Organic Nitrogen Compounds: Amines, Amides, Amino Acids and Proteins

Q: What is the order of basicity of amines?

Primary aliphatic amines ammonia primary aromatic amines. Alkyl groups are electron-donating, increasing N lone pair availability. Aromatic groups delocalise the N lone pair into the benzene ring, reducing availability.

Edexcel International A-Level Chemistry· Unit 5 (WCH15) Spec 19.1–19.12· 45 min read

1. Structure, Nomenclature and Basicity of Amines★★☆☆☆⏱ 10 min

📘 Definition

Primary Amine

An organic compound where a nitrogen atom is covalently bonded to one alkyl/aryl group and two hydrogen atoms, with the general formula $R N H_{2}$

Amines are classified as primary, secondary, or tertiary based on the number of alkyl/aryl groups bonded to the nitrogen atom. IUPAC nomenclature uses the suffix -amine for the parent chain, with substituents numbered relative to the amine group. Aromatic amines have nitrogen directly bonded to a benzene ring, e.g. phenylamine ( $C_{6} H_{5} N H_{2}$ ).

Amine basicity depends on the availability of the lone pair of electrons on the nitrogen atom to accept a proton from an acid: primary aliphatic amines have electron-donating alkyl groups that increase lone pair availability (stronger base than ammonia), while primary aromatic amines have the N lone pair delocalised into the benzene ring (weaker base than ammonia).

📐 Worked Example

Place the following compounds in order of increasing basicity: ammonia, butylamine, phenylamine. Explain your answer.

1
First, classify each compound: butylamine = primary aliphatic amine, phenylamine = primary aromatic amine.
2
Apply the basicity trend: aromatic amines < ammonia < aliphatic amines.
3
Final order: phenylamine < ammonia < butylamine. Explanation: Butylamine's alkyl group increases N lone pair availability, phenylamine's delocalised N lone pair reduces availability.

Exam tip:

Always link basicity directly to nitrogen lone pair availability, do not just state the trend without reasoning to gain full marks.

2. Reactions and Preparation of Amines★★★☆☆⏱ 12 min

Reaction with water: Forms weakly alkaline solutions, e.g. $C H_{3} C H_{2} C H_{2} C H_{2} N H_{2} + H_{2} O ⇌ C H_{3} C H_{2} C H_{2} C H_{2} N H_{3}^{+} + O H^{-}$
Reaction with acids: Forms ionic ammonium salts, e.g. butylamine + HCl → butylammonium chloride
Reaction with halogenoalkanes: Forms secondary/tertiary amines via nucleophilic substitution
Reaction with ethanoyl chloride: Forms an amide via elimination of HCl
Reaction with $C u^{2 +}$ ions: Forms deep blue complex ions, similar to ammonia

Key preparation methods for amines: Aliphatic primary amines are made by heating a halogenoalkane with excess ammonia in ethanol, or reducing nitriles with $L i A l H_{4}$ in dry ether. Aromatic primary amines are made by reducing nitro-compounds with tin and concentrated HCl, followed by neutralisation with NaOH to liberate the free amine.

📐 Worked Example

State the reagents and conditions required to prepare phenylamine from nitrobenzene.

1
First reagent: Tin (Sn) metal and concentrated hydrochloric acid, heated under reflux.
2
Second reagent: Excess sodium hydroxide solution, to neutralise the intermediate phenylammonium salt and release free phenylamine.

Exam tip:

Do not forget the neutralisation step for aromatic amine preparation, it is a common marking point. Full reaction mechanisms are not required for any amine reactions.

3. Azo Dyes, Amides and Polymer Synthesis★★★☆☆⏱ 10 min

📘 Definition

Diazotisation

The reaction of a primary aromatic amine with nitrous acid (formed in situ from NaNO₂ and dilute HCl) at temperatures below 5°C to form a stable benzenediazonium ion salt.

Benzenediazonium ions undergo coupling reactions with phenols in alkaline conditions to form brightly coloured azo dyes, used in textiles and food colouring. Amides are formed when acyl chlorides react with amines, eliminating HCl as a byproduct.

Addition polymers: Form from alkene monomers, no byproducts, e.g. poly(propenamide), poly(ethenol)
Condensation polymers: Form from difunctional monomers, eliminate small molecules like $H_{2} O$ or HCl, e.g. nylon, proteins

📐 Worked Example

Draw the repeat unit of nylon-6,6, formed from 1,6-diaminohexane and hexanedioic acid.

1
Each amide bond forms by eliminating one $H_{2} O$ molecule between an amine and carboxylic acid group.
2
$-NH-(CH_2)_6-NH-CO-(CH_2)_4-CO-$
3
Add trailing bonds at each end of the repeat unit, and write the letter n outside brackets to indicate multiple repeat units.

Exam tip:

Poly(ethenol) is water soluble due to hydroxyl groups that form hydrogen bonds with water molecules, making it ideal for soluble hospital laundry bags.

4. Amino Acids, Zwitterions and Proteins★★☆☆☆⏱ 8 min

📘 Definition

Zwitterion

A dipolar ion with both a positive ammonium group ( $- N H_{3}^{+}$ ) and negative carboxylate group ( $- C O O^{-}$ ), with overall neutral charge, formed by amino acids at their isoelectric point.

Amino acids have the general formula $H_{2} N - C H (R) - C O O H$ , where R is a variable side chain. All naturally occurring amino acids (except glycine) have a chiral central carbon atom, so they rotate the plane of plane-polarised light. Amino acids are amphoteric: the carboxyl group can donate a proton, and the amine group can accept a proton.

