Last updated: 22 August 2026
A halogenoalkane can look deceptively simple: take an alkane and replace one hydrogen atom by fluorine, chlorine, bromine or iodine. Yet that one substitution changes the chemistry of the molecule dramatically because the carbon–halogen bond is polar. The carbon bonded to the halogen becomes electron-poor, creating a natural site for attack by electron-rich species.
This chapter therefore connects several important Organic Chemistry ideas. We will first see why different halogenoalkanes react at different rates, then follow their substitution and elimination reactions, and finally use molecular structure to understand the SN1 and SN2 mechanisms, stereochemistry and the unusual unreactivity of halogenoarenes.
Try these five questions before reading the chapter. Use the structures and reaction schemes carefully — Organic Chemistry often becomes much easier once you learn to read what is happening to the molecule.
- 5 diagnostic questions
- Structures and reaction schemes where useful
- Instant explanation for every answer option
The carbon–halogen bond controls halogenoalkane reactivity
Halogenoalkanes have the general form R–X, where X is F, Cl, Br or I. Because halogens are more electronegative than carbon, the bonding electron pair is attracted towards the halogen. The carbon therefore carries a partial positive charge, δ⁺, while the halogen carries a partial negative charge, δ⁻.
This makes the carbon atom an electron-poor centre. A nucleophile — a species that donates a lone pair of electrons — can attack this carbon and eventually replace the halogen.
The C–X bond is polar, but bond polarity alone does not determine how rapidly a halogenoalkane reacts. The strength of the C–X bond is crucial.
Why the most polar carbon–halogen bond is not the most reactive
It is tempting to predict that fluoroalkanes should be the most reactive because fluorine is the most electronegative halogen and the C–F bond is highly polar. Experimentally, however, fluoroalkanes are the least reactive towards hydrolysis.
Bond polarity: determines how electron-poor the carbon atom is and therefore helps make nucleophilic attack possible.
Bond strength: determines how difficult it is to break the carbon–halogen bond. For the hydrolysis trend down Group 17, this is the dominant factor.
Carbon–halogen bond strength down Group 17
Going from F to Cl to Br to I, the halogen atom becomes larger and the C–X bond becomes longer. Orbital overlap between carbon and the halogen becomes less effective, so the bond becomes weaker.
| Bond | Relative bond strength | Relative ease of hydrolysis |
|---|---|---|
| C–F | Strongest | Slowest |
| C–Cl | Strong | Slower |
| C–Br | Weaker | Faster |
| C–I | Weakest | Fastest |
If an examination question asks why iodoethane hydrolyses faster than bromoethane, do not simply say that iodine is less electronegative. State that the C–I bond is weaker than the C–Br bond and therefore requires less energy to break.
Nucleophilic substitution in halogenoalkanes
In nucleophilic substitution, a nucleophile replaces the halogen atom. The nucleophile supplies an electron pair to the electron-deficient carbon, while the carbon–halogen bond eventually breaks and the halogen leaves as a halide ion.
The 9476 syllabus uses bromoethane to illustrate three particularly important substitutions: formation of an alcohol, formation of a nitrile and formation of a primary amine.
Hydrolysis: forming an alcohol
Bromoethane is heated with aqueous sodium hydroxide. Hydroxide ion acts as the nucleophile and replaces Br, producing ethanol.
This reaction is called hydrolysis because the halogenoalkane is converted into an alcohol under aqueous conditions.
Formation of a nitrile: adding one carbon atom
Heating bromoethane with potassium cyanide in ethanol replaces Br by CN. The carbon atom in CN becomes part of the organic molecule, so the carbon chain increases by one carbon atom.
CN⁻ is especially useful in organic synthesis because replacing X by CN increases the carbon chain by one carbon atom.
Formation of a primary amine
Bromoethane reacts with ammonia in ethanol when heated under pressure. Ammonia acts as a nucleophile and ultimately gives ethylamine.
In practice, excess ammonia helps favour formation of the primary amine because the amine product is itself nucleophilic and can otherwise undergo further alkylation.
