Organic Chemistry: Hydrocarbons — A-Level Chemistry Study Guide

For: A-Level Chemistry candidates sitting A-Level Chemistry.

Covers: alkane substitution and combustion, alkene addition and polymerisation, free radical and electrophilic addition mechanisms, structural and stereoisomerism, and standard hydrocarbon distinguishing tests

You should already know: IGCSE Chemistry, basic algebra.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the A-Level Chemistry style for educational use. They are not reproductions of past Cambridge International examination papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official Cambridge mark schemes for grading conventions.

1. What Is Organic Chemistry: Hydrocarbons?

Hydrocarbons are organic compounds consisting exclusively of carbon and hydrogen atoms, the simplest class of organic molecules that form the backbone of all more complex organic structures. They are split into two broad groups: aliphatic (straight, branched, or non-aromatic ring structures) and aromatic (ring structures with delocalised pi electrons). This guide focuses on the aliphatic subgroups alkanes (saturated, containing only single C-C bonds) and alkenes (unsaturated, containing at least one C=C double bond), as specified for core assessment in the A-Level Chemistry syllabus.

2. Alkanes — substitution and combustion

Alkanes are saturated hydrocarbons with the general formula $C_n{H}_{2n+2}$ for non-cyclic structures, and $C_n{H}_{2n}$ for cyclic alkanes. Their low reactivity stems from the strong, non-polar nature of C-C and C-H bonds, so reactions only occur under high-energy conditions.

Combustion

Alkanes are widely used as fuels due to their highly exothermic combustion reactions. Complete combustion in excess oxygen produces only carbon dioxide and water: $$C_n{H}_{2n+2}+\frac{3n+1}{2}{O}_2\to nC{O}_2+(n+1)H_{2}O$$ For example, complete combustion of propane: $C_3{H}_{8(g)}+5O_{2(g)}\to 3C{O}_{2(g)}+4H_2{O}_{(g)}$, with an enthalpy change of ~-2220 kJ mol⁻¹. Incomplete combustion in limited oxygen produces toxic carbon monoxide or solid carbon (soot) instead of CO2.

Free Radical Substitution

Alkanes react with halogens (Cl2, Br2) in the presence of ultraviolet (UV) light via substitution, where one hydrogen atom is replaced by a halogen atom to form a halogenoalkane and hydrogen halide. Multiple substitution can occur if excess halogen is present, producing a mixture of products.

Worked Example

Write the balanced equation for the monosubstitution of butane with chlorine, including reaction conditions.

Answer: $C_4{H}_{10}+C{l}_2\stackrel{UV}{\to }{C}_4{H}_{9}Cl+HCl$ Exam note: Examiners will deduct a mark if you omit the UV light condition for this reaction.

3. Alkenes — addition reactions and polymerisation

Alkenes are unsaturated hydrocarbons with the general formula $C_n{H}_{2n}$ for non-cyclic structures, containing at least one C=C double bond. The high electron density of the pi bond in the C=C group makes alkenes far more reactive than alkanes, and they primarily undergo addition reactions, where the pi bond breaks and two new single covalent bonds form.

Key Addition Reactions

1. Hydrogenation: Addition of H2 gas with a nickel catalyst at 150°C, producing a saturated alkane. Used industrially to turn liquid vegetable oils into solid margarine.
2. Halogenation: Addition of halogens at room temperature (no catalyst required), producing a dihalogenoalkane.
3. Hydrohalogenation: Addition of hydrogen halides (HCl, HBr, HI) at room temperature, following Markovnikov’s rule: the hydrogen atom adds to the carbon in the double bond with the most hydrogen atoms already attached, forming the more stable carbocation intermediate and a major substituted product. For example, propene + HBr produces 90% 2-bromopropane and 10% 1-bromopropane.
4. Hydration: Addition of steam with a concentrated H3PO4 catalyst at 300°C and 60 atm, producing an alcohol. Used industrially to manufacture ethanol from ethene.

Addition Polymerisation

Thousands of alkene monomers can undergo addition reactions to form long, saturated polymer chains. The pi bond in each monomer breaks, and the monomers link together via single C-C bonds. For example, ethene monomers form poly(ethene), and chloroethene monomers form poly(vinyl chloride) (PVC).

Worked Example

Draw the repeating unit of poly(propene).

Answer: ${\left[-CH(C{H}_3)-C{H}_2-\right]}_n$ Exam note: Never include a double bond in the polymer repeating unit, as the pi bond breaks during polymerisation.

4. Mechanisms — free radical, electrophilic addition

Reaction mechanisms describe the step-by-step movement of electrons during a reaction, and are a high-weight assessment component in A-Level Chemistry papers.

