Describe the reactions of alcohols including oxidation, esterification, dehydration and the tri-iodomethane test, classify primary, secondary and tertiary alcohols, and explain the greater acidity of phenol and its ease of ring substitution

What this dot point is asking

SEAB wants you to classify alcohols (primary, secondary, tertiary), describe their oxidation, esterification, dehydration and the tri-iodomethane test, and explain why phenol is both a stronger acid than an aliphatic alcohol and more reactive than benzene toward the ring. The oxidation outcomes and the phenol comparisons are recurring exam questions.

The answer

Classifying alcohols

• Primary ($1^\circ$): the OH carbon is bonded to one other carbon (or none), e.g. ethanol.
• Secondary ($2^\circ$): the OH carbon is bonded to two other carbons, e.g. propan-2-ol.
• Tertiary ($3^\circ$): the OH carbon is bonded to three other carbons, e.g. 2-methylpropan-2-ol.

Oxidation of alcohols

Using acidified potassium dichromate(VI) (orange to green, Cr$^{+6}$ to Cr$^{+3}$):

• Primary alcohol: to an aldehyde (distil off immediately), then to a carboxylic acid (reflux with excess oxidant).
• Secondary alcohol: to a ketone (no further oxidation).
• Tertiary alcohol: not oxidised (the OH carbon has no hydrogen to remove; oxidation would need to break a C-C bond, which does not happen). The dichromate stays orange, a useful distinguishing test.

Esterification

Alcohols react with carboxylic acids (concentrated sulfuric acid catalyst, reflux) to form esters and water:

Dehydration

Heating an alcohol with concentrated sulfuric or phosphoric acid (or passing over hot $\text{Al}_2\text{O}_3$) removes water to give an alkene: $\text{C}_2\text{H}_5\text{OH} \rightarrow \text{CH}_2\text{=CH}_2 + \text{H}_2\text{O}$.

The tri-iodomethane (iodoform) test

Warming with iodine and sodium hydroxide gives a pale yellow precipitate of tri-iodomethane ($\text{CHI}_3$) for compounds containing the $\text{CH}_3\text{CH(OH)}-$ group (or the $\text{CH}_3\text{CO}-$ group). So ethanol and propan-2-ol give a positive test, but methanol and propan-1-ol do not. This pinpoints a methyl-carbinol structure.

Phenol: acidity

Phenol is a stronger acid than ethanol. When phenol loses $\text{H}^+$, the resulting phenoxide ion is stabilised because the negative charge on oxygen is delocalised into the benzene ring. The ethoxide ion from ethanol has no such delocalisation, so its charge is concentrated. The greater stability of the phenoxide ion means phenol gives up $\text{H}^+$ more readily.

Phenol: ring reactivity

A lone pair on the phenol oxygen is partly delocalised into the ring, raising its electron density. This makes the ring more attractive to electrophiles, so phenol reacts with bromine water at room temperature without a catalyst, giving a white precipitate of 2,4,6-tribromophenol, unlike benzene which needs a halogen carrier.

Examples in context

Example 1. Making esters for flavours and fragrances. Refluxing an alcohol with a carboxylic acid and a little concentrated sulfuric acid produces a sweet-smelling ester, the basis of artificial fruit flavourings. SEAB pairs the esterification equation with the idea of a reversible reaction, asking how the position of equilibrium can be shifted to improve the ester yield.

Example 2. Distinguishing functional groups in an unknown. Faced with an unknown organic liquid, candidates use the dichromate oxidation and the tri-iodomethane test together to narrow down whether it is a primary, secondary or tertiary alcohol (or a methyl ketone). This combination of tests is exactly the reasoning rewarded in qualitative-analysis questions.

Try this

Q1. State the product and conditions when propan-1-ol is oxidised to a carboxylic acid. [2 marks]

• Cue. Propanoic acid; reflux with excess acidified potassium dichromate(VI).

Q2. Explain why phenol reacts with bromine water without a catalyst but benzene does not. [2 marks]

• Cue. An oxygen lone pair delocalises into the ring, raising its electron density and making it more attractive to the electrophilic bromine.

Q3. Name the organic product of dehydrating butan-2-ol, and the conditions. [2 marks]

• Cue. But-2-ene (major) and but-1-ene; concentrated sulfuric or phosphoric acid, heat (or hot $\text{Al}_2\text{O}_3$).