CHAPTER 12: ALKANES

Question 1

What are alkanes?

Answer 1

• Saturated hydrocarbons containing single C-C and C-H bonds σ-bonds.
• σ-bonds are overlaps of orbitals directly between bonding atoms.
• There is free rotation of the σ-bond.
• Each C atom has 4 electron pairs in 4 σ-bonds.
• Repulsion between pairs gives 3D tetrahedral shape.
• Each bond angle is 109.5°.

Question 2

Why are there variations in boiling points of alkanes?

Answer 2

• Boiling points increase as chain length increases.
• Increase in chain length increases Mr & SA.
• This introduces more London forces.
• There’s more surface contact between molecules.
• More energy is needed to overcome them.
• Branched isomers of alkanes have lower boiling points.
• There is less surface contact between molecules.
• This results in fewer London forces.
• Branches also prevent molecules from being as compact as straight-chain molecules.
• This further decreases IMF’s & boiling point.

Question 3

What are the properties of alkanes?

Answer 3

High Bond Enthalpy:
• Two s-orbitals overlap in σ-bonds.
• They overlap in a straight line.
• Gives highest electron density between nuclei.
• This means strong electrostatic attraction between nuclei & shared pair of electrons.
• This gives σ-bonds high bond enthalpy.
• Makes bonds difficult to break, so have low reactivity.

Low Polarity:
• Nucleophiles are electron rich species which donate pairs of electrons to create covalent bonds.
• They’re attracted to + or δ+ species.
• Electrophiles are electron deficient species which accept pairs of electrons to create covalent bonds.
• They’re attracted to - or δ- species.
• C-C bonds are non-polar.
• Electronegativity of C & H is very similar.
• The C-H bond is considered as non-polar.
• They don’t attract nucleophiles or electrophiles.
• Therefore, lack of polarity gives them low reactivity.

Question 4

What happens in combustion of alkanes?

Answer 4

Complete Combustion:
• In plentiful supplies of O₂ they give CO₂ & H₂O.
• CₓHᵧ + (x + ʸ/₄)O₂ → xCO₂ + ʸ/₂(H₂O).

Incomplete Combustion
• Occurs in limited O₂ supply.
• Each consecutive alkane molecule needs 1.5 more O₂ molecules for complete combustion.
• Hydrogen atoms are always oxidised to water.
• But, combustion of carbon may form CO or C.
• CO is colourless, odourless & slightly toxic.
• It’s better at binding to haemoglobin than O₂.
• This could lead to suffocation.

Question 5

How do alkanes react with halogens?

Answer 5

• They react in photochemical reactions.
• This requires ultraviolet light to begin.
• A hydrogen atom is substituted by Cl or Br.
• This is known as a free-radical substitution reaction.
• It has 3 stages: Initiation, Propagation, Termination.

Question 6

What is the initiation stage of alkanes reacting with halogens?

Answer 6

• Covalent bond in Br/Cl is broken by homolytic fission.
• Each atom in Br/Cl takes one electron from the pair.
• This forms two highly reactive Br/Cl radicals.
• Br-Br → Br• + •Br or Cl-Cl → Cl• + •Cl.
• Radicals are shown with dots to represent electrons.

Question 7

What is the propagation stage of alkanes reacting with halogens?

Answer 7

• Reaction Propagates through 2 propagation steps.
1) CH₄ +Br• → •CH₃ + HBr or CH₄ +Cl• → •CH₃ + HCl.
2) •CH₃ + Br₂ → CH₃Br + Br• or •CH₃ + Cl₂ → CH₃Cl + Cl•.
• In step one, a Br/Cl radical reacts with a C-H bond in methane.
• This forms a methyl radical & a molecule of HBr/HCl.
• In step two, each methyl radical reacts with another Br/Cl molecule.
• Forms organic product: bromomethane (or chloromethane) with a new Br/Cl radical.
• The new Br/Cl Radical reacts with another CH₄.
• The 2 steps cycle in a chain reaction.
• Propagation is terminated if 2 radicals collide.

Question 8

What is the termination stage of alkanes reacting with halogens?

Answer 8

• Two radicals collide.
• This forms a molecule with paired electrons.
• There are various termination steps:
1) Br• + •Br → Br₂ (or Cl• + •Cl → Cl₂).
2) •CH₃ + •CH₃ → C₂H₆.
3) •CH₃ + •Br → CH₃Br (or •CH₃ + •Cl → CH₃Cl).
• Both radicals are removed after collision.
• This terminates the reaction.

Question 9

Why is further substitution a limit of radical substitution?

Answer 9

• CH₃Br/Cl is formed in 2nd propagation step.
• Another Br/Cl radical can collide with CH₃Br/Cl.
• Substitutes another H atom & forms CH₂Br₂/Cl₂.
• Substitution can continue until there are no H atoms.
• This gives a mixture of: CH₃Br/Cl, CH₂Br₂/Cl₂, CHBr₃/Cl₃ & CBr₄/Cl₄.
• This lowers the yield of desired product.
• This makes the reaction inefficient.

Question 10

Why is substitution at different points in the chain a limit of radical substitution?

Answer 10

• If carbon chain is longer, you can get mixture of monosubstituted isomers by substitution at different positions in the carbon chain.
• E.g. Pentane can form 3 monosubstituted isomers.
• This gives a mixture of different products.
• This lowers the yield of desired product.
• This makes the reaction inefficient.