ChemistryISC Class 12 · Aldehydes, Ketones and Carboxylic Acids

Oxidation, Reduction and Tests for Aldehydes and Ketones (ISC Class 12)

Updated 15 October 2026 · Fact-checked

Aldehydes are oxidised easily to carboxylic acids, so they reduce Tollens' reagent (silver mirror) and Fehling's solution (red Cu₂O); ketones do not. Both are reduced to alcohols by NaBH₄ or LiAlH₄, and to hydrocarbons by Clemmensen (Zn–Hg/HCl) or Wolff–Kishner (hydrazine, KOH, glycol) reduction. Methyl ketones give iodoform.

Only aldehydes are oxidised by mild reagents, so only they give the silver mirror (Tollens') and red Cu₂O (Fehling's, aliphatic); ketones show no change.

Illustration drawn for QuestionBuddy.

Understand Oxidation, Reduction and Tests for Aldehydes and Ketones

Why aldehydes are easy to oxidise. An aldehyde has a hydrogen atom on the carbonyl carbon. Even mild oxidising agents remove it and give a carboxylic acid: $C H_{3} C H O + [O] \to C H_{3} C O O H$ . Ketones have no such hydrogen; they are oxidised only by strong oxidants (conc. HNO₃, KMnO₄/H₂SO₄, K₂Cr₂O₇/H₂SO₄) at high temperature, with cleavage of carbon-carbon bonds next to the carbonyl group, giving a mixture of acids with fewer carbon atoms. When ethanal is made by oxidising ethanol, it must be distilled off as soon as it forms; otherwise it is oxidised further to ethanoic acid.

Tests based on oxidation (aldehydes only).
• Tollens' test: ammoniacal silver nitrate, $[A g (N H_{3})_{2}]^{+}$ , warmed with an aldehyde gives a silver mirror (or black precipitate): $C H_{3} C H O + 2 [A g (N H_{3})_{2}]^{+} + 3 O H^{-} \to C H_{3} C O O^{-} + 2 A g + 2 H_{2} O + 4 N H_{3}$ . Aliphatic and aromatic aldehydes both respond.
• Fehling's test: Fehling's solution A (aqueous CuSO₄) and B (alkaline sodium potassium tartrate, Rochelle salt), mixed and warmed with an aliphatic aldehyde, gives a reddish-brown precipitate of Cu₂O: $R C H O + 2 C u^{2 +} + 5 O H^{-} \to R C O O^{-} + C u_{2} O + 3 H_{2} O$ . Aromatic aldehydes (benzaldehyde) do not respond. Benedict's solution behaves similarly.
• Schiff's test: aldehydes restore the pink colour of Schiff's reagent (decolourised rosaniline); ketones generally do not.

The iodoform test. Methyl ketones (CH₃CO–), ethanal, and alcohols with CH₃CH(OH)– give a yellow precipitate of iodoform with I₂ and NaOH: $C H_{3} C O C H_{3} + 3 I_{2} + 4 N a O H \to C H I_{3} + C H_{3} C O O N a + 3 N a I + 3 H_{2} O$ . It separates methyl ketones from other ketones, and ethanal from other aldehydes (methanal and propanal do not respond).

Reduction to alcohols. Sodium borohydride (NaBH₄), lithium aluminium hydride (LiAlH₄) or catalytic hydrogenation (H₂ with Ni, Pt or Pd) reduce aldehydes to primary alcohols and ketones to secondary alcohols: $C H_{3} C H_{2} C H O + 2 [H] \to C H_{3} C H_{2} C H_{2} O H$ ; $C H_{3} C O C H_{3} + 2 [H] \to C H_{3} C H (O H) C H_{3}$ .

Reduction to hydrocarbons (C=O → CH₂).
• Clemmensen reduction: zinc amalgam and concentrated HCl: $C H_{3} C O C H_{3} + 4 [H] \to C H_{3} C H_{2} C H_{3} + H_{2} O$ . Suitable for compounds stable to acid.
• Wolff–Kishner reduction: the carbonyl compound forms a hydrazone with hydrazine, which is heated with KOH (or sodium ethoxide) in a high-boiling solvent such as ethylene glycol; nitrogen is lost: $C H_{3} C H O N H_{2} N H_{2} C H_{3} C H = N N H_{2} K O H, glycol, heat C H_{3} C H_{3} + N_{2}$ . Overall: $C H_{3} C H O + N_{2} H_{4} \to C_{2} H_{6} + N_{2} + H_{2} O$ . Suitable for compounds sensitive to acid.
• Red phosphorus and HI also reduce C=O to CH₂: $C H_{3} C H O + 4 H I \to C_{2} H_{6} + H_{2} O + 2 I_{2}$ .

