Ch 1 · Primary and secondary amines · AS & A Level Chemistry 9701

Topic 1 of 4 in Nitrogen compounds — AS & A Level Chemistry 9701 — 57 lessons.

Introduction

AS chemistry introduced primary amines from halogenoalkanes. A2 adds two further preparative routes, reduction of nitriles and reduction of amides, and asks for a more quantitative comparison of base strength between primary amines, secondary amines and ammonia, building on the AS idea that an alkyl group's electron-donating inductive effect makes amines basic.

Key ideas — Preparing amines by reduction of nitriles

A nitrile, RCN, is reduced by lithium aluminium hydride in dry ether (or by hydrogen gas with a nickel catalyst) to a primary amine, extending the carbon chain by one carbon and retaining the nitrogen atom: CH_3CN + 4[H] arrow CH_3CH_2NH_2

🧑‍🏫 Teacher's explanation

Think of a nitrile group (-C≡ N) as a carbon atom that is 'hungry' for electrons because it is triple-bonded to nitrogen. To turn this into an amine, we need to 'stuff' extra hydrogen atoms into that triple bond to saturate it. It is like taking a tightly wound spring and slowly letting it unwind by adding small pieces of metal to it.

📖 Definitions

🔍 Worked example — Easy

Write the balanced equation for the reduction of ethanenitrile, ceCH3CN, to ethanamine, ceCH3CH2NH2, using [H] to represent hydrogen atoms.

🔍 Worked example — Standard

A student reduces propanenitrile, ceCH3CH2CN, using LiAlH_4 in dry ether. Calculate the molar mass of the resulting primary amine.

🔍 Worked example — Exam-style

How many moles of propan-1-amine are produced from 0.50 moles of propanenitrile if the reduction is 100\% efficient?

⚠️ Common errors

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🎯 Exam question

Key ideas — Preparing amines by reduction of amides

An unsubstituted amide, RCONH_2, is reduced by lithium aluminium hydride to a primary amine with the same number of carbon atoms, since the reduction converts the carbonyl carbon into a CH_2 group: CH_3CONH_2 + 4[H] arrow CH_3CH_2NH_2 + H_2O Reducing an N-substituted amide, RCONHR', in the same way gives a secondary amine, RCH_2NHR', since the substituent already present on nitrogen remains attached.

🧑‍🏫 Teacher's explanation

Think of an amide as a molecule with a 'functional group' that acts like a handle. This handle is the carbonyl group (C=O). When we use a powerful reducing agent like lithium aluminium hydride (LiAlH_4), we are essentially performing a chemical 'surgery' where we strip away the oxygen atoms and replace them with hydrogen atoms.

📖 Definitions

🔍 Worked example — Easy

Calculate the number of carbon atoms in the product formed when propanamide, CH_3 extCH_2 extCONH_2, is reduced by LiAlH_4.

🔍 Worked example — Standard

Write the balanced equation for the reduction of ethanamide, CH_3 extCONH_2, to ethanamine, CH_3 extCH_2 extNH_2, using [H] to represent the reducing equivalents.

🔍 Worked example — Exam-style

A secondary amide, CH_3 extCONHCH_3 (N-methylacetamide), is reduced using LiAlH_4. Determine the IUPAC name of the resulting amine.

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🎯 Exam question

(a) Write the structural formula for X. [1] (b) Compound X is reduced using LiAlH_4 to form compound Y. Write the structural formula for Y. [2] (c) Explain why compound Y is a base. [1] [4 marks]

Key ideas — Forming amides by condensation with an acyl chloride

Ammonia or a primary/secondary amine reacts with an acyl chloride in a condensation reaction, releasing hydrogen chloride, which is neutralised by a second molecule of the nitrogen nucleophile: CH_3COCl + 2NH_3 arrow CH_3CONH_2 + NH_4Cl CH_3COCl + 2CH_3NH_2 arrow CH_3CONHCH_3 + CH_3NH_3Cl

