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.
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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) is sp³-hybridised: three bonding pairs (one N–C, two N–H) and one lone pair occupy near-tetrahedral positions around N. Bond angles ~107° (compressed from 109.5° 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.
R–NH₂
1 alkyl group on N
2 H atoms on N
H-bond donor ✓✓
e.g. ethylamine CH₃CH₂NH₂
R–NH–R'
2 alkyl groups on N
1 H atom on N
H-bond donor ✓
e.g. dimethylamine (CH₃)₂NH
R–N(R')R''
3 alkyl groups on N
0 H atoms on N
No H-bond donation
e.g. trimethylamine (CH₃)₃N
Simple primary amines: [alkyl]amine — methylamine, ethylamine, propylamine. With locants: propan-1-amine (NH₂ at C1), propan-2-amine (NH₂ at C2), butan-2-amine. Secondary/tertiary: N- prefix for substituents on N — N-methylethanamine (ethyl chain, methyl on N); N,N-dimethylpropan-1-amine (propyl chain, two methyl groups on N).
Amine vs amide — lone pair availability determines basicity. In ethanamine (left) the N lone pair sits in a free sp³ orbital and readily accepts H⁺. In ethanamide (right) the lone pair is delocalised into the C=O pi system (resonance, dashed arrow ↔) — it cannot be donated to accept a proton.
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 (polar C–N bond) 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.
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. Aryl amines (aniline): Kb ~ 10⁻¹⁰ — lone pair partially delocalised into benzene ring.
Properties of ammonium salts: ionic; typically solid at room temperature; non-volatile (no free lone pair to interact with olfactory receptors → no fishy odour); water-soluble. This is why treating amines with acid eliminates the smell.
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 nitrogen is approximately sp²-hybridised because its lone pair participates in resonance with the adjacent C=O pi system:
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⁺.
IUPAC naming: suffix –amide; –CONH₂ carbon always C1.
| Name | Formula | BP |
|---|---|---|
| Methanamide | HCONH₂ | 193°C |
| Ethanamide | CH₃CONH₂ | 220°C |
| Propanamide | CH₃CH₂CONH₂ | 213°C |
| Butanamide | CH₃CH₂CH₂CONH₂ | 216°C |
The lone pair on N in an amide is delocalised into the C=O — it participates in the pi system and is no longer sitting in a pure non-bonding orbital freely available to accept H⁺. Dissolving ethanamide in water gives pH ≈ 7 (neutral). Kb ~ 10⁻¹⁵ — negligibly small. Compare: ethylamine Kb ~ 4 × 10⁻⁴ → noticeably basic.
Amides have BOTH: (1) N–H bonds → H-bond donors; (2) C=O group → strong H-bond acceptor (C=O is a stronger acceptor than alcohol O). 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.
"Classify the amine by the number of carbons in the chain." Classification counts alkyl groups on N, not chain length. Propan-2-amine has –NH₂ on a secondary carbon but is a PRIMARY amine (one alkyl group on N). Always count at the nitrogen.
"Amides are basic because they contain N–H bonds like amines." Having N–H bonds does not confer basicity if the lone pair is delocalised. In amides, the lone pair on N is tied up in the amide resonance — it cannot accept H⁺ from water. Kb ~ 10⁻¹⁵ — essentially neutral despite the N–H bonds.
"Amine + carboxylic acid produces an ester and water." This reaction produces an ionic ammonium carboxylate salt at room temperature — no water is produced and no covalent ester bond forms. Esterification requires an O nucleophile (alcohol), a catalyst (conc. H₂SO₄), and reflux — none of these apply to amine + acid neutralisation.
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⁺ from HCl; Cl⁻ becomes the counterion of the ammonium salt.
N is bonded to: one (CH₃)₂CH– group (isopropyl) + two H atoms. ONE alkyl group on N → PRIMARY amine (propan-2-amine).
N is bonded to: two ethyl groups + one H. TWO alkyl groups on N → SECONDARY amine (diethylamine).
