Why is ethylamine a stronger base than phenylamine in water, and why is an amide much less basic than either? In ethylamine, CH₃CH₂NH₂, the nitrogen lone pair is relatively available to accept H⁺ and the electron-donating alkyl group supports that basic behaviour. In phenylamine, C₆H₅NH₂, the nitrogen lone pair can be delocalised into the aromatic ring, making it less available to accept a proton. The familiar qualitative ordering of aqueous basicities is generally ethylamine greater than ammonia greater than phenylamine under comparable school conditions. In an amide such as ethanamide, CH₃CONH₂, the nitrogen lone pair is delocalised toward the carbonyl group, making its nitrogen far less basic in the ordinary school comparison. A structural formula therefore provides the best starting point for predicting acid–base behaviour.
For Hougang JC1 and JC2 families searching for H2 Chemistry tuition, amine basicity explanations, phenylamine bromine water observations, ethylamine and ammonia comparisons, amide hydrolysis or the 2027 SEAB 9476 syllabus, the immediate answer is to locate the nitrogen lone pair and explain what makes it more or less available. This reference guide includes worked acid–base and salt formation equations, the aqueous versus gaseous basicity distinction, formation of phenylamine from nitrobenzene, the reaction of phenylamine with aqueous bromine, and a comparison of amines with amides. It follows the examinable content and links to the eduKate Hougang Chemistry learning route rather than presenting disconnected memorisation tables.
Amine Versus Amide Basicity: Where Is the Nitrogen Lone Pair?
Ethylamine is a typical alkylamine electron-pair donor
In ethylamine, CH₃CH₂NH₂, the nitrogen atom has a lone pair that can accept a proton. A familiar acid–base reaction is CH₃CH₂NH₂ + HCl → CH₃CH₂NH₃⁺Cl⁻ in an appropriate representation. The ethyl group can donate electron density inductively, often increasing basic strength relative to ammonia under the comparison convention used in the school course. That conclusion is about the availability of the nitrogen lone pair and the solution equilibria, not simply the fact that nitrogen occurs in the formula.
Phenylamine has a lone pair involved in the aromatic system
In phenylamine, C₆H₅NH₂, nitrogen is attached directly to a benzene ring. Its lone pair can be delocalised into the ring, leaving it relatively less available for protonation compared with a typical alkylamine. That helps explain why phenylamine is a weaker base than ethylamine in familiar aqueous comparisons. The tutor should draw the actual N attachment and explain the relevant electron distribution rather than assert ‘all aromatic chemicals are weak bases’.
An amide is not merely an amine with a different name
In an amide such as CH₃CONH₂, the nitrogen lone pair is delocalised toward the adjacent carbonyl group. This reduces ordinary nitrogen basicity compared with a simple amine. An amide also has a characteristic –C(=O)–N– connection, distinct from a primary amine R–NH₂ without the adjacent acyl group. Students should recognise the functional-group structure before choosing an acid–base explanation or hydrolysis reaction.
A three-structure independent practice question
Show ethylamine, phenylamine and ethanamide. Ask the learner to circle the nitrogen lone pair, identify what molecular feature affects its availability, and explain the relative tendency to accept H⁺. Then change ethylamine to a different alkylamine or phenylamine to a ring-substituted analogue where the syllabus supports that comparison. A student who can transfer the electronic reason is better prepared than one who recalls a ranked list without understanding it.
Keep the route ownership precise
This page owns amine versus amide structure and basicity. The acyl chloride and amide reaction guide owns amide formation and hydrolysis; phenol versus ethanol covers a different acidity comparison. For the full H2 progression, use the JC Chemistry hero and Hougang Chemistry hub.
For current course and three-student lesson options, confirm the actual H2 syllabus, teaching venue and availability directly. No Hougang Chemistry teaching location is implied by the local guide.
A tiny electron pair can change an entire chemical class
A student draws NH₃, CH₃CH₂NH₂ and C₆H₅NH₂. All have nitrogen capable of bearing a lone pair, so the student assumes they must have exactly the same basic strength.
