Alkanes

Alkanes are saturated hydrocarbons with the general formula $\mathrm{C}_n\mathrm{H}_{2n+2}$. "Saturated" means every carbon atom is bonded to the maximum possible number of hydrogen atoms -- there are no C=C or $\mathrm{C}\equiv\mathrm{C}$ bonds. All carbon-carbon and carbon-hydrogen bonds are $\sigma$ bonds, formed by head-on overlap of hybridised orbitals ($sp^3$ for carbon).

Structure and Bonding

Each carbon in an alkane is $sp^3$ hybridised with tetrahedral geometry (bond angles approximately $109.5^\circ$). The C--C bond length is approximately $154\,\mathrm{pm}$, and the C--H bond length is approximately $109\,\mathrm{pm}$.

Free rotation about C--C $\sigma$ bonds produces an infinite set of molecular conformations. The three staggered conformations of ethane (gauche and anti) are energy minima, while the eclipsed conformations are energy maxima. The rotational energy barrier in ethane is approximately $12\,\mathrm{kJ/mol}$.

Physical Properties

State at Room Temperature

The state depends on chain length. Methane to butane ($\mathrm{C}_1$--$\mathrm{C}_4$) are gases, pentane to hexadecane ($\mathrm{C}_5$--$\mathrm{C}_{16}$) are liquids, and heptadecane and above are waxy solids.

Boiling Points

Boiling point increases with chain length due to increasing surface area and therefore stronger London (dispersion) forces. The relationship is approximately linear for the first ~20 members, with each additional $-\mathrm{CH}_2-$ group contributing roughly 20--25 K to the boiling point.

Branched alkanes have lower boiling points than their straight-chain isomers because branching reduces the surface area for intermolecular contact. For example, 2,2-dimethylpropane (b.p. $9.5^\circ\mathrm{C}$) vs pentane (b.p. $36^\circ\mathrm{C}$) vs 2-methylbutane (b.p. $28^\circ\mathrm{C}$).

Solubility

Alkanes are non-polar and are insoluble in water. They dissolve in non-polar organic solvents (the general principle of "like dissolves like"). Their density is less than that of water.

Free Radical Substitution

Alkanes are generally unreactive. The C--C and C--H bonds are strong and non-polar. However, under UV light or at high temperatures, alkanes undergo substitution reactions via a radical chain mechanism.

The Chlorination of Methane

The reaction between methane and chlorine under UV light is the canonical example.

$$\mathrm{CH}_4 + \mathrm{Cl}_2 \xrightarrow{\mathrm{UV}} \mathrm{CH}_3\mathrm{Cl} + \mathrm{HCl}$$

Mechanism: Step-by-Step

Initiation: Homolytic fission of the $\mathrm{Cl-Cl}$ bond, requiring UV light to supply the bond dissociation energy ($\mathrm{BDE} = 242\,\mathrm{kJ/mol}$):

$$\mathrm{Cl}_2 \xrightarrow{h\nu} 2\mathrm{Cl}^\bullet$$

Propagation -- Step 1: A chlorine radical abstracts a hydrogen atom from methane, forming hydrogen chloride and a methyl radical:

$$\mathrm{Cl}^\bullet + \mathrm{CH}_4 \to \mathrm{HCl} + \mathrm{CH}_3^\bullet$$

This step has $\Delta H \approx +4\,\mathrm{kJ/mol}$ (slightly endothermic). The H--Cl bond formed ($431\,\mathrm{kJ/mol}$) nearly compensates the C--H bond broken ($435\,\mathrm{kJ/mol}$).

Propagation -- Step 2: The methyl radical reacts with a chlorine molecule, forming chloromethane and regenerating the chlorine radical:

$$\mathrm{CH}_3^\bullet + \mathrm{Cl}_2 \to \mathrm{CH}_3\mathrm{Cl} + \mathrm{Cl}^\bullet$$

This step has $\Delta H \approx -108\,\mathrm{kJ/mol}$ (exothermic). The C--Cl bond formed ($349\,\mathrm{kJ/mol}$) and the $\mathrm{Cl-Cl}$ bond broken ($242\,\mathrm{kJ/mol}$) give a net release.

The chlorine radical is regenerated in step 2, which is why this is a chain reaction. A single initiation event can produce thousands of product molecules before termination.

Termination: Radical-radical combination reactions that remove reactive species from the chain:

$$2\mathrm{Cl}^\bullet \to \mathrm{Cl}_2$$ $$\mathrm{Cl}^\bullet + \mathrm{CH}_3^\bullet \to \mathrm{CH}_3\mathrm{Cl}$$ $$2\mathrm{CH}_3^\bullet \to \mathrm{C}_2\mathrm{H}_6$$

Termination becomes significant only when radical concentrations are high enough to make bimolecular encounters probable.

Energetics of the Chain Reaction

The overall reaction is exothermic: $\Delta H \approx -104\,\mathrm{kJ/mol}$. However, the initiation step requires significant energy input ($242\,\mathrm{kJ/mol}$). The propagation steps provide the driving force: the first propagation step is slightly endothermic, and the second is strongly exothermic.

Multiple Substitution

The chloromethane product can itself undergo further substitution, producing dichloromethane ($\mathrm{CH}_2\mathrm{Cl}_2$), trichloromethane/chloroform ($\mathrm{CHCl}_3$), and tetrachloromethane/carbon tetrachloride ($\mathrm{CCl}_4$). The product mixture is statistical, not selective, which limits the synthetic utility of radical chlorination.

