Organic Chemistry

Learn and apply Organic Chemistry in the published Chemistry course sequence.

GCE A-Level H2 Chemistry 9476-2027

Learning goals

Representations and Nomenclature (A Level Organic)
Isomerism and Stereochemistry (A Level Organic)
Alkanes and Free-radical Substitution
Alkenes and Electrophilic Addition
Halogenoalkanes: SN1/SN2 and Elimination
Alcohols: Oxidation and Dehydration
Organic Mechanisms: Curly Arrows, Electrophiles, Nucleophiles
Arenes and Electrophilic Substitution
Carbonyls: Nucleophilic Addition and Tests
Carboxylic Acids and Derivatives
Amines, Amides, Amino Acids
Polymers: Addition and Condensation

What You’ll Learn

Read structures, name compounds and distinguish constitutional, geometrical and optical isomers.
Use electron movement, reactive species and electronic effects to explain organic mechanisms.
Predict reactions from the functional group, reagent and essential conditions rather than from isolated memory cues.
Join reactions into synthesis routes and use qualitative evidence without claiming more than the observations support.

This route is the H2 Chemistry (9476) organic sequence. H1 Chemistry (8873) has separately composed lessons that retain its specified reactions without importing H2 curly-arrow mechanisms, detailed stereochemistry or the complete H2 functional-group range. H3 Chemistry (9813) is a separate syllabus, not a successor course.

Prerequisites

Chemical Bonding: electronegativity, bond polarity, sigma and pi bonds, orbital overlap and intermolecular forces.
Acids and Bases: electron-pair donor/acceptor language and acid–base equilibria.
Energetics: bond-energy and stability comparisons.
Reaction Kinetics: activation energy, rate equations and mechanistic evidence.

Before synthesis work, be able to count carbon atoms from skeletal formulae, conserve charge and atoms, and identify the functional group that actually reacts.

Lessons (Recommended Order)

Foundations and mechanism language

Representations and Nomenclature — translate formulae and name the prescribed classes.
Isomerism and Stereochemistry — distinguish constitutional, cis–trans and optical isomerism.
Mechanisms and Curly Arrows — build the electron-flow toolkit used throughout the course.

Hydrocarbons and core transformations

Carbonyl, carboxylic, nitrogen and polymer chemistry

Follow the order on a first pass. During revision, enter at the lesson matching the reaction family or evidence gap in your last question.

Quick Reference

Evidence or condition	Reaction family	First reasoning move
UV light with a halogen	free-radical substitution	use homolytic fission and single-headed arrows
electron-rich C = C with Br₂ or HX	electrophilic addition	polarise the reagent and form the more stable intermediate where relevant
benzene with Cl₂/AlCl₃ or Br₂/AlBr₃	electrophilic substitution	generate a stronger electrophile and restore aromaticity
warm aqueous nucleophile with a halogenoalkane	nucleophilic substitution	compare substrate, nucleophile and solvent for SN1/SN2
hot ethanolic base with a halogenoalkane	elimination	remove H⁺ and the leaving group to form C = C
aldehyde or ketone with HCN	nucleophilic addition	attack the δ + carbonyl carbon
acyl chloride with water, alcohol or ammonia	nucleophilic acyl substitution	addition is followed by elimination

Use this sequence in every unfamiliar problem: functional group → reagent role → essential condition → reaction family → product → evidence or mechanism.

Common Exam Traps

Drawing a curly arrow from an electrophile or positive charge instead of from an electron pair.
Treating every bromine reaction alike: alkene addition, arene ring substitution and methylbenzene side-chain substitution require different conditions and mechanisms.
Giving a reagent but omitting the solvent, temperature, catalyst or distillation/reflux choice that determines the product.
Calling 2,4-DNPH an aldehyde test; it detects aldehydes and ketones.
Forcing one mechanism label from the substrate alone when solvent, nucleophile and kinetic evidence also matter.
Showing the right product with impossible charges, missing atoms or an unregenerated catalyst.
Memorising a route in only one direction and then failing a synthesis question that begins from the product.
Importing H2 mechanism requirements into H1 8873.

