Organic Chemistry

1. Introduction to Organic Chemistry

What is Organic Chemistry?

Organic chemistry is the study of carbon compounds. The definition traditionally excludes simple carbon compounds such as carbon monoxide ($\mathrm{CO}$), carbon dioxide ($\mathrm{CO}_2$), carbonates ($\mathrm{CO}_3^{2-}$), bicarbonates ($\mathrm{HCO}_3^-$), and carbides, as these are classified as inorganic.

Carbon is unique in its ability to form a vast number of compounds. There are over ten million known organic compounds, far exceeding the number of known inorganic compounds.

Why Carbon?

Carbon ($Z = 6$, electron configuration $1s^2\, 2s^2\, 2p^2$) has four valence electrons and can form four covalent bonds. Several properties make carbon exceptionally suited as the backbone of organic molecules:

Property	Description
Tetravalency	Carbon has four valence electrons and forms four covalent bonds, giving a wide variety of bonding patterns
Catenation	Carbon atoms can bond to other carbon atoms to form chains and rings of varying lengths
Multiple bonding	Carbon can form single ($\sigma$), double ($\sigma + \pi$), and triple ($\sigma + 2\pi$) bonds
Isomerism	The same molecular formula can represent different compounds with different structures
Bond strength	C-C and C-H bonds are relatively strong, making organic compounds generally stable

Functional Groups

A functional group is an atom or group of atoms that determines the characteristic chemical reactions of a molecule. Organic compounds are classified primarily by their functional groups.

Functional Group	Structure	Suffix / Prefix	Example
Alkane	C-C single bonds only	-ane	$\mathrm{CH}_4$ (methane)
Alkene	C=C double bond	-ene	$\mathrm{C}_2\mathrm{H}_4$ (ethene)
Alkyne	C$\equiv$C triple bond	-yne	$\mathrm{C}_2\mathrm{H}_2$ (ethyne)
Haloalkane	C-X (X = F, Cl, Br, I)	-halo-	$\mathrm{CH}_3\mathrm{Cl}$ (chloromethane)
Alcohol	-OH	-ol	$\mathrm{C}_2\mathrm{H}_5\mathrm{OH}$ (ethanol)
Aldehyde	-CHO	-al	$\mathrm{CH}_3\mathrm{CHO}$ (ethanal)
Ketone	>C=O	-one	$\mathrm{CH}_3\mathrm{COCH}_3$ (propanone)
Carboxylic acid	-COOH	-oic acid	$\mathrm{CH}_3\mathrm{COOH}$ (ethanoic acid)
Ester	-COO-	-oate	$\mathrm{CH}_3\mathrm{COOCH}_3$ (methyl ethanoate)
Amine	-NH$_2$	-amine	$\mathrm{CH}_3\mathrm{NH}_2$ (methylamine)
Amide	-CONH$_2$	-amide	$\mathrm{CH}_3\mathrm{CONH}_2$ (ethanamide)

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2. Hydrocarbons

Hydrocarbons are compounds containing only carbon and hydrogen. They are the simplest organic compounds and form the basis of all other organic chemistry.

2.1 Alkanes ($\mathrm{C}_n\mathrm{H}_{2n+2}$)

General Properties

Alkanes are saturated hydrocarbons containing only single C-C and C-H bonds. They contain the maximum possible number of hydrogen atoms per carbon atom.

Naming

The first ten alkanes must be memorised:

Prefix	Formula	Name
Meth-	$\mathrm{CH}_4$	Methane
Eth-	$\mathrm{C}_2\mathrm{H}_6$	Ethane
Prop-	$\mathrm{C}_3\mathrm{H}_8$	Propane
But-	$\mathrm{C}_4\mathrm{H}_{10}$	Butane
Pent-	$\mathrm{C}_5\mathrm{H}_{12}$	Pentane
Hex-	$\mathrm{C}_6\mathrm{H}_{14}$	Hexane
Hept-	$\mathrm{C}_7\mathrm{H}_{16}$	Heptane
Oct-	$\mathrm{C}_8\mathrm{H}_{18}$	Octane
Non-	$\mathrm{C}_9\mathrm{H}_{20}$	Nonane
Dec-	$\mathrm{C}_{10}\mathrm{H}_{22}$	Decane

Structural Isomerism

Structural isomers have the same molecular formula but different structural formulae. There are three main types:

Chain isomerism: Different arrangements of the carbon skeleton.

For $\mathrm{C}_4\mathrm{H}_{10}$:

• Butane (straight chain): $\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{CH}_3$
• 2-methylpropane (branched): $\mathrm{CH}_3\mathrm{CH}(\mathrm{CH}_3)\mathrm{CH}_3$

Position isomerism: The functional group or substituent is attached at different positions on the carbon chain.

For $\mathrm{C}_3\mathrm{H}_7\mathrm{Cl}$:

• 1-chloropropane: $\mathrm{CH}_2\mathrm{ClCH}_2\mathrm{CH}_3$
• 2-chloropropane: $\mathrm{CH}_3\mathrm{CHClCH}_3$

Functional group isomerism: Molecules with the same molecular formula but different functional groups.

For $\mathrm{C}_2\mathrm{H}_6\mathrm{O}$:

• Ethanol ($\mathrm{CH}_3\mathrm{CH}_2\mathrm{OH}$) -- alcohol
• Methoxymethane ($\mathrm{CH}_3\mathrm{OCH}_3$) -- ether

Physical Properties

Property	Trend / Value
Boiling point	Increases with chain length (more van der Waals forces)
State at r.t.p.	$\mathrm{C}_1$-$\mathrm{C}_4$: gas; $\mathrm{C}_5$-$\mathrm{C}_{17}$: liquid; $\gt{}\mathrm{C}_{17}$: solid
Solubility	Insoluble in water, soluble in organic solvents
Density	Less dense than water

Branched isomers have lower boiling points than their straight-chain counterparts because branching reduces the surface area of contact between molecules, weakening van der Waals forces.

Chemical Properties

Alkanes are relatively unreactive because C-C and C-H bonds are strong and non-polar. Their main reactions are:

Combustion:

Complete combustion (in excess oxygen):

$\mathrm{C}_n\mathrm{H}_{2n+2} + \frac{3n+1}{2}\mathrm{O}_2 \to n\mathrm{CO}_2 + (n+1)\mathrm{H}_2\mathrm{O}$

Incomplete combustion (in limited oxygen):

$2\mathrm{C}_n\mathrm{H}_{2n+2} + (2n+1)\mathrm{O}_2 \to 2n\mathrm{CO} + (2n+2)\mathrm{H}_2\mathrm{O}$

Carbon (soot) may also be produced.

Halogenation (Free Radical Substitution):

Alkanes react with halogens ($\mathrm{Cl}_2$ or $\mathrm{Br}_2$) in the presence of UV light to form haloalkanes.

$\mathrm{CH}_4 + \mathrm{Cl}_2 \xrightarrow{\mathrm{UV}} \mathrm{CH}_3\mathrm{Cl} + \mathrm{HCl}$

This is a substitution reaction because a hydrogen atom is replaced by a chlorine atom.

Free Radical Substitution Mechanism

The reaction proceeds via a free radical chain mechanism with three stages:

Initiation: UV light provides energy to break the Cl-Cl bond homolytically (each atom gets one electron), generating chlorine free radicals.

$\mathrm{Cl}_2 \xrightarrow{\mathrm{UV}} 2\mathrm{Cl}^\bullet$

Propagation: The chlorine radical attacks a methane molecule, abstracting a hydrogen atom and forming HCl and a methyl radical. The methyl radical then attacks another chlorine molecule.

$\mathrm{Cl}^\bullet + \mathrm{CH}_4 \to \mathrm{HCl} + \mathrm{CH}_3^\bullet$

$\mathrm{CH}_3^\bullet + \mathrm{Cl}_2 \to \mathrm{CH}_3\mathrm{Cl} + \mathrm{Cl}^\bullet$

The chain continues because a new chlorine radical is generated in each propagation step.

Termination: Two radicals combine, ending the chain. Any two radicals can combine.

$\mathrm{Cl}^\bullet + \mathrm{Cl}^\bullet \to \mathrm{Cl}_2$

$\mathrm{CH}_3^\bullet + \mathrm{CH}_3^\bullet \to \mathrm{C}_2\mathrm{H}_6$

$\mathrm{CH}_3^\bullet + \mathrm{Cl}^\bullet \to \mathrm{CH}_3\mathrm{Cl}$

warning

warning $\mathrm{CH}_3\mathrm{Cl}$, $\mathrm{CH}_2\mathrm{Cl}_2$, $\mathrm{CHCl}_3$, and $\mathrm{CCl}_4$ because the substitution can continue on the same carbon atom. Controlling the ratio of methane to chlorine influences the proportion of products.

Cracking of Hydrocarbons

Cracking is the process of breaking down large hydrocarbon molecules into smaller, more useful ones. It is important because the demand for shorter-chain hydrocarbons (e.g., petrol) exceeds the supply from fractional distillation.

Thermal cracking:

• Uses high temperatures ($400$--$700^\circ\mathrm{C}$) and high pressures
• Produces a mixture of alkanes and alkenes
• Produces shorter-chain alkanes as fuels

$\mathrm{C}_{14}\mathrm{H}_{30} \xrightarrow{\Delta} \mathrm{C}_7\mathrm{H}_{16} + \mathrm{C}_7\mathrm{H}_{14}$

Catalytic cracking:

• Uses a zeolite catalyst at slightly lower temperatures ($450^\circ\mathrm{C}$)
• Produces branched alkanes, cycloalkanes, and aromatic compounds
• Preferred industrially because it gives higher-quality products and operates at lower temperatures

2.2 Alkenes ($\mathrm{C}_n\mathrm{H}_{2n}$)

General Properties

Alkenes are unsaturated hydrocarbons containing at least one carbon-carbon double bond (C=C). 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, and is weaker than the $\sigma$ bond.

Naming

Alkenes are named by replacing the -ane suffix with -ene. The position of the double bond is indicated by the lower-numbered carbon atom involved.

$\mathrm{CH}_2=\mathrm{CHCH}_2\mathrm{CH}_3 \quad \mathrm{but-1-ene}$

$\mathrm{CH}_3\mathrm{CH}=\mathrm{CHCH}_3 \quad \mathrm{but-2-ene}$

E/Z (Cis/Trans) Isomerism

The C=C double bond prevents free rotation. When each carbon of the double bond carries two different substituents, two distinct isomers exist:

• Cis isomer: The two largest substituents are on the same side
• Trans isomer: The two largest substituents are on opposite sides

The modern CIP (Cahn-Ingold-Prelof) priority rules assign E (entgegen, opposite) or Z (zusammen, together) based on the atomic number of substituents:

1. Assign priority to each substituent on each carbon (higher atomic number = higher priority)
2. If the two highest-priority groups are on the same side: Z
3. If the two highest-priority groups are on opposite sides: E

info

E/Z isomerism is a type of stereoisomerism. Stereoisomers have the same structural formula but different spatial arrangements of atoms. For E/Z isomerism to occur, each carbon of the C=C must have two different substituents.

