S5 Chemistry · Unit 1

Introduction to Organic Chemistry

Classification · Formulas · Functional Groups · IUPAC Nomenclature

1.1

Classification of Organic Compounds

What is Organic Chemistry?

Organic chemistry is the branch of chemistry that deals with the study of carbon-containing compounds. The word "organic" originally referred to substances obtained from living organisms, but today millions of synthetic organic compounds are known.

Carbon is unique because it can form 4 covalent bonds and can bond to itself to form long chains, rings, and branched structures — a property called catenation.

Definition An organic compound is any compound that contains carbon atoms covalently bonded to hydrogen, and often to oxygen, nitrogen, sulfur, halogens, or other elements.
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Exceptions to the carbon rule Some carbon-containing compounds are NOT classified as organic: CO₂ (carbon dioxide), CO (carbon monoxide), carbonates (e.g. CaCO₃), bicarbonates, and cyanides (CN⁻) are considered inorganic.

A: Classification by Carbon Skeleton (Chain Structure)

Organic compounds are first grouped by the shape of their carbon backbone:

ClassCarbon SkeletonExampleNotes
Acyclic (open-chain)No ring; carbon atoms form a chainButane, CH₃CH₂CH₂CH₃Also called aliphatic
CyclicContains at least one ring of carbon atomsCyclohexane, benzeneIncludes aromatic and non-aromatic rings
AlicyclicCyclic but NOT aromaticCyclohexane C₆H₁₂Behave like aliphatic compounds
AromaticContains a benzene ring (C₆H₆)Benzene, toluene, phenolHighly stable due to delocalized π electrons
HeterocyclicRing contains atoms other than C (e.g. N, O, S)Pyridine (N in ring), furan (O in ring)Very common in medicines and DNA bases

B: Classification by Degree of Unsaturation

Organic compounds are also classified by the type of bonds between carbon atoms:

ClassBondsGeneral FormulaExamples
SaturatedC–C single bonds onlyCₙH₂ₙ₊₂ (alkanes)Methane CH₄, ethane C₂H₆
UnsaturatedC=C double or C≡C triple bonds presentCₙH₂ₙ (alkenes), CₙH₂ₙ₋₂ (alkynes)Ethene C₂H₄, ethyne C₂H₂
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Quick Memory Tip Saturated = no more hydrogen can be added (like saturated fat — full!). Unsaturated = still has room for more hydrogen (can react with H₂ in addition reactions).
1.2

Types of Formulas for Organic Compounds

There are four main ways to represent an organic molecule. Each gives different levels of information.

Type of FormulaWhat it showsExample: Butane
Molecular (Molecular Formula) The total number of each type of atom in one molecule C₄H₁₀
Empirical Formula The simplest whole-number ratio of atoms C₂H₅
Structural Formula Shows how atoms are connected (all bonds shown) CH₃–CH₂–CH₂–CH₃
Displayed (Full Structural) Formula Shows every atom and every bond explicitly All H atoms drawn out around each C
Skeletal Formula Zigzag lines; C atoms at each corner/end; H atoms implied A zigzag with 3 line segments
Condensed Structural Formula Groups H atoms with each carbon, reading left to right CH₃CH₂CH₂CH₃
// Butane: Different representations compared

Molecular: C₄H₁₀
Empirical: C₂H₅
Condensed: CH₃–CH₂–CH₂–CH₃
IUPAC name: butane

Relationship between Molecular and Empirical Formulas

The molecular formula is always a whole-number multiple of the empirical formula.

If empirical formula is CH₂ and the molar mass is 56 g/mol:

Mass of CH₂ unit = 12 + 2 = 14 g/mol  →   n = 56 ÷ 14 = 4

So molecular formula = (CH₂)₄ = C₄H₈ (this is butene or cyclobutane)

1.3

Functional Groups and Homologous Series

Functional Group A functional group is an atom or group of atoms within a molecule that is responsible for its characteristic chemical reactions. The rest of the molecule (usually a carbon chain) is called the alkyl group (R–).
Homologous Series A homologous series is a family of organic compounds that (i) have the same functional group, (ii) differ by a –CH₂– unit from one member to the next, (iii) have the same general formula, (iv) show a gradual change in physical properties, and (v) have similar chemical properties.
FamilyFunctional GroupGeneral FormulaExample (n=2)
AlkanesC–C (single bond, no FG)CₙH₂ₙ₊₂Ethane C₂H₆
AlkenesC=CCₙH₂ₙEthene C₂H₄
AlkynesC≡CCₙH₂ₙ₋₂Ethyne C₂H₂
Alcohols–OH (hydroxyl)CₙH₂ₙ₊₁OHEthanol C₂H₅OH
Halogenoalkanes–X (X = F,Cl,Br,I)CₙH₂ₙ₊₁XChloroethane C₂H₅Cl
Aldehydes–CHOCₙH₂ₙOEthanal CH₃CHO
Ketones–CO–CₙH₂ₙOPropanone CH₃COCH₃
Carboxylic acids–COOHCₙH₂ₙO₂Ethanoic acid CH₃COOH
Esters–COO–CₙH₂ₙO₂Methyl ethanoate CH₃COOCH₃
Amines–NH₂CₙH₂ₙ₊₁NH₂Ethylamine C₂H₅NH₂
Amides–CONH₂Ethanamide CH₃CONH₂
Nitriles–CNCₙH₂ₙ₊₁CNEthanenitrile CH₃CN

