IB Chemistry HLTopic 6 — Electron Pair SharingPaper 1 & 2Organic~12 min read
Nucleophilic Substitution in Halogenoalkanes
Same reactants, same product, two completely different routes. Which one a halogenoalkane takes depends on how crowded the carbon is — and that single idea explains the rate equation, the mechanism and the shape of the energy profile.
📚 What you need to know
There are two mechanisms: SN1 and SN2.
S = substitution, N = nucleophilic. The number is how many species are in the slow step.
Tertiary halogenoalkanes go by SN1. Primary go by SN2. Secondary can do either.
SN1 is two steps with a carbocation intermediate. Rate = k[halogenoalkane].
SN2 is one step through a transition state. Rate = k[halogenoalkane][nucleophile].
SN1 is unimolecular, SN2 is bimolecular (molecularity of the slow step).
SN2 gives inversion of configuration because the attack comes from behind.
First, a reminder about nucleophile strength
Before the mechanisms, one quick point that both share. A nucleophile works better when its lone pair is easy to hand over.
Charge: a negative species beats its neutral version. OH– is stronger than H2O.
Electronegativity: with the same charge, the less electronegative atom grips its pair more loosely. NH3 is stronger than H2O.
Order of nucleophile strength
CN– > OH– > NH3 > H2O
SN1: the two-step route
A tertiary halogenoalkane has three alkyl groups packed around the carbon holding the halogen. There is simply no room for a nucleophile to squeeze in. So the molecule does something else: it lets the halogen go first.
🧩 The SN1 mechanism
Step 1 (slow). The C–X bond breaks heterolytically. The halogen leaves as X– taking both electrons.
This leaves a tertiary carbocation intermediate with a positive carbon.
This step is rate-determining because it is the hard one.
Step 2 (fast). The nucleophile attacks the positive carbon and donates a lone pair.
Only the halogenoalkane appears in the slow step, so the rate does not depend on the nucleophile.
The carbocation is a real, if short-lived, species. That is what makes this an intermediate rather than a transition state.
Rate equation for SN1
rate = k[halogenoalkane]
SN2: the one-step route
A primary halogenoalkane has only one alkyl group in the way, so there is a clear path to the carbon. The nucleophile does not wait for the halogen to leave — it comes in from the opposite side and pushes it out.
Everything happens at once. The new C–Nu bond is forming while the old C–X bond is breaking. There is no intermediate, only a high-energy arrangement called a transition state where both bonds are partly formed.
Rate equation for SN2
rate = k[halogenoalkane][nucleophile]
The number in SN1 and SN2 is not the number of steps. It is the molecularity of the slow step — how many species have to collide for the rate-determining step to happen. SN1 has two steps but a molecularity of one.
The energy profiles
You can tell the two mechanisms apart from the shape of the energy diagram alone. Count the humps.
A transition state sits at the top of a hump and cannot be isolated. An intermediate sits in a dip and exists, briefly, as a real species.
Inversion of configuration
In SN2 the nucleophile cannot come in on the same side as the halogen — that side is blocked, which is called steric hindrance. So it attacks from directly behind, opposite the leaving group.
As the new bond forms, the three other groups on the carbon are pushed through and end up pointing the other way. The molecule is turned inside out.
Drawing tip: use a wedge and a dashed bond to show the 3D arrangement before and after, and dotted bonds in the transition state itself.
Side by side
Feature
SN1
SN2
Typical halogenoalkane
Tertiary
Primary
Number of steps
Two
One
Slow step involves
Halogenoalkane only
Halogenoalkane and nucleophile
Rate equation
rate = k[RX]
rate = k[RX][Nu]
Molecularity of slow step
Unimolecular
Bimolecular
Key species in the middle
Carbocation intermediate
Transition state
Energy profile
Two peaks with a dip
One peak
Effect on shape
Attack from either face
Inversion of configuration
Secondary halogenoalkanes sit in the middle. They can go either way, and which route wins depends on the solvent, the temperature and how strong the nucleophile is. If an exam question uses a secondary halogenoalkane, read the conditions carefully.
Worked examples
WORKED EXAMPLE
Doubling the concentration of OH– has no effect on the rate of hydrolysis of a certain halogenoalkane. Doubling the halogenoalkane concentration doubles the rate. Deduce the mechanism and the class of halogenoalkane.
Read the nucleophile result
No effect means the nucleophile is not in the slow step. Its order is zero.
Read the halogenoalkane result
Doubling doubles the rate, so it is first order.
Write the rate equationrate = k[halogenoalkane]Match it to a mechanism
One species in the slow step means unimolecular.
SN1, so the halogenoalkane is tertiarythe slow step must be the C–X bond breaking to give a carbocation
WORKED EXAMPLE
Explain why bromoethane reacts with hydroxide by SN2 while 2-bromo-2-methylpropane reacts by SN1.
Look at the crowding
Bromoethane is primary: one alkyl group, so the carbon is reachable.
So the nucleophile attacks directly
It comes in from behind the C–Br bond in one step. No carbocation is needed.
Now the tertiary case
2-bromo-2-methylpropane has three alkyl groups blocking the carbon, so a direct attack is not possible.
What makes SN1 workable
Those same three groups push electron density towards the positive carbon, making the tertiary carbocation stable enough to form.
Steric hindrance blocks SN2; carbocation stability allows SN1two reasons are wanted here — crowding and carbocation stability
💡 Exam tip
Label the slow (rate-determining) step on any mechanism you draw. It is almost always a marking point.
Use double-headed curly arrows throughout, starting from a lone pair or a bond.
Put the transition state in square brackets with dotted bonds to both the nucleophile and the leaving group.
Show wedge and dashed bonds when a question mentions inversion. Flat drawings cannot show it.
Rate data is a gift: zero order in the nucleophile means SN1, first order means SN2.
Never call the transition state an intermediate. It cannot be isolated.
⚠️ Common mix-up
Thinking the 1 and 2 mean the number of steps. They mean molecularity of the slow step. SN1 has two steps.
Confusing intermediate and transition state. Intermediate = a dip, a real species. Transition state = a peak, not isolable.
Putting the nucleophile in the SN1 rate equation. It is not in the slow step, so it does not appear.
Drawing SN2 attack from the same side as the halogen. That is exactly what steric hindrance prevents.
Assigning tertiary to SN2 and primary to SN1. It is the other way round. Crowded carbon means SN1.
Forgetting the halide ion in the products. It always comes out as X–.
Up next: Relative Rates of Nucleophilic Substitution — the three things that decide how fast any of this actually happens.
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