Hydrocarbons

Question 0

What is homolysis?

Answer 0

This is even splitting of a covalent bond and gives free radicals.

Question 1

What is hetrolysis?

Answer 1

Hetrolysis is “uneven” splitting of a covalent bond and produces ions.

Question 2

What is the first step in purifying hydrocarbons?

Answer 2

The hydrocarbons in crude oil are separated out into simpler mixtures through fractional distillation.

Question 3

Outline the process of cracking.

Answer 3

Cracking is the process by which long chain hydrocarbons are broken up into smaller and unsaturated products which are more useful. This process can be carried out in a thermal cracker (run at 750-850°C) or catalytic cracker (run at 400-500°C).

Question 4

What is the catylist used in a catalytic cracker?

Answer 4

A fluidized bed of zeolites (aluminosilicates).

Question 5

What can the products of cracking be used for?

Answer 5

These may be used to make other chemicals or can be added to gasoline to increase the octane number (more branched chains).

Question 6

Give a typical cracking equation.

Answer 6

C₁₀H₂₂ → C₅H₆ + C₇H₁₆

Question 7

What are the conditions necessary for the photo chlorination of methane?

Answer 7

The presence of light. (U.V required)

Question 8

What is the first step in the photochlorination of methane?

Answer 8

This is the initiation step.

Homolysis of the Cl-Cl bond occurs when a photon of light is absorbed. An attack on the methane by the Cl⋅ radicals formed then occurs.

Question 9

What is the second step in the photochlorination of methane?

Answer 9

This is propagation and occurs through:

Cl⋅ +CH₄ → CH₃⋅ + HCl
CH₃⋅ + Cl₂ → CH₃Cl + Cl⋅

This step is a chain reaction so that once a chlorine molecule has been split light does not need to be absorbed for the reaction to continue so thousands of molecules can be formed for each photon of light absorbed.

Question 11

What is the third and final stage in the photochlorination of methane?

Answer 11

This is the termination step. There are three possible versions:

CH₃⋅ + CH₃⋅ → C₂H₆
CH₃ + Cl⋅ → CH₃Cl
Cl⋅ + Cl⋅ → Cl₂

Question 12

What are the three methods of producing Alkenes?

Answer 12

1. ) Industrially - in the cracking process
2. ) Dehydration of alcohols through an elimination reaction.
3. ) from halogeno alkanes (elimination reaction)

Question 13

Outline the possible process of creating an Alkene through the dehydration of alcohol.

Answer 13

This is the removal of the elements of water achieved in three ways:

1. ) the alcohol vapour is passed over aluminum oxide at 300-350°C
2. ) the alcohol vapour is passed over phosphoric acid at 200-250°C
3. ) the alcohol is mixed with excess concentrated sulphuric acid and heated to 170°C

Question 14

How would you produce an Alkene from a halogeno Alkene?

Answer 14

By heating the halogeno Alkene with potassium of sodium hydroxide dissolved in ethanol.

Question 15

What do the bond energy terms of the C-C bond and the C=C bond show?

Answer 15

That the C=C(610KJmol⁻¹) double bond is less than twice as strong as the C-C (346KJmol⁻¹) single bond.

Question 16

Why might it be expected that when carbon atoms bond, the four bonds will have slightly different values?

Answer 16

Because the valance electrons of carbon are made out of two 2S electrons and two 2P electrons. However the bonds are found to be all the same.

Question 17

Why do the four Bonds for carbon in ethene have the same bond energy terms?

Answer 17

It is assumed that one of the 2S electrons is promoted into a 2P orbital to form four SP³ hybrid orbitals and all the bonds are the same.

Question 18

Describe the strength of an SP³ hybrid orbital.

Answer 18

The SP³ orbital is more concentrated in direction than a P orbital or S orbital so it is able to overlap more and form a stronger bond.

Question 19

How does each carbon atom produce three SP² orbitals in ethene?

Answer 19

They use a 2S orbital and two of the 2P orbitals to form three SP³ hybrid orbitals.

Question 20

What bond type is formed when two P orbitals (I.e SP² orbitals in ethene) overlap endways?

Answer 20

They form a sigma bond.

Question 21

Describe the bonding structure in ethene.