N is bonded to: three methyl groups, no H. THREE alkyl groups on N → TERTIARY amine (trimethylamine). 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) Primary; (CH₃)₂CHNH₂ + HCl → [(CH₃)₂CHNH₃⁺][Cl⁻] | (b) Secondary; (C₂H₅)₂NH + HCl → [(C₂H₅)₂NH₂⁺][Cl⁻] | (c) Tertiary; (CH₃)₃N + HCl → [(CH₃)₃NH⁺][Cl⁻]
(a) Predict the order of increasing boiling point for: ethanamine (CH₃CH₂NH₂, primary amine, MW 45), dimethylamine ((CH₃)₂NH, secondary amine, MW 45), ethanamide (CH₃CONH₂, amide, MW 59). Justify each step. (b) Ethanamide dissolves in water to give pH ≈ 7, while ethanamine gives pH > 7. Explain this difference.
Dimethylamine (secondary, 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 of the three.
Ethanamine (primary, 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 (each molecule donates and accepts simultaneously). Also higher MW (59 vs 45) → stronger dispersion. Dramatically higher BP.
Ethanamine (pH > 7): N lone pair in a free sp³ orbital → freely accepts H⁺ from water: CH₃CH₂NH₂ + H₂O ⇌ CH₃CH₂NH₃⁺ + OH⁻. OH⁻ produced → basic solution. Kb ~ 4 × 10⁻⁴ (weak base, but measurably basic).
Ethanamide (pH ≈ 7): N lone pair is delocalised into the adjacent C=O by resonance (CH₃–C(=O)–NH₂ ↔ CH₃–C(–O⁻)=NH₂⁺). The lone pair participates in the pi system — it cannot be donated to accept H⁺ from water. No OH⁻ produced → neutral. Kb ~ 10⁻¹⁵.
(a) Increasing BP: dimethylamine (7.4°C) < ethanamine (16.6°C) < ethanamide (220°C). Dimethylamine < ethanamine: secondary amine has 1 N–H vs 2 N–H in primary → weaker H-bonding. Ethanamine < ethanamide: amide adds C=O as a strong H-bond acceptor to the N–H donor, creating a stronger cooperative network; also higher MW. (b) Ethanamine: free N lone pair accepts H⁺ → OH⁻ → basic. Ethanamide: N lone pair delocalised into C=O → unavailable → neutral.
Glycylglycine is the simplest dipeptide — two glycine amino acids (H₂N–CH₂–COOH) linked by one peptide/amide bond. (a) Draw the amide bond in glycylglycine and identify which atoms came from the –COOH and which from the –NH₂ of the two glycine molecules. (b) Write the condensation equation. (c) Explain why glycylglycine does not dissolve in hexane. (d) Explain why the amide bond has "restricted rotation" and its biological significance for protein structure.
Two glycines link: –COOH of first + –NH₂ of second → –CO–NH– bond. The structural region around the bond: …–CH₂–C(=O)–NH–CH₂–…
Glycylglycine contains multiple highly polar, H-bonding groups: N–H donors, C=O acceptor, free –COOH (H-bond donor + acceptor), free –NH₂ (H-bond acceptor, donor). The molecule is extensively H-bonded between its own molecules. Hexane is non-polar — only dispersion forces. For glycylglycine to dissolve, H-bonds between glycylglycine molecules must be disrupted and new glycylglycine–hexane interactions would form — but these would be only dispersion forces, far weaker than the H-bonds disrupted. The energy balance is highly unfavourable → insoluble in hexane. In water, by contrast, both O–H and N–H of water can interact with all polar groups in glycylglycine → favourable dissolution.
The amide bond has resonance: R–C(=O)–NH–R' ↔ R–C(–O⁻)=N⁺H–R'. The C–N bond has partial double-bond character → rotation around C–N requires breaking the pi contribution → energetically unfavourable → the amide bond is planar with restricted rotation. The C, O, N, and their bonded atoms are all coplanar.
