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9701 Chemistry · Topics 14 & 28 · AS + A Level

Alkanes, Alkenes and Arenes Cheat Sheet — A Level Chemistry 9701

Hydrocarbons carry two of the three mechanisms 9701 asks you to reproduce in full. These sheets cover free radical substitution step by step, electrophilic addition to alkenes with Markovnikov’s rule and carbocation stability, then the A2 material on benzene’s delocalisation, why it resists addition, and the electrophilic substitution reactions — nitration, halogenation and Friedel–Crafts — with the electrophile-generating step for each.

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What’s on this cheat sheet

Sheet 1 of 2 — 9701 Chemistry · Topic 14 · AS Level
Alkanes and Alkenes

01 · Alkanes — structure

CnH2n+2, all σ bonds, sp³ carbon, tetrahedral at 109.5°. Free rotation about every C–C bond.

Non‑polar, so only London forces: boiling point rises with chain length and falls with branching, and they are insoluble in water.

02 · Why alkanes are unreactive

C–C and C–H bonds are strong and almost non‑polar, so there is no δ+ carbon for a nucleophile to attack and no electron‑rich site for an electrophile. Only radicals and combustion get past that.

03 · Combustion

Complete: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O. Incomplete: limited oxygen gives CO (toxic, binds haemoglobin) and carbon soot.

Other pollutants: NOx from the high engine temperature, SO₂ from sulfur impurities, and unburnt hydrocarbons — all removed or reduced by a catalytic converter.

04 · Free‑radical substitution

CH₄ + Cl₂ → CH₃Cl + HCl, UV light

Initiation Cl₂ → 2Cl• (homolysis by UV)
Propagation Cl• + CH₄ → •CH₃ + HCl
•CH₃ + Cl₂ → CH₃Cl + Cl•
Termination Cl• + •CH₃ → CH₃Cl
2•CH₃ → C₂H₆

Poor for synthesis: further substitution gives CH₂Cl₂, CHCl₃ and CCl₄, and termination gives a mixture.

05 · Cracking

Long alkanes are broken into shorter, more useful ones plus an alkene.

Catalytic — zeolite, ~450 °C, moderate pressure; gives branched alkanes and arenes for fuels. Thermal — higher temperature and pressure; gives more alkenes for polymers.

06 · Alkenes — structure

CnH2n. The C=C is one σ plus one π bond; each carbon is sp², trigonal planar at 120°.

The π electrons sit above and below the plane, exposed and easily attacked — so alkenes react with electrophiles. The π bond also blocks rotation, giving cis–trans isomerism.

07 · Addition reactions of alkenes

Reagent Conditions Product
H₂ Ni, 150 °C alkane
Br₂ room temp dibromoalkane
HBr room temp bromoalkane
steam H₃PO₄, 300 °C, 60 atm alcohol
cold dilute KMnO₄ diol

Test for a C=C: bromine water is decolourised from orange to colourless at room temperature.

08 · Electrophilic addition mechanism

1 The π electrons attack the δ+ end of HBr (or induce a dipole in Br₂).
2 The H–Br bond breaks heterolytically, giving Br⁻ and a carbocation.
3 Br⁻ attacks the carbocation to give the product.

Curly arrow 1 from the C=C to the H; arrow 2 from the H–Br bond to Br; arrow 3 from the lone pair on Br⁻ to the positive carbon.

09 · Markovnikov’s rule

The hydrogen adds to the carbon that already carries more hydrogens, because that route goes through the more stable carbocation.

Stability order tertiary > secondary > primary: alkyl groups are electron‑releasing and spread the positive charge. So propene + HBr gives mainly 2‑bromopropane.

10 · Oxidation of alkenes

Cold dilute acidified KMnO₄ → diol; purple decolourises.

Hot concentrated acidified KMnO₄ → the C=C is cleaved. A terminal =CH₂ gives CO₂; =CHR gives a carboxylic acid; =CR₂ gives a ketone — which lets you deduce where the double bond was.

11 · Worked example — locate the C=C

An alkene C₅H₁₀ gives propanone and ethanoic acid on vigorous oxidation. Identify it.

Propanone ← (CH₃)₂C=
Ethanoic acid ← =CHCH₃
Join the fragments: (CH₃)₂C=CHCH₃, 2‑methylbut‑2‑ene.

12 · Addition polymerisation

n CH₂=CH₂ → –(CH₂–CH₂)n–. Draw the repeat unit in brackets with bonds through them and n outside.

Poly(alkenes) are inert and non‑biodegradable, so disposal means landfill, incineration with energy recovery, or recycling. PVC releases HCl on burning.

13 · Worked example — Markovnikov

Predict the major product of but‑1‑ene + HBr.

H can add to C1 or C2.
Adding to C1 gives a secondary carbocation on C2.
Adding to C2 gives a primary carbocation on C1.

The secondary cation is the more stable, so the major product is 2‑bromobutane; 1‑bromobutane forms in small amounts.

14 · Alkanes against alkenes

Alkane Alkene
bonding all σ σ + π
attacked by radicals electrophiles
typical reaction substitution addition
Br₂(aq) no change decolourised
KMnO₄ no change decolourised

Marks lost here

— Leaving out a termination step, or writing initiation without UV.

— Forgetting the induced dipole when Br₂ attacks a C=C.

— Predicting the minor product by ignoring carbocation stability.

— Drawing a polymer repeat unit still containing a double bond.

Sheet 2 of 2 — 9701 Chemistry · Topic 28 · A Level
Arenes — Benzene Chemistry

15 · Structure of benzene

C₆H₆, planar, regular hexagon, all bond angles 120°, every carbon sp². Each carbon’s remaining p orbital overlaps sideways into a delocalised π system above and below the ring.