Answer 21

Each carbon forms 3 sigma bonds using SP² orbitals two with hydrogen atoms and one carbon atom, with each bond at an angle of 120°. These bonds are planar. The unhybridized P orbital of each carbon is at right angles to the plane of the SP² orbitals. These P orbitals on adjacent carbon atoms are close enough to overlap sideways. The overlapping takes place above and below the plane of the SP² bonds and is called a π bond. π bonds are not as strong as Σ bonds because there is less overlapping.

Question 22

What are all the reactions of alkenes?

Answer 22

They are all electrophillic addition reactions and since there is a high concentration of electrons in the double bonds the attack will be by electrophiles.

Question 24

What reaction can the second step of the first photochlorination of methane give rise to?

Answer 24

A “secondary chlorination”:

CH₃Cl⋅ → ⋅CH₂Cl + HCl
⋅CH₂Cl + Cl₂ → CH₂Cl₂ + Cl⋅

And so forth.

Question 25

What is catalytic Hydrogenation (addition of hydrogen)?

Answer 25

This is when hydrogen adds across a double bond in an to form a saturated bond. It is an example of heterogenous catalysis, the hydrogen and double bond compound are adsorbed onto the surface of the catalyst using the “D” orbitals.

Question 26

What is the use of catylitic hydrogenation?

Answer 26

This is used in the food industry to produce margarines by hydrogenating oils. The softness of the margarines can be controlled by the amount of hydrogenation, the more hydrogenation, the harder the margarine.

Question 27

What are the two possible products created from the addition of hydrobromic acid (HBr) to Propene?

Answer 27

1-bromopropane and 2-bromopropane.

Question 28

What is the main product produced in the addition of hydrobromic acid to propene and why!

Answer 28

It is found that the main product is 2-Bromopropane. This is because of Markovnikov’s rule.

Question 29

What is Markovnikov’s rule?

Answer 29

This is a rule governing the addition of Hydrogen compounds across a double bond:-

When a hydrogen compound adds on across a double bond the hydrogen atom always adds on to the carbon atom which already has the most Hydrogen atoms in it.

Question 30

Explain Markovnikov’s rule.

Answer 30

The carbocation can form on either the C¹ or C² carbon in prop-1-ene.

When the Carbocation forms on the C² it is a secondary carbocation (only one Hydrogen atom attached to the reacting carbon atom).

When the carbocation forms on the C¹ carbon it is a primary carbocation (two hydrogen atoms attached to the reacting Carbon atom).

The secondary Carbocation is more stable because It is next to a CH₃ group which is electron releasing and will stabilize the ion. So 2-bromopropane production is favoured

Question 31

What happens when you add bromine liquid or bromine dissolved In a a non polar solvent to Ethene?

Answer 31

1,2-dibromoethane is formed

Question 32

What happens when you add bromine water to Ethene?

Answer 32

It forms 2-bromoethanol. This acts as a test for a double bond as shaking an Alkene with bromine water causes the bromine water to be decolourised.

Question 33

Outline the mechanism of the addition of Bromine.

Answer 33

The bromine molecule becomes polarized which is induced by approaching the ethene molecule. The partial positive charge on the bromine will then attack the double bond.

Question 34

How can the direct addition of water to ethene be achieved and what does this produce?

Answer 34

By passing ethene and steam at 300°C and 60atm over phosphoric acid absorbed on silica pellets. This process is used industrially and produces ethanol.

Question 35

What kind of polymerization do alkenes undergo?

Answer 35

Addition polymerization.

Question 36

What is addition polymerization and how is it achieved?

Answer 36

This is when molecules join up together to form long chains. The first polymerizations involved very high pressures (100atm) and used oxygen as a catylist. Modern methods use more powerful catylist and take place at atmospheric pressure.

Question 37

What is the difference between plastics formed at normal pressures and high pressures?

Answer 37

Plastics formed at high pressures is a low density, very pliable material.

Plastics formed at normal pressures have a higher density and are tougher.

Question 38

Give some examples of plastics.

Answer 38

PVC (polyvinylchloride) from CH₂=CHCl
Teflon (polytetraflouroethene) from F₂C=CF₂
Polystirene from CH=CH₂