Subject: Chemistry | Level: A-Level | Exam Board: AQA
Master the reactions and mechanisms of Halogenoalkanes! This guide covers everything from nucleophilic substitution and elimination to the environmental impact of CFCs on the ozone layer.
Revision Notes & Key Concepts
Revision Podcast Transcript
Hello and welcome to Chemistry Unlocked — the revision podcast that gets you exam-ready. I'm your host, and today we're diving deep into one of the most fascinating and frequently examined topics in A-Level Chemistry: Halogenoalkanes. Whether you're revising for the first time or doing a final polish before your exam, this episode has everything you need. We'll cover the core chemistry, the mechanisms, the environmental impact of CFCs, and I'll give you my top exam tips to help you pick up every available mark. So grab a pen, get comfortable, and let's get into it. Let's start with the basics. What exactly is a halogenoalkane? Simply put, it's an alkane — a hydrocarbon — where one or more hydrogen atoms have been replaced by a halogen atom. The halogens we're talking about are fluorine, chlorine, bromine, and iodine. So if you take ethane, which is CH3CH3, and replace one hydrogen with bromine, you get bromoethane — CH3CH2Br. That's your classic halogenoalkane. Now, the key thing that makes halogenoalkanes so interesting — and so reactive — is the carbon-halogen bond. Because halogens are much more electronegative than carbon, the bond is polar. The carbon carries a partial positive charge — we write that as delta-plus — and the halogen carries a partial negative charge — delta-minus. This polarity is what makes the carbon atom susceptible to attack by nucleophiles. Think of the carbon as a target — it's electron-deficient, and nucleophiles are attracted to it like a magnet. Now let's talk about bond enthalpy, because this is where the reactivity differences between the different halogenoalkanes come from. Bond enthalpy is the energy required to break one mole of a bond in the gaseous phase. The stronger the bond, the more energy needed to break it, and therefore the less reactive the compound. Here are the values you need to know. The carbon-fluorine bond has an enthalpy of approximately 485 kilojoules per mole — that's the strongest. Carbon-chlorine is around 339 kilojoules per mole. Carbon-bromine is about 284 kilojoules per mole. And carbon-iodine is the weakest at approximately 218 kilojoules per mole. So what does this mean for reactivity? Fluoroalkanes are the least reactive — that C-F bond is incredibly strong and very hard to break. Iodoalkanes are the most reactive — the C-I bond is weak and breaks easily. The order of reactivity is: iodo greater than bromo greater than chloro greater than fluoro. Remember this — it comes up all the time in exam questions. Now let's move on to the reactions themselves. There are two main types of reaction that halogenoalkanes undergo: nucleophilic substitution and elimination. Let's take them one at a time. In nucleophilic substitution, a nucleophile attacks the electron-deficient carbon atom and replaces — or substitutes — the halogen, which leaves as a halide ion. The halide ion is called the leaving group. There are three key nucleophiles you need to know for your exam. First: hydroxide ions, OH minus. When a halogenoalkane reacts with aqueous sodium hydroxide — that's NaOH dissolved in water — the hydroxide ion acts as a nucleophile. It donates its lone pair to the carbon, the C-halogen bond breaks, and an alcohol is produced. So bromoethane plus aqueous NaOH gives ethanol plus bromide ions. The conditions are aqueous NaOH, heated under reflux. Second: cyanide ions, CN minus. When a halogenoalkane reacts with potassium cyanide dissolved in ethanol — KCN in ethanol — the cyanide ion acts as the nucleophile. It attacks the carbon, the halide leaves, and a nitrile is produced. Crucially, this reaction adds a carbon atom to the chain — so it's used in synthesis to extend carbon chains. This is an important application that examiners love to test. Third: ammonia, NH3. When a halogenoalkane is heated with excess concentrated ammonia in a sealed tube, the ammonia molecule acts as a nucleophile through its nitrogen lone pair. The product is an amine. So bromoethane plus ammonia gives ethylamine plus hydrogen bromide. Note that the HBr can react with more ammonia to form ammonium bromide, which is why excess ammonia is used. Now let's look at the mechanism for nucleophilic substitution. This is where many candidates lose marks, so pay close attention. The mechanism involves curly arrows. A