Alkanes | B.Sc. 1st Year Chemistry

1. Introduction to Alkanes

Alkanes are saturated hydrocarbons containing only carbon and hydrogen atoms. In an open-chain alkane, all carbon-carbon bonds are single bonds.

Because the carbon atoms are joined only by single sigma bonds, alkanes are called saturated hydrocarbons.

General Formula

CnH2n+2

This formula is applicable to open-chain saturated hydrocarbons containing only carbon and hydrogen.

Examples

Methane CH4 Ethane C2H6 Propane C3H8 Butane C4H10 Pentane C5H12 Hexane C6H14

Why are alkanes saturated?

Each carbon atom has four valencies. In an alkane, carbon uses all available valencies through single bonds with carbon or hydrogen. Therefore, hydrogen cannot be added without breaking an existing bond.

Remember: Alkane = saturated hydrocarbon = only C-C and C-H single bonds.

2. Structure of Alkanes

Hybridization of Carbon

The carbon atoms in alkanes are generally sp³ hybridized. One s orbital and three p orbitals combine to form four equivalent sp³ hybrid orbitals.

Carbon hybridization = sp3 Number of sigma bonds around carbon = 4 Approximate bond angle = 109.5°

Shape around carbon

The four sp³ orbitals arrange themselves approximately in a tetrahedral arrangement to minimize repulsion between electron pairs.

Bonding in methane

CH4 Carbon = sp3 hybridized 4 C-H sigma bonds

C-C bond in ethane

CH3-CH3 C-C = sigma bond C-H = sigma bonds

Rotation around a carbon-carbon single bond is possible because a sigma bond has cylindrical symmetry. This gives rise to different conformations.

Exam point: Alkanes contain only sigma bonds. The carbon atoms are sp³ hybridized and have approximately tetrahedral geometry.

3. Nomenclature of Alkanes

The IUPAC name of an alkane is obtained by selecting the longest continuous carbon chain and naming the substituents attached to it.

Basic steps

Find the longest continuous carbon chain.
Give the parent chain the corresponding alkane name.
Number the chain from the end nearest the substituent.
Identify the substituent or substituents.
Give their positions using numbers.
Arrange different substituents alphabetically.

Parent chain names

Number of carbon atoms	Parent name
1	Methane
2	Ethane
3	Propane
4	Butane
5	Pentane
6	Hexane
7	Heptane
8	Octane

Example

CH3-CH(CH3)-CH2-CH3

Longest chain = 4 carbon atoms = butane.

There is a methyl group on carbon 2.

Name = 2-methylbutane

Memory tip: First find the longest chain. Then number it. Do not start numbering before deciding the parent chain.

4. Isomerism in Alkanes

Isomers are compounds having the same molecular formula but different arrangements of atoms.

Chain isomerism

Alkanes show chain isomerism because the carbon skeleton can be arranged in different ways.

Example: C4H10

CH3-CH2-CH2-CH3 n-butane

CH3-CH(CH3)-CH3 2-methylpropane

Both have molecular formula C4H10 but different carbon skeletons.

Important point

Chain isomerism in simple open-chain alkanes begins with butane.

Remember: Methane, ethane and propane do not have chain isomers. Butane is the first alkane showing chain isomerism.

5. Conformation of Alkanes

Conformations are different spatial arrangements produced by rotation around a carbon-carbon single bond without breaking the bond.

Ethane

CH3-CH3

Rotation around the C-C sigma bond gives different conformations.

Staggered conformation

In the staggered arrangement, C-H bonds on one carbon are positioned between the C-H bonds on the other carbon.

It has lower torsional strain and is more stable.

Eclipsed conformation

In the eclipsed arrangement, the C-H bonds of one carbon are directly behind the C-H bonds of the other carbon.

Electron cloud repulsion is greater, so this conformation is less stable.

Remember: Staggered = more stable. Eclipsed = less stable.

6. Preparation of Alkanes

6.1 Hydrogenation of Alkenes

Alkenes can be converted into alkanes by adding hydrogen in the presence of a suitable catalyst.

Overall reaction

CH2=CH2 + H2 --Ni / Pt / Pd--> CH3-CH3

Step 1: Hydrogen and the alkene interact with the metal catalyst.

Step 2: The H-H bond is activated on the catalyst surface.

