Difficulty: Intermediate | Prerequisites: Basic alkene structure, Lewis acids/bases, arrow-pushing notation
Alkenes are the first major functional group where you move from "identify the structure" to "predict what happens." These three reactions, halogenation, halohydrin formation, and hydrogenation, are the entry point to addition chemistry. Each one adds atoms across the C=C double bond, but they differ in what gets added, from which face, and through what intermediate. If you can map out the mechanism for each, you have the tools to predict products and stereochemistry for most of the addition reactions in the course. You should already be comfortable drawing carbocations, recognising nucleophiles and electrophiles, and pushing curved arrows.
Alkenes react as nucleophiles, donating their pi electrons to electrophiles. Halogenation adds two halogen atoms anti to each other through a cyclic halonium ion. Halohydrin formation is the same mechanism but with water as the nucleophile, so you get a halogen and an OH on opposite faces. Hydrogenation delivers two hydrogens to the same face (syn addition) using a metal catalyst.
Alkene
A hydrocarbon containing at least one carbon-carbon double bond (C=C). The double bond consists of one sigma bond and one pi bond. Think of it as the reactive handle of the molecule: the pi electrons sit above and below the plane and are available to donate to electrophiles.
Electrophilic addition
The general reaction type for alkenes: an electrophile attacks the electron-rich pi bond, forming a new sigma bond, followed by a nucleophile completing the addition. In simple terms, this means the double bond breaks open and two new groups attach.
Halonium ion (bromonium ion)
A three-membered cyclic intermediate formed when a halogen (e.g. Br2) reacts with an alkene. The halogen bridges both carbons of the former double bond, carrying a positive charge. Think of it as a tiny, strained ring that forces the incoming nucleophile to attack from the opposite face.
Anti addition
An addition where the two new groups end up on opposite faces of the original double bond. In simple terms, one group comes from above the plane and the other from below.
Syn addition
An addition where both new groups attach to the same face of the double bond. Think of it as both atoms landing on the same side of the molecule at once.
Halohydrin
The product of adding a halogen and a hydroxyl group (OH) across a double bond. The halogen and OH end up anti to each other. In simple terms, it is a molecule that has both a halogen and an alcohol group on adjacent carbons.
Markovnikov regioselectivity
In additions to unsymmetrical alkenes, the nucleophile (e.g. OH) adds to the more substituted carbon, and the electrophilic atom (e.g. Br) adds to the less substituted carbon. Think of it as "the richer carbon gets richer" in terms of substituents.
Hydrogenation
The addition of molecular hydrogen (H2) across a double bond using a metal catalyst such as Pt, Pd, or Ni. This is a syn addition. In simple terms, you are saturating the alkene by converting the double bond into a single bond.
Catalytic surface (heterogeneous catalysis)
The metal catalyst (Pt, Pd, Ni) provides a surface on which both the alkene and H2 adsorb. Because both sit on the same face of the metal, both hydrogens are delivered from the same side. Think of it as both the alkene and the hydrogen lying flat on a table, so the hydrogens can only land on the top face.
Reagents: Br2 (or Cl2), no solvent nucleophile needed.
Product: A vicinal dihalide, with the two halogen atoms on adjacent carbons, anti to each other.
Mechanism (two steps):
The pi electrons of the alkene attack Br2, displacing bromide (Br-) and forming a cyclic bromonium ion. This is the electrophilic attack / pi-transfer (PT) step. The bromonium ion is a three-membered ring with a positive charge on bromine.
The displaced Br- acts as a nucleophile and attacks one of the carbons of the bromonium ion from the back side (anti to the bridging bromine). This opens the ring and gives the vicinal dibromide.
Stereochemistry: Anti addition. Because the bromonium ion blocks one face, the nucleophilic bromide must attack from the opposite face. On a cyclohexane ring, this results in a trans-diaxial product.
Key details:
The reaction works with Br2 or Cl2. Fluorine is too reactive (explosive), iodine is too unreactive.
