Welcome to Hydrocarbons: Alkenes
Welcome to your study guide for Alkenes, a core topic in Unit A2 2: Organic Chemistry for CCEA A Level Life and Health Sciences. If you have ever wondered how vegetable oils are turned into spreads, how everyday plastics like poly(ethene) are manufactured, or how chemists test for unsaturated molecules in the lab, this chapter holds the answers!
Don't worry if organic chemistry mechanisms feel a bit daunting at first. We will break down every structure, reaction, and exam technique step by step so you feel fully confident heading into your exam.
---1. Structure and Bonding in Alkenes
What is an Alkene?
An alkene is an unsaturated hydrocarbon that contains at least one carbon–carbon double covalent bond (\(\text{C}=\text{C}\)). The term unsaturated means the molecule contains double (or multiple) bonds, so it does not contain the maximum possible number of hydrogen atoms.
For straight-chain and branched acyclic alkenes with one double bond, the general molecular formula is:
\(\text{C}_n\text{H}_{2n}\)
For example, if an alkene has \(4\) carbon atoms (\(n = 4\)), its molecular formula is \(\text{C}_4\text{H}_8\).
The Nature of the \(\text{C}=\text{C}\) Double Bond
The carbon–carbon double bond is not just two identical single bonds. It is composed of two distinctly different types of covalent bonds:
1. The \(\sigma\) (sigma) bond: Formed by the direct, end-on overlap of atomic orbitals along the internuclear axis between the two carbon atoms. This is a strong, stable covalent bond.
2. The \(\pi\) (pi) bond: Formed by the sideways overlap of adjacent, parallel \(p\) orbitals above and below the plane of the carbon atoms.
Why are alkenes so much more reactive than alkanes?
The electrons in the \(\pi\) bond form a region of high electron density located above and below the internuclear axis. Because these electrons are more exposed and held further from the carbon nuclei than \(\sigma\) electrons, the \(\pi\) bond is easily attacked by electron-seeking species (electrophiles). Consequently, the \(\pi\) bond breaks relatively easily during chemical reactions.
Restricted Rotation:
Unlike single \(\text{C}–\text{C}\) bonds which can rotate freely, the sideways overlap of the \(\pi\) bond locks the carbon atoms in place. The double bond prevents free rotation around the \(\text{C}=\text{C}\) axis at room temperature, because rotating the bond would require breaking the \(\pi\) bond.
Key Takeaway: The \(\text{C}=\text{C}\) double bond consists of one strong \(\sigma\) bond and one more exposed \(\pi\) bond. The high electron density of the \(\pi\) bond makes alkenes reactive, while its geometry prevents free rotation.
---2. Nomenclature and Isomerism
IUPAC Rules for Naming Alkenes
Naming alkenes follows a systematic set of rules:
1. Identify the longest continuous carbon chain containing the \(\text{C}=\text{C}\) double bond. Replace the suffix -ane with -ene.
2. Number the carbon chain from the end that gives the \(\text{C}=\text{C}\) bond the lowest possible locant number. For example, \(\text{CH}_3\text{CH}_2\text{CH}=\text{CH}_2\) is but-1-ene, not but-3-ene.
3. Identify and number any branched alkyl groups or halogen substituents. List substituents alphabetically with their respective position numbers (e.g., 2-methylbut-2-ene or 1,2-dibromoethene).
Positional and Structural Isomerism
Alkenes can exhibit positional isomerism, where molecules have the same molecular formula and carbon skeleton but differ in the position of the \(\text{C}=\text{C}\) double bond along the chain. For example:
• But-1-ene: \(\text{CH}_2=\text{CH}–\text{CH}_2–\text{CH}_3\)
• But-2-ene: \(\text{CH}_3–\text{CH}=\text{CH}–\text{CH}_3\)
Geometric (cis / trans or \(E/Z\)) Isomerism
Geometric isomerism is a form of stereoisomerism where molecules have the same structural formula but a different 3D arrangement of atoms in space. For an alkene to show geometric isomerism, two conditions must be met:
1. There must be restricted rotation around the \(\text{C}=\text{C}\) double bond.
2. Each carbon atom of the \(\text{C}=\text{C}\) double bond must be attached to two different atoms or groups.
Understanding cis vs trans:
• cis-isomer: Matching or higher-priority groups are located on the same side of the double bond.
