S5 Chemistry – Unit 9: Amines and Amino Acids

9.1

Nomenclature and Classification

Amines — Definition Amines are organic compounds derived from ammonia (NH₃) by replacing one, two, or three hydrogen atoms with alkyl or aryl groups. They contain a nitrogen atom with a lone pair. Named with the suffix –amine (for aliphatic amines) or as N-substituted amines.

Classification by Degree of Substitution

Class	Structure	Definition	Example	Name
Primary (1°)	RNH₂	N bonded to 1 alkyl group	CH₃NH₂	Methylamine
Secondary (2°)	R₂NH	N bonded to 2 alkyl groups	(CH₃)₂NH	Dimethylamine
Tertiary (3°)	R₃N	N bonded to 3 alkyl groups	(CH₃)₃N	Trimethylamine
Quaternary (4°) salt	R₄N⁺X⁻	N bonded to 4 alkyl groups (cation)	(CH₃)₄N⁺I⁻	Tetramethylammonium iodide

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Important: Amine Classification vs Alcohol Classification For alcohols, the classification (1°/2°/3°) refers to the carbon bearing –OH. For amines, the classification refers to the nitrogen atom directly — how many carbon atoms are bonded to N. E.g. (CH₃)₃CNH₂ has a tertiary carbon but is a primary amine (N bonded to 1 C).

IUPAC Nomenclature

Simple amines: Name the alkyl group(s) attached to N, then add –amine. E.g. CH₃NH₂ = methylamine; (C₂H₅)₂NH = diethylamine.
IUPAC systematic: Name the longest chain attached to N as the parent alkane, replace –e with –amine, and give the position of –NH₂. E.g. CH₃CH₂CH₂NH₂ = propan-1-amine.
N-substituents: Use N– prefix for substituents on nitrogen. E.g. CH₃NHCH₂CH₃ = N-methylethanamine (or methyl ethylamine).
Aromatic amines: Phenylamine (aniline) = C₆H₅NH₂.

Name	Formula	Class	Smell
Methylamine	CH₃NH₂	1°	Fishy/ammonia
Ethylamine	C₂H₅NH₂	1°	Fishy
Propylamine	C₃H₇NH₂	1°	Fishy/putrid
Dimethylamine	(CH₃)₂NH	2°	Fishy/ammonia
Trimethylamine	(CH₃)₃N	3°	Fishy (dead fish)
Phenylamine (aniline)	C₆H₅NH₂	1° (aromatic)	Unpleasant
Putrescine (1,4-diaminobutane)	H₂N(CH₂)₄NH₂	1° diamine	Rotting flesh

Example 1

Naming and Classifying Amines

Name and classify: (a) CH₃CH₂NHCH₃ (b) (C₂H₅)₃N (c) (CH₃)₃CNH₂

N bonded to CH₃ and C₂H₅ → 2 groups on N → secondary. Name: N-methylethanamine (or methyl ethylamine).

N bonded to 3 ethyl groups → tertiary. Name: triethylamine.

N bonded to 1 group (the C(CH₃)₃ group) → primary amine (the carbon bearing N is tertiary but the amine itself is primary). Name: 2-methylpropan-2-amine (or tert-butylamine).

9.2

Physical Properties

Boiling Points and Hydrogen Bonding

Primary and secondary amines can form N–H···N hydrogen bonds between molecules, but these are weaker than O–H···O bonds in alcohols (N is less electronegative than O). Therefore amines have higher boiling points than alkanes of similar M_r, but lower than alcohols.

Tertiary amines cannot form N–H···N H-bonds (no N–H). They have even lower boiling points — similar to ethers.

Solubility in Water

Short-chain amines (C₁–C₆) are soluble in water: the N atom accepts H-bonds from water (N···H–O), and N–H can donate H-bonds to water. As chain length increases, solubility decreases. All amines are basic in water (produce OH⁻).

Amine	M_r	B.P. (°C)	Compare to alcohol
Methylamine (1°)	31	−6	Methanol B.P. = 65°C
Ethylamine (1°)	45	+17	Ethanol B.P. = 78°C
Propylamine (1°)	59	+48	Propan-1-ol B.P. = 97°C
Dimethylamine (2°)	45	+7	—
Trimethylamine (3°)	59	+3	No N–H bonds
Phenylamine (aromatic)	93	+184	High B.P. due to ring London forces

9.3

Basicity of Amines

Basicity Amines act as Brønsted-Lowry bases by accepting a proton (H⁺) using the lone pair on nitrogen: RNH₂ + H₂O ⇌ RNH₃⁺ + OH⁻. They also act as Lewis bases (lone pair donors). The strength of a base is measured by pK_b; lower pK_b = stronger base.

