NEET

CHEMISTRY
MODULE - 4
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 INDEX
S. No. Chapter Name Page No.

1. Classification & Nomenclature

Theory and Exercise 1.1 to 1.5 1-31

Exercise 2 32-35

Exercise 3 36-38

Exercise 4 (Previous Year’s Questions) 39-40

Answer Key 41
2. Isomerism (Structural, Conformational & Geometrical Isomerism)

Theory and Exercise 1.1 to 1.3 43-62

Exercise 2 63-66

Exercise 3 67-69
Exercise 4 (Previous Year’s Questions) 70

Answer Key 71
3. General Organic Chemistry

Theory and Exercise 1.1 to 1.6 73-102

Exercise 2 103-109
Exercise 3 110-113

Exercise 4 (Previous Year’s Questions) 114-116

Answer Key 117

4. Hydrocarbons

Theory and Exercise 1.1 to 1.3 119-163

Exercise 2 164-169

Exercise 3 170-174

Exercise 4 (Previous Year’s Questions) 175-180

Answer Key 181
5. Purification & Analysis of Organic Compounds

Theory and Exercise 1.1 to 1.3 183-203

Exercise 2 204-206

Exercise 3 207-208
Exercise 4 (Previous Year’s Questions) 209

Answer Key 210

Classification & Nomenclature CHEMISTRY

Chapter CLASSIFICATION
&
1
NOMENCLATURE

Chapter Summary Introduction

Around the year 1780, chemists began to distinguish between
• Introduction
organic compounds obtained from plants and animals and
• Classification of organic inorganic compounds prepared from mineral sources. Berzilius,
compound a Swedish chemist proposed that a ‘vital force’ was responsible
for the formation of organic compounds. However, this notion
• Homologous Series
was rejected in 1828 when F. Wohler synthesised an organic
• Common Name compound, urea from an inorganic compound, ammonium
• Derived System cyanate.
• IUPAC Name NH4CNO Heat
 →NH2CONH2
Ammonium cyanate Urea
• Nomenclature of open
The pioneering synthesis of acetic acid by Kolbe (1845) and that
chain compounds
of methane by Berthelot (1856) showed conclusively that
• Nomenclature of Alicyclic
organic compounds could be synthesised from inorganic
compounds sources in a laboratory,
• Nomenclature of The development of electronic theory of covalent bonding
Aromatic Compounds ushered organic chemistry into its modern shape.
(a) Tetravalence of carbon : Shapes of organic compounds
Carbon contains four electrons in outermost shell, so need to
make 4 bonds to get nearest noble gas configuration.
Spectroscopic studies and properties of methane have revealed
that the four C-H bonds in methane are identical and
symmetrically disposed at an angle of 109° 28' to each other.
This may be accounted for on the basis of Hybridization.
i.e. Carbon (ground state) 1S2 2S2 2Px1 2Py1 2Pz0
1s 2s 2p
Carbon (excited state)
sp hybridization
3

The SP3 hybrid orbitals of carbon overlap with four 1s atomic

orbitals of four H atoms.

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CHEMISTRY Classification & Nomenclature
1s of H
H
sp3 orbitals of C

C
1s of H
1s of H

1s of H H H H
The four valencies of carbon atom can be represented by following way
Hybridisation
Structure σ bonds π bonds [No. of σ bond Shape Bond Angle No. of Bond angles
+ No. of l.p.]
Tetrahedral
C 4 0 sp3 109°28' 6
(Non planar)
Planar
C 3 1 sp2 120° 3
(Trigonal)
C 2 2 sp Linear 180° 1
C 2 2 sp Linear 180° 1

Note: Double bond between carbon atoms is known as olefinic bond and triple bond between carbon
atoms is known as Acetylenic bond.
(b) Catenation: Self linking property of C-atom is known as catenation. It is responsible for the variety and
large number of organic compounds. It may also give rise to open chain and closed chain nature of
compounds. Bond energy for catenation of C is maximum.
Bond energy (in Kcal:) C—C Si—Si N—N P—P
85 54 39 50
SP2 SP2 SP3 SP SP σ bonds = 13(σ e– = 26)
↓ ↓ ↓ ↓ ↓
Ex. CH2 = CH — CH — C ≡ CH π bonds = 3(π e– = 6)
CH3 Olefinic bond = 1
↑
SP3 Acetylenic bond = 1
(c) Tetrahedral shape: The four covalent bonds are directed towards the four corners of a regular
tetrahedron

109°28’
C

Note: Hybridisation influences the bond length and bond enthalpy (strength) in compounds. The sp hybrid
orbital contains more s character and hence it is closer to its nucleus and forms shorter and stronger
bonds than the sp3 hybrid orbital. The sp2 hybrid orbital is intermediate in s character between sp and
sp3 and, hence, the length and enthalpy of the bonds it forms, are also intermediate between them. The
change in hybridisation affects the electronegativity of carbon. The greater the s character of the hybrid
orbitals, the greater is the electronegativity. Thus, a carbon atom having an sp hybrid orbital with 50%
s character is more electronegative than that possessing sp2 or sp3 hybridised orbitals.

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Classification & Nomenclature CHEMISTRY
Classification of Carbon

There are four types of carbon present in organic compounds. The carbon which is attached to no other

carbon atom or directly attached with one, two, three and four carbon atoms are known as primary,

secondary, tertiary and quarternary carbon atom respectively.

On the basis of carbon atom, hydrogen atoms bonded with 1o, 2o and 3o C-atoms are named as primary,

secondary and tertiary hydrogen atom respectively.

C C
C C 3°
C C C C C
1° 3° C C 4°

[Note: 4°H is not possible because 4°C is attached with four carbons]

Ex.

1°C 2°C 3°C 4°C 1°H 2°H 3°H Csp3 Csp2 Csp

CH3
CH3 – CH – CH2 – C – CH3 5 1 1 1 15 2 1 8 0 0
CH3 CH3

1 7 3 0 3 14 1 9 2 0

Classification of organic compound

The existing large number of organic compounds and their ever-increasing number has made it necessary to

classify them on the basis of their structures. organic compounds are broadly classified as fallow:

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CHEMISTRY

ORGANIC COMPOUND

Open -chain or acyclic

Closed – Chain or
Or aliphatic compounds
Cyclic compound or Ring compounds

Alicyclic Aromatic
Saturated Unsaturated compounds compound
(Single bond between (double or triple bond
carbon-carbon atom) between carbon-carbon atom)

CH3–CH2–CH3 Non-Benzenoid

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Homocyclic Heterocyclic Benzenoid Heterocyclic
Propane compound
compounds compounds compounds compound
O CH3–CH2=CH2 HC ≡ CH
propene Ethyne Cyclo NH2
CH3–C–H O
O hexane O
Acetaldehyde Tetrahydrofuran Furan
CH2=CH–C–OH (THF) O
O Cyclo Tropone
Benzene Aniline N
Acrylic acid butene |
CH3–C–OH S H
O N Thiophene Pyrrole
Acetic acid |
CH2=CH–C–H H
CH3C ≡ N
Acrylaldehyde Naphthalene N
Acetonitrile Pyridine
Pyrrolidine
Classification & Nomenclature
Classification & Nomenclature CHEMISTRY
Aliphatic or Open chain compounds: Those compounds which consists of unbranched (straight) or
branched chains.
CH3
Ex. CH3 CH2 CH2 CH3, CH2–CH2–CH2–CH3, CH3 CH2 CH CH3 , CH3 CH CH2 OH
OH CH3
(Unbranched) (Branched)
These compounds are further categorised as :

Saturated compounds
(a) In such type, carbon are bonded with single bonds and do not have any carbon-carbon multiple bond.
O O
Ex. - CH3–CH2–CH3 , CH3–CH2–C≡N, CH3 CH2 C CH3 , CH3 CH2 C OH

Unsaturated Compounds
(a) These compounds have at least one carbon-carbon multiple bond (may be double bond or triple bond)
CH2 = CH – CH3 Propene
O
CH ≡ C – CH3 Propyne , CH2=CH–C≡N, CH ≡ C C H
(b) Due to presence of π bonds these are more reactive.

Closed chain compounds

In these compounds first & last carbon are attached with each other to form a ring.
CH2 H2C CH2
Ex. H2C CH2 , O
Cyclopropane Epoxyethane

Alicyclic compounds: These are the compounds having the properties like aliphatic compounds.
These may be saturated or unsaturated like aliphatic compounds.
These are further classified as :

(a) Homocyclic compounds

Those compounds in which the ring is formed by one type of atoms only. When ring is composed of C
atoms then compound is called carbocyclic.

cyclopentane cyclobutene cyclohexane

(b) Heterocyclic Compounds

These are cyclic compounds having ring composed of more than one type of atoms.

N
O
H
Tetrahydro furan (THF) Pyrrolidine
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CHEMISTRY Classification & Nomenclature
Aromatic compounds Compounds which follow Huck’s rule are called aromatic compound.
Huck’s rule :
(i) Compound should be cyclic.
(ii) Compound should have planer ring.
(iii) There should be cyclic resonance in ring.
(iv) Number of electrons involved in cyclic resonance should be = (4n+2)π eΘ . Where
n = 0, 1, 2, 3, . . .
n=0 2π electrons or 1 pair
n=1 6π electrons or 3 pairs
n=2 10π electrons or 5 pairs
n=3 14π electrons or 7 pairs

(a) Benzoid compounds: These compounds are composed of Benzene ring.

Naphthalene Benzene Anthracene Phenanthrene

NH2

Aniline
(b) Non benzoid compounds: These compounds are composed of other than benzene ring.

O
Tropone

(C) Heterocyclic Compounds: These are cyclic compounds having ring composed of more than one type of
atoms.

O S N N
H
Furan Thiophene Pyrrole Pyridine
Note: Atoms other than carbon in the ring are known as hetero atoms. Like O. S & N in above compound.
Homologous Series
A series of compounds in which compounds having same functional group are arranged in their increasing
molecular weight as called homologous series.
Characterstics of Homologous series
(i) Member of a homologous series are called homologous of each other.
(ii) All the members of a series have same general formula, general methods of preparation and similar
chemical properties due to same functional group.

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Classification & Nomenclature CHEMISTRY
(iii) The homologues show difference in physical properties due to change in molecular mass and structural
arrangement of molecule.
(iv) Members of a homologous series differ by n(–CH2–) in their molecular formula and by n(14) in their
molecular weight.
(v) More than one different homologous series can have same general formula.
(vi) Homologues can never be isomers and isomers can never be homologues since isomers always have
same molecular formula while homologues have different molecular formula.

Some Standard Homologous Series Are

S.No. Name of Series General Formula I-homologue II-homologue
1. Alkane CnH2n+2 CH4 CH3–CH3
2. Alkene CnH2n CH2=CH2 CH2=CH–CH3
3. Alkyne CnH2n–2 HC≡CH HC≡C–CH3
4. Halo alkane CnH2n+1X CH3–X CH3–CH2–X
5. Alcohol CnH2n+2O CH3–OH CH3–CH2–OH
6. Ether CnH2n+2O CH3–O–CH3 CH3–O–CH2–CH3
7. Aldehyde CnH2nO H–CHO CH3–CHO
8. Ketone CnH2nO CH3 C CH3 CH3 C CH2 CH3
O O
9. Carboxylic acid CnH2nO2 H–COOH CH3–COOH
10. Ester CnH2nO2 H C O CH3 H C O CH2CH3
O O
or CH3 C O CH3
O
11. Anhydride CnH2n–2O3 O O O O
H C C H CH3 C O C H
Unstable
12. Amide CnH2n+1NO H–CONH2 CH3–CONH2
13. Acid halide CnH2n–1XO(X=Cl,-Br,–I,–F) O O
H C Cl CH3 C Cl
(Unstable)
14. Nitro alkane CnH2n+1NO2 O O
CH3–N CH3–CH2–N
O O
15. Amine CnH2n+3N CH3–NH2 CH3–CH2–NH2
16. Alkenyne CnH2n–4 CH2=CH–C≡CH CH3–CH=CH–C≡CH
17. Alkadiene CnH2n–2 CH2=C=CH2 CH3–CH=C=CH2

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CHEMISTRY Classification & Nomenclature
Exercise 1.1
1. Ratio of 1°, 2° & 3° C atoms in 6. Which one is not correct for a homologous
CH3 series -
(1) All members have same general
(mesitylene) respectively formula
CH3 CH3 (2) All members have same chemical
is:- properties
(1) 3: 2: 1 (2) 1: 2: 3 (3) All members have same physical
(3) 1: 1: 1 (4) 3: 1: 1 properties
(4) All members have same functional
2. Which of the following have only one type
group
of H-atoms?
7. Number of C-atoms in 3rd member of
(1) (2)
ketone homologous series are -
(1) 5 (2) 3
(3) (4) (1) and (3) both
(3) 4 (4) 6

8. Moleculer formula of 2nd member of

3. Which of the following is/are unsaturated
homologous series of 1º amine is:
hydrocarbon?
(1) CH3 − CH2 − C − CH2 − CH3 (1) C2H5N (2) C2H7N
|| (3) C2H6N (4) C3H9N
O
9. Number of π bonds in
(2) CH3 –CH– C ≡ N
| CH2=CH–CH=CH–C≡CH is:
CH3
(1) 2 (2) 3
(3) CH3 – CH = CH − CH − CH3 (3) 4 (4) 5
|
CH3
10. The total number of secondary H–atoms in
(4) All the structure given below are:
4. Which of the following is Homocyclic CH3 − CH − CH2 − CH3
|
compound:- CH3
(1) Pyrrole (2) Nephthaline
(1) 1 (2) 4
(3) Thiophene (4) Pyridine
(3) 3 (4) 2
5. Maximum number of hetero atoms are
11. No. of hetero atoms present in the following
present in the ring.
heterocyclic compound is:
O
N

(1) (2) O
N
N H
O (1) 3 (2) 2
NO2 (3) 1 (4) zero
O
(3) (4) 12. Select the molecule which has only one π− bond.
N (1) CH≡CH (2) CH2=CHCHO
| O 2N NO2
H (3) CH3CH=CH2 (4) CH3–CN
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Classification & Nomenclature CHEMISTRY
13. Which of the following statement is 17. Which of the following is homocyclic
incorrect regarding alkenes: compound
(1) There will be a double bond between NH2
any two carbon atoms. (1) (2)
(2) General formula is CnH2n–2
(3) These are called olefins because they
(3) (4) All
reacts with halogens to form oily
substances Olefins
18. Which of the following pair represents
(4) Due to presence of π bonds these are homologues
more reactive than alkanes. (1) CH3–CH2–OH, CH3–CH2–CH2–CH2–OH

14. Number of C-atoms in 2nd member of ester (2) CH3–CN, CH3–CH2–NH2

(3) CH3–NH2, CH3–NH–CH3
homologous series are:
(4) All
(1) 4 (2) 3
(3) 5 (4) 6 19. Incorrect statement is:
(1) Homologues differ by ‘CH3’ in their
15. How many sp C-atoms are present in given
molecular formula
compound:
(2) Homologues can be isomers
CH ≡ C – CH2 – CH = CH – CN
(3) Different homologous series always
(1) 2 (2) 4
have different general formula
(3) 3 (4) 5
(4) All

16. Minimum C required for a open chain 20. Molecular formula of third member of ether
compound to have two 3°C and one 2°C: family (Alkoxy alkane) is:
(1) 7 (2) 5 (1) C3H6O (2) C3H8O
(3) 8 (4) 6 (3) C4H10O (4) C4H8O

Nomenclature Of Organic Compounds

Organic compounds are named mainly in three systems :

(a) Common Name or Trivial Name System

(b) Derived Name System

(c) IUPAC Name or Geneva Name System

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CHEMISTRY Classification & Nomenclature
Common Name

Some Common Names Based on Source :

S.No. Compound Common Name Source
1. CH4 Marsh gas (Fire damp) Marshy Area
2. CH3OH Wood spirit (Carbinol) Destructive
distillation of wood
3. CH3CH2OH Grain alcohol Grain
4. NH2 C NH2 Urea (Carbamide) Urine
O
5. HCOOH Formic acid Formica (Red ants)
6. CH3COOH Acetic acid Acetum (Vinegar)
7. HOOC–COOH Oxalic acid Oxalis plant
8. CH3 CH COOH Lactic acid Lactum (Milk)
OH
9. CH3CH2CH2COOH Butyric acid Butter
10. HO CH COOH Tartaric acid Tamarind
HO CH COOH
11. HO CH COOH Malic acid Malum (Apple)
CH2 COOH
12. CH2 COOH Citric acid Citrus (Lemon)
HO C COOH Fruits

CH2 COOH

Common Names Of Hydrocarbon Radicals

(A) For saturated hydrocarbons:
-H
Alkane  Alkyl (monovalent radical)
→
( CnH2n+2 ) ( CnH2n+1− )

Ex. 1 (n =1) CH4 → –CH3 Methyl Ex. 2 (n =2) C2H6 → –C2H5 Ethyl
Ex. 3 (n =3)CH3–CH2–CH3 (i) CH3–CH2–CH2– n-Propyl
–H
Propane (ii) CH3–CH–CH3 Iso-Propyl

Note: Radicals having same molecular formula but different structures are differentiated in their
common names by using prefix as n, iso, neo, secondary, tertiary, active etc.
Ex.4 (n = 4)
(a) –H (i) CH3–CH2–CH2– CH2– n-butyl
CH3–CH2–CH2–CH3
n-Butane (ii) CH3–CH2–CH–CH3 sec–butyl /Active–sec butyl

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Classification & Nomenclature CHEMISTRY

(b) CH3–CH–CH3 –H (i) CH3–CH–CH2– Isobutyl

CH3 CH3
Isobutane
(ii) CH3–C–CH3 tert–butyl
CH3

Ex.5 (n = 5)
(i) CH3–CH2– CH2–CH2–CH2– n-Pentyl

(a) CH3–CH2– CH2–CH2–CH3 –H (ii) CH3–CH2–CH2–CH–CH3 Active sec. pentyl

n-Pentane
(iii) CH3–CH2–CH–CH2–CH3 sec. pentyl

(i) CH3–CH–CH2–CH2– Iso-Pentyl

–H CH3
(b) CH3–CH–CH2–CH3 Active iso pentyl
(ii) CH3–CH–CH–CH3
CH3
Iso Pentane CH3

(iii) CH3–C–CH2–CH3 tert pentyl or tert amyl

CH3

(iv) –CH2–CH–CH2–CH3 Active pentyl or active amyl

CH3
CH3 CH3

(C) CH3—C—CH3 → CH3—C—CH2– neo-pentyl

CH3 CH3
neo-pentane

Note: Pentyl is also called amyl group.

Bivalent Radical
–H
Alkane Alkyl removal of H from
Alkylidene
the same carbon

removal of H from
adjacent/vicinal carbon
Alkylene
Ex.(i) –H
CH4 CH3– –H CH2
Methane Methyl Methylidene
Ex.(ii) CH3–CH3 –H CH3–CH2–
–H
CH3–CH
ethane ethyl (ethylidene)
–H
CH2–CH2
(ethylene)

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CHEMISTRY Classification & Nomenclature

Ex.(iii) CH3– CH2– CH3 –H –H

CH3–CH2–CH2– CH3– CH2–CH
n-propane n-propyl (n-propylidene)
–H
CH3–CH–CH2
(propylene)
(B) For unsaturated hydrocarbons:
(a) −H
Alkene  → Alkenyl
Ex.(i) CH2 = CH2 → CH2= CH– Ethenyl (vinyl)
ethene

Ex.(ii) CH –CH=CH –H –CH2–CH=CH2 2-Propenyl (Allyl)

3 2
Propene –H
CH3–C=CH2 ISO-Propenyl

–H Propenyl
CH3–CH=CH–
(b) –H
Alkyne → Alkynyl
−H
Ex.(i) H − C ≡ C − H →H − C ≡ C − ethynyl
ethyne

Ex.(ii) CH3–CH2≡CH –H
CH3–C≡C– Propynyl
Propyne –H
–CH2–C≡CH 2-Propynyl (Propargyl)
Aryl Radical

–H
–C6H5 or
(i)
Phene/Benzene Phenyl

CH2– CH C

–H –H
CH3
(ii) Benzyl Benzal or Benzo
Benzylidene (Benzylidyne)

Toluene

CH3 CH3 CH3

O–Tollyl m–Tollyl
p–Tollyl
CH2OH CH2Cl CHCl2 CCl3

Ex.

Benzyl alcohol Benzyl chloride Benzal chloride Benzo chloride

Note: Aromatic hydrocarbons are called Arenes and respective radicals are called aryl radicals.
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Classification & Nomenclature CHEMISTRY
1. Common Names of Hydrocarbon Derivatives:
Common names of hydrocarbons derivatives are given in two different systems.
System – I [Used for naming of those compounds which have functional group without carbon]
Format : Name of hydrocarbon radical (R–) + Name of functional group (–Z)

R–Z

S.NO Functional Group Name of Functional

Group
1 –SO3H Sulphonic acid
2 –OH Alcohol
3 –SH Thioalcohol
4 –NH2/–NH–/–N– Amine
|
5 –O– Ether
6 –S– Thioether
7 –X Halide
8 C Ketone
O
9 –C≡N Cyanide
10 –N C isocyanide

Ex. CH3–CH2–OH CH3–CH2–CH2–OH

ethyl alcohol n-propyl alcohol

CH3–CH–CH3
OH CH3–CH2–CH2–CH2–OH
Iso propyl alcohol n-butyl alcohol
CH3–CH–CH2–CH3 CH3–CH–CH2–OH
OH CH3
Secondary butyl alcohol iso butyl alcohol

CH3
CH3 C OH CH3–NH2
CH3 Methyl amine
t-butyl alcohol

CH3–CH2–NH2 CH3–NH–CH3
ethyl amine dimethyl amine

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CHEMISTRY Classification & Nomenclature
CH3–N–CH3
Et3N or (CH3–CH2)3 N
CH3 Triethyl amine
Trimethyl amine
CH3–CH2–N–CH2–CH3 CH3 O CH2CH3
CH3 Ethyl methyl ether
Diethyl methyl amine
CH3 CH SO3H CH3 C CH CH3
CH3 O CH3
Isopropyl sulphonic acid Isopropyl methyl ketone

CH2 Br

Benzyl bromide
Exception : although ketone, cyanide, isocyanide are functional group with carbon can be given name
as system-I.
System II:
Format : Prefix + Suffix

On the basis of total no. It is suggested for

of carbon in compound functional group
including C of present in the
functional group compound

If total carbon ⇒ One Two Three Four Five

Prefix ⇒ Form Acet Propion Butyr Valer
And suffix is given by following table
S.No. Functional Group Suffix
1. –COOH ic acid
2. O O ic anhydride
C O C
3. C OR Alkyl ……. ate
O
4. –COX yl halide
5. –CONH2 amide
6. –CHO aldehyde
7. –C≡N –o nitrile
8. N C –o isonitrile
Ex. H2C 0 CH3CN CH3–NC CH3–CHO
Formaldehyde, Acetonitrile, Acetoisonitrile, Acetaldehyde

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Classification & Nomenclature CHEMISTRY

H–COOH CH3–COOH CH3–C–Cl CH3–CH–CHO H–C–NH2 CH3–C–NH2

Formic acid Acetic acid O O O
CH3
Acetyl chloride Isobutyraldehyde Formamide Acetamide

O O
CH3 C O CH3 CH3 C CH3 C CH3 CH C O C2H5
O O
O CH3 C CH3 CH2 C CH3 O

O O
Methyl acetate Acetic anhydride Acetic propionic anhydride Ethyl isobutyrate
• Prefix “Acryl” is used for the compounds which have total three carbon atoms and double bond is on
2nd carbon.
Ex. CH2 = CH–COOH Acrylic acid
CH2 = CH–CHO Acryl aldehyde [Acrolein]
• Prefix “Croton” is used for the compounds which have total four carbon atoms and double bond is on
2nd carbon
Ex. CH3–CH = CH–COOH Crotonic acid
CH3–CH = CH – CHO Croton aldehyde

Derived System
According to this system name of any compound is given by using name of representative compound of the
homologous series.
Series Name of Homologous Name of Representative Structure of group
series compound
1. Alkane Methane
C
2. Alkene Ethylene >C C<
3. Alkyne Acetylene C C
4. Alcohol Carbinol
C OH

