Arenes

Question 1

What is the structure of benzene?

Answer 1

Benzene consists of a hexagonal planer arrangement of 6 carbon atoms, with bond angle between each carbon being 120°. Each carbon atom makes 4 bonds. 3 σ-bonds, of which 2 are to adjacent carbon atoms and 1 to a hydrogen atom. This leaves a valent outer shell electron in a p-orbital above and below plane of hexagonal planer.

Question 2

What is the electron delocalisation model of benzene?

Answer 2

The valent electron in the p-orbital above and below plane of the carbon ring overlap with the electrons in adjacent p-orbitals on the 2 adjacent carbon atoms. This occurs with every p-orbital in every carbon atom of the carbon ring, which results in a system of π-bonds distributed across the whole carbon ring, resulting in the formation of ring shaped bonding regions above and below the plane of the carbon ring.

Question 3

What is the kekelé model of benzene?

Answer 3

The kekelé model of benzene suggests that only pairs of p-orbital electrons overlap. This results in 3 sets of C=C double bonds forming across the benzene ring, each consisting of a σ-bond and π-bond.

Question 4

What are the delocalised π-bond electrons in benzene?

Answer 4

In benzene, the electrons from the valent p-orbitals overlap side-on to form ringed bonding regions. This results in the π-bonds formed being distributed over the whole benzene ring. The electrons from each p-orbital of each carbon atom are also distributed over the whole ring and areshared between all 6 carbons. Becasue these electrons are shared between more than 2 carbon atoms and are found in π-bonds, they are called delocaised π-bond electrons.

Question 5

What is the effect of the delocalisation model of electrons on the properties of benzene?

Answer 5

• The bond lengths between each carbon atom in the benzene rings are equal, resulting in benzene being a regular hexagon in shape. The kekelé structure would result in the C=C double bonds being shorter than the C-C bonds.
• Following the trend of an increase in enthalpies of hydrogenation between cyclohexene and cyclohexadiene, it would be logical for cyclehexa-1,3,5-triene (kekelé structure of benzene) to have a higher enthalpy of hydrogenation the 2 previous. However, benzene actually has a lower enthalpy of hydrogenation than cyclohexadiene. This suggests benzene has more stable bonding structure than kekelé structure, hence the electron delocalisation model.
• Because benzene is more stable than kekele structure, it does not take part in addition reactions with bromine, and thus doesn’t decolourise bromine water as cyclohexene and cyclohexadiene. This is further evidence that benzene does not contain C=C bonds.
• Addition reactions of benzene would result formation of less stable products (due to electrons from delocalised π-bond system being used) and thus are not favourable and do not take place under normal conditions. Instead, benzene takes place in substitution reactions whereby the hydrogen atoms are replaced by other groups. These preserve the delocalised structure of electrons in benzene and are thus more favourable

Question 6

What types of reactions does benzene take part in?

Answer 6

• Nitration.
• Halogenation.
• Methylation.

Question 7

What is the process of nitration of benzene?

Answer 7

Conditions: Benzene is added to a mixture of conc. nitric (HNO₃) and conc. sulfuric (H₂SO₄) acid. The sulfuric acid acts as the catalyst. The mixture is kept at 50°C to avoid multiple substitutions.

Reactants: HNO₃, benzene.

Products: Nitrobenzene, H₂O.

Name of mechanism: Electrophilic substitution.

Equation:

C₆H₆ + HNO₃ → C₆H₅NO₂ + H₂O

Question 8

What is the mechanism for nitration of benzene?

Answer 8

1. Formation of electrophile - Conc. H₂SO₄ and HNO₃ are mixed and cooled to form the nitronium (NO₂⁺) electrophile in the reaction:

H₂SO₄ + HNO₃ ⇔ HSO₄^- + NO₂⁺+ H₂O

2. Electrophilic substitution - First, an electron from the delocalised rings in benzene is donated to electrophile to form new C-N bond, forming an unstable C₆H₆NO₂⁺ intermediate. The C-H bond breaks by heterolytic fission to form a H⁺ ion, with both bonding electrons going to C atom and back into electron cloud to re-stabalise benzene ring.
3. Regeneration of catalyst - H⁺ combines with HSO₄^- in equation:

HSO₄^- + H⁺ → H₂SO₄

to regenerate catalyst.

