Amines smell like fish and rotting flesh, and yet the same functional group — slightly modified — forms every protein in your body, every nylon fibre in your clothing, and every painkiller in your medicine cabinet.
Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.
Putrescine (butane-1,4-diamine) and cadaverine (pentane-1,5-diamine) are the compounds responsible for the smell of rotting flesh — both are diamines produced when bacteria decompose amino acids in dead tissue. The same class of compounds, amines, includes adrenaline (the fight-or-flight hormone), dopamine (the reward neurotransmitter), and all local anaesthetics from lidocaine to cocaine.
They are weak bases, they smell bad in small molecules, and they are essential to life in large ones.
Before you read on: Write down what structural feature you think amines share that makes them weak bases. What does "weak base" mean at the molecular level — what is actually happening in solution?
The nitrogen atom in an amine has a lone pair that is freely available for proton acceptance — unlike the lone pair on the nitrogen of an amide, which is locked into the adjacent carbonyl — and this single structural difference separates the most reactive organic bases from some of the most inert.
Amines contain a nitrogen atom bonded to one, two, or three alkyl groups, with remaining bonds to hydrogen. The nitrogen in a primary amine (R–NH₂) has three bonding pairs (one N–C, two N–H) and one lone pair in near-tetrahedral positions around N — bond angles ~107° (compressed slightly by the lone pair's greater repulsion). The lone pair is the defining reactive feature: it is the source of the amine's basicity and H-bond accepting ability, and it is freely available because it is not involved in any adjacent pi bond.
R–NH₂R–NH–R'R–N(R')R''Simple primary amines: [alkyl]amine — methylamine, ethylamine, propylamine. With locants: propan-1-amine (NH₂ at C1), propan-2-amine (NH₂ at C2). Secondary/tertiary: N- prefix for substituents on N — N-methylethanamine; N,N-dimethylpropan-1-amine.
Amine vs amide — lone pair availability determines basicity. In ethanamine (left) the N lone pair is freely available and readily accepts H⁺. In ethanamide (right) the lone pair is delocalised into the C=O pi system (resonance ↔) — it cannot be donated to accept a proton.
Which of the following compounds is a secondary amine?
Amines sit between alkanes and alcohols in the boiling point ranking — they form hydrogen bonds, but weaker ones than alcohols, because nitrogen is less electronegative than oxygen, making the N–H bond a weaker H-bond donor.
Primary and secondary amines have N–H bonds — H-bond donors. The lone pair on N is an H-bond acceptor. N–H···N hydrogen bonds form between amine molecules — weaker than O–H···O bonds because nitrogen (EN 3.0) is less electronegative than oxygen (EN 3.5), making the N–H bond less polar and the N–H hydrogen less δ⁺.
Tertiary amines have NO N–H bonds — they cannot donate H-bonds between their own molecules. IMF limited to dipole-dipole and dispersion → lower BP than primary or secondary amines of same size.
Short-chain amines (C1–C4) are miscible with water via N–H donation to water and lone pair acceptance from water's O–H. Tertiary amines dissolve too — the lone pair on N accepts H-bonds from water even without N–H to donate. The fishy smell of fish is primarily trimethylamine. Squeezing lemon juice converts it to an ionic trimethylammonium salt (non-volatile, odourless) — acid-base chemistry solving a sensory problem.
True or False: Trimethylamine (a tertiary amine) is insoluble in water because it has no N–H bonds to donate hydrogen bonds.
Every amine reaction follows from one principle: the lone pair on nitrogen attacks a proton donor and forms a new N–H bond — whether the proton comes from water (giving basic solution) or from an acid (giving an ammonium salt).
Alkyl amines: Kb ~ 10⁻⁴ — stronger bases than NH₃ (Kb = 1.8 × 10⁻⁵) because alkyl groups donate electron density to N inductively, increasing lone pair availability.
Properties of ammonium salts: ionic; typically solid at room temperature; non-volatile; water-soluble. This is why treating amines with acid eliminates the smell.
Complete the cloze: When methylamine (CH₃NH₂) reacts with ethanoic acid at room temperature, the product is an ionic ____, specifically methylammonium ethanoate. No ____ is produced (unlike esterification). The reaction requires no ____ and proceeds at room temperature.
Adding a carbonyl group directly next to the nitrogen of an amine transforms it from one of the most reactive organic bases into a compound that is essentially neutral in water — one structural change, reactivity reversed completely.
An amide contains the –CONH₂ group: a carbonyl carbon (C=O) directly bonded to a nitrogen bearing one or two H atoms. The lone pair on nitrogen participates in resonance with the adjacent C=O pi system — it is delocalised rather than freely available:
R–C(=O)–NH₂ ↔ R–C(–O⁻)=NH₂⁺
Consequences: (1) the C–N bond has significant double-bond character → restricted rotation around C–N; (2) the lone pair on N is delocalised into the C=O pi system and is NOT freely available to accept H⁺.