The relative rates of substitution at different hydrogen positions follow the order:

$$3^\circ\,\mathrm{H} \gt 2^\circ\,\mathrm{H} \gt 1^\circ\,\mathrm{H}$$

This is because the stability of the radical intermediate increases with substitution: tertiary radicals are stabilised by hyperconjugation and inductive effects from neighbouring alkyl groups.

Worked Example. For the monochlorination of propane, calculate the ratio of 1-chloropropane to 2-chloropropane.

There are six primary hydrogens and two secondary hydrogens. The relative reactivity of $2^\circ$ vs $1^\circ$ hydrogens is approximately 4:1.

$$\frac◆LB◆[2\mathrm{-chloropropane}]◆RB◆◆LB◆[1\mathrm{-chloropropane}]◆RB◆ = \frac◆LB◆2 \times 4◆RB◆◆LB◆6 \times 1◆RB◆ = \frac{8}{6} = \frac{4}{3}$$

The product ratio is 4:3 in favour of 2-chloropropane.

Comparison with Bromination

Bromine radicals are less reactive but more selective than chlorine radicals. The H--Br bond formed ($366\,\mathrm{kJ/mol}$) is weaker than the H--Cl bond ($431\,\mathrm{kJ/mol}$), making the first propagation step more endothermic. By the Hammond postulate, the transition state for hydrogen abstraction by $\mathrm{Br}^\bullet$ resembles the products (radical character is more developed), so the stability of the radical product has a larger effect on the activation energy.

Bromination is therefore far more selective: $3^\circ\,\mathrm{H} : 2^\circ\,\mathrm{H} : 1^\circ\,\mathrm{H} \approx 1600 : 82 : 1$.

Combustion

Complete Combustion

In excess oxygen, alkanes burn to produce carbon dioxide and water:

$$\mathrm{C}_n\mathrm{H}_{2n+2} + \frac{3n+1}{2}\mathrm{O}_2 \to n\mathrm{CO}_2 + (n+1)\mathrm{H}_2\mathrm{O}$$

All combustion reactions are highly exothermic. For methane:

$$\mathrm{CH}_4 + 2\mathrm{O}_2 \to \mathrm{CO}_2 + 2\mathrm{H}_2\mathrm{O} \quad \Delta H_c^\circ = -890\,\mathrm{kJ/mol}$$

Incomplete Combustion

When oxygen is limited, carbon monoxide and/or carbon (soot) are produced:

$$2\mathrm{C}_n\mathrm{H}_{2n+2} + (2n+1)\mathrm{O}_2 \to 2n\mathrm{CO} + (2n+2)\mathrm{H}_2\mathrm{O}$$ $$\mathrm{C}_n\mathrm{H}_{2n+2} + \frac{n+1}{2}\mathrm{O}_2 \to n\mathrm{C} + (n+1)\mathrm{H}_2\mathrm{O}$$

Carbon monoxide is a toxic, colourless, odourless gas that binds irreversibly to haemoglobin ($K \approx 200\times$ that of $\mathrm{O}_2$), reducing the blood's oxygen-carrying capacity. Incomplete combustion of alkanes in poorly ventilated spaces is a major cause of CO poisoning.

Calculating Enthalpy of Combustion from Bond Enthalpies

$$\Delta H_c = \sum \mathrm{BE}(\mathrm{reactants}) - \sum \mathrm{BE}(\mathrm{products})$$

This uses mean bond enthalpies and introduces systematic error because the actual bond enthalpies in the specific molecules differ from the mean values. For precise work, experimental enthalpies (from calorimetry) must be used.

Worked Example. Estimate $\Delta H_c$ for methane using mean bond enthalpies: C--H = $413\,\mathrm{kJ/mol}$, O=O = $498\,\mathrm{kJ/mol}$, C=O (in $\mathrm{CO}_2$) = $805\,\mathrm{kJ/mol}$, O--H = $464\,\mathrm{kJ/mol}$.

Bonds broken: $4 \times 413 + 2 \times 498 = 1652 + 996 = 2648\,\mathrm{kJ/mol}$.

Bonds formed: $2 \times (2 \times 805) + 4 \times 464 = 3220 + 1856 = 5076\,\mathrm{kJ/mol}$.

$$\Delta H_c = 2648 - 5076 = -2428\,\mathrm{kJ/mol}$$

This overestimates the experimental value ($-890\,\mathrm{kJ/mol}$ per mole of methane, which is $-890\,\mathrm{kJ/mol}$ for the balanced equation as written with 1 mol methane). The discrepancy arises because mean bond enthalpies are averages and do not account for the specific molecular environment. Note: the bond enthalpy calculation above is per mole of methane as written.

Environmental Impact of Alkanes

Fossil Fuels and the Carbon Cycle

Alkanes are the primary components of fossil fuels: natural gas (mostly methane), petroleum (mix of liquid alkanes $\mathrm{C}_5$--$\mathrm{C}_{12}$), and coal (complex hydrocarbon structures).

Combustion of fossil fuels releases $\mathrm{CO}_2$, which contributes to the enhanced greenhouse effect. The atmospheric concentration of $\mathrm{CO}_2$ has increased from approximately $280\,\mathrm{ppm}$ in pre-industrial times to over $420\,\mathrm{ppm}$ as of 2025.

Methane as a Greenhouse Gas

Methane is a far more potent greenhouse gas than $\mathrm{CO}_2$ on a per-molecule basis. Its global warming potential (GWP) over 100 years is approximately 28--36 times that of $\mathrm{CO}_2$. Major anthropogenic sources include natural gas leaks, livestock (enteric fermentation), rice paddies, and landfill decomposition.