Physical Properties

Alkenes have physical properties similar to alkanes of comparable molecular mass:

• Boiling points increase with chain length
• Lower boiling points than the corresponding alkanes (slightly less surface contact due to the double bond restricting rotation)
• Insoluble in water, soluble in organic solvents

Chemical Properties

The C=C double bond is the reactive site of alkenes. The $\pi$ bond is relatively weak and readily breaks to allow addition reactions, where atoms or groups are added across the double bond.

Test for unsaturation (Bromine water test):

When bromine water (orange-brown) is added to an alkene, the bromine adds across the double bond and the solution is decolourised.

$\mathrm{CH}_2=\mathrm{CH}_2 + \mathrm{Br}_2 \to \mathrm{CH}_2\mathrm{BrCH}_2\mathrm{Br}$

This is a useful test to distinguish alkenes from alkanes (alkanes do not decolourise bromine water without UV light).

Hydrogenation:

$\mathrm{CH}_2=\mathrm{CH}_2 + \mathrm{H}_2 \xrightarrow{\mathrm{Ni catalyst, } 150^\circ\mathrm{C}} \mathrm{CH}_3\mathrm{CH}_3$

Used industrially to convert vegetable oils to margarine (hardening).

Halogenation:

$\mathrm{CH}_2=\mathrm{CH}_2 + \mathrm{Cl}_2 \to \mathrm{CH}_2\mathrm{ClCH}_2\mathrm{Cl}$

Halogenation of alkenes does not require UV light (unlike alkanes).

Hydration:

$\mathrm{CH}_2=\mathrm{CH}_2 + \mathrm{H}_2\mathrm{O} \xrightarrow{\mathrm{H}_3\mathrm{PO}_4, 300^\circ\mathrm{C}, \mathrm{high pressure}} \mathrm{CH}_3\mathrm{CH}_2\mathrm{OH}$

Hydrohalogenation:

$\mathrm{CH}_2=\mathrm{CH}_2 + \mathrm{HBr} \to \mathrm{CH}_3\mathrm{CH}_2\mathrm{Br}$

Markovnikov's Rule:

When HX adds to an unsymmetrical alkene, the hydrogen atom attaches to the carbon that already has more hydrogen atoms (the less substituted carbon).

$\mathrm{CH}_3\mathrm{CH}=\mathrm{CH}_2 + \mathrm{HBr} \to \mathrm{CH}_3\mathrm{CHBrCH}_3 \quad \mathrm{(2-bromopropane, major product)}$

The minor product, 1-bromopropane ($\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{Br}$), is formed in smaller amounts.

tip

Markovnikov's Rule can be remembered as "the rich get richer": the carbon with more H atoms gets the additional H.

Industrial Importance of Alkenes

Alkenes are the raw materials for the polymer industry:

• Polyethylene (polythene): From ethene. Used for plastic bags, bottles, insulation
• Polypropene: From propene. Used for ropes, carpets, plastic containers
• Polyvinyl chloride (PVC): From chloroethene (vinyl chloride). Used for pipes, window frames
• PTFE (Teflon): From tetrafluoroethene. Used for non-stick coatings

2.3 Alkynes ($\mathrm{C}_n\mathrm{H}_{2n-2}$)

General Properties

Alkynes contain at least one carbon-carbon triple bond (C$\equiv$C), which consists of one $\sigma$ bond and two $\pi$ bonds. The simplest alkyne is ethyne ($\mathrm{HC}\equiv\mathrm{CH}$), commonly known as acetylene.

Naming

Alkynes are named by replacing the -ane suffix with -yne.

$\mathrm{HC}\equiv\mathrm{CH} \quad \mathrm{ethyne}$

$\mathrm{CH}_3\mathrm{C}\equiv\mathrm{CH} \quad \mathrm{propyne}$

Physical Properties

Similar to alkanes and alkenes: low boiling points for small molecules, increasing with chain length. Ethyne is a colourless gas at room temperature.

Chemical Properties

Alkynes undergo addition reactions similar to alkenes, but can accept two moles of reagent because the triple bond has two $\pi$ bonds.

Hydrogenation (stepwise):

$\mathrm{HC}\equiv\mathrm{CH} + \mathrm{H}_2 \xrightarrow{\mathrm{Pd/C}} \mathrm{H}_2\mathrm{C}=\mathrm{CH}_2 \xrightarrow{\mathrm{H}_2} \mathrm{CH}_3\mathrm{CH}_3$

Halogenation (stepwise):

$\mathrm{HC}\equiv\mathrm{CH} + \mathrm{Br}_2 \to \mathrm{BrHC}=\mathrm{CHBr} \xrightarrow{\mathrm{Br}_2} \mathrm{Br}_2\mathrm{HC}-\mathrm{CHBr}_2$

Hydration:

Ethyne reacts with water in the presence of $\mathrm{HgSO}_4$ and dilute $\mathrm{H}_2\mathrm{SO}_4$ catalyst to form ethanal (acetaldehyde):

$\mathrm{HC}\equiv\mathrm{CH} + \mathrm{H}_2\mathrm{O} \xrightarrow{\mathrm{HgSO}_4, \mathrm{H}^+} \mathrm{CH}_3\mathrm{CHO}$

Higher alkynes give ketones.

Worked Example: Alkyne Addition

Ethyne reacts with excess bromine water. Write the equation for the overall reaction and state the type of reaction.

Solution

Since ethyne has two $\pi$ bonds, it can react with two moles of $\mathrm{Br_2}$:

$\mathrm{HC}\equiv\mathrm{CH} + 2\mathrm{Br_2} \to \mathrm{Br_2HC}-\mathrm{CHBr_2}$

The product is 1,1,2,2-tetrabromoethane. This is an addition reaction (bromine adds across both $\pi$ bonds stepwise). The bromine water is decolourised.

Worked Example: Comparing Alkane, Alkene, and Alkyne Reactivity

A student has three unlabelled gas jars containing ethane, ethene, and ethyne. Describe a simple chemical test to distinguish between them.

Solution

Add bromine water to each gas jar:

1. Ethane: No reaction. Bromine water remains orange-brown (no C=C or C$\equiv$C to react).
2. Ethene: Decolourisation of bromine water (C=C undergoes electrophilic addition).
3. Ethyne: Decolourisation of bromine water (C$\equiv$C undergoes electrophilic addition).

This distinguishes ethane from the other two, but ethene and ethyne both decolourise bromine water.

To distinguish ethene from ethyne, use ammoniacal $\mathrm{CuCl}$: ethyne forms a red precipitate of copper(I) acetylide ($\mathrm{Cu_2C_2}$), while ethene does not react.

2.4 Arenes (Benzene)

Benzene Structure

Benzene has the molecular formula $\mathrm{C}_6\mathrm{H}_6$. Two models describe its structure:

Kekule model (1865):

• Alternating single and double bonds in a hexagonal ring
• Could not explain why benzene does not undergo addition reactions like alkenes
• Could not explain why all C-C bonds have identical lengths ($0.140$ nm, intermediate between C-C single bond at $0.154$ nm and C=C double bond at $0.134$ nm)

Delocalised model (modern):

• Each carbon atom is $sp^2$ hybridised, forming three $\sigma$ bonds (two to adjacent carbons, one to hydrogen)
• The remaining unhybridised $p$ orbital on each carbon overlaps with those on neighbouring carbons, forming a delocalised $\pi$ system above and below the ring
• All six $\pi$ electrons are delocalised over the entire ring
• This delocalisation gives benzene extra stability (resonance energy of approximately $150$ kJ/mol)
• Represented as a hexagon with a circle inside, or alternating double bonds with understanding that the $\pi$ electrons are delocalised

$\mathrm{Bond length in benzene} = 0.140 \mathrm{ nm}$

$\mathrm{C-C single bond} = 0.154 \mathrm{ nm}, \quad \mathrm{C=C double bond} = 0.134 \mathrm{ nm}$

The intermediate bond length confirms the delocalised model.

Electrophilic Aromatic Substitution (EAS)

Benzene undergoes substitution reactions rather than addition reactions because substitution preserves the stable delocalised $\pi$ system. The mechanism is electrophilic aromatic substitution.

General mechanism:

1. An electrophile ($\mathrm{E}^+$) attacks the benzene ring
2. A temporary positively charged intermediate (arenium ion / sigma complex) forms
3. The intermediate loses $\mathrm{H}^+$ to restore the delocalised $\pi$ system

Nitration:

Reagents: concentrated $\mathrm{HNO}_3$ with concentrated $\mathrm{H}_2\mathrm{SO}_4$ (catalyst) at $50$--$60^\circ\mathrm{C}$.

Generation of electrophile:

$\mathrm{HNO}_3 + \mathrm{H}_2\mathrm{SO}_4 \to \mathrm{NO}_2^+ + \mathrm{HSO}_4^- + \mathrm{H}_2\mathrm{O}$

Substitution:

$\mathrm{C}_6\mathrm{H}_6 + \mathrm{NO}_2^+ \to \mathrm{C}_6\mathrm{H}_5\mathrm{NO}_2 + \mathrm{H}^+$

Product: nitrobenzene.

Halogenation (with a halogen carrier):

Reagents: $\mathrm{Cl}_2$ or $\mathrm{Br}_2$ with a Lewis acid catalyst such as $\mathrm{AlCl}_3$ or $\mathrm{FeBr}_3$.

$\mathrm{Cl}_2 + \mathrm{AlCl}_3 \to \mathrm{Cl}^+ + \mathrm{AlCl}_4^-$

$\mathrm{C}_6\mathrm{H}_6 + \mathrm{Cl}^+ \to \mathrm{C}_6\mathrm{H}_5\mathrm{Cl} + \mathrm{H}^+$

Friedel-Crafts Alkylation:

Reagents: haloalkane with $\mathrm{AlCl}_3$ catalyst.

$\mathrm{CH}_3\mathrm{Cl} + \mathrm{AlCl}_3 \to \mathrm{CH}_3^+ + \mathrm{AlCl}_4^-$

$\mathrm{C}_6\mathrm{H}_6 + \mathrm{CH}_3^+ \to \mathrm{C}_6\mathrm{H}_5\mathrm{CH}_3 + \mathrm{H}^+$

Product: methylbenzene (toluene).

Friedel-Crafts Acylation:

Reagents: acyl chloride with $\mathrm{AlCl}_3$ catalyst.

$\mathrm{CH}_3\mathrm{COCl} + \mathrm{AlCl}_3 \to \mathrm{CH}_3\mathrm{CO}^+ + \mathrm{AlCl}_4^-$

$\mathrm{C}_6\mathrm{H}_6 + \mathrm{CH}_3\mathrm{CO}^+ \to \mathrm{C}_6\mathrm{H}_5\mathrm{COCH}_3 + \mathrm{H}^+$

Product: phenylethanone (acetophenone).

info

Friedel-Crafts reactions cannot be used on benzene rings that already bear strongly electron-withdrawing groups (e.g., $-\mathrm{NO}_2$) because these deactivate the ring towards electrophilic attack.

