Addition to Alkenes Overview

Introduction

An alkene contains a carbon–carbon sigma bond and a more exposed pi bond. The pi electrons can donate to an electrophile, making the double bond a useful starting point for building alcohols, halides, dihalides and other molecules. The phrase "addition to an alkene" describes a change in connectivity, but several mechanisms can produce it. Product orientation and stereochemistry reveal which electron path is likely.

Core explanation

In a typical addition, the alkene C=C loses its pi component and gains two new sigma bonds, one at each carbon. The carbon framework remains connected by a C–C single bond. One new group may be H, Br, OH or another substituent depending on reagent. A balanced structure must place one added fragment on each former double-bond carbon and account for all atoms and charges. A triple bond behaves differently and will be treated separately.

The alkene pi bond is relatively electron-rich and extends above and below the plane of the bonded carbons. It can therefore donate an electron pair to an electron-poor species. In acid addition, the electrophile may be H⁺ from a hydrogen halide or acid-catalysed water chemistry. The pi bond forms a C–H bond and leaves a carbocation on the other carbon; a nucleophile then captures it. The more stable possible carbocation often controls Markovnikov regiochemistry , placing H on the carbon that already has more hydrogens and the incoming nucleophile on the more substituted carbon. This familiar wording assumes a simple unsymmetrical alkene and a cationic mechanism.

Other additions do not pass through a free carbocation. Bromine or chlorine addition forms a three-membered halonium ion, then a nucleophile opens it from the opposite face, often giving anti addition . Hydroboration adds B–H across the alkene in a concerted syn step and, after oxidation, yields an alcohol with OH at the less substituted carbon. Catalytic hydrogenation transfers hydrogen atoms at a metal surface, typically giving syn addition. These pathways show that neither Markovnikov orientation nor free rotation is universal for all alkene additions.

Two separate questions must be asked about an unsymmetrical alkene. Regiochemistry: which alkene carbon receives which fragment? Stereochemistry: do the two new groups arrive from the same face (syn), opposite faces (anti), or as a mixture? A planar carbocation may be attacked from either face, whereas a bridged ion restricts attack. If the product has no new stereocentre, syn and anti language may not distinguish products even though the mechanism has a geometric preference.

Conditions matter as much as reagents. HBr normally adds through a polar electrophilic pathway, but radical initiation with peroxides can change it to an anti-Markovnikov radical chain reaction. Water in acid gives a different route from water with bromine, which makes a halohydrin. A reaction scheme must specify all relevant reagents and conditions before predicting product position. Rearrangements can occur in carbocation-forming additions but are not expected from a concerted hydroboration or a bridged halonium path.

Addition often competes with polymerisation or oxidation when reactive intermediates are present. In a synthesis problem, first identify the reagent class and likely intermediate, then draw the major product. Avoid applying one rule to every alkene reaction simply because all end by replacing C=C with C–C.

Step-by-step reasoning

Mark the two alkene carbons and count their attached carbon groups and hydrogens. Identify the reagent's electrophilic and nucleophilic parts. Decide whether the likely intermediate is a carbocation, halonium ion, radical, surface-bound species or concerted borane addition. Use that intermediate to place groups and predict possible rearrangement. Finally check syn/anti geometry, stereocentres, valence and atom balance.

Visual explanation

Draw a horizontal C=C with the pi cloud above and below. An electrophile approaches the cloud from above, receiving a curved arrow from the pi bond. Then split the drawing into several pathway panels: an open carbocation captured by nucleophile, a bridged bromonium opened from below, and concerted B–H addition from one face. The panels have the same starting alkene but different product rules.

Real-world analogy

An open two-seat bench can receive two new occupants, but how they sit depends on the entrance route. A doorway from one side may force opposite-side placement, while a shared vehicle may deliver both from one side. The alkene's two carbons are the seats; the mechanism is the entrance route. Knowing only that both seats end occupied does not reveal the route.

Real-world example

Ethene is converted on a large scale into useful materials through addition chemistry, including hydration to ethanol and halogenation to 1,2-dihalo compounds used as further feedstocks. In laboratory synthesis, a chemist may instead use hydroboration–oxidation of a terminal alkene to put OH on the terminal carbon, choosing a mechanism for its regioselectivity rather than merely seeking any alcohol.

Why?

Why is the pi bond the reactive part of an alkene? Its side-on overlap is weaker and more exposed than the C–C sigma bond, so its electron density can interact with an approaching electrophile while the sigma framework remains intact. Converting one pi bond into two new sigma bonds can be energetically favourable, but the activation barrier and product distribution depend on reagent and pathway.

Common misconception

"All additions to unsymmetrical alkenes obey Markovnikov's rule." That rule describes the common regiochemistry of certain polar additions through cationic character. Hydroboration–oxidation and peroxide-initiated HBr addition give anti-Markovnikov products by different mechanisms, while addition of identical halogens has no H-versus-X orientation question.

Worked example

Question: Compare the expected location of OH when propene is converted to an alcohol by (a) acid-catalysed hydration and (b) hydroboration followed by oxidation.

Reasoning: In (a), protonation that creates the more stable secondary carbocation allows water capture at C2. In (b), boron attaches to the less substituted C1 during concerted addition, and oxidation replaces B with OH.

Answer: Acid hydration gives mainly propan-2-ol; hydroboration–oxidation gives propan-1-ol. The two reagents use different pathways and regiochemical rules.

Quick check

1. Does bromine addition to an alkene normally require a free carbocation intermediate? Answer: No. It commonly proceeds through a bridged bromonium ion that is opened by bromide.

Exam focus

State reagent conditions before predicting an alkene addition product. Show where both incoming fragments end up and explain orientation using the actual intermediate. Distinguish regiochemistry from syn/anti stereochemistry. When a carbocation forms, check for rearrangement; when a halonium ion forms, expect backside opening and anti addition.

Advanced insight

The same alkene can be a nucleophile toward an electrophile and a substrate for radical or metal-catalysed pathways. Frontier-orbital interactions, solvent stabilisation and catalyst surfaces determine different transition states. Product patterns are therefore evidence about the pathway, but identical connectivity can sometimes arise from multiple mechanisms, making stereochemical or kinetic experiments valuable.

Summary

Alkene addition consumes the pi bond and places new sigma-bonded fragments on its two carbons. The pi electrons commonly attack an electrophile, but the resulting pathway may involve a carbocation, halonium ion, radical, metal surface or concerted borane transfer. Regiochemistry and stereochemistry depend on that mechanism. Identify all reagents and conditions before applying Markovnikov, anti-Markovnikov, syn or anti labels.

Practice questions

1. What happens to the alkene pi bond in a typical addition reaction? Answer: Its electrons are used to form new sigma bonds while the C–C sigma bond remains. 2. What two questions should be separated for an unsymmetrical alkene addition? Answer: Regiochemistry asks which carbon receives each group; stereochemistry asks the relative faces or spatial arrangement of addition. 3. Why can HBr and Br₂ give different kinds of intermediates? Answer: Polar HBr addition can protonate the alkene to a carbocation, while Br₂ commonly forms a bridged bromonium ion. 4. Which hydration route places OH at C1 of propene? Answer: Hydroboration–oxidation, because boron adds to the less substituted carbon and is then replaced by OH.