Photochemical Smog

Incomplete combustion of alkanes (particularly from internal combustion engines) releases unburnt hydrocarbons and nitrogen oxides ($\mathrm{NO}_x$). In sunlight, these undergo a series of free radical reactions to produce ozone ($\mathrm{O}_3$), peroxyacyl nitrates (PANs), and other secondary pollutants at ground level. This mixture constitutes photochemical smog, which causes respiratory problems and damages vegetation.

Cracking

The demand for shorter-chain hydrocarbons (as fuels and chemical feedstocks) exceeds their natural abundance in crude oil. Cracking is the thermal decomposition of long-chain alkanes into shorter molecules. Two types:

Thermal cracking: High temperature ($400$--$900^\circ\mathrm{C}$) and pressure. Produces a mixture of alkenes and smaller alkanes. Free radical mechanism. Example:

$$\mathrm{C}_{10}\mathrm{H}_{22} \to \mathrm{C}_5\mathrm{H}_{10} + \mathrm{C}_5\mathrm{H}_{12}$$

Catalytic cracking: Lower temperature ($450^\circ\mathrm{C}$) and moderate pressure, using a zeolite catalyst. Produces branched alkanes, cycloalkanes, and aromatic compounds. The catalyst provides an alternative reaction pathway with lower activation energy (carbocation mechanism). Catalytic cracking also produces higher-quality gasoline (higher octane rating due to branching).

Common Pitfalls

1. Assuming radical substitution gives a single product. The product is always a mixture. Chloromethane reacts further to give dichloromethane, chloroform, and carbon tetrachloride. This limits the synthetic utility of radical halogenation.

2. Writing the wrong termination reactions. Termination requires two radicals. A radical combining with a non-radical molecule is not termination -- it is a propagation step.

3. Confusing the energetics of propagation steps. The first propagation step (hydrogen abstraction) is typically slightly endothermic for chlorination, while the second (halogen attack) is strongly exothermic. For bromination, the first step is much more endothermic, which accounts for the greater selectivity.

4. Forgetting the UV initiation requirement. Without UV light, the $\mathrm{Cl-Cl}$ bond does not undergo homolytic fission at room temperature. The reaction does not proceed in the dark.

5. Incorrect stoichiometry in combustion equations. Always balance the equation fully. For an alkane $\mathrm{C}_n\mathrm{H}_{2n+2}$, the stoichiometric oxygen requirement is $\frac{3n+1}{2}$ moles per mole of alkane.

6. Confusing thermal and catalytic cracking. Thermal cracking uses higher temperatures without a catalyst and produces alkenes via free radical chemistry. Catalytic cracking uses lower temperatures with a zeolite catalyst and produces more branched and cyclic products.

7. Assuming all C--H bonds are equally reactive. In radical halogenation, tertiary C--H bonds are weaker and more reactive than secondary, which are more reactive than primary. This selectivity is much more pronounced for bromine than for chlorine.

Radical Halogenation: Selectivity and Relative Rates

Reactivity of Different C--H Bonds

C--H bond type	Bond enthalpy ($\mathrm{kJ/mol}$)	Relative reactivity (Cl)	Relative reactivity (Br)
Primary ($1^\circ$)	410	1	1
Secondary ($2^\circ$)	395	3.8	82
Tertiary ($3^\circ$)	380	5.0	1600

Chlorine is relatively unselective (only a 5:1 preference for tertiary over primary) because the first propagation step is only slightly endothermic, so the transition state is early (reactant-like) and the energy difference between C--H bond types is small.

Bromine is highly selective (1600:1 for tertiary over primary) because the first propagation step is significantly endothermic, so the transition state is late (product-like) and the full energy difference between C--H bond types is expressed.

Calculating Product Distributions

Worked Example. Predict the relative amounts of 1-chloropropane and 2-chloropropane from the chlorination of propane.

Propane has 6 primary hydrogens and 2 secondary hydrogens.

Relative amount of 1-chloropropane $= 6 \times 1 = 6$

Relative amount of 2-chloropropane $= 2 \times 3.8 = 7.6$

Percentage of 1-chloropropane $= 6 / (6 + 7.6) \times 100 = 44\%$

Percentage of 2-chloropropane $= 7.6 / (6 + 7.6) \times 100 = 56\%$

The secondary product predominates despite having fewer hydrogens, because each secondary hydrogen is nearly 4 times more reactive.

Detailed Fractional Distillation of Crude Oil

Fraction Composition

Fraction	Carbon range	Boiling range ($^\circ\mathrm{C}$)	Uses
Refinery gas	$\mathrm{C}_1$--$\mathrm{C}_4$	Below 25	Bottled gas, fuel
Petrol (gasoline)	$\mathrm{C}_5$--$\mathrm{C}_{10}$	$30$--$150$	Motor fuel
Naphtha	$\mathrm{C}_6$--$\mathrm{C}_{14}$	$100$--$200$	Petrochemical feedstock
Kerosene (paraffin)	$\mathrm{C}_{10}$--$\mathrm{C}_{16}$	$150$--$250$	Jet fuel, heating
Diesel oil	$\mathrm{C}_{14}$--$\mathrm{C}_{20}$	$220$--$350$	Diesel engines
Fuel oil	$\mathrm{C}_{18}$--$\mathrm{C}_{30}$	$350$--$500$	Ship fuel, power stations
Bitumen	$\mathrm{C}_{30}+$	Above 500	Road surfacing, roofing

Why Demand Exceeds Supply

Modern demand favours lighter fractions (petrol, diesel, naphtha) over heavier fractions (fuel oil, bitumen). Cracking and reforming convert surplus heavy fractions into more valuable lighter products.

