Edexcel International A Level Chemistry: Organic nitrogen compounds: amines, amides, amino acids and proteins (YCH11) – Revision Notes

These notes condense Topic 19: Organic Nitrogen Compounds: Amines, Amides, Amino Acids and Proteins of the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry specification (YCH11), Issue 1, September 2017, outcomes 19.1 to 19.12 including Core Practical 15. It is examined as IA2 material, within Unit 5: Transition Metals and Organic Nitrogen Chemistry. The Topic 19 study guide has the full teaching and worked examples.

Use these notes with the course page for Edexcel IAL Chemistry, a topic-by-topic printable checklist and the Topic 19 practice questions. For a quick check of your weak areas, take a free diagnostic.

Names and formulae (19.1)

Family	Group	Example	Name
Primary amine	–NH₂ on carbon	CH₃CH₂CH₂NH₂	propylamine
Secondary amine	–NH– between two carbons	(CH₃CH₂)₂NH	diethylamine
Aromatic amine	–NH₂ on a ring	C₆H₅NH₂	phenylamine
Primary amide	–CONH₂	CH₃CH₂CH₂CONH₂	butanamide
N-substituted amide	–CONH–R	HCONHCH₃	N-methylmethanamide
2-amino acid	H₂N–CH(R)–COOH	H₂NCH(CH₃)COOH	2-aminopropanoic acid

• An “N-” locant means the group is on nitrogen, not on a carbon.
• Skeletal formulae: draw every N and O, and the H atoms on them.

Amine reactions (19.2)

Learn these for butylamine (C₄H₉NH₂) and phenylamine (C₆H₅NH₂).

Reagent	Type	Product / observation
Water	Base (proton acceptor)	RNH₃⁺ + OH⁻, alkaline solution
HCl(aq)	Acid–base	RNH₃⁺Cl⁻ salt; reversed by NaOH
Halogenoalkane	Nucleophilic substitution	Secondary amine, then tertiary amine, then quaternary ammonium salt
Ethanoyl chloride	Forms amide (room temperature, fast)	N-substituted ethanamide + RNH₃⁺Cl⁻
Cu²⁺(aq)	Base, then ligand	Butylamine: pale blue precipitate, deep blue solution in excess

General equations to adapt:

• RNH₂ + R′Br → RNHR′ + HBr
• CH₃COCl + 2RNH₂ → CH₃CONHR + RNH₃⁺Cl⁻
• [Cu(H₂O)₆]²⁺ + 4RNH₂ → [Cu(RNH₂)₄(H₂O)₂]²⁺ + 4H₂O (in excess amine)

Solubility and base strength (19.3)

• Miscible with water: N–H and the N lone pair hydrogen bond with water. A big hydrocarbon part (the ring in phenylamine) lowers solubility.
• Base strength depends on lone-pair availability.

Base	Effect on N lone pair	Strength
Phenylamine	Delocalised into the ring π system	Weakest
Ammonia	No effect	Middle
Butylamine	Alkyl group releases electron density	Strongest

Making amines (19.4, 19.5)

Target	Start from	Reagents and conditions	Note
Primary aliphatic amine	Halogenoalkane	Excess conc. NH₃ in ethanol, heat in a sealed tube	Excess NH₃ reduces multiple substitution
Primary aliphatic amine	Nitrile	LiAlH₄ in dry ether (or H₂ / Ni)	Only primary amine; RCN + 4[H] → RCH₂NH₂
Aromatic amine	Nitro-compound	Sn + conc. HCl, heat under reflux; then NaOH(aq)	ArNO₂ + 6[H] → ArNH₂ + 2H₂O

Method in steps: nitrobenzene to phenylamine

1. Reflux nitrobenzene with tin and concentrated hydrochloric acid.
2. Cool, then add sodium hydroxide solution until alkaline: this frees the amine from C₆H₅NH₃⁺.
3. Separate the phenylamine (for example by steam distillation, then solvent extraction).

