Last updated: 21 August 2026
Alkanes can seem chemically quiet because their molecules contain only relatively strong C–C and C–H σ bonds. Introduce a carbon–carbon double bond, however, and the behaviour changes dramatically. The second bond in C=C is a π bond whose electrons lie above and below the plane of the carbon atoms, making them much more accessible to approaching reagents.
This single structural feature explains most of the chemistry in this chapter. We will begin with ethene, CH₂=CH₂, and then extend the same ideas to substituted alkenes such as propene and but-2-ene. Once you can connect the structure of the π bond to electrophilic addition, stereochemistry and oxidation, alkene reactions become a set of related patterns rather than a list to memorise.
Try these five questions before reading the chapter. Use the structures and reaction schemes carefully — Organic Chemistry often becomes much easier once you learn to read what is happening to the molecule.
- 5 diagnostic questions
- Structures and reaction schemes where useful
- Instant explanation for every answer option
Structure and bonding in alkenes
Alkenes are hydrocarbons containing a carbon–carbon double bond. They are described as unsaturated because the C=C bond can undergo addition reactions in which new atoms or groups become attached to the two carbon atoms.
For an open-chain alkene containing one C=C bond, the general molecular formula is CₙH₂ₙ. Ethene, for example, has the formula C₂H₄, compared with C₂H₆ for the corresponding alkane, ethane.
The C=C double bond contains one σ bond and one π bond. The exposed, electron-rich π bond is the part that is usually broken during an addition reaction.
The σ bond and π bond
Each carbon atom involved in the double bond is approximately sp² hybridised. Three sp² orbitals are arranged in a trigonal planar geometry and form σ bonds. One unhybridised p orbital remains on each carbon atom.
These two parallel p orbitals overlap sideways. The resulting electron density lies above and below the plane containing the carbon atoms and forms the π bond.
σ bond: formed by head-on orbital overlap along the internuclear axis. Electron density is concentrated directly between the bonded nuclei.
π bond: formed by sideways overlap of parallel p orbitals. Its electron density lies above and below the plane of the σ-bond framework and is more exposed to approaching reagents.
Shape and bond angles around C=C
Because each double-bonded carbon is sp² hybridised, the arrangement around each carbon is approximately trigonal planar. Bond angles are therefore about 120°.
The two carbon atoms of the C=C bond and the atoms directly attached to them lie approximately in the same plane.
In ethene, each carbon forms three σ bonds: two C–H σ bonds and one C–C σ bond. The remaining unhybridised p orbital on each carbon overlaps sideways to form the π bond. Thus the complete C=C bond contains 1 σ + 1 π.
Restricted rotation and cis-trans isomerism
Rotation around an ordinary C–C single bond can occur without destroying the σ bond. A C=C double bond is different because the two p orbitals must remain parallel for sideways overlap to be maintained.
Rotating one end of a C=C bond relative to the other would destroy the sideways p-orbital overlap and therefore break the π bond. The energy required is too large for free rotation under ordinary conditions.
Why restricted rotation produces stereoisomers
Because the groups attached to the two double-bonded carbons are locked in position, different spatial arrangements can become distinct compounds. But restricted rotation alone is not enough: each carbon of the double bond must have two different substituents.
| Alkene | Cis-trans isomerism? | Reason |
|---|---|---|
| Propene | No | The terminal carbon of C=C has two H atoms. |
| But-1-ene | No | The terminal carbon of C=C has two H atoms. |
| But-2-ene | Yes | Each carbon of C=C has one H and one CH₃ group. |
| 2-methylpropene | No | One double-bonded carbon is attached to two identical CH₃ groups. |
Why alkenes undergo electrophilic addition
Most of the characteristic reactions of alkenes can be understood as electrophilic addition. An electrophile is an electron-deficient species that accepts an electron pair.
The π bond contains a region of relatively high electron density above and below the plane of the molecule. This makes it attractive to electrophiles. During an addition reaction, the π bond is broken and two new σ bonds are formed.
