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Test the C=O.

Six original Cambridge-style questions on nucleophilic addition and identifying carbonyl compounds.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
Keep these straight

Mechanism first, then three tests in order.

01
Explain
[4 marks]

Describe the mechanism of the reaction between propanone and HCN in the presence of KCN. Name the product.

  • The C=O is polar; the carbon is δ+. ✓
  • A lone pair on the carbon of CN− attacks the δ+ carbon; the π electrons of C=O move onto the oxygen, forming an alkoxide intermediate. ✓
  • The O− takes H+ from HCN, regenerating CN−. ✓
  • Product: 2-hydroxy-2-methylpropanenitrile, (CH3)2C(OH)CN. ✓
02
Explain
[3 marks]

Butanone reacts with HCN to give a product that does not rotate plane-polarised light. Explain why, even though the product contains a chiral centre.

  • The product CH3C(OH)(CN)CH2CH3 has four different groups on one carbon, so it is chiral. ✓
  • The C=O carbon in butanone is trigonal planar, so CN− attacks from above or below with equal probability. ✓
  • Equal amounts of the two enantiomers form (a racemic mixture), and their rotations cancel. ✓
03
Recall
[3 marks]

Describe a test that distinguishes propanal from propanone. Give the reagent, conditions and observations for each compound.

  • Warm each with Tollens' reagent (or Fehling's solution). ✓
  • Propanal: silver mirror (or blue to brick-red precipitate with Fehling's). ✓
  • Propanone: no change; ketones are not oxidised. ✓
04
Analysis
[3 marks]

Compound R, C4H8O, gives an orange precipitate with 2,4-DNPH, does not react with Fehling's solution, and gives a yellow precipitate with alkaline iodine. Identify R, explaining each result.

  • 2,4-DNPH positive: R is an aldehyde or ketone. ✓
  • No reaction with Fehling's: R is a ketone. ✓
  • CHI3 forms: R has CH3CO–. R is butanone, CH3COCH2CH3. ✓
05
Calculation
[2 marks]

Excess Tollens' reagent is warmed with 0.880 g of ethanal (Mr 44.0). Each mole of aldehyde reduces two moles of Ag+. Calculate the mass of silver deposited (Ar Ag = 107.9).

  • n(ethanal) = 0.880 ÷ 44.0 = 0.0200 mol, so n(Ag) = 0.0400 mol. ✓
  • Mass = 0.0400 × 107.9 = 4.32 g. ✓
06
Analysis
[2 marks]

Which of ethanal, propanal, propanone and pentan-3-one give a positive tri-iodomethane test? Explain.

  • Ethanal and propanone: both contain CH3CO–. ✓
  • Propanal (CH3CH2CO–) and pentan-3-one (two ethyl groups) have no CH3 on the C=O carbon, so are negative. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.