Factors Affecting Amine Basicity

1. Alkyl groups (electron-donating): Alkyl groups donate electrons to N, increasing the electron density on N and making it a better H⁺ acceptor. Thus: alkylamines are stronger bases than NH₃.

2. Aromatic ring (electron-withdrawing by resonance): In phenylamine (aniline), the lone pair on N is delocalised into the benzene ring (resonance), making it less available for protonation. Phenylamine is therefore a much weaker base than aliphatic amines.

3. Order of basicity: 2° alkylamine > 1° alkylamine > NH₃ > phenylamine (aromatic) > amides.

Base	pK_b	Relative Strength	Reason
Diethylamine (2°)	3.0	Strongest (aliphatic)	Two electron-donating ethyl groups
Ethylamine (1°)	3.4	Strong	One ethyl group donates electrons to N
Methylamine (1°)	3.4	Strong	Methyl group donates electrons to N
Ammonia (NH₃)	4.7	Moderate	No alkyl groups; reference
Phenylamine (C₆H₅NH₂)	9.4	Weak	Lone pair delocalised into benzene ring
Ethanamide (CH₃CONH₂)	~15	Extremely weak	Lone pair delocalised into C=O; barely basic

Reactions as Bases

With water (weak base equilibrium): RNH2 + H2O <==> RNH3+ + OH- With acids (strong base reaction, forms salt): RNH2 + HCl --> RNH3+Cl- (alkylammonium chloride salt) RNH2 + H2SO4 --> (RNH3)2SO4 e.g. CH3NH2 + HCl --> CH3NH3+Cl- (methylamine) (methylammonium chloride)

Example 2

Comparing Basicities

Arrange in order of increasing base strength: phenylamine, ethylamine, ammonia, diethylamine. Explain the trend.

Phenylamine is weakest: lone pair on N delocalised into benzene ring → less available for H⁺ → pK_b = 9.4.

Ammonia: no alkyl groups, no resonance → pK_b = 4.7.

Ethylamine: one electron-donating ethyl group increases e⁻ density on N → pK_b = 3.4.

Diethylamine: two ethyl groups → most electron-dense N → pK_b = 3.0 → strongest base.

✓

Order: phenylamine < NH₃ < ethylamine < diethylamine.

9.4

Preparation of Amines

Method 1: Reduction of Nitriles (LiAlH₄)

R-CN + 4[H] --LiAlH4, dry ether--> R-CH2-NH2 (primary amine) e.g. CH3CN + 4[H] --> CH3CH2NH2 (ethylamine)

Chain is extended by 1C from the nitrile carbon. This is an excellent route to primary amines.

Method 2: Reduction of Amides (LiAlH₄)

RCONH2 + 4[H] --LiAlH4--> RCH2NH2 (primary amine) RCONHR' + 4[H] --LiAlH4--> RCH2NHR' (secondary amine) RCONR'2 + 4[H] --LiAlH4--> RCH2NR'2 (tertiary amine)

Method 3: Nucleophilic Substitution of Halogenoalkanes with Excess NH₃

A halogenoalkane reacts with excess concentrated ammonia in ethanol in a sealed tube. The product is a mixture, but using excess NH₃ favours the primary amine.

R-X + NH3 (excess, alc.) --> R-NH2 + HX (primary amine, major) R-NH2 + R-X --> R2NH + HX (secondary, minor) R2NH + R-X --> R3N + HX (tertiary, minor) R3N + R-X --> R4N+X- (quaternary salt)

The reaction produces a mixture; excess NH₃ minimises secondary and tertiary formation.

Method 4: Reduction of Nitrobenzene (for Phenylamine)

C6H5NO2 + 6[H] --Sn/HCl, then NaOH--> C6H5NH2 + 2H2O (nitrobenzene) (phenylamine/aniline)

Tin (Sn) and concentrated HCl are used as the reducing agent. The phenylamine salt formed is treated with NaOH to liberate free phenylamine.

