Alkanes

In a nutshell

Alkanes are saturated hydrocarbons: compounds of carbon and hydrogen only, held together entirely by single C-C and C-H σ-bonds.

This subtopic explains their shape, why their boiling points change with chain length and branching, why they are so unreactive, and the two reactions they do have: combustion (as fuels) and radical substitution with chlorine or bromine in UV light.

Assumed knowledge: Basic concepts of organic chemistry: naming, formulae, homolytic fission and radicals; Bonding and structure: electron pair repulsion and London forces.

Core content

Properties of alkanes: σ-bonds and free rotation

Every bond in an alkane is a single bond, so alkanes are saturated hydrocarbons containing single C-C and C-H bonds as σ-bonds.

A σ-bond forms by the overlap of orbitals directly between the bonding atoms. The shared pair of electrons sits on the line joining the two nuclei.

Because the overlap lies along that line, one end of the bond can spin without breaking the overlap. So there is free rotation about a σ-bond: the two CH₃ groups in ethane can twist relative to each other.

Properties of alkanes: tetrahedral shape and 109.5°

Each carbon atom in an alkane is surrounded by four bonded pairs of electrons and no lone pairs.

The explanation, step by step:

1. The four bonded pairs repel each other equally.
2. They spread out to get as far apart as possible.
3. That arrangement is tetrahedral, with a bond angle of 109.5° around each carbon.

The same is true of every carbon in the chain, which is why a straight-chain alkane is really a zig-zag, not a straight line. That zig-zag is what a skeletal formula draws.

Properties of alkanes: boiling points and chain length

Alkane molecules are held to each other only by induced dipole-dipole interactions (London forces). To boil an alkane you overcome these forces between molecules; the covalent bonds inside the molecules do not break.

Boiling point rises as the carbon chain gets longer:

The causal chain the marks follow:

1. A longer chain has more electrons.
2. More electrons give stronger induced dipole-dipole interactions (London forces) between molecules.
3. More energy is needed to overcome these forces, so the boiling point is higher.

Reading the graph:

• Methane to butane (C₁ to C₄) boil below room temperature, so they are gases at room temperature; pentane to decane are liquids.
• The rise is steady but gets slightly smaller with each extra carbon (73 °C from C₁ to C₂, but 23 °C from C₉ to C₁₀).

Properties of alkanes: boiling points and branching

Branched isomers have lower boiling points than their straight-chain isomer. The three isomers of C₅H₁₂ show it clearly:

These are isomers, so they have the same number of electrons. The difference comes from shape:

1. Branching makes a molecule more compact (closer to a sphere).
2. Compact molecules cannot get as close to each other, so there are fewer points of contact / less surface contact between molecules.
3. This gives weaker London forces between molecules.
4. Less energy is needed to overcome them, so the boiling point is lower.

Still don't get it? · why branching lowers the boiling point

Think of two pieces of Velcro tape. Press two long strips flat against each other and they grip all along their length; it takes real effort to pull them apart. Now scrunch each strip into a ball. Two balls only touch at one spot, so they fall apart easily, even though there is exactly the same amount of Velcro.

Now rebuild it for molecules:

1. Pentane and 2,2-dimethylpropane are both C₅H₁₂, so they have the same number of electrons. That rules out "more electrons" as the reason.
2. London forces only act where molecules are close together.
3. Long, straight pentane molecules can lie alongside each other, close together over their whole length: lots of contact.
4. Compact, ball-shaped 2,2-dimethylpropane molecules only touch at a few points.
5. Less contact means weaker London forces, so less energy is needed to separate the molecules and the substance boils at a lower temperature.

Exam wording: "branched isomers have fewer surface points of contact, so weaker London forces (induced dipole-dipole interactions), so less energy is needed to overcome the intermolecular forces."

Reactions of alkanes: why alkanes are unreactive

Alkanes do not react with most reagents (acids, bases, oxidising agents at room temperature). There are two reasons, and you need both:

• High bond enthalpy: C-C and C-H bonds are strong, so a lot of energy is needed to break them.
• Very low polarity of the σ-bonds: carbon and hydrogen have very similar electronegativities, so the C-H bonds are almost non-polar and C-C bonds are non-polar. There are no δ+ or δ- sites to attract polar or charged reagents.

