Edexcel International A Level Chemistry: Introductory organic chemistry and alkanes (YCH11)

This guide teaches Topic 4, Introductory Organic Chemistry and Alkanes, of the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry (XCH11/YCH11) specification, Issue 1, September 2017. It covers outcomes 4.1 to 4.18 (parts 4A and 4B). The topic sits in Unit 1: Structure, Bonding and Introduction to Organic Chemistry, an International AS (IAS) unit.

The revision notes condense this page, and the practice questions let you test yourself. The course hub is Edexcel International A Level Chemistry, the printable checklist lists every specification point, and the free 10-minute diagnostics point to your weak areas.

What this topic covers

Spec outcome	You need to know or do
4.1–4.3	Hazard versus risk; risk assessments; ways to reduce risk
4.4	Homologous series and functional group
4.5	IUPAC names; structural, displayed and skeletal formulae (prefixes up to C₁₀)
4.6	Classify reactions: addition, substitution, oxidation, reduction, polymerisation
4.7–4.8	Homolytic and heterolytic fission; free radical and electrophile
4.9	General formulae; hydrocarbon and saturated
4.10–4.11	Structural isomers of alkanes and cycloalkanes up to six carbons
4.12	Fractional distillation, cracking and reforming, with equations
4.13–4.14	Pollutants; CO toxicity; acidic oxides of nitrogen and sulfur
4.15–4.16	Alternative fuels; carbon neutrality of petrol, bioethanol and hydrogen
4.17–4.18	Combustion, halogenation and the free radical substitution mechanism

There is no core practical in this topic. The specification gives one further suggested practical: cracking alkanes by thermal decomposition, including liquid paraffin over aluminium oxide.

Hazard, risk and working safely (4.1–4.3)

A hazard is anything with the potential to cause harm, such as a flammable, toxic or corrosive substance. A risk is the chance that the hazard actually causes harm, given how, and how much of, the substance is used. Hazard is a property of the substance; risk depends on what you do with it.

Many organic compounds are volatile and flammable, and some are toxic. Their vapours spread and can ignite away from the liquid. So you carry out a risk assessment before any practical: identify each hazard, judge the risk and decide how to control it. The specification names three ways to reduce risk:

1. Work on a smaller scale: less vapour, less heat, a smaller fire if one starts.
2. Take precautions specific to the hazard: no naked flames for flammable liquids, a fume cupboard for toxic vapour, gloves and goggles for corrosives.
3. Use an alternative method with less hazardous substances, such as a dilute solution of a reagent.

Worked example: planning a halogenation

A student plans to shake cyclohexane with pure liquid bromine in bright light. Cyclohexane is highly flammable; bromine is toxic and corrosive. Use a few cm³ in a test tube (smaller scale); work in a fume cupboard away from flames, with gloves and goggles (hazard-specific); use a dilute solution of bromine in an organic solvent instead of pure bromine (less hazardous alternative).

Homologous series and functional groups (4.4)

In a homologous series, every member shares one functional group and one general formula, and each member differs from the next by CH₂. Members show a gradual trend in physical properties and have similar chemical properties.

The functional group is the part of the molecule responsible for its characteristic reactions: C=C in alkenes, –OH in alcohols.

Naming and drawing compounds (4.5)

The stem shows how many carbons the longest chain has: meth- C₁, eth- C₂, prop- C₃, but- C₄, pent- C₅, hex- C₆, hept- C₇, oct- C₈, non- C₉, dec- C₁₀. The ending shows the series: -ane for alkanes.

Rules for branched alkanes

1. Find the longest continuous carbon chain. It does not have to be drawn in a straight line.
2. Number the chain from the end that gives the branches the lowest numbers.
3. Name each branch (methyl, ethyl) with its position; use di-, tri- for repeats.
4. List different branches alphabetically, ignoring di- and tri-. Commas go between numbers, hyphens between numbers and words.

Formula types. A structural formula shows the arrangement unambiguously, e.g. CH₃CH(CH₃)CH₂CH₃. A displayed formula draws every atom and bond. A skeletal formula is a zigzag in which each corner and line end is a carbon and hydrogens on carbon are not drawn: pentane is a zigzag of four lines, cyclohexane a hexagon.

Worked example: a hidden longest chain

Name CH₃CH(CH₃)CH₂CH(CH₂CH₃)CH₃.