Free Radical Substitution (Alkane Halogenation)

This mechanism occurs in three stages, with homolytic bond breaking (each atom retains one electron from the broken bond) forming free radicals (species with an unpaired electron):

1. Initiation: UV light breaks the weak Cl-Cl bond to form two chlorine free radicals: $C{l}_2\stackrel{UV}{\to }2Cl\bullet$
2. Propagation: A chain reaction where free radicals react with stable molecules to form new free radicals: $Cl\bullet +C{H}_4\to C{H}_3\bullet +HCl$ $C{H}_3\bullet +C{l}_2\to C{H}_{3}Cl+Cl\bullet$
3. Termination: Two free radicals combine to form a stable molecule, ending the chain: $Cl\bullet +Cl\bullet \to C{l}_2$, $C{H}_3\bullet +Cl\bullet \to C{H}_{3}Cl$, $C{H}_3\bullet +C{H}_3\bullet \to C_2{H}_6$

Exam tip: Always draw a single dot next to the atom with the unpaired electron for all free radical species, missing dots will lose marks.

Electrophilic Addition (Alkene Addition Reactions)

Electrophiles are electron-deficient species that accept a pair of electrons to form a new covalent bond. The mechanism for addition of HBr to ethene is as follows:

1. The electron-rich pi bond in the C=C double bond attacks the partially positive H atom of polar HBr, breaking the H-Br bond heterolytically (both electrons go to the more electronegative Br atom) to form a positively charged carbocation intermediate and a Br⁻ ion.
2. The Br⁻ ion donates a lone pair of electrons to the carbocation, forming the stable bromoethane product.

Worked Example

Explain why 2-bromopropane is the major product of the reaction between propene and HBr.

Answer: The secondary carbocation intermediate (positive charge on the central carbon, bonded to two alkyl groups) formed when H adds to the terminal CH2 group is more stable than the primary carbocation (positive charge on the terminal carbon, bonded to one alkyl group) that would form if H added to the central carbon. More stable intermediates form faster, so the product derived from the secondary carbocation is dominant.

5. Isomerism — structural and stereoisomerism

Isomers are compounds with the same molecular formula but different arrangements of atoms, leading to different physical and/or chemical properties. There are two broad classes of isomerism for hydrocarbons:

Structural Isomerism

Compounds with the same molecular formula but different structural formula (different order of bonded atoms):

1. Chain isomerism: Different carbon chain arrangements, e.g. butane ($C{H}_{3}C{H}_{2}C{H}_{2}C{H}_3$) and 2-methylpropane ($(C{H}_3{)}_{3}CH$)
2. Position isomerism: Same functional group, different position on the carbon chain, e.g. 1-chloropropane ($C{H}_{3}C{H}_{2}C{H}_{2}Cl$) and 2-chloropropane ($C{H}_{3}CHClC{H}_3$)
3. Functional group isomerism: Different functional groups, e.g. propene ($C_3{H}_6$, alkene) and cyclopropane ($C_3{H}_6$, cyclic alkane)

Stereoisomerism

Compounds with the same structural formula but different spatial arrangement of atoms:

1. Cis-trans (geometric) isomerism: Occurs in alkenes where each carbon in the C=C double bond is bonded to two different groups. The cis isomer has identical groups on the same side of the double bond, while the trans isomer has identical groups on opposite sides. For example, cis-but-2-ene has a higher boiling point than trans-but-2-ene due to its higher polarity.
2. Optical isomerism: Occurs when a molecule has a chiral centre (a carbon atom bonded to four distinct groups), forming two non-superimposable mirror images called enantiomers.

Worked Example

How many structural isomers have the molecular formula $C_4{H}_8$?

Answer: 4: but-1-ene, but-2-ene, 2-methylpropene, and cyclobutane. Exam note: Always include cyclic isomers when counting isomers for formulas matching $C_n{H}_{2n}$.

6. Distinguishing tests — bromine water, alkaline KMnO₄

These qualitative tests are used to differentiate between saturated alkanes and unsaturated alkenes, and are frequently tested in both theory and practical assessment papers.

Bromine Water Test

Bromine water is an orange-brown aqueous solution of Br2. Procedure: Add 2 cm³ of bromine water to 2 cm³ of the unknown hydrocarbon, shake gently at room temperature with no UV light exposure.

• Alkane result: No reaction, orange-brown colour remains (alkanes only react with bromine under UV light, which is excluded from the standard test)
• Alkene result: Immediate decolourisation, solution turns colourless, as Br2 adds across the C=C double bond to form a colourless dihalogenoalkane.

Exam tip: If you state that alkanes decolourise bromine water without specifying UV light, you will lose marks.

Alkaline Potassium Manganate(VII) (KMnO4) Test

Dilute alkaline KMnO4 is a purple oxidising agent. Procedure: Add 3 drops of 0.01 mol dm⁻³ alkaline KMnO4 to 2 cm³ of the unknown hydrocarbon, warm gently.

• Alkane result: No reaction, purple colour remains
• Alkene result: Purple colour fades to colourless, with a brown precipitate of manganese(IV) oxide (MnO2) forming, as the C=C double bond is oxidised to a diol (two OH groups added across the bond).

Worked Example

A student has two unlabelled test tubes containing cyclopentane and cyclopentene. Describe a test to identify each compound, including observations.

