The big idea: In a halogenoalkane (e.g. CH3CH2Br), the carbon–halogen bond is polar: the halogen is more electronegative than carbon, so it pulls the bonding electrons towards itself. This leaves the carbon slightly positive (δ+) and the halogen slightly negative (δ−).
That δ+ carbon is an easy target for an electron-rich species. A nucleophile is an electron-pair donor — it brings a lone pair to the δ+ carbon, forms a new bond, and kicks the halogen out. The halogen leaves with the C–X bonding pair as a halide ion (the leaving group).
The hydroxide ion's lone pair attacks the δ+ carbon; at the same time the C–X bonding pair shifts onto X, which leaves as a halide ion.
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Key terms: - Nucleophile (Nu⁻) — an electron-pair donor; it has a lone pair and is attracted to a δ+ centre (e.g. OH⁻, CN⁻, NH3, H2O). - Electrophile — the opposite: an electron-pair acceptor (here, the δ+ carbon). - Leaving group — the atom/ion that departs with the bonding pair (the halide ion, X⁻). - Substitution — one group is replaced by another; the carbon skeleton is unchanged.
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A curly arrow shows the movement of a pair of electrons. The tail starts at the electrons that move (a lone pair or a bond) and the head points to where that pair ends up (a new bond, or onto an atom). In nucleophilic substitution there are two curly arrows.
The two arrows
- Arrow 1 — from the nucleophile to the carbon. The tail is on the lone pair of the nucleophile (e.g. on the O of OH⁻); the head points at the δ+ carbon, forming the new bond.
- Arrow 2 — from the C–X bond to the halogen. The tail is on the C–X bonding pair; the head points onto the halogen, which leaves as X⁻ (taking both electrons).
The hydroxide ion's lone pair attacks the δ+ carbon; at the same time the C–X bonding pair shifts onto X, which leaves as a halide ion.
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Common arrow mistakes: - Drawing the arrow from the carbon to the nucleophile (backwards) — it must start at the electron pair that moves, i.e. the nucleophile's lone pair. - Forgetting the second arrow — the C–X bond must break, so the bonding pair has to go onto the halogen. - Forgetting the charges: OH⁻ becomes neutral once it bonds, and the halogen leaves as X⁻ (a negative ion).
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The product depends on the nucleophile. The most-tested case is warm aqueous hydroxide (NaOH or KOH dissolved in water): the OH⁻ replaces the halogen and the product is an alcohol.
Hydroxide → alcohol: Bromoethane reacted with warm aqueous NaOH gives ethanol:
CH3CH2Br + OH⁻ → CH3CH2OH + Br⁻
The –Br is replaced by –OH; a bromide ion is released. The conditions are warm (gentle heat) and the hydroxide is aqueous (in water).
| Nucleophile | Conditions | Organic product |
|---|---|---|
| hydroxide, OH⁻ | warm aqueous NaOH/KOH | an alcohol (–OH) |
| water, H2O | warm; slower than OH⁻ | an alcohol (–OH) |
| cyanide, CN⁻ | warm KCN in ethanol | a nitrile (–CN), adds a carbon |
| ammonia, NH3 | excess NH3, heated, sealed | an amine (–NH2) |
Why the rate changes down the group: Iodoalkanes react fastest, fluoroalkanes slowest. The deciding factor is the carbon–halogen bond strength: the C–I bond is the weakest, so it breaks most easily; the C–F bond is the strongest. (Bond strength matters more here than how polar the bond is.)
How this is tested: R3.4 shows up two ways:
- Paper 1A (MCQ): identify the organic product of a halogenoalkane + aqueous hydroxide (it is the alcohol). - Paper 2: 'use curly arrows to show the electron-pair movement' in the nucleophilic substitution — a near-guaranteed mechanism mark.
The markers want the two arrows in the right places and the correct products (alcohol + halide ion).
Scoring the mechanism: Show the lone pair on the nucleophile and start arrow 1 there (not at the carbon). Always draw the second arrow from the C–X bond onto the halogen, and label the leaving group as X⁻.
IB-style question — substitution of 1-bromopropane (a)
1-bromopropane, CH3CH2CH2Br, is warmed with aqueous sodium hydroxide. (a) Use curly arrows to describe the electron-pair movement in this reaction, and identify the leaving group. [2]
How to score the marks
- Mark 1 — arrow to the carbon. A curly arrow starts at the lone pair on the hydroxide oxygen and points to the δ+ carbon (the one bonded to Br), forming the new C–O bond.
- Mark 2 — arrow breaking C–Br + leaving group. A second curly arrow goes from the C–Br bonding pair onto the bromine; the leaving group is the bromide ion, Br⁻.
Final answer
Arrow 1: O lone pair (of OH⁻) → δ+ carbon. Arrow 2: C–Br bond → Br. Leaving group = Br⁻ (shown below).
The hydroxide ion's lone pair attacks the δ+ carbon; at the same time the C–X bonding pair shifts onto X, which leaves as a halide ion.
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IB-style question — the organic product (b)
(b) Identify the organic product formed, and state the two conditions needed. [2]
How to score the marks
- Mark 1 — product. The –Br is replaced by –OH, so the organic product is the alcohol propan-1-ol, CH3CH2CH2OH (a bromide ion is also released).
- Mark 2 — conditions. The sodium hydroxide must be aqueous (dissolved in water) and the mixture is warmed (gently heated).
Final answer
Product: propan-1-ol, CH3CH2CH2OH. Conditions: warm, aqueous NaOH.