Chapter 12: Properties of the alkanes

Question 1

What are alkanes?

Answer 1

• The main components of natural gas and crude oil
• Among the most stable organic compounds
• Their lack of reactivity has allowed crude oil deposits to remain in the earth for many millions of years
• They are mainly used in fuels

Question 2

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 joined to 4 other atoms by single covalent bonds
• These are a type of covalent bond called a sigma bond

Question 3

What is a sigma bond?

Answer 3

• 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

Question 4

Each carbon atom in an alkane has how many sigma bonds

Answer 4

4 sigma bonds

• Either C-C or C-H
• Each sigma bond acts as axes around which the atoms can rotate freely, these shapes are not rigid

Question 5

The shape of alkanes

Answer 5

• Each carbon atom is surrounded by four electron pairs in four sigma bonds
• Repulsion between these electron pairs results in 3D tetrahedral arrangement around each carbon atom
• Each bond is approximately 109.5 degrees

Question 6

Briefly describe fractional distillation

Answer 6

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

Question 7

Why does boiling point of alkanes increase?

Answer 7

• The answer lies with the weak intermolecular forces called London forces
• These forces hold molecules together in solids and liquids but, once broken, the molecules move apart from each other and the alkane becomes a gas
• The greater the intermolecular forces, the higher the boiling point

Question 8

Effect of chain length on boiling point

Answer 8

• 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 9

The effect of branching on boiling point

Answer 9

• Isomers of alkanes have the same molecular mass
• If you compare the boiling points of branched isomers with straight-chain isomers, you find that the branched isomers have lower boiling points

Question 10

Explain the effect on boiling point

Answer 10

• The reason for this difference lies again London forces
• There are fewer surface points of contact between molecules of the branched alkanes, giving fewer London forces
• Another factor lies with the shape of the molecules
• The branches get in the way and prevent the branched molecules getting as close together as straight-chained molecules, decreasing the intermolecular forces further

Question 11

Reactivity of alkanes

Answer 11

• Alkanes do not react with most common reagents. The reasons for their lack of reactivity are:
• C-C and C-H sigma 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 12

Combustion of alkanes

Answer 12

• Despite their low reactivity, all alkanes react with a plentiful supply of oxygen to produce carbon dioxide and water
• This reaction is called combustion
• All combustion processes give out heat, and alkanes are used as fuels because they are readily available, easy to transport

Question 13

In a plentiful supply of oxygen, alkanes burn completely. What does this produce

Answer 13

Carbon dioxide + water

Question 14

Incomplete combustion of alkanes

Answer 14

• In a limited supply of oxygen, there is not enough oxygen for complete combustion
• Carbon monoxide is produced

Question 15

Reactions of alkanes with halogens

Answer 15

• In the presence of sunlight, alkanes react with halogens
• The high-energy ultraviolet radiation present in sunlight provides the initial energy for a reaction to take place
• This is a substitution reaction

Question 16

The mechanism for the bromination of methane is an example of what?

Answer 16

radical substitution

Question 17

Mechanism for bromination of alkanes: Step 1 (Initiation)

Answer 17

• The covalent bond in the bromine molecule is broken by homolytic fission
• Each bromine atom takes one electron from the pair, forming two highly reactive bromine radicals
• The energy for this bond fission is provided by UV radiation

Question 18

Mechanism for bromination of alkanes: Step 2 Propogation

Answer 18

• In the first propagation step, a bromine radical, reacts with a C-H bond in the methane, forming a methyl radical and a molecule of HBr
• In the second propogation each methyl radical reacts with another bromine molecule, forming the organic product bromoethane, CH3Br, together with a new bromine radical

Question 19

When is propogation terminated?

Answer 19

When two radicals collide

Question 20

Describe termination

Answer 20

• Two radicals collide, forming a molecule with all electrons
  paired
• When two radicals collide and react, both radicals are removed removed from the reaction mixture, stopping the reaction

Question 21

Limitations of radical substitution in organic synthesis

Answer 21

• Although radical substation gives us a way of making haloalkanes this reaction has problems that limit it’s importance for synthesis of just one organic compound

Question 22

Further substitution

Answer 22

• In the mechanism above, bromoethane, CH3Br, was formed in the second propogation step
• Another bromine radical can collide with a bromoethane molecule, substituting a further hydrogen atom to form dibromoethane, CH2Br
• Further substitution can continue until all hydrogen atoms have been substituted
• The result is a mixture of CH3Br, CH2Br, CHBr3 and CBbr4

Question 23

Substitution at different points in the carbon chain

Answer 23

• For methane, all four hydrogen atoms are bonded to the same carbon atom, so only one monobromo compound, CH3Br is possible
• With ethane, similarly only one monosaturated C2H5Br is possible
• If the carbon chain is longer, we will get a mixture of monosubstituted isomers by substitution at different positions in the carbon chain