After drawing any mechanism, audit four things: electron source, electron destination, charge conservation and regeneration of any catalyst.

Hub Quiz and Structured Practice

Start with the Organic Chemistry hub quiz for retrieval across the full sequence. Then use structured Organic Chemistry practice for mechanisms, deductions and multi-step synthesis.

For each error, record whether it was a structure error, condition error, electron-flow error or evidence error. That classification tells you which lesson to revisit.

Practise

Work through questions with marking and feedback as you learn.

About 10 minutes

Questions are picked at random each time you start. You'll see the answer after each question. It's for practice only and doesn't count towards mastery.

Recent attempts

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This check needs JavaScript to record answers. The reviewed questions are listed below for study.

Which option names all five compounds correctly and classifies the alcohol and the carbonyl compound correctly? (i) CH₂=C(CH₃)CH₂CH₃ (ii) CH₃CBr(CH₃)CH₂CH₃ (iii) CH₃CH(OH)CH₂CH₃ (iv) CH₃CH₂CH₂CHO (v) CH₃CH₂CH(CH₃)COOH

(i) 2-methylbut-1-ene; (ii) 2-bromo-2-methylbutane; (iii) butan-2-ol, a secondary alcohol; (iv) butanal, an aldehyde; (v) 2-methylbutanoic acid
(i) 3-methylbut-3-ene; (ii) 2-bromo-2-methylbutane; (iii) butan-2-ol, a secondary alcohol; (iv) butanal, an aldehyde; (v) 2-methylbutanoic acid
(i) 2-methylbut-1-ene; (ii) 2-bromo-2-methylbutane; (iii) butan-2-ol, a primary alcohol; (iv) butanone, a ketone; (v) 2-methylbutanoic acid
(i) 2-methylbut-1-ene; (ii) 3-bromo-3-methylbutane; (iii) butan-2-ol, a secondary alcohol; (iv) butanal, an aldehyde; (v) 3-methylbutanoic acid

Which option names all four compounds correctly? (i) CH₃CH₂COOCH₃ (ii) CH₃CH₂CH₂CH₂NH₂ (iii) CH₃CH₂CH₂CONH₂ (iv) H₂NCH₂COOH

(i) methyl propanoate; (ii) butylamine; (iii) butanamide; (iv) aminoethanoic acid
(i) propyl methanoate; (ii) butylamine; (iii) butanamide; (iv) aminoethanoic acid
(i) methyl propanoate; (ii) butylamine; (iii) butanamide; (iv) 2-aminopropanoic acid
(i) methyl propanoate; (ii) butylamine; (iii) propanamide; (iv) aminoethanoic acid

Which statements are correct? (1) Ethene reacts with hydrogen over a nickel catalyst in an addition reaction to form ethane. (2) Ethane and chlorine react rapidly by substitution in the dark at room temperature. (3) Heating bromoethane with NaOH(aq) replaces Br by OH in a substitution reaction, forming ethanol.

(1) and (3) only
(1), (2) and (3)
(3) only
(1) and (2) only

Which statements are correct? (1) Ethanol burns completely in oxygen to form carbon dioxide and water. (2) Ethanol heated with concentrated H₃PO₄ undergoes elimination of water to form ethene. (3) Propanone is oxidised to propanoic acid when heated with acidified K₂Cr₂O₇. (4) Ethanal is reduced to ethanol by LiAlH₄ in dry ether.

(1), (2) and (4) only
(1), (2), (3) and (4)
(2) and (4) only
(1) and (3) only

Which statements are correct? (1) Poly(propene) is a condensation polymer, because each propene molecule loses a hydrogen atom as it joins the chain. (2) A polyester formed from ethane-1,2-diol and hexanedioic acid is a condensation polymer, because water is eliminated as each ester link forms. (3) The dipeptide H₂NCH₂CONHCH(CH₃)COOH contains one peptide bond, formed between two α-amino acids. (4) Heating this dipeptide with NaOH(aq) hydrolyses the peptide bond, giving H₂NCH₂COO⁻Na⁺ and H₂NCH(CH₃)COO⁻Na⁺.

