Unit 2 · Organic Chemistry

The Alkanes

Saturated hydrocarbons — nomenclature, isomerism, occurrence, preparation, properties, and uses.

2.1

Nomenclature of Alkanes

Definition Alkanes are saturated hydrocarbons — organic compounds containing only carbon and hydrogen atoms joined exclusively by single C–C and C–H bonds. They belong to the homologous series with general formula CnH2n+2.

The Homologous Series

Each successive member differs by –CH2– and shows a gradual change in physical properties while retaining similar chemical behaviour. All alkanes are colourless, relatively unreactive, and non-polar.

nNameMolecular FormulaCondensed FormulaState (25°C)
1MethaneCH4CH4Gas
2EthaneC2H6CH3CH3Gas
3PropaneC3H8CH3CH2CH3Gas
4ButaneC4H10CH3(CH2)2CH3Gas
5PentaneC5H12CH3(CH2)3CH3Liquid
6HexaneC6H14CH3(CH2)4CH3Liquid
7HeptaneC7H16CH3(CH2)5CH3Liquid
8OctaneC8H18CH3(CH2)6CH3Liquid
10DecaneC10H22CH3(CH2)8CH3Liquid
18+Octadecane…C18+H38+—Solid (wax)

IUPAC Rules for Naming Branched Alkanes

  1. Longest chain — identify the longest continuous carbon chain to give the parent name.
  2. Lowest locants — number the chain from the end nearest the first branch point so substituents get the lowest possible numbers.
  3. Name substituents — alkyl groups: methyl (–CH3), ethyl (–C2H5), propyl (–C3H7), isopropyl (–CH(CH3)2), etc.
  4. Alphabetical order — list substituents alphabetically (ignoring multiplying prefixes di–, tri–) before the parent chain name.
  5. Identical substituents — use di–, tri–, tetra– with separate locants for each instance.
  6. Punctuation — numbers separated by commas; numbers and letters separated by hyphens. No space in the final name.
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Memory Aid — First Four Names My Excellent Pupils Bring → Methane, Ethane, Propane, Butane; then 5=pent, 6=hex, 7=hept, 8=oct, 9=non, 10=dec.
Example 1

Naming a Branched Alkane

Name: CH3–CH(CH3)–CH2–CH3

1
Longest chain: C1–C2–C3–C4 = 4 carbons → parent: butane.
2
Number from nearest branch: branch is on C2 from either end → locant = 2.
3
Substituent: –CH3 = methyl.
✓
Name: 2-methylbutane.
Example 2

Two Identical Branches

Name: CH3–CH(CH3)–CH2–CH(CH3)–CH3

1
Longest chain: 5 C → pentane.
2
Number from left: branches at C2, C4. From right: also C2, C4. Lowest set = 2,4.
3
Two methyl groups → prefix di.
✓
Name: 2,4-dimethylpentane.
Example 3

Choosing the Correct Longest Chain

Name the compound where a central carbon has four CH3 groups attached: C(CH3)4

1
Expand: the central C plus any one CH3 branch gives a 2-C chain. But we can pick a 3-C chain by taking central C + one CH3 as C1, central as C2, another CH3 as C3 → propane parent? No — we still have two CH3 branches hanging off C2.
2
Longest chain = 3 C (pick any three carbons in a line: one CH3–C–CH3). Parent: propane. Remaining two CH3 are both on C2.
✓
Name: 2,2-dimethylpropane.
2.2

Isomerism in Alkanes

Definition Structural (chain) isomers are compounds with the same molecular formula but a different connectivity (arrangement) of atoms. In alkanes, isomerism arises from different branching of the carbon skeleton.
Molecular FormulaNo. of IsomersNames
CH41Methane
C2H61Ethane
C3H81Propane
C4H102Butane; 2-methylpropane
C5H123Pentane; 2-methylbutane; 2,2-dimethylpropane
C6H145Hexane; 2-methylpentane; 3-methylpentane; 2,2-dimethylbutane; 2,3-dimethylbutane
C7H169—
C10H2275—
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Systematic Approach to Drawing Isomers Start with the maximum-length straight chain. Then reduce chain by one carbon and attach it as a branch, moving the branch along. Reduce by two carbons next, and so on. Always check no two drawn structures are identical (can be renumbered to the same name).
Example 4

