Alkanes

Question 1

Characteristics of alkanes (hydrocarbons, saturated/ unsaturated, Mr, C atom and molecular geometry)

Answer 1

Hydrocarbons because only contain carbon and hydrogen

Saturated because they contain only carbon-carbon and carbon-hydrogen single bonds

CnH2n+2

Each C atom is sp3 hybridised and has a tetrahedral molecular geometry, bond angle is 109.5

Question 2

Classification of hydrogen and carbon atoms

Answer 2

1°C atom is bonded to one other C atom
2°C atom is bonded to two other C atoms
etc…

1°H atom is bonded to a C atom that is bonded to 1 other C atom
2°H atom is bonded to a C atom that is bonded to 2 other C atoms

Question 3

Conditions for isomerism

Answer 3

From butane onwards, alkanes have constitutional isomers due to branching of hydrocarbon chains

Branched-chain alkanes display enantiomerism

Cycloalkanes display cis-trans isomerism

Question 4

Physical properties (polarity, bp mp, solubility, density)

Answer 4

C-H is essentially non-polar due to negligible difference in electronegativity
Hence alkanes are non-polar, with IMF being relatively weak id-id

bp mp increase with more carbon atoms
- No. of electrons per alkane molecule increase
- Size of electron cloud increases
- Ease of polarisation of electron cloud increases
- Stronger id-id resulting in more energy required to overcome

bp mp decrease with increasing degree of branching
- Molecule becomes more spherical/ compact
- Surface area available for intermolecular interactions decrease
- Extent of contact between neighbouring molecules reduces, weakening id-id that can be easily overcome

Solubility
- Non-polar, so they are soluble in non-polar solvents
- Dissolved compounds of low polarity and not compounds of high polarity

Density
- Increase with increasing molecular size but tends to level off at 0.8g cm-3, less dense than water

Question 5

Preparation of alkanes

Answer 5

Reduction from alkenes:
H2(g), Ni, heat
- Nickel is least active and requires elevated temperature and pressure
H2(g), Pt or Pd

Reduction of alkynes: Pd and heat
Reduction of alkene to cycloalkane: Ni and high temp, pressure

Question 6

Unreactivity

Answer 6

No effect, whether hot or cold, two immiscible layers observed
- NaOH, HCL, H2SO4, KMnO4

Unaffected by polar reagants due to it being saturated and non-polar

Because:
1. Non-polar
- Do not contain any region of high electron density and do not attract electrophilic reagants
- Do not contain any electron deficient sites to attract nucleophilic reagents
2. Relatively strong C-C and C-H bonds which do not break under normal conditions

Question 7

Combustion

Answer 7

Burn readily in air or oxygen when ignited, exothermic reaction

IN excess oxygen
Burn with non-sooty flame, produce CO2 and H2O

IN limited oxygen
Burn with sooty flame, forms CO (g) and C (as soot) in addition to CO2 and H2O

Question 8

Free radicals

Homolytic fission

Answer 8

Atom/ group of atoms that has unpaired electrons
Highly reactive and intermediates, have equal numbers of protons and electrons and electrically neutral
More alkyl groups attached to carbon with unpaired electron, more stable the alkyl radical as alkyl groups are electron donating

Bond breaks in a way that each of atom involved in forming bond acquires one of 2 bonding electrons thus forming free radicals (curly arrow with half a head represents one electron movement)

Question 9

Essential points in Free-radical substitution

Answer 9

1. Name of mechanism: Free-radical substitution
2. 3 Steps: Initiation, propagation, termination
3. Conditions: UV light or heat
4. Half arrows from Cl-Cl bond in initiation step
5. Repeating part (a) and part (b) in propagation step
6. Termination steps include reaction forming required product
7. Placing of dots on correct atoms in steps

Question 10

Initiation step

Answer 10

Chlorine is supplied with energy to split into free radicals via homolytic fission of Cl-Cl bond

Involved bond breaking and not proceed in dark or at room temp
Easier to break Cl-Cl bond than C-H bond or the C-C bond

Question 11

Propagation step

Answer 11

Hydrogen abstraction- Highly reactive chlorine radical collide with methane molecule and abstracts one hydrogen atom from it, forming hydrogen chloride and methyl radical

Halogen abstraction- Methyl radical react with chlorine molecule, leading to formation of cholromethane and chlorine free radical

Cycle repeats

Hydrogen radical never formed
- Formation of H-Cl bond more exothermic than formation of C-Cl bond

Question 12

Termination step

Answer 12

2 free radicals collide and combine to form a stable product