Part 3 · Chapter 34

Practical Organic Chemistry

The detective work of the laboratory — a systematic set of colour, smell and precipitate tests that reveal an unknown compound's functional group and identity

Fundamentals of Chemistry Prof. Mithun Mondal Reading time ≈ 55 min
i What you'll learn
  • How a preliminary examination — state, colour, odour and the ignition test — narrows down an unknown.
  • How to detect unsaturation with bromine water and Baeyer's reagent.
  • The laboratory tests for alcohols and phenols, including the Lucas test.
  • The tests for aldehydes and ketones — 2,4-DNP, Tollens', Fehling's and the iodoform test.
  • The tests for carboxylic acids and amines (carbylamine, Hinsberg).
  • The classic distinguishing tests and a systematic scheme for naming an unknown.
Section 34-1

The Detective's Method

The whole of organic chemistry now comes together at the bench. Faced with an unknown compound, the chemist works like a detective: gather clues from how it looks and burns, find which elements it holds (Chapter 33), then run targeted tests for each functional group. Every test in this chapter is simply a reaction you already know — oxidation, addition, acid–base — chosen because it produces an unmistakable colour, smell, gas or precipitate.

Section 34-2

Preliminary Examination

Before any reagent, simple observation pays. Physical state and colour hint at the class; a characteristic odour (the fruity smell of an ester, the sharp smell of an acid, the fishy smell of an amine) is a strong clue. The ignition (flame) test is especially useful.

clean blue flame aliphatic / saturated sooty luminous flame aromatic / unsaturated
The ignition test — a sooty flame betrays an aromatic ring
Why the flame tells tales. A high carbon-to-hydrogen ratio — found in aromatic rings and unsaturated chains — burns with incomplete combustion, throwing out unburnt carbon as black soot in a luminous yellow flame. Saturated aliphatic compounds, richer in hydrogen, burn cleanly blue. One match narrows the field before a single reagent is opened.
Section 34-3

Detecting Unsaturation

A carbon–carbon double or triple bond is revealed by two reagents that it consumes.

ReagentObservationReaction type
Bromine in \(\ce{CCl4}\) (or bromine water)reddish-brown colour disappearsaddition across the multiple bond
Baeyer's reagent (cold dilute alkaline \(\ce{KMnO4}\))purple fades; brown \(\ce{MnO2}\) appearsoxidation to a vicinal diol
A caution. Baeyer's test is decolourised by anything readily oxidised — aldehydes, formic acid, phenols — not just by double bonds. Confirm unsaturation with bromine and read the preliminary clues together; no single test is proof on its own.
Section 34-4

Alcohols & Phenols

Both carry an \(\ce{-OH}\) group, but their chemistry differs sharply: phenols are weakly acidic and give strong colours with iron(III), while alcohols do not.

GroupTestPositive result
Alcoholceric ammonium nitrateyellow reagent turns red/wine
Alcoholester test (acid + conc. \(\ce{H2SO4}\))fruity-smelling ester
1°/2°/3° alcoholLucas reagent (conc. \(\ce{HCl}\)+\(\ce{ZnCl2}\))turbidity: 3° at once, 2° in ~5 min, 1° none (cold)
Phenolneutral \(\ce{FeCl3}\)violet / blue / green colour
Phenolbromine waterwhite precipitate (2,4,6-tribromophenol)
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The Lucas test logic
turbidity speed follows carbocation stability: 3° > 2° > 1°

The test forms an insoluble alkyl chloride (the turbidity) by an \(S_N1\) route, whose rate tracks carbocation stability. A tertiary alcohol reacts instantly, a secondary one in a few minutes, and a primary one not at all in the cold — so the time to cloudiness names the class.

Section 34-5

Aldehydes & Ketones

The carbonyl group \(\ce{>C=O}\) is found first, then aldehyde is told from ketone by tests that exploit the aldehyde's easy oxidation.

