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

This study guide covers 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. It covers all three points: 3.3.1.1 Nomenclature, 3.3.1.2 Reaction mechanisms and 3.3.1.3 Isomerism. Everything in it 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.

For recall, use the revision notes. To test yourself, try the practice questions. See the OxfordAQA A-Level Chemistry hub, the printable checklist and a free diagnostic.

What this unit covers

Spec point	What you need to do	Stage
3.3.1.1 Nomenclature	Six kinds of formula; homologous series; drawing structural, displayed and skeletal forms; IUPAC names both ways, no chain or ring above six carbons	International AS
3.3.1.2 Reaction mechanisms	Radical dots; a balanced equation per radical step; curly arrows elsewhere	International AS
3.3.1.3 Isomerism	Define structural isomer and stereoisomer; three kinds of structural isomer; E and Z forms; Cahn–Ingold–Prelog (CIP) rules	International AS

Section 3.3.1 lists no required practical.

3.3.1.1 Six ways to write a formula

The specification names six formula types. Here they are for 3-methylbut-1-ene.

Formula type	What it shows	3-methylbut-1-ene
Empirical	Lowest whole-number ratio of the atoms present	CH₂
Molecular	True count of each kind of atom in a molecule	C₅H₁₀
General	An algebraic formula that fits every member of the series	CₙH₂ₙ (alkenes)
Structural	Each carbon with its attached atoms, written in order	CH₂=CHCH(CH₃)CH₃
Displayed	Every atom and every bond	below
Skeletal	Bonds as lines, carbons and C–H hydrogens left out	below

The displayed formula shows all ten hydrogen atoms and the double bond:

                H
                |
            H - C - H
    H   H       |   H
    |   |       |   |
H - C = C ----- C - C - H
                |   |
                H   H

Skeletal formula rules. Each line is a bond, and each line end or corner is a carbon unless another atom is written there. Leave out H atoms on carbon; write in every other atom with its hydrogens (–OH, not –O). For 3-methylbut-1-ene: a double line (C1=C2), a single line to C3, then two lines from C3, to C4 and to the methyl carbon.

Worked example 1: from composition to formulas

A compound contains 24.2% carbon, 4.0% hydrogen and 71.7% chlorine by mass. Its Mr is 99.0. Find its empirical and molecular formulas. (Ar: C = 12.0, H = 1.0, Cl = 35.5.)

Moles in 100 g:   C = 24.2 / 12.0 = 2.017   H = 4.0 / 1.0 = 4.0   Cl = 71.7 / 35.5 = 2.020
Divide by 2.017:  C = 1.00                  H = 1.98              Cl = 1.00
Empirical formula:          CH₂Cl     (empirical formula mass = 12.0 + 2.0 + 35.5 = 49.5)
Units per molecule:         99.0 / 49.5 = 2
Molecular formula:          C₂H₄Cl₂

Two structures fit: CH₃CHCl₂ and CH₂ClCH₂Cl. The method is from 3.1.2; see amount of substance.

Homologous series

The specification describes a homologous series as a series of compounds containing the same functional group. Members:

• share a general formula;
• differ from one member to the next by CH₂;
• have similar chemical reactions, because the functional group is the same;
• show a gradual change in physical properties, such as boiling point, as the chain lengthens.

Worked example 2: using a general formula

An unbranched alcohol has Mr = 102.0. Find its molecular formula and name it, given that the –OH is on the end carbon.

General formula of alcohols:   CₙH₂ₙ₊₁OH
Mr = 12.0n + 1.0(2n + 2) + 16.0 = 14n + 18
14n + 18 = 102  →  n = 6
Molecular formula: C₆H₁₃OH (C₆H₁₄O)    Name: hexan-1-ol

3.3.1.1 IUPAC naming

The limit for naming and for drawing is six carbon atoms per chain and per ring.

