Chemistry • Year 12 • Module 7 • Lesson 16
Amines & Amides: Structure, Properties & Reactions
Apply amine/amide knowledge to real data, graph interpretation, a comparison table, and a cause-and-effect Australian context scenario.
1. Interpret a boiling-point data table — C₃ functional groups
The table below lists measured boiling points and dominant intermolecular forces for six C₃ compounds. Use it to answer the questions that follow. 9 marks
| Compound | Formula | Class | BP (°C) | Key IMF | N–H donors? |
|---|---|---|---|---|---|
| Propane | CH₃CH₂CH₃ | Alkane | −42 | Dispersion only | No |
| Trimethylamine | (CH₃)₃N | Tertiary amine | 3 | Dipole–dipole; dispersion | No |
| Propan-1-amine | CH₃CH₂CH₂NH₂ | Primary amine | 48 | N–H H-bonding; dipole; dispersion | Yes (2) |
| Propan-1-ol | CH₃CH₂CH₂OH | Alcohol | 97 | O–H H-bonding; dipole; dispersion | No |
| Propanoic acid | CH₃CH₂COOH | Carboxylic acid | 141 | O–H H-bonding (dimer); dipole; dispersion | No |
| Propanamide | CH₃CH₂CONH₂ | Amide | 213 | N–H donor + C=O acceptor cooperative network | Yes (2) |
Data sourced from NIST WebBook standard reference.
1.1 Explain, using electronegativity values, why propan-1-amine (BP 48°C) has a lower boiling point than propan-1-ol (BP 97°C) despite both forming hydrogen bonds. 3 marks
1.2 Trimethylamine (BP 3°C) has a higher boiling point than propane (BP −42°C) even though trimethylamine also cannot donate H-bonds. Explain this difference in BP using IMF reasoning. 2 marks
1.3 Propanamide has a higher BP than propanoic acid despite having a slightly lower molecular mass (73 vs 74 g mol⁻¹). Use the data in the table to explain this counterintuitive result. 2 marks
1.4 Based on the table, predict the approximate boiling point range and key IMF for N-methylpropan-1-amine (CH₃CH₂CH₂NH(CH₃)), a secondary amine with four carbons total (MW 73). Justify your prediction by comparison to the C₃ compounds in the table. 2 marks
2. Graph interpretation — basicity of amines across substitution classes
The bar chart below shows the pKb values (negative log of Kb) for a series of amines and one amide. A lower pKb means a stronger base. 7 marks
Figure 2.1. pKb values for selected nitrogen compounds. Lower pKb = stronger base. Data from NIST Chemistry WebBook and Clayden Organic Chemistry 2nd edn.
2.1 Which compound in the chart is the strongest base? State the pKb value and identify whether it is a primary, secondary, or tertiary amine. 2 marks
2.2 Trimethylamine (tertiary) is a weaker base than dimethylamine (secondary). Using the data and your lesson knowledge, suggest a reason why adding a third alkyl group does not continue to increase basicity. 2 marks
2.3 Ethanamide has a pKb of 15.1, far higher than any amine on the chart. Explain, using resonance, why the amide is essentially not basic. 3 marks
3. Compare and contrast — amines vs amides
Complete the two-column comparison table below. All entries must be based on lesson content. 8 marks (1 per row)
| Feature | Amines (e.g. propan-1-amine) | Amides (e.g. propanamide) |
|---|---|---|
| Functional group structure | ||
| Basicity / pH in water | ||
| Lone pair availability on N | ||
| Reaction with HCl | ||
| Boiling point (C₃, °C) | ||
| H-bond donor ability | ||
| How formed | ||
| Australian/industrial example |
4. Cause-and-effect chain — trimethylamine in Australian seafood
When fresh Australian barramundi is left at room temperature, the naturally occurring trimethylamine N-oxide (TMAO) in its tissue is broken down by bacteria, releasing trimethylamine ((CH₃)₃N). Adding a squeeze of lemon juice before serving is traditional culinary practice to reduce the odour. Trace the cause-and-effect chain below. 5 marks
