Chapter 12: Alkanes

Question 1

1. What is the general formula for alkanes?
2. Give four examples of alkanes used in everyday life.

Answer 1

1. The general formula of alkanes is CⁿH²ⁿ⁺².
2. Image below.

Question 2

Describe the type of bonding in alkanes.

Answer 2

• Alkanes are saturated hydrocarbons, containing only carbon and hydrogen atoms joined together by single covalent bonds.
• Each carbon atom in an alkane is has four (covalent) sigma bonds.
  
  • A covalent bond is defined as a shared pair of electrons. A sigma bond is the result of the overlap of two orbitals, one from each bonding atom.
  • Each overlapping orbital contains one electron, so the sigma bond has two electrons that are shared between the bonding atoms.
  • A sigma bond is positioned on a line directly between bonding atoms.

Question 3

Describe the shape of alkanes.

Answer 3

• Each carbon atom is surrounded by four electron pairs in four σ-bonds.
• Repulsion between these electron pairs results in a 3D tetrahedral arrangement around each carbon atom.
• Each bond angle is approximately 109.5˚.
• The σ-bonds act as axes around which the atoms can rotate freely, so these shapes are not rigid.

Question 4

How do oil refineries separate crude oil? Explain why this is possible.

Answer 4

• Oil refineries separate crude oil into fractions by fractional distillation in a distillation tower.
  
  • Each fraction contains a range of alkanes.
• Separation like this is possible because the boiling points of the alkanes are different, increasing as their chain length increases.

Question 5

Explain why the boiling point of alkanes increases as chain length increases.

Answer 5

• The boiling point increases because of the London forces. These forces act in liquids and solids but, once broken, the molecules move apart from each other and the alkane becomes a gas.
• The London forces act between molecules that are in close surface contact. As the chain length increases, the molecules have a larger surface area so more surface contact is possible between molecules. The London forces between the molecules will be greater and so more energy is required to overcome the forces.

Question 6

Explain the effect of branching on the boiling point of alkanes.

Answer 6

• Isomers of alkanes have the same molecular mass. Branched isomers have lower boiling points.
• The reason for this difference is because there are fewer surface points of contact between molecules of branched alkanes, giving fewer London forces.
• Another factor lies with the shape of the molecules. The branches prevent the branched molecules getting as close together as straight-chain molecules, decreasing the intermolecular forces further.

Question 7

Alkanes do not react with the most common reagents. Explain why.

Answer 7

The reasons for alkanes’ lack of reactivity are:

• C–C and C–H σ-bonds are strong
• C–C bonds are non-polar
• the electronegativity of carbon and hydrogen is so similar that the C–H bond can be considered to be non-polar.

Question 8

Describe the complete combustion of alkanes. Give an example.

Answer 8

• In a plentiful supply of oxygen, alkanes burn completely to produce carbon dioxide and water.

Question 9

Describe the incomplete combustion of alkanes. Give and example.

Answer 9

• In a limited supply of oxygen, incomplete combustion takes place.
• In incomplete combustion, the hydrogen atoms in the alkane are always oxidised to water, but combustion of the carbon may be incomplete, forming the toxic gas carbon monoxide [CO] or even carbon itself as soot.

Question 10

Describe the reaction of alkanes with halogens. Give an example.

Answer 10

• In the presence of sunlight, alkanes react with halogens. The high-energy UV radiation present in sunlight provides the initial energy for a reaction to take place.
• For example, methane reacts with bromine to form bromomethane.
  
  • This is a substitution reaction, as a hydrogen atom in the alkane has been substituted by a halogen atom.

Question 11

Describe the reaction mechanism for the first stage of the bromination of methane.

EXTENSION: What type of reaction is the bromination of methane?

Answer 11

• In the initiation stage, the reaction is started when the covalent bond in a bromine molecule is broken by homolytic fission.
• Each bromine atom takes one electron from the pair, forming to highly reactive bromine radicals.
• The energy for this bond fission is provided by UV radiation.

EXTENSION: The bromination of methane is a radical substitution reaction.

Question 12

Describe the reaction mechanism for the second stage of the bromination of methane.

Answer 12

• In the propagation stage, the reaction propagates through two propagation steps, a chain reaction.

Question 13

Describe the reaction mechanism for the third stage of the bromination of methane.

Answer 13

Question 14

Although radical substitution gives us a way of making haloalkanes, this reaction has problems that limit its important for organic synthesis. Describe these limitations.

Answer 14

• ​Further substitution. Bromomethane [CH₃Br] is formed in the second propagation step. Another bromine radical can collide with a bromomethane molecule, substituing a further hydrogen atom to form dibromoethane [CH₂Br₂]. Further substitution can continue until all hydrogen atoms have been substituted. The result is a mixture of CH3Br, CH₂Br₂, CHBr₃ and CBr₄.
• Substitution at different positions in a carbon chain. For methane, all four hydrogen atoms are bonded to the same carbon atom, so only one monobromo compound [CH₃Br] is possible. Similarly, only one monosubstituted product is possible with ethane [C₂H₅Br].
  
  • If the carbon chain is longer, we will get a mixture of monosubstituted isomers by substitution at different positions in the carbon chain.