Worked Example: Predicting EAS Products

Predict the products when benzene undergoes (a) chlorination with $\mathrm{Cl_2}/\mathrm{AlCl_3}$, and (b) Friedel-Crafts alkylation with $\mathrm{CH_3CH_2Cl}/\mathrm{AlCl_3}$.

Solution

(a) Chlorination:

$\mathrm{Cl_2} + \mathrm{AlCl_3} \to \mathrm{Cl^+} + \mathrm{AlCl_4^-}$

$\mathrm{C_6H_6} + \mathrm{Cl^+} \to \mathrm{C_6H_5Cl} + \mathrm{H^+}$

Product: chlorobenzene ($\mathrm{C_6H_5Cl}$)

(b) Friedel-Crafts alkylation:

\mathrm{CH_3CH_2Cl} + \mathrm{AlCl_3} \to \mathrm{CH_3CH_2^+ + \mathrm{AlCl_4^-}

\mathrm{C_6H_6} + \mathrm{CH_3CH_2^+} \to \mathrm{C_6H_5CH_2CH_3 + \mathrm{H^+}

Product: ethylbenzene ($\mathrm{C_6H_5CH_2CH_3}$)

Worked Example: Why Benzene Does Not Undergo Addition

Explain why benzene does not decolourise bromine water under normal conditions, whereas cyclohexene does.

Solution

Cyclohexene has an isolated C=C double bond. The $\pi$ bond is relatively weak and readily undergoes electrophilic addition with $\mathrm{Br_2}$, decolourising bromine water.

Benzene has a delocalised $\pi$ system of 6 electrons spread over the entire ring. This delocalisation provides approximately $150 \mathrm{ kJ/mol}$ of extra stability (resonance energy). An addition reaction would destroy this stable delocalised system, requiring a large input of energy. Instead, benzene undergoes electrophilic substitution, which preserves the delocalised $\pi$ system. Substitution does not require breaking the aromatic system, so the reaction is energetically favourable, but it does not decolourise bromine water because substitution (not addition) occurs.

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3. Functional Groups

3.1 Alcohols

Classification

Alcohols contain the hydroxyl group (-OH) attached to an $sp^3$ hybridised carbon atom.

Type	Structure	Example
Primary (1$^\circ$)	-OH attached to a carbon bonded to at most one other carbon	$\mathrm{CH}_3\mathrm{CH}_2\mathrm{OH}$ (ethanol)
Secondary (2$^\circ$)	-OH attached to a carbon bonded to two other carbons	$\mathrm{CH}_3\mathrm{CH(OH)CH}_3$ (propan-2-ol)
Tertiary (3$^\circ$)	-OH attached to a carbon bonded to three other carbons	$(\mathrm{CH}_3)_3\mathrm{COH}$ (2-methylpropan-2-ol)

Naming

Alcohols are named using the -ol suffix. The position of the -OH group is indicated by a number.

$\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{OH} \quad \mathrm{propan-1-ol}$

$\mathrm{CH}_3\mathrm{CH(OH)CH}_3 \quad \mathrm{propan-2-ol}$

Physical Properties

Property	Description
Boiling point	Higher than corresponding alkanes due to hydrogen bonding between -OH groups
Solubility	Small alcohols (C1--C3) are miscible with water; solubility decreases with increasing chain length
Volatility	Lower than corresponding alkanes (hydrogen bonding must be overcome)

Oxidation of Alcohols

Different classes of alcohols give different products upon oxidation.

Primary alcohols:

Oxidised first to aldehydes, then to carboxylic acids.

$\mathrm{RCH}_2\mathrm{OH} \xrightarrow{[\mathrm{O}]} \mathrm{RCHO} \xrightarrow{[\mathrm{O}]} \mathrm{RCOOH}$

To obtain the aldehyde (without further oxidation to the carboxylic acid), distil the product out as it forms.

Oxidising agents: acidified potassium dichromate(VI) ($\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7 / \mathrm{H}^+$), which changes colour from orange to green upon reduction.

Secondary alcohols:

Oxidised to ketones (which are resistant to further oxidation).

$\mathrm{R}_2\mathrm{CHOH} \xrightarrow{[\mathrm{O}]} \mathrm{R}_2\mathrm{C=O}$

Tertiary alcohols:

Not oxidised by common oxidising agents because the carbon bearing the -OH has no hydrogen atoms to remove.

Worked Example: Predicting Oxidation Products

Predict the products when the following alcohols are heated with acidified potassium dichromate(VI) under reflux: (a) butan-1-ol, (b) butan-2-ol, (c) 2-methylpropan-2-ol.

Solution

(a) Butan-1-ol ($\mathrm{CH_3CH_2CH_2CH_2OH}$) is primary. Under reflux (strong oxidation), it is fully oxidised to butanoic acid:

$\mathrm{CH_3CH_2CH_2CH_2OH} \xrightarrow{[\mathrm{O}],\ \mathrm{reflux}} \mathrm{CH_3CH_2CH_2COOH}$

The colour change is orange to green.

(b) Butan-2-ol ($\mathrm{CH_3CH_2CH(OH)CH_3}$) is secondary. It oxidises to butanone:

$\mathrm{CH_3CH_2CH(OH)CH_3} \xrightarrow{[\mathrm{O}]} \mathrm{CH_3CH_2COCH_3}$

(c) 2-Methylpropan-2-ol ($\mathrm{(CH_3)_3COH}$) is tertiary. No oxidation occurs; the orange colour of dichromate persists.

3.2 Carboxylic Acids

Naming

Carboxylic acids contain the -COOH group and are named with the -oic acid suffix.

$\mathrm{HCOOH} \quad \mathrm{methanoic acid}$

$\mathrm{CH}_3\mathrm{COOH} \quad \mathrm{ethanoic acid}$

$\mathrm{CH}_3\mathrm{CH}_2\mathrm{COOH} \quad \mathrm{propanoic acid}$

Properties

Acidity:

Carboxylic acids are weak acids. They partially dissociate in water:

$\mathrm{CH}_3\mathrm{COOH} \rightleftharpoons \mathrm{CH}_3\mathrm{COO}^- + \mathrm{H}^+$

$K_a = \frac{[\mathrm{CH}_3\mathrm{COO}^-][\mathrm{H}^+]}{[\mathrm{CH}_3\mathrm{COOH}]} = 1.8 \times 10^{-5}$

$\mathrm{p}K_a = -\log_{10}(1.8 \times 10^{-5}) = 4.74$

Carboxylic acids are stronger acids than alcohols (phenol has $\mathrm{p}K_a \approx 10$) but weaker than mineral acids. The carboxylate ion ($\mathrm{RCOO}^-$) is stabilised by resonance delocalisation of the negative charge over the two oxygen atoms.

Reactions:

• With metals: $2\mathrm{CH}_3\mathrm{COOH} + 2\mathrm{Na} \to 2\mathrm{CH}_3\mathrm{COONa} + \mathrm{H}_2$
• With bases: $\mathrm{CH}_3\mathrm{COOH} + \mathrm{NaOH} \to \mathrm{CH}_3\mathrm{COONa} + \mathrm{H}_2\mathrm{O}$
• With carbonates: $2\mathrm{CH}_3\mathrm{COOH} + \mathrm{Na}_2\mathrm{CO}_3 \to 2\mathrm{CH}_3\mathrm{COONa} + \mathrm{H}_2\mathrm{O} + \mathrm{CO}_2$

Esterification

Carboxylic acids react with alcohols in the presence of a strong acid catalyst (concentrated $\mathrm{H}_2\mathrm{SO}_4$) to form esters:

$\mathrm{RCOOH} + \mathrm{R}'\mathrm{OH} \rightleftharpoons \mathrm{RCOOR}' + \mathrm{H}_2\mathrm{O}$

This is a reversible, condensation reaction (two molecules join with the elimination of a small molecule, in this case water). The equilibrium can be shifted towards the ester by using an excess of one reactant or by removing water.

3.3 Esters

Naming

Esters are named with the alkyl group from the alcohol first, followed by the -oate from the carboxylic acid.

$\mathrm{CH}_3\mathrm{COOCH}_2\mathrm{CH}_3 \quad \mathrm{ethyl ethanoate}$

Properties

• Pleasant, fruity odours (used in flavourings and perfumes)
• Lower boiling points than carboxylic acids or alcohols of comparable mass (no hydrogen bonding between ester molecules)
• Insoluble in water

Hydrolysis of Esters

Acid hydrolysis: Reversible reaction; the ester is heated with dilute acid.

$\mathrm{RCOOR}' + \mathrm{H}_2\mathrm{O} \xrightarrow{\mathrm{H}^+} \mathrm{RCOOH} + \mathrm{R}'\mathrm{OH}$

Alkaline hydrolysis (saponification): Irreversible reaction; the ester is heated with aqueous NaOH or KOH.

$\mathrm{RCOOR}' + \mathrm{NaOH} \to \mathrm{RCOONa} + \mathrm{R}'\mathrm{OH}$

The carboxylate salt produced is the soap (if a long-chain ester is used). This is the basis of soap making:

1. A fat or oil (triglyceride ester) is heated with concentrated NaOH solution
2. The ester bonds are hydrolysed, producing glycerol and sodium salts of fatty acids (soap)
3. Soap is precipitated by adding saturated NaCl solution (salting out)

Worked Example: Comparing Ester Hydrolysis Methods

Compare the products of acid hydrolysis and alkaline hydrolysis of methyl propanoate ($\mathrm{CH_3CH_2COOCH_3}$). Explain why alkaline hydrolysis is preferred for soap making.

Solution

Acid hydrolysis:

$\mathrm{CH_3CH_2COOCH_3} + \mathrm{H_2O} \xrightarrow{\mathrm{H^+}} \mathrm{CH_3CH_2COOH} + \mathrm{CH_3OH}$

Products: propanoic acid and methanol. This is reversible.

Alkaline hydrolysis (saponification):

$\mathrm{CH_3CH_2COOCH_3} + \mathrm{NaOH} \to \mathrm{CH_3CH_2COONa} + \mathrm{CH_3OH}$

Products: sodium propanoate and methanol. This is irreversible.

Why alkaline is preferred for soap: The product is the sodium carboxylate salt, which is the soap molecule (surfactant). The carboxylate ion has a long hydrophobic tail and a hydrophilic head, allowing it to emulsify grease. In acid hydrolysis, the product is the carboxylic acid, which does not have the ionic character needed for soap action. Furthermore, alkaline hydrolysis goes to completion (irreversible), giving a higher yield.

3.4 Aldehydes and Ketones

Both contain the carbonyl group (>C=O).

Aldehydes: The carbonyl group is at the end of the carbon chain (bonded to at least one H).

$\mathrm{RCHO}$

Naming: -al suffix. The -al carbon is always carbon 1.

$\mathrm{CH}_3\mathrm{CHO} \quad \mathrm{ethanal}$

Ketones: The carbonyl group is within the carbon chain (bonded to two carbons).

$\mathrm{R}_2\mathrm{C=O}$

Naming: -one suffix. The position is indicated by a number.

