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How are alcohols and carboxylic acids made and how do they react?

Classification and oxidation of alcohols, dehydration and esterification, the reactions of carboxylic acids, and the preparation and hydrolysis of esters.

A focused answer to WJEC A-Level Chemistry Unit 2, covering primary, secondary and tertiary alcohols and their oxidation, dehydration to alkenes, esterification, the acidity and reactions of carboxylic acids, and ester hydrolysis.

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What this dot point is asking

WJEC wants you to classify alcohols, oxidise them appropriately, dehydrate them to alkenes, carry out esterification, and describe the acidity and reactions of carboxylic acids and the formation and hydrolysis of esters.

The answer

Classifying and oxidising alcohols

The oxidant is acidified potassium dichromate(VI), which turns from orange to green. To isolate an aldehyde, distil it off; to reach the acid, reflux with excess oxidant.

Dehydration and esterification

Carboxylic acids

Carboxylic acids are weak acids that release H+\text{H}^+ from the COOH\text{COOH} group. They react with reactive metals, carbonates (effervescence of CO2\text{CO}_2, a test) and bases to form salts.

Esters and their hydrolysis

Esters form by esterification and can be hydrolysed back: acid hydrolysis (dilute acid, reversible) gives the acid and alcohol; base hydrolysis (saponification, irreversible) gives the carboxylate salt and alcohol.

Controlling the oxidation product

The key practical skill is controlling how far a primary alcohol is oxidised. Using a limited amount of oxidant and distilling the product off as it forms isolates the aldehyde, because removing it from the flask stops it being oxidised further. Refluxing with excess oxidant instead returns the vapour to the flask, so the aldehyde stays in contact with the oxidant and is oxidised all the way to the carboxylic acid. The same reagent therefore gives two different products depending purely on the apparatus and the amount of oxidant, a distinction examiners test directly.

Why esterification needs concentrated acid

Fischer esterification is a slow, reversible reaction, and concentrated sulfuric acid plays two roles. It catalyses the reaction by protonating the carbonyl, making it more open to attack, and it absorbs the water produced, shifting the equilibrium toward the ester by Le Chatelier. Using dilute acid would fail on both counts: it would not remove water and would dilute the mixture. This is why the condition must be stated as concentrated, not just sulfuric acid.

Examples in context

Esters as flavours and fragrances. Many fruity smells (ethyl ethanoate, pentyl ethanoate) are esters made by Fischer esterification, used in foods, perfumes and solvents. Biodiesel by transesterification. Vegetable-oil esters are converted to methyl esters (biodiesel), an industrial-scale ester reaction linked to the renewable-fuels theme of Unit 2's wider impact content.

Try this

Q1. State the colour change when a primary alcohol is oxidised by acidified potassium dichromate. [1 mark]

  • Cue. Orange to green.

Q2. Name the ester formed from methanol and propanoic acid. [1 mark]

  • Cue. Methyl propanoate.

Q3. Give the reagents and conditions to dehydrate ethanol to ethene. [1 mark]

  • Cue. Heat with concentrated sulfuric acid (or pass over hot aluminium oxide).

Q4. State how you would isolate an aldehyde rather than a carboxylic acid when oxidising a primary alcohol. [1 mark]

  • Cue. Distil the aldehyde off as it forms, using a limited amount of oxidant.

Q5. State the two roles of concentrated sulfuric acid in esterification. [2 marks]

  • Cue. It catalyses the reaction and removes water, shifting the equilibrium toward the ester.

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 20184 marksDescribe how you would oxidise propan-1-ol to propanoic acid, and explain how the conditions differ from those used to obtain propanal.
Show worked answer β†’

Reflux propan-1-ol with excess acidified potassium dichromate(VI). The orange dichromate is reduced to green chromium(III).

CH3CH2CH2OH+2[O]β†’CH3CH2COOH+H2O\text{CH}_3\text{CH}_2\text{CH}_2\text{OH} + 2[\text{O}] \rightarrow \text{CH}_3\text{CH}_2\text{COOH} + \text{H}_2\text{O}.

To stop at the aldehyde (propanal), distil the product off as it forms using a limited amount of oxidant, so it is removed before further oxidation.

To reach the acid you reflux with excess oxidant so the aldehyde is oxidised further.

Markers reward reflux with excess acidified dichromate, the colour change, the equation, and distillation to isolate the aldehyde.

WJEC 20214 marksCalculate the maximum mass of ethyl ethanoate that could be made from 23.0 g of ethanol reacting with excess ethanoic acid. The molar mass of ethanol is 46.0 g mol-1 and of ethyl ethanoate is 88.0 g mol-1.
Show worked answer β†’

The esterification is CH3COOH+C2H5OHβ‡ŒCH3COOC2H5+H2O\text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH} \rightleftharpoons \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O}, a 1:11:1 ratio.

Moles of ethanol =23.0/46.0=0.500= 23.0 / 46.0 = 0.500 mol.

Since ethanoic acid is in excess, ethanol is limiting, so maximum moles of ester =0.500= 0.500 mol.

Maximum mass of ester =0.500Γ—88.0=44.0= 0.500 \times 88.0 = 44.0 g.

Markers reward the limiting reagent, the 1:11:1 ratio, and the mass with units (this is the theoretical maximum, ignoring the equilibrium).

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