Arenes Flashcards
Explain the relative resistance to bromination of benzene compared with alkenes. (4)
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
Explain the evidence that led scientists to doubt the model proposed by Kekule. (3)
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
What is an arene?
An aromatic hydrocarbon containing one or more benzene rings
Problems with Kekule’s benzene structure
- 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
Describe the delocalised model of benzene
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.
Under normal conditions benzene does not…
react with strong acids
react with halogens
decolourise bromine water
Nitration of benzene
- type of reaction
- reagents
- catalyst
- conditions
- equation
- Electrophilic sub
- Conc. nitric acid
- Conc. sulfuric acid
- 50 degrees (low)
- Benzene + HNO3 -> nitrobenzene + H2O
Nitration of methylbenzene
Can lead to the formation of 2,4,6 trimethylbenzene TNT
Halogenation of benzene
Reacts in the presence of halogen carrier which catalyzes reaction such AlCl3, Fe, FeCl3
Happens at RTP
Electrocphilic sub
HCl produced if Cl
Catalyst reactions for nitration
sulfuric acid generates the NO2+ electrophile:
HNO3 + H2SO4 -> NO2+ + HSO4- + H2O
H+ + HSO4 -> H2SO4
Catalyst reactions for halogenation
Halogen carrier generates either bromonium, chloronium or iodonium ion
Br2 + FeBr3 -> Br+ + FeBr4 -
H+ + FeBr4- -> FeBr3 + HBr
What happens when phenol is dissolved in water?
It forms a weak acidic solution by losing H+ from the OH group
How is the salt phenoxide formed?
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
Phenol reaction with bromine
RTP
No catalyst
white precipitate of 2,4,6 tribromophenol is formed
Phenol + 3Br2 -> 2,4,6 tribromophenol + 3HBr
Comparison of reactions of bromine with benzene and phenol
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
Why are phenols acidic?
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
How can an arylamine NOT be prepared?
Nucleophilic substitution between bromobenzene and ammonia because electron density in pi bond of benzene ring will repel nucleophile.
How are arylamines prepared?
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
Fuller equations for each step of preparing arylamine
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
Reaction of nitrobenzene with tin (IV) and hydrochloric acid
Forms phenylamine, water and tin chloride
(o)-NO2 + 3/2 Sn + 6HCl -> (o)-NH2 + 2H2O + 3/2 SnCl4
How to prepare nitrobenzene from benzene
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
Preparation of primary aliphatic amines
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
What are the two steps in synthesis of dyes from phenylamine?
Diazotisation and coupling
What is diazotisation?
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
