# Detection of Functional Groups

> Functional group detection for JEE Chemistry: Lassaigne's test, iodoform, Tollens, Fehling, 2,4-DNP, Lucas, Hinsberg and carbylamine tests with colour outcomes.

- Canonical URL: https://prepelephant.com/topics/jee/chemistry/detection-of-functional-groups
- Exam / course: JEE · Subject: Chemistry
- Publisher: PrepElephant (https://prepelephant.com) — Prepared and reviewed by the PrepElephant Academic Review Team
- First published: 2026-10-02
- Last updated: 2026-10-02
- How to cite: "Detection of Functional Groups", PrepElephant, https://prepelephant.com/topics/jee/chemistry/detection-of-functional-groups

## Direct answer

Iodoform's yellow crystals of CHI3 appear only when the system contains a methyl carbonyl, CH3CO-, or the CH3CH(OH)- group that oxidises into one — which is why the test fingers methyl ketones, acetaldehyde, ethanol and secondary alcohols of that shape while ignoring ethanol's homologues. The same specificity-by-mechanism logic runs the whole identification kit: Lassaigne's sodium fusion converts bonded N, S and halogens into NaCN, Na2S and NaX for detection (Prussian blue for nitrogen, violet with sodium nitroprusside for sulphur); Tollens reagent mirrors aldehydes but not ketones; Fehling solution accepts aliphatic aldehydes only; and the amine tier is separated by Hinsberg's reagent, the carbylamine stink test and the azo-dye coupling. Every test is a tiny mechanism statement, which is how the exam frames it.

## What you must remember

- **Lassaigne ledger:** N gives Prussian blue Fe4[Fe(CN)6]3 with ferric chloride; S gives a violet colour with sodium nitroprusside; both together give blood-red ferric thiocyanate; halides precipitate as AgCl (white, soluble in dilute NH4OH), AgBr (pale yellow, soluble in concentrated NH4OH), AgI (yellow, insoluble).
- **Carbonyl split:** 2,4-DNP gives orange-red hydrazones with both aldehydes and ketones (not with carboxylic acids or esters); Tollens silver mirror flags aldehydes and formic acid; Fehling's red Cu2O flags aliphatic aldehydes — benzaldehyde refuses it.
- **Iodoform club:** CH3CO- compounds (acetone, acetophenone), CH3CHO, ethanol and CH3CH(OH)- secondary alcohols — the only primary alcohol in the club is ethanol.
- **Acid fizz:** carboxylic acids liberate CO2 briskly from sodium bicarbonate; phenols do not, separating the two acidities at a stroke.
- **Phenol signatures:** violet colouration with neutral FeCl3 and a white precipitate of 2,4,6-tribromophenol from bromine water.
- **Amine tier tests:** Hinsberg separates cleanly — primary amines form sulphonamides soluble in alkali, secondary form insoluble ones, tertiary refuse to react; carbylamine's foul isocyanide odour flags primary amines only; diazonium coupling with beta-naphthol gives a red dye for aromatic primary amines.
- **Lucas ladder:** anhydrous ZnCl2 in concentrated HCl clouds tertiary alcohols immediately, secondary within about five minutes, primary not at room temperature — reactivity riding on carbocation stability.
- **Unsaturation pair:** bromine in CCl4 decolourises and Baeyer's alkaline KMnO4 turns brown for alkenes and alkynes, with the caveat that sulphur dioxide and other reductants also decolourise permanganate.

## Separating three carbonyl liquids

Three unlabelled bottles hold acetaldehyde, benzaldehyde and acetophenone. One reagent, 2,4-DNP, merely confirms all three carry C=O — orange precipitates across the board. Now Tollens: acetaldehyde and benzaldehyde silver the tube, acetophenone does not, so the ketone is pinned. Finish with Fehling: acetaldehyde's red Cu2O appears, benzaldehyde stays clear — the aromatic aldehyde lacks the reducing power. Three steps, complete separation, and the same matrix solves every "distinguish between" question by building a truth table of outcomes. The exam's craft is picking pairs that need exactly one discriminating test — aniline versus N-methylaniline (carbylamine positive only for the first), phenol versus benzoic acid (FeCl3 versus bicarbonate fizz), propan-2-ol versus propan-1-ol (Lucas at room temperature).

## How the exam frames it

Detection of elements and functional groups sits inside the Purification and Characterisation unit of the current JEE Main syllabus, delivered mostly as match-the-test or identify-the-compound questions — pure reward marks if the colour outcomes are banked. JEE Advanced occasionally asks why a test works, expecting the mechanism sentence: iodoform needs the methyl carbonyl for exhaustive halogenation, Tollens works because aldehydes oxidise to carboxylates while ketones cannot without C-C cleavage. The traps: expecting Fehling to answer for benzaldehyde, forgetting formic acid mirrors an aldehyde in both Tollens and Fehling, and reading a negative bromine-water test as proof of saturation when a phenol may simply have precipitated instead.

## Frequently asked questions

### Which test distinguishes aromatic from aliphatic aldehydes?

Fehling's solution — aliphatic aldehydes reduce the blue copper tartrate complex to red copper(I) oxide, while benzaldehyde and other aromatic aldehydes leave it unchanged.

### Why does formic acid give a positive Tollens test?

Its structure carries an aldehydic hydrogen on the carboxyl carbon, so it oxidises to carbonic species while reducing silver just as an aldehyde does.

### Why is iodoform positive with ethanol but not propanol?

Ethanol oxidises in situ to acetaldehyde, which has the methyl carbonyl motif the exhaustive iodination requires; propanol's oxidation product lacks it.

### How does the Hinsberg test separate the three amine classes?

Primary amines give sulphonamides with an acidic N-H that dissolves in alkali, secondary amines give N,N-disubstituted sulphonamides that cannot, and tertiary amines simply do not react with the reagent.

### What colour signals nitrogen in Lassaigne's test?

Prussian blue — the deep-blue ferric ferrocyanide Fe4[Fe(CN)6]3 precipitate formed when the fused sodium cyanide is first converted to sodium ferrocyanide with ferrous sulphate, then oxidised with ferric chloride.