Organic tests are quick chemical reactions that distinguish one functional group from another. CBSE Class 12 and NEET both test these directly — expect one to two questions a year on tests.
Core Concepts
Tollen’s test
Ag(NH3)2+ reagent. Aldehydes reduce Ag+ to Ag metal, giving a shiny silver mirror on the tube. Ketones do not react. Distinguishes aldehydes from ketones.
The chemistry: the aldehyde is oxidised to a carboxylate, and Ag+ is reduced to metallic silver. The reaction requires a warm water bath and a clean glass tube.
All aldehydes give Tollen’s test — both aliphatic and aromatic. This is different from Fehling’s test, which fails with aromatic aldehydes.
Fehling’s test
Fehling’s A (CuSO4) and B (alkaline tartrate). Aldehydes reduce Cu(II) to Cu(I), giving a red precipitate of Cu2O. Aromatic aldehydes (like benzaldehyde) do not react.
Why does benzaldehyde fail Fehling’s test? The tartrate complex in Fehling’s B is bulky. Benzaldehyde, with its large phenyl group, has steric difficulty entering the coordination sphere. With Tollen’s reagent (which is a smaller complex), benzaldehyde works fine.
Iodoform test
I2 + NaOH with methyl ketones (CH3CO-) or ethanol or compounds with CH3CH(OH)- gives yellow crystals of iodoform (CHI3) with a characteristic smell. Identifies the methyl ketone group.
Compounds that give a positive iodoform test:
- Methyl ketones: acetone, acetophenone (CH3COC6H5)
- Ethanol: CH3CH2OH (oxidised in situ to CH3CHO, then to CH3COOH which has the CH3CO- group)
- Secondary methyl carbinols: CH3CH(OH)R (oxidised in situ to methyl ketone)
- Acetaldehyde: CH3CHO
A classic NEET question: “Which of the following gives iodoform test?” The answer is any compound with a CH3CO- group or a CH3CH(OH)- group. Ethanol gives it; methanol does not.
Lucas test
ZnCl2 + HCl. Tertiary alcohols give immediate turbidity; secondary in 5-10 minutes; primary do not react at room temperature. Distinguishes alcohol classes.
The turbidity is due to formation of the insoluble alkyl chloride. The mechanism is SN1 — tertiary alcohols form the most stable carbocations and react fastest.
| Alcohol Class | Time for Turbidity | Reason |
|---|---|---|
| Tertiary | Immediate | Stable 3° carbocation |
| Secondary | 5-10 minutes | Less stable 2° carbocation |
| Primary | No reaction at RT | 1° carbocation too unstable |
Baeyer’s test
Dilute alkaline KMnO4. Unsaturated compounds (alkenes, alkynes) decolourise the purple solution. Distinguishes saturated from unsaturated hydrocarbons.
The purple KMnO4 is reduced to brown MnO2, causing decolourisation. The alkene is oxidised to a diol. Saturated hydrocarbons do not react, so the purple colour persists.
Test for amines
1° amines give carbylamine (foul smell) with CHCl3 + KOH. 2° amines give no reaction. 3° amines give no reaction. 1° and 2° react with Hinsberg’s reagent differently.
Carbylamine (isocyanide) test:
The foul-smelling isocyanide is unmistakable. This test works only for primary amines.
Hinsberg’s test (benzenesulphonyl chloride, C6H5SO2Cl):
- 1° amine → sulphonamide soluble in alkali (acidic NH, dissolves in NaOH)
- 2° amine → sulphonamide insoluble in alkali (no acidic H)
- 3° amine → no reaction with Hinsberg’s reagent
Test for carboxylic acids
Sodium bicarbonate test: carboxylic acids produce brisk effervescence (CO2) with NaHCO3. Phenols do not react with NaHCO3 (too weak an acid). This distinguishes carboxylic acids from phenols.
Master Reference Table
| Test | Reagent | Positive Result | Identifies |
|---|---|---|---|
| Tollen’s | Ag(NH3)2+ | Silver mirror | Aldehydes (all) |
| Fehling’s | CuSO4 + alkaline tartrate | Red precipitate (Cu2O) | Aldehydes (not aromatic) |
| Iodoform | I2 + NaOH | Yellow crystals (CHI3) | CH3CO- or CH3CH(OH)- |
| Lucas | ZnCl2 + conc. HCl | Turbidity (time varies) | Alcohol class (1°, 2°, 3°) |
| Baeyer’s | Dilute KMnO4 | Decolourisation | Unsaturation (C=C, C≡C) |
| Carbylamine | CHCl3 + KOH | Foul smell | Primary amines |
| NaHCO3 | Sodium bicarbonate | Effervescence (CO2) | Carboxylic acids |
| FeCl3 | Neutral FeCl3 | Violet colour | Phenols |
Worked Examples
Tollen’s test — acetaldehyde gives silver mirror, acetone does not. Or iodoform — both give positive test (acetone has CH3CO-), so iodoform alone is not enough.
Iodoform test — ethanol gives yellow precipitate (CH3CH(OH)- group), methanol does not. Definitive test.
FeCl3 test — phenol gives violet colour, ethanol gives no colour. Or NaOH test — phenol dissolves in NaOH (forms sodium phenoxide), ethanol does not react.
Both are carboxylic acids. But formic acid (HCOOH) has an aldehyde-like H-CO group. It gives a positive Tollen’s test (silver mirror). Acetic acid does not. Formic acid also reduces Fehling’s solution.
You have three unlabelled bottles containing ethanol, ethanal, and ethanoic acid. How do you identify each?
Test all three with NaHCO3 → ethanoic acid gives effervescence (CO2). The other two do not react.
Test remaining two with Tollen’s reagent → ethanal gives silver mirror. Ethanol does not react.
The last bottle is ethanol by elimination. Three tests, three identifications.
Common Mistakes
Saying benzaldehyde gives Fehling’s test. It does not — aromatic aldehydes fail Fehling’s.
Confusing Tollen’s and Fehling’s. Tollen’s gives silver mirror; Fehling’s gives red precipitate.
Writing that iodoform is given only by methyl ketones. Ethanol and CH3CH(OH)- compounds also give it.
Saying the Lucas test works for all alcohols. It works best for small-molecule alcohols. Large alcohols may be insoluble in the reagent, giving misleading results.
Forgetting that formic acid gives Tollen’s test. It is the only carboxylic acid that does, because it has an aldehyde-like structure (H-COOH has H directly bonded to C=O).
Exam Weightage and Revision
Distinguishing tests carry 1-2 questions in NEET every year and 3-5 marks in CBSE Class 12. The questions follow a predictable pattern: “How will you distinguish between X and Y?” or “Which of the following gives a positive test with reagent Z?”
| Test | NEET Frequency | Typical Question Format |
|---|---|---|
| Tollen’s/Fehling’s | Every year | Distinguish aldehyde from ketone |
| Iodoform | Most years | Which compound gives iodoform? |
| Lucas | Every 2 years | Arrange alcohols by Lucas test reactivity |
| NaHCO3 | Occasional | Distinguish acid from phenol |
| Carbylamine | Occasional | Identify primary amine |
The single highest-yield preparation is the master reference table above. Memorise it — every row is a potential NEET question.
Practice Questions
Q1. An organic compound gives a positive iodoform test and a positive Tollen’s test. What could it be?
Positive iodoform → has CH3CO- or CH3CH(OH)- group. Positive Tollen’s → is an aldehyde. The compound that satisfies both is acetaldehyde (CH3CHO). It has the CH3CO- group (iodoform positive) and is an aldehyde (Tollen’s positive).
Q2. How would you distinguish between 1-propanol, 2-propanol, and 2-methyl-2-propanol using one reagent?
