Edexcel A-Level Chemistry: Organic Foundations — Complete Revision Guide (9CH0)

Organic foundations is the first half of the A-Level organic syllabus and the topic where most candidates either gain confidence or lose it. The mechanisms, the homologous series, the curly arrows and the IUPAC names look like a lot to remember, but in fact the entire section reduces to about a dozen reactions and four mechanism types. Once you can name a compound, identify its functional group, and recognise the type of reaction in front of you, the choice of mechanism follows almost automatically.

Organic foundations also sets up everything in organic advanced. Carbonyls are simply alcohol oxidation products. Carboxylic acids are simply oxidised aldehydes. Ester and amide formation use mechanisms that are direct extensions of nucleophilic substitution. So the time you invest here returns multiple times over the course of the qualification.

This guide walks through the organic foundations content in 9CH0 topic by topic. It covers IUPAC nomenclature; structural, E/Z and optical isomerism; the alkanes (combustion and free radical substitution); the alkenes (electrophilic addition and Markovnikov's rule); halogenoalkanes (SN1, SN2, E1, E2 mechanisms); and alcohols (oxidation, dehydration). For each topic you will find the core ideas, common pitfalls, a worked example or mechanism summary, and a link into the LearningBro Organic Foundations course.

What the Edexcel 9CH0 Specification Covers

Edexcel A-Level Chemistry (9CH0) is examined through Paper 1 (Inorganic and Physical, 1h45, 90 marks), Paper 2 (Organic and Physical, 1h45, 90 marks) and Paper 3 (General and Practical, 2h30, 120 marks). Organic foundations is examined predominantly on Paper 2, with practical organic questions on Paper 3.

Sub-topic	Spec area	Typical Paper 2 weight
IUPAC nomenclature	Topic 6	2-4 marks
Structural and E/Z isomerism	Topic 6	3-5 marks
Optical isomerism	Topic 18	3-5 marks
Alkanes: combustion and free radical	Topic 6	3-5 marks
Alkenes: electrophilic addition	Topic 6	4-6 marks
Markovnikov's rule	Topic 6	3-5 marks
Halogenoalkanes: SN1 / SN2	Topic 6	4-6 marks
Halogenoalkanes: E1 / E2 elimination	Topic 6	3-5 marks
Alcohols: oxidation and dehydration	Topic 7	4-6 marks
Reaction mechanism summary	Topics 6, 7, 18	4-8 marks

These weights are estimates, modelled on the Edexcel 9CH0 paper format. What is reliable is that a multi-step organic synthesis question — usually six to ten marks — appears on essentially every Paper 2.

IUPAC Nomenclature

IUPAC names follow a strict structural recipe: identify the longest carbon chain containing the principal functional group, number it to give the lowest locants to substituents and the functional group, name substituents alphabetically with locants, and use the appropriate suffix and prefix conventions for functional groups.

Functional group	Suffix	Example
Alkane	-ane	Propane
Alkene	-ene	Propene
Alcohol	-ol	Propan-1-ol
Halogenoalkane	(prefix halo-)	2-Bromopropane
Aldehyde	-al	Propanal
Ketone	-one	Propan-2-one
Carboxylic acid	-oic acid	Propanoic acid
Ester	-oate	Methyl propanoate
Amine	-amine (or amino-)	Propan-1-amine
Amide	-amide	Propanamide

A common pitfall is to forget to number from the end nearest the functional group; another is to use the wrong locant set when two equally valid numberings exist (use the one giving the lowest set of locants overall). See the nomenclature lesson for naming drills.

Isomerism: Structural, E/Z and Optical

Structural isomers have the same molecular formula but different structural formulae. The three sub-types are chain isomerism (different carbon skeletons), positional isomerism (functional group in a different position), and functional group isomerism (different functional groups, e.g. propan-1-ol and methoxyethane).

E/Z isomerism (geometric isomerism) occurs at C=C double bonds where each carbon of the double bond carries two different groups. The bond cannot rotate, so the two groups on each carbon are locked in either E (entgegen, opposite sides) or Z (zusammen, same side) configurations. Use Cahn-Ingold-Prelog priority rules to assign E or Z: rank the two groups on each carbon by atomic number, and check whether the higher-priority groups are on the same side (Z) or opposite sides (E).