Catalytic Reforming

Catalytic reforming converts straight-chain alkanes into branched alkanes and cycloalkanes (which have higher octane ratings) using a platinum catalyst at $500^\circ\mathrm{C}$. This is the primary industrial source of aromatic hydrocarbons (benzene, methylbenzene) which are produced by the dehydrogenation of cycloalkanes.

Greenhouse Effect and Carbon Footprint

Carbon Dioxide as a Greenhouse Gas

$\mathrm{CO}_2$ absorbs infrared radiation emitted by the Earth's surface and re-radiates it back towards the surface, warming the atmosphere. The concentration of atmospheric $\mathrm{CO}_2$ has increased from approximately $280\,\mathrm{ppm}$ (pre-industrial) to over $420\,\mathrm{ppm}$ (2024), primarily due to combustion of fossil fuels.

Global Warming Potential (GWP)

Greenhouse gas	GWP (100-year)	Atmospheric lifetime	Primary source
$\mathrm{CO}_2$	1	100--1000 years	Fossil fuel combustion
$\mathrm{CH}_4$	28	12 years	Natural gas leaks, agriculture
$\mathrm{N}_2\mathrm{O}$	265	114 years	Fertiliser use, industrial
$\mathrm{CF}_4$ (CFC-14)	6,630	50,000 years	Electronics industry

Carbon Capture and Storage (CCS)

CCS involves capturing $\mathrm{CO}_2$ from point sources (power stations, cement works) and storing it underground in depleted oil/gas reservoirs or deep saline aquifers. Pre-combustion capture (gasification), post-combustion capture (amine scrubbing), and oxy-fuel combustion are the three main approaches.

Practice Problems

Problem 1

Write the mechanism for the radical bromination of ethane to give bromoethane. Calculate $\Delta H$ for each propagation step given: C--H bond enthalpy in ethane $= 420\,\mathrm{kJ/mol}$, H--Br bond enthalpy $= 366\,\mathrm{kJ/mol}$, C--Br bond enthalpy in bromoethane $= 285\,\mathrm{kJ/mol}$, Br--Br bond enthalpy $= 193\,\mathrm{kJ/mol}$.

Solution:

Initiation: $\mathrm{Br}_2 \xrightarrow{h\nu} 2\mathrm{Br}^\bullet \quad \Delta H = +193\,\mathrm{kJ/mol}$

Propagation step 1: $\mathrm{Br}^\bullet + \mathrm{CH}_3\mathrm{CH}_3 \to \mathrm{HBr} + \mathrm{CH}_3\mathrm{CH}_2^\bullet$

$$\Delta H_1 = \mathrm{BE}(\mathrm{C{-}H}) - \mathrm{BE}(\mathrm{H{-}Br}) = 420 - 366 = +54\,\mathrm{kJ/mol}$$

Propagation step 2: $\mathrm{CH}_3\mathrm{CH}_2^\bullet + \mathrm{Br}_2 \to \mathrm{CH}_3\mathrm{CH}_2\mathrm{Br} + \mathrm{Br}^\bullet$

$$\Delta H_2 = \mathrm{BE}(\mathrm{Br{-}Br}) - \mathrm{BE}(\mathrm{C{-}Br}) = 193 - 285 = -92\,\mathrm{kJ/mol}$$

Overall: $\Delta H = +54 + (-92) = -38\,\mathrm{kJ/mol}$ (exothermic).

The endothermic first propagation step ($+54\,\mathrm{kJ/mol}$) is much less favourable than for chlorination ($+4\,\mathrm{kJ/mol}$), which is why bromination is slower but more selective.

Problem 2

Explain why the radical chlorination of 2-methylpropane gives predominantly 2-chloro-2-methylpropane, even though there are nine primary hydrogens and only one tertiary hydrogen.

Solution:

Although the ratio of primary to tertiary hydrogens is 9:1, the relative reactivity of tertiary vs primary hydrogens toward $\mathrm{Cl}^\bullet$ is approximately 5:1. The expected ratio is:

$$\frac◆LB◆[2\mathrm{-chloro-2-methylpropane}]◆RB◆◆LB◆[1\mathrm{-chloro-2-methylpropane}]◆RB◆ = \frac◆LB◆1 \times 5◆RB◆◆LB◆9 \times 1◆RB◆ = \frac{5}{9}$$

In practice, the tertiary product predominates even more than this ratio predicts because the statistical calculation assumes all primary positions are equivalent, but in 2-methylpropane there are two different primary environments ($-\mathrm{CH}_3$ on C-1 vs $-\mathrm{CH}_3$ on C-3). The tertiary radical $(\mathrm{CH}_3)_3\mathrm{C}^\bullet$ is stabilised by hyperconjugation from the nine neighbouring C--H bonds and the inductive effect of three methyl groups, making it significantly lower in energy than the primary radical $\mathrm{CH}_3\mathrm{CH}^\bullet(\mathrm{CH}_3)\mathrm{CH}_3$.

Problem 3

Write equations for the complete and incomplete combustion of butane ($\mathrm{C}_4\mathrm{H}_{10}$). Calculate the enthalpy change for complete combustion using bond enthalpies, and explain why incomplete combustion occurs in limited oxygen.