In an electrophilic addition mechanism, electron movement begins from the electron-rich π bond and moves towards an electrophile.
The general pattern
Although the precise intermediate depends on the reagent, the underlying idea is the same. An electrophilic part of the reagent is attracted to the π electrons. The π bond reacts, and the two atoms of the reagent ultimately add across the two carbons that originally formed C=C.
When drawing a curly-arrow mechanism, always ask: where are the electrons now, and where are they going? A curly arrow represents movement of an electron pair, so an arrow involving the alkene normally begins at the π bond rather than at a positively charged or electron-deficient atom.
Addition of bromine to alkenes
Bromine provides a useful example of electrophilic addition. As a Br₂ molecule approaches the electron-rich π bond, its electron cloud is polarised. A temporary dipole is produced, making the nearer bromine atom electron-deficient.
The reaction removes the C=C double bond because the π bond is replaced by two new C–Br σ bonds. This is why the reaction is classified as addition: the two reactant molecules combine to produce one organic product.
What happens in the mechanism?
The approaching Br₂ molecule becomes polarised as Brδ+–Brδ−. The π electrons are attracted towards the electron-deficient bromine atom. At the same time, the Br–Br bond undergoes heterolytic fission, producing Br⁻. Subsequent attack by Br⁻ completes addition across the original double bond.
In a curly-arrow diagram, the important examination habit is to show arrows beginning at electron pairs or bonds. The first arrow therefore starts from the C=C π bond.
Bromine as a test for unsaturation
Bromine is also used as a qualitative test for a carbon–carbon double bond. An alkene rapidly decolourises bromine at room temperature because the bromine is consumed in an addition reaction.
Alkene present: the orange-brown colour of bromine is decolourised.
Alkane under ordinary test conditions: no rapid decolourisation occurs because an alkane does not possess the electron-rich π bond required for electrophilic addition.
Addition of hydrogen halides and Markovnikov orientation
Hydrogen halides such as HCl, HBr and HI also add across C=C. With a symmetrical alkene such as ethene, there is only one possible structural product. With an unsymmetrical alkene such as propene, however, two orientations appear possible.
Why is one product favoured?
In the first stage of addition of HBr, the π bond attacks Hδ+. Protonation can occur in two directions. One direction produces a primary carbocation, whereas the other produces a secondary carbocation.
The pathway involving the more stable carbocation is favoured. Alkyl groups help stabilise the electron-deficient positively charged carbon, giving the general stability order:
For propene, the favoured pathway produces a secondary carbocation. Br⁻ then attacks this carbocation, giving 2-bromopropane as the major product.
Markovnikov's rule is a useful shortcut: during addition of HX to an unsymmetrical alkene, H usually becomes attached to the carbon that already carries more hydrogen atoms. The deeper reason is the relative stability of the possible carbocation intermediates.
Do not use the rule without understanding the mechanism
Markovnikov's rule should not be treated as an isolated fact to memorise. If you can identify the two possible carbocations and compare their stability, you can predict the major product even when the alkene looks unfamiliar.
For CH₃CH=CH₂ reacting with HBr, placing H on the terminal CH₂ carbon leaves the positive charge on the middle carbon. This gives a secondary carbocation, which is more stable than the alternative primary carbocation. Br⁻ therefore attacks the secondary carbocation to give CH₃CHBrCH₃ as the major product.
Hydrogenation of alkenes
Alkenes react with hydrogen in the presence of a metal catalyst. Hydrogen adds across the C=C bond, converting the alkene into an alkane.
This reaction is called hydrogenation or reduction. One hydrogen atom becomes bonded to each carbon that originally belonged to the C=C bond.
Hydration of alkenes with steam
An alkene can also be converted into an alcohol by addition of water across the double bond. Industrially, this is carried out using steam in the presence of a phosphoric acid catalyst at elevated temperature and pressure.
