ChemistryCBSE Class 12 · Aldehydes, Ketones and Carboxylic Acids

Oxidation, Reduction and Tests for Aldehydes and Ketones

Updated 15 October 2026 · Fact-checked

Aldehydes and ketones are reduced to alcohols by $NaBH_4$ or $LiAlH_4$, and to hydrocarbons by Clemmensen (Zn–Hg/HCl) or Wolff–Kishner (hydrazine, then KOH in ethylene glycol) reduction. Aldehydes are easily oxidised, so they give a silver mirror with Tollens' reagent and red Cu₂O with Fehling's solution; ketones do not. Methyl ketones give yellow iodoform with NaOI.

Understand Oxidation, Reduction and Tests for Aldehydes and Ketones

1. Reduction.

*To alcohols.* Sodium borohydride ( $N a B H_{4}$ ), lithium aluminium hydride ( $L i A l H_{4}$ ) or catalytic hydrogenation reduces aldehydes to primary alcohols and ketones to secondary alcohols: $C H_{3} C H_{2} C H O \to C H_{3} C H_{2} C H_{2} O H$ ; $C H_{3} C O C H_{3} \to C H_{3} C H (O H) C H_{3}$ .

*To hydrocarbons.* The C=O group can be reduced right down to $C H_{2}$ . In Clemmensen reduction the compound is treated with zinc amalgam and concentrated HCl. In Wolff–Kishner reduction it is first converted to its hydrazone with hydrazine, and the hydrazone is heated with KOH (or sodium ethoxide) in a high-boiling solvent such as ethylene glycol, releasing nitrogen: $> C = O N H_{2} N H_{2} > C = N N H_{2} K O H, ethylene glycol, Δ > C H_{2} + N_{2}$ . Clemmensen uses strong acid, so it suits compounds that are stable to acid; Wolff–Kishner uses strong base, so it suits compounds that are acid-sensitive.

2. Oxidation. Aldehydes differ from ketones here. An aldehyde has an H attached to the carbonyl carbon, so it is easily oxidised to a carboxylic acid with the same number of carbons, even by mild oxidants. Common strong oxidants (nitric acid, $K M n O_{4}$ , $K_{2} C r_{2} O_{7}$ ) work, and so do the mild reagents used in tests. Ketones are oxidised only by strong oxidising agents at higher temperature, and the reaction breaks a C–C bond, giving a mixture of acids with fewer carbon atoms.

3. Tests that separate aldehydes from ketones.

*Tollens' test.* Tollens' reagent is freshly prepared ammoniacal silver nitrate. Warming an aldehyde with it produces a bright silver mirror on the inside of the tube, because the aldehyde is oxidised to the carboxylate anion and $A g^{+}$ is reduced to metallic silver: $R C H O + 2 [A g (N H_{3})_{2}]^{+} + 3 O H^{-} \to R C O O^{-} + 2 A g + 2 H_{2} O + 4 N H_{3}$ . Both aliphatic and aromatic aldehydes respond.

*Fehling's test.* Fehling's solution is a mixture of Fehling A (aqueous copper sulphate) and Fehling B (alkaline sodium potassium tartrate, Rochelle salt). Heating an aldehyde with it gives a reddish-brown precipitate of copper(I) oxide: $R C H O + 2 C u^{2 +} + 5 O H^{-} \to R C O O^{-} + C u_{2} O + 3 H_{2} O$ . Aromatic aldehydes (such as benzaldehyde) do not respond to Fehling's test, which lets you tell them from aliphatic aldehydes.

Ketones give neither test.

4. Haloform (iodoform) reaction. Aldehydes and ketones with at least one methyl group next to the carbonyl, $C H_{3} C O -$ (methyl ketones and ethanal), react with sodium hypohalite ( $N a O X$ , from $X_{2}$ and NaOH). The three hydrogens of the methyl group are replaced by halogen, and the $C X_{3}$ group is then cleaved, giving the sodium salt of an acid with one fewer carbon and a haloform. With iodine and NaOH, iodoform ( $C H I_{3}$ ), a yellow solid with a characteristic smell, separates: $R C O C H_{3} + 3 N a O I \to R C O O N a + C H I_{3} + 2 N a O H$ . The C=C bond, if present, is not attacked. Alcohols with the $C H_{3} C H (O H) -$ group (ethanol, propan-2-ol, butan-2-ol) also give the test, because NaOI first oxidises them to $C H_{3} C O -$ compounds.

Together these tests answer most 'distinguish between' questions in this chapter.