Biological significance: Every peptide bond in a protein is an amide bond with restricted rotation. Only the bonds to the alpha-carbon (Cα) rotate freely. This constrains the backbone into specific repeating patterns — the alpha helix and beta sheet — which are the secondary structures that define a protein's shape. A protein's shape determines its function (enzyme active site geometry, antibody binding surface, structural rigidity). Without the restricted rotation of the peptide bond, proteins could adopt any conformation and would lose their specific three-dimensional architecture entirely.
(a) –C(=O)–NH– bond; C=O from –COOH; –NH– from –NH₂; H₂O = –OH + H. (b) Condensation equation as above. (c) Glycylglycine has multiple polar H-bonding groups; hexane is non-polar — new hexane interactions would be only dispersion forces, far weaker than the H-bonds disrupted → insoluble. (d) Resonance gives C–N partial double-bond character → restricted rotation → planar peptide bonds. In proteins: constrains backbone into alpha helix and beta sheet secondary structures, defining 3D shape and function.
Amine classification: Count alkyl groups on N. 1 = primary (R–NH₂); 2 = secondary (R–NH–R'); 3 = tertiary (R–N(R')R''). NOT the number of carbons in the chain.
Amines as weak bases: R–NH₂ + H₂O ⇌ R–NH₃⁺ + OH⁻. Kb ≪ 1. Alkyl amines Kb ~ 10⁻⁴. Lone pair on N (sp³, freely available) accepts H⁺.
Amine + acid → ammonium salt: R–NH₂ + HCl → [R–NH₃⁺][Cl⁻]. Room temperature; no catalyst; ionic product; no water. R–NH₂ + R'COOH → [R–NH₃⁺][R'COO⁻].
Boiling points (C3): Propane (−42°C) < trimethylamine (3°C) < propan-1-amine (48°C) < propan-1-ol (97°C) < propanoic acid (141°C) < propanamide (213°C).
N–H vs O–H H-bonding: O (EN 3.5) > N (EN 3.0) → O–H more polar → stronger H-bond donor → alcohol BP > amine BP at same chain length.
Amide structure: –CONH₂. N lone pair delocalised into C=O (resonance) → NOT basic (Kb ~ 10⁻¹⁵). Has N–H + C=O → strongest IMF → highest BP of all functional groups.
Amide bond formation: R–COOH + H₂N–R' → R–CO–NH–R' + H₂O (condensation). Same as peptide bond in proteins; same as nylon linkage.
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 | ? | ? | ? |
Rank the following five compounds from lowest to highest boiling point, and explain each step using IMF reasoning. Compound data:
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, forms N–H···N H-bonds). Option D is wrong — larger alkyl groups increase dispersion forces and would tend to raise BP; the cause of lower BP is absence of H-bond donors, not steric bulk.
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. Option C is wrong — having N–H bonds does not confer basicity if the lone pair is delocalised. Option D is wrong — N in an amide does have a lone pair, but it is delocalised, not absent.
Q3 — B. Amine + carboxylic acid at room temperature → ionic ammonium carboxylate salt. The amine lone pair accepts H⁺ from the carboxylic acid's O–H. Product: propylammonium ethanoate [CH₃CH₂CH₂NH₃⁺][CH₃COO⁻]. No water is produced (this is neutralisation, not esterification). Option A is wrong — esterification requires an alcohol (O nucleophile) + conc. H₂SO₄ + reflux. Option C (amide formation) can occur on prolonged heating but not at room temperature under these conditions.
Q4 — C. The key is electronegativity: O (3.5) > N (3.0) → O–H bond is more polar → the 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. Option A counts oxygen atoms — not valid reasoning. Option B is not correct as stated — both propan-1-amine and propan-1-ol form H-bonds with multiple neighbours.