Once again, the overall pattern is addition across C=C. An H atom and an OH group become attached to the two carbons that originally formed the double bond.
Mild oxidation of alkenes to diols
Alkenes can be oxidised by cold, dilute alkaline potassium manganate(VII), KMnO₄. Under these mild conditions, two hydroxyl groups are introduced across the C=C bond to form a vicinal diol, meaning that the two –OH groups are on neighbouring carbon atoms.
In alkaline solution, the purple colour of manganate(VII) is replaced by the brown colour of manganese(IV) oxide, MnO₂. This provides another chemical test for unsaturation.
Recognising the change in the carbon skeleton
Under mild oxidation conditions, the carbon–carbon bond itself is not cleaved. The two carbon atoms that formed C=C remain bonded to one another. The key structural change is simply:
Cold, dilute oxidation keeps the carbon skeleton together and converts the C=C region into a diol.
Strong oxidation and oxidative cleavage
Hot acidified potassium manganate(VII) causes much more extensive oxidation. Instead of merely adding oxygen-containing groups across C=C, the double bond is cleaved. Each carbon of the original double bond becomes part of a new oxygen-containing product.
The product formed from each alkene carbon depends on how many hydrogen atoms were originally attached to that carbon.
To predict oxidative-cleavage products, inspect the two carbon atoms of C=C separately. Ask how many hydrogen atoms each one carries before oxidation.
| Carbon in the original C=C | Typical oxidation product | Structural idea |
|---|---|---|
| =CH₂ | CO₂ and H₂O | A terminal carbon carrying two H atoms is oxidised completely. |
| =CHR | Carboxylic acid, RCOOH | The alkene carbon carries one H atom. |
| =CR₂ | Ketone | The alkene carbon carries no H atoms. |
Example: oxidation of propene
Consider propene, CH₃CH=CH₂. The middle alkene carbon is =CHR and therefore gives a carboxylic acid. The terminal alkene carbon is =CH₂ and is oxidised completely to carbon dioxide.
Example: a ketone and a carboxylic acid
In 2-methylbut-2-ene, one carbon of the double bond carries no hydrogen atoms and therefore forms a ketone. The other carries one hydrogen atom and therefore forms a carboxylic acid.
Using oxidation products to identify an unknown alkene
Oxidative cleavage can be used in reverse. If the products are known, identify the carbonyl carbon of each product, remove the oxygen-containing part conceptually, and reconnect those two carbons with a C=C bond.
Suppose strong oxidation produces propanone and ethanoic acid. Propanone tells you that one alkene carbon originally carried two CH₃ groups and no H. Ethanoic acid tells you that the other alkene carbon originally carried one CH₃ group and one H. Reconnecting those carbons gives (CH₃)₂C=CHCH₃, which is 2-methylbut-2-ene.
Reaction summary for alkenes
The individual reactions are much easier to remember when they are all viewed as consequences of the C=C bond. Most reactions replace the π bond by new σ bonds, while strong oxidation goes further and breaks the carbon–carbon framework at the double bond.
| Reagent and conditions | Type of reaction | Main organic result |
|---|---|---|
| H₂, Ni catalyst, heat | Hydrogenation / reduction | Alkane |
| Br₂, room temperature | Electrophilic addition | Dibromoalkane |
| HCl, HBr or HI | Electrophilic addition | Halogenoalkane |
| Steam, H₃PO₄ catalyst, high temperature and pressure | Hydration | Alcohol |
| Cold, dilute alkaline KMnO₄ | Mild oxidation | Vicinal diol |
| Hot, acidified KMnO₄ | Strong oxidation | Oxidative cleavage products |
Do not memorise the reactions as six unrelated equations. Start by locating C=C, identify the reagent, and then decide what happens to the π bond.