Method 5: Gabriel Synthesis (Pure Primary Amines)

The Gabriel synthesis gives pure primary amines without contamination by secondary or tertiary amines. Potassium phthalimide reacts with a halogenoalkane; hydrolysis liberates the primary amine.

Phthalimide-K+ + R-X --> N-alkylphthalimide + KX N-alkylphthalimide + H2O/H+ (or hydrazine) --> R-NH2 + phthalic acid

Example 3

Preparing Propylamine from 1-Bromopropane

Describe two methods to prepare propylamine (CH₃CH₂CH₂NH₂) from 1-bromopropane.

Method A (via NH₃):
CH₃CH₂CH₂Br + excess NH₃(alc.) → CH₃CH₂CH₂NH₂ + HBr
(Gives mixture of 1°, 2°, 3°; use excess NH₃ to favour 1°)

Method B (via nitrile, purer):
Step 1: CH₃CH₂CH₂Br + KCN → CH₃CH₂CH₂CN + KBr (butanenitrile)
Step 2: CH₃CH₂CH₂CN + 4[H] →^LiAlH₄ CH₃CH₂CH₂CH₂NH₂ (butan-1-amine, 4C)
Note: this gives butan-1-amine (4C), not propylamine (3C). For propylamine via nitrile: start from ethyl bromide → propanenitrile → propylamine.

9.5

Reactions of Amines

Reaction 1: As Bases (with acids)

RNH2 + HCl --> RNH3+Cl- (alkylammonium chloride) RNH2 + H2SO4 --> (RNH3)2SO4 (C6H5)NH2 + HCl --> (C6H5)NH3+Cl- (phenylammonium chloride)

Reaction 2: Acylation with Acyl Chlorides → Amide

RNH2 + R'COCl --> R'CONHR + HCl (N-substituted amide) e.g. CH3NH2 + CH3COCl --> CH3CONHCH3 + HCl (methylamine) (ethanoyl chloride) (N-methylethanamide)

This is an acylation reaction — introduces an acyl group (–COR') onto N. Used to make paracetamol (acylation of 4-aminophenol with ethanoic anhydride).

Reaction 3: Acylation with Acid Anhydrides → Amide

RNH2 + (R'CO)2O --> R'CONHR + R'COOH e.g. C6H5NH2 + (CH3CO)2O --> CH3CONHC6H5 + CH3COOH (phenylamine) (ethanoic anhydride) (N-phenylethanamide/acetanilide)

Reaction 4: Reaction with Halogenoalkanes → Higher Amines / Quaternary Salt

RNH2 + R'X --> RNHR' + HX (secondary amine) RNHR' + R'X --> RNR'2 + HX (tertiary amine) RNR'2 + R'X --> RN+(R')3 X- (quaternary ammonium salt)

Reaction 5: Diazotisation (Phenylamine only)

Phenylamine reacts with NaNO₂ + HCl at 0–5°C to form a diazonium salt. This reaction must be kept cold because diazonium salts decompose above 5°C.

C6H5NH2 + NaNO2 + 2HCl --0-5 degC--> C6H5N2+Cl- + NaCl + 2H2O (phenylamine) (benzenediazonium chloride)

Diazonium salts undergo coupling reactions with activated aromatic compounds (e.g. phenol, naphthol) to give intensely coloured azo dyes:

C6H5N2+Cl- + C6H5OH --NaOH (alkaline)--> C6H5-N=N-C6H4OH + HCl (benzenediazonium) (phenol) (yellow/orange azo dye)

Example 4

Synthesis of Paracetamol

Paracetamol is made by acylation of 4-aminophenol (H₂N–C₆H₄–OH) with ethanoic anhydride. Write the equation and name the reaction type.

4-aminophenol has –NH₂ group that reacts with ethanoic anhydride.

H₂N–C₆H₄–OH + (CH₃CO)₂O → CH₃CO–NH–C₆H₄–OH + CH₃COOH

Product: paracetamol (N-(4-hydroxyphenyl)ethanamide). Reaction type: acylation (nucleophilic acyl substitution at the –NH₂ group).

✓

Ethanoic anhydride is preferred over ethanoyl chloride: no corrosive HCl; by-product is ethanoic acid (recyclable).

