A halogenoalkane has a weakness built into it. One of its carbon atoms is short of electrons, and anything carrying a spare pair will go straight for it — pushing the halogen out on the way in.
📚 What you need to know
A nucleophile is an electron-rich species that donates a pair of electrons. The name means “nucleus loving”, so it heads for positive charge.
The C–X bond is polar because the halogen is more electronegative, which leaves the carbon δ+.
In nucleophilic substitution the nucleophile bonds to that carbon and the halogen departs as a halide ion — the leaving group.
The reaction is fastest for iodoalkanes and slowest for fluoroalkanes, because the rate follows bond strength, not polarity.
Charged nucleophiles (OH–, CN–) give a neutral product in one step.
Neutral nucleophiles (H2O, NH3) give a positively charged product that then loses H+.
Warm aqueous NaOH or KOH hydrolyses a halogenoalkane to an alcohol.
What makes something a nucleophile
Two things, and only two. It must have a lone pair to hand over, and it must be willing to hand it over. That is all “electron-rich” means in practice.
Because of that, some nucleophiles are better than others. A hydroxide ion carries a full negative charge on its oxygen; a water molecule carries only a partial one. Same element, same lone pairs, but the ion is far more eager to donate — so OH– is the stronger nucleophile of the two.
Charged nucleophiles
Neutral nucleophiles
OH– hydroxide
H2O water
CN– cyanide
NH3 ammonia
Cl– chloride
ROH alcohols
R– carbanions
RNH2 amines
Every one of those neutral species has a lone pair on an oxygen or a nitrogen. If you can spot the lone pair, you can spot the nucleophile — you do not need to memorise the list.
The polar C–X bond
Halogens are more electronegative than carbon, so in a C–X bond the shared pair is pulled towards the halogen. The halogen ends up δ– and the carbon δ+. That partially positive carbon is the target: it is the one place in the molecule that a lone pair can be attracted to.
Notice where the new bond forms. It has to be at the carbon that was holding the leaving group — nowhere else in the molecule is short of electrons.
What actually happens
Hydrolysis is the standard example. Warm a halogenoalkane with aqueous sodium or potassium hydroxide and the hydroxide ion substitutes for the halogen, giving an alcohol.
Two practical details examiners like. The mixture is warmed because the reaction is slow at room temperature, and ethanol is added as a solvent because the halogenoalkane will not mix with water on its own. Without it the two layers barely touch and almost nothing happens.
🧩 Reading any substitution question
Find the nucleophile — the species with a lone pair, usually written with a negative charge or a colon.
Find the δ+ carbon — the one attached to the halogen.
Find the leaving group — the halogen, which departs as X–.
Write the product by swapping the nucleophile in for the halogen, and put X– on the right-hand side.
If the nucleophile was neutral, add a second step in which the product loses H+.
Why the halogen decides the rate
All four halogenoalkanes react the same way, but at wildly different speeds. The bond that has to break is the C–X bond, so the weaker that bond, the faster the reaction.
Bond enthalpies from the data booklet. The gap between C–F and C–Cl is enormous, which is why fluoroalkanes are treated as inert.
This is the bit worth understanding rather than learning. C–F is the most polar of the four, so the carbon there is the most δ+ — and yet it is the slowest by miles. Polarity decides where the nucleophile attacks; bond strength decides how fast anything happens.
Neutral nucleophiles need an extra step
If the nucleophile arrives with a lone pair but no negative charge, it still donates that pair — but it keeps the positive charge that donating leaves behind. The first product is therefore a cation, which then throws off a proton to become neutral.
Ammonia behaves the same way: it attacks as NH3, gives a positively charged alkylammonium ion, and loses H+ to finish as an amine.
WORKED EXAMPLE
1-bromopropane is warmed with aqueous potassium cyanide. Identify the nucleophile and the leaving group, and write the equation for the reaction.
Step 1 — the nucleophileThe cyanide ion has a lone pair on carbon and a full negative charge.CN⁻Step 2 — the leaving groupThe C–Br bond breaks and bromine takes both electrons.Br⁻Step 3 — swap them overCH₃CH₂CH₂Br + CN⁻ → CH₃CH₂CH₂CN + Br⁻The product is butanenitrile. Note the carbon chain grew by one — that is why cyanide is such a useful nucleophile.
WORKED EXAMPLE
1-iodobutane is hydrolysed much faster than 1-chlorobutane, even though the C–Cl bond is more polar. Explain why.
What has to happenIn both molecules the C–X bond must break before the reaction can finish.Compare the bondsC–I = 228 kJ mol⁻¹, C–Cl = 324 kJ mol⁻¹Iodine is a much larger atom, so the bonding pair is further from both nuclei and held far more weakly.the weaker C–I bond breaks more easily, so the rate is higherThen deal with the polarity point directly: the greater δ+ on carbon in C–Cl does not compensate for the much stronger bond.
WORKED EXAMPLE
Chloroethane reacts with water rather than with hydroxide ions. State the two steps that occur and explain why a second step is needed.
Step 1 — the attackWater donates a lone pair on oxygen to the δ+ carbon and the chloride ion leaves.CH₃CH₂Cl + H₂O → CH₃CH₂OH₂⁺ + Cl⁻Step 2 — the tidy-upThe oxygen donated a pair without having a negative charge to lose, so it is left positive.CH₃CH₂OH₂⁺ → CH₃CH₂OH + H⁺the cation loses H⁺ to give the neutral alcoholWater is also a weaker nucleophile than OH⁻, so this route is slower as well as longer.
💡 Exam tip
Define a nucleophile as an electron-rich species that donates a pair of electrons — not just “attracted to positive charge”.
Always name both the nucleophile and the leaving group; questions ask for them separately.
Explain the polarity by electronegativity difference, then say the carbon is δ+.
For rate questions, quote bond enthalpy values and say the weaker bond breaks more easily.
Remember the practical conditions: warm, aqueous NaOH or KOH, with ethanol as a solvent.
If the nucleophile is neutral, write both steps. Losing the mark for the deprotonation is a common way to drop a point.
⚠️ Common mix-up
Predicting that fluoroalkanes react fastest because the C–F bond is the most polar. Bond strength decides the rate.
Writing the halogen leaving as a neutral atom. It takes the bonding pair with it, so it leaves as X–.
Attacking the wrong carbon. Only the carbon bonded to the halogen carries a δ+ charge.
Forgetting the second step with H2O or NH3, and writing a neutral product straight away.
Calling the nucleophile an electrophile because it attacks. It attacks by giving electrons, which makes it a nucleophile.
Up next: Heterolytic Fission — we have been saying the halogen “leaves with both electrons” without naming it. That way of breaking a bond has a name, and it is where nucleophiles and electrophiles come from in the first place.
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