Question 9

What is the process of halogenation of benzene?

Answer 9

Conditions: Benzene is mixed with halogen and a halogen carrier (e.g. AlCl₃).

Reactants: Benzene, halogen.

Products: Halogenobenzene, Hydrogen halide.

Name of mechanism: Electrophilic substitution.

Equation:

C₆H₆ + A₂ → C₆H₅A + HA

“A” being any halogen.

Question 10

What is the mechanism for bromination (halogenation) of benzene?

Answer 10

1. Generation of electrophile - Br-Br bond breaks by heterolytic fission to form Cl^- and Cl⁺ions. Cl^- reacts with FeBr₃ to form FeBr₄^-. Overall reaction:

FeBr₃ + Br₂ → Br⁺ + FeBr₄^-

2. Electrophilic substitution - First, an electron from the delocalised rings in benzene is donated to electrophile to form new C-Br bond, forming an unstable C₆H₆Br⁺ intermediate. The C-H bond breaks by heterolytic fission to form a H+ ion, with both bonding electrons going to C atom and back into electron cloud to re-stabalise benzene ring.
3. Regenerating catalyst - H⁺ reacts with FeBr₄^- in the reaction:

H⁺ + FeBr₄^- → FeBr₃ + HBr

to regenerate catalyst.

Question 11

What is the process of methylation of benzene?

Answer 11

Conditions: Benzene is mixed with chloromethane (CH₃Cl) and a halogen carrier (e.g. AlCl₃).

Reactants: Benzene, chloromethane.

Products: Methylbenzene, Hydroge chloride.

Name of mechanism: Electrophilic substitution.

Equation:

C₆H₆ + CH₃Cl → C₆H₅CH₃ + HCl

Question 12

What is the mechanism for methylation of benzene?

Answer 12

1. Generation of electrophile - C-Cl bond breaks by heterolytic fission to form Cl^- and CH₃⁺ ions. Cl- reacts with AlCl₃ to form AlCl₄^-. Overall reaction:

AlCl₃ + CH₃Cl → CH₃⁺ + AlCl₄^-

2. Electrophilic substitution - First, an electron from the delocalised rings in benzene is donated to electrophile to form new C-C bond, forming an unstable C₆H₆CH₃⁺ intermediate. The C-H bond breaks by heterolytic fission to form a H+ ion, with both bonding electrons going to C atom and back into electron cloud to re-stabalise benzene ring.
3. Regenerating catalyst - H+ reacts with AlCl₄^- in the reaction:

H⁺ + AlCl₄^- → AlCl₃ + Hcl

to regenerate catalyst.

Question 13

Why are phenols more acidic than alcohols?

Answer 13

Delocalisation of a lone pair of electrons on the O in -OH reduces the electron density of O and thus its attraction towards the H⁺ ions when they dissociate. This means that it is easier for protons to dissociate from phenols than alcohols, making phenols more acidic.

Question 14

What is the general formula for the reaction of phenol with an alkali (e.g. NaOH)?

Answer 14

Phenol + alkali → Phenoxide salt + water

E.g.:

C₆H₅OH + NaOH → C₆H₅O^-Na⁺ + H₂O

Question 15

What is the general formula for the reaction of phenol with a metal?

Answer 15

Phenol + Metal → Phenoxide salt + Hydrogen

E.g.:

2C₆H₅OH + 2Na → 2C₆H₅O^-Na⁺ + H₂

Question 16

What is the process of bromination of phenol?

Answer 16

Conditions: RTP, no catalysts required.

Reactants: Phenol, bromine.

Products: 2,4,6-tribromophenol, hydrogen bromide.

Observations: The orange bromine water is decolourised and a white precipitate is formed.

Equation:

Phenol + 3Br₂→ 2,4,6-tribromophenol + 3HBr

Question 17

What are some common uses for phenols?