Properties of a Homologous Series — illustrated with Alkanes

As the number of carbon atoms increases by 1 (–CH₂–), the following properties change gradually:

AlkaneFormulaMrBoiling Point (°C)State at 25°C
MethaneCH₄16−162Gas
EthaneC₂H₆30−89Gas
PropaneC₃H₈44−42Gas
ButaneC₄H₁₀58−1Gas
PentaneC₅H₁₂72+36Liquid
HexaneC₆H₁₄86+69Liquid
DecaneC₁₀H₂₂142+174Liquid
IcosaneC₂₀H₄₂282+343Solid (wax)
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Why do boiling points increase? Larger molecules have more electrons → stronger van der Waals (London dispersion) forces between molecules → more energy needed to separate them → higher boiling point.
1.4

General Rules of IUPAC Nomenclature

What is IUPAC?

IUPAC stands for the International Union of Pure and Applied Chemistry. Their naming system provides a unique, systematic name for every organic compound that conveys its structure.

The Three Building Blocks of an IUPAC Name

Every IUPAC name has up to three parts: PREFIX + ROOT/STEM + SUFFIX

PartTells you…Example
PrefixNames and positions of substituents (branches, halogens)2-methyl, 3-chloro
Root/StemThe length of the longest carbon chainmeth(1), eth(2), prop(3), but(4), pent(5), hex(6), hept(7), oct(8), non(9), dec(10)
SuffixThe main functional group (class of compound)-ane, -ene, -yne, -ol, -al, -one, -oic acid

Step-by-Step IUPAC Naming Rules

Rule 1 — Find the longest chain: Identify the longest continuous chain of carbon atoms that contains the principal functional group. This gives the root name (stem).

Rule 2 — Number the chain: Number the carbon atoms from the end closest to the principal functional group (or the branch, if no functional group). The functional group / substituents must get the lowest possible locants (numbers).

Rule 3 — Name substituents: Identify all substituents attached to the main chain (branches = alkyl groups; halogens, etc.). Use prefixes: methyl (–CH₃), ethyl (–C₂H₅), propyl (–C₃H₇), fluoro (–F), chloro (–Cl), bromo (–Br), iodo (–I).

Rule 4 — Multiple identical substituents: Use di-, tri-, tetra- when the same substituent appears 2, 3, 4 times. List their positions separately: 2,2-dimethyl (not 2-dimethyl).

Rule 5 — Alphabetical order: When there are two or more different substituents, list them alphabetically (ignore di-, tri-, etc. when alphabetising).

Rule 6 — Suffix for class: End the name with the correct suffix: –ane (alkane), –ene (alkene), –yne (alkyne), –ol (alcohol), –al (aldehyde), –one (ketone), –oic acid (carboxylic acid), –oate (ester), –amine (amine).

Chain lengthPrefix (stem)Example (alkane)Formula
1 carbonmeth-methaneCH₄
2 carbonseth-ethaneC₂H₆
3 carbonsprop-propaneC₃H₈
4 carbonsbut-butaneC₄H₁₀
5 carbonspent-pentaneC₅H₁₂
6 carbonshex-hexaneC₆H₁₄
7 carbonshept-heptaneC₇H₁₆
8 carbonsoct-octaneC₈H₁₈
9 carbonsnon-nonaneC₉H₂₀
10 carbonsdec-decaneC₁₀H₂₂

Worked Examples

Worked Example 1 — Naming a Branched Alkane

Question: Name the following compound: CH₃–CH(CH₃)–CH₂–CH₃

1
Find the longest chain. Count the main chain: C1–C2–C3–C4 = 4 carbons. Root = but-
2
Identify the substituent. Carbon 2 has an extra –CH₃ (methyl group) attached.
3
Number the chain. Number from the end closest to the branch: the branch is on C2 from the left OR C3 from the right. C2 is lower → number from left. Branch position = 2.
4
Write the name. Substituent (2-methyl) + root (but) + suffix (-ane).
Answer 2-methylbutane
Worked Example 2 — Drawing the Structure from a Name

Question: Draw the condensed structural formula of 3-ethyl-2-methylpentane.