All Isomers of C5H12

1
Longest chain = 5: CH3CH2CH2CH2CH3 → pentane.
2
Chain = 4 + one methyl: CH3CH(CH3)CH2CH3 → 2-methylbutane. (Branch at C3 from left = C2 from right → same compound.)
3
Chain = 3 + two methyls both on C2: C(CH3)4 as propane core → 2,2-dimethylpropane.
✓
Total: 3 isomers.
2.3

Occurrence of Alkanes

Natural Gas

Predominantly methane (85–95%) with smaller amounts of ethane, propane, and butane. Found trapped in porous rock above petroleum deposits, formed by anaerobic decomposition of organic matter over millions of years.

Petroleum (Crude Oil)

A complex mixture of alkanes (C5–C40+), cycloalkanes, and aromatic hydrocarbons. Formed from ancient marine organisms buried under sediment and subjected to heat and pressure. Separated into useful fractions by fractional distillation.

FractionCarbon RangeB.P. Range (°C)Use
Refinery gasC1–C4<30Fuel; LPG
Gasoline (petrol)C5–C1030–200Motor fuel
Kerosene (paraffin)C10–C16150–270Jet fuel; heating
Diesel/gas oilC14–C20200–350Diesel engines
Lubricating oilC20–C50300–370Lubricants; waxes
Residue (bitumen)C50+>370Road surfacing

Other Sources

Methane is produced in biogas (anaerobic digestion of organic waste), in coal seams (firedamp), from landfill sites, from the digestive systems of ruminants, and as methane clathrates on the ocean floor.

2.4

Laboratory Preparation of Alkanes

Method 1: Decarboxylation (Kolbe’s Method)

Sodium salt of a carboxylic acid is heated strongly with soda lime (NaOH + CaO). The product alkane has one fewer carbon than the carboxylate.

RCOONa + NaOH --CaO, heat--> R-H + Na2CO3 e.g. CH3COONa + NaOH --> CH4 + Na2CO3 (sodium ethanoate) (methane)

Method 2: Wurtz Reaction

An alkyl halide reacts with sodium metal in dry ether. Best when a single alkyl halide is used (otherwise a mixture of three products forms).

2 R-X + 2 Na --dry ether--> R-R + 2 NaX e.g. 2 CH3Br + 2 Na --> CH3-CH3 + 2 NaBr (ethane)

Method 3: Catalytic Hydrogenation of Alkenes

R-CH=CH2 + H2 --Ni/Pt, 150 degC--> R-CH2-CH3 e.g. CH2=CH2 + H2 --> CH3CH3 (ethane)

Method 4: Reduction of Alkyl Halide

R-X + Zn + HCl --> R-H + ZnXCl or R-X + LiAlH4 (dry ether) --> R-H + ...
Example 5

Preparation of Butane via Decarboxylation

Write the equation for preparing butane by decarboxylation.

1
Butane = C4H10 (4 carbons). The carboxylate must have 4+1 = 5 carbons → sodium pentanoate: CH3CH2CH2CH2COONa.
2
CH3CH2CH2CH2COONa + NaOH → CaO, Δ CH3CH2CH2CH3 + Na2CO3
✓
Product: butane ✔
2.5

Physical Properties

States at Room Temperature

C1–C4: gases  |  C5–C17: liquids  |  C18+: solids (waxes)

Boiling and Melting Points

Both increase steadily with chain length because longer chains have greater surface area, leading to stronger London (van der Waals) dispersion forces between molecules.

Branched isomers have lower boiling points than straight-chain isomers of the same molecular formula because branching makes the molecule more compact, reducing surface area and weakening dispersion forces.

