Every alkene contains at least one C=C double bond — this is the functional group that gives alkenes their characteristic chemistry. Alkanes (single C-C bonds only) are relatively unreactive because their bonds are strong and non-polar. Alkenes, on the other hand, are reactive precisely because of that double bond.

Think of the double bond like a loaded spring sitting inside the molecule — full of energy (electron density) and just waiting for something to come along and react with it.

They're called unsaturated hydrocarbons because:

• They contain carbon-carbon double bonds (not "saturated" with the maximum possible hydrogens)
• They're made of hydrogen and carbon atoms only

Naming Alkenes

Alkenes are named with the suffix -ene (alk + ene). For chains of 4 or more carbons, you must number the chain (starting from the end closest to the double bond) and state the position of the double bond using the lower-numbered carbon involved.

This is the concept that everything else in the chapter hangs off, so let's really slow down here.

Each carbon atom has 4 outer-shell electrons (configuration 1s²2s²2p²). To get a full outer shell, carbon needs to form 4 covalent bonds. When two carbon atoms form a double bond, they still only share 4 electrons total between them across that double bond — but those 4 electrons form two different kinds of bond, not one strong bond and one weak one that are somehow "the same."

σ (Sigma) Bonds — the "head-on" bond

A sigma bond forms when two atomic orbitals overlap end-to-end, directly along the line joining the two nuclei. Picture two hands clasping palm-to-palm — that's the overlap. The electron density sits symmetrically between the two nuclei, and this direct overlap is what makes σ bonds strong.

Every single covalent bond you've met before (C-H, C-C in alkanes, etc.) is a σ bond. In ethene, each carbon uses three of its four electrons to form three σ bonds: two to hydrogen atoms, and one to the other carbon atom.

π (Pi) Bonds — the "sideways" bond

After forming those three σ bonds, each carbon has one electron left over sitting in an unhybridised p orbital, sticking up above and below the plane of the molecule. When two carbons are next to each other, these p orbitals can overlap sideways (not end-to-end) — this sideways overlap is the π bond.

Because the π bond's electron density sits above and below the plane of the molecule (not directly between the nuclei), it's weaker and more exposed than a σ bond. This exposed electron cloud is exactly what makes the C=C bond so attractive to electrophiles — it's basically "unprotected" electron density sticking out where things can attack it.

In a single C-C σ bond, the two carbon atoms can spin freely relative to each other — like two coins joined by a pin, they can rotate without breaking anything. But in a C=C double bond, that sideways-overlapping π bond gets in the way. If you tried to rotate one carbon relative to the other, you'd have to break the π bond's overlap — which takes serious energy. So the atoms attached to a C=C bond are locked in position, fixed either "up" or "down" relative to the double bond.

This restricted rotation is the entire reason geometric isomerism exists in alkenes.

For cis/trans (or E/Z) isomers to be possible at all, you need two different groups on each carbon of the double bond. 2-methylpropene, for example, has two identical CH₃ groups on the same carbon — so no matter how you look at it, there's no "different arrangement" possible. But-2-ene, however, has an H and a CH₃ on each carbon — giving genuine cis and trans forms.

When cis/trans breaks down: E/Z naming

The cis/trans system only works cleanly when two groups are identical. Once you have four different groups attached to the C=C bond, you need the more rigorous E/Z system, based on Cahn-Ingold-Prelog (CIP) priority rules.

1. Step 1: On each carbon of the double bond, compare the two attached atoms/groups. Look at the atomic number of the first atom attached — higher atomic number = higher priority.
2. Step 2: If the first atoms tie (e.g. both carbon), move outward to the next atoms attached and compare again.
3. Step 3: If the two highest-priority groups (one per carbon) are on the same side → Z isomer (from German "zusammen" = together). If on opposite sides → E isomer (from German "entgegen" = opposite).

Worked example — 1-bromo-1-propen-2-ol: This molecule has Br/H on one carbon and OH/CH₃ on the other — four different groups, so we need E/Z.

• Carbon 1: Br (atomic number 35) beats H (atomic number 1) → Br has priority
• Carbon 2: O (atomic number 8) beats C in CH₃ (atomic number 6) → OH has priority
• If Br and OH are on opposite sides → E-1-bromo-1-propen-2-ol
• If Br and OH are on the same side → Z-1-bromo-1-propen-2-ol

Because the C=C bond is electron-rich, it's a target for electrophiles — species that "love electrons" (usually because they're positively charged or have a partial positive charge). In an electrophilic addition reaction, the electrophile adds across the double bond, the π bond breaks, and two new σ bonds form — one to each carbon.

Four key types of electrophilic addition to know:

Notice the pattern: in every case, a molecule "splits" across the double bond, with one part attaching to each carbon. Hydrogenation is how margarine is made industrially (converting C=C bonds in vegetable oils into C-C bonds using a nickel catalyst) — a nice real-world hook to remember it by.

Oxidation with Cold Dilute KMnO₄

Alkenes can also be oxidised by cold, dilute, acidified potassium manganate(VII) — a powerful oxidising agent. You can think of this as two steps bolted together: an oxygen atom is delivered by the KMnO₄ (oxidation), then water in the solution adds an -OH and an -H across the (now broken) double bond.

This is one of the classic "distinguish two organic compounds" tests, and it comes up constantly in practical/observation-based exam questions.