Reactions of alkanes: combustion

Alkanes burn in oxygen in a highly exothermic reaction, which is why they are used as fuels (petrol, diesel, natural gas, camping gas).

Complete combustion happens in a plentiful supply of oxygen. The only products are carbon dioxide and water:

• CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)
• C₈H₁₈(l) + 12½O₂(g) → 8CO₂(g) + 9H₂O(l)

Incomplete combustion happens in a limited supply of oxygen. Carbon monoxide (and sometimes carbon, as soot) forms instead of carbon dioxide:

• CH₄(g) + 1½O₂(g) → CO(g) + 2H₂O(l)
• CH₄(g) + O₂(g) → C(s) + 2H₂O(l)

The danger from CO: carbon monoxide is toxic. It binds to haemoglobin in red blood cells more strongly than oxygen does, so the blood carries less oxygen around the body. It is colourless and odourless, so a build-up (for example from a faulty gas boiler in a poorly ventilated room) goes unnoticed and can kill.

Reactions of alkanes: radical substitution with chlorine and bromine

In ultraviolet (UV) radiation, alkanes react with chlorine or bromine. A hydrogen atom on the alkane is replaced by a halogen atom, giving a haloalkane and a hydrogen halide:

CH₄ + Cl₂ → CH₃Cl + HCl

The mechanism is radical substitution. It runs through three stages. OCR writes it as a sequence of equations, with a dot, •, on every radical to show its unpaired electron. You do not need half curly arrows.

1. Initiation: making radicals

Cl₂ → 2Cl•

• UV radiation supplies the energy to break the Cl-Cl bond.
• The bond breaks by homolytic fission: each bonding atom receives one electron from the bonded pair, forming two radicals.
• A radical is a species with an unpaired electron.

2. Propagation: the chain reaction

CH₄ + Cl• → CH₃• + HCl

CH₃• + Cl₂ → CH₃Cl + Cl•

• The Cl• removes a hydrogen atom from methane, making HCl and a methyl radical.
• The methyl radical takes a Cl atom from Cl₂, making the product and regenerating Cl•.
• Each propagation step has one radical on each side, so the radical count never drops. The regenerated Cl• goes round again: a chain reaction.

3. Termination: removing radicals

Cl• + Cl• → Cl₂

CH₃• + Cl• → CH₃Cl

CH₃• + CH₃• → C₂H₆

• Two radicals combine to form a molecule with no unpaired electron, so the chain stops.

Bromine reacts in exactly the same way: swap every Cl for Br (Br₂ → 2Br•, and HBr is formed in propagation).

Still don't get it? · the radical chain reaction

Picture a game of tag. One player is "it". When "it" tags someone, that person becomes "it" and the first player is free. The game never runs out of an "it", so it keeps going round and round. It only stops when two "it" players bump into each other and both drop out.

Now rebuild it with chlorine:

1. Initiation creates the first players who are "it". UV radiation splits one Cl₂ into two Cl•, each with one unpaired electron. An unpaired electron makes a species very reactive.
2. Propagation step 1: a Cl• grabs an H atom from CH₄. The Cl is now paired up (in HCl), but the carbon has been left with an unpaired electron: CH₃• is now "it".
3. Propagation step 2: CH₃• grabs a Cl atom from a Cl₂ molecule. The carbon is now paired up (in CH₃Cl), but the other Cl atom is left unpaired: a new Cl• is "it".
4. That new Cl• starts step 1 again. One initiation can lead to thousands of propagation cycles.
5. Termination is the two "it" players colliding: any two radicals join, their unpaired electrons pair up into a bond, and two radicals vanish.

Exam check: initiation has radicals only on the right; propagation has one radical on each side; termination has radicals only on the left.

Reactions of alkanes: limitations of radical substitution in synthesis

Radical substitution is a poor way to make one particular haloalkane, because it gives a mixture of organic products. There are two reasons.

1. Further substitution. The product can react again. Cl• can remove an H from CH₃Cl just as it did from CH₄, so more H atoms get replaced:

• CH₃Cl + Cl₂ → CH₂Cl₂ + HCl
• CH₂Cl₂ + Cl₂ → CHCl₃ + HCl
• CHCl₃ + Cl₂ → CCl₄ + HCl

The propagation steps for the first of these are:

CH₃Cl + Cl• → CH₂Cl• + HCl

CH₂Cl• + Cl₂ → CH₂Cl₂ + Cl•

2. Substitution at different positions in the carbon chain. In a longer alkane, the radical can remove an H from any carbon. Propane gives two different monosubstituted products:

Termination steps add to the mixture too, by joining two organic radicals into a longer alkane (for example ethane from two CH₃•).

