This chemistry grades 11-12 lesson on halogenoalkanes notes gives the key ideas first, then worked examples and diagrams. After that, practise with the worksheet on halogenoalkanes and check yourself with the quiz.
Quick answer: Alkanes react with halogens in UV light by free radical substitution, a chain reaction of initiation, propagation and termination. Halogenoalkanes react with nucleophiles in nucleophilic substitution, by SN2 for primary and SN1 for tertiary compounds. Weaker carbon-halogen bonds react faster, so iodoalkanes are the most reactive.
Level: Chemistry grades 11-12 (A Level · IB · Class 11-12) · Curriculum: IB R3.4
Halogenoalkanes notes: the key ideas
What are halogenoalkanes and how are they classified?
Halogenoalkanes are alkanes with one or more hydrogen atoms replaced by a halogen. The carbon–halogen bond is polar because the halogen is more electronegative, so the carbon carries a partial positive charge (δ+) that attracts nucleophiles. They are classified by the number of carbon atoms attached to the carbon bearing the halogen: primary (1), secondary (2) and tertiary (3).
A nucleophile is an electron-pair donor. It attacks an electron-deficient (δ+) carbon atom and forms a new covalent bond using its lone pair.
How do alkanes react with halogens?
Alkanes are unreactive, but in UV light they react with chlorine or bromine by free radical substitution. A radical is a species with an unpaired electron, formed by homolytic fission, where each atom of the broken bond keeps one electron.
Further substitution gives CH₂Cl₂, CHCl₃ and CCl₄, so the product is a mixture. Using excess alkane makes mono-substitution more likely.
The Cl–Cl bond enthalpy is 242 kJ mol⁻¹. Calculate the longest wavelength of light that can split one molecule (h = 6.63 × 10⁻³⁴ J s, c = 3.00 × 10⁸ m s⁻¹, NA = 6.02 × 10²³ mol⁻¹).
Energy per bond = 242 000 ÷ 6.02 × 10²³ = 4.02 × 10⁻¹⁹ J. λ = hc ÷ E = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) ÷ 4.02 × 10⁻¹⁹ = 4.95 × 10⁻⁷ m = 495 nm.
How does nucleophilic substitution work?
In nucleophilic substitution the nucleophile replaces the halogen, which leaves as a halide ion. The conditions decide the product.
| Nucleophile | Conditions | Product (from bromoethane) |
|---|---|---|
| OH⁻ | aqueous NaOH, warm | ethanol, CH₃CH₂OH |
| CN⁻ | KCN in ethanol, reflux | propanenitrile, CH₃CH₂CN (chain longer by one carbon) |
| NH₃ | excess ammonia in ethanol, sealed tube | ethylamine, CH₃CH₂NH₂ |
With hot ethanolic potassium hydroxide the halogenoalkane loses HBr instead, forming an alkene: this is elimination, and the hydroxide ion acts as a base.
What is the difference between SN1 and SN2?
Primary halogenoalkanes react by SN2: one step, with the nucleophile attacking as the halide leaves. Tertiary halogenoalkanes react by SN1: the halide leaves first to form a planar carbocation, which is then attacked. Secondary ones can follow either route.
| Feature | SN2 | SN1 |
|---|---|---|
| Steps | one | two (carbocation intermediate) |
| Rate equation | rate = k[RX][Nu⁻] | rate = k[RX] |
| Typical halogenoalkane | primary | tertiary |
| Stereochemistry | inversion of configuration | racemic mixture |
The rate of hydrolysis depends on the C–X bond strength: C–I > C–Br > C–Cl in speed, because weaker bonds break more easily. The C–F bond is so strong that fluoroalkanes hardly react.
Kofi in Accra warms 5.00 g of 1-bromobutane (Mr = 136.9) with aqueous sodium hydroxide and obtains 1.96 g of butan-1-ol (Mr = 74.0). Calculate the percentage yield.
n = 5.00 ÷ 136.9 = 0.03652 mol; theoretical mass = 0.03652 × 74.0 = 2.703 g. Yield = 1.96 ÷ 2.703 × 100 = 72.5 %.
How can you compare the rates of hydrolysis?
Warm the halogenoalkane with ethanolic silver nitrate. Water slowly hydrolyses it, releasing halide ions that precipitate at once as AgCl (white), AgBr (cream) or AgI (yellow). The shorter the time to a precipitate, the faster the hydrolysis. Iodoalkanes give a precipitate first, then bromo, then chloro.
Why do CFCs damage the ozone layer?
In the stratosphere UV breaks the C–Cl bond in CFCs to give Cl• radicals, which catalyse ozone destruction: Cl• + O₃ → ClO• + O₂ and ClO• + O → Cl• + O₂. The chlorine radical is regenerated, so one radical can destroy thousands of ozone molecules.
Do not confuse the roles of hydroxide. With aqueous NaOH it is a nucleophile and gives an alcohol; with ethanolic KOH and heat it is a base and gives an alkene. The solvent changes the product.
Draw curly arrows from a lone pair or bond to where electrons end up. The nucleophile arrow starts at the lone pair and points to the δ+ carbon; the second arrow starts at the C–X bond and points to the halogen.
- Free radical substitution: initiation, propagation, termination; needs UV light
- Nucleophile = electron-pair donor: OH⁻ gives alcohols, CN⁻ nitriles, NH₃ amines
- SN2: one step, rate = k[RX][Nu], inversion; SN1: carbocation, rate = k[RX], racemic
- C–X bond enthalpy: C–F 467, C–Cl 346, C–Br 290, C–I 228 kJ mol⁻¹
- Ethanolic KOH and heat causes elimination to an alkene
Radical and nucleophilic substitution reactions: frequently asked questions
Why do iodoalkanes react faster than chloroalkanes?
The C–I bond is longer and weaker (228 kJ mol⁻¹) than the C–Cl bond (346 kJ mol⁻¹). Less energy is needed to break it, so substitution is faster.
What is a free radical?
It is a species with an unpaired electron, such as Cl•. It forms when a covalent bond splits evenly in homolytic fission, usually in UV light.
How do SN1 and SN2 mechanisms differ?
SN2 happens in one step and its rate depends on both reactants, while SN1 has a carbocation intermediate and its rate depends only on the halogenoalkane. SN2 inverts the configuration and SN1 gives a racemic mixture.
Why is ethanol used as the solvent for the cyanide reaction?
Water would also act as a nucleophile and form an alcohol as a by-product. Ethanol keeps the nucleophile as the cyanide ion only.
Why is excess ammonia used to make amines?
The amine product is also a nucleophile and could react further with the halogenoalkane. A large excess of ammonia makes the primary amine the main product.
