Unit 5- Topic 18

Every aromatic compound has benzene - every aliphatic compound is those who do are not aromatic Benzene is an arene Is a colourless liquid Boiling temperature of 80 It is insoluble in water Toxicity - carcinogen Molecular formula: C6H6 it is highly unsaturated

-the three C=C should mean that benzene decolourises bromine water in an addition reaction, however it doesn’t. a substitution reaction occurs this suggests that there are no C=C present. -in dibromobenzene, four isomers should exist in Kekule structures but only three were known to exist. This suggests that the bonds between the carbon atoms in the benzene ring are the same, not different -X-ray diffraction show a value of C-C in benzene as an intermeadiate between the C-C and C=C in cyclohexene. this suggests all the carbon carbon bonds are all the same -thermochemical data about enthalpy changes show benzene has a lower enthalpy change than expected -infra-red spectra shows absoptions which are typical of an aromatic C=C stretch

the enthalpy change of cyclohexene (-120) and the enthalpy change of cyclohexa-1, 4-diene (-239) indicate that the enthalpy change for adding 1 mol of H2 to 1 mol of C=C bond is -120. -Kekule’s structures would be the treble of cyclohexene. so -120 x 3, -360 -the actual benzene value is much lower.

after the sigma bonds are formed, each carbon atom has one electron in a p-orbital, there is a formation of one large pi-bond made up of all six electrons. these six electrons form a delocalised pi-bond. this means that: there are no individual C=C bonds, so there is no addition reaction with bromine. There are three isomers because there is no difference in the arrangement of electrons when the bromine atoms are on adjacent carbon atoms. When charge is spread around in a species, there is more stability.

because substitution preserves the stability of the delocalised electrons in the pi bond

benzene + hydrogen -> cyclohexane (C6H12) conditions: heating under pressure with nickel catalyst

benzene + oxygen -> carbon dioxide + water observation: smoky flame

benzene + bromine -> bromobenzene + hydrogen bromide conditions: heated under reflux with halogen carrier catalyst (AlCl3, AlBr3, FeBr3 (iron (III) fillings are suitable since it reacts to form iron (III) bromide) reaction: substitution

benzene + concentrated nitric acid -> nitrobenzene + water conditions: warming with concentrated sulfuric acid as a catalyst. Carried out between 50º and 60ºC. At lower temps the reaction is too slow and at higher it can form dinitrobenzene and trinitrobenzene reaction: substitution

-using a reagent represented by XY, one of the hydrogen atoms in benzene is substituted by Y, other product is HX -a catalyst is needed- aluminium chloride, iron(III) chloride and iron(III) bromide can also work -anhydrous conditions are needed because water would react with the catalyst and sometimes also with the organic product

reaction: substitution of one hydrogen atoms of benzene by an alkyl group benzene + halogenoalkane -> alkylbenzene (methylbenzene) + hydrogen chloride conditions: anhydrous conditions, aluminium chloride catalyst

reaction: substitution of one of the hydrogen atoms in benzene by an acyl group` benzene + acyl chloride -> ketone (phenylketone) + hydrogen chloride conditions: anhydrous conditions, aluminium chloride catalyst

reaction: replacement of a hydrogen atom by an -SO3H group benzene + sulfuric acid -> benzenesulfonic acid + water conditions: warmed with fuming sulfuric acid at 40ºC for 20 to 30 minutes if the question asks about sulfuric acid use H2SO4 if it asks about SO3 as the electrophile just substitute SO3 in the sulfuric acid REACTANT

made by dissolving sulfur trioxide in concentrated sulfuric acid

-the benzene ring is electron rich due to the delocalised electrons in the pi bond allows it to attract electrophiles. -using Y+ to represent electrophile. as Y+ approaches the delocalised pi bond, it attracts two of the six electrons, formig a covalent bond so giving an intermeadiate species with a positive charge- one carbon is joined to both H and Y -this intermeadiate lacks stability , so in the next step the H leaves as H+ and the electrons join the four to restore the delocalised pi-bopnd

Step 1: formation of electrophile= AlBr3 + Br2 -> Br+ + AlBr4 - step 2: electrophilic attack step 3:formation of aromatic product step 4: regeneration of the catalyst= AlBr4 - + H+ -> AlBr3 + HBr

Step 1: formation of electrophile (nitric acid acts as a base) = HNO3 + H2SO4 -> NO2+ + HSO4 - + H2O step 2: electrophilic attack step 3:formation of aromatic product step 4: regeneration of the catalyst= HSO4 - + H+ -> H2SO4

Step 1: formation of electrophile= AlCl3 + CH3Cl -> CH3+ + AlCl4 - step 2: electrophilic attack step 3:formation of aromatic product step 4: regeneration of the catalyst= AlCl4 - + H+ -> AlCl3 + HCl

Step 1: formation of electrophile= AlCl3 + CH3COCl -> CH3CO+ + AlCl4 - step 2: electrophilic attack step 3:formation of aromatic product step 4: regeneration of the catalyst= AlCl4 - + H+ -> AlCl3 + HCl

a hydroxyl group joined to the benzene ring. You can expect it to undergo electrophilic substitutions reaction due to the benzene ring, it has the physical properties of alcohols

phenol + bromine water -> 2,4,6-tribromophenol + hydrogen bromide conditions: at room temperature without a catalyst observations: bromine water decolourises and organic white precipitate is formed. reaction: substitution reaction occurs three times as there are three bromine atoms in the organic product

-the lone pair of electrons in the pz-orbital on the oxygen atom of the OH group interacts with the delocalised pi electrons in the benzene ring in phenol. -the electron density above and below the ring of atoms is increased making the molecule much more reactive towards electrophiles -bromine molecules are polarised as they approach the benzene ring in phenol. eventually the Br-Br bond breaks and the Br+ electrophile attacks the benzene ring in phenol