6.1.1 Aromatic Compounds Flashcards
Summarise the nitration of benzene (5)
- electrophilic substitution with H2SO4 catalyst and dilute nitric acid.
1. Formation of electrophile: nitric acid (HNO3) + sulphuric acid = nitronium ion (electrophile) and HSO4 and water
- Nitronium ion breaks pi system and bonds with benzene
- H+ ion is released - reformation of catalyst:
H+ + HSO4- = H2SO4
Overall equation: under 50 degrees
Benzene + nitric acid = nitrobenzene and water
Summarise the bromination of benzene
- electrophilic substitution with iron bromide catalyst and bromine.
1. Formation of electrophile:
Br2 + FeBr3 = Br+ + FeBr4
- Benzene attacks Br+ and releases H+
- Reformation of catalyst:
H+ FeBr4 = FeBr3 + HBr
Overall:
Benzene + Br2 = Bromobenze and H+
Why is the Kekule’s model of benzene incorrect? (3)
- Bond Length: Kekule’s model shows 3 double bonds and 3 single bonds that vary in length (134nm or 153nm). However through X Ray diffraction it was revealed that all bonds in benzene are the same length: 0.139nm
- Hydrogenation enthalpy: Kekule’s model infers that benzene has triple the enthalpy of cyclohexene however it is 208 KJmol-1 instead of 360KJmol-1
- Reactivity: Benzene does not react like alkenes as it does not react with bromine without a catalyst.
Describe the bonding in Benzene (4)
- 6 carbons and 6 Hydrogens in a delocalised pi system
- The P orbitals with an electron overlap sideways and above and below the carbons so the electron density is shared across all 6 carbons
- The double bonds contain sigma bonds through the overlapping of S orbitals and pi bonds
- The C-H bond is a localised sigma bond
What are the conditions needed for the chlorination of benzene?(3)
- Electrophilic substitution
- Cl2
- Halogen carrier: AlCl3 (catalyst)
- Overall equation:
Benzene + Cl2 = chlorobenzene + HCl
Describe the Alkylation of benzene? (3)
- Electrophilic substitution
- Catalyst: AlCl3
- Reactants: Haloalkane and benzene
- overall equation:
Benzene + Haloalkane = RBenzene + HCl
Describe the acylation of benzene (5)
- Electrophilic substitution
- Reactants:
Benzene and Acyl halogen - Catalyst: AlCl3
- electrophile formed: COCH3+
- overall equation:
Benzene + Acyl chloride = Phenyl + HCl
Describe the reactivity of a phenol (4)
- More reactive than benzene
- Reacts with metals and metal hydroxides but not carbonates
- Is 2,4 directing
- Reacts with bromine without catalyst therefore decolourises
Why is phenol more reactive than benzene? (2)
- The O atom donates a lone pair of electrons to the pi system which increases the electron density of the aromatic ring
- This polarises the aromatic ring allowing it to induce dipoles in molecules like Br2 to undergo Electrophilic substitution without a catalyst
Describe the bromination of phenol (5)
- No catalyst needed
- Phenol is 2,4 directing so if 3Br2 is used then it will form 2,4,6- tribromophenol
- Observations:
Bromine is decolourised
White precipitate of 2,4,6 tribromophenol is formed at room temperature - Overall reaction:
Phenol + 3Br2 = 2,4,6 tribromophenol + 3HBr
Describe the nitration of phenol.
- No catalyst
- Reactants:
Phenol + dilute HNO3 at room temperature - Forms 2- nitrophenol and 4- nitrophenol as Phenol is 2,4 directing
How would you separate 2-nitrophenol and 4-nitrophenol
- Fractional distillation
- One will have a higher boiling point than the other
What directing group is phenylamine?
- 2,4 directing
- An activating group so no catalyst is needed in reactions
What directing group is Nitrophenol?
- 3, directing
- A deactivating group
How do you test for phenols? (3)
- Neutral iron chloride test.
- Pale yellow solution of FeCl3 is added which turns violet in the presence of phenol.
- FeCl2 replaces H in the OH group of phenol and HCl is made
