Organic Chemistry

:::info Board Coverage AQA Paper 2 | Edexcel Paper 2 | OCR A Gateway C5 & C6 | WJEC C4 :::

1. Introduction to Organic Chemistry

1.1 What Is Organic Chemistry?

Organic chemistry is the study of carbon compounds. Carbon is unique because each carbon atom Can form four covalent bonds and can bond with other carbon atoms to form chains and rings of Virtually unlimited length. This property, called catenation, gives rise to millions of known Organic compounds.

Carbon forms four covalent bonds because it has four electrons in its outer shell and needs four More to achieve a stable configuration. The strength and versatility of the C-C bond (about 348 KJ/mol) makes long chains stable, while the C-H bond (about 412 kJ/mol) provides a convenient way to Satisfy the remaining valences.

1.2 Hydrocarbons

Hydrocarbons are compounds containing only carbon and hydrogen atoms. They are the simplest Organic compounds and are the main components of crude oil and natural gas.

Hydrocarbons are classified into two broad families: alkanes (saturated, single bonds only) and alkenes (unsaturated, containing at least one C=C double bond).

1.3 Homologous Series

A homologous series is a family of organic compounds with the same general formula, similar Chemical properties, and successive members differing by CH $_2$ .

Properties of a homologous series:

Same functional group
Similar chemical properties (the functional group determines the reactions)
Gradual change in physical properties (e.g. Boiling point increases with chain length)
Each successive member differs by CH $_2$ (relative molecular mass increases by 14)

The regular change in physical properties is a consequence of increasing molecular size. Longer Chains have stronger London dispersion forces between molecules, leading to higher boiling points, Higher melting points, and lower volatility.

1.4 Naming Organic Compounds

The first four members of each homologous series have historical names:

Number of carbons	Prefix
1	Meth-
2	Eth-
3	Prop-
4	But-
5	Pent-
6	Hex-
7	Hept-
8	Oct-

These prefixes are combined with the suffix that identifies the homologous series: -ane for alkanes, -ene for alkenes, -ol for alcohols, -oic acid for carboxylic acids.

2. Alkanes

2.1 General Properties

Alkanes are saturated hydrocarbons — all carbon-carbon bonds are single bonds. The general Formula is:

$\mathrm{C_n\mathrm{H_{2n+2}$

The formula is derived from the fact that each carbon atom forms four bonds. In a straight chain, The two end carbons are bonded to three hydrogens each, and the remaining $(n-2)$ carbons are bonded To two hydrogens each. Total hydrogens: $2 \times 3 + (n-2) \times 2 = 6 + 2n - 4 = 2n + 2$ .

Name	Formula	Structure	Boiling Point ( $^{\circ}$ C)
Methane	CH $_4$	1 carbon	-162
Ethane	C $_2$ H $_6$	2 carbons	-89
Propane	C $_3$ H $_8$	3 carbons	-42
Butane	C $_4$ H $_{10}$	4 carbons	-0.5
Pentane	C $_5$ H $_{12}$	5 carbons	36
Hexane	C $_6$ H $_{14}$	6 carbons	69
Heptane	C $_7$ H $_{16}$	7 carbons	98
Octane	C $_8$ H $_{18}$	8 carbons	126

Trend: Boiling point increases with chain length due to stronger London forces (more surface Area for intermolecular attraction, more electrons to polarise).

2.2 Combustion of Alkanes

Complete combustion (plenty of oxygen):

$\mathrm{C_n\mathrm{H_{2n+2} + \frac{3n+1}{2}\mathrm{O_2 \to n\mathrm{CO_2 + (n+1)\mathrm{H_2\mathrm{O$

Example: $\mathrm{CH_4 + 2\mathrm{O_2 \to \mathrm{CO_2 + 2\mathrm{H_2\mathrm{O$

Example: $2\mathrm{C_4\mathrm{H_{10} + 13\mathrm{O_2 \to 8\mathrm{CO_2 + 10\mathrm{H_2\mathrm{O$

Incomplete combustion (limited oxygen) produces carbon monoxide (CO) and/or carbon (soot):

$2\mathrm{CH_4 + 3\mathrm{O_2 \to 2\mathrm{CO + 4\mathrm{H_2\mathrm{O$ $\mathrm{CH_4 + \mathrm{O_2 \to \mathrm{C + 2\mathrm{H_2\mathrm{O$

:::caution Carbon monoxide is a toxic, colourless, odourless gas that binds to haemoglobin more Strongly than oxygen, preventing oxygen transport in the blood. Incomplete combustion is dangerous Because CO is produced without any visible or olfactory warning.

