Addition Reactions to Alkenes: JEE Main Guide

Alkene addition reactions are the backbone of JEE Main Organic Chemistry and appear in three to five questions per session. The key is knowing which addition pathway — electrophilic, free-radical, or hydroboration — applies under which conditions, and then applying the correct regiochemistry (Markovnikov or anti-Markovnikov) and stereochemistry (syn or anti). Master these rules and alkene questions become mechanical.

Electrophilic Addition: The Standard Pathway

Alkenes are electron-rich (π bond is nucleophilic). Electrophilic addition proceeds via: (1) electrophile attack on π bond → forms carbocation intermediate (Markovnikov step — electrophile adds to the carbon bearing fewer Hs, forming the more stable carbocation); (2) nucleophile attacks carbocation → product. Markovnikov's rule: in HX addition to an unsymmetrical alkene, H adds to the carbon with more hydrogens (forms the more stable 2° or 3° carbocation). Example: CH₂=CHCH₃ + HBr → CH₃CHBrCH₃ (not CH₂BrCH₂CH₃). The anti-Markovnikov product CH₂BrCH₂CH₃ is the minor product in ionic conditions.

Acid-catalysed hydration (H₂O/H⁺): follows Markovnikov — OH adds to the more substituted carbon. Mechanism: H⁺ adds to less substituted carbon → carbocation → water attack → oxonium ion → proton loss → alcohol. Hydroboration-oxidation (BH₃/THF then H₂O₂/NaOH): gives anti-Markovnikov alcohol (OH adds to less substituted carbon) with syn stereochemistry (both B and H add to the same face). This contrast (acid hydration → Markovnikov; hydroboration → anti-Markovnikov, syn) is tested in every third session. Take a free alkene reactions mock. See our alkenes and alkynes guide.

Halogenation and Halohydrin Formation

Addition of Cl₂ or Br₂ in inert solvent (CCl₄): anti addition (the two halogens add to opposite faces of the double bond — anti stereochemistry). Mechanism: halogen molecule approaches π bond → forms cyclic halonium ion (bridged) → nucleophilic backside attack by X⁻ on the more substituted carbon of the halonium → anti product. This anti addition is diagnostic: meso compound from cis-alkene + Br₂; racemic mixture from trans-alkene + Br₂. JEE tests these stereochemical outcomes.

Halohydrin formation (Cl₂ or Br₂ in water): halonium ion intermediate + water (nucleophile) attacks the more substituted carbon → Markovnikov alcohol with halogen on the less substituted carbon, anti stereochemistry. Product: β-halohydrin. Test: Br₂/water vs Br₂/CCl₄ — the water changes the regiochemistry of nucleophile addition but not the anti stereochemistry of the halonium mechanism.

Free Radical Addition: Anti-Markovnikov

In the presence of peroxides (ROOR) or UV light, HBr adds to alkenes by a free-radical chain mechanism: Br• adds to the less substituted carbon (gives the more stable secondary radical) → Markovnikov radical → H• from HBr → anti-Markovnikov product. The peroxide effect applies only to HBr (not HCl or HI — bond dissociation energies make the chain process unfavourable for HCl and HI). No peroxide → ionic mechanism → Markovnikov product. Peroxide → radical mechanism → anti-Markovnikov product. This switch is the single most tested regiochemistry distinction in JEE alkene problems.

Ozonolysis: Cleave the Double Bond

Ozonolysis (O₃ then reductive workup with Zn/H₂O or Me₂S): cleaves the C=C bond completely, giving two carbonyl fragments. Each carbon of the double bond becomes a carbonyl group — if it had one H it gives an aldehyde; if it had no H it gives a ketone. Reductive workup: aldehydes are stable. Oxidative workup (H₂O₂): aldehydes are further oxidised to carboxylic acids; ketones remain. JEE question type: given the ozonolysis products, identify the starting alkene. Work backward: each carbonyl carbon was one end of the double bond. For the complete organic synthesis strategy see our organic synthesis guide and for reaction mechanisms see our reaction mechanisms guide.