Year 12 Chemistry Module 7, Organic Chemistry ⏱ ~45 min Lesson 16 of 23 Core

Amines & Amides, Structure, Properties & Reactions

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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Today's hook: The smell of rotting fish and the strength of nylon come from the same family of nitrogen compounds. How does structure drive both?

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Connect: nitrogen compounds in everyday chemistry

Warm up first

Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.

Worksheets

Practise this lesson

Four printable worksheets that build from the foundations up to exam-style questions, start at whatever level suits you.

Build Foundations & guided practice Apply Application practice Master Mastery challenge Exam Exam-style questions

Before You Read

The Smell of Decay, and the Basis of Life

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?

Learning Intentions

goals

Know

Amine classification: primary (1 alkyl on N), secondary (2), tertiary (3)
Amines are weak bases: R–NH₂ + H₂O ⇌ R–NH₃⁺ + OH⁻; Kb ≪ 1
Amines + acids → ionic ammonium salts; room temperature, no catalyst
Amide structure: –CONH₂; N lone pair delocalised into C=O → not basic
BP ranking: amide > carboxylic acid > alcohol > amine > alkane

Understand

Why N–H H-bonds are weaker than O–H H-bonds (electronegativity difference)
Why tertiary amines have lower BPs than primary amines (no N–H donors)
Why amides are NOT basic, N lone pair delocalised into C=O by resonance
Why amides have the highest BP, N–H donor + C=O acceptor network

Can Do

Classify amines as primary, secondary, or tertiary from structure
Write equations for amine reactions with HCl and carboxylic acids
Explain why amides are neutral using resonance/delocalisation
Rank compound BPs using IMF arguments at the same chain length

Scan these before reading

vocab

Amine (–NH₂)A functional group derived from ammonia where one or more H atoms are replaced by organic groups; acts as a weak base.

Primary amineOne carbon group attached to nitrogen (RNH₂); e.g., methylamine (CH₃NH₂).

Amide (–CONH₂)A functional group with a carbonyl group bonded to a nitrogen; formed by condensation of carboxylic acid with amine.

Basicity of aminesAmines accept protons from water: RNH₂ + H₂O ⇌ RNH₃⁺ + OH⁻; Kb depends on electron-donating/withdrawing groups.

Amide bond formationCarboxylic acid + amine → amide + water (condensation); requires heating or activation.

Peptide bondAn amide linkage (–CO–NH–) between amino acids in proteins; formed by condensation polymerisation.

📚 Core Content

Explain: classify and name amines

Amine Structure, The Nitrogen Lone Pair and Classification

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.

Structure of the Amine Group

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.

Classification, Count Alkyl Groups on Nitrogen

Primary (1°)
R–NH₂
1 alkyl group on N · 2 H atoms on N · H-bond donor ✓✓
e.g. ethylamine CH₃CH₂NH₂

Secondary (2°)
R–NH–R'
2 alkyl groups on N · 1 H atom on N · H-bond donor ✓
e.g. dimethylamine (CH₃)₂NH

Tertiary (3°)
R–N(R')R''
3 alkyl groups on N · 0 H atoms on N · No H-bond donation
e.g. trimethylamine (CH₃)₃N

IUPAC Naming

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.

Comparison of ethanamine and ethanamide lone pair availability and basicity

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.

Must Do: Amine classification is determined by counting alkyl groups on nitrogen not carbons adjacent to the N, not total chain length. Ethylamine: one ethyl group on N → primary. N-methylethanamine: methyl + ethyl on N → secondary. Always locate N first, count its alkyl groups directly.

Common Error: Confusing amine classification (count alkyl groups on N) with alcohol classification (count carbons bonded to C–OH). Propan-2-amine (CH₃CH(NH₂)CH₃) has –NH₂ on a secondary carbon, but the AMINE is primary (only one alkyl group bonded to N). Applying the alcohol classification logic to an amine always produces wrong answers.

Which of the following compounds is a secondary amine?

Represent: connect bonding to properties

Physical Properties of Amines, N–H Bonding vs O–H Bonding

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.

Intermolecular Forces in Amines

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.

Full Boiling Point Ranking at C3

Propane

−42°C

Dispersion only

Trimethylamine (3°)

3°C

Dipole-dipole; no N–H donors

Propan-1-amine (1°)

48°C

N–H H-bonding (weaker donor)

Propan-1-ol

97°C

O–H H-bonding (stronger donor)

Propanoic acid

141°C

O–H + dimer (2 H-bonds/pair)

Propanamide

213°C

N–H + C=O network (strongest)

Key Comparisons to Memorise

Primary amine vs alcohol (same chain): Amine BP < Alcohol BP. Reason: O (EN 3.5) > N (EN 3.0) → O–H more polar → stronger H-bond donor → more energy to break O–H···O bonds.
Primary vs secondary amine (same formula): Primary BP > secondary BP. Primary has 2 N–H bonds (2 donors); secondary has only 1 N–H bond → weaker overall H-bonding. Example: ethylamine (1°, BP 16.6°C) vs dimethylamine (2°, BP 7.4°C), both C₂H₇N.
Tertiary amine vs primary amine: Tertiary BP < primary BP. No N–H bonds in tertiary → dipole-dipole only.