Q5 — A. Propanamide has both N–H bonds (donors) and a C=O group (strong acceptor). This combination creates a cooperative H-bond network: each molecule simultaneously donates (via N–H) and accepts (via C=O) H-bonds from neighbours. The network is more extensive and stronger than alcohol H-bonding (one O–H per molecule) or carboxylic acid dimerisation (two H-bonds per pair, but limited to pairs). Option B is wrong — primary amide has 2 N–H bonds, not three. Option C is wrong — amides are NOT basic. Option D is wrong — propanamide (MW 73) actually has LOWER mass than propanoic acid (MW 74); dispersion forces argument fails.
Q6 (4 marks): (a) Trimethylamine is a tertiary amine — three methyl groups on N, no H atoms on N. It cannot donate hydrogen bonds between its own molecules because it has no N–H bonds (H-bond donors require an N–H or O–H bond). IMF in liquid trimethylamine: dipole-dipole and dispersion only. (1 mark for classification, 1 mark for H-bonding explanation). (b) (CH₃)₃N + HCl → [(CH₃)₃NH⁺][Cl⁻] (1 mark). (c) The product is an ionic ammonium salt [(CH₃)₃NH⁺][Cl⁻]. The lone pair on N is now used in the N–H bond formed when H⁺ was accepted — there is no free lone pair available to interact with olfactory receptors. The salt is also ionic and non-volatile (high melting point, does not readily evaporate at room temperature) → no molecules reach the nose → odourless. (1 mark)
Q7 (5 marks) — sample: Order: dimethylamine (7.4°C) < ethanamine (16.6°C) < ethanamide (220°C) (1 mark).
Dimethylamine vs ethanamine (same formula C₂H₇N): both are amines with N–H bonds, but 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 intermolecular N–H···N hydrogen bonding in ethanamine = higher BP. (2 marks)
Ethanamine vs ethanamide: both have N–H bonds, but 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 to a neighbour's C=O) and accepts H-bonds (into its own C=O from a neighbour's N–H). This networked H-bonding is far stronger than the N–H···N H-bonding in ethanamine, and ethanamide also has higher molecular mass (59 vs 45) → stronger dispersion forces. Combined effects produce the dramatically higher BP of 220°C. (2 marks)
Q8 (6 marks) — sample: The student's claim is incorrect (1 mark). Not all nitrogen-containing compounds produce basic solutions. Ethanamide does NOT produce a basic solution in water; pH ≈ 7 (neutral). The reason lies in the resonance structure of the amide group (1 mark): CH₃–C(=O)–NH₂ ↔ CH₃–C(–O⁻)=NH₂⁺. The lone pair on N participates in this resonance — it is delocalised into the C=O pi system and occupies a p-type orbital involved in the pi bond rather than a purely non-bonding sp³ orbital (1 mark). As a result, this lone pair is not freely available to accept H⁺ from water, so the equilibrium R–CONH₂ + H₂O ⇌ R–CONH₃⁺ + OH⁻ lies extremely far to the left. Kb (ethanamide) ~ 10⁻¹⁵ — negligibly small; pH ≈ 7 (1 mark). By contrast, ethanamine (CH₃CH₂NH₂) has an N lone pair in a free sp³ orbital with no adjacent carbonyl to delocalise it. The lone pair freely accepts H⁺ from water → R–NH₂ + H₂O ⇌ R–NH₃⁺ + OH⁻ — Kb ~ 4 × 10⁻⁴ → clearly basic solution (1 mark). Biological significance: the amide (peptide) bond (–CO–NH–) in proteins has this same restricted lone pair. The partial C=N double-bond character from resonance makes the peptide bond planar with restricted rotation. This planarity constrains the protein backbone into specific secondary structures (alpha helix, beta sheet) that define the three-dimensional shape of each protein — and therefore its biological function as an enzyme, structural protein, or signalling molecule (1 mark).
The structural feature that makes all amines weak bases is the lone pair on nitrogen — sitting in a free sp³ orbital, 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.
Putrescine and cadaverine are small, volatile diamines — the lone pairs on their two N atoms make them smell strongly and act as weak bases. Dopamine and adrenaline are larger, with the same N lone pairs but additional OH groups and aromatic rings that make them less volatile and more biologically targeted. 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.
Amines & Amides