Hydrogen halides like HBr are polar molecules — bromine is more electronegative than hydrogen, so it pulls the bonding electrons toward itself, giving H a slight positive charge (δ+) and Br a slight negative charge (δ-).

Here's the step-by-step mechanism, using curly arrows to track electron movement:

1. The C=C π bond acts as a nucleophile, attacking the δ+ hydrogen of H-Br. A curly arrow goes from the middle of the double bond to the H atom.
2. The H-Br bond breaks heterolytically (both electrons go to Br) — a curly arrow goes from the H-Br bond to the Br atom, forming a Br⁻ ion.
3. This leaves a positively charged carbocation intermediate on the carbon that did NOT bond to hydrogen.
4. The Br⁻ ion (now acting as a nucleophile) attacks the positive carbocation, forming the final C-Br bond.

Worked example — propene + HBr: Propene (CH₃-CH=CH₂) can form two possible carbocations when H⁺ adds:

• If H adds to the terminal CH₂, a secondary carbocation forms on the middle carbon (more stable, because two alkyl groups help stabilise the positive charge) → leads to 2-bromopropane (major product)
• If H adds to the middle carbon, a primary carbocation forms on the end carbon (less stable, only one alkyl group nearby) → leads to 1-bromopropane (minor product)

Mechanism with Halogens (X₂) — the "induced dipole" trick

Br₂ is a non-polar molecule (both atoms have equal electronegativity, so the electron pair is shared equally) — so how does it act as an electrophile? The answer: the double bond induces a temporary dipole in the approaching Br₂ molecule.

Once this induced dipole forms, the mechanism proceeds exactly like the HX mechanism: the π bond attacks the δ+ bromine, the Br-Br bond breaks heterolytically to release Br⁻, a carbocation intermediate forms, and the Br⁻ ion attacks the carbocation to complete the addition.

Addition polymerisation is really just electrophilic addition happening thousands of times in a row — except instead of adding a small molecule like Br₂ across the double bond, alkene molecules add to each other, over and over, forming a long chain. The π bond breaks in each monomer, and new C-C single bonds link the monomers together.

Deducing Repeat Units — the trap students fall into

This is the single most commonly mis-answered part of this topic, so pay close attention.

Worked example: If you're shown a section of a polymer chain and asked to identify the monomer, follow this method:

1. Find the smallest repeating pattern in the chain (usually 2 carbons in the backbone).
2. Draw that repeating section with square brackets and a subscript n.
3. Change the C-C single bond in the repeat unit back into a C=C double bond.
4. That's your monomer!

For example, a polymer chain with the repeat unit -CH(OH)-CH₂- came from the monomer CH(OH)=CH₂ (ethenol). A repeat unit of -CH₂-CH(CO₂H)- came from the monomer CH₂=CH(CO₂H) (prop-2-enoic acid).

Because polyalkenes (like poly(ethene) and PVC) are made entirely of unreactive C-C and C-H bonds, they are non-biodegradable — microorganisms simply can't break them down. This creates a genuine environmental challenge, and exam questions love to test whether you understand the trade-offs of each disposal method.

Biodegradable and Compostable Polymers

Chemists have designed alternative polymers that avoid the landfill problem entirely:

• Biodegradable polymers (like polyesters and polyamides) can be broken down by hydrolysis reactions — this is a major advantage over polyalkenes, which have no reactive functional group for microorganisms to attack.
• Compostable polymers are usually plant-based (e.g. plant starch in biodegradable bin liners, sugar cane fibres replacing polystyrene in disposable cups/plates) and degrade naturally, leaving no harmful residues.

• I can state the general formula of alkenes and explain what "unsaturated" means
• I can name alkenes correctly, including specifying the position of the double bond
• I can explain the difference between σ and π bonds, including how each forms
• I can explain why the π bond restricts rotation around the C=C bond
• I can identify whether a molecule can show cis/trans isomerism
• I can apply CIP priority rules to assign E/Z isomers, including cases with tied first atoms
• I can list the four electrophilic addition reactions of alkenes (H₂, H₂O, HX, X₂) with correct conditions
• I can describe the electrophilic addition mechanism with HX, including curly arrows
• I can explain how a non-polar Br₂ molecule becomes an electrophile (induced dipole)
• I can apply Markovnikov's rule to predict major/minor products using carbocation stability
• I can describe the bromine water saturation test and predict observations
• I can describe the cold dilute KMnO₄ oxidation test and name the product (diol)
• I can explain addition polymerisation and draw general/displayed formulae
• I can deduce a monomer from a given repeat unit (and vice versa)
• I can compare landfill, incineration, and recycling as plastic disposal methods
• I can explain why PVC incineration requires gas scrubbing, with the neutralisation equation
• I can explain why polyesters/polyamides are biodegradable but polyalkenes are not

What examiners are really checking for in this topic:

• Can you connect the bonding (σ/π) to the reactivity (electrophilic addition)? This link is tested constantly.
• Can you correctly apply CIP priority rules even in tricky cases where first atoms tie?
• Can you draw full curly-arrow mechanisms with correct arrow start/end points (arrows start from a bond or lone pair, never from an atom alone)?
• Can you justify Markovnikov's rule using carbocation stability, not just state the answer?
• Can you evaluate environmental trade-offs (disposal methods) with balanced, specific reasoning rather than vague statements?