$\mathrm{CH}_3\mathrm{COCH}_3 \quad \mathrm{propanone}$

Properties

Property	Aldehydes	Ketones
Boiling point	Lower than corresponding alcohols	Similar to aldehydes
Odour	Pungent (e.g., ethanal)	Pleasant (e.g., propanone = nail polish remover)
Oxidation	Easily oxidised to carboxylic acids	Resistant to oxidation

Nucleophilic Addition

The carbonyl carbon is electron-deficient (partially positive due to the electronegative oxygen), making it susceptible to attack by nucleophiles.

Reaction with HCN (cyanohydrin formation):

$\mathrm{RCHO} + \mathrm{HCN} \to \mathrm{RCH(OH)CN}$

The mechanism:

1. The nucleophile $\mathrm{CN}^-$ attacks the carbonyl carbon
2. A tetrahedral intermediate forms
3. The intermediate picks up a proton from $\mathrm{HCN}$

3.5 Amines

Classification and Naming

Amines contain the amino group (-NH$_2$). They are classified by the number of alkyl groups attached to the nitrogen:

Type	Structure	Example
Primary	$\mathrm{RNH}_2$	$\mathrm{CH}_3\mathrm{NH}_2$ (methylamine)
Secondary	$\mathrm{R}_2\mathrm{NH}$	$(\mathrm{CH}_3)_2\mathrm{NH}$ (dimethylamine)
Tertiary	$\mathrm{R}_3\mathrm{N}$	$(\mathrm{CH}_3)_3\mathrm{N}$ (trimethylamine)

Basicity

Amines are weak bases. The lone pair on nitrogen can accept a proton:

$\mathrm{RNH}_2 + \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{RNH}_3^+ + \mathrm{OH}^-$

Amines form salts with acids:

$\mathrm{CH}_3\mathrm{NH}_2 + \mathrm{HCl} \to \mathrm{CH}_3\mathrm{NH}_3^+\mathrm{Cl}^-$

The basicity of amines depends on the availability of the lone pair. Electron-donating groups increase basicity; electron-withdrawing groups decrease it.

Hydrogen Bonding

Primary and secondary amines can form hydrogen bonds (N-H), giving them higher boiling points than comparable hydrocarbons. Tertiary amines cannot form N-H hydrogen bonds but can accept H-bonds.

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4. Polymerisation

4.1 Addition Polymerisation

In addition polymerisation, monomers with C=C double bonds join together without the loss of any small molecule. The double bond opens to form single bonds linking monomers into a long chain.

General reaction:

$n\mathrm{CH}_2=\mathrm{CHX} \to \left[\mathrm{--CH}_2\mathrm{--CHX--}\right]_n$

Polymer	Monomer	Repeating Unit	Uses
Polyethylene	Ethene	$--\mathrm{CH}_2\mathrm{--CH}_2--$	Bags, bottles, films
Polypropene	Propene	$--\mathrm{CH}_2\mathrm{--CH}(\mathrm{CH}_3)--$	Ropes, carpets, containers
PVC	Chloroethene	$--\mathrm{CH}_2\mathrm{--CHCl}--$	Pipes, cable insulation
Polystyrene	Phenylethene (styrene)	$--\mathrm{CH}_2\mathrm{--CH}(\mathrm{C}_6\mathrm{H}_5)--$	Packaging, insulation
PTFE (Teflon)	Tetrafluoroethene	$--\mathrm{CF}_2\mathrm{--CF}_2--$	Non-stick coatings

tip

When drawing addition polymers, always show the repeating unit in square brackets with n outside. Include enough of the structure to show the pattern clearly (typically at least two repeating units).

4.2 Condensation Polymerisation

In condensation polymerisation, monomers join together with the elimination of a small molecule (usually water or HCl). The monomers must have two functional groups each.

Polyesters:

Formed from a dicarboxylic acid and a diol.

$n\mathrm{HOOC}\mathrm{--R}\mathrm{--COOH} + n\mathrm{HO}\mathrm{--R}'\mathrm{--OH} \to \left[\mathrm{--OC}\mathrm{--R}\mathrm{--COO}\mathrm{--R}'\mathrm{--O--}\right]_n + 2n\mathrm{H}_2\mathrm{O}$

Example: Terylene (PET) from benzene-1,4-dicarboxylic acid and ethane-1,2-diol.

$\mathrm{HOOC}\mathrm{--C}_6\mathrm{H}_4\mathrm{--COOH} + \mathrm{HOCH}_2\mathrm{CH}_2\mathrm{OH} \to \left[\mathrm{--OC}\mathrm{--C}_6\mathrm{H}_4\mathrm{--COOCH}_2\mathrm{CH}_2\mathrm{O--}\right]_n + 2n\mathrm{H}_2\mathrm{O}$

Polyamides (Nylons):

Formed from a dicarboxylic acid and a diamine.

$n\mathrm{HOOC}\mathrm{--R}\mathrm{--COOH} + n\mathrm{H}_2\mathrm{N}\mathrm{--R}'\mathrm{--NH}_2 \to \left[\mathrm{--OC}\mathrm{--R}\mathrm{--CONH}\mathrm{--R}'\mathrm{--NH--}\right]_n + 2n\mathrm{H}_2\mathrm{O}$

Example: Nylon-6,6 from hexanedioic acid and hexane-1,6-diamine.

Worked Example: Condensation Polymer from Monomers

Nylon-6,6 is formed from hexanedioic acid ($\mathrm{HOOC(CH_2)_4COOH}$) and hexane-1,6-diamine ($\mathrm{H_2N(CH_2)_6NH_2}$). (a) Write the equation for the formation of the repeating unit. (b) What small molecule is eliminated?

Solution

(a)

$n\mathrm{HOOC(CH_2)_4COOH} + n\mathrm{H_2N(CH_2)_6NH_2} \to \left[\mathrm{--OC(CH_2)_4CONH(CH_2)_6NH--}\right]_n + 2n\mathrm{H_2O}$

(b) Water ($\mathrm{H_2O}$) is eliminated at each amide bond formed. For each repeating unit, one molecule of water is eliminated (from the -COOH of the acid and one -NH$_2$ of the diamine).

Worked Example: Identifying Biodegradable Polymers

Explain why poly(lactic acid) (PLA) is biodegradable but polyethylene is not.

Solution

PLA contains ester bonds ($-\mathrm{COO-}$) in the polymer backbone. Ester bonds can be hydrolysed by water and by enzymes produced by microorganisms, breaking the polymer into small molecules that can be absorbed by the environment.

Polyethylene contains only C-C and C-H bonds in the polymer backbone. These are very strong and non-polar; microorganisms do not produce enzymes that can break them. Polyethylene persists in the environment for hundreds of years.

Polyurethane:

Formed from a diisocyanate and a diol. The reaction produces a urethane linkage (no small molecule is eliminated, but it is still classified as a step-growth polymer).

4.3 Biodegradable Polymers

Conventional polymers are non-biodegradable because the strong covalent bonds in the polymer backbone cannot be broken by microorganisms.

Poly(lactic acid) (PLA):

• Made from corn starch or sugarcane (renewable resource)
• Ester bonds in the polymer backbone can be hydrolysed by microorganisms
• Used for packaging, medical sutures, disposable cups

Poly(hydroxybutyrate) (PHB):

• Produced by bacterial fermentation (e.g., Alcaligenes eutrophus)
• Completely biodegradable
• Used for bottles, plastic bags, medical implants

4.4 Problems with Plastic Waste

Problem	Description
Non-biodegradability	Most plastics persist in the environment for hundreds of years
Landfill accumulation	Plastics take up significant landfill space
Marine pollution	Microplastics in oceans harm marine life and enter the food chain
Toxic additives	Plasticisers, flame retardants, and stabilisers can leach into the environment
Incineration issues	Burning plastics releases toxic gases (e.g., HCl from PVC, dioxins)

Solutions include recycling, biodegradable alternatives, reduced usage, and improved waste management.

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5. Reactions and Mechanisms

5.1 Oxidation of Alcohols

The oxidation pathway depends on the class of alcohol:

$\mathrm{Primary alcohol} \xrightarrow{[\mathrm{O}]} \mathrm{Aldehyde} \xrightarrow{[\mathrm{O}]} \mathrm{Carboxylic acid} \xrightarrow{[\mathrm{O}]} \mathrm{CO}_2 + \mathrm{H}_2\mathrm{O}$

$\mathrm{Secondary alcohol} \xrightarrow{[\mathrm{O}]} \mathrm{Ketone} \quad \mathrm{(no further oxidation)}$

$\mathrm{Tertiary alcohol} \quad \mathrm{(no oxidation with common reagents)}$

Oxidising agents used:

• Acidified potassium dichromate(VI): $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7 / \mathrm{H}_2\mathrm{SO}_4$ (orange to green)
• Acidified potassium manganate(VII): $\mathrm{KMnO}_4 / \mathrm{H}_2\mathrm{SO}_4$ (purple to colourless)

5.2 Reduction Reactions

Reduction is the addition of hydrogen or removal of oxygen.

• Carboxylic acid to primary alcohol: $\mathrm{RCOOH} \xrightarrow{\mathrm{LiAlH}_4} \mathrm{RCH}_2\mathrm{OH}$
• Aldehyde to primary alcohol: $\mathrm{RCHO} \xrightarrow{\mathrm{NaBH}_4} \mathrm{RCH}_2\mathrm{OH}$
• Ketone to secondary alcohol: $\mathrm{R}_2\mathrm{C=O} \xrightarrow{\mathrm{NaBH}_4} \mathrm{R}_2\mathrm{CHOH}$

info

info reduce aldehydes and ketones but not carboxylic acids. $\mathrm{LiAlH}_4$ is more reactive and can reduce carboxylic acids, esters, and amides.

5.3 Hydrolysis of Esters

See Section 3.3 for detailed mechanisms.