5. Aldehyde Acetaldehyde
C CHO

6. Ketone Acetone
C C C
O
7. Carboxylic acid Acetic acid
C COOH

Note: Representative member of a homologous series is that simplest member of the series which
represents all the characteristics of series.
H CH3 CH3
Ex. CH3–CH2–C–CH3 CH3–CH– C – CH3 CH3
CH3 CH3 CH3–CH– COOH
Ethyl dimethyl methane Isopropyltrimethylmethane Dimethylacetic acid

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CHEMISTRY Classification & Nomenclature
H CH3 O
CH3–CH2–C–OH CH3–C___C–H
CH3 CH3
Ethylmethyl carbinol Trimethyl acetaldehyde
Exercise 1.2
1. Which of the following is tertiary radical:- 7. Which of the following has wrong common
(1) CH3–CH– (2) CH2 = CH – name:
CH3 (1) CH3 − CH − CH2 − OH Isobutyl alcohol
(3) CH3–CH2 –CH2– (4) (CH3)3C – |
CH3
CH3 (2) CH3 –CH–CH3 sec. propyl alcohol
2. Common name of CH3–C–Cl is:- |
OH
CH3
CH3
(1) Neobutyl chloride
(3) CH3–C–CH2–OH Neopentyl alcohol
(2) Neopentyl chloride
(3) Tert. butyl chloride CH3
(4) Isobutyl chloride (4) CH3 –OH methyl alcohol
3. Common name of CH2 = CH – CN
8. The structure of Acetonitrile is:
(1) Vinyl Cyanide (2) Acrylonitrile
(1) H–CN
(3) Ethyl cyanide (4) Both (1) & (2)
(2) CH3–CH2–CN
4. Structure of Isopropyl acetate is:-
(3) CH3–NC
(1) CH3–CH– C –O–CH3
(4) CH3–CN
CH3 O
(2) H3C–C–O–CH–CH3
9 Common name of given compound:
O CH3 CH3–C–O–CH=CH2
(3) CH3–CH– C –O–H O
CH3 O (1) Ethenyl ethanoate
(4) H–C–O–CH–CH3 (2) Vinyl ethanoate
O CH3 (3) Vinyl acetate
(4) Ethenyl acetate
5. Identify primery amine:
(1) Trimethyl amine
(2) sec. Butyl methyl amine 10. Incorrect common name is
(3) Ethyl methyl amine (1) (CH3)3C–OH tert. butyl alcohol
(4) Tert. butyl amine (2) (CH3)2CH–OH Isopropyl alcohol
(3) CH2=CH–CH2–OH Acryl alcohol
6. Identify secondary radical:
(4) CH3COOH Acetic acid
(1) CH2 = CH – CH2 –
CH3 − CH − CH2 − CH3
(2) 11. Which is true for Isobutyl alcohol
|
(3) CH3 − CH − CH2 − (1) It is 1° alcohol
| (2) It has three 1° carbon atoms
CH3 (3) It has one 3° carbon atom
(4) All
(4) CH2

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Classification & Nomenclature CHEMISTRY
2
12. sp carbon atoms present in Acryl aldehyde 15. Which of the following is not correctly
is- matched:
(1) 0 (2) 3 (1) Acetonitrile CH2 = CHCN
(3) 2 (4) 1
(2) Allyl chloride CH2 = CH − CH2Cl
13. Which is correctly named (3) s-Butyl group CH3–CH–C2H5
(1) CH2 = CH – CH = O (Acryl aldehyde)
(2) CH2 = CH – CH2 – OH (Acryl alcohol)
(4) Ethylene chloride CH2Cl –CH2Cl
(3) CH2 = CH – CH2 – COOH (Crotonic acid)
(4) All
14. Structural formula of isopropyl methanoate 16. Common name of the given structure is
is: CH2–OH
(1) CH3–C–O–CH–CH3
CH2–OH
O CH3
(2) H–C–O–CH2–CH–CH3 (1) Ethylene Glycol

O CH3 (2) Ethene dialcohol

(3) CH3–C–O–CH2–CH2 (3) Glycerol
O CH3 (4) Ethylene alcohol
(4) H–C–O–CH–CH3
O CH3

IUPAC Name or Geneva Name System

(International union of pure and applied chemistry)
1. FORMAT :
Secondary prefix (P–2) + Primary Prefix (P–1) + Word Root (Alk) + Primary Suffix (S–1) + Secondary Suffix (S–2)

Names of “Cyclo” word is Number of carbon It represents saturation or Suffix of principal

substituents in used if principal atoms in selected unsaturation in principle functional group
alphabetical order carbon chain is principle carbon carbon chain
with position (If cyclic chain ane (C – C)
required)
ene (C = C) (With
yne (C ≡ C ) Position
if required)

No. of carbon in PCC Word root No. of carbon in PCC Word root
1C  → Meth 11C  → Undec
2C  → Eth 12C  → Dodec
3C  → Prop 13C  → Tridec
4C  → But 14C  → Tetradec
5C  → Pent 15C  → Pentadec
6C  → Hex 16C  → Hexadec
7C  → Hept 17C  → Heptadec
8C  → Oct 18C  → Octadec
9C  → Non 19C  → Nonadec
10C 
→ Dec 20C 
→ Icos
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CHEMISTRY Classification & Nomenclature
2. Selection of principal functional group (PFG) : If compound has only one functional group then it is
considered as principal functional group.
Note: Following functional group are always considered as substituent.
Functional group Prefix name
–X Halo
– OR Alkoxy
–C–C–
Epoxy
O
– NO2 Nitro
– NO Nitroso

• If compound has more than one functional groups then according to IUPAC priority table, Higher
priority functional group is considered as principal functional group (PFG) & rest of the functional
groups are considered as substituent. (suffix is used for PFG and preffix is used for substituent)
S. Functional group Prefix Suffix
No.
1. –(C) OOH (carboxylic acid) × oic acid
– COOH carboxy carboxylic acid
2. – SO3H (sulphonic acid) sulpho sulphonic acid
3. O × oic anhydride
–(C)
O(anhydride)
–(C)
O
O × Carboxylic anhydride
–C
O
–C
O
4 – (C)OOR (ester) × alkyl ........... oate
–COOR alkoxy carbonyl / Alkanoyloxy alkyl...... carboxylate
O O
–C–OR –O–C–R
5. O × oyl halide
–(C)–X(acid halide)
O Halo carbonyl/ halo formyl carbonyl halide
–C–X
6. O × amide
–(C)–N–(amide)
O carbamoyl / akanoyl amino Carboxamide
–C–N– O O
–C–N– –N–C–R
7 –(C)N (cyanide) × Nitrile
–CN cyano Carbonitrile

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Classification & Nomenclature CHEMISTRY

8. –N C(isocyanide) isocyano/carbyl amino Isonitrile

9. –(C)HO (aldehyde) oxo Al

–CHO formyl Carbaldehyde
10. –(C)–(Ketone) keto/oxo One
O
11. –OH (alcohol) hydroxy Ol
12. – SH (thio alcohol) mercapto Thiol
13. –N–(amine) amino amine

Note: (C) atom written in brackets means it has been included in the parent chain
Ex.
1. CH2 = CH – CH2 – Cl
No. of F.G. = 2
P.F.G. = Not Present (Cl is a substituent)

2. CH2 = CH – CH2 – OH
No. of F.G. = 2
P.F.G. = –OH
Cl CN

3. CH2 = CH –CH– CH2–CH–OH

No. of F.G. = 4
P.F.G. = –CN [–Cl, – OH are substituent]

4. CH3–CH–C ≡ C – CH–C–Cl
N=0 CN O
No. of F.G. = 4
P.F.G. = –COCl [–N=0, –CN are substituents]

Note: C= C & C ≡ C are also considered as functional group

3. Selection of principal carbon chain (PCC)
• Select the longest continuous chain of carbon atoms which have maximum number of PFG, multiple
bonds and substituents
Priority order : PFG > Maximum no. of multiple bond > Maximum no. of carbon atoms > Maximum
no. of substituents (Chain length)
Case – I : If P.F.G. has carbon then its carbon is included in principal carbon chain (P.C.C)
NO2

Ex. 1. CH2 = CH –CH– CH2–CH– C–NH2

P.F.G.
CH2–OH O

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CHEMISTRY Classification & Nomenclature

CN P.F.G.

2. CH3 – CH2 –C– CH–CH2– CH2– OH

CH2
Case – II : If P.F.G. does not have carbon then including the carbon to which P.F.G. is directly attached,
C atoms are convered on its both side to get free terminals.
CH3
Ex. 1. CH –CH – C – OH P.F.G
3

CH2 CH2–Cl
CH3
CH2 – CH3
2. CH3 – CH2 – C – CH2 – C – SO3H P.F.G.
CH2 CH3
CH3 O P.F.G.

3. CH3 – CH – C
CH2– CH3

4. Numbering of principal carbon chain (PCC)

• Selected principle carbon chain is numbered from the side where P.F.G, multiple bonds and
substituents is getting lowest possible number
Priority order : P.F.G. > Multiple bond (= > ≡ ) > Lowest set of locants of Substitutents >
Alphabatical order
4 3 2 1 1 2 3 3 2 1
Ex. CH3–CH2–CH=CH2 , CH2=CH–CH2–NO2 , CH2=CH–CH2–OH
6 5 4 3 2 1
CH ≡ C–CH2–CH = CH– CH3 , CH3–C ≡ C–CH2–CH = CH2
1 2 3 4 5 6

1 2 3 4 5 1 2 3 4 5 4 3 2 1
CH2=CH–CH2–C ≡ CH , CH3–CH–CH=CH2 , CH2–CH2–CH–CH=CH2
OH Cl OH
2 3 1 2 3 4 5 4 3 2 1
CH3–CH2–CH – C – CH2–CH3 ,CH ,CH3–CH–CH2–CH– CH3
3–C–CH2–CH3
C≡N CH2 O Cl Br
1 4
Chloro Bromo
1 2 3 4 5 6 7 6 5 4 3 2 1
CH3–CH2–CH–CH2– CH–CH2–CH3 , CH3–CH2–CH–CH2– CH–CH3
C2H5 CH3 Br CH3
Ethyl Methyl
6 5 4 3 2 1 Br 4 3 2 1 Cl
CH3 – CH2 – 3CH – CH – CH – CH3
, CH – CH2 – CH2 – C — Cl
1 2 6 4 5 Br Cl
Br CH3 Cl
2, 3, 4 right numbering
3, 4, 5 wrong numbering
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Classification & Nomenclature CHEMISTRY
5. Use of numerical prefixs : These are used to show multiplicity of more than one identical substituents
1 2 3 4
• For simple substituents we use di, tri, tetra Ex. CH3–CH–CH–CH3 2,3-dimethyl butane
CH3 CH3
CH2Cl
1 2 3 4 5
• For complex substituents we use bis, tris, Ex. CH2=CH–C–CH2–CH3 3,3-bis (chloro methyl)
CH2Cl pent-1-ene

6. Alphabatical order of substituents :

Prefix cyclo, iso & Neo are considered in alphabetical order but other prefixes like n, sec, tert, di, tri …
etc are not considered in alfabatical order Ex. dimethyl, isopropyl, isobutyl, sec. butyl
Note: If two or more than two double bonds/triple bonds are present in parent chain then "a" is added
to word root (alka).
1 2 3 4 1 2 3 4 5
Ex. (i) CH2 = CH–CH= CH2 Buta -1, 3-diene =
(iii) CH2 C–CH2 –C ≡ CH Pent-1-en-4yne
1 2 3 4 5
(ii) CH ≡ C–CH2 –C ≡ CH Penta -1,4-diyne

2. If carbon containing F.G. is present as a susbstituent in a compund (except aldehyde & ketone) then its
carbon is not included in p.c.c.
Substituent
O O OH O
Ex.
H2N – C – CH2 – CH – CH2 – C – OH NC–CH2– CH2–CH–CH=CH–C–NH2
↑ ↑
Substituent CH3 (P.F.G.) Substituent P.F.G

3. If aldehyde or ketone is present as substituent in a compound and if possible then include its carbon in
p.c.c. but if not possible (as per p.c.c. selection rule), it is excluded.

4. Two prefixes [formyl and oxo] are suggested for –CH=O and their use is conditional
• Formyl → When its C is not included in p.c.c.
• Oxo → When its C is included in p.c.c.

5. If more than two carbon containing identical P.F.G are directly attached to principal carbon chain then
their carbon is not included in principal carbon chain.
• When C of P.F.G. is out side the p.c.c. its additional suffix is used as secondary suffix.
OH
3 2 1
H2C – C – CH2 P.F.G.=3 2-Hydroxypropane-1,2,3-tricarboxylic acid
HOOC COOH COOH
O O
1 2 3 4 5
HO – C – CH2 – CH – CH2 – C – OH P.F.G. = 2 3-(Carboxy methyl) pentane dioic acid
CH2 – COOH

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CHEMISTRY Classification & Nomenclature
IUPAC name of open chain compounds
1 3 5 2 3
2 4 6 1 4 6
1. 2. 3. 5
3-Methyl pentane 2,3-dimethyl hexane Hex-3-ene

6 3
4 1
5 3
4. 2 3,4,5-Trimethyl hex-5-en-2-ol 5. 4 2
1,3-Butadiene
1 OH
1
3
4 2 2
5
3 1 4 O
3
2 1
6.
6
7 7. 8. 4 OC2H5
6 5
5-Methyl hept-4-en-1-yne Ethyl but-2-enoate
7
4,4-diethyl heptane
O O
4 2
4 3 2 1 5 1
HO OH Me Me
3
9. 10. Me Me
10
OH OH 6 8
7 9
2,3-Dihydroxy-1,4-butanedioic acid
(Tartaric acid) 5,6-Diethyl-7-methyl-3-decene
O
4 5
5 3 2 4
1
11. 6 12. 3 3,3-Diethyl-2,4-dimethyl pentane
7 1 2
3-Ethyl-4,5-dimethyl-5-hepten-2-one
Exercise 1.3
1. Principal functional group in the compound 3. Which of the following has correct
numbering according to IUPAC.
CH3–CH–CH–CHO is:- 1 2 3 4 5
CONH2 (1) CH2=CH–CH2–CH2–CN
(1) Aldehyde (2) Amine 2 3 4 5 6 7

(3) Amide (4) Ketone (2) CH–CH2–CH=CH–C≡CH

1CH2
2. Which of the following selected principal 2 3
1 4 5
carbon chain (PCC) is correct:- (3) CH≡C–CH–CH=CH2
CH3 OH
2 3 4
(1) CH3 – C – OH (4) CH3–CH2–CH–CH2–CH3
CH3 1CN

(2) CH2 — CH2 4. What is the correct IUPAC name of the

COOH COOH following?
CH3–CH=CH–CH–CH2–CH3
(3) CH3 – CH2 – CH – CH2– CH3
COOH
CN (1) 4-Carboxy hex-2-ene
CH3 – CH2 – C — C – CH2– CH3 (2) Hex-4-ene-3-carboxylic acid
(4)
CH2 CH2 (3) 2-Ethyl pent-3-enoic acid
(4) 2-(prop-1-enyl) butanoic acid
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Classification & Nomenclature CHEMISTRY
5. Select the incorrect match of IUPAC name 10. IUPAC name of CH3–CH=CH–C≡CH is
with following structure:_ (1) Pent-2-en-4-yne
Me Et (2) Pent-3-en-1-yne
(3) Pent-3-yne-1-en
(1) (2-ethyl-4-methylpentane)
O (4) Pent-2-yne-1-en
(2) H
(2-ethyl-pent-3-enal) 11. IUPAC name of the (CH3)2CHCH(CH3)2 is
(1) 1,1,2,3-Tetramethylethane
(2) 1,2-Di-isopropylethane
(3) CH3COOMe (methyl ethanoate)
(3) 2,3-Dimethylbutane
(4) HC≡C–CH=CH–CH3 (pent-3-en-1-yne)
(4) 2,3,3-Trimethylbutane
6. The IUPAC name of
12. The IUPAC name of CH3–CH– CH3 is
Cl
NH2
CH3–C–CH2–CH=CHCH3 :
(1) Dimethyl amine
CHO
(2) 2-Amino propane
(1) 5-chloro-5-Formyl hex-2-ene
(3) Isopropyl amine
(2) 2-methyl-2-chloro hex-4-en-1-al
(4) 2-Propanamine
(3) 2-chloro-2- methyl hex-4-en-1-carbaldehyde
(4) 2-chloro-2-methyl hex-4-en-1-al 13. The IUPAC name of the compound
OCH3
O
CH3 CH3–CH2– CH– CH2–OH
7. The IUPAC name of CH3 – C –CH is:
CH3 (1) 2-Methoxy-1-butanol
(1) 2-Methylbutan-3-one (2) 3-Methoxy-1-butanol
(2) 3-Methylbutan-2-al (3) 3-Methoxy-4-butanol
(3) 2-Methylbutan-3-al (4) 1-Hydroxy-2-methoxy-butane
(4) 3-Methylbutan-2-one
14. The IUPAC name of the compound is
8. IUPAC name of the given compound is
O CN
CH3–CH=C–CH3
CH2–CH3 CH3–C–CH2–C–CH3
(1) 2-Ethyl-2-butene CH3
(2) 3-Ethyl-2-butene (1) 4–Cyano–4–methyl–2–oxopentane
(2) 2-Methyl-4-oxo pentanenitrile
(3) 3-Methyl-3-pentene (3) 2,2-Dimethyl-4-oxo pentanenitrile
(4) 4-Cyano-4-methylpentan-2-one
(4) 3-Methyl-2-pentene
15. The IUPAC name of the following compound
9. The IUPAC name
of the compound
CH CH –CH–CH –CH is
3 2 2 3
CH3–CH– CH2– CH2–OH is
CH3CH2–CH–CH2–CH3
CH3
(1) 1-Pentanol (1) 3,4-Dimethyloctane
(2) Pentanol (2) 3-sec. Pentylpentane
(3) 2-Methyl-4-butanol (3) 3,4-Diethylhexane
(4) 3-Methyl-1-butanol (4) 3,4-Dimethylhexane

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CHEMISTRY Classification & Nomenclature
16. The IUPAC name of CH3CH2NHCHO is 22. The correct IUPAC name of
(1) N-formylethanamine CH3–CH2–C–COOH is :
(2) Ethylaminomethanal CH2
(3) N-Ethylmethanamide
(1) 2-Methyl butanoic acid
(4) N-Ethylmethanol
(2) 2-Ethylprop-2-enoic acid
17. The IUPAC name of (3) 2- Carboxy-1- butene
O O (4) None of the above

23. The IUPAC name of compound

O H HO–C=O CH3
is:
CH3 – C = C – C–H
(1) Acetic anhydride NH2 Cl
(2) Formyl anhydride
(1) 2-amino-3-chloro-2-methyl-2-
(3) Butane–2,4–dione
pentenoic acid
(4) Ethanoic methanoic anhydride
(2) 3-amino-4-chloro-2-methyl-2-
18. The correct IUPAC name of: pentenoic acid
CH3 – CH2 – CH – CH2COCl
(3) 4-amino-3-chloro-2-methyl-2-
CH3
pentenoic acid
(1) 3–Methylpentanoyl chloride
(2) 3–Methylbutanoyl chloride (4) All of the above
(3) 1–Chloro–3–ethylbutanone
F Cl
(4) 1–Chloro–3–methyl pentanone
24. CH3–C – CH2 – C – CH3
19. The IUPAC name of N≡C–CH2–CH2–OH is CH2CH3 CH2CH3
(1) 3–Hydroxy propanecarbonitrile
The IUPAC name of this compound is :
(2) 3–Hydroxy propanenitrile
(1) 2-choro-2,4-diethyl-4-fluoro pentane
(3) 2–Hydroxy ethyl cyanide
(2)3-fluoro-5-chloro-3-methyl-5-ethylhexane
(4) 1–Hydroxy–2–cyanoethane
(3)3-chloro-5-fluoro-3,5-dimethylheptane
20. The IUPAC name for the given compound is: (4)3,5-dimethyl-5-fluoro-3-chloroheptane

O 25. The I.U.P.A.C. name of

(1) 2-Acetyl prop-1-ene (C2H5)2CH.CH2OH is :-
(2) Pent-1-en-4-one (1) 2-Ethyl butan-1-ol
(3) Pent-4-en-2-one
(2) 2-Methyl pentan-1-ol
(4) Formyl propene
(3) 2-Ethyl pentan-1-ol
21. The IUPAC name of the given compound is (4) 3-Ethyl butan-1-ol
Br
Cl–C–CHO 26. The correct IUPAC name of 2-ethyl-
F 3-pentyne is :
(1) 2-Bromo-2-chloro-2-fluoroethanal (1) 3-methyl hex -4-yne
(2) 1-Bromo-1-chloro-2-fluoroethanal (2) 4-ethyl pent-2-yne
(3) 2-Fluoro-2-chloro-2-bromoethanal (3) 4-methyl hex-2-yne
(4) 1-Fluoro-1-chloro-1-bromoethanal (4) None of these

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Classification & Nomenclature CHEMISTRY
27. The IUPAC name of the given structure is :- 34. The IUPAC name of
H2N–CH– CH–CHO C6H5CH = CH – COOH is :
HOOC COOH (1) Cinnamic acid
(1) 3-Amino-2-formyl butane-1, 4-dioic acid (2) 1-Phenyl-2-carboxy ethane
(2) 3-Amino-2, 3-dicarboxy propanal (3) 3-Phenyl prop-2-enoic acid
(3) 2-Amino-3-formyl butane-1, 4-dioic acid (4) Dihydroxy-3-phenyl propionic acid
(4) 1-Amino-2-formyl succinic acid
35. The IUPAC name of C6H5—CH—CH2—CCl3
28. I.U.P.A.C. name of
(CH3)2CH−CH2−CH2Br is :- C6H5
(1) 1-bromopentane is
(2) 2-methyl-4-bromopentane (1)1,1,1-Trichloro-3,3-diphenyl propane
(3) 1-bromo-3-methylbutane (2)1,1-Diphenyl-3,3,3-trichloro propane
(4) 2-methyl-3-bromopropane (3) Both (1) and (2)
(4) 3,3,3-trichloro-1,1-diphenyl propane
29. N
36. Which one of the following I.U.P.A.C.
IUPAC name of above compound is :- names is correct ?
(1) N, 1– di methyl propanamine (1) 2-Methyl-3-ethyl pentane
(2) N-methyl–2–butanamine (2) 2-Ethyl-3-methyl pentane
(3) Sec. butyl methyl amine (3) 3-Ethyl-2-methyl pentane
(4) 1–methyl–1–methylamino propane (4) 3-methyl-2-ethyl pentane
30. The correct name of 3,3-dimethyl
37. The IUPAC name of
propanamide is
(1) 2-Methyl butanamide CH3–C– O– CH2–C–OH is
(2) 3-Methyl butanamide O O
(3) Iso-propyl ethanamide (1) 1-acetoxy acetic acid
(4) Iso propyl acetamide (2) 2-acetoxy ethanoic acid
(3) 2-ethanoyl oxyacetic acid
31. IUPAC name will be CH2–CH– CH2
(4) 2-ethanoyl oxyethanoic acid
CN CN CN
(1) 1,2,3-Tricyano propane 38. CH3–O–C–CH2–COOH
(2) Propane 1,2,3- trinitrile O
(3) 1,2,3-cyano propane The correct IUPAC systematic name
(4) Propane-1,2,3-tricarbonitrile of the above compound is :
32. The IUPAC name of compound- (1) 2-Acetoxy ethanoic acid
HOOC–CH–COOH (2) 2-Methoxy carbonyl ethanoic acid
COOH (3) 3-Methoxy formyl ethanoic acid
(1) Tricarboxy methane (4) 2-Methoxy formyl acetic acid
(2) Propane trioic acid 39. The IUPAC name of
(3) Tributanoic acid CH3
(4) Methanetricarboxylic acid CH3CH(OH)CH2–C–OH is
33. The IUPAC name of is – CH3
Br (1) 1, 1-Dimethyl-1, 3-butanediol
(1) 2-bromo-4-isopropylpentane (2) 4-Methyl-2, 4-pentanediol
(2) 2, 3-dimethyl-5-bromohexane (3) 1, 3, 3-Trimethyl-1, 3-propanediol
(3) 2-bromo-4, 5-dimethylhexane (4) 2-Methyl-2, 4-pentanediol
(4) 5-bromo-2, 3-dimethylhexane

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CHEMISTRY Classification & Nomenclature

IUPAC names of Alicyclic compounds

1. Format : P–2 + P–1(cylco) + alk + S–1 + S–2

2. If compound is composed of ring and open chain then principal carbon chain is selected according to
given priority order.
Part containing (P.F.G.) > part containing (M.B.) > part containing more carbon in chain > part
containing substituent.
If every thing is same in both part then priority is given to cyclic part.

