Question
Given an unknown organic compound, how do we identify its functional group using chemical tests? Summarise the key tests for alcohols, aldehydes, ketones, carboxylic acids, and amines.
Solution — Step by Step
Tollens test (silver mirror test): Add ammoniacal AgNO3. Aldehydes reduce to metallic silver, forming a mirror on the test tube wall.
Fehling test: Add Fehling solution (Cu2+ in alkaline tartrate). Aldehydes give a red precipitate of Cu2O.
Ketones do NOT respond to either test (they cannot be easily oxidised).
Since both are carbonyl compounds, use:
- Tollens/Fehling: positive = aldehyde, negative = ketone
- 2,4-DNP test: Both aldehydes AND ketones give a yellow-orange precipitate with 2,4-dinitrophenylhydrazine. This confirms the C=O group but does not distinguish between them.
- Iodoform test: Methyl ketones () and acetaldehyde () give a yellow precipitate of CHI3 with I2/NaOH.
Carboxylic acids: Add NaHCO3 solution — brisk effervescence (CO2 gas) confirms -COOH group.
Alcohols: No effervescence with NaHCO3. Confirm with:
- Lucas test (for classifying 1-degree/2-degree/3-degree): ZnCl2/conc. HCl. 3-degree gives immediate turbidity, 2-degree takes 5 minutes, 1-degree shows no reaction at room temperature.
- Ceric ammonium nitrate test: gives a red colour with alcohols.
Carbylamine test (isocyanide test): Heat with CHCl3 + alcoholic KOH. Primary amines give a foul-smelling isocyanide.
Hinsberg test: React with benzenesulphonyl chloride (C6H5SO2Cl).
- 1-degree amine: soluble product (dissolves in NaOH)
- 2-degree amine: insoluble product
- 3-degree amine: no reaction
graph TD
A[Unknown compound] --> B{Effervescence with NaHCO3?}
B -->|Yes| C[Carboxylic acid]
B -->|No| D{2,4-DNP test positive?}
D -->|Yes: C=O present| E{Tollens test?}
D -->|No| F{Lucas test / Carbylamine test}
E -->|Silver mirror| G[Aldehyde]
E -->|No reaction| H[Ketone]
F -->|Turbidity with Lucas| I[Alcohol]
F -->|Foul smell with CHCl3/KOH| J[Primary amine]Why This Works
Each functional group has a characteristic reactivity that we are exploiting:
- Aldehydes are easily oxidised (reducing agents) — they reduce Ag+ and Cu2+
- Carboxylic acids are strong enough acids to decompose NaHCO3
- Alcohols undergo nucleophilic substitution with Lucas reagent (rate depends on carbocation stability)
- Primary amines have a lone pair on N that reacts with CHCl3 to form isocyanide
The 2,4-DNP test is a “screening test” for the carbonyl group. Once you confirm C=O is present, Tollens/Fehling narrows it down to aldehyde or ketone.
Alternative Method
A quick “which test for which group” reference:
| Functional Group | Test | Positive Result |
|---|---|---|
| Aldehyde | Tollens | Silver mirror |
| Aldehyde | Fehling | Red ppt (Cu2O) |
| Ketone (methyl) | Iodoform | Yellow ppt (CHI3) |
| Any C=O | 2,4-DNP | Orange-yellow ppt |
| -COOH | NaHCO3 | Effervescence (CO2) |
| -OH (alcohol) | Lucas | Turbidity (3-degree instant) |
| 1-degree amine | Carbylamine | Foul smell |
| 1-degree/2-degree/3-degree amine | Hinsberg | Soluble/Insoluble/No reaction |
Common Mistake
Students often forget that the iodoform test is positive for acetaldehyde () too — not just methyl ketones. Also, secondary alcohols with the group (like 2-propanol) give a positive iodoform test because they are first oxidised to methyl ketones in alkaline I2 conditions. This is a classic JEE Main trap.