Class 12 Aldehydes and Ketones — The Tests That Identify Them
"Distinguish between the following pairs of compounds" is one of the most predictable question types in the Class 12 organic paper, and this chapter supplies most of the answers. The difficulty is that the tests overlap in confusing ways: one test detects both aldehydes and ketones, another detects only aldehydes, a third detects a particular group that can sit in either. This article sorts them out, gives a balanced equation for each, and — most usefully — explains with oxidation numbers why aldehydes and ketones part company.
The two structures, and the one difference that matters
Ketone: R–CO–R′ (carbonyl carbon carries two carbon groups)
- R, R′ — alkyl or aryl groups. In methanal (HCHO) both positions on the carbonyl carbon are hydrogen.
- The carbonyl group >C=O is polar: the carbon is δ+ and the oxygen δ−. That polarity drives every addition reaction in this chapter.
- The hydrogen on the aldehyde carbon is the whole story for the oxidation tests. A ketone has no such hydrogen, so it cannot be oxidised without breaking a C–C bond.
The oxidation-number argument
This is the cleanest way to justify the difference in an answer, and it takes three lines. Assign oxidation numbers to the carbonyl carbon using the usual rules: a bond to a more electronegative atom counts −1 for that atom's partner, a bond to hydrogen gives the carbon −1, a C–C bond counts zero.
| Compound | Bonds on the carbonyl carbon | Oxidation number of that carbon |
|---|---|---|
| Ethanal, CH₃CHO | =O (+2), one H (−1), one C (0) | +1 |
| Propanone, CH₃COCH₃ | =O (+2), two C (0, 0) | +2 |
| Ethanoic acid, CH₃COOH | =O (+2), –OH (+1), one C (0) | +3 |
Read the table as a journey. An aldehyde carbon sits at +1 and can climb to +3 simply by swapping its hydrogen for an –OH — a two-electron oxidation that a mild reagent can supply. A ketone carbon is already at +2 with no hydrogen left to give, so the only way up is to break a carbon–carbon bond, which needs harsh conditions. That single fact explains why Tollens' and Fehling's reagents work on aldehydes and ignore ketones.
The five tests, in one table
| Test | Reagent | Positive result | What it actually detects |
|---|---|---|---|
| 2,4-DNP (Brady's) | 2,4-dinitrophenylhydrazine | Orange to red precipitate | Any carbonyl — aldehyde or ketone |
| Tollens' | Ammoniacal AgNO₃, [Ag(NH₃)₂]⁺ | Silver mirror on the tube wall | Aldehydes (aliphatic and aromatic) |
| Fehling's | Alkaline Cu²⁺–tartrate complex | Red-brown precipitate of Cu₂O | Aliphatic aldehydes only |
| Iodoform | I₂ + NaOH (or NaOI) | Yellow precipitate of CHI₃ | CH₃CO– group, or CH₃CH(OH)– group |
| Schiff's | Decolourised fuchsin | Pink colour returns | Aldehydes (ketones do not, or only very slowly) |
Notice what each row is really claiming. Only Tollens', Fehling's and Schiff's separate aldehyde from ketone. 2,4-DNP proves a carbonyl exists but says nothing about which kind. The iodoform test does not test for "aldehyde" at all — it tests for a methyl group attached to a carbonyl (or to a CH–OH that gets oxidised to one).
The equations, balanced and checked
Tollens' test (silver mirror).
Count: left has 2 C, 18 H, 3 O, 4 N, 2 Ag. Right has 2 C (in the ammonium ethanoate), 7 + 9 + 2 = 18 H, 2 + 1 = 3 O, 1 + 3 = 4 N, 2 Ag. It balances.
Fehling's test.
Count: left 2 C, 4 + 5 = 9 H, 1 + 5 = 6 O, 2 Cu, net charge +4 − 5 = −1. Right 2 C, 3 + 6 = 9 H, 2 + 1 + 3 = 6 O, 2 Cu, net charge −1. Balanced in atoms and charge — check both when an equation contains ions.
Iodoform test. For ethanal and for propanone:
CH₃COCH₃ + 3 I₂ + 4 NaOH → CHI₃↓ + CH₃COONa + 3 NaI + 3 H₂O
Check the second: left 3 C, 6 + 4 = 10 H, 1 + 4 = 5 O, 6 I, 4 Na. Right 1 + 2 = 3 C, 1 + 3 + 6 = 10 H, 2 + 3 = 5 O, 3 + 3 = 6 I, 1 + 3 = 4 Na. Balanced.
Both give iodoform, which is exactly why the test cannot tell an aldehyde from a ketone. What it can do is tell ethanal from propanal: propanal (CH₃CH₂CHO) has no CH₃CO– group, so it is negative.
Worked example 1 — how much silver is in a silver mirror?
Question. 2.20 g of ethanal is warmed with excess Tollens' reagent. Calculate the maximum mass of silver deposited.
Step 1 — the ratio. From the balanced equation, 1 mol of ethanal gives 2 mol of Ag.