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

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.
Clemmensen reduction: $> C = O Z n - H g, conc. H C l > C H_{2} + H_{2} O$; Acidic conditions.
Wolff–Kishner reduction: $> C = O N H_{2} N H_{2} > C = N N H_{2} K O H, ethylene glycol, Δ > C H_{2} + N_{2}$; Basic conditions.
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; all 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 precipitate; aliphatic aldehydes only.
Iodoform reaction: $R C O C H_{3} + 3 N a O I \to R C O O N a + C H I_{3} ↓ + 2 N a O H$; For $C H_{3} C O -$ and $C H_{3} C H (O H) -$ compounds; yellow precipitate.

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

Most questions in this topic are 'distinguish between' pairs, 'complete the reaction', or identify-the-unknown puzzles. Use a fixed decision path.

1Classify each compound: aldehyde (aliphatic or aromatic) or ketone; does it have $C H_{3} C O -$ (or $C H_{3} C H (O H) -$ )?
2To separate an aldehyde from a ketone, use Tollens' (silver mirror) or Fehling's (red precipitate) test.
3To separate an aliphatic aldehyde from an aromatic one, use Fehling's test (aromatic aldehydes do not respond).
4To pick out a methyl ketone (or ethanal, or a $C H_{3} C H (O H) -$ alcohol), use the iodoform test.
5For each test, write the reagent, what you observe with each compound, and the equation if asked.
6For reductions, decide whether the target is an alcohol ( $N a B H_{4}$ / $L i A l H_{4}$ ) or a hydrocarbon (Clemmensen or Wolff–Kishner), and choose acid or base conditions according to the rest of the molecule.
7In identification puzzles, list every observation as a constraint and test candidate structures against all of them.

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.

Tollens' reagent deposits silver

Suppose a craft teacher wants to silver the inside of a small glass ornament using Tollens' reagent and plans to use 0.22 g of ethanal as the reducing agent.

Write the ionic equation and calculate the maximum mass of silver that can be deposited. (Ethanal 44 g mol⁻¹, Ag 108 g mol⁻¹)

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 $= 0.22/44 = 0.0050$ mol.
Each mole of aldehyde deposits 2 mol of Ag: $0.010$ mol Ag.
Mass $= 0.010 \times 108 = 1.08$ g.

Answer: Up to about 1.08 g of silver.

Memory hook: One aldehyde, two silvers: the aldehyde's easy oxidation is what makes the mirror.

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.

Tollens for all, Fehling for the aliphatic
All aldehydes pass Tollens; benzaldehyde fails Fehling; ketones fail both.
Use it when: Choosing the right test.
C for Clemmensen and Concentrated HCl; K for Kishner and KOH
Clemmensen: Zn–Hg with conc. HCl (acid). Wolff–Kishner: hydrazine, then KOH (base). Both end at $C H_{2}$ .
Use it when: Matching reagents to named reductions.
Iodoform needs CH₃–C=O or CH₃–CH–OH
Look for a methyl next to C=O (or next to CHOH). Methanal and propanal fail; ethanal passes.
Use it when: Iodoform test questions.

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: Which gives iodoform? · 1 mark · MCQBoard level

Which of the following gives the iodoform test?
(a) Propanal (b) Pentan-3-one (c) Propanone (d) Benzaldehyde

Try it first, then show the solutionSolution

Only propanone, $C H_{3} C O C H_{3}$ , has a $C H_{3} C O -$ group.

Answer: (c) Propanone.

Example 2: Two ways to make propane from propanone · 2 marks · Very short answerBoard level

Write the reagents and the reactions for converting propanone into propane by (a) Clemmensen reduction and (b) Wolff–Kishner reduction.

Try it first, then show the solutionSolution

(a) $C H_{3} C O C H_{3} Z n - H g, conc. H C l C H_{3} C H_{2} C H_{3} + H_{2} O$ (1 mark).
(b) $C H_{3} C O C H_{3} N H_{2} N H_{2} (C H_{3})_{2} C = N N H_{2} K O H, ethylene glycol, Δ C H_{3} C H_{2} C H_{3} + N_{2}$ (1 mark).

Answer: (a) Zn–Hg/conc. HCl. (b) $N H_{2} N H_{2}$ , then KOH in ethylene glycol on heating.

Example 3: Distinguishing pairs · 3 marks · Short answerHarder board level

Give a simple chemical test to distinguish between: (a) propanal and propanone (b) acetophenone and benzophenone (c) ethanal and propanal.

Try it first, then show the solutionSolution

(a) Tollens' test: propanal gives a silver mirror; propanone does not. (Fehling's test also works.) (1 mark)
(b) Iodoform test: acetophenone ( $C_{6} H_{5} C O C H_{3}$ ) gives a yellow precipitate of $C H I_{3}$ with $I_{2}$ /NaOH; benzophenone ( $C_{6} H_{5} C O C_{6} H_{5}$ ) does not (1 mark).
(c) Iodoform test: ethanal ( $C H_{3} C H O$ ) gives yellow iodoform; propanal does not (1 mark).