AlkaneMrB.P. (°C)M.P. (°C)Notes
Methane16−162−183Main component of natural gas
Ethane30−89−183—
Propane44−42−188LPG component
Butane58−1−138LPG; lighter fuel
Pentane72+36−130Liquid at room temp
Hexane86+69−95Common lab solvent
Octane114+126−57Petrol component; octane rating
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Effect of Branching on B.P. (C5H12 isomers) Pentane: 36°C  >  2-methylbutane: 28°C  >  2,2-dimethylpropane: 9°C. Increasing branching = lower B.P.

Solubility and Polarity

Alkanes are non-polar molecules. They are insoluble in water (polar solvent) but dissolve freely in non-polar solvents such as hexane, benzene, CCl4, and ether (like dissolves like). All alkanes are less dense than water.

2.6

Chemical Properties and Reactions

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General Reactivity Alkanes are relatively unreactive. Strong, non-polar C–C (347 kJ/mol) and C–H (413 kJ/mol) bonds resist attack by polar reagents, acids, bases, and oxidising agents. Their main reactions are combustion and halogenation by free-radical substitution.

Reaction 1: Combustion

Complete Combustion (excess O2)

CnH(2n+2) + (3n+1)/2 O2 --> n CO2 + (n+1) H2O e.g. CH4 + 2O2 --> CO2 + 2H2O (DeltaH = -890 kJ/mol) C3H8 + 5O2 --> 3CO2 + 4H2O

Incomplete Combustion (limited O2)

2CH4 + 3O2 --> 2CO + 4H2O (limited oxygen) CH4 + O2 --> C + 2H2O (very limited oxygen)

Carbon monoxide (CO) is a colourless, odourless, toxic gas that binds to haemoglobin ~240× more strongly than O2, causing asphyxiation.

Reaction 2: Halogenation — Free-Radical Substitution

Alkanes react with Cl2 or Br2 in UV light or high temperature (not in dark). The reaction is not selective and gives a mixture of all possible substitution products.

CH4 + Cl2 --(hv)--> CH3Cl + HCl (chloromethane) CH3Cl + Cl2 --(hv)--> CH2Cl2 + HCl (dichloromethane) CH2Cl2+ Cl2 --(hv)--> CHCl3 + HCl (chloroform) CHCl3 + Cl2 --(hv)--> CCl4 + HCl (tetrachloromethane)

Free-Radical Mechanism (3 Stages)

INITIATION (UV light causes homolytic fission): Cl2 --(hv)--> 2 Cl* PROPAGATION (chain reaction): Cl* + CH4 --> *CH3 + HCl *CH3 + Cl2 --> CH3Cl + Cl* (Cl* regenerated) TERMINATION (two radicals combine): Cl* + Cl* --> Cl2 *CH3 + Cl* --> CH3Cl *CH3 + *CH3 --> C2H6
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Why UV light is needed The Cl–Cl bond requires ~243 kJ/mol to break homolytically. UV photons provide sufficient energy to generate Cl radicals that initiate the chain reaction. The reaction does not occur in the dark.

Reaction 3: Thermal Cracking

Thermal cracking (400–900°C, high pressure) breaks long-chain alkanes into shorter alkanes and alkenes. Catalytic cracking (450–550°C, zeolite catalyst) is more controlled, producing more branched alkanes (higher octane rating) and useful alkenes.

C16H34 --> C8H18 + C8H16 (example thermal cracking) C10H22 --> C4H10 + C3H6 + C3H6 (example)
Example 6

Balanced Equation — Combustion of Butane

Write the balanced equation for complete combustion of butane (C4H10).

