Alkenes

Q: What is the molecular geometry and approximate bond angle around a double-bonded carbon atom in an alkene?

The arrangement is trigonal planar with a bond angle of approximately \( 120^\circ \) .

Q: What two conditions must be met for an alkene to exhibit \( E/Z \) stereoisomerism?

There must be restricted rotation around the \( \text{C}=\text{C} \) double bond, and two different groups must be attached to each carbon atom of the double bond.

Q: What is the chemical test and positive observation used to confirm unsaturation in an organic compound?

Add bromine water ; a positive result is a colour change from orange-brown to colourless (decolourisation).

Q: Why are tertiary carbocations more stable than secondary and primary carbocations?

Alkyl groups exert an electron-donating positive inductive effect , which pushes electrons toward the positively charged carbon and disperses the charge.

Q: What reagents and conditions are required for the catalytic hydrogenation of alkenes to alkanes?

Hydrogen gas (\( \text{H}_2 \)) , a nickel (\( \text{Ni} \)) catalyst , and a temperature of around \( 150^\circ\text{C} \) .

Q: What reagents and conditions are required for the industrial hydration of alkenes to form alcohols?

Steam (\( \text{H}_2\text{O}\text{(g)} \)) , a concentrated phosphoric acid (\( \text{H}_3\text{PO}_4 \)) catalyst , a temperature of \( \approx 300^\circ\text{C} \), and a pressure of \( 60\text{--}70\text{ atm} \).

Chemistry 1110 · CCEA A Level · 10 min read · Updated: 31 Aug 2026

Welcome to Alkenes!

Welcome to one of the most exciting and dynamic topics in AS Chemistry! While alkanes are relatively unreactive, alkenes are the chemical world's real go-getters. From producing the plastics in your phone case to making synthetic fabrics and ripening fruit, alkenes are at the heart of the modern chemical industry.

Don't worry if organic mechanisms have felt intimidating in the past. We will break every concept down into simple, step-by-step bites so that you can tackle your exams with total confidence.

1. Structure and Bonding in Alkenes

Alkenes are unsaturated hydrocarbons. The word unsaturated simply means they contain at least one carbon-to-carbon double bond (\( \text{C}=\text{C} \)), while hydrocarbon means they are composed solely of carbon and hydrogen atoms.

For an alkene with just one double bond, the general molecular formula is:
\( \text{C}_n\text{H}_{2n} \)

The Nature of the Double Bond: Sigma (\( \sigma \)) and Pi (\( \pi \)) Bonds

A single covalent bond is made of one sigma (\( \sigma \)) bond. A double bond, however, consists of one \( \sigma \)-bond and one \( \pi \)-bond:

• Sigma (\( \sigma \)) Bond: Formed by the direct, end-on overlap of atomic orbitals. The electron density is concentrated along the line directly between the two carbon nuclei. This is a strong, stable covalent bond.
• Pi (\( \pi \)) Bond: Formed by the sideways overlap of adjacent unhybridised p-orbitals. The electron density is concentrated in two lobes: one above and one below the plane of the carbon atoms.

Analogy: Imagine shaking hands firmly with someone directly in front of you—that is an end-on \( \sigma \)-bond. Now imagine two people giving each other high-fives above and below their handshake at the same time—that sideways interaction represents the \( \pi \)-bond!

Bond Angles and Geometry

Each carbon atom involved in the double bond has three regions of electron density (two single bonds and one double bond). These electron clouds repel each other equally according to VSEPR theory:

• The arrangement around each double-bonded carbon is trigonal planar.
• The bond angle is approximately \( 120^\circ \).

Quick Takeaway: The \( \text{C}=\text{C} \) double bond is composed of one strong \( \sigma \)-bond and one exposed \( \pi \)-bond with an electron cloud situated above and below the planar molecule (bond angle \( \approx 120^\circ \)).

2. Stereoisomerism: \( E/Z \) Isomerism

Stereoisomers are molecules with the same structural formula, but their atoms are arranged differently in 3D space.

Why does \( E/Z \) Isomerism Occur?

In alkanes, carbon atoms can freely spin around single \( \sigma \)-bonds. In alkenes, however, the sideways overlap of the \( \pi \)-bond locks the carbon atoms firmly in place. Rotating around the double bond would require breaking the \( \pi \)-bond, which takes a lot of energy.

For a compound to exhibit \( E/Z \) isomerism, two conditions MUST be met:
1. Restricted rotation around the \( \text{C}=\text{C} \) double bond.
2. Two different groups attached to each carbon atom of the double bond.