curly arrow represents the movement of a pair of electrons. Here are the golden rules: curly arrows must start from a lone pair or from a bond. They must end at an atom or at a bond being formed. Never draw a curly arrow starting from a positive charge or ending in mid-air. For the reaction of bromoethane with hydroxide ions: Step one — draw the hydroxide ion with a lone pair shown on the oxygen. Draw a curly arrow from that lone pair to the carbon atom of the C-Br bond. Step two — simultaneously, draw a second curly arrow from the C-Br bond to the bromine atom. This shows the bond breaking heterolytically — both electrons going to the bromine. Step three — the products are ethanol and a bromide ion. The carbon in the transition state is surrounded by five groups — it's trigonal bipyramidal. The hydroxide attacks from the opposite side to the bromine — this is called backside attack. The bromine leaves from the other side. This is the SN2 mechanism — substitution, nucleophilic, bimolecular. Now let's talk about elimination. This is the second major reaction type. In elimination, the hydroxide ion acts not as a nucleophile but as a base. Instead of attacking the carbon bearing the halogen, it removes a hydrogen atom from an adjacent carbon. This causes the C-H bond and the C-X bond to break simultaneously, and a carbon-carbon double bond forms — giving an alkene. The conditions for elimination are: NaOH or KOH dissolved in ethanol — not water — and heated. The key distinction is the solvent. Aqueous NaOH gives substitution. Ethanolic NaOH gives elimination. This is one of the most commonly tested points in the entire topic, so burn it into your memory: water gives alcohol, ethanol gives alkene. Right, let's talk exam technique. I'm going to walk you through the most common mistakes I see, and how to avoid them. Mistake number one: Curly arrow errors. This is the single biggest source of lost marks in mechanism questions. The most common errors are: drawing an arrow from a bond to a lone pair when it should be the other way around; starting an arrow from a partial positive charge rather than a lone pair; and forgetting to show the lone pair on the nucleophile at all. My advice: always draw the lone pairs on your nucleophile before you draw any arrows. It forces you to think about where the arrow starts. Mistake number two: Confusing substitution and elimination conditions. Candidates frequently write NaOH without specifying the solvent, and lose marks as a result. The conditions are everything. Always state: aqueous NaOH for substitution, ethanolic NaOH for elimination. And always state the temperature — heated under reflux. Mistake number three: Not linking bond enthalpy to reaction rate. If a question asks you to explain why iodoalkanes react faster than chloroalkanes, you must mention bond enthalpy explicitly. Say: The C-I bond has a lower bond enthalpy than the C-Cl bond, so less energy is required to break it, meaning the activation energy is lower and the reaction proceeds faster. That's a three-mark answer right there. Mistake number four: The ozone depletion equations. Many candidates can write one of the two equations but not both. You need both. Equation one: Cl radical plus O3 gives ClO radical plus O2. Equation two: ClO radical plus O3 gives 2O2 plus Cl radical. Notice that the chlorine radical is regenerated in equation two — this is what makes it a catalyst. It's not consumed in the overall reaction. Examiners will specifically ask you to explain why chlorine acts as a catalyst, and the answer is that it is regenerated. Mistake number five: Forgetting that the rate of hydrolysis is tested experimentally using silver nitrate solution. When a halogenoalkane is hydrolysed in ethanol and then silver nitrate solution is added, a precipitate of silver halide forms. The faster the precipitate forms, the faster the hydrolysis. Iodoalkanes give a yellow precipitate of silver iodide most quickly. Chloroalkanes give a white precipitate of silver chloride most slowly. This is a classic required practical context. Time for a quick-fire quiz! I'll ask a question, give you a few seconds to think, then give the answer. Question one: What is the order of reactivity of halogenoalkanes, from most to least reactive? ... Answer: Iodo, bromo, chloro, fluoro. Question two: What product is formed when bromoethane reacts with aqueous NaOH? ... Answer: Ethanol. Question three: What product is formed when bromoethane reacts with ethanolic NaOH? ... Answer: Ethene — an alkene. Question four: Write the two equations for the catalytic destruction