AQA A-Level Chemistry: Introduction to organic chemistry (7405)

This guide teaches section 3.3.1 Introduction to organic chemistry of the AQA AS and A-level Chemistry specification (7404/7405), version 1.2, July 2026, for AS and A-level exams June 2016 onwards. It covers 3.3.1.1 Nomenclature, 3.3.1.2 Reaction mechanisms and 3.3.1.3 Isomerism. All of it is AS content, so nothing here is A-level only, and the section lists no required practical. It is examined on Paper 2 at AS and A-level, and A-level Paper 3 can test any content.

Links to the revision notes, practice questions, course hub, checklist and free 10-minute diagnostics are at the end of this guide.

What this section covers

Specification point	What you must be able to do	AS / A-level
3.3.1.1 Nomenclature	Six formula types; homologous series; draw structural, displayed and skeletal formulas; IUPAC names to and from structures (chains and rings up to six carbons each)	AS
3.3.1.2 Reaction mechanisms	Radical dot notation; balanced equations for free-radical steps; curly arrows for electron-pair movement	AS
3.3.1.3 Isomerism	Define structural isomer and stereoisomer; chain, position and functional group isomers; E–Z isomers; Cahn–Ingold–Prelog (CIP) priority rules	AS

3.3.1.1 Representing organic compounds

Organic compounds can be written in six ways.

Formula type	What it shows	Butan-1-ol
Empirical	Simplest whole-number ratio of atoms of each element	C₄H₁₀O
Molecular	Actual number of atoms of each element in one molecule	C₄H₁₀O
General	Algebraic formula for every member of a homologous series	CₙH₂ₙ₊₁OH
Structural	Arrangement of atoms, carbon by carbon, without showing every bond	CH₃CH₂CH₂CH₂OH
Displayed	Every atom and every bond	see below
Skeletal	Carbon skeleton as lines; functional groups shown	see below

For butane, C₄H₁₀, the empirical formula differs from the molecular formula: it is C₂H₅.

A displayed formula shows every bond as a line, including the O–H bond:

    H   H   H   H
    |   |   |   |
H - C - C - C - C - O - H
    |   |   |   |
    H   H   H   H

A skeletal formula follows three rules:

1. Each line is a bond; each end and each corner is a carbon atom unless another atom is written there.
2. Hydrogen atoms on carbon are not drawn (each carbon makes four bonds).
3. Every other atom is drawn, with any hydrogen on it (so –OH is written in full).

So butan-1-ol is a zigzag of three C–C lines plus a fourth line from the end carbon to OH, and a plain hexagon is cyclohexane, C₆H₁₂.

Worked example 1: from percentage composition to formulas

A hydrocarbon contains 85.7% carbon by mass. Its Mr is 56.0. Find its empirical, molecular and general formula. (Ar: C = 12.0, H = 1.0.)

Mass of H in 100 g:   100 − 85.7 = 14.3 g
Moles:                C = 85.7 / 12.0 = 7.14     H = 14.3 / 1.0 = 14.3
Ratio (÷ 7.14):       C = 1.00                   H = 2.00
Empirical formula:    CH₂   (empirical formula mass = 14.0)
Number of CH₂ units:  56.0 / 14.0 = 4
Molecular formula:    C₄H₈

C₄H₈ fits CₙH₂ₙ, shared by alkenes and cycloalkanes, so the data cannot tell you which family it is in (see Worked example 6). For the amount-of-substance background, see the mole revision notes.

Homologous series

A homologous series is a series of compounds containing the same functional group. Its members:

• have the same general formula
• differ from the next member by CH₂
• have similar chemical properties, because they have the same functional group
• show a gradual trend in physical properties, such as boiling point rising as the chain gets longer.

The functional group is the atom or group of atoms that gives the series its typical reactions.

3.3.1.1 IUPAC naming

You must name compounds and draw them from names for chains and rings of up to six carbon atoms each.

