Topic 17.4 Benzene

Ring compounds in which there are delocalised electrons (based on benzene)

The bonding in a compound which has delocalised electrons forming pi bonding in a hydrocarbon ring

Molecules have chains atoms and with rings that are not aromatic (openchain)

-Hydrogenation Heat under pressure with a nickel catalyst C6H6 +3H2 –> C6H12

-Benzene burns with a smokey flame C6H6 + 7.5O2 –> 6CO2 + 3H2O

Heated under reflux with a catalyst (a halogen carrier)

Reagents- concentrated nitric acid + sulfuric acid Conditions- warm (not about 50°C)

Organic product is methylbenzene (touline) Adding alkyl group

-Delocalised electrons in ring act as nucleophile -Attack acyl chlorides -Reactive intermediate needed -AlCl catalyst: R-COCl + AlCl3 –> R-CO^+ + AlCl4^- (Anhydrous conditions (dry) as water would react with the catalyst)

An electron pair acceptor – an atom or group of atoms that is attracted to a region of high electron density and can accept a pair of electrons.

An electron-rich species that can donate a pair of electrons

A reaction in which a functional group in a compound is replaced by (or is substituted by) another.

Chemical reaction wherein two or more reactants come together to form a larger single product.

Benzene isn’t very reactive, so if Kekulé’s model were to be correct, there would be 3 double bonds in benzene, and so it would react with bromine and decolourise it –> Benzene doesn’t decolourise bromine

The bond lengths in benzene are all the same length. If Kekulé’s model was correct, there would be 2 different bond lengths, corresponding to the double bonds and the single bonds in the ring.

The hydrogenation enthalpy of benzene is expected to be three times that of cyclohexene, if it were Kekulé’s model. However, it is smaller, and less energy is produced than expected, meaning that the actual model of benzene is a lot more stable than Kekulé’s model.

Should be 4 isomers however only 3 are known to exist

Step 1: Formation of the electrophile Bromine reacts with the catalyst to form Br+: AlCl3 + Br2 –> Br+ + AlCl3Br- Step 2: Electrophilic attack Step 3: Formation of the aromatic product Step 4: Formation of the inorganic products AlCl3Br- + H+ –> AlCl3 + HBr

-NO2 + ion = electrophile Step 1: formation of electrophile (conc.) nitric acid and (conc.) sulphuric acid to form NO2^+: HNO3 + H2SO4 –> NO2^+ + HSO4^- + H2O Step 2: Electrophilic attack Step 3: Formation of aromatic product Step 4: Formation of inorganic products HSO4^- + H^+ –> H2SO4

-Alkalyation means the substitution of one of the hydrogen atoms of benzene by an alkyl group. -The reagent is a halogenoalkane, and the products are an alklybenzene (methylbenzene) and hydrogen chloride. -The equation for the reaction with chloromethane:

Because that would involve breaking the delocalisation and losing that stability.

In the absence of a catalyst (cold conditions) reagents: Aqueous bromine (?)

The increased reactivity compared to benzene is due to the lone pair of p-orbital electrons on the oxygen atom in the phenol group.