H-H -> H* + H*

Step 3: Hydrogen atoms are transferred to the two carbon atoms of the C=C bond.

CH2=CH2 + H* + H* -> CH3-CH3

Step 4: Ethane leaves the catalyst surface.

Remember: Alkene + H2 + metal catalyst → alkane.

6.2 Reduction of Alkyl Halides

Alkyl halides can be reduced to the corresponding alkanes.

Example

CH3-CH2-Cl + 2[H] -> CH3-CH3 + HCl

Step 1: The alkyl halide accepts hydrogen/reducing equivalents.

CH3-CH2-Cl + [H] -> CH3-CH2• + HCl

Step 2: The carbon-halogen bond is replaced by a carbon-hydrogen bond.

CH3-CH2• + [H] -> CH3-CH3

In actual laboratory preparations, the exact reducing reagent may vary depending on the syllabus and reaction.

6.3 Wurtz Reaction

Alkyl halides react with sodium metal in dry ether to form a higher alkane.

Overall reaction

2R-X + 2Na --dry ether--> R-R + 2NaX

Step 1: Sodium transfers an electron to the alkyl halide.

R-X + e⁻ -> R• + X⁻

Step 2: The alkyl radical can receive another electron and form an organosodium species.

R• + Na -> R-Na

Step 3: The organosodium species reacts with another alkyl halide.

R-Na + R-X -> R-R + NaX

This is a simplified textbook representation. The actual sodium-metal reaction can involve several metal-mediated pathways.

Memory tip: Wurtz joins two alkyl groups. R-X + R-X → R-R.

6.4 Decarboxylation with Soda Lime

Sodium salts of carboxylic acids give alkanes containing one carbon atom less when heated with soda lime.

General reaction

R-COONa + NaOH --CaO, heat--> R-H + Na2CO3

Step 1: The carboxylate salt is heated with sodium hydroxide in the presence of calcium oxide.

R-COONa + NaOH --heat--> intermediate

Step 2: The carboxyl carbon is removed as the carbonate part.

R-COONa + NaOH -> R-H + Na2CO3

Example

CH3COONa + NaOH --CaO, heat--> CH4 + Na2CO3

Sodium acetate has two carbon atoms, but methane has only one carbon atom.

Remember: Soda-lime decarboxylation gives an alkane with one carbon less.

6.5 Kolbe Electrolysis

Electrolysis of an aqueous solution of sodium or potassium salts of carboxylic acids can produce alkanes through radical coupling.

Anode reaction

RCOO⁻ -> R• + CO2 + e⁻

Step 1: The carboxylate ion loses an electron at the anode.

RCOO⁻ -> RCOO• + e⁻

Step 2: The carboxyl radical loses carbon dioxide.

RCOO• -> R• + CO2

Step 3: Two alkyl radicals combine.

R• + •R -> R-R

Cathode reaction

2H2O + 2e⁻ -> H2 + 2OH⁻

Overall reaction

2RCOONa + 2H2O -> R-R + 2CO2 + H2 + 2NaOH

Memory tip: Kolbe electrolysis removes CO2 from two carboxylate ions and then joins the two remaining alkyl radicals.

7. Physical Properties of Alkanes

Physical state

The lower members such as methane, ethane, propane and butane are gases at room temperature. Higher members occur as liquids or solids depending on molecular size.

Solubility

Alkanes are non-polar molecules and are practically insoluble in water.

They dissolve better in non-polar organic solvents.

Density

Most alkanes have density lower than water.

Boiling point

Boiling point generally increases as molecular mass and surface area increase.

Branching usually decreases boiling point because branching reduces the effective surface area for intermolecular attraction.

Polarity

C-C and C-H bonds have relatively small electronegativity differences, so alkanes are considered non-polar overall.

Remember: More carbon atoms → generally higher boiling point. More branching → generally lower boiling point.

8. Chemical Reactions of Alkanes

Alkanes are comparatively less reactive than alkenes and alkynes because they contain strong C-C and C-H sigma bonds.

However, they undergo important reactions under suitable conditions.

Important reactions

Free-radical halogenation
Combustion
Cracking or pyrolysis
Isomerization
Controlled oxidation under special conditions

Among these, free-radical substitution is especially important for examination questions.

9. Free Radical Halogenation of Alkanes

Alkanes react with chlorine or bromine in the presence of light or heat to give haloalkanes.