Incomplete combustion occurs when the oxygen supply is insufficient. The extent of incompleteness Depends on the oxygen-to-fuel ratio. With very limited oxygen, solid carbon (soot) is produced; with Moderately limited oxygen, carbon monoxide is produced.

2.3 Halogenation of Alkanes

Alkanes react with halogens in the presence of UV light (substitution reaction):

$\mathrm{CH_4 + \mathrm{Cl_2 \xrightarrow{\mathrm{UV} \mathrm{CH_3\mathrm{Cl + \mathrm{HCl$

This is a free radical substitution reaction. The reaction can continue to produce Dichloromethane, trichloromethane, and tetrachloromethane.

The mechanism involves three stages:

Initiation: UV light provides enough energy to break the Cl-Cl bond homolytically (each atom gets one electron), producing chlorine radicals.
Propagation: A chlorine radical abstracts a hydrogen from methane, producing HCl and a methyl radical. The methyl radical then reacts with Cl $_2$ Producing chloromethane and another chlorine radical. This is a chain reaction.
Termination: Two radicals combine, ending the chain.

2.4 Higher Tier: Free Radical Substitution Mechanism

Initiation:

$\mathrm{Cl_2 \xrightarrow{\mathrm{UV} 2\mathrm{Cl^\bullet$

Propagation:

$\mathrm{CH_4 + \mathrm{Cl^\bullet \to \mathrm{CH_3^\bullet + \mathrm{HCl$ $\mathrm{CH_3^\bullet + \mathrm{Cl_2 \to \mathrm{CH_3\mathrm{Cl + \mathrm{Cl^\bullet$

Termination:

$\mathrm{Cl^\bullet + \mathrm{Cl^\bullet \to \mathrm{Cl_2$ $\mathrm{CH_3^\bullet + \mathrm{Cl^\bullet \to \mathrm{CH_3\mathrm{Cl$ $\mathrm{CH_3^\bullet + \mathrm{CH_3^\bullet \to \mathrm{C_2\mathrm{H_6$

The propagation step is self-sustaining: each time a chlorine radical is consumed, another is Produced. This is why UV light is needed only to start the reaction, not to sustain it.

2.5 Worked Examples on Alkane Reactions

Worked Example. Write the balanced equation for the complete combustion of pentane (C $_5$ H $_{12}$ ).

$\mathrm{C_5\mathrm{H_{12} + 8\mathrm{O_2 \to 5\mathrm{CO_2 + 6\mathrm{H_2\mathrm{O$

Check: C: 5 = 5, H: 12 = 12, O: 16 = 10 + 6 = 16.

Worked Example. Write the balanced equation for the incomplete combustion of propane producing Carbon monoxide.

$2\mathrm{C_3\mathrm{H_8 + 7\mathrm{O_2 \to 6\mathrm{CO + 8\mathrm{H_2\mathrm{O$

3. Alkenes

3.1 General Properties

Alkenes are unsaturated hydrocarbons — they contain at least one carbon-carbon double bond (C=C). The general formula is:

$\mathrm{C_n\mathrm{H_{2n}$

Name	Formula
Ethene	C $_2$ H $_4$
Propene	C $_3$ H $_6$
Butene	C $_4$ H $_8$

The double bond consists of one sigma bond and one pi bond. The pi bond is formed by the sideways Overlap of p orbitals above and below the plane of the molecule. This pi bond is weaker than the Sigma bond and is what makes alkenes more reactive than alkanes.

3.2 Test for Unsaturation

Alkenes decolourise bromine water (orange to colourless). This is because the double bond opens And bromine adds across it:

$\mathrm{C_2\mathrm{H_4 + \mathrm{Br_2 \to \mathrm{C_2\mathrm{H_4\mathrm{Br_2$

Alkanes do NOT decolourise bromine water (no double bond to react with).

3.3 Addition Reactions

Alkenes undergo addition reactions because the double bond can open up, allowing new atoms to Bond to the carbon atoms.

