2 7 .

8 | Practical Organic Chemistr y

(iii) Beilstein test: The organic compound is heated on a clean copper wire in Bunsen flame. A green coloured
flash indicates chlorine, blue-green of bromine and blue of iodine. Fluorine is not detected in this test.
Exception: Urea (NH2CONH2) and thiourea (NH2CSNH2) do not contain halogens but still give this test.

(d) Test for phosphorous: The organic compound is heated with an oxidizing agent (sodium peroxide).
Phosphorous is oxidized to phosphate. The solution is then boiled with conc. HNO3 and then treated with
ammonium molybdate. A canary yellow ppt. confirms the presence of phosphorous.
Na3PO 4 + 3HNO3 → H3PO 4 + 3NaNO3

H3PO 4 + 12(NH4 )2 MoO 4 + 21HNO3 → (NH4 )3 PO 4 .12MoO3 + 21NH4NO3 + 12H2O

Ammonium Ammonium phospho
molybdate molybdate

Exception: If the organic compound contains arsenic (As), then it also gives the same test forming ammonium
arsenomolybdate forming a canary yellow ppt.

2.3 Detection of Organic Groups

(a) Hydroxyl group (alcohol and phenol)

1. Lucas’s reagent (ZnCl2; conc. HCl): This reagent converts alcohols into the corresponding alkyl chlorides.
Zinc chloride (a Lewis acid) increases the reactivity of alcohols towards acid. The test depends on the rate of
reaction of primary, secondary, and tertiary alcohols with the reagent at room temperature.
(i) RCH2OH → No reaction at room temperature.
(ii) R 2CHOH → R 2CHCl + H2O (1 hour or maybe longer)
(iii) R3COH → R3CCl + H2O (immediately)
To 1 mL of the alcohol in a small test tube add 6 mL of Lucas’ reagent at room temperature. Close the
tube with a cork, shake and allow to stand.
•• Primary alcohols - The aqueous phase remains clear (except allyl alcohol - droplets after 7 minutes).
No turbidity or cloudiness is observed.
•• Secondary alcohols - Very slow reaction (~ 1 hour or maybe longer) when droplets of alkyl chloride
may be seen. Turbidity or cloudiness is observed.
•• Tertiary alcohols - Very fast reaction and droplets of the alkyl chloride formed almost immediately.
Turbidity or cloudiness is observed.

2. Victor meyer’s test: This test consists of the following steps:

(i) The given alcohol is first converted into its alkyl iodide by treating it with P & Iodine.
(ii) The alkyl iodide is then treated with silver nitrite (AgNO2) to convert it into a corresponding nitro alkane.
(iii) The nitroalkane is then treated with nitrous acid (HNO2), i.e., NaNO2 + HCl.
(iv) The resulting solution is finally made alkaline with aqueous NaOH or KOH. If blood red colouration
appears it is a primary alcohol. If blue colouration appears it is a secondary alcohol. If the solution
remains colourless, indicates a tertiary alcohol.
RCH2OH + P / I2 → RCH2 − I + AgNO2 → RCH2NO2 + HONO →

RC( = NOH) − NO2 + NaOH → Blood Red colour
R 2CH − OH + P / I2 → R 2CH − I + AgNO2 → R 2CHNO2 + HONO →

R 2C(N =O) − NO2 + NaOH → Blue colour
R C − OH + P / I2 → R3C − I + AgNO2 → R3C − NO2 + HONO →
3
No Reaction + NaOH → Colourless
 Chemi str y | 27.9

 
 
3. Iodoform test: Iodoform test is given by alcohols having  CH3 − C − OH  group. In Iodoform test, the
 
|
 H 
compound reacts with I2 / NaOH reagent to give iodoform (CHl3) which is pale yellow in colour. It is insoluble
in water and has an antiseptic smell.

(b) Phenol detection

(i) Bromine water: Phenols are generally highly reactive towards electrophilic reagents and are readily
brominated by bromine water. e.g.
OH OH
Br Br
+ 3Br₂

Br
Dissolve or suspend about 0.05 g of the compound in 2 mL of dilute hydrochloric acid and add bromine
water dropwise until the bromine colour remains. A white precipitate of the bromophenol may form.
Solid bromophenol derivatives can be used for the confirmation of the structure of a phenol.

(ii) Ferric chloride test: Most phenols react with iron (III) chloride to form coloured complexes. The colours
vary - red, purple, blue or green - depending on various factors, e.g. the phenolic compound used, the
solvent, concentration. Since some phenols do not give colours, a negative test must not be taken as
significant without supporting information.

(c) Aldehyde detection

(i) Fehling’s solution: Aldehydes reduce Fehling’s solution to yellow or red copper (I) oxide.
Preparation of the reagent: Mix equal volumes of Fehling’s solution I (aqueous alkaline potassium
tartrate) and Fehling’s solution II (copper sulphate solution).
Exception: Only aliphatic aldehydes give this test. Aromatic aldehydes don’t give this test.