5.4 Test Tube Reactions for Functional Group Identification

A systematic approach to identifying functional groups:

Test	Reagent	Positive Result	Functional Group Detected
Bromine water	$\mathrm{Br}_2$ (aq)	Decolourisation (orange to colourless)	Alkene, alkyne (C=C, C$\equiv$C)
2,4-DNPH test	2,4-dinitrophenylhydrazine	Orange/yellow precipitate	Aldehyde or ketone (C=O)
Fehling's test	Fehling's A + B (blue)	Brick-red precipitate	Aldehyde (but not ketone)
Tollens' test	Tollens' reagent (colourless)	Silver mirror	Aldehyde (but not ketone)
Acidified dichromate	$\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7 / \mathrm{H}^+$ (orange)	Orange to green	Primary or secondary alcohol, aldehyde
$\mathrm{PCl}_5$ test	Phosphorus(V) chloride	White fumes of HCl; effervescence	Alcohol (-OH) or carboxylic acid (-COOH)
$\mathrm{NaOH}$ (aq)	Sodium hydroxide solution	Acid reacts; CO$_2$ produced with carboxylic acid	Carboxylic acid
Iodine test	$\mathrm{I}_2$ solution	Blue-black colour	Starch (not a standard organic functional group)

Distinguishing aldehydes from ketones:

• Fehling's solution: aldehydes give a brick-red precipitate ($\mathrm{Cu}_2\mathrm{O}$); ketones do not
• Tollens' reagent: aldehydes give a silver mirror ($\mathrm{Ag}$); ketones do not
• 2,4-DNPH: both give a precipitate (confirms C=O), so use Fehling's or Tollens' to distinguish

5.5 Reaction Pathway Tables

Key interconversions between functional groups:

From	To	Reagent / Conditions
Alkene	Alkane	$\mathrm{H}_2$ / Ni catalyst
Alkene	Haloalkane	$\mathrm{HX}$ (e.g., HBr)
Alkene	Alcohol	$\mathrm{H}_2\mathrm{O}$ / $\mathrm{H}_3\mathrm{PO}_4$, $300^\circ\mathrm{C}$
Alkene	Diol	Cold, dilute, alkaline $\mathrm{KMnO}_4$
Alkene	Halogenoalkane (dihalide)	$\mathrm{Br}_2$ or $\mathrm{Cl}_2$
Haloalkane	Alcohol	NaOH (aq), heat
Haloalkane	Amine	Excess concentrated $\mathrm{NH}_3$ (alcoholic), heat
Alcohol	Aldehyde	Mild oxidation ($\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$, distil)
Alcohol	Carboxylic acid	Strong oxidation ($\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$, reflux)
Alcohol	Ester	Carboxylic acid + conc. $\mathrm{H}_2\mathrm{SO}_4$, heat
Carboxylic acid	Ester	Alcohol + conc. $\mathrm{H}_2\mathrm{SO}_4$, heat
Carboxylic acid	Salt	Base (e.g., NaOH, $\mathrm{Na}_2\mathrm{CO}_3$)
Ester	Carboxylic acid + alcohol	Acid hydrolysis (dilute $\mathrm{H}_2\mathrm{SO}_4$, heat)
Ester	Carboxylate salt + alcohol	Alkaline hydrolysis (NaOH, heat)
Aldehyde / Ketone	Cyanohydrin	HCN (trace NaOH catalyst)

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6. Worked Examples

Worked Example 1

Draw all the structural isomers of $\mathrm{C}_4\mathrm{H}_9\mathrm{Cl}$ and name them.

Solution:

There are four structural isomers:

1. 1-chlorobutane: $\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{CH}_2\mathrm{Cl}$ (primary)
2. 2-chlorobutane: $\mathrm{CH}_3\mathrm{CH}_2\mathrm{CHClCH}_3$ (secondary)
3. 1-chloro-2-methylpropane: $\mathrm{CH}_3\mathrm{CH}(\mathrm{CH}_3)\mathrm{CH}_2\mathrm{Cl}$ (primary)
4. 2-chloro-2-methylpropane: $(\mathrm{CH}_3)_3\mathrm{CCl}$ (tertiary)

Worked Example 2

A hydrocarbon $\mathrm{A}$ with molecular formula $\mathrm{C}_4\mathrm{H}_8$ decolourises bromine water and reacts with $\mathrm{HBr}$ to give two products, $\mathrm{B}$ (major) and $\mathrm{C}$ (minor). Identify $\mathrm{A}$, $\mathrm{B}$, and $\mathrm{C}$.

Solution:

$\mathrm{C}_4\mathrm{H}_8$ has one degree of unsaturation ($\mathrm{C}_n\mathrm{H}_{2n}$), so it is an alkene. Since it gives two products with HBr, it is unsymmetrical.

$\mathrm{A}$ = but-1-ene: $\mathrm{CH}_2=\mathrm{CHCH}_2\mathrm{CH}_3$

Reaction with HBr (Markovnikov addition):

$\mathrm{B}$ (major): 2-bromobutane: $\mathrm{CH}_3\mathrm{CHBrCH}_2\mathrm{CH}_3$

$\mathrm{C}$ (minor): 1-bromobutane: $\mathrm{CH}_2\mathrm{BrCH}_2\mathrm{CH}_2\mathrm{CH}_3$

Worked Example 3

Calculate the volume of $\mathrm{CO}_2$ produced (at r.t.p.) when $5.8 \mathrm{ g}$ of ethanol ($\mathrm{C}_2\mathrm{H}_5\mathrm{OH}$) is completely combusted.

Solution:

$\mathrm{C}_2\mathrm{H}_5\mathrm{OH} + 3\mathrm{O}_2 \to 2\mathrm{CO}_2 + 3\mathrm{H}_2\mathrm{O}$

Molar mass of $\mathrm{C}_2\mathrm{H}_5\mathrm{OH} = 2(12) + 6(1) + 16 = 46 \mathrm{ g/mol}$

Moles of ethanol $= \frac{5.8}{46} = 0.126 \mathrm{ mol}$

From the equation: 1 mol ethanol produces 2 mol $\mathrm{CO}_2$.

Moles of $\mathrm{CO}_2 = 2 \times 0.126 = 0.252 \mathrm{ mol}$

Volume at r.t.p. $= 0.252 \times 24.0 = 6.05 \mathrm{ dm}^3$

Worked Example 4

Explain why the boiling point of propan-1-ol ($97^\circ\mathrm{C}$) is much higher than that of propane ($-42^\circ\mathrm{C}$) even though they have similar molar masses.

Solution:

Propan-1-ol molecules can form hydrogen bonds with each other through the -OH group. Hydrogen bonding is a much stronger intermolecular force than the van der Waals forces between propane molecules. A significantly greater amount of energy is required to overcome hydrogen bonds, resulting in a much higher boiling point for propan-1-ol.

Worked Example 5

A compound $\mathrm{D}$ with molecular formula $\mathrm{C}_3\mathrm{H}_8\mathrm{O}$ gives the following results:

• Reacts with $\mathrm{PCl}_5$ to produce white fumes
• Reacts with acidified $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$ (orange to green) to give compound $\mathrm{E}$
• Compound $\mathrm{E}$ gives a silver mirror with Tollens' reagent

Identify $\mathrm{D}$ and $\mathrm{E}$.

Solution:

$\mathrm{C}_3\mathrm{H}_8\mathrm{O}$ has possible structures: propan-1-ol, propan-2-ol, or methoxyethane.

Reaction with $\mathrm{PCl}_5$ (white fumes = HCl) confirms the presence of -OH, so $\mathrm{D}$ is an alcohol (not methoxyethane).

Oxidation of $\mathrm{D}$ gives $\mathrm{E}$, which gives a silver mirror with Tollens' reagent, confirming $\mathrm{E}$ is an aldehyde. Only primary alcohols oxidise to aldehydes.

$\mathrm{D}$ = propan-1-ol: $\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{OH}$

$\mathrm{E}$ = propanal: $\mathrm{CH}_3\mathrm{CH}_2\mathrm{CHO}$

Worked Example 6

Draw the repeating unit of the polymer formed from propene and name the polymer.

Solution:

Propene: $\mathrm{CH}_2=\mathrm{CHCH}_3$

The double bond opens during addition polymerisation:

$n\mathrm{CH}_2=\mathrm{CHCH}_3 \to \left[\mathrm{--CH}_2\mathrm{--CH}(\mathrm{CH}_3)--\right]_n$

Repeating unit: $--\mathrm{CH}_2\mathrm{--CH}(\mathrm{CH}_3)--$

Name: polypropene.

Worked Example 7

Ethanoic acid reacts with ethanol in the presence of concentrated $\mathrm{H}_2\mathrm{SO}_4$ to form an ester. Calculate the percentage yield if $6.0 \mathrm{ g}$ of ethanoic acid produces $6.5 \mathrm{ g}$ of ethyl ethanoate.

Solution:

$\mathrm{CH}_3\mathrm{COOH} + \mathrm{CH}_3\mathrm{CH}_2\mathrm{OH} \rightleftharpoons \mathrm{CH}_3\mathrm{COOCH}_2\mathrm{CH}_3 + \mathrm{H}_2\mathrm{O}$

Molar mass of $\mathrm{CH}_3\mathrm{COOH} = 60 \mathrm{ g/mol}$

Moles of $\mathrm{CH}_3\mathrm{COOH} = \frac{6.0}{60} = 0.10 \mathrm{ mol}$

Theoretical moles of ester $= 0.10 \mathrm{ mol}$ (1:1 ratio)

Molar mass of $\mathrm{CH}_3\mathrm{COOCH}_2\mathrm{CH}_3 = 88 \mathrm{ g/mol}$

Theoretical mass of ester $= 0.10 \times 88 = 8.8 \mathrm{ g}$

$\mathrm{Percentage yield} = \frac{6.5}{8.8} \times 100\% = 73.9\%$

Worked Example 8

Write a balanced equation for the nitration of benzene and describe the role of concentrated $\mathrm{H}_2\mathrm{SO}_4$ in the reaction.

Solution:

$\mathrm{C}_6\mathrm{H}_6 + \mathrm{HNO}_3 \xrightarrow{\mathrm{conc. } \mathrm{H}_2\mathrm{SO}_4, 50\mathrm{--}60^\circ\mathrm{C}} \mathrm{C}_6\mathrm{H}_5\mathrm{NO}_2 + \mathrm{H}_2\mathrm{O}$

Concentrated $\mathrm{H}_2\mathrm{SO}_4$ acts as a catalyst and as a dehydrating agent. It protonates $\mathrm{HNO}_3$ to generate the nitronium ion ($\mathrm{NO}_2^+$), which is the electrophile that attacks the benzene ring:

$\mathrm{HNO}_3 + \mathrm{H}_2\mathrm{SO}_4 \to \mathrm{NO}_2^+ + \mathrm{HSO}_4^- + \mathrm{H}_2\mathrm{O}$

Worked Example 9

Explain why the hydrolysis of ethyl ethanoate with aqueous NaOH is irreversible, whereas hydrolysis with aqueous $\mathrm{H}_2\mathrm{SO}_4$ is reversible.

Solution:

In alkaline hydrolysis, the product is the carboxylate ion ($\mathrm{CH}_3\mathrm{COO}^-$), which is a very weak acid and does not react with the alcohol to re-form the ester. The equilibrium lies completely to the right.

In acid hydrolysis, the products are the carboxylic acid ($\mathrm{CH}_3\mathrm{COOH}$) and the alcohol, which can re-react in the presence of the acid catalyst to form the ester. The equilibrium is reversible and both forward and backward reactions occur simultaneously.

Worked Example 10

$10.0 \mathrm{ g}$ of a triglyceride ester (molar mass $885 \mathrm{ g/mol}$) is completely hydrolysed with excess NaOH. Calculate the mass of soap (sodium carboxylate, molar mass $306 \mathrm{ g/mol}$) produced.

Solution:

A triglyceride has three ester groups, so it produces 3 moles of soap per mole of triglyceride.

Moles of triglyceride $= \frac{10.0}{885} = 0.0113 \mathrm{ mol}$

Moles of soap $= 3 \times 0.0113 = 0.0339 \mathrm{ mol}$

Mass of soap $= 0.0339 \times 306 = 10.37 \mathrm{ g}$

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Additional Worked Examples

Worked Example: Combustion Stoichiometry

Calculate the volume of oxygen (at r.t.p.) needed for the complete combustion of $2.0 \mathrm{ dm^3}$ of butane gas ($\mathrm{C_4H_{10}}$).