Azo dyes (19.6)

1. Diazotisation: phenylamine + NaNO₂ + HCl, below 10 °C → benzenediazonium chloride, C₆H₅N₂⁺Cl⁻.
2. Coupling: add to phenol in NaOH(aq) → yellow-orange azo dye, C₆H₅–N=N–C₆H₄OH.

Too warm, and the diazonium ion breaks down to N₂ gas and phenol.

Amides (19.7)

Acyl chloride + NH₃ → primary amide + NH₄Cl. Acyl chloride + amine → N-substituted amide. Both are fast at room temperature.

Polymers (19.8 to 19.10)

Polymer	Type	Monomer(s)	Repeat unit	Lost
Nylon-6,6	Condensation	H₂N(CH₂)₆NH₂ + HOOC(CH₂)₄COOH (or the diacyl dichloride)	–[NH(CH₂)₆NHCO(CH₂)₄CO]–	H₂O (or HCl)
Protein	Condensation	2-amino acids	–[NH–CH(R)–CO]–	H₂O
Poly(propenamide)	Addition	CH₂=CHCONH₂	–[CH₂–CH(CONH₂)]–	nothing
Poly(ethenol)	Addition (made via poly(ethenyl ethanoate))	notional monomer CH₂=CHOH	–[CH₂–CH(OH)]–	nothing

Properties from hydrogen bonding

• Polyamides: N–H···O=C hydrogen bonds between chains give strength and fairly high melting temperatures (good fibres).
• Poly(ethenol): many –OH groups hydrogen bond with water, so it dissolves. Used for soluble hospital laundry bags and for liquid-detergent capsules (liquitabs).

Method in steps: drawing a condensation repeat unit

1. Write the two monomers with their reacting groups at the ends.
2. Remove –OH (or –Cl) from the acid end and –H from the amine end.
3. Join with –CONH–, add brackets with extension bonds, and add n.

Amino acids (19.11)

Condition	Form of 2-aminopropanoic acid
Solid / neutral solution	Zwitterion, H₃N⁺CH(CH₃)COO⁻
In acid	H₃N⁺CH(CH₃)COOH
In alkali	H₂NCH(CH₃)COO⁻

• Zwitterions explain the high melting temperature and solubility in water.
• Plane-polarised light: a single enantiomer rotates the plane in a polarimeter. Glycine (no chiral carbon) and racemic mixtures do not.
• Peptide bond: –COOH of one + –NH₂ of the next → –CONH– + H₂O. Two different amino acids give two dipeptides.

Core Practical 15: tests for unknowns (19.12)

Test	Result	Shows
Warm with NaOH(aq); test gas with damp red litmus	Litmus goes blue	NH₄⁺
Acidified BaCl₂(aq)	White precipitate	SO₄²⁻
Dilute acid; limewater	Milky	CO₃²⁻ or HCO₃⁻
HNO₃ + AgNO₃(aq)	White, cream, yellow	Cl⁻, Br⁻, I⁻
Shake with bromine water	Orange colour lost	C=C
PCl₅	Steamy fumes	–OH
Warm with H⁺ / Cr₂O₇²⁻	Orange turns green	Aldehyde, or primary or secondary alcohol
Fehling’s / Benedict’s	Brick-red precipitate	Aldehyde
2,4-DNPH	Yellow-orange precipitate	C=O
I₂ + NaOH(aq)	Pale yellow precipitate	CH₃CO– (or CH₃CH(OH)–)
Na₂CO₃ or NaHCO₃	Fizzing	–COOH

Worked reminders

Titre for an amine. What volume of 0.0500 mol dm⁻³ HCl neutralises 20.0 cm³ of 0.0800 mol dm⁻³ ethylamine?

1. n(amine) = 0.0800 × 20.0 ÷ 1000 = 0.00160 mol.
2. Amine : HCl = 1 : 1, so n(HCl) = 0.00160 mol.
3. Volume = 0.00160 ÷ 0.0500 = 0.0320 dm³ = 32.0 cm³. Use methyl orange (weak base, strong acid).

