You are viewing a free preview of this lesson.
Subscribe to unlock all 10 lessons in this course and every other course on LearningBro.
Halogenoalkanes (also called haloalkanes or alkyl halides) are organic compounds in which one or more hydrogen atoms of an alkane have been replaced by halogen atoms (F, Cl, Br, or I). They have the general formula CₙH₂ₙ₊₁X (for monohalogenoalkanes), where X is a halogen.
Halogenoalkanes are far more reactive than alkanes because the C-X bond is polar — the halogen is more electronegative than carbon, creating a δ+ carbon that is susceptible to attack by nucleophiles.
Halogenoalkanes are classified as primary, secondary, or tertiary depending on how many carbon atoms are bonded to the carbon that carries the halogen:
| Classification | Carbon Bearing the Halogen | Example |
|---|---|---|
| Primary (1°) | Bonded to one other carbon (or none, as in CH₃Cl) | CH₃CH₂Br (bromoethane) |
| Secondary (2°) | Bonded to two other carbons | (CH₃)₂CHBr (2-bromopropane) |
| Tertiary (3°) | Bonded to three other carbons | (CH₃)₃CBr (2-bromo-2-methylpropane) |
This classification is critically important because it determines which reaction mechanism occurs (SN1 vs SN2) and whether elimination competes with substitution.
The C-X bond is polar, giving halogenoalkanes a permanent dipole. However, they cannot form hydrogen bonds with each other (no O-H or N-H), so their intermolecular forces are permanent dipole-dipole interactions and van der Waals forces.
The polar C-X bond makes the carbon electrophilic (δ+), and the halogen can leave as a halide ion (X⁻). This makes halogenoalkanes susceptible to two main reaction types:
graph TD
A["Halogenoalkane (R-X)"] -->|"NaOH(aq), reflux"| B["Alcohol (R-OH)"]
A -->|"KCN in ethanol/water, reflux"| C["Nitrile (R-CN)"]
A -->|"Excess NH₃ in ethanol, sealed tube, heat"| D["Amine (R-NH₂)"]
A -->|"NaOH in ethanol, reflux"| E["Alkene"]
B -->|"Conc. H₂SO₄, 170 °C"| E
E -->|"HX, room temp"| A
A nucleophile (electron-pair donor) attacks the δ+ carbon, replacing the halogen:
R-X + Nu⁻ → R-Nu + X⁻
Common nucleophilic substitution reactions:
| Nucleophile | Reagent and Conditions | Product | Product Type |
|---|---|---|---|
| OH⁻ | NaOH(aq), heat under reflux | R-OH | Alcohol |
| CN⁻ | KCN in ethanol/water, heat under reflux | R-CN | Nitrile |
| NH₃ | Excess concentrated NH₃ in ethanol, sealed tube, heat | R-NH₂ | Amine |
The reaction with OH⁻ is called hydrolysis. The reaction with CN⁻ is important because it increases the carbon chain length by one — a key synthetic tool.
Why is CN⁻ nucleophilic? The carbon atom in CN⁻ has a lone pair of electrons available for donation. It attacks the δ+ carbon of the halogenoalkane, forming a new C-C bond. This is one of the few simple reactions that creates a new carbon-carbon bond.
A strong base removes a hydrogen atom from a carbon adjacent to the C-X carbon. The halogen leaves as X⁻, and a C=C double bond forms:
R-CH₂-CHX-R' + base → R-CH=CH-R' + HX
The distinction between substitution and elimination depends on the conditions:
Exam tip: This aqueous vs ethanolic distinction is frequently tested. The same reagent (NaOH) gives different products depending on the solvent. Aqueous = substitution. Ethanolic = elimination.
The type of halogenoalkane determines whether nucleophilic substitution proceeds by an SN2 or SN1 mechanism (detailed in the next lesson):
Other factors include:
The rate of hydrolysis of halogenoalkanes depends on the strength of the C-X bond:
| Bond | Bond Enthalpy (kJ mol⁻¹) | Rate of Hydrolysis |
|---|---|---|
| C-F | 484 | Extremely slow (essentially unreactive) |
| C-Cl | 338 | Slow |
| C-Br | 276 | Moderate |
| C-I | 238 | Fast |
The C-I bond is the weakest and breaks most easily, making iodoalkanes the most reactive. The C-F bond is the strongest, making fluoroalkanes essentially inert to hydrolysis.
Important distinction: Reactivity is determined by bond strength (enthalpy), not bond polarity. C-F is the most polar bond but is the least reactive because it is the strongest. This is a common exam question designed to test whether you understand that bond enthalpy determines reactivity for halogenoalkanes.
The relative hydrolysis rates of 1-chlorobutane, 1-bromobutane, and 1-iodobutane can be compared experimentally:
The halide ions produced by hydrolysis react with Ag⁺ to form insoluble silver halide precipitates:
The precipitate appears fastest for the iodoalkane (weakest C-I bond) and slowest for the chloroalkane (strongest C-Cl bond among the three tested).
Chlorofluorocarbons (CFCs) such as CCl₃F and CCl₂F₂ were widely used as refrigerants, aerosol propellants, and foam-blowing agents because of their low toxicity and chemical stability. However, their very stability is the problem — they persist in the atmosphere and eventually reach the stratosphere.