Number of C atoms	1	2	3	4	5	6
Name stem	meth-	eth-	prop-	but-	pent-	hex-

Families you meet in the International AS organic sections, 3.3.2 to 3.3.6:

Family	Group	How it shows in the name
Alkane	C–C and C–H only	-ane
Alkene	C=C	-ene, with a locant
Halogenoalkane	–F, –Cl, –Br, –I	fluoro-, chloro-, bromo-, iodo-
Alcohol	–OH	-ol
Aldehyde	–CHO	-al (the CHO carbon is C1)
Ketone	C=O within the chain	-one
Carboxylic acid	–COOH	-oic acid (the COOH carbon is C1)
Nitrile	–C≡N	-nitrile (the CN carbon is counted and is C1)

Rings take cyclo- before the stem (cyclopentane).

Method

1. Pick the longest carbon chain that includes the functional group. Its length sets the stem.
2. Number from the end that gives the functional group the lowest locant. In an alkane, number so the set of substituent locants is as low as possible.
3. Write substituents as prefixes in alphabetical order (bromo before chloro before methyl), using di- and tri- for repeats. The multiplier does not count in the alphabetical order.
4. Put commas between numbers and hyphens between numbers and letters.

Worked example 3: naming

Name CH₃CH(CH₃)CHClCH₂CH₃.

Longest chain:          5 carbons → pentane
Numbering left to right:  CH₃ on C2, Cl on C3 → locant set {2, 3}
Numbering right to left:  Cl on C3, CH₃ on C4 → locant set {3, 4}
Lower set wins:         {2, 3}
Alphabetical order:     chloro before methyl
Name:                   3-chloro-2-methylpentane

Worked example 4: drawing from names

(a) 2-methylcyclopentan-1-ol. Skeletal: a pentagon with OH on one corner (C1) and a short line for CH₃ on the next corner (C2). Molecular formula C₆H₁₂O.

(b) hex-4-en-1-ol. Six-carbon chain; –OH on C1; C=C between C4 and C5.

HOCH₂CH₂CH₂CH=CHCH₃        (C₆H₁₂O)

The –OH takes the suffix and the lowest number; the C=C is numbered from the same end.

3.3.1.2 Reaction mechanisms

Free-radical mechanisms

• A radical has an unpaired electron, shown by a dot: Br•, CH₃CH₂•.
• No curly arrows are needed in radical steps.
• Every step needs its own balanced equation.

Worked example 5: steps for a radical reaction

2,2-Dimethylpropane, C(CH₃)₄, reacts with chlorine in ultraviolet light. Every hydrogen atom in the molecule is equivalent, so only one monochloro product forms: 1-chloro-2,2-dimethylpropane. Write equations for the steps.

Cl₂ → 2Cl•                                    (UV light breaks the Cl–Cl bond)
Cl• + C(CH₃)₄ → (CH₃)₃CCH₂• + HCl             (radical in, radical out)
(CH₃)₃CCH₂• + Cl₂ → (CH₃)₃CCH₂Cl + Cl•
Cl• + Cl• → Cl₂
(CH₃)₃CCH₂• + Cl• → (CH₃)₃CCH₂Cl
2(CH₃)₃CCH₂• → (CH₃)₃CCH₂CH₂C(CH₃)₃           (C₁₀H₂₂)

Check that atoms balance in every line. One radical in and one out keeps the chain going; two in and none out stops it. Section 3.3.2.4 names these steps initiation, propagation and termination, using methane.

Curly arrows in other mechanisms

• A curly arrow shows the movement of an electron pair.
• A new covalent bond: the arrow begins at a lone pair or at an existing covalent bond.
• A broken covalent bond: the arrow begins on that bond and ends on the atom that takes both electrons.
• To outline a mechanism, draw the structures of every species involved and add the arrows.

Worked example 6: two arrows, no carbon

Ammonia reacts with hydrogen chloride: NH₃ + HCl → NH₄⁺ + Cl⁻. Describe the arrows.

• Arrow 1 starts on the lone pair on nitrogen and points to the H of H–Cl. This forms the new N–H bond.
• Arrow 2 starts on the H–Cl bond and points to Cl. The bond breaks and Cl leaves as Cl⁻ with both electrons.

