Which amine class has three alkyl/aryl groups on nitrogen?
A — Tertiary amine: N bonded to three carbon groups (R₃N). Primary = one R (RNH₂); secondary = two R (R₂NH); quaternary = four R groups, permanent positive charge (R₄N⁺). McMurry & Ballantine 8e Ch 24 §24.1
Question 2
Correct basicity order of amines?
B — Aliphatic > NH₃ > aromatic amines. Alkyl groups donate electrons by induction, making the lone pair on N more available (more basic) than in NH₃. In aromatic amines, the lone pair is delocalised into the π ring → far less available → least basic. McMurry 8e §24.3; pKₐ(conjugate acid): aliphatic ~10–11; NH₄⁺ 9.26; anilinium 4.6
Question 3
Why is aniline (C₆H₅–NH₂) less basic than methylamine?
C — Resonance delocalisation. In aniline, the –NH₂ nitrogen lone pair overlaps with the aromatic π system (resonance donation). This reduces electron density on N, making it a weaker base. Methylamine has no such resonance; the lone pair is fully available. McMurry 8e §24.3
Question 4
pKₐ(conjugate acid) of aniline is ~4.6; of methylamine is ~10.7. This means:
D — Higher pKₐ(conjugate acid) = stronger base. Methylamine (pKₐ 10.7) is a stronger base than aniline (pKₐ 4.6). A higher pKₐ means the conjugate acid is a weaker acid, i.e. it holds the proton more tightly → the amine accepted the proton more willingly → stronger base. McMurry 8e §24.3
Question 5
Diazotisation requires:
A — Aromatic primary amine + NaNO₂/HCl at 0–5°C. NaNO₂ + HCl generates nitrous acid (HNO₂) in situ, which reacts with Ar–NH₂ to form the diazonium salt ArN₂⁺Cl⁻. Temperature is kept near 0°C to prevent decomposition of the unstable diazonium salt. McMurry 8e §24.8
Question 6
A diazonium salt (ArN₂⁺) is:
B — Unstable; used immediately. Diazonium salts decompose above ~5°C (losing N₂ gas). They are highly reactive intermediates: the N₂⁺ group is an excellent leaving group. In synthetic practice the diazonium is generated and immediately converted to the target aryl compound without isolation. McMurry 8e §24.8
Question 7
Sandmeyer reaction: ArN₂⁺ + CuCl →?
C — ArN₂⁺ + CuCl → ArCl + N₂. The Sandmeyer reaction uses Cu(I) halide or Cu(I) cyanide as the source of the replacing group. CuCl gives ArCl; CuBr gives ArBr; CuCN gives ArCN. Cu(I) acts as a single-electron transfer (radical) mediator. McMurry 8e §24.8
Question 8
To convert ArN₂⁺ → ArBr, the Sandmeyer reagent is:
B — CuBr (or HBr/Cu) replaces –N₂⁺ with –Br to give the aryl bromide. D (KI) is used for aryl iodides via direct iodination without a copper catalyst (Balz-Schiemann variant is different — KI works directly for ArI). McMurry 8e §24.8
Question 9
To convert ArN₂⁺ → ArCN:
A — CuCN in the Sandmeyer reaction converts the diazonium to an aryl nitrile (ArCN). This is synthetically valuable: ArCN can be hydrolysed to ArCOOH or reduced to ArCH₂NH₂, extending the carbon chain. McMurry 8e §24.8
Question 10
Direct replacement ArN₂⁺ → ArI (no copper catalyst) uses:
B — KI directly. Aryl iodides are uniquely made from diazonium salts by simple addition of KI (no Cu catalyst needed) because I⁻ is a good nucleophile and ArI is relatively stable. ArN₂⁺ + KI → ArI + N₂ + KCl. McMurry 8e §24.8
Question 11
Hydrolysis of diazonium salt (ArN₂⁺ + H₂O):
C — ArOH (phenol). Warming the diazonium salt in aqueous acid causes the N₂⁺ to leave as N₂ and water acts as a nucleophile, giving a phenol. This is how phenols are made from anilines in synthesis. McMurry 8e §24.8
