Preparing phenol and substituting its ring

The antiseptic in the cupboard

Phenol was the first antiseptic used in surgery. It is a colourless solid that melts near $41\ ^\circ\text{C}$ and turns pink in air, because a little of it oxidises. A phenol is a compound with OH on a benzene ring. The methylphenols are the cresols. The benzene rings with two OH groups are catechol (1,2), resorcinol (1,3) and hydroquinone (1,4).

Making phenol

Four methods are in the syllabus.

1. Cumene process. Benzene and propene give cumene (isopropylbenzene). Air oxidises cumene to cumene hydroperoxide. Dilute acid splits that into phenol and propanone. Both products are useful, which makes the method economical.
2. From a haloarene. Chlorobenzene is heated with aqueous sodium hydroxide at about $623\ \text{K}$ under $300\ \text{atm}$. Sodium phenoxide forms, and acid frees the phenol.
3. From benzenesulphonic acid. The acid is fused with sodium hydroxide, which gives sodium phenoxide. Acid then frees the phenol.
4. From a diazonium salt. Benzenediazonium chloride warmed with water gives phenol and nitrogen: $\text{C}_{6}\text{H}_{5}\text{N}_{2}\text{Cl} + \text{H}_{2}\text{O} \to \text{C}_{6}\text{H}_{5}\text{OH} + \text{N}_{2} + \text{HCl}$.

Why the ring is so reactive

The oxygen has lone pairs. One of them overlaps with the ring and pushes electron density into it, especially onto the ortho and para carbons. The ring is therefore much more reactive to electrophiles than benzene. OH is activating and ortho, para directing. In the phenoxide ion the effect is even stronger.

Substitution on the ring

Bromination. Bromine water needs no catalyst. It gives a white precipitate of 2,4,6-tribromophenol at once, with three substitutions: $\text{C}_{6}\text{H}_{5}\text{OH} + 3 \text{Br}_{2} \to \text{C}_{6}\text{H}_{2}\text{Br}_{3}\text{OH} + 3 \text{HBr}$. This is a test for phenol. In a solvent of low polarity such as carbon disulphide at low temperature, only one bromine goes in. The product is a mixture of o-bromophenol and p-bromophenol, with the para isomer in larger amount.

Nitration. Dilute nitric acid at low temperature, about $298\ \text{K}$, gives a mixture of o-nitrophenol and p-nitrophenol. They can be separated by steam distillation. The ortho isomer has a hydrogen bond inside the molecule, so it does not bond to its neighbours, and it carries over with the steam. The para isomer makes hydrogen bonds between molecules and stays behind. Concentrated nitric acid oxidises phenol and gives a poor yield of 2,4,6-trinitrophenol (picric acid). Picric acid is made better by sulphonating first and then nitrating.

Two reactions that put a group on the ortho carbon

Kolbe reaction. Sodium phenoxide reacts with carbon dioxide at about $400\ \text{K}$ and $4$ to $7\ \text{atm}$. Acid then gives salicylic acid (2-hydroxybenzoic acid). Carbon dioxide is a weak electrophile, so the very reactive phenoxide ion is needed. Salicylic acid and ethanoic anhydride give aspirin.

Reimer-Tiemann reaction. Phenol is heated with chloroform and aqueous sodium hydroxide at about $340\ \text{K}$. Acid then gives salicylaldehyde (2-hydroxybenzaldehyde). The electrophile is dichlorocarbene, $:\text{CCl}_{2}$, formed from chloroform and the base. With carbon tetrachloride in place of chloroform, the product is salicylic acid.

Other reactions

• Reduction. Distilled with zinc dust, phenol loses its OH and gives benzene.
• Oxidation. Chromic acid oxidises phenol to p-benzoquinone, a yellow ring with two C=O groups.
• Ferric chloride test. Neutral ferric chloride gives a violet colour with phenol. Alcohols give no colour.

Phenol (3 g, $0.032\ \text{mol}$) needs at least $15\ \text{g}$ of bromine. This is three times the moles, and about $9.5\ \text{g}$ of product forms at $90\ \%$.

Practise with new numbers

Try it in Python

The program finds which reagent runs out first when phenol meets a limited amount of bromine, and the mass of white solid formed.

Common mistakes

• Using a Lewis acid catalyst with bromine water. The ring is so reactive that none is needed, and the product has three bromine atoms.
• Expecting only one nitro group from concentrated nitric acid. It oxidises phenol and gives picric acid in low yield.
• Writing the Kolbe product as the para acid. The main product is the ortho acid, salicylic acid.
• Forgetting the second step. The Kolbe and the Reimer-Tiemann reactions end with an acid.
• Mixing up the product of the cumene process. It gives phenol and propanone, not benzene.

Remember these

• Phenol: cumene process, haloarene with NaOH at 623 K and 300 atm, sulphonic acid fusion, diazonium salt with water.
• The ring is activated: bromine water gives 2,4,6-tribromophenol at once.
• Dilute nitric acid gives o- and p-nitrophenol. Steam distillation takes the ortho isomer.
• Kolbe: phenoxide with $\text{CO}_{2}$ gives salicylic acid. Reimer-Tiemann: $\text{CHCl}_{3}$ with NaOH gives salicylaldehyde.
• Violet colour with neutral ferric chloride.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. What are the two products of the cumene process?

2. Why can o-nitrophenol be separated from p-nitrophenol by steam distillation?

3. Write the reagents for turning phenol into salicylaldehyde.

4. Why does phenol react with bromine water without a catalyst, while benzene does not?

References

• NCERT Chemistry Part II (Class 12), Chapter 7, the sections on phenol.
• Morrison and Boyd, Organic Chemistry, the chapter on phenols.
• OpenStax Chemistry 2e, Chapter 20, section 20.2 (the structure of phenols).