How SN2 and SN1 substitution differ
The same overall substitution — replacing X by a nucleophile — can occur by different molecular pathways. At A-Level, these are described as SN2 and SN1 mechanisms.
The important point is not merely to memorise the names. You should be able to connect each mechanism to the structure of the halogenoalkane and explain the stereochemical outcome.
The SN2 mechanism: nucleophile and leaving group move together
In an SN2 mechanism, nucleophilic attack and departure of the halide occur in the same step. The nucleophile approaches the δ⁺ carbon from the side opposite the C–X bond. As the new bond forms, the C–X bond breaks.
For an SN2 mechanism, draw a curly arrow from the nucleophile's lone pair to the carbon attached to X, and another curly arrow from the C–X bond to X.
There is no carbocation intermediate. Instead, bond formation and bond breaking occur together through a transition state.
Why primary halogenoalkanes favour SN2
Backside attack requires the nucleophile to reach the carbon attached to the halogen. In a primary halogenoalkane, relatively few alkyl groups surround this carbon, so there is less steric hindrance.
As the carbon becomes more substituted, surrounding alkyl groups obstruct the approaching nucleophile. SN2 therefore becomes increasingly difficult, particularly for tertiary halogenoalkanes.
Why SN2 gives inversion of configuration
If the reacting carbon is a chiral centre, attack from the side opposite the leaving group turns the tetrahedral arrangement inside out. The result is called inversion of configuration.
Think of SN2 as a one-step replacement from the back. The nucleophile enters from one side as the leaving group departs from the other, so a chiral centre undergoes inversion.
The SN1 mechanism: a carbocation forms first
An SN1 reaction proceeds in stages. First, the C–X bond breaks heterolytically and X leaves with the bonding electron pair. This produces a carbocation intermediate.
The nucleophile then donates an electron pair to the positively charged carbon of the carbocation.
Why tertiary halogenoalkanes favour SN1
Forming a carbocation is much more favourable when that carbocation is relatively stable. Alkyl groups have an electron-donating effect and help disperse the positive charge. A tertiary carbocation is therefore more stable than a secondary carbocation, which is more stable than a primary carbocation.
This explains why tertiary halogenoalkanes have a greater tendency to undergo SN1 substitution, while primary halogenoalkanes favour SN2. Secondary substrates lie between these extremes and their behaviour depends more strongly on the reaction conditions.
Why SN1 gives racemisation
A carbocation centre is planar. If the original molecule was optically active, the nucleophile can attack the planar carbocation from either side. Both configurations are therefore formed, producing racemisation.
| Feature | SN2 | SN1 |
|---|---|---|
| Basic pathway | One-step nucleophilic attack and C–X bond breaking | Carbocation forms first, then nucleophile attacks |
| Structural factor | Favoured by low steric hindrance | Favoured when a stable carbocation can form |
| Typical trend | Primary halogenoalkanes favoured | Tertiary halogenoalkanes favoured |
| Stereochemical outcome | Inversion of configuration | Racemisation |
Elimination competes with substitution
Hydroxide ion does not always act as a nucleophile. Under ethanolic conditions it can behave as a base and remove a hydrogen atom from a carbon adjacent to the carbon bearing the halogen.
The hydrogen and halogen are eliminated and a carbon–carbon double bond forms. The product is therefore an alkene.
NaOH(aq), heat: OH⁻ behaves mainly as a nucleophile and substitution gives an alcohol.
NaOH in ethanol, heat: OH⁻ behaves mainly as a base and elimination gives an alkene.
Do not identify the reaction from the reagent alone. The solvent is part of the reaction conditions: aqueous NaOH and ethanolic NaOH lead to different major pathways.
How to distinguish different halogenoalkanes
A useful test begins by hydrolysing the halogenoalkane so that its halogen atom is released as a halide ion. The halide ion can then be identified using silver nitrate.
Step 1: hydrolyse the halogenoalkane
Heat the halogenoalkane with aqueous NaOH. If hydrolysis occurs, X is released as X⁻.
Step 2: acidify the mixture
Add dilute nitric acid. This removes excess OH⁻ that could otherwise interfere with the subsequent silver-ion test.