Solution:

Complete combustion:

$$2\mathrm{C}_4\mathrm{H}_{10}(g) + 13\mathrm{O}_2(g) \to 8\mathrm{CO}_2(g) + 10\mathrm{H}_2\mathrm{O}(l)$$

Incomplete combustion (producing CO):

$$2\mathrm{C}_4\mathrm{H}_{10}(g) + 9\mathrm{O}_2(g) \to 8\mathrm{CO}(g) + 10\mathrm{H}_2\mathrm{O}(l)$$

Further incomplete combustion (producing C):

$$2\mathrm{C}_4\mathrm{H}_{10}(g) + 5\mathrm{O}_2(g) \to 8\mathrm{C}(s) + 10\mathrm{H}_2\mathrm{O}(l)$$

Incomplete combustion occurs when the supply of oxygen is insufficient for all carbon to be fully oxidised to $\mathrm{CO}_2$. The products are a mixture of $\mathrm{CO}_2$, $\mathrm{CO}$, $\mathrm{C}$ (soot), and unburnt hydrocarbons. $\mathrm{CO}$ is toxic because it binds irreversibly to haemoglobin, reducing the blood's oxygen-carrying capacity.

Enthalpy of complete combustion using bond enthalpies (per mole of butane):

Bonds broken: $10 \times \mathrm{C-H}$ ($10 \times 413 = 4130$) + $3 \times \mathrm{C-C}$ ($3 \times 347 = 1041$) + $\frac{13}{2} \times \mathrm{O=O}$ ($6.5 \times 498 = 3237$) = $8408\,\mathrm{kJ/mol}$

Bonds formed: $4 \times 2 \times \mathrm{C=O}$ ($8 \times 805 = 6440$) + $5 \times 2 \times \mathrm{O-H}$ ($10 \times 464 = 4640$) = $11080\,\mathrm{kJ/mol}$

$$\Delta H_c = 8408 - 11080 = -2672\,\mathrm{kJ/mol}$$

The literature value is $-2877\,\mathrm{kJ/mol}$. The discrepancy arises from using mean bond enthalpies rather than specific values for butane and its products.

Conformations of Alkanes

Free Rotation About Sigma Bonds

Rotation about single ($\sigma$) bonds is relatively free, giving alkanes a range of conformations. For ethane, the two methyl groups can rotate relative to each other:

• Eclipsed conformation: H atoms on the front and back carbons are aligned. Highest energy due to torsional strain and steric repulsion between electron clouds.
• Staggered conformation: H atoms on the front carbon are positioned between those on the back carbon. Lowest energy (most stable).
• The torsional barrier is approximately $12\,\mathrm{kJ/mol}$ for ethane, small enough to allow rapid rotation at room temperature.

Conformations of Butane

Butane has two important staggered conformations:

• Anti conformation: The two methyl groups are maximally separated ($180^\circ$ dihedral angle). Lowest energy.
• Gauche conformation: The two methyl groups are $60^\circ$ apart. Higher energy than anti by approximately $3.7\,\mathrm{kJ/mol}$ due to van der Waals repulsion between the methyl groups.

The energy difference between anti and gauche is small, so both conformations are populated at room temperature, with the anti conformation being more abundant.

Environmental and Industrial Significance

Crude Oil and Fractional Distillation

Crude oil is a mixture of hydrocarbons that is separated by fractional distillation based on boiling point:

Fraction	Carbon range	Boiling range ($^\circ\mathrm{C}$)	Uses
Refinery gas	$\mathrm{C}_1$--$\mathrm{C}_4$	Below 40	Fuel gas, LPG
Gasoline (petrol)	$\mathrm{C}_5$--$\mathrm{C}_{10}$	40--170	Motor fuel
Naphtha	$\mathrm{C}_6$--$\mathrm{C}_{12}$	60--200	Petrochemical feedstock
Kerosene	$\mathrm{C}_{10}$--$\mathrm{C}_{16}$	170--250	Jet fuel, heating
Diesel	$\mathrm{C}_{14}$--$\mathrm{C}_{20}$	250--340	Diesel engines
Fuel oil	$\mathrm{C}_{20}$--$\mathrm{C}_{40}$	340--500	Ship fuel, power stations
Bitumen	$>\mathrm{C}_{50}$	Above 500	Roads, roofing

Cracking

Larger hydrocarbons (high boiling fractions) are in less demand than smaller ones. Cracking breaks large molecules into smaller, more useful ones:

Thermal cracking: High temperature ($400$--$900^\circ\mathrm{C}$) and pressure. Produces a mixture of alkanes and alkenes. Free radical mechanism. Used to produce ethene.

$$\mathrm{C}_{10}\mathrm{H}_{22} \xrightarrow{\Delta} \mathrm{C}_8\mathrm{H}_{18} + \mathrm{C}_2\mathrm{H}_4$$

Catalytic cracking: Lower temperature ($450^\circ\mathrm{C}$) with a zeolite catalyst. Produces branched alkanes and cycloalkanes (better for petrol, higher octane rating).

Octane Rating

The octane rating measures the resistance of a fuel to premature detonation (knocking) in a petrol engine:

• 2,2,4-trimethylpentane (iso-octane): octane rating = 100 (very resistant to knocking).
• Heptane: octane rating = 0 (knocks severely).
• Branched and cyclic alkanes have higher octane ratings than straight-chain alkanes.

Catalytic cracking increases octane rating by producing more branched molecules. Tetraethyl lead ($\mathrm{Pb}(\mathrm{C}_2\mathrm{H}_5)_4$) was historically used as an anti-knock additive but is now banned due to lead toxicity.

Additional Practice Problems

Problem 3

Write the mechanism for the radical chlorination of propane, showing all propagation and termination steps. Explain why 1-chloropropane and 2-chloropropane are both formed, and predict which is the major product.