No carbocation rearrangement occurs because the intermediate is a bromonium ion, not an open carbocation.
Br2 in CCl4 is a classic "test for unsaturation": the brown Br2 colour disappears when an alkene is present.
Reagents: Br2 (or Cl2) in H2O.
Product: A halohydrin, with OH and the halogen on adjacent carbons, anti to each other.
Mechanism (three steps):
Pi electrons of the alkene attack Br2, forming the cyclic bromonium ion and releasing Br-. (Same first step as halogenation.)
Water (the solvent, present in large excess) acts as the nucleophile instead of Br-. Water attacks the more substituted carbon of the bromonium ion from the anti face. This is because the more substituted carbon bears more of the positive charge and is more electrophilic.
A proton transfer (PT) step deprotonates the oxonium ion to give the neutral halohydrin product.
Regioselectivity:
The halogen ends up on the less substituted carbon.
The OH ends up on the more substituted carbon.
This follows from the bromonium ion's charge distribution: the more substituted carbon is better able to stabilise positive charge, so the nucleophile (water) preferentially attacks there.
Stereochemistry: Anti addition, for the same reason as halogenation. The bromonium ion blocks one face, so water attacks from the opposite side.
Key details:
The difference between halogenation and halohydrin formation is simply the solvent. In an inert solvent, Br- is the nucleophile (giving a dihalide). In water, H2O is the nucleophile (giving a halohydrin).
If the alkene is symmetrical, there is no regioselectivity question, as both carbons are equivalent.
Reagents: H2 with a metal catalyst (Pt, Pd, or Ni).
Product: An alkane. The double bond is fully reduced to a single bond.
Mechanism:
Both the alkene and H2 adsorb onto the surface of the metal catalyst.
The H-H bond breaks on the metal surface, generating metal-bound hydrogen atoms.
Both hydrogen atoms are delivered to the same face of the alkene (the face touching the catalyst surface).
The product desorbs from the catalyst surface as an alkane.
Stereochemistry: Syn addition. Both hydrogens are delivered from the same face because they come from the catalyst surface. On a disubstituted cycloalkene, this produces a cis product.
Key details:
This is a heterogeneous catalysis reaction: the catalyst is a solid and the reactants are in solution or gas phase.
The reaction is exothermic. The heat released (heat of hydrogenation) can be used to compare the relative stability of alkene isomers: a more stable alkene releases less heat upon hydrogenation.
No carbocation intermediate is formed, so there is no possibility of rearrangement.
The catalyst is not consumed, it is regenerated after each cycle.
Reaction | Reagents | Product | Stereochemistry |
|---|---|---|---|
Halogenation | Br2 (or Cl2) | Vicinal dihalide | Anti addition |
Halohydrin | Br2 / H2O | Halohydrin (Br + OH) | Anti addition |
Hydrogenation | H2, Pt (or Pd, Ni) | Alkane | Syn addition |
Halogenation is used industrially to produce intermediates for pharmaceuticals, pesticides, and flame retardants. Ethylene dibromide, made by bromination of ethylene, was historically added to leaded petrol as a lead scavenger.
Hydrogenation is the basis of food manufacturing processes: margarine is made by partially hydrogenating vegetable oils (converting some C=C double bonds to single bonds), which raises the melting point and turns a liquid oil into a solid spread.
Students often think the bromonium ion is an open carbocation. It is not. The bromine bridges both carbons, which is precisely why anti addition is enforced and why carbocation rearrangements do not occur in halogenation.
Students confuse which atom goes where in halohydrin formation. The halogen goes to the less substituted carbon. The OH goes to the more substituted carbon. The nucleophile attacks the carbon bearing more positive charge, which is the more substituted one.
Students sometimes assume hydrogenation gives anti addition because "the two hydrogens come from two different H atoms in H2." Both hydrogens are delivered from the same catalyst surface, so it is syn.