🧑‍🏫 Teacher's explanation

Think of an acyl chloride as a highly reactive 'activated' version of a carboxylic acid. In a carboxylic acid, the -OH group is a poor leaving group, making the carbon less eager to react. In an acyl chloride, the -Cl group is an excellent leaving group, making the carbonyl carbon extremely electrophilic (hungry for electrons).\n\nWhen an amine approaches, the nitrogen's lone pair attacks the carbon. This is a condensation reaction because two molecules combine to form one larger molecule, with a small molecule (HCl) being expelled. However, because HCl is a strong acid and amines are bases, the HCl immediately reacts with another molecule of the amine to form a salt. This neutralization is what actually drives the reaction forward to completion.\n\nAnalogy: Imagine a high-speed collision between a magnet and a piece of metal. The metal is so attracted to the magnet that it snaps on instantly. The acyl chloride is the magnet, and the amine is the metal. The HCl is like the sparks flying off during the impact that get caught by a nearby sponge (the second amine molecule).

📖 Definitions

🔍 Worked example — Easy

Write the balanced equation for the reaction between ethanoyl chloride and ammonia.

🔍 Worked example — Standard

Calculate the mass of propanoyl chloride (M_r = 92.5) required to react completely with 0.10 moles of methylamine (CH_3NH_2) to produce a secondary amide, assuming 2 moles of amine are used per mole of acyl chloride.

🔍 Worked example — Exam-style

A student reacts 0.50 moles of acetyl chloride (CH_3COCl) with an excess of ethylamine (CH_3CH_2NH_2). Calculate the total moles of salt produced in the reaction.

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🎯 Exam question

Key ideas — Basicity of amines in aqueous solution

The nitrogen atom of an amine carries a lone pair of electrons and, just like ammonia, accepts a proton from water to form a substituted ammonium ion and hydroxide ion, giving an alkaline solution: CH_3CH_2NH_2 + H_2O leftharpoons CH_3CH_2NH_3^+ + OH^- The strength of this equilibrium, and hence the base strength of the amine, depends on how readily the lone pair on nitrogen is able to accept a proton.

🧑‍🏫 Teacher's explanation

Imagine a nitrogen atom is like a person holding a single, precious ball (the lone pair of electrons). To be a 'base', this person must be willing to hand that ball over to a passing water molecule (a proton). If the nitrogen is surrounded by groups that push more electron density toward it, it becomes 'greedier' and more willing to share, making it a stronger base.

📖 Definitions

🔍 Worked example — Easy

Write the balanced equation for the reaction of ammonia with water to show why it is basic.

🔍 Worked example — Standard

Predict whether ethylamine (CH_3CH_2NH_2) is a stronger or weaker base than ammonia (NH_3) based on the inductive effect of the ethyl group.

🔍 Worked example — Exam-style

Explain why phenylamine (C_6H_5NH_2) is a weaker base than ammonia (NH_3).

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🎯 Exam question

Key ideas — Comparing the basicity of primary and secondary amines with ammonia

Alkyl groups are electron-donating by induction, so they increase the electron density on the nitrogen atom relative to ammonia, making the lone pair more available to bind a proton. A primary amine, such as ethylamine, is therefore a stronger base than ammonia. A secondary amine has two electron-donating alkyl groups on nitrogen, so it is generally a stronger base still than the corresponding primary amine, since electron density on nitrogen is increased further. For simple aliphatic amines, the overall trend in base strength is therefore secondary amine > primary amine > ammonia.

🧑‍🏫 Teacher's explanation

Imagine the nitrogen atom in an amine is like a person trying to hand out a gift (the lone pair of electrons) to a stranger (a proton, H^+). Ammonia is like a person with nothing but their own hands; they can give, but they don't have much extra 'stuff' to offer. However, an amine has 'alkyl groups' attached to the nitrogen. Think of these alkyl groups as extra pockets full of cash (electrons) that they keep pushing toward the person's hands.

Ch 1 · Primary and secondary amines · AS & A Level Chemistry 9701

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