Alkanes — OCR A-Level Chemistry

This row is a course heading, not an assessed answer target: it tells you that the alkane topic in section 4.1.2 covers the characteristic properties of alkanes. Read it as a map of what to study rather than a question to answer. Work through your specification and textbook treatment of alkanes and build a short summary covering their general formula and homologous series, their saturation (only single C–C and C–H bonds), the σ-bond description of those bonds, free rotation about σ-bonds, the non-polar nature of the molecules, and the resulting physical and chemical behaviour such as low reactivity, increasing boiling points with chain length and combustion. Link each property to a reason, for example why non-polar alkanes do not dissolve well in water. Use the heading to check you can explain, not just list, these properties before moving to the named sub-statements.

Alkanes are saturated hydrocarbons: each carbon forms four single bonds, so the molecules contain only single C–C and C–H bonds with no C=C or C≡C. Each of these single bonds is a σ-bond, formed when orbitals on the two bonding atoms overlap directly along the line joining the nuclei (end-on overlap), giving a region of electron density concentrated between the atoms. Because the σ-bond electron density is symmetrical about the bond axis, rotation about a C–C σ-bond does not break the bond, so the groups can rotate freely relative to one another. In a multiple-choice question, expect options that test saturation, the identity of the bonds, the meaning of direct orbital overlap, or the consequence of free rotation. Read each option against all three ideas before choosing.

Each carbon atom in an alkane is bonded to four other atoms by four single covalent bonds, and those four bonding pairs repel one another as far apart as possible. Electron pair repulsion therefore pushes the four bonds towards the corners of a tetrahedron, giving a bond angle of about 109.5°. In ethane, CH₃CH₃, each carbon is bonded to three hydrogen atoms and to the other carbon, so the H–C–H and H–C–C angles are all close to 109.5°. The shape is described as tetrahedral about each carbon. The repulsion acts between bonding pairs of electrons, not between the nuclei or the atoms themselves, and the four pairs are equivalent because all the bonds are single bonds.

Alkanes are non-polar molecules held together in the liquid by induced dipole–dipole interactions, also called London forces. These arise from temporary fluctuations in electron distribution that induce dipoles in neighbouring molecules. As carbon-chain length increases, the molecule has more electrons and a larger surface area, so the induced dipoles are stronger and more energy is needed to overcome them, raising the boiling point. Branching has the opposite effect: a more branched isomer has a smaller surface area for contact between molecules, so the London forces are weaker and the boiling point is lower than that of its straight-chain isomer. For example, pentane boils higher than its branched isomer 2,2-dimethylpropane.

This row is a study heading, not an assessed answer target: it tells you which reactions of alkanes to read about and learn within 4.1.2. Work through the specification sub-points that sit under it, so you can describe how alkanes burn and how they react with halogens. For combustion, learn that alkanes burn in a plentiful supply of oxygen to give carbon dioxide and water, and that a limited supply gives carbon monoxide or carbon, which is why ventilation matters. For halogenation, learn that alkanes react with halogens such as chlorine or bromine in the presence of ultraviolet light, by a free-radical substitution mechanism. Use the heading to organise your notes into two columns, combustion and halogenation, and check each column against the sub-points.

Alkanes are relatively unreactive because their C–C and C–H bonds are strong σ-bonds with high bond enthalpies, so a lot of energy is needed to break them. The bonds are also essentially non-polar: carbon and hydrogen have similar electronegativities, so the σ-bond electrons are shared fairly evenly and there is no significant δ⁺ or δ⁻ centre. Many reagents, such as polar reagents and nucleophiles, attack regions of partial charge, so they find little to react with on an alkane. This explains why alkanes mainly undergo combustion and free-radical substitution, where radicals or high-energy conditions are involved, rather than addition or substitution by polar reagents. Link this to the bonding work in 2.2.2 j on σ-bonds and bond polarity.

Alkanes burn exothermically and are used as fuels. In a plentiful supply of oxygen, complete combustion occurs: the alkane reacts with O₂ to give carbon dioxide and water only, for example CH₄ + 2O₂ → CO₂ + 2H₂O. In a limited supply of oxygen, incomplete combustion occurs, producing carbon monoxide and/or carbon (soot) alongside water, for example 2CH₄ + 3O₂ → 2CO + 4H₂O. Carbon monoxide is colourless, odourless and toxic because it binds to haemoglobin in the blood, reducing oxygen transport, so faulty or poorly ventilated appliances are dangerous. Carbon (soot) also causes respiratory irritation and deposits. A question may ask you to balance an equation, identify the products, or explain why CO is hazardous.

Alkanes react with chlorine or bromine in the presence of ultraviolet radiation by radical substitution. UV light supplies energy to break the halogen molecule by homolytic fission, forming two halogen radicals; this is initiation. In propagation, a halogen radical abstracts a hydrogen atom from the alkane to give a hydrogen halide and an alkyl radical, then the alkyl radical reacts with a halogen molecule to give the halogenoalkane and a new halogen radical. Termination combines two radicals to end the chain, for example two halogen radicals, two alkyl radicals, or an alkyl and a halogen radical. A question may ask you to identify the type of fission, classify a step, or spot the termination step.

Radical substitution converts an alkane into a halogenoalkane, but it is a poor way to make one pure product. Once some halogenoalkane has formed, the halogen can substitute again at the same carbon or elsewhere, giving di- and poly-substituted products such as dichloromethane alongside chloromethane when methane reacts with chlorine. Substitution can also occur at different positions along a carbon chain, so a longer alkane such as propane yields a mixture of 1-chloropropane and 2-chloropropane. Separating these similar products is difficult, so the yield of the single desired compound is low. A multiple-choice question tests whether you can identify both limitations: further substitution and substitution at different chain positions.