Carbons	1	2	3	4	5	6
Stem	meth-	eth-	prop-	but-	pent-	hex-

Family	Functional group	Name ending or prefix	Example
Alkane	C–C only	-ane	propane
Cycloalkane	ring of C	cyclo- … -ane	cyclopentane
Alkene	C=C	-ene	but-1-ene
Halogenoalkane	–F, –Cl, –Br, –I	fluoro-, chloro-, bromo-, iodo-	2-chloropropane
Alcohol	–OH	-ol	ethanol
Aldehyde	–CHO	-al	propanal
Ketone	C=O inside the chain	-one	butanone
Carboxylic acid	–COOH	-oic acid	ethanoic acid
Nitrile	–C≡N	-nitrile	propanenitrile

These are the families met in AS organic chemistry (3.3.1 to 3.3.6). Other families come later.

Method in steps

1. Find the longest carbon chain that contains the functional group. This gives the stem.
2. Number it so the main functional group gets the lowest number (for an alkane, the substituents do).
3. In aldehydes, carboxylic acids and nitriles the functional-group carbon is carbon 1, so no number is needed (propanal, not propan-1-al).
4. Add side chains and halogens as prefixes in alphabetical order, using di-, tri- for repeats (2,2-dimethyl).
5. Commas go between numbers; hyphens go between numbers and words.

When a molecule has both an –OH group and a C=C, the –OH takes the suffix and gets the lowest number: CH₂=CHCH₂CH₂OH is but-3-en-1-ol.

Worked example 2: naming from a structural formula

Name CH₃CH₂C(CH₃)₂CH(OH)CH₃.

Longest chain with the C–OH:   CH₃–CH₂–C–CH(OH)–CH₃  → 5 carbons → pentan-…-ol
Number from the right (OH end): C1 = CH₃, C2 = CH(OH), C3 = C(CH₃)₂, C4 = CH₂, C5 = CH₃
Suffix:                        -2-ol
Side chains:                   two methyl groups, both on C3 → 3,3-dimethyl
Name:                          3,3-dimethylpentan-2-ol

Worked example 3: drawing from a name

Draw the structural formula of 4-methylpentan-2-one.

pentan → 5-carbon chain:           C–C–C–C–C
-2-one → C=O on carbon 2:           CH₃–CO–C–C–C
4-methyl → CH₃ on carbon 4:         CH₃COCH₂CH(CH₃)CH₃
Check: C₆H₁₂O, every carbon has four bonds.

3.3.1.2 Reaction mechanisms

A mechanism shows how a reaction happens step by step. The specification sets two sets of conventions.

Free-radical mechanisms

• A radical has an unpaired electron. You show it with a dot: Cl•, CH₃•.
• You do not need curly arrows for radical mechanisms.
• You must write a balanced equation for each step.

Worked example 4: steps for a free-radical reaction

Ethane reacts with chlorine in ultraviolet light to give chloroethane and hydrogen chloride. Write equations for the steps.

Cl₂ → 2Cl•                         (UV light splits the Cl–Cl bond)
Cl• + C₂H₆ → C₂H₅• + HCl
C₂H₅• + Cl₂ → C₂H₅Cl + Cl•
Cl• + Cl• → Cl₂
C₂H₅• + Cl• → C₂H₅Cl
C₂H₅• + C₂H₅• → C₄H₁₀

In the second and third steps one radical goes in and one comes out, so the reaction continues. When two radicals meet, both dots disappear. These steps are named initiation, propagation and termination in 3.3.2.4, where the reaction of methane with chlorine is the example in the specification.

Curly arrows in other mechanisms

In other mechanisms, curly arrows show the movement of electron pairs.

• To form a covalent bond, the arrow starts from a lone pair of electrons or from another covalent bond, and points to where the new bond forms.
• To break a covalent bond, the arrow starts from the bond and points to the atom that takes the electron pair.

To outline a mechanism, draw the structures of the species involved and add the curly arrows.

Worked example 5: reading a two-arrow step

A hydroxide ion reacts with bromoethane, CH₃CH₂Br. The C–Br bond is polar, with C δ+ and Br δ−.