9.6

Amino Acids

Amino Acids — Definition Amino acids are compounds containing both an amino group (–NH₂) and a carboxyl group (–COOH) in the same molecule. In α-amino acids (the biologically important ones), both groups are on the same carbon (α-carbon). General formula: H₂N–CHR–COOH where R is the side chain.

Amphoteric Nature of Amino Acids

Amino acids are amphoteric — they can act as both an acid and a base:

As a base (accepts H+): H2N-CHR-COOH + H+ --> +H3N-CHR-COOH (cation, acidic solution) As an acid (donates H+): H2N-CHR-COOH + OH- --> H2N-CHR-COO- + H2O (anion, alkaline solution) At the isoelectric point (zwitterion form): H2N-CHR-COOH <==> +H3N-CHR-COO- (neutral molecule) (zwitterion -- internal salt)

The zwitterion (dipolar ion) is the predominant form at the isoelectric point (pI) — when the amino acid carries no net charge. This is the form that exists as a solid.

Amino Acid	Abbreviation	R Group	Essential?	Character
Glycine	Gly (G)	–H	No	Simplest; no chiral centre
Alanine	Ala (A)	–CH₃	No	Simplest chiral amino acid
Valine	Val (V)	–CH(CH₃)₂	Yes	Branched non-polar
Serine	Ser (S)	–CH₂OH	No	Polar, –OH side chain
Cysteine	Cys (C)	–CH₂SH	No	Forms disulfide bridges
Phenylalanine	Phe (F)	–CH₂C₆H₅	Yes	Aromatic, non-polar
Lysine	Lys (K)	–(CH₂)₄NH₂	Yes	Basic side chain
Aspartic acid	Asp (D)	–CH₂COOH	No	Acidic side chain

9.7

Optical Isomerism

Optical Isomerism Optical isomers (enantiomers) are non-superimposable mirror images of each other. They arise when a molecule contains a chiral centre — a carbon atom bonded to four different groups. The two isomers rotate plane-polarised light in opposite directions: the (+) or (d) form rotates it clockwise; the (−) or (l) form rotates it anticlockwise.

Chirality in Amino Acids

All α-amino acids (except glycine) have a chiral α-carbon bonded to: –H, –NH₂, –COOH, and –R (side chain, 4 different groups). They therefore exist as two enantiomers.

Almost all natural amino acids have the L configuration (same spatial arrangement as L-glyceraldehyde) and are biologically active. The D-enantiomers are rare in nature but occur in some bacterial cell walls.

Racemic Mixture

A racemic mixture (racemate) contains equal amounts of both enantiomers. It has no net optical activity (the rotations cancel). Racemic mixtures are produced in laboratory synthesis (because attack occurs equally from both faces of the molecule) but not in biological systems (enzymes are stereospecific).

Conditions for Optical Isomerism

The molecule must have at least one chiral centre (asymmetric carbon — C bonded to 4 different groups).
The molecule must be non-superimposable on its mirror image.
Example: 2-hydroxypropanoic acid (lactic acid): CH₃CH(OH)COOH — C2 is bonded to H, OH, CH₃, and COOH → all 4 different → chiral.

Example 5

Identifying Chiral Centres

Which of these compounds has a chiral centre? (a) CH₃CH₂OH (b) CH₃CH(OH)CH₃ (c) CH₃CH(Br)COOH (d) Glycine (H₂NCH₂COOH)

CH₃CH₂OH: C1 = CH₃ (no); C2 = C bonded to OH, H, H, CH₃ — two identical H → no chiral centre.

CH₃CH(OH)CH₃: C2 bonded to OH, H, CH₃, CH₃ — two identical CH₃ → no chiral centre.

CH₃CH(Br)COOH: C2 bonded to Br, H, CH₃, COOH — all 4 different → chiral centre ✔. Shows optical isomerism.

Glycine H₂NCH₂COOH: α-C bonded to NH₂, H, H, COOH — two identical H → no chiral centre. Glycine is the only achiral amino acid.

9.8

Peptides and Proteins

Peptide Bond A peptide bond is an amide bond (–CO–NH–) formed between the –COOH of one amino acid and the –NH₂ of another, with the loss of water (condensation reaction). Dipeptide = 2 amino acids linked; polypeptide = many; protein = large polypeptide(s) with specific 3D structure.