(2), (3) and (4) only
(1), (2), (3) and (4)
(2) and (3) only
(4) only

Which statements are correct? (1) Benzene is a planar hexagonal molecule with C–C–C bond angles of 120°, and its π electrons are delocalised above and below the ring. (2) In propene, CH₂=CHCH₃, every bond angle is about 120°, because the molecule contains a C=C bond. (3) 1,2-dichloroethene shows cis–trans isomerism but 1,1-dichloroethene does not, because in 1,1-dichloroethene one carbon of the C=C bond carries two identical atoms. (4) C₃H₈O has three constitutional isomers: propan-1-ol, propan-2-ol and methoxyethane.

(1), (3) and (4) only
(1), (2), (3) and (4)
(1) and (3) only
(4) only

Which of these compounds has a chiral centre?

CH₃CHClCH₂CH₃
CH₃CH₂CH₂CH₂Cl
(CH₃)₂CHCH₂Cl
(CH₃)₃CCl

Which option classifies all four species correctly? NH₃, NO₂⁺, HO• and CN⁻

NH₃: nucleophile; NO₂⁺: electrophile; HO•: free radical, formed by homolytic fission; CN⁻: nucleophile
NH₃: electrophile, because it has three δ+ H atoms; NO₂⁺: electrophile; HO•: free radical; CN⁻: nucleophile
NH₃: nucleophile; NO₂⁺: electrophile; HO•: formed by heterolytic fission; CN⁻: nucleophile
NH₃: nucleophile; NO₂⁺: nucleophile, because its O atoms have lone pairs; HO•: free radical; CN⁻: nucleophile

Methylbenzene is treated separately with (P) hot acidified KMnO₄ and (Q) Cl₂ with an AlCl₃ catalyst in the dark. Which option gives the main organic products?

P: benzoic acid, because the side chain is oxidised and the ring is unchanged; Q: a mixture of 1-chloro-2-methylbenzene and 1-chloro-4-methylbenzene, by electrophilic substitution of the ring.
P: no reaction, because the benzene ring resists oxidation; Q: a mixture of 1-chloro-2-methylbenzene and 1-chloro-4-methylbenzene.
P: benzoic acid; Q: 1-chloro-3-methylbenzene.
P: benzoic acid; Q: (chloromethyl)benzene, C₆H₅CH₂Cl.

Propanal, benzaldehyde and propanone are warmed separately with Fehling’s solution. Which compounds give a red-brown precipitate?

Propanal only
Propanal and benzaldehyde
Propanal, benzaldehyde and propanone
Propanal and propanone

Which statements are correct? (1) Ethylamine is a stronger base than ammonia, because the electron-donating ethyl group makes the N lone pair more available. (2) Ethanamide is a weaker base than ammonia, because its N lone pair is delocalised into the C=O group. (3) Propanoic acid is a stronger acid than 2-chloropropanoic acid.

(1) and (2) only
(1), (2) and (3)
(2) only
(1) only

Which statements are correct? (1) Poly(propene) waste persists in landfill because its non-polar C–C and C–H bonds are chemically inert. (2) Polyesters cannot biodegrade, because their ester links cannot be broken by water. (3) Plastics are made mainly from crude oil, which is a finite resource.

(1) and (3) only
(1), (2) and (3)
(3) only
(1) only

Practise after feedback

After a check, practise the skills it showed you need to work on.

About 10 minutes

Questions are picked at random each time you start. You'll see the answer after each question. It's for practice only and doesn't count towards mastery.

Recent attempts

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This check needs JavaScript to record answers. The reviewed questions are listed below for study.