Step 3: add aqueous silver nitrate
Ag⁺ reacts with Cl⁻, Br⁻ or I⁻ to form differently coloured silver halide precipitates.
| Halide ion | Silver halide formed | Observation |
|---|---|---|
| Cl⁻ | AgCl | White precipitate |
| Br⁻ | AgBr | Pale cream precipitate |
| I⁻ | AgI | Yellow precipitate |
The speed at which halide ions are released can also provide information about relative hydrolysis rates. Iodoalkanes generally release I⁻ more readily than comparable bromoalkanes or chloroalkanes because C–I is the weakest bond.
Why halogenoarenes are unusually unreactive
Chlorobenzene contains a C–Cl bond just as chloroethane does, but it does not readily undergo the nucleophilic substitution reactions typical of a chloroalkane. The benzene ring fundamentally changes the nature of the C–Cl bond and the geometry of nucleophilic attack.
Delocalisation strengthens the C–Cl bond
One of chlorine's lone pairs can overlap with the delocalised π system of the benzene ring. Electron density from this lone pair becomes partially delocalised into the ring.
As a result, the bond between the ring carbon and chlorine gains partial double-bond character. It is therefore stronger and more difficult to break than the ordinary C–Cl single bond in a chloroalkane.
Steric hindrance also makes nucleophilic attack difficult
The carbon bearing chlorine is part of the planar aromatic ring. The geometry of the benzene ring obstructs the type of backside attack needed for an SN2 pathway, making nucleophilic substitution much less favourable.
For 9476, explain the unreactivity of chlorobenzene using both delocalisation of a chlorine lone pair into the aromatic ring and steric hindrance.
Distinguishing a halogenoalkane from a halogenoarene
The difference in hydrolysis behaviour provides a characteristic test. Heat both compounds with aqueous NaOH, acidify with dilute nitric acid and then add aqueous silver nitrate.
Halogenoalkane: hydrolysis releases halide ions, so an appropriate silver halide precipitate can be obtained.
Halogenoarene: does not readily hydrolyse under these conditions, so little or no halide ion is released and the expected silver halide precipitate is absent.
Fluorinated compounds: useful because they are unreactive
The exceptionally strong C–F bond makes many fluoroalkanes and fluorohalogenoalkanes chemically stable. This relative inertness has made fluorinated compounds useful in applications where a substance must resist unwanted chemical reaction.
Historically, fluorohalogenoalkanes such as chlorofluorocarbons were used in applications including refrigeration and aerosol systems partly because they were stable under ordinary conditions.
Chemical inertness can be both useful and environmentally problematic: a compound that does not react readily near Earth's surface can remain in the atmosphere for a long time.
CFCs and the ozone layer
Chlorofluorocarbons, or CFCs, are associated with depletion of the stratospheric ozone layer. Their stability allows them to persist long enough to reach the upper atmosphere, where they can ultimately lead to destruction of ozone.
For this syllabus, you need to recognise the environmental effect. The detailed reaction mechanism by which CFCs deplete ozone is not required.
HFCs and HCFCs are not environmentally harmless
Hydrofluorocarbons, HFCs, and hydrochlorofluorocarbons, HCFCs, were introduced as replacements for CFCs in various applications, but they are not environmentally consequence-free.