• The row as written has five carbons, but through the ethyl branch the chain is six carbons long, so the parent is hexane.
• Two methyl branches remain. Numbering from the nearer end puts them on carbons 2 and 4, not 3 and 5.
• Name: 2,4-dimethylhexane, C₈H₁₈.

Classifying reactions (4.6)

Type	What happens	Example
Addition	Two molecules join to make one product	C₂H₄ + H₂ → C₂H₆
Substitution	One atom or group is replaced by another	C₂H₆ + Br₂ → C₂H₅Br + HBr
Oxidation	Gain of oxygen or loss of hydrogen	Combustion of an alkane
Reduction	Gain of hydrogen or loss of oxygen	Hydrogen added to an alkene
Polymerisation	Many small monomers join into a long chain	n C₂H₄ → poly(ethene)

One reaction can fit two labels. Adding hydrogen to an alkene is an addition and also a reduction.

Bond breaking, free radicals and electrophiles (4.7–4.8)

• Homolytic fission: each atom takes one electron of the pair. The products are free radicals. Cl–Cl → 2Cl•
• Heterolytic fission: one atom takes both electrons. The products are ions. H–Br → H⁺ + Br⁻

A free radical is a species with an unpaired electron, shown by a single dot. An electrophile is a species that accepts a pair of electrons to form a new covalent bond, such as H⁺ attacking an alkene in Topic 5.

Alkanes and cycloalkanes (4.9)

Alkanes and cycloalkanes are hydrocarbons (compounds of carbon and hydrogen only) that are saturated (single bonds only).

• Alkanes: CₙH₂ₙ₊₂
• Cycloalkanes: CₙH₂ₙ (closing the ring removes two hydrogens)

Worked example: identifying from relative molecular mass

An alkane has Mᵣ = 114. Each CH₂ unit adds 14 and the two extra hydrogens add 2, so 14n + 2 = 114, giving n = 8: C₈H₁₈. A cycloalkane with Mᵣ = 84 gives 14n = 84, so n = 6: C₆H₁₂.

Structural isomerism (4.10–4.11)

Structural isomers have the same molecular formula but different structural formulae. For cycloalkanes, vary ring size as well as branch positions.

Worked example: the isomers of C₆H₁₄

Structural formula	Name
CH₃CH₂CH₂CH₂CH₂CH₃	hexane
CH₃CH(CH₃)CH₂CH₂CH₃	2-methylpentane
CH₃CH₂CH(CH₃)CH₂CH₃	3-methylpentane
CH₃C(CH₃)₂CH₂CH₃	2,2-dimethylbutane
CH₃CH(CH₃)CH(CH₃)CH₃	2,3-dimethylbutane

That is five isomers. An ethyl on carbon 2 of butane is not new: the longest chain becomes five carbons (3-methylpentane). For cycloalkanes, C₄H₈ gives cyclobutane and methylcyclopropane.

Fuels from crude oil (4.12)

Crude oil is a mixture of hydrocarbons, mostly alkanes. Alkanes are used as fuels.

• Fractional distillation separates crude oil into fractions by boiling temperature in a column that is cooler towards the top. It is a physical separation.
• Cracking breaks long-chain alkanes into shorter alkanes (for petrol) and alkenes (for polymers). Thermal cracking uses high temperature; catalytic cracking uses a zeolite catalyst.
• Reforming turns straight-chain alkanes into branched and cyclic hydrocarbons, which burn more smoothly in petrol engines.

Worked example: writing the equations

Heptadecane, C₁₇H₃₆, cracks to nonane and butene only.

• Carbon left after nonane: 17 − 9 = 8, so two butene molecules.
• Check hydrogen: 20 + 2 × 8 = 36.
• C₁₇H₃₆ → C₉H₂₀ + 2C₄H₈

Reforming heptane into methylcyclohexane: C₇H₁₆ → C₇H₁₄ + H₂.

Further suggested practical: cracking liquid paraffin

Liquid paraffin on mineral wool is heated with aluminium oxide as catalyst, and the gas is collected over water. It decolourises bromine water, showing an alkene. Lift the delivery tube out of the water before you stop heating, so cold water is not sucked back into the hot tube.