Amides have BOTH: (1) N–H bonds → H-bond donors; (2) C=O group → strong H-bond acceptor. Together these create an extensive, multi-directional H-bond network. Each molecule simultaneously donates H-bonds via N–H to neighbours' C=O groups AND accepts H-bonds into its own C=O. This cooperative network is stronger than alcohol H-bonding, carboxylic acid dimerisation, or amine N–H bonding → amide BP is highest of all functional group classes at the same chain length.
Amide bond formation by condensation. The –OH from the carboxylic acid and one H from the amine combine as water. The product –CO–NH– bond is the amide bond — identical to the peptide bond linking amino acids in proteins and the linkage in nylon.
Why does ethanamide (CH₃CONH₂) dissolve in water to give pH ≈ 7 (neutral), while ethylamine (CH₃CH₂NH₂) gives pH > 7?
For each compound, classify the amine as primary, secondary, or tertiary, and write the equation for its reaction with HCl: (a) (CH₃)₂CHNH₂ (b) (C₂H₅)₂NH (c) (CH₃)₃N
Locate N in each structure. Count the number of alkyl groups bonded directly to N. 1 = primary; 2 = secondary; 3 = tertiary. For the HCl reaction: lone pair on N accepts H⁺; Cl⁻ becomes the counterion.
(CH₃)₂CHNH₂: N bonded to one isopropyl group + two H atoms → ONE alkyl group on N → PRIMARY amine.
(C₂H₅)₂NH: N bonded to two ethyl groups + one H → TWO alkyl groups on N → SECONDARY amine.
(CH₃)₃N: N bonded to three methyl groups, no H → THREE alkyl groups → TERTIARY amine. Note: tertiary amine still reacts with HCl — the lone pair on N accepts H⁺ even though there are no N–H bonds in the starting material.
(a) Predict the order of increasing boiling point for: ethanamine (CH₃CH₂NH₂, 1°, MW 45), dimethylamine ((CH₃)₂NH, 2°, MW 45), ethanamide (CH₃CONH₂, amide, MW 59). (b) Ethanamide dissolves in water to give pH ≈ 7, while ethanamine gives pH > 7. Explain.
Dimethylamine (2°, 1 N–H, BP 7.4°C) — only one N–H bond per molecule → fewer H-bond donors than ethanamine → weaker intermolecular N–H···N bonding → lowest BP.
Ethanamine (1°, 2 N–H, BP 16.6°C) — two N–H bonds per molecule → more H-bond donors → stronger overall N–H···N bonding → higher BP than dimethylamine.
Ethanamide (amide, 2 N–H + C=O, BP 220°C) — N–H donor AND C=O strong acceptor → extensive cooperative H-bond network. Also higher MW (59 vs 45) → stronger dispersion. Dramatically higher BP.
Ethanamine (pH > 7): N lone pair is freely available → accepts H⁺ from water: CH₃CH₂NH₂ + H₂O ⇌ CH₃CH₂NH₃⁺ + OH⁻. OH⁻ produced → basic solution.
Ethanamide (pH ≈ 7): N lone pair is delocalised into the adjacent C=O by resonance. The lone pair participates in the pi system — it cannot be donated to accept H⁺ from water. No OH⁻ produced → neutral.
Complete the microtasks in the Learn phase to unlock Practice. You need -- XP.
0 / -- XP earned
For each compound below: (i) classify as primary, secondary, or tertiary amine; (ii) write the equation for its reaction with HCl; (iii) write the IUPAC name of the salt produced.
| Compound | (i) Class | (ii) Equation with HCl | (iii) Salt name |
|---|---|---|---|
| CH₃NH₂ | ? | ? | ? |
| (CH₃)₂NH | ? | ? | ? |
| C₂H₅N(CH₃)₂ | ? | ? | ? |
1. Which of the following correctly classifies N,N-diethylpropan-1-amine and explains why it has a lower boiling point than propan-1-amine?
2. A student dissolves ethanamide (CH₃CONH₂) in water and measures pH ≈ 7. Which explanation is correct?
3. A student reacts propan-1-amine with ethanoic acid at room temperature. What is the product and what type of reaction has occurred?
4. Which statement correctly explains why propan-1-amine (BP 48°C) has a lower boiling point than propan-1-ol (BP 97°C), despite both having similar molecular masses and H-bonding groups?
5. The boiling point of propanamide (213°C) is much higher than both propan-1-ol (97°C) and propanoic acid (141°C), despite propanamide having a lower molecular mass than propanoic acid. Which explanation is correct?
Question 6 (4 marks) — The "fishy" odour of a seafood counter is primarily trimethylamine, (CH₃)₃N. (a) Classify trimethylamine and explain whether it can donate hydrogen bonds between its own molecules. (b) Write the equation for the reaction of trimethylamine with hydrochloric acid. (c) Explain why the product of this reaction is odourless.
Question 7 (5 marks) — Compare the boiling points of the following three C₂ compounds and explain the trend using IMF reasoning: ethanamine (CH₃CH₂NH₂, BP 16.6°C), dimethylamine ((CH₃)₂NH, BP 7.4°C), ethanamide (CH₃CONH₂, BP 220°C). In your response, refer to the specific H-bonding differences between these compounds.