Solution

$2\mathrm{C_4H_{10}} + 13\mathrm{O_2} \to 8\mathrm{CO_2} + 10\mathrm{H_2O}$

From the equation: 2 moles of butane react with 13 moles of $\mathrm{O_2}$.

Since volumes of gases at the same temperature and pressure are proportional to moles:

$V(\mathrm{O_2}) = \frac{13}{2} \times V(\mathrm{C_4H_{10}}) = \frac{13}{2} \times 2.0 = 13.0 \mathrm{ dm^3}$

Worked Example: Electrophilic Addition Mechanism

Describe the mechanism for the reaction of propene with $\mathrm{HBr}$, explaining why 2-bromopropane is the major product.

Solution

Step 1: The $\mathrm{HBr}$ molecule approaches the $\pi$ electron cloud of the C=C double bond. The electron-rich double bond polarises $\mathrm{HBr}$.

Step 2: The $\pi$ bond breaks and forms a bond with the $\mathrm{H}$ atom from $\mathrm{HBr}$. The H adds to C-1 (the less substituted carbon, following Markovnikov's rule), and a secondary carbocation intermediate forms at C-2:

$\mathrm{CH_2}=\mathrm{CHCH_3} + \mathrm{HBr} \to \mathrm{CH_3}-\mathrm{C^+HCH_3} + \mathrm{Br^-}$

Step 3: The bromide ion attacks the carbocation:

$\mathrm{CH_3}-\mathrm{C^+HCH_3} + \mathrm{Br^-} \to \mathrm{CH_3CHBrCH_3}$

The major product is 2-bromopropane because the secondary carbocation intermediate is more stable than the primary carbocation that would form if H added to C-2.

Worked Example: Identifying a Polymer

A polymer has the repeating unit $--\mathrm{CH_2--CHCl}--$. Name the monomer, the polymer, and state one use.

Solution

To find the monomer, remove the bonds between repeating units and add a double bond:

Monomer: $\mathrm{CH_2}=\mathrm{CHCl}$ (chloroethene, also called vinyl chloride)

Polymer name: polyvinyl chloride (PVC)

Uses: pipes, window frames, cable insulation, flooring.

Worked Example: Ester Formation and Hydrolysis

Ethanol ($\mathrm{CH_3CH_2OH}$) reacts with propanoic acid ($\mathrm{CH_3CH_2COOH}$) in the presence of concentrated $\mathrm{H_2SO_4}$. (a) Name the ester formed. (b) Write the equation. (c) Explain why the ester is immiscible with water.

Solution

(a) The ester is named using the alkyl group from the alcohol (ethyl) and the -oate from the acid (propanoate): ethyl propanoate.

(b) $\mathrm{CH_3CH_2COOH} + \mathrm{CH_3CH_2OH} \rightleftharpoons \mathrm{CH_3CH_2COOCH_2CH_3} + \mathrm{H_2O}$

Conditions: concentrated $\mathrm{H_2SO_4}$ catalyst, heat under reflux.

(c) The ester molecule lacks any -OH or other strongly polar group that can form hydrogen bonds with water. Although the C=O bond is polar, the overall molecule is much less polar than the parent alcohol or acid. Without hydrogen bonding to water, the ester is immiscible.

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7. Practice Questions

Exam-Style Practice Questions

Question 1: Draw and name all the structural isomers of $\mathrm{C}_5\mathrm{H}_{12}$ (pentane).

There are three structural isomers of pentane:

1. Pentane (n-pentane): $\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{CH}_2\mathrm{CH}_3$
2. 2-methylbutane (isopentane): $\mathrm{CH}_3\mathrm{CH}(\mathrm{CH}_3)\mathrm{CH}_2\mathrm{CH}_3$
3. 2,2-dimethylpropane (neopentane): $\mathrm{C}(\mathrm{CH}_3)_4$

The boiling points decrease in the order: pentane ($36^\circ\mathrm{C}$) > 2-methylbutane ($28^\circ\mathrm{C}$) > 2,2-dimethylpropane ($9.5^\circ\mathrm{C}$), because branching reduces surface area and weakens van der Waals forces.

Question 2: A compound $\mathrm{X}$ has the molecular formula $\mathrm{C}_3\mathrm{H}_6\mathrm{O}$. It decolourises bromine water, gives a positive result with 2,4-DNPH, but does not react with Tollens' reagent. Identify $\mathrm{X}$ and explain the observations.

$\mathrm{C}_3\mathrm{H}_6\mathrm{O}$: degree of unsaturation = $\frac{2(3)+2-6}{2} = 1$.

Decolourises bromine water: contains a C=C double bond.

Positive 2,4-DNPH: contains a C=O carbonyl group.

Since there is one degree of unsaturation for C=C and the molecular formula has both C=C and C=O suggested, but the degree of unsaturation is only 1, we must reconsider. The compound is propanone ($\mathrm{CH}_3\mathrm{COCH}_3$), a ketone.

2,4-DNPH positive: propanone has C=O (confirms carbonyl).

Tollens' negative: propanone is a ketone, not an aldehyde.

Bromine water: propanone does NOT decolourise bromine water under normal conditions. This question contains a contradiction; the correct answer is propanal ($\mathrm{CH}_3\mathrm{CH}_2\mathrm{CHO}$), which is an aldehyde (positive 2,4-DNPH and positive Tollens'), or the compound may be prop-2-en-1-ol ($\mathrm{CH}_2=\mathrm{CHCH}_2\mathrm{OH}$), which decolourises bromine water but would not give a positive 2,4-DNPH test.

Revised answer: The compound $\mathrm{X}$ is propanone ($\mathrm{CH}_3\mathrm{COCH}_3$). Positive 2,4-DNPH confirms the carbonyl group. Negative Tollens' confirms it is a ketone, not an aldehyde. Propanone does not decolourise bromine water under standard conditions; if the question states it does, the compound would more likely be prop-2-enal ($\mathrm{CH}_2=\mathrm{CHCHO}$), which has both C=C and C=O (degree of unsaturation = 2, which requires reconsideration of the formula).

Question 3: Describe the mechanism for the reaction of ethene with bromine, including the intermediate formed.

The reaction is electrophilic addition.

Step 1: The $\mathrm{Br}_2$ molecule approaches the $\pi$ electron cloud of the C=C bond. The electron-rich double bond polarises the $\mathrm{Br}_2$ molecule ($\mathrm{Br}^\delta{+}$--$\mathrm{Br}^\delta{-}$).

Step 2: The $\pi$ bond breaks and a new $\sigma$ bond forms between one carbon and the electrophilic bromine atom. The $\mathrm{Br}$-$\mathrm{Br}$ bond breaks heterolytically, producing a bromide ion ($\mathrm{Br}^-$). A carbocation intermediate is formed:

$\mathrm{CH}_2=\mathrm{CH}_2 + \mathrm{Br}_2 \to \mathrm{CH}_2\mathrm{Br}-\mathrm{CH}_2^+ + \mathrm{Br}^-$

Step 3: The bromide ion attacks the carbocation from the opposite side, forming the product 1,2-dibromoethane:

$\mathrm{CH}_2\mathrm{Br}-\mathrm{CH}_2^+ + \mathrm{Br}^- \to \mathrm{CH}_2\mathrm{BrCH}_2\mathrm{Br}$

Question 4: Explain why benzene undergoes substitution reactions rather than addition reactions with bromine, despite having a high electron density.

Benzene has a delocalised $\pi$ system of 6 electrons spread over the entire ring. This delocalisation gives the molecule significant extra stability (resonance energy of approximately $150$ kJ/mol). An addition reaction would destroy this delocalised system, requiring a large input of energy. A substitution reaction replaces one hydrogen atom while preserving the delocalised $\pi$ system, so the product retains the aromatic stability. Therefore, substitution is energetically favoured over addition.

Question 5: State and explain the trend in boiling points for the following compounds: methane ($-162^\circ\mathrm{C}$), ethane ($-89^\circ\mathrm{C}$), propane ($-42^\circ\mathrm{C}$), butane ($0^\circ\mathrm{C}$).

Boiling point increases with increasing chain length. As the number of carbon atoms increases, the molecular mass increases, and there are more electrons in the molecule. More electrons lead to stronger van der Waals (London dispersion) forces between molecules. Stronger intermolecular forces require more energy to overcome, resulting in higher boiling points. Additionally, larger molecules have greater surface area for intermolecular contact.

Question 6: Compound $\mathrm{Y}$ ($\mathrm{C}_4\mathrm{H}_8\mathrm{O}_2$) is hydrolysed with aqueous NaOH to give ethanol and the sodium salt of ethanoic acid. Deduce the structure of $\mathrm{Y}$ and write an equation for its formation from ethanoic acid and ethanol.

The products of hydrolysis are ethanol ($\mathrm{CH}_3\mathrm{CH}_2\mathrm{OH}$) and sodium ethanoate ($\mathrm{CH}_3\mathrm{COONa}$). The original compound $\mathrm{Y}$ is therefore the ester formed from ethanoic acid and ethanol.

$\mathrm{Y}$ = ethyl ethanoate: $\mathrm{CH}_3\mathrm{COOCH}_2\mathrm{CH}_3$

Formation equation:

$\mathrm{CH}_3\mathrm{COOH} + \mathrm{CH}_3\mathrm{CH}_2\mathrm{OH} \rightleftharpoons \mathrm{CH}_3\mathrm{COOCH}_2\mathrm{CH}_3 + \mathrm{H}_2\mathrm{O}$

Conditions: concentrated $\mathrm{H}_2\mathrm{SO}_4$ catalyst, heat under reflux.

Question 7: Describe a chemical test to distinguish between propan-1-ol and propan-2-ol.

Both propan-1-ol and propan-2-ol react with $\mathrm{PCl}_5$ to give white fumes (both contain -OH), so this test does not distinguish them.

Use acidified potassium dichromate(VI) followed by a confirmatory test:

1. Add acidified $\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$ (orange) and warm gently. Both will turn green (both oxidise).
2. Distil the product from the reaction mixture.
3. Test the distillate with Tollens' reagent.

Propan-1-ol oxidises to propanal (aldehyde), which gives a silver mirror with Tollens' reagent.

Propan-2-ol oxidises to propanone (ketone), which does not give a silver mirror with Tollens' reagent.

Alternatively, if the oxidation products can be isolated, 2,4-DNPH gives an orange precipitate for both (both produce carbonyl compounds), but only the aldehyde product (from propan-1-ol) gives a positive Fehling's test.

Question 8: Draw the structure of Nylon-6,6 and identify the two monomers from which it is made.

Nylon-6,6 is made from hexanedioic acid (adipic acid) and hexane-1,6-diamine.

Hexanedioic acid: $\mathrm{HOOC}(\mathrm{CH}_2)_4\mathrm{COOH}$

Hexane-1,6-diamine: $\mathrm{H}_2\mathrm{N}(\mathrm{CH}_2)_6\mathrm{NH}_2$

Repeating unit of Nylon-6,6:

$\left[\mathrm{--OC}(\mathrm{CH}_2)_4\mathrm{CONH}(\mathrm{CH}_2)_6\mathrm{NH--}\right]_n$

The condensation polymerisation eliminates water molecules at each amide bond formed.