Formation of a Dipeptide

H2N-CHR1-COOH + H2N-CHR2-COOH --> H2N-CHR1-CO-NH-CHR2-COOH + H2O ^peptide bond^ (dipeptide -- R1 at N-terminus, R2 at C-terminus)

Note that two different dipeptides are possible from two different amino acids: R1–R2 and R2–R1 (different N- and C-terminus arrangements).

Hydrolysis of Peptides

Acid hydrolysis (breaks all peptide bonds): Polypeptide + nH2O --6M HCl, 110 degC, 24h--> mixture of amino acids Enzymatic hydrolysis (specific): Trypsin, pepsin, etc. cleave specific peptide bonds selectively.

Levels of Protein Structure

Primary structure: the sequence of amino acids (determined by DNA); held by peptide bonds.
Secondary structure: local folding patterns — α-helix and β-pleated sheet; held by H-bonds between C=O and N–H of the backbone.
Tertiary structure: overall 3D shape; held by H-bonds, disulfide bridges (–S–S–), ionic interactions, van der Waals forces, and hydrophobic interactions.
Quaternary structure: arrangement of multiple polypeptide subunits (e.g. haemoglobin = 4 subunits).

Example 6

Drawing Dipeptide Structures

Write the structural formula of the dipeptide formed between glycine (H₂NCH₂COOH) and alanine (H₂NCHCH₃COOH). How many different dipeptides are possible from these two amino acids?

Gly–Ala: H₂N–CH₂–CO–NH–CH(CH₃)–COOH (glycine at N-terminus, alanine at C-terminus)

Ala–Gly: H₂N–CH(CH₃)–CO–NH–CH₂–COOH (alanine at N-terminus, glycine at C-terminus)

✓

Two different dipeptides are possible: Gly–Ala and Ala–Gly. They are different compounds with different properties.

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Exercises

Classify each amine as primary, secondary, or tertiary and give its IUPAC name:
(a) CH₃CH₂NH₂ (b) (CH₃)₂NCH₂CH₃ (c) (CH₃CH₂)₂NH (d) C₆H₅NH₂

(a) Primary; ethanamine (ethylamine)
(b) Tertiary; N,N-dimethylethanamine (dimethylethylamine); N bonded to 3 carbons
(c) Secondary; diethylamine (N-ethylethanamine); N bonded to 2 C groups
(d) Primary (aromatic); phenylamine (aniline)
Explain why phenylamine is a much weaker base than ethylamine, even though both are primary amines.

In ethylamine, the ethyl group donates electrons to N (inductive effect), increasing electron density on N and making the lone pair more available for protonation → stronger base (pK_b ≈ 3.4).

In phenylamine, the lone pair on N is delocalised by resonance into the benzene ring (overlap of N lone pair with the π system of the ring). This reduces the electron density on N and makes the lone pair much less available for protonation → much weaker base (pK_b ≈ 9.4).
Write equations for the reactions of methylamine with: (a) HCl; (b) ethanoyl chloride; (c) bromoethane (1 mole).

(a) CH₃NH₂ + HCl → CH₃NH₃⁺Cl⁻ (methylammonium chloride — salt)
(b) CH₃NH₂ + CH₃COCl → CH₃CONHCH₃ + HCl (N-methylethanamide — secondary amide)
(c) CH₃NH₂ + C₂H₅Br → CH₃NHC₂H₅ + HBr (N-methylethanamine — secondary amine)
Explain what a zwitterion is and describe the behaviour of an amino acid in acidic, neutral, and alkaline solution.

A zwitterion is a molecule that carries both a positive and a negative charge simultaneously but has no net overall charge. In amino acids, at the isoelectric point: the –COOH has lost H⁺ (→ –COO⁻) and the –NH₂ has gained H⁺ (→ –NH₃⁺).