The similarities are important. But so are the attachments to nitrogen. A lone pair surrounded by an electron-donating ethyl group is not in quite the same electronic environment as one able to interact with a delocalised aromatic π system.
The purpose of learning relative basicity is to make that structure–property connection. Chemistry becomes more meaningful when the student can explain a difference instead of memorising an ordering as a mysterious fact.
The key is not the nitrogen label alone, but what happens to its electrons when the molecule interacts with a proton and the solvent.
What is an amine?
An amine is an organic nitrogen compound derived in a broad structural sense from ammonia by replacing one or more hydrogen atoms with suitable carbon-containing groups.
A primary amine has one organic group attached to nitrogen, a secondary amine has two, and a tertiary amine has three in the standard classification of neutral amines.
Examples include:
- Ethylamine (ethanamine), CH₃CH₂NH₂: a primary aliphatic amine.
- Dimethylamine, (CH₃)₂NH: a secondary aliphatic amine.
- Trimethylamine, (CH₃)₃N: a tertiary aliphatic amine.
- Phenylamine (aniline), C₆H₅NH₂: a primary aromatic amine, with –NH₂ attached directly to a benzene ring.
Counting the hydrogens in the entire molecule does not determine the amine’s class. Count how many organic groups are attached to nitrogen.
For example, ethylamine is primary even though the ethyl group contains two carbon atoms.
Why amines act as bases
Nitrogen in a typical amine has a lone pair of electrons. That pair can accept H⁺, producing an ammonium-type ion.
For ethylamine:
CH₃CH₂NH₂ + H⁺ ⇌ CH₃CH₂NH₃⁺.
In water the proton-transfer equilibrium can also be represented as:
CH₃CH₂NH₂ + H₂O ⇌ CH₃CH₂NH₃⁺ + OH⁻.
This describes ethylamine as a weak base under ordinary aqueous conditions.
The lone pair also qualifies the amine as a Lewis base, an electron-pair donor. The Brønsted–Lowry viewpoint describes its proton acceptance.
A student who says a base must already contain the OH⁻ group would incorrectly exclude ethylamine. Hydroxide is produced through its interaction with water in the equilibrium shown.
Worked example 1 — ethylamine reacting with hydrochloric acid
Ethylamine accepts a proton from a strong acid such as hydrochloric acid, forming an ethylammonium salt.
A suitable overall equation is:
CH₃CH₂NH₂ + HCl → [CH₃CH₂NH₃]⁺Cl⁻.
The product is ethylammonium chloride. The carbon skeleton remains unchanged; nitrogen has accepted a proton and acquired a positive formal charge.
Do not call this a new amide or suggest that the carbon–nitrogen bond has been destroyed. It is an acid–base reaction involving the lone pair.
It is also not necessary to invent a separate new covalent nitrogen–chlorine bond. The chloride ion is the counter-ion to the positively charged ethylammonium species.
Why ethylamine is more basic than ammonia in water
Ammonia, NH₃, and ethylamine both contain a nitrogen lone pair. But the ethyl group is electron donating by inductive effects relative to hydrogen.
This can increase electron density at nitrogen and favour proton acceptance in the ordinary aqueous school comparison. Consequently ethylamine is generally more basic than ammonia under comparable conditions.
That does not mean the +I effect is the only possible determinant of the basicity of every amine. In aqueous solutions, hydration of the protonated and unprotonated forms, steric effects and molecular structure also matter.
For the specific syllabus comparison ethylamine versus ammonia, the electron-donating ethyl group is an important explanatory feature.
A useful answer explicitly identifies the proton-accepting lone pair and the group influencing its availability.
Why phenylamine is less basic than ammonia
Phenylamine has its amino group attached directly to an aromatic ring. The nitrogen lone pair can interact with the ring’s delocalised π system.
This delocalisation makes the lone pair less available for attachment to H⁺ than the more localised lone pair in ammonia. Phenylamine is therefore a weaker base than ammonia in the familiar aqueous comparison.
The important comparison is not that benzene has “more protons” or that nitrogen has lost its electron pair. The pair still exists in the molecule, but it participates in an electronic system that changes its ability to accept a proton.
This distinction builds directly on the concept of delocalisation learnt earlier in bonding and benzene chemistry.