Worked examples

1. Balancing a complete and an incomplete combustion equation

Task: write balanced equations for the complete combustion of propane, C₃H₈, and for its incomplete combustion to carbon monoxide.

Balance in the order C, then H, then O last (oxygen appears on its own as O₂, so it is easiest to fix at the end).

Complete combustion

1. 3 C atoms give 3CO₂.
2. 8 H atoms give 4H₂O.
3. Count O atoms on the right: $3 \times 2 + 4 \times 1 = 10$, so you need 5O₂.

C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l)

Incomplete combustion to CO

1. 3 C atoms give 3CO.
2. 8 H atoms give 4H₂O.
3. O atoms on the right: $3 \times 1 + 4 \times 1 = 7$, so you need $\tfrac{7}{2}$ = 3½O₂.

C₃H₈(g) + 3½O₂(g) → 3CO(g) + 4H₂O(l)

A half in front of O₂ is fine. If you prefer whole numbers, double everything: 2C₃H₈ + 7O₂ → 6CO + 8H₂O.

Shortcut for any alkane, C_nH_2n+2, burning completely:

$\text{C}_n\text{H}_{2n+2} + \frac{3n+1}{2}\,\text{O}_2 \longrightarrow n\,\text{CO}_2 + (n+1)\,\text{H}_2\text{O}$

Check with octane ($n = 8$): $\frac{3(8)+1}{2} = \frac{25}{2} = 12\tfrac{1}{2}$ O₂, giving 8CO₂ and 9H₂O, as in Core content.

Where marks go: miscounting the O atoms in water (each H₂O has one O), or writing CO₂ as the carbon product when the question says the oxygen supply is limited.

2. Writing the mechanism for a new alkane and halogen

Task: propane reacts with bromine in UV radiation. Write the mechanism that forms 2-bromopropane, CH₃CHBrCH₃.

The pattern is identical to methane and chlorine. The only new skill is putting the radical on the right carbon: for 2-bromopropane, the H must come off the middle carbon.

Initiation

Br₂ → 2Br•

Propagation

CH₃CH₂CH₃ + Br• → CH₃CH•CH₃ + HBr

CH₃CH•CH₃ + Br₂ → CH₃CHBrCH₃ + Br•

Termination (any one is usually asked; know all three types)

Br• + Br• → Br₂

CH₃CH•CH₃ + Br• → CH₃CHBrCH₃

CH₃CH•CH₃ + CH₃CH•CH₃ → (CH₃)₂CHCH(CH₃)₂

Checks before you move on:

• UV radiation stated for initiation.
• Every radical has its dot, and in CH₃CH•CH₃ the dot is on the middle carbon, the one that lost its H.
• The first propagation step makes HBr, not H• and not H₂.
• Each propagation step has one radical on each side.

Where marks go: writing CH₃CH₂CH₃ + Br• → CH₃CH₂CH₂Br + H•. That makes the wrong isomer and produces H•, which is never formed in this mechanism.

3. Overall equations for further substitution

Task: write the overall equation for the conversion of methane into tetrachloromethane, CCl₄, and for ethane into hexachloroethane, C₂Cl₆.

Each H atom replaced uses one Cl₂ and makes one HCl.

1. CH₄ has 4 H atoms to replace, so 4Cl₂ in and 4HCl out:

CH₄ + 4Cl₂ → CCl₄ + 4HCl

2. C₂H₆ has 6 H atoms to replace, so 6Cl₂ in and 6HCl out:

C₂H₆ + 6Cl₂ → C₂Cl₆ + 6HCl

Check the Cl atoms for ethane: left $6 \times 2 = 12$; right $6 + 6 = 12$. Balanced.

Where marks go: writing H₂ as the by-product. The hydrogen always leaves as the hydrogen halide.