Hydrogenation: $\mathrm{C_2\mathrm{H_4 + \mathrm{H_2 \to \mathrm{C_2\mathrm{H_6$ (with nickel Catalyst)

Hydration: $\mathrm{C_2\mathrm{H_4 + \mathrm{H_2\mathrm{O \to \mathrm{C_2\mathrm{H_5\mathrm{OH$ (with Phosphoric acid catalyst)

Reaction with steam: This is used industrially to make ethanol.

Reaction with hydrogen bromide: $\mathrm{C_2\mathrm{H_4 + \mathrm{HBr \to \mathrm{C_2\mathrm{H_5\mathrm{Br$

Reaction with halogens: $\mathrm{C_2\mathrm{H_4 + \mathrm{Br_2 \to \mathrm{C_2\mathrm{H_4\mathrm{Br_2$

In all addition reactions, the pi bond breaks and two new sigma bonds form. The carbon atoms go from Sp $^2$ hybridised (trigonal planar) to sp $^3$ hybridised (tetrahedral).

3.4 Addition Polymers

Alkenes can undergo addition polymerisation to form long-chain molecules called polymers (plastics).

The double bond in each alkene molecule (monomer) opens and links to form a polymer chain.

Example: Poly(ethene) from ethene

$n\mathrm{C_2\mathrm{H_4 \to \mathrm{(-CH_2\mathrm{-CH_2\mathrm{-)_n$

Example: Poly(propene) from propene

$n\mathrm{C_3\mathrm{H_6 \to \mathrm{(-CH_2\mathrm{-CH(CH_3\mathrm{)-)_n$

Example: Poly(chloroethene) / PVC from chloroethene

$n\mathrm{C_2\mathrm{H_3\mathrm{Cl \to \mathrm{(-CH_2\mathrm{-CHCl-)_n$

3.5 Problems with Polymers

Most polymers are not biodegradable. They persist in landfill sites and can cause environmental Problems. Disposal by incineration can release toxic gases if chlorine-containing polymers (like PVC) are burned.

Solutions:

Recycling: Sorting and reprocessing waste plastics. Different polymers must be separated because they cannot be recycled together.
Biodegradable polymers: Made from plant materials (e.g. Corn starch), which can be broken down by microorganisms.
Incineration: Burning plastics for energy recovery (but releases CO $_2$ and potentially toxic gases).

3.6 Summary: Alkanes vs. Alkenes

Property	Alkanes	Alkenes
General formula	C $_n$ H $_{2n+2}$	C $_n$ H $_{2n}$
Bond type	Single bonds only	Contains C=C double bond
Reactivity	Less reactive	More reactive
Bromine water test	No reaction	Decolourises
Combustion	Complete and incomplete	Same

4. Alcohols

4.1 General Properties

Alcohols contain the hydroxyl group (-OH). The general formula is:

$\mathrm{C_n\mathrm{H_{2n+1}\mathrm{OH$

Name	Formula	Boiling Point ( $^{\circ}$ C)
Methanol	CH $_3$ OH	65
Ethanol	C $_2$ H $_5$ OH	78
Propanol	C $_3$ H $_7$ OH	97
Butanol	C $_4$ H $_9$ OH	118

Physical properties:

Soluble in water (the -OH group forms hydrogen bonds with water)
Boiling points higher than corresponding alkanes (hydrogen bonding between molecules)
Ethanol is miscible with water in all proportions

4.2 Reactions of Alcohols

Combustion:

$\mathrm{C_2\mathrm{H_5\mathrm{OH + 3\mathrm{O_2 \to 2\mathrm{CO_2 + 3\mathrm{H_2\mathrm{O$

Oxidation: Ethanol can be oxidised to ethanoic acid using an oxidising agent (e.g. Potassium Dichromate(VI) with dilute sulfuric acid).

$\mathrm{C_2\mathrm{H_5\mathrm{OH \to \mathrm{CH_3\mathrm{COOH$

The oxidation proceeds in two stages: ethanol is first oxidised to ethanal (an aldehyde), which is Then further oxidised to ethanoic acid (a carboxylic acid).

Dehydration: Ethanol can be dehydrated to ethene using aluminium oxide catalyst at high Temperature.

$\mathrm{C_2\mathrm{H_5\mathrm{OH \to \mathrm{C_2\mathrm{H_4 + \mathrm{H_2\mathrm{O$

Reaction with sodium:

$2\mathrm{C_2\mathrm{H_5\mathrm{OH + 2\mathrm{Na \to 2\mathrm{C_2\mathrm{H_5\mathrm{ONa + \mathrm{H_2$

This reaction is similar to the reaction of sodium with water but is less vigorous. It confirms that Alcohols contain the -OH group.