For the aromatic-ring foundation, use Benzene versus Alkenes: Hougang H2 Chemistry.
Worked example 2 — phenylamine salt formation
Phenylamine can accept a proton from hydrochloric acid:
C₆H₅NH₂ + HCl → [C₆H₅NH₃]⁺Cl⁻.
The product is phenylammonium chloride.
The fact that phenylamine is weaker than ethylamine does not mean phenylamine can never form salts with acids. It means the equilibrium tendency to accept a proton differs under comparable conditions.
This illustrates an exam-friendly principle: the terms stronger base and weaker base describe relative tendencies, not absolute categories of reactive versus inert molecules.
A student should be able to explain why an acid can protonate both ethylamine and phenylamine while their basicities remain different.
What about primary, secondary and tertiary amines?
For gaseous-phase basicity, the electron-donating effects of alkyl groups can influence how strongly an isolated amine accepts a proton. Comparing primary, secondary and tertiary structures therefore involves the number and nature of alkyl substituents and their electronic effects.
For aqueous-phase basicity, solvation is also important. Protonated amines and unprotonated amines interact differently with water, so the simple rule “tertiary is always the strongest base” is not dependable as a universal aqueous ranking.
The 2027 H2 syllabus deliberately distinguishes the gaseous-phase discussion of primary, secondary and tertiary amines as Lewis bases from the specific aqueous comparison of ammonia, ethylamine and phenylamine.
That division is useful for students. It prevents them from applying an oversimplified gas-phase trend to an aqueous calculation with different physical effects.
Why an amide is not simply another amine
An amide has a nitrogen attached to a carbonyl-containing group. A typical primary amide can be written as R–C(=O)–NH₂.
Ethanamide has structural formula CH₃C(=O)NH₂.
The nitrogen lone pair can delocalise towards the neighbouring carbonyl group. This strongly reduces its availability for simple proton acceptance at nitrogen and gives the amide linkage partial double-bond character.
Amides are therefore treated as essentially neutral in the syllabus’s familiar acid–base comparison, unlike the readily protonated amines.
Do not confuse amide with amine simply because both contain the letters “am” and nitrogen. Their carbonyl connectivity is a decisive chemical difference.
Worked example 3 — recognise an amine and an amide
Compound A: CH₃CH₂NH₂.
Compound B: CH₃CONH₂.
Compound A is ethylamine. It has no carbonyl carbon directly attached to nitrogen and acts as a typical weak base in water.
Compound B is ethanamide. Its nitrogen bonds directly to a carbonyl carbon, allowing the lone pair to interact with C=O.
If asked which is more basic at its nitrogen in the simple school comparison, ethylamine is the correct choice. The answer should mention nitrogen lone-pair availability and delocalisation, not just quote a name.
A student who writes “both have NH₂ so both are equally basic” has noticed the shared group but ignored the surrounding chemical structure.
Phenylamine and aqueous bromine: an aromatic-ring reaction
Phenylamine’s –NH₂ group can donate electron density into the benzene ring. This activates the ring towards electrophilic substitution.
In the standard H2 qualitative reaction, phenylamine reacts with aqueous bromine to form 2,4,6-tribromophenylamine, typically observed as a white precipitate as the bromine reagent is decolourised.
A simplified net equation is:
C₆H₅NH₂ + 3Br₂ → C₆H₂Br₃NH₂ + 3HBr.
The product still contains the –NH₂ group attached to the aromatic ring. Three ring hydrogens have been replaced with bromine at the 2-, 4- and 6-positions.
This is related to, but not identical with, the familiar phenol bromination test. Phenol contains –OH while phenylamine contains –NH₂. Their different functional groups must remain correctly identified in every drawing.
Worked example 4 — distinguish phenylamine and ethylamine by structure
Consider a question that provides only the structural formulas C₆H₅NH₂ and CH₃CH₂NH₂, then asks which is the stronger base in aqueous medium.
First identify both as primary amines: nitrogen is attached to one organic group in each.
Next identify their different attached groups. In phenylamine, nitrogen attaches directly to a benzene ring and its lone pair can delocalise into the π system. In ethylamine, the ethyl group has an electron-donating inductive effect.