Ex. open chain → P.C.C. , Ring → substituent

OH
P.F.G
Ring : P.C.C. , Open chain : Substituent

Ring : P.C.C., Open chain : Substituent

3. If carbon containing functional group is directly attached with cyclic ring then it is considered as part
of parent chain but carbon of functional group is not included in parent chain.
4 3 O
Ex. (Wrong numbering) (Alk → Hept ×)
5
2 C1
– OH
6
7 Cl
HO
3 2 O 3 2 O
4 C – OH (Alk → Hex) 4 C – OH
1 1
5 5
6 Cl 6 Cl
HO HO
6-chloro-5-hydroxy cyclo hex-2-ene carboxylic acid
P-2 P-1 alk S-1 S-2
2 1
CH2–C≡N
Cyclohexyl ethane nitrile
Note : Rest of the rules are same as open chain compound.
CN CHO O
2 COOC2H5
1

Ex. Cyclohexane Carbonitrile Cyclo propane carbaldehyde Ethyl 2-oxo cyclohexane carboxylate

Some other examples

NH2
OH H3C 6
1 NO2
1 2 3
4 2 4 2
5 3
3 4 1
5 CH3 5 CH3
6 NO2 6
3-Methylcyclohexanol 6-Methyl-4-nitrocyclohex-2-en-1 amine 1-Methyl-3-nitrocyclohexene

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Classification & Nomenclature CHEMISTRY

Exercise 1.4
1. The IUPAC name for the compound CH3
O 5. The IUPAC name of is:
H2C—CH–CH3 OH
(1) 3–Methylcyclo–1–buten–2–ol
(1) Propylene oxide
(2) 1,2-Oxopropane (2) 4–Methylcyclo–2–buten–1–ol
(3) 1,2-Epoxypropane (3) 4–Methylcyclo–1–buten–3–ol
(4) Methyloxirane
(4) 2–Methylcyclo–3–buten–1–ol.
2. What is the structure of 3-Phenylpentane?
(1) CH3–CH–CH2–CH2–CH2–CH3 6. Which is correct IUPAC name:
(1) CH3–C–CH2–Br 1-Bromopropanone
O

(2) CH3–CH2–CH–CH2–CH3 O
O

(2) O O Cyclohexane-1,3,5-
(3) CH3–CH2–CH–CH2–CH3
trione
O
CHO
OH
(4) CH3–CH2–CH–CH2–CH3 (3)
2-Ketocylopentanecarbaldehyde
(4) All are correct

3. Correct IUPAC name of is:

C = CH– CH3
(1) 1–Ethyl–3,3–dimethylcyclohexane 7. The IUPAC name of H5C2 is
(2) 3–Ethyl–1,1–dimethylcyclohexane
(1) 3-Cyclopropyl-3-ethyl-2-propene
(3) 1,1–Dimethyl–3–ethylcyclohexane
(2) 1-Cyclopropyl-1-ethylpropene
(4) 3,3–Dimethyl–1,1–ethylcyclohexane
(3) 3-Cyclopropyl-2-pentene
4. The IUPAC name of the compound (4) (1-Ethyl-1-propenyl) cyclopropane
HO C≡CH
Cl is
Br 8. The IUPAC name of is

(1) 1-Chloro-2-bromocylohexane
(1) 1-Ethynyl-1-hydroxycyclohexane
(2) 1,2-Bromochlorocyclohexane
(2) 1-(Hydroxycyclohexyl) ethyne
(3) 4-Bromo-3-chlorocyclohexane
(3) 1-Ethynylcyclohexanol
(4) 1-Bromo-2-chlorocyclohexane
(4) 1-Acetylenyl-1-hydroxycyclohexane

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CHEMISTRY Classification & Nomenclature

Nomenclature of Aromatic Compounds

1. If one principal functional group is present on bezene ring, its common name is accepted in IUPAC
system.
Ex.
COOH CN CH = O
OH

Benzoic acid Benzonitrile Benzaldehyde

Phenol
(Benzene carboxylic acid) (Benzene carbonitrile) (Benzene carbaldehyde)
O
NH2 CONH2 C – O– R

Aniline Benzamide Alkyl benzoate

(Benzenamine) (Benzene carboxamide) (Alkyl benzene carboxylate)

2. If more than one principal functional groups are present on ring then they are named as derivative of
benzene and suffix for P.F.G. is used as per IUPAC rules.
Ex.
` COOH OH CH=O
COOH

CH=O
Benzene-1,2-dicarboxylic acid OH
(Phthalic acid) Benzene-1,4-diol Benzene-1,3-dicarbaldehyde
(Quinol)
3. In aromatic hydrocarbon if side chain is also present then except methyl benzene, ethyl benzene,
ethenyl benzene, isopropyl benzene, ethynyl benzene, side chain is considered as main part and ring
is considered as a substituent and is named as phenyl.
Ex.
4 3 2 1
1 2 3
CH3 – CH2 – CH – CH3 4 5
1 2 3
CH2 = CH – CH2
2-Phenylbutane 3-Phenylpropane 2-Methyl-4-phenyl pentane
Exception
CH3 CH2–CH3 CH=CH2 C≡CH CH3–CH–CH3

Methyl Ethyl Ethenyl Ethynyl Isopropyl

benzene benzene benzene benzene benzene
(Toluene) (Styrene) (Cumene)
Note: Toluene is also valid in IUPAC

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Classification & Nomenclature CHEMISTRY
4. In Hydrocarbon derivative, the part having secondary functional group [Ring or side chain] is
considered as main part and other is considered as substituent.
Ex.
1 2 3 4
CH2 – CH2 – CH – CH3 CH2 – CH2 – CH2 – CH3
4 2
CH2 – CH2 – CH3
1
Br 3 5 CH2 – CH2 – OH 3
4 2
2 6 5 1
1 OH
3-Bromo-1-phenyl butane 2-phenyl ethan-1-ol 6
Br
1-Bromo-4-butylbenzene 3-propylphenol

5. If functional group is present in both parts the part having principal functional group is considered as
the main part.
Ex.
2
OH
2 1 1
3 1 CH2 – CH2 – OH 2 6

4 6 3 5
Br 5 4

2-[4-bromophenyl] CH2 – CH2 – Br

1 2
ethanol 4-[2-Bromoethyl]
phenol
Ex.
CH3 CHCl2 CH2 OH
1
CH3
2

1,2-Dimethyl benzene Phenyldicholoromethane Phenyl methanol

(o-Xylene) (Benzal chloride) (Benzyl alcohol)

OCH3 O–C6H5
CHO

Methoxy benzene Phenoxy benzene

(Anisol) (Diphenyl ether) Benzaldehyde
1 2
CO CH3
CHO
1 OH CH2–CH2–CHO
2

2-hydroxy benzaldehyde 1–phenyl ethenone

(salicylaldehyde) 3-phenylpropanal (Acetophenone)

COC6H5 NO2
1
NO2 2

3 NO2
Diphenylmethanone 1,3- Dinitrobenzene
(Benzophenone) Nitrobenzene (m-dinitrobenzene)

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CHEMISTRY Classification & Nomenclature

NH2 NH2 CH2NH2

NH2

Benzenamine Benzene-1,2-diamine Phenyl methanamine

(Aniline) (O-Phenylenediamine) (Benzylamine)
Ex.
CH3

CH2–CH2–NH2
O O O–C–CH3
O
C6H5–C–O–C–C6H5 (4-Methyl Phenyl)
2-Phenyl ethanamine Benzoic anhydride ethanoate

Exercise 1.5
1. Which of the following name will be 3. The correct IUPAC name of following
incorrect? compound is:
OH
(1) 3, 6-Dimethyl cyclohexene Cl Br

(1) 2-Bromo -6-chloro phenol

(2) 1, 6-Dimethyl cyclohexene (2) 1-Bromo -2-hydroxy -3-chloro benzene
(3) 2-chloro -6- bromo phenol
(3) 6, 6-Dimethyl cyclohexene (4) 1-chloro -2- hydroxy -3- bromo benzene

4. The correct IUPAC name of the following

(4) 1, 5-Dimethyl cyclohexene
compound is
OH
2. The correct IUPAC name of

COOH
CN
Br
H3C–O–C O–C–CH3 (1) 4-Bromo-3-cyano phenol
O O (2) 6-Bromo-3-hydroxy benzo nitrile
(3) 2-cyano-4-hydroxy bromo benzene
(1) 3-methoxy carbonyl-5-ethanoyloxy
(4) 2-Bromo-5-hydroxy benzo nitrile
benezene carboxylic acid
(2) 5-methoxy carbonyl-3- ethanoyloxy O OH
benezene carboxylic acid 5. The IUPAC name of is
(3) 5-ethanoyloxy -3- methoxy carbonyl- (1) 4-hydroxy -2-pentanone
benezene carboxylic acid (2) 4–oxo-2-Pentanal
(4) 3-ethanoyloxy -5- methoxy carbonyl- (3) Pentane-4-ol-2-one
benezene carboxylic acid (4) Pentane-2-one-4-ol

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Classification & Nomenclature CHEMISTRY
6. Correct IUPAC name of following 9. The IUPAC name of
compound is: OH O
CH3 OH

(1) 2-Carboxyphenol
(2) 2-Hydroxybenzoic acid
(3) 1-Carboxy-2-hydroxybenzene
Cl (4) (2–Hydroxyphenyl) methanoic acid
(1) 1–Methyl–4–chlorobenzene
10. Give IUPAC name of the following:
(2) 1–Chloro–4–methylbenzene OC2H5
O
(3) 4–Chlorotoluene C
(4) Both (2) & (3)

7. Correct IUPAC name of following

compound is: Br
Cl (1) Ethyl–1–bromo 4–benzoate
(2) Ethyl 4–bromobenzoate

NO2 (3) Bromobenzoate

O2 N
(1) 4–Chloro–1,3–dinitrobenzene (4) Bromobenzene ester
(2) 1–Chloro–2,4–dinitrobenzene
11. IUPAC name of given compound is :-
(3) 2–Chloro–1,5–dinitrobenzene
Cl
(4) 2,4–Dinitro–1–chlorobenzene
COOH

8. The IUPAC name of O2N CHO

COOH
OCH3
(1) 4–chloro benzene–1,3–dicarboxylic acid
(1) 2-Methoxy-4-nitrobenzaldehyde
(2) 5–carboxy–2–chloro benzoic acid
(2) 4-Nitro-2-methoxybenzaldehyde
(3) 4–chloro benzene–1,3–dioic acid
(3) 3-Methoxy-4-formylnitrobenzene
(4) 2-Formyl-4-nitroanisole (4) Both (1) and (3)

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CHEMISTRY Classification & Nomenclature
Exercise7.
2 The IUPAC name of following
1. Which of the following has more than one polyfunctional compound is:
type of H-atom: O
OHC COOH
(1) (2)
H H
(3) CH3 (4) CH2
(1) 2, 4-Dioxo cyclohexanoic acid
2. Number of hetero atoms in the ring is: (2) 2,4-Dioxocycloheptanoic acid
CO–CH2 (3) 4-Formyl-2-oxocyclohexane-1-
NH O
CO–CH2 carboxylic acid
(1) 1 (2) 2 (4) 2,4-Dioxocyclohexane-1-carboxylic
(3) 4 (4) 3 acid
3. The homologue of phenol is:
O O
CH2OH OH
CH3 8. H OH IUPAC name is:
(1) (2)

OH (1) 4-Ethyl-5-oxopentanoic acid

OH
OH (2) 4-Ethyl-4-formylbutanoic acid
(3) (4)
(3) 4-Formylhexanoic acid
4. Number of π electrons present in (4) 4-Oxohexanoic acid
naphthalene is
(1) 4 (2) 5 9. Incorrect match is:
(3) 10 (4) 14 Br
5. Carbon atoms in the compound (1) 2-Bromobut-3-ene
NC CN are
C=C Me
NC CN (2) O 1-Methoxyprop-2-ene
(1) sp hybridized CH3
(2) sp2 hybridized (3) CH3 –C–OH 1,1-Dimethylethanol
(3) sp and sp2 hybridized
(4) sp, sp2 and sp3 hybridized CH3

(4) All
6. Structure formula of 4–Ethyl–1–fluoro–2–
nitrobenzene is: 10. The IUPAC name of Cl–C–OC2H5 is:-
F C2H5 O
O2 N
(1) (2) F (1) Ethoxyformyl chloride
C2H5 NO2 (2) Ethoxymethanoyl chloride
F NO2 (3) Ethylchloromethanoate

(3) (4)
F (4) Ethoxycarbonyl chloride

O2 N C2H5 C2H5

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Classification & Nomenclature CHEMISTRY

Cl 15. The IUPAC name of

CH3 CH2-CHO
11. has the IUPAC name:-
OHC–CH2– CH2– CH–CH2–CHO
C2H5 (1) 4, 4-di(formylmethyl) butanal
(1) 3–Chloro–1–ethyl–2–methyl
(2) 2-(oxoethyl) butane-1, 4-
cyclopentane
(2) 1–Chloro–3–ethyl–2–methyl dicarbaldehyde
cyclopentane (3) 3-Acetal hexane-1-6-dial
(3) 4–Chloro–1–ethyl–5–methyl (4) 3-(2-oxoethyl) hexane-1, 6-dial
cyclopentane
(4) 2–Chloro–5–ethyl–1–methyl
cyclopentane 16. The IUPAC name of :
12. IUPAC name of the following compound: HO
O CH2–CH3
O O
C
O (1) Ethoxy-2-methyl hydroxy-2butenoate
C
(2) Ethyl-2-Hydroxy methyl-2-butenoate
O
(1) Cyclohexane-1,2-dicarboxylic acid (3) Ethyl-2-hydroxy methyl-2-butene
(2) Cyclohexane-1,2-dicarboxylic carboxylate
anhydride (4) Ethoxy-2-Methyl hydroxy-2-butene
(3) Cyclohexane ethanoic anhydride
Carboxylate
(4) Cyclohexanebutanoic anhydride
F
17. The correct IUPAC name of the
13. The IUPAC name of is:
following compound is :
I Cl
Br O=C–CH2– CH–CHO
(1) 1-Bromo-2-chloro-3-fluoro-6-iodo OH H–C=O
benzene
(1) 3,3-diformyl propanoic acid
(2) 2-Bromo-1-chloro-5-fluoro-3-iodo
benzene (2) 3-formyl-4-oxo-butanoic acid
(3) 4-Bromo-2-chloro-5-iodo-1-fluoro (3) 3,3-dioxo propanoic acid
benzene
(4) 3,3-dicarbaldehyde propanoic acid
(4) 2-Bromo-3-chloro-1-iodo-5-fluoro
benzene O
CH3 18. The IUPAC name of C–CH3 is
14. The IUPAC name of CH–CHO is:
(1) Phenyl ethanone
(1) 2-Phenylpropan-3-al
(2) Methyl phenyl ketone
(2) Formylethylbenzene
(3) 2-Phenylpropanal (3) Acetophenone
(4) Ethylformylbenzene (4) Phenyl methyl ketone

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CHEMISTRY Classification & Nomenclature
19. IUPAC name of the given compound is COOH
22. The IUPAC name of
CH3
COOH
CH3–C–COOC2H5
(1) 6-carboxy-2,3-dimethyl-4-heptenoic acid
(2) 2, 3, 6- trimethyl-4-heptene-1,7-dioic
acid
NO2 (3) 2,5,6-trimethyl-3-heptene-1,7-dioic
(1) Ethyl-2-methyl-2-(m-nitro phenyl) acid
(4) 6-carboxy-2-5-dimethyl-3-heptenoic
propanoate acid
(2) Ethyl-2-methyl-2-(o-nitro phenyl)
O
propanoate 23. C

(3) Ethyl-2-methyl-2-(3-nitro phenyl) H O Cl

IUPAC name this compound is :-
propanoate (1) 1–chloro–3–methyl butane–1, 4–dione
(4) Ethyl-2-methyl-2-(3-nitro phenyl) (2) 3–formyl–butanoyl chloride
(3) 3–methyl–4–oxo–butanoyl chloride
propanoic acid (4) None

24. Write the correct IUPAC name of the

20. IUPAC name of given compound following bond line formula?
CHO
N
Cl OH (1) N-Methyl N–ethyl propan-1-amine
(1) 2-hydroxy-3-chloro-cyclopent-4-enal (2) N-Ethyl N–methyl propan-1-amine
(3) N-Methyl N–propyl ethan-1-amine
(2) 3-Chloro-2-hydroxy-cyclopent-4-enal (4) N-Ethyl N–propyl N-methyl amine
(3) 4-Chloro-5-hydroxy-cyclopent-2-
enecarbaldehyde Me
Me Me
(4) 4-Chloro-5-hydroxy cyclohex-4–enal 25. The IUPAC name
Me Me
this compound is
21. IUPAC name of given compound is (1) 2, 4 ,5–Triethyl–3–nonene
(2) 5, 6–Diethyl–3–methyl–4–decene
COCl
COOH (3) 2, 4 ,5–Triethyl–3–octene
(4) 3–Heptyl–5–methyl–3–heptane
NH2
(1) 6-amino-2-Chlorocarbonyl benzene 26. The IUPAC name of
Carboxylic acid BrCH2–CH–CO–CH2–CH2CH3 is
CONH2
(2) 6-amino-2-Chloroformyl benzoic acid
(1) 2–(Bromomethyl)-3–oxohexanamide
(3) 1-amino-2-carboxy-3-Chloroformly
(2) 1–Bromo–2–amino–3–oxohexane
benzene (3) 1–Bromo–2-amino–n–propyl ketone
(4) 2-amino-6-Chlorocarbonyl-benzene (4) 3–Bromo–2-propyl–propenamide
Carboxylic acid

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Classification & Nomenclature CHEMISTRY
27. Which of the following compound is 28. The IUPAC name of given compound is
wrongly named? O
(1) CH3CH2CH2CHCOOH ; (CH3)2 CH(CH2)2–C–N(CH3)2
Cl
2-chloro pentanoic acid (1) N, N, 4–Trimethylpentanamide
(2) CH3C≡CCHCOOH ; (2) Dimethylamino–4–methylpentanone
CH3
2-Methyl hex–3–enoic acid (3) N, N–Dimethylamino–4–methyl
(3) CH3CH2CH =CHCOCH3;
pentan amide
Hex–3-en–2-one
(4) CH3–CHCH2CH2CHO ; (4) 2–Methyl–5–oxodimethylpentanamide
CH3
4-Methyl pentanal

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CHEMISTRY Classification & Nomenclature

Exercise 3

1. Which of the following statements is

4. Assertion : The IUPAC name for
incorrect?
CH≡ C—CH2 —CH=CH2 is pent-4-en-1-yne
(1) Aromatic compounds may have hetero Reason : A tripe bond is always given
atom in the ring a lower locant than a double bond.
(2) Tropolone is a non-benzenoid compound (1) If both Assertion & Reason are True &
(3) Pyridine is a heterocyclic aromatic the Reason is a correct explanation of
compound the Assertion.
(2) If both Assertion & Reason are True but
(4) Aniline is a non-benzenoid compound
Reason is not a correct explanation of
the Assertion.
2. Consider the following statements : (3) If Assertion is True but the Reason is
I. The presence of functional group False.
disables systematisation of organic (4) If both Assertion & Reason are false.
compound into different classes.
II. In polyfunctional compounds, one of 5. Assertion : The correct IUPAC name
CH3
the functional group is chosen as the CH CH
of compound 3
principal functional group.
III. The order of priority for functional [1–Methylethyl]cyclohexane.
group is Reason : It is derivative of cyclohexane
—COCl > —COOR > —SO3H > —COOH because the no. of C–atoms in the ring is
IV. CH2(OH)CH2 (OH) is named as Ethane- more than the side chain.
1,2-diol. (1) If both Assertion & Reason are True &
the Reason is a correct explanation of
Read the given statements and choose the
the Assertion.
correct option. (2) If both Assertion & Reason are True but
(1) Both I and II (2) Both II and IV Reason is not a correct explanation of
(3) I, II and III (4) All of these the Assertion.
(3) If Assertion is True but the Reason is
3. Assertion : CH3–CH–CH2–CH–CHO is False.
(4) If both Assertion & Reason are false.
Cl CH3
2–chloro-4-methyl- pentanal. 6. Assertion : NC–CH2CH2COOH is called 3-
Reason : Number is given according to the cyano propanoic acid.
alphabetical order of substituents. Reason : While naming polyfunctional
compounds, –COOH group gets preference
(1) If both Assertion & Reason are True &
over cyano for selecting principal
the Reason is a correct explanation of functional group.
the Assertion. (1) If both Assertion & Reason are True & the
(2) If both Assertion & Reason are True but Reason is a correct explanation of the
Reason is not a correct explanation of Assertion.
the Assertion. (2) If both Assertion & Reason are True but
Reason is not a correct explanation of
(3) If Assertion is True but the Reason is the Assertion.
False. (3) If Assertion is True but the Reason is
(4) If both Assertion & Reason are false. False.
(4) If both Assertion & Reason are false.

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Classification & Nomenclature CHEMISTRY
7. Assertion : The IUPAC name of citric acid is 9. Assertion : The IUPAC name of
2-hydroxy-propane-1,2,3- CH3–CH–C–CH–CH3
tricarboxylic acid. CH3–O O OCH2CH3
COOH 2-Ethoxy-4- methoxy pentan -3- one.
Reason : large substituents are assigned
HOOC COOH
OH lower locants.
Citric acid (1) If both Assertion & Reason are True &
Reason : When an unbranched C Chain is the Reason is a correct explanation of
directly linked to more than two C- the Assertion.
(2) If both Assertion & Reason are True but
containing principal functional groups, then it
Reason is not a correct explanation of
is named as a derivative of the parent alkane
the Assertion.
which does not include the C atoms of the (3) If Assertion is True but the Reason is
functional groups. False.
(1) If both Assertion & Reason are True & (4) If both Assertion & Reason are false.
the Reason is a correct explanation of
the Assertion. 10. Which is correctly matched

(2) If both Assertion & Reason are True but

Reason is not a correct explanation of (1) Phenanthrene
the Assertion. (2) Anthracene
(3) If Assertion is True but the Reason is
False. (3) Naphthalene
(4) If both Assertion & Reason are false.

8. Assertion : IUPAC name of urea is (4) Benzene

aminomethanamide.
Reason : In NH2–C–NH2
11. Assertion: CH2=CH–CH2–NO2 is named as
O
1-Nitroprop-2-ene in IUPAC.
principal functional group is “amide” Reason: –NO2 group being secondary
and “amino behave as substituent. functional group gets priority over multiple
(1) If both Assertion & Reason are True & bonds in numbering.
the Reason is a correct explanation of (1) If both Assertion & Reason are True &
the Reason is a correct explanation of
the Assertion.
the Assertion.
(2) If both Assertion & Reason are True but (2) If both Assertion & Reason are True but
Reason is not a correct explanation of Reason is not a correct explanation of
the Assertion.
the Assertion.
(3) If Assertion is True but the Reason is
(3) If Assertion is True but the Reason is False.
False. (4) If both Assertion & Reason are false.
(4) If both Assertion & Reason are false.

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CHEMISTRY Classification & Nomenclature
12. Assertion: Benzene is an aromatic 13. Assertion: Homologues can never be
compound. isomers.
Reason: Benzene is homocyclic compound. Reason: Homologues always have different
(1) If both Assertion & Reason are True & molecular formula while isomers always
the Reason is a correct explanation of have same molecular formula.
(1) If both Assertion & Reason are True &
the Assertion.
the Reason is a correct explanation of
(2) If both Assertion & Reason are True but
the Assertion.
Reason is not a correct explanation of (2) If both Assertion & Reason are True but
the Assertion. Reason is not a correct explanation of
(3) If Assertion is True but the Reason is the Assertion.
False. (3) If Assertion is True but the Reason is
False.
(4) If both Assertion & Reason are false.
(4) If both Assertion & Reason are false.