Using the tests together. Tollens' or Fehling's separates aldehydes from ketones; Fehling's separates aliphatic from aromatic aldehydes; iodoform picks out CH₃CO– compounds; neutral FeCl₃ or NaHCO₃ separates phenols and acids from carbonyl compounds. Always state the result for both compounds in a pair.

Oxidation, Reduction and Tests for Aldehydes and Ketones: formulas and key facts

Tollens' test: $R C H O + 2 [A g (N H_{3})_{2}]^{+} + 3 O H^{-} \to R C O O^{-} + 2 A g + 2 H_{2} O + 4 N H_{3}$; Silver mirror; aliphatic and aromatic aldehydes.
Fehling's test: $R C H O + 2 C u^{2 +} + 5 O H^{-} \to R C O O^{-} + C u_{2} O + 3 H_{2} O$; Red-brown Cu₂O; aliphatic aldehydes only.
Clemmensen reduction: $R_{2} C = O Z n - H g, conc. HCl R_{2} C H_{2} + H_{2} O$; Acidic conditions.
Wolff–Kishner reduction: $R_{2} C = O N H_{2} N H_{2} R_{2} C = N N H_{2} K O H, ethylene glycol, heat R_{2} C H_{2} + N_{2}$; Basic conditions; ethanal gives ethane.
Reduction to alcohols: $R C H O N a B H_{4} or L i A l H_{4} R C H_{2} O H; R_{2} C O \to R_{2} C H O H$; Aldehyde → 1° alcohol; ketone → 2° alcohol.

How to solve Oxidation, Reduction and Tests for Aldehydes and Ketones questions

Questions here are of three types: name the reagent or product, distinguish a pair, and identify unknowns from test results. Use a decision table.

1Classify each compound: aldehyde (aliphatic or aromatic), methyl ketone, other ketone, or alcohol.
2To separate aldehyde from ketone: Tollens' (silver mirror) or Fehling's (red precipitate) — positive for the aldehyde only.
3To separate aliphatic from aromatic aldehyde: Fehling's — positive only for the aliphatic one.
4To separate CH₃CO– (or CH₃CH(OH)–) compounds from others: iodoform test (yellow precipitate).
5For reduction, decide the product: alcohol (NaBH₄, LiAlH₄, H₂/Ni) or hydrocarbon (Clemmensen in acid, Wolff–Kishner in base, red P/HI).
6Write each test as reagent → observation for compound 1 → observation for compound 2.

Oxidation, Reduction and Tests for Aldehydes and Ketones in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

The silver mirror is a quantitative oxidation

Suppose a craftsperson silvers the inside of a small decorative glass bulb by warming Tollens' reagent with 0.44 g of ethanal (M = 44) until all the ethanal has reacted. (Ag = 107.9)

Write the equation, calculate the maximum mass of silver deposited, and explain why propanone could not be used instead.

Try it first, then show the solutionSolution

$C H_{3} C H O + 2 [A g (N H_{3})_{2}]^{+} + 3 O H^{-} \to C H_{3} C O O^{-} + 2 A g + 2 H_{2} O + 4 N H_{3}$ .
Moles of ethanal $= \frac{0.44}{44} = 0.010$ mol, so silver $= 2 \times 0.010 = 0.020$ mol.
Mass of silver $= 0.020 \times 107.9 = 2.16$ g.
Propanone has no H on its carbonyl carbon, so mild Tollens' reagent cannot oxidise it and no silver forms.

Answer: At most 2.16 g of silver. Propanone (a ketone) does not reduce Tollens' reagent.

Memory hook: One aldehyde, two silvers: the aldehyde H is what reduces Ag⁺.

Tricks to remember Oxidation, Reduction and Tests for Aldehydes and Ketones

Short memory aids. Say each one aloud once, then use it on a question the same day.