Alkane + X2 --hv / heat--> Alkyl halide + HX

The reaction proceeds through free radicals and is therefore called a free-radical chain reaction.

Three main stages

Initiation
Propagation
Termination

Very important: In a mechanism question, do not write only the overall reaction. Show initiation, propagation and termination separately.

10. Chlorination of Methane: Complete Free-Radical Mechanism

Methane reacts with chlorine in the presence of ultraviolet light to form chloromethane and hydrogen chloride.

Overall reaction

CH4 + Cl2 --hv--> CH3Cl + HCl

Step 1: Initiation

Ultraviolet light breaks the chlorine-chlorine bond homolytically. Each chlorine atom takes one electron from the bonding pair.

Cl-Cl --hv--> Cl• + •Cl

Cl2 --hv--> 2Cl•

The chlorine radicals are highly reactive and start the chain reaction.

Step 2: Propagation 1

A chlorine radical abstracts a hydrogen atom from methane.

Cl• + H-CH3 -> H-Cl + •CH3

Cl• + CH4 -> HCl + CH3•

A methyl radical is produced.

Step 3: Propagation 2

The methyl radical attacks chlorine.

CH3• + Cl-Cl -> CH3-Cl + Cl•

Chloromethane is formed and a new chlorine radical is produced.

This new chlorine radical can react with another methane molecule.

Step 4: Chain continuation

Cl• + CH4 -> HCl + CH3•

CH3• + Cl2 -> CH3Cl + Cl•

Cl• + CH4 -> HCl + CH3•

CH3• + Cl2 -> CH3Cl + Cl•

The chlorine radical is regenerated, so the chain can continue through many cycles.

Step 5: Termination

The chain ends when two radicals combine to form a stable molecule.

Cl• + •Cl -> Cl2

CH3• + •Cl -> CH3Cl

CH3• + •CH3 -> CH3-CH3

Mechanism in one sequence

1. Cl2 --hv--> 2Cl•

2. Cl• + CH4 -> HCl + CH3•

3. CH3• + Cl2 -> CH3Cl + Cl•

4. Cl• + CH4 -> HCl + CH3•

5. CH3• + Cl2 -> CH3Cl + Cl•

6. Cl• + Cl• -> Cl2

7. CH3• + Cl• -> CH3Cl

8. CH3• + CH3• -> C2H6

Memory tip: Initiation makes radicals. Propagation uses radicals and makes new radicals. Termination removes radicals.

Common mistake: Do not write CH4 + Cl2 → CH3Cl + HCl and call that the mechanism. That is only the overall reaction.

11. Chlorination of Ethane

Overall reaction

C2H6 + Cl2 --hv--> C2H5Cl + HCl

Initiation

Cl2 --hv--> 2Cl•

Propagation Step 1

Cl• + CH3-CH3 -> HCl + CH3-CH2•

A hydrogen atom is removed from ethane and an ethyl radical is formed.

Propagation Step 2

CH3-CH2• + Cl2 -> CH3-CH2Cl + Cl•

The ethyl radical reacts with chlorine and regenerates the chlorine radical.

Chain continuation

Cl• + CH3-CH3 -> HCl + CH3-CH2•

CH3-CH2• + Cl2 -> CH3-CH2Cl + Cl•

Termination

Cl• + Cl• -> Cl2

CH3-CH2• + Cl• -> CH3-CH2Cl

CH3-CH2• + •CH2-CH3 -> CH3-CH2-CH2-CH3

Important: Termination by radical combination is possible, but the major chain reaction proceeds through propagation steps.

12. Bromination of Propane

Bromination is useful for understanding radical stability and product selectivity.

Overall reaction

CH3-CH2-CH3 + Br2 --hv--> CH3-CHBr-CH3 + HBr

2-bromopropane is favored because abstraction of the secondary hydrogen produces a more stable secondary radical than abstraction of a primary hydrogen.

Initiation

Br-Br --hv--> Br• + •Br

Propagation Step 1

Bromine radical abstracts a secondary hydrogen from propane.

Br• + CH3-CH2-CH3 -> HBr + CH3-CH•-CH3

Propagation Step 2

CH3-CH•-CH3 + Br2 -> CH3-CHBr-CH3 + Br•

The bromine radical is regenerated, so the chain continues.