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Classification & Nomenclature CHEMISTRY

Exercise 4 (Previous Year's Questions)

1. Which nomenclature is not according to 4. The total number of π−bond electrons in the
IUPAC system [AIPMT-2012] following structure is: [AIPMT-2015]
H H H
(1) CH3 – CH – CH – CH2–CH3 H3C
CH3
CH3
H3C
2-Methyl-3 phenylpentane
H2C H CH3
(2) CH3–C–CH2–CH2–CH2COOH
O (1) 8 (2) 12
(3) 16 (4) 4
5-oxohexanoic acid
(3) Br–CH2–CH=CH2 1-Bromo-prop-2-ene 5. The enolic form of ethyl acetoacetate as
CH3 below has: [AIPMT-2015]
(4) CH3–CH2–C–CH2–CHCH3 H3C
CH H3C CH2
Br
O O
CH3 C C
C C
4-Bromo,2,4- dimethylhexane
OH OC2H5 O OC2H5
2. Structure of the compound whose IUPAC Enolic Ketonic
name is 3-Ethyl-2-hydroxy-4-methylhex -3- (1) 16 sigma bonds and 1 pi-bond
(2) 9 sigma bonds and 2 pi-bonds
en-5-ynoic acid is: [NEET UG-2013]
(3) 9 sigma bonds and 1 pi-bond
COOH (4) 18 sigma bonds and 2 pi-bonds
(1)
OH 6. The pair of electrons in the given carbanion,
OH CH3C≡CΘ, is present in which of the
following orbital? [NEET-I-2016]
(2) COOH (1) 2p (2) sp3
2
(3) sp (4) sp
OH 7. The IUPAC name of the compound
O O
(3) COOH
H–C
is: [NEET-UG-2017]
OH
(4) COOH
(1) 5-Formylhex-2-en-3-one
(2) 5-Methyl-4-oxohex-2-en-5-al
3. Which of the following organic compound (3) 3-Keto-2-methylhex-5-enal
has same hybridization as its combustion (4) 3-Keto-2-methylhex-4-enal
product CO2? [AIPMT-2014] 8. Which of the following molecules represents
(1) Ethane the order of hybridisation sp2, sp2, sp, sp from
(2) Ethyne left to right atoms? [NEET-UG-2018]
(1) HC≡C–C≡CH (2) CH2=CH–C≡CH
(3) Ethene
(3) CH2=CH–CH=CH2 (4)CH3–CH=CH–CH3
(4) Ethanol
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CHEMISTRY Classification & Nomenclature

9. The number of sigma (σ) and pi (π) bonds in 12. The correct IUPAC name of the following
pent–2–en–4–yne is: [NEET-UG-2019] compound is: [NEET-UG-2022]
(1) 10σ bonds and 3π bonds Cl OH
(2) 8σ bonds and 5π bonds
Br
(3) 11σ bonds and 2π bonds
(4) 13σ bonds and no π bonds (1) 1-bromo-5-chloro-4-methylhexan-3-ol
(2) 6-bromo-2-chloro-4-methylhexan-4-ol
10. How many (i) sp2hybridised carbon atoms (3) 1-bromo-4-methyl-5-chlorohexan-3-ol
and (ii) π bonds are present in the following (4) 6-bromo-4-methyl-2-chlorohexan-4-ol
compound? [NEET-UG-2020 (COVID-19)]
13. A compound with a molecular formula of
C≡C–COOCH3 C6H14 has two tertiary carbons. Its IUPAC
name is: [NEET-2024]
(1) 7,5 (2) 8, 6
(3) 7, 6 (4) 8, 5 (1) 2,3-dimethylbutane
(2) 2,2-dimethylbutane
11. The correct structure of 2,6–Dimethyl dec– (3) n-hexane
4–ene is: [NEET-UG-2021] (4) 2-methylpentane

14. The correct IUPAC name of the compound

(1)
[RE-NEET-2024]
NO2
F
(2)
C2H5is :

(3) (1) 4-Ethyl-1-fluoro-2-nitrobenzene

(2) 4-Ethyl-1-fluoro-6-nitrobenzene
(3) 3-Ethyl-6-fluoro-1-nitrobenzene
(4) (4) 1-Ethyl-4-fluoro-3-nitrobenzene

40 Sarvam Career Institute

Classification & Nomenclature CHEMISTRY
ANSWER KEY

Exercise 1.1
Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans. 3 4 3 2 3 3 1 2 3 4 2 3 2 2 3 1 4 1 4 3

Exercise 1.2
Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Ans. 4 3 4 2 4 2 2 4 3 3 4 2 1 4 1 1

Exercise 1.3
Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans. 3 4 4 3 1 4 4 4 4 2 3 4 1 3 3 3 4 1 2 3
Que. 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39
Ans. 1 2 2 3 1 3 3 3 2 2 4 4 4 3 1 3 4 2 4

Exercise 1.4
Q ue. 1 2 3 4 5 6 7 8
Ans. 3 4 2 4 2 4 3 3

Exercise 1.5
Que. 1 2 3 4 5 6 7 8 9 10 11
Ans. 3 4 1 4 1 4 2 1 2 2 1

Exercise 2
Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans. 3 2 2 3 3 4 3 3 4 3 2 2 2 3 4 2 2 1 3 3
Que. 21 22 23 24 25 26 27 28
Ans. 4 3 3 2 2 1 2 1

Exercise 3
Que. 1 2 3 4 5 6 7 8 9 10 11 12 13
Ans. 4 2 4 4 1 1 1 1 3 4 4 2 1

Exercise 4 (Previous Year's Questions)

Q ue. 1 2 3 4 5 6 7 8 9 10 11 12 13 14
Ans. 3 3 2 1 4 4 4 2 1 3 1 1 1 1

Sarvam Career Institute 41

 Notes

Sarvam Career Institute

Isomerism CHEMISTRY

Chapter ISOMERISM
STRUCTURAL,
2
GEOMETRICAL
AND
CONFORMATIONAL
Chapter Summary Introduction

• Introduction Two or more than two organic compounds having the same
molecular formula but different physical, chemical or biological
• Structural Isomerism
properties are called isomers and this phenomenon is called
• Stereo Isomerism
isomerism. The name was given by Berzelius.
• Geometrical Isomerism
• Configuration of
Classification of Isomers:
geometrical Isomers
Isomers
• Conformational Isomerism
M.F= Same
• Gauche effect M.F= Same
S.F. = Different S.F. = Same
Structural isomers Stereoisomers S.C.F.= Different
or
Constitutional isomers
Configurational Conformational/
Chain isomers
isomers Rotational isomers
(non-interconvertible) (interconvertible)
Position isomers

Metamers

Functional Geometrical Optical

group isomers isomers isomers

Ring chain isomers

Tautomers
M.F. = molecular formula
S.F. = structural formula
SCF = stereochemical formula

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CHEMISTRY Isomerism
Structural Isomerism
1. Chain Isomerism:
• Compounds belong to same homologous series (same functional group)
• Compounds have different number of carbon atoms in principal carbon chain or in side chain or both.
CH3
Ex. (i) CH3 – CH2–CH2 – CH3 and CH3–CH–CH3
Butane (P.C.C. = 4C) 2–Methylpropane (P.C.C. = 3C)
C2H5 CH3
(ii) CH3 (Difference in number of C atoms in side chain)

2. Position Isomerism:
• Compounds belong to same homologous series (same functional group)
• Compounds have difference in position of principal functional group, multiple bond or substituent in
same carbon chain.
4 3 2 1 4 3 2 1
Ex. (i) CH3–CH2–CH=CH2 & CH3–CH=CH–CH3(P.C.C.= 4C)
CH3 CH3
1 2 3 3 2 1
(ii) CH3–C–CH3 & CH3–CH–CH2–OH
OH Isobutylalcohol
t-Butylalcohol (P.C.C. = 3C)
(P.C.C. = 3C)
CH3
CH3–CH2–CH=CH2 CH3–CH=CH–CH3 CH3–C=CH2
(iii) (I) (II) (III)
I & II→ Position isomers II & III→ Chain isomers I & III→ Chain isomers

3. Metamerism:
• Compounds have same poly valent function group (P.V.F.G.)
• Compounds have difference in units (groups) attached with P.V.F.G.
Polyvalent Functional group [More than one valency] are:
O
O
–O–, –S–, –C–, –C–O–, –C–NH–, –C–N–, –NH–, –N–, –C
O ,–S–O–
O O O O –C
O
O
Ex. A. (i) CH3–CH2–CH2–O–CH3 I & II → Metamers
(ii) CH3–CH2–O–CH2–CH3 I & III → Position isomers (priority) & metamers
CH3 − CH − O − CH3
|
(iii) CH3 II & III → Metamers
O O
CH3–C–NH–CH3 H–C–NH–CH2CH3
B. N–Methylethanamide & N–Ethylmethanamide
groups–CH3, –CH3 groups–H, –CH2CH3

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Isomerism CHEMISTRY
4. Functional Isomerism:
• Compounds having same molecular formula but different functional groups are called functional
isomers.
• Compounds belong to different homologous series.
• Members of those different homologous series show functional isomerism, which have same general
formula.
Alcohol and ether
Ex. (i) → CH3—CH2—OH and CH3—O—CH3
(G.F. → CnH2n+2 O)

(ii) Aldehydes and ketones → CH3 − CH2 − C − H and CH3 − C − CH3
(G.F. → CnH2n O) || ||
O O

(iii) Acids and ester → CH3 − C − OH and H − C − O − CH3

(G.F. → CnH2nO2) || ||
O O

(iv) Cyanide and isocyanide → CH3—CH2—CH2—CN and CH3—CH2—CH2—NC

(G.F. → CnH2n–1 N)

Nitro and Nitrite → O

(v) CH3–CH2–N and CH3—CH2—O—N=O
(G.F. → CnH2n+1O2N) O

(vi) 1°, 2°, 3° amines (i) CH3—CH2—CH2—NH2 (ii) CH3—NH—CH2—CH3 (iii) CH3
|
(G.F. → CnH2n+3N) CH3 − N − CH3

(vii) Amide and Oxime → CH3 − C − NH2 and CH3—CH=N–OH

(G.F. → CnH2n+1ON) ||
O
OH
CH2OH CH3
(viii) Aromatic Alcohol and phenolic compounds: and
(CnH2n–6O) (Aromatic alcohol) (phenol)
Note: Alkyl halides do not show Functional isomerism.

5. Ring Chain Isomerism (RCI):

• Compounds having same M.F. in which one has open chain and other has closed chain structure are
called R.C.I.
• Compounds belong to different homologous series.
Me
Ex. (I) CH3–C≡C–CH3 (II) CH2=CH–CH=CH2 (III) (IV)

I & II → F.G.I.

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CHEMISTRY Isomerism
I & III → F.G.I. & R.C.I.
I & IV → F.G.I. & R.C.I.
II & III → F.G.I. & R.C.I.
II & IV → F.G.I. & R.C.I.
III & IV → Chain isomers
Note: All R.C.I. are F.G.I. but converse is not true.
6. Tautomerism:
• Two compounds have different functional groups due to shifting of acidic H-atom and change in
position of multiple bonds.
• The two isomers exist in dynamic equilibrium.
• The two forms in tautomeric equilibrium are called tautomers of each other.
• The interconvertibility of tautomers is a chemical reaction which can be categorised as acid base
reaction.
O OH
Ex. CH3–C–CH3 CH2=C–CH3
(keto) (Enol)

Solved Example:
Ex. 1 Structural isomeric aldehydes & ketones with molecular formula C4H8O.
Ans. 3
O O O
(i) CH3–CH2–CH2–C–H (ii) CH3–CH–C–H (iii) CH3–C–CH2–CH3
CH3
C4H8O (2 aldehyde and 1 ketone) total 3 isomers are possible.
Ex. 2 Benzenoid isomers of C7H8O.
Ans. 5
CH3
CH2–OH CH3 CH3 O
–CH3
OH

(a) (b) (c) OH (d) OH (e)

• a, b – Functional isomers • b, c – Position isomers
• c, d – Position isomers • a, d – Functional isomers
• a, e – Functional isomers • b, e – Functional isomers
Note: Alcoholic and phenolic groups are Functional isomers.

Exercise 1.1
1. 1-Chlorobutane & 2-Chlorobutane are: 2. & are called is :

(1) Position isomers (1) Position isomers

(2) Chain isomers (2) Chain isomers
(3) Geometrical isomers (3) Functional isomers
(4) Ring-chain isomers (4) Ring chain isomers

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Isomerism CHEMISTRY
3. Possible number of disubstituted benzene 12. How many chain isomers are possible for
isomers are : C6H14 :
(1) 1 (2) 2 (1) 5 (2) 4
(3) 3 (4) 4 (3) 3 (4) 2
13. How many benzenoid isomeric phenols are
4. Which of the following are isomers:
possible with the molecular Formula
(1) Ethanol and ethoxyethane
C7H8O:
(2) Methanol and methoxymethane
(1) 3 (2) 4
(3) Propanoic acid and ethylacetate
(3) 2 (4) 5
(4) Propionaldehyde and acetone
14. A pair of position isomers is:
5. Functional group isomer of CH3COOCH3 is:
(1) CH3CH2COOH (2) HOCH2–CH2CHO (1) &
(3) HCOOC2H5 (4) Both (1) & (2) O &
(2) O
CH3
6. & are:
(1) Chain isomer (2) Functional isomer (3) &
(3) Position isomer (4) Metamers OH HO CH3

Me C2H5
7. CH3CONH2 & HCONHCH3 are called: (4) &
(1) Position isomer (2) Chain isomer
(3) Metamer (4) Functional isomer
15. Given compounds are:
O
8. The formula C4H8O2 represents:
(1) Only an carboxylic acid Cl–CH2 C–O–CH3 and
(2) Only an ester
(3) Only an alcohol O
(4) Both carboxylic acid and ester C–O–CH2–CH2–Cl

9. Propene and cyclopropane are: (1) Chain isomers

(1) Chain isomers (2) Metamers
(2) Position isomers
(3) Position isomers
(3) Geometrical isomers
(4) Ring chain isomers
(4) Ring chain isomers

10. The minimum number of carbon atoms

Me Me Me H
present in an organic compound to show N N
chain isomerism are:
16. and are:
(1) 2 (2) 3
(3) 5 (4) 4
Me
11. The minimum number of carbon atoms (1) Functional isomers
present in an organic compound to be able (2) Metamers
to show position isomerism are:
(3) Position isomers
(1) 3 (2) 4
(4) Chain isomers
(3) 2 (4) 5

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CHEMISTRY Isomerism
O O 23. Given compound shows which type of
OH OH
17. and H are isomerism
O O O O
(1) Homologues (2) Chain isomers
S O and S O
(3) Functional isomers (4) not a isomer
O O
18. Which of the following pairs of compounds (1) Chain isomerism
are functional group isomers? (2) Positional isomerism
(3) Metamerism
(1) and O
OH (4) Functional group isomerism
O
O
(2) and 24. The compound which is not isomeric with
diethyl ether is
O
(3) OH and (1) methyl n-propyl ether.
(2) 1-butanol.
O O
(4) and (3) 2-methyl propan-2-ol.
O O
(4) butanone.
19. CH3 – CH2 –CHO & CH2 = CH – CH2OH are:
(1) Functional group isomers 25. Isomers have essentially identical.
(2) Chain isomers (1) Structural formula
(3) Position isomers (2) Chemical properties
(4) Metamers
(3) Molecular formula
20. Alkynes are isomers of- (4) Physical properties
(1) Cycloalkane
(2) Alkadiene 26. What is the relation between
(3) Alkene 3-Ethylpentane and 3-Methylhexane ?
(4) All of the above
(1) Chain isomers
(2) Position isomers
21. C7H7Cl shows how many benzenoid
aromatic isomers? (3) Functional isomers
(1) 4 (2) 3 (4) Homologues
(3) 5 (4) 6

27. CH3–CH2–NH– CHO; CH3–CH–CHO

22. Choose the homologues-
CH3
NH2
Ι ΙΙ
(1) &
Which type of isomerism is observed
OH between I and II ?
CH2OH (1) Chain isomers
(2) &
(2) Position isomers
(3) CH3NH2 & (CH3)2NH (3) Functional isomers
(4) All the above (4) Metamers

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Isomerism CHEMISTRY
28. The correct relationship among the 34. How many structural isomers of all the
following pairs of given compounds is secondary alcohols with molecular formula
O O C5H12O.
O O (1) 2 (2) 3
(I) (3) 4 (4) 5
O
O 35. Relation between given compounds is:
O
O (II) ,
O
O
O (1) Position isomers (2) Chain isomers
O (3) Identical (4) Functional isomers
(III)
36. Which type of isomerism is observed
(I, II) (II, III) between I and II ?
(1) Functional Metamers CH3–CH2–C–OCH3
Isomers O
(2) Metamers Functional (I)
Isomers CH3–C–CH2–O–CH3
(3) Metamers Metamers O
(4) Functional Functional (II)
Isomers Isomers (1) Functional isomerism
(2) Metamerism
29. Molecular formula C4H10O represent (3) Position isomerism
(1) Two primary alcohol (4) Stereoisomerism
(2) One secondary alcohol
(3) One tertiary alcohol 37. & show isomerism –
(4) All of these
(1) Chain (2) Position
(3) Functional (4) None of these
30. How many positional isomers are possible
for dimethyl cyclohexane. 38. The total number of cyclic compounds
(1) 3 (2) 4 (neglecting stereoisomers) with the
(3) 5 (4) 6 molecular formula C5H10 is
(1) 4 (2) 5
31. How many aromatic isomers are possible (3) 6 (4) 7
for trichlorobenzene (C6H3Cl3)?
39. Which of the following pairs of compounds
(1) 2 (2) 3 are not isomers -
(3) 4 (4) 5 (1) Propyne and cyclopropane
(2) Propyne and propadiene
32. The number of ether isomers represented (3) Propene and cyclopropane
by formula C4H10O is (only structural) (4) 1-Propanol and methoxyethane
(1) 4 (2) 3
(3) 2 (4) 1 40. What is true for 1,2– pentadiene?
(1) It is functional isomer of Pentyne
33. Total number of 2° amine isomers of C4H11N (2) It is position isomer of Pentyne
would be (only structural) (3) It is chain isomer of 3–Methyl –1–butyne
(1) 4 (2) 3 (4) It is metamer of cyclopentene
(3) 5 (4) 2
Sarvam Career Institute 49
CHEMISTRY Isomerism
41. Functional isomerism is shown by − 45. Tautomerism is not shown by -
(1) o-Nitrophenol and p-nitrophenol (1) O (2) CH NOH
O
(2) Dimethyl ether and ethanol
O
(3) 2-Pentenoic acid and 3-pentenoic acid
(3) O (4) O
(4) Acetaldehyde and acetone
O
42. The minimum number of carbon atoms in 46. Which will not show tautomerism -
ketone to show position isomerism - O
O OH
(1) 3 (2) 4
(3) 5 (4) 6 (1) (2)
O
43. The minimum number of carbon atoms in N–OH
ketone to show metamerism: O
O
(1) 3 (2) 4
(3) (4)
(3) 5 (4) 6

44. Which can’t show Tautomerism? 47. Tautomerism will be shown by

O O (1) CH2==CH—CHO
O (2) CH3—CH == CH—CHO
(1) (2) OH
OH OH (3) CH2==CH—CH2
OH O
(3) (4) O
(4) C6H5−C−H

48. Identify the type of isomerism

(i) CH3NHCH2CH2CH3 and CH3CH2NHCH2CH3 ______________________

CH3 − N − CH2CH3
|
(ii) CH3NHCH2CH2CH3 and CH3 ______________________

and and
(iii) ______________________ (iv) ______________________
OH
O CN
CN
(v) and (vi) and
______________________ ______________________

CH2OH CH3
O
OH
(vii) and (viii) and
O H ______________________ ______________________

Me

(ix) and ______________________

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Isomerism CHEMISTRY

(x) CH3–CH2–CH2–CH2–OH, CH3–CH2–CH–OH ______________________

CH3
O O
(xi) H–C–NH–CH2–CH3 & CH3–C–NHCH3 ______________________

(xii) CH3–CH2–CH–CH2–CH3 & CH3– CH2– CH– CH3 ______________________

CH3 C2H5

(xiii) and ______________________

(xiv) CH3 CH2–CH2–CH3 & CH3–CH2 CH2–CH3 _______________________

O O
(xv) CH3–CH2– C–OH & CH3 –CH–C–H ______________________

OH
Cl Br CH3
C2H5 CH3
(xvi) and ______________________ (xvii) and _____________________

Br Cl

(xviii) CH3–CH2–CH=0 & CH2–CH– CH2 ______________________

O

(xix) CH2=CH–CH2–OH & CH3–C– CH3 ______________________

O
CH3
O–CH2–CH3 OCH3
(xx) and ______________________

Stereoisomerism
Stereoisomers: compounds having same structural formula & molecular formula but different stereochemical
formula are called stereo isomer. These isomers have same connectivity of atoms and groups (same structural
formula) but different orientation of groups about a stereo centre (system).

Ex.

I, III → position Isomers same M.F. but diff. S.F.

II, III → position Isomers same M.F. but diff. S.F.
I, II → stereoisomers same M.F. & S.F. but different S.C.F.

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CHEMISTRY Isomerism
Stereoisomers

Configurational Conformational/
isomers Rotational isomers
(non-interconvertible) (interconvertible)

Geometrical isomers Optical isomers

Configurational Isomers: It arises due to different orientations along a stereo centre and these isomers
can be separated and these isomers do not convert into one-another at room temperature therefore,
they are true isomers.
Ex. cis-2-Butene
& trans-2-Butene
Geometrical Isomerism
The compound having different 3-D arrangement which are not interconvertible by free rotation about any set
of bonds are called geometrical isomers and the phenomenon is called geometrical isomerism.
(I) This isomerism arises due to restricted rotation about a system that is a set of two atoms where rotation
of one atom with respect to other atom is not free.
• Restricted rotation is observed due to
(i) double bond [ –N=N–]
,
(ii) Ring structure
• Due to the restricted rotation about double bond or the ring structure the molecules exist in two or
more orientations.

Ex. and (Restricted Rotation)

(II) Different groups should be attached at each atom of the system. (Set of two atoms where rotation is
restricted)
a a a d
Ex. C=C , C=C
b d b e
Case (a): Geometrical Isomers in Alkenes
• Alkenes show G.I. due to restricted rotation about double bond
• Only those alkenes show G. I. in which "Each double bonded carbon individually have different atoms
or groups"
a a a d
C=C C=C
b b b c
Show G.I Show G.I
a b a d
C=C C=C
a b b d
Does not Show G.I Does not Show G.I

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Isomerism CHEMISTRY
Case (b): Geometrical Isomers in Oximes [>C = N–OH]
• Oximes show G. I. due to restricted rotation about double bond.
• Only those oximes show Geometrical isomerism in which double bonded carbon atom has two different
units.
Ex. Acetaldoximes has two Geometrical isomers –
CH3 − C − H CH3 − C − H
|| ||
N − OH HO− N
syn anti
When H and OH are on the same side When H and OH are on the opposite side
Case (c): Geometrical Isomers in Azo Compounds (–N ·· ·· )
= N–

Ex. Ph–N=N–Ph (Azo benzene)

•• Ph •• Ph
N N
N N
•• Ph Ph ••
(syn) (Anti)
Case (d): Geometrical Isomers in Cycloalkanes Cycloalkanes show Geometrical isomers due to
restricted rotation about single bond. Only those cyclo alkanes show Geometrical isomers in which at
least two different sp3 carbons of the ring individually have two different units.
Two Geometrical isomers
Me Me Me Me
H H cis
H H
Me H trans
H Me
Ex. Which of the following show Geometrical isomerism–
Cl
H
N
(1)H H (2) (3) (4)
Cl Cl Cl Cl Br
Ans.(1), (2) and (4)

Case (e): Geometrical Isomers in Cycloalkene

Condition - Minimum 8 membered ring is required to show G.I. w.r.t. double bond in the ring
Ex. Which of the following show G.I.
G.I. wrt d.b. in the ring = × G.I. wrt d.b. in the ring = √
(1) No G.I. (2) Show G.I.
G.I. wrt single bond of ring = × G.I. wrt single bond of ring = ×

Case (f): Geometrical Isomers in Allenes

• In Allenes, continuous π bonds are perpendicular to each other.
• If no. of π bonds are even then terminal valencies will be in ⊥ plane and compound will never show G.I.
• If number of π bonds are odd then terminal valencies will be in same plane and compound can show
G.I. if both terminals have different units.

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CHEMISTRY Isomerism
Ex. Which of the following show G.I.
(1) H3C C = C = C CH3 No G.I.
H H

(2) H3C C = C = C = C CH3 Show G.I.