Cause: bacteria decompose TMAO in barramundi tissue →
Effect 1: _______________________________________________________ (name the product and its class) 1 mark
→ so …
Effect 2: _______________________________________________________ (state the sensory consequence and give the chemistry reason why this compound smells) 1 mark
→ adding lemon juice (citric acid) →
Effect 3: _______________________________________________________ (write the type of reaction that occurs and the type of product formed) 1 mark
→ so …
Effect 4: _______________________________________________________ (explain why the smell is eliminated at the molecular level) 1 mark
Overall outcome: ________________________________________________ 1 mark
5. Predict and justify — amine drug delivery
Many Australian pharmaceuticals manufactured at CSL Limited contain amine functional groups. For example, the analgesic paracetamol (acetaminophen) contains an amide group (–NH–CO–CH₃). Paracetamol’s water solubility is critical for its absorption as a tablet. 4 marks
5.1 Predict whether paracetamol would be more soluble in water or in hexane (non-polar solvent), and justify your prediction in terms of the intermolecular forces involved in each solvent. 2 marks
5.2 Despite containing an –NH– group, paracetamol does not produce a basic solution when dissolved in water. Using your knowledge of amides, explain why. 2 marks
Q1.1 — Amine vs alcohol BP (3 marks)
Oxygen (EN = 3.5) is more electronegative than nitrogen (EN = 3.0) [1 mark]. The O–H bond in propan-1-ol is more polar than the N–H bond in propan-1-amine, making the H in O–H more strongly δ⁺ [1 mark]. Therefore O–H···O hydrogen bonds are stronger and require more energy to break than N–H···N hydrogen bonds → propan-1-ol has a higher BP [1 mark].
Q1.2 — Trimethylamine vs propane BP (2 marks)
Both trimethylamine and propane are incapable of H-bond donation, but trimethylamine has a significant dipole moment (the N is more electronegative than C; lone pair on N creates a molecular dipole) giving dipole–dipole interactions absent in the symmetric non-polar propane [1]. These additional dipole–dipole forces raise the BP of trimethylamine above that of propane despite similar MW (59 vs 44 g mol⁻¹) [1].
Q1.3 — Propanamide higher BP than propanoic acid (2 marks)
Propanoic acid has only O–H donor + C=O acceptor, and forms dimers (2 H-bonds per pair of molecules). Propanamide has N–H donors AND a C=O acceptor in a cooperative network: each molecule simultaneously donates (via N–H) and accepts (via C=O) H-bonds from multiple neighbours [1]. This multi-directional cooperative network is stronger overall than the two-molecule dimer of a carboxylic acid, requiring more energy to disrupt → higher BP despite lower MW [1].
Q1.4 — Predict BP for N-methylpropan-1-amine (2 marks)
N-methylpropan-1-amine is a secondary amine (two alkyl groups on N: methyl + propyl), MW 73. It has one N–H bond (H-bond donor) plus dipole–dipole and dispersion forces [1]. Predicted BP: approximately 60–75°C — higher than primary propan-1-amine (48°C) due to greater MW and dispersion forces, but lower than propan-1-ol (97°C) because N–H bonding is weaker than O–H bonding. Accept any value in the range 55–80°C with valid reasoning [1]. (Actual BP of N-methylpropan-1-amine is 63°C.)
Q2.1 — Strongest base (2 marks)
Dimethylamine is the strongest base with pKb = 3.27 (lowest pKb value = highest Kb = strongest base) [1]. It is a secondary amine (two methyl groups on N) [1].