Answer: Add bromine water to each test tube, shake at room temperature. The test tube where the orange-brown colour immediately decolourises contains cyclopentene. The test tube where the orange-brown colour remains contains cyclopentane.

7. Common Pitfalls (and how to avoid them)

• Wrong move: Omitting UV light as a condition for alkane-halogen substitution. Why: Students confuse alkane substitution conditions with alkene halogenation, which occurs at room temperature with no UV. Correct move: Always write "UV light" above the reaction arrow for free radical substitution, and specify "room temperature, dark" for alkene halogenation if conditions are requested.
• Wrong move: Drawing double bonds in polymer repeating units. Why: Students copy the monomer structure directly instead of recalling the pi bond breaks during polymerisation. Correct move: The repeating unit only contains single bonds between the two carbons that formed the monomer double bond, with extending bonds outside the square brackets to show the continuous chain.
• Wrong move: Claiming all alkenes have cis-trans isomers. Why: Students forget the requirement that each carbon in the C=C double bond must be bonded to two distinct groups. Correct move: Check if either carbon in the double bond has two identical groups (e.g. propene $C{H}_{3}CH=C{H}_2$ has one carbon bonded to two H atoms, so no cis-trans isomerism).
• Wrong move: Forgetting the unpaired electron dot on free radical species. Why: Students rush through mechanism drawings and skip required notation. Correct move: Add a single dot next to the atom with the unpaired electron for every free radical in initiation, propagation and termination steps.
• Wrong move: Stating that alkanes decolourise bromine water in standard test conditions. Why: Students mix up the UV-dependent substitution reaction with the dark addition reaction of alkenes. Correct move: Alkanes only react with bromine water if exposed to UV light, so the standard dark test shows no colour change for alkanes.

8. Practice Questions (A-Level Chemistry Style)

Question 1

Methane reacts with chlorine gas in the presence of UV light to form a mixture of products. (a) Write the balanced equation for the initiation step of this reaction. [1 mark] (b) Write two propagation step equations for the formation of chloromethane. [2 marks] (c) Write one termination step equation that forms ethane. [1 mark]

Worked Solution

(a) $C{l}_2\stackrel{UV}{\to }2Cl\bullet$ (1 mark for correct equation, including UV condition and free radical dots) (b) $Cl\bullet +C{H}_4\to C{H}_3\bullet +HCl$ (1 mark); $C{H}_3\bullet +C{l}_2\to C{H}_{3}Cl+Cl\bullet$ (1 mark) (c) $C{H}_3\bullet +C{H}_3\bullet \to C_2{H}_6$ (1 mark)

Question 2

But-2-ene reacts with hydrogen bromide to form a single major product. (a) Name the mechanism for this reaction. [1 mark] (b) Draw the structure of the major product. [1 mark] (c) Explain why but-2-ene exhibits cis-trans isomerism, but but-1-ene does not. [2 marks]

Worked Solution

(a) Electrophilic addition (1 mark) (b) $C{H}_{3}CHBrC{H}_{2}C{H}_3$ (2-bromobutane, 1 mark) (c) But-2-ene has each carbon in the C=C double bond bonded to two distinct groups (one CH3 and one H) (1 mark). But-1-ene has one carbon in the C=C double bond bonded to two identical H atoms, so no cis-trans isomerism is possible (1 mark).

Question 3

A 0.1 mol sample of a gaseous hydrocarbon undergoes complete combustion, producing 0.4 mol of CO2 and 0.5 mol of H2O. (a) Calculate the molecular formula of the hydrocarbon. [2 marks] (b) State whether this hydrocarbon is an alkane or alkene, and give a reason for your answer. [1 mark]

Worked Solution

(a) Moles of C in hydrocarbon = moles of CO2 = 0.4 mol, so 0.1 mol hydrocarbon contains 0.4 mol C → 1 mol hydrocarbon contains 4 C atoms (1 mark). Moles of H in hydrocarbon = 2 × moles of H2O = 1.0 mol, so 1 mol hydrocarbon contains 10 H atoms. Molecular formula = $C_4{H}_{10}$ (1 mark). (b) Alkane, as it matches the general formula $C_n{H}_{2n+2}$ for alkanes (1 mark).

9. Quick Reference Cheatsheet

10. What's Next

This hydrocarbons topic is the foundational building block for all subsequent organic chemistry content in the A-Level Chemistry syllabus. The reaction mechanisms you learned here (free radical substitution, electrophilic addition) will be extended to more complex functional groups including halogenoalkanes, alcohols, carboxylic acids, and aromatic compounds later in the course. Isomerism is a recurring concept that will appear in every organic chemistry topic, so mastering it now will save you significant time when studying more complex molecules later. Distinguishing tests are a core component of practical organic chemistry assessment, and you will encounter many more test procedures for other functional groups in upcoming modules.

If you have any questions about reaction mechanisms, isomer counting, or test observations as you study, you can ask Ollie our AI tutor for step-by-step explanations and extra practice questions at any time on the homepage. You can also access more A-Level Chemistry study guides, past paper walkthroughs, and practical assessment tips to help you prepare for your A-Level Chemistry exams.

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