Students mix up the solvent's role: Br2 alone gives a dihalide, Br2 in water gives a halohydrin. The solvent changes the nucleophile, not the electrophile.
⚠️ You will almost certainly be asked to draw the bromonium ion intermediate and show why the product is anti.
⚠️ Predicting the regiochemistry of halohydrin formation (which carbon gets OH, which gets Br) is a very common exam question.
⚠️ Be prepared to compare syn vs. anti addition across multiple reaction types. A table or flowchart is useful revision here.
⚠️ Hydrogenation stereochemistry (syn addition on a ring) is frequently tested with cyclohexene substrates: you need to draw both hydrogens on the same face.
True or false: Halogenation of an alkene proceeds through an open carbocation intermediate.
False. It proceeds through a cyclic halonium (bromonium) ion.
In halohydrin formation with Br2/H2O, the OH group ends up on the ______ substituted carbon.
More substituted.
True or false: Hydrogenation is an example of anti addition.
False. Hydrogenation is syn addition.
What change in solvent converts a halogenation into a halohydrin formation?
Using water as the solvent instead of an inert solvent like CCl4.
True or false: Carbocation rearrangements can occur during bromination of an alkene.
False. The bromonium ion intermediate prevents rearrangement.
Q: Draw the product of the reaction of cyclohexene with Br2. What is the stereochemistry and why?
A: The product is trans-1,2-dibromocyclohexane. The stereochemistry is anti because the bromonium ion intermediate blocks one face of the ring, forcing the nucleophilic Br- to attack from the opposite face.
Q: When 2-methylpropene reacts with Br2/H2O, which carbon gets the OH and which gets the Br? Explain.
A: The OH goes to the more substituted carbon (the one bearing the methyl groups) and the Br goes to the less substituted carbon. Water, the nucleophile, preferentially attacks the carbon of the bromonium ion that bears more positive charge, which is the more substituted carbon.
Q: Explain why hydrogenation is a syn addition.
A: Both the alkene and H2 adsorb onto the same face of the metal catalyst surface. Because both hydrogen atoms are delivered from the catalyst surface, they must add to the same face of the double bond.
Q: A student claims that halohydrin formation follows Markovnikov's rule. Is this correct?
A: It depends on perspective. In Markovnikov's original sense (H goes to the carbon with more H's), the rule was stated for HX additions. For halohydrin formation, the result is consistent in that the nucleophile (OH) ends up on the more substituted carbon, but the reasoning is about charge distribution in the bromonium ion rather than carbocation stability in the classical Markovnikov sense.
Q: What product do you expect from hydrogenation of 1-methylcyclohexene with H2/Pt?
A: cis-1-Methylcyclohexane (the methyl and the new H on C1 are on the same face, and the H on C2 is also delivered from the same face as the H on C1). The product is methylcyclohexane with both new hydrogens cis to each other.
These three reactions connect directly to the later alkene reactions (syn-dihydroxylation, anti-dihydroxylation, ozonolysis, hydroboration-oxidation) covered in Part 2 of these notes. The syn vs. anti framework you build here carries through every addition reaction in the course.
Halohydrin formation also connects to epoxide chemistry: treating a halohydrin with base gives an epoxide via intramolecular SN2, bridging alkene reactions to the substitution and ring-opening material.
Hydrogenation connects to thermodynamics and stability: heats of hydrogenation are used to compare alkene stability (more substituted alkenes are more stable and release less heat).
alkene addition reactions, halogenation mechanism, bromonium ion, halonium ion, anti addition, syn addition, halohydrin formation, Markovnikov regioselectivity, catalytic hydrogenation, vicinal dihalide, Br2 addition to alkenes, Br2/H2O, electrophilic addition, pi bond reactivity, cyclohexene bromination, trans-dihalide, heterogeneous catalysis, Pt Pd Ni catalyst, heat of hydrogenation, organic chemistry study notes