1
Apply CnH2n+2 formula with n=4: need 4 CO2 and 5 H2O.
2
Balance O: right side has 4×2 + 5×1 = 13 O atoms → need 13/2 O2.
3
C4H10 + 13/2 O2 → 4 CO2 + 5 H2O. Multiply by 2:
✓
2 C4H10 + 13 O2 → 8 CO2 + 10 H2O ✔
2.7

Uses of Alkanes

Alkane(s)ApplicationReason / Notes
MethaneNatural gas fuel; synthesis gas (H2 + CO) productionHigh energy density; clean burning
Propane & ButaneLPG — cooking, heating, portable fuelLiquefied easily under moderate pressure
C5–C10Petrol / gasoline — motor fuelCorrect volatility for combustion engines
C10–C16Kerosene — jet fuel, paraffin lampsHigh energy; right boiling range
C14–C20Diesel fuelCompression-ignition engines
C20–C50Lubricating oils, greases, paraffin waxHigh viscosity; thermal stability
C50+Bitumen — roads, waterproofingThermoplastic; excellent waterproofing
Alkanes generallyFeedstock for petrochemical industryRaw material for plastics, detergents, solvents
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Environmental Impact Combustion of alkanes releases CO2 (greenhouse gas) and, in limited oxygen, toxic CO and particulates. Methane itself is a potent greenhouse gas (~25× CO2 over 100 years). Steps to reduce impact include catalytic converters, cleaner engine technology, and transition to renewable energy.
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Exercises

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Multiple Choice Quiz — 25 Questions

Unit 2 Quiz — Alkanes

25 Questions · Select the best answer · Click Submit to see results
Q1

The general molecular formula for alkanes is:

Q2

How many structural isomers does C₄H₁₀ have?

Q3

What process is used to separate petroleum into fractions?

Q4

The IUPAC name of (CH₃)₃CH is:

Q5

Halogenation of alkanes requires:

Q6

Which termination step produces ethane during chlorination of methane?

Q7

Decarboxylation of sodium propanoate (CH₃CH₂COONa) with NaOH/CaO produces:

Q8

Why are alkanes relatively unreactive with most chemical reagents?

Q9

Which petroleum fraction is used as jet fuel?

Q10

The purpose of catalytic cracking in petroleum refining is:

Q11

The reaction of methane with excess chlorine under UV light gives as the final product:

Q12

Cyclohexane (C₆H₁₂) is a cycloalkane. Its general formula is:

Q13

Which statement about the free-radical chain reaction of methane with chlorine is correct?

Q14

Natural gas is a fossil fuel formed from:

Q15

The reason alkanes are immiscible with water is:

Q16

Reforming of naphtha using a platinum catalyst converts it to:

Q17

The boiling point of hexane (C₆H₁₄) compared to pentane (C₅H₁₂) is:

Q18

Thermal cracking uses:

Q19

Which best describes the empirical formula of alkanes with n > 1?

Q20

Which alkane would be used as a component of diesel fuel?

Q21

When propane burns in insufficient oxygen, which products can form?

Q22

Why does 2-methylbutane have a lower boiling point than n-pentane despite having the same molecular formula?

Q23

The IUPAC name of CH₃CH₂CH(CH₃)CH₂CH₃ is:

Q24

A mixture of branched alkane isomers in petrol is preferable to n-hexane because:

Q25

Methane (CH₄) has a tetrahedral structure because:

📝 Go to Unit Test →
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Unit Test

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Instructions Total: 50 marks  |  Time: 50 minutes  |  Attempt all questions  |  Show all working.

Section A — Short Answer

30 marks
Q1 [3 marks]

Give the IUPAC name for each compound:

(a) CH3CH2CH(CH3)CH3   (b) (CH3)2CHCH(CH3)CH3   (c) CH3C(CH3)2CH2CH3

(a) 2-methylbutane
(b) 2,3-dimethylbutane
(c) 2,2-dimethylbutane
Q2 [4 marks]

Draw the structural formulae and give IUPAC names for all 3 isomers of C5H12.

1. Pentane: CH3CH2CH2CH2CH3
2. 2-methylbutane: CH3CH(CH3)CH2CH3
3. 2,2-dimethylpropane: C(CH3)4
Q3 [4 marks]

Write balanced equations for: (a) complete combustion of propane; (b) incomplete combustion of ethane producing CO.

(a) C3H8 + 5O2 → 3CO2 + 4H2O
(b) 2C2H6 + 5O2 → 4CO + 6H2O
Q4 [6 marks]

Write the full free-radical mechanism for the monochlorination of ethane (C2H6) by Cl2 under UV light. Include all three stages.