Acidic solution: excess H⁺ protonates –COO⁻ → ⁺H₃N–CHR–COOH (cation, positive charge)
Neutral/isoelectric pH: ⁺H₃N–CHR–COO⁻ (zwitterion, no net charge)
Alkaline solution: OH⁻ removes H⁺ from –NH₃⁺ → H₂N–CHR–COO⁻ (anion, negative charge)
Which of the following has a chiral centre? For those that do, draw/describe the two enantiomers.
(a) CH₃CHBrCH₃ (b) CH₃CH(NH₂)COOH (c) CH₃COCH₃ (d) CHClBrI

(a) CH₃CHBrCH₃: C2 bonded to Br, H, CH₃, CH₃ — two identical CH₃ → no chiral centre.
(b) CH₃CH(NH₂)COOH: C2 bonded to NH₂, H, CH₃, COOH — all different → chiral centre ✔. Enantiomers: L-alanine (natural) and D-alanine (mirror image).
(c) CH₃COCH₃: carbonyl carbon bonded to two CH₃ → no chiral centre.
(d) CHClBrI: C bonded to H, Cl, Br, I — all different → chiral centre ✔. Two mirror-image enantiomers exist.
Write the structural formula of the dipeptide Ala–Gly (alanine at N-terminus, glycine at C-terminus). How many water molecules are lost when a tripeptide forms from three amino acids?

Ala–Gly: H₂N–CH(CH₃)–CO–NH–CH₂–COOH
The CO–NH is the peptide bond.

For a tripeptide from 3 amino acids: 2 water molecules are lost (one per peptide bond formed; n amino acids → n−1 peptide bonds → n−1 water molecules).

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Unit Test

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Instructions Total: 50 marks | Time: 55 minutes | Attempt all questions | Show all working.

Section A — Short Answer

30 marks

Q1 [4 marks]

Name and classify each as primary, secondary, or tertiary:
(a) CH₃CH₂CH₂NH₂ (b) (CH₃CH₂)₂NH (c) (CH₃)₃N (d) C₆H₅NHCH₃

(a) Propan-1-amine (propylamine); primary
(b) Diethylamine (N-ethylethanamine); secondary
(c) Trimethylamine; tertiary
(d) N-methylphenylamine; secondary (N bonded to phenyl and methyl)

Q2 [5 marks]

Explain with reference to electron donation/withdrawal why the base strength follows this order:
diethylamine > ethylamine > NH₃ > phenylamine > ethanamide.

Base strength depends on availability of lone pair on N for H⁺ acceptance.

Diethylamine: two ethyl groups donate electrons to N (inductive) → highest e⁻ density → strongest base.
Ethylamine: one ethyl group donates → less e⁻ donation than diethylamine.
Ammonia: no alkyl groups; bare lone pair, moderate availability.
Phenylamine: lone pair delocalised into benzene π system by resonance → much less available → weak base (pK_b 9.4).
Ethanamide: lone pair delocalised into C=O (amide resonance) → almost completely unavailable → extremely weak base (barely basic).

Q3 [6 marks]

Write equations for the following reactions:
(a) Ethylamine + HCl (b) Ethylamine + ethanoyl chloride (c) Ethylamine + bromoethane (excess) (d) Phenylamine + NaNO₂/HCl at 0°C (e) Phenylamine + ethanoic anhydride (f) The coupling of benzenediazonium chloride with phenol
Name all organic products.

(a) C₂H₅NH₂ + HCl → C₂H₅NH₃⁺Cl⁻ — ethylammonium chloride
(b) C₂H₅NH₂ + CH₃COCl → CH₃CONHC₂H₅ + HCl — N-ethylethanamide
(c) C₂H₅NH₂ + C₂H₅Br → (C₂H₅)₂NH + HBr — diethylamine (then further to triethylamine etc. with excess)
(d) C₆H₅NH₂ + NaNO₂ + 2HCl →^0–5°C C₆H₅N₂⁺Cl⁻ + NaCl + 2H₂O — benzenediazonium chloride
(e) C₆H₅NH₂ + (CH₃CO)₂O → CH₃CONHC₆H₅ + CH₃COOH — N-phenylethanamide (acetanilide)
(f) C₆H₅N₂⁺Cl⁻ + C₆H₅OH →^NaOH C₆H₅–N=N–C₆H₄OH + HCl — an orange/yellow azo dye

Q4 [5 marks]

Describe the amphoteric behaviour of amino acids. Using alanine (CH₃CH(NH₂)COOH) as your example, write equations showing its behaviour in: (a) acidic solution; (b) alkaline solution; (c) at its isoelectric point.

Amino acids are amphoteric — they have both acidic (–COOH) and basic (–NH₂) groups and can react with both acids and bases.

(a) Acidic solution (excess H⁺):
⁺H₃N–CH(CH₃)–COO⁻ + H⁺ → ⁺H₃N–CH(CH₃)–COOH (cation, net +1 charge)

(b) Alkaline solution (excess OH⁻):
⁺H₃N–CH(CH₃)–COO⁻ + OH⁻ → H₂N–CH(CH₃)–COO⁻ + H₂O (anion, net −1 charge)

(c) Isoelectric point (zwitterion):
H₂N–CH(CH₃)–COOH ⇌ ⁺H₃N–CH(CH₃)–COO⁻ (no net charge; this is the predominant solid form)

Q5 [4 marks]

Explain what is meant by optical isomerism. State the condition required for a molecule to show optical isomerism and identify which of these has a chiral centre: (a) CH₃CHClCH₃; (b) CH₃CHClCH₂CH₃; (c) glycine.

Optical isomerism: Two molecules with the same structural formula that are non-superimposable mirror images of each other. They rotate plane-polarised light in opposite directions.

Condition: At least one chiral centre — a carbon bonded to four different groups.

(a) CH₃CHClCH₃: C2 bonded to Cl, H, CH₃, CH₃ — two identical CH₃ → no chiral centre.
(b) CH₃CHClCH₂CH₃: C2 bonded to Cl, H, CH₃, CH₂CH₃ — all four different → chiral centre ✔.
(c) Glycine H₂NCH₂COOH: α-C bonded to H, H, NH₂, COOH — two identical H → no chiral centre.

Q6 [6 marks]

Write the structural formula of the dipeptide Val–Ala (valine at N-terminus, alanine at C-terminus). Using the general structure H₂N–CHR–COOH:
(a) Mark the peptide bond clearly. [2]
(b) Show what happens when this dipeptide is hydrolysed with 6M HCl. [2]
(c) How many peptide bonds are in a protein containing 200 amino acid residues? [2]

Val–Ala dipeptide:
H₂N–CH(CH(CH₃)₂)–CO–NH–CH(CH₃)–COOH
(underlined = peptide bond)

(a) The peptide bond is the –CO–NH– linkage between the two amino acid residues. It is a covalent amide bond formed by condensation (loss of H₂O) from the –COOH of valine and –NH₂ of alanine.

(b) Hydrolysis:
Val–Ala + H₂O →^{6M HCl, 110°C} H₂N–CH(iPr)–COOH + H₂N–CH(CH₃)–COOH
(valine + alanine as free amino acids)

(c) A polypeptide with n amino acids has n−1 peptide bonds. For 200 amino acids: 199 peptide bonds.

Section B — Extended Response

20 marks

Q7 [10 marks]

(a) Describe the preparation of phenylamine from benzene in two steps: nitration of benzene to give nitrobenzene, then reduction to phenylamine. Write equations and state conditions for each step. [4 marks]

(b) Phenylamine can be converted into an azo dye. Describe the two-stage process (diazotisation then coupling), giving equations, conditions, and why low temperature is essential. [4 marks]

(c) Compare phenylamine and ethylamine as bases, explaining the difference in terms of electronic structure. [2 marks]

(a)
Step 1 — Nitration: C₆H₆ + HNO₃ →^{conc. H₂SO₄, 50°C} C₆H₅NO₂ + H₂O (electrophilic substitution; NO₂⁺ electrophile)

Step 2 — Reduction: C₆H₅NO₂ + 6[H] →^{Sn/conc. HCl} C₆H₅NH₃⁺Cl⁻ →^NaOH C₆H₅NH₂ + NaCl + H₂O
(phenylammonium chloride first; NaOH liberates free phenylamine)

(b)
Stage 1 — Diazotisation (0–5°C):
C₆H₅NH₂ + NaNO₂ + 2HCl →^0–5°C C₆H₅N₂⁺Cl⁻ + NaCl + 2H₂O
Why cold? Diazonium salts (C₆H₅N₂⁺) are unstable and decompose above 5°C: C₆H₅N₂⁺ + H₂O → C₆H₅OH + N₂.

Stage 2 — Coupling (alkaline conditions):
C₆H₅N₂⁺Cl⁻ + C₆H₅OH →^NaOH C₆H₅–N=N–C₆H₄OH + HCl
Gives intensely coloured azo dye (orange/yellow) due to extended conjugated π system.

(c)
Ethylamine (pK_b 3.4): ethyl group donates electrons inductively to N → lone pair electron density increased → stronger base.
Phenylamine (pK_b 9.4): lone pair on N delocalised into benzene ring by resonance → significantly reduced availability for H⁺ acceptance → much weaker base. Difference of ~6 pK_b units means phenylamine is ~10⁶ times weaker than ethylamine.

Q8 [10 marks]

(a) Amino acids show optical isomerism. Explain what this means, why almost all natural amino acids are L-form, and what a racemic mixture is. [4 marks]

(b) Draw the structures of all possible tripeptides that can be formed from one molecule of glycine (Gly) and two molecules of alanine (Ala). How many distinct tripeptides are possible? [3 marks]

(c) Describe the four levels of protein structure, stating the type of interaction that maintains each level. [3 marks]

(a)
Amino acids (except glycine) have a chiral α-carbon bonded to 4 different groups. Two non-superimposable mirror image forms (enantiomers) exist, rotating plane-polarised light in opposite directions.
L-form in nature: Enzymes that synthesise proteins are themselves chiral (made of L-amino acids) and are stereospecific — they only incorporate L-amino acids into polypeptides. L-amino acids are recognised and used by ribosomes; D-amino acids are not (with rare exceptions in bacteria).
Racemic mixture: Equal amounts of L- and D-enantiomers with no net optical rotation. Formed in laboratory synthesis (equal probability of attack from both faces).

(b) Tripeptides from 1 Gly + 2 Ala:
Positions for G and A,A in a tripeptide (3 positions):
1. Gly–Ala–Ala: H₂N–CH₂–CO–NH–CH(CH₃)–CO–NH–CH(CH₃)–COOH
2. Ala–Gly–Ala: H₂N–CH(CH₃)–CO–NH–CH₂–CO–NH–CH(CH₃)–COOH
3. Ala–Ala–Gly: H₂N–CH(CH₃)–CO–NH–CH(CH₃)–CO–NH–CH₂–COOH
Total: 3 distinct tripeptides.

(c) Four levels of protein structure:
1° Primary: Sequence of amino acids; maintained by covalent peptide bonds.
2° Secondary: Local folding (α-helix, β-sheet); maintained by hydrogen bonds between backbone C=O and N–H.
3° Tertiary: Overall 3D shape; maintained by H-bonds, disulfide bridges, ionic interactions, and hydrophobic interactions between R groups.
4° Quaternary: Assembly of multiple polypeptide subunits; maintained by same interactions as tertiary (no new bond type; same non-covalent forces between subunits).

← Unit 8: Esters, Amides & Nitriles Unit 10: Phase Diagrams →

Nomenclature and Classification

Classification by Degree of Substitution

IUPAC Nomenclature

Naming and Classifying Amines

Physical Properties

Boiling Points and Hydrogen Bonding

Solubility in Water

Basicity of Amines

Factors Affecting Amine Basicity

Reactions as Bases

Comparing Basicities

Preparation of Amines

Method 1: Reduction of Nitriles (LiAlH4)

Method 2: Reduction of Amides (LiAlH4)

Method 3: Nucleophilic Substitution of Halogenoalkanes with Excess NH3

Method 4: Reduction of Nitrobenzene (for Phenylamine)

Method 5: Gabriel Synthesis (Pure Primary Amines)

Preparing Propylamine from 1-Bromopropane

Reactions of Amines

Reaction 1: As Bases (with acids)

Reaction 2: Acylation with Acyl Chlorides → Amide

Reaction 3: Acylation with Acid Anhydrides → Amide

Reaction 4: Reaction with Halogenoalkanes → Higher Amines / Quaternary Salt

Reaction 5: Diazotisation (Phenylamine only)

Synthesis of Paracetamol

Amino Acids

Amphoteric Nature of Amino Acids

Optical Isomerism

Chirality in Amino Acids

Racemic Mixture

Conditions for Optical Isomerism

Identifying Chiral Centres

Peptides and Proteins

Formation of a Dipeptide

Hydrolysis of Peptides

Levels of Protein Structure

Drawing Dipeptide Structures

Exercises

Interactive Quiz

Unit 9 Quiz — Amines & Amino Acids

Unit Test

Section A — Short Answer

Section B — Extended Response

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Method 1: Reduction of Nitriles (LiAlH₄)

Method 2: Reduction of Amides (LiAlH₄)

Method 3: Nucleophilic Substitution of Halogenoalkanes with Excess NH₃