Alkanes Revision Notes
Subject: Chemistry | Level: A-Level | Exam Board: AQA
Master AQA A-Level Chemistry's core topic, Alkanes (3.2). This guide breaks down everything from boiling point trends governed by van der Waals forces to the critical mechanisms of combustion and free-radical substitution, providing examiner insights to help you secure top marks.
Revision Notes & Key Concepts
Revision Podcast Transcript
AQA A-Level Chemistry — Alkanes, Topic 3.2 A Study Guide Podcast Episode Duration: approximately 10 minutes --- INTRO (approximately 1 minute) --- Hello and welcome! I'm your chemistry tutor, and today we're diving deep into one of the most important topics in AQA A-Level Chemistry — Alkanes, covered in Topic 3.2 of the specification. Now I know what some of you might be thinking — alkanes? They're just boring saturated hydrocarbons, right? Wrong! Alkanes are absolutely central to understanding organic chemistry, and more importantly, they come up in almost every A-Level exam paper in some form. Whether it's a question on boiling point trends, combustion equations, cracking conditions, or the mechanism of free-radical substitution — examiners love this topic because it tests both recall and application. So in the next ten minutes, I'm going to walk you through everything you need to know. We'll cover the core concepts clearly, then I'll give you my top exam tips, highlight the most common mistakes I see candidates make, run you through a quick-fire recall quiz, and finish with a summary of the absolute must-know points. Grab a pen, because you'll want to take notes. Let's go! --- CORE CONCEPTS (approximately 5 minutes) --- Let's start with the basics. Alkanes are saturated hydrocarbons — that means they contain only carbon and hydrogen atoms, and all the carbon-carbon bonds are single bonds. The general formula is C-n H-2n-plus-2. So methane is CH4, ethane is C2H6, propane is C3H8, and so on. They form a homologous series, meaning each successive member differs by a CH2 unit. Now, the first major concept you need to nail is PHYSICAL PROPERTIES — specifically boiling points. Here's the key idea: alkanes are non-polar molecules. The only intermolecular forces acting between them are van der Waals forces — also called London dispersion forces. These are temporary dipole-induced dipole interactions caused by the movement of electrons creating momentary uneven charge distributions. Here's what examiners want you to say: as the chain length increases, the molecules have more electrons and a greater surface area for contact. This means the van der Waals forces between molecules are STRONGER. And crucially — this is the mark-winning phrase — MORE ENERGY IS REQUIRED TO OVERCOME these forces. That's why boiling points increase as chain length increases. A common mistake? Candidates say the forces are "stronger" but forget to say more energy is needed to overcome them. That second part is what earns the mark. Remember: state the force, then state the consequence. Branching also matters. Branched-chain alkanes have lower boiling points than their straight-chain isomers. Why? Because branched molecules are more spherical — they have a smaller surface area for contact, so the van der Waals forces are weaker, and less energy is needed to separate them. Moving on to COMBUSTION. Alkanes burn in oxygen. Complete combustion produces carbon dioxide and water. For example: methane plus two moles of oxygen gives carbon dioxide plus two moles of water. Incomplete combustion — when there's insufficient oxygen — produces carbon monoxide and water, or even carbon particles as soot. Carbon monoxide is toxic because it binds irreversibly to haemoglobin, preventing oxygen transport. This is why catalytic converters in cars are so important. They contain platinum and rhodium catalysts and convert harmful gases. The key equation AQA loves to ask about is: 2NO plus 2CO gives N2 plus 2CO2. Learn that equation — it comes up as a recall item regularly. Another environmental concern is sulfur dioxide from burning fossil fuels containing sulfur impurities. This contributes to acid rain. The solution? Flue gas desulfurisation. Two equations to memorise here. First: calcium oxide reacts with sulfur dioxide to give calcium sulfite — CaO plus SO2 gives CaSO3. Second: calcium carbonate reacts with sulfur dioxide and oxygen to give calcium sulfate and carbon dioxide — this is the wet scrubbing process. AQA frequently asks for these as recall items, so commit them to memory. Now let's talk about CRACKING — breaking large alkane molecules into smaller, more useful ones. There are two types: thermal cracking and catalytic cracking. Thermal cracking uses very high temperatures — between 700 and 1200 degrees Celsius — and high pressures up to 70 atmospheres. There is no catalyst. The products are mainly alkenes, like ethene and propene, plus hydrogen. These alkenes are valuable feedstocks for making polymers and other chemicals. Catalytic cracking uses lower temperatures — around 450 to 500 degrees Celsius — slight pressure, and a zeolite catalyst. The zeolite has a porous structure that provides a large surface area. The products are mainly aromatic hydrocarbons like benzene and toluene, plus branched alkanes. These are used in high-octane motor fuels. The classic exam trap here is mixing up the two. Thermal cracking equals high temperature, high pressure, no catalyst, makes alkenes. Catalytic cracking equals moderate temperature, slight pressure, zeolite catalyst, makes aromatics and branched alkanes. Drill that distinction until it's automatic. Now for the big one — FREE-RADICAL SUBSTITUTION. This is the mechanism by which alkanes react with halogens like chlorine in the presence of ultraviolet light. The mechanism has three stages: Initiation, Propagation, and Termination. INITIATION: UV light provides energy to break the chlorine-chlorine bond by HOMOLYTIC FISSION. This means each atom gets one electron from the shared pair, forming two chlorine radicals. Write it as: Cl2 gives 2 Cl dot. The dot represents the unpaired electron. This is a crucial mark point — if you write the dot on the wrong atom, you lose the mark. PROPAGATION: This is a two-step chain reaction. Step one: a chlorine radical attacks a methane molecule. The chlorine radical takes a hydrogen atom, forming hydrogen chloride, and leaving a methyl radical. Write: Cl dot plus CH4 gives HCl plus dot-CH3. Step two: the methyl radical attacks a chlorine molecule, taking one chlorine atom to form chloromethane and regenerating a chlorine radical. Write: dot-CH3 plus Cl2 gives CH3Cl plus Cl dot. Notice how the radical produced in step one — the methyl radical — is the reactant in step two. And the radical produced in step two — the chlorine radical — is the reactant in step one. This is what makes it a CHAIN REACTION. The radical is never consumed overall; it just keeps cycling. Examiners specifically look for this linkage between the two propagation steps. A critical mistake I see all the time: candidates write the propagation steps producing H2 instead of HCl. Remember — the chlorine radical takes a hydrogen from methane to form HCl, not H2. The chlorine ends up in the HCl molecule. TERMINATION: The chain reaction ends when two radicals collide and combine to form a stable molecule. There are three possible termination reactions: Cl dot plus Cl dot gives Cl2. Dot-CH3 plus dot-CH3 gives C2H6 — ethane. And Cl dot plus dot-CH3 gives CH3Cl — chloromethane. Any combination of two radicals counts as termination. Note that chloromethane can be produced in both propagation and termination — this is fine and expected. --- EXAM TIPS AND COMMON MISTAKES (approximately 2 minutes) --- Right, let's talk exam strategy. Here are my top tips for this topic. Tip one: In boiling point questions, always use a two-part answer. State that van der Waals forces are stronger due to more electrons and greater surface area, THEN state that more energy is required to overcome these forces. Both parts are needed for the mark. Tip two: When writing free-radical substitution mechanisms, always check that your radical dot is on the CARBON atom in the methyl radical — dot-CH3 — not on the hydrogen. Placing the dot on the wrong atom is one of the most penalised errors in this topic. Tip three: For cracking questions, the command word matters. If the question says "State the conditions", you need to give temperature, pressure, and catalyst — all three. If it says "Explain why a zeolite catalyst is used", you need to mention its porous structure and large surface area. Tip four: Memorise the catalytic converter equation — 2NO plus 2CO gives N2 plus 2CO2 — and both flue gas desulfurisation equations. These appear as one-mark recall items and are easy marks to bank. Tip five: In propagation steps, always show the chain reaction linkage. Examiners award a mark specifically for showing that the radical produced in step one is consumed in step two, and vice versa. Common mistake alert: Do NOT say that covalent bonds break during fractional distillation. Fractional distillation separates alkanes by boiling point — only the INTERMOLECULAR FORCES are overcome, not the covalent bonds within the molecules. This is a classic error that costs candidates marks every year. --- QUICK-FIRE RECALL QUIZ (approximately 1 minute) --- Okay, quick-fire time! I'll ask the question, give you three seconds to think, then I'll give the answer. Question one: What type of bond fission occurs in the initiation step of free-radical substitution? Answer: Homolytic fission. Question two: What catalyst is used in catalytic cracking? Answer: Zeolite. Question three: What are the two propagation steps in the chlorination of methane? Give the equations. Answer: Step one — Cl dot plus CH4 gives HCl plus dot-CH3. Step two — dot-CH3 plus Cl2 gives CH3Cl plus Cl dot. Question four: Why do branched alkanes have lower boiling points than straight-chain isomers of the same molecular formula? Answer: Branched alkanes have a smaller surface area for contact, so van der Waals forces are weaker, and less energy is required to overcome them. Question five: Write the equation for the catalytic converter reaction removing nitrogen oxides and carbon monoxide. Answer: 2NO plus 2CO gives N2 plus 2CO2. How did you do? If you stumbled on any of those, go back and review that section — those are exactly the kinds of questions that appear in AQA papers. --- SUMMARY AND SIGN-OFF (approximately 1 minute) --- Let's wrap up with the absolute must-know points for Alkanes. One: Alkanes are saturated hydrocarbons with the general formula CnH2n+2. Their only intermolecular forces are van der Waals forces. Two: Boiling points increase with chain length because van der Waals forces are stronger, requiring more energy to overcome. Branching decreases boiling points. Three: Complete combustion gives CO2 and water. Incomplete combustion gives CO and soot. Catalytic converters convert 2NO plus 2CO to N2 plus 2CO2. Four: Thermal cracking uses high temperature, high pressure, no catalyst, and produces alkenes. Catalytic cracking uses moderate temperature, slight pressure, zeolite catalyst, and produces aromatics and branched alkanes. Five: Free-radical substitution has three stages — Initiation (homolytic fission by UV), Propagation (two-step chain reaction), and Termination (two radicals combine). Always put the radical dot on the carbon in dot-CH3. That's everything for Alkanes, Topic 3.2. You've got this! Keep practising those mechanism equations, and remember — in the exam, be explicit, be precise, and always link cause to effect. Good luck, and I'll see you in the next episode!
Key Terms & Definitions
- Saturated Hydrocarbon
- A compound containing only hydrogen and carbon atoms, where all the carbon-carbon bonds are single bonds.
- Homologous Series
- A series of organic compounds with the same functional group and similar chemical properties, in which successive members differ by a CH₂ group.
- Homolytic Fission
- The breaking of a covalent bond where one electron from the bonding pair goes to each atom, forming two radicals.
- Free Radical
- A species with an unpaired electron, making it highly reactive.
- Cracking
- The breaking down of long-chain hydrocarbon molecules into smaller, more useful hydrocarbon molecules.
- Isomers
- Molecules that have the same molecular formula but a different structural arrangement of atoms.
Worked Examples
Worked Example
Question: Explain why the boiling point of dodecane (C₁₂H₂₆) is significantly higher than that of butane (C₄H₁₀). (4 marks)
Solution: Step 1: Identify the intermolecular forces present in both molecules. Both dodecane and butane are non-polar molecules, so the only intermolecular forces are van der Waals forces. Step 2: Compare the structure of the two molecules. Dodecane is a much larger molecule with a longer carbon chain than butane. Step 3: Relate the size of the molecule to the strength of the intermolecular forces. Dodecane has more electrons and a larger surface area of contact between its molecules compared to butane. This results in stronger van der Waals forces between dodecane molecules. Step 4: Link the strength of the forces to the energy required to overcome them. More energy is required to overcome the stronger van der Waals forces in dodecane, hence it has a higher boiling point.
Worked Example
Question: The reaction of methane with chlorine in the presence of UV light produces chloromethane. Outline the mechanism for this free-radical substitution reaction. (5 marks)
Solution: Step 1: Initiation. The UV light provides the energy for the homolytic fission of the Cl-Cl bond, forming two chlorine radicals. Equation: Cl₂ --(UV light)--> 2Cl• Step 2: Propagation (Step 1). A chlorine radical reacts with a methane molecule, forming hydrogen chloride and a methyl radical. Equation: Cl• + CH₄ → HCl + •CH₃ Step 3: Propagation (Step 2). The methyl radical is reactive and attacks a chlorine molecule, forming chloromethane and regenerating a chlorine radical, which can continue the chain reaction. Equation: •CH₃ + Cl₂ → CH₃Cl + Cl• Step 4: Termination. The reaction is terminated when two radicals collide and combine. Any one of the following is acceptable. Equations: Cl• + Cl• → Cl₂ OR •CH₃ + •CH₃ → C₂H₆ OR Cl• + •CH₃ → CH₃Cl
Worked Example
Question: Thermal cracking of the alkane C₁₅H₃₂ can produce ethene, propene, and octane (C₈H₁₈) in one possible reaction. Construct a balanced equation for this reaction. (3 marks)
Solution: Step 1: Write the reactant and the known products in an unbalanced equation. C₁₅H₃₂ → C₂H₄ + C₃H₆ + C₈H₁₈ Step 2: Balance the carbon atoms. The question implies that these are the only products, but a check is needed. 15 carbons in the reactant. On the product side, we have 2 + 3 + 8 = 13 carbons. This means another product with 2 carbons is needed. Since thermal cracking produces alkenes, another mole of ethene is a likely product. Step 3: Construct the full balanced equation. Let's try with 2 moles of ethene. Reactant: C₁₅H₃₂ Products: 2 x C₂H₄ (ethene) + 1 x C₃H₆ (propene) + 1 x C₈H₁₈ (octane) Carbon check: (2 x 2) + 3 + 8 = 4 + 3 + 8 = 15. (Balanced) Hydrogen check: (2 x 4) + 6 + 18 = 8 + 6 + 18 = 32. (Balanced) Final Answer: C₁₅H₃₂ → 2C₂H₄ + C₃H₆ + C₈H₁₈
Practice Questions
Question: State the general formula for an alkane and the shape and bond angle around each carbon atom. (3 marks)
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Question: Explain why catalytic cracking is an important industrial process. (4 marks)
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Question: Further substitution can occur during the chlorination of methane, forming dichloromethane (CH₂Cl₂). Outline the two propagation steps that would lead to the formation of a •CHCl₂ radical. (2 marks)
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Question: Write a balanced equation for the incomplete combustion of pentane (C₅H₁₂) to form carbon monoxide and water. (2 marks)
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Question: A student states that 'butane has a higher boiling point than 2-methylpropane because it has a longer chain'. Criticise this statement. (3 marks)
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