• trans-isomer: Matching or higher-priority groups are located on opposite sides (across) the double bond.
Memory Trick: Remember that trans means across (like a transatlantic flight across the ocean), while in cis, the matching groups are on the "C-ame S-ide".
Quick Check: Does propene show geometric isomerism? Look at \(\text{C}_1\): it has two hydrogen atoms attached to it (\(\text{H}\) and \(\text{H}\)). Because both groups on that carbon are identical, propene cannot exist as cis/trans isomers.
Key Takeaway: Always number from the end closest to the \(\text{C}=\text{C}\) bond. Geometric (cis/trans) isomerism requires both restricted rotation and two distinct groups attached to each double-bonded carbon.
---3. Chemical Reactions and Mechanisms
The Electrophilic Addition Mechanism
The characteristic reaction of alkenes is electrophilic addition. An electrophile is an electron-deficient species (an atom, positive ion, or molecule with a \(\delta+\) region) that can accept a pair of electrons.
How it works step-by-step:
1. The electron-rich \(\pi\) bond of the alkene attacks the electron-deficient (\(\delta+\)) part of the electrophile.
2. A pair of electrons from the \(\pi\) bond forms a new covalent bond with the electrophile, breaking the \(\pi\) bond.
3. This creates a temporary, positively charged intermediate called a carbocation.
4. The remaining negatively charged species (nucleophile) attacks the positively charged carbocation, supplying a lone pair of electrons to form the final addition product.
1. Halogenation (Addition of Bromine — Chemical Test for Unsaturation)
When an alkene reacts with bromine (\(\text{Br}_2\)), the bromine molecule adds across the double bond to form a dihalogenoalkane:
\(\text{R–CH=CH–R'} + \text{Br}_2 \rightarrow \text{R–CHBr–CHBr–R'}\)
• Reagents and Conditions: Bromine water or \(\text{Br}_2\) in an organic solvent at room temperature.
• Observation: The orange/brown solution turns colourless rapidly.
• Lab Application: This decolourisation reaction serves as the definitive laboratory test to distinguish unsaturated hydrocarbons (alkenes) from saturated hydrocarbons (alkanes).
2. Hydrogenation (Addition of \(\text{H}_2\))
Hydrogen gas adds across the \(\text{C}=\text{C}\) double bond to convert an alkene into an alkane:
\(\text{CH}_2=\text{CH}_2 + \text{H}_2 \rightarrow \text{CH}_3–\text{CH}_3\)
• Reagents and Conditions: Hydrogen gas (\(\text{H}_2\)) with a finely divided Nickel (\(\text{Ni}\)) catalyst at elevated temperature (typically \(\sim 150\ ^\circ\text{C}\)).
• Industrial Application: Used in the "hardening" of liquid unsaturated vegetable oils to manufacture solid spreads and margarine.
3. Hydration (Addition of Steam)
Gaseous water (steam) adds across the double bond to produce an alcohol:
\(\text{CH}_2=\text{CH}_2 + \text{H}_2\text{O}\text{(g)} \rightarrow \text{CH}_3\text{CH}_2\text{OH}\)
• Reagents and Conditions: Steam (\(\text{H}_2\text{O}\text{(g)}\)) in the presence of an acid catalyst (concentrated \(\text{H}_3\text{PO}_4\) or \(\text{H}_2\text{SO}_4\)) under high temperature and pressure.
• Industrial Application: Direct synthesis of industrial ethanol from ethene.
4. Addition of Hydrogen Halides (\(\text{HX}\)) & Markovnikov's Rule
Hydrogen halides (such as \(\text{HBr}\) or \(\text{HCl}\)) add across the double bond to yield halogenoalkanes:
\(\text{CH}_2=\text{CH}_2 + \text{HBr} \rightarrow \text{CH}_3\text{CH}_2\text{Br}\)
What happens with unsymmetrical alkenes?
When \(\text{HX}\) reacts with an unsymmetrical alkene (such as propene), two different products are possible. Markovnikov’s Rule predicts the major product:
"The hydrogen atom adds to the carbon of the double bond that already has the greater number of hydrogen atoms attached to it."