1
Identify root: pent → 5 carbons in main chain.
C1–C2–C3–C4–C5
2
Identify substituents: 2-methyl (–CH₃ on C2), 3-ethyl (–C₂H₅ on C3).
3
Attach substituents to the main chain:
CH₃–CH(CH₃)–CH(C₂H₅)–CH₂–CH₃
4
Verify: count all carbons: 5 (chain) + 1 (methyl) + 2 (ethyl) = 8 carbons total → molecular formula C₈H₁₈.
Answer CH₃–CH(CH₃)–CH(C₂H₅)–CH₂–CH₃  |  C₈H₁₈
Worked Example 3 — Empirical vs Molecular Formula

Question: A hydrocarbon contains 85.7% carbon and 14.3% hydrogen by mass. Its molar mass is 56 g/mol. Determine the empirical and molecular formulas.

1
Find mole ratio. Assume 100 g sample:
Moles C = 85.7 ÷ 12 = 7.14 mol
Moles H = 14.3 ÷ 1 = 14.3 mol
2
Divide by smallest value (7.14):
C : H = 7.14/7.14 : 14.3/7.14 = 1 : 2
3
Empirical formula: CH₂
Empirical formula mass = 12 + 2 = 14 g/mol
4
Find n: n = 56 ÷ 14 = 4
Answer Empirical formula: CH₂  |  Molecular formula: C₄H₈
Worked Example 4 — Identifying the Functional Group & Homologous Series

Question: For each compound below, identify the functional group and state the homologous series it belongs to:
(a) CH₃CH₂OH    (b) CH₃COOH    (c) CH₃CHO    (d) CH₃CH₂NH₂

a
CH₃CH₂OH → functional group: –OH (hydroxyl) → Alcohol (ethanol)
b
CH₃COOH → functional group: –COOH (carboxyl) → Carboxylic acid (ethanoic acid)
c
CH₃CHO → functional group: –CHO (aldehyde) → Aldehyde (ethanal)
d
CH₃CH₂NH₂ → functional group: –NH₂ (amino) → Amine (ethylamine)
Summary (a) Alcohol  ·  (b) Carboxylic acid  ·  (c) Aldehyde  ·  (d) Amine
Worked Example 5 — Naming a Compound with Multiple Substituents

Question: Name: CH₃–C(CH₃)₂–CH₂–CH₂–CH₃

1
Longest chain: C1–C2–C3–C4–C5 = 5 carbons → pent-
2
C2 has two methyl groups attached → 2,2-dimethyl
3
Numbering: from left, branches at C2; from right, at C4. Left gives lower numbers → use left.
Answer 2,2-dimethylpentane

Exercises with Answers

Exercise 1

Classify the following as saturated or unsaturated, and as aliphatic, aromatic, or alicyclic:
(a) C₆H₆    (b) C₄H₈    (c) C₃H₈    (d) Cyclohexane C₆H₁₂

(a) C₆H₆ → unsaturated, aromatic (benzene — contains a benzene ring with delocalized electrons)
(b) C₄H₈ → unsaturated, aliphatic (contains C=C double bond; open chain)
(c) C₃H₈ → saturated, aliphatic (propane — only C–C single bonds; open chain)
(d) C₆H₁₂ → saturated, alicyclic (cyclohexane — ring structure but no benzene ring)
Exercise 2

Give the IUPAC name for each compound:
(a) CH₃–CH₂–CH₂–CH₂–CH₃
(b) CH₃–CH(Cl)–CH₃
(c) CH₃–CH₂–OH
(d) CH₃–CH₂–COOH

(a) 5-carbon chain, no substituents, –ane suffix → pentane
(b) 3-carbon chain, Cl on C2, –ane suffix → 2-chloropropane
(c) 2-carbon chain, –OH group, –ol suffix → ethanol
(d) 3-carbon chain (including the carboxyl C), –oic acid suffix → propanoic acid
Exercise 3

A compound has the molecular formula C₄H₈O. It contains the functional group –CHO.
(a) What homologous series does it belong to?
(b) Write its IUPAC name.

(a) The –CHO functional group → Aldehydes
(b) 4-carbon chain + aldehyde suffix: butanal (CH₃CH₂CH₂CHO)
Exercise 4

A hydrocarbon has the empirical formula CH. Its molar mass is 78 g/mol. Find the molecular formula and identify the compound.

