Electrophilic Addition to Alkenes

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 electrons grab the proton.

The alkene is the nucleophile here. The proton adds to the carbon that gives the more stable cation — that is all Markovnikov’s rule is. Slow step, so it decides the outcome.

Step 2 — bromide traps the cation.

Fast, and it does not affect which product forms — step 1 already decided that. 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.

How each reagent works — the arrow pushing

Electron flow only. Follow the arrows; the structures do the talking.

HBr / HX — Markovnikov, via a carbocation2 steps

Why it works · The alkene π bond is electron-rich — it is the nucleophile. In H–Br the H is δ+ (Br pulls the bonding pair), so the π electrons grab that proton and the H lands so as to leave the more stable (more-substituted) carbocation — that is Markovnikov. Bromide then traps the cation. Electrophilic addition; not redox.

H₂O / H₂SO₄ — acid-catalyzed hydration2 steps

Why it works · Same electrophilic addition, but the electrophile is H⁺ from the acid and the nucleophile is water. The π bond grabs H⁺ to give the more-substituted carbocation (Markovnikov); water adds, then loses a proton to give the Markovnikov alcohol. Acid is a catalyst — it is returned at the end.

Br₂ — bromonium ion, anti addition2 steps

Why it works · Br–Br has no permanent dipole, but the electron-rich π bond polarizes it — the near Br becomes δ+. The π attacks it and, because the resulting Br still has lone pairs, it bridges both carbons as a bromonium ion (no open carbocation → no rearrangement). Bromide then attacks the opposite face, so the two Br’s add anti.

Br₂ / H₂O — halohydrin1 step

Why it works · The same bromonium forms, but now water is the nucleophile that opens it. In an unsymmetrical bromonium the more-substituted carbon carries more δ+ (it better stabilizes positive charge), so water attacks there — anti to bromine. Result: OH on the more-substituted carbon, Br on the other (Markovnikov halohydrin).

CH₂I₂/Zn or :CCl₂ — carbene cyclopropanation1 step

Why it works · A carbene (Simmons–Smith carbenoid from CH₂I₂/Zn, or dichlorocarbene from CHCl₃/base) has a carbon with both a lone pair and an empty orbital. It adds across the alkene π in one concerted [2+1] step → a cyclopropane, stereospecifically (cis stays cis).

Quiz yourself

Tap a question to reveal the answer — free, no login.

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.

Draw this on the whiteboard

Open the OChem Board whiteboard — benzene rings, curved arrows, wedge/dash bonds and a clickable periodic table built in. No account needed.

Open the whiteboard →