Q2.2 — Why trimethylamine weaker than dimethylamine (2 marks)
Although three alkyl groups donate more electron density to N inductively (which would strengthen basicity), in trimethylamine the three methyl groups create steric hindrance around the nitrogen lone pair [1]. Water molecules find it harder to approach and protonate the lone pair on N; also in aqueous solution, the trimethylammonium cation cannot stabilise solvation as effectively as the dimethylammonium cation (which has more available N–H bonds for H-bonding with water). The steric effect partially counteracts the inductive effect, reducing the observed basicity [1].
Q2.3 — Why ethanamide is not basic (3 marks)
In ethanamide, the nitrogen lone pair is not in a free sp³ orbital. It participates in resonance with the adjacent C=O group: CH₃C(=O)NH₂ ↔ CH₃C(⁻O)=NH₂⁺ [1]. The lone pair is delocalised into the C=O pi system, occupying a p-type orbital involved in the pi bond rather than a non-bonding sp³ orbital [1]. It is therefore not freely available to accept H⁺ from water. Kb ∼ 10⁻¹⁵ — negligibly small; pH ≈ 7; the compound is essentially neutral [1].
Q3 — Compare and contrast table
| Feature | Amines | Amides |
|---|---|---|
| Functional group | –NH₂ (primary), etc. | –CO–NH₂ |
| Basicity / pH | Weak base; pH > 7; Kb ∼ 10⁻⁴ | Neutral; pH ≈ 7; Kb ∼ 10⁻¹⁵ |
| Lone pair on N | Freely available (sp³) | Delocalised into C=O (not freely available) |
| Reaction with HCl | Forms ionic ammonium salt at room temp | No reaction (neutral; Kb too small) |
| BP (C₃, °C) | Propan-1-amine: 48°C | Propanamide: 213°C |
| H-bond donor | Primary: 2 N–H; Secondary: 1 N–H; Tertiary: none | 2 N–H bonds AND strong C=O acceptor |
| How formed | Reductive amination, natural biosynthesis | Condensation of carboxylic acid + amine (+ heat) |
| Australian example | Trimethylamine (fishy smell in Australian seafood) | Nylon polyamide (Pacific Brands; surgical sutures) |
Q4 — Cause-and-effect chain
Effect 1: Bacteria break down TMAO and release trimethylamine ((CH₃)₃N), a tertiary amine.
Effect 2: The barramundi develops a strong fishy odour. Trimethylamine is volatile (low BP = 3°C, no H-bond donors to self-associate) and the lone pair on N interacts with olfactory receptors.
Effect 3: An acid–base neutralisation occurs (amine acts as base, citric acid as acid), forming an ionic ammonium citrate salt: (CH₃)₃N + H⁺ → [(CH₃)₃NH⁺][citrate⁻].
Effect 4: The trimethylammonium salt is ionic and non-volatile (no free lone pair; much higher melting point → no evaporation at room temperature); no amine molecules reach olfactory receptors.
Overall outcome: The perceived fishy smell is eliminated because acid–base chemistry converts the volatile amine to a non-volatile ionic salt — a practical application of amine acid–base chemistry in the Australian food industry.
Q5.1 — Paracetamol solubility (2 marks)
More soluble in water. Paracetamol has an amide N–H and a C=O acceptor that can both form H-bonds with water molecules (H-bond donation via N–H to water’s lone pairs; H-bond acceptance via C=O from water’s O–H) [1]. Hexane is non-polar and can only form dispersion interactions; disrupting the strong H-bonds within paracetamol–paracetamol and paracetamol–water interactions to dissolve in hexane would be energetically unfavourable [1].
Q5.2 — Paracetamol not basic (2 marks)
The nitrogen in paracetamol is part of an amide group (–NH–CO–). The lone pair on this N is delocalised into the adjacent C=O by resonance [1], making it unavailable to accept H⁺ from water. Kb ∼ 10⁻¹⁵ — effectively neutral. The same structural explanation applies to all amides: having N–H bonds does not confer basicity if the lone pair is delocalised into an adjacent carbonyl [1].