(ii) Tollen’s reagent (Ammoniacal silver nitrate solution): Aldehydes are readily oxidized to carboxylic
acids and will reduce Tollen’s reagent to produce a silver mirror on the inside of a clean test tube.
Exception: α -hydroxy ketones also give this test but other ketones do not. Hemiacetals and formic acid
also give this test. But acetals do not.

(d) Aldehyde or Ketone detection

(i) Brady’s reagent (2,4-Dinitrophenylhydrazine): A test for the carbonyl group (C=O) in aldehydes and
ketones. 2,4-Dinitrophenylhydrazine gives sparingly soluble yellow or red 2,4-dinitrophenylhydrazones
with aldehydes and ketones.
R R H
C O + H₂N N NO₂ C N N NO₂
R R
H
NO₂ NO₂
R,R’ = hydrogen or alkyl

(ii) Iodoform Test: Iodoform test is given by simple methyl ketones and acetaldehyde only i.e. compounds
having (CH3 − C = O−) group. In Iodoform test, the compound reacts with I2 / NaOH reagent to give
iodoform (CHI3 ) which is pale yellow in colour. It is insoluble in water and has an antiseptic smell.
2 7 . 1 0 | Practical Organic Chemistr y

(e) Carboxylic acid: Test with 5% aq. NaHCO3

R − CO2H + NaHCO3 → R − CO2 − Na+ + CO2 + H2O
Sodium hydrogen carbonate reacts with carboxylic acids to give the sodium salt of the acid and liberates carbon
dioxide. If the acid is insoluble in water and the reaction is sluggish, dissolve the acid in methanol and add
carefully to a saturated sodium hydrogen carbonate solution, when a vigorous effervescence will be observed.

(f) Distinction between alcohol, phenol and carboxylic acid

(i) All alcohols react with sodium metal (Na) to evolve H2 gas. But do not react with NaOH or NaHCO3.

2CH3CH2OH + 2Na → 2CH3CH2ONa + H2

(ii) All phenols react with sodium metal (Na) and NaOH. But do not react with NaHCO3.
_
OH O Na+

+ NaOH + H₂O

Sodium phenoxide

(iii) All carboxylic acids react with sodium metal (Na), NaOH and NaHCO3.

(g) Amines
(i) Reaction with nitrous acid:
AMINE REACTION
1° aliphatic → N2 evolved
RNH2 + HNO2 → ROH + N2 + H2O
1° aromatic → Diazonium salt is formed.
ArNH2 + HNO2 → ArN ≡ N+

Add the cold diazonium solution and with swirling to a cold solution of 2-naphthol (0.2 g) in 5% NaOH
solution (2mL). An orange-red azo dye is formed.
2° aliphatic → Yellow oily nitrosamines are generally formed.
2° aromatic → R NH + HNO → R N − NO
2 2 2

3° aliphatic → No visible reaction

3° aromatic → Dialkylanilines yield green solid p-nitroso compounds

(if p-position unsubstituted).
(ii) Hinsberg reaction: In this test amine is shaken well with Hinsberg reagent (benzene sulphonyl chloride)
in the presence of aqueous alkali (either NaOH or KOH). A primary amine will form a soluble sulfonamide
salt which precipitates after addition of diluted hydrochloric acid. A secondary amine in the same reaction
will directly form an insoluble sulfonamide. A tertiary amine will not react with the sulfonamide but is
insoluble. After adding a dilute acid this insoluble amine is converted to a soluble ammonium salt. In this
way the reaction can distinguish between the three types of amines.
 Chemi str y | 27.11

O O H
H HCl
Ph S Cl + H N Ph S N R
R
O O
N
N-Alkylbenzene
sulphonamide (insoluble)

HCl NaOH

O H 
Ph S N R Cl
O H
(Soluble)

A precipitate which is an insoluble compound (which is insoluble in NaOH solution also) indicates 2°
amines.
O O R
R
Ph S Cl + H N Ph S N R + HCl
R
O O
N,N-Dialkyl benzene sulphonamide
(insoluble in NaOH)

3° amines do not react with the reagent.

(iii) Reaction with Carbon disulphide (CS2): Primary amine reacts with CS2 and further reacts with base due
to acidic hydrogen present on nitrogen.
R – NH2 + CS2 → R – NH – CSSH → reacts with base
Secondary amine reacts with CS2 but doesn’t react with a base due to the absence of acidic hydrogen.
R 2NH + CS2 → R 2`N − CSSH does not react with base.
Tertiary amine do not react with CS2 .

