Arenes

Question 1

Explain the relative resistance to bromination of benzene compared with alkenes. (4)

Answer 1

In benzene, electrons are delocalised.
In alkenes, π-electrons are localised/between two carbons
benzene has a lower electron density
alkene/C=C has a higher electron density
benzene polarises bromine / Br2 LESS
OR benzene attracts bromine / Br2 LESS
OR benzene induces a weaker dipole in bromine / Br2

Question 2

Explain the evidence that led scientists to doubt the model proposed by Kekule. (3)

Answer 2

All bond lengths between the carbons have the same lengths, whereas double and single bonds have different lengths
Delta H hydrogenation less exothermic than expected when compared to Delta H hydrogenation for cyclohexene
Resistant to electrophilic attack under normal conditions, does not decolourise bromine water

Question 3

What is an arene?

Answer 3

An aromatic hydrocarbon containing one or more benzene rings

Question 4

Problems with Kekule’s benzene structure

Answer 4

• Benzene has lower reactivity than the alkenes, each c=c bond would be expected to react with bromine water and decolourise it, however this doesn’t happen. (Led to equilibrium model)
• Carbon - carbon bond lengths are all the same
• Enthalpy change of hydrogenation of benzene is higher than the expected (3x cyclohexene), actual structure has much less energy than Kekule’s structure

Question 5

Describe the delocalised model of benzene

Answer 5

Each carbon is bonded to two other carbons and one hydrogen, forming sigma bonds and a trigonal planar shape around the C atom. The fourth outer electron is in a 2p - orbital above and below the plane of the molecule, and this sideways overlaps with other electrons in p-orbitals on the carbons forming a pi cloud ring of electron density above and below the plane of the carbon atoms, delocalised.

Question 6

Under normal conditions benzene does not…

Answer 6

react with strong acids
react with halogens
decolourise bromine water

Question 7

Nitration of benzene

• type of reaction
• reagents
• catalyst
• conditions
• equation

Answer 7

1. Electrophilic sub
2. Conc. nitric acid
3. Conc. sulfuric acid
4. 50 degrees (low)
5. Benzene + HNO3 -> nitrobenzene + H2O

Question 8

Nitration of methylbenzene

Answer 8

Can lead to the formation of 2,4,6 trimethylbenzene TNT

Question 9

Halogenation of benzene

Answer 9

Reacts in the presence of halogen carrier which catalyzes reaction such AlCl3, Fe, FeCl3
Happens at RTP
Electrocphilic sub
HCl produced if Cl

Question 10

Catalyst reactions for nitration

Answer 10

sulfuric acid generates the NO2+ electrophile:
HNO3 + H2SO4 -> NO2+ + HSO4- + H2O
H+ + HSO4 -> H2SO4

Question 11

Catalyst reactions for halogenation

Answer 11

Halogen carrier generates either bromonium, chloronium or iodonium ion
Br2 + FeBr3 -> Br+ + FeBr4 -
H+ + FeBr4- -> FeBr3 + HBr

Question 12

What happens when phenol is dissolved in water?

Answer 12

It forms a weak acidic solution by losing H+ from the OH group

Question 13

How is the salt phenoxide formed?

Answer 13

Acidic solution of phenol (aq) is neutralised by sodium hydroxide (aq) 
Phenol + NaOH -> (o)-O-Na+ + H2O
OR
Phenol reaction with sodium metal
2 Phenol + 2Na -> 2 (o)- O-Na+ + H2

Question 14

Phenol reaction with bromine

Answer 14

RTP
No catalyst
white precipitate of 2,4,6 tribromophenol is formed
Phenol + 3Br2 -> 2,4,6 tribromophenol + 3HBr

Question 15

Comparison of reactions of bromine with benzene and phenol

Answer 15

Reaction with phenol occurs more readily because:

• lone pair of electrons on oxygen atom on the phenol group is drawn in benzene ring
• this creates high electron density in ring structure
• increased density polarizes the bromine molecules which are then attracted

Question 16

Why are phenols acidic?