Therefore ethylamine is more basic than phenylamine in the familiar aqueous school comparison.
The right explanation is not “secondary beats primary”; both of these compounds are primary. The difference comes from their aromatic versus alkyl environment.
Making ethylamine: nitrile reduction
The H2 syllabus gives the reduction of nitriles as an example of forming primary amines.
For instance, ethanenitrile, CH₃CN, can be reduced under an appropriate chemical system to ethylamine, CH₃CH₂NH₂.
The net organic transformation may be represented as:
CH₃CN + 4[H] → CH₃CH₂NH₂.
Suitable reducing systems include LiAlH₄ in the proper controlled conditions or catalytic hydrogenation in specified cases. The exact reagent and conditions should match the question.
The nitrile carbon becomes the carbon bearing the amine group in the reduced product. Students who lose a carbon atom may have failed to track the carbon skeleton.
This is a structural question, not a preparation procedure for unsupervised experimentation.
Making phenylamine: from nitrobenzene
The 2027 H2 syllabus explicitly includes reduction of nitrobenzene to phenylamine using tin and concentrated hydrochloric acid with heating, followed by treatment with aqueous sodium hydroxide.
Conceptually, the nitro group –NO₂ on the aromatic ring becomes –NH₂ after reduction and work-up.
A simplified net reduction representation is:
C₆H₅NO₂ + 6[H] → C₆H₅NH₂ + 2H₂O.
The symbol [H] describes reducing equivalents in the school equation, not an instruction to use isolated free hydrogen atoms.
Under acidic reaction conditions the amine product can initially be protonated; adding base liberates free phenylamine. That is why the sequence includes a work-up step.
An exam question may deliberately provide both reagents. Explain what each contributes rather than treating the two conditions as redundant.
Amide formation: connect the earlier acyl chloride chapter
Ethanoyl chloride can react with an appropriate primary amine to form an amide.
For example:
CH₃COCl + CH₃CH₂NH₂ → CH₃CONHCH₂CH₃ + HCl in the simplified net organic equation.
The product is N-ethylethanamide. Nitrogen is now directly linked to the carbonyl carbon through the characteristic amide group.
In an actual reaction mixture, hydrogen chloride can protonate excess amine, and reaction conditions and stoichiometry matter. The displayed net equation communicates the key bond change.
For more detailed structural practice, see Acyl Chlorides, Esters and Amides: Hougang H2 Guide.
The useful contrast is that the same nitrogen-containing starting amine can be basic and also participate as a nucleophile in forming an amide.
What does amide hydrolysis produce?
An amide can be hydrolysed under appropriate aqueous acidic or alkaline conditions and heating.
For a simple primary amide, acidic hydrolysis ultimately leads to a carboxylic acid together with an ammonium-type nitrogen product in acidic solution. Alkaline hydrolysis gives a carboxylate salt and ammonia for an unsubstituted primary amide under the standard net model.
The specific products depend on whether nitrogen has organic substituents and which work-up conditions are applied.
For ethanamide in alkaline aqueous conditions, a simplified equation is:
CH₃CONH₂ + OH⁻ → CH₃COO⁻ + NH₃.
Check that the atoms and charges balance, and recognise that the carboxylate—not neutral ethanoic acid—is the expected main carboxyl-derived species in the alkaline mixture.
The earlier acid–base principles return here. Organic reaction products cannot be named correctly if the solution’s protonation conditions are ignored.
Amine versus amide: a five-question diagnostic
Give the learner ethylamine, phenylamine and ethanamide structures. Then ask:
- Which nitrogen lone pair is most readily available for proton acceptance in the usual aqueous comparison?
- Which amine’s lone pair can delocalise into a benzene ring?
- Which nitrogen lone pair is delocalised into a neighbouring carbonyl group?
- Which compound is expected to form an ordinary ammonium salt with dilute strong acid?
- Which compound can form a white aromatic tribromo product with aqueous bromine under the familiar syllabus conditions?
The answers connect physical structure, acid–base chemistry and reaction evidence. If one fails, identify the reason before assigning another full chapter of practice.