Evidence: all six C–C bonds are the same length, between a single and a double bond; and the enthalpy of hydrogenation is about 150 kJ mol⁻¹ less exothermic than three times that of cyclohexene, so benzene is more stable than the Kekulé structure predicts.

16 · Why substitution, not addition

Addition would destroy the delocalised system and its extra stability. Substitution keeps the ring intact, so benzene undergoes electrophilic substitution — and it needs stronger conditions than an alkene because the π electrons are spread out and less available.

17 · The four substitutions

Reaction Reagents Electrophile
nitration conc. HNO₃ / conc. H₂SO₄, 55 °C NO₂⁺
halogenation Cl₂ or Br₂ with AlCl₃ / FeBr₃ Cl⁺ or Br⁺
Friedel–Crafts alkylation RCl / AlCl₃ R⁺
Friedel–Crafts acylation RCOCl / AlCl₃ RCO⁺

AlCl₃ is a halogen carrier — it polarises the reagent to generate the electrophile and is regenerated at the end.

18 · The mechanism in three steps

1 · Generate the electrophile
HNO₃ + 2H₂SO₄ → NO₂⁺ + H₃O⁺ + 2HSO₄⁻

2 · Attack — two π electrons from the ring form a bond to the electrophile, giving an unstable intermediate with a broken delocalised system and a positive charge.

3 · Restore aromaticity — H⁺ is lost from that carbon and the ring’s delocalisation is re‑formed.

19 · Side‑chain reactions

UV light favours free‑radical substitution in the side chain: methylbenzene + Cl₂ → chloromethylbenzene.

AlCl₃ and no light favours substitution in the ring instead. The condition decides the site.

Hot alkaline KMnO₄ oxidises any alkyl side chain all the way to –COOH: methylbenzene → benzoic acid.

20 · Directing effects

Group Effect Directs to
–OH, –NH₂, –CH₃ activating 2‑ and 4‑
–NO₂, –COOH, –CHO deactivating 3‑

Electron‑releasing groups push charge into the ring, making it more reactive than benzene; electron‑withdrawing groups do the reverse. Phenol reacts with bromine water alone, without a catalyst.

21 · Worked example — plan a synthesis

Make 3‑nitrobenzoic acid from methylbenzene.

Nitrate first and you get the 2‑ and 4‑ isomers, since –CH₃ directs there.

So oxidise first:
1 · hot alkaline KMnO₄, then H⁺ → benzoic acid
2 · conc. HNO₃ / conc. H₂SO₄, 55 °C → the –COOH group directs to the 3‑ position.

22 · Comparing benzene with alkenes

Benzene Alkene
π electrons delocalised localised
Br₂(aq) no reaction decolourised
typical reaction substitution addition

This test distinguishes them in seconds and is the standard exam question.

23 · Equations to know

C₆H₆ + HNO₃ → C₆H₅NO₂ + H₂O
C₆H₆ + Br₂ → C₆H₅Br + HBr
C₆H₆ + CH₃Cl → C₆H₅CH₃ + HCl
C₆H₆ + CH₃COCl → C₆H₅COCH₃ + HCl

C₆H₅CH₃ + 3[O] → C₆H₅COOH + H₂O
C₆H₅NO₂ + 6[H] → C₆H₅NH₂ + 2H₂O

24 · Benzene, methylbenzene, phenol

Reactivity to electrophiles rises in the order benzene < methylbenzene < phenol.

–CH₃ releases electrons inductively; –OH donates a whole lone pair into the delocalised system. The more electron‑rich the ring, the more readily it attracts an electrophile — which is why phenol needs only bromine water while benzene needs a halogen carrier.

25 · Worked example — identify the product

Methylbenzene is refluxed with Cl₂ in the presence of AlCl₃, then the product is treated with hot alkaline KMnO₄.

AlCl₃, no UV → ring substitution, and –CH₃ directs to 2‑ and 4‑ → 4‑chloromethylbenzene
Hot alkaline KMnO₄ → the side chain is oxidised
4‑chlorobenzoic acid

26 · Why the conditions matter

Nitration above about 60 °C gives dinitro products; below 50 °C the reaction is impractically slow. The quoted 55 °C is the working compromise.

Friedel–Crafts must be run in dry conditions, since AlCl₃ is hydrolysed by water and the catalyst is destroyed.

Marks lost here

— Drawing benzene with three localised double bonds when asked about delocalisation.

— Omitting the step that regenerates the aromatic ring by losing H⁺.

— Saying benzene decolourises bromine water.

— Nitrating before oxidising when the directing effects demand the reverse.

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Written and reviewed by Fahad H. AhmadChemistry tutor at Mega Lecture · 10M+ lecture views · Book a free trial class

Alkanes, Alkenes and Arenes — Frequently Asked Questions

Why does benzene undergo substitution rather than addition?

Addition would destroy the delocalised π system and lose the stabilisation of roughly 150 kJ mol⁻¹ shown by the enthalpy of hydrogenation. Substitution restores the ring, so it is energetically favoured.

What is Markovnikov’s rule and why does it work?

When an unsymmetrical molecule adds across an unsymmetrical alkene, the hydrogen adds to the carbon that already has more hydrogens. This is because it forms the more stable carbocation — tertiary is more stable than secondary, which is more stable than primary, thanks to the electron-donating alkyl groups.

What are the three steps of free radical substitution?

Initiation, in which UV light homolytically splits the halogen into radicals; propagation, a two-step chain that consumes and regenerates radicals; and termination, in which two radicals combine. Examiners expect the correct step to be named and the dot to be shown on every radical.

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