STRUCTURAL ISOMERISM, STRUCTURAL

IDENTIFICATION & POC

CONTENTS
Particulars Page No.

Theory 01 – 15

Exercise - 1 16 – 26
Part - I : Numerical Type Questions
Part - II : Only One option correct Type
Part - III : One or more than one correct options

Exercise - 2 26 – 28
Part - I : JEE(ADVANCED)/IIT-JEE Problems (Previous Years)
Part - II : JEE(MAIN) / AIEEE Problems (Previous Years)

Answer Key 29

Self Assessment Paper (SAP) 30 – 36

Part - 1 : Paper JEE (main) pattern
Part - 2 : Paper JEE (advanced) pattern

Answer Key 37

RRP Solutions 37 – 42

JEE (Advanced) Syllabus

Concepts : Structural identification

Catalytic hydrogenation, Monochlorination & ozonolysis reactions.
Detection of elements (N, S, halogens); Detection and identification of the following functional groups:
hydroxyl (alcoholic and phenolic), carbonyl (aldehyde and ketone), carboxyl, amino and nitro.

JEE (Main) Syllabus

General introduction, qualitative analysis of organic compounds & Lab test of functional groups.
 JEE (Adv.)-Chemistry Structural isomerism, Structural identification & POC

1. Structural Isomerism
Isomerism
The phenomenon of existence of two or more compounds possessing the same molecular formula but different
physical or chemical or both properties is known as isomerism. Such compounds are known as isomers.

Classification of isomerism

Isomerism

Structural (constitutional) isomerism Stereoisomerism

Configurational

Position Chain Ring-chain Functional Metamerism Tautomerism

isomerism isomerism isomerism isomerism
Conformational

1 Structural isomerism
When two or more organic compounds have same molecular formula but different structural formula,
(i.e., they differ in connectivity of atoms) they are called structural isomers and the phenomenon is called
structural isomerism.

CH3 - CH = CH - CH3 , CH3CH2CH = CH2 ù

But - 2 - ene But - 1 - ene úû structural Isomers

CH3
|
CH3 - CH - CH2CH3 ù
CH - CH2 - CH3 , | ú Identical compounds
| CH3 ú (not isomers)
CH3 ú
2 - Methylbutane û
2 - Methyl bu tan e

1.1 Chain isomerism

Compounds having same molecular formula but different carbon skeletons (either difference in main chain or
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side chain) are known as chain isomers & phenomenon is known as chain isomerism.
Condition : They should have same nature of locants.

(i)

Pentane Isopentane Neopentane

(Main chain of 5C) (Main chain of 4 C) (Main chain of 3C)
All the above are chain isomers.

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(ii)

Size of main chain = 3 Size of main chain = 3

Size of longest Side chain = 2 Size of longest side chain = 1
Both are chain isomers due to difference in number of carbon atoms in side chain.

(iii) &

2-Ethylbutanenitrile 2-Methylpentanenitrile
Both are chain isomers due to difference in number of carbon atoms in parent chain.
1.2 Position isomerism
Compounds having same carbon skeleton along with same nature of locants but having different position of
locants are known as position isomers & phenomenon is position isomerism.
(i) &

Difference only in position of –OH group so these are positional isomers.

(ii) , &

o-Xylene m-Xylene p-Xylene

Difference only in position of –CH3 group so these are positional isomers.
1.3 Functional isomerism
Compounds having same molecular formula but different functional groups are known as functional isomers &
phenomenon is functional isomerism.
(i) CH3CH2–NH2 & CH3–NH–CH3 (ii) &

Ethanamine Dimethylamine Propanal Propanone

1.4 Ring-chain isomerism

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CH3–CH=CH–CH3 (But-2-ene) & Cyclobutane

Sometimes it is also known as functional isomerism.

Note : (1) 1º, 2º, 3º amines are functional isomers.
(2) 1º, 2º, 3º amides are functional isomers.
(3) Alcohol attached to sp2 C is chemically different from alcohol attached to sp3 C.
(4) Alchol and enol are functional isomers.

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& are functional isomers.

(5) Following compounds do not exist at room temperature therefore should not be considered as structural
isomers.

–
(i) – C = C –OH (enol) (ii) –CºC–OH (ynol) (iii) – C – OH (gemdiol)
–
–

–
OH
–

–
(iv) – C – OH (hemiacetal) (v) – C – O– C = C – (vi) –O–O– (peroxy compound)

–
–
–

–
OR OH

1.5 Metamerism
It arises due to different alkyl chains on either side of the functional group.
(Polyvalent hetro atomic functional group must be present in the compounds).

(i) CH3–CH2–O–CH2–CH2 & CH3–O–CH2–CH2–CH3

Ethoxy ethane Methoxy propane
[Ethyl groups on either sides of O.] [Methyl & propyl groups on either sides of O.]

(ii) are metamers.

are metamers.

Que. Identify relationship between the given pair of compounds.

(1) (i) CH3 – CH2CH2 – CH3 (ii)

Butane 2–Methylpropane
Size of main chain = 4 Size of main chain = 3
Size of side chain = 0 Size of side chain = 1
Structure (i) & (ii) are chain isomers.
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(2) (i) (ii)

Size of main chain = 6 Size of main chain = 6

Size of side chain = 2 Size of side chain 1 = 1
Size of side chain 2 = 1
Structure (i) & (ii) are chain isomers.
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(3) (i) (ii)

Size of main chain = 6 Size of main chain = 3

Size of side chain = 0 Size of side chain 1 = 1
Size of side chain 2 = 1
Size of side chain 3 = 1
Structure (i) & (ii) are chain isomers.
H3 C - CH2 - CH = CH2 (but - 1 - ene) ù
(4) ú position isomers
H3 C - CH = CH - CH2 (but - 2 - ene)û

(5) HC º C - CH2 - CH2 - CH3 (pent - 1 - yne) ù

H3 C - C º C - CH2 - CH3 ú position isomers
(pent - 2 - yne) û

(6) (i) CH3 – CH2OH (ii) CH3 – O – CH3

Ethanol Methoxymethane
Functional group – OH Functional group – O – (Ether)
Structure (i) & (ii) are functional isomers.

(7) (i) (ii)

Functional groups – COOH Functional groups (Ester)

Structure (i) & (ii) are functional isomers.

(8) (i) C2H5 – O – C2H5 (ii) C3H7 – O – CH3

Diethyl ether Methyl propyl ether
Alkyl groups –C2H5 &–C2H5 Alkyl groups –C3H7 & – CH3
Structure (i) & (ii) are metamers.

O O
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(9) (i) (ii)

Struture (i) and (ii) have different alkyl groups but same functional groups, so these are metamers.

(10) (i) CH3–O–CH2–CH2–CH=O (ii) CH3–CH2–CH2–O–CH=O

Functional groups-ether and aldehyde Functional group-Ester
Struture (i) and (ii) have different functional groups, so these are functional isomers.

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1.6. Degree of unsaturation (DU)

The presence of double bonds or rings within a molecule is indicated by a quantity called degree of
unsaturation.
Applications : To identify the no. of p bonds or rings and also helpful in determining the structure of the
molecule.
Definition : Deficiency of 2H atoms with respect to fully saturated acyclic hydrocarbon is equivalent to
one DU. It is also known as Index of Hydrogen Deficiency (IHD) or Double Bond Equivalence (DBE)

–2H –2H
H3 C – H2C – CH3 ¾¾ ¾® ¾¾ ¾® CH3–CºCH or CH2=C=CH2 or
(DU = O)

Where n = number of carbon atoms in the molecule

Note : Total no. of cyclic rings + double bonds will give us degree of unsaturation.
One double bond = one DU
One ring = one DU
One triple bond = two DU

(2 ´ 2 + 2) – 4
(i) CH2=CH2 DU= =1 (ii) DU = 2
2

(iii) DU = 4 (iv) DU = 7
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(2 ´ 2 + 2) – 4
(v) C2FClBrI DU= =1
2

(2 ´ 15 + 2) – (28 - 2)
(vi) C15H28O2N2 DU= =3
2

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2. Structural identification
Introduction
The main objective of an organic chemist is the determination of the structure of a new organic compound
which has been obtained in pure state either from a natural source or synthesised in the laboratory.
In order to establish the correct structure of an organic compound, it is necessary to detect skeleton of
compound, elements and functional groups present in the organic compound.

2.1. Catalytic hydrogenation

Alkenes, Alkynes, polyenes or polyynes can be hydrogenated by using catalysts Ni/Pt/Pd at room temperature.
All Carbon–Carbon p bonds(C=C, CºC) get hydrogenate. The reaction can’t be stopped at any
intermediate stage.
Steps in the hydrogenation of a C=C double bond at a catalyst surface (for example Ni or Pt)
(1) The reactants are adsorbed on the catalyst surface and H2 dissociates.
(2) A H atom bonds to one C atom. The other C atom is still attached to the surface.
(3) A second C atom bonds to a H atom. The molecule leaves the surface.