Answer: (a) Tollens'. (b) Iodoform. (c) Iodoform.

Example 4: Tollens' and Fehling's tests · 3 marks · Short answerHarder board level

Write the ionic equations for the reactions of ethanal with Tollens' reagent and with Fehling's solution, giving the observation in each case. Which of these tests would benzaldehyde fail?

Try it first, then show the solutionSolution

Tollens': $C H_{3} C H O + 2 [A g (N H_{3})_{2}]^{+} + 3 O H^{-} \to C H_{3} C O O^{-} + 2 A g + 2 H_{2} O + 4 N H_{3}$ ; a silver mirror forms (1 mark).
Fehling's: $C H_{3} C H O + 2 C u^{2 +} + 5 O H^{-} \to C H_{3} C O O^{-} + C u_{2} O + 3 H_{2} O$ ; a reddish-brown precipitate forms (1 mark).
Benzaldehyde gives Tollens' test but not Fehling's test (1 mark).

Answer: Silver mirror; reddish-brown $C u_{2} O$ ; benzaldehyde fails Fehling's test.

Example 5: Four C₅H₁₀O carbonyl compounds · 5 marks · Long answerHOTS / competency

Four carbonyl compounds A, B, C and D have the formula $C_{5} H_{10} O$ . A gives the iodoform test but not Tollens' test. B gives neither test. C and D give Tollens' test; C has no α-hydrogen, and D has an unbranched chain. Identify A to D, and name the product of Clemmensen reduction of B.

Try it first, then show the solutionSolution

A: a methyl ketone (iodoform +, Tollens' −): pentan-2-one, $C H_{3} C O C H_{2} C H_{2} C H_{3}$ (or 3-methylbutan-2-one; the unbranched answer is pentan-2-one) (1 mark).
B: a ketone without $C H_{3} C O -$ : pentan-3-one, $C H_{3} C H_{2} C O C H_{2} C H_{3}$ (1 mark).
C: an aldehyde with no α-hydrogen: 2,2-dimethylpropanal, $(C H_{3})_{3} C C H O$ (1 mark).
D: straight-chain aldehyde: pentanal, $C H_{3} C H_{2} C H_{2} C H_{2} C H O$ (1 mark).
Clemmensen reduction of B gives pentane, $C H_{3} C H_{2} C H_{2} C H_{2} C H_{3}$ (1 mark).

Answer: A pentan-2-one; B pentan-3-one; C 2,2-dimethylpropanal; D pentanal; B → pentane.

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: Saying ketones give Tollens' test
Do this: Only aldehydes (and a few special compounds) reduce Tollens' reagent; ordinary ketones do not.
Usually costs: 1 mark.
Mistake: Using Fehling's test for benzaldehyde
Do this: Aromatic aldehydes do not respond to Fehling's; use Tollens' for them.
Usually costs: 1 mark.
Mistake: Expecting iodoform from propanal
Do this: Propanal is $C H_{3} C H_{2} C H O$ : no $C H_{3} C O -$ group. Only ethanal among aldehydes gives the test.
Usually costs: 1 mark.
Mistake: Giving an alcohol for Clemmensen reduction
Do this: Clemmensen and Wolff–Kishner give hydrocarbons ( $C H_{2}$ ); $N a B H_{4}$ / $L i A l H_{4}$ give alcohols.
Usually costs: 1 mark.

Exam tips

For each 'distinguish' pair, write: reagent, observation with compound 1, observation with compound 2.
Name Fehling A and B and Tollens' reagent correctly if asked how they are prepared.
In the iodoform equation, remember the acid salt has one carbon fewer than the ketone.
Mention the yellow colour of iodoform and its smell as the observation.
For identification puzzles, write the structure, not just the name.