H H

(3) H3C C = C = C = C H No G.I.

H H

Configuration of geometrical isomers

(A) cis–trans system If same groups are present on same side of the system then cis and if same groups are
present on opposite of the system then it is called trans.
a a b a
C=C C=C
b b a b
[Same groups, same side] [Same groups opposite side]
cis trans
Ex. Cl C = C Cl Cl F
H H Br C = C Cl
cis trans
(B) syn-anti system
Condition Minimum one terminal should have lone pair.
Ex. (1) Oxime
(i) Aldoxime Oxime of aldehyde other than formaldehyde will show G.I. and exist in two forms Syn & Anti.
R •
•
H •
•

C=N C=N
H OH R OH
syn anti
Note: H & OH same side → syn
H & OH opposite side → anti
(ii) Ketoxime
• Ketoximes of symmetrical ketones do not show G.I.
• Ketoximes of unsymmetrical ketones show G.I.
•• C2H5 ••
CH3
C=N C=N
C2H5 OH CH3 OH
Syn-ethyl methyl ketoxime Anti-ethyl methyl ketoxime
(2) Azo Compounds
•• •• ••
CH3
N=N N=N
••
CH3 CH3 CH3
syn anti
Note: lp-lp same side → syn
lp-lp opposite side → anti

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Isomerism CHEMISTRY
(C) E – Z System : Main units are selected on both side of the system
When main units are on opposite side→ E[Entgegen i.e. opposite]
When main units are on same side → Z[Zusammen i.e. together]
a b
C let
main unit
C
d C
‘Z’

CIP sequence Rules (Given by Cahn, Ingold & Prelog) for selecting main unit:
Rule I: Unit having 1st atom with higher atomic number gets priority.
Br F CI CH3
C=C C=C
I Cl Cl C N H2
‘Z’ Cl Cl ‘E’

Rule II: If 1st atoms of two units are isotopes then heavier isotope gets priority.
H NO2
C=C
D CH3
‘E’
Rule III: If 1st atom of two units are same then sets of next atoms (arranged in decreasing atomic
number) are compared and at the first difference, atom with higher atomic number gets priority.
[H,H,H] H3C CF2CL [Cl F F]
C=C
[ C H H] H2C–H2C CCl2F [Cl Cl F]
‘Z’
Rule IV: If multiple bond is present in unit then consider them as:-
C C
C = C → –C – C– –C ≡ C– → –C – C–
C C C C
N C
C=O→ C–O –C ≡ N– → –C – N
O C N C

O [O,O, H] CH3 H CH3 CH3

Ex. Ex. C=C Ex. C=C
N≡C C—H Cl CH3 H Cl
H N C=C ‘Z’ ‘E’
C—OH
H ‘Z’ [O,O, O]
O

[C,H,H] CH3 – CH2 CH2Cl [Cl,H,H]

Ex. C=C
[C,C,C] CH ≡ C COOH[O,O,O]
‘E’
Number of Geometrical Isomers (G.I.) :
(a) If compound is unsymmetrical (in respect of IUPAC numbering) or compound with dissimilar ends then,
number of G.I. = 2n [n = number of double bonds or system where G.I. is possible]

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CHEMISTRY Isomerism
Ex. CH3–CH=CH–CH=CH–CH=CH–CH2CH3
As n = 3 → number of G.I = 23 = 8
(b) If compound is symmetrical or compound with similar ends then

(n = even number) (n = odd number)

n –1 n-1
n-1
G.I.= 2 + 22 G.I.= 2n-1+ 2 2

Ex. CH3–CH=CH–CH=CH–CH=CH–CH3
n = 3 (odd number)
n −1
No of G.I. = 2n−1 + 2 2

3 −1
= 23−1 + 2 2
= 22 + 21
=4+2
= 6 Ans.
Physical properties of Geometrical isomers
Physical Br Br H Br Remarks
properties C = C C=C
H Ι H Br ΙΙ H
Dipole moment I > II Cis-isomer has resultant of
dipoles while in trans isomer
dipole moments cancel out
Boiling point I > II Molecules having higher dipole
moment have higher boiling
point due to larger
intermolecular force of
attraction
Solubility I > II More polar molecules are more
(in H2O) soluble in H2O (Polar solvent)
Melting point II > I More symmetric isomers have
higher melting points due to
better packing in crystalline
lattice & trans isomers are more
symmetric than cis.
Stability II > I The molecule having more
vander waal strain are less
stable. In cis isomer the bulky
groups are closer so, they have
larger vander waals strain.

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Table
Physical properties H3C H H2C Br
C=C C=C
H Ι Br H ΙΙ H
Dipole moment I > II
Boiling point I > II
Solubility (in H2O) I > II
Melting point I > II
Stability I > II

Exercise 1.2
1. Fumaric acid shows geometrical isomerism 5. Configuration of the given molecule is:
with: ⊕
H3N NH2
(1) Malonic acid (2) Maleic acid C
(3) Malic acid (4) Tartaric acid C
H2N C≡N
2. Which of the following will show
(1) E (2) Z
geometrical isomerism:
(3) Trans (4) 2 & 3 both
(1) 1-Butene
(2) 1, 2-Dibromoethene
(3) But-2-yne 6. Incorrect match is:
(4) Isobutylene M.F. No. of isomers
(1) C4H8 3 alkenes
3. Which of the following pair of compounds (2) C2H7N 2
shows correct configuration: (3) C2H2Br2 3
Ph Ph Ph (4) C5H12 3
(1) N=N & N=N
Syn-azobenzene anti-azobenzene Ph
7. Which of the following compound can show
CH3 CH3 OH geometrical isomerism:
(2) H C = N OH & H C = N (1) (CH3)2C = CH – C2H5
syn-acetaldoxime anti-acetaldoxime (2) H2C = CBr2
D
(3) C6H5 – CH = CH – CH3
CH2CH2CH3 CICH2 Br
C=C C=C (4) CH3 – CH = CCl2
(3) H CH3 & BrCH2 CH3
Z-isomer
E-isomer 8. Which of the following is not true regarding
(4) All of these hex-2-ene:
(1) Boiling point of cis is higher
4. Which is incorrect statement: (2) Trans isomer has zero dipole moment
F Cl F Br (3) Trans is more stable than cis
C C (4) Can show position isomerism
(I) C (II) C 9. Which of the following compounds will
Br I I CI show geometrical isomerism:
(1) I has Z configuration (1) 2-Butene
(2) II has E configuration (2) 1-Chloro-1-butene
(3) I & II are geometrical isomers of each other
(3) 1-Phenylpropene
(4) None
(4) All of above
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CHEMISTRY Isomerism
10. Which is a pair of geometrical isomers: 16. Which of the following has Z-configuration:
Cl Br Cl Br H3C C2H5
I. C=C II. C=C (1) C=C
H Br H CH3 H H
Cl CH3 H Br Br CH(CH3)2
III. C=C IV. C=C (2) HOCH C = C CH2–CH3
Br H Cl CH3 2

(1) I and II (2) I and III Cl H

(3) II and IV (4) III and IV (3) Br C = C D

(4) All the above

11. But-2-ene exhibits cis-trans isomerism due
to:
(1) Rotation around C3–C4 sigma bond 17. Which of the following is/are cis isomer?
(2) Restricted rotation around C=C bond Cl Cl
(1) C=C
(3) Rotation around C1–C2 bond H H
(4) Rotation around C2–C3 double bond H3C CH3
(2) C=C
H Br
12. The number of geometrical isomers of
CH3CH=CHCH=CH–CH=CHCl are: Cl Cl
(1) 2 (2) 4 (3)
(3) 6 (4) 8 H H
(4) All of these
13. Which of the following functional group
containing compounds can exhibit 18. Which of the following cannot show
geometrical isomerism: geometrical isomerism?
(1) > C = C < (2) > C = N – (1) CH3CH = CHC2H5
(3) – N = N – (4) All of these (2) (CH3)2 C = CHC2H5
14. Geometrical isomerism is shown by (3) CH3CH = CHCH3
H I (4) All the three
(1) H C = C Br

H 19. Which of the following will not show cis-

I
(2) H C C = C Br trans isomerism?
3
(1) Butanedioic acid
H3C CI
(3) C=C (2) 1,2-dimethyl cyclopropane
H3C Br (3) 1,2-dichloroethene
H CI (4) Butenedioic acid
(4) C=C
H3C Cl
20. CH3–C– CI is
15. Among the following compounds, the one
CI–C–Br
which can exhibit geometrical isomerism is:
(1) 1,3-Butadiene (1) Trans
(2) 1,2-Butadiene (2) Z
(3) 1,3-Pentadiene (3) Both Z & trans
(4) 1,4-Pentadiene (4) cis

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Conformational Isomerism
• The isomerism arises due to free rotation of atoms with respect to second atom in a molecule so it is
also called rotational isomerism.
• Due to free rotation about single bond, a molecule has several 3-D temporary arrangements which are
called conformers or rotamers or conformations.
• Infinite number of conformations are possible for the compound.
• It is to remember that C–C single bond is not completely free it is hindered by small barrier of 1-20
kJ/mol. It is due to
(i) Repulsive interaction between adjacent bonds called torsional strain.
(ii) Repulsive interaction between adjacent atoms called steric strain.
H
H H
H H
C θ C
C H H C
H
H H H H
Here θ(60°) is dihedral angle, angle between two planes.

Ex. Which of the following show conformational isomerism.

H H
O O
(1) C (2) C (3)
H H H
H H H H H
H H H H H
O O
(4) (5) C C (6) C C C
H H
H H H H H H H H H
Ans.2, 4, 5, 6
In general conformations are represented by sawhorse and Newman projection.

1. Sawhorse projection:
The central C–C bond as a somewhat longer straight line. Upper end of the line is slightly tilted towards
right or left hand side. The front carbon is shown at the lower end of the line, whereas the rear carbon
is shown at the upper end. Each carbon has three lines attached to it corresponding to three hydrogen
atoms. The lines are inclined at an angle of 120° to each other.
H H H
H H H

H H
H H H H
(i) Eclipsed (ii) Staggered
Fig. Sawhorse projection of ethane

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CHEMISTRY Isomerism
2. Newman Projection:
In this projection, the molecule is viewed at the C-C bond head on. The carbon atom nearer to the eye
is represented by a point. Three hydrogen atoms attached to the front carbon atom are shown by three
lines drawn at an angle of 120 to each other. The rear carbon atom (the carbon away from the eye) is
represented by circle and the three hydrogen atoms are shown attached to it by the shorter lines drawn
at an angle of 120 to each other.
HH H
H H
H H H
H
HH H
(i) Eclipsed (ii) Staggered
Fig. Newman’s projection of ethane
Conformational Analysis of Ethane
Ethane molecule contains a carbon-carbon σ bond and each carbon is further attached to 3 H-atoms.
It exists in two extreme conformations are:
(i) Eclipsed Conformation [H-atoms attached with two carbons are as closed together as possible]
(ii) Staggered Conformation [H-atoms attached with two carbons are as far apart as possible]

60 rotation
12.5kJ/mol

The potential energy barrier between the two extreme conformations of ethane is about 12.5 kJ/mol.
The potential energy of ethane molecule is minimum for the staggered conformation, increase with
rotation and reaches at maximum for the eclipsed conformation. Most of ethane molecules naturally
exist in the most stable staggered conformation. Any other intermediate conformations or
conformation other than eclipsed & staggered are called skew conformations

Energy Profile of Eclipsed and Staggered Forms of Ethane

Stability Order Staggered > Skew > Eclipsed

Conformational Analysis of Butane
1 2 3 4
CH3–CH2–CH2–CH3

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If we consider rotation about the C2 - C3 bond of butane, we find that there are four important
conformations shown as:

60º 60º 60º

60º

Fully eclipsed
conformation
60º

The stability order → Anti > Gauche > Partially eclipsed > Fully eclipsed
(iv) (ii = vi) (iii = v) (i)
Order of potential energy → Anti < Gauche < Partially eclipsed < Fully eclipsed
(iv) (ii = vi) (iii = v) (i)
Note: In all conformations, the bond angles, the bond lengths remain same.

Gauche Effect
Gauche conformation, in some cases, becomes even more stable than anti when intra-molecular hydrogen-
bonding takes place.
• Cases where gauche is more stable than anti :
X Y X Y
| | –OH –OH
−C − C − –NH2 NH2
| | –OH NH2
⊕
–COOΘ − NH3
–COOH –COOH
–F –OH

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Exercise 1.3
1. CH3 – CH2 – CH2 – CH3. There is free rotation 3. How many conformations does ethane
about (C2 σ C3) bond. The most stable form have:
is obtained after rotation of __________, (1) 1 (2) 2
starting from fully eclipsed conformation. (3) 3 (4) Infinite
(1) 60° (2) 120°
(3) 240° (4) 180° 4. The eclipsed and staggered conformation of
ethane is due to:
2. Which one of the following is the most (1) Free rotation about C–C single bond
stable conformation of 2,3-butanediol: (2) Restricted rotation about C–C single
OH OH bond
H OH H CH3 (3) Absence of rotation about C–C bond
(1) H (2) (4) None of the above
CH3 H CH3
CH3 OH
5. Dihedral angle required to get maximum
OH OH stable conformer from minimum stable
HO CH3 HO CH3 conformer in C2–C3 bond of n-butane is:
(3) CH3 (4) H (1) 360° (2) 180°
H CH3
H H (3) 120° (4) 240°

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Exercise 2
1. Assign double bond configurations to the 6. For geometrical isomers of 3–hexene:
following: H C2H5 H H
COOH C = C & C = C
NC C2H5 H C2H5 C2H5
CH2OH
(1) M.P. is high and dipole moment high for

H2N–H2C CN trans
(1) E, Z (2) Z, E (2) M.P. is low and dipole moment low for
(3) E, E (4) Z, Z trans
(3) M.P. is high and dipole moment low for
2. HO – CH2– CH2 – F
trans
Which conformer of above compound is most
stable across C – C? (4) M.P. is low and dipole moment high for
(1) Anti staggered (2) Partially eclipsed trans
(3) Gauche (4) Fully eclipsed
7. Which of the following pairs of compounds are
3. The geometrical isomerism is shown by:
CH2 not positional isomers?
CH2
(1) (2) HO CH3 OH
(1) and
CHCl CH3
CHCl
(3) (4) CH3
(2) and
CH3 OH OH
4. What are the correct designations for the
structure below? CH2OH OCH3
(3) and

(4) All of these

(1) E, E
(2) Z, E 8. CH3–CH=C – C=CH–CH3
(3) E, Z Br Br
(4) Geometrical isomerism not possible.
How many geometrical isomers are possible

5. Which of the following will not show for the above compound:
geometrical isomerism? (1) 2 (2) 3
(1) CH3–C=CH–CH2–CH3 (3) 4 (4) 6
CH3
9. CH3–CH=C – C=CH–CH3
(2) CH3–CH–CH=CH–CH2–CH3
Br Cl
CH3
How many geometrical isomers are possible
(3) CH3– CH=CH–CH3 for this compound?
(4) CH3– CH2–CH=CH–CH2–CH3 (1) 2 (2) 3
(3) 4 (4) 6
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CHEMISTRY Isomerism
10. CH2=CH–CH=CH–CH=CH2 14. Which of the following has incorrect relation?
Me
How many geometrical isomers are possible
Pr Pr
for this compound? (1) & identical
Me
(1) 2 (2) 3
(3) 4 (4) 8 Me Et
(2) positional
&
Pr Pr isomers

11. Which of the following compound can show

(3) Et Et
positional
geometrical isomerism? &
isomers
Pr
Br Cl Pr
(1) C=C Me
I Cl
Me Et
CH3 (4) & Homologues
(2) Pr
C
CH3
15. Which of the following order is correct-
F Et
(3) C=C COOH
Cl Et OCOH
CH3 & functional isomer
CH3
(4) C OMe OEt
CH3 CH3 Et
& metamers
Me
12. Which of the following is correct set of CH2–CH2–OH CH2–O–CH3

physical properties of the geometrical metamers

&
isomers?
Me–N–Me CH2–NH–CH3
CH3 H CH3 F
C=C & C=C functional
H F H &
H isomers
(I) (II)
(1) TFTF (2) FTTF
Dipole Boiling Melting Stability (3) TTFT (4) TFFT
moment point point
16. Which of the following cannot be written in an
(1) I > II I < II II > I I > II isomeric form?
(2) II > I II > I II > I II > I (1) CH3–CH(OH)–CH2–CH3
(2) CH3–CHO
(3) I > II I > II I > II I > II
(3) CH2=CH–Cl
(4) II > I II > I I > II I > II (4) Cl–CH2CH2–Cl

17. Only two isomeric monochloro derivatives are

13. An alkane forms isomers if it possesses possible for (excluding stereo)
(1) n-butane
minimum no. of carbon atoms-
(2) 2,2-dimethylpentane
(1) 1 (2) 2 (3) benzene
(3) 3 (4) 4 (4) neopentane

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18. An alkene exists as geometrical Isomers. The 24. Which of the following group of compounds
minimum number of carbon atoms in the are not functional isomers?
molecule is -
(1) 1 (2) 2 (1) Nitro alkane and alkyl nitrite.
(3) 3 (4) 4 (2) Aldehydes and ketones.

19. Which of the following is not true for maleic (3) Alcohols and aldehydes.
acid and fumaric acid.? (4) Alcohols and ether (both saturated).
(1) Configurational isomers
(2) Stereo isomers
(3) Z and E isomers 25. The correct structure of trans-hex-2-enal is
(4) Optical isomers CHO
(1)
20. Functional isomers of CH3–C–O–H is - CHO
|| (2)
O
(1) Amide (2) Ester
(3) CHO
(3) Anhydride (4) All
CHO
OH CH2OH (4)

21. Compound and are-

26. Which is a pair of geometrical isomers –
CH3
(1) CI Me and Br H
(1) Functional isomers C=C C=C
Br H Cl Me
(2) metamers
(3) Tautomers
(4) Ring chain isomers (2) Ph OH and Me ••
C=N C=N
Me •• Ph OH
22. Which will not show metamerism?
(1) CH3 O Et (3)
Br Br and Br Br
(2) CH3 − O − CH3
O
Ph
(3) CH3 CH2 C O CH3 (4) Me C=C
Ph
and
H
C=C
H Me Me Me
(4) CH3–CH2–NH–CH2–CH3

23. The functional isomers of 1−butyne are

27. The correct stereochemical name of –
(I) CH3−C≡C−CH3 (II) CH3−C=CH2
CH3 H H
CH3 C=C CH3
H CH2 C = C
(III) CH2=C=CH−CH3 COOCH3
(IV) CH2=CH−CH=CH2
(1) Methyl 2–methylhepta (2E, 5E) dienoate
(1) (I), (III) and (IV) only
(2) (II) and (III) only (2) Methyl 2–methylhepta (2Z, 5Z) dienoate
(3) (II) and (IV) only (3) Methyl 2–methylhepta (2E, 5Z) dienoate
(4) (III) and (IV) only (4) Methyl 2–methylhepta (2Z, 5E) dienoate
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28. Which of the following represents the structure 29. The IUPAC name of compound
H Me
having cis arrangement around each double bond? Me
O is
H
Me C2H5
(1)
(1) (Z)-4, 6, 7- Trimethyl hept-4-en-3-one
H
(2) (E)-4, 6-Dimethyl oct-4-en-3-one
(3) (Z)-4, 6-Dimethyl oct-4-en-3-one
(2)
(4) (E)-4,6-Dimethyl oct-4-en-6-one
30. The number of cis-trans isomer possible for
(3)
the following compound

(4) (1) 2 (2) 4

H H
(3) 6 (4) 8

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Exercise 3
1. Assertion : Trans-1-chloropropene has higher 4. Assertion : Benzaldehyde forms two oximes
dipole moment than cis-1-chloropropene. on reacting with NH2OH.
Reason : The resultant of the two vectors in Reason : The two oximes arise due to
trans-1-chloropropene is more than in geometrical isomerism around C=N bond.
cis-1-chloropropene. (1) If both Assertion & Reason are True & the
(1) If both Assertion & Reason are True & the Reason is a correct explanation of the
Reason is a correct explanation of the Assertion
Assertion (2) If both Assertion & Reason are True but
(2) If both Assertion & Reason are True but Reason is not a correct explanation of
Reason is not a correct explanation of the Assertion
the Assertion (3) If Assertion is True but the Reason is
(3) If Assertion is True but the Reason is False
False (4) If both Assertion & Reason are False
(4) If both Assertion & Reason are False
5. Assertion : The boiling point of cis 1,
2. Assertion : All the carbon atoms in but-2-ene 2-dichloro ethene is higher than that of
lie in one plane. corresponding trans isomer.
Reason : Carbon atoms in it are sp2 and Reason : Cis-1,2-dichloro ethene has higher
sp3-hybridized. dipole moment as compared to that of the
(1) If both Assertion & Reason are True & trans-isomer.
the Reason is a correct explanation of (1) If both Assertion & Reason are True & the
the Assertion Reason is a correct explanation of the
(2) If both Assertion & Reason are True but Assertion
Reason is not a correct explanation of (2) If both Assertion & Reason are True but
the Assertion Reason is not a correct explanation of
(3) If Assertion is True but the Reason is the Assertion
False
(3) If Assertion is True but the Reason is
(4) If both Assertion & Reason are False
False
(4) If both Assertion & Reason are False
3. Assertion : is planar molecule but
6. Assertion : Ethanol cannot show position
cannot show G.I. isomerism.
Reason : In styrene every carbon is sp2 Reason : Ethanol cannot show isomerism.
hybridised. (1) If both Assertion & Reason are True & the
(1) If both Assertion & Reason are True & the
Reason is a correct explanation of the
Reason is a correct explanation of the
Assertion
Assertion
(2) If both Assertion & Reason are True but
(2) If both Assertion & Reason are True but
Reason is not a correct explanation of Reason is not a correct explanation of
the Assertion the Assertion
(3) If Assertion is True but the Reason is (3) If Assertion is True but the Reason is
False False
(4) If both Assertion & Reason are False (4) If both Assertion & Reason are False

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7. Assertion : CH3 – CH2 – Br and CH3 – CH2 – I 10. Match the column:
are functional group isomers. Column–I Column–II
Reason : They have same molecular formula. (Compound) (Type of
(1) If both Assertion & Reason are True & the structural
Reason is a correct explanation of the isomers)
Assertion (a) n-Butane, isobutane (p) Ring chain
(2) If both Assertion & Reason are True but
(b) Ethanol, (q) Position
Reason is not a correct explanation of
methoxymethane
the Assertion
(c) But–1–ene, (r) Chain
(3) If Assertion is True but the Reason is
Cyclobutane isomerism
False
(d) 1–Propanol,2– (s) Functional
(4) If both Assertion & Reason are False
propanol,

8. Assertion : Boiling point of cis isomers are (1) (a)–(p), (b)–(r), (c)–(s), (d)–(q)
higher than trans-isomers. (2) (a)–(q), (b)–(p), (c)–(s), (d)–(r)
Reason : dipole moments of cis isomers are (3) (a)–(r), (b)–(s), (c)–(p), (d)–(q)
higher than trans-isomers. (4) (a)–(p), (b)–(r), (c)–(q), (d)–(s)
(1) If both Assertion & Reason are True & the
Reason is a correct explanation of the 11. Match the column:
Assertion Column–I Column–II
(2) If both Assertion & Reason are True but (Conformation of (Property)
Reason is not a correct explanation of n-butane)
the Assertion (a) Fully eclipsed (p) Has two methyl-
(3) If Assertion is True but the Reason is hydrogen
False interactions
(4) If both Assertion & Reason are False (b) Partially (q) Dihedral angle
eclipsed between two
9. Assertion: Acetone cannot show position
methyl group is
isomerism.
60°
Reason: Acetone cannot show isomerism.
(c) Gauche (r) Has least energy
(1) If both Assertion & Reason are True & the
(d) Anti (s) Has one methyl-
Reason is a correct explanation of the
Assertion. methyl interaction

(2) If both Assertion & Reason are True but (1) (a)–(s), (b)–(p), (c)–(q), (d)–(r)
Reason is not a correct explanation of
(2) (a)–(p), (b)–(q), (c)–(r), (d)–(s)
the Assertion.
(3) If Assertion is True but the Reason is (3) (a)–(q), (b)–(p), (c)–(s), (d)–(r)
False.
(4) (a)–(r), (b)–(p), (c)–(q), (d)–(s)
(4) If both Assertion & Reason are false.