| Type | Important feature | Environmental point |
|---|---|---|
| CFCs | Contain carbon, fluorine and chlorine | Associated with depletion of the ozone layer |
| HCFCs | Contain hydrogen as well as fluorine and chlorine | Reduced but non-zero ozone impact and can contribute to climate change |
| HFCs | Contain hydrogen and fluorine but no chlorine | Do not have the same chlorine-based ozone problem, but many are significant greenhouse gases |
Common Mistakes
| Common mistake | Why it is wrong | Better approach |
|---|---|---|
| Saying fluoroalkanes react fastest because C–F is the most polar bond. | The C–F bond is exceptionally strong, so it is difficult to break. | Compare C–X bond strengths. For similar molecules, C–I is weakest and iodoalkanes hydrolyse fastest. |
| Writing only “NaOH” for both substitution and elimination. | The solvent determines which pathway is favoured. | State NaOH(aq), heat for hydrolysis and NaOH in ethanol, heat for elimination. |
| Saying every primary halogenoalkane must react by SN2 and every tertiary one must react by SN1. | These are mechanistic trends rather than absolute labels independent of conditions. | Explain primary substrates in terms of lower steric hindrance and tertiary substrates in terms of greater carbocation stability. |
| Saying chlorobenzene is unreactive simply because the C–Cl bond is non-polar. | The C–Cl bond is polar. Its low reactivity arises from features of the aromatic structure. | Explain delocalisation of a chlorine lone pair, partial double-bond character and steric hindrance. |
| Adding AgNO₃ immediately after heating with aqueous NaOH. | Excess OH⁻ can interfere with the silver-ion test. | Acidify with dilute HNO₃ first, then add AgNO₃(aq). |
| Assuming HFCs and HCFCs have no environmental disadvantages. | Replacing CFCs does not automatically remove all environmental impact. | Recognise that HFCs can be significant greenhouse gases and HCFCs can still have ozone and climate impacts. |
What you should be able to do
Explain halogenoalkane reactivity
Relate hydrolysis rates to the relative strengths of the C–F, C–Cl, C–Br and C–I bonds.
Recall the three key substitutions
Convert bromoethane into an alcohol, nitrile or primary amine using the correct reagents and conditions.
Explain SN1 and SN2
Use carbocation stability and steric hindrance to explain the preferred substitution pathway.
Predict stereochemical outcomes
Recognise inversion of configuration for SN2 and racemisation for SN1.
Distinguish substitution from elimination
Use aqueous versus ethanolic NaOH conditions to predict whether an alcohol or alkene is formed.
Test and compare halogen derivatives
Use hydrolysis followed by acidification and AgNO₃ testing, and explain why halogenoarenes behave differently.
These ten questions test whether you can apply the chapter to structures, reactions and unfamiliar examples.
- 10 mastery questions
- Structures and reaction schemes where useful
- Instant explanation for every answer option
Frequently asked questions
Questions students often ask about halogenoalkanes
Halogenoalkanes become much easier once you separate three ideas: carbon–halogen bond strength, the substitution mechanism, and the reaction conditions. These questions bring those ideas together.
Why is iodoethane more reactive than fluoroethane even though the C–F bond is more polar?
The important factor in the hydrolysis trend is the strength of the carbon–halogen bond. C–F is very strong, whereas C–I is much weaker. Breaking C–I therefore requires less energy, so iodoethane hydrolyses more readily despite the greater polarity of C–F.
How do I remember the difference between aqueous and ethanolic NaOH?
Think about the product you want. NaOH(aq) supplies OH⁻ under aqueous conditions for substitution, replacing X and giving an alcohol. NaOH in ethanol with heat favours elimination, removing H and X and producing an alkene.
Why does SN2 cause inversion but SN1 causes racemisation?
In SN2, the nucleophile must attack from the side opposite the leaving group, so the arrangement around a chiral carbon is inverted. In SN1, a planar carbocation forms first. The nucleophile can then approach from either side, giving both configurations and hence racemisation.
Why must nitric acid be added before silver nitrate in the halide test?
The halogenoalkane was hydrolysed using aqueous NaOH, so excess OH⁻ may remain. Dilute nitric acid neutralises this OH⁻ before AgNO₃ is added, preventing unwanted silver-containing precipitates from interfering with identification of the silver halide.
Do I need to learn the detailed mechanism by which CFCs destroy ozone?
No. For syllabus 9476, you should recognise the effect of CFCs on the ozone layer and understand that alternatives such as HFCs and HCFCs can also have significant environmental impacts, but the detailed ozone-depletion mechanism is not required.
Halogenoalkane chemistry is really a story about what happens at one polar carbon–halogen bond. Once you connect bond strength to reactivity, steric hindrance to SN2, carbocation stability to SN1, and solvent conditions to substitution versus elimination, the reactions stop looking like separate facts and become variations of the same underlying chemical reasoning.