4.3 Uses of Alcohols

Ethanol: Solvents, fuels, alcoholic drinks
Methanol: Industrial solvent, fuel (but toxic)
Ethanol as a biofuel: Produced by fermentation of sugars, can be used as a renewable fuel

4.4 Fermentation

Fermentation uses yeast to convert sugars into ethanol and carbon dioxide:

$\mathrm{C_6\mathrm{H_{12}\mathrm{O_6 \to 2\mathrm{C_2\mathrm{H_5\mathrm{OH + 2\mathrm{CO_2$

Conditions:

Temperature: 30—40 $^{\circ}$ C (optimal for yeast enzymes)
Anaerobic (absence of oxygen)
pH: slightly acidic

The ethanol produced can be purified by distillation. Fermentation produces ethanol Concentrations of up to about 15%, beyond which the ethanol itself begins to inhibit the yeast.

Worked Example. Calculate the maximum mass of ethanol that can be produced from 180 g of glucose By fermentation.

$M_r(\mathrm{glucose) = 180, \quad M_r(\mathrm{ethanol) = 46$

$n(\mathrm{glucose) = \frac{180}{180} = 1 \mathrm{ mol$

From the equation, 1 mol glucose produces 2 mol ethanol.

$m(\mathrm{ethanol) = 2 \times 46 = 92 \mathrm{ g$

5. Carboxylic Acids

5.1 General Properties

Carboxylic acids contain the carboxyl group (-COOH). The general formula is:

$\mathrm{C_n\mathrm{H_{2n+1}\mathrm{COOH$

Name	Formula
Methanoic acid	HCOOH
Ethanoic acid	CH $_3$ COOH
Propanoic acid	C $_2$ H $_5$ COOH
Butanoic acid	C $_3$ H $_7$ COOH

5.2 Properties

React with carbonates to produce a salt, water, and CO $_2$
React with metals to produce a salt and hydrogen
Form salts called carboxylates (e.g. Sodium ethanoate)
Weak acids (partially dissociate in solution)
Higher boiling points than alkanes and alcohols of similar size (form dimers via hydrogen bonding)

Carboxylic acids are weak acids because the O-H bond in the -COOH group is only partially Dissociated. The equilibrium lies far to the left: most carboxylic acid molecules remain Undissociated in solution.

5.3 Reactions

With sodium carbonate:

$2\mathrm{CH_3\mathrm{COOH + \mathrm{Na_2\mathrm{CO_3 \to 2\mathrm{CH_3\mathrm{COONa + \mathrm{H_2\mathrm{O + \mathrm{CO_2$

With metals:

$2\mathrm{CH_3\mathrm{COOH + 2\mathrm{Na \to 2\mathrm{CH_3\mathrm{COONa + \mathrm{H_2$

With alcohols (esterification):

$\mathrm{CH_3\mathrm{COOH + \mathrm{C_2\mathrm{H_5\mathrm{OH \rightleftharpoons \mathrm{CH_3\mathrm{COOC_2\mathrm{H_5 + \mathrm{H_2\mathrm{O$

5.4 Esters

Esters are formed when a carboxylic acid reacts with an alcohol (esterification). This is a condensation reaction (water is eliminated).

The reaction is catalysed by a strong acid (e.g. Concentrated sulfuric acid). The reaction is Reversible, so the yield can be improved by using an excess of one reactant or by removing the water As it forms.

Esters have distinctive sweet/fruity smells and are used in:

Flavourings and fragrances
Solvents
Plasticisers (added to plastics to make them flexible)

6. Crude Oil and Fractional Distillation

6.1 Formation of Crude Oil

Crude oil is formed from the remains of microscopic marine organisms that died millions of years Ago. Under high pressure and temperature, these remains were converted into a mixture of Hydrocarbons. This process takes millions of years, making crude oil a non-renewable resource.

6.2 Fractional Distillation

Crude oil is a mixture of hydrocarbons with different chain lengths. These are separated by fractional distillation in a fractionating column.