Question 9: Explain the mechanism of free radical substitution for the reaction of methane with chlorine under UV light. Why is a mixture of chlorinated products obtained?

See Section 2.1 for the full mechanism (initiation, propagation, termination).

A mixture of products ($\mathrm{CH}_3\mathrm{Cl}$, $\mathrm{CH}_2\mathrm{Cl}_2$, $\mathrm{CHCl}_3$, $\mathrm{CCl}_4$) is obtained because once a chlorine radical abstracts a hydrogen from $\mathrm{CH}_3\mathrm{Cl}$ (the product), a new $\mathrm{CH}_2\mathrm{Cl}^\bullet$ radical is formed in the propagation step. This radical can then react with more $\mathrm{Cl}_2$ to form $\mathrm{CH}_2\mathrm{Cl}_2$, and the process continues. Each successive chlorination makes the remaining C-H bonds slightly weaker (due to the electron-withdrawing effect of Cl), so further substitution is progressively easier.

Question 10: A student claims that poly(ethene) is harmful to the environment because it cannot be degraded by microorganisms. Evaluate this claim and suggest alternatives.

The claim is correct. Poly(ethene) consists of very long chains of carbon-carbon bonds that are chemically inert and not recognised by enzymes in microorganisms. This means poly(ethene persists in landfill sites and the natural environment for hundreds of years. It also breaks down into microplastics that accumulate in marine ecosystems.

Alternatives include:

1. Biodegradable polymers such as PLA (poly(lactic acid)) or PHB (poly(hydroxybutyrate)), which have ester or other bonds that can be hydrolysed by microorganisms
2. Recycling poly(ethene) mechanically (melting and reforming) or chemically (breaking down into monomers for repolymerisation)
3. Reducing usage through reusable bags and containers
4. Using biopolymers derived from renewable resources (e.g., starch-based polymers, cellulose)

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Summary Table

Topic	Key Concept	Example
Alkanes	Saturated, $\mathrm{C}_n\mathrm{H}_{2n+2}$	Free radical substitution with $\mathrm{Cl}_2$
Alkenes	Unsaturated, $\mathrm{C}_n\mathrm{H}_{2n}$, addition reactions	Bromine water decolourisation test
Alkynes	Triple bond, $\mathrm{C}_n\mathrm{H}_{2n-2}$	Hydration to carbonyl compounds
Benzene	Delocalised $\pi$ system, EAS	Nitration, Friedel-Crafts
Alcohols	-OH group, classified as 1$^\circ$, 2$^\circ$, 3$^\circ$	Oxidation to aldehydes/ketones
Carboxylic acids	-COOH, weak acids, $\mathrm{p}K_a \approx 4.7$	Esterification
Esters	-COO-, fruity odours	Saponification for soap
Aldehydes / Ketones	C=O, distinguished by Fehling's and Tollens'	Nucleophilic addition with HCN
Addition polymerisation	Alkene monomers, no small molecule lost	Polyethylene, PVC, PTFE
Condensation polymerisation	Two functional groups, small molecule eliminated	Nylon-6,6, PET (Terylene)

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Exam Tips

• When drawing organic mechanisms, always show curly arrows representing the movement of electron pairs (from nucleophile to electrophile, or from bond to atom).
• For isomerism questions, systematically consider chain, position, and functional group isomerism.
• When asked to distinguish between two compounds, choose a test that gives a clearly different observable result for each compound.
• In polymer questions, always show the repeating unit correctly and include the bracket notation with $n$.
• For benzene questions, remember that benzene undergoes substitution (not addition) because the delocalised $\pi$ system must be preserved.
• Markovnikov's Rule applies to addition of HX to unsymmetrical alkenes: the H adds to the carbon with more H atoms.
• When writing equations for esterification, always include the acid catalyst and remember the reaction is reversible.
• For cracking, specify whether it is thermal or catalytic and state the conditions required.
• Remember that primary alcohols give aldehydes then carboxylic acids on oxidation; secondary give ketones; tertiary give nothing.

Additional Exam-Style Problems

Question 11: Write balanced equations for the complete combustion of (a) propane and (b) propene.

(a) $\mathrm{C}_3\mathrm{H}_8 + 5\mathrm{O}_2 \to 3\mathrm{CO}_2 + 4\mathrm{H}_2\mathrm{O}$

(b) $\mathrm{C}_3\mathrm{H}_6 + \frac{9}{2}\mathrm{O}_2 \to 3\mathrm{CO}_2 + 3\mathrm{H}_2\mathrm{O}$

Or: $2\mathrm{C}_3\mathrm{H}_6 + 9\mathrm{O}_2 \to 6\mathrm{CO}_2 + 6\mathrm{H}_2\mathrm{O}$

Question 12: Describe the difference between thermoplastic and thermosetting polymers. Give one example of each.

Thermoplastic polymers can be melted and reshaped repeatedly. They consist of separate polymer chains held together by weak intermolecular forces. When heated, these forces are overcome and the polymer softens. Examples: polyethylene, polypropene, PVC.

Thermosetting polymers cannot be remelted once set. They have cross-links between polymer chains that form a rigid 3D network. Heating causes decomposition rather than melting. Examples: Bakelite, melamine, urea-methanal resin.

Question 13: Compound $\mathrm{Z}$ ($\mathrm{C}_4\mathrm{H}_8\mathrm{O}_2$) is a sweet-smelling liquid that is immiscible with water. When heated with aqueous NaOH, it produces methanoic acid and propan-1-ol. Identify $\mathrm{Z}$ and write equations for its hydrolysis.

$\mathrm{Z}$ = propyl methanoate: $\mathrm{HCOOCH}_2\mathrm{CH}_2\mathrm{CH}_3$

Acid hydrolysis: $\mathrm{HCOOCH}_2\mathrm{CH}_2\mathrm{CH}_3 + \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{HCOOH} + \mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{OH}$

Alkaline hydrolysis: $\mathrm{HCOOCH}_2\mathrm{CH}_2\mathrm{CH}_3 + \mathrm{NaOH} \to \mathrm{HCOONa} + \mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{OH}$

Question 14: Explain why the $\pi$ bond in ethene is weaker than the $\sigma$ bond.

The $\sigma$ bond is formed by the end-on (head-on) overlap of $sp^2$ hybrid orbitals, which is efficient and concentrates electron density directly between the two nuclei. The $\pi$ bond is formed by the sideways overlap of unhybridised $p$ orbitals above and below the plane of the molecule. This sideways overlap is less effective, and the $\pi$ electron density is further from the nuclei and more exposed. The $\pi$ bond therefore has a lower bond energy (approximately $270$ kJ/mol for C=C $\pi$ component vs $350$ kJ/mol for C-C $\sigma$ component) and is more easily broken in addition reactions.

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Problem Set

Problem 1: Name the following compound: $\mathrm{CH_3CH(CH_3)CH_2CH_2Cl}$

If you get this wrong, revise: Naming (Section 2.1)

Solution

The longest carbon chain containing the Cl is 5 carbons (pentane). Number from the end nearer Cl.

Cl is on C-1. The methyl substituent is on C-2.

Name: 1-chloro-2-methylpentane

Problem 2: Draw and name all structural isomers of $\mathrm{C_4H_9Cl}$.

If you get this wrong, revise: Structural Isomerism (Section 2.1)

Solution

1. 1-chlorobutane: $\mathrm{CH_3CH_2CH_2CH_2Cl}$ (primary)
2. 2-chlorobutane: $\mathrm{CH_3CH_2CHClCH_3}$ (secondary)
3. 1-chloro-2-methylpropane: $\mathrm{CH_3CH(CH_3)CH_2Cl}$ (primary)
4. 2-chloro-2-methylpropane: $\mathrm{(CH_3)_3CCl}$ (tertiary)

Problem 3: Write the initiation, propagation (one step), and one termination step for the free radical chlorination of methane.

If you get this wrong, revise: Free Radical Substitution Mechanism (Section 2.1)

Solution

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

Propagation: $\mathrm{Cl^\bullet} + \mathrm{CH_4} \to \mathrm{HCl} + \mathrm{CH_3^\bullet}$

Termination: $\mathrm{Cl^\bullet} + \mathrm{Cl^\bullet} \to \mathrm{Cl_2}$

Problem 4: State the reagents and conditions for converting propene to propan-1-ol. Name the type of reaction and write the equation.

If you get this wrong, revise: Chemical Properties of Alkenes (Section 2.2)

Solution

Reagents: $\mathrm{H_2O}$ with $\mathrm{H_3PO_4}$ catalyst at $300^\circ\mathrm{C}$ under high pressure.

Type: hydration (addition reaction).

$\mathrm{CH_3CH=CH_2} + \mathrm{H_2O} \xrightarrow{\mathrm{H_3PO_4}} \mathrm{CH_3CH_2CH_2OH}$

Note: The major product follows Markovnikov's rule, giving propan-2-ol. To obtain propan-1-ol specifically, an indirect route via hydroboration-oxidation would be needed (beyond DSE scope).

Problem 5: A compound has the molecular formula $\mathrm{C_4H_8}$. It does not decolourise bromine water. Draw and name two possible structures.

If you get this wrong, revise: Structural Isomerism (Section 2.1)

Solution

$\mathrm{C_4H_8}$ corresponds to one degree of unsaturation ($\mathrm{C}_n\mathrm{H}_{2n}$). If it does not decolourise bromine water, it is NOT an alkene. It must be a cycloalkane.

1. Cyclobutane: a four-membered ring ($\mathrm{C_4H_8}$)
2. Methylcyclopropane: a three-membered ring with a methyl substituent ($\mathrm{C_4H_8}$)

Both have the formula $\mathrm{C_4H_8}$ but are cyclic, so the C=C test is negative.

Problem 6: Write the equation for the nitration of methylbenzene (toluene) and state the conditions.

If you get this wrong, revise: Electrophilic Aromatic Substitution (Section 2.4)

Solution

Conditions: concentrated $\mathrm{HNO_3}$ with concentrated $\mathrm{H_2SO_4}$ (catalyst) at $50$--$60^\circ\mathrm{C}$.

$\mathrm{C_6H_5CH_3} + \mathrm{HNO_3} \xrightarrow{\mathrm{conc.\ H_2SO_4}} \mathrm{C_6H_4(CH_3)(NO_2)} + \mathrm{H_2O}$

The product is a mixture of 2-nitrotoluene (minor) and 4-nitrotoluene (major), because the $-\mathrm{CH_3}$ group directs substitution to the ortho and para positions.

Problem 7: Classify the following alcohols as primary, secondary, or tertiary: (a) butan-2-ol, (b) 2-methylpropan-2-ol, (c) pentan-1-ol. State the oxidation product of each.

If you get this wrong, revise: Classification and Oxidation of Alcohols (Section 3.1)

Solution

(a) $\mathrm{CH_3CH_2CH(OH)CH_3}$ -- secondary -- oxidises to butanone ($\mathrm{CH_3CH_2COCH_3}$)

(b) $\mathrm{(CH_3)_3COH}$ -- tertiary -- not oxidised by common reagents

(c) $\mathrm{CH_3CH_2CH_2CH_2CH_2OH}$ -- primary -- oxidises to pentanal ($\mathrm{CH_3CH_2CH_2CH_2CHO}$), then to pentanoic acid ($\mathrm{CH_3CH_2CH_2CH_2COOH}$)

Problem 8: Ethanoic acid reacts with methanol to form an ester. (a) Name the ester. (b) Write the equation. (c) State the role of the concentrated sulphuric acid.

