Part 3: Phenol, acidity and directing effects

Explain why phenol behaves differently from benzene and use substituents to plan where the next group will enter the ring.

The OH group changes both acid–base and ring chemistry

A phenol has OH directly attached to an aromatic ring. This differs from an aromatic molecule with a side-chain alcohol, such as C₆H₅CH₂OH. Phenol is a weak acid and reacts with NaOH to form phenoxide and water, but does not react with carbonate in the standard qualitative test.

Phenol is sufficiently acidic for neutralisation by hydroxide but much weaker than a carboxylic acid in this comparison. Do not use “weak” to mean no reaction with any base. The negative charge in phenoxide can be stabilised by interaction with the ring; this is a useful explanation, while the required test distinction is NaOH reaction without carbonate effervescence.

Electron donation makes the ring easier to attack

A lone pair on oxygen can interact with the aromatic π system, increasing ring electron density and its susceptibility to electrophiles. Explain donation from an oxygen p orbital; do not say the whole OH group leaves or that oxygen donates an electron as a redox step.

Phenol reacts readily with bromine water without the halogen carrier needed for benzene. The solution decolourises and a white precipitate of 2,4,6-tribromophenol forms. The OH group remains attached; three ring H atoms are replaced by Br, producing three HBr molecules.

Phenol reacts with dilute nitric acid to form a mixture of 2-nitrophenol and 4-nitrophenol. Concentrated sulfuric acid is not needed for this specified phenol reaction. The difference in conditions is evidence of the activated ring, not permission to apply benzene’s nitration conditions to every aromatic molecule.

Locate positions relative to the group already present

OH and NH₂ are electron-donating 2- and 4-directing groups in the required scope; the equivalent 6-position is also adjacent to the existing group. NO₂ is electron-withdrawing and 3-directing, with positions 3 and 5 equivalent in a monosubstituted ring.

Number the carbon carrying the existing group as 1. Positions 2 and 6 are adjacent, 3 and 5 are one further round, and 4 is opposite. When the starting molecule has several substituents, inspect whether their preferences reinforce or conflict, and use any supplied additional directing information. OCR may give data for groups beyond OH, NH₂ and NO₂.

Worked planning example: changing a group changes the next step

Suppose an original synthesis problem needs a new substituent in position 3 relative to a nitro group. Introducing NO₂ before that substitution supplies the required 3-directing influence. If the nitro group is first reduced to NH₂, the directing preference changes to positions 2 and 4, so the order of steps affects the product mixture.

This is a planning principle, not a guarantee that every imagined sequence works under every condition. Check that the reagent is compatible with all groups present and that the required aromatic reaction is within the supplied toolkit. In an answer, draw the intermediate structure and state why its existing group directs the next substitution.

For phenol bromination, 2,4,6 substitution and the precipitate distinguish it from simple benzene behaviour. For a comparison question, name the electron-density change in phenol and the relative difficulty in benzene; describing only one molecule leaves the comparison incomplete.

Quick checks

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

Sources