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12. Match the column: 14. Match the column:
Column–I Column–II (No. of
(Compound) Geometrical Column–I Column–II
isomers)
(a) 1,2- (p) 6 (Compound) (Isomeric form)
dichloroethene
(a) H H (p) E
(b) Chlorobromo- (q) 4 C=C
Fluoroiodoethene HOOC COOH
(c) 2,4–Hexadiene (r) 3
(d) 2,4– Heptadiene (s) 2 (b) F NH2 (q) Z
C=C
(1) (a)–(s), (b)–(q), (c)–(r), (d)–(p) Cl Br
(2) (a)–(s), (b)–(p), (c)–(r), (d)–(q)
(3) (a)–(p), (b)–(q), (c)–(r), (d)–(s) (c) CH3 D (r) Trans
(4) (a)–(p), (b)–(r), (c)–(q), (d)–(s) C=C
O 2N H
13. Match the column: (d) (s) Cis
H CH3
Column–I Column–II (Type C=C
(Compound) of isomers) CH3 H
(a) NC CN (p) Metamerism
(1) (a)–(r), (b)–(p), (c)–(q), (d)–(s)

and (2) (a)–(s), (b)–(q), (c)–(p), (d)–(r)

CN CN (3) (a)–(p), (b)–(s), (c)–(r), (d)–(q)

(4) (a)–(q), (b)–(r), (c)–(s), (d)–(p)

(b) O (q) Chain
Cl isomerism

and
Cl
O
(c) NH (r) Ring chain
isomerism
O
and

NH
O
(d) (s) Functional
isomerism
O
and

(1) (a)–(q), (b)–(s), (c)–(p), (d)–(r)

(2) (a)–(q), (b)–(s), (c)–(r), (d)–(p)
(3) (a)–(p), (b)–(q), (c)–(s), (d)–(r)
(4) (a)–(q), (b)–(p), (c)–(s), (d)–(r)
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Exercise 4 (Previous Year's Questions)
1. The number of structural isomers possible 4. With respect to the conformers of ethane,
from the molecular formula C3H9N is : which of the following statements is true?
[AIPMT PRE-2015] [NEET-2017]
(1) 2 (2) 3 (1) Bond angle remains same but bond
(3) 4 (4) 5 length changes
(2) Bond angle changes but bond length
2. In which of the following molecules, all remains same
atoms are coplanar? [NEET-2016] (3) Both bond angle and bond length
change
(4) Both bond angles and bond length
(1) (2) remains same

5. Dihedral angle of least stable conformer of

(3) (4)
ethane is : [NEET(UG) 2021]

3. The correct statement regarding the (1) 120° (2) 180°

comparison of staggered and eclipsed (3) 60° (4) 0°
conformation of ethane, is:- [NEET-I 2016]
(1) The staggered conformation of ethane
is less stable than eclipsed 6. The compound which shows metamerism is:
conformation, because staggered
conformation has torsional strain [NEET(UG) 2021]
(2) The eclipsed conformation of ethane is (1) C5H12 (2) C3H8O
more stable than staggered
conformation, because eclipsed (3) C3H6O (4) C4H10O
conformation has no torsional strain
(3) The eclipsed conformation of ethane is
more stable than staggered conformation 7. Which one of the following compounds can
even through the eclipsed conformation exist as cis-trans isomers? [NEET-2025]
has torsional strain (1) 1, 1-Dimethylcyclopropane
(4) The staggered conformation of ethane (2) 1, 2-Dimethylcyclohexane
is more stable than eclipsed (3) Pent-1-ene
conformation, because staggered (4) 2-Methylhex-2-ene
conformation has no torsional strain.

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ANSWER KEYS

Exercise 1.1
Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans. 1 1 3 4 4 2 4 4 4 4 3 1 1 4 2 1 4 2 1 2
Que. 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
Ans. 1 1 3 4 3 1 3 3 4 2 2 2 2 2 3 1 3 2 1 1
Que. 41 42 43 44 45 46 47 48
Ans. 2 3 3 2 1 4 2 i Metamer vi Chain xi Metamer xvi identical
ii FGI vii Not isomer xii identical xvii Chain
iii Position viii FGI xiii identical xviii RCI
iv FGI ix Chain xiv Chain xix FGI
v FGI x Position xv FGI xx Metamer

Exercise 1.2
Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans. 2 2 4 3 4 1 3 2 4 3 2 4 4 2 3 4 4 2 1 3

Exercise 1.3
Que. 1 2 3 4 5
Ans. 4 3 4 1 2

Exercise 2
Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans. 3 3 4 4 1 3 3 2 3 1 4 3 4 2 3 3 1 4 4 2
Que. 21 22 23 24 25 26 27 28 29 30
Ans. 1 2 4 3 2 4 4 2 3 1

Exercise 3
Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14
Ans. 1 2 2 1 1 3 4 1 3 3 1 2 1 2

Exercise 4 (Previous Year's Questions)

Q ue. 1 2 3 4 5 6 7
Ans. 3 2 4 4 4 4 2

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Chapter GENERAL
ORGANIC
3 CHEMISTRY

Chemical Reaction
Chapter Summary
Chemical reaction involve breaking of existing bonds and
• Chemical Reaction formations of new bonds.
hν
• Reaction intermediates H3C—H+Cl—Cl H3C—Cl+H—Cl
(New bond)
• Inductive effect (existing bond)
A sequential account of each step, describing details of electron
• Application of I-effect
movement, energetics during bond cleavage and bond
• Resonance formation, and the rates of transformation of reactants into
• Application of Resonance products (kinetics) is referred to as reaction mechanism.
Organic Molecule + Attacking Reagent –→ [Intermediate] –→
effect
Product (s) + By Products (s)
• Hyperconjugation effect Substrate & Reagent : The reactant which supplies carbon to the
• Application of H-effect new bond is called substrate and the other reactant is called
reagent.
• Electromeric effect
Concepts to Understand Reaction Mechanism:
• Aromatic, Anti Aromatic & (1) Fission of covalent bond (2) Reagent
Non Aromatic compounds. (3) Intermediate (4) Electronic effect
Type of covalent bond fission :
(a) Heterolytic Cleavage/Fission: The bond fission in which
unequal distribution of electrons takes place is known as
heterolytic cleavage. due to unequal distribution of electrons,
ions are formed, that's why it is also known as ionic bond
cleavage.
⊕ Θ
Ex. CH3 • • Br H3 C+Br
(b) Homolytic Cleavage/Fission: The bond fission in which equal
distribution of electrons takes place is known as homolytic
cleavage.
Due to equal distribution of electrons, neutral species is formed,
which is known as free radical and cleavage is known as non
ionic bond cleavage/homolytical fission.

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• •
Ex. CH3 • • H H3 C+H
⊕ Θ
–C + ו Z Ionic cleavage
Or
Θ ⊕
Heterolytic fission
–C • • Z –C • × + Z

–C • + × Z Non ionic cleavage

or homolytic fission
Reaction
intermediate

Types of Attacking Reagents: There are two types of regents:

(a) Electrophilic Reagent or Electrophiles:
The reagents which accept eΘ to form covalent bond are called electrophiles. Electrophiles may be
positively charged or neutral.
(i) Positively Charged Electrophiles:
⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕
H, SO3H, NO, NO2, X, R, R–C, C6H5–N2
O
Ex. etc.
(ii) Neutral Electrophiles: central atom is e– deficient
(a) All Lewis acids as:
Ex. BF3, AlCl3, ZnCl2, BeCl2, FeCl3, SnCl2, SnCl4 etc.
(b) Free radicals, carbenes and nitrenes act as electrophiles.
• • ••
Ex. H , R, : CH2 , R–C–N•• etc.
O
Note: Vaccant orbital is required for a species to act as electrophile.
(b) Nucleophilic Reagent or Nucleophiles:
The reagent which donate electron to form covalent bond are called nucleophile.
Nucleophiles may be negatively charged ions or neutral & possess a lone pair or π electrons.
(i) Negatively charged nucleophiles.
Θ Θ Θ Θ Θ Θ Θ Θ Θ
Ex. H, OH, OR, CN, X, R, R –COO, NH2 , SH etc.
(ii) All Lewis base which contains lone pairs
•• •• •• •• •• ••
Ex. H2 O , R O H, R– O –R, N H3, R– N H2, R3 N etc.
•• •• ••

(iii) π- electron donor:

Ex. CH2=CH2, CH≡CH,

Ambident Nucleophile: species which have two different electron rich centre or species in which two
or more different atoms bear a lone pair.

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.. •• Θ ..
Θ
Ex. O–N = O , N H2– O H, C ≡ N etc.
••

Ambiphile: Can act as both electrophile and nucleophile.

O
Ex. –C≡N, –C–H etc.

Reaction Intermediates:
Chemical species formed in the intermediate steps of a reaction are called reaction intermediates
1. Carbocation:
Empty unhybridized
'p' orbital

120° C +
sp2 hybridized carbon
(Trigonal planer structure)
• Positively charged carbon with three bond pairs is called carbocation.
• Due to electron deficiency it acts as an electrophile and always attack on electron rich site.
• It has incomplete octet.

2. Carbanions:
:

C–
Sp3
hybridized
Trigonal pyramidal structure
• Negatively charged carbon with three bond pairs and one lone pair is called carbanion.
• It has eight electron so has complete octet.
• Due to presence of non bonded electrons it acts as a nucleophile.

3. Free Radical:
Unhybridized P orbital
containing odd electron
.
120° C
sp2 hybridized carbon
Trigonal Planar structure
• Electrically neutral species in which carbon has three bond pairs and one unpaired eΘ is known as
carbon free radical.
• It has seven electron or odd electron in outermost shell i.e. 7e¯ species.
• It has incomplete octet, hence it is electron deficient species.

Electron displacement effect in covalent bonds:

There are four effects which affect the properties of molecules.
(1) Inductive effect [I-effect] (2) Resonance
(3) Hyperconjugation effect [H-effect] (4) Electromeric effect or polarizability effect [E-effect]

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Inductive Effect:
• Polarisation of σ-bond caused by polarisation of adjacent σ-bond is called inductive effect.
• It causes permanent polarisation in molecule, hence it is a permanent effect.
• Partial displacement of σ e– density takes place.
• The I– effect always operate through σ bond, it does not involve π bond electron.
• Its magnitude decreases with distance and it is negligible after 3rd carbon atom hence, it is distance
1
dependant effect. I effect ∝
distance
Types of Inductive Effect:
(a) –I Effect:
The groups having electron withdrawing power more than hydrogen.
⊕ ⊕
–NR3 > –NH3 >–NO2>–CN>–COOH>–F>–Cl>–Br>–I>–OR>–OH>–C≡CH>–NH2>–C6H5>–CH=CH2.
(b) +I Effect: The groups having electron releasing power more than hydrogen.
Θ Θ Θ
–NH > – O > –CO O > –C(CH3)3 > –CH(CH3)2 > –CH2 – CH3 > – CH3 > –D
The hydrogen atom is reference for +I & –I series. The inductive effect of hydrogen is assumed to be zero.

Application of I-Effect:
(1) Stability of Carbocation (C⊕) or Carbonium Ion. Electron releasing group increases stability by
decreasing positive charge on carbon.
1 +I
Stability of C⊕ ∝ ∝
Charge –I
CH3
⊕
Ex. (i) CH3 C⊕ > CH3 CH > CH3
⊕
CH2 > CH3
⊕ (+I-effect ∝ number of +I groups)

CH3 CH3
3° 2° 1°
⊕ ⊕ ⊕ ⊕
(ii) CH3 CH2 > Br CH2–CH2 > Cl CH2–CH2 > F CH2–CH2

–I effect ↑, Stability of C⊕ ↓

(2) Stability of Carbanion (CΘ): Electron withdrawing group increases stability by decreasing negative
charge on carbon.
 Θ 1 –I 
Stability of C ∝ Charge ∝ +I 
 
CH3
Θ Θ Θ Θ
Ex. (i) CH3–C < CH–CH3 < CH3–CH2 < CH3

CH3 CH3
3º 2° 1°
Θ
Number of +I group ↓ +I effect ↓ Stability of C ↑
Θ Θ Θ
(ii) CH2–CH2 F > CH2–CH2 OH > CH2–CH2 NH2

Θ
–I effect ↓ Stability of C ↓
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Θ Θ Θ
(iii) CH2–CH2 → F CH2–CH2–CH–CH3 CH2–CH2–CH2–CH2–F

F
Distance of –I group ↑ –I effect ↓ Stability ↓

(3) Acidic Strength:

Ka
H − 

A 
 AΘ + H+
 
acid Conjugate base

 1 
 Acidic strength ∝ Ka ∝ pKa ∝ Stability of conjugate base(Anion)
 
Ex. Order of acidic strength:
(i) CH3–C–OH < Cl–CH2–C–OH
O O
D –H⊕ D –H⊕

Θ < Θ
CH3→ C–O Cl ← CH2 – C–O (Conjugate base stability)
O O

+I of –CH3 –I of –Cl
 −I 
 Acidic strength ∝ Stability of conjugatebase(anion) ∝ +I 

−I
acidic strength ∝
+I

(ii) CH3–CH2–COOH < Cl–CH2–COOH

<
F–CH2–COOH
–H ⊕
–H ⊕ –H⊕
Θ Θ Θ
CH3–CH2–COO < Cl-CH2COO < F-CH2COO (Conjugate base stability)
(+I of –CH3) (- I of Cl) (- I of F)

(iii) CH3SH > CH3–OH [ S – H bond is weaker than O – H bond]

–H⊕ –H⊕
Θ Θ
> CH3–O
CH3S
(Conjugate base stability)
(iv) CH3–CH2–CH–COOH > CH3–CH–CH2–COOH > CH2–CH2–CH2–COOH
F F F
(v) CCl3—COOH > Cl2CH—COOH > ClCH2—COOH

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Exercise 1.1
1. Which is most acidic compound: 8. The correct order of increasing acidic
(1) C2H5COOH strength of the compounds:
(2) CH3CH2CH2COOH
(a) CH3CO2H (b) MeOCH2CO2H
(3) CH3COOH
(4) HCOOH Me
(c) CF3CO2H (d) Me CO2H
2. Maximum –I effect is exerted by the group
(1) d < a < c < b (2) d < a < b < c
(1) –F (2) –OCH3
(3) –Cl (4) –NO2 (3) a < d < c < b (4) b < d < a < c

3. Which of the following is most stable: 9. Which of the following groups has the
Θ Θ
(1) CH3 (2) CH2= CH highest +I effect:
Θ Θ (1) CH3— (2) CH3CH2—
(3) CH≡ C (4) CH3–C≡ C
(3) (CH3)2 CH— (4) (CH3)3 C—
4. Which of the following substituted
carboxylic acids has the highest Ka value: 10. Pair of groups exerting (–I) effect is:
(1) CH3–CH2–CH–COOH (1) –NO2 and –CH3 (2) –NO2 and –Cl
Cl (3) –Cl and –CH3 (4) –CH3 and –C2H5
(2) CH3–CH–CH2–COOH
Cl
11. Correct order of acidic strength is:
(3) CH2–CH2–CH2–COOH
(1) HCOOH > CH3COOH > C2H5COOH
Cl
(2) C2H5COOH > CH3COOH > HCOOH
(4) CH3–CH–CH2–COOH
(3) HCOOH > C2H5COOH > CH3COOH
Br
(4) CH3COOH > HCOOH > C2H5COOH
5. Which of the following is electrophile
(1) SO3 (2) Na+ (3) FΘ 12. The number of electrons present in the
(4) NH3 ⊕
valence shell of carbon of CH3 CH2 ion
6. Which of the following is nucleophile bearing +ve charge:
(1) H2O (2) CH2=CH2 (1) 8 (2) 7
(3) Both 1 & 2 (4) None of these (3) 6 (4) 4
7. The inductive effect:
(1) Implies the atom’s ability to cause bond 13. The hybridisation of C of methylcarbanion is:
polarization
(1) sp3 (2) sp2
(2) Increases with increase of distance
(3) Implies the transfer of lone pair of (3) sp (4) All of these
electrons from more electronegative
atom to the lesser electronegative atom
14. Carbanion of CH3CHΘ
2 has shape:
in a molecule
(4) Implies the transfer of lone pair of (1) Planar
electrons from lesser electronegative (2) Pyramidal
atom to the more electronegative atom in (3) Squar bipyramidal
a molecule
(4) All of the above

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15. Which of the following presents the correct 17. Express in decreasing order of (+I) -
order of the acidic strength in the given (a) CH3CH2–CH2–
compounds: (b) CH3–
CH3
(1) FCH2COOH > CH3COOH > BrCH2COOH >
(c) CH3–C–CH2–CH3
ClCH2COOH
(2) BrCH2COOH> ClCH2COOH > FCH2COOH > CH3
CH3COOH CH3–C–
(d) CH3
(3) FCH2COOH >ClCH2COOH > BrCH2COOH >
CH3–CH–CH2-
CH3COOH (e)
CH2CH3
(4) CH3COOH > BrCH2COOH > ClCH2COOH >
Correct answer is –
FCH2COOH (1) (c) > (d) > (e) > (a) > (b)
(2) (d) > (a) > (b) > (c) > (e)
16. The correct order of decreasing acidic (3) (a) > (b) > (c) > (d) > (e)
strength of trichloroacetic acid (A), (4) (a) > (b) > (c) > (e) > (d)
trifluoroacetic acid (B), acetic acid (C) and
formic acid (D) is: 18. Which of the following has maximum pKa:
(1) B > A > D > C (2) B > D > C > A (1) CH2FCOOH (2) CH2ClCOOH
(3) CH3COOH (4) HCOOH
(3) A > B > C > D (4) A > C > B > D

Resonance:
It involves complete delocalization of π e– / eΘ of lp /unpaired eΘ is called as resonance or complete
transfer of πe– from one shell to another shell is called as Resonance.

Conditions For Resonance:

1. When π bonds are in conjugated system then e– of one π bond are transferred towards another π bond.
⊕ Θ Θ ⊕
Ex. (i) CH2=CH–CH=CH2 CH2–CH=CH–CH2 CH2–CH=CH–CH2

⊕
(ii) CH2=CH–C–H CH2–CH=C–H
O Θ
O
2. If lone pair and π-bond are in conjugated system then e– of lone pair are transferred towards π bond.
Θ ⊕
Ex. (i) CH2=CH–OH CH2–CH=OH

Θ Θ
(ii) CH2=CH–CH2 CH2–CH=CH2

3. If positive charge (vacant orbital) and π bond are in conjugated system then e– of π bond are transferred
towards positive charge (vacant orbital).
⊕ ⊕
Ex. CH2= CH–CH2 CH2– CH=CH2

4. If unpaired eΘ and π bond are in conjugated system.

• •
Ex. CH2 = CH — CH2 CH2 – CH = CH2

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5. If lone pair and positive charge (vacant orbital) are in conjugated system then e– of lone pair are
transferred towards positive charge (vacant orbital).
⊕ •• ⊕
Ex. (i) CH2—OH CH2 = OH
••
Cl—C—Cl Cl—C =Cl
(ii)
Cl Cl
The Rules to draw resonating structure:
(i) The positions of the nuclei of the atoms must remain the same in all resonating structures.
(ii) All canonical forms must have the same number of unpaired electron.
(iii) All resonating structure must have same net charge/Sum of charge.
(iv) All resonating structure must be according to lewis dot formula.
Draw Resonating Structures:
⊕ ⊕ ⊕
CH2 CH2 CH2 CH2 CH2
1. ⊕ ⊕
(I) (II) (III) (IV) (V)
5-Resonating structure

••
OH ⊕ ⊕ Θ ⊕
••
OH
OH OH OH
2.
Θ Θ
(I) (II) (III) (IV) (V)
5-Resonating structure

Θ Θ
Θ
CH2 CH2 CH2 CH2 CH2
3.
Θ Θ

(I) (II) (III) (IV) (V)

5-Resonating structure

⊕
C–H C–H C–H C–H C–H
4.
O ⊕ OΘ ⊕ OΘ OΘ O

(I) (II) (III) (IV) (V)

5-Resonating structure

Resonance Energy: The difference in energy between the resonance hybrid and the most stable
canonical structure is referred as the resonance energy. It is consumed in stabilising the molecule.
The resonance energy of a resonance hybrid is the difference between the theoritical and experimental
heats of hydrogenation of the compound.
Catalyst
Ex. + H2 → + 28.6 kcal/mol

Cyclohexene

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Accordingly,
Catalyst
+ 3H2  → + 3 × 28.6 = 85.8 Kcal/mole. (Theoretical)

But practical value of heat of hydrogenation is 49.8 Kcal/mol

Therefore, benzene has 85.8 – 49.8 Kcal/mol less energy than expected for a typical compound with
three double bonds. Hence resonance energy of benzene molecule is 85.8 – 49.8 = 36.0 Kcal/mole.
Types of Mesomeric Effect:
(a) +M effect (b) –M effect

(a) Positive Mesomeric Effect (+M Effect): When the group releases electron to the conjugated system it
shows +M effect.
+M Group: Lone pair containing group like
Θ Θ O O
      ••
− NH , –O , −NH2 , −OH , −OR , −NR2 , −NHR , − X , –NH–C–CH3 , –O–C–CH
••
•• 3
••
NH2←(+M) group

Ex.

Θ Θ O O
     ••
order of +M Group: − NH > –O > −NH2 > −NHR > −NR2 > −OH > −OR > –NH–C–CH3 > –O–C–CH
••
•• 3>

−F  > −Br
 > −Cl  > −I

(b) Negative Mesomeric Effect (–M Effect): When the group withdraws electron from the conjugated
system it shows –M effect.
–M group: —CHO, —COOH, —COOR, —COR, —NO2, —CN, —CONH2, —SO3H
O ←N= O ←(–M) group

Ex.

order of –M Group: –NO2 > —CN > —SO3H > —CHO > —COR > —COOR > —COOH > —CONH2
M-effect of a group on benzene ring :
•• ⊕ ⊕ ⊕ •• δ⊕
NH2 NH2 NH2 NH2 NH2 NH2
Θ Θ δΘ
+M •• •• δΘ
≡
••
Θ δΘ
H
Θ Θ Θ δΘ
O←N=O O←N–O O←N–O O←N–O O←N=O O←N O
⊕ δ⊕ δ⊕
–M ⊕
≡
⊕ ⊕ δ⊕

• M-effect of a group affects ortho and para position while meta position remains unaffected.
• +M group increases π eΘ density at ortho and para position and -M group decrease π eΘ density at ortho
and para position in both the cases meta position remains neutral.

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• M-effect of a group affects ortho and para position equally while I-effect depends on distance so it is
maximum at ortho and minimum at para.
• Comparison of M and I-effect of a group.
In general M-effect of a group dominates over its I-effect but in the following groups their I-effect
dominates over their M-effect.
O
|| Θ
(I) —X [–I > +M] Net effect ⇒ e withdrawing
Θ
(II) −C − O [+I > –M] net effect → eΘ releasing
Comparision of stability of resonating structure:
(i) Resonating structure having complete octate is more stable
(ii) Neutral resonating structure is more stable than charged resonating structure
Or
Resonating structure having more covalent bond is more stable
(iii) Resonating structure having negative charge on more EN atom and positive charge on less EN atom is
more stable
(iv) Resonating structure having opposite charge closer and same charge away is more stable

Ex. Arrange the following in correct order of stability.

Θ ⊕ Θ ⊕
Ex. (i) CH= CH − Cl > CH2–CH=Cl > CH2–CH–Cl
2
Complete octet & Neutral complete octet & charged incomplete octet
⊕ ⊕ Θ
(ii) R–C–OH > R–C=O–H > R–C–OH > R–C–OH
O OΘ OΘ O⊕
Complete octet Complete octet Carbon with oxygen with
and neutral and charged incomplete octet incomplete octet

Exercise 1.2
1. Examine the following two structures for 2. Find the stability order of given resonating
the anilinium ion and choose the correct structures:
statement from the ones given below.
⊕ O
⊕ (I) N
NH3 NH3 OΘ
⊕
⊕
⊕ O
(II) Θ N
OΘ
I II
Θ
(1) II is not an acceptable canonical ⊕ O
structure because carbonium ions are ⊕ N
(III)
less stable than ammonium ion. O
Θ
(2) II is not an acceptable canonical
structure because it is non-aromatic.
Θ
⊕ O
(IV) ⊕ N
(3) II is not an acceptable canonical OΘ
structure because the nitrogen has 10 (1) I > II > III > IV (2) I > III > II > IV
valence electrons.
(3) I > III > IV > II (4) IV > I > III > II
(4) II is an acceptable canonical structure.

