IB Chemistry HLTopic 6 — Electron Pair SharingPaper 1 & 2Trends~11 min read
Relative Rates of Nucleophilic Substitution
Three things decide how fast a halogenoalkane reacts: who is attacking, which halogen is leaving, and how crowded the carbon is. Learn them as a checklist and rate questions stop being guesswork.
📚 What you need to know
Three factors matter, whether the mechanism is SN1 or SN2.
1. The nucleophile: stronger means faster. Charge and electronegativity decide strength.
2. The halogen: the weaker the C–X bond, the faster the reaction.
Bond strength order: C–F > C–Cl > C–Br > C–I, so reactivity is the reverse.
Iodoalkanes react fastest; fluoroalkanes barely react at all.
3. The class: primary goes SN2, tertiary goes SN1, secondary can do either.
Carbocation stability rises with more alkyl groups because of the positive inductive effect.
Silver nitrate gives a precipitate whose colour and speed identify the halide.
Factor 1: how good is the nucleophile?
A nucleophile has to push a lone pair into a δ+ carbon. Anything that makes that pair more available makes the reaction faster.
Charge. A negative species has an extra electron and shares more freely than the neutral version. OH– beats H2O every time.
Electronegativity. With the same charge, the less electronegative donor atom holds its lone pair more loosely. Nitrogen is less electronegative than oxygen, so NH3 beats H2O.
Strongest to weakest
CN– > OH– > NH3 > H2O
A shortcut worth knowing: a conjugate base is always a better nucleophile than its conjugate acid. That is why OH– beats H2O and why NH2– would beat NH3.
Factor 2: which halogen is leaving?
Whatever the mechanism, at some point the C–X bond has to break. The energy needed to do that is the bond enthalpy, and it drops sharply as you go down group 17.
Notice how much bigger the C–F bar is than the rest. That gap is why fluoroalkanes are treated as unreactive rather than just slow.
Reactivity of halogenoalkanes
iodo > bromo > chloro > fluoro
Careful with the logic here. Iodine is the least electronegative halogen, so the C–I carbon is the leastδ+. You might expect iodoalkanes to be slowest. They are the fastest, because bond strength matters far more than bond polarity.
The silver nitrate test
You can watch this trend happen on the bench. Warm each halogenoalkane with aqueous silver nitrate in ethanol. As halide ions are released they meet Ag+ and drop out as a precipitate.
Two things tell you which halogen it was: the colour of the precipitate and how quickly it appears.
The order the precipitates appear in matches the bond strength trend exactly. That is the point of the experiment.
Halide released
Precipitate
Colour
How quickly
F–
none (AgF is soluble)
—
no reaction
Cl–
AgCl
white
slowly
Br–
AgBr
cream
moderately
I–
AgI
yellow
quickly
Factor 3: how crowded is the carbon?
The number of alkyl groups on the carbon holding the halogen changes both which mechanism runs and how fast it goes.
Primary (one alkyl group): plenty of room, so the nucleophile attacks directly. SN2, and no carbocation is ever formed.
Tertiary (three alkyl groups): the carbon is blocked, but the carbocation it would form is stable. SN1.
Secondary: either, depending on the solvent, the temperature and the nucleophile.
Why tertiary carbocations are more stable
Alkyl groups push electron density towards whatever they are attached to. This is the positive inductive effect. Pushing electrons towards a positive carbon spreads the charge out and lowers the charge density, which makes the ion more stable.
Three alkyl groups push three times. One alkyl group pushes once. That is the whole explanation.
The arrowheads on the bonds are not curly arrows. They just show the direction electron density is being pushed.
🧩 Answering any “which reacts faster” question
Are the nucleophiles different? If so, compare charge first, then electronegativity.
Are the halogens different? If so, compare C–X bond enthalpy. Weaker wins.
Are the classes different? If so, decide the mechanism for each one.
Only one thing should be changing. Identify it and ignore the rest.
State the reason, not just the answer. “Weaker C–I bond needs less energy to break.”
Check the direction. Weaker bond means faster; more stable carbocation means faster SN1.
Worked examples
WORKED EXAMPLE
1-chlorobutane, 1-bromobutane and 1-iodobutane are each warmed with aqueous silver nitrate. Predict the order in which precipitates appear and explain your reasoning.
Spot what is changing
Same chain, same class, same nucleophile. Only the halogen differs.
Compare bond enthalpiesC–Cl 324 > C–Br 285 > C–I 228 kJ mol–1Turn that into a rate
The weakest bond breaks first, so the iodide is released first.
Yellow AgI first, then cream AgBr, then white AgClgive the colours as well as the order — the question is testing both
WORKED EXAMPLE
2-bromo-2-methylpropane hydrolyses much faster than 1-bromobutane under the same conditions. Explain why, referring to the mechanism in each case.
Identify the classes
2-bromo-2-methylpropane is tertiary; 1-bromobutane is primary.
Tertiary route
It goes SN1. The slow step forms a tertiary carbocation, which three alkyl groups stabilise by the positive inductive effect.
Why that is fast
A more stable carbocation means a lower activation energy for the rate-determining step.
Primary route
It goes SN2, and the nucleophile and the halogenoalkane must collide correctly in one step, which is a bigger ask.
The stable tertiary carbocation lowers the activation energy of the slow stepboth bromides, both hydroxide — so class must be the reason
💡 Exam tip
Learn the bond enthalpy order even if you cannot recall the exact numbers. The order earns the mark.
Always say weaker bond, less energy needed, faster reaction as a chain. One-word answers lose marks.
Give precipitate colours precisely: white, cream, yellow. “Off-white” and “pale” are risky.
Fluoroalkanes give no precipitate for two reasons: the C–F bond is too strong and AgF is soluble.
When explaining carbocation stability, use the phrase positive inductive effect.
Check what the question is actually varying before you start writing.
⚠️ Common mix-up
Using electronegativity to predict the halogen trend. That gives the wrong answer. Use bond enthalpy.
Saying stronger bond means faster. It is the reverse. Strong bonds are hard to break.
Swapping cream and yellow. Bromide is cream, iodide is yellow.
Claiming primary halogenoalkanes form carbocations. They do not — that is the point of SN2.
Muddling inductive effect with electronegativity. Alkyl groups push electrons; electronegative atoms pull them.
Comparing two things that differ in more than one way. Then no single factor explains it.
Up next: Mechanisms of Electrophilic Addition — back to alkenes, now with full curly arrows and the carbocation making a return appearance.
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