In the stratosphere, UV radiation breaks C-Cl bonds homolytically, releasing chlorine radicals:
CCl₃F → CCl₂F• + Cl•
These chlorine radicals catalyse the destruction of ozone (O₃):
Propagation step 1: Cl• + O₃ → ClO• + O₂ Propagation step 2: ClO• + O → Cl• + O₂
The chlorine radical is regenerated — it is a catalyst. A single chlorine radical can destroy thousands of ozone molecules before being removed by a termination reaction.
The Montreal Protocol (1987) led to the global phase-out of CFCs. Replacement compounds include HFCs (hydrofluorocarbons), which do not contain chlorine and do not deplete ozone (though they are potent greenhouse gases, so further alternatives are being sought).
Note the mechanism similarity: CFC ozone destruction follows the same radical chain reaction pattern (initiation, propagation, termination) as free radical substitution of alkanes. The chlorine radical acts as a catalyst because it is consumed in step 1 and regenerated in step 2.
| Starting Material | Reagent & Conditions | Product | Reaction Type |
|---|---|---|---|
| R-X | NaOH(aq), reflux | R-OH | Nucleophilic substitution |
| R-X | KCN, ethanol/water, reflux | R-CN | Nucleophilic substitution |
| R-X | Excess NH₃, ethanol, sealed tube, heat | R-NH₂ | Nucleophilic substitution |
| R-X | NaOH in ethanol, reflux | Alkene + HX | Elimination |
Edexcel 9CH0 specification Topic 7 — Organic Chemistry I, sub-topic 7.1 covers halogenoalkanes: classification as primary, secondary or tertiary; the polarity of the C-X bond and the relative reactivity F < Cl < Br < I in nucleophilic substitution (where bond strength dominates over polarity); reactions with hydroxide ion (substitution → alcohol), cyanide ion (substitution → nitrile, extending the carbon chain by one), ammonia (substitution → primary amine, then secondary etc.), and elimination with KOH in ethanol (E2 → alkene + HX) — refer to the official specification document for exact wording. Sub-topic 7.1 is examined directly on Paper 2 (mechanism, products, observation in CP5 hydrolysis) and Paper 3 (synoptic with kinetics, NMR, organic synthesis routes). The data booklet provides bond enthalpies for C-F (467), C-Cl (340), C-Br (290), C-I (228) kJ mol−1 — these set the reactivity ordering.
Question (8 marks):
(a) Predict and name the products of the reaction of 1-bromopropane with (i) NaOH dissolved in water, heated under reflux, and (ii) NaOH dissolved in ethanol, heated under reflux. (4)
(b) Write the equations for both reactions, classify each as substitution or elimination, and explain why the solvent determines the product. (4)
Solution with mark scheme:
(a)(i) Aqueous NaOH — substitution: OH⁻ attacks the C-Br carbon, replacing Br⁻.
M1 — substitution mechanism identified.
A1 — product is propan-1-ol (CH3−CH2−CH2−OH).
(a)(ii) Ethanolic NaOH — elimination: OH⁻ acts as a base, abstracting H from the carbon adjacent to C-Br (β-hydrogen). The C=C forms; HBr is lost.
M1 — elimination mechanism identified.
A1 — product is propene (CH2=CH−CH3) + H2O + NaBr.
(b) Step 1 — equations.
(i) CH3−CH2−CH2−Br+OH−→CH3−CH2−CH2−OH+Br− (substitution).
(ii) CH3−CH2−CH2−Br+OH−→CH2=CH−CH3+H2O+Br− (elimination).
M1 — substitution equation correct.
M1 — elimination equation correct (with H2O and Br⁻ as products).
Step 2 — explanation of solvent dependence.
M1 — in water, OH⁻ is heavily solvated and acts as a nucleophile (attacks the electrophilic C); in ethanol, OH⁻ is less solvated and acts as a stronger base (abstracts the β-H instead).
A1 — additionally, the more polar solvent (water) favours the substitution mechanism by stabilising the polar transition state of nucleophilic attack; the less polar solvent (ethanol) destabilises charged species, favouring the concerted elimination (E2) pathway.
Total: 8 marks (M5 A3).
Question (6 marks): A student carries out CP5 (rate of hydrolysis of haloalkanes) by adding equal volumes of 1-bromobutane, 1-chlorobutane and 1-iodobutane to ethanol, then adding silver nitrate solution and timing how long until the precipitate forms.
(a) Write the equation for the reaction of 1-bromobutane with water (or hydroxide), and explain why ethanol is needed as a solvent. (2)
(b) Predict the order of precipitate formation (fastest to slowest) for the three substrates and justify the order using bond enthalpy data. (3)
(c) State the colour of the precipitate for each halide. (1)
Mark scheme decomposition by AO:
(a)
(b)
(c)
Total: 6 marks split AO1 = 3, AO2 = 3.
Connects to:
Topic 7.2 — Alcohols (next lesson): halogenoalkanes are a key intermediate in the synthesis of alcohols (RX+OH−→ROH+X−). The reverse reaction (alcohol → halogenoalkane) uses HX or PX3 reagents.
Topic 9 — Kinetics: the rate of hydrolysis depends on the C-X bond enthalpy (bond breaking is the rate-limiting step). The order of reactivity F < Cl < Br < I is set by bond strength, not polarity (which would predict the opposite, since C-F is most polar).
Subscribe to continue reading
Get full access to this lesson and all 10 lessons in this course.