Dipeptide mass. Glycine (Mr 75.0) + alanine (Mr 89.0) → dipeptide + H₂O. Mr = 75.0 + 89.0 − 18.0 = 146.0. The two orders, Gly–Ala and Ala–Gly, have the same Mr but are different compounds: in one the free –NH₂ is on the glycine end, in the other on the alanine end.

Amide equation. Propanoyl chloride with methylamine: CH₃CH₂COCl + 2CH₃NH₂ → CH₃CH₂CONHCH₃ + CH₃NH₃⁺Cl⁻. The product is N-methylpropanamide.

Must-know distinctions

• Addition vs condensation polymer: addition opens C=C and loses nothing; condensation joins two different groups and loses a small molecule.
• Amine vs amide: amines are bases; the –CONH– group in amides is not usefully basic, because the C=O draws the lone pair away.
• Diazotisation vs coupling: the first needs cold acid; the second needs alkaline phenol.
• Halogenoalkane vs nitrile route: the nitrile route gives one product and an extra carbon.

Quick self-test

1. Name CH₃CH₂CH₂CONH₂.
2. Put in order of increasing base strength: ammonia, ethylamine, phenylamine.
3. Give the reagent and conditions to turn propanenitrile into propylamine.
4. What mass of ethanoyl chloride (Mr 78.5) is needed to make 0.0400 mol of an N-substituted ethanamide?
5. Calculate the percentage by mass of oxygen in the poly(ethenol) repeat unit, C₂H₄O.
6. Why must the diazotisation mixture stay below 10 °C?
7. Write the zwitterion of 2-aminopropanoic acid.
8. Explain why a glycine solution has no effect on plane-polarised light.
9. Calculate the percentage by mass of nitrogen in the poly(propenamide) repeat unit, C₃H₅NO.
10. Give the Mr of N-butylethanamide, C₆H₁₃NO.
11. State what you see when butylamine is added drop by drop, then in excess, to copper(II) sulfate solution.

Answers

1. Butanamide.
2. Phenylamine < ammonia < ethylamine.
3. LiAlH₄ in dry ether (or H₂ gas over a Ni catalyst).
4. 0.0400 × 78.5 = 3.14 g.
5. Mr = 44.0; 16.0 ÷ 44.0 × 100 = 36.4%.
6. The benzenediazonium ion decomposes when warmer, giving nitrogen and phenol.
7. H₃N⁺CH(CH₃)COO⁻.
8. Glycine has two H atoms on its central carbon, so it has no chiral centre.
9. Mr = 71.0; 14.0 ÷ 71.0 × 100 = 19.7%.
10. 115.0.
11. A pale blue precipitate, which then disappears, giving a deep blue solution, once butylamine is in excess.

Where marks are usually lost

• Saying phenylamine is basic “because of the benzene ring” without explaining delocalisation of the lone pair.
• Using one mole of amine with ethanoyl chloride when two are needed.
• Omitting “excess ammonia” or “sealed tube / ethanol” in the halogenoalkane route.
• Writing aqueous LiAlH₄; it must be in dry ether.
• Leaving out the NaOH step after Sn / HCl.
• Giving “room temperature” for diazotisation, or coupling in acid.
• Drawing the repeat unit of poly(ethenol) with a C=C, or a condensation repeat unit with –OH still on the end.
• Explaining poly(ethenol) solubility without naming hydrogen bonds with water.
• Forgetting that two different amino acids can make two dipeptides.
• Recording only the final colour in a test.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry (XCH11/YCH11) specification, Issue 1, September 2017, Pearson Education Limited – Unit 5, Topic 19: Organic Nitrogen Compounds: Amines, Amides, Amino Acids and Proteins, outcomes 19.1 to 19.12 (Core Practical 15).