Alkanes are the simplest family of organic compounds. They are saturated hydrocarbons, meaning they contain only carbon and hydrogen atoms connected by single covalent bonds. Alkanes serve as the backbone of organic chemistry and are major components of crude oil and petroleum fuels.
🔑 Key Principle
Alkanes are non-polar, saturated hydrocarbons. Their physical properties, such as boiling point trends, are determined by weak intermolecular London dispersion forces. Their chemical stability arises from the high bond enthalpies of their non-polar C-C and C-H single bonds.
1. Saturated Hydrocarbons and Bonding
Every carbon atom in an alkane forms four single covalent bonds. These single bonds are known as sigma (σ) bonds, which are formed by the end-on overlap of atomic orbitals.
A compound containing only carbon and hydrogen atoms, in which all carbon-to-carbon bonds are single covalent bonds.
A single covalent bond formed by the direct, end-on overlap of atomic orbitals, with the electron density concentrated along the internuclear axis between the bonding nuclei.
Orbital Overlap and Geometry
In alkanes, each carbon atom is \( \text{sp}^3 \) hybridised. The four \( \text{sp}^3 \) hybrid orbitals overlap end-on with either a \( 1\text{s} \) orbital from a hydrogen atom, or another \( \text{sp}^3 \) hybrid orbital from an adjacent carbon atom. This end-on overlap provides maximum orbital overlap, making the resulting \( \sigma \) bonds strong and stable.
Because the four bonding electron pairs repel each other equally to minimise repulsion, they adopt a tetrahedral arrangement around each carbon atom, as predicted by Valence Shell Electron Pair Repulsion (VSEPR) theory. This structure has the following characteristics:
- A bond angle of \( 109.5^\circ \).
- Free rotation of the bonded atoms around the \( \sigma \) bond axis. This is because the end-on overlap has cylindrical symmetry along the line connecting the two nuclei, meaning rotation does not disrupt the orbital overlap.
Students often define saturated as, having the maximum possible number of hydrogen atoms bonded to carbon. In exams, you must state that all carbon-carbon bonds are single covalent bonds to receive full credit.
2. Physical Properties: Boiling Point Trends
Alkanes are non-polar molecules. The electronegativity values of carbon (2.5) and hydrogen (2.1) are very similar, meaning that the C-H bonds are virtually non-polar. Consequently, the only intermolecular forces acting between alkane molecules are weak London dispersion forces (temporary dipole-induced dipole interactions).
The boiling points of alkanes are determined by the strength of these London dispersion forces, which depend on two factors:
Effect of Chain Length
As the carbon chain length of straight-chain alkanes increases:
- The boiling point increases.
- This is because longer molecules have more electrons and a larger molecular surface area.
- This leads to more contact points between adjacent molecules, increasing the strength of the temporary dipoles and the overall London dispersion forces.
- Consequently, more thermal energy is required to overcome these intermolecular forces and separate the molecules into the gas phase.
Effect of Branching
For isomers (alkanes with the same molecular formula but different structural formulas), branching has a significant impact on boiling points:
- Increased branching decreases the boiling point.
- Branched molecules are more compact and spherical in shape compared to their straight-chain isomers.
- This shape reduces the surface area in contact between adjacent molecules, preventing them from packing closely together.
- As a result, the London dispersion forces are weaker, and less thermal energy is required to overcome them and boil the compound.
When discussing physical properties in exams, avoid the generic term, van der Waals forces, if the syllabus or question refers specifically to London forces. Always refer to them as London (dispersion) forces. Additionally, always make clear that boiling involves breaking intermolecular forces, not covalent bonds.
Solution:
- Both molecules are isomers with the same molecular formula, \(\text{C}_5\text{H}_{12}\), and therefore have the same number of electrons.
- Pentane is a straight-chain alkane. It has a larger surface area and can pack closely with neighbouring pentane molecules, resulting in stronger London dispersion forces.
- 2,2-dimethylpropane is a highly branched, compact molecule. Its spherical shape reduces the contact surface area between molecules, meaning they cannot pack as closely.
- Therefore, the London dispersion forces between 2,2-dimethylpropane molecules are weaker than those between pentane molecules.
- Consequently, less thermal energy is needed to separate 2,2-dimethylpropane molecules, giving it a lower boiling point than pentane.
3. Chemical Reactivity of Alkanes
Alkanes are generally unreactive and do not react with common reagents like acids, alkalis, or oxidising agents under standard laboratory conditions. Their low reactivity is due to two structural features:
- Strong C-C and C-H bonds: The covalent bonds in alkanes have high bond enthalpies (C-C is approximately \( 347\text{ kJ mol}^{-1} \) and C-H is approximately \( 413\text{ kJ mol}^{-1} \)). A large amount of activation energy is required to break these bonds.
- Non-polar bonds: Carbon and hydrogen have very similar electronegativities. The C-H and C-C bonds are non-polar, which means alkanes lack areas of high or low electron density. Consequently, they do not attract nucleophiles or electrophiles.
The only main reactions alkanes undergo are:
- Combustion: Reacting with oxygen to produce energy (as they are highly exothermic).
- Free Radical Substitution: Reacting with halogens in the presence of ultraviolet (UV) light.
Solution:
- The general formula for acyclic alkanes is \( \text{C}_n\text{H}_{2n+2} \).
- Given \( n = 14 \), the number of hydrogen atoms is: \[ 2n + 2 = 2(14) + 2 = 30 \]
- Therefore, the molecular formula of the alkane is \( \text{C}_{14}\text{H}_{30} \).
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