Common exam mistakes

• Explaining boiling points with "bigger mass" or "bigger M_r". The cause is more electrons, giving stronger London forces.
• Saying boiling breaks C-C or C-H bonds. Boiling overcomes the intermolecular London forces; the covalent bonds stay intact.
• Explaining branching with "fewer electrons". Isomers have the same number of electrons. The mark is for fewer points of (surface) contact, so weaker London forces.
• Stopping the chain one link short. "Stronger London forces" alone is not enough: finish with "more energy is needed to overcome them".
• Giving 109°, 120° or 90° for the angle in an alkane. It is 109.5°, and the explanation needs four bonded pairs repelling equally / as far apart as possible.
• Saying alkanes are unreactive "because they are saturated". The spec's reasons are high bond enthalpy and very low polarity of the σ-bonds. Give both.
• Describing a σ-bond as just "a shared pair of electrons". Say overlap of orbitals directly between the bonding atoms. Do not bring in sp³ hybridisation: it is not required.
• Saying CO is dangerous because it is a greenhouse gas. The danger is that CO is toxic: it binds to haemoglobin and reduces oxygen transport.
• Forgetting UV radiation in initiation, or saying UV breaks a C-H bond. UV breaks the halogen-halogen bond.
• Writing H• or H₂ in propagation. The first propagation step makes the hydrogen halide (HCl or HBr) and an alkyl radical.
• Missing dots on radicals, or putting the dot on the wrong carbon (for 2-bromopropane, it belongs on the middle carbon: CH₃CH•CH₃).
• Defining homolytic fission as "a bond breaking evenly". Say each bonding atom receives one electron from the bonded pair, forming two radicals.
• Giving only one limitation of radical substitution. Name both: further substitution and substitution at different positions in the chain.

Key definitions

• Hydrocarbon: a compound containing carbon and hydrogen only.
• Saturated: containing single carbon-carbon bonds only.
• Alkane: a saturated hydrocarbon containing single C-C and C-H bonds as σ-bonds (general formula C_nH_2n+2).
• σ-bond: a bond formed by the overlap of orbitals directly between the bonding atoms (free rotation is possible about it).
• Induced dipole-dipole interactions (London forces): attractions between an instantaneous dipole in one molecule and the dipole it induces in a neighbouring molecule.
• Homolytic fission: bond breaking in which each bonding atom receives one electron from the bonded pair, forming two radicals.
• Radical: a species with an unpaired electron.
• Radical substitution: a reaction, initiated by UV radiation, in which a hydrogen atom in an alkane is replaced by a halogen atom via a radical mechanism (initiation, propagation, termination).
• Initiation: the step that forms radicals from a molecule (for example Cl₂ → 2Cl• in UV radiation).
• Propagation: steps in which a radical reacts with a molecule to form a product and another radical, so the chain continues.
• Termination: a step in which two radicals combine to form a molecule, removing radicals and ending the chain.

Specification

• I can describe alkanes as saturated hydrocarbons with C-C and C-H σ-bonds, explain a σ-bond as the overlap of orbitals directly between the bonding atoms, and state that there is free rotation about it. (4.1.2 a)
• I can explain the tetrahedral shape and 109.5° bond angle around each carbon in terms of electron pair repulsion, and draw a 3-D diagram. (4.1.2 b)
• I can explain how chain length and branching affect the boiling points of alkanes in terms of London forces (induced dipole-dipole interactions), and read the trend from a graph or table. (4.1.2 c)
• I can explain the low reactivity of alkanes in terms of high bond enthalpy and very low polarity of the σ-bonds. (4.1.2 d)
• I can write balanced equations for the complete and incomplete combustion of alkanes and explain the danger of carbon monoxide. (4.1.2 e)
• I can write the radical substitution mechanism for an alkane with chlorine or bromine in UV radiation, labelling initiation, propagation and termination and showing radicals with a dot. (4.1.2 f)
• I can explain why radical substitution gives a mixture of products, in terms of further substitution and substitution at different positions in the chain. (4.1.2 g)

Related notes

• Basic concepts of organic chemistry: naming alkanes, formulae, homolytic fission and radicals
• Bonding and structure: electron pair repulsion, polarity and London forces
• Enthalpy changes: enthalpy of combustion and average bond enthalpies
• Alkenes: the C=C double bond, a π-bond that prevents rotation and makes alkenes more reactive
• Haloalkanes: the products of radical substitution, and chlorine radicals breaking down ozone