Solution:

Initiation:

$$\mathrm{Cl}_2 \xrightarrow{\mathrm{UV}} 2\mathrm{Cl}\cdot$$

Propagation (pathway to 1-chloropropane):

$$\mathrm{Cl}\cdot + \mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_3 \to \mathrm{HCl} + \mathrm{CH}_2\cdot\mathrm{CH}_2\mathrm{CH}_3 \quad (\text{abstracts a primary H})$$$$\mathrm{CH}_2\cdot\mathrm{CH}_2\mathrm{CH}_3 + \mathrm{Cl}_2 \to \mathrm{CH}_2\mathrm{ClCH}_2\mathrm{CH}_3 + \mathrm{Cl}\cdot$$

Propagation (pathway to 2-chloropropane):

$$\mathrm{Cl}\cdot + \mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_3 \to \mathrm{HCl} + \mathrm{CH}_3\mathrm{CH}\cdot\mathrm{CH}_3 \quad (\text{abstracts a secondary H})$$$$\mathrm{CH}_3\mathrm{CH}\cdot\mathrm{CH}_3 + \mathrm{Cl}_2 \to \mathrm{CH}_3\mathrm{CHClCH}_3 + \mathrm{Cl}\cdot$$

Termination:

$$\mathrm{Cl}\cdot + \mathrm{Cl}\cdot \to \mathrm{Cl}_2$$$$\mathrm{Cl}\cdot + \mathrm{R}\cdot \to \mathrm{RCl}$$$$\mathrm{R}\cdot + \mathrm{R}'\cdot \to \mathrm{RR}'$$

(where R and R' are any organic radicals formed)

Major product: 1-chloropropane predominates statistically because propane has 6 primary hydrogens vs 2 secondary hydrogens. However, the selectivity factor for chlorination is only about 3.8:1 (secondary:primary), so:

• 1-chloropropane: $6 \times 1 = 6$ (relative yield)
• 2-chloropropane: $2 \times 3.8 = 7.6$ (relative yield)

Percentage of 2-chloropropane $= 7.6/(6 + 7.6) = 56\%$. The secondary product (2-chloropropane) is the major product despite fewer secondary hydrogens.

Problem 4

$10.0\,\mathrm{g}$ of an alkane is completely burned in excess oxygen. The $\mathrm{CO}_2$ produced is absorbed by $\mathrm{NaOH}$ solution, causing an increase in mass of $30.8\,\mathrm{g}$. Identify the alkane.

Solution:

Mass of $\mathrm{CO}_2$ absorbed $= 30.8\,\mathrm{g}$.

$$n(\mathrm{CO}_2) = \frac{30.8}{44.0} = 0.700\,\mathrm{mol}$$

If the alkane has formula $\mathrm{C}_n\mathrm{H}_{2n+2}$, combustion produces $n$ moles of $\mathrm{CO}_2$:

$$n = 0.700$$

Molar mass of alkane $= 10.0/0.700 \times (1/n)$... Wait: $n(\text{alkane}) = n(\mathrm{CO}_2)/n_\mathrm{C}$.

If 1 mol alkane produces $n$ mol $\mathrm{CO}_2$: $n(\text{alkane}) = 0.700/n$.

$M = 10.0/n(\text{alkane}) = 10.0n/0.700 = 14.3n$.

For $\mathrm{C}_5\mathrm{H}_{12}$: $M = 60 + 12 = 72$. $14.3n = 72 \implies n = 5.03 \approx 5$.

The alkane is pentane ($\mathrm{C}_5\mathrm{H}_{12}$, $M = 72.0\,\mathrm{g/mol}$).

Check: $n(\text{pentane}) = 10.0/72.0 = 0.139\,\mathrm{mol}$. $n(\mathrm{CO}_2) = 5 \times 0.139 = 0.694\,\mathrm{mol}$. $m(\mathrm{CO}_2) = 0.694 \times 44.0 = 30.5\,\mathrm{g}$ (close to $30.8\,\mathrm{g}$; the small discrepancy is due to rounding).

Advanced Alkanes

Combustion Calculations: Incomplete Combustion and Environmental Impact

Incomplete combustion occurs when oxygen supply is limited, producing carbon monoxide and/or carbon (soot) in addition to (or instead of) carbon dioxide.

$\mathrm{C}_n\mathrm{H}_{2n+2} + \frac{3n+1}{2}\mathrm{O}_2 \to n\mathrm{CO} + (n+1)\mathrm{H}_2\mathrm{O} \quad \text{(partial)}$

$\mathrm{C}_n\mathrm{H}_{2n+2} + \frac{n+1}{2}\mathrm{O}_2 \to n\mathrm{C} + (n+1)\mathrm{H}_2\mathrm{O} \quad \text{(very limited)}$

Environmental impact of combustion products:

Product	Environmental Effect
$\mathrm{CO}_2$	Greenhouse gas, contributes to global warming
$\mathrm{CO}$	Toxic (binds to haemoglobin, reducing $\mathrm{O}_2$ transport)
$\mathrm{C}$ (soot/particulates)	Respiratory problems, global dimming
$\mathrm{NO}_x$	Acid rain, photochemical smog (formed from $\mathrm{N}_2$ in air at high temperature)
$\mathrm{SO}_2$	Acid rain (from sulphur impurities in fuel)

Free Radical Substitution: Advanced Mechanism

The mechanism proceeds through three stages. Using ethane and chlorine as an example:

Stage 1 -- Initiation: $\mathrm{Cl}_2 \xrightarrow{h\nu} 2\mathrm{Cl}^\bullet$

The Cl--Cl bond ($243\,\mathrm{kJ/mol}$) is homolytically cleaved by UV light. The bond dissociation energy must be supplied by a photon with energy $E \geq 243\,\mathrm{kJ/mol}$, corresponding to $\lambda \leq 493\,\mathrm{nm}$ (visible blue-green light).