Optical isomerism occurs when a molecule contains a chiral centre — a carbon bonded to four different groups. The two arrangements are non-superimposable mirror images, called enantiomers. Enantiomers rotate plane-polarised light in equal and opposite directions. A 1:1 mixture of enantiomers is a racemate and shows no net rotation.

Worked example. Identify the type of isomerism in (a) butane and 2-methylpropane, (b) but-2-ene cis and trans, (c) 2-bromobutane (single compound). (a) Chain isomers — same molecular formula C4H10, different skeletons. (b) E/Z isomers — same C=C bond connectivity but different geometric arrangement. (c) Optical isomers exist because C2 carries four different groups (H, CH3, C2H5, Br).

A common pitfall is to call any isomer "structural" when E/Z and optical have specific names. Another is to label as chiral a carbon that has two of the same substituent (e.g. a methyl carbon in 2-methylpropane). See the isomerism lesson.

Alkanes: Combustion and Free Radical Substitution

Alkanes (CnH2n+2) are saturated hydrocarbons with only single C-C and C-H bonds. They are unreactive toward most polar reagents because the bonds are non-polar and strong. Their two main reactions are combustion (with oxygen) and free radical substitution (with halogens in UV light).

Complete combustion gives CO2 and H2O. Incomplete combustion (limited O2) gives CO and/or C and H2O. The CO is a toxic product of vehicle exhaust and a cause of poisoning in faulty domestic appliances.

Free radical substitution with chlorine in UV light proceeds in three stages:

• Initiation: Cl2 → 2Cl• (UV homolysis)
• Propagation: Cl• + CH4 → HCl + •CH3, then •CH3 + Cl2 → CH3Cl + Cl•
• Termination: 2Cl• → Cl2, or •CH3 + Cl• → CH3Cl, or 2•CH3 → C2H6

The mechanism produces a mixture of products: CH3Cl, CH2Cl2, CHCl3, CCl4, and longer chains via termination. This makes free radical substitution a poor laboratory synthesis but an important industrial process.

Worked example. Calculate the percentage of methane converted to chloromethane in a typical free-radical substitution if 80 percent forms CH3Cl, 10 percent CH2Cl2, 5 percent CHCl3, 3 percent CCl4 and 2 percent ethane. Just 80 percent — and the rest is mixed products that must be separated. The "lab-poor / industry-significant" nature of free radical substitution is a recurring exam point.

A common pitfall is to draw curly arrows for free radical mechanisms — these mechanisms use single-headed (fish-hook) arrows for one-electron movements. Another is to forget initiation and go straight to propagation. See the alkane lesson.

Alkenes: Electrophilic Addition and Markovnikov's Rule

Alkenes (CnH2n) contain a C=C double bond. The pi bond is a region of high electron density that attracts electrophiles (electron-deficient species). The standard alkene reaction is electrophilic addition.

The mechanism for HBr + propene: (1) the pi electrons attack the δ+ end of the H-Br dipole; (2) the C-H bond forms while the H-Br bond breaks heterolytically, leaving Br-; (3) Br- attacks the carbocation to form the product. Curly arrows show two-electron movements.

Markovnikov's rule says that when an asymmetric reagent (like HBr or H2O) adds across an asymmetric alkene, the H attaches to the C with more H atoms already (so the X attaches to the C with more substituents). This is because the intermediate carbocation must be the most stable possible — and tertiary carbocations are more stable than secondary, which are more stable than primary. The carbocation forms at the more substituted carbon, so the nucleophile ends up there.

Worked example. Predict the major product of HBr + 2-methylpropene. The two possible carbocations are tertiary (2-methylpropan-2-yl carbocation) and primary (2-methylpropyl carbocation). The tertiary is much more stable, so HBr adds to give 2-bromo-2-methylpropane as the major product.