Aldehydes have a 'handle' H
The H on the carbonyl carbon is a handle that oxidants grab; ketones have no handle, so Tollens' and Fehling's leave them alone.
Use it when: Explaining why only aldehydes respond.
Fehling fails for benzene rings
Fehling's: Fails with aromatic aldehydes (benzaldehyde); Tollens' works for both.
Use it when: Separating aliphatic from aromatic aldehydes.
Clemmensen = Clean acid, Wolff–Kishner = Knocks off N₂ in base
Clemmensen uses Zn–Hg/HCl (acid); Wolff–Kishner uses hydrazine and KOH (base) and gives off N₂.
Use it when: Choosing the reduction to CH₂.

Oxidation, Reduction and Tests for Aldehydes and Ketones: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Reagents for Clemmensen reduction · 1 mark · MCQBoard level

The Clemmensen reduction converts a carbonyl group into a CH₂ group using:
(a) NaBH₄ in ethanol
(b) zinc amalgam and concentrated HCl
(c) hydrazine and KOH in ethylene glycol
(d) H₂ with Pd/BaSO₄

Try it first, then show the solutionSolution

NaBH₄ gives alcohols; (c) is Wolff–Kishner; (d) is the Rosenmund catalyst.
Clemmensen uses Zn–Hg and conc. HCl.

Answer: (b).

Example 2: Tollens' test on propanal · 2 marks · Short answerBoard level

Describe Tollens' test for propanal with an equation and the observation. What would be seen with propanone?

Try it first, then show the solutionSolution

Warm propanal with Tollens' reagent (ammoniacal silver nitrate): a silver mirror forms on the tube (½ mark).
$C H_{3} C H_{2} C H O + 2 [A g (N H_{3})_{2}]^{+} + 3 O H^{-} \to C H_{3} C H_{2} C O O^{-} + 2 A g + 2 H_{2} O + 4 N H_{3}$ (1 mark).
Propanone gives no silver mirror (½ mark).

Answer: Silver mirror with propanal; no change with propanone.

Example 3: Three pairs, three tests · 3 marks · Short answerHarder board level

Give one chemical test each to distinguish: (a) methanal and ethanal; (b) ethanal and benzaldehyde; (c) propanal and propanone.

Try it first, then show the solutionSolution

(a) Iodoform test (I₂ + NaOH, warm): ethanal gives a yellow precipitate of CHI₃; methanal does not (1 mark).
(b) Fehling's solution, warm: ethanal gives a reddish-brown precipitate of Cu₂O; benzaldehyde does not (1 mark).
(c) Tollens' reagent, warm: propanal gives a silver mirror; propanone does not (or Fehling's: red precipitate with propanal only) (1 mark).

Answer: (a) Iodoform. (b) Fehling's. (c) Tollens' (or Fehling's).

Example 4: Three ways to reduce butanone · 3 marks · Short answerHarder board level

Give the product and equation when butanone is treated with (a) NaBH₄, (b) zinc amalgam and conc. HCl, (c) hydrazine followed by KOH in ethylene glycol on heating. Name reactions (b) and (c).

Try it first, then show the solutionSolution

(a) $C H_{3} C O C H_{2} C H_{3} + 2 [H] \to C H_{3} C H (O H) C H_{2} C H_{3}$ : butan-2-ol (1 mark).
(b) $C H_{3} C O C H_{2} C H_{3} + 4 [H] \to C H_{3} C H_{2} C H_{2} C H_{3} + H_{2} O$ : butane; Clemmensen reduction (1 mark).
(c) Hydrazone, then loss of N₂: $C H_{3} C O C H_{2} C H_{3} + N_{2} H_{4} \to C_{4} H_{10} + N_{2} + H_{2} O$ overall: butane; Wolff–Kishner reduction (1 mark).

Answer: (a) Butan-2-ol. (b) Butane (Clemmensen). (c) Butane (Wolff–Kishner).

Example 5: Two C₃H₆O isomers · 5 marks · Long answerHOTS / competency

Two isomeric compounds X and Y have the formula C₃H₆O. X gives a silver mirror with Tollens' reagent and a red precipitate with Fehling's solution. Y gives neither, but gives a yellow precipitate with I₂ and NaOH. Both are converted into the same hydrocarbon Z by Wolff–Kishner reduction.
(a) Identify X, Y and Z. [2]
(b) Write the equation for the reaction of Y with I₂ and NaOH. [1]
(c) What alcohols are formed when X and Y are reduced with NaBH₄? Classify each. [2]

Try it first, then show the solutionSolution

(a) X is an aliphatic aldehyde: propanal, CH₃CH₂CHO. Y is a methyl ketone: propanone, CH₃COCH₃. Z = propane, CH₃CH₂CH₃ (2 marks).
(b) $C H_{3} C O C H_{3} + 3 I_{2} + 4 N a O H \to C H I_{3} + C H_{3} C O O N a + 3 N a I + 3 H_{2} O$ (1 mark).
(c) X gives propan-1-ol, a primary alcohol (1 mark); Y gives propan-2-ol, a secondary alcohol (1 mark).