Process:

Crude oil is heated to about $400^{\circ}\mathrm{C$ to vaporise it
The vapour enters the fractionating column at the bottom
The column is hotter at the bottom and cooler at the top
Hydrocarbons condense at different heights depending on their boiling point
Shorter chains (lower boiling points) condense near the top
Longer chains (higher boiling points) condense near the bottom

Fractions and their uses:

Fraction	Carbon Chain	Use
Refinery gas	C $_1$ —C $_4$	Bottled gas, fuel
Gasoline (petrol)	C $_5$ —C $_10$	Car fuel
Naphtha	C $_6$ —C $_12$	Chemical feedstock
Kerosene	C $_11$ —C $_15$	Jet fuel, paraffin
Diesel oil	C $_15$ —C $_19$	Diesel engines
Fuel oil	C $_20$ —C $_30$	Ship fuel, power stations
Bitumen	C $_{30+}$	Road surfacing

The trend in properties down the fractionating column: shorter chains are more volatile, have lower Boiling points, are less viscous, and burn more cleanly. Longer chains are less volatile, have Higher boiling points, are more viscous, and burn with more soot.

6.3 Cracking

Longer-chain hydrocarbons are in greater supply than demand, while shorter chains are in high Demand. Cracking breaks long-chain alkanes into shorter, more useful alkanes and alkenes.

Thermal cracking: High temperature and pressure. Produces a mixture of shorter alkanes and Alkenes. Used to produce ethene for polymer manufacture.

Catalytic cracking: Lower temperature, uses a zeolite catalyst. Produces more branched-chain Alkanes and aromatic compounds. Produces higher-quality petrol.

Example:

$\mathrm{C_{10}\mathrm{H_{22} \to \mathrm{C_8\mathrm{H_{18} + \mathrm{C_2\mathrm{H_4$

(decane $\to$ octane + ethene)

Cracking is a thermal decomposition reaction. The C-C bonds in long-chain alkanes break when heated, Producing shorter fragments. Some of these fragments contain a C=C double bond (alkenes), which is Why cracking is a useful source of alkenes for the chemical industry.

Worked Example. A hydrocarbon with formula C $_{12}$ H $_{26}$ undergoes cracking to produce C $_8$ H $_{18}$ and another product. Identify the other product.

The atoms must balance: 12 C and 26 H on the left. The known product has 8 C and 18 H, so the other Product has $12 - 8 = 4$ C and $26 - 18 = 8$ H. The formula is C $_4$ H $_8$ Which is butene (an Alkene).

$\mathrm{C_{12}\mathrm{H_{26} \to \mathrm{C_8\mathrm{H_{18} + \mathrm{C_4\mathrm{H_8$

7. Biodegradable Polymers and Natural Polymers

7.1 Natural Polymers

Polymer	Monomer	Links
Starch	Glucose	Glycosidic bonds
Cellulose	Glucose	Glycosidic bonds
Protein	Amino acids	Peptide bonds
DNA	Nucleotides	Phosphodiester bonds

Natural polymers are formed by condensation polymerisation, where monomers join with the elimination Of a small molecule ( water).

7.2 Biodegradable Polymers

Polymers made from renewable resources (e.g. Corn starch, sugar cane) that can be broken down by Microorganisms. Examples include PLA (polylactic acid) and PHA (polyhydroxyalkanoates).

Advantages: Reduce landfill waste, made from renewable resources, lower carbon footprint.

Disadvantages: More expensive than conventional plastics, may require specific composting Conditions, may contaminate recycling streams if mixed with conventional plastics.

7.3 Addition vs. Condensation Polymerisation

Feature	Addition Polymerisation	Condensation Polymerisation
Monomer type	Alkene (C=C)	Two different functional groups
Small molecule lost	None	Water (or other)
Example	Poly(ethene)	Nylon, polyester
By-products	None	Water

8. Alcohols vs. Carboxylic Acids vs. Esters: A Comparison

Property	Alcohols	Carboxylic Acids	Esters
Functional group	-OH	-COOH	-COO-
Boiling point trend	Higher than alkanes	Higher than alcohols	Lower than both
Acidity	Very weak	Weak	Not acidic
Hydrogen bonding	Yes	Yes (stronger, dimers)	Accept but do not donate
Typical reaction	Oxidation, dehydration	Esterification	Hydrolysis
Odour	Characteristic	Pungent	Sweet/fruity