(i) 2-methylbut-1-ene; (ii) 2-bromo-2-methylbutane; (iii) butan-2-ol, a secondary alcohol; (iv) butanal, an aldehyde; (v) 2-methylbutanoic acid
(i) 3-methylbut-3-ene; (ii) 2-bromo-2-methylbutane; (iii) butan-2-ol, a secondary alcohol; (iv) butanal, an aldehyde; (v) 2-methylbutanoic acid
(i) 2-methylbut-1-ene; (ii) 2-bromo-2-methylbutane; (iii) butan-2-ol, a primary alcohol; (iv) butanone, a ketone; (v) 2-methylbutanoic acid
(i) 2-methylbut-1-ene; (ii) 3-bromo-3-methylbutane; (iii) butan-2-ol, a secondary alcohol; (iv) butanal, an aldehyde; (v) 3-methylbutanoic acid

Which option names all four compounds correctly? (i) CH₃CH₂COOCH₃ (ii) CH₃CH₂CH₂CH₂NH₂ (iii) CH₃CH₂CH₂CONH₂ (iv) H₂NCH₂COOH

(i) methyl propanoate; (ii) butylamine; (iii) butanamide; (iv) aminoethanoic acid
(i) propyl methanoate; (ii) butylamine; (iii) butanamide; (iv) aminoethanoic acid
(i) methyl propanoate; (ii) butylamine; (iii) butanamide; (iv) 2-aminopropanoic acid
(i) methyl propanoate; (ii) butylamine; (iii) propanamide; (iv) aminoethanoic acid

(1) and (3) only
(1), (2) and (3)
(3) only
(1) and (2) only

(1), (2) and (4) only
(1), (2), (3) and (4)
(2) and (4) only
(1) and (3) only

(2), (3) and (4) only
(1), (2), (3) and (4)
(2) and (3) only
(4) only

(1), (3) and (4) only
(1), (2), (3) and (4)
(1) and (3) only
(4) only

Which of these compounds has a chiral centre?

CH₃CHClCH₂CH₃
CH₃CH₂CH₂CH₂Cl
(CH₃)₂CHCH₂Cl
(CH₃)₃CCl

Which option classifies all four species correctly? NH₃, NO₂⁺, HO• and CN⁻

NH₃: nucleophile; NO₂⁺: electrophile; HO•: free radical, formed by homolytic fission; CN⁻: nucleophile
NH₃: electrophile, because it has three δ+ H atoms; NO₂⁺: electrophile; HO•: free radical; CN⁻: nucleophile
NH₃: nucleophile; NO₂⁺: electrophile; HO•: formed by heterolytic fission; CN⁻: nucleophile
NH₃: nucleophile; NO₂⁺: nucleophile, because its O atoms have lone pairs; HO•: free radical; CN⁻: nucleophile

Methylbenzene is treated separately with (P) hot acidified KMnO₄ and (Q) Cl₂ with an AlCl₃ catalyst in the dark. Which option gives the main organic products?

P: benzoic acid, because the side chain is oxidised and the ring is unchanged; Q: a mixture of 1-chloro-2-methylbenzene and 1-chloro-4-methylbenzene, by electrophilic substitution of the ring.
P: no reaction, because the benzene ring resists oxidation; Q: a mixture of 1-chloro-2-methylbenzene and 1-chloro-4-methylbenzene.
P: benzoic acid; Q: 1-chloro-3-methylbenzene.
P: benzoic acid; Q: (chloromethyl)benzene, C₆H₅CH₂Cl.

Propanal, benzaldehyde and propanone are warmed separately with Fehling’s solution. Which compounds give a red-brown precipitate?

Propanal only
Propanal and benzaldehyde
Propanal, benzaldehyde and propanone
Propanal and propanone

(1) and (2) only
(1), (2) and (3)
(2) only
(1) only

(1) and (3) only
(1), (2) and (3)
(3) only
(1) only

Check what I know

Start here to see which parts you already know.

About 8 minutes

Answer 10 short questions. It shows what to work on next and doesn't count towards mastery.

Recent attempts

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This check needs JavaScript to record answers. The reviewed questions are listed below for study.