Answer: (a) X propanal, Y propanone, Z propane. (b) Iodoform reaction as shown. (c) Propan-1-ol (1°) and propan-2-ol (2°).

Where students lose marks in Oxidation, Reduction and Tests for Aldehydes and Ketones

The slips examiners see most often, and what to write instead.

Mistake: Wrong observation for Tollens' test
ISC examiners reported 'white precipitate' instead of a silver mirror (2024).
Do this: Tollens' test: silver mirror (or a black/grey precipitate of silver). Fehling's: reddish-brown precipitate of Cu₂O.
Usually costs: 1 mark.
Mistake: Saying ethanal is removed because it is poisonous
Examiners noted 'poisonous' or 'explosive' given instead of 'oxidised further to ethanoic acid' (2024).
Do this: The reason is chemical: aldehydes are easily oxidised to carboxylic acids.
Usually costs: 1 mark.
Mistake: Using an unsuitable test for a pair
Examiners reported the Lucas test used to distinguish ethanol from propanal, and tests described without observations for both compounds (2025).
Do this: Choose a test that one compound passes and the other fails (iodoform for ethanol/propanal; Tollens' or Schiff's would also work), and state both observations.
Usually costs: 1 mark per pair.
Mistake: Naming the wrong reduction
In a 2025 fill-in, 'Stephen' and 'propane' were written instead of 'Wolff–Kishner' and 'ethane'.
Do this: Hydrazine + KOH + glycol = Wolff–Kishner; ethanal gives ethane (same carbon count).
Usually costs: 1 mark.

Exam tips

Learn the composition of each reagent: Tollens' (ammoniacal AgNO₃), Fehling's (CuSO₄ + alkaline Rochelle salt), Schiff's (decolourised rosaniline), iodoform (I₂ + NaOH).
In identification problems, a silver mirror means –CHO; a red Fehling precipitate means aliphatic –CHO; a yellow iodoform precipitate means CH₃CO– or CH₃CH(OH)–.
Write IUPAC names for answers (ethanal, propanone) as examiners asked for them in 2024.
For reductions, the carbon count does not change: a C₄ ketone gives a C₄ alkane or alcohol.
For each distinguishing test, finish with the negative result for the other compound.