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3. The order of stability of the following 8. Which of the following is wrong about
resonating structures is: resonance:
Θ ⊕ (1) Most stable resonating structures
O O O contribute the most to resonance
⊕ Θ
CH2=CH–CH CH2–CH=CH CH2–CH=CH hybrid.
(I) (II) (III)
(2) All resonating structures have same
(1) II > I > III (2) I > III > II
number of unpaired electron.
(3) I > II > III (4) III > II > I
(3) All resonating structures have same
number of bond pair electrons.
4. Mesomeric effect is the resonance of:
(1) π- electrons and lone pair (4) All resonating structures have same
(2) σ- electrons only amount of net charge.
(3) π and σ-electrons both
(4) π- electrons only 9. Which of the following contain + M & - I
effect.
5. Which of the following resonating (1) O = CH – (2) – NO2
structures contributes the most to (3) – Cl (4) CH3 –
resonance hybrid ?
OCH3 OCH3
10. Which of the following is not a pair of
resonating structures?
(1) ⊕ (2) ⊕ Θ
•• •
(1) Θ O and
••
O
••
•
⊕
OCH3 OCH3
⊕ (2) and
(3) (4) N
•• N
••

Θ
•• ••
O
•• O
••
6. Which of the following group has (+M) (3) CH3–C •• and CH3–C ⊕
OH
•• OH
effect on benzene ring ? ••
O O and
S–OH C–H (4)
(1) O (2)
Θ
• •• ••
NH–CH3 CH3 O••
• O
• •
•• ⊕
(3) (4) 11. (I) H–C–OH
•• (II) H–C=OH
••

Θ ⊕
•• •• •
O
• • • O••
•• ••
Θ (III) H–C–OH (IV) H–C–OH
••
7. CH2–C–CH3 and CH2=C–CH3 are: ••
⊕ ••
Θ
• •
O •O•
••
Increasing order of stability is -
Θ
(1) I < III < II < IV
(1) Resonating structures
(2) IV < III < II < I
(2) Tautomers
(3) III < IV < II < I
(3) Geometrical isomers
(4) II < IV < III < I
(4) Optical isomers

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Applications of Resonance Effect

1. Stability of Carbocation:
• More resonance in carbocation makes it more stable.
1
• Stability ∝ +M / +I ∝
–M / –I
Ex. Stability orders of Carbocations:-
⊕ ⊕ ⊕
(i) CH2=CH–CH2 > CH3 CH2–CH2
> CH2=CH CH2–CH2
stable by +I of Alkyl group –I of Vinyl group
resonance
⊕ ⊕ ⊕
(ii) CH3–C–CH=CH2 > CH3–CH–CH=CH2 > CH2–CH=CH2
CH3
Resonance and +I Resonance and +I Resonance
of 2–CH3 of 1–CH3 only

⊕ ⊕ ⊕
(iii) (C6H5)3 C > (C6H5)2 CH > C6H5 C H2
Resonance Resonance Resonance
with 3 rings with 2 rings with 1 ring
⊕ ⊕ ⊕
(iv) CH2 > CH2 > CH2
Resonance with Resonance with Resonance with
3π bond 2π bond 1π bond

⊕ ⊕
(v) ⊕ > CH3 > CH3
Resonance +I effect –I effect

⊕ ⊕ ⊕

(vi) , ,
(a) (b) (c)
Resonance more resonance no resonance
stability b > a > c

⊕ ⊕
(vii) CH2–C–H < CH2–CH2–C–H
O O
more –I of —CHO less –I of —CHO

2. Stability of Carbanion:
• More resonance in carbanion makes it more stable.
1
• Stability ∝ –M / –I ∝
+M / +I

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Ex. Give stability order of following compounds

Θ Θ Θ
(i) CH2=CH–CH2 , CH2=CH, CH3—CH2
charge on sp2C & resonance charge on sp2C charge on sp3C
stability order: I > II > III

Θ Θ Θ Θ Θ Θ Θ
(ii) (C6H5)3C > (C6H5)2CH > C6H5CH2 > CH2═CH—CH2 > CH2=CH > CH3 > CH3—CH2
Θ Θ
> (CH3)2—CH > (CH3)3C

Θ Θ Θ
(iii) , ,

(a) (b) (c)

Resonance with 1 π resonance with 2π no resonance
stability order: b > a > c
Θ Θ Θ
(iv) CH2—NO2 CH2—CH2—NO2 CH3— CH—NO2
(a) (b) (c)
Resonance No resonance Resonance but +I effect
stability order: a > c > b
Θ Θ
(v) Θ, CH2 , CH3

(a) (b) (c)

Resonance and +I effect Resonance Resonance and more +I effect
stability order: b > a > c
O
Θ
(vi) CH3–C–O Θ
•• O

(a) (b)
Resonance with oxygen Resonance with carbon
(Equivalent R.S.) (Non equivalent R.S.)
stability order: a > b
3. Stability of free radical :
• Comparison similar to carbocation
• Stability ∝ Extent of resonance
1
∝ +M / +I ∝
–M / –I
• •

Ex. (i)
• •
(a) (b) (c) (d)
Resonance and Less +I effect More +I effect Resonance and
more +I effect less +I effect
Ans. a>d>c>b

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•
• • •

Ex. (ii)
(a) (b) (c) (d)
No resonance &
Resonance Resonance No resonance
unpaired e– on
with 1π bond with 2π bond +I –effect
unsaturated C
–I effect
Ans. b>a>d>c

Question: Give stability order of following compounds

⊕
CH2
⊕ ⊕ ⊕
(i) CH2 CH2 CH2

NO2
NO2 NO2
–M and M=0 –M and (Reference)
more –I only –I less –I
stability order is IV > II > III > I
⊕ ⊕ ⊕
CH2 CH2 CH2 ⊕
CH2
OCH3
(ii)
(I) OCH3
(II) OCH3 (IV)
(III)
+M and M=0 +M and (Reference)
more –I only –I less –I
Stability order III > I > IV > II
Θ
Θ CH2 Θ Θ
CH2 CH2 CH2

NO2
(iii)
NO2
NO2
–M and M=0 –M and (Reference)
more –I only –I less –I
stability order I > III > II > IV

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Exercise 1.3
1. Most stable carbocation is: 6. Which of the following is most stable:
⊕ ⊕ ⊕
(1) CH2=CH (2) CH2=CH–CH2 (1) CH3
⊕
⊕ (2) CH3– CH 2
(3) CH2 (4) ⊕ CH3 ⊕
(3) CH3– CH –CH3
⊕
(4) CH2=CH– CH 2
2. Least stable carbonium ion in the following
will be:
7. Which of the following is least stable
(1) ⊕ (2) ⊕ carbocations:
⊕ ⊕
⊕ (1) CH2= CH (2) CH3– CH 2
(3) CH2 (4) ⊕
⊕ ⊕
(3) (Ph)2 C H (4) Ph CH 2
3. Arrange stability of the given carbocations
in decreasing order: 8. Which one of the carbanions is most stable:
Θ Θ
⊕ ⊕ ⊕ ⊕ CH2 CH2
CH2 CH2 CH2 CH2
NO2
(1) (2)

NO2
OCH3 OH NH2 Cl
Θ Θ
(I) (II) (III) (IV) CH2 CH2
(1) I > II > III > IV (2) III > II > I > IV CN
(3) IV > I > II > III (4) III > I > II > IV (3) (4)

Θ Θ CN
4. Arrange the carbanions, ( CH3 ) 3 C, CCl3 ,
9. Choose the most stable Carbocation:
Θ Θ
( CH3 )2 CH, C6H5 CH2 in order of their (1) CH3–CH=CH– CH2
⊕

decreasing stability: ⊕
Θ Θ Θ Θ (2) CH2=CH– CH –CH3
(1) ( CH3 )2 CH > CCl3 > C6H5 CH2 > ( CH3 )3 C ⊕
(3) CH2= C –CH2–CH3
Θ Θ Θ Θ
(2) CCl3 > C6H5 CH2 > (CH3 )2 CH > ( CH3 )3 C (4) CH2=CH–CH2 – CH2
⊕

Θ Θ Θ Θ
(3) ( CH3 )3 C > (CH3 )2 CH > C6H5 CH2 > CCl3
10. Most stable carbonium ion in the following
Θ Θ Θ Θ will be:
(4) C6H5 CH2 > CCl3 > (CH3 )3 C > (CH3 )2 CH
(1) ⊕ (2) ⊕
5. The increasing order of stability of the ⊕
(3) CH2 (4) ⊕
following free radicals is:
• • • •
(1) (C6H5)3 C <(C6H5)2 CH <(CH3)3 C <(CH3)2 CH
11. Which free radical is the most stable:
• • • • • •
(2) (C6H5)2 CH <(C6H5)3 C <(CH3)3 C <(CH3)2 CH (1) C6H5–CH2 (2) CH2=CH–CH2
• • • •
(3) (CH3)2 CH <(CH3)3 C <(C6H5)3 C <(C6H5)2 CH • •
(3) CH3–CH–CH3 (4) CH3–C–CH3
• • • •
(4) (CH3)2 CH <(CH3)3 C <(C6H5)2 CH <(C6H5)3 C CH3

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12. Most stable carbanion is :- 15. Correct order of stability is –
Θ Θ ⊕ ⊕ ⊕ ⊕
(1) HC ≡ C (2) H2C=CH CH2 CH2 CH2 CH2
CH3 Θ
Θ
(3) CH3–C–CH2
(4) CH3–C=CH
CH3 OCH3
CH3 OCH3 NH2
Ι ΙΙ ΙΙΙ ΙV
13. The order of stability of the following (1) I > II > III > IV (2) III > II > I > IV
carbocations is- (3) IV > I > II > III (4) II > III > I > IV
⊕
(I) CH=
2 CH − CH2
16. Consider the following carbanions :
⊕
Θ
(II) CH3 − CH2 − CH2 ••
(I) H3CO CH2
⊕
CH2 Θ
••
(II) O2N CH2
(III) Θ
••
(III) CH2
(1) I > II > III (2) III > I > II
Θ
••
(3) III > II > I (4) II > III > I (IV) H3C CH2
Correct decreasing order of stability is-
14. The order of decreasing stability of the
(1) II > III > IV > I (2) III > IV > I > II
Cations is-
(3) IV > I > II > III (4) I > II > III > IV
⊕ ⊕
CH3 CHCH3 CH3 CHOCH3
Ι ΙΙ 17. Which one is most stable free radical?
⊕ Me
CH3 CHCOCH3 •
•
ΙΙΙ (1) (2)
(1) III > II > I (2) I > II > III
(3) II > I > III (4) I > III > II • •
(3) CH=
2 CH − CH2 (4) CH3 − C− CH3
|
CH3

4. Acidic Strength comparison:

Acidic strength ∝ Stability of conjugatebase
• Carboxylic acids are more acidic than phenols
OH
R–C–OH
O
–H⊕ –H⊕
Θ
R–C–O R–C=O OΘ O
Θ Θ
O O
Two equivalent resonating structures 5, unequal R.S.
make the anion stable less stable anion
so corresponding acid is more acidic.

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• Phenol is more acidic than alcohols:

Ph–OH R–OH
⊕ ⊕
–H –H
Ph–OΘ R–OΘ
Conjugate base stable by resonance no resonance
So, it is more acidic.

Question: Give acidic strength order for following compound

O OH
O
(i) CH3–C–OH CH3–S–OH CH3–CH2–OH
O
(I) (II) (III) (IV)
Ans. Acidic strength α Stability of conjugate base
Θ
O O
Θ
O CH3–S–O Θ
Θ CH3–CH2–O
CH3–C–O O
(Reso. with (Reso. with (Reso.with C)
oxygen) more oxygen) (No Reso.)
Acidic strength: II > I > IV > III

(ii) OH C≡CH OH OH

(I) (II) (III) (IV)

Θ Θ Θ Θ
Ans. O C≡C O O

negative on negative on negative on more negative on

more EN Less EN (sp‘C’) EN and more Resonance more EN & Less Resonance
Acidic order — III > IV > I > II

OH OH OH OH

NO2
(iii)
NO2
NO2
–M and M=0 –M and (Referance)
more –I only –I less –I
but intra molecular
H-bonding
Acidic strength → III > I > II > IV

Note: Due to intramolecular H-bonding in ortho nitrophenol it is less acidic than para nitrophenol.

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CH O= N→O
OH
O
N
O
O←N=O H–O
intramolecular H-bonding intermolecular H-bonding
OH OH OH OH

NO2 NO2 NO2 NO2

(iv)

NO2 NO2
3 (–M) 2 (–M) 1 (–M) (Referance)
Acidic strength → 1 > 2 > 3 > 4
COOH COOH COOH COOH

NO2
(v)
NO2
NO2
ortho effect M=0 –M and (Referance)
only –I less –I
Acidic strength → 1 > 3 > 2 > 4
⊕
5. Basic Strength Comparison:- B.S. ∝ H accepting tendency ∝ lone pair donating tendency.
1
• Basic strength ∝
Resonance(l.P.)
1
• Basic strength ∝
%Scharacter of N
1
• Basic strength ∝ +M, + I ∝
−M, − I

Question: Give basic strength order:

•• •• ••
NH2 CH2–NH2 NH–CH3
(i)
l.P. is in l.p. is not in l.p. is in resonance
resonance resonance and +I of CH3
Basic strength → II > III > I
••
NH2
(ii) ••
N
••

N
H
l.p. in resonance l.p. is not in l.p. is not in
resonance & Nsp2 resonance & Nsp3
Basic strength→ III > II > I

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••
(iii) NH2
N ••
N
H
distribution of distribution of l.p. l.p. is not in
l.p. among N & four C among N & three C resonance
(l.p. is part of aromaticity) (lp is not the part of aromaticity)
Basic strength→ III > II > I
••
O •• ••
N N
(iv) ••
N ••
N H H
l.p. is not in l.p. is not in resonance l.p. is not in resonance distribution of
2 3
resonance & Nsp Nsp & –I effect of & +I-effect l.p. among N & four C
oxygen (l.p. is part of aromaticity)

Basic strength→ III > II > I > IV

Comparison of basic strength in aliphatic amines

(i) CH3–NH2, (CH3)2NH, (CH3)3N, NH3
B.S. in aqueous medium 2° > 1° > 3° > ammonia
In gas phase : 3° > 2° > 1° > ammonia
(ii) C2H5–NH2 , (C2H5)2NH, (C2H5)3N, NH3
B.S. in aqueous medium 2° > 3° > 1° > ammonia
In gas phase : 3° > 2° > 1° > ammonia
Note: if medium is not given then consider aqueous medium

Exercise 1.4
1. Acidic strength order is:- 3. Amongst the following, the most basic
COOH COOH compound is:
(1) C6H5NH2
(1) (2)
(2) p-NO2 – C6H4NH2
CMe3 (3) m-NO2 – C6H4NH2
CMe3
(4) C6H5CH2NH2
COOH
COOH

(3) CMe3 (4) 4. Amongst the following the most basic

compound is:
(1) aniline (2) benzylamine
(1) 4 > 1 > 2 > 3 (2) 3 > 1 > 2 > 4
(3) 4 > 3 > 1 > 2 (4) 3 > 4 > 1 > 2 (3) p–nitroaniline (4) acetanilide

2. The correct order of acidic strength of the 5. Correct acidic nature is :-

following compounds is: (1) CH3OH < CH3SH < CH3NH2
A. Phenol B. p-Cresol (2) CH3NH2 < CH3OH < CH3SH
C. m-Nitrophenol D. p-Nitrophenol (3) CH3SH < CH3OH < CH3NH2
(1) D > C > A >B (2) B > D > A >C (4) None of these
(3) A > B > D >C (4) C > B > A >D
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6. Arrange basic strength of the given 12. Which nitrogen in LSD (Lysergic acid and
compounds in decreasing order: diethylamide) is most basic:
(a) CH3–CH2–NH2
O
(b) CH2=CH–NH2 Ι ΙΙΙ
(c) C6H5–NH2 HN C–N(C2H5)2
ΙΙ
(1) a > b > c (2) a > c > b N
(3) c > b > a (4) b > c > a
CH3
7. Which of the following is the strongest (1) I
base: (2) II
(1) NH2 (2) NHCH3 (3) III
(4) All are equally basic
NH2
(3) (4) CH2NH2 13. (A) (B) (C)
CH3
N N
H NH2
8. Which of the following substituents will
decrease the acidity of phenol: Choose the incorrect statement:
(1) –NO2 (2) –CN (1) A is more basic than C
(3) –CH3 (4) –CHO (2) B is more basic than A
(3) B is more basic than C
9. Consider the following compounds: (4) All are aromatic bases
NH2 NH2 NH2 NH2

14. (I) NH2 (II) CH3O NH2

NO2 Cl CH3 (III) NO2 NH2

(a) (b) (c) (d)
Arrange these compounds in decreasing (IV) NH2
order of their basicity:
NO2
(1) a > b > c > d (2) b > c > a > d
(3) d > a > c > b (4) d > a > b > c The correct order of decreasing basicity of
the above compound is:
10. Which nitrogen is protonated readily in the (1) I > II > III > IV
guanidine?
2 (2) II > I > IV > III
NH2 (3) III > IV > II > I
NH = C
1 NH2
3 (4) II > I > III > IV
(1) 1
(2) 2 15. Which one of the following is least basic in
(3) 3 character ?
(4) All with equal rate

11. Which of the following shows the correct (1) N (2) N N–H
order of decreasing acidity:
H
(1) PhCO2H > PhSO3H > PhCH2OH > PhOH
(2) PhSO3H > PhOH > PhCO2H > PhCH2OH (3) (4)
(3) PhCO2H >PhOH > PhCH2OH > PhSO3H N N
(4) PhSO3H > PhCO2H > PhOH > PhCH2OH H H

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16. Which one of the following compounds is 18. Arrange the following in correct order of
most acidic: acidic strength :
OH OH (I) C2H5SH
(1) (2) (II) C2H5OH
NO2 CH3 (III) C2H5NH2
OH (IV) CH3CH2CH3
(3) (4) ClCH2CH2OH (1) IV > II > I > III (2) IV > II > III > I
(3) I > II > III > IV (4) III > I > IV > II

17. Which one of the following compounds has 19. Among the following the pka is minimum for :–
the most acidic nature (1) C6H5OH (2) HCOOH
(3) C2H5OH (4) CH3C ≡ CH
CH2OH OH
(1) (2) 20. The correct order of increasing basic nature
of the bases NH3, CH3NH2 and (CH3)2 NH is -
OH
(1) NH3 < CH3NH2 < (CH3)2NH
OH CH
(3) (4) (2) CH3NH2 < (CH3)2NH < NH3
(3) CH3NH2 < NH3 < (CH3)2NH
(4) (CH3)2NH < NH3 < CH3NH2

6. Bond length comparison:

B.L. > > >
(I) CH2=CH–Cl(C Cl) (Due to resonance)
(II) CH3–CH2–Cl (C–Cl) (No resonance)
B.L.→ II > I
(I) CH3–CH=O C=O (No resonance)
(II) CH2 = CH – CH = O (C O) (Due to resonance)

B.L. → II > I

• Resonance
Single bond (–)  → Acquired double bond charecters so B.L. ↓ ( )

• Resonance
Double bond (=)  → Acquired single bond charecters so B.L. ↑ ( )

Ex. (i) Compare C–C bond length

, CH3 – CH3, CH2=CH2, CH≡CH

(a) (b) (c) (d)

Ans. b > a > c > d

(C–C) (C C) (C=C) (C≡C)

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Ex. (ii) Compare C–O bond length
O O O OH

(a) (b) (c) (d)

Resonance More Resonance No Resonance No Resonance
(C O) (C O) (C O) (C–O)
Single bond More Single bond
characters characters
Ans. d > b > a > c
Hyperconjugation Effect (H-Effect)

H-Effect: Complete delocalisation of e– of C–H σ bond towards carbon-carbon π bond or positive charge or
free radical is called as H-effect.
• It is also called as no bond resonance and Nathen & Baker effect.
• It is permanent effect
• It is distance independent effect

Conditions of H-Effect:
1. When C–H σ bond of sp3 ‘C’ is in conjugation with +ve charge in carbocation.
Orbital overlap here

H
2e Vacant p orbital

C C
⊕ H
SP3
H
H
H
α-carbon H H⊕ H H Hδ⊕
⊕ ⊕ δ⊕ δ⊕
H C CH2 H C CH2 H C CH2 H C CH2 ≡ H C CH2
H H H H⊕ Hδ⊕
αH

2. When C–H σ bond of sp3 ‘C’ is in conjugation with unpaired eΘ in free radical.
•
H H H H
• •
H C CH2 H C CH2 H C CH2 H C CH2
•
H H H H
αH

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3. When C–H σ bond of sp3 ‘C’ is in conjugation with carbon-carbon double bond in alkene.
H H⊕ H H
Θ Θ Θ
H–C–CH=CH2 H–C=CH–CH2 H–C=CH–CH2 H⊕ C=CH–CH2
1 2 3 1 2 3 1 2 3 1 2 3

H H H ⊕
H
αH
Note: In alkyl carbanion, H-effect is not observed.

4. When C–X σ bond of sp3 ‘C’ is in conjugation with carbon-carbon double bond then it is called reverse
hyper conjugation (–H effect)
Θ
X X X X
⊕ Θ ⊕ ⊕
X–C–CH = CH2 ↔ X–C = CH–CH2 ↔ X C = CH–CH2 ↔ X–C= CH – CH2
X X X X
Θ

5. H-effect of a group on benzene ring:

+H-effect:
δ⊕
H H H H H
δ⊕ δ⊕
H–C–H H⊕ C–H H⊕ C–H H–C–H H….C….H
Θ δΘ δΘ
•• Θ
••
..........≡
•• δΘ
Θ
–H-effect:
δΘ
X X X X X
Θ Θ Θ δΘ δΘ
X–C–X X C–X X C–X X C–X X….C….X
⊕ δ⊕ δ⊕
⊕
..........≡

⊕ δ⊕

(i) +H-effect of a group increases eΘ density at o- & p-position.

(ii) –H-effect of a group decreases eΘ density at o-&p-position.
(iii) H-effect does not affect meta position.
(iv) H-effect is equal at ortho and para position.

Order of +H-effect of a group on benzene ring.