Question 12
Reduction of diazonium salt with H₃PO₂:
D — ArH. Hypophosphorous acid (H₃PO₂) reduces the diazonium salt, replacing –N₂⁺ with –H. This deamination reaction is useful for removing an amino group after it has served as a directing group in aromatic substitution. McMurry 8e §24.8
Question 13
Azo coupling (ArN₂⁺ + phenol/ArNH₂) produces:
A — Azo compound (Ar–N=N–Ar'). The diazonium cation acts as an electrophile in EAS on the activated ring (phenol or arylamino compound). The –N=N– chromophore absorbs visible light, producing highly coloured azo dyes. Used in textile dyes and diagnostic reagents (e.g., azo dye tests for bile). McMurry 8e §24.9
Question 14
Sulfonamides inhibit bacterial growth by:
B — Competitive inhibition of dihydropteroate synthase. Sulfonamides are structural analogues of PABA (p-aminobenzoic acid). They compete with PABA for the enzyme dihydropteroate synthase, blocking the conversion of PABA to dihydrofolic acid. Without folate, bacteria cannot synthesise nucleotides. McMurry 8e §24.10; Rang & Dale Pharmacology Ch 50
Question 15
The PABA that sulfonamides mimic is used by bacteria to synthesise:
C — Folate (dihydrofolic acid). Bacteria must synthesise folate de novo from PABA because they cannot absorb preformed folate. Human cells get folate from the diet (dietary folate); they lack the PABA-utilising enzyme. This is the basis of selectivity — sulfonamides harm bacteria but not host cells. McMurry 8e §24.10; Rang & Dale Ch 50
Question 16
Acylation of a primary amine (Schotten-Baumann) stops at:
B — Amide. Once the amide (RCONH–R') is formed, the nitrogen lone pair is heavily delocalised into the carbonyl C=O; the amide nitrogen is a very poor nucleophile (pKₐ ~25 for conjugate acid). Further acylation does not occur under normal conditions. This selectivity is exploited in protecting-group chemistry. McMurry 8e §21.7, §24.6
Question 17
The product of alkylating a primary amine can over-alkylate to form:
B — A mixture of all degrees. NH₃ + RX → RNH₂ → R₂NH → R₃N → R₄N⁺X⁻. Each product is still nucleophilic and can react again. This is why direct alkylation is synthetically unselective; reductive amination or Gabriel synthesis are preferred for clean 1° or 2° amine preparation. McMurry 8e §24.5
Question 18
Quaternary ammonium salts differ from tertiary amines in that they:
B — Permanent positive charge. All four bonds on nitrogen are to carbon; there is no lone pair and the positive charge cannot be neutralised by deprotonation. Quaternary ammonium compounds (e.g. benzalkonium chloride) are quaternary regardless of pH. They act as antiseptics (cationic surfactants disrupting bacterial membranes). McMurry 8e §24.1
Question 19
The local anaesthetic procaine is an ester of PABA with a diethylaminoethanol. At physiological pH 7.4, the tertiary amine group is largely:
C — Protonated (ammonium form). Aliphatic tertiary amines have pKₐ(conjugate acid) ~9–10. At pH 7.4, which is well below pKₐ, the Henderson-Hasselbalch equation indicates the majority (>99%) of the amine is in the protonated form. This ionic form improves water solubility for injection formulation, while the small free-base fraction penetrates neuronal membranes. McMurry 8e §24.3; Katzung Pharmacology §Local Anaesthetics
Question 20
What is the IUPAC name for C₆H₅–NH₂?
D — Both “phenylamine” and “aniline” are acceptable IUPAC names. IUPAC 2013 recommendations retain “aniline” as a preferred retained name. “Phenylamine” is the systematic substitutive name. Benzylamine (B) is C₆H₅CH₂NH₂ — note the extra –CH₂– spacer. McMurry 8e §24.1; IUPAC 2013 Blue Book
D-1: Primary amine▼
A compound in which nitrogen is bonded to one carbon group and two hydrogen atoms (R–NH₂). Primary amines are the most nucleophilic and among the most basic of the amine classes. They are prepared by: (1) alkylation of NH₃ (though gives mixtures), (2) reduction of a nitro compound (ArNO₂ + [H] → ArNH₂), (3) reduction of an amide (RCONH₂ + LiAlH₄ → RCH₂NH₂), or (4) Gabriel synthesis (clean 1° product). McMurry 8e §24.1, §24.5
D-2: Diazotisation▼
The conversion of an aromatic primary amine to a diazonium salt (ArN₂⁺Cl⁻) using NaNO₂ + HCl at 0–5°C. HCl + NaNO₂ → HNO₂ (nitrous acid) in situ; HNO₂ nitrosates –NH₂ through the nitrosonium ion (NO⁺) to give –N₂⁺. Temperature is maintained near 0°C because diazonium salts are thermally unstable and decompose (releasing N₂) above ~5°C. Aliphatic 1° amines also form diazonium ions but these are too unstable to use synthetically — they decompose immediately to alcohols + N₂. McMurry 8e §24.8
D-3: Sandmeyer reaction▼
Replacement of the diazonium group (N₂⁺) of an arenediazonium salt with a halide or cyanide using a Cu(I) salt as catalyst. The Cu(I) mediates single-electron transfer (radical mechanism): ArN₂⁺ + CuX → Ar• + N₂ + Cu(II)X → ArX + Cu(I). Products: CuCl → ArCl; CuBr → ArBr; CuCN → ArCN. Exception: ArI requires only KI (no Cu needed). McMurry 8e §24.8
D-4: Azo coupling▼
An electrophilic aromatic substitution in which a diazonium cation (ArN₂⁺) acts as a weak electrophile and attacks an activated arene (phenol at pH 8–9, or arylamino compound) at the para position. Product: an azo compound (Ar–N=N–Ar'). The extended conjugation of the –N=N– chromophore absorbs visible light, producing intensely coloured compounds. Commercial use: azo dyes (>50% of all synthetic dyes). Clinical/diagnostic use: diazo tests for bilirubin, Sudan III for fat staining. McMurry 8e §24.9
D-5: Sulfonamide▼
A class of antibacterial compounds derived from sulfanilamide (4-aminobenzenesulfonamide), which is a structural analogue of PABA (p-aminobenzoic acid). Mechanism: competitive inhibition of dihydropteroate synthase → blocks bacterial de novo folate synthesis → bacteria cannot synthesise purines/thymidine → growth arrest. Selective for bacteria because human cells use dietary folate (lack dihydropteroate synthase). First discovered by Gerhard Domagk (Prontosil, 1935). Still used: sulfamethoxazole (co-trimoxazole with trimethoprim). McMurry 8e §24.10
D-6: Acylation — Schotten-Baumann▼
Reaction of a primary or secondary amine with an acid chloride or acid anhydride in the presence of aqueous base (NaOH or NaHCO₃, the Schotten-Baumann conditions) to give an amide. R–NH₂ + R'COCl → R'CONH–R + HCl. The base neutralises HCl as it forms. Reaction stops at the amide because nitrogen in an amide is a very poor nucleophile (lone pair delocalised into C=O). This selectivity contrasts with alkylation, which over-alkylates. McMurry 8e §21.7
E1. Explain why aliphatic amines are stronger bases than aromatic amines, using methylamine vs aniline. Include pKₐ values and resonance argument. (6 marks)Essay
Key pKₐ values (conjugate acid):
Methylamine (CH₃NH₂): pKₐ ~10.7 → stronger base
Aniline (C₆H₅NH₂): pKₐ ~4.6 → much weaker base
Why aliphatic > aromatic:
In methylamine, the methyl group donates electrons by induction, increasing electron density on N → lone pair more available for donation to H⁺.
In aniline, the nitrogen lone pair participates in resonance with the aromatic π system. Four resonance structures show electron density shifted from N onto the ring → lone pair is partially delocalised and less available to bind H⁺.
Evidence: aniline is a better nucleophile at carbon than nitrogen (N lone pair occupied) relative to methylamine.
Consequence: anilinium ion (C₆H₅NH₃⁺, pKₐ 4.6) is a much stronger acid than methylammonium (pKₐ 10.7), confirming aniline holds a proton weakly → weak base.