9. Summary of Organic Reactions

9.1 Reaction Types Summary

Reaction Type	Description	Example
Combustion	Burning in oxygen	$\mathrm{CH_4 + 2\mathrm{O_2 \to \mathrm{CO_2 + 2\mathrm{H_2\mathrm{O$
Addition	Atoms added across C=C double bond	$\mathrm{C_2\mathrm{H_4 + \mathrm{Br_2 \to \mathrm{C_2\mathrm{H_4\mathrm{Br_2$
Substitution	Atom replaced by another atom	$\mathrm{CH_4 + \mathrm{Cl_2 \to \mathrm{CH_3\mathrm{Cl + \mathrm{HCl$
Polymerisation	Monomers join to form long chains	$n\mathrm{C_2\mathrm{H_4 \to \mathrm{poly(ethene)$
Esterification	Acid + alcohol $\to$ ester + water	$\mathrm{CH_3\mathrm{COOH + \mathrm{C_2\mathrm{H_5\mathrm{OH \to \mathrm{ester + \mathrm{H_2\mathrm{O$
Fermentation	Sugar $\to$ ethanol + carbon dioxide	$\mathrm{C_6\mathrm{H_{12}\mathrm{O_6 \to 2\mathrm{C_2\mathrm{H_5\mathrm{OH + 2\mathrm{CO_2$

Common Pitfalls

Confusing alkanes and alkenes. Alkanes are saturated (single bonds only); alkenes are unsaturated (contain C=C double bond).
Writing the wrong general formula. Alkanes: C $_n$ H $_{2n+2}$ ; alkenes: C $_n$ H $_{2n}$ .
Forgetting that incomplete combustion produces carbon monoxide (toxic) or carbon (soot), not just CO $_2$ and H $_2$ O.
Confusing the test for saturation with other tests. Bromine water decolourises with alkenes (addition) but not with alkanes.
Writing polymer equations incorrectly. Make sure the repeating unit is correct and show the $n$ subscript.
Confusing condensation polymerisation with addition polymerisation. Addition: no small molecule eliminated; condensation: water (or other small molecule) eliminated.
Stating that ethanol is produced by hydration of ethene and fermentation under the same conditions. Hydration uses high temperature and a phosphoric acid catalyst; fermentation uses yeast at 30—40 $^{\circ}$ C.
Forgetting that carboxylic acids are weak acids. They dissociate partially, not completely. They still react with carbonates and metals.
Confusing the terms “volatile” and “flammable”. Volatile means evaporates; flammable means catches fire . Shorter-chain hydrocarbons are more volatile.
Not balancing combustion equations. Remember to count both C and H atoms on both sides, and check that O atoms balance too.

Practice Questions

Write the balanced equation for the complete combustion of butane (C $_4$ H $_{10}$ ).
Explain why the boiling point of alkanes increases with chain length.
Draw the displayed formula and write the balanced equation for the polymerisation of propene.
Describe the test to distinguish between hexane and hexene.
Ethanol can be produced by fermentation of glucose or by hydration of ethene. Describe both methods, including the conditions and equations.
Describe how fractional distillation separates crude oil into fractions.
Write the equation for the reaction of propanoic acid with sodium carbonate.
500 g of octane (C $_8$ H $_{18}$ ) undergoes complete combustion. Calculate the mass of CO $_2$ produced.
Explain why poly(ethene) causes environmental problems and describe two ways to reduce these problems.
Cracking of C $_{15}$ H $_{32}$ produces C $_{10}$ H $_{22}$ and one other product. Write the balanced equation and name the other product.
Explain the mechanism of the free radical substitution of methane with chlorine, including all three stages.
Describe the similarities and differences between the reactions of alcohols with sodium and the reactions of carboxylic acids with sodium.
Explain why the boiling point of ethanoic acid is higher than that of ethanol, despite both having similar molecular masses.
Describe how biodegradable polymers differ from conventional polymers, and explain why they are considered more environmentally friendly.
Write the balanced equation for the esterification of propanoic acid with ethanol, and name the ester produced.
Calculate the maximum volume of CO $_2$ produced at RTP when 360 g of glucose undergoes fermentation.
Explain why the products of cracking always include at least one alkene.
Compare the environmental impact of disposing of poly(ethene) by landfill and by incineration.
Name the ester formed from methanol and ethanoic acid, and write the balanced equation for its formation.
Explain why crude oil is classified as a non-renewable resource.