Stage 2 -- Propagation: $\mathrm{Cl}^\bullet + \mathrm{CH}_3\mathrm{CH}_3 \to \mathrm{CH}_3\mathrm{CH}_2^\bullet + \mathrm{HCl}$ $\mathrm{CH}_3\mathrm{CH}_2^\bullet + \mathrm{Cl}_2 \to \mathrm{CH}_3\mathrm{CH}_2\mathrm{Cl} + \mathrm{Cl}^\bullet$

The chain carrier ($\mathrm{Cl}^\bullet$) is regenerated, allowing the chain to continue.

Stage 3 -- Termination: $2\mathrm{Cl}^\bullet \to \mathrm{Cl}_2$ $2\mathrm{CH}_3\mathrm{CH}_2^\bullet \to \mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{CH}_3 \quad \text{(butane)}$ $\mathrm{CH}_3\mathrm{CH}_2^\bullet + \mathrm{Cl}^\bullet \to \mathrm{CH}_3\mathrm{CH}_2\mathrm{Cl}$

Termination produces a mixture of products, which is why free radical substitution is not a useful synthetic method for preparing pure haloalkanes.

Relative Reactivity of Halogens in Free Radical Substitution

The reactivity of halogens in free radical substitution with alkanes follows the trend:

$\mathrm{F}_2 \gg \mathrm{Cl}_2 > \mathrm{Br}_2 \gg \mathrm{I}_2$

Halogen	Bond dissociation energy (kJ/mol)	Reactivity	Selectivity
$\mathrm{F}_2$	158	Extremely high (explosive)	Very low (random substitution)
$\mathrm{Cl}_2$	243	Moderate	Moderate (3:1 primary:secondary)
$\mathrm{Br}_2$	193	Low (requires heat or UV)	High (1600:1 primary:secondary)
$\mathrm{I}_2$	151	Negligible (reaction is endothermic)	--

Fluorine reacts too violently to be useful. Iodine does not react because the H-abstraction step is endothermic (the H--I bond formed, $297\,\mathrm{kJ/mol}$, is weaker than the C--H bond broken, $\approx 410\,\mathrm{kJ/mol}$ for primary C--H).

Bond Dissociation Energies and Radical Stability

C--H Bond	Bond dissociation energy (kJ/mol)	Radical formed
$\mathrm{CH}_3$--H (primary)	439	$\mathrm{CH}_3^\bullet$ (least stable)
$\mathrm{CH}_3\mathrm{CH}_2$--H (primary)	423	$\mathrm{CH}_3\mathrm{CH}_2^\bullet$
$(\mathrm{CH}_3)_2\mathrm{CH}$--H (secondary)	413	$(\mathrm{CH}_3)_2\mathrm{CH}^\bullet$
$(\mathrm{CH}_3)_3\mathrm{C}$--H (tertiary)	404	$(\mathrm{CH}_3)_3\mathrm{C}^\bullet$ (most stable)
$\mathrm{H}_2\mathrm{C}=\mathrm{CH}$--H (vinylic)	465	$\mathrm{H}_2\mathrm{C}=\mathrm{CH}^\bullet$ (very unstable)
$\mathrm{Ph}$--H (benzylic C--H)	473 (Ph--H)	$\mathrm{Ph}^\bullet$
$\mathrm{PhCH}_2$--H (benzylic)	375	$\mathrm{PhCH}_2^\bullet$ (very stable)

Radical stability: benzylic $\approx$ tertiary $>$ secondary $>$ primary $>$ methyl $>$ vinylic

Weaker bonds break more easily, producing more stable radicals. The stability of tertiary radicals arises from hyperconjugation (electron donation from adjacent C--H and C--C sigma bonds into the half-filled $p$ orbital) and inductive effects (electron-donating alkyl groups stabilise the electron-deficient radical centre).

Isomer Counting for Alkanes

Carbon atoms	Number of structural isomers
1	1 (methane)
2	1 (ethane)
3	1 (propane)
4	2 (butane, 2-methylpropane)
5	3 (pentane, 2-methylbutane, 2,2-dimethylpropane)
6	5
7	9
8	18
9	35
10	75

Environmental Chemistry: Catalytic Converters

Catalytic converters reduce the emissions from internal combustion engines:

• Reduction catalyst (platinum, palladium): Converts $\mathrm{NO}_x$ to $\mathrm{N}_2$ and $\mathrm{O}_2$: $2\mathrm{NO} \to \mathrm{N}_2 + \mathrm{O}_2$ $2\mathrm{NO}_2 \to \mathrm{N}_2 + 2\mathrm{O}_2$

• Oxidation catalyst (platinum, rhodium): Converts $\mathrm{CO}$ to $\mathrm{CO}_2$ and unburnt hydrocarbons to $\mathrm{CO}_2$ and $\mathrm{H}_2\mathrm{O}$: $2\mathrm{CO} + \mathrm{O}_2 \to 2\mathrm{CO}_2$ $\mathrm{C}_n\mathrm{H}_{2n+2} + \frac{3n+1}{2}\mathrm{O}_2 \to n\mathrm{CO}_2 + (n+1)\mathrm{H}_2\mathrm{O}$

Limitations:

• Catalytic converters only work at high temperatures ($> 300^\circ\mathrm{C}$), so they are ineffective during cold starts.
• Lead compounds (from leaded petrol) poison the catalyst by coating the active sites.
• Sulphur compounds in fuel can produce $\mathrm{SO}_2$ and $\mathrm{SO}_3$ which can deactivate the catalyst.

