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Electrophilic Addition to Alkenes

How a C=C double bond reacts with HX, water, and halogens through carbocations or bridged ions.

Quick answer An alkene's π bond attacks an electrophile to form a cation — an open carbocation (HX, acid/water) or a bridged halonium ion (Br2, halohydrins). A nucleophile then adds to the more substituted carbon (Markovnikov's rule).
Mechanism · Electrophilic Addition (HBr)2 steps
Step 1 — the π bond grabs the proton.
HHCH3HHBrslowCH3CH3H+2° carbocationBr
The alkene's π electrons attack H–Br. The proton adds to the less-substituted carbon, so the + charge lands on the more-substituted carbon — the more stable (2°) cation. This slow step sets Markovnikov regiochemistry.
Step 2 — bromide traps the carbocation.
CH3CH3H+BrfastBrCH3CH3H2-bromopropane
The bromide ion's lone pair attacks the empty orbital of the carbocation, forming the C–Br bond. This step is fast. Net result: H and Br add across the double bond, Markovnikov.

Nearly every alkene reaction is the same two-step move — π bond attacks an electrophile, a cation forms, a nucleophile traps it — differing only in the intermediate.

The one alkene, three destinations

Propene (the nucleophile)
+ HBr → 2-bromopropane
+ H₂O/H⁺ → 2-propanol
+ Br₂ → 1,2-dibromopropane

1. The π Bond Is the Nucleophile — It Attacks the Electrophile

The π electrons form a new σ bond to an electrophile, leaving the other carbon a cation that a nucleophile then captures.

Electron-rich π bond
Secondary carbocation intermediate

2. Adding HX Runs Through the More Stable Carbocation — That Is Markovnikov's Rule

The π bond grabs the proton of HX to give the more stable carbocation (3° > 2° > 1°), so X lands on the more substituted carbon.

Markovnikov addition of HBr to propene.

The nucleophilic π bond reaches for the proton of H–Br.
Protonation at the terminal carbon gives the more stable 2° carbocation; Br⁻ leaves.
Bromide traps the cation to give 2-bromopropane.

Watch for rearrangements: a free carbocation can undergo a 1,2-hydride or alkyl shift to upgrade a 2° cation to 3°.

3. Acid-Catalyzed Hydration Adds H–OH the Same Markovnikov Way

Water plus catalytic acid runs the same mechanism with water as nucleophile — OH on the more substituted carbon, and it can rearrange too.

Acid-catalyzed hydration puts OH on the more substituted carbon.

4. Halogenation Goes Through a Bridged Halonium Ion — Forcing Anti Addition

With Br2 or Cl2 the halogen bridges both carbons as a cyclic halonium ion, blocking one face so X must attack from the back — giving anti vicinal dihalides with no rearrangement.

Anti addition of Br₂ gives 1,2-dibromopropane.

5. Halohydrins Form When Water Opens the Halonium at the More Substituted Carbon

Run halogenation in water and water opens the halonium from the back face (anti) at the more substituted carbon, giving a halohydrin — OH on the more substituted carbon, X on the neighbor.

Br₂ in H₂O
Bromohydrin: OH on more substituted C

6. The Syn Additions (Hydroboration, Hydrogenation) Break This Pattern on Purpose

Hydroboration–oxidation is concerted, so it is syn, never rearranges, and puts OH on the less substituted carbon (anti-Markovnikov); hydrogenation (H2, Pd) adds two H syn to reduce the alkene.

Hydrogenation: a concerted syn delivery of two H atoms.

7. Summary

Open carbocation (HX, hydration): Markovnikov · non-stereospecific · can rearrange · Bridged halonium (Br2, halohydrins): anti · no rearrangement · nucleophile opens at more substituted C · Concerted (hydroboration, hydrogenation): syn · no rearrangement · hydroboration is anti-Markovnikov.

Ask "what is the intermediate?" first, and the regiochemistry and stereochemistry follow.

Quiz yourself

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2-Bromopropane (CC(C)Br). Protonating the terminal CH2 gives a secondary carbocation, which is more stable than the primary one; bromide then traps that cation. This is Markovnikov's rule — Br ends up on the more substituted carbon.

Because the intermediate is a bridged bromonium ion, not an open carbocation. The bromine bridges both carbons and blocks one face, so the incoming Br is forced to attack from the opposite face — anti addition. The same bridging is why halogenation never rearranges.

Water opens the bromonium ion at the more substituted carbon (more partial positive charge), so OH lands there and Br on the less substituted carbon — a bromohydrin (CC(O)CBr) — added anti.

Only the open-carbocation pathways — HX addition and acid-catalyzed hydration — can undergo 1,2-hydride or alkyl shifts. Halogenation, halohydrin formation, hydroboration, and hydrogenation all avoid a free carbocation, so they don't rearrange. A product whose skeleton implies a shifted, more-stable cation is the tell.

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