Peptide bonds ( $- C O N H -$ ) form via condensation reactions between two amino acids, eliminating one molecule of water. Long chains of amino acids linked by peptide bonds are called polypeptides, or proteins when they have a functional biological structure.

📐 Worked Example

Draw the zwitterion structure of alanine, where R = $C H_{3}$ .

1
The amine group accepts a proton to become $- N H_{3}^{+}$ .
2
The carboxylic acid group donates a proton to become $- C O O^{-}$ .
3
$H_3N^+-CH(CH_3)-COO^-$
4
Confirm the overall charge of the ion is zero.

Exam tip:

When drawing zwitterions, always show full charges on the nitrogen and oxygen atoms, do not omit them or you will lose marks.

5. Practical Identification of Organic Nitrogen Compounds★★★★☆⏱ 5 min

Amines: Turn damp red litmus paper blue (alkaline), form deep blue complex with $C u^{2 +}$ ions
Amides: Hydrolyse with warm NaOH to release ammonia gas, which turns damp red litmus blue
Amino acids: High melting points, soluble in water due to zwitterion formation

✓ Quick check

Which test would you use to distinguish between butylamine and butanamide at room temperature?
- Add dilute HCl
- Test with damp red litmus paper
- Heat with NaOH and test gas output
Reveal answer
1 —
Butylamine is alkaline so will turn damp red litmus blue immediately, butanamide is neutral so no change at room temperature.

Exam tip:

Use IR data provided in the exam to identify amide bonds (strong ~1650 cm⁻¹ absorption) and amine groups (broad ~3300-3500 cm⁻¹ absorption).

6. Common Pitfalls

Wrong move:

Stating aromatic amines are stronger bases than ammonia

Why:

Forgetting delocalisation of the N lone pair into the benzene ring reduces proton acceptance ability

Correct move:

Link basicity to N lone pair availability, placing aromatic amines below ammonia in basicity order

Wrong move:

Drawing zwitterions with uncharged $- N H_{2}$ and $- C O O H$ groups

Why:

Zwitterions require a protonated amine and deprotonated carboxyl group for neutral dipolar structure

Correct move:

Always draw zwitterions with $- N H_{3}^{+}$ and $- C O O^{-}$ groups, showing full charges

Wrong move:

Including $H_{2} O$ or HCl in condensation polymer repeat units

Why:

Small molecules are eliminated during condensation, so they are not part of the polymer chain

Correct move:

Only include atoms remaining in the polymer chain in repeat units, with trailing bonds and n notation

Wrong move:

Forgetting to keep diazotisation reactions below 5°C

Why:

Benzenediazonium ions decompose above 5°C to form phenol and nitrogen gas, so no azo dye forms

Correct move:

Always state diazotisation is carried out at 0-5°C using an ice bath

Wrong move:

Confusing addition and condensation polymer formation

Why:

Addition polymers form from alkenes with no byproducts, condensation forms from difunctional monomers with small molecule elimination

Correct move:

Check for double bonds in monomers to identify addition polymers, check for two distinct functional groups for condensation polymers

7. Quick Reference Cheatsheet

Compound Class	Key Properties/Reactions	Key Reagents/Conditions
Primary Aliphatic Amine	Stronger base than ammonia, forms salts/amides/complexes	Excess ammonia/ethanol, $L i A l H_{4}$ for nitrile reduction
Primary Aromatic Amine	Weaker base than ammonia, forms diazonium ions	Sn/conc HCl, heat, NaOH neutralisation
Amide	Forms via acylation, hydrolyses to amine + carboxylic acid	Acyl chloride + amine, warm NaOH for hydrolysis
Amino Acid	Amphoteric, forms zwitterions, peptide bonds	Condensation polymerisation to form proteins
Polymer	Addition (poly(propenamide)), condensation (nylon/proteins)	Alkene monomers (addition), diamine + diacid (condensation)
Azo Dye	Formed via diazotisation + phenol coupling	0-5°C, alkaline coupling conditions

8. Frequently Asked

What is the order of basicity of amines?

Primary aliphatic amines > ammonia > primary aromatic amines. Alkyl groups are electron-donating, increasing N lone pair availability. Aromatic groups delocalise the N lone pair into the benzene ring, reducing availability.

How do I distinguish between addition and condensation polymers?

Addition polymers form from alkene monomers, retain all monomer atoms in the repeat unit. Condensation polymers form from difunctional monomers, eliminate a small molecule (e.g. $H_{2} O$ , HCl) during polymerisation.

What is a zwitterion?

A zwitterion is a dipolar ion with both positive and negative functional groups, and overall neutral charge. Amino acids form zwitterions at their isoelectric point, with $- N H_{3}^{+}$ and $- C O O^{-}$ groups.

Going deeper

What's Next

Now that you have mastered organic nitrogen compounds for Edexcel IAL Chemistry Unit 5, you are ready to revise full Unit 5 content including transition metals and practical assessment skills to prepare for your WCH15 exam. This topic is frequently tested in both multiple choice and structured long answer questions, so practice recalling reaction conditions and drawing structures accurately. Work through past paper questions on this topic to familiarise yourself with exam phrasing and marking scheme expectations, especially for questions asking you to explain basicity trends or draw polymer repeat units. Connect this content to earlier organic chemistry topics like functional group identification to prepare for practical unknown identification questions.