(i) –CH3 > – CH2– CH3 > – CH(CH3)2 > – C(CH3)3
(3αH) (2 αH) (1 αH) (No Hα)
(No H-effect)
(ii) – CH3 > – CD3

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Application of H-Effect

1. Stability of Carbocation:
• Stability of carbocation can be explained by M, I and H-effect.
• Stability ∝ H-effect ∝ No. of αH
Ex. Give stability order for:-
CH3
⊕ ⊕ ⊕
(i) CH3–C⊕ > CH3–CH > CH3–CH2 > CH3
CH3 CH3
9 αH 6 αH 3 αH Zero αH
CH3 ⊕
CH2
⊕
⊕
(ii)

7αH 1αH 4αH

Stability order 1 > 3 > 2

2. Stability of Carbon Free Radicals:

• Stability ∝ H-effect ∝ No. of αH
Ex. Give stability order for:
CH3
. . CH3
.
CH2
.
CH3
(i) CH3 C CH3 CH

CH3 CH3
9 αH 6 αH 3 αH Zero αH
Stability order I > II > III > IV
• • • •
(ii) CH2= CH–CH2 CH3– CH–CH = CH2 CH3 CH3–CH – CH3
Resonance Resonance & 3αH Resonance & 5αH 6αH
Stability order III > II > I > IV

3. Stability of Alkenes:
• Stability ∝ Resonance
• Stability ∝ H-effect ∝ No. of αH
Ex. (i) CH3—CH═CH2 > CH2═CH2
3 αH Zero αH

CH3 CH3
(ii) CH3 C = C CH3 > C=C > CH3
C=C
H >
CH3 CH3 CH3 H CH3 H
(12αH) (9αH) (6αH)

CH3 H CH3 CH3

C=C C=C
H CH3 > H H > CH3 H > H H
(Trans) (Cis) C=C C=C
H H H H
(6αH) (6 αH) (3αH) (No αH)

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4. Heat of Hydrogenation(HOH):
R—CH═CH2 + H2 → R—CH2—CH3 + ∆H (HOH)
Heat evolved when one mole of any unsaturated hydrocarbon are hydrogenated is called heat of
hydrogenation (HOH) If alkene is more reactive towards hydrogen then it will evolve more HOH.
HOH ∝ Number of π bonds
1
∝ (If π bonds are equal )
stability of alkene

Question: Compare HOH

CH3 CH3 H CH3 CH3–CH2
C = CH2 CH3 H
CH3 C = C C=C C=C
H CH3 H H H H
(I) (II) (III) (IV)
Stability: I > II >III > IV
H.O.H. I < II < III < IV

Question: Give stability order of following compounds

⊕ ⊕
CH2 ⊕ ⊕ CH2
CH2 CH2
(i)

NO2 OCH3
Cl CH3
–M +M
(–Ι>+M) +H
So stability order IV > III > II > I

Θ Θ
CH2 Θ Θ CH2
CH2 CH2
(ii)

NO2 OCH3
Cl CH3
–M +M
(–Ι>+M) +H
stability order I > II > III > IV

Question: Give acidic strength order for following compounds

OH OH OH OH

NO2 Cl CH3 OCH3

–M (–I>+M) +H +M
Acidic strength→ I > II > III > IV

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Question: Give basic strength order for following compounds
NH2 NH2 NH2 NH2

NO2 Cl CH3 OCH3

(–M) (–I>+M) (+H) (+M)
basic strength →IV > III > II > I
Electromeric Effect
Electromeric Effect (E-effect):- The complete transfer of the shared pair of π electron of a multiple bond to one
of the shared atoms in the presence of an attacking reagent is called as E-effect.
It is a temporary effect and comes into play only when a reagent attacks on substrate having π
electrons.
(a) +E-effect:- When electrons of π bonds are transferred to atom of the multiple bond to which the reagent
gets attached, the effect is called +E-effect.
H⊕ ⊕
Ex. C═C C—C
H
(b) –E-Effect:- When electrons of π bond are transferred to atom of the multiple bond to which
the attacking reagent is not attached, the effect is called as –E-effect.
Θ
C═O C≡N Θ
Ex. C—O
CN

Exercise 1.5
1. Which of the following is most stable (3) ⊕
carbocation?
⊕ ⊕
CH2 CH2 (4) ⊕

(1) (2)
3. Among the following alkenes
CH3 CH2–CH3 (I) 1-butene
⊕ (II) trans-2-butene
⊕ CH2
CH2 (III) cis-2-butene
(IV) Isobutylene
(3) (4)
the order of decreasing stability is:
CH3–CH–CH3 CH3–C–CH3 (1) II > I > III > IV
CH3
(2) III > IV > I > II
(3) IV > I > II > III
(4) IV > II > III > I
2. Select the most stable carbocation among
the following:
4. Correct order of stability is:
⊕
(1) (1) CH2=CH2> CH3–CH=CH2> (CH3)2C=CH2
(2) CH2=CH2< CH3–CH=CH2 < (CH3)2C=CH2
(2) ⊕ (3) CH2=CH2< (CH3)2C=CH2 < CH3–CH=CH2
(4) CH3–CH=CH2 < CH2=CH2< (CH3)2C=CH2

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5. Arrange the Stability of following: 9. Electromeric effect:

(1) Comes into play at the demand of

I II III attacking reagent.

(1) I < II < III (2) II < I < III
(2) involves displacement of electrons in a
(3) I < III < II (4) II < III < I
sigma bond.
6. Which of the following alkene is most
(3) Comes into play in the molecule when
stable:
at least one atom has unshared pair of
(1) (2) electrons.

(4) is permanent effect.

(3) (4)

7. Give the decreasing order of hyperconjugative 10. Which statement is correct for electromeric
effect of R in R–CH = CH2, where R is: effect:
(1) It is a temporary effect
I. Me – II. Et–
(2) It is the property of π bond
III. Me2CH– IV. Me3C– (3) It takes place in presence of reagent,
(1) I > II > III > IV (2) IV > III > II > I i.e., electrophile or nucleophile
(3) II > I > III > IV (4) IV > III > I > II (4) All are correct

8. Which amongst the following is the most 11. Which is wrong electromeric effect?
⊕ Θ
stable carbocation: (1) C=O C–O
⊕ ⊕
(1) CH3 CH2 (2) CH3 ⊕ Θ
(2) CH3–CH=CH2 CH3–CH–CH2
CH3
⊕
(3) CH3–C⊕ (4) CH3–CH Θ ⊕
(3) −C ≡ N → − C =N
CH3 CH3 Θ ⊕
(4) CH ≡ CH → CH =
CH

Aromatic, Anti Aromatic and Non Aromatic Compounds

Conditions for a compound to be aromatic:

(i) Cyclic

(ii) Planar ring

(iii) Cyclic resonance

(iv) According to Huckel’s rule

Number of π eΘ in cyclic resonance = (4n + 2)π eΘ

Where n = 0, 1, 2, 4, . . . . . .

π eΘ = 2, 6, 10, 14, . . . . . . .

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Condition for a compound to be anti aromatic:

(i) Cyclic

(ii) Planar ring

(iii) Cyclic resonance

(iv) Number of π eΘ in cyclic resonance = 4π eΘ

Where n = 1, 2, 3, 4, . . . . .

π eΘ = 4, 8, 12, 16, . . . . .

Note: Compounds other than aromatic and antiaromatic are known as nonaromatic compounds

Stability : Aromatic > Non aromatic > Anti Aromatic

S.No. Compound Cyclic Planar Cyclic resonance π Electron in cyclic Aromatic (A)/
Resonance Anti-aromatic(AA)/
Non- Aromatic(NA)
1. ⊕    2πe – A

2. Θ    4πe– AA

3. ⊕    4πe– AA

4. O    2πe– A

5. Θ    6πe– A

6. O    4πe– AA

7.    6πe– A

8.    4πe– NA

9.    4πe– AA

10. ⊕    6πe– A

11.    6πe– A
••
O
••

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12.    6πe– A
••
S
••

13.    6πe– A
••
N
H
14. N    6πe– A
N
H
15. NH2    6πe– A

16.    6πe– A
••
N
17.    6πe– A

18.    10πe– A

19.    14πe– A

20. Θ    6πe– NA

21.    6πe– +6πe– A

22.     NA

Exercise 1.6
1. An aromatic molecule will not: 3. Which of these cyclopropene systems is
(1) have 4n π electrons aromatic
(2) have (4n + 2) π electrons (I) (II) (III)
(3) be planar
Θ ⊕
(4) be cyclic (1) I (2) II
(3) III (4) all of these
2. Which of the following molecules, in pure
form, is unstable at room temperature ? 4. Which is aromatic compound among the
following?
(1) (2) ⊕ Θ

(1) (2)
O
(3) (4)
(3) (4) All the above
N

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5. Which of the following is most stable 8. Tautomerism is due to :-
carbocation? (1) Delocalization of sigma electrons
⊕ (2) Delocalization of pi electrons
(1) (2) ⊕
(3) Migration of active–H atom
⊕ (4) None is correct
CH2 ⊕
(3) (4) CH=
2 CH − CH2
9. Which of the following does not show
6. Which of the following compounds can tautomerism ?
exhibit tautomerism (1) C6H5COCH3
CHO COCH3 (2) CH3CHO
(1) (2) (3) CH3COCH3
(4) C6H5COC(CH3)3
NO2 CH3
(3) (4) CH3–C–CH3
10. The number of π electrons in benzene are:
NO2
(1) 8πe– (2) 3πe–
(3) 6πe– (4) 2πe–
7. Tautomerism is exhibited by
11. Which of the following is heterocyclic
(1) CH=CH–OH
aromatic species ?
(2) O
(1) (2)
O
O Θ

(3) O
(3) (4)
O N
O
H
(4) All the above

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Exercise 2
1. Which of the following is most stable: 6. Which is most stable carbanion:
Θ Θ Θ
(1) CH3 (2) CH3–C–CH3 CH2
Θ
CH3
(1) (2)
Θ Θ
(3) F–C–F (4) Cl–C–Cl
F F (3) Θ

2. In the anion HCOO–, the two carbon- (4) CH2=CH– CH –CH=CH2

oxygen bonds are found to be equal length. 7. Arrange the Carbocations in decreasing
What is the reason for it: order of stability:
(1) the C=O bond is weaker than the C–O OH ⊕
⊕ OH
bond ⊕ OH
(2) the anion HCOO– has two equivalent
resonating structures I II III
(3) the electronic orbitals of carbon atom (1) I > II > III (2) I > III > II
are hybridized (3) II > III > I (4) III > II > I
(4) the anion is obtained by removal of
proton from the acid molecule 8. Consider the following compounds
NH2
3. Rank the following radicals in order of O
Decreasing stability: CH3–CH2–NH2 CH3–C–NH2
•
I II III
•
Correct order of their basic strength is:
I II (1) I < II < III (2) II > I > III
•
• (3) III > II < I (4) II < III < I

9. Arrange the following in decreasing of

III IV their basic nature:
(1) III > II > I > IV (2) III > II < I < IV NH2
(3) II > III > II > IV (4) III < II < I < IV

4. Which is the decreasing order of acidity in, N N

(I) HCOOH (II) CH3COOH H
(III) CH3CH2COOH (IV) C6H5COOH I II III
(1) I > II > III > IV (2) IV > III > II > I (1) I > II > III (2) I > III > II
(3) IV > I > II > III (4) I > IV > II > III (3) III < II > I (4) II > I > III
5. Which is maximum acidic compound:
10. Give the correct order of increasing acidity
COOH COOH
NO2 of the following compounds:
(1) (2) (I) OH (II) OH

COOH COOH (III) COOH (IV) C ≡CH

(3) (4) (1) II < I < IV < III (2) IV < II < I < III
(3) I < II < IV < III (4) IV < I < II < III
NO2 OCH3
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11. Which of the following pairs does not O
represent canonical structures ? Ha
+ •–• • + • • •–• • 15. What is the order of acidic
(1) CH3–C ≡ N–O
• ••
and CH3–C = N–O
• ••
Hb
Hc
–• •• ••
+ O •–• + O •• strength of the labelled H atoms:
• ••
(2) CH2 = N •–• and CH2 – N •–•
O
• ••
•
O
• ••
•
(1) Ha > Hc > Hb (2) Hb > Hc > Ha
+ •–• •–• + (3) Hc > Ha > Hb (4) Ha > Hb > Hc
(3) CH2=N=•N• •• and CH2–N≡N••

•• + 16. Consider the following bases:

O•• O–H
(4) C6H5 – C + and C6H5 – C •• •• (I) o-nitroaniline (II) m-nitroaniline
NH3 NH2 (III) p-nitroaniline
Order of the base strength is:
12. Which is aromatic species:
(1) II > III > I (2) II > I > III
(1) –2 (2) + (3) I > II > III (4) I > III > II

17. Increasing order of basic strength of the

(3) (4) All the above
following is:
I. CH3NH2 II. (CH3)2NH
13. The hybridization states of the nitrogen III. (CH3)3N IV. C6H5NH2
atoms in pyridine, piperidine and pyrrole
V. C6H5CH2NH2
are respectively:
(1) IV < V < I < II < III (2) IV < V < III < I < II
•• •• (3) IV < III < V < I < II (4) IV < V < I < III < II
N
••
N N
H H
Pyridine Piperidine Pyrrole 18. Which of the following compound is not
are respectively: resonance stabilized ?
(1) sp2, sp3 and sp2 • •• Θ
O•
(2) sp2, sp3 and sp3 (1) (2)
(3) sp3, sp3 and sp3 O
(4) sp2, sp2 and sp2 Θ
(3) (4)
Θ O O
14. Select an acid from each pair having
stronger conjugate base:
(A) H2O H2 S 19. Which of the following ketonic compound
(I) (II)
is the least stable ?
(B) CH 3 – C ≡ CH CH 3 – CH = CH 2
(I) (II) O
O
O
(C) O (1) (2)
NH
O
NH
O O
(I) (II) O
(1) I, I, II (2) II, I, I (3) (4)
(3) I, II, I (4) I, II, II
O O O

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20. The correct order of stability for the 25. Among the following the dissociation
following species is: constant (Ka) is highest for:
⊕ ⊕ (1) C6H5OH (2) C6H5CH2OH
O (3) CH3C≡CH (4) CH3NH3+Cl–
(I) (II)
⊕
O ⊕
26. Among the following the aromatic
(III) (IV) molecule is:
(1) II > IV > I > III (2) I > II > III > IV
(1) (2)
(3) II > I > IV > III (4) I > III > II > IV
⊕
21. Which of the following represent the ⊕
decreasing order of Ka values ?
(3) (4)
COOH
(ii) Θ

27. Pyridine is less basic than triethylamine

NO2 because:
(1) Pyridine has aromatic character
OH
(iii) (2) Nitrogen in pyridine is sp2 hybridized
HO (3) Pyridine is a cyclic system
(i) (4) In pyridine, lone pair of nitrogen is
C
delocalized
(iv) CH
(1) (ii) > (i) > (iii) > (iv) 28. Among the following the weakest base is:
(2) (ii) > (iii) > (i) > (iv) (1) C6H5CH2NH2
(3) (i) > (ii) > (iii) > (iv) (2) C6H5CH2NHCH3
(4) (ii) > (iv) > (i) > (iii) (3) O2NCH2NH2
(4) CH3NHCHO
22. Which of the following compounds will not
give effervescence with sodium bicarbonate ? 29. Which of the following orders is correct for
OH the energy required for homolytic cleavage
O 2N NO2 of indicated C–H bonds?
(1) C6H5CO2H (2) H3

NO2 H2
SO3H H1
(3) C6H5OH (4) (1) H1 > H2 > H3 (2) H3 > H1 > H2
(3) H2 > H1 > H3 (4) H3 > H2 > H1
23. Among the following strongest acid is:
(1) CH3COOH (2) C6H5COOH 30. The decreasing order of stability of
(3) m-CH3OC6H4COOH (4) p-CH3OC6H4COOH following cations is
⊕ ⊕ ⊕ ⊕
24. The strongest base among the following is: CH2 CH2 CH2 CH2

(1) (2)
N N
H OH NH2 CH3 Cl

(3) (4) NH2 (P) (Q) (R) (S)

N (1) P > Q > R > S (2) Q > S > R > P
H (3) Q > P > S > R (4) Q > P > R > S

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31. The decreasing order of stability of 37. Which of the following will not be soluble
following anions is in sodium hydrogen carbonate?
Θ Θ Θ Θ (1) 2, 4, 6-trinitrophenol
CH2 CH2 CH2 CH2
(2) Benzoic acid
(3) o-nitrophenol
(4) Benzenesulphonic acid
NO2 CN CH3
38. In benzene C–C bond length between all
(P) (Q) (R) (S) carbons are equal because of:
(1) P > Q > R > S (2) S > R > Q > P (1) Tautomerism (2) sp2 hybridization
(3) Q > P > R > S (4) P > Q > S > R (3) Isomerism (4) Resonance
32. Polarization in acrolein as:
+δ −δ
39. The order of acidic strength of the
=
(1) C H2 C H − CHO hydrogen atoms (Hα, Hβ, Hγ) in the given
−δ +δ molecule is:
=
(2) C H2 C H − CHO O O
−δ +δ
(3) C H=
2 CH − CH O H3C C C
+δ −δ CH CH CH2
(4) C H=
2 CH − CH O
Hγ Hβ Hα
33. Which of the following order of acidic (1) Hγ > Hα > Hβ (2) Hγ > Hβ > Hα
strength is correct: (3) Hβ > Hγ > Hα (4) Hβ > Hα > Hγ
(1) RCOOH > ROH > HOH > HC≡CH
(2) RCOOH > HOH > ROH > HC≡CH 40. Among the following the compound having
(3) RCOOH > HOH > HC≡CH > ROH the most acidic alpha-hydrogen is:
(4) RCOOH > HC≡CH > HOH > ROH (1) CH3CHO
(2) CH3COCH3
34. Which of the following is more basic than
aniline: (3) CH3–C–CH2CHO
(1) Diphenyl amine (2) Triphenyl amine O
(3) p-nitro aniline (4) Benzyl amine (4) CH3–CO–CH2–CO2CH3

35. The stability of carbanions in the 41. Compare heat of hydrogenation of the
following: following:
Θ
Θ
(a) RC ≡ C (b) (i) (ii)
Θ Θ
(c) R2C = CH (d) R3C − CH2
is in the order of: (iii)
(1) (d)>(b)>(c)>(a) (2) (a)>(c)>(b)>(d) (1) i > ii > iii (2) iii > ii > i
(3) (a)>(b)>(c)>(d) (4) (b)>(c)>(d)>(a) (3) ii > iii > i (4) ii > i > iii

36. Given are cyclohexanol (I), acetic acid (II) 42. Correct order of stability is:
2,4, 6-trinitrophenol (III) and phenol (IV).
(1) 1-butene > trans-2-butene > cis-2-butene
In these the order of decreasing acidic
(2) trans-2-butene > 1-butene > cis-2-butene
character will be:
(1) III > II > IV > I (2) II > III > I > IV (3) trans-2-butene > cis-2-butene > 1-butene
(3) II > III > IV > I (4) III > IV > II > I (4) cis-2-butene > trans-2-butene > 1-butene

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Θ 47. The correct order of stability of the
43. CH3 is less stable than :-
following carbocations is –
Θ ⊕ ⊕
(1) CH3 − CH2 (1) CH3CH2 CH2 > CH2 =CH − CH2 >
Θ ⊕ ⊕
(2) CH3 − CH2 − CH3 = CH2 CH C ( CH3 )2
CH2 CH CHCH3 >=
Θ ⊕ ⊕
(3) CH2 − NO2 (2) CH2 =
CH CH2 > CH3CH2 CH2 >
Θ ⊕ ⊕
(4) CH3 − CH − C2H5 CH2 CH C ( CH3 )2=
= > CH2 CH CHCH3
⊕
(3) CH2 CH C ( CH3 )2 >
=
44. Arrange Increasing order of their Ka
⊕ ⊕
value. CH2 =CH CHCH3 > CH2 =CH − CH2 >
COOH ⊕
I oxalic acid COOH CH3CH2 CH2
⊕
II succinic acid, (4) CH2 =CH − C HCH3 > CH2 =
(HOOC − CH2 − CH2 − COOH) ⊕ ⊕
CH C ( CH3 )2 > CH3CH2 − CH2 >
III malonic acid (HOOC − CH2 − COOH) ⊕
CH= CH − CH2
IV adipic acid (HOOC − ( CH2 )4 − COOH) 2

(1) III < II < I < IV (2) II < III > I > IV 48. (a) CH2 = CH – CH = CH2
(3) I > III > II > IV (4) II > I > III < IV (b) CH3 – CH = CH2
(c) CH3 – CH3
single bond length would be
45. Which of the following pairs represent
(1) a (146 pm) b (151 pm) c (153 pm)
resonating structures - (2) a (151 pm) b (146 pm) c (153 pm)
(1) CICH2CH = CHCH3 and (3) a (146 pm) b (153 pm) c (151 pm)
(4) a (153 pm) b (146 pm) c (151 pm)
CH2 = CH − CH − CH3
|
Cl 49. Which of the following molecule has
longest C = C bond length ?
Θ ⊕ ⊕ Θ
(2) :CH2 − N ≡ N: and CH2 = N = N (1) CH2 = C = CH2
(2) CH3–CH = CH2
→ CH3
(3) CH3 − C ≡ N and CH3 − N =
C
(4) All the above (3) CH3–C–CH=CH2
CH3
46. Select the correct option for stability (4) CH3–C=CH2
of following resonance structures :- CH3
Θ ⊕
O CH2, O CH2 COOH
X Y
50.
Θ ⊕
O CH2
X
Z pka value of the compound decreases
(1) x > y > z (2) z > y > x if X is:
(3) y > x > z (4) y > z > x (1) –NO2 (2) –NH2
(3) –OH (4) – OCH3

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51. The correct order of increasing 56. Which one of the following compounds is
dissociation constant [Ka] of the following non-aromatic ?
compound is-
(1) (2)
OH OH
O
NO2 ⊕

(3) (4)
⊕
NO2 NO2 57. The correct order of acidic character of the
(I) (II) following compounds is :
OH OH OH COOH COOH COOH
O2N NO2

NO2 NO2 CH3

NO2
(III) (IV) (I) (II) (III) (IV)
(1) II < IV < I < III (2) IV < III < I < II (1) II > III > IV > I (2) III > II > I > IV
(3) IV > III > II > I (4) I > II > III > IV
(3) IV < II < I < III (4) IV < I < II < III
58. A. Phenyl methanamine
52. Which type of hybrid orbitals do C and N B. N, N-Dimethylaniline
C. N-Methyl aniline
use in the formation of C–N bond in
D. Benzenamine
pyrrole? Choose the correct order of basic nature of
H the above amines.
N (1) D > C > B > A (2) D > B > C > A
(3) A > C > B > D (4) A > B > C > D
59. Among the following, the aromatic
(1) sp2 and sp2 (2) sp2 and sp3 compounds are:
(3) sp and sp3 (4) sp and sp2
(A) O (B)
53. Give the correct order of decreasing
basicity of the following compounds,
I. C6H5NH2 II. (C6H5)2 NH (C) (D)
III. (C6H5)3 N IV. C6H11NH2 Θ ⊕
(1) I > II > III > IV (2) IV > III > II > I Choose the correct answer from the
(3) IV > I > II > III (4) III > II > I > IV following options:
(1) (A) and (B) only
54. Correct order of basic strength is– (2) (A), (B) and (C) only
NH2 NHCH3 N(CH3)2 (3) (B), (C) and (D) only
(4) (B) and (C) only
⊕ ⊕
CH2 CH2 ⊕
(I) (II) (III) CH2 ⊕
CH
(1) II < III < I (2) I > II > III 60.
(3) II > III> I (4) I < II < III
O
H
55. Which of the following species is anti (A) (B) (C) (D)
aromatic?
⊕ Among the given species the Resonance
stabilised carbocations are:
(1) (2) (1) (C) and (D) only
(2) (A), (B) and (C) only
Θ (3) (A), (B) and (D) only
(3) (4) (4) (A) and (B) only
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61. Which of the following structures are 66. The functional group that shows negative
aromatic in nature ? resonance effect is :
Θ (1) –OH (2) –OR
⊕ Θ
(3) –COOH (4) –NH2
67. The interaction between π bond and lone
(A) (B) (C) (D) pair of electrons present on an adjacent
(1) A, B, C and D (2) Only A and B atom is responsible for
(3) Only A and C (4) Only B, C and D (1) Resonance effect (2)Electromeric effect
62. Which of the following carbocations is (3) Inductive effect (4) Hyperconjugation
most stable? 68. Arrange the following in increasing order
OCH OCH3 of basicity
(1) (2) NH NH
⊕ ⊕

(3) H CO ⊕
3 H2N NH2 H2N CH3
⊕
(4) H3CO
(I) (II)
63. Arrange the following in decreasing acidic NH NH
strength.
OH OH OH OH
H2N OH H3C CH3
(III) (IV)
(1) IV < II < III < I (2) IV < II < I < III
NO2 OCH3
NO2 (3) IV < III < II < I (4) III < IV < II < I
OCH3
(A) (B) (C) 69. For the given compounds, the correct order
(D)
(1) A > B > C > D (2) B > A > C > D of increasing pka value
(3) D > C > A > B (4) D > C > B > A (A) OH (B) O2N OH

64. Increasing order of stability of the OCH3

resonance structure is :
Θ
OHC (C)HO (D) NO2
NH2
A. ⊕
⊕ OH
OHC
NH2 (E) OH
B. Θ OCH3
⊕
O (1) (E) < (D) < (C) < (B) < (A)
Θ
H (2) (D) < (E) < (C) < (B) < (A)
C. NH2
(3) (E) < (D) < (B) < (A) < (C)
Θ ⊕
NH2 (4) (B) < (D) < (C) < (A) < (E)
D. OHC
(1) C < D < B < A (2) C < A < B < D 70. Which among the following is incorrect
(3) D < C < A < B (4) D < C < B < A statement?
(1) Electromeric effect dominates over
65. The increasing order of ka for the following
inductive effect
phenols is
(2) The electromeric effect is, temporary
(a) 2, 4-Dinitrophenol
effect
(b) 4 - Nitrophenol
(c) 2, 4, 5- Trimethylphenol (3) The organic compound shows
(d) Phenol electromeric effect in the presