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

This guide teaches 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. Learning outcomes 19.1 to 19.12 are covered, including Core Practical 15. The topic belongs to International A2 (IA2), within Unit 5: Transition Metals and Organic Nitrogen Chemistry; its practical skills also feed into Unit 6: Practical Skills in Chemistry II.

Course links: the course page for Edexcel IAL Chemistry and a topic-by-topic printable checklist. Addition polymer basics are in the alkenes guide. A free diagnostic shows weak spots quickly.

Topic 19 at a glance

Outcome	You should be able to
19.1	Name and draw amines, amides and amino acids
19.2	Reactions of butylamine and phenylamine (five reagents)
19.3	Miscibility with water; basicity order
19.4	Primary aliphatic amines from halogenoalkanes and nitriles
19.5	Aromatic amines from nitro-compounds (Sn / conc. HCl)
19.6	Diazotisation and coupling with phenol
19.7	Amides from acyl chlorides
19.8	Nylon and proteins; poly(propenamide) and poly(ethenol)
19.9	Repeat units of these polymers
19.10	Properties linked to hydrogen bonding
19.11	Amino acid experiments
19.12	Core Practical 15: analysis of unknowns

19.1 Naming and drawing

A primary amine has one carbon on the nitrogen (R–NH₂), a secondary amine two, a tertiary amine three. The specification uses common names: butylamine is CH₃CH₂CH₂CH₂NH₂ and phenylamine is C₆H₅NH₂. A secondary amine such as CH₃CH₂NHCH₃ is named N-methylethylamine; the “N-” shows the methyl group sits on nitrogen.

Amides contain –CONH–. A primary amide ends –CONH₂: CH₃CONH₂ is ethanamide. When a carbon group replaces one hydrogen on nitrogen, the name gains an N- prefix: CH₃CH₂CONHCH₃ is N-methylpropanamide.

Amino acids have an –NH₂ and a –COOH group. In the 2-amino acids that build proteins both groups sit on the same carbon: H₂NCH(R)COOH. With R = CH₂CH₃ the systematic name is 2-aminobutanoic acid.

Worked example 1: three formulae for one amide

Draw N-ethylethanamide three ways.

1. Structural: CH₃CONHCH₂CH₃.
2. Displayed: every bond shown.

    H   O       H   H
    |   ‖       |   |
H — C — C — N — C — C — H
    |       |   |   |
    H       H   H   H

3. Skeletal: a zigzag line with N in the chain, =O on the second carbon and the H on nitrogen drawn in.

19.2 and 19.3 Reactions and basicity of amines

Why amines are bases

The nitrogen lone pair accepts a proton; the more available it is, the stronger the base.

• Butylamine is stronger than ammonia. The alkyl group pushes electron density onto nitrogen (positive inductive effect).
• Phenylamine is weaker than ammonia. The lone pair overlaps with the ring’s delocalised π system, so it is less available.

Order of base strength: phenylamine < ammonia < butylamine.

Why short amines mix with water

N–H bonds and the nitrogen lone pair hydrogen bond with water, so small amines such as butylamine are miscible. Phenylamine dissolves only slightly: the large ring cannot hydrogen bond.

The five reactions in 19.2

Reagent	Butylamine	Phenylamine
Water	Equilibrium: C₄H₉NH₂ + H₂O ⇌ C₄H₉NH₃⁺ + OH⁻; alkaline solution	Same equilibrium; only weakly alkaline
Acid (HCl)	C₄H₉NH₂ + HCl → C₄H₉NH₃⁺Cl⁻ (butylammonium chloride)	C₆H₅NH₂ + HCl → C₆H₅NH₃⁺Cl⁻; the oily amine dissolves, and reappears with alkali
Halogenoalkane, e.g. bromoethane	Nucleophilic substitution: C₄H₉NH₂ + C₂H₅Br → C₄H₉NHC₂H₅ + HBr; further substitution gives tertiary amines and quaternary ammonium salts	Same type of substitution
Ethanoyl chloride	Fast at room temperature: CH₃COCl + 2C₄H₉NH₂ → CH₃CONHC₄H₉ + C₄H₉NH₃⁺Cl⁻ (N-butylethanamide)	CH₃COCl + 2C₆H₅NH₂ → CH₃CONHC₆H₅ + C₆H₅NH₃⁺Cl⁻ (N-phenylethanamide)
Copper(II) ions	Pale blue precipitate of Cu(OH)₂, then in excess a deep blue solution, [Cu(C₄H₉NH₂)₄(H₂O)₂]²⁺	Also acts as a ligand through its lone pair

The copper reaction mirrors aqueous ammonia (17.22): deprotonation first, then ligand exchange in excess. In the ethanoyl chloride equation the second mole of amine removes the HCl formed; without excess amine you see steamy fumes.

