AS Chemistry: Alkanes and Alkenes — Revision Notes

Condensed for the final weeks. For the full explanation, use the Hydrocarbons: Alkanes and Alkenes study guide.

Free-radical substitution (alkanes)

Requires UV light. Three stages, all of which must be shown:

INITIATION    Cl2  --UV-->  2Cl.               homolytic fission
PROPAGATION   Cl. + CH4  ->  .CH3 + HCl
              .CH3 + Cl2 ->  CH3Cl + Cl.       radical regenerated
TERMINATION   Cl. + Cl.  ->  Cl2
              .CH3 + Cl. ->  CH3Cl
              .CH3 + .CH3 -> C2H6              any two radicals

• Homolytic fission: one electron to each atom, shown with single-headed (fishhook) arrows, in contrast to the double-headed curly arrows used for heterolytic fission elsewhere in organic chemistry.
• Propagation steps consume a radical and produce one — this is why the chain continues.
• The reaction is poorly controlled: further substitution gives CH₂Cl₂, CHCl₃, CCl₄.

Electrophilic addition (alkenes)

The C=C is a region of high electron density, which attracts electrophiles.

Step 1  The pi bond breaks; electrons attack the electrophile
        An INDUCED DIPOLE forms in a non-polar molecule such as Br2
Step 2  A CARBOCATION intermediate forms
Step 3  The nucleophile (e.g. Br-) attacks the carbocation

Heterolytic fission — both electrons to one atom, shown with double-headed curly arrows.

The reactions of alkenes

Reagent	Conditions	Product
H₂	Ni catalyst, 150 °C	Alkane
Br₂	Room temperature	Dibromoalkane — decolourises bromine water
HBr	Room temperature	Halogenoalkane
H₂O(g)	H₃PO₄ catalyst, 300 °C, 60 atm	Alcohol
Cold dilute acidified KMnO₄	—	Diol; purple → colourless
O₂	—	Combustion

The test for unsaturation: bromine water, orange-brown → colourless.

Locating a double bond by oxidative cleavage

Hot, concentrated acidified KMnO₄ (unlike the cold, dilute acidified version used for the diol test) breaks the C=C bond completely. The fragments produced tell you exactly where the double bond was:

terminal   =CH2   ->  oxidised all the way to CO2 + H2O
internal   =CH-   ->  oxidised to a CARBOXYLIC ACID
disubstituted =CR2 ->  oxidised to a KETONE

Worked example. An alkene is treated with hot concentrated acidified KMnO₄ and gives propanone, (CH₃)₂C=O, plus CO₂ and water. Deduce the structure of the alkene.

Propanone comes from a fully-substituted =CR2 carbon: (CH3)2C=
CO2 + H2O comes from a terminal =CH2 carbon

Combining the two fragments:  (CH3)2C=CH2   (2-methylpropene)

Working backwards from the oxidation products to the alkene’s structure this way is a standard structure-determination question, and is worth distinguishing clearly from the cold dilute acidified KMnO₄ diol test — the two use the same reagent but very different conditions, and give completely different types of information.

Markovnikov’s rule

When HBr adds to an unsymmetrical alkene, the hydrogen adds to the carbon already bearing more hydrogens, because this route forms the more stable carbocation.

Carbocation stability:   tertiary > secondary > primary

Alkyl groups are electron-donating, so they spread the positive charge and stabilise the cation. Explaining the major product requires naming the carbocation, not just quoting the rule.

Exam traps

• Omitting UV light from the initiation step.
• Using double-headed arrows in a radical mechanism — radicals need single-headed arrows.
• Forgetting the induced dipole when a non-polar molecule such as Br₂ approaches the C=C.
• Saying bromine water goes “clear” — it goes colourless.
• Quoting Markovnikov without explaining carbocation stability.
• Writing termination as a single step — any two radicals combining counts.
• Using cold dilute acidified KMnO₄ conditions when a question specifically requires C=C bond cleavage (which needs hot, concentrated acidified KMnO₄) — the two conditions test entirely different things.
• Forgetting that a fully-substituted =CR₂ carbon gives a ketone on oxidative cleavage, not a carboxylic acid — only a =CH– carbon (with one hydrogen still attached) can be oxidised as far as an acid.

Self-test

1. Name the three stages of free-radical substitution and the type of bond fission.
2. Why does a propagation step keep the chain going?
3. What is observed when an alkene is added to bromine water?
4. Give the conditions for converting an alkene to an alcohol.
5. Explain, using carbocation stability, the major product when HBr adds to propene.
6. An alkene is cleaved by hot concentrated acidified KMnO₄ to give ethanoic acid and propanone only. Deduce the structure of the original alkene.
7. What condition distinguishes the diol test from the oxidative-cleavage reaction, given that both use acidified KMnO₄?

Answers: 1. Initiation, propagation, termination; homolytic fission. 2. Each step uses one radical and generates another, so the reactive species is continuously regenerated. 3. The orange-brown colour disappears — the solution becomes colourless. 4. Steam, phosphoric acid catalyst, about 300 °C and 60 atm. 5. H adds to the carbon with more hydrogens, giving a secondary carbocation rather than a primary one; the two electron-donating alkyl groups spread the positive charge, so 2-bromopropane is the major product. 6. Ethanoic acid comes from a =CH– carbon bearing a methyl group (CH₃CH=), and propanone comes from a =CR₂ carbon bearing two methyl groups ((CH₃)₂C=), so the alkene is CH₃CH=C(CH₃)₂ (2-methylbut-2-ene). 7. Temperature and concentration: cold, dilute acidified KMnO₄ gives the diol (the C=C bond stays intact); hot, concentrated acidified KMnO₄ cleaves the C=C bond completely.

For the full mechanisms and further worked examples, see the Hydrocarbons: Alkanes and Alkenes study guide; for exam-style practice with full worked answers, see the Alkanes and Alkenes practice questions.