Part 2: Nitration, Friedel–Crafts reactions and phenol

Reuse the electrophilic-substitution pattern with correctly generated electrophiles, then explain why phenol brominates under milder conditions.

Three stages organise every aromatic substitution

Before drawing any aromatic mechanism, identify the electrophile and show how the reagents produce it. A full-headed curly arrow then moves a ring π pair towards the electrophile, forming a C–E bond and a positively charged intermediate. Finally loss of H⁺ with the C–H bond pair returned to the ring restores delocalisation.

At the carbon attacked, show both E and H in the intermediate. In the product, H has been replaced by E. The second curly arrow starts at the C–H bond, not at the H⁺ symbol. Catalyst-regeneration equations explain why the catalyst is absent from the overall equation.

Nitronium is generated by the acid mixture

Warm benzene with concentrated nitric and sulfuric acids under controlled conditions, commonly around 50–60°C for mononitration. Sulfuric acid enables formation of the nitronium ion, NO₂⁺, from nitric acid. A convenient balanced generation equation is HNO₃ + H₂SO₄ → NO₂⁺ + HSO₄⁻ + H₂O.

The ring attacks the nitrogen of NO₂⁺, not one of its oxygens. After the intermediate forms, HSO₄⁻ accepts the proton lost from the ring, regenerating H₂SO₄. The product is nitrobenzene, C₆H₅NO₂, and overall the reaction forms water.

Temperature control limits further substitution and moderates an exothermic reaction. More vigorous conditions may introduce additional nitro groups, so do not write a universal condition independent of the substrate. Nitration is a route to nitroarenes, which can later be reduced to aromatic amines.

Add a carbon group through an alkyl electrophile

An alkyl halide with anhydrous AlCl₃ can alkylate benzene. In the simple A-Level model, RCl + AlCl₃ → R⁺ + AlCl₄⁻ shows generation of the alkyl electrophile. The ring bonds to the carbon carrying the electrophilic character; loss of H⁺ then restores aromaticity.

For chloroethane, the added group is CH₂CH₃, so the product is ethylbenzene. Anhydrous conditions preserve the Lewis-acid catalyst, which otherwise reacts with water. Suitable warming or reflux may be used according to the supplied procedure.

Friedel–Crafts alkylation may give further substitution because alkyl groups activate the ring; some substrates can also rearrange. These limitations explain why a clean single product should not be assumed for every unfamiliar halogenoalkane. The required mechanism is the simple electrophilic-substitution model, with complete structures checked.

Keep C=O in the newly attached group

An acyl chloride with anhydrous AlCl₃ produces an acylium electrophile, RCO⁺. The ring attacks its carbonyl carbon. Acylation therefore gives an aryl ketone, C₆H₅COR, whereas alkylation gives C₆H₅R. The carbonyl is retained throughout the substitution.

Ethanoyl chloride, CH₃COCl, gives C₆H₅COCH₃, phenylethanone. The acyl electrophile can be represented CH₃–C⁺=O with the charge on the electrophilic carbon, or an appropriate resonance form. A free CH₃⁺ would give methylbenzene and is the wrong electrophile.

This route introduces a carbonyl that can be reduced to a secondary alcohol or undergo cyanide addition. Write the intermediate with the complete acyl group bonded through carbon, then draw C–H bond electrons returning to the ring. The AlCl₃ is regenerated in the simplified catalytic account; laboratory work-up also releases product from aluminium-containing complexes.

A ring-bound OH group changes ring reactivity

Phenol has OH directly attached to an aromatic carbon. C₆H₅CH₂OH is a benzyl alcohol, not a phenol, because CH₂ separates OH from the ring. An oxygen lone pair in phenol can overlap with the ring π system, increasing electron density and activating the ring towards electrophiles.

Phenol decolourises bromine water at room temperature without a halogen-carrier catalyst and forms a white precipitate of 2,4,6-tribromophenol. Donation of electron density helps polarise Br₂ and lowers the barrier to substitution compared with benzene. This is substitution, despite the loss of bromine colour.

The three Br atoms occupy the two positions next to OH and the position opposite it. Start with carbon 1 attached to OH and place Br at 2, 4 and 6; there remain two ring hydrogens plus the OH hydrogen. The equation consumes three Br₂ and forms three HBr.

Use the observation together with the candidate structures

Bromine-water decolourisation is not unique to alkenes. A phenol can also remove the colour, but the characteristic white tribromophenol precipitate and other structural evidence distinguish the case. Within a simple comparison, benzene shows no rapid decolourisation without its catalyst.

As a useful acid–base connection, phenol is a weak acid and forms phenoxide with NaOH. It is generally too weak to release CO₂ readily from carbonate under the standard qualitative comparison with carboxylic acids. These connections help plan CP15 tests; do not use smell or a single colour change as a complete identification.

Quick checks

Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official Edexcel mark allocations.

Sources