Combustion and pollutants (4.13–4.14, 4.17)

Complete combustion gives carbon dioxide and water. For hexane:

2C₆H₁₄ + 19O₂ → 12CO₂ + 14H₂O

In a limited supply of oxygen, incomplete combustion gives carbon monoxide or carbon particulates:

2C₆H₁₄ + 13O₂ → 12CO + 14H₂O

The pollutants emitted when alkane fuels burn are carbon monoxide, oxides of nitrogen, oxides of sulfur, carbon particulates and unburned hydrocarbons.

• Carbon monoxide is toxic because it binds to haemoglobin more strongly than oxygen does, so the blood carries less oxygen to the cells.
• Oxides of nitrogen form because the very high temperature inside an engine makes atmospheric nitrogen combine with oxygen (N₂ + O₂ → 2NO). NO is oxidised to NO₂, which forms nitric acid in rain: 4NO₂ + O₂ + 2H₂O → 4HNO₃.
• Oxides of sulfur come from sulfur impurities in the fuel (S + O₂ → SO₂), and SO₂ + H₂O → H₂SO₃.

These acidic oxides cause acid rain.

Alternative fuels and carbon neutrality (4.15–4.16)

Crude oil is finite, so its fuels are not sustainable, and burning them releases CO₂, a greenhouse gas linked to climate change. A fuel is carbon-neutral if there is no net release of CO₂ over its whole life cycle: production, transport and use.

• Petrol is not carbon-neutral: its carbon was locked underground for millions of years.
• Bioethanol is made by fermenting sugars from plants. In theory the CO₂ absorbed by photosynthesis (6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂) equals the CO₂ released by fermentation (C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂) and by combustion (2C₂H₅OH + 6O₂ → 4CO₂ + 6H₂O): six in, six out. In practice, farm machinery, fertiliser, distillation and transport release extra CO₂.
• Hydrogen burns to water only (2H₂ + O₂ → 2H₂O), but its footprint depends on how it is made. Hydrogen from methane and steam (CH₄ + H₂O → CO + 3H₂) releases carbon oxides; hydrogen from electrolysis is close to carbon-neutral only if the electricity is renewable or nuclear.

Worked example: CO₂ from two fuels

Calculate the mass of CO₂ released when 1.00 kg of each fuel burns completely (C = 12.0, H = 1.0, O = 16.0).

• Hexane: Mᵣ = 86.0, so 1000 ÷ 86.0 = 11.63 mol. Each mole gives 6 mol CO₂: 69.77 mol × 44.0 g mol⁻¹ = 3.07 kg CO₂.
• Ethanol: Mᵣ = 46.0, so 1000 ÷ 46.0 = 21.74 mol. Each mole gives 2 mol CO₂: 43.48 mol × 44.0 g mol⁻¹ = 1.91 kg CO₂.

All the hexane CO₂ is fossil carbon; the ethanol’s carbon came from the air within one growing season.

Halogenation: free radical substitution (4.17–4.18)

In ultraviolet light, a halogen atom replaces a hydrogen atom of the alkane. For ethane and bromine:

C₂H₆ + Br₂ → C₂H₅Br + HBr

Initiation. UV light breaks Br–Br homolytically: two curly half-arrows run from the middle of the bond, one to each bromine atom.

Br₂ → 2Br•

Propagation. Each step uses one radical and makes another.

Br• + C₂H₆ → C₂H₅• + HBr

C₂H₅• + Br₂ → C₂H₅Br + Br•

Termination. Two radicals join.

Br• + Br• → Br₂

C₂H₅• + Br• → C₂H₅Br

C₂H₅• + C₂H₅• → C₄H₁₀

Limited use in synthesis. Bromoethane can itself be attacked by Br•, giving C₂H₄Br₂ and then more highly substituted products, and termination adds butane. The mixture must be separated, so the yield of any one product is low.

Common errors

• Defining risk as “how dangerous the substance is”. That is the hazard.
• Taking the row on the page as the longest chain without checking the branches.
• Using two-headed curly arrows in the initiation step instead of half-arrows.
• Making H• in propagation. The first step makes HBr and an alkyl radical.
• Saying CO “stops you breathing” instead of explaining its binding to haemoglobin.
• Calling bioethanol fully carbon-neutral, or hydrogen carbon-neutral whatever its source.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry (XCH11/YCH11) specification, Issue 1, September 2017, Pearson Education Limited: Unit 1, Topic 4: Introductory Organic Chemistry and Alkanes, outcomes 4.1 to 4.18.