The H–Cl bond is polar (H δ+), so the lone pair is drawn to H; see shapes, polarity and intermolecular forces.

Worked example 7: an arrow from a bond

In the first step of the reaction of hex-3-ene with sulfuric acid, the C=C forms a bond to hydrogen.

• Arrow 1 starts on the C=C double bond and points to an H atom of H–OSO₃H.
• Arrow 2 starts on that O–H bond and points to O, giving HSO₄⁻.
• Products: the carbocation CH₃CH₂CH₂C⁺HCH₂CH₃ and HSO₄⁻. You meet the complete mechanisms in sections 3.3.3, 3.3.4 and 3.3.5.

3.3.1.3 Isomerism

Structural isomers

Structural isomers are compounds with the same molecular formula but different structural formulas.

Type	What differs	Example pair
Chain	Branching of the carbon skeleton	pentan-1-ol and 3-methylbutan-1-ol, C₅H₁₂O
Position	Where the functional group sits on the same skeleton	CH₃CHCl₂ and CH₂ClCH₂Cl, C₂H₄Cl₂
Functional group	The functional group itself	pentanal and cyclopentanol, C₅H₁₀O

Stereoisomerism and E–Z isomers

Stereoisomers share a structural formula but differ in how their atoms are arranged in space.

E–Z isomerism is a form of stereoisomerism caused by restricted rotation about the planar C=C double bond. The doubly bonded carbons and their four attached atoms sit in one plane and cannot twist. E–Z isomers need:

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

Applying the CIP rules

1. Take each C=C carbon in turn and compare the two atoms bonded directly to it. Higher atomic number means higher priority.
2. If they are the same, list the atoms attached to each, highest atomic number first, and compare them in order. Stop at the first point of difference.
3. Higher-priority groups on the same side: Z. On opposite sides: E.

Worked example 8: all the structural isomers of C₃H₅Cl

CH₂=CHCH₂Cl     3-chloroprop-1-ene    C1 carries H and H → no E–Z
ClCH=CHCH₃      1-chloroprop-1-ene    C1: Cl and H; C2: CH₃ and H → E–Z isomers
CH₂=CClCH₃      2-chloroprop-1-ene    C1 carries H and H → no E–Z
chlorocyclopropane (three-membered ring with one Cl) → no C=C, no E–Z

The three chloropropenes are position isomers. Chlorocyclopropane is a functional group isomer of each, because a ring replaces the C=C. The stereoisomers of 1-chloroprop-1-ene:

    Cl       H                Cl       CH₃
      \     /                   \     /
       C = C                     C = C
      /     \                   /     \
     H       CH₃               H       H

  (E)-1-chloroprop-1-ene    (Z)-1-chloroprop-1-ene

Cl (17) outranks H (1) on the left carbon; CH₃ (C, 6) outranks H on the right.

Worked example 9: looking beyond the first atom

Decide whether this alkene is E or Z.

    ClCH₂       CH₃
         \     /
          C = C
         /     \
  (CH₃)₂CH       H

• Left carbon: CH₂Cl and CH(CH₃)₂ both attach through C, a tie.
• Next atoms: CH₂Cl has (Cl, H, H); CH(CH₃)₂ has (C, C, H). Highest first: Cl (17) beats C (6), so CH₂Cl wins, though the other group has more carbons.
• Right carbon: CH₃ beats H.
• CH₂Cl and CH₃ are both on the upper side: Z.

Common errors

• Counting only the corners of a skeletal formula and missing the line ends.
• Ordering prefixes by locant or by “di-” rather than alphabetically.
• Drawing curly arrows on radical equations, or leaving the dot off a radical.
• Beginning a curly arrow at a charge sign or a bare atom instead of at a lone pair or bond.
• Ranking CIP groups by size or by the number of atoms rather than by atomic number at the first difference.

Next steps

Use the revision notes to condense this guide, then try the practice questions.

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.