Assigning \( E \) and \( Z \): The Cahn-Ingold-Prelog (CIP) Priority Rules

To determine whether an isomer is \( E \) or \( Z \), follow these simple steps:

Step 1: Split the double bond vertically in half and look at one carbon atom at a time.
Step 2: Assign priority to the two attached atoms based on their atomic number (\( Z \)). The atom with the higher atomic number gets high priority, and the other gets low priority.
Note: If the directly attached atoms are identical (for example, two carbon chains), move along the chains atom-by-atom until you find the first point of difference.
Step 3: Compare the positions of the two high-priority groups across the double bond:
• \( Z \) Isomer: The high-priority groups are on the same side (both top or both bottom).
• \( E \) Isomer: The high-priority groups are on opposite sides (diagonally across).

Memory Trick:
• \( Z \) = "Zame Zide" (Same Side)
• \( E \) = "Enemies" (Opposite Sides / Across from each other)

What about cis and trans?

Cis/trans notation is a special sub-category of \( E/Z \) isomerism used when each double-bonded carbon has an identical group (often a hydrogen atom) attached to it:
• cis: Identical groups are on the same side (\( Z \)).
• trans: Identical groups are on opposite sides (\( E \)).

Quick Takeaway: Check both carbons of the double bond. If each has two different groups, apply CIP priority rules based on atomic number: high priorities on the "zame zide" = \( Z \); on opposite sides = \( E \).

3. Reactivity and Electrophilic Addition

Why are alkenes so much more reactive than alkanes? The answer lies in the \( \pi \)-bond!

The \( \pi \)-electron cloud sits above and below the plane of the carbon atoms, making it exposed and easily accessible. This creates a region of high electron density that readily attacks and attracts electron-deficient species called electrophiles.

• Electrophile: An electron-pair acceptor (often carrying a positive charge \( + \) or a partial positive charge \( \delta^+ \)).

Reaction 1: Addition of Halogens (e.g., \( \text{Br}_2 \)) — The Test for Unsaturation

When bromine water is added to an alkene and shaken, the solution rapidly turns from orange-brown to colourless. This is the standard diagnostic test for unsaturation.

Equation for ethene with bromine:
\( \text{CH}_2=\text{CH}_2 + \text{Br}_2 \rightarrow \text{CH}_2\text{Br}\text{CH}_2\text{Br} \) (1,2-dibromoethane)

Step-by-Step Electrophilic Addition Mechanism for \( \text{Br}_2 \):

1. Induced Dipole: Although \( \text{Br}_2 \) is non-polar, as it approaches the dense \( \pi \)-electron cloud of the alkene, the electrons in the \( \text{Br}-\text{Br} \) bond are repelled. This induces a temporary dipole: \( \text{Br}^{\delta+} - \text{Br}^{\delta-} \).
2. Attack by the \( \pi \)-bond: A curly arrow goes from the \( \text{C}=\text{C} \) double bond to the \( \text{Br}^{\delta+} \). At the same time, the \( \text{Br}-\text{Br} \) bond breaks heterolytically, with the electron pair moving onto the second bromine atom to form a bromide ion (\( \text{Br}^- \)).
3. Carbocation Intermediate: One carbon forms a bond with bromine, leaving the adjacent carbon with an empty orbital and a positive charge (a carbocation).
4. Nucleophilic Attack: A curly arrow goes from a lone pair on the bromide ion (\( \text{:\!Br}^- \)) to the positively charged carbon atom of the carbocation, yielding the dihalogenoalkane product.

Reaction 2: Addition of Hydrogen Halides (e.g., \( \text{HBr} \))

Hydrogen halides are already permanently polar (\( \text{H}^{\delta+} - \text{Br}^{\delta-} \)).

Equation with ethene:
\( \text{CH}_2=\text{CH}_2 + \text{HBr} \rightarrow \text{CH}_3\text{CH}_2\text{Br} \) (bromoethane)

• The \( \pi \)-electrons attack the \( \text{H}^{\delta+} \) atom, forming a \( \text{C}-\text{H} \) bond.
• The \( \text{H}-\text{Br} \) bond breaks heterolytically to produce \( \text{:\!Br}^- \).
• The \( \text{:\!Br}^- \) attacks the carbocation intermediate to give the halogenoalkane.

Unsymmetrical Alkenes and Markovnikov's Rule

When an unsymmetrical hydrogen halide (like \( \text{HBr} \)) reacts with an unsymmetrical alkene (like propene, \( \text{CH}_3\text{CH}=\text{CH}_2 \)), two different products can form: a major product and a minor product.

Carbocation Stability:

Alkyl groups (such as \( -\text{CH}_3 \)) push electrons away from themselves towards the positively charged carbon. This is known as the positive inductive effect. It helps disperse and stabilise the positive charge:

• Primary (\( 1^\circ \)) Carbocation: The positive carbon is bonded to only one alkyl group (least stable).
• Secondary (\( 2^\circ \)) Carbocation: The positive carbon is bonded to two alkyl groups (more stable).
• Tertiary (\( 3^\circ \)) Carbocation: The positive carbon is bonded to three alkyl groups (most stable).

Stability Order: Tertiary (\( 3^\circ \)) \( > \) Secondary (\( 2^\circ \)) \( > \) Primary (\( 1^\circ \))

Markovnikov's Rule Simplified:

"The rich get richer!" In the addition of \( \text{HX} \) to an unsymmetrical alkene, the hydrogen atom attaches to the double-bonded carbon that already has the greater number of hydrogen atoms. This pathway proceeds via the more stable carbocation intermediate, yielding the major product.

Example with Propene:
• Addition of \( \text{H}^+ \) to carbon-1 creates a secondary carbocation (\( \text{CH}_3\text{C}^+\text{HCH}_3 \)) \( \rightarrow \) leading to 2-bromopropane (major product).
• Addition of \( \text{H}^+ \) to carbon-2 creates a primary carbocation (\( \text{CH}_3\text{CH}_2\text{C}^+\text{H}_2 \)) \( \rightarrow \) leading to 1-bromopropane (minor product).

Reaction 3: Catalytic Hydrogenation (Addition of \( \text{H}_2 \))

Alkenes react with hydrogen gas to form alkanes.
• Reagents and Conditions: Hydrogen gas (\( \text{H}_2 \)), Nickel (\( \text{Ni} \)) catalyst, temperature around \( 150^\circ\text{C} \).
• Equation: \( \text{CH}_2=\text{CH}_2 + \text{H}_2 \rightarrow \text{CH}_3\text{CH}_3 \)
• Application: Used industrially in the hydrogenation of liquid vegetable oils to produce solid spreads such as margarine.

Reaction 4: Hydration (Addition of Steam, \( \text{H}_2\text{O}\text{(g)} \))

Alkenes react with steam to produce alcohols.
• Reagents and Conditions: Steam (\( \text{H}_2\text{O}\text{(g)} \)), concentrated phosphoric acid (\( \text{H}_3\text{PO}_4 \)) catalyst, high temperature (\( \approx 300^\circ\text{C} \)) and pressure (\( \approx 60\text{--}70\text{ atm} \)).
• Equation: \( \text{CH}_2=\text{CH}_2 + \text{H}_2\text{O} \rightarrow \text{CH}_3\text{CH}_2\text{OH} \) (ethanol)

Quick Takeaway: Alkenes undergo electrophilic addition due to their high electron density. Unsymmetrical alkenes form major products via the most stable carbocation intermediate (tertiary \( > \) secondary \( > \) primary).

4. Addition Polymerisation

Alkenes can join together end-to-end in large numbers to form long-chain molecules known as addition polymers.

• Monomer: The individual small alkene molecule containing a \( \text{C}=\text{C} \) double bond.
• Polymer: The long-chain molecule made up of thousands of repeating units joined by single \( \text{C}-\text{C} \) bonds.
• Repeating Unit: The specific arrangement of atoms that repeats over and over again along the polymer chain.

How to Draw Repeating Units:

1. Draw the monomer in an 'H' or 'X' shape, placing all side-groups vertically above or below the two double-bonded carbons.
2. Change the \( \text{C}=\text{C} \) double bond into a single \( \text{C}-\text{C} \) bond.
3. Extend open bonds beyond the square brackets on both sides.
4. Place square brackets around the unit with a subscript '\( n \)' on the bottom right.

Example: Ethene (\( n\text{CH}_2=\text{CH}_2 \)) polymerises into poly(ethene): \( \text{--[--CH}_2\text{--CH}_2\text{--]--}_n \)

Environmental Considerations of Polymers

Because poly(alkenes) are saturated hydrocarbons containing only strong, non-polar \( \text{C}-\text{C} \) and \( \text{C}-\text{H} \) single bonds, they are chemically unreactive and non-biodegradable. Methods for handling polymer waste include:

• Recycling: Sorting polymers by type, melting them down, and reshaping them into new products. This conserves finite crude oil reserves.
• Incineration (Energy Recovery): Burning polymer waste to generate electricity. However, this releases greenhouse gases (such as \( \text{CO}_2 \)) and potentially toxic gases (e.g., toxic \( \text{HCl} \) from chlorinated polymers like PVC, which must be neutralised using scrubbers with bases like \( \text{CaO} \)).
• Feedstock Recycling: Chemically breaking polymers back down into monomers and basic hydrocarbons to use as raw chemical feedstock.

