9701 Chemistry · Topics 15 & 29 · AS + A Level
Halogenoalkanes Cheat Sheet — A Level Chemistry 9701
Halogenoalkanes are the crossroads of the organic syllabus — almost every synthesis route passes through one. This sheet covers the Sᴺ1 and Sᴺ2 mechanisms and which substrate takes which route, the competition between substitution and elimination, why the C–I bond hydrolyses fastest despite iodine’s low electronegativity, and the environmental chemistry of CFCs.
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What’s on this cheat sheet
Halogen Compounds
01 · Why they react
The halogen is more electronegative than carbon, so the C–X bond is polar with a δ+ carbon — the site every nucleophile attacks.
Rate of hydrolysis is decided by bond enthalpy, not polarity: C–I is weakest so iodoalkanes react fastest; C–F is strongest and barely reacts at all.
02 · Nucleophilic substitution
| Reagent | Conditions | Product |
|---|---|---|
| NaOH(aq) | warm, aqueous | alcohol |
| KCN in ethanol | reflux | nitrile (chain +1 C) |
| excess NH₃ in ethanol | heat, sealed tube | amine |
| H₂O | slow, warm | alcohol |
The cyanide route is the standard way to lengthen a carbon chain; the nitrile can then be hydrolysed to an acid or reduced to an amine.
03 · SN2 mechanism
The nucleophile attacks the δ+ carbon from the side opposite the halogen while the C–X bond breaks. The transition state has five groups round the carbon.
rate = k[RX][Nu⁻] — second order overall.
The configuration is inverted.
04 · SN1 mechanism
Slow C–X breaks heterolytically → a tertiary carbocation, stabilised by three electron‑releasing alkyl groups.
Fast the nucleophile attacks the carbocation.
rate = k[RX] — first order; the nucleophile is absent from the rate equation.
Attack from either face gives a racemic mixture.
05 · Choosing between them
Primary → SN2 (little steric hindrance, no stable carbocation available). Tertiary → SN1 (crowded carbon, but a very stable carbocation). Secondary goes by both.
The kinetics tell you which: if the nucleophile appears in the rate equation it is SN2.
06 · Elimination
Hot ethanolic KOH gives an alkene: OH⁻ acts as a base, removing H from the carbon next to the C–X, and HX is lost.
The competition is set by the conditions: aqueous NaOH → substitution, ethanolic KOH, hot → elimination. Tertiary halogenoalkanes eliminate most readily, and unsymmetrical ones can give more than one alkene.
07 · Comparing hydrolysis rates
RI → yellow precipitate first
RBr → cream precipitate next
RCl → white precipitate slowest
Ethanol is the common solvent; the silver halide precipitates as soon as the halide ion is released.
08 · Equations to know
CH₃CH₂Br + CN⁻ → CH₃CH₂CN + Br⁻
CH₃CH₂Br + 2NH₃ → CH₃CH₂NH₂ + NH₄Br
CH₃CH₂Br + KOH(ethanol) → CH₂=CH₂ + KBr + H₂O
09 · Uses and hazards
Halogenoalkanes are versatile intermediates — the –X can be swapped for –OH, –CN, –NH₂ and more, which is why they sit at the centre of synthesis routes.
Also used as solvents, refrigerants, anaesthetics and in PTFE. CFCs are inert enough to reach the stratosphere, where UV homolysis releases Cl• radicals that catalyse ozone breakdown.
10 · Ozone depletion
Cl• + O₃ → ClO• + O₂
ClO• + O → Cl• + O₂
overall O₃ + O → 2O₂
The Cl• is regenerated, so it acts as a catalyst and one radical destroys many ozone molecules. Replacements: HFCs, which carry no chlorine.
11 · Worked example — deduce the mechanism
The nucleophile is absent from the rate equation, so it takes part after the slow step.
Slow step = C–Br breaking → tertiary carbocation.
Mechanism is SN1, and the product from an optically active substrate would be racemic.
12 · Making halogenoalkanes
| From | Reagent |
|---|---|
| alkane | Cl₂ or Br₂ with UV |
| alkene | HBr, or Br₂ |
| alcohol | PCl₅, or HCl / ZnCl₂ |
The alcohol route is the cleanest for synthesis; the UV route gives a mixture and is rarely useful.
13 · Worked example — predict the products
(a) substitution → butan‑2‑ol
(b) elimination → but‑1‑ene and but‑2‑ene, since H can be removed from either neighbouring carbon; but‑2‑ene is the major product and shows cis–trans isomerism.
Conditions decide the product
NaOH(aq), warm → alcohol (substitution)
KOH in ethanol, hot → alkene (elimination)
KCN in ethanol, reflux → nitrile
Excess NH₃, sealed tube → primary amine
Marks lost here
— Explaining reactivity by bond polarity; it is bond enthalpy that decides the rate.
— Giving aqueous conditions for elimination or ethanolic for substitution.
— Drawing SN2 in two steps or SN1 in one.
— Forgetting excess ammonia, without which further substitution gives secondary and tertiary amines.
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Halogenoalkanes — Frequently Asked Questions
Why does the C–I bond hydrolyse fastest?
Rate of hydrolysis depends on bond enthalpy, not polarity. C–I is the weakest carbon–halogen bond at about 228 kJ mol⁻¹, so it breaks most easily. C–F is the most polar but by far the strongest, so fluoroalkanes hydrolyse most slowly.
When does elimination happen instead of substitution?
Hot ethanolic potassium hydroxide with the hydroxide acting as a base favours elimination to form an alkene. Warm aqueous potassium hydroxide with the hydroxide acting as a nucleophile favours substitution to form an alcohol. The solvent is the clue examiners plant in the question.
Which halogenoalkanes follow Sᴺ1 and which follow Sᴺ2?
Tertiary halogenoalkanes go by Sᴺ1 through a carbocation intermediate stabilised by three alkyl groups. Primary halogenoalkanes go by Sᴺ2 in one concerted step with inversion of configuration. Secondary halogenoalkanes can do either.