Reagent	Conditions	Product
Br2	Bromine water, room temp	1,2-dibromoalkane (decolourises Br2)
HBr	Room temp	Bromoalkane (Markovnikov)
H2	Ni catalyst, ~150 °C	Alkane
H2O (steam)	Conc H3PO4 catalyst, 300 °C, 60 atm	Alcohol (Markovnikov)
KMnO4 (cold dilute)	Aqueous, alkaline	Diol
O3 then H2O / Zn	Ozonolysis	Two carbonyls

The Br2/water test is the classic alkene identifier — orange-brown bromine water decolourises immediately on shaking with an alkene.

A common pitfall is to predict the anti-Markovnikov product for HBr addition without realising it. Another is to draw double-headed arrows for the wrong bond. See the alkene lesson for full mechanism diagrams.

Halogenoalkanes: SN1, SN2, E1, E2

Halogenoalkanes (CnH2n+1X) contain a polar C-X bond. The δ+ carbon is attacked by nucleophiles (electron-rich species) in substitution reactions, or attacked at the β-hydrogen by bases in elimination reactions. The four possible mechanisms are SN1, SN2, E1, E2.

SN2 (bimolecular substitution) — concerted single-step mechanism where the nucleophile attacks from the back face as the leaving group departs. Rate = k [RX][Nu-]. Inversion of configuration (Walden inversion) at chiral centres. Favoured by primary halogenoalkanes (less steric hindrance) and strong nucleophiles.

SN1 (unimolecular substitution) — two-step mechanism: ionisation to a carbocation, then nucleophilic attack. Rate = k [RX]. Racemisation at chiral centres because the planar carbocation can be attacked from either face. Favoured by tertiary halogenoalkanes (stable carbocation) and polar protic solvents.

E2 (bimolecular elimination) — concerted single-step mechanism where the base removes a β-hydrogen as the leaving group departs and the C=C double bond forms. Rate = k [RX][base]. Anti-periplanar geometry required.

E1 (unimolecular elimination) — two-step mechanism: ionisation to a carbocation, then loss of a β-hydrogen. Rate = k [RX]. Favoured by tertiary halogenoalkanes, weak bases.

Substrate	Substitution	Elimination
Primary (1°)	SN2 dominates	E2 (with strong bulky base)
Secondary (2°)	SN1 / SN2 mixture	E1 / E2 mixture
Tertiary (3°)	SN1 dominates	E1 / E2 dominates with strong base

In practice, the choice between substitution and elimination is also governed by reagent and conditions: aqueous KOH favours substitution (water is the nucleophile); ethanolic KOH favours elimination (ethoxide is a stronger base); the reflux temperature also pushes toward elimination.

Worked example. Predict the major product of 2-bromo-2-methylpropane heated with ethanolic KOH. Tertiary substrate plus strong base in alcoholic solvent — E1 or E2 elimination dominates. Product: 2-methylpropene + KBr + ethanol.

A common pitfall is to draw the SN1 mechanism with the carbocation and nucleophile in the same step — they are separate steps. Another is to forget that SN1 produces racemisation while SN2 produces inversion. See the halogenoalkane lesson for full mechanism diagrams.

Alcohols: Oxidation and Dehydration

Alcohols (R-OH) classify as primary, secondary or tertiary based on the number of carbons attached to the carbinol carbon. Oxidation behaviour depends on classification.

• Primary alcohols → aldehydes (distil out as soon as formed) → carboxylic acids (reflux with excess oxidising agent).
• Secondary alcohols → ketones (further oxidation requires C-C bond breaking, which acidified dichromate cannot achieve under standard conditions).
• Tertiary alcohols → no oxidation under standard conditions.

Standard oxidising agents are acidified potassium dichromate (K2Cr2O7 / H2SO4), which turns from orange (Cr2O7^2-) to green (Cr^3+) on reaction. Acidified KMnO4 also works but is less commonly used because it is too aggressive.

Worked example. Distinguish between propan-1-ol, propan-2-ol and 2-methylpropan-2-ol using acidified dichromate and Tollens' reagent. (a) All three: dichromate goes orange → green for the first two, no change for the tertiary. (b) Tollens' on the dichromate-oxidised products: the propan-1-ol product (propanal, an aldehyde) gives a silver mirror; the propan-2-ol product (propanone, a ketone) does not.

