Skip to main content
ScotlandChemistrySyllabus dot point

What chemistry gives food its flavour and aroma?

Flavour molecules, the effect of cooking on flavour, the difference between aldehydes and ketones, oxidation of aldehydes, and the volatility and solubility of flavour and aroma compounds.

An SQA Higher Chemistry answer on the chemistry of cooking, covering flavour molecules, how cooking changes flavour, the difference between aldehydes and ketones and their carbonyl group, the oxidation of aldehydes to carboxylic acids, and the volatility and solubility of aroma compounds.

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. Flavour and aroma molecules
  3. Aldehydes and ketones
  4. Oxidation of aldehydes
  5. Worked example: moles of a flavour aldehyde
  6. Volatility and solubility
  7. Examples in context
  8. Try this

What this key area is asking

The SQA wants you to describe flavour molecules, explain how cooking can change flavour, distinguish aldehydes from ketones by their structure, describe the oxidation of aldehydes, and link the volatility and solubility of molecules to how we perceive flavour and aroma. The aldehyde-versus-ketone test is a near-certain exam item, so the structural reason and the test results must be precise.

Flavour and aroma molecules

The taste and smell of food come from small organic molecules. We detect flavour on the tongue and aroma through volatile molecules reaching the nose. Cooking changes flavour because heat can break down, oxidise or release flavour molecules, and can also create new ones (for example, the browning reactions that give roasted food its taste).

Aldehydes and ketones

Oxidation of aldehydes

The reason is structural: the aldehyde's carbonyl carbon carries a hydrogen that can be replaced by an OH-OH to form the acid, whereas the ketone's carbonyl carbon has no such hydrogen.

Worked example: moles of a flavour aldehyde

Volatility and solubility

A molecule's behaviour depends on its size and the forces between molecules:

  • Volatility: smaller molecules and those with weaker intermolecular forces vaporise more easily, so they are detected as aroma.
  • Solubility: molecules that can hydrogen bond dissolve in water; non-polar molecules dissolve in oil. This is why some flavours wash out in water while others stay in the fat.

Examples in context

The chemistry of cooking explains why a steak smells so good. Heating drives the volatile aldehydes and other small carbonyl molecules off the surface so they reach the nose, while browning reactions between sugars and amino acids generate hundreds of new aroma compounds. Vanilla flavour comes from vanillin, an aldehyde that gives a positive Fehling's test, which is why food chemists can confirm its presence chemically. The contrast between water-soluble and fat-soluble flavours is why a cook adds fat to a sauce: many flavour molecules are non-polar and dissolve in the oil, carrying taste through the dish, whereas water-soluble flavours such as those in a stock are extracted into the cooking water.

Try this

Q1. State the test that distinguishes an aldehyde from a ketone. [2 marks]

  • Cue. A mild oxidising agent (Fehling's blue to brick-red, Tollens' silver mirror, or acidified dichromate orange to green) reacts with the aldehyde but not the ketone.

Q2. Why must an aroma molecule be volatile? [1 mark]

  • Cue. It must vaporise easily so that it can travel to and be detected by the nose.

Q3. Calculate the number of moles in 5.80 g5.80 \text{ g} of the aldehyde ethanal (GFM=44.0 gGFM = 44.0 \text{ g}). [1 mark]

  • Cue. n=5.80/44.0=0.132 moln = 5.80 / 44.0 = 0.132 \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 20193 marksBoth propanal and propanone have the molecular formula C3H6OC_3H_6O and contain a carbonyl group. (a) Name the homologous series each belongs to. (b) Describe a chemical test, including the result, that would distinguish between them.
Show worked answer →

Markers reward correctly classifying each compound and a valid test with the distinguishing results.

(a) Propanal is an aldehyde (carbonyl at the end of the chain, bonded to a hydrogen). Propanone is a ketone (carbonyl within the chain, bonded to two carbons).

(b) Use a mild oxidising agent such as Fehling's solution. Warm each compound with the reagent: the aldehyde (propanal) is oxidised to a carboxylic acid and gives a positive result (the blue Fehling's solution turns to a brick-red precipitate of copper(I) oxide). The ketone (propanone) is not oxidised, so the solution stays blue.

Acidified dichromate (orange to green) or Tollens' reagent (silver mirror) are equally acceptable, provided the result for each compound is stated.

SQA Higher 20223 marksA flavour molecule, the aldehyde butanal (C4H8OC_4H_8O, GFM=72.0 gGFM = 72.0 \text{ g}), is oxidised to butanoic acid. (a) Calculate the number of moles in 3.60 g3.60 \text{ g} of butanal. (b) State why butanal can be oxidised but a ketone of the same formula cannot.
Show worked answer →

Part (a) is a mole calculation; part (b) tests the aldehyde-versus-ketone distinction.

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

n=3.6072.0=0.0500 moln = \frac{3.60}{72.0} = 0.0500 \text{ mol}

(b) An aldehyde has its carbonyl carbon bonded to at least one hydrogen atom, so it can be oxidised to a carboxylic acid by removing that hydrogen and adding oxygen. In a ketone the carbonyl carbon is bonded to two other carbon atoms and has no hydrogen to remove, so it cannot be oxidised in the same way.

Markers reward the correct mole value with a unit and the structural reason linked to the hydrogen on the carbonyl carbon.

Related dot points

Sources & how we know this