H-atoms attached to the Alkene attached to the

Metal surface Metal surface One H-atom is
H H transferred to alkene
H H C–C H H
H H H H C–C
(1) (2)
H H
Metal surface

H Second H atom is
H transferred to alkene H H Two H-atoms have added to the same
face of the double bond (syn-addition).
(3) H H (4) H C–C H
The product is a saturated alkane.
H H H

Note : (1) Aromatic p bonds are stable at room temperature but can be hydrogenate at high temperature.
(2) It can be concluded that the hydrogenation product of an alkene or alkyne or any unsaturated
compound is always a saturated compound.
(3) The no. of moles of H2 consumed by 1 mole of compound is equal to the no. of p bonds.
(4) During catalytic hydrogenation carbon skeleton does not change.
Application : This reaction gives an information about molecule that the molecule, is saturated or
unsaturated.
Ni
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(i) R–CH=CH–R + H2 ¾¾®

¾ R–CH2–CH2–R
Ni / Pt / Pd
(ii) R–CºC–R + 2H2 ¾¾ ¾ ¾ ¾ ¾® R–CH2–CH2–R
H2

R – CH = CH – R ¾H
¾¾2
® R – CH2 – CH2 – R
(Not isolated )

2H2 / Ni
(iii) CH2=CH–CH=CH2 ¾¾ ¾ ¾¾® CH3–CH2–CH2–CH3

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CH = CH2 CH2 – CH3

H2 / Ni

–
(iv) ¾¾ ¾ ¾ ¾ ¾ ¾¾®
room temperature

CH2 – CH3
H 2/Ni

–
(100 – 150ºC)

2.2. Monohalogenation
When an alkane or a cycloalkane is treated with halogen a photochemical reaction takes place, in which
a C–H bond cleaves and a C–X bond is formed. In such reactions if one H-atom is substituted by one
halogen atom, then this is known as monohalogenation reaction.

Application : If a molecule has more than one type of H-atoms, then on monochlorination, it forms a
mixture of monochloroisomers. All these products (structures) are position isomers.
Conclusion : Hence, it can be concluded that the total no. of position isomers (structural) of monochloro
compounds is equal to the number of different types of H-atoms present in the reactant. The different type
of H-atoms are also known as non-identical hydrogens or non-equivalent hydrogens or chemically different
hydrogens.

Note : In aromatic hydrocarbons, the hydrogen atoms of the side-chain are chlorinated, but H-atoms of benzene
ring are stable.

(i) CH4 CH3Cl + HCl ¼1–Monochloroproduct½

Monochlorination

Cl
–

(ii) + HCl ¼1–Monochloroproduct½

Monochlorination

CH3 CH2Cl
–

(iii) ¼1–Monochloroproduct½
Monochlorination

Note : Only one monochloro product is formed because aromatic H atoms are inert towards this reaction.

(iv) CH3–CH2–CH2–CH3 2 Products (structural isomers)

Monochlorination
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(v) CH3–CH2–CH2–CH2–CH3 3 Products (structural isomers)

Monochlorination

(vi) 4 Products (structural isomers)

Monochlorination
CH3
–

(vii) 5 Products (structural isomers)

Monochlorination

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2.3. Ozonolysis
Ozonolysis reaction is used to determine the position of C=C, CºC in a molecule. In this reaction alkene,
alkyne and polyalkene on ozonolysis undergo oxidative cleavage. It is of two types.
(i) Reductive ozonolysis
Reagents are : (1) O3 (ozone) (2) Zn or (CH3)2S and H2O or CH3COOH
The products are carbonyl compounds (aldehydes, ketones).
(ii) Oxidative ozonolysis
Reagents are : (1) O3 (ozone) (2) H2O2 or H2O
The products are ketones and/or acids.
Note : (i) Ozonolysis does not interfere with other functional groups.
(ii) At higher temperature, the aromatic double bonds can also undergo ozonolysis reaction.

(a) Reductive ozonolysis

(1) O3
R–CH C–R R–CH=O + O=C–R + ZnO + H2O
(2) Zn/H2O
R R
(1) O3
R–CºC–H R–C–C–H + ZnO + H2O
(2) Zn/H2O
OO
(b) Oxidative ozonolysis
(1) O3
R–CH C–R (2) H O R–COOH + O=C–R + H2O
2 2

R R
(1) O3
R–CºC–H (2) H O R–COOH + HCOOH
2 2

H2O+CO2
(1) O3
Ex. (i) CH2=CH2 CH2=O + CH2=O
(2) Zn/H2O
(1) O3
(ii) CH3–CH2–CH=CH2 CH3–CH2–CH=O + O=CH2
(2) Zn/H2O
(1) O3
(iii) CH2=CH–CH2–CH=CH–CH3 CH2=O + O=CH–CH2–CH=O + O=CH–CH3
(2) Zn/H2O

(1) O3
(iv) 2 OHC–CH2–CHO (Propanedial)
(2) Zn/H2O

(1) O3
(v) O=CH–CH2– C –CH2–CH=O + O=CH2
(2) Zn/H2O
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O ( D)
¾¾3¾¾®
(vi) Zn,H O or (glyoxal)
2

(i)O3
CH = CH – CH3 + CH3CH=O
(ii)Zn, H2O
–

(vii)
2 O=CH–CH=O + O=CH–C–CH=O + CH3CH=O
(ii)Zn, H2O
O
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3. Practical Organic Chemistry (POC)

While looking for the information regarding the functional nature of the compound, we go for the following lab
tests.

3.1. Test for acidic/active hydrogen

When any compound release H2 gas after reaction with sodium or potassium or alkali metals or sodamide
then this reflects the presence of acidic hydrogen.
Active H : The H atoms which are attached with more electronegative atoms like O,N,S,X, Csp.

– + 1
General reaction : Z – H + Na ® Z Na + H2 ­
2
S.No Reactant Reagent (Na metal) Product
1 R - NH2 ¾¾
Na
® 1
R - NHNa + H 2 ­
2
2 R - SH ¾¾
Na
® 1
RSNa + H 2 ­
2
3 R - COOH ¾¾
Na
® 1
R - COONa + H 2 ­
2
4 PhOH ¾¾
Na
® 1
PhONa + H 2 ­
2
5 R—OH ¾¾
Na
® 1
R— O–Na+ + H2 ­
2
6 R—SO3H ¾¾
Na
® 1
R—SO3–Na+ + H2 ­
2
7 R—CºCH ¾¾
Na
® 1
R—CºC–Na+ + H2 ­
2
8 R — CH = CH2 ¾¾
Na
® No reaction.
9 R — CH2—O —CH3 ¾¾
Na
® No reaction.
10 HCl ¾¾®
Na
1
NaCl + H 2 ­
2
11 H ¾¾
Na
® Å
– Na
1
+ H2
2

3.2 Test for unsaturation

(a) Bromine water test (Br2 + H2O, Red-brown solution)
This is used to distinguish between saturated (alkane) and unsaturated (alkene/alkyne) hydrocabon.
S.No Reactant Reagent Product Observation
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1 R–CH2–CH3 Bromine water – No reaction

(Br2 + H2O)
2 R–CH=CH2 Bromine water Red-brown colour
(Br2 + H2O) disappears

3 R–CºCR Bromine water Red-brown colour

(Br2 + H2O) disappears

Note : Benzene does not give this test, whereas phenol and aniline give this test.
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(b) Baeyer reagent (Cold dil. alkaline KMnO4 Pink/purple solution)
This is also used to distinguish between saturated (alkane) and unsaturated (alkene/alkyne) compounds.
S.No Reactant Reagent Product Observation
1 R–CH2–CH3 Cold dil. alkaline KMnO4 – No reaction

2 R–CH=CH2 Cold dil. alkaline KMnO4 Purple colour disappears

3 R–CºCR Cold dil. alkaline KMnO4 Purple colour disappears

3.3 Test for terminal alkyne

S.No Reactant Reagent Product Observation

1 RCºCH Tollen’s Reagent [AgNO3 + R–CºCAg White precipitate
NH OH or {Ag(NH ) }+ OH ] +
–
4 3 2
NH4NO3
2 RCºCH Ammonical cuprous chloride R–C º CCu Red precipitate
(Cu2Cl2 + NH4OH) ¯ + 2NH4Cl

3.4. Test for alcohols

(a) Lucas reagent test (Conc. HCl + anhydrous ZnCl2)
* It gives white turbidity or cloudiness with alcohols (OH groups attached with sp3 hybridised carbon).
* Lucas reagent is also used to distinguish between 1°, 2°, 3° alcohol because 1°, 2°, 3° alcohols react with
different rates.

(i) 1° alcohol R–CH2–OH

(does not give appreciable reaction or gives white turbidity in 30 min.)

(ii) 2° alcohol R2CH –OH (gives white turbidity in 5 min.½

(iii) 3° alcohol R3C – OH R3C–Cl (gives white turbidity immediately)

* Phenols and enols do not give Lucas test.

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(b) Victor Mayer test

NO2
P+I2 AgNO2 HNO2 base
1° Alcohol : RCH2–OH RCH2–I RCH2–NO2 R–C= N–OH Blood Red colour

2° Alcohol %

3° Alcohol %

(c) Cerric Ammonium Address

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It is group reagent for alcohols.
3.5. Test for phenol or enol
Neutral FeCl3 test :
It forms violet purple complex with phenol or enol (OH groups attached with sp2 carbon).
3-
éæ ö ù
6 PhOH + FeCl3 ¾¾® êçç PhO ÷÷ Feú + 3HCl + 3H+
êëè ø6 úû
(Coloured complex)
* It does not give positive test with alcohols.

3.6. Test for carbonyl compounds

2, 4-DNP ¼2,4-Dinitrophenyl hydrazine+½Test :
Carbonyl compounds (all aldehydes and ketones) give yellow–orange precipitate with 2,4–DNP. It is also
known as Brady’s reagent.

C=O + H2N – NH NO2 C=N – NH NO2

NO2 NO2
2, 4-DNP (yellow–orange precipitate of Hydrazone)

3.7. Test for aldehydes

Q
(a) Tollen’s reagent [AgNO3 + NH4OH or {Ag(NH3)2}+ OH ] :
Tollen’s reagent gives silver mirror or Black precipitate with aldehydes.
O
||
AgNO 3 +NH4OH
R–CH = O ¾¾ ¾ ¾¾ ¾¾® R – C – OQ + Ag ¯
silver mirror
Note : HCOOH also gives this test.