Answer 16

The lone pair on the oxygen becomes partially delocalised into the pi bond on the benzene ring
This increases partial positive charge on phenol hydrogen atom and therefore it is more likely to be lost as an H+
The delocalisation also means that the negative charge on the phenoxide is spread over several atoms and it is more stable and less likely to recombine with H+ ion

Question 17

How can an arylamine NOT be prepared?

Answer 17

Nucleophilic substitution between bromobenzene and ammonia because electron density in pi bond of benzene ring will repel nucleophile.

Question 18

How are arylamines prepared?

Answer 18

Reduction of nitrobenzene using a mixture of tin and conc. HCl heated under reflux. Conc. HCl is then neutralized (using sodium hydroxide solution)

(o)-NO2 + 6[H] -> (o)-NH2 + 2H2O

Question 19

Fuller equations for each step of preparing arylamine

Answer 19

1. (o)-NO2 + 7H+ + 6e- -> (o)-NH3+ + 2H2O
Electrons provided through:
Sn -> Sn2+ + 2e-
Sn2+ -> Sn4+ + 2e-
2. sodium hydroxide solution added
(o)-NH3+ + OH- -> (o)-NH2 + H2O

Question 20

Reaction of nitrobenzene with tin (IV) and hydrochloric acid

Answer 20

Forms phenylamine, water and tin chloride

(o)-NO2 + 3/2 Sn + 6HCl -> (o)-NH2 + 2H2O + 3/2 SnCl4

Question 21

How to prepare nitrobenzene from benzene

Answer 21

Treat benzene with a mixture of conc. nitric acid with conc. sulfuric acid catalyst at 50 degrees. Yellow oily nitrobenzene will be formed.

(o) + HNO3 -> (o)-NO2 + H2O

Question 22

Preparation of primary aliphatic amines

Answer 22

Warm a halogenoalkane with excess ammonia using ethanol as solvent as halogenoalkane is insoluble in water, in nucleophilic sub.
1. A salt is formed:
CH3CH2Cl + NH3 _> CH3CH2NH2 + HCl
2. Further ammonia reacts with the hydrogen chloride formed:
HCl + NH3 -> NH4+Cl-
3. As a nucelophile, ammonia has lone pair of electrons and attacks the delta + carbon atom in the polar carbon-halogen bond . However, propylamine has a lone pair of electrons which can attack another molecule of 1-chloroethane causing further substitution:
CH3CH2Cl + CH3CH2NH2 -> (CH3CH2)2NNH + HCl

Question 23

What are the two steps in synthesis of dyes from phenylamine?

Answer 23

Diazotisation and coupling

Question 24

What is diazotisation?

Answer 24

First step in which the diazonium ion is formed. Mixture of phenylamine and nitrous acid is kept at below 10 degrees and diazonium salt is formed. HNO2 (nitrous acid) is generated in the reaction mixture by reacting together sodium nitrate, NaNO2, and excess HCl:
NaNO2 + HCl -> HNO2 + NaCl
(o)-NH2 + HNO2 + HCl -> (o)-N+Cl-=–N +2H2O

Question 25

What is coupling?

Answer 25

When diazonium salt is reacted with a phenol or other aromatic compound under alkaline conditions.
azo functional group: -N=N-
diazonium salt + (o)-OH + NaOH -> (o)-N=N-(o)-OH + NaCl + H2O

Question 26

During the conversion of phenylamine to a benzenediazonium ion, why must temp. not be lower or higher than 10 degrees c. ?

Answer 26

If lower then reaction rate will be far too low, higher will result in thermal decomposition of benzenediazonium cation

Question 27

Main feature of methyl orange which makes it water-soluble?

Answer 27

Ionic which makes it soluble in water because polar water molecules interact with ions and separate them

Question 28

Nitration of phenol only requires dilute acid, why are different conditions employed?

Answer 28

Phenol is activated towards electrophilic attack because one the lone pairs of electrons on oxygen atom can delocalised into the ring of electrons in the benzene ring, increasing the electron density, therefore no need for catalyst