Water Solubility and Odour

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.

Must Do: In any IMF comparison between an amine and an alcohol at the same chain length, name the electronegativity values (O = 3.5, N = 3.0), explain that the O–H bond is more polar than N–H, connect this to H-bond donation strength, and conclude with the BP consequence. An answer that only says "O–H H-bonds are stronger than N–H H-bonds" without the electronegativity reasoning earns half marks at best.

Common Error: "Tertiary amines are insoluble in water because they have no N–H bonds." Tertiary amines DO dissolve in water (short chain), the lone pair on N accepts H-bonds from water's O–H groups. The absence of N–H means they cannot donate H-bonds between their own molecules (hence lower BP), but they can still interact with water as H-bond acceptors. Trimethylamine is fully miscible with water.

True or False: Trimethylamine (a tertiary amine) is insoluble in water because it has no N–H bonds to donate hydrogen bonds.

Respond: write weak-base and salt equations

Amines as Weak Bases, Reactions with Acids

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).

Ionisation in Water

Amine as weak base R–NH₂ + H₂O ⇌ R–NH₃⁺ + OH⁻
Kb = [R–NH₃⁺][OH⁻] / [R–NH₂] ≪ 1
Solution is basic (pH > 7) but Kb ≪ 1, most molecules remain un-protonated

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.

Reactions with Acids, Ammonium Salt Formation

With HCl (strong acid, goes to completion) R–NH₂ + HCl → [R–NH₃⁺][Cl⁻]
e.g. CH₃CH₂NH₂ + HCl → [CH₃CH₂NH₃⁺][Cl⁻] (ethylammonium chloride)
Room temperature; no catalyst; exothermic; ionic salt formed; no gas produced

With carboxylic acid (weak acid) R–NH₂ + R'COOH → [R–NH₃⁺][R'COO⁻]
e.g. CH₃NH₂ + CH₃COOH → [CH₃NH₃⁺][CH₃COO⁻] (methylammonium ethanoate)
Room temperature; ionic ammonium carboxylate salt; no water produced

Regenerating the free amine (add NaOH) [R–NH₃⁺][Cl⁻] + NaOH → R–NH₂ + NaCl + H₂O
NaOH deprotonates the ammonium cation; free amine released (fishy smell returns)

Properties of ammonium salts: ionic; typically solid at room temperature; non-volatile; water-soluble. This is why treating amines with acid eliminates the smell.

Must Do: Write the product of amine + acid as an ionic salt with both cation and anion shown: [R–NH₃⁺][Cl⁻] or R–NH₃⁺Cl⁻. Do NOT write "R–NHCl" (wrong formula, H is added to N, not Cl) or "R–NH₂·HCl" (dot notation without charges).

Common Error: Confusing amine + acid (→ ammonium salt) with esterification (carboxylic acid + alcohol → ester + water). Key differences: (1) amine + acid is instantaneous at room temperature, no catalyst; (2) esterification needs conc. H₂SO₄ and reflux; (3) amine + acid gives an IONIC SALT with no water produced; (4) esterification gives a COVALENT ester + water.

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.

Feedback: explain amide resonance and non-basicity

Amides, Structure, Properties and Why They Are Not Basic

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.

Amide Structure and Resonance

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⁺.

Why Amides Often Have Very High Boiling or Melting Points

Primary and secondary amides can combine N–H hydrogen-bond donation with strong C=O acceptance, creating extensive, multi-directional networks. This often gives amides very high boiling or melting points. Do not use a universal functional-group ranking: the exact structure, substitution, association, molar mass and whether a compound melts or decomposes all matter, while carboxylic-acid dimers complicate simple comparisons.

Condensation reaction of propanoic acid and methylamine forming N-methylpropanamide and water, migrated to a sequence layout.

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.

Biomolecule link, protein hydrolysis: Because peptide bonds are amide bonds, the reverse reaction applies directly to proteins. Hydrolysis of a protein, adding water across each peptide (amide) bond, breaks it down into its amino acid monomers. This is catalysed in the body by proteases (and in the lab by hot acid or base) and is the same hydrolysis pattern as triglycerides → glycerol + fatty acids and polysaccharides → glucose. See Lesson 23 for the three biomolecule classes together.

Must Do: The explanation for why amides are not basic MUST include "delocalised" or "resonance" and describe the mechanism: the lone pair on N is delocalised into the C=O pi system, reducing its availability for proton acceptance.