(iv) Carbylamine test: 1° aliphatic and aromatic amines on heating with CHCl3 and KOH give an offensive
smell due to the formation of carbylamine.

RNH2 + CHCl3 + 3KOH → R − NC +3KCl + 3H2O

Alkyl isocyanide
or
Alkyl carbyl amine

PhNH2 + CHCl3 + 3KOH → Ph − N ≡ C+ 3KCl + 3H2O

Aniline

2° and 3° aliphatic and aromatic amines do not give this test.

(v) Azo-dye test: 1° aromatic amines on heating with NaNO2 and HCl give stable diazonium salts which
couple with alkaline solution of β -naphthol to give orange-red azo dye.
NaNO₂ + HCl
PhNH₂ + HNO₂ Ph N N Cl
273 K
Benzene diazonium chloride
OH OH

Ph N N
pH = 9-10
-Naphthol

Orange-red dye
2 7 . 1 2 | Practical Organic Chemistr y

(h) Amides
(i) Simple primary amides can be decomposed by boiling with alkali and thereby evolving ammonia which
has pungent smell. E.g. CH3 − CO − NH2 + NaOH → CH3 − CO2 − Na+ + NH3

(ii) Hofmann rearrangement: Amides show Hofmann rearrangement to form a primary amine when
treated with Br2 / NaOH
O O
Br₂ C H₂O
R R NH₂
NaOH N CO₂
R NH₂

This test confirms presence of amide.

(i) Ester detection

(i) Hydroxamic acid test: R – CO – OR’ + H2N – OH → R – CO – NH – OH + R’ – OH
Esters react with hydroxylamine in the presence of sodium hydroxide to form the sodium salt of the
corresponding hydroxamic acid. On acidification and addition of ferric chloride the magenta-coloured
iron (III) complex of the hydroxamic acid is formed.
(ii) Hydrolysis test: Esters undergo saponification reaction i.e. hydrolysed with NaOH. The compound is
heated with NaOH with a small amount of phenolphthalein. The initial pink colour of the solution is
discharged due to the formation of acid, since phenolphthalein shows pink colour in basic medium and
is colourless in acidic medium.
O O
∆
R – C – O – R' + NaOH R – C – ONa + R'OH

(j) Nitro group detection

(i) Mulliken’s test: The nitro compound is dissolved in alcohol and is reduced with Zn / NH4 Cl or CaCl2
solution and Zn dust, and boiled. It is then filtered, cooled, and heated with ammoniacal AgNO3 , white
to grey and the black precipitate confirms the presence of nitro group.
Zn
RNO2 + 4[H]  → RNHOH + H2O
NH4 Cl

(ii) Azo dye test: Aromatic nitro compounds are first reduced to −NH2 group with SnCl2+conc. HCl which
on reaction with nitrous acid forms diazonium salts. The stable diazonium salt forms orange red dye with
alkaline solution of β -naphthol.
[H]
PhNO₂ Sn + HCl PhNH₂
Nitro
benzene HNO₂

OH
Orange-red Ph N N Cl
azo dye -Naphthol

(k) Sugars, Carbohydrates detection

(i) Molisch’s test: This is a general test for carbohydrates. Dissolve 20 - 30 mg of the compound in 2 mL
water and add 0.5 mL of the reagent (a 20% solution of 2-naphthol in ethanol). Pour 2 mL of concentrated
sulphuric acid from a dropper carefully down the side of the tube so that the acid forms a layer beneath
the aqueous solution without mixing with it. A red colouration, changing to dark purple forms at the
interface.
(ii) Fehling’s test: This forms the reduction test of carbohydrates. Fehling’s solution contains blue alkaline
cupric hydroxide solution, heated with reducing sugars gets reduced to yellow or red cuprous oxide
and is precipitated. Hence, formation of the yellow or brownish-red coloured precipitate helps in the
detection of reducing sugars in the test solution.
 Chemi str y | 27.13

(iii) Osazone Test: The ketoses and aldoses react with phenylhydrazine to produce a phenylhydrazone
which further reacts with another two molecules of phenylhydrazine to yield osazone. Needle-
shaped yellow osazone crystals are produced by glucose, fructose and mannose, whereas
lactosazone produces mushroom shaped crystals. Crystals of different shapes will be shown by
different osazones. Flower-shaped crystals are produced by maltose.