Worked example 2: finding the concentration of butylamine

A 25.0 cm³ portion of aqueous butylamine needs 18.40 cm³ of 0.150 mol dm⁻³ hydrochloric acid, using methyl orange. Find the concentration in mol dm⁻³ and g dm⁻³ (Mr of butylamine = 73.0).

1. n(HCl) = 0.150 × 18.40 ÷ 1000 = 0.00276 mol.
2. Ratio 1 : 1, so n(C₄H₉NH₂) = 0.00276 mol.
3. Concentration = 0.00276 ÷ 0.0250 = 0.110 mol dm⁻³.
4. Mass concentration = 0.1104 × 73.0 = 8.06 g dm⁻³.

Methyl orange suits a weak base–strong acid titration (end point below pH 7).

19.4 and 19.5 Making amines

Primary aliphatic amines (19.4)

From halogenoalkanes. Heat the halogenoalkane with excess concentrated ammonia in ethanol, in a sealed tube (under pressure):

CH₃(CH₂)₃Cl + 2NH₃ → CH₃(CH₂)₃NH₂ + NH₄Cl

Excess ammonia cuts down further substitution to secondary and tertiary amines.

By reducing nitriles. Use lithium tetrahydridoaluminate(III) in dry ether, or hydrogen gas over nickel:

CH₃CH₂CH₂CN + 4[H] → CH₃CH₂CH₂CH₂NH₂

This gives only the primary amine, and (via a nitrile made from a halogenoalkane and cyanide) adds one carbon.

Aromatic amines (19.5)

Reflux the nitro-compound with concentrated hydrochloric acid and tin. The product is the salt of the amine, so add sodium hydroxide solution to release the free amine:

C₆H₅NO₂ + 6[H] → 2H₂O + C₆H₅NH₂

Worked example 3: yield of a substituted phenylamine

7.05 g of 1-fluoro-4-nitrobenzene (C₆H₄FNO₂, Mr 141.0) is reduced. After purification, 3.94 g of 4-fluorophenylamine (C₆H₆FN, Mr 111.0) is collected. Find the percentage yield.

1. n(nitro-compound) = 7.05 ÷ 141.0 = 0.0500 mol.
2. One mole of nitro-compound gives one mole of amine: theoretical mass = 0.0500 × 111.0 = 5.55 g.
3. Yield = 3.94 ÷ 5.55 × 100 = 71.0%.

19.6 Azo dyes

Step 1: diazotisation. Nitrous acid, HNO₂, is unstable, so make it in the flask from sodium nitrite and hydrochloric acid, below 10 °C (ice bath):

C₆H₅NH₂ + HCl + HNO₂ → C₆H₅N₂⁺Cl⁻ + 2H₂O

The product is benzenediazonium chloride. Above about 10 °C it decomposes, giving off nitrogen gas and forming phenol.

Step 2: coupling. Add the cold diazonium solution to phenol dissolved in sodium hydroxide solution. A yellow-orange solid forms: an azo dye.

C₆H₅N₂⁺ + C₆H₅OH → C₆H₅N=NC₆H₄OH + H⁺

The –N=N– azo group links two benzene rings; delocalisation across the molecule makes it coloured.

19.7 Amides from acyl chlorides

Acyl chlorides react quickly at room temperature with ammonia to give primary amides (C₆H₅COCl + 2NH₃ → C₆H₅CONH₂ + NH₄Cl gives benzamide) and with amines to give N-substituted amides (19.2 table).

19.8 and 19.9 Polymers containing nitrogen or hydrogen-bonding groups

Condensation polymers (19.8 i)

In condensation polymerisation, monomers join and a small molecule (water or hydrogen chloride) is lost at each link.

Nylon-6,10 forms from hexane-1,6-diamine, H₂N(CH₂)₆NH₂, and decanedioyl dichloride, ClOC(CH₂)₈COCl. Each amide link releases HCl. Repeat unit:

–[NH(CH₂)₆NHCO(CH₂)₈CO]–

With the dicarboxylic acid instead, water is lost. The numbers give the carbons in the diamine, then in the acid part.

Proteins are condensation polymers of 2-amino acids. The –COOH of one amino acid and the –NH₂ of the next form a peptide link (an amide link, –CONH–), releasing water.

Addition polymers (19.8 ii)

These form by opening a C=C; nothing is lost.

• Poly(propenamide) from propenamide, CH₂=CHCONH₂. Repeat unit: –[CH₂–CH(CONH₂)]–.
• Poly(ethenol): repeat unit –[CH₂–CH(OH)]–. The notional monomer, ethenol, is unstable, so the polymer is made from poly(ethenyl ethanoate) by replacing its ethanoate groups with –OH.

Show extension bonds through the brackets and leave no C=C.

19.10 Properties and hydrogen bonding

Polyamides. N–H on one chain hydrogen bonds to C=O on the next. With London forces along the chains, this gives strong fibres with fairly high melting temperatures.

Poly(ethenol). Every second carbon carries an –OH group. These form hydrogen bonds with water, so the polymer dissolves. Uses:

• soluble laundry bags: hospital linen goes into the washing machine still in its bag, so staff do not handle it;
• liquid-detergent capsules (liquitabs): the film dissolves in the wash and releases the detergent.

Leaving some ethanoate groups in the chain changes the hydrogen bonding, which lets manufacturers adjust how readily the film dissolves.

19.11 Experiments with amino acids

Acidity, basicity and zwitterions (19.11 i)

In the solid and in neutral solution, an amino acid exists mainly as a zwitterion: the –COOH has given its proton to the –NH₂ in the same molecule.

H₃N⁺–CH(R)–COO⁻

• Add acid: the –COO⁻ is protonated, giving H₃N⁺CH(R)COOH.
• Add alkali: the –NH₃⁺ loses a proton, giving H₂NCH(R)COO⁻.

Experiments: show that the amino acid neutralises both dilute hydrochloric acid and sodium hydroxide; note its high melting temperature (an ionic lattice of zwitterions).

Worked example 4: identifying an amino acid

0.721 g of an amino acid H₂NCH(R)COOH reacts with exactly 28.00 cm³ of 0.250 mol dm⁻³ sodium hydroxide (1 : 1). Identify R.

1. n(NaOH) = 0.250 × 28.00 ÷ 1000 = 0.00700 mol.
2. Mr = 0.721 ÷ 0.00700 = 103.0.
3. H₂NCH(R)COOH minus R is C₂H₄NO₂ (74.0), so R = 29.0, which is C₂H₅.
4. The acid is 2-aminobutanoic acid.

Plane-polarised light (19.11 ii)

Pass plane-polarised monochromatic light through the solution in a polarimeter. Every amino acid found in proteins except glycine has a chiral carbon. One enantiomer on its own turns the plane of polarisation. A glycine solution, or a racemic mixture, causes no overall rotation.

Peptide bond formation (19.11 iii)

Two amino acids can join in two orders, so glycine and 2-aminobutanoic acid give two dipeptides, each with Mr = 75.0 + 103.0 − 18.0 = 160.0. Heating with hydrochloric acid breaks peptide links again.

19.12 Core Practical 15: analysis of unknowns

The specification gives only the title, “Analysis of some inorganic and organic unknowns”. Unit 6 may ask you to recall and interpret tests for ions, gases and organic functional groups from Units 1 and 2:

Looking for	Test	Positive result
CO₃²⁻ / HCO₃⁻	Dilute acid; limewater	Gas turns limewater milky
SO₄²⁻	Acidified barium chloride	White precipitate
NH₄⁺	Warm with NaOH(aq)	Ammonia given off: damp red litmus goes blue
Cl⁻, Br⁻, I⁻	HNO₃ then AgNO₃(aq); then ammonia	White / cream / yellow precipitates
Group 1 and 2 cations	Flame test	Characteristic colour
Transition metal ions	NaOH(aq), NH₃(aq), then excess	Coloured precipitates; some dissolve
C=C	Shake with bromine water	Orange colour disappears
–OH (dry sample)	PCl₅	Steamy HCl fumes
Aldehyde	Fehling’s or Benedict’s; Tollens’	Brick-red precipitate; silver mirror
C=O	2,4-DNPH	Yellow-orange precipitate
CH₃CO– group	Iodine in alkali	Pale yellow precipitate of CHI₃
–COOH	Sodium carbonate or hydrogencarbonate	Fizzing; CO₂
Amine	Smell; litmus; Cu²⁺(aq)	Fishy; alkaline; deep blue in excess

Record each observation in full (start and end colours), then deduce.

Common errors

• Ranking phenylamine above ammonia as a base.
• Forgetting the sodium hydroxide step after tin and hydrochloric acid.
• Letting the diazonium solution warm above 10 °C, or coupling with phenol in acid.
• Leaving a C=C in an addition repeat unit, or omitting the water or HCl lost in condensation.
• Saying glycine rotates plane-polarised light.

Where next

Condense this guide using the revision notes, test yourself on the practice questions, and see how to prepare for each Edexcel IAL Chemistry unit.

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).