Chlorohydrin Formation of Alkenes with Cl₂ and H₂O (Anti Addition via Chloronium Ion)

Alkenes treated with chlorine in water give chlorohydrins. The alkene polarizes Cl₂, forming a three-membered chloronium ion and Cl⁻. Water attacks the more substituted carbon from the backside (anti), delivering Markovnikov OH and placing Cl on the less substituted carbon. Deprotonation furnishes the neutral chlorohydrin. Because no free carbocation forms, rearrangements do not occur and stereochemical outcomes are predictable.

Introduction

Chlorohydrin formation is an electrophilic addition that proceeds through a bridged chloronium ion. The π bond donates to one chlorine of Cl₂ as the Cl–Cl σ bond heterolyzes to Cl⁻. Simultaneously, the bound chlorine donates a lone pair into the adjacent carbon, closing the three-membered ring. Water, present in large excess, attacks anti at the more substituted carbon where the bridged cation is better stabilized, giving Markovnikov OH placement. Proton transfer to solvent delivers the neutral anti chlorohydrin.

Quick Summary

• Reagents: Cl₂/H₂O (Cl₂ + H₂O ⇌ HOCl + HCl); practical lab alternatives include NaOCl (HOCl) or N-chlorosuccinimide (NCS) in water.
• Outcome: Anti addition of Cl and OH; OH installs at the more substituted carbon (Markovnikov), Cl on the less substituted carbon.
• Mechanism: Chloronium ion formation → H₂O anti attack at the more substituted carbon (SN2-like) → deprotonation to the chlorohydrin.
• Stereochemistry: Anti; when two stereocenters form, products are enantiomeric pairs (no meso because Cl ≠ OH).
• Rearrangements: None; the bridged intermediate prevents carbocation shifts.

Mechanism (Chlorohydrin Formation)

The alkene π bond attacks one chlorine atom as the Cl–Cl σ bond shifts to the distal chlorine, forming Cl⁻. The bound chlorine donates a lone pair to the adjacent carbon, closing the bridged chloronium ion.

Water approaches anti to the bridge and attacks the more substituted carbon (greater positive character), opening the chloronium to give an oxonium intermediate.

A base (H₂O or Cl⁻) removes the proton from oxygen, giving the neutral chlorohydrin with Cl and OH anti across the former double bond.

The final product places OH on the more substituted carbon and Cl on the less substituted carbon with an anti relationship.

Mechanistic Checklist (Exam Focus)

• Draw the chloronium ion explicitly; avoid free carbocations.
• Show water attacking anti at the more substituted carbon and the bridge opening.
• Include the deprotonation that neutralizes the oxonium.
• Emphasize Markovnikov OH placement and anti stereochemistry in the product sketch.

Worked Example — Chlorohydrin from Styrene

• Substrate: Styrene (C=CC1=CC=CC=C1).
• Reagents: Cl₂ (aq), 0–25 °C, subdued light.
• Pathway: Chloronium formation across the vinyl–aryl C=C; water attacks anti at the benzylic carbon (more substituted), then deprotonation.
• Outcome: 2-chloro-1-phenylethanol (chlorine on the terminal carbon; benzylic OH), formed as enantiomeric pairs when both carbons become stereocenters.

When Multiple Alkenes Are Present

Reaction is fastest at the double bond that forms the more stabilized chloronium (greater substitution, allylic/benzylic stabilization) and that is more accessible to backside attack by H₂O. If candidate alkenes are similar in electronics and sterics, mixtures can form.

Practical Tips & Pitfalls

• Medium: Maintain aqueous conditions; without sufficient water, the reaction furnishes vicinal dichlorides via Cl⁻ opening.
• Light & temperature: Subdued light and 0–25 °C minimize radical pathways and side reactions.
• Alternatives: NaOCl (HOCl) or NCS/H₂O can replace Cl₂/H₂O in many laboratory settings.
• Quench: Destroy excess halogen with aqueous Na₂S₂O₃; handle HCl/HOCl with proper PPE.
• Common errors: Omitting the chloronium, drawing syn addition, or placing OH on the less substituted carbon.

Exam-Style Summary

• Cl₂/H₂O adds anti via a chloronium intermediate.
• OH installs at the more substituted carbon (Markovnikov); Cl at the less substituted carbon.
• Products are anti; enantiomeric pairs result when two stereocenters form.
• No rearrangements (no free carbocation).

Interactive Toolbox

• Practice halohydrin additions in the Reaction Solver and compare chlorohydrin vs. bromohydrin outcomes.
• Use the Mechanism Solver to export the stepwise curved-arrow diagrams for flashcards.

FAQ / Exam Notes

• Why does water attack the more substituted carbon? That carbon bears greater positive character in the chloronium; backside attack there is favored.
• Can other nucleophiles participate? Yes—ROH, Cl⁻, carboxylates, etc., open the chloronium to give anti products bearing the new nucleophile. For a closely related reaction, see the bromohydrin formation guide.
• What prevents rearrangements? The bridged intermediate keeps both carbons engaged; no free carbocation forms.