Electrophilic Addition to Alkenes

Introduction

The alkene π electron cloud is exposed above and below its sigma framework and can interact with an electron-pair acceptor. In electrophilic addition, the π component is consumed while new sigma bonds attach groups across the former C=C. Understanding this broad pattern helps interpret hydrogen-halide, halogen, and hydration reactions without treating them as identical mechanisms.

Core explanation

An ordinary C=C double bond has one strong sigma framework and one pi component. In a polar addition such as HBr addition, pi electrons form a bond to electrophilic hydrogen. The H–Br bond breaks heterolytically, and a carbocation can form on the other alkene carbon. Bromide then donates an electron pair to that cation, making the second new sigma bond. The net reaction adds H and Br to the two former double-bond carbons. Both atoms remain in the product, unlike substitution where one group replaces another.

Electron-flow arrows must begin at an electron source. In the first step, a two-electron curved arrow goes from the π bond toward H, not from H toward the alkene. A second arrow shows the H–Br electron pair moving to Br. In the next step, bromide's lone pair points to the carbocation carbon. These diagrams describe a polar mechanism. Other alkene additions have different intermediates: halogen addition often involves a bridged halonium ion rather than a freely rotating carbocation, and hydrogenation involves a catalyst surface. Do not force every addition into the HBr template.

The regiochemistry of addition to an unsymmetrical alkene depends on which atom attaches to which double-bond carbon. Carbocation stability often helps predict HBr or HCl orientation under normal polar conditions. Stereochemistry also matters: a planar carbocation may be attacked from either face, whereas a bridged intermediate can constrain the relative positions of added groups. Product drawings must therefore specify when a stereochemical conclusion is justified by the mechanism.

Addition generally reduces the number of π bonds and raises the number of sigma bonds. This can release energy, but it does not mean every electrophile reacts rapidly with every alkene. Substituents, solvent, temperature, concentration, and competing mechanisms affect rate and outcome. A reliable problem solution identifies reagent and conditions, marks the original C=C carbons, and tracks atoms into the product. A conservation check catches missing hydrogen or halogen atoms.

Step-by-step reasoning

1. Locate the electron-rich alkene π bond and incoming electrophile. 2. Draw the first new bond using a curved arrow from electrons. 3. Identify the resulting intermediate under the stated reagent conditions. 4. Complete capture by a nucleophile and verify all atoms in the product.

Visual explanation

Draw C=C with a cloud above it and an incoming H–Br molecule. Show H bonding to one carbon, a temporary cation on the other, and Br⁻ attaching there.

Real-world analogy

A flexible two-part connector can redirect one of its links to grasp an incoming object, leaving a second attachment point to be filled. The molecular mechanism is governed by electrons, not mechanical hooks.

Real-world example

Addition of HBr to ethene produces bromoethane under suitable conditions. This transforms a π-bond-containing feedstock into a saturated haloalkane with new C–H and C–Br bonds.

Why?

Why can the π bond act as an electron donor? Its electron density extends beyond the carbon-carbon axis and can form a new sigma bond to an electron-deficient reagent.

Common misconception

“The C=C bond disappears completely during addition.” Its pi component is consumed, but the carbon-carbon sigma bond remains as the product's C–C single bond.

Worked example

Predict HCl addition to ethene. The π bond bonds to H from HCl, and the H–Cl bond supplies Cl⁻. The initially symmetric ethene gives no regioisomer choice. Chloride attaches to the other former double-bond carbon, yielding CH₃CH₂Cl. Count atoms: C₂H₄ + HCl gives C₂H₅Cl. The carbon-carbon connection remains, now as a single bond.

Quick check

1. Which alkene bond component is consumed in ordinary addition? Answer: The pi component; the carbon-carbon sigma connection remains.

Exam focus

Start curved arrows at electron pairs. Identify whether the specific reagent follows a carbocation, halonium, radical, or catalyst-surface pathway before predicting stereochemistry.

Advanced insight

Reaction-coordinate diagrams can show a high-energy intermediate between two transition states for a stepwise carbocation pathway. A concerted pathway instead has no isolable intermediate along that coordinate.

Summary

Electrophilic addition uses alkene π electrons to form new sigma bonds across C=C. Reagent identity and reaction conditions determine the intermediate, regioselectivity, and possible stereochemistry of products.

Practice questions

1. Does HBr addition retain both H and Br in the organic product? Answer: Yes, one adds to each former double-bond carbon in the simple case. 2. Where should a curved arrow start for attack on H⁺? Answer: At the π-bond electron pair. 3. Must Br₂ addition use the same free-carbocation pathway as HBr addition? Answer: No. It commonly proceeds through a bridged bromonium intermediate.