M4 Alkanes Flashcards
What are alkanes?
Saturated hydrocarbons containing single C-C and C-H bonds as sigma bonds, with free rotation of the sigma bond.
What are sigma bonds?
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.
Each carbon atom in an alkane has four sigma bonds, either C-C or C-H.
What is the shape of alkanes?
- Each carbon atom is surrounded by four electron pairs in four sigma bonds.
- Repulsion between these electron pairs results in a 3D tetrahedral arrangement around each carbon atom, with a bond angle of 109.5°.
- The sigma bond acts as axes around which the atoms can rotate freely, so these shapes are not rigid.
How do boiling points of alkanes differ with length and branching?
- London forces act between molecules that are close in surface contact.
- As chain length increases, molecules have a larger surface area so more surface contact is possible between molecules. The London forces between molecules will be greater and so more energy is required to overcome the forces.
- The more branched an alkane the lower the boiling point as there are fewer surface points of contact between molecules of the branched alkanes, giving fewer surface points of contact between molecules of the branched alkanes, giving fewer London forces. The branches also get in the way and prevent branched molecules getting as close together as straight-chain molecules, decreasing intermolecular forces further.
Why does boiling point increase?
- London forces are weak intermolecular forces that hold molecules together in solids and liquids.
- Once broken the molecules move apart and the alkane becomes a gas.
- The greater the intermolecular forces, the higher the boiling point.
Why are alkanes unreactive?
- C-H and C-C sigma bonds are strong
- C-C bonds are non-polar
- The electronegativity of carbon and hydrogen is similar so C-H bond can be considered non-polar.
- Due to the lack of polar bonds, alkanes have high bond enthalpies which require lots of energy to overcome.
Describe complete combustion of alkanes
- In a plentiful supply of oxygen, alkanes buen completely to produce carbon dioxide and water.
eg. complete combustion of propane:
C3H8 + 5 O2 = 3 CO2 + 4 H2O
(g) (g) (g) (l)
I
Describe incomplete combustion of alkanes
- In a limited supply of oxygen, there is not enough oxygen for complete combustion. When oxygen is limited, the hydrogen atoms in the alkane are always oxidised in water, forming the toxic gas carbon monoxide, or even carbon as soot.
eg. C7H16 + 7.5 O2 = 7 CO + 8 H2O
(l) (g) (g) (l)
Describe the reaction of alkanes with halogens
- In 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.
eg. CH4 (g) + Br2 (l) = CH3Br (g) + HBr(g)
methane bromomethane
This is a substitution reaction as a hydrogen atom in the alkane has been substituted by a halogen atom.
What is homolytic fission?
A covalent bond breaks and each atom retains one of the shared pairs of electrons. Each fragment is known as a free radical.
What is a radical?
- A species with an unpaired electron.
- Radicals are most commonly formed when the bond being broken has electrons that are fairly equally shared
- Radicals are so reactive as it has an impaired electron. They have no overall charge
What is the mechanism for bromination of alkanes?
- The mechanism for the bromination of methane is an example of radical substitution.
- The mechanism takes place in three stages, called initiation, propagation and termination.
Describe initiation
- In the initiation stage, the reaction is started when the covalent bond in a bromine molecule is broken by homolytic fission.
- Each bromine takes one electron from the pair, forming two highly reactive bromine radicals. This energy for bond fission is provided by UV radiation.
Br-Br —> Br• + Br•
Describe propagation
- The reaction propagates through two propagation steps, a chain reaction
- In the first step a bromine radical (Br•) reacts with a C-H bond in the methane, forming a methyl radical, •CH3 and a molecules of hydrogen bromide HBr
CH4 + Br• —> •CH3 + HBr - In the second propagation step, each methyl radical reacts with another bromine molecule, forming the organic product bromomethane, with a new bromine radical
•CH3 + Br2 —> Br•
The new bromine radical then reacts with another CH4 molecule as in the first propagation step, and the two steps can continue to cycle through in a chain reaction
The propagation steps could continue until all the reactants have been used up, propagation is terminated whenever two radicals collide.
Describe termination
- In the termination stage, two radicals collide, forming a molecule with all electrons paired. There are a number of possible termination steps.
Br• + Br• —> Br2
•CH3 + •CH3 —> C2H6
•CH3 + •Br —> CH3Br