Which option names all five compounds correctly and classifies the alcohol and the carbonyl compound correctly? (i) CH₃CH₂CH=CHCH₃ (ii) CH₃CHClCH₂CH₃ (iii) (CH₃)₃COH (iv) CH₃CH₂COCH₂CH₃ (v) CH₃CH(CH₃)CH₂COOH

(i) pent-2-ene; (ii) 2-chlorobutane; (iii) 2-methylpropan-2-ol, a tertiary alcohol; (iv) pentan-3-one, a ketone; (v) 3-methylbutanoic acid
(i) pent-3-ene; (ii) 2-chlorobutane; (iii) 2-methylpropan-2-ol, a tertiary alcohol; (iv) pentan-3-one, a ketone; (v) 3-methylbutanoic acid
(i) pent-2-ene; (ii) 2-chlorobutane; (iii) 2-methylpropan-2-ol, a secondary alcohol; (iv) pentan-3-one, an aldehyde; (v) 3-methylbutanoic acid
(i) pent-2-ene; (ii) 3-chlorobutane; (iii) 2-methylpropan-2-ol, a tertiary alcohol; (iv) pentan-3-one, a ketone; (v) 2-methylbutanoic acid

Which option names all four compounds correctly? (i) CH₃COOCH₂CH₂CH₃ (ii) CH₃CH₂CH₂NH₂ (iii) CH₃CH₂CONH₂ (iv) H₂NCH(CH₃)COOH

(i) propyl ethanoate; (ii) propylamine; (iii) propanamide; (iv) 2-aminopropanoic acid
(i) ethyl propanoate; (ii) propylamine; (iii) propanamide; (iv) 2-aminopropanoic acid
(i) propyl ethanoate; (ii) propanamide; (iii) propylamine; (iv) 2-aminopropanoic acid
(i) propyl ethanoate; (ii) propylamine; (iii) propanamide; (iv) 3-aminopropanoic acid

Which statements are correct? (1) Ethane reacts with chlorine in ultraviolet light at room temperature in a substitution reaction, forming chloroethane and hydrogen chloride. (2) Ethene reacts with bromine dissolved in tetrachloromethane in an addition reaction, forming 1,2-dibromoethane. (3) Bromoethane heated with sodium hydroxide dissolved in ethanol undergoes substitution, forming ethanol.

(1) and (2) only
(1), (2) and (3)
(2) only
(1) and (3) only

Which statements are correct? (1) Ethanol heated under reflux with excess acidified K₂Cr₂O₇ is oxidised to ethanoic acid, and the solution turns from orange to green. (2) Propanone is reduced by LiAlH₄ in dry ether to propan-1-ol. (3) Ethanal is oxidised to ethanoic acid by hot acidified KMnO₄, and it is reduced to ethanol by hydrogen with a nickel catalyst.

(1) and (3) only
(1), (2) and (3)
(3) only
(1) and (2) only

Which statements are correct? (1) Poly(ethene) is an addition polymer: each molecule is a macromolecule of thousands of repeat units formed from ethene with no other product. (2) The polymer formed from hexane-1,6-diamine and hexanedioic acid is an addition polymer, because its monomers join end to end into long chains. (3) Proteins are condensation polymers of α-amino acids joined by peptide (amide) bonds. (4) Heating a protein with aqueous acid hydrolyses its peptide bonds, releasing its amino acids (as their protonated forms).

(1), (3) and (4) only
(1), (2), (3) and (4)
(1) and (3) only
(4) only

Which statements are correct? (1) In but-2-ene, the bond angles around each carbon atom of the C=C bond are about 120°, and those two carbons and the four atoms bonded to them lie in one plane. (2) Cis- and trans-but-2-ene exist because the π bond, formed by sideways overlap of p orbitals above and below the plane, prevents rotation about the C=C bond. (3) C₄H₈ has four alkene isomers: but-1-ene, cis-but-2-ene, trans-but-2-ene and 2-methylpropene; but-1-ene, but-2-ene and 2-methylpropene are constitutional isomers, with different connectivity. (4) Each carbon atom in ethane has its four σ bonds arranged at 90° to one another in a plane.

(1), (2) and (3) only
(1), (2), (3) and (4)
(2) and (3) only
(1) only

Propene reacts with hydrogen bromide. Which statement describes the first step of the mechanism correctly?