of ozone by chlorine radicals. ... Answer: Cl radical plus O3 gives ClO radical plus O2. Then ClO radical plus O3 gives 2O2 plus Cl radical. Question five: Why does the C-F bond make fluoroalkanes less reactive than iodoalkanes? ... Answer: The C-F bond has a higher bond enthalpy — it is stronger and requires more energy to break, so the activation energy for reaction is higher. Question six: What reagent and conditions are needed to convert a halogenoalkane into a nitrile? ... Answer: Potassium cyanide, KCN, dissolved in ethanol, heated. Let's bring it all together. Here are your six key takeaways from today's episode. One: Halogenoalkanes contain a polar C-X bond. The carbon is delta-plus and is susceptible to nucleophilic attack. Two: Reactivity increases down Group 7 — iodoalkanes are most reactive, fluoroalkanes are least reactive — because bond enthalpy decreases. Three: Nucleophilic substitution occurs with aqueous NaOH, KCN in ethanol, or excess ammonia, producing alcohols, nitriles, or amines respectively. Four: Elimination occurs with ethanolic NaOH and produces alkenes. The solvent is the key difference from substitution. Five: Curly arrows must start from a lone pair or bond and end at an atom or bond being formed. Never start from a charge. Six: CFCs deplete the ozone layer via a catalytic radical chain. Chlorine radicals are regenerated — that's what makes them catalysts. Know both equations. That's it for today's episode of Chemistry Unlocked. You've covered the full halogenoalkanes topic — mechanisms, conditions, bond enthalpy, and environmental chemistry. If you found this useful, go back and test yourself on the quick-fire questions without looking at your notes. Active recall is the most powerful revision technique there is. Good luck in your exams — you've got this. See you next time.
Key Terms & Definitions
- Halogenoalkane
- An alkane in which one or more hydrogen atoms have been replaced by a halogen atom.
- Nucleophile
- An electron pair donor.
- Substitution Reaction
- A reaction in which an atom or group of atoms is replaced by a different atom or group of atoms.
- Elimination Reaction
- A reaction in which a small molecule (like HBr or $H_2O$) is removed from a larger molecule, leaving a double bond.
- Bond Enthalpy
- The energy required to break one mole of a specific bond in the gaseous state.
- Radical
- A species with an unpaired electron.
Worked Examples
Worked Example
Question: 1-bromopropane reacts with aqueous sodium hydroxide. Draw the mechanism for this reaction and state the role of the hydroxide ion. (4 marks)
Solution: Step 1: Draw the structure of 1-bromopropane. Step 2: Draw the $OH^-$ ion with a lone pair and a negative charge. Step 3: Draw a curly arrow from the lone pair on the oxygen to the $\delta+$ carbon atom attached to the bromine. Step 4: Draw a curly arrow from the middle of the C-Br bond to the bromine atom. Step 5: Draw the products: propan-1-ol and a $Br^-$ ion. Role: The hydroxide ion acts as a nucleophile.
Worked Example
Question: Explain why 1-iodopropane reacts faster than 1-chloropropane when heated with aqueous silver nitrate. (3 marks)
Solution: Step 1: The C-I bond is weaker than the C-Cl bond (or has a lower bond enthalpy). Step 2: Therefore, the C-I bond requires less energy to break. Step 3: This means the activation energy for the reaction of 1-iodopropane is lower, so the reaction proceeds faster.
Worked Example
Question: Write two equations to show how chlorine radicals catalyse the decomposition of ozone. (2 marks)
Solution: Equation 1: $Cl\bullet + O_3 \rightarrow ClO\bullet + O_2$ Equation 2: $ClO\bullet + O_3 \rightarrow 2O_2 + Cl\bullet$
Practice Questions
Question: State the reagent and conditions required to convert 1-bromobutane into butanenitrile.
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Question: Explain why chlorofluorocarbons (CFCs) are damaging to the ozone layer. Include relevant equations in your answer.
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Question: A student reacts 2-bromo-2-methylpropane with hot ethanolic potassium hydroxide. Name the mechanism, state the role of the hydroxide ion, and draw the structure of the organic product.
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Question: Compare the rates of hydrolysis of 1-chlorobutane, 1-bromobutane, and 1-iodobutane. Explain your answer.
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Question: When 1-bromopropane reacts with ammonia, the product is propan-1-amine. Why must excess ammonia be used?
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