Alkenes represent a major family of hydrocarbons that are defined by the presence of at least one carbon-to-carbon double bond. Unlike alkanes, which are saturated and relatively unreactive, the double bond in alkenes makes them highly valuable starting materials in synthesis.
🔑 Key Principle
Alkenes are unsaturated hydrocarbons because they contain at least one \( \text{C}=\text{C} \) double bond, meaning they contain fewer hydrogen atoms than the corresponding alkane with the same number of carbon atoms. The general formula for aliphatic alkenes with one double bond is \( \text{C}_n\text{H}_{2n} \).
The Nature of the C=C Double Bond
The carbon-carbon double bond is not simply two identical single bonds. It is composed of two distinct types of covalent bonds: a sigma (\( \sigma \)) bond and a pi (\( \pi \)) bond.
- The Sigma (\( \sigma \)) Bond: This is formed by the end-on overlap of orbitals directly between the two carbon nuclei. It is a strong bond with high electron density along the axis connecting the nuclei.
- The Pi (\( \pi \)) Bond: This is formed by the sideways overlap of two adjacent, parallel p-orbitals. The electron density is concentrated in two lobes: one above and one below the plane of the sigma framework.
A compound containing only carbon and hydrogen atoms, which features at least one carbon-carbon multiple bond (such as a double or triple bond) or an aromatic ring.
A covalent bond formed by the sideways overlap of adjacent p-orbitals, resulting in electron density being concentrated above and below the nodal plane of the bonding atoms.
🔑 Key Principle
The sideways overlap of p-orbitals in a pi bond prevents the carbon atoms from rotating relative to each other around the double bond axis. Any attempt to rotate around the \( \text{C}=\text{C} \) bond would require the pi bond to be broken, which requires a significant amount of energy.
1. Formation of the Sigma (\( \sigma \)) bond: Each carbon atom uses three valence electrons to form three strong sigma bonds (one with the other carbon atom and two with hydrogen atoms). These are formed by the direct end-on overlap of hybridised orbitals along the line between the nuclei.
2. Formation of the Pi (\( \pi \)) bond: Each carbon atom has one remaining unhybridised p-orbital containing one electron. These p-orbitals project perpendicularly above and below the plane of the sigma framework. They overlap sideways with each other, sharing their electrons to form a pi bond cloud above and below the nodal plane.
3. Bond Strength: The pi bond is weaker than the sigma bond because sideways overlap is less effective than the direct end-on overlap of a sigma bond. The orbital overlap is less concentrated between the carbon nuclei, making it easier to break.
Bond Angles and Shape
Each carbon atom in the double bond is surrounded by three areas of electron density (two single bonds and one double bond). According to Valence Shell Electron Pair Repulsion (VSEPR) theory, these electron regions repel each other to get as far apart as possible.
This results in a trigonal planar geometry around each carbon atom, with bond angles of approximately 120° (specifically, \( 121.3^\circ \) for \( \text{H}-\text{C}-\text{C} \) and \( 117.4^\circ \) for \( \text{H}-\text{C}-\text{H} \) in ethene due to the slightly greater repulsion from the double bond).
Compounds with the same structural formula but a different arrangement of their atoms in three-dimensional space.
E/Z Stereoisomerism
Because there is restricted rotation around the \( \text{C}=\text{C} \) double bond, groups attached to the double-bonded carbon atoms are locked in place. If both carbon atoms in the double bond have two different groups attached to them, stereoisomerism arises.
The IUPAC system uses the Cahn-Ingold-Prelog (CIP) priority rules to assign names to these stereoisomers:
- Examine the two atoms attached to one carbon of the double bond. Assign priority based on their atomic numbers (the atom with the higher atomic number receives higher priority).
- Examine the two atoms attached to the second carbon of the double bond and assign priority in the same way.
- Compare the positions of the two high-priority groups:
- If the high-priority groups are on the same side of the double bond, it is the Z-isomer (from German zusammen, together).
- If the high-priority groups are on opposite sides of the double bond, it is the E-isomer (from German entgegen, opposite).
To recall which isomer is which, a simple memory trick is: Z-isomer has the high-priority groups on the 'Zame Side'. The C=C bond cannot rotate because the pi bond would have to break, which prevents the isomers from interconverting at room temperature.
For E/Z stereoisomerism to occur, two criteria must be satisfied:
- There must be restricted rotation around a carbon-carbon double bond. This is satisfied in both compounds by the presence of the pi bond.
- Each carbon atom in the double bond must be attached to two different groups.
Let us check the groups on each carbon:
- In but-2-ene (\( \text{CH}_3\text{CH}=\text{CHCH}_3 \)):
- Carbon 2 is attached to a hydrogen atom (\( -\text{H} \)) and a methyl group (\( -\text{CH}_3 \)). These are different.
- Carbon 3 is attached to a hydrogen atom (\( -\text{H} \)) and a methyl group (\( -\text{CH}_3 \)). These are also different.
- Therefore, but-2-ene meets both criteria and exhibits E/Z stereoisomerism.
- In but-1-ene (\( \text{CH}_2=\text{CHCH}_2\text{CH}_3 \)):
- Carbon 1 is attached to two identical hydrogen atoms.
- Because one of the carbons in the double bond is attached to two identical groups, but-1-ene does not meet the second criterion and cannot exhibit E/Z stereoisomerism.
Testing for Unsaturation: The Bromine Water Test
Because the pi bond is relatively weak and has a high concentration of negative charge above and below the nodal plane, it is easily attacked by electron-deficient species (electrophiles).
This is the basis of the classic laboratory test for alkenes (and unsaturation in general):
- Reagent: Bromine water, which is an orange-brown solution.
- Observation: When shaken with an alkene, the orange-brown colour disappears to leave a colourless solution.
- Reaction: An addition reaction takes place, where bromine adds across the double bond to form a colourless dihalogenoalkane.
When writing about the bromine water test, always state the starting colour of the reagent (orange, orange-brown, or brown) and the final colour (colourless). Avoid writing 'clear' instead of colourless. Remember: the decolourisation of bromine water is an addition reaction, not a substitution.