Common Mistakes
| Common mistake | Why it is wrong | Better approach |
|---|---|---|
| Describing C=C as two σ bonds. | A double bond consists of one σ bond from head-on overlap and one π bond from sideways p-orbital overlap. | Remember C=C = 1 σ + 1 π, and connect the π bond to alkene reactivity. |
| Assuming every alkene can exhibit cis-trans isomerism. | Restricted rotation is not enough. Each carbon of C=C must also have two different groups attached. | Inspect the substituents on each double-bonded carbon separately before deciding whether cis-trans isomerism is possible. |
| Applying Markovnikov's rule without considering carbocation stability. | The rule is a consequence of the preferred reaction pathway, not an arbitrary numbering rule. | Draw or imagine both possible carbocations and identify which is more stable. |
| Confusing mild and strong oxidation by KMnO₄. | Cold, dilute oxidation forms a diol without breaking the carbon skeleton, whereas hot strong oxidation cleaves C=C. | Check the reaction conditions before predicting the product. |
What you should be able to do
Explain alkene bonding
Describe a C=C bond as one σ bond and one π bond and explain how the orbitals overlap.
Explain alkene shape
Relate sp² hybridisation to trigonal planar geometry and bond angles of approximately 120°.
Recognise cis-trans isomerism
Decide whether an alkene can form cis-trans isomers by examining the groups attached to each carbon of C=C.
Explain electrophilic addition
Use the electron-rich π bond to explain reactions with bromine and hydrogen halides.
Predict alkene reactions
Recall appropriate reagents and conditions for hydrogenation, hydration, halogenation and mild oxidation.
Interpret oxidative cleavage
Predict strong-oxidation products and work backwards from those products to determine the original alkene.
These ten questions test whether you can apply the chapter to structures, reactions and unfamiliar examples.
- 10 mastery questions
- Structures and reaction schemes where useful
- Instant explanation for every answer option
Frequently asked questions
Common Questions About Alkenes
The same few ideas — π bonding, restricted rotation and electron-rich C=C bonds — explain many questions that initially seem unrelated.
Why are alkenes more reactive than alkanes?
Alkenes contain a π bond in addition to the C–C σ bond. The π electron cloud lies above and below the molecular plane and is more exposed to approaching reagents. Electrophiles can therefore interact with these electrons relatively easily. Alkanes contain only σ bonds and do not possess the same electron-rich region.
Why can the carbon atoms of a C=C bond not rotate freely?
The π bond is formed by sideways overlap of two parallel p orbitals. Rotation about the carbon–carbon axis would make those orbitals cease to be parallel and would destroy their overlap. Free rotation would therefore require the π bond to be broken, which requires substantial energy.
Does every alkene show cis-trans isomerism?
No. Although every ordinary C=C bond has restricted rotation, cis-trans isomerism is possible only when each carbon of the double bond is attached to two different groups. Propene therefore does not show cis-trans isomerism because its terminal carbon is bonded to two identical hydrogen atoms.
Why is 2-bromopropane the major product when HBr reacts with propene?
Protonation of propene can in principle produce either a primary or a secondary carbocation. The secondary carbocation is more stable, so the pathway leading through it is favoured. Br⁻ then attacks that carbocation, producing 2-bromopropane as the major product. Markovnikov's rule is a useful description of this outcome, but carbocation stability explains why it occurs.
How can I predict the products of strong oxidation of an alkene?
First locate the C=C bond and examine each of its carbon atoms separately. A double-bonded carbon with no H normally gives a ketone. A double-bonded carbon with one H gives a carboxylic acid. A terminal =CH₂ carbon is oxidised completely to CO₂ and H₂O. Treating the two alkene carbons independently makes even unfamiliar oxidative-cleavage questions much easier.
Alkene chemistry becomes much more manageable once you stop seeing each reaction as a separate fact. Start with the C=C bond: its π electrons explain electrophilic addition, its orbital arrangement explains restricted rotation, and its position tells you what happens during oxidation. Follow the electrons and the carbon skeleton, and most alkene questions become predictable rather than memorised.