Empirical formula mass = 12 + 1 = 13 g/mol
n = 78 ÷ 13 = 6
Molecular formula = (CH)₆ = C₆H₆
This compound is benzene — an aromatic hydrocarbon with a ring structure and delocalized electrons.
Exercise 5

Write the condensed structural formula for:
(a) 2-methylpropane    (b) 3-methylhexane    (c) 2,3-dimethylbutane

(a) CH₃CH(CH₃)CH₃ — 4 carbons total, methyl branch on C2 of a 3-carbon chain (but IUPAC: rename — the longest chain is actually 4 → 2-methylpropane = isobutane)
(b) CH₃CH₂CH(CH₃)CH₂CH₂CH₃ — hexane chain (6C) + methyl on C3
(c) CH₃CH(CH₃)CH(CH₃)CH₃ — butane chain (4C) + methyl on C2 and C3
Exercise 6 — Extended

State four characteristics of a homologous series and illustrate each with an example from the alcohol series (CₙH₂ₙ₊₁OH).

1. Same general formula: All alcohols fit CₙH₂ₙ₊₁OH (e.g. methanol CH₃OH, ethanol C₂H₅OH, propanol C₃H₇OH)

2. Same functional group: All have the –OH (hydroxyl) group responsible for their chemical behaviour

3. Differ by –CH₂– unit: Methanol → ethanol → propanol each adds one –CH₂– (mass difference of 14 g/mol)

4. Gradual change in physical properties: Boiling points rise steadily: methanol 65°C, ethanol 78°C, propanol 97°C, butanol 118°C

5. Similar chemical properties: All alcohols undergo oxidation, esterification, and dehydration reactions
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Quick Quiz — Unit 1

Multiple Choice Quiz

10 Questions · Select best answer

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Unit 1 Test

Section A — Short Answer (30 marks)

A1. (2 marks) Define the term organic compound and give two examples of carbon compounds that are NOT classified as organic.

A2. (3 marks) Distinguish between: (a) saturated and unsaturated hydrocarbons; (b) aliphatic and aromatic compounds.

A3. (4 marks) Explain the meaning of the term homologous series. State four characteristics of a homologous series.

A4. (6 marks) Give the IUPAC name for each of the following:
(i) CH₃CH₂CH₂CH₂OH    (ii) CH₃CH₂COCH₃    (iii) (CH₃)₃CH
(iv) CH₃CH(Br)CH₂CH₃    (v) CH₃CH₂NH₂    (vi) CH₃COOC₂H₅

A5. (4 marks) Draw the full structural (displayed) formula for: (a) propan-1-ol, (b) propanal, (c) propanone, (d) propanoic acid.

A6. (4 marks) A compound contains 40.0% C, 6.7% H, and 53.3% O by mass. The molar mass is 60 g/mol. Find the empirical and molecular formula. Name the compound.

A7. (7 marks) Complete the table:

CompoundMolecular FormulaFunctional GroupHomologous SeriesIUPAC Name
CH₃OH
CH₃CHO
C₃H₇ClHalogenoalkane1-chloropropane
CH₃COCH₃
C₂H₅COOH

Section B — Structured Questions (20 marks)

B1. (8 marks) Consider the following compounds:

(I) CH₄    (II) C₂H₄    (III) C₂H₂    (IV) C₂H₅OH    (V) C₆H₆

(a) Which compound belongs to the alkyne series? Name it. (2)

(b) Which compound is aromatic? Explain why it is classified as aromatic. (2)

(c) Write the general formula for the series to which compound (II) belongs. (1)

(d) Why does compound (I) have a much lower boiling point than compound (IV) despite (IV) having only 2 carbons? (3)

B2. (12 marks) (a) Define the term functional group. (2)

(b) For each functional group listed, name the homologous series, write the general formula, and give one named example: –OH, –COOH, –CHO, –CO–, –NH₂, –COO–. (6 × 1 mark each)

(c) Explain why the boiling points increase steadily within a homologous series. (2)

(d) State and explain TWO differences between the molecular formula and the structural formula of an organic compound. (2)

📋 Model Answers — Unit Test

A4 answers: (i) butan-1-ol  |  (ii) butan-2-one  |  (iii) 2-methylpropane  |  (iv) 2-bromobutane  |  (v) ethylamine (ethanamine)  |  (vi) ethyl ethanoate

A6 answer: Moles: C = 40/12 = 3.33; H = 6.7/1 = 6.7; O = 53.3/16 = 3.33 → Ratio 1:2:1 → empirical CH₂O (mass 30). n = 60/30 = 2 → molecular formula C₂H₄O₂ethanoic acid (acetic acid) CH₃COOH

B1(d): Compound (IV) ethanol has –OH group capable of forming hydrogen bonds (due to high electronegativity of O and presence of O–H). Methane only has weak van der Waals forces. Hydrogen bonds are much stronger intermolecular forces → much more energy needed to separate ethanol molecules → higher boiling point.

Next Unit → Unit 2: Alkanes