OxfordAQA A-Level Chemistry: Introduction to organic chemistry (9620) – Revision Notes

These notes summarise section 3.3.1 Introduction to organic chemistry of the OxfordAQA International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards: 3.3.1.1 Nomenclature, 3.3.1.2 Reaction mechanisms and 3.3.1.3 Isomerism. Every point is International AS content; none is International A-level only. Unit 2 (Organic 1 and Physical 1) assesses it, and Unit 5 (practical and synoptic) assesses all content.

The full explanations and worked examples are in the study guide. Then test yourself with the practice questions. The OxfordAQA A-Level Chemistry hub and printable checklist span the whole course; a free diagnostic shows where to begin.

3.3.1.1 Nomenclature

Key definitions

Term	Meaning
Empirical formula	Lowest whole-number ratio of the atoms of each element present
Molecular formula	True count of each kind of atom in one molecule
General formula	Formula in terms of n that fits the whole series, e.g. alkanes CₙH₂ₙ₊₂
Structural formula	Groups written carbon by carbon, e.g. CH₃CH₂CHO
Displayed formula	Every atom and every bond drawn
Skeletal formula	Lines for bonds; carbons at ends and corners; no C–H hydrogens drawn
Homologous series	Family of compounds that all contain the same functional group
Functional group	The part of the molecule that decides the family’s typical reactions

Homologous series, four features: same general formula; successive members differ by CH₂; similar chemical reactions; gradual trend in physical properties.

Empirical vs molecular: cyclohexane is C₆H₁₂, so its empirical formula is CH₂. They are the same when the ratio cannot be simplified (C₃H₈O stays C₃H₈O).

Naming in four steps

1. Longest chain that contains the functional group → stem (meth, eth, prop, but, pent, hex).
2. Number from the end that gives the functional group the lowest locant; for alkanes, the lowest set of substituent locants.
3. Prefixes in alphabetical order (bromo, chloro, ethyl, iodo, methyl); di-/tri- do not count for order.
4. Commas between numbers, hyphens between numbers and words.

Group	Suffix or prefix	Numbering note
C=C	-ene	lowest locant for the first carbon of the C=C
–OH	-ol	takes the suffix over C=C: pent-4-en-2-ol
–CHO	-al	always C1, no locant
C=O in chain	-one	locant needed once there are five or more carbons (pentan-2-one)
–COOH	-oic acid	always C1
–C≡N	-nitrile	CN carbon is C1 and counts in the stem
halogen	fluoro, chloro, bromo, iodo	prefix only
ring	cyclo-	cyclohexane, cyclopentanol

Worked reminders

BrCH₂CH₂CH(CH₃)CH₃     → Br on C1, CH₃ on C3     → 1-bromo-3-methylbutane
CH₃COCH(CH₃)CH₂CH₃     → C=O on C2, CH₃ on C3    → 3-methylpentan-2-one

Drawing from a name: method in steps

1. Read the stem and suffix first: they give the chain length and the main group.
2. Draw the bare chain and number it from the end the suffix locant needs.
3. Add each prefix group at its locant.
4. Fill every carbon up to four bonds with H, then check the molecular formula.

2,4-dimethylhexan-3-one:
hexan-3-one        → C–C–C(=O)–C–C–C
2,4-dimethyl       → CH₃ on C2 and on C4
structural formula → CH₃CH(CH₃)COCH(CH₃)CH₂CH₃     (C₈H₁₆O)

Reading skeletal formulas. A zigzag of four lines has five carbons (three corners plus two ends). With Cl on the second carbon from one end, it is 2-chloropentane, C₅H₁₁Cl.

3.3.1.2 Reaction mechanisms

Free-radical notation

• Radical = species with an unpaired electron, shown by a dot (I•, C₂H₅•).
• No curly arrows needed for radical mechanisms.
• Write a balanced equation for each step. Pattern for a halogen X₂ and an alkane RH:

X₂ → 2X•                  splits the X–X bond
X• + RH → R• + HX         one radical in, one out
R• + X₂ → RX + X•         one radical in, one out
R• + X• → RX              two radicals in, none out

Where you use this notation next

Section	Mechanism	Notation
3.3.2.4	Free-radical substitution of methane by chlorine (initiation, propagation, termination)	dots and equations
3.3.3	Nucleophilic substitution by OH⁻, CN⁻ and NH₃; elimination with KOH	curly arrows
3.3.4	Electrophilic addition of HBr, H₂SO₄ and Br₂ to alkenes	curly arrows
3.3.5	Acid-catalysed elimination of water from alcohols	curly arrows

Curly arrows (all other mechanisms)

Arrow	Starts at	Ends at
Forms a bond	a lone pair, or another covalent bond	the atom that gains the bond
Breaks a bond	the middle of that bond	the atom that keeps the pair

One arrow = one pair of electrons. “Outline a mechanism” means draw each species’ structure with its lone pairs and charges, plus every arrow.

