Cconcentrated HNO3 and concentrated H2SO4, 10 °C[1]
Nitration of benzene requires the strong electrophile, NO2+, formed when nitric acid is protonated by the stronger sulfuric acid, followed by loss of water: HNO3 + 2H2SO4 → H2O + NO2+ + 2HSO4−
The nitration reaction is highly exothermic. Mononitration occurs below 55 °C, further nitration to dinitrobenzene can occur above 55 °C and trinitration occurs above 100 °C. Hence the reaction mixture must be cooled and maintained below 55 °C for mononitration.
CIt donates protons to the nitric acid, thus forming NO2+ ions.[1]
The function of sulfuric acid is to aid in the generation of the strong electrophile, NO2+, by first, protonation of nitric acid (sulfuric acid is a stronger Brønsted–Lowry acid).
CCH3Cl, AlCl3 → KMnO4, H2SO4 → Cl2, AlCl3[1]
Calkylation, nitration, free radical substitution[1]
Alkylation gives C6H5CH3. Nitration (–CH3 is 2,4-directing) gives 4-nitromethylbenzene. Free radical substitution then gives ClCH2C6H4NO2.
In B, –NO2 is 3-directing.
I reagent: KMnO4 with dilute H2SO4. Condition: heat under reflux. Type of reaction: side-chain oxidation[1]
II reagent: Br2, excess methylbenzene. Condition: uv light. Type of reaction: free radical (side-chain) substitution[1]
III reagent: Br2, anhydrous FeBr3 catalyst. Condition: room temperature. Type of reaction: electrophilic (aromatic) substitution[1]
The positional isomers are and [1]
4-bromomethylbenzene is more likely to be formed along with A.[1]
B reacts with Cl2 in the absence of uv light to give dense white fumes (of HCl) with a suitable catalyst ⇒ electrophilic substitution ⇒ presence of benzene ring (since it requires a catalyst to react with Cl2).
B can be oxidised by KMnO4 to form C ⇒ presence of alkyl side-chain on benzene.
C is exactly neutralised by one mole of NaOH ⇒ C only has 1 –CO2H group after side-chain oxidation ⇒ B only has 1 alkyl substituent.
B: [1]
C: [1]
C6H5CH2CH3 + 6[O] → C6H5CO2H + 2H2O + CO2[1]
The –NO2 group is a highly deactivating substituent. It reduces the electron density of the benzene ring such that it becomes less susceptible to an electrophilic attack by a rather unreactive electrophile like CH3CO+.[1]
The presence of a second, strongly activating group like –OCH3 increases the electron density of the benzene ring sufficiently for it to be susceptible to electrophilic attack by CH3CO+.[1]
Each of the 10 carbon atoms in naphthalene is sp2 hybridised, with one singly filled 2p orbital perpendicular to the plane of the molecule. A C–C σ bond is formed with neighbouring C atoms via sp2–sp2 overlap.[1]
All 10 p orbitals with one electron each overlap sideways, leading to a delocalised electron cloud of 10 π electrons covering all 10 C atoms. The bonds between the carbon atoms are in between a single C–C and a double C=C bond.[1]