• Arrow 1: from a lone pair on the O of :OH⁻ to the δ+ carbon. It forms the C–O bond.
• Arrow 2: from the C–Br bond to Br. It breaks the bond; Br leaves as Br⁻ with the pair.
• Products: CH₃CH₂OH and Br⁻.

An arrow can also start from a bond to make a new bond, such as from a C=C to another atom. Full mechanisms come in 3.3.3 to 3.3.5; bond polarity is in the bonding study guide.

3.3.1.3 Isomerism

Structural isomers

Structural isomers have the same molecular formula but a different structural formula. There are three types:

Type	What changes	Example pair
Chain	The carbon skeleton (branching)	pentane and 2,2-dimethylpropane, C₅H₁₂
Position	Position of the functional group on the same skeleton	propan-1-ol and propan-2-ol, C₃H₈O
Functional group	The functional group itself	propanal and propanone, C₃H₆O

Stereoisomers and E–Z isomerism

Stereoisomers have the same structural formula but a different arrangement of atoms in space.

E–Z isomerism is one type. It happens because there is restricted rotation about the planar C=C double bond. The two carbons of the C=C and the four atoms joined to them lie in one plane, and the groups cannot swap sides by rotating. So E–Z isomers exist when:

1. there is a C=C double bond, and
2. each carbon of the C=C carries two different groups.

CIP priority rules

1. On each carbon of the C=C, look at the two atoms joined directly to it. The atom with the higher atomic number has the higher priority.
2. If the two atoms are the same, move one bond further out and compare the sets of atoms there. Stop at the first point of difference.
3. Z: the two higher-priority groups are on the same side of the double bond. E: they are on opposite sides.

Worked example 6: C₄H₈ — structural and E–Z isomers

Find all the alkene isomers of C₄H₈ and show which have E–Z isomers.

CH₂=CHCH₂CH₃      but-1-ene          C1 has H and H → no E–Z
CH₃CH=CHCH₃       but-2-ene          each C has H and CH₃ → E–Z isomers
(CH₃)₂C=CH₂       2-methylpropene    C1 has H and H → no E–Z

But-1-ene and but-2-ene are position isomers. But-1-ene and 2-methylpropene are chain isomers. Cyclobutane and methylcyclopropane are functional group isomers of all three, because a ring replaces the C=C.

The two stereoisomers of but-2-ene:

   H₃C       CH₃              H₃C       H
      \     /                    \     /
       C = C                      C = C
      /     \                    /     \
     H       H                  H       CH₃

  (Z)-but-2-ene              (E)-but-2-ene

On each carbon CH₃ (C, atomic number 6) beats H (1). Both CH₃ on the same side gives Z.

Worked example 7: first point of difference

Assign E or Z to this isomer of 3-methylpent-2-ene:

   H₃C       CH₃
      \     /
       C = C
      /     \
     H       CH₂CH₃

• Left carbon: CH₃ beats H.
• Right carbon: CH₃ and CH₂CH₃ both join through C, so compare the next atoms. CH₃ has (H, H, H); CH₂CH₃ has (C, H, H). C beats H, so CH₂CH₃ has the higher priority.
• The higher-priority groups, CH₃ (top left) and CH₂CH₃ (bottom right), are on opposite sides: E.

Common errors

• Missing the ends of a skeletal zigzag as carbons, so the chain is too short.
• Drawing H on carbon in a skeletal formula, or leaving the H off –OH.
• Numbering an alkene the wrong way (pent-3-ene instead of pent-2-ene).
• Drawing curly arrows for a radical step, or a radical with no dot.
• Starting a curly arrow on a charge sign rather than on a lone pair.
• Claiming E–Z isomers when one carbon of the C=C carries two identical groups.
• Using group size, not atomic number, for CIP priority.

Where to go next

• Introduction to organic chemistry revision notes
• Introduction to organic chemistry practice questions
• AQA A-Level Chemistry hub and printable checklist
• All free 10-minute diagnostics

Official syllabus

AQA, AS and A-level Chemistry specification (7404/7405), version 1.2, July 2026, for AS and A-level exams June 2016 onwards: section 3.3.1 Introduction to organic chemistry (3.3.1.1 to 3.3.1.3).