(l) Proteins
(i) Ninhydrin test: Amino acids also react with ninhydrin at pH = 4. The reduction product obtained from
ninhydrin then reacts with NH3 and excess ninhydrin to yield a blue colored substance. This test is given
by only amino acids and proteins which contain free −NH2 groups in their structure. Proline doesn’t give
this test due to absence of free −NH2 group.
(ii) Xanthoproteic test: Some amino acids contain aromatic groups that are derivatives of benzene. These
aromatic groups can undergo reactions that are characteristics of benzene and benzene derivatives.
One such reaction is the nitration of a benzene ring with nitric acid. The amino acids that have activated
benzene ring can readily undergo nitration. This nitration reaction, in the presence of activated benzene
ring, forms yellow product.
Example: tyrosine, tryptophan, phenylalanine and glutamic acid.
(iii) Millon’s Test: Millon’s test is specific to phenol containing structures (tyrosine is the only common
phenolic amino acid). Millon’s reagent is concentrated HNO3 , in which mercury is dissolved. As a result
of the reaction a red precipitate or a red solution is considered as positive test. A yellow precipitate of
HgO is NOT a positive reaction but usually indicates that the solution is too alkaline.
Example: tyrosine, phenylalanine, glycine and β -naphtol.
(iv) Biuret Test: The Biuret Test positively identifies the presence of proteins (not less than two Peptides). The
reaction in this test involves the complex formation of the proteins with Cu2+ ions in a strongly alkaline
solution.
Example: gelatin, casein and albumin.

(m) Alkene and alkyne

(i) Br2- water test: The decoloration of a solution of bromine in water with dichloromethylene as catalyst
is an analytical test for the presence of alkenes:
CH2 CH2 + Br2 → BrCH2 − CH2Br
=
It is also used as a quantitative test of unsaturation.
(ii) Bromine in CCl4: Bromine will add to the carbon-carbon double bond of alkenes to produce
dibromoalkanes and with alkynes to produce tetrabromoalkanes. When this reaction occurs, molecular
bromine is consumed, and its characteristic dark redbrown color disappears if bromine is not added in
excess. The rapid disappearance of the bromine color is a positive test for unsaturation.

C=C + Br₂ CCl₄ Br C C Br

Alkene Red-Brown
Colorless

Br Br

C ≡C + 2Br₂ CCl₄ C C
Alkyne Red-Brown
Br Br
Colorless
2 7 . 1 4 | Practical Organic Chemistr y

(iii) Baeyer’s reagent: (Alkene only): Baeyer’s reagent is an alkaline solution of cold potassium permanganate,
which is a powerful oxidant making this a redox reaction. Reaction with double or triple bonds (–C = C –
or – C ≡ C –) in an organic material causes the color to fade from purplish-pink to brown. It is a syn addition
reaction.
2KMnO₄ + H₂O 2KOH + 2MnO₂ + 3[O]
Purple Brownish
colour red ppt.

R³ R⁴
1.2KMnO₄
2.2H₂O
R³ R⁴OH
2
+ MnO₂+K₂MnO₄
R¹ R² R¹ OH
R²

HC CH + H₂O + 4[O] HOOC COOH

(n) Detection of terminal alkyne

(i) AgNO3 in ammonia: The reactions of alkynes with certain metal cations, e.g. Ag+ gives acetylides. Thus,
few drops of diammine silver(I) hydroxide Ag(NH3)2OH reacts with terminal alkynes signaled by the
formation of a silver white precipitate of the silver acetylide.
RC ≡ C − H + Ag(NH3 )2+ (in NH4 OH) → RC ≡ C − Ag (insoluble) + NH3 + NH+4

(ii) Acidic nature of terminal alkyne: Terminal alkyne being acidic reacts with strong bases like NaNH2 to
form sodium acetylide.
H − C ≡ C − H + NaNH2 (in ammonia or ether) → H − C ≡ C − Na (sodium acetylide)+NH3

(iii) With Grignard reagent: Terminal alkynes react with Grignard reagent to form alkanes.
RC ≡ C − H + C2H5MgBr (in ether) → RC ≡ C − MgBr + C2H6

(o) Detection of alkoxy group

(i) Hot Hydroiodic Acid (HI) test: The relatively unreactive C – O bonds in an ether can be cleaved by
treatment with conc. HI to give two alkyl iodides and water. The alkyl iodides produced can react with
mercuric nitrate, Hg(NO3)2 , to give the orange-red coloured mercuric iodide (HgI2).

ROR '+ 2HI (Hot) → RI + R 'I + H2O

2RI + Hg(NO3 )2 → HgI2 + 2R − O − NO2
Vermillion coloured

(ii) Cold HI Test: When ether is treated with cold HI, the smaller alkyl group in ether forms iodide while
bigger one forms alcohol in case of asymmetric ether.
CH3 − CH2 − O − CH3 + cold HI → CH3 I + CH3 − CH2 − OH

MASTERJEE CONCEPTS

Exceptions:
•• Lassaigne’s test is not given by compounds which do not contain C atoms but contain N atoms.
Example: NH2NH2 (hydrazine), NH2OH (hydroxyl amine).
Reason: Since, these compounds do not contain carbon, in sodium fusion extract, CN− ion will not form
and Prussian blue colour is not observed.