Reactions of Hydrocarbons

In short: Alkenes react by addition: hydrogen, halogens, hydrogen halides or water add across the C=C double bond. Alkenes decolourise bromine water quickly and alkanes do not, which tells them apart, and alkanes react with halogens only by substitution in UV light.

Two kinds of reaction

Reaction	Hydrocarbon	What happens
Addition	Alkenes and alkynes (unsaturated)	Atoms add across the double or triple bond. One product.
Substitution	Alkanes (saturated)	A hydrogen atom is swapped for another atom. Two products.

The C=C double bond has one strong bond and one weaker bond. The weaker bond breaks easily, so alkenes are much more reactive than alkanes.

Addition reactions of alkenes

The double bond opens and one new atom joins each carbon.

Adding	Name	Conditions	Product from ethene
$H_{2}$	Hydrogenation	Nickel or Pd/Pt catalyst, heat	Ethane
$B r_{2}$ or $C l_{2}$	Halogenation	Room temperature, no UV needed	1,2-dibromoethane
$HCl$ or $HBr$	Hydrohalogenation	Room temperature	Chloroethane
$H_{2} O$	Hydration	Dilute $H_{2} S O_{4}$ or $H_{3} P O_{4}$ catalyst, heat	Ethanol

$C H_{2} = C H_{2} + B r_{2} \to C H_{2} BrC H_{2} Br$

$C H_{2} = C H_{2} + H_{2} O \to C H_{3} C H_{2} OH$

The bromine water test

Bromine water is orange-brown.

Shake it with an alkene: it turns colourless immediately, because the bromine adds across the double bond.
Shake it with an alkane in the dark: no change. (In UV light it fades only slowly, by substitution.)

This is the standard test for unsaturation. Potassium permanganate solution (purple) is also decolourised by alkenes.

Markovnikov's rule

When HX or water adds to an unsymmetrical alkene like propene, two products are possible:

$C H_{2} = CHC H_{3} + HBr \to C H_{3} CHBrC H_{3} (major) or C H_{2} BrC H_{2} C H_{3} (minor)$

Markovnikov's rule: the H goes to the carbon of the double bond that already has more hydrogens. "The rich get richer." So the Br goes to the middle carbon, and 2-bromopropane is the main product.

For a symmetrical alkene, like but-2-ene, both carbons are the same, so only one product forms (2-bromobutane).

Adding $H_{2}$ or $B r_{2}$ never gives two products, because the same atom goes to each carbon.

Substitution reactions of alkanes

Alkanes do not react with halogens in the dark. In ultraviolet (UV) light, a hydrogen atom is replaced by a halogen atom:

$C H_{4} + C l_{2} UV light C H_{3} Cl + HCl$

Notice there are two products, one of them a hydrogen halide. With more chlorine, more hydrogens are replaced: $C H_{2} C l_{2}$ , $CHC l_{3}$ and $CC l_{4}$ form, giving a mixture.

Substitution is slow and gives mixtures, so making a haloalkane from an alkene (addition, one product) is usually better.

Comparing addition and substitution

	Addition (alkene + $B r_{2}$ )	Substitution (alkane + $B r_{2}$ )
Speed	Fast	Slow
Conditions	Room temperature, dark is fine	Needs UV light
Products	One	Two (or more), including HBr

Reaction pathways

Exam questions often ask you to link reactions in a flow chart:

$ethane C l_{2}, UV chloroethane$

$ethene H_{2} O, acid ethanol oxidise ethanoic acid$

Identify the reagent, conditions and type of reaction for each arrow.

Hydrocarbon reactions

Addition (alkenes)

H2: Ni catalyst
Br2 or Cl2
HX
H2O: acid catalyst
One product

Bromine water test

Alkene: colourless fast
Alkane: no change in dark

Markovnikov

H to carbon with more H
Unsymmetrical alkenes only

Substitution (alkanes)

Halogen + UV light
Two products
Mixture forms

Year 12Harder pathways and isomer counts

How many products? But-1-ene + HCl gives two products (1-chlorobutane minor, 2-chlorobutane major). But-2-ene + HCl gives only 2-chlorobutane.

Alkynes add twice. Ethyne + 2 $H_{2}$ (Ni) gives ethane. With one mole of $B r_{2}$ , ethyne gives 1,2-dibromoethene; with two moles, 1,1,2,2-tetrabromoethane.

Monosubstitution products. The number of different monochloro products equals the number of different hydrogen environments. Propane has 2 (1-chloropropane and 2-chloropropane). 2-methylpropane has 2. Butane has 2.

Working backwards. A compound adds HCl to give 2-chloro-2-methylpentane as the main product. The Cl is on C2, which had fewer H, so the starting alkene had its double bond at C2: 2-methylpent-2-ene or 2-methylpent-1-ene.

Markovnikov with water. 2-methylbut-2-ene, $(C H_{3})_{2} C = CHC H_{3}$ , adds water so the H goes to the CH carbon and the OH to the carbon with no H: the main product is 2-methylbutan-2-ol.

Planning two steps. To make a target, first use addition to put atoms on an alkene (one product), then substitution in UV light to swap one more H. For $C F_{3} CHClBr$ : $C F_{2} = CHCl$ adds HF to give $C F_{3} C H_{2} Cl$ , then $B r_{2}$ in UV light replaces one H.

Quick summary

Alkenes undergo addition: $H_{2}$ (catalyst), halogens, hydrogen halides, water (acid catalyst).
Bromine water: decolourised fast by alkenes, not by alkanes in the dark.
Markovnikov: H adds to the carbon that already has more H.
Alkanes undergo substitution with halogens in UV light, giving two products.

Quiz

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Alkenes mainly undergo:
1. A Addition reactions
2. B Substitution reactions
3. C Condensation reactions
4. D No reactions
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A: Addition reactions
The double bond opens to add atoms.
Alkanes react with chlorine by:
1. A Addition
2. B Substitution
3. C Hydration
4. D Polymerisation
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B: Substitution
A hydrogen is swapped for a chlorine.
What do alkanes need to react with halogens?
1. A Dark conditions
2. B UV light
3. C Water
4. D An acid catalyst
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B: UV light
UV light provides the energy to start the reaction.