Common Error: "Amides are acidic because they contain a C=O group like carboxylic acids." Amides are NOT acidic. The C=O in an amide is not associated with an ionisable –OH group, there is no O–H bond in an amide to donate to water. Classifying amides as acidic is always wrong.

Insight: The planarity of the amide bond (partial C=N double-bond character from resonance) has profound biological consequences. In proteins, the peptide bond (–CO–NH–) is planar with restricted rotation, each peptide unit is a rigid flat structure. This constrains the polypeptide backbone into specific repeating patterns: the alpha helix and beta sheet. These secondary structures arise directly from the planarity of the peptide bond.

Why does ethanamide (CH₃CONH₂) dissolve in water to give pH ≈ 7 (neutral), while ethylamine (CH₃CH₂NH₂) gives pH > 7?

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Key Equations, Amines & Amides (Consolidation)

Amine structures

R–NH₂ (1°) R–NH–R' (2°) R–N(R')R'' (3°)

Classification: count alkyl groups on N (not carbons in chain)

Amine as weak base

R–NH₂ + H₂O ⇌ R–NH₃⁺ + OH⁻

Kb = [R–NH₃⁺][OH⁻] / [R–NH₂] Kb ≪ 1 → weak base

Amine + strong acid

R–NH₂ + HCl → [R–NH₃⁺][Cl⁻]

Alkylammonium chloride (ionic salt); room temperature

Amine + carboxylic acid

R–NH₂ + R'COOH → [R–NH₃⁺][R'COO⁻]

Alkylammonium carboxylate salt; room temperature

Ammonium salt → free amine

[R–NH₃⁺][Cl⁻] + NaOH → R–NH₂ + NaCl + H₂O

NaOH deprotonates the ammonium cation

Amide formation (condensation)

R–COOH + H₂N–R' → R–CO–NH–R' + H₂O

Acid + amine → amide + water; same bond as peptide bond

BP ranking (same chain length)

AMIDE > ACID > ALCOHOL > AMINE > ALKANE

Amide highest: N–H donor + C=O acceptor → strongest H-bond network

Apply: compare amines and amides

📋 Worked Examples

Classifying Amines and Writing Acid Reactions

GIVEN

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

METHOD

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.

1, (a)

(CH₃)₂CHNH₂: N bonded to one isopropyl group + two H atoms → ONE alkyl group on N → PRIMARY amine.

(CH₃)₂CHNH₂ + HCl → [(CH₃)₂CHNH₃⁺][Cl⁻] (propan-2-ammonium chloride)

2, (b)

(C₂H₅)₂NH: N bonded to two ethyl groups + one H → TWO alkyl groups on N → SECONDARY amine.

(C₂H₅)₂NH + HCl → [(C₂H₅)₂NH₂⁺][Cl⁻] (diethylammonium chloride)

3, (c)

(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.

(CH₃)₃N + HCl → [(CH₃)₃NH⁺][Cl⁻] (trimethylammonium chloride)

(a) Primary → [(CH₃)₂CHNH₃⁺][Cl⁻] | (b) Secondary → [(C₂H₅)₂NH₂⁺][Cl⁻] | (c) Tertiary → [(CH₃)₃NH⁺][Cl⁻]

Comparing BP and Solubility: Amine vs Amide vs Alcohol

GIVEN

(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.

STEP 1, BP ranking

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.

STEP 2, pH comparison

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.

Increasing BP: dimethylamine (7.4°C) < ethanamine (16.6°C) < ethanamide (220°C). (b) Ethanamine: free N lone pair → basic. Ethanamide: N lone pair delocalised into C=O → neutral (Kb ~10⁻¹⁵).

Why is an amide much less basic than a comparable amine?

Practice: classify, compare and explain

🔀 Sort the Steps +7 XP

To distinguish between an amine and an amide using chemical tests, arrange these steps in the correct order:

Add NaOH solution and warm, amides hydrolyse slowly releasing NH₃ (detected with damp red litmus turning blue)

Add dilute HCl, amines react readily forming a soluble ammonium salt; amides react slowly

Confirm amine by characteristic fishy odour and amine salt precipitation with HCl

Add the unknown compound to water and test pH with universal indicator

If solution is basic (pH > 7), the compound is likely an amine (lone pair on N acts as base)

Practice Activity

ACTIVITY, AMINE CLASSIFICATION AND REACTIONS

Amine Classification and Reactions

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	?	?	?

❓ Check Your Understanding

Short Answer

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.

Reveal Answers

Multiple Choice Answers

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.

Short Answer Sample Answers

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 with negligible volatility under these conditions, so far fewer molecules enter the gas phase and reach the nose.

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.

Revisit: Putrescine, Dopamine, and the Lone Pair

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.

Review: retrieve nitrogen chemistry

Check what actually stuck

🏆 Review, Key Ideas

Quick Recall, Amines & Amides

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?

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