10. Polymers in Detail

10.1 Identifying Monomers from Polymers

To identify the monomer, remove the brackets and the subscript $n$ :

Polymer	Repeating Unit	Monomer
Poly(ethene)	$-$ CH $_2$ -CH $_2$ - $-$ _n$	Ethene (C $_2$ H $_4$ )
Poly(propene)	$-$ CH $_2$ -CH(CH $_3$ )- $-$ _n$	Propene (C $_3$ H $_6$ )
PVC	$-$ CH $_2$ -CHCl- $-$ _n$	Chloroethene (C $_2$ H $_3$ Cl)
Poly(tetrafluoroethene)	$-$ CF $_2$ -CF $_2$ - $-$ _n$	Tetrafluoroethene (C $_2$ F $_4$ )

10.2 Problems with Disposal

Disposal Method	Advantages	Disadvantages
Landfill	Cheap, simple	Leaches chemicals, never decomposes
Incineration	Energy recovery	Can release toxic gases (e.g. HCl from PVC)
Recycling	Conserves resources	Sorting is difficult and expensive

10.3 Biodegradable Alternatives

PLA (polylactic acid) is made from fermented corn starch. It breaks down in composting conditions Within months. PHB (polyhydroxybutyrate) is produced by bacterial fermentation and is fully Biodegradable.

10.4 Worked Examples

Worked Example. A polymer has the repeating unit $-$ CH $_2$ -CHCl- $-$ _n$. (a) Identify the Monomer. (b) Draw the displayed formula of the monomer. (c) Name the polymer.

(a) The monomer is chloroethene, C $_2$ H $_3$ Cl.

(b) H $_2$ C=CHCl

Worked Example. Write the equation for the polymerisation of but-2-ene and name the polymer.

$n\mathrm{CH_3\mathrm{CH=\mathrm{CHCH_3 \to \mathrm{(-CH(CH_3\mathrm{)-CH_2\mathrm{-)_n$

The polymer is poly(but-2-ene), also called polybutene.

11. Cracking in More Detail

11.1 Why Cracking Is Necessary

The demand for shorter-chain hydrocarbons (for petrol, alkenes for polymers) is much greater than The supply from fractional distillation. Cracking converts less useful long-chain alkanes into more Valuable shorter-chain alkanes and alkenes.

11.2 Types of Cracking

Type	Conditions	Products
Thermal cracking	High temperature (400—900 $^{\circ}$ C), high pressure	Shorter alkanes + alkenes
Catalytic cracking	Moderate temperature, zeolite catalyst	Branched alkanes + cycloalkanes + aromatic compounds

11.3 Worked Examples

Worked Example. C $_{14}$ H $_{30}$ undergoes thermal cracking. One product is C $_8$ H $_{18}$ . Identify the other product.

Atom balance: C: $14 - 8 = 6$ H: $30 - 18 = 12$ . The other product is C $_6$ H $_{12}$ Which is Cyclohexane.

$\mathrm{C_{14}\mathrm{H_{30} \to \mathrm{C_8\mathrm{H_{18} + \mathrm{C_6\mathrm{H_{12}$

Worked Example. Explain why the product of cracking always contains at least one alkene.

In thermal cracking, the C-C bonds break randomly. When a bond breaks and the resulting fragments Rearrange, at least one fragment will have an unsaturated (double) bond because there are too few Hydrogen atoms to fully saturate all the carbon atoms. This is a consequence of the fact that Alkanes have a general formula of C $_n$ H $_{2n+2}$ — splitting a chain necessarily creates fragments With fewer hydrogen atoms relative to their carbon atoms.

12. Additional Practice Questions

Calculate the theoretical yield of CO $_2$ when 25 g of CaCO $_3$ reacts with excess HCl.
What mass of NaCl is produced when 11.7 g of sodium reacts with excess chlorine gas?
A student performs a titration and obtains the following results: 24.8, 24.6, 24.7 cm $^3$ . What average titre should they use?
Calculate the atom economy for the reaction: $2\mathrm{Na{} + \mathrm{Cl{}_2 \to 2\mathrm{NaCl{}$ .
What is the percentage yield if 5.0 g of CaCO $_3$ produces only 2.0 g of CO $_2$ ?
Calculate the mass of water of crystallisation in 12.5 g of MgSO $_4 \cdot 7$ H $_2$ O.
What volume of gas is produced at RTP when 0.200 mol of Zn reacts with excess H $_2$ SO $_4$ ?
A solution contains 20 g of NaOH in 500 cm $^3$ . What is the molar concentration?
Calculate the atom economy for the Haber process: $\mathrm{N{}_2 + 3\mathrm{H{}_2 \rightleftharpoons 2\mathrm{NH{}_3$ , where ammonia is the desired product.
5.0 g of a mixture of Mg and Zn reacts with excess acid to produce 2.4 dm $^3$ of H $_2$ at RTP. Calculate the percentage of Mg in the mixture if the Zn is in excess. (Hint: assume only Mg reacts.)