If you get this wrong, revise: Esterification (Section 3.2)

Solution

(a) Methyl ethanoate

(b) $\mathrm{CH_3COOH} + \mathrm{CH_3OH} \rightleftharpoons \mathrm{CH_3COOCH_3} + \mathrm{H_2O}$

(c) Concentrated $\mathrm{H_2SO_4}$ acts as a catalyst and as a dehydrating agent, removing water and shifting the equilibrium towards the ester product.

Problem 9: Draw the repeating unit of the addition polymer formed from chloroethene ($\mathrm{CH_2=CHCl}$) and name the polymer.

If you get this wrong, revise: Addition Polymerisation (Section 4.1)

Solution

$n\mathrm{CH_2=CHCl} \to \left[\mathrm{--CH_2--CHCl--}\right]_n$

Repeating unit: $--\mathrm{CH_2--CHCl}--$

Polymer name: polyvinyl chloride (PVC)

Problem 10: Compound $\mathrm{A}$ ($\mathrm{C_3H_6O}$) gives the following results: (i) positive 2,4-DNPH test, (ii) no reaction with Tollens' reagent, (iii) no reaction with $\mathrm{PCl_5}$. Identify $\mathrm{A}$.

If you get this wrong, revise: Test Tube Reactions (Section 5.4)

Solution

Positive 2,4-DNPH: contains C=O (carbonyl group).

No reaction with Tollens': not an aldehyde (it is a ketone).

No reaction with $\mathrm{PCl_5}$: no -OH group (not an alcohol or carboxylic acid).

$\mathrm{A}$ = propanone ($\mathrm{CH_3COCH_3}$), which is a ketone with a C=O group, consistent with all observations.

Problem 11: Describe how to convert ethanol to ethanoic acid in two steps, giving reagents and conditions for each step.

If you get this wrong, revise: Reaction Pathway Tables (Section 5.5)

Solution

Step 1: Ethanol to ethanal (mild oxidation)

Reagent: acidified $\mathrm{K_2Cr_2O_7}$

Conditions: heat, distil the product out as it forms (to prevent further oxidation)

$\mathrm{CH_3CH_2OH} \xrightarrow{[\mathrm{O}],\ \mathrm{distil}} \mathrm{CH_3CHO}$

Step 2: Ethanal to ethanoic acid (strong oxidation)

Reagent: acidified $\mathrm{K_2Cr_2O_7}$

Conditions: heat under reflux (ensures complete oxidation)

$\mathrm{CH_3CHO} \xrightarrow{[\mathrm{O}],\ \mathrm{reflux}} \mathrm{CH_3COOH}$

Alternatively, a single step of oxidation under reflux converts ethanol directly to ethanoic acid.

Problem 12: $6.0 \mathrm{ g}$ of impure calcium carbonate is reacted with excess hydrochloric acid, producing $1.15 \mathrm{ dm^3}$ of $\mathrm{CO_2}$ at r.t.p. Calculate the percentage purity of the calcium carbonate.

If you get this wrong, revise: Combustion and Stoichiometry (Section 2.1)

Solution

$\mathrm{CaCO_3} + 2\mathrm{HCl} \to \mathrm{CaCl_2} + \mathrm{H_2O} + \mathrm{CO_2}$

$n(\mathrm{CO_2}) = \frac{1.15}{24.0} = 0.0479 \mathrm{ mol}$

$n(\mathrm{CaCO_3}) = 0.0479 \mathrm{ mol}$

$m(\mathrm{pure\ CaCO_3}) = 0.0479 \times 100 = 4.79 \mathrm{ g}$

$\mathrm{Percentage\ purity} = \frac{4.79}{6.0} \times 100\% = 79.8\%$

Problem 13: Write equations for the complete combustion of methane and for the incomplete combustion of methane producing carbon monoxide.

If you get this wrong, revise: Chemical Properties of Alkanes (Section 2.1)

Solution

Complete combustion (excess $\mathrm{O_2}$):

$\mathrm{CH_4} + 2\mathrm{O_2} \to \mathrm{CO_2} + 2\mathrm{H_2O}$

Incomplete combustion (limited $\mathrm{O_2}$):

$2\mathrm{CH_4} + 3\mathrm{O_2} \to 2\mathrm{CO} + 4\mathrm{H_2O}$

With even less oxygen, carbon (soot) may also be produced: $\mathrm{CH_4} + \mathrm{O_2} \to \mathrm{C} + 2\mathrm{H_2O}$

Problem 14: Explain what is meant by the term "electrophile" and explain why $\mathrm{Br_2}$ acts as an electrophile when it reacts with ethene.

If you get this wrong, revise: Chemical Properties of Alkenes (Section 2.2)

Solution

An electrophile is an electron-deficient species that accepts a pair of electrons from a nucleophile (an electron-rich species).

When $\mathrm{Br_2}$ approaches the electron-rich $\pi$ bond of ethene, the $\pi$ electrons induce a dipole in the $\mathrm{Br_2}$ molecule, making one bromine atom partially positive ($\mathrm{Br}^{\delta+}$). This partially positive bromine acts as the electrophile and is attracted to the $\pi$ bond. The $\pi$ bond then donates electrons to form a new $\sigma$ bond with $\mathrm{Br}$.

Problem 15: Describe a chemical test to distinguish between cyclohexane and cyclohexene.

If you get this wrong, revise: Test for Unsaturation (Section 2.2)

Solution

Add bromine water (orange-brown) to each compound separately.

• Cyclohexane: No reaction; bromine water remains orange-brown. Cyclohexane is a saturated cycloalkane with no C=C bond.
• Cyclohexene: Bromine water is decolourised (orange-brown to colourless). Cyclohexene has a C=C bond that undergoes electrophilic addition with $\mathrm{Br_2}$.

Problem 16: Draw the repeating unit of Terylene (PET), formed from benzene-1,4-dicarboxylic acid and ethane-1,2-diol. State the type of polymerisation.

If you get this wrong, revise: Condensation Polymerisation (Section 4.2)

Solution

Type: condensation polymerisation (water is eliminated).

Repeating unit:

$\left[\mathrm{--OC\mathrm{-}C_6H_4\mathrm{-}COOCH_2CH_2O--}\right]_n$

The benzene-1,4-dicarboxylic acid provides the $-\mathrm{COO-}$ linkage and the ethane-1,2-diol provides the $-\mathrm{OCH_2CH_2O-}$ linkage.

Problem 17: Compound $\mathrm{P}$ ($\mathrm{C_4H_8O_2}$) is hydrolysed with aqueous NaOH to give ethanol and the sodium salt of propanoic acid. (a) Identify $\mathrm{P}$. (b) Write an equation for the formation of $\mathrm{P}$.

If you get this wrong, revise: Hydrolysis of Esters (Section 3.3)

Solution

(a) The hydrolysis products are ethanol ($\mathrm{CH_3CH_2OH}$) and sodium propanoate ($\mathrm{CH_3CH_2COONa}$). Therefore, $\mathrm{P}$ is the ester of propanoic acid and ethanol.

$\mathrm{P}$ = ethyl propanoate: $\mathrm{CH_3CH_2COOCH_2CH_3}$

(b) Formation (esterification):

$\mathrm{CH_3CH_2COOH} + \mathrm{CH_3CH_2OH} \rightleftharpoons \mathrm{CH_3CH_2COOCH_2CH_3} + \mathrm{H_2O}$

Conditions: concentrated $\mathrm{H_2SO_4}$ catalyst, heat under reflux.

Problem 18: Explain the meaning of the term "saturated" as applied to hydrocarbons, and explain why alkenes are described as "unsaturated."

If you get this wrong, revise: General Properties of Alkanes and Alkenes (Sections 2.1 and 2.2)

Solution

Saturated hydrocarbons contain only single C-C and C-H bonds. They have the maximum possible number of hydrogen atoms for their carbon skeleton (general formula $\mathrm{C_nH_{2n+2}}$ for alkanes). They cannot undergo addition reactions.

Unsaturated hydrocarbons contain at least one multiple bond (C=C or C$\equiv$C). They have fewer hydrogen atoms than the corresponding saturated compound (general formula $\mathrm{C_nH_{2n}}$ for alkenes). The multiple bond can be "saturated" by adding atoms across it in addition reactions, such as with $\mathrm{Br_2}$ or $\mathrm{H_2}$.

Problem 19: Write an equation for the cracking of $\mathrm{C_{14}H_{30}}$ to produce ethene and another product. State the type of cracking and the conditions.

If you get this wrong, revise: Cracking of Hydrocarbons (Section 2.1)

Solution

$\mathrm{C_{14}H_{30}} \to \mathrm{C_2H_4} + \mathrm{C_{12}H_{26}}$

Type: thermal cracking (produces a mixture of alkanes and alkenes).

Conditions: high temperature ($400$--$700^\circ\mathrm{C}$) and high pressure.

Note: Catalytic cracking (using zeolite at $\sim 450^\circ\mathrm{C}$) could also produce ethene but typically gives more branched products.

Problem 20: Compound $\mathrm{Q}$ has the molecular formula $\mathrm{C_4H_{10}O}$. It does not react with $\mathrm{PCl_5}$, does not decolourise bromine water, and does not react with acidified $\mathrm{K_2Cr_2O_7}$. Identify $\mathrm{Q}$ and explain the observations.

If you get this wrong, revise: Functional Group Identification (Section 5.4)

Solution

Does not react with $\mathrm{PCl_5}$: no -OH group (not an alcohol or carboxylic acid).

Does not decolourise bromine water: no C=C or C$\equiv$C (not an alkene or alkyne).

Does not react with acidified $\mathrm{K_2Cr_2O_7}$: consistent with no -OH group.

The only remaining possibility for $\mathrm{C_4H_{10}O}$ without -OH is an ether.

$\mathrm{Q}$ = diethyl ether ($\mathrm{CH_3CH_2OCH_2CH_3}$) or methyl propyl ether ($\mathrm{CH_3OCH_2CH_2CH_3}$).

Problem 21: Calculate the percentage yield when $5.0 \mathrm{ g}$ of propanoic acid ($M = 74.0 \mathrm{ g/mol}$) reacts with excess ethanol to produce $4.8 \mathrm{ g}$ of ethyl propanoate ($M = 102 \mathrm{ g/mol}$).

If you get this wrong, revise: Esterification (Section 3.2)

Solution

$n(\mathrm{propanoic\ acid}) = \frac{5.0}{74.0} = 0.0676 \mathrm{ mol}$

Theoretical moles of ester $= 0.0676 \mathrm{ mol}$ (1:1 ratio)

$\mathrm{Theoretical\ mass} = 0.0676 \times 102 = 6.90 \mathrm{ g}$

$\mathrm{Percentage\ yield} = \frac{4.8}{6.90} \times 100\% = 69.6\%$