Practice questions on Oxidation, Reduction and Tests for Aldehydes and Ketones

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
Which compound gives a reddish-brown precipitate with Fehling's solution?
(a) benzaldehyde
(b) propanone
(c) propanal
(d) acetophenone
Hint
Aliphatic aldehydes only.
Answer
(c) propanal.
Q2 · 2 marks · Short answerBoard level
In the preparation of ethanal by oxidising ethanol with acidified dichromate, ethanal is distilled off as soon as it forms. Why?
Hint
What happens to an aldehyde left with an oxidant?
Answer
Ethanal is easily oxidised; if it stays in the flask with the oxidising agent it is oxidised further to ethanoic acid: $C H_{3} C H O + [O] \to C H_{3} C O O H$ . Ethanal boils at a low temperature, so distilling it off at once prevents this.
Q3 · 2 marks · Short answerHarder board level
Write the ionic equation for the reaction of ethanal with Fehling's solution and state the change in oxidation state of copper.
Hint
Cu²⁺ is reduced to copper(I) oxide.
Answer
$C H_{3} C H O + 2 C u^{2 +} + 5 O H^{-} \to C H_{3} C O O^{-} + C u_{2} O + 3 H_{2} O$ . Copper changes from +2 to +1.
Q4 · 3 marks · Short answerHarder board level
How will you convert (a) propanone into propane, (b) ethanal into ethane without using acid, and (c) propanal into propan-1-ol? Give reagents and equations.
Hint
Choose Clemmensen or Wolff–Kishner by the conditions allowed.
Answer
(a) Clemmensen: Zn–Hg/conc. HCl, $C H_{3} C O C H_{3} + 4 [H] \to C H_{3} C H_{2} C H_{3} + H_{2} O$ . (b) Wolff–Kishner: NH₂NH₂, then KOH in ethylene glycol, heat: $C H_{3} C H O + N_{2} H_{4} \to C_{2} H_{6} + N_{2} + H_{2} O$ . (c) NaBH₄ (or LiAlH₄, or H₂/Ni): $C H_{3} C H_{2} C H O + 2 [H] \to C H_{3} C H_{2} C H_{2} O H$ .
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): Propanone does not give a silver mirror with Tollens' reagent.
Reason (R): Ketones have no hydrogen atom on the carbonyl carbon, so they are oxidised only by strong oxidants with cleavage of C–C bonds.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Tollens' reagent is a mild oxidant.
Answer
(a).
Q6 · 3 marks · Short answerHarder board level
Give two chemical tests to distinguish ethanal from benzaldehyde, stating the observation with each compound.
Hint
One test is negative for the aromatic aldehyde; another depends on the CH₃CO– group.
Answer
(i) Fehling's solution: ethanal gives a reddish-brown precipitate of Cu₂O; benzaldehyde gives none. (ii) Iodoform test: ethanal gives a yellow precipitate of CHI₃; benzaldehyde does not. (Tollens' does not help: both give a silver mirror.)
Q7 · 3 marks · Short answerHOTS / competency
0.58 g of propanal (M = 58) is warmed with excess Fehling's solution until reaction is complete. Calculate the mass of copper(I) oxide formed (Cu₂O: M = 143). What colour is it?
Hint
1 mol aldehyde gives 1 mol Cu₂O.
Answer
Moles of propanal $= \frac{0.58}{58} = 0.010$ ; Cu₂O $= 0.010$ mol $= 0.010 \times 143 = 1.43$ g. It is a reddish-brown precipitate.
Q8 · 3 marks · Case-basedHOTS / competency
Read and answer.
Suppose a flavour chemist is given an 'almond oil' sample that should contain benzaldehyde but may have been mixed with acetophenone (C₆H₅COCH₃). Three reagents are on her bench: ammoniacal silver nitrate, a copper(II) tartrate complex in alkali, and iodine in sodium hydroxide.
(i) Which reagent shows the presence of benzaldehyde, and what is seen? [1]
(ii) Which reagent shows the presence of acetophenone, and what is seen? [1]
(iii) Why is Fehling's solution of no use here? [1]
Hint
One is an aromatic aldehyde, the other a methyl ketone.
Answer
(i) Tollens' reagent: a silver mirror (benzaldehyde is oxidised; acetophenone is not). (ii) I₂ and NaOH: a yellow precipitate of iodoform from the CH₃CO– group of acetophenone. (iii) Fehling's solution does not react with aromatic aldehydes or with ketones, so both compounds give a negative result.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Product of an aldehyde with Tollens' reagent?Show answer
Carboxylate ion, with a silver mirror.
2.Which aldehydes do not give Fehling's test?Show answer
Aromatic aldehydes such as benzaldehyde.
3.Reagents for Wolff–Kishner reduction?Show answer
Hydrazine, then KOH (or NaOEt) in ethylene glycol on heating.
4.Product of propanone with NaBH₄?Show answer
Propan-2-ol.
5.What does Schiff's reagent show?Show answer
Aldehydes restore its pink colour.
6.Why are ketones resistant to oxidation?Show answer
No H on the carbonyl carbon; oxidation needs C–C bond cleavage.

Oxidation, Reduction and Tests for Aldehydes and Ketones: frequently asked questions

How do you distinguish an aldehyde from a ketone?

Warm with Tollens' reagent: an aldehyde gives a silver mirror, a ketone does not. Fehling's solution gives a red precipitate with aliphatic aldehydes only.

What is the difference between Clemmensen and Wolff–Kishner reduction?

Both convert C=O to CH₂. Clemmensen uses zinc amalgam and concentrated HCl (acidic); Wolff–Kishner uses hydrazine and KOH in ethylene glycol (basic).

Why does benzaldehyde not give Fehling's test?

It is an aromatic aldehyde and is not oxidised by the mild copper(II) complex in Fehling's solution, although it does give Tollens' test.

Which carbonyl compounds give the iodoform test?

Ethanal and methyl ketones (compounds with CH₃CO–); alcohols with CH₃CH(OH)– also respond.

What happens when ketones are oxidised strongly?

C–C bonds next to the carbonyl group break and a mixture of carboxylic acids with fewer carbon atoms forms.