Question 8 (6 marks) — A student states: "Ethanamide dissolves in water to give a basic solution because it contains nitrogen, like all amines." Evaluate this claim, using structural and electronic reasoning to explain whether the student is correct. In your response, include the resonance structure of ethanamide, compare its Kb with that of ethanamine, and explain the biological significance of the structural feature responsible for the amide's behaviour.
Q1 — C. N,N-diethylpropan-1-amine: N has two ethyl groups + one propyl group = THREE alkyl groups → tertiary amine. No N–H bonds → cannot donate H-bonds → only dipole-dipole and dispersion forces → lower BP than propan-1-amine (primary, 2 N–H bonds).
Q2 — B. The lone pair on N in ethanamide is delocalised into the C=O pi system (resonance: CH₃–C(=O)–NH₂ ↔ CH₃–C(–O⁻)=NH₂⁺). This lone pair cannot be donated to accept H⁺ from water. Kb ~10⁻¹⁵ → negligible basicity → pH ≈ 7.
Q3 — B. Amine + carboxylic acid at room temperature → ionic ammonium carboxylate salt. Product: propylammonium ethanoate [CH₃CH₂CH₂NH₃⁺][CH₃COO⁻]. No water is produced.
Q4 — C. O (3.5) > N (3.0) → O–H bond is more polar → H in O–H is more δ⁺ → O–H is a stronger H-bond donor → O–H···O bonds require more energy to break than N–H···N bonds → alcohol has higher BP.
Q5 — A. Propanamide has both N–H bonds (donors) and a C=O group (strong acceptor). This combination creates a cooperative H-bond network where each molecule simultaneously donates (via N–H) and accepts (via C=O) H-bonds. More extensive than alcohol or carboxylic acid H-bonding.
Q6 (4 marks): (a) Tertiary amine — three methyl groups on N, no H atoms on N. Cannot donate hydrogen bonds between its own molecules (no N–H bonds = no H-bond donors). IMF: dipole-dipole and dispersion only. (b) (CH₃)₃N + HCl → [(CH₃)₃NH⁺][Cl⁻]. (c) The product is an ionic ammonium salt. The lone pair on N is now used in the N–H bond formed when H⁺ was accepted — no free lone pair to interact with olfactory receptors. The salt is also ionic and non-volatile → odourless.
Q7 (5 marks): Order: dimethylamine (7.4°C) < ethanamine (16.6°C) < ethanamide (220°C). Dimethylamine vs ethanamine: dimethylamine (secondary) has only ONE N–H bond per molecule while ethanamine (primary) has TWO. More N–H bonds = more H-bond donors = stronger overall N–H···N H-bonding in ethanamine = higher BP. Ethanamine vs ethanamide: ethanamide additionally has a highly polar C=O group that is a strong H-bond acceptor. The combination of N–H donors AND C=O acceptors creates a cooperative H-bond network: each ethanamide molecule simultaneously donates H-bonds (via N–H) and accepts H-bonds (into its C=O). This networked H-bonding is far stronger than N–H···N H-bonding alone, producing the dramatically higher BP of 220°C.
Q8 (6 marks): The student's claim is incorrect. Not all nitrogen-containing compounds produce basic solutions. Ethanamide does NOT produce a basic solution; pH ≈ 7 (neutral). The reason lies in the resonance structure: CH₃–C(=O)–NH₂ ↔ CH₃–C(–O⁻)=NH₂⁺. The lone pair on N is delocalised into the C=O pi system — it is not freely available to accept H⁺ from water, so no OH⁻ is produced. Kb (ethanamide) ~10⁻¹⁵ — negligibly small; pH ≈ 7. By contrast, ethanamine (CH₃CH₂NH₂) has a freely available lone pair on N — it accepts H⁺ from water → Kb ~4 × 10⁻⁴ → clearly basic. Biological significance: the amide (peptide) bond (–CO–NH–) in proteins has the same restricted lone pair. Partial C=N double-bond character makes the peptide bond planar with restricted rotation, constraining the protein backbone into alpha helix and beta sheet secondary structures that define 3D shape and function.
The structural feature that makes all amines weak bases is the lone pair on nitrogen — freely available to accept H⁺ from water (R–NH₂ + H₂O ⇌ R–NH₃⁺ + OH⁻). "Weak base" means Kb ≪ 1 — the equilibrium lies left; most molecules remain un-protonated, but enough OH⁻ is produced to raise pH above 7. The amide modification (adding C=O adjacent to N) locks the lone pair into resonance — turning a reactive base into a neutral, structurally rigid bond that holds every protein together.
1. What structural feature makes amines weak bases?
2. Why does propan-1-amine have a lower BP than propan-1-ol?
3. Why is ethanamide NOT basic?
4. Write the equation and conditions for amine + HCl. What type of product forms?
5. Why does the amide (peptide) bond have biological significance?
Work through this topic 1-on-1 with an experienced HSC tutor.
Book a free session →