2.5- Hydrocarbons

Question 1

(a)

products when a hydrocarbon fuel undergoes complete combustion

Answer 1

Water (H2O) and carbon dioxide (CO2 ).

Question 2

(a)

balanced chemical equation for the complete combustion of ethane

Answer 2

2C2H6 + 7O2 → 4CO2 + 6H2O

Question 3

(a)

balanced symbol equation for the incomplete combustion of methane to form carbon monoxide and water

Answer 3

2CH4 + 3O2 → 2CO + 4H2O

Question 4

(a)

balanced symbol equation for the incomplete combustion of methane to form carbon particulates and water

Answer 4

CH4 + O2 → C + 2H2O

Question 5

(a)

benefits related to the burning of fossil fuels for energy

Answer 5

Benefits:
- Produces large amounts of energy.
- Currently relatively easily available.

Question 6

(a)

drawbacks related to the burning of fossil fuels for energy

Answer 6

Drawbacks:
- Uses finite resources
- Produces carbon dioxide which contributes to global warming.
- Produces carbon monoxide which is toxic.
- Produces acidic gases which form acid rain.

Question 7

(c)

Mechanism of radical substitution

Answer 7

photochlorination of alkanes

Question 8

(c)

Photochlorination of alkanes

Answer 8

A photochlorination reaction occurs when one of the hydrogen atoms on an alkane is replaced by a chlorine atom. The reaction is initiated by UV light.

Question 9

(c)

Halogenation

Answer 9

Alkanes react with chlorine in uv light to form a halogenoalkane.

Question 10

(c)

Radical halogenation of alkanes

Answer 10

(1) Initiation
Cl2 → 2Cl.
(2a) Propagation
Cl. + CH4 → .CH3 + HCl
(2b) Propagation
CH3 + Cl2 → CH3Cl + Cl.
(3) Termination
2Cl. → Cl2
Cl. + . CH3 → CH3Cl
2 .CH3 → CH3CH3

Question 11

(c)

Radical halogenation of alkanes explaination

Answer 11

(1) Initiation:
UV light provide energy to split the chlorine molecules into free radicals….
(2a) Propagation:
A chlorine radical reacts with a methane molecule. It produces another radical.
(2b) Propagation:
The methyl radical can collide with chloride molecules to produce a chlorine radical.

Question 12

(d)

difference in reactivity of alkanes and alkenes

Answer 12

Alkenes have a high electron density in the carbon-carbon double bond, C=C. This makes them more susceptible to attacks from electrophiles, compared to alkanes, making alkenes more reactive.

Question 13

(g)

Electrophilic addition

Answer 13

Halogens react with alkenes to form dihalogenoalkanes.

H2C=CH2 + Br2 → CH2BrCH2Br

Question 14

(h)

Bromine/bromine water and potassium manganate(VII) tests for alkenes

Answer 14

Add bromine water: The orange solution decolourises if the alkene C=C bond is present.
Add acidified potassium dichromate: The purple solution decolourises if the alkene C=C bond is present.

Question 15

(i)

Explain why the electrophilic addition of bromine to propene can produce a major and a minor product

Answer 15

This reaction can produce two products — a major and a minor product. Two types of intermediates can form — a primary and a secondary carbocation. The secondary carbocation is more stable than the primary one. The more stable secondary carbocation leads to the major product, 2-bromopropane. The primary carbocation results in the minor product, 1-bromopropane.

Question 16

(j)

Hydrogenation

Answer 16

Ethene reacts with hydrogen and a nickel catalyst at 150˚C to produce ethane.

H2C=CH2 + H2 → CH3CH3

Question 17

(j)

Importance of hydrogenation

Answer 17

This is commercially important since unsaturated vegetable oils are reacted with hydrogen to form margarine.

Question 18

(k)

Addition polymerisation

Answer 18

The double bonds in alkenes and substituted alkenes (monomers) can open up and join together to form long chains called polymers.

Question 19

(k)

Poly(propene)

Answer 19

Used in textiles, automotive parts, and packaging materials.

Question 20

(k)

Poly(chloroethene) or PVC

Answer 20

Insulation for electrical wires, windows, gutters, pipes

Question 21

(k)

Poly(phenylethene) or
polystyrene

Answer 21

Drinking cups, make disposable plastic cutlery and household appliances