McMurry 8e §24.3
E2. Describe diazotisation of aniline. State conditions, why temperature must be 0–5°C, and list four Sandmeyer products with reagents. (8 marks)Essay
Reagents & conditions: NaNO₂ + HCl (excess), 0–5°C. NaNO₂ + HCl → HNO₂ (nitrous acid, unstable) → NO⁺ (nitrosonium ion, the actual electrophile). NO⁺ attacks the –NH₂ nitrogen of aniline → N-nitrosoaniline → tautomerises and loses H₂O → ArN₂⁺Cl⁻ (benzenediazonium chloride).
Why 0–5°C? The diazonium salt is thermally unstable; above ~5°C it loses N₂ spontaneously (“decomposition”): ArN₂⁺ + H₂O → ArOH + N₂ + H⁺. At 0–5°C decomposition is slow enough to permit use of ArN₂⁺ as a synthetic intermediate.
E3. Explain the mechanism of sulfonamide antibacterial action. Why are sulfonamides selective for bacteria and not human cells? (6 marks)Essay
Target enzyme: Dihydropteroate synthase (DHPS) in bacteria.
Normal pathway: Bacteria synthesise folate de novo: PABA + pteridine → dihydropteroate → dihydrofolic acid → tetrahydrofolic acid (THF). THF is the cofactor for thymidylate synthesis and purine synthesis (nucleotides for DNA replication).
Mechanism of inhibition: Sulfonamides (e.g. sulfamethoxazole) are structural analogues of PABA. They enter the DHPS active site (competitive inhibition) and block PABA binding → dihydropteroate cannot form → no folate → bacteria cannot make thymidine or purines → DNA synthesis arrested → bacteriostasis.
Selectivity:
Bacteria must synthesise folate de novo from PABA because they lack folate transporters (cannot import preformed folate from the environment).
Human (eukaryotic) cells lack dihydropteroate synthase entirely — they obtain folate from the diet via folate transporters. No DHPS → no sulfonamide target → no toxicity to host.
McMurry 8e §24.10; Rang & Dale Pharmacology Ch 50
E4. Compare alkylation vs acylation of primary amines. Explain why acylation is the preferred route to a secondary amide in synthesis. (6 marks)Essay
Alkylation (R–NH₂ + R'X):
Mechanism: S₂ — amine acts as nucleophile attacking alkyl halide.
Problem: over-alkylation. Each successive product (2°, 3° amine) is still nucleophilic. Result: a mixture of 1°, 2°, 3° amines and quaternary ammonium salt.
Impractical for clean synthesis of a specific degree of amine without special methods (e.g., Leuckart, Gabriel, reductive amination).
Stops at the amide: the amide nitrogen lone pair is delocalised into the carbonyl (resonance: –N–C=O ↔ –N=C–O⁻), making it essentially non-nucleophilic. No further acylation.
Advantage: clean, single product; acid chloride or anhydride is highly electrophilic → fast and high-yielding.
In protecting-group strategy: acylation temporarily converts a 1° amine to an amide (less reactive), a step used repeatedly in peptide synthesis.
McMurry 8e §21.7, §24.5, §24.6
E5. Azo dyes are made by coupling a diazonium salt with an activated aromatic ring. Explain the chemical steps from aniline to an azo dye. State one clinical use of azo chemistry. (7 marks)Essay
Step 2 — Azo coupling (electrophilic aromatic substitution): PhN₂⁺ is a weak electrophile. It attacks an electron-rich (activated) arene at the para position. Two main coupling partners:
N,N-dimethylaniline: PhN₂⁺ + C₆H₅N(CH₃)₂ → methyl orange (yellow-orange, used as pH indicator)
Chromophore: The –N=N– azo group conjugated with the two arene rings creates an extended π system. This absorbs visible light (~430–530 nm), yielding yellow/orange/red colours. Electron-withdrawing or -donating substituents shift the absorption wavelength (colour tuning).
One clinical use:Van den Bergh (diazo) reaction for serum bilirubin. Bilirubin (a tetrapyrrole) is cleaved by diazonium reagent (diazotised sulfanilic acid) into two azodipyrroles that absorb at 540 nm. Distinguishes direct (conjugated) from indirect (unconjugated) bilirubin clinically. Also: Sudan III/IV (azo dyes) used histologically to stain lipids/fat droplets in frozen sections.
McMurry 8e §24.8–24.9; Tietz Textbook of Clinical Chemistry §Bilirubin