Use Lucas reagent (ZnCl2 + conc. HCl). 2-Methyl-2-propanol (tertiary) → immediate turbidity. 2-Propanol (secondary) → turbidity in 5-10 minutes. 1-Propanol (primary) → no turbidity at room temperature. One reagent, three different responses.
Q3. Compound A (C2H4O2) gives effervescence with NaHCO3 and a silver mirror with Tollen’s reagent. Identify A.
Effervescence with NaHCO3 → carboxylic acid. Silver mirror with Tollen’s → aldehyde-like group. C2H4O2 with both properties = formic acid (HCOOH). It is the only carboxylic acid that gives Tollen’s test because of its H-C=O group.
Q4. Why does acetophenone give a positive iodoform test but benzophenone does not?
Acetophenone (C6H5COCH3) has a CH3CO- group — the methyl group adjacent to the carbonyl is required for iodoform. Benzophenone (C6H5COC6H5) has two phenyl groups on either side of the carbonyl — no methyl group, no iodoform.
FAQs
Can we use these tests in any order?
Strategy matters. Always start with the test that eliminates the most possibilities. NaHCO3 test first separates acids from non-acids. Then Tollen’s separates aldehydes from the rest. Plan your testing sequence before starting.
Why are there multiple tests for the same functional group?
Different tests have different selectivities and sensitivities. Tollen’s works for all aldehydes; Fehling’s fails for aromatic aldehydes. Having multiple tests lets you cross-verify and handle edge cases.
Are these tests used in modern chemistry?
Classical tests are still taught and used in teaching labs. Modern analytical labs use NMR, IR, and mass spectrometry instead — they are faster, more informative, and need smaller samples. But the logic of distinguishing tests remains the same.
Make a table of tests and what they identify. Six rows. NEET picks from this table almost every year.
Systematic Approach to “Identify the Unknown” Problems
NEET and CBSE sometimes give problems where you need to identify an unknown compound using tests. Here is a systematic flowchart:
Step 1: Litmus or pH test. Turns blue litmus red → acidic (carboxylic acid or mineral acid). Turns red litmus blue → basic (amine).
Step 2: NaHCO3 test. Effervescence (CO2) → carboxylic acid. No effervescence → not a carboxylic acid.
Step 3: 2,4-DNP test. Yellow/orange precipitate → carbonyl (aldehyde or ketone). No precipitate → no C=O.
Step 4: Tollen’s test. Silver mirror → aldehyde. No mirror → ketone (if 2,4-DNP was positive) or other compound.
Step 5: Iodoform test. Yellow precipitate → CH3CO- group (methyl ketone) or CH3CH(OH)- group (secondary methyl carbinol) or ethanol.
Step 6: FeCl3 test. Violet colour → phenol. Other colours possible for enols and hydroxamic acids.
Step 7: Lucas test. For alcohols — immediate turbidity (3°), delayed (2°), no reaction (1°).
This sequence tests from most to least common. Most exam problems can be solved in 3-4 steps.
Why Chemistry Labs Still Teach These Tests
In modern research labs, NMR and mass spectrometry have replaced classical wet tests. But the thinking behind the tests — “what functional group reacts with this reagent, and what would I observe?” — is the same logic used when interpreting spectra. The tests also reinforce reaction chemistry: Tollen’s test is a redox reaction, Lucas test is an SN1 reaction, and iodoform is a halogenation-oxidation sequence. Understanding the test means understanding the mechanism.
Phenol vs Enol — A Subtle Distinction
Both phenols and enols have -OH groups. Both give colours with FeCl3. But phenols are stable (the -OH is on a stable aromatic ring), while enols are usually unstable and exist in equilibrium with their keto form (keto-enol tautomerism). A positive FeCl3 test in a carbonyl compound means significant enol content — this happens with beta-diketones like acetylacetone and ethyl acetoacetate.
Tests are the practical layer under organic theory. They are memorisable and high-yield.