Step 2 — molar mass of ethanal (C 12.011, H 1.008, O 15.999):
2(12.011) + 4(1.008) + 15.999 = 24.022 + 4.032 + 15.999 = 44.05 g/mol
Step 3 — moles.
n(CH₃CHO) = 2.20 ÷ 44.05 = 0.0499 mol
n(Ag) = 2 × 0.0499 = 0.0999 mol
Step 4 — mass of silver (Ag = 107.868 g/mol):
0.0999 × 107.868 = 10.77 g
Reality check: a real test tube uses milligrams, not grams — this is a paper calculation to test the 1 : 2 ratio, and the ratio is the mark.
Worked example 2 — mass of iodoform
Question. 0.58 g of propanone gives a positive iodoform test. What is the theoretical mass of CHI₃?
Step 1 — the ratio. 1 mol propanone → 1 mol CHI₃.
Step 2 — molar masses (I = 126.904 g/mol):
M(CH₃COCH₃) = 3(12.011) + 6(1.008) + 15.999 = 36.033 + 6.048 + 15.999 = 58.08 g/mol
M(CHI₃) = 12.011 + 1.008 + 3(126.904) = 12.011 + 1.008 + 380.712 = 393.73 g/mol
Step 3 — moles and mass.
n = 0.58 ÷ 58.08 = 0.00999 mol
mass = 0.00999 × 393.73 = 3.93 g
The product weighs nearly seven times the starting material — that is what putting three iodine atoms on one carbon does, and it is why the yellow precipitate is so easy to see.
Worked example 3 — separating four look-alike compounds
Question. Four bottles contain propanal, propanone, benzaldehyde and propan-1-ol. Give a scheme to identify each.
Test 1 — 2,4-DNP. Propan-1-ol gives no precipitate; the other three do. Propan-1-ol is identified.
Test 2 — Tollens' on the remaining three. Propanal and benzaldehyde give the silver mirror; propanone does not. Propanone is identified.
Test 3 — Fehling's on the last two. Propanal gives the red-brown Cu₂O precipitate; benzaldehyde does not. Both are now identified.
Confirm with the iodoform test. Propanone is positive; propanal, benzaldehyde and propan-1-ol are all negative. (Careful — ethanal and propan-2-ol would both have been positive, which is why iodoform is a confirmation here and not the first step.)
The logic to carry into the exam: start with the test that splits the set most evenly, then narrow down. Writing four unrelated tests earns fewer marks than a scheme that eliminates in order.
Laboratory cautions
These are genuine hazards, and they are fair viva questions.
- Tollens' reagent must be prepared fresh and never stored. On standing it can deposit silver nitride and related solids that are shock-sensitive and explosive. Wash the tube out immediately after the test.
- Fehling's solution is prepared by mixing A and B only when needed — the copper(II) tartrate complex does not keep well.
- 2,4-dinitrophenylhydrazine is kept damp because the dry solid is shock-sensitive. Handle it under a teacher's supervision only.
- Iodoform has a strong, persistent smell; do the test in a well-ventilated space.
Common mistakes that cost marks
- Saying benzaldehyde gives a positive Fehling's test. It does not. Aromatic aldehydes fail Fehling's but pass Tollens' — this exact pair is a favourite question.
- Calling the iodoform test an aldehyde test. It detects the CH₃CO– group. Propanone (a ketone), ethanol and propan-2-ol (alcohols) are all positive; propanal is negative.
- Using 2,4-DNP to distinguish aldehyde from ketone. Both give the orange precipitate. It only proves a carbonyl group is present.
- Forgetting that methanoic acid (HCOOH) reduces Tollens' reagent. It has an H attached to the carbonyl carbon, so it behaves like an aldehyde — the one carboxylic acid that does.
- Writing "Ag₂O" as the silver product. The visible mirror is metallic silver, Ag.
- Leaving equations unbalanced in charge. The Fehling's equation is written in ions; the charges must match on both sides, not just the atoms.
- Assuming a ketone cannot be oxidised at all. It can, under vigorous conditions, but the C–C bond breaks and you get a mixture of shorter acids — which is why it is useless as a test.
Why this chapter matters
| Where it appears | What is actually tested |
|---|---|
| CBSE / ICSE Class 12 organic unit | "Distinguish between" pairs, reasons for reactivity, named reactions |
| Practical examination | Functional group identification in an unknown organic sample |
| Viva | Why Tollens' is made fresh, why benzaldehyde fails Fehling's |
| NEET / JEE foundation | Aldol and Cannizzaro conditions, nucleophilic addition order |
| IIT-JAM / CUET-PG later on | Mechanism of addition, protecting groups, selective oxidation |
Question patterns are revised from session to session, so check your own board's current syllabus document rather than assuming last year's pattern repeats.
Want to see the oxidation-number argument for yourself? Feed CH3CHO, CH3COCH3 and CH3COOH into the Oxidation Number calculator and watch the carbonyl carbon move from +1 to +2 to +3. Once you can generate those numbers yourself, "why do aldehydes reduce Tollens' reagent but ketones do not" becomes a question you can answer from first principles.
Open the Oxidation Number Calculator →Struggling to keep the organic tests straight before the practical exam? ABC Chemistry teaches Class 11–12 chemistry at the Gurugram centre and runs online classes for students across India — details at abcchemistry.in.