18. Oragnic III (ARENES)
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Give the 2 names of the models for the structure of benzene
Kekule model: contains double bonds between every other carbon molecule Delocalised model: one electron from the p orbital of each carbon atom is delocalised so the p orbitals overlap to form a delocalised p system
Give the pieces of evidence used to provide evidence for the delocalised model of benzene
Physical factor: C-C bond lengths are all equal in length in benzene, so there can be no double bonds as suggested by kekule model, as double bonds would be shorter Thermodynamic factor: If there were alternating double bonds like the kekule model suggests, the enthalpy change of hydrogenation would be expected to be 3x as exothermic as that of cyclohexene (-360). However, actual enthalpy change of hydrogenation of benzene is less exothermic (-208) than that value, so it’s more thermodynamically stable and does not contain alternating hydrogen bonds. Chemical factor: benzene does not react with bromine without a catalyst, signifying it is less charge dense
Describe how the orbitals of carbon atoms overlap to form the pi bonds and sigma bonds in the delocalized model of the benzene ring
One electron from the p orbital of each carbon atom is delocalized to form the delocalized pi SYSTEM Pi bonds form from the sideWAYS overlap of p orbitals Sigma bond form from the head on overlap of orbitals: s+s s+p p+p
Describe the combustion reaction of benzene
C6H6 + 7.5O2 —> 6CO2 + 3H2O
Benzene typically goes through …… reactions
Electrophillic substitution NOT additional reactions as it is Kinetically stable ie: its electrons are delocalised
Describe the reaction between benzene and bromine
C6H6 + Br2 —> C6H5Br Forms bromobenzene Electrophillic substitution Requires a halogen carrier catalyst Heat under reflux 1) Generate the electrophile : Br2 + FeBr3 —> FeBr4- + Br+ 2) Mechanism: Curly arrow benzene circle to Br+ —> extend the lines from the carbon to have a H and Br. Note, the bromine circle is broken and is positive (+) —> form bromobenzene and H+ 3) Regenerate catalyst: H+ + FeBr4- —> HBr + FeBr3
Describe the reaction of benzene with concentrated nitric acid
C6H6 + HNO3 —> C6H5NO + H2O Forms nitrobenzene Need concentrated nitric acid Need concentrated sulphuric acid catalyst 50 degrees Celsius 1) Generate the electrophile : H2SO4 + HNO3 —> NO2+ + HSO4- + H2O 2) Mechanism: Curly arrow benzene circle to NO2+ —> extend the lines from the carbon to have a H and NO2. Note, the bromine circle is broken and is positive (+) —> form nitrobenzene and H+ 3) Regenerate catalyst: H+ + HSO4- —> H2SO4
Describe the freidal craft ALKYLATION reaction of benzene
1) Generate the electrophile : R-Cl + AlCl3 —> R+ + AlCl4- 2) Mechanism: Curly arrow benzene circle to R+ —> extend the lines from the carbon to have a H and R. Note, the bromine circle is broken and is positive (+) —> form methylbenzene and H+ 3) Regenerate catalyst: AlCl4- + H+ —> AlCl3 + HCl
Describe the freidal craft ACYLATION reaction of benzene
1) Generate the electrophile : CH3COCl + AlCl3 —> CH3CO+ + AlCl4- 2) Mechanism: Curly arrow benzene circle to CH3O+ —> extend the lines from the carbon to have a H and CH3CO+ Note, the bromine circle is broken and is positive (+) —> form aryl ketones and H+ 3) Regenerate catalyst: AlCl4- + H+ —> AlCl3 + HCl
What catalysts can be used to generate the electrophile + are they heterogenous or homogenous
AlCl3 or FeBr3 - they heterogenous catalysts as they are gasses and benzene is a liquid at room temperature
Why is phenol more reactive with bromine than a normal benzene ring
The lone pair of electrons on oxygen on the OH group can donate into the pi system which activates the ring and makes it more electron dense, making phenol more susceptible to an electrophilic attack as phenol acts as an electrophile and is is able to polarise Br2 molecule more. Unlike benzene, phenol reacts readily with bromine without a catalyst, producing a white precipitate of 2,4,6-tribromophenol.
Describe the bromination of phenol
Produces 2,4,6 tribromophenol (a white precipitate) Requires 3Br2 —> forms 2,4,6 tribromophenol + 3HBr Under room temperature, not heated under reflux like benzene
Explain why 2-hydroxybenzoic acid has a lower melting point than 4-hydroxybenzoic acid
In 2-hydroxybenzoic acid, the hydroxyl and carboxylic acid groups are adjacent, allowing for the formation of intramolecular hydrogen bonds (hydrogen bonding within the same molecule). This “locks” the functional groups together, reducing the molecule’s ability to form strong intermolecular hydrogen bonds (bonds between different molecules). In contrast, the spatial arrangement of 4-hydroxybenzoic acid prevents intramolecular bonding, allowing molecules to form extensive, strong intermolecular hydrogen bonds with one another instead. Because it requires more energy to break these intermolecular forces, 4-hydroxybenzoic acid has a higher melting point.
Explain why phenol is more acidic than alkyl alchohol - methanol, ethanol
Phenol is more acidic than alkyl alchohol as the lone pair on the oxygen of OH can donate into the pi system. This weakens the O-H bond and so the H+ is easier to remove. H+ forms oxonium ion (H3O+). The delocalised pi system therefore stabilises the conjugate base anion
Explain the bonding in benzene
Each carbon atom in the benzene ring forms three sigma bonds due to head on overlap of orbitals. Therefore, each cabron atom has 1 remaining electron in a p orbital. These 6 p orbitals overlap side on with their adjacent p orbitals to produce a delocalised pi system
Phenol + strong base Phenol + weak base
Phenol + strong base (NaOH) —> C6H5O-Na+ + H2O Phenol + weak base (Na2CO3) —> to weak of a weak acid to react so don’t produce bubbles ld CO2 gas like expected