3.3.1.3 Isomerism

Two definitions

• Structural isomers: same molecular formula, different structural formula.
• Stereoisomers: structural formula identical, but the atoms are arranged differently in 3-D space.

Three types of structural isomer (all C₆H₁₂O)

Type	Pair
Chain	hexanal and 2-methylpentanal
Position	hexan-2-one and hexan-3-one
Functional group	hexanal and hexan-3-one

E–Z isomers: method in steps

1. Find the C=C. Restricted rotation about this planar bond is the cause of E–Z isomerism.
2. Check that each carbon of the C=C carries two different groups. If not, stop: no E–Z.
3. On each carbon, rank the two groups by the atomic number of the atom attached directly.
4. If tied, list the next atoms in each group, highest first, and compare at the first point of difference.
5. Higher priorities same side = Z; opposite sides = E.

Atomic numbers you need most: H 1, C 6, N 7, O 8, F 9, Cl 17, Br 35, I 53.

Drawing reminder. For 1-bromo-2-chloroprop-1-ene, BrCH=C(Cl)CH₃, Br beats H on one carbon and Cl beats CH₃ on the other. Br and Cl on opposite sides gives the E isomer:

    Br       CH₃
      \     /
       C = C
      /     \
     H       Cl
  (E)-1-bromo-2-chloroprop-1-ene

Keep the 120° layout around each C=C carbon, so it is obvious which side each group is on.

Worked reminder. Rank –CH₂CH₂Cl against –CH(CH₃)₂. Both attach through C. Next atoms: (C, H, H) against (C, C, H). First C ties C; then H loses to C. So –CH(CH₃)₂ wins. The Cl further out never gets compared, because the decision is made one bond earlier.

Must-know distinctions

• Empirical formula is a ratio; molecular formula is the real count.
• Structural formula (CH₃CH₂OH) is not a molecular formula (C₂H₆O).
• Structural isomers differ in how the atoms are connected; stereoisomers differ only in 3-D arrangement.
• Chain isomers change the skeleton; position isomers move the group; functional group isomers change the group.
• Radical steps: dots and equations. Other mechanisms: curly arrows from lone pairs or bonds.
• CIP uses atomic number, not group size or mass of the whole group.

Quick self-test

1. Give the empirical formula of cyclobutane.
2. Give the general formula of the alkenes and the molecular formula of the alkene with six carbons.
3. Name CH₃CHBrCHBrCH₃.
4. Name CH₂=C(CH₃)CH₂CH₂CH₃.
5. Give the structural formula of 3-chloropentanal.
6. State the type of isomerism shown by 1-chloropentane and 1-chloro-3-methylbutane.
7. Does 2,3-dimethylbut-2-ene have E–Z isomers? Explain.
8. Which group has the higher CIP priority: –CH₂Br or –CCl₃?
9. Write the equation for the step in which a cyclopentyl radical, C₅H₉•, reacts with chlorine.
10. Where does a curly arrow that breaks a covalent bond start?
11. State the cause of E–Z isomerism.

Answers

1. CH₂ (C₄H₈ ÷ 4).
2. CₙH₂ₙ; C₆H₁₂.
3. 2,3-dibromobutane.
4. 2-methylpent-1-ene (C=C at C1, methyl at C2).
5. CH₃CH₂CHClCH₂CHO (the CHO carbon is C1, so Cl is on C3).
6. Chain isomers (same functional group position on C1, different carbon skeleton).
7. No. Each carbon of the C=C carries two identical CH₃ groups.
8. –CH₂Br. Both attach through C; next atoms (Br, H, H) against (Cl, Cl, Cl); Br (35) beats Cl (17) at the first comparison, so the three chlorines do not matter.
9. C₅H₉• + Cl₂ → C₅H₉Cl + Cl•
10. From the middle of the bond being broken.
11. Restricted rotation about the planar C=C double bond.

Where marks are usually lost

• Defining structural isomers without naming the molecular formula as the shared one.
• Counting only corners, not line ends, in a skeletal formula.
• Leaving off the H of –OH in a skeletal formula, or drawing hydrogens on carbon.
• Giving a locant to -al or -oic acid (butan-1-al is wrong).
• Choosing a longest chain that misses the functional group carbon.
• Missing dots on radicals, or unbalanced radical equations.
• Starting a curly arrow on a negative charge or a nucleus instead of a lone pair or bond.
• Ranking CIP groups by counting atoms instead of comparing the highest atomic number first.
• Swapping E and Z: Z means the higher priorities are on the same side.

Official syllabus

OxfordAQA, International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards: section 3.3.1 Introduction to organic chemistry (International AS), points 3.3.1.1 to 3.3.1.3.