and concentrated hydrochloric acid (HCl).
  • The reaction mixture is heated.
  • The initial product is the phenylammonium salt due to the acidic conditions.
  • A strong base like NaOH is added at the end to liberate the free phenylamine.

Quick check

  1. Name the two reagents required to reduce nitrobenzene.1 mark
  2. What is the role of sodium hydroxide in the preparation of phenylamine?1 mark

End-of-chapter exercise

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

  1. Draw the skeletal formula for (a) propan-1-amine and (b) N-methylethanamine. Classify each as primary, secondary or tertiary.4 marks
  2. Explain, with reference to the electronic structure of the molecules, why ethylamine is a stronger base than ammonia.3 marks
  3. Phenylamine is a very weak base. Explain why it is significantly less basic than ammonia.3 marks
  4. Write a balanced chemical equation for the reaction of propylamine (CH₃CH₂CH₂NH₂) with dilute nitric acid (HNO₃). Name the organic product.2 marks
  5. Describe the reagents and conditions needed to convert nitrobenzene into phenylamine.3 marks
  6. The compound C₄H₁₁N has several structural isomers that are amines. Draw the displayed formula for one primary, one secondary, and one tertiary isomer.3 marks
  7. Outline a two-step synthesis of ethylamine from bromoethane that results in a good yield of the primary amine and avoids significant formation of secondary amines. Give reagents and conditions for both steps.5 marks
  8. When bromoethane is heated with a limited amount of ethanolic ammonia, a mixture of products is formed. Give the structure of the secondary amine formed in this reaction and explain why it is formed.3 marks
  9. Predict, with reasoning, whether trimethylamine, (CH₃)₃N, is a stronger or weaker base than dimethylamine, (CH₃)₂NH, in the gaseous phase.3 marks
  10. A student plans to synthesise butan-1-amine. Two possible starting materials are 1-chloropropane and 1-chlorobutane. Explain which starting material should be used for a two-step synthesis via a nitrile, and outline the reaction pathway.4 marks

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