A curly arrow goes from the C=C π bond to the δ+ H of H–Br and another from the H–Br bond to Br; the bond breaks heterolytically, forming Br⁻ and the secondary carbocation CH₃C⁺HCH₃.
Br⁻ acts as a nucleophile and attacks a carbon atom of the C=C bond first.
The H–Br bond breaks homolytically into H• and Br• radicals, which add to the C=C bond.
The π electrons attack H to form the primary carbocation CH₃CH₂CH₂⁺, which is more stable because the charge is at the end of the chain.

Benzene is nitrated with a mixture of concentrated nitric and sulfuric acids, keeping the mixture at about 50 °C. Which statement is correct?

H₂SO₄ protonates HNO₃ to generate the electrophile NO₂⁺; benzene undergoes electrophilic substitution, so the delocalised π system is restored in nitrobenzene.
NO₂⁺ adds across one C=C bond of benzene, in the same way that bromine adds to ethene.
The nucleophile NO₃⁻ attacks the benzene ring.
H₂SO₄ is used up because it supplies the nitro group.

Butanal and butanone are tested separately with 2,4-dinitrophenylhydrazine (2,4-DNPH), with warm Tollens’ reagent and with warm alkaline aqueous iodine. Which set of results is correct?

2,4-DNPH: orange precipitate with both; Tollens’ reagent: silver mirror with butanal only; alkaline iodine: yellow precipitate with butanone only.
2,4-DNPH: orange precipitate with butanal only; Tollens’ reagent: silver mirror with butanal only; alkaline iodine: yellow precipitate with butanone only.
2,4-DNPH: orange precipitate with both; Tollens’ reagent: silver mirror with both; alkaline iodine: yellow precipitate with butanone only.
2,4-DNPH: orange precipitate with both; Tollens’ reagent: silver mirror with butanal only; alkaline iodine: yellow precipitate with butanal only.

Which option correctly identifies the stronger acid of ethanoic acid and chloroethanoic acid, and the stronger base of ethylamine and ethanamide, with correct reasons?

Chloroethanoic acid, because electron-withdrawing Cl disperses the negative charge of its carboxylate ion; ethylamine, because in ethanamide the N lone pair is delocalised into the C=O group.
Ethanoic acid, because the Cl atom makes the O–H bond harder to break; ethylamine, because the amide lone pair is delocalised.
Chloroethanoic acid, because Cl disperses the carboxylate charge; ethanamide, because its C=O group donates electrons to N.
Chloroethanoic acid, because Cl disperses the carboxylate charge; the two bases are equally strong, because both contain an –NH₂ group.

Check my progress

When you feel ready, answer on your own to show what you can do.

About 10 minutes

Answer 12 questions. You'll see your score, the answers and explanations at the end. Your result can count towards your course progress.

Recent attempts

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Check again

After practising what your progress check showed, check those skills again.

About 10 minutes

Answer 12 questions. You'll see your score, the answers and explanations at the end. Your result can count towards your course progress.

Recent attempts

History is stored only in this browser.

No completed attempts are saved yet.

Review

Come back later to see whether your learning has lasted.

About 10 minutes

Answer 12 questions. You'll see your score, the answers and explanations at the end. A scheduled review counts towards your course progress only when it is due.

Recent attempts

History is stored only in this browser.

No completed attempts are saved yet.

Organic revision loop

For an efficient organic revision loop:

redraw one reaction map from memory;
explain one mechanism aloud from electron source to destination;
solve one unfamiliar synthesis or deduction;
check reagents, conditions, observations and stereochemical consequences separately.

Use Practical and Qualitative Analysis to connect functional-group reactions with the supplied Paper 4 evidence tables.

Organic Chemistry

What You’ll Learn

Prerequisites

Quick Reference

Hub Quiz and Structured Practice

Practise

Practice complete

Your answers

Practise after feedback

Practice complete

Your answers

Check what I know

Starting check complete

Your answers

Check my progress

Progress check complete

Your answers

Check again

Progress check complete

Your answers

Review

Review complete

Your answers

Organic revision loop