TestPositive resultDetects
2,4-DNP (Brady's reagent)orange / yellow precipitateany carbonyl (aldehyde & ketone)
Tollens' reagentbright silver mirroraldehydes only
Fehling's solutionbrick-red \(\ce{Cu2O}\) precipitatealiphatic aldehydes
Schiff's reagentpink/magenta colouraldehydes
Iodoform (\(\ce{I2}\)+\(\ce{NaOH}\))yellow \(\ce{CHI3}\) precipitatemethyl ketones & \(\ce{CH3CH(OH)-}\)
The iodoform test reaches beyond carbonyls. A yellow precipitate of iodoform appears not only for methyl ketones (\(\ce{CH3CO-}\)) but also for ethanal and for any alcohol with the \(\ce{CH3CH(OH)-}\) unit — including ethanol and propan-2-ol — because these are first oxidised to the necessary methyl-carbonyl by the reagent. So it is a test for that structural fragment, not for a single functional group.
Section 34-6

Carboxylic Acids

Carboxylic acids are the strongest common organic acids and announce themselves clearly: they turn blue litmus red and, decisively, liberate carbon dioxide from sodium hydrogen carbonate.

The bicarbonate test
\[ \ce{RCOOH + NaHCO3 -> RCOONa + H2O + CO2 ^} \]
Bicarbonate separates acid from phenol. Both phenols and carboxylic acids dissolve in sodium hydroxide, so \(\ce{NaOH}\) cannot tell them apart. But only the stronger carboxylic acid is acidic enough to release \(\ce{CO2}\) from the weaker \(\ce{HCO3-}\); a phenol gives no effervescence. The brisk fizz is the cleanest single distinction between the two.
Section 34-7

Amines

Amines are basic and often fishy-smelling. The most useful bench tests both identify an amine and tell its class.

TestResultIdentifies
Carbylamine (\(\ce{CHCl3}\)+alc. \(\ce{KOH}\))extremely foul (isocyanide) smell1° amines only (aliphatic & aromatic)
Diazotisation + couplingorange-red azo dyearomatic 1° amines
Hinsberg's (benzenesulphonyl chloride)see belowdistinguishes 1°, 2°, 3°
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Hinsberg's test logic
the number of N–H bonds decides the outcome

A amine forms a sulphonamide with one acidic N–H, so it dissolves in \(\ce{KOH}\). A amine forms a sulphonamide with no N–H, which stays insoluble in \(\ce{KOH}\). A amine has no N–H to react and gives no product at all. Solubility in alkali therefore reads off the class.

Section 34-8

Classic Distinguishing Tests

Examiners love the question "how would you distinguish A from B?" The trick is to pick one reagent that gives a clear positive with one and nothing with the other.

DistinguishReagentWho responds
Ethanol vs phenolneutral \(\ce{FeCl3}\)phenol → violet; ethanol → none
Phenol vs benzoic acid\(\ce{NaHCO3}\)benzoic acid → \(\ce{CO2}\); phenol → none
Aldehyde vs ketoneTollens' / Fehling'saldehyde → mirror / red ppt
Ethanal vs propanaliodoformethanal → yellow ppt
1° vs 2° vs 3° alcoholLucas reagentturbidity timing differs
Formic vs acetic acidTollens' reagentformic acid → silver mirror (it reduces)
Section 34-9

A Systematic Scheme

Put together, the tests follow a logical order — broad clues first, specific confirmation last — so that each step narrows the possibilities.

1 · observe state, colour, odour 2 · ignition (flame) test 3 · detect elements (Lassaigne) 4 · find the functional group 5 · confirmatory test → identity
The systematic scheme — from observation to identity
The closing idea of the book. Every test here is just a reaction from an earlier chapter put to a diagnostic use: oxidation gives Tollens' its mirror, acid–base chemistry gives the bicarbonate its fizz, carbocation stability gives Lucas its timing. Practical organic chemistry is not a new subject — it is everything you have learned, asked to introduce itself.
Worked Examples

Putting It to Work

1 Read the flame

Problem. An unknown burns with a sooty, luminous yellow flame. What does this suggest?