(b) Fehling’s solution

It is an alkaline solution of cupric ion complexed with sodium potassium tartrate.
There are two solutions in Fehling solution
Solution (A) : CuSO4 solution
Solution (B) : Alkaline solution of sodium potassiumtartrate.
When these two solutions are mixed we get deep blue coloured solution.

CuSO4 + 2NaOH ¾¾® Cu(OH)2 + Na2SO4

Cu(OH)2 + ¾¾®
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Equal volume of both the solutions are heated with aldehyde to give red brown precipitate of cuprous oxide
(Cu2O) which confirms the presence of aldehyde.
R – CHO + 2CuO ¾¾® RCOOH + Cu2O (Red ppt) ¯
Blue

RCHO + 2Cu2+ + ¾¾® RCOO + ¯ + 2H2O

(c) Benedict solution

It also consists of two solutions.
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Solution (A) : CuSO4 solution
Solution (B) : Alkaline solution of sodium citrate.
CuSO4 + 2NaOH ¾¾® Cu(OH)2 + Na2SO4

Cu(OH)2 + ¾¾®

Aldehyde gives positive test with Benedict solution.

RCH = O + + ¾¾® +

(d) Schiff’s reagent

It is dilute solution of rosaniline hydrochloride whose pink colour has been discharged by passing SO 2.
Aldehyde restores pink colour when treated with Schiff’s reagent (Magenta solution in H2SO3).
* Aromatic aldehydes (Benzaldehyde) do not give Fehling and Benedict tests.

NH
CH3

Schiff reagent

H2N NH2

3.8. Iodoform Test

Reagents : I2 + NaOH or NaOI ¼Where R = H, alkyl, aryl group½

Acetaldehyde, all methyl ketones & ethyl alcohol give Iodoform test.

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3.9. Test for acids/esters/amides

(a) Sodium bicarbonate test (NaHCO3)
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All the acids which are stronger than H2CO3 give CO2 gas with NaHCO3.
HCl + NaHCO3 ¾® NaCl + H2CO3 ¾® H2O + CO2 ­
RCOOH + NaHCO3 ¾® RCO2Na + H2CO3 ¾® H2O + CO2 ­
RSO3H + NaHCO3 ¾® RSO3Na + H2CO3¾® H2O + CO2 ­

Note : If electron withdrawing group (NO2) is present at ortho or para position of phenol then it gives positive test with
sodium bicarbonate.

(b) Litmus test % Acid converts blue litmus into red litmus while base converts red litmus into blue.
(c) Acid amide gives smell of ammonia when heated with alkali.

(d) Esters are sweet (fruity) smelling liquids.

Esters when react with NaOH & phenolphthalein, pink colour disappears on heating.
D
RCOOR’ + NaOH + Phenolphthalein (Pink) ¾¾® R COOH + R’ OH (Colourless solution)

3.10. Test for amines

(i) NaNO2 + aqueous HCl test % it is used to distinguish between 1º, 2º and 3º amines and also distinguish
between aliphatic and aromatic primary amines.

1° Aliphatic amine % R–CH2–NH2 R–CH2–OH + N2 ­

1° Aromatic amine : ¼diazonium salt)

2° Amine % R – NH – R (N -Nitroso amine) yellow oily liquid

3° Amine : R3N unstable nitrite

(ii) Carbyl amine test (CHCl3 + KOH)

1° Amine (Aliphatic and Aromatic)
CHCl +KOH
R–CH2–NH2 ¾¾ ¾3 ¾ ¾® R–CH2–N º C + 3KCl + 3H2O
(unpleasent smell of isocyanide)
CHCl 3 +KOH
BOBO-BA\Reliable\Adv.\Organic Chemistry\Structural isomerism, Structural identification & POC\Theory

Ph – NH2 ¾¾ ¾ ¾ ¾ ® Ph – N º C

* 2° Amines and 3° Amines do not give this test.

(iii) Hofmann mustard oil test
It is a test for 1º amine and aniline. Primary amine reacts with carbon disulphide to form dithioalkyl carbamic
acids which decompose on heating with mercuric chloride (HgCl2) to give alkyl isothiocyanate having smell
like mustard oil.
SH
HgCl2
¾¾ ¾
¾® R – N = C = S + HgS + 2HCl
|

(a) R – NH2 + S=C=S ¾¾® R – NH – C D

S
||

Dithioethylcarbamic acid Alkylisothiocyanate

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D
(b) CH3CH2NH2 + S=C=S ¾ ¾ ® HgCl2
¾¾ ¾¾ ® CH3CH2–N = C=S + HgS + 2HCl
D
(1º amine) Ethyl isothiocyanate
Similarly aniline gives phenlyisothiocyanate.
(i) S=C=S
(c) C6H5NH2 C6H5N=C=S
(ii) HgCl2/
Phenylisothiocyanate
* 2° Amines and 3° Amines do not give this test.

(iv) Hinsberg reagent (C6H5SO2Cl)

This is used to distinguish between 1°,2° & 3° amines.

Pyridine Base
1° Amine % R–NH2 + PhSO2Cl R–NH–SO2–Ph Compound is soluble in base.

Pyridine Base
2° Amine % R2NH + PhSO2Cl R2N–SO2–Ph Compound is insoluble in base.

Pyridine
3° Amine : R3N + PhSO2Cl No reaction.

3.11. Test of nitro group (Mulliken- Barker Test)

Nitroalkane & nitrobenzene gives black ppt on reduction with Zn and ammonium chloride followed by
treating with Tollen’s reagent. This is also called Mulliken’s test.

(a) R–NO2 R–NHOH Ag(black ppt)¯

(b) + Zn + NH4Cl ¾® AgNO +NH OH

3 ¾¾
4 ¾® ¯
¾¾ ¾ ¾

4. Elements detection
4.1. Detection of carbon and hydrogen
Carbon and hydrogen are detected by heating the compound with copper(II) oxide. Carbon present in the
compound is oxidised to carbon dioxide (tested with lime-water, which develops turbidity) and hydrogen
to water (tested with anhydrous copper sulphate, which turns blue).
D
C + 2CuO ¾¾® 2Cu + CO2
BOBO-BA\Reliable\Adv.\Organic Chemistry\Structural isomerism, Structural identification & POC\Theory

D
2H + CuO ¾¾® Cu + H2O
D
CO2 + Ca(OH)2 ¾¾® CaCO3 ¯+ H2O
D
5H2O + CuSO4 ¾¾® CuSO4.5H2O
White Blue
4.2. Detection of other elements
Nitrogen, sulphur, halogens and phosphorus present in an organic compound are detected by “Lassaigne’s
test”. The elements present in the compound are converted from their covalent form to their ionic form by
fusing the organic compound with sodium metal.
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Following reactions take place: C, N, S and X come from organic compound.
Cyanide, sulphide and halide of sodium formed by sodium fusion are extracted from the fused mass by
boiling it with distilled water. This extract is known as sodium fusion extract or Lassaigne solution.
D D
Na + C + N ¾¾® NaCN ; 2Na + S ¾¾® Na2S
D D
Na + X ¾¾® NaX ; Na + C + N + S ¾¾® NaSCN
(i) Test for nitrogen
The sodium fusion extract is boiled with iron(II) sulphate and then acidified with dilute sulphuric acid. The
formation of Prussian blue or green colour confirms the presence of nitrogen. Alternatively FeCl3 and dil.
HCl may be added.
6NaCN + FeSO4 ¾¾® Na4[Fe(CN)6]
Na4[Fe(CN)6] + 4Fe3+ ¾¾ ¾® Fe4[Fe(CN)6]3. xH2O
xH 2O

Ferric ferrocyanide (Prussian blue)

(ii) Test for sulphur
(a) The sodium fusion extract is acidified with acetic acid and lead acetate is added to it. A black
precipitate of lead sulphide indicates the presence of sulphur.
Na2S + (CH3COO)2Pb ¾¾® PbS (Black)
(b) On treating sodium fusion extract with sodium nitroprusside, appearance of a violet colour indicates
the presence of sulphur.
Na2S + Na2[Fe(CN)5 NO] ¾¾® Na4[Fe(CN)5NOS]
Sodium nitroprusside Sodium thionitroprusside (Violet/Purple)
(c) In case, nitrogen and sulphur both are present in an organic compound, then sodium thiocyanate is
formed. It gives blood red colour with neutral FeCl 3.