Initiation: Cl2 →(hv) 2 Cl•

Propagation:
Cl• + C2H6 → •C2H5 + HCl
•C2H5 + Cl2 → C2H5Cl + Cl•

Termination:
Cl• + Cl• → Cl2
•C2H5 + Cl• → C2H5Cl
•C2H5 + •C2H5 → C4H10
Main product: chloroethane
Q5 [4 marks]

Give two laboratory methods for preparing methane, with a balanced equation for each.

1. Decarboxylation:
CH3COONa + NaOH →CaO, Δ CH4 + Na2CO3

2. Reduction of iodomethane:
CH3I + Zn + HCl → CH4 + ZnICl
Q6 [5 marks]

Explain the trend in boiling points of straight-chain alkanes from methane to octane. Why do branched isomers have lower boiling points than straight-chain isomers of the same molecular formula?

Trend: B.P. increases from −162°C (methane) to +126°C (octane) as chain length increases.
Reason: Alkanes are non-polar; only London dispersion forces act between molecules. Longer chains have greater surface area → stronger dispersion forces → more energy needed to separate molecules → higher B.P.
Branching: Branched isomers are more compact/spherical → less surface contact between molecules → weaker dispersion forces → lower B.P. than straight-chain isomer with same Mr.
Q7 [4 marks]

State four uses of alkanes, naming the specific fraction or compound used in each case and linking to a relevant property.

1. Methane — fuel: high energy density, clean burning.
2. Propane/butane (LPG) — portable cooking fuel: liquefied under moderate pressure, easily stored.
3. Petrol (C5–C10) — motor fuel: correct volatility for combustion engines.
4. Bitumen (C50+) — road surfacing: thermoplastic, waterproof, durable.

Section B — Extended Response

20 marks
Q8 [10 marks]

(a) Describe the fractional distillation of petroleum: explain how the fractionating column works, what principle governs separation, and name four fractions with their uses. [6 marks]

(b) Define cracking, write one equation for thermal cracking of a long-chain alkane, and state one reason why cracking is important in the petroleum industry. [4 marks]

(a) Crude oil is heated in a furnace and fed into the base of a fractionating column, which is hot at the bottom (~400°C) and cool at the top (~40°C). Vapours rise and condense when they reach the level where temperature equals their boiling point; fractions are drawn off at different levels. Separation is based on differences in boiling point, which increases with chain length due to stronger van der Waals forces.
Fractions: Refinery gas (C1–C4, fuel); Petrol (C5–C10, motor fuel); Kerosene (C10–C16, jet fuel); Bitumen (C50+, road surfacing).

(b) Cracking is the thermal or catalytic decomposition of large alkane molecules into smaller, more useful hydrocarbons.
Equation: C16H34 → C8H18 + C8H16
Importance: Converts low-demand heavy fractions into high-demand petrol and alkenes (monomers for plastics).
Q9 [10 marks]

(a) Write the structural formulae and IUPAC names for all five isomers of hexane (C6H14). [5 marks]

(b) Arrange the five isomers in order of increasing boiling point and justify your answer using intermolecular force theory. [5 marks]

(a)
1. Hexane: CH3(CH2)4CH3
2. 2-methylpentane: CH3CH(CH3)CH2CH2CH3
3. 3-methylpentane: CH3CH2CH(CH3)CH2CH3
4. 2,2-dimethylbutane: CH3C(CH3)2CH2CH3
5. 2,3-dimethylbutane: CH3CH(CH3)CH(CH3)CH3

(b) Increasing B.P.:
2,2-dimethylbutane (50°C) < 2,3-dimethylbutane (58°C) < 2-methylpentane (60°C) < 3-methylpentane (63°C) < hexane (69°C)

Justification: All isomers have the same Mr = 86 and are non-polar; only London dispersion forces differ. Hexane (straight chain) has the greatest surface area → strongest forces → highest B.P. Increasing branching reduces surface contact area → weaker forces → lower B.P. 2,2-dimethylbutane is most compact → lowest B.P.