A common examination trap: basicity and nucleophilicity are related but different
Basicity concerns proton acceptance in the relevant equilibrium setting. Nucleophilicity concerns electron-pair donation towards an electrophilic centre and is strongly affected by reaction context and kinetics.
A compound can act as a base and as a nucleophile in different reactions, but the two labels are not interchangeable measures of strength across all solvents and substrates.
This distinction is useful when ethylamine accepts H⁺ in one question yet reacts with an acyl chloride in another. The same nitrogen lone pair participates, but the reacting partner and product differ.
A strong student identifies which reaction is taking place before applying the vocabulary.
What does SEAB H2 Chemistry 9476 require?
The official 2027 SEAB H2 Chemistry 9476 syllabus includes preparation of ethylamine from reduced amides and nitriles, reduction of nitrobenzene to phenylamine, formation of amine salts, gaseous-phase basicity of primary, secondary and tertiary amines, and relative aqueous basicities of ammonia, ethylamine and phenylamine.
It also includes phenylamine’s reaction with aqueous bromine, amide formation, the relative neutrality of amides due to nitrogen lone-pair delocalisation, amide hydrolysis and reduction. The SEAB 2027 A-Level syllabuses distinguish H2 9476 from H1 8873.
A learner needs the correct depth for the actual subject, not every possible amine pKa table or an entire university course in nitrogen chemistry.
A seven-day Hougang JC Chemistry plan
Day 1: identify primary, secondary and tertiary amines and distinguish amines from amides.
Day 2: draw protonation reactions for ammonia, ethylamine and phenylamine.
Day 3: explain the aqueous basicity comparison by identifying alkyl inductive effects and aromatic lone-pair delocalisation.
Day 4: contrast gaseous and aqueous amine basicity and explain why solvation matters in water.
Day 5: write the phenylamine bromination product and distinguish it from phenol’s familiar bromine test.
Day 6: map ethylamine formation from a nitrile, phenylamine formation from nitrobenzene, and an amide formed from an acyl chloride.
Day 7: solve a mixed unknown-structure question without seeing the worked solution, then explain why the nitrogen environment decides each answer.
The objective is learning continuity: the acid–base model, reaction product and structural explanation should remain consistent across new questions.
How a three-student Chemistry lesson can help
One student may memorise the basicity ranking without understanding the electron pair. Another may understand delocalisation but confuse ethylamine and ethanamide structures. A third may name the right bromination product yet misread the observed precipitate.
A carefully managed three-student tutorial can reveal those different starting positions. The tutor should inspect each learner’s structural reasoning, correct the first wrong move and test an unfamiliar nitrogen compound.
For parent decisions, begin with Hougang JC Chemistry Tuition, Hougang H1/H2 Syllabus Library and Amino Acids and Peptide Bonds. Confirm subject fit, class venue, timetable and availability rather than assuming the Hougang named in a learning guide is a physical teaching location.
Frequently asked questions
Is phenylamine an amide?
No. Phenylamine has –NH₂ attached directly to a benzene ring. An amide has nitrogen attached to a carbonyl carbon within an –C(=O)–N– arrangement.
Why is ethylamine more basic than phenylamine?
The ethyl group donates electron density towards nitrogen, while the phenylamine nitrogen lone pair can delocalise into the aromatic ring and is less available for proton acceptance.
Does phenylamine form salts with strong acids?
Yes. It is a weak base, not a non-base; suitable strong acids can protonate it to give phenylammonium salts.
Why are amides comparatively neutral?
Their nitrogen lone pair is strongly delocalised towards a neighbouring carbonyl group, making it much less available as a base in the usual comparison.
Why does phenylamine react with bromine water?
Its –NH₂ group activates the aromatic ring towards electrophilic substitution. The familiar school result includes white 2,4,6-tribromophenylamine precipitate.
Which topics follow?
Connect amines and amides to Phenol versus Ethanol Acidity, SN1/SN2 Nucleophilic Substitution, and eduKate Sengkang Science Hub. When the student can track one nitrogen lone pair through several contexts, the chapter becomes a coherent system instead of a set of exceptions.