Practice Problems

Question 1: Homologous series

Explain what is meant by a homologous series. Describe two trends in physical properties as you go down the alkane series.

Answer

A homologous series is a family of organic compounds with the same general formula, similar chemical properties, and successive members differing by $-\mathrm{CH{}_2-$ .

Trends in alkanes: (1) boiling point increases down the series because longer molecules have stronger London forces between them. (2) Viscosity increases because longer chains become more entangled. (3) Flammability generally decreases as molecules get larger.

Question 2: Alkane reactions

Describe the complete combustion of butane ( $\mathrm{C{}_4\mathrm{H{}_{10}$ ). Write the balanced equation and explain why incomplete combustion is dangerous.

Answer

Complete combustion: $2\mathrm{C{}_4\mathrm{H{}_{10} + 13\mathrm{O{}_2 \to 8\mathrm{CO{}_2 + 10\mathrm{H{}_2\mathrm{O{}$ .

Incomplete combustion occurs when there is insufficient oxygen, producing carbon monoxide ( $\mathrm{CO{}$ ) and/or carbon (soot) instead of $\mathrm{CO{}_2$ . Carbon monoxide is dangerous because it is a colourless, odourless gas that binds to haemoglobin in red blood cells, reducing their capacity to carry oxygen. This can lead to carbon monoxide poisoning and death.

Question 3: Alkenes and addition reactions

Describe the test to distinguish between an alkane and an alkene. Write an equation for the reaction of ethene with bromine water.

Answer

Add bromine water (orange-brown) to each hydrocarbon. Alkanes do not react with bromine water at room temperature — the colour remains. Alkenes decolourise bromine water because the $\mathrm{C=C{}$ double bond undergoes an addition reaction.

$\mathrm{CH{}_2=\mathrm{CH{}_2 + \mathrm{Br{}_2 \to \mathrm{CH{}_2\mathrm{BrCH{}_2\mathrm{Br{}$ (1,2-dibromoethane).

Question 4: Alcohols

Describe the reaction of ethanol with sodium. Write the equation and name the type of reaction.

Answer

Ethanol reacts with sodium, producing sodium ethoxide and hydrogen gas. Effervescence is observed (bubbles of $\mathrm{H{}_2$ ).

$2\mathrm{CH{}_3\mathrm{CH{}_2\mathrm{OH{} + 2\mathrm{Na{} \to 2\mathrm{CH{}_3\mathrm{CH{}_2\mathrm{ONa{} + \mathrm{H{}_2$

This is a substitution reaction (the -OH hydrogen is replaced by sodium).

Question 5: Polymers

Explain how addition polymers are formed from monomers, using poly(ethene) as an example. Include the displayed formula of the repeating unit.

Answer

Addition polymers are formed when many small alkene monomers join together, opening their $\mathrm{C=C{}$ double bonds. No other product is formed.

For poly(ethene): many ethene molecules ( $\mathrm{CH{}_2=\mathrm{CH{}_2$ ) join together. The double bond opens and single bonds form between adjacent monomers.

Repeating unit: $-\mathrm{CH{}_2-\mathrm{CH{}_2-$ (displayed as a bracket with the repeating unit inside and n outside).

Worked Examples

Example 1: pH calculation

Calculate the pH of a $0.050\,\text{mol\,dm}^{-3}$ solution of HCl.

Solution:

HCl is a strong acid, so $[\text{H}^+] = 0.050\,\text{mol\,dm}^{-3}$ .

$\text{pH} = -\log_{10}[\text{H}^+] = -\log_{10}(0.050) = 1.30$

Summary

This topic covers the key concepts of Organic Chemistry for GCSE Chemistry. Focus on understanding the fundamental principles, practising with exam-style questions, and applying your knowledge to unfamiliar contexts.

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