Answer 17

Phenol:

• Synthesis of dyes.
• Synthesis of drugs such as aspirin.

Alkylphenols:

• Used in detergents.

Chlorophenols:

• Used in antiseptics and disinfectants.

Salicylic acid:

• Used to manufacture analgesics such as aspirin.

Question 18

What is the order of reactivity between cyclehexene, benzene and phenol?

Answer 18

Cyclehexene is the most reactive, followed by phenol, followed by benzene.

Question 19

In terms of bromination, why is cyclohexene the most reactive?

Answer 19

Cyclohexene contains a C=C double bond made from a σ and π bond. The π consists of localised areas of electron density, above and below σ bond. This means that it has the highest electron density out of all the π bonds. It is able to induce a dipole across Br₂ molecules strong enough for it to readily act as an electrophile, having the strongest attraction towards the areas of electron density and taking part in electrophilic addition reaction with cyclohexene.

Question 20

In terms of bromination, why is benzene the least reactive?

Answer 20

Because the π-bonds in benzene are distributed across the whole carbon ring and the delocalised π-bond electrons are shared between all 6 carbon atoms, benzene ‘electron rings’ have the lowest electron density out of the regions of electron density in the 3 compounds. It is unable to induce a dipole across a Br₂ molecule strong enough for the molecule to act as an electrophile, so there is the weakest attraction between the Br₂ molecule and areas of electron density in the molecule. This is why halogen carriers are required to generate bromonium ions which are stronger electrophiles in order to get benzene to react with bromine.

Question 21

In terms of bromination, why is phenol the second most reactive?

Answer 21

Due to phenol also having the delocalised π-bond clouds, its electron density is not as great as that of cyclohexene. However, a lone pair of p-orbital electrons from the O in -OH delocalises into the benzene electron cloud, which increases electron density in the benzene ‘electron rings’. It is able to induce a dipole across Br₂ molecules strong enough for the molecules to act as electrophiles. They have a great enough attraction towards areas of electron density in phenol to take part in electrophilic substitution reactions at RTP without a catalyst.

Question 22

How are aromatic amines prepared?

Answer 22

Nitrobenzene and other nitroarenes can be reduced to phenylamines. They are heated with a mixture of tin and concentrated HCl under reflux (both make up reducing agent). The excess HCl in the mixture is then neutralised.

Question 23

What is the equation for the reduction of nitrobenzene?

Answer 23

Nitrobenzene + 6[H] → Phenylamine + 2H₂O

(C₆H₅)NO₂ + 6[H] → (C₆H₅)NH₂ + 2H₂O

Question 24

What is the process of diazotisation?

Answer 24

Conditions: Temperature kept between 0-10°C.

Reactants: Phenylamine, Nitrous acid (HNO₂), HCl.

Products: Benzenediazonium chloride, H₂O.

Equation:

C₆H₅NH₂ + HNO₂ + HCl → C₆H₅N₂⁺Cl^- + 2H₂O

Question 25

What are the steps involved in Diazotisation?

Answer 25

1. Nirous acid is generated by mixing together sodium nitrate (NaNO₂) and HCl in reaction:

NaNO₂ + HCl → HNO₂ + NaCl

2. The mixture is cooled to between 0 and 10°C.
3. Phenylamine is added to the mixture and reacts to give the diazonium salt.

Question 26

Why does the mixture have to be kept between 0 and 10°C?

Answer 26

• If temperature drops below 0°C, reaction mixture freezes, which prevents reactants from colliding with each other and thus stops reaction from taking place.
• Temperature kept below 10°C because above 10°C, the diazonium ions decompose and the azo groups are released as N₂ gas, which decreases yield of reaction.

Question 27

What is the process of coupling?

Answer 27

Conditions: Reaction mixture has to be alkaline.

Reactants: Benzodiazonium chloride, phenol, NaOH.

Products: Azo dye, NaCl, H₂O.

Equation:

C₆H₅N₂⁺Cl^- + C₆H₅OH + NaOH → C₆H₅N=NC₆H₅OH + NaCl + H₂O