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How do alcohols and phenols differ in their reactions?

The reactions of alcohols, oxidation and dehydration, the acidity and characteristic reactions of phenols, and tests to distinguish them.

A focused answer to WJEC A-Level Chemistry Unit 4, covering the reactions of alcohols including oxidation and dehydration, the weak acidity and ring reactions of phenols, and chemical tests that distinguish alcohols from phenols.

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What this dot point is asking

WJEC wants you to describe the reactions of alcohols (oxidation and dehydration), explain the weak acidity and ring reactions of phenols, and use chemical tests to distinguish alcohols from phenols.

The answer

Reactions of alcohols

Phenols

Distinguishing tests

Why phenol differs from an aliphatic alcohol

In an ordinary alcohol the oxygen lone pairs stay localised on the oxygen. In phenol, one oxygen lone pair overlaps with the delocalised ring system, which has two consequences. First, the negative charge of the phenoxide ion (formed when phenol loses a proton) is delocalised into the ring, stabilising it, so phenol is acidic enough to react with sodium hydroxide. Second, the extra electron density pushed into the ring activates it toward electrophiles, so phenol brominates at room temperature without a catalyst, substituting at the 2, 4 and 6 positions. An aliphatic alcohol such as ethanol does neither: it is neutral and has no aromatic ring to activate.

Summary of distinguishing tests

Reagent Phenol Ethanol
Bromine water Decolourised, white precipitate No reaction
Sodium hydroxide Dissolves (acidic) No reaction
Neutral iron(III) chloride Violet colour No colour
Sodium metal Hydrogen gas Hydrogen gas

Note that sodium reacts with both because both have an Oβˆ’H\text{O}-\text{H} bond, so it cannot distinguish them; the first three tests can.

Examples in context

Antiseptics and resins. Phenol's reactivity underpins antiseptics and phenol-formaldehyde resins (Bakelite), while its weak acidity is exploited in separating it from neutral organics. Breathalysers and oxidation. The orange-to-green oxidation of ethanol by acidified dichromate, the same reaction tested here, was the chemistry behind early roadside breathalysers.

Try this

Q1. State the reagent that distinguishes phenol from ethanol and the observation for phenol. [1 mark]

  • Cue. Bromine water; phenol decolourises it and forms a white precipitate.

Q2. Name the organic product when a secondary alcohol is oxidised. [1 mark]

  • Cue. A ketone.

Q3. Explain why phenol is acidic. [1 mark]

  • Cue. The phenoxide ion is stabilised by delocalisation of the negative charge into the ring.

Q4. Explain why phenol reacts with bromine water without a catalyst but benzene does not. [2 marks]

  • Cue. The oxygen lone pair feeds electron density into the ring, activating it, so phenol reacts at room temperature; benzene needs a halogen carrier to generate a strong enough electrophile.

Q5. Name the organic product when a tertiary alcohol is warmed with acidified dichromate. [1 mark]

  • Cue. No reaction (tertiary alcohols are not oxidised; the solution stays orange).

Exam-style practice questions

Practice questions written in the style of WJEC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

WJEC 20204 marksDescribe how you could use chemical tests to distinguish between separate samples of ethanol and phenol. Give the reagent and the observation for each.
Show worked answer β†’

Add bromine water to each. Phenol decolourises bromine water at room temperature and forms a white precipitate of 2,4,6-tribromophenol; ethanol does not react.

Alternatively, add sodium hydroxide solution. Phenol is weakly acidic and dissolves; ethanol does not.

Or add neutral iron(III) chloride: phenol gives a violet colour, ethanol gives none.

Markers reward at least one valid reagent with the correct observation for both substances, clearly distinguishing them.

WJEC 20184 marksCalculate the mass of propanal that could be obtained by the controlled oxidation of 12.0 g of propan-1-ol, assuming a 70.0 percent yield. Molar masses: propan-1-ol 60.0, propanal 58.0 g mol-1.
Show worked answer β†’

Moles of propan-1-ol =12.0/60.0=0.200= 12.0 / 60.0 = 0.200 mol.

The oxidation CH3CH2CH2OH+[O]β†’CH3CH2CHO+H2O\text{CH}_3\text{CH}_2\text{CH}_2\text{OH} + [\text{O}] \rightarrow \text{CH}_3\text{CH}_2\text{CHO} + \text{H}_2\text{O} is 1:11:1, so theoretical moles of propanal =0.200= 0.200 mol.

Theoretical mass =0.200Γ—58.0=11.6= 0.200 \times 58.0 = 11.6 g.

At 70.0 percent yield, actual mass =0.700Γ—11.6=8.12= 0.700 \times 11.6 = 8.12 g.

Markers reward the moles of alcohol, the 1:11:1 ratio, the theoretical mass, and applying the yield (distillation is needed to stop at the aldehyde).

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