An alkene has the opposite problem to a halogenoalkane. There is too much electron density in the C=C, not too little — so instead of attracting things that carry spare electrons, it attracts things that are short of them.
📚 What you need to know
An electrophile is an electron-deficient species that accepts a pair of electrons to form a covalent bond.
Electrophiles carry a full or partial positive charge: H+, NO2+, R+, and neutral ones such as HX, X2 and H2O.
The C=C double bond is one σ bond plus one weaker π bond, and the π electrons sit outside the plane of the molecule.
In electrophilic addition the π bond breaks and two new σ bonds form — one to each carbon.
Hydration: steam at about 300 °C and 60 atm with a H3PO4 or H2SO4 catalyst gives an alcohol.
Halogenation: X2 at room temperature gives a dihalogenoalkane. Bromine water going from orange to colourless is the test for unsaturation.
Hydrohalogenation: HX at room temperature gives a halogenoalkane, with rate HI > HBr > HCl.
What an electrophile is
Straight mirror image of a nucleophile. A nucleophile has a pair to give away; an electrophile has a gap to fill. It forms its bond by accepting a pair, which is why it is drawn to regions of high electron density.
Positively charged electrophiles
Neutral electrophiles
H+ hydrogen ion
HX hydrogen halides
NO2+ nitronium ion
X2 halogens
NO+ nitrosonium ion
H2O water
R+ carbocations
RX halogenoalkanes
A neutral molecule can be an electrophile because of polarity, not charge. In H–Br the hydrogen is δ+, and that is enough. Notice that HX and RX appear on this list and on the nucleophile-target list from the substitution page — the same polar bond makes one end an electrophile and gives the other end a leaving group.
Why the C=C is a target
A double bond is not two identical bonds. The first is a σ bond along the line joining the carbons, strong and buried between them. The second is a π bond, formed sideways, with its electron density spread above and below the plane of the molecule.
That matters for two reasons. The π electrons are exposed, sitting on the outside of the molecule where an approaching species meets them first; and the π bond is weaker than a σ bond, so it is the one that gives way.
The bond angle of 120° is a consequence of the same picture: three regions of electron density round each carbon, pushed as far apart as a flat arrangement allows.
What “addition” means here
Nothing leaves. The whole of the attacking molecule ends up in the product, split between the two carbons that used to be double bonded. That is the difference between addition and substitution, and it is worth stating in exactly those terms.
Alkanes cannot do this. They have no π bond to break and no region of high electron density, so an electrophile has nothing to attack.
The three reactions to know
Addition of steam (hydration)
Pass ethene and steam over an acid catalyst under pressure and water adds across the double bond to give ethanol. This is how industrial ethanol is made, and it is faster and more efficient than fermentation.
Hydration of ethene
CH2=CH2 + H2O → CH3CH2OH 300 °C, 60 atm, H3PO4 catalyst
The mechanism goes through an intermediate in which H+ and HSO4– add across the double bond. Water then hydrolyses that intermediate, releasing the alcohol and regenerating the acid — which is exactly why sulfuric acid counts as a catalyst here rather than a reactant.
Addition of halogens (halogenation)
No heat, no catalyst, no pressure. Shake an alkene with a halogen at room temperature and the π bond breaks, giving a dihalogenoalkane with one halogen on each carbon.
Because the reaction is so easy, it makes a perfect test. Bromine water is orange; the dibromo product is colourless. Shake an unknown with bromine water and watch what happens to the colour.
Say “decolourised”, not “clear”. Bromine water is already clear in the sense of see-through — what changes is the colour, and examiners mark the distinction.
Addition of hydrogen halides (hydrohalogenation)
Also room temperature, also rapid. The H–X bond is polar, so the δ+ hydrogen is the electrophilic end, and the product is a halogenoalkane.
Hydrohalogenation of ethene
CH2=CH2 + HBr → CH3CH2Br rate: HI > HBr > HCl
That rate order should look familiar. It is the same argument as the halogenoalkanes on the substitution page — the weaker H–X bond breaks more readily, so HI reacts fastest. Once you have the “weakest bond breaks first” idea, it pays for itself across the whole topic.
Reagent
Conditions
Product type
Example product
H2O as steam
300 °C, 60 atm, H3PO4
alcohol
ethanol
X2, e.g. Br2
room temperature
dihalogenoalkane
1,2-dibromoethane
HX, e.g. HBr
room temperature
halogenoalkane
bromoethane
WORKED EXAMPLE
Propene is shaken with bromine at room temperature. State the type of reaction, write the equation and name the product.
Type of reactionBromine is electron-deficient once it is polarised by the π cloud, and nothing leaves the alkene.electrophilic additionThe equationCH₃CH=CH₂ + Br₂ → CH₃CHBrCH₂BrThe nameOne bromine on each of the two carbons that were double bonded — carbons 1 and 2.1,2-dibromopropaneCheck the atoms balance: C₃H₆ + Br₂ gives C₃H₆Br₂. Nothing is left over, which is the signature of an addition.
WORKED EXAMPLE
Two unlabelled samples are hexane and hex-1-ene. Describe a test that distinguishes them and state the observation for each.
The testShake each sample with bromine water at room temperature.Hex-1-eneIt contains a C=C, so electrophilic addition occurs and a colourless dibromo compound forms.orange to colourlessHexaneSaturated, no π bond, so there is no reaction under these conditions.stays orangeGive the observation for both samples. A test with only one stated outcome does not distinguish anything.
WORKED EXAMPLE
Explain why alkenes undergo addition reactions but alkanes do not, and why HI adds to ethene faster than HCl does.
Alkenes against alkanesAn alkene has a π bond above and below the plane: a region of high electron density that attracts electrophiles. It is also weaker than a σ bond, so it can break and be replaced by two stronger σ bonds.alkanes are saturated, with no π bond to breakHI against HClBoth add the same way, so the difference is how easily the H–X bond breaks. H–I is the longer, weaker bond.the weaker H–I bond breaks more readily, so HI reacts fastestFull order: HI > HBr > HCl.
💡 Exam tip
Define an electrophile as an electron-deficient species that accepts a pair of electrons.
Say the C=C is a region of high electron density — that phrase is what earns the mark.
State that the π bond breaks and two σ bonds form, so the product is saturated.
Learn the hydration conditions properly: 300 °C, 60 atm, phosphoric or sulfuric acid catalyst.
For the bromine water test, always give both observations and use the word decolourised.
Name products with locants: 1,2-dibromoethane, not just “dibromoethane”.
⚠️ Common mix-up
Calling it nucleophilic addition. The alkene supplies the electrons, so the attacking species is the electrophile.
Putting both halogens on the same carbon. One goes to each of the two carbons of the old double bond.
Saying the solution goes “clear”. It goes colourless; it was clear all along.
Breaking the σ bond instead of the π bond. The σ bond survives and becomes the C–C single bond.
Writing a leaving group into an addition equation. Nothing leaves — that is what makes it addition.
That closes the electron-pair sharing chain: a nucleophile giving a pair to an electron-poor carbon, a bond breaking heterolytically to create those two species in the first place, and an electrophile taking a pair from a π bond. Every mechanism in this topic is one of those three moves, or a combination of them.
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