Solution. A high C:H ratio burns incompletely:

Working
\[ \text{sooty flame} \Rightarrow \textbf{aromatic or highly unsaturated} \]
2 Alcohol or phenol?

Problem. A liquid with an \(\ce{-OH}\) group gives a violet colour with neutral \(\ce{FeCl3}\). Which is it?

Solution. Only phenols give the iron(III) colour:

Working
\[ \text{violet with } \ce{FeCl3} \Rightarrow \textbf{phenol} \]
3 Aldehyde vs ketone

Problem. Two carbonyl compounds both give a 2,4-DNP precipitate. One also gives a silver mirror. Identify each.

Solution. Tollens' responds only to the aldehyde:

Working
\[ \text{silver mirror} \Rightarrow \textbf{aldehyde};\quad \text{no mirror} \Rightarrow \textbf{ketone} \]
4 Phenol vs benzoic acid

Problem. Both dissolve in \(\ce{NaOH}\). One reagent tells them apart — which, and how?

Solution. Only the stronger acid frees \(\ce{CO2}\) from bicarbonate:

Working
\[ \ce{NaHCO3}:\ \text{benzoic acid} \to \ce{CO2 ^};\ \text{phenol} \to \text{no fizz} \]
5 The iodoform clue

Problem. Which of ethanol and propan-1-ol gives a positive iodoform test, and why?

Solution. Only ethanol carries the \(\ce{CH3CH(OH)-}\) unit:

Working
\[ \ce{CH3CH2OH} \to \textbf{yellow CHI3};\quad \ce{CH3CH2CH2OH} \to \text{none} \]
6 Name the amine class

Problem. In Hinsberg's test an amine forms a product insoluble in \(\ce{KOH}\). Which class is it?

Solution. An insoluble sulphonamide has no acidic N–H:

Working
\[ \text{insoluble in } \ce{KOH} \Rightarrow \textbf{secondary amine} \]
Review

Chapter Summary

Preliminary tests

State, colour, odour and the ignition test (sooty = aromatic) narrow the field first.

Unsaturation

Bromine water and Baeyer's reagent are decolourised by C=C / C≡C bonds.

–OH groups

Phenol → violet with FeCl₃; alcohols → Lucas timing gives 1°/2°/3°.

Carbonyls

2,4-DNP finds the carbonyl; Tollens'/Fehling's pick out the aldehyde; iodoform the CH₃CO– unit.

Acids & amines

NaHCO₃ fizz = acid (not phenol); carbylamine = 1° amine; Hinsberg names the class.

The scheme

Observe → ignite → detect elements → find the group → confirm the identity.

Practice

Problems

For each item, name the reagent, the expected observation, and the reaction behind it. Difficulty rises down the list.

  1. What does a sooty flame indicate, and why does it occur?
  2. Name two tests for unsaturation and the change observed in each.
  3. How would you distinguish ethanol from phenol?
  4. Explain the Lucas test and how it tells 1°, 2° and 3° alcohols apart.
  5. Which test detects all carbonyls, and which singles out aldehydes?
  6. Explain the iodoform test and list two compounds that give it positively.
  7. How do you distinguish a carboxylic acid from a phenol in one test?
  8. Describe the carbylamine test and state which amines respond.
  9. Explain Hinsberg's test and the result for each class of amine.
  10. How would you distinguish formic acid from acetic acid?
  11. An unknown is neutral, gives a silver mirror and a yellow iodoform precipitate. Suggest its identity.
  12. Outline the systematic scheme for identifying an unknown organic compound.
Tip: read every test as a known reaction wearing a diagnostic disguise. The silver mirror is just an aldehyde being oxidised; the bicarbonate fizz is just an acid–base reaction; the Lucas timing is just carbocation stability. Ask "what reaction is this reagent provoking, and what visible thing does it make?" — and the unknown gives up its name. With that, you have travelled from the mole in Chapter 1 to identifying a compound with your own hands. The course is complete.