WJEC A-Level Chemistry Unit 4, June 2025: Question 2

📌 What This Question Tests

This question tests recall and application of key organic functional group reactions for secondary alcohols:

The Triiodomethane (Iodoform) Reaction: Identifying the positive observation for methyl secondary alcohols ( CH₃–CH(OH)– group).
Oxidation of Alcohols: Predicting the product and drawing the structural formula when a secondary alcohol is oxidised using acidified potassium dichromate(VI).

Yellow solid or yellow precipitate

Mark Scheme: [1 Mark] for stating "yellow solid" or "yellow precipitate".

Alkaline iodine (I₂ in NaOH(aq)) is the reagent for the triiodomethane (iodoform) test.
It gives a positive result with compounds containing either:
- A methyl carbonyl group: CH₃–C=O
- A methyl hydroxy group: CH₃–CH(OH)–
Propan-2-ol has the structure CH₃–CH(OH)–CH₃ , which contains this group.
The yellow precipitate formed is triiodomethane (CHI₃), which possesses a characteristic medicinal smell.

When asked "State what is seen...", you must describe an observation (colour change, precipitate, effervescence), not the name of a chemical. Writing only "triiodomethane" scores 0 marks.
Always state both the colour and the state (e.g. "yellow precipitate" or "yellow solid").

Writing "nothing" or "no reaction", confusing secondary alcohols with tertiary alcohols.
Stating "yellow solution" instead of solid or precipitate.
Naming the product (e.g. "triiodomethane formed") without giving an observation.

Any clearly drawn representation of propanone showing connectivity:

H₃C \ C = O / H₃C

Acceptable structural representations: CH₃COCH₃ or displayed/skeletal formulae of propanone.

Mark Scheme: [1 Mark] for the correct structure of propanone.

Acidified potassium dichromate(VI), K₂Cr₂O₇ / H⁺ , is a common oxidising agent.
Primary alcohols oxidise to aldehydes, then to carboxylic acids.
Secondary alcohols oxidise exclusively to ketones:
CH₃–CH(OH)–CH₃ + [O] → CH₃–CO–CH₃ + H₂O
Tertiary alcohols resist oxidation under standard conditions.
The orange dichromate(VI) ions (Cr₂O₇²⁻) are reduced to green chromium(III) ions (Cr³⁺).

The question specifically asks to "Give the structure...". Writing just the name "propanone" will not gain credit.
Make sure all bonds connect to the correct atoms. Ensure the C=O double bond originates from the central carbon atom, not an oxygen bonded to hydrogen.
Count your carbons: propan-2-ol has 3 carbons, so the ketone must also have 3 carbons.

Drawing propanoic acid ( CH₃CH₂COOH ) or propanal ( CH₃CH₂CHO ) by mistakenly treating propan-2-ol as a primary alcohol (propan-1-ol).
Leaving the hydroxyl hydrogen attached while forming a double bond, which gives carbon 5 bonds.
Writing the molecular formula ( C₃H₆O ) rather than a structural formula.

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