Dehydration of alcohols (loss of water to give an alkene) uses concentrated sulfuric or phosphoric acid as catalyst. The mechanism is E1: protonation of the OH group, loss of water to give a carbocation, then loss of a β-hydrogen to form the alkene. Tertiary alcohols dehydrate most readily; primary alcohols require more forcing conditions.

A common pitfall is to over-oxidise primary alcohols and miss the aldehyde step (you must distil out as soon as formed); another is to write a tertiary alcohol oxidation product. See the alcohols lesson.

Reaction Mechanism Summary

By the end of organic foundations you should have a small library of mechanisms automated. Drawing each one quickly and accurately is the single highest-leverage skill on Paper 2.

Mechanism	Substrate	Reagent	Product	Arrows
Free radical substitution	Alkane	Cl2 / Br2, UV	Halogenoalkane	Single-headed (fish-hook)
Electrophilic addition	Alkene	HBr, Br2, H2O/H+	Saturated product	Curly (two-electron)
SN2 nucleophilic substitution	Primary RX	OH-, CN-, NH3	Alcohol/nitrile/amine	Curly, concerted
SN1 nucleophilic substitution	Tertiary RX	H2O, ethanol	Alcohol/ether	Curly, two steps
E2 elimination	Secondary/tertiary RX	KOH/ethanol, hot	Alkene	Curly, concerted
E1 elimination	Tertiary RX	Heat, weak base	Alkene	Curly, two steps
Acid-catalysed dehydration	Alcohol	Conc H2SO4/H3PO4, heat	Alkene	Curly, E1-like

A common pitfall is to use the wrong arrow type — single-headed arrows for radical mechanisms only, double-headed for everything else. See the mechanism summary lesson.

Common Mark-Loss Patterns

• Forgetting to number from the end nearest the functional group in IUPAC names.
• Failing to assign E/Z by priority rules; assuming "cis" and "trans" still work in all cases.
• Calling a non-chiral carbon a chiral centre.
• Drawing curly arrows for free radical mechanisms.
• Predicting anti-Markovnikov products for HBr addition.
• Confusing SN1 and SN2 mechanism diagrams (one step vs two).
• Over-oxidising primary alcohols when the question wants the aldehyde stage.
• Drawing the dehydration mechanism without the protonation first step.
• Forgetting curly arrow direction (always from electron source to electron sink).
• Drawing dot-and-cross diagrams when curly arrows are wanted.

How to Revise This Topic

• Practise IUPAC names every day for two weeks. Twenty names a day. By the end you should not need to count.
• Build a mechanism deck. One mechanism per card, drawn cleanly with all arrows. Drill until automatic.
• Drill Markovnikov problems by predicting both possible carbocations for any HX + alkene and choosing the more stable.
• Practise the substrate-mechanism table until you can predict SN1/SN2/E1/E2 from substrate + reagent in five seconds.
• Use the LearningBro practice quizzes to test under timed conditions.

Linking to Other Topics

Organic foundations is heavily synoptic. Bond polarity from bonding explains why electrophiles attack pi bonds and nucleophiles attack δ+ carbons. Reaction profiles from kinetics and equilibrium underpin SN1 versus SN2 and Markovnikov reasoning. The organic advanced topic builds directly on these mechanisms — carbonyl chemistry uses electrophilic addition; ester formation uses nucleophilic attack on a δ+ carbon; condensation polymerisation uses the same mechanisms repeatedly. The analytical chemistry topic uses NMR and IR to identify the products of these reactions.

Final Word

Organic foundations is the gateway to the second half of A-Level Chemistry. Once you can name compounds, recognise functional groups and draw the standard mechanisms cleanly, the rest of organic becomes a matter of applying these tools. The full LearningBro Organic Foundations course walks through every reaction with worked examples, mechanism drills and AI tutor feedback. Get this section fluent and Paper 2 becomes much shorter.