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
Fehling's solution is reduced by:
(a) benzaldehyde (b) propanone (c) ethanal (d) acetophenone
Hint
Aliphatic aldehydes only.
Answer
(c) Ethanal.
Q2 · 2 marks · Very short answerBoard level
Name the products when butanal and butanone are each reduced with sodium borohydride.
Hint
Aldehyde → 1°, ketone → 2°.
Answer
Butanal → butan-1-ol (primary). Butanone → butan-2-ol (secondary).
Q3 · 2 marks · Very short answerHarder board level
0.29 g of propanal (58 g mol⁻¹) is warmed with excess Fehling's solution. Calculate the mass of copper(I) oxide formed. ( $C u_{2} O$ = 143 g mol⁻¹)
Hint
1 mol aldehyde gives 1 mol $C u_{2} O$ .
Answer
Moles of propanal $= 0.29/58 = 0.0050$ ; $C u_{2} O$ formed $= 0.0050$ mol; mass $= 0.0050 \times 143 = 0.715$ g.
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): Ketones are not oxidised by Tollens' reagent.
Reason (R): Oxidation of a ketone requires the breaking of a carbon–carbon bond.
(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
No H on the carbonyl carbon.
Answer
(a).
Q5 · 3 marks · Short answerHarder board level
Write the equation for the iodoform reaction of acetophenone. Name two alcohols that also give a positive iodoform test, and explain why.
Hint
$C H_{3} C H (O H) -$ group.
Answer
$C_{6} H_{5} C O C H_{3} + 3 N a O I \to C_{6} H_{5} C O O N a + C H I_{3} + 2 N a O H$ . Ethanol and propan-2-ol: NaOI oxidises $C H_{3} C H (O H) -$ to $C H_{3} C O -$ (ethanal, propanone), which then gives iodoform.
Q6 · 2 marks · Very short answerHarder board level
When would you choose Wolff–Kishner reduction rather than Clemmensen reduction to convert a carbonyl group into $C H_{2}$ ? Give reagents for both.
Hint
Acid versus base.
Answer
Clemmensen uses Zn–Hg and conc. HCl (acidic); Wolff–Kishner uses $N H_{2} N H_{2}$ then KOH in ethylene glycol (basic). Choose Wolff–Kishner when the molecule contains groups that are damaged by strong acid; choose Clemmensen when it has base-sensitive groups.
Q7 · 3 marks · Short answerHOTS / competency
Give one chemical test each to distinguish: (a) benzaldehyde and ethanal (b) pentan-2-one and pentan-3-one (c) propanal and propan-1-ol.
Hint
Fehling's, iodoform, Tollens'.
Answer
(a) Fehling's test: ethanal gives a red-brown precipitate; benzaldehyde does not (iodoform also works: ethanal positive). (b) Iodoform: pentan-2-one gives yellow $C H I_{3}$ ; pentan-3-one does not. (c) Tollens': propanal gives a silver mirror; propan-1-ol does not.
Q8 · 4 marks · Case-basedHOTS / competency
Read and answer.
Suppose a lab assistant finds three unlabelled bottles known to contain ethanal, propanone and benzaldehyde. She tests a sample from each:
Bottle X: silver mirror with Tollens', red precipitate with Fehling's, yellow precipitate with $I_{2}$ /NaOH.
Bottle Y: no change with Tollens' or Fehling's, yellow precipitate with $I_{2}$ /NaOH.
Bottle Z: silver mirror with Tollens', no change with Fehling's or $I_{2}$ /NaOH.
(a) Identify X, Y and Z. (1 mark)
(b) Write the equation for the reaction of Y with $I_{2}$ /NaOH (as NaOI). (1 mark)
(c) Why does Z fail Fehling's test but pass Tollens'? (1 mark)
(d) Which bottle's contents would give propan-2-ol with $N a B H_{4}$ ? (1 mark)
Hint
Match each test to a structural feature.
Answer
(a) X ethanal, Y propanone, Z benzaldehyde. (b) $C H_{3} C O C H_{3} + 3 N a O I \to C H_{3} C O O N a + C H I_{3} + 2 N a O H$ . (c) Aromatic aldehydes do not reduce Fehling's solution, but they are oxidised by Tollens' reagent. (d) Y (propanone).

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.Reagents for Clemmensen reduction?Show answer
Zinc amalgam and concentrated HCl.
2.What is Tollens' reagent?Show answer
Ammoniacal silver nitrate solution.
3.Precipitate in Fehling's test?Show answer
Reddish-brown copper(I) oxide, $C u_{2} O$ .
4.Which aldehyde gives the iodoform test?Show answer
Ethanal.
5.Product of reducing a ketone with $L i A l H_{4}$ ?Show answer
A secondary alcohol.
6.Do aromatic aldehydes give Fehling's test?Show answer
No.

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

How do you distinguish an aldehyde from a ketone in Class 12?

Use Tollens' reagent (silver mirror with aldehydes) or Fehling's solution (red-brown precipitate with aliphatic aldehydes). Ketones give neither.

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

Both convert C=O into CH_2. Clemmensen uses zinc amalgam with concentrated HCl (acidic); Wolff–Kishner uses hydrazine, then KOH in ethylene glycol (basic).

Which compounds give the iodoform test?

Compounds with a CH_3CO\text{–} group (methyl ketones and ethanal) and alcohols with a CH_3CH(OH)\text{–} group.

Why are aldehydes easily oxidised but ketones are not?

Aldehydes have a hydrogen on the carbonyl carbon that can be removed to form an acid; ketones would have to break a C–C bond.

Why doesn't benzaldehyde give Fehling's test?

Aromatic aldehydes are not oxidised by Fehling's solution under the test conditions, though they do reduce Tollens' reagent.