IB Chemistry Electrophilic Addition Reactions: The Complete FAQ

You need to know that alkenes react with electrophiles like Br2, HBr and H2O across the C=C double bond, forming a saturated product. At HL you must also draw the full curly-arrow mechanism, including the carbocation intermediate, and apply Markovnikov's rule to unsymmetrical alkenes.

Bromine's non-polar bond becomes polarised as it approaches the electron-rich C=C double bond, making one Br atom slightly positive. The pi bond attacks this Br, the Br–Br bond breaks heterolytically, and a carbocation intermediate forms. A bromide ion then attacks the positive carbon, giving the dibromoalkane.

Alkenes have a localised, reactive pi bond that's easily attacked by electrophiles, so addition happens readily and destroys the double bond. Arenes like benzene have a delocalised ring of six pi electrons that gives extra stability, so they resist addition and undergo electrophilic substitution instead, which keeps the ring intact.

Markovnikov's rule says that when H–X adds across an unsymmetrical alkene, the hydrogen attaches to the carbon that already has more hydrogens, giving the more stable carbocation and the major product. You need it whenever a question asks for the major organic product of an unsymmetrical alkene reacting with HBr, HCl or H2O.

Both levels know that alkenes undergo addition reactions and can identify them using the bromine-water decolourisation test. Only HL students are examined on the full curly-arrow mechanism, the carbocation intermediate and Markovnikov's rule — this extra depth appears as AHL-only content in the IB Chemistry guide.

Expect "describe the mechanism", "outline" or "deduce", all of which require curly arrows showing electron movement, correct partial charges, and any intermediate drawn explicitly. "State" or "identify" questions, by contrast, just want the reagent, product or reaction type named — no mechanism needed.

A curly arrow always starts from a bond or a lone pair — never from an atom on its own — and points to where the electron pair ends up. For electrophilic addition, that means one arrow from the pi bond to the electrophile, and a second from the breaking bond to the leaving group.

The most common mistake is drawing the curly arrow starting from an atom (like Br) instead of from the bond or lone pair, which examiners mark as a fundamental error even if the final product is correct. The second most common: forgetting to draw the carbocation intermediate entirely.

Addition happens to alkenes because the pi bond breaks and nothing leaves the molecule — two things join together across the double bond. Substitution happens to arenes because the delocalised ring stays intact and a hydrogen atom is replaced by something else, so one group swaps for another.

Typical questions ask you to draw the mechanism for an alkene reacting with HBr or Br2, predict the major product using Markovnikov's rule, explain why benzene resists addition, or identify an unknown compound from a positive bromine-water test in a practical-based question.

Electrophilic addition happens to alkenes, where an electron-poor species attacks the electron-rich C=C bond and adds across it. Nucleophilic substitution happens to haloalkanes, where an electron-rich species replaces a halogen leaving group. Look at the starting functional group — a double bond means addition, a C–halogen bond means substitution.

Yes — at HL, electrophilic addition adds a full curly-arrow mechanism, carbocation stability arguments and Markovnikov's rule on top of the SL requirement to simply recognise the reaction and its test. Students moving from SL-level organic chemistry often underestimate how much extra precision the HL mechanism demands.