Cracking: Thermal vs Catalytic

Feature	Thermal cracking	Catalytic cracking
Temperature	$700$--$900^\circ\mathrm{C}$	$450$--$500^\circ\mathrm{C}$
Pressure	High ($\approx 7000\,\mathrm{kPa}$)	Slightly above atmospheric
Catalyst	None	Zeolite (aluminosilicate)
Products	High proportion of alkenes, some shorter alkanes	Branched alkanes, cycloalkanes, aromatic hydrocarbons
Carbon chain length	Shorter chains	Shorter chains
Process	Free radical mechanism	Carbocation (ionic) mechanism

Common Pitfalls

1. Writing balanced combustion equations: For $\mathrm{C}_n\mathrm{H}_{2n+2}$, the coefficient of $\mathrm{O}_2$ is $\frac{3n+1}{2}$, not $n + \frac{n+1}{2}$ (these are the same, but students often get the algebra wrong). For odd $n$, $\frac{3n+1}{2}$ is not an integer, so double the entire equation.

2. Confusing substitution and addition: Alkanes undergo substitution (one atom replaces another). Alkenes undergo addition (atoms add across the double bond). Students sometimes write addition reactions for alkanes.

3. Forgetting that free radical substitution produces a mixture: The product mixture includes all possible isomers plus side products from termination steps. This is why free radical substitution is not used for synthesis.

4. Bond dissociation energy vs mean bond enthalpy: Bond dissociation energy is for breaking a specific bond in a specific molecule. Mean bond enthalpy is an average across different molecules. Use bond dissociation energies for radical stability arguments, not mean bond enthalpies.

Exam-Style Questions with Full Mark Schemes

Q1 (5 marks)

Write an equation for the incomplete combustion of butane ($\mathrm{C}_4\mathrm{H}_{10}$) that produces carbon monoxide. State two environmental problems caused by incomplete combustion.

Mark Scheme:

$2\mathrm{C}_4\mathrm{H}_{10} + 9\mathrm{O}_2 \to 8\mathrm{CO} + 10\mathrm{H}_2\mathrm{O}$ (2 marks for correct balanced equation).

Two from (1 mark each): Carbon monoxide is toxic because it binds irreversibly to haemoglobin in red blood cells, reducing the blood's oxygen-carrying capacity. Carbon monoxide is a colourless, odourless gas so it is difficult to detect. Soot/particulates cause respiratory problems and contribute to global dimming.

Q2 (6 marks)

Explain the term homolytic fission. Use the reaction between methane and chlorine to describe the mechanism of free radical substitution, including all three stages. (6 marks)

Mark Scheme:

Homolytic fission: the breaking of a covalent bond in which each bonding electron is taken by one of the atoms, producing two free radicals, each with an unpaired electron (1 mark).

Initiation: $\mathrm{Cl}_2 \xrightarrow{h\nu} 2\mathrm{Cl}^\bullet$ (1 mark).

Propagation: $\mathrm{Cl}^\bullet + \mathrm{CH}_4 \to \mathrm{CH}_3^\bullet + \mathrm{HCl}$ and $\mathrm{CH}_3^\bullet + \mathrm{Cl}_2 \to \mathrm{CH}_3\mathrm{Cl} + \mathrm{Cl}^\bullet$ (2 marks for both propagation steps).

Termination: any two radicals combine, e.g. $2\mathrm{Cl}^\bullet \to \mathrm{Cl}_2$ or $\mathrm{CH}_3^\bullet + \mathrm{Cl}^\bullet \to \mathrm{CH}_3\mathrm{Cl}$ (1 mark).

The chain carrier ($\mathrm{Cl}^\bullet$) is regenerated in the propagation step, allowing the chain reaction to continue (1 mark).

Q3 (4 marks)

Explain why free radical substitution of propane with bromine is more selective than with chlorine, and predict the major organic product.

Mark Scheme:

The H-abstraction step with bromine is endothermic (or only slightly exothermic), so the transition state resembles the radical product (1 mark). This means the stability of the radical product strongly influences the activation energy: the more stable radical (secondary) has a lower activation energy, giving greater selectivity (1 mark).

With chlorine, the H-abstraction step is highly exothermic, so the transition state resembles the reactant and radical stability has less influence on the rate, giving lower selectivity (1 mark).

The major product is 2-bromopropane because the secondary hydrogen is abstracted preferentially over the primary hydrogens (1 mark).

Q4 (5 marks)

A student cracks a long-chain alkane $\mathrm{C}_{16}\mathrm{H}_{34}$ and obtains a mixture including propene ($\mathrm{C}_3\mathrm{H}_6$) and an alkane. Write a balanced equation for this cracking reaction and state the type of cracking most likely to produce propene.

Mark Scheme:

$\mathrm{C}_{16}\mathrm{H}_{34} \to \mathrm{C}_3\mathrm{H}_6 + \mathrm{C}_{13}\mathrm{H}_{28}$ (2 marks for correct balanced equation).

The cracking that produces an alkene (propene) is thermal cracking (1 mark), which operates at high temperature ($700$--$900^\circ\mathrm{C}$) and produces a high proportion of alkenes via a free radical mechanism (1 mark).

Catalytic cracking produces mainly branched and cyclic alkanes/aromatics, not primarily alkenes (1 mark).

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tip

Diagnostic Test Ready to test your understanding of Alkanes? The diagnostic test contains the hardest questions within the A-Level specification for this topic, each with a full worked solution.

Unit tests probe edge cases and common misconceptions. Integration tests combine Alkanes with other chemistry topics to test synthesis under exam conditions.

See Diagnostic Guide for instructions on self-marking and building a personal test matrix.