Quick Takeaway: Addition polymerisation involves opening up \( \text{C}=\text{C} \) double bonds to form saturated, durable, non-biodegradable long chains.

Summary and Common Pitfalls to Avoid

• Pitfall 1: Forgetting double-headed vs single-headed curly arrows. In electrophilic addition mechanisms, arrows represent the movement of a pair of electrons, so always use double-headed arrows originating from a bond or lone pair.
• Pitfall 2: Drawing repeating units with double bonds. Remember, addition polymers contain only single bonds along their carbon backbone.
• Pitfall 3: Confusing \( E/Z \) assignments. Always compare atomic numbers directly attached to each carbon of the double bond first before looking further down the chain.

Quick check

Can you answer these now?

Open each question to check the key ideas from this chapter.

What is the general molecular formula for an alkene containing one double bond?

\( \text{C}_n\text{H}_{2n} \)

How do the formation and electron distributions of sigma (\( \sigma \)) and pi (\( \pi \)) bonds differ in a \( \text{C}=\text{C} \) double bond?

A \( \sigma \)-bond forms by the end-on overlap of orbitals with electron density concentrated between the nuclei, whereas a \( \pi \)-bond forms by the sideways overlap of unhybridised p-orbitals with electron density concentrated in lobes above and below the plane of the carbon atoms.

What is the molecular geometry and approximate bond angle around a double-bonded carbon atom in an alkene?

The arrangement is trigonal planar with a bond angle of approximately \( 120^\circ \).

What two conditions must be met for an alkene to exhibit \( E/Z \) stereoisomerism?

There must be restricted rotation around the \( \text{C}=\text{C} \) double bond, and two different groups must be attached to each carbon atom of the double bond.

How are the Cahn-Ingold-Prelog (CIP) priority rules applied to assign \( E \) or \( Z \) configurations to alkenes?

Priorities are assigned to the two groups on each double-bonded carbon based on higher atomic number (\( Z \)); if the high-priority groups are on the same side it is the \( Z \) isomer, and if on opposite sides it is the \( E \) isomer.

What is the chemical test and positive observation used to confirm unsaturation in an organic compound?

Add bromine water; a positive result is a colour change from orange-brown to colourless (decolourisation).

How does a non-polar halogen molecule (such as \( \text{Br}_2 \)) react with an alkene during electrophilic addition?

The high electron density of the \( \pi \)-bond induces a dipole in the \( \text{Br}_2 \) molecule (\( \text{Br}^{\delta+}-\text{Br}^{\delta-} \)), allowing the \( \text{Br}^{\delta+} \) to act as an electrophile.

Why are tertiary carbocations more stable than secondary and primary carbocations?

Alkyl groups exert an electron-donating positive inductive effect, which pushes electrons toward the positively charged carbon and disperses the charge.

State Markovnikov's rule and explain why it governs the major product in the addition of \( \text{HX} \) to an unsymmetrical alkene.

The hydrogen atom attaches to the double-bonded carbon with the greater number of hydrogen atoms because this pathway proceeds via the more stable carbocation intermediate.

What reagents and conditions are required for the catalytic hydrogenation of alkenes to alkanes?

Hydrogen gas (\( \text{H}_2 \)), a nickel (\( \text{Ni} \)) catalyst, and a temperature of around \( 150^\circ\text{C} \).

What reagents and conditions are required for the industrial hydration of alkenes to form alcohols?

Steam (\( \text{H}_2\text{O}\text{(g)} \)), a concentrated phosphoric acid (\( \text{H}_3\text{PO}_4 \)) catalyst, a temperature of \( \approx 300^\circ\text{C} \), and a pressure of \( 60\text{--}70\text{ atm} \).

What is addition polymerisation and how is a repeating unit drawn from an alkene monomer?

It is the joining of many alkene monomers by opening \( \text{C}=\text{C} \) double bonds into single bonds; the repeating unit is drawn with a \( \text{C}-\text{C} \) single bond, extended open bonds through square brackets, and a subscript \( n \).

Why are poly(alkenes) non-biodegradable, and what environmental risk is associated with incinerating chlorinated polymers like PVC?

They contain only strong, unreactive, non-polar \( \text{C}-\text{C} \) and \( \text{C}-\text{H} \) bonds; incinerating chlorinated polymers releases toxic \( \text{HCl} \) gas, which requires neutralisation with base scrubbers (e.g., \( \text{CaO} \)).

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