Alternative primary pathway

Br• + CH3-CH2-CH3 -> HBr + CH3-CH2-CH2•

CH3-CH2-CH2• + Br2 -> CH3-CH2-CH2Br + Br•

This gives 1-bromopropane, but the secondary pathway is favored.

Termination

Br• + Br• -> Br2

CH3-CH•-CH3 + Br• -> CH3-CHBr-CH3

CH3-CH•-CH3 + •CH-CH3 -> coupled radical product

The exact radical-radical coupling products depend on which radicals meet.

Remember: Bromination is generally more selective than chlorination. Secondary and tertiary hydrogen abstraction is therefore strongly favored when the corresponding more stable radical can form.

13. Combustion of Alkanes

Alkanes burn in oxygen and release heat. Complete combustion produces carbon dioxide and water.

Combustion of methane

Overall reaction

CH4 + 2O2 -> CO2 + 2H2O + heat

Step 1: Methane reacts with oxygen.

CH4 + O2 -> CO2 + H2O

Step 2: Balance carbon.

CH4 + O2 -> CO2 + H2O

Step 3: Balance hydrogen.

CH4 + O2 -> CO2 + 2H2O

Step 4: Balance oxygen.

CH4 + 2O2 -> CO2 + 2H2O

General combustion equation

CnH2n+2 + (3n+1)/2 O2 -> nCO2 + (n+1)H2O

Example: propane

C3H8 + O2 -> CO2 + H2O

C3H8 + O2 -> 3CO2 + 4H2O

C3H8 + 5O2 -> 3CO2 + 4H2O

Incomplete combustion

When oxygen supply is insufficient, carbon monoxide or carbon particles may form.

2CH4 + 3O2 -> 2CO + 4H2O

CH4 + O2 -> C + 2H2O

Common mistake: Do not assume every combustion reaction produces only CO2. Limited oxygen can produce CO or carbon.

14. Cracking of Alkanes

Cracking is the breaking of large hydrocarbon molecules into smaller hydrocarbons by heat, pressure and/or suitable catalysts.

Depending on the conditions, cracking can produce smaller alkanes, alkenes and other hydrocarbons.

Example

C10H22 -> C5H12 + C5H10

Here the products are pentane and pentene.

General idea of bond breaking

C-C bond --heat--> smaller hydrocarbon fragments

In radical cracking, homolytic bond cleavage can produce radicals that undergo further reactions.

R-R --heat--> R• + •R

R• -> smaller radical + alkene

Remember: Cracking converts large hydrocarbons into smaller, more useful hydrocarbon molecules.

15. Wurtz Reaction in Detail

Wurtz reaction is mainly used to prepare symmetrical higher alkanes from alkyl halides.

Example: preparation of ethane

2CH3Cl + 2Na --dry ether--> CH3-CH3 + 2NaCl

Step 1: Sodium transfers an electron to methyl chloride.

CH3-Cl + e⁻ -> CH3• + Cl⁻

Step 2: The methyl radical can interact with sodium.

CH3• + Na -> CH3Na

Step 3: Methyl sodium reacts with another methyl halide.

CH3Na + CH3Cl -> CH3-CH3 + NaCl

The above is a simplified mechanism used for understanding the reaction. Sodium-metal coupling can involve more complicated pathways.

Memory tip: Two R groups join together: R-X + R-X → R-R.

16. Decarboxylation: Detailed Reaction

Sodium salts of carboxylic acids are heated with soda lime to prepare alkanes containing one carbon atom fewer.

Example: sodium acetate

CH3COONa + NaOH --CaO, heat--> CH4 + Na2CO3

Step 1: Sodium acetate is heated with sodium hydroxide and calcium oxide.

CH3COONa + NaOH --CaO, Δ--> CH4 + Na2CO3

Step 2: The carboxyl carbon is removed into carbonate.

CH3-COO⁻ -> CH3-H

Step 3: The remaining alkyl group receives hydrogen.

CH3• + H -> CH4

The simple reaction equation is the safest form to write in an examination unless a detailed mechanism is specifically requested.

Key point: Decarboxylation decreases the carbon number by one.

17. Kolbe Electrolysis: Step-by-Step

Kolbe electrolysis is an electrochemical method of preparing alkanes from salts of carboxylic acids.

