IB Chemistry HLTopic 6 — Electron Pair SharingPaper 1 & 2Organic~12 min read
The Mechanisms of Electrophilic Addition
You already know what comes out. Now you have to show the working — two steps, a carbocation in the middle, and curly arrows that start and finish in exactly the right places.
📚 What you need to know
Every electrophilic addition is two steps with a carbocation intermediate.
Step 1 (slow): the π electrons attack the electrophile. Heterolytic fission gives a carbocation.
Step 2 (fast): the negative ion donates a lone pair to the carbocation.
Curly arrows must be double-headed, start from a lone pair or the C=C, and point to a δ+ atom or a positive charge.
HX is permanently polar, so the electrophile is ready-made.
X2 is non-polar and only becomes an electrophile through an induced dipole.
Water needs an acid catalyst; H3O+ is the electrophile and is regenerated at the end.
Hydrogenation is assumed knowledge — no mechanism required.
The curly arrow rules
Examiner reports complain about curly arrows more than almost anything else in organic chemistry. Three rules cover it.
Double-headed. A pair of electrons is moving, not one.
Start from electrons. Either a lone pair, or the middle of the C=C. Never from an atom letter and never from a positive charge.
End where the electrons land. On a δ+ atom, on a positive carbon, or onto an atom that is taking a bond with it.
One extra habit. When a bond breaks heterolytically, the arrow starts on that bond and ends on the atom keeping the electrons. Draw that arrow and the bromide ion in your product suddenly makes sense.
Adding a hydrogen halide
HBr is polar all the time: bromine is more electronegative, so the hydrogen carries a permanent δ+. That makes it an electrophile before it even meets the alkene.
Count your arrows against this before you hand a mechanism in. Missing the second arrow in step 1 is the most common slip.
🧩 Drawing the HX mechanism from scratch
Draw the alkene with the double bond clearly showing, and the H–X molecule nearby.
Mark δ+ on the hydrogen and δ– on the halogen.
Arrow 1: from the middle of the C=C to the δ+ hydrogen.
Arrow 2: from the H–X bond to the halogen. This is the heterolytic fission.
Draw the carbocation with the + on the correct carbon, and the halide ion with its lone pairs.
Arrow 3: from a lone pair on the halide to the positive carbon. Draw the product.
Adding a halogen
Br2 is a problem at first glance. Both atoms are identical, so the molecule is non-polar. There is no δ+ end for the alkene to attack.
What saves the reaction is that the π cloud is a big pile of negative charge. As the bromine molecule drifts close, that charge repels the electrons in the Br–Br bond, pushing them towards the far bromine. A dipole is induced where none existed before.
This is the only real difference between the halogen and hydrogen halide mechanisms. Everything after the dipole forms is the same.
Bromine and ethene
CH2=CH2 + Br2 → CH2BrCH2Br
If a question asks you to compare the HBr and Br2 mechanisms, the answer is one sentence: hydrogen halides are permanently polar, halogens are non-polar and rely on a temporary induced dipole. Everything else matches.
Adding water — the acid catalyst version
Water is a weak electrophile. Its δ+ hydrogens are not positive enough to be attacked by an alkene on their own, so nothing happens without help.
Add a strong acid and the picture changes. H3O+ carries a full positive charge and is a proper electrophile. The alkene attacks that instead.
Step 3 is easy to forget. Without it you are left with a positively charged product, which cannot be the answer to “name the organic product”.
All four side by side
Reagent
Electrophile
Where it comes from
Mechanism needed?
HX
δ+ hydrogen
permanent polarity
Yes
X2
δ+ halogen
induced dipole
Yes
H2O with acid
H3O+
the acid catalyst
Yes
H2 with Ni or Pt
not applicable
surface catalysis
No
Worked examples
WORKED EXAMPLE
Describe, with curly arrows, the mechanism for the reaction of ethene with HCl. State how many arrows are needed and where each one starts and finishes.
Set up the polarity
Chlorine is more electronegative, so the H is δ+ and the Cl is δ–.
Step 1 needs two arrows
Arrow 1 from the middle of the C=C to the δ+ H. Arrow 2 from the H–Cl bond to the Cl.
What that leavesCH3CH2+ and :Cl–Step 2 needs one arrow
From a lone pair on Cl– to the positive carbon.
Three arrows in total; the product is chloroethanestep 1 is slow and rate-determining; step 2 is fast
WORKED EXAMPLE
Br2 has no permanent dipole. Explain how it can still act as an electrophile towards ethene.
Start with why there is no dipole
Two identical atoms means no electronegativity difference, so the bonding pair is shared evenly.
Bring the alkene into it
The C=C π bond is a region of high electron density.
What that does to the bromine
As Br2 approaches, the π electrons repel the Br–Br bonding pair towards the far atom.
Result
The near bromine becomes δ+ and can accept the pair of π electrons.
A temporary induced dipole makes the near Br the electrophilethe word “temporary” matters — the dipole vanishes if the alkene moves away
💡 Exam tip
Every arrow must be double-headed. A single-headed arrow here says radicals and loses the mark.
Draw the arrow from the C=C starting on the bond, not on one of the carbon letters.
Show the lone pairs on your halide ion. The arrow in step 2 has to come from one.
Label step 1 as slow and step 2 as fast if the question mentions rate at all.
Put the + on the correct carbon of the carbocation, and keep every hydrogen accounted for.
For hydration, finish with the deprotonation step and say the catalyst is regenerated.
⚠️ Common mix-up
Forgetting the second arrow in step 1. Without it the H–X bond never breaks and no halide ion appears.
Starting an arrow at a positive charge. Arrows start where electrons are, not where they are missing.
Saying Br2 is polar. It is not. Say induced or temporary dipole.
Using water directly as the electrophile. It is too weak; the acid catalyst supplies H3O+.
Stopping at the oxonium ion. The organic product is the neutral alcohol.
Drawing a mechanism for hydrogenation. It is not required and wastes time.
Up next: Addition to Unsymmetrical Alkenes — what happens when the two carbons of the double bond are not the same, and how to work out which product wins.
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