• Why does this happen? Reaction pathways proceed via the more stable carbocation intermediate. Alkyl groups are electron-donating (inductive effect), so stability increases in the order:
tertiary (\(3^\circ\)) > secondary (\(2^\circ\)) > primary (\(1^\circ\))
• For example, when \(\text{HBr}\) reacts with propene (\(\text{CH}_3\text{CH}=\text{CH}_2\)), the secondary carbocation (\(\text{CH}_3\text{CH}^+\text{CH}_3\)) forms much more readily than the primary carbocation (\(\text{CH}_3\text{CH}_2\text{CH}_2^+\)). Thus, 2-bromopropane is the major product.
5. Oxidation with Acidified Potassium Manganate(VII)
Alkenes are oxidised by acidified potassium manganate(VII) to form diols (compounds with two \(-\text{OH}\) groups):
\(\text{CH}_2=\text{CH}_2 + [\text{O}] + \text{H}_2\text{O} \rightarrow \text{CH}_2(\text{OH})\text{CH}_2(\text{OH})\)
• Product: Ethane-1,2-diol (from ethene).
• Observation: The purple solution turns colourless.
• Equation representation: In equations, the oxidising agent is represented as \([\text{O}]\), reacting together with water (\(\text{H}_2\text{O}\)) to add two \(-\text{OH}\) groups across the double bond.
Key Takeaway: Alkenes undergo electrophilic addition due to their electron-rich \(\pi\) bond. For unsymmetrical alkenes adding \(\text{HX}\), Markovnikov's rule guides you to the major product via the most stable carbocation intermediate.
---4. Industrial and Environmental Context
Addition Polymerisation
Alkenes act as monomers that join together to form extremely long saturated polymer chains called polyalkenes:
• Under heat, high pressure, and suitable catalysts, the weak \(\pi\) bond of each alkene monomer breaks.
• The newly freed electrons form new single covalent \(\sigma\) bonds linking adjacent monomer units together.
• Key Feature: There are no by-products formed during addition polymerisation; the polymer is the sole product.
• Example: \(n(\text{CH}_2=\text{CH}_2) \rightarrow -[\text{CH}_2–\text{CH}_2]_n-\) (ethene forms poly(ethene)).
Cracking of Crude Oil Fractions
Crude oil contains a surplus of long-chain alkanes, but society has a much higher demand for shorter petrol fractions and reactive alkene feedstocks. Cracking (both thermal and catalytic) breaks strong \(\text{C}–\text{C}\) bonds in long-chain alkanes to produce smaller alkanes and valuable alkenes (such as ethene and propene) to supply the chemical industry.
---5. Examiner Pitfalls & Revision Checklist
Common Exam Mistakes to Avoid:
• "Turns clear" vs "Turns colourless": Never write that bromine water or acidified \(\text{KMnO}_4\) "turns clear". Clear means transparent (bromine water is already clear orange). The correct mark scheme term is always "turns colourless" or "decolourises".
• Inaccurate Curly Arrows: When drawing mechanisms, always start your curly arrow directly from the electron pair (the \(\text{C}=\text{C}\) \(\pi\) bond line or a lone pair) and point it accurately towards the specific atom receiving the electrons.
• Incomplete cis/trans explanations: Do not just write "because of the double bond". You must explicitly state both factors: (1) restricted rotation around the \(\text{C}=\text{C}\) bond, and (2) two different groups attached to each carbon of the double bond.
• Oxidation Equations: Do not attempt to balance full manganate redox formulas in organic equations unless specifically asked; use \([\text{O}] + \text{H}_2\text{O}\) to represent the addition of two \(-\text{OH}\) groups to form a diol.
Quick Summary Review:
• Bonding: \(1\ \sigma\) bond + \(1\ \pi\) bond (restricted rotation, high electron density).
• Test for unsaturation: Bromine water goes from orange/brown to colourless.
• Hydration: Steam + acid catalyst (\(\text{H}_3\text{PO}_4\)) \(\rightarrow\) Alcohol.
• Hydrogenation: \(\text{H}_2\) + \(\text{Ni}\) catalyst (\(\sim 150\ ^\circ\text{C}\)) \(\rightarrow\) Alkane.
• Oxidation: Acidified \(\text{KMnO}_4\) (purple \(\rightarrow\) colourless) \(\rightarrow\) Diol.
• Markovnikov: \(\text{H}\) goes to the carbon with more hydrogens (\(3^\circ > 2^\circ > 1^\circ\) carbocation stability).