Example: sodium acetate

Step 1: Ion formation

CH3COONa -> CH3COO⁻ + Na⁺

Step 2: Oxidation at anode

CH3COO⁻ -> CH3COO• + e⁻

Step 3: Loss of carbon dioxide

CH3COO• -> CH3• + CO2

Step 4: Radical coupling

CH3• + •CH3 -> CH3-CH3

Step 5: Cathode reaction

2H2O + 2e⁻ -> H2 + 2OH⁻

Overall reaction

2CH3COONa + 2H2O -> C2H6 + 2CO2 + H2 + 2NaOH

Memory sequence: Carboxylate → radical → CO2 loss → radical coupling → alkane.

18. Oxidation of Alkanes

Alkanes are generally resistant to mild oxidation, but under suitable conditions they can undergo oxidation. Complete oxidation is combustion.

Complete oxidation

CH4 + 2O2 -> CO2 + 2H2O

Controlled oxidation

Under special catalytic and controlled conditions, some alkanes can form oxygen-containing products. The exact product depends on the alkane and reaction conditions.

For basic B.Sc. examination answers, distinguish ordinary combustion from selective catalytic oxidation.

Do not write: "All alkanes easily undergo oxidation." Alkanes are relatively unreactive toward mild oxidizing agents because their C-C and C-H sigma bonds are strong.

19. Alkane vs Alkene vs Alkyne

Feature	Alkane	Alkene	Alkyne
General formula	CnH2n+2	CnH2n	CnH2n-2
Main bond	C-C single	C=C double	C≡C triple
Hybridization	sp3	sp2	sp
Approximate bond angle	109.5°	120°	180°
Saturation	Saturated	Unsaturated	Unsaturated
Typical reaction	Substitution	Addition	Addition

Exam shortcut: Alkane = single bond = sp3. Alkene = double bond = sp2. Alkyne = triple bond = sp.

20. Common Mistakes

Writing CnH2n for an alkane. The open-chain alkane formula is CnH2n+2.
Calling alkanes unsaturated.
Forgetting that methane is CH4.
Numbering an alkane from the wrong end.
Calling chain isomerism possible in propane.
Writing only the overall chlorination reaction when a free-radical mechanism is asked.
Forgetting the initiation step.
Forgetting that propagation regenerates a radical.
Putting radical-radical reactions under propagation. They belong to termination.
Writing Cl+ instead of Cl• when describing a chlorine radical.
Forgetting the condition hv for photochemical chlorination or bromination.
Saying soda-lime decarboxylation gives an alkane with the same number of carbon atoms.
Forgetting that Kolbe electrolysis involves loss of CO2 and coupling of two radicals.

21. Worked Problems

Problem 1: Molecular formula of an alkane

Find the molecular formula of an open-chain alkane containing 7 carbon atoms.

General formula = CnH2n+2 n = 7 H = 2(7) + 2 = 14 + 2 = 16 Formula = C7H16

Therefore, the molecular formula is C7H16.

Problem 2: Number of hydrogen atoms

Find the molecular formula of an alkane containing 10 carbon atoms.

CnH2n+2 n = 10 H = 2(10) + 2 = 22 Formula = C10H22

Problem 3: Decarboxylation

What alkane is obtained by heating sodium propionate with soda lime?

CH3CH2COONa + NaOH --CaO, heat--> CH3CH3 + Na2CO3

Sodium propionate contains three carbon atoms. The product contains two carbon atoms.

Product = ethane.

Problem 4: Wurtz reaction

What product is obtained when ethyl bromide reacts with sodium in dry ether?

2CH3CH2Br + 2Na --dry ether--> CH3CH2CH2CH3 + 2NaBr

Product = butane.

Problem 5: Chlorination mechanism

Write the propagation steps for chlorination of methane.

Step 1: Cl• + CH4 -> HCl + CH3• Step 2: CH3• + Cl2 -> CH3Cl + Cl• Chain continues: Cl• + CH4 -> HCl + CH3• CH3• + Cl2 -> CH3Cl + Cl•

22. TU Examination Preparation

2-mark questions

Define alkane.
Write the general formula of open-chain alkanes.
Why are alkanes called saturated hydrocarbons?
What is chain isomerism?
Write the first three members of the alkane series.
What is Wurtz reaction?
What is decarboxylation?
What is Kolbe electrolysis?
What is a free radical?
What is the role of UV light in chlorination?

5-mark questions

Explain the structure and hybridization of alkanes.
Explain chain isomerism with suitable examples.
Explain the Wurtz reaction with an example.
Explain preparation of alkanes by decarboxylation.
Explain Kolbe electrolysis.
Explain the physical properties of alkanes.
Explain combustion and cracking of alkanes.

10-mark questions

Explain the free-radical mechanism of chlorination of methane in detail.
Discuss the preparation and important reactions of alkanes.
Explain free-radical halogenation with initiation, propagation and termination steps.
Describe Wurtz reaction, decarboxylation and Kolbe electrolysis with equations.
Discuss the structure, physical properties and chemical reactions of alkanes.

How to write a 10-mark free-radical mechanism answer

Write the overall reaction.
Write the reaction conditions.
Explain initiation with the radical equation.
Write propagation step 1 as an equation.
Write propagation step 2 as an equation.
Show chain continuation if required.
Write termination equations.
Explain why the chain reaction continues.
Write a short memory point or conclusion.

Best exam structure: Do not fill a mechanism answer with only paragraphs. The important reaction steps should be visible as actual chemical equations.

23. Quick Revision

Alkanes are saturated hydrocarbons.
Open-chain alkane formula = CnH2n+2.
Carbon is sp3 hybridized.
Approximate bond angle = 109.5°.
Only sigma bonds are present.
Chain isomerism starts from butane.
Staggered ethane is more stable than eclipsed ethane.
Alkanes are generally non-polar.
Alkanes are practically insoluble in water.
Wurtz reaction joins two alkyl groups.
Soda-lime decarboxylation gives an alkane with one carbon less.
Kolbe electrolysis proceeds through alkyl radicals.
Free-radical halogenation is a substitution reaction.
Initiation produces radicals.
Propagation consumes and regenerates radicals.
Termination removes radicals.
Complete combustion produces CO2 and H2O.
Incomplete combustion can produce CO or carbon.
Cracking converts large hydrocarbons into smaller hydrocarbons.

Most important equations

CnH2n+2

2R-X + 2Na --dry ether--> R-R + 2NaX

R-COONa + NaOH --CaO, heat--> R-H + Na2CO3

RCOO⁻ -> R• + CO2 + e⁻

R• + •R -> R-R

CH4 + Cl2 --hv--> CH3Cl + HCl

Cl2 --hv--> 2Cl•

Cl• + CH4 -> HCl + CH3•

CH3• + Cl2 -> CH3Cl + Cl•

CH4 + 2O2 -> CO2 + 2H2O + heat

24. Quick Check

1. What is the general formula of an open-chain alkane?

CnH2n+2.

2. What type of hybridization is present in alkane carbon?

sp3 hybridization.

3. Which is more stable in ethane, staggered or eclipsed?

Staggered conformation is more stable.

4. What are the three stages of a free-radical chain reaction?

Initiation, propagation and termination.

5. What happens in the initiation step of methane chlorination?

Chlorine molecules undergo homolytic cleavage under UV light to form chlorine radicals: Cl2 --hv--> 2Cl•.

6. Write the first propagation step of methane chlorination.

Cl• + CH4 → HCl + CH3•.

7. What is the product of soda-lime decarboxylation of sodium acetate?

Methane, CH4.

8. What is the main idea of the Wurtz reaction?

Two alkyl groups are joined together to form a higher alkane.

9. What are the main products of complete combustion of an alkane?

Carbon dioxide and water, with release of heat.

10. Why does bromination show greater selectivity than chlorination?

Bromination has a more selective hydrogen-abstraction step, so formation of the more stable radical is strongly favored.

Final Revision Note

For alkanes, the most important part to understand is the relationship between their saturated structure and their reactions. The strong C-C and C-H sigma bonds make them relatively unreactive, but under suitable conditions they undergo substitution, combustion and cracking.

For free-radical halogenation, learn the actual sequence:

Cl2 --hv--> 2Cl•

Cl• + CH4 -> HCl + CH3•

CH3• + Cl2 -> CH3Cl + Cl•

Cl• + CH4 -> HCl + CH3•

CH3• + Cl2 -> CH3Cl + Cl•

Cl• + Cl• -> Cl2

CH3• + Cl• -> CH3Cl

CH3• + CH3• -> C2H6

If you can write and explain these equations correctly, the mechanism portion of an alkane question becomes much easier to remember and reproduce in the examination.