You already know from GCSE bonding that a bond between atoms of different electronegativity is polar, and from Alkanes that halogenoalkanes are made when an alkane meets a halogen in UV light. This page is where they come alive: that polar C–X bond makes the carbon open to attack, so halogenoalkanes react by nucleophilic substitution (with the curly arrows you met in the introduction) and by elimination — and their most famous cousins, the CFCs, tore a hole in the ozone layer.
The polar C–X bond
A halogen is more electronegative than carbon, so the carbon–halogen bond is polar: the carbon carries a small positive charge (δ+) and the halogen a small negative charge (δ−). That δ+ carbon is a target — it attracts species with electrons to spare.
Recall electronegativity and what makes a bond polar from Bonding (3.1.3).
A nucleophile is an electron-pair donor — a species with a lone pair (and often a negative charge) that it uses to form a new bond to a δ+ carbon.
In nucleophilic substitution a nucleophile swaps places with the halogen. Three are examined at A-level — meet them and their conditions first, then we’ll draw the two-arrow mechanism they all share.
Three nucleophiles: OH−, CN−, NH3
Three nucleophiles are examined, and each gives a different product under its own conditions. The single most valuable thing to memorise is this grid — because AQA often marks the reagent and the conditions separately.
| Nucleophile | Reagent | Conditions | Product |
|---|---|---|---|
| OH− (hydrolysis) | NaOH or KOH | aqueous, warm, reflux | an alcohol (R–OH) |
| CN− | KCN | in ethanol (aqueous ethanol), reflux | a nitrile (R–CN) — adds a carbon |
| NH3 | excess ammonia | in ethanol, sealed tube (heat) | a primary amine (R–NH2) |
The cyanide route is special because it lengthens the carbon chain by one — a nitrile that can later be reduced to an amine or hydrolysed to a carboxylic acid. The ammonia route needs excess ammonia to stop at the primary amine; with excess halogenoalkane the amine keeps reacting to give secondary, tertiary and quaternary products.
That build-up of secondary, tertiary and quaternary products is exactly how amines are made — you’ll cover it in full in Amines (3.3.11), an A-level (Year 13) topic.
The CN group brings its own carbon, so the nitrile always has one more carbon than the halogenoalkane. Name it from the new, longer chain:
- bromoethane (2 C) → propanenitrile (3 C)
- 1-bromopropane (3 C) → butanenitrile (4 C)
Naming the nitrile from the precursor’s carbon count — calling the 1-bromopropane product ‘propanenitrile’ — is wrong: always count the extra carbon the CN adds.
Interactive — reagent + conditions, as a pair
Reagent
Conditions
Organic product
- Quote the reagent AND the conditions. Examiner reports award these as separate marks — “KCN” alone is not enough; say KCN in ethanol, reflux.
- Cyanide is ethanolic; hydroxide (for substitution) is aqueous. The solvent is the detail students most often drop.
- Ammonia must be in excess to give the primary amine; otherwise further substitution occurs.
🧪 Exam-style questions
1-Bromopropane reacts with a solution of potassium cyanide in aqueous ethanol. What is the organic product? Tick (✓) one box.
Source: AQA A-Level Chemistry past papers.
The mechanism
All three nucleophiles do the very same thing: a lone pair (or a negative charge) attacks the δ+ carbon, and the C–Br bonding pair leaves with the halogen as Br−. Two curly arrows — here they are for each nucleophile in turn.
Cyanide is the one to watch: it attacks through its carbon, because that is where the lone pair and the negative charge sit. Start the arrow there — and the new C–C bond is what lengthens the chain by one.
Ammonia carries no charge at all, so watch what conservation of charge forces to happen — a positive intermediate, and a second ammonia needed to reach the amine.
- Charge is conserved in every step. Add up the charges on the left and on the right — they must match. Two neutral species in (NH3 + the halogenoalkane), so a + ion and a − ion out.
- Every arrow starts where the electrons are — a lone pair or a bond — and ends where they are going: an atom, or the gap where a new bond forms. Never from empty space, and never from a positive charge.
- A neutral nucleophile makes a positive intermediate. NH3 has no charge to donate, so the ethylammonium ion comes out + as the Br− leaves — that is charge being conserved, not a slip.
Interactive — place the arrows
- The carbon is δ+ because the halogen is more electronegative — that polarity is why the nucleophile attacks the carbon (not the halogen).
- The charge and the lone pair go on the atom that attacks. On the O in OH−, on the C in CN− (never the N) — that is the atom that forms the new bond.
- Arrows start on a lone pair and on the C–X bond. One curly arrow from the nucleophile’s lone pair to the carbon; one from the C–X bond to the halogen — never from empty space.
- The halogen leaves as X− (it takes both bonding electrons).
Recall the curly-arrow rules from Introduction to organic chemistry.
🧪 Exam-style questions
Which compound is formed from bromoethane in a nucleophilic substitution reaction? Tick (✓) one box.
Source: AQA A-Level Chemistry past papers.
Rate & the C–X bond
Different halogenoalkanes hydrolyse at very different rates — and the reason is not the obvious one.
You might expect the most polar bond to react fastest: fluorine is the most electronegative, so C–F has the biggest δ+ carbon. But it is the bond enthalpy that controls the rate, not the polarity. The C–I bond is the weakest, so it breaks most easily and iodoalkanes hydrolyse fastest; the C–F bond is the strongest, so fluoroalkanes hydrolyse slowest:
rate: iodo > bromo > chloro > fluoro (weakest C–X bond → fastest)
You can watch the trend in a test tube. Warm each halogenoalkane with silver nitrate solution in ethanol: the water acts as the nucleophile (a slow hydrolysis), and the instant a halide ion is released, Ag+ catches it as a precipitate of AgX. Yellow AgI appears first, cream AgBr next, white AgCl last — the order the precipitates appear is the rate order. Those precipitate colours are the same ones you meet in the silver nitrate test for halide ions in Group 7.
the cause — bond enthalpy
the weakest bond breaks most easily — so the C–I bond hydrolyses fastest, strong C–F barely at all
the evidence — the silver nitrate test
the water is the nucleophile; Ag+ just catches the halide the instant it leaves, as insoluble AgX
- Bond enthalpy, not polarity, controls the rate. The C–I bond is weakest, so it breaks most easily and reacts fastest — even though C–F is the most polar.
- The trend is iodo > bromo > chloro > fluoro. In the AgNO3 test, the iodoalkane gives its precipitate (yellow AgI) first.
🧪 Exam-style questions
1-Bromobutane and 1-iodobutane are hydrolysed by sodium hydroxide under the same conditions, and the halide-ion concentration is monitored. The iodobutane reacts faster. State how a concentration–time graph would show this, and explain why the rates differ.
Show answer
The graph for 1-iodobutane rises more steeply / reaches its final halide concentration sooner. 1 mark
The C–I bond is weaker (has a lower bond enthalpy) than the C–Br bond, so it breaks more easily. 1 mark
Source: AQA A-Level Chemistry past papers.
Substitution vs elimination
Warm a halogenoalkane with potassium hydroxide and the same OH− can do two different jobs — and the conditions decide which wins. Both happen at once; you just shift the balance.
| Substitution | Elimination | |
|---|---|---|
| OH− acts as a… | nucleophile | base |
| Conditions | aqueous KOH, warm | ethanolic KOH, hot (reflux) |
| Product | an alcohol | an alkene (+ H2O + KX) |
Both routes use the same reagent (OH−, from KOH), so all you need to fix is conditions → product:
- COWS — Cold (just warm, not hot), OH−, Water (aqueous) → Substitution (an alcohol).
- HEAT — Hot, Ethanolic → an Alkene. Elimination is the one that needs real heat, so the word is the reminder.
In elimination, the OH− (as a base) pulls an H+ off the carbon next to the C–Br; that C–H bonding pair becomes the C=C double bond, and the C–Br bond breaks to release Br−. Because the hydrogen can leave from more than one neighbouring carbon, you often get a mixture of alkenes (for example 2-bromobutane gives but-1-ene and but-2-ene).
aqueous KOH · warm — substitution
OH− as a nucleophile — its lone pair attacks the δ+ carbon
ethanolic KOH · hot — elimination
OH− as a base — it takes an H from the carbon next to the C–Br
Interactive — you set the solvent
- Aqueous → substitution; ethanolic → elimination. The solvent flips the outcome — and it is a common lost mark.
- OH− is a nucleophile in substitution, a base in elimination. Name the role the question asks about.
- Elimination can give more than one alkene (a mixture), because the H can be removed from different neighbouring carbons.
🧪 Exam-style questions
When 2-bromobutane is warmed with potassium hydroxide, substitution and elimination both occur. What is the role of the hydroxide ions in the elimination reaction? Tick (✓) one box.
When 2-bromobutane is warmed with potassium hydroxide, substitution and elimination both occur. Which of these compounds is not produced? Tick (✓) one box.
Which compound is not formed by reacting 3-bromo-3-methylhexane with warm, ethanolic potassium hydroxide? Tick (✓) one box.
Source: AQA A-Level Chemistry past papers.
Ozone depletion
CFCs (chlorofluorocarbons) were prized as refrigerants and solvents precisely because they are so unreactive — but that stability let them drift, intact, all the way up to the ozone layer, where they did real damage.
Ozone (O3) in the upper atmosphere is beneficial: it absorbs harmful ultraviolet radiation that would otherwise reach living organisms. High up, UV is energetic enough to break a C–Cl bond in a CFC, making a chlorine radical, Cl• — and a single Cl• then catalyses the destruction of ozone, over and over:
Cl• + O3 → ClO• + O2
ClO• + O3 → 2O2 + Cl•
Add the two steps and the ClO• and one Cl• cancel: overall 2O3 → 3O2, with the Cl• regenerated. Because it is not used up, one chlorine radical destroys thousands of ozone molecules — the essence of a catalyst.
The evidence, gathered by different research groups (notably Rowland and Molina), persuaded governments to ban CFCs as solvents and refrigerants, and chemists developed chlorine-free alternatives — a case study in how science drives legislation.
Interactive — run the cycle
- Chlorine atoms (radicals), not molecules, catalyse the decomposition — give both steps so the Cl• is shown being regenerated.
- Ozone is beneficial because it absorbs UV. That is the reason its loss matters — not global warming.
- UV breaks the C–Cl bond (the weakest in a CFC) to make the radical.
🧪 Exam-style questions
Most scientists believe the concentration of ozone in the upper atmosphere should not be allowed to decrease. Which is a correct reason for this belief? Tick (✓) one box.
State why the ozone layer is beneficial for living organisms.
Show answer
It absorbs (harmful) ultraviolet radiation. 1 mark
Source: AQA A-Level Chemistry past papers.
Capstone: reagent + condition
Halogenoalkanes are the crossroads of organic synthesis: from one C–X you can reach an alcohol, a nitrile (and, from there, an amine or a carboxylic acid), an amine, or an alkene — and the conditions choose the destination. This is the grid to know cold:
| Want… | Reagent | Conditions | OH− / role |
|---|---|---|---|
| an alcohol | NaOH / KOH | aqueous, warm, reflux | nucleophile (substitution) |
| a nitrile (+1 C) | KCN | ethanolic, reflux | — |
| a primary amine | excess NH3 | ethanolic, sealed tube | — |
| an alkene | KOH | ethanolic, hot (reflux) | base (elimination) |
Read the question for the product it wants, then quote the reagent and the conditions together — because on this topic, AQA marks them as a pair. The aqueous-versus-ethanolic call is the one to get right.
Capstone — build the full recipe
Reagent
Solvent
Temperature
- The C–X bond is polar (δ+ carbon); halogenoalkanes react by nucleophilic substitution — a nucleophile’s lone pair attacks the δ+ carbon (one arrow) and the C–X bond breaks, X leaving as X− (second arrow).
- OH− (NaOH/KOH, aqueous, warm) → alcohol. CN− (KCN, ethanolic, reflux) → nitrile (adds a carbon). NH3 (excess, ethanolic, sealed tube) → primary amine. Quote the reagent and the conditions.
- Rate of hydrolysis follows the C–X bond enthalpy, not polarity: C–I is the weakest bond → fastest; C–F strongest → slowest. So iodo > bromo > chloro > fluoro.
- Substitution vs elimination: with OH−, aqueous favours substitution (OH− as a nucleophile → alcohol); ethanolic, hot favours elimination (OH− as a base → alkene). Both occur together.
- Ozone in the upper atmosphere absorbs harmful UV. UV breaks a C–Cl bond in a CFC → Cl•, which catalyses ozone decomposition: Cl• + O3 → ClO• + O2; ClO• + O3 → 2O2 + Cl• (Cl• regenerated).