Na + C + N + S ¾¾® NaSCN ; Neutral FeCl3 + NaSCN ¾¾® Fe(SCN)3(Blood red)

(iii) Test for halogens
The sodium fusion extract is acidified with nitric acid and then treated with silver nitrate. A white precipitate,
soluble in ammonium hydroxide shows the presence of chlorine, a yellowish precipitate, sparingly soluble
in ammonium hydroxide shows the presence of bromine and a yellow precipitate, insoluble in ammonium
hydroxide shows the presence of iodine.
Note The sodium fusion extract is first boiled with concentrated nitric acid to decompose cyanide or sulphide
of sodium formed during Lassaigne’s test. These ions would otherwise interfere with silver nitrate test for
halogens.
NaX + AgNO3 ¾¾® AgX ¯
(iv) Test for phosphorus
The compound is heated with an oxidising agent (sodium peroxide). The phosphorus present in the
compound is oxidised to phosphate. The solution is boiled with nitric acid and then treated with ammonium
molybdate. A yellow colouration or precipitate indicates the presence of phosphorus.
BOBO-BA\Reliable\Adv.\Organic Chemistry\Structural isomerism, Structural identification & POC\Theory

Na3PO4 + 3HNO3 ¾¾® H3PO4 + 3NaNO3

H3PO4 + 12(NH4)2MoO4 + 21 HNO3 ¾¾® (NH4)3PO4. 12 MoO3 + 21 NH4NO3 + 12H2O

Ammonium Ammonium
molybdate Phosphomolybdate

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JEE (Adv.)-Chemistry Structural isomerism, Structural identification & POC

PART - I : NUMERICAL TYPE QUESTIONS

Section (A) : Structural isomerism
A-1. How many number of all structurally isomeric dienes with molecular formula C5H8 are possible ?
A-2. How many cyclic structurally isomeric amines of molecular formula C3H7N are possible?
A-3. How many structurally isomeric ethers with molecular formula C5H12O are possible?
A-4. How many structurally isomeric esters with molecular formula C5H10O2 are possible?
A-5. How many structurally isomeric ketones with molecular formula C6H12O are possible?
A-6. How many number of all aldehydes (structurally isomeric) with molecular formula C5H10O are possible?
A-7. How many benzenoid structural isomers are possible for C7H8O ?
A-8. Total number of benzenoid isomers of molecular formula C9H12 would be-
A-9. How many structural alkenes of formula C2FClBrI are possible ?

A-10. How many structural isomers can be obtained by the replacement of one hydrogen atom of propene with
chlorine ?
Section (B) : Structural Identification
B-1. How many alkenes, alkynes and alkadienes can be hydrogenate to form Isopentane (Including all structural
isomers)?
B-2. Find the number of structural isomers of fully saturated cycloalkane of molecular formula C6H12 which give two
monochloro structural products.
H2/Ni Cl2/hv
(a) (b)
No. of No. of
CH3 products monochloro
B-3.
C=C structural products
H
O3/Zn
H2O
(c)
No. of
ozonolysis
products
Calculate sum of number of products formed in the reaction a, b and c.
BOBO-BA\Reliable\Adv.\Organic Chemistry\Structural isomerism, Structural identification & POC\Exercise

Section (C) : POC (Lab tests and Element detection)

C-1. How many of the following compounds decolorise Br2 water solution ?

(I) (II) (III) Me – C º C – Me (IV)

(V) (VII) ` (VIII) Me – CH = CH – Et (IX)

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C-2. Structure of Ascorbic acid is represented as follows.

(Ascorbic acid)
How many of the following reagents can give positive test with ascorbic acid?
2,4-DNP Na Metal HCl + ZnCl2 FeCl 3
(II) (III) (IV) (V)

NaOH + Phenolthalein dil.KMnO4 Br2/H2O AgNO3 + NH4OH I2 + NaOH

(VI) (VII) (VIII) (XI) (X)

C-3. Among the following the number of compounds which react with Fehling’s solution is :

O O
|| ||
, H3C – C – H , H C – CH – C – H , ,
3 2

, HCHO , HCOOH , CH3COCH3 .

C-4.@ Observe the the following compounds.

O Me
OH CH3 CH3 Ph
CH3–C–OH C
O
CH3 O
(III) (IV) (V)
(I) (II)

O Ph
CH3 O CH3
C H3–CH–CH 2–CH 3 C C Ph–CH–CH 3
OH OH
O O
(VII) (VIII) (IX)
(VI)
BOBO-BA\Reliable\Adv.\Organic Chemistry\Structural isomerism, Structural identification & POC\Exercise

Number of compounds which can give positive haloform test = (x)

Number of compounds which can give positive Lucas reagent test = (y)
Report your answer (x + y)

C-5. In the Lassaigne’s test, one of the organic compound X gives blood red colour with FeCl3. Compound X, when
fused with sodium metal forms compound Y. Molecular mass of compound Y is

C-6. How many oxygen atoms are present in 1 molecule of ammonium phosphomolybdate ?

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JEE (Adv.)-Chemistry Structural isomerism, Structural identification & POC

PART - II : ONLY ONE OPTION CORRECT TYPE

Section (A) : Structural isomerism

A-1. Relation between compounds & is:

(A) Position isomers (B) Chain isomers (C) Identical (D) Functional isomers

A-2. Which type of isomerism is observed between I and II ?

(I) CH3 – CH2 – C – OCH3 , (II)

||
O

(A) Functional isomerism (B) Metamerism

(C) Position isomerism (D) Stereoisomerism

A-3. Which of the following is a pair of structural isomers?

(A) and (C) and

(C) and (D) and

A-4. Which one of the following compound is not isomer of others?

(A) (B) (C) (D)

A-5. What is the relation between 3-Ethylpentane and 3-Methylhexane ?

BOBO-BA\Reliable\Adv.\Organic Chemistry\Structural isomerism, Structural identification & POC\Exercise

(A) Chain isomers (B) Position isomers

(C) Functional isomers (D) No relation

A-6. Which type of isomerism is observed between I and II?

(I) CH3 – CH2 – NH – CHO (II) CH3–CH–CHO

NH2
(A) Chain isomers (B) Position isomers
(C) Functional isomers (D) Metamers

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A-7. Which of the following pair is identical ?

(A) (B)

(C) (D)

A-8. Degree of unsaturation (DU) & total number of different functional groups in given compound are?

CHO OH O
NC NH

NH
NC
O CH2OH

(A) 8, 7 (B) 9, 8 (C) 12, 8 (D) 12, 7

A-9. How many positional isomers are possible for dimethylcyclohexane?

(A) 3 (B) 4 (C) 5 (D) 6

A-10. How many aromatic isomers are possible for trichlorobenzene (C6H3Cl3) ?
(A) 2 (B) 3 (C) 4 (D) 5

A-11. The number of ether isomers represented by formula C4H10O is (only structural)?
(A) 4 (B) 3 (C) 2 (D) 1

A-12. Total number of 2° amine isomers of C4H11N would be (only structural)?

(A) 4 (B) 3 (C) 5 (D) 2

A-13. Find the total number of structurally isomeric 1° amides with molecular formula C5H11NO?
(A) 1 (B) 3 (C) 2 (D) 4

A-14. How many structural isomers of all the tertiary alcohols with molecular formula C6H14O are ?
(A) 2 (B) 3 (C) 4 (D) 5
BOBO-BA\Reliable\Adv.\Organic Chemistry\Structural isomerism, Structural identification & POC\Exercise

A-15. Total number of structural isomers for C5H10 would be?

(A) 8 (B) 6 (C) 9 (D) 10

Section (B) : Structural Identification

B-1. Compound A (C6H12) does not absorb H2 in presence of Ni. It forms two monochloro structural isomers on
photochemical chlorination. Its structure can be :

(A) (B) (C) (D)

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JEE (Adv.)-Chemistry Structural isomerism, Structural identification & POC
B-2. Which alkyne will give 3-Ethylheptane on catalytic hydrogenation?

(A) (B) (C) (D)

B-3.@ Which of the following hydrocarbons give same product on hydrogenation?

(A) 2-Methylhex-1-ene & 3-Methylhex-3-ene
(B) 3-Ethylhex-1-en-4-yne & 2-Methylhept-2-en-4-yne
(C) 3-Ethylcycloprop-1-ene & 1,2-Dimethylcycloprop-1-ene
(D) 2-Methylbut-2-ene & 3-Methylbut-1-ene

B-4. How many number of moles of hydrogen will be required for complete hydrogenation of one mole of the
following compound ?

(A) 6 (B) 7 (C) 5 (D) 3

B-5. Only two structural isomeric monochloro derivatives are possible for :-
(A) n-Pentane (B) 2,4-Dimethyl pentane
(C) Toluene (D) 2,3-Dimethyl butane

B-6. The number of possible monochloro derivatives of 2,2,3,3-Tetramethylbutane is -

(A) 2 (B) 3 (C) 4 (D) 1

B-7.@ Which of the following alkene gives four monochloro (structural isomer) products after hydrogenation ?
(A) Pent-2-ene (B) 2-Methylbut-2-ene (C) 3-Methylhex-2-ene (D) 2, 3-Dimethylbut-2-ene

B-8. Which of the following compounds will give four monochloro (structural) product on monochlorination?

(A) (B) (C) (D)

BOBO-BA\Reliable\Adv.\Organic Chemistry\Structural isomerism, Structural identification & POC\Exercise

O / Zn
B-9.@ X ¾¾3 ¾
¾® +

Y..
The IUPAC name of compound Y is :
(A) 2-Cyclohexylbutane (B) 1-Methylpropylcyclohexane
(C) Butylcyclohexane (D)1-Cyclohexylbutane
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 JEE (Adv.)-Chemistry Structural isomerism, Structural identification & POC
B-10. An alkene give two moles of HCHO, one mole of CO2 and one mole of on ozonolysis.

What is its structure?

(A) (B)

(C) (D)

B-11. An unknown compound on ozonolysis gives acid C3H6O2 and a ketone C4H8O. From this information, identifiy
structure of unknown compound.
CH3

(A) (CH3)2C = CHCH2 – CH2CH3 (B) CH3CH2 – C = CHCH2CH3

(C) (D) CH3CH2CH2CH = CHCH2CH3

Section (C) : POC (Lab tests and Element detection)

C-1. When one mole of the given compound reacts with sodium metal then how many moles of H2 gas will be
released?

O
||
OH (1mole)

(A) 1 mole (B) 1.5 mole (C) 2 mole (D) 0.5 mole

C-2.@

Identify X :
(A) CH3 – CH2 – C º C – CH2 – CH3 (B) CH3 – C º C – CH2 – CH2 – CH3

(C) (D)

C-3. Compound 'A' gives red precipitate with Cu2Cl2 / NH4OH solution and decolourises bromine water. The compound
'A' can be :
BOBO-BA\Reliable\Adv.\Organic Chemistry\Structural isomerism, Structural identification & POC\Exercise

(A) CH2 = CH - C - CH3 (B) CH2 = CH - C - H (C) CH3–CºCH (D) PhCHO

|| ||
O O

C-4. An organic compound does not react appreciably with Lucas reagent but gives white precipitate with Tollen’s
reagent. What is the possible structure of compound ?
(A) (B)
OH
(C) (D) CH2=C=CH–CH2–OH

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C-5. The group reagent for the test of alcohols is :
(A) Cerric ammonium nitrate (B) Schiff’s reagent
(C) Molisch’s reagent (D) Bromine water
C-6.@ The following two compounds I and II can be distinguished by using reagent.

(I) (II)

(a) aq. NaHCO3 (b) Neutral FeCl3 (c) Fehling solution (d) Na metal
(A) a or c (B) b or c (C) c or d (D) b or d
C-7. Which of the following compounds will not react with I2 /OH– ?

(A) (B) (C) (D) CH3 – CHO

C-8. The compound A gives following reactions.

Na metal
H2 gas
A(C6H 8O2) 2, 4-DNP
Yellow orange ppt
O3
Zn/H2O
B(C6H8O4)

Its structure can be

(A) (B) OHC – (CH2)2 – CH = CH – COOH

(C) (D)

C-9.@ An organic compound X (C4H8O2) gives positive test with NaOH and phenolphthalein. Structure of X will be :
(A) CH3 - CH2 - CH2 - C - OH (B) CH3 - C - C - CH3
|| || ||
O O O

(C) CH3 - C - O - C2H5 (D) CH3 - C - OCH3

|| ||
BOBO-BA\Reliable\Adv.\Organic Chemistry\Structural isomerism, Structural identification & POC\Exercise

O O

C-10. Which of the following compound will give smell of NH3 with conc. NaOH ?
(A) CH3 - CH2 - C - NH2 (B) CH3 - C - CH2 - NH2
|| ||
O O

(C) (D) CH3 - CH2 - C - OH

||
O

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 JEE (Adv.)-Chemistry Structural isomerism, Structural identification & POC
C-11. Which of the following will not give positive test with CHCl 3 / KOH ?
(A) CH3–CH2–NH–CH3 (B) CH3–CH2–CH2–NH2

CH3
|
(C) (D) CH3 - CH - NH2

C-12. A positive carbylamine test is given by :

(A) N,N–dimethylaniline (B) 2, 4-dimethylaniline
(C) N-methyl-o-methylaniline (D) N-methylaniline

C-13. A research scholar get a mixture of three product during an experiment with ammonia. In product I only one
H of ammonia is replaced by ethyl group and in II two H atoms of ammonia are replaced by ethyl groups and
in III all the H-atoms are replaced by ethyl groups . Which test he should use to distinguish or separate the
products ?
(A) Carbyl amine test (B) Iodoform test (C) Fehling solution test (D) Hinsberg test

C-14._ Which of the following would produce effervescence with sodium bicarbonate?

(A) (B) (C) (D) All of these

C-15. A compound is heated with zinc dust and ammonium chloride followed by addition of the Tollen's reagent.
Formation of silver mirror indicates the presence of following group

(A) –CHO (B) (C) –NO2 (D) –NH2

C-16. In the Lassaigne’s test, one of the organic compound gives red colour with FeCl 3. Compound can be :
(A) Na2S (B) NH2CSNH2 (C) C6H5Cl (D) NaCN

C-17. Lassaigne’s test is used in qualitative analysis to detect

(A) Nitrogen (B) Sulphur (C) Chlorine (D) All of these
BOBO-BA\Reliable\Adv.\Organic Chemistry\Structural isomerism, Structural identification & POC\Exercise

C-18. The compound that does not give a blue colour in Lassaigne’s test is
(A) C6H5–NH2 (B) CH3CONH2 (C) NH2–NH2 (D) C6H5–NO2

C-19. Nitrogen containing organic compound when fused with sodium metal then it forms:
(A) NaNO2 (B) NaCN (C) NaNH2 (D) NaN3

C-20. The sodium extract of an organic compound on acidification with acetic acid and addition of lead acetate
solution gives a black precipitate. The organic compound contains
(A) Nitrogen (B) Halogen (C) Sulphur (D) Phosphorus

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JEE (Adv.)-Chemistry Structural isomerism, Structural identification & POC
PART - III : ONE OR MORE THAN ONE OPTIONS CORRECT TYPE
Section (A) : Structural isomerism
A-1. Which of the following pairs of structures represent the constitutional isomers ?

(A) CH2=CHCH2CH3 and

(B) CH3OCH2CH3 and

(C) (CH3)3CCH2CH2CH2OH and (CH3)2 CHCH2OCH2CH2CH3

(D) CH2ClCHClCH2CHO and CHCl2CH=CH–CH2–OH
A-2. Which of the following is/are the correct relationship ?
NH2
NHCH3 CH2 – CH2 – NH2 CH2 – NHC2H5
CH3 CH3

CH3 CH3 CH3

CH3
I II III IV

(A) I & II are functional isomers. (B) II & IV are metamers.

(C) I & IV are position isomers. (D) I & III are chain isomers.
A-3. Which of the following are functional isomers of methyl ethanoate ?
O
||
(A) CH3 – CH2 – COOH (B) CH3 C
CH H
|
OH

O O
|| ||
(C) CH3 – O – CH 2 – C – H (D) CH2 – C – CH3
|
OH

A-4. Which of the following can be the isomer(s) of C8H8O ?

BOBO-BA\Reliable\Adv.\Organic Chemistry\Structural isomerism, Structural identification & POC\Exercise

(A) (B) (C) (D)

Section (B) : Structural Identification

B-1.@ Which of the following compound on reductive ozonolysis give glyoxal as one of the product?

(A) (B) (C) (D)

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 JEE (Adv.)-Chemistry Structural isomerism, Structural identification & POC

CHO
(1) O3
CH3–CH
(2) Zn/H2O
B-2.@ C8H12 CHO
(X) H2/Ni
Y

True statements is/are

(A) Structure of X is

(B) Structure of X is

(C) Y on monochlorination produce 3 monochloro structural products.

COOH
(D) Oxidative ozonolysis product of X is CH 3–CH
COOH

B-3. Which of the following compounds give 1,4-Dimethyl cyclohexane when undergo catalytic hydrogenation?

CH2

(A) (B) (C) (D)

CH3

Section (C) : POC (Lab tests and Element detection)

C-1. Which of the following compounds on reaction with Na metal, liberate hydrogen gas ?

(A) CH3–OH (B) CH3–CºCH (C) Ph–OH (D)

C-2. Which of the following will perform iodoform reaction with I2/OH– ?
(A) CH3COCH2CH3 (B) CH3CONH2 (C) C6H5COCH3 (D) CH3CHO

O3 / Zn, H2O
C-3._@ ¾¾ ¾¾¾® (X) + (Y)
Compound (X) and (Y) can be distinguish by
(A) Tollen’s reagent (B) Fehling solution (C) Haloform test (D) 2,4-DNP Test
BOBO-BA\Reliable\Adv.\Organic Chemistry\Structural isomerism, Structural identification & POC\Exercise

C-4.@ Formic acid and acetaldehyde can be distinguish by-

(A) I2 + NaOH (B) Tollen’s reagent (C) Fehling solution (D) 2,4-DNP test

C-5.@ An organic compound having molecular formula C3H4, reacts with sodium metal to give a colourless and
odourless gas. Select the correct statements about this organic compound.
(A) It gives bromine water test
(B) It reacts with Bayer’s reagent
(C) It reacts with Tollen’s reagent
(D) It reacts with ammonical cuprous chloride.

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JEE (Adv.)-Chemistry Structural isomerism, Structural identification & POC
C-6. Compound P Liberates H2 gas with Na metal. P gives white precipitate with Tollen’s reagent, there is no
response towards Lucas reagent and compound Q gives instant turbidity with anhydrous ZnCl 2 / HCl, and
with sodium metal 1 mole of compound Q liberates 11.2 litre H2 gas at STP. Find the structural formula of
compound P and Q.
O CH3
|| |
(A) P is CH2 = CH - C - H (B) Q is CH3 - C - CH2 - O - CH3
|
OH
CH3
|
(C) P is CH3 – O – C º C – H (D) Q is CH3 - C - CH - CH3
| |
OH OH

C-7. Phenol can be distinguished from ethanol by

(A) Cerric ammonium nitrate (B) Neutral FeCl3
(C) Br2, H2O (D) Blue litmus

PART - I : JEE (ADVANCED) / IIT-JEE PROBLEMS (PREVIOUS YEARS)

* Marked Questions may have more than one correct option.
1. Four isomeric para-disubstituted aromatic compounds A to D with molecular formula C8H8O2 were given for
identification. Based on the following observations, give structures of the compounds. [JEE 2002, 5/60]
(i) Both A and B form a silver mirror with Tollen’s reagent; also B gives a positive test with FeCl 3
solution.
(ii) C gives positive iodoform test.
(iii) D is readily extracted in aqueous NaHCO3 solution.
2. Identify a reagent from the following list which can easily distinguish between 1-butyne and 2-butyne ?
[IIT-JEE-2002(S), 3/90]
(A) bromine, CCl4 (B) H2, Lindlar catalyst
(C) dilute H2SO4, HgSO4 (D) Ammonical Cu2Cl2 solution
3. In conversion of 2-butanone to propanoic acid which reagent is used ? [JEE 2005, 3/84]

(A) NaOH, NaI / (B) Fehling solution (C) NaOH, I2 / (D) Tollen's reagent
BOBO-BA\Reliable\Adv.\Organic Chemistry\Structural isomerism, Structural identification & POC\Exercise

4. The total number of cyclic isomers possible for a hydrocarbon molecular formula C4H6 is / are :
[IIT-JEE 2010]
5. In allene (C3H4), the type(s) of hybridisation of the carbon atoms is (are) : [IIT-JEE 2012]
(A) sp and sp3
(B) sp and sp 2
(C) only sp3
(D) sp and sp
2 3

6. The IUPAC name(s) of the following compound is(are) : [JEE. Adv 2017]

H3 C Cl
(A) 4-methylchlorobenzene (B) 4-chlorotoluene
(C) 1-chloro-4-methylbenzene (D) 1-methyl-4-chlorobenzene

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 JEE (Adv.)-Chemistry Structural isomerism, Structural identification & POC

PART - II : JEE(MAIN) / AIEEE PROBLEMS (PREVIOUS YEARS)

1. The general formula CnH2nO2 could be for open chain [AIEEE- 2003]
(1) diketones (2) carboxylic acids (3) diols (4) dialdehydes.

2. Which of the following reagents may be used to distinguish between phenol and benzoic acid ?
[AIEEE 2011, 4/120]
(1) Aqueous NaOH (2) Tollen's reagent (3) Molisch reagent (4) Neutral FeCl3

3.* Silver mirror test is given by which one of the following compounds? [AIEEE 2011, 4/120]
(1) Acetaldehyde (2) Acetone (3) Formaldehyde (4) Benzophenone

4. Ozonolysis of an organic compound 'A' produces acetone and propionaldehyde in equimolar mixture. Identify
'A' from the following compounds : [AIEEE 2011, 4/120]
(1) 1-Pentene (2) 2-Pentene
(3) 2-Methyl-2-pentene (4) 2-Methyl-1-pentene

5. Which branched chain isomer of the hydrocarbon with molecular mass 72u gives only one isomer of mono
substituted alkyl halide ? [AIEEE 2012, 4/120]
(1) Tertiary butyl chloride (2) Neopentane
(3) Isohexane (4) Neohexane

6. Iodoform can be prepared from all except : [AIEEE 2012, 4/120]

(1) Ethyl methyl ketone (2) Isopropyl alcohol
(3) 3–Methyl–2–butanone (4) Isobutyl alcohol

7. On heating an aliphatic primary amine with chloroform and ethanolic potassium hydroxide, the organic
compound formed is : [JEE MAINS -2014, 4/120]
(1) an alkanol (2) an alkanediol (3) an alkyl cyanide (4) an alkyl isocyanide

8. In the Victor-Meyer’s test, the colour given by 1°, 2° and 3° alcohols are respectively :
[JEE (Main) 2014 Online (20-04-14), 4/120]
(1) Red, colourless, blue (2) Red, blue, colourless
(3) Colourless, red, blue (4) Red, blue, violet
BOBO-BA\Reliable\Adv.\Organic Chemistry\Structural isomerism, Structural identification & POC\Exercise

9. Match the organic compounds in column-I with lassaigne’s test results in column-II appropriately :
[JEE(Main) 2015 Online (11-04-15), 4/120]
Column-I Column-II
(A) Aniline (i) Red colour with FeCl3
(B) Benzene sulfonic acid (ii) Violet colour with sodium nitroprusside
(C) Thiourea (iii) Blue colour with hot and acidic solution of FeSO4
(1) (A)-(ii); (B)-(iii); (C)-(i) (2) (A)-(iii); (B)-(i); (C)-(ii)
(3) (A)-(iii); (B)-(ii); (C)-(i) (4) (A)-(ii); (B)-(i); (C)-(iii)

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JEE (Adv.)-Chemistry Structural isomerism, Structural identification & POC
10. Which compound would give 5-keto-2-methyl hexanal upon ozonolysis ? [JEE Main 2015, 4/120]

CH3 CH3
CH3
H3C
(1) (2) (3) (4)
CH3 CH3

11. The test to distinguish primary, secondary and tertiary amine is :

[JEE (Main) 2016 Online (09-04-16), 4/120]
(1) Mustard oil test (2) C6H5SO2Cl
(3) Sandmeyer’s reaction (4) Carbylamine reaction
12. The tests performed on compound X and their inferences are: [JEE Main 2019, Online]
Test Inference
(a) 2,4 - DNP test Coloured precipitate
(b) Iodoform test Yellow precipitate
(c) Azo-dye test No dye formation
Compound 'X' is:

NH2 OH H3C CH3 H 3C CH3

N NH2 N
CH3 COCH3
CHO CHO
(1) (2) (3) (4)

13. An unsaturated hydrocarbon X absorbs two hydrogen molecules on catalytic hydrogenation, and also gives
following reaction : [JEE Main 2020, Online]

3 O [Ag(NH ) ]+
X ¾¾¾¾ ¾
® A ¾¾¾¾¾¾
3 2
® B(3 - oxo - hexanedicarboxylic acid)
Zn/H2 O

X will be :-
BOBO-BA\Reliable\Adv.\Organic Chemistry\Structural isomerism, Structural identification & POC\Exercise

(1) (2) (3) (4)

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 JEE (Adv.)-Chemistry Structural isomerism, Structural identification & POC

EXERCISE # 1
PART - I
A-1. 6 A-2. 4 A-3. 6 A-4. 9 A-5. 6
A-6. 4 A-7. 5 A-8. 8 A-9. 3 A-10. 3
B-1. 6 B-2. 1 B-3. 9 C-1. 5 C-2. 6
C-3. 4 C-4. 8 C-5. 81 C-6. 40

PART - II
A-1. (C) A-2. (A) A-3. (C) A-4. (D) A-5. (A)
A-6. (C) A-7. (D) A-8. (C) A-9. (B) A-10. (B)
A-11. (B) A-12. (B) A-13. (D) A-14. (B) A-15. (D)
B-1. (C) B-2. (B) B-3. (D) B-4. (C) B-5. (D)
B-6. (D) B-7. (B) B-8. (D) B-9. (B) B-10. (B)
B-11. (B) C-1. (D) C-2. (A) C-3. (C) C-4. (C)
C-5. (A) C-6. (B) C-7. (C) C-8. (C) C-9. (C)
C-10. (A) C-11. (A) C-12. (B) C-13. (D) C-14. (D)
C-15. (C) C-16. (B) C-17. (D) C-18. (C) C-19. (B)
C-20. (C)

PART - III
A-1. (ACD) A-2. (ABD) A-3. (ABCD) A-4. (BCD) B-1. (BCD)
B-2. (BCD) B-3. (ABC) C-1. (ABC) C-2. (ACD) C-3. (AC)
C-4. (AD) C-5. (ABCD) C-6. (BC) C-7. (ABCD)

EXERCISE # 2
PART - I

1. (A) (B) (C) (D)

BOBO-BA\Reliable\Adv.\Organic Chemistry\Structural isomerism, Structural identification & POC\Exercise

2. (D) 3. (C) 4. 5 5. (B) 6. (BC)

PART - II
1. (2) 2. (4) 3. (1,3) 4. (3) 5. (2)
6. (4) 7. (4) 8. (2) 9. (3) 10. (2)
11. (2) 12. (2) 13. (1)

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JEE (Adv.)-Chemistry Structural isomerism, Structural identification & POC

This Section is not meant for classroom discussion. It is being given to promote self-study and self
testing amongst the Reliable students.

PART- 1 : PAPER JEE (MAIN) PATTERN

SECTION–I : (Maximum Marks : 80)
˜ This section contains TWENTY questions.
˜ Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options is correct.
˜ For each question, darken the bubble corresponding to the correct option in the ORS.
˜ For each question, marks will be awarded in one of the following categories :
Full Marks : +4If only the bubble corresponding to the correct option is darkened.
Zero Marks : 0 If none of the bubbles is darkened.
Negative Marks : –1 In all other cases
1. On mixing a certain alkane with chlorine and irradiating it with ultraviolet light, it forms only one
monochloroalkane, this alkane could be :
(A) propane (B) pentane (C) isopentane (D) neopentane.

2. The prussian blue colour obtained during the test of nitrogen by Lassaigne’s test is due to the formation of :
(A) Fe4[Fe(CN)6]3 (B) Na3[Fe(CN)6] (C) Fe(CN)3 (D) Na4(Fe(CN)5NOS]

3. Of the five isomeric hexanes,the isomer which can give two monochlorinated compounds is ?
(A) n–Hexane (B) 2,3-Dimethylbutane (C) 2,2-Dimethylbutane (D) 2-Methylpentane

4. Among the following the one that gives positive iodoform test upon reaction with I2 and NaOH is ?
(A) CH3CH2CH(OH)CH2CH3 (B) C6H5CH2CH2OH
(C) CH3 - CH - CH3 (D) PhCHOHCH3
|
CH2 - OH

5. In the following sequence of reactions, the alkene affords the compound 'B'
O3 H O
CH3CH = CHCH3 ¾¾¾® A ¾¾2¾® B, compound B is
Zn
BOBO-BA\Reliable\Adv.\Organic Chemistry\Structural isomerism, Structural identification & POC\SAP

(A) CH3CH3CHO (B) CH3COCH3 (C) CH3CH2COCH3 (D) CH3CHO

6. If 1 mole H2 is reacted with 1 mole of the following compound.

Which double bond will be hydrogenated ?

(A) c (B) b (C) a (D) d

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 JEE (Adv.)-Chemistry Structural isomerism, Structural identification & POC
7. Identify the hydrocarbon having molecular formula C5H6 which gives white ppt with ammonical AgNO3 ?
(A) (B) (C) (D)
8. Which of the following compounds will give a positive iodoform test ?
(A) Methanol (B) 2,2-Dimethylpropanol
(C) Ethanol (D) Methanal

9. The following two compounds I and II can be distinguished by using reagent

(I) (II)

(I) (II)
(a) aq. NaHCO3 (b) Neutral FeCl3
(c) Blue litmus solution (d) Na metal (e) HCl/ZnCl2 anhydrous
(A) a or c (B) b or e (C) d or e (D) c or d

10. Which of the following compound cannot give Iodoform when react with IO –(hypoiodite)?
(A) (B)

(C) (D)

11. Which of the following statement is incorrect ?

(A) Phenol gives positive bromine water test.
(B) Aniline gives foul smelling compound on reaction with CHCl3 + KOH.
(C) Formic acid gives positive Tollen's test.
(D) Nitrobenzene gives positive Tollen's test.

O3
12. (x) C7H12 P+Q
Me2S
Compound P responds to Tollen’s test and iodoform test but Q does not respond with both the reagents.
Structure of compound (x) is :

(A) (B) (C) (D)

13. Yellow precipitate obtained during the test of halogen by lassaigne’s test is due to the formation of
(A) AgF (B) AgCl (C) AgBr (D) None of these
BOBO-BA\Reliable\Adv.\Organic Chemistry\Structural isomerism, Structural identification & POC\SAP

14. The Hinsberg's method is used for :

(A) preparation of primary amines (B) preparation of secondary amines
(C) preparation of tertiary amines (D) separation of amine mixtures

15. Which is incorrect match with respect to the reagent used for lab test ?
(A) Carbonyl ® 2,4–DNP
(B) Nitro ethane ® Zn, NH4Cl and AgNO3 (Mulliken Barker test)
(C) Phenol ® Anhydrous ZnCl2 + Conc. HCl (Lucas Reagent)
(D) Benzoic acid ® NaHCO3

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JEE (Adv.)-Chemistry Structural isomerism, Structural identification & POC
16. On oxidative ozonolysis of 3-Methylhex-3-ene, two products A & B are formed. A gives CO2 gas with sodium
bicarbonate, but B does not. The structures of A & B are respectively :

(A) & CH3–CH2–COOH (B) CH3–CH2–COOH & CH3–CH2–CH=O

(C) CH3–CH2–COOH & (D) CH3–CH2–CH2–COOH &

17. NH2 and can be differentiated by :

(A) Carbylamine reaction (B) Iodoform test

(C) Cold KMnO4 (D) Br2–H2O

18. Which of the following compound gives azo dye test ?

(A) (B) (C) (D)

19. Which of the following reagent can be used to distinguish the given compounds I & II ?

&
BOBO-BA\Reliable\Adv.\Organic Chemistry\Structural isomerism, Structural identification & POC\SAP

(A) Na metal (B) NaHCO3 (C) Lucas Reagent (D) 2, 4-D.N.P

20. A compound (P) on reaction with “Q” in basic medium (KOH) gives a bad smelling compound (CH3CH2NC).
Compound Q can be prepare by reaction of acetone with calciumhypochlorite (Ca(OCl)2]. P and Q can be
(A) CH3–CH2–NH2 & CHCl3 (B) CH3–CH2–NO2 & CH3Cl
(C) CH3–CH2–NH–CH3 & COCl2 (D) (CH3–CH2) 3N & Cl2

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 JEE (Adv.)-Chemistry Structural isomerism, Structural identification & POC
SECTION-II : (Maximum Marks: 20)
˜ This section contains FIVE questions.
˜ The answer to each question is a NUMERICAL VALUE.
˜ For each question, enter the correct numerical value (If the numerical value has more than two decimal
places, truncate/round-off the value to TWO decimal places; e.g. 6.25, 7.00, –0.33, –.30, 30.27, –127.30,
if answer is 11.36777..... then both 11.36 and 11.37 will be correct) by darken the corresponding bubbles
in the ORS.
For Example : If answer is –77.25, 5.2 then fill the bubbles as follows.
˜ Answer to each question will be evaluated according to the following marking scheme:
Full Marks : +4 If ONLY the correct numerical value is entered as answer.

21. How many structural isomeric ketones having molecular formula (C5H10O) give iodoform test ?

H / Ni Cl2 / hn
22. ¾¾2 ¾
¾® P ¾¾ ¾¾® Q (Total number of monochloro structural products).

23. How many alcohols give immediate turbidity with Lucas reagent having molecular formula (C 5H12O)?

24. How many hydrocarbons having molecular mass 68 can give white precipitate with Tollen's reagent ?

25. How many isomeric structural alkene on catalytic hydrogenation gives 2-Methylhexane?

PART 2 : PAPER JEE (ADVANCED) PATTERN

SECTION-I : (Maximum Marks : 12)
˜ This section contains FOUR questions.
˜ Each question has FOUR options (A), (B), (C) and (D). ONLY ONE of these four options is correct.
˜ For each question, darken the bubble corresponding to the correct option in the ORS.
˜ For each question, marks will be awarded in one of the following categories :
Full Marks : +3 If only the bubble corresponding to the correct option is darkened.
Zero Marks : 0 If none of the bubbles is darkened.
Negative Marks : –1 In all other cases

1. An organic compound “A” of molecular weight 120, gives Tollen’s reagent test and 2,4-DNP test but no
Iodoform with I2/OHQ. The compound “A” may be –
(A) Benzoic acid (B) Phenyl methyl ketone
(C) 2-Phenyl ethanal (D) 1-Phenyl ethanol

2. ‘X’ compound (C4H8O) decolorises bromine water & gives instant turbidity with lucas reagent. When ‘X’ react
with I2 & NaOH it give yellow ppt Identify ‘X’.
BOBO-BA\Reliable\Adv.\Organic Chemistry\Structural isomerism, Structural identification & POC\SAP

O CH3
(A) CH3–C–CH2–CH3 (B) CH 3–CH–CH=CH 2 (C) CH 3–C–CH 3 (D) CH 3–HC=CH–CH 2
OH OH OH

3. Compounds I and II can be distinguished by using reagent.

(I) (II)
4–Hydroxy–4–methylpent–2–enoic acid 5–Hydroxypent–2–ynoic acid
(A) NaHCO3 (B) Br2 / H2O
(C) HCl / ZnCl2 (anhydrous) (D) Cu2Cl2 / NH4OH
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JEE (Adv.)-Chemistry Structural isomerism, Structural identification & POC
4. Test to differentiate between ethanol (CH3CH2OH) and phenol (Ph–OH) is/are :
(A) NaHCO3 test (B) Neutral FeCl3 (C) Sodium metal test (D) All of these

SECTION-II : (Maximum Marks: 32)

˜ This section contains EIGHT questions.
˜ Each question has FOUR options for correct answer(s). ONE OR MORE THAN ONE of these four option(s)
is (are) correct option(s).
˜ For each question, choose the correct option(s) to answer the question.
˜ Answer to each question will be evaluated according to the following marking scheme:
Full Marks : +4 If only (all) the correct option(s) is (are) chosen.
Partial Marks : +3 If all the four options are correct but ONLY three options are chosen.
Partial Marks : +2 If three or more options are correct but ONLY two options are chosen,
both of which are correct options.
Partial Marks : +1 If two or more options are correct but ONLY one option is chosen
and it is a correct option.
Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered).
Negative Marks : –1 In all other cases.
˜ For Example : If first, third and fourth are the ONLY three correct options for a question with second
option being an incorrect option; selecting only all the three correct options will result in +4 marks. Selecting
only two of the three correct options (e.g. the first and fourth options), without selecting any incorrect
option (second option in this case), will result in +2 marks. Selecting only one of the three correct options
(either first or third or fourth option), without selecting any incorrect option (second option in this case),
will result in +1 marks. Selecting any incorrect option(s) (second option in this case), with or without
selection of any correct option(s) will result in –1 marks.

5. Which of the following compounds after complete hydrogenation will form three monochloro structural iso-
meric products ?

(A) (B)

C º CH
|
(C) (D) HC º C – CH – C º CH
BOBO-BA\Reliable\Adv.\Organic Chemistry\Structural isomerism, Structural identification & POC\SAP

6. Correct statment(s) about is /are :

3
(A) librate mole of H2 on treatment with Na. (B) Positive test with FeCl 3
2

(C) Positive test with NaHCO3 (D) Positive test with Tollen's reagent

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 JEE (Adv.)-Chemistry Structural isomerism, Structural identification & POC

7. Correct statement(s) about is/are

(A) It gives coloured solution with neutral FeCl3 solution.

(B) It liberates H2 gas with Na metal.
(C) It gives positive Iodoform test.
(D) It forms sweet smelling compound with alcohols.

8. Which of the following lab tests are given by Ph–CH=CH–COOH (cinnamic acid) ?
(A) Br2 water test (B) Neutral FeCl3 (C) NaHCO3 test (D) Tollen’s test

9. Which of the following tests will be given by (Squaric acid)?

(A) Br2 water test (B) 2, 4-DNP test (C) Neutral FeCl3 (D) Tollen’s test

O / Zn
10._@ ¾¾¾¾
3
® Products
H2O

Which of the following product(s) is/are formed in above reaction?

O
(A) (B) O O (C) H–C–H (D) CH3–C–H
O O

O / Zn
11._ ¾¾¾¾
3
® Products
H2O

Which of the following product(s) is/are formed in above reaction ?

BOBO-BA\Reliable\Adv.\Organic Chemistry\Structural isomerism, Structural identification & POC\SAP

(A) H–C–H (B) OHC–C–CH2–CHO

O O

(C) OHC–CH2–C–CH2 CHO (D) H–C–C–H

O O O

12. Acetaldehyde and propyne can be distinguish by :

(A) Tollen’s reagent (B) I2/NaOH (C) 2,4-DNP test (D) neutral FeCl3
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JEE (Adv.)-Chemistry Structural isomerism, Structural identification & POC
SECTION-III : (Maximum Marks: 18)
˜ This section contains SIX questions.
˜ The answer to each question is a NUMERICAL VALUE.
˜ For each question, enter the correct numerical value (in decimal notation, truncated/rounded-off to the
second decimal place; e.g. 6.25, 7.00, –0.33, –.30, 30.27, –127.30, if answer is 11.36777..... then both
11.36 and 11.37 will be correct) by darken the corresponding bubbles in the ORS.
For Example : If answer is –77.25, 5.2 then fill the bubbles as follows.
˜ Answer to each question will be evaluated according to the following marking scheme:
Full Marks : +3 If ONLY the correct numerical value is entered as answer.
Zero Marks : 0 In all other cases.

13. In how many reactions CO2 gas is released after reaction ?

(1) + NaHCO3 ¾¾® (2) + NaHCO3 ¾¾®

(3) + NaHCO3 ¾¾® (4) CH3–CH2–OH + NaHCO3 ¾¾®

(5) CH3–C º C–H + NaHCO3 ¾¾® (6) HCOOC2H5 + NaHCO3 ¾¾®

14._@ How many alkadienes form Isopentane after hydrogenation reaction (Including all structural isomers)?

15.@ How many acyclic structural isomeric compounds having molecular formula C6H12O can give haloform test
and 2,4-DNP test?

16. If sodium fusion extract of halogen is acidified with nitric acid followed by addition of AgNO 3. Find out the
correct statements.
(a) A white precipitate soluble in ammonium hydroxide confirm the presence of chloride ion.
(b) A yellow precipitate soluble in ammonium hydroxide confirm the presence of bromide ion.
(c) A yelllow precipitate insoluble in ammonium hydroxide confirm the presence of iodine ion.
(d) This sodium fusion extract of halogen is first boiled with conc. HNO 3 to decompose cyanide or sulphide
ions formed during lassaigne’s test, that interfere with AgNO 3 test for halogens.
BOBO-BA\Reliable\Adv.\Organic Chemistry\Structural isomerism, Structural identification & POC\SAP

17.@ How many no. of active hydrogen atoms are present in a compound (mol.mass 90) 0.45 g of which when
treated with Na metal liberates 112 ml of the H2 gas at STP ?

18._ ‘n’ number of alkenes yield 2,2,3,4,4-pentamethyl-pentane on catalytic hydrogenation and ‘m’ number of
monochloro structural isomers are possible for this compound.
Report your answer as (n + m).

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 JEE (Adv.)-Chemistry Structural isomerism, Structural identification & POC

PART - 1

1. (D) 2. (A) 3. (B) 4. (D) 5. (D)

6. (D) 7. (A) 8. (C) 9. (B) 10. (A)

11. (D) 12. (C) 13. (C) 14. (D) 15. (C)

16. (C) 17. (A) 18. (B) 19. (C) 20. (A)
21. 2 22. 2 23. 1 24. 2 25. 5

PART - 2
1. (C) 2. (B) 3. (C) 4. (B) 5. (CD)
6. (ABCD) 7. (ABC) 8. (AC) 9. (ABC) 10. (ABC)
11. (ABC) 12. (ABC) 13. 2 14. 2 15. 4

16. 4 17. 2 18. 4

Solution
PART - 1
1. The number of monohalogenation products obtained from any alkane depends upon the number of different
types of hydrogen it contains. Compound containing only one type of hydrogen gives only one monohalogenation
product.
CH3CH2CH3 — two types of hydrogen
propane (two monohalogenation structural products)
CH3CH2CH2CH2CH3 — three types of hydrogen
pentane (three monohalogenation structural products)

— four types of hydrogen (four monohalogenation structural products)

BOBO-BA\Reliable\Adv.\Organic Chemistry\Structural isomerism, Structural identification & POC\SAP

— one types of hydrogen (one monohalogenation structural product)

Thus the given alkane should be neopentane.

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JEE (Adv.)-Chemistry Structural isomerism, Structural identification & POC

2. 6NaCN + FeSO4 ¾¾® Na4[Fe(CN)6]

Na4[Fe(CN)6] + 4Fe3+ ¾¾ ¾® Fe4[Fe(CN)6]3. xH2O

xH 2O

Ferric ferrocyanide (Prussian blue)

3. 2,3-Dimethylbutane has two chemically different hydrogen atoms so it can give two monochlorinated
structural compounds.

4. For positive iodoform test, alcohol molecules must have CH3 - CH - group
|
OH

I 2 + NaOH –
Ph - CH - CH3 ¾¾ ¾ ¾¾® CHI3 + Ph – COO .
|
OH

O / Zn
CH3CH = CHCH3 ¾¾H¾
3 ¾®
5. O
2CH3CHO.
2

6. Aromatic p bonds are stable and cannot hydrogenate at room temperature.

7. Terminal alkyne can react with ammonical AgNO3 and compound have 3 DU.

CH3–CH
8. , | groups give positive iodoform test.
OH

9. (a) Both give the test with aq. NaHCO3 because both have –COOH group and acidic hydrogen.
(b) II give + ve test with neutral FeCl3 due to presence of phenolic –OH group, but (I) does not.
(c) In (I) and (II) acidic hydrogen atom is present so both give + ve test with blue litmus paper.
(d) In (I) and (II) acidic hydrogen atom is present so both give + ve test with Na metal.
(e) (I) give + ve test with HCl/ZnCl 2 due to presence of aliphatic alcoholic group, but (II) does not.

CH3–CH
10. , | groups give positive iodoform test.
OH
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11. Nitrobenzene does not give positive Tollen's test.

O3
12. CH3CHO +
Me2S

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 JEE (Adv.)-Chemistry Structural isomerism, Structural identification & POC
13. NaBr + AgNO3 ¾¾® AgBr (yellow)

14. Hinsberg's reagent is used to separation of amines mixtures.

15. Phenol does not gives Lucas reagent test.

16. O /H O
¾® CH3–CH2–COOH +
¾¾3¾ 2¾

17. 1º and 2º amine can be differentiated by Carbylamine test.

18. Aromatic 1º amine gives positive azo dye test.

19. Lucas reagent is used to distinguish between alcohol and phenol.

20. CH3–CH2–NH2 + CHCl3 + KOH ¾® CH3CH2NC

CH3–CO–CH3 + Ca(OCl)2 ¾® CHCl3 + (CH3COO)2Ca

21. ,

H / Ni Cl / hn
22. ¾¾2 ¾
¾® P ¾¾2¾¾® +

23.
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24. C5H8 , (Molecular Mass = 68)

25. , , , ,

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JEE (Adv.)-Chemistry Structural isomerism, Structural identification & POC
PART - 2
1. The compound gives Tollen’s reagent and 2,4-DNP test i.e. the compound is aldehyde. Further it gives no
Iodoform test. Expected molecular formula is C8H8O. Therefore the correct answer is (C).

2. CH 3–CH–CH=CH 2 gives bromine water test, lucas reagent test and haloform test.
OH

CH3

3.
HC = C – C – CH3 HOOC –C C – CH2 – CH2
HOOC H OH OH
(I) (II)
(I) gives immediate turbidit with by Lucas reagent and (II) does not give turbidity appreciably.

4. Ethanol can not give neutral FeCl3 test but phenol gives this test.

5. H / Ni s
¾¾2 ¾
¾®

C º CH CH2 – CH 3
| H2 / Ni |
HC º C – CH – C º CH ¾¾ ¾ ¾® H3 C – CH2 – CH – CH2 – CH3 3 monochloro products

Both structures give three monochloro structural isomeric products.

6.

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7.

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 JEE (Adv.)-Chemistry Structural isomerism, Structural identification & POC
9. Generally terminal alkyne and –CH=O group gives positive test with tollens reagent.

reductive
10. ¾¾¾¾® O+O O + H–C–H
ozonolysis

reductive
11. ¾¾¾¾® 2HCHO + H–C–C–CH2–C–H + H–C–CH2–C–CH2–C–H
ozonolysis
O O O O O O

12. Acetaldehyde and propyne can be distinguish by Tollen's reagent and Iodoform test.

13. (1) + NaHCO3 ¾¾® + H2O + CO2 ­

(2) + NaHCO3 ¾¾® + H2O + CO2 ­

(3) + NaHCO3 ¾¾® No reaction

(4) CH3–CH2–OH + NaHCO3 ¾¾® No reaction

(5) CH3–C º C–H + NaHCO3 ¾¾® No reaction

(6) HCOOC2H5 + NaHCO3 ¾¾® No reaction

CH3 CH3
14.
H 2C =C–C =CH 2 CH3 –C=C=CH2
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CH3–C–CH2–CH2–CH2–CH3 CH3–C–CH–CH2–CH3
15.
O O CH3

CH3
CH3–C–CH2–CH–CH3 CH3–C–C–CH3
O CH3 O CH3

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JEE (Adv.)-Chemistry Structural isomerism, Structural identification & POC

17. 122 ml of H2 is obtained from 0.45 g

0.45 ´ 22400
22400 ml of H2 is obtained from = 90 g
112

90 g compound gives one mole H2 gas

i.e. 2H obtained from 1 mole of compound.
Ans. No. of active H = 2

18. Only one alkene

Three monochloro isomers are possible as it has three different types of 'H' atoms.

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