The big idea: In a halogenoalkane (R–X), the carbon–halogen bond is polar: the halogen is more electronegative than carbon, so the carbon carries a partial positive charge (δ+) and the halogen a partial negative charge (δ−).
That δ+ carbon attracts electron-rich species called nucleophiles. A nucleophile uses a lone pair to bond to the δ+ carbon and replaces the halogen. The halogen leaves as a halide ion — this is nucleophilic substitution.
Three terms to nail: - Nucleophile (Nu) — an electron-pair donor that attacks the δ+ carbon (e.g. OH⁻, CN⁻, NH3; it has a lone pair). - Leaving group — the halide (Cl⁻, Br⁻, I⁻) that departs, taking both bonding electrons. - Substrate — the halogenoalkane being attacked.
Overall: R–X + Nu⁻ → R–Nu + X⁻.
There are two mechanisms by which this happens — SN2 and SN1. They give the same overall product but proceed in different ways, and which one operates depends on the halogenoalkane. The next two sections take each in turn.
SN2 stands for Substitution, Nucleophilic, bimolecular (the 2 means two species are involved in the rate-determining step). It happens in a single, concerted step — bond-making and bond-breaking occur at the same time.
Nucleophilic substitution: the nucleophile's lone pair attacks the δ+ carbon while the C–halogen bond breaks heterolytically, the halide leaving with both bonding electrons.
Interactive diagram
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What happens in the one step: - The nucleophile attacks the δ+ carbon from the side opposite to the leaving group (back-side attack). - As the new Nu–C bond starts to form, the C–halogen bond starts to break. - At the midpoint there is a transition state with partial bonds to both the nucleophile and the leaving group. - Because the attack is from the back, the other three groups are pushed through — the carbon's configuration is inverted (like an umbrella turning inside-out).
Because both the halogenoalkane and the nucleophile are present in the single (rate-determining) step, both concentrations affect the rate:
- rate of reaction (mol dm⁻³ s⁻¹)
- rate constant
- concentration of the halogenoalkane (mol dm⁻³)
- concentration of the nucleophile (mol dm⁻³)
Why primary halogenoalkanes prefer SN2: A primary carbon (one carbon attached) is relatively uncrowded, so the nucleophile can reach it easily for back-side attack. Bulky groups on a tertiary carbon would block this approach — so primary halogenoalkanes react by SN2.
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SN1 stands for Substitution, Nucleophilic, unimolecular (the 1 means one species is involved in the rate-determining step). It happens in two steps via a carbocation intermediate.
| Step | What happens | Speed |
|---|---|---|
| Step 1 | The C–halogen bond breaks heterolytically — the halide leaves with both bonding electrons, forming a positively charged carbocation. | slow (rate-determining) |
| Step 2 | The nucleophile's lone pair attacks the carbocation, forming the new C–Nu bond. | fast |
Only the slow step sets the rate: The slow first step (heterolysis) is the rate-determining step, and it involves only the halogenoalkane — the nucleophile joins in the later fast step. So the rate depends on only one concentration:
- rate of reaction (mol dm⁻³ s⁻¹)
- rate constant
- concentration of the halogenoalkane (mol dm⁻³)
Why tertiary halogenoalkanes prefer SN1: SN1 needs the carbocation to form first. A tertiary carbocation is the most stable (the three attached alkyl groups push electron density onto the positive carbon — a positive inductive effect — spreading out the charge). Because the tertiary carbocation forms readily, tertiary halogenoalkanes react by SN1.
A primary carbocation is very unstable, so primary substrates avoid SN1 and go by SN2 instead.
How this is tested: R3.4 (HL) appears as a Paper 1A MCQ ('which substrate reacts fastest / by which mechanism?') and as a short Paper 2 deduce / explain question.
The two classic Paper 2 asks are: state the mechanism and rate equation for a given halogenoalkane (justified by it being 1°/2°/3°), and explain the order of reactivity of the halogenoalkanes down a group (C–I fastest).
Decide by the substrate. Count the carbons attached to the carbon bonded to the halogen:
SN2
- One concerted step (no intermediate).
- Nucleophile attacks the carbon from the opposite side to the leaving group.
- Goes through a transition state with partial bonds to both Nu and the leaving group.
- rate = k[halogenoalkane][Nu⁻] — second order.
- Favoured by primary (1°) halogenoalkanes (little steric hindrance).
SN1
- Two steps via a carbocation intermediate.
- Step 1 (slow): the C–halogen bond breaks heterolytically.
- Step 2 (fast): the nucleophile attacks the carbocation.
- rate = k[halogenoalkane] — first order (Nu⁻ absent).
- Favoured by tertiary (3°) halogenoalkanes (stable carbocation).
The carbon–halogen reactivity trend: For the same carbon skeleton, the rate of substitution depends on the carbon–halogen bond strength — and the weaker the bond, the better the leaving group, so the faster the reaction:
C–I (fastest) > C–Br > C–Cl > C–F (slowest)
The C–I bond is the longest and weakest (iodine's bonding electrons are far from its nucleus), so it breaks most easily and iodide is the best leaving group. The C–F bond is the shortest and strongest, so fluoroalkanes react slowest of all.
Two-mark mechanism answers: Name the mechanism (SN1 or SN2), give the rate equation, and justify with the substrate class (1° → SN2; 3° → SN1) or the carbocation/steric reason. For the halide trend, anchor it to bond strength → leaving-group ability, not electronegativity.
IB-style question — which mechanism? (a)
2-bromo-2-methylpropane, (CH3)3CBr, is warmed with aqueous sodium hydroxide. (a) State, with a reason, whether the reaction proceeds by SN1 or SN2, and write the rate equation. [3]
How to score the marks
- Mark 1 — classify the substrate. The carbon bonded to Br is attached to three other carbons, so (CH3)3CBr is a tertiary halogenoalkane.
- Mark 2 — choose the mechanism with a reason. A tertiary substrate forms a stable tertiary carbocation, so the reaction proceeds by SN1.
- Mark 3 — rate equation. The rate-determining step involves only the halogenoalkane: rate = k[(CH3)3CBr] (first order overall).
Final answer
Tertiary substrate → SN1 (stable carbocation); rate = k[(CH3)3CBr].
IB-style question — explain the halide order (b)
(b) 1-chlorobutane, 1-bromobutane and 1-iodobutane are each warmed with aqueous sodium hydroxide. Explain which reacts fastest. [2]
How to score the marks
- Mark 1 — identify the fastest. 1-iodobutane reacts fastest because the C–I bond is the weakest (longest) carbon–halogen bond.
- Mark 2 — link to leaving group. The weak C–I bond breaks most easily, so iodide is the best leaving group, giving the fastest substitution (order C–I > C–Br > C–Cl).
Final answer
1-iodobutane is fastest: the C–I bond is weakest, so it breaks most easily and iodide is the best leaving group.