The problems are on the oxidation of alcohols worksheet. Try them on your own first, then compare with the answers below.
Quick answer: Primary alcohols are oxidised to aldehydes and then carboxylic acids, secondary alcohols to ketones, and tertiary alcohols resist oxidation; acidified dichromate turns from orange to green. These worked answers correct the 20 problems of the Chemistry 11-12 worksheet on oxidation of alcohols, from classify the alcohols to silver from ethanal: each solution shows the formula, the substitution and the final answer with its unit.
Each answer gives the reasoning before the final result in bold. Atomic masses used: H 1.0, C 12.0, O 16.0, Ag 107.9. When an equation uses [O] or [H], the symbols stand for the oxidising or reducing agent, not free atoms.
Level: Chemistry grades 11-12 (A Level · IB · Class 11-12) · Curriculum: A Level 3.3.5
Problem 1 Classify the alcohols
1. Butan-1-ol: primary. Butan-2-ol: secondary. 2-methylpropan-1-ol: primary. 2-methylpropan-2-ol: tertiary.2. 2-methylpropan-2-ol.
3. Count the carbon atoms bonded to the C–OH carbon: one, two, one and three respectively.
Problem 2 Colour of dichromate
1. Orange to green.2. Orange: Cr₂O₇²⁻; green: Cr³⁺.
3. A tertiary alcohol has no hydrogen on the C–OH carbon, so the mixture stays orange.
Problem 3 Oxidising propan-1-ol
1. Propanal.2. Propanoic acid.
3. CH₃CH₂CH₂OH + [O] → CH₃CH₂CHO + H₂O; CH₃CH₂CHO + [O] → CH₃CH₂COOH. Both equations balance for atoms.
Problem 4 Oxidising propan-2-ol
1. Propanone, a ketone (C=O inside the chain).2. CH₃CH(OH)CH₃ + [O] → CH₃COCH₃ + H₂O.
3. The carbonyl carbon has no hydrogen left, so a further oxidation would need a C–C bond to break.
Problem 5 Reading the carbonyl tests
1. Ethanal: silver mirror. Propanone: no change.2. Both give an orange precipitate.
3. Tollens’ reagent (or Fehling’s solution) separates them; 2,4-DNPH does not.
Problem 6 Reducing carbonyls
1. Propan-1-ol.2. Butan-2-ol.
3. Propan-1-ol is primary; butan-2-ol is secondary.
Problem 7 Yield of a ketone
1. n = 8.80 / 88.0 = 0.100 mol.2. Ratio 1 : 1, so mass = 0.100 × 86.0 = 8.60 g.
3. Yield = 6.45 / 8.60 × 100 = 75.0 %.
Problem 8 Distilling an aldehyde
1. Distillation.2. Propanal boils at 49 °C, much lower than propan-1-ol at 97 °C, so it leaves the flask as it forms and cannot be oxidised to the acid.
3. They give small bubbles so that the liquid boils smoothly instead of bumping.
Problem 9 Boiling points and hydrogen bonds
1. Propan-1-ol: hydrogen bonds. Propanal: permanent dipole-dipole forces. Propanoic acid: hydrogen bonds (often in dimers).2. Propanal has no O–H group, so its molecules cannot hydrogen-bond to each other.
3. The acid has both C=O and O–H, so molecules pair up through two hydrogen bonds, which need more energy to separate.
Problem 10 Two isomers of C₃H₆O
1. P is propanal; Q is propanone.2. P: aldehyde; Q: ketone.
3. Q has no hydrogen on the carbonyl carbon, so it cannot be oxidised by the mild oxidant.
Problem 11 Naming hydroxynitriles
1. CH₃CH(OH)CN: 2-hydroxypropanenitrile.2. (CH₃)₂C(OH)CN: 2-hydroxy-2-methylpropanenitrile.
3. The first product is chiral (four different groups: H, OH, CN, CH₃); the second has two CH₃ groups so it is not.
Problem 12 Why C=O is attacked by nucleophiles
1. Oxygen is more electronegative than carbon, so the shared electrons are pulled to oxygen.2. The carbon carries δ+ and nucleophiles seek positive centres.
3. In an alkene the C=C is electron-rich and non-polar; in C=O the carbon is electron-poor, so the type of attacking species reverses.
Problem 13 Tertiary or secondary
1. Warm each with acidified potassium dichromate(VI).2. Pentan-2-ol turns green; 2-methylbutan-2-ol stays orange.
3. Pentan-2-one.
Problem 14 Racemic mixtures
1. 2-hydroxypropanenitrile.2. Two enantiomers in equal amounts (1 : 1).
3. The planar C=O is attacked equally from both faces; the two enantiomers rotate light by equal and opposite amounts, so the net rotation is zero.
Problem 15 Identifying by melting point
1. Propanal.2. Yes: propanal is an aldehyde, so it gives a silver mirror.
3. Impurities lower and widen the melting range, so a pure derivative is needed for a valid comparison.
Problem 16 Alcohol in wine
1. Left: 6 C, 18 H + 16 H = 34 H, 3 + 14 = 17 O, 4 Cr; charge −4 + 16 = +12. Right: 6 C, 12 + 22 = 34 H, 6 + 11 = 17 O, 4 Cr; charge +12. Both sides match.2. n(Cr₂O₇²⁻) = 0.0200 × 0.0184 = 3.68 × 10⁻⁴ mol; n(ethanol) = 3.68 × 10⁻⁴ × 3/2 = 5.52 × 10⁻⁴ mol.
3. In 250 cm³: 5.52 × 10⁻³ mol; mass = 0.2539 g; volume = 0.2539 / 0.789 = 0.322 cm³ in 5.00 cm³ of wine: 6.4 % by volume.
Problem 17 Four butanols
1. W: butan-1-ol; X: butan-2-ol; Y: 2-methylpropan-2-ol; Z: 2-methylpropan-1-ol.2. W: butanal (then butanoic acid); X: butanone; Z: 2-methylpropanal.
3. A mirror means an aldehyde, so the alcohols are primary (W and Z); the chain shape separates them. A ketone means secondary (X only); no reaction means tertiary (Y).
Problem 18 Find the error in the mechanism
1. The cyanide ion has attacked the oxygen, but oxygen carries δ−.2. The correct intermediate is CH₃CH(O⁻)CN, which gains H⁺ to give CH₃CH(OH)CN.
3. CN⁻ is a nucleophile (it donates a lone pair from carbon) and it attacks the δ+ carbon of C=O.
Problem 19 Identify a compound by data
1. m/z 72 confirms C₄H₈O (4(12) + 8 + 16 = 72). The 1715 cm⁻¹ peak shows C=O; the lack of a broad peak shows no O–H. The DNPH result confirms a carbonyl; no mirror means a ketone.2. Butanone, CH₃COCH₂CH₃.
3. CH₃C(OH)(CN)CH₂CH₃, 2-hydroxy-2-methylbutanenitrile; it is chiral (CH₃, C₂H₅, OH and CN are all different).
Problem 20 Silver from ethanal
1. n = 2.20 / 44.0 = 0.0500 mol.2. n(Ag) = 2 × 0.0500 = 0.100 mol; mass = 0.100 × 107.9 = 10.8 g.
3. Volume = 0.100 / 0.200 = 0.500 dm³ = 500 cm³.
