Alkanes

Question 1

Alkanes are non-polar

Answer 1

Electronegativity difference between H and C atoms are negligible, the C-H bond is essentially non-polar.

Question 2

Boiling point of alkanes increases as number of C atom increases

Answer 2

1. Number of C atom increases
2. Number of electrons per alkane molecules increases
3. Size of electron cloud increases
4. Ease of polarisation increases
5. Strength of ID-ID attraction increases
6. More energy required to overcome stronger ID-ID attraction between molecules.

Question 3

Boiling point of alkanes decreases as degree of branching increases

Answer 3

1. Increase in branching, molecules become more spherical
2. Surface area for intermolecular forces of attraction decreases
3. Extent of contact with neighbouring molecule decreases
4. Less energy required to overcome weaker ID-ID attraction between molecules

Question 4

Melting point of alkanes differ when the number of carbon is odd/even

Answer 4

1. Alkanes with an even number of carbon atoms are packed more closely in the crystalline state
2. Attractive forces between individual molecules are greater
3. Melting points are higher for alkanes with an even number of carbon atoms.

Question 5

Solubility of alkanes

Answer 5

In non-polar solvent, alkanes are soluble.
In water and highly polar solvent, alkanes are insoluble.

Question 6

Preparation of alkanes: Reduction of alkenes

Answer 6

Reaction and Reagent: H2(g), Ni and heat/ H2(g), Pt or Pd

Question 7

Lack of reactivity of alkanes

Answer 7

1. Alkanes are non-polar, therefore unable to attract charged species.
2. Alkanes are fully saturated, they do not contain region of high electron density, thus unable to attract electrophiles.
3. Alkanes contain relatively strong C-C bond and C-H bonds that do not break under normal conditions.

Question 8

Halogenation of alkanes

Answer 8

Reaction: Free radical substitution
Reagent: UV light, heat, sunlight and X2

Question 9

Free radical substitution

Answer 9

Step 1: Initiation
X2 is supplied energy from UV/heat/sunlight to undergo homolytic fission, forming 2X.

Step 2: Propagation
Hydrogen abstraction: Hydrocarbon reacts with reactive halogen radical.
Halogen abstraction: Hydrocarbon radical reacts with with halogen molecule to form the halogeno-compound

Step 3: Termination
Any two free radicals collides and combines to form a stable product