Skip to main content
ScotlandChemistrySyllabus dot point

Why does food go off, and how do antioxidants protect it?

Oxidation as the loss of electrons, the oxidation of alcohols and aldehydes, the rancidity of edible oils caused by the oxidation of carbon-to-carbon double bonds, and the action of antioxidants in preventing oxidation.

An SQA Higher Chemistry answer on the oxidation of food, covering oxidation as electron loss, the oxidation of alcohols to aldehydes and carboxylic acids, the rancidity caused by oxidation of double bonds in edible oils, and how antioxidants protect food by being oxidised themselves.

Generated by Claude Opus 4.810 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

Jump to a section
  1. What this key area is asking
  2. Oxidation in terms of electrons
  3. Oxidation of alcohols and aldehydes
  4. Rancidity of oils
  5. Antioxidants
  6. Worked example: moles and the oxidation sequence
  7. Examples in context
  8. Try this

What this key area is asking

The SQA wants you to define oxidation in terms of electrons, describe the oxidation of alcohols and aldehydes, explain why edible oils turn rancid, and explain how antioxidants prevent oxidation. Identifying oxidation from an ion-electron equation and explaining the antioxidant mechanism are dependable exam earners.

Oxidation in terms of electrons

Oxidation of alcohols and aldehydes

Functional groups can be oxidised in a clear sequence:

  • A primary alcohol is oxidised to an aldehyde, then further to a carboxylic acid.
  • A secondary alcohol is oxidised to a ketone.
  • An aldehyde is oxidised to a carboxylic acid.

Tertiary alcohols and ketones resist this kind of oxidation, because their relevant carbon has no hydrogen that can be removed. The usual oxidising agent in the lab is acidified dichromate, Cr2O72Cr_2O_7^{2-}, which changes from orange to green as the chromium is reduced from Cr6+Cr^{6+} to Cr3+Cr^{3+} while the organic molecule is oxidised. In each oxidation step the molecule either gains an oxygen atom or loses two hydrogen atoms, raising its oxygen-to-hydrogen ratio.

Rancidity of oils

Antioxidants

Antioxidants are molecules that prevent the oxidation of food. They work by being oxidised in preference to the food molecules, so the antioxidant is sacrificed and the food is protected.

You can show this with an ion-electron half-equation: the antioxidant is oxidised, for example C6H8O6C6H6O6+2H++2eC_6H_8O_6 \rightarrow C_6H_6O_6 + 2H^+ + 2e^-, losing electrons so that oxygen reacts with the antioxidant rather than the oil. Because the electrons appear on the right (product) side, the equation confirms this is an oxidation. The antioxidant is sacrificed first and the fat is spared, which is why a small amount of antioxidant can protect a much larger quantity of food.

Worked example: moles and the oxidation sequence

Examples in context

The oxidation of food is why a bottle of cooking oil left open slowly develops a stale, "off" smell, and why crisps and biscuits are sealed in nitrogen-flushed packets to keep oxygen away from their fats. Antioxidants such as vitamin C (ascorbic acid) and vitamin E are added to processed foods, often listed as E-numbers, precisely because they are oxidised in preference to the fats and so extend shelf life. The same chemistry runs in the body: dietary antioxidants are thought to mop up the reactive oxygen species generated by metabolism. The orange-to-green dichromate test used in the lab to follow alcohol oxidation is the basis of the original breathalyser, which detected ethanol in breath by the colour change as the alcohol was oxidised.

Try this

Q1. Name the product when a primary alcohol is oxidised twice. [1 mark]

  • Cue. A carboxylic acid (via an aldehyde).

Q2. Explain how an antioxidant protects food from oxidation. [2 marks]

  • Cue. It is oxidised in preference to the food, losing electrons, so the food molecules are not oxidised.

Q3. Calculate the number of moles in 7.40 g7.40 \text{ g} of butan-1-ol (GFM=74.0 gGFM = 74.0 \text{ g}). [1 mark]

  • Cue. n=7.40/74.0=0.100 moln = 7.40 / 74.0 = 0.100 \text{ mol}.

Exam-style practice questions

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

SQA Higher 20183 marksVitamin C (C6H8O6C_6H_8O_6) is an antioxidant. Its action can be shown by the ion-electron equation C6H8O6C6H6O6+2H++2eC_6H_8O_6 \rightarrow C_6H_6O_6 + 2H^+ + 2e^-. (a) State whether this represents oxidation or reduction, with a reason. (b) Explain how an antioxidant protects food.
Show worked answer →

Markers reward the correct identification with a reason and a clear explanation of the protective mechanism.

(a) This represents oxidation, because electrons are lost (the 2e2e^- appear on the right-hand, product side of the equation).

(b) An antioxidant protects food by being oxidised in preference to the food molecules. The antioxidant loses electrons (is oxidised) so that oxygen or other oxidising species react with it rather than with the food, so the food is not oxidised and does not spoil.

A common loss is saying reduction (gain of electrons) or claiming the antioxidant blocks oxygen rather than being oxidised itself.

SQA Higher 20213 marksEthanol can be oxidised in two stages. (a) Name the product of the first oxidation and the product of the second oxidation. (b) Calculate the number of moles in 9.20 g9.20 \text{ g} of ethanol (GFM=46.0 gGFM = 46.0 \text{ g}). (c) State the type of compound formed when a secondary alcohol is oxidised.
Show worked answer →

Parts (a) and (c) test the oxidation sequence; part (b) is a mole calculation.

(a) The first oxidation of ethanol (a primary alcohol) gives the aldehyde ethanal; the second oxidation gives the carboxylic acid ethanoic acid.

(b) Using n=m/GFMn = m / GFM:

n=9.2046.0=0.200 moln = \frac{9.20}{46.0} = 0.200 \text{ mol}

(c) Oxidising a secondary alcohol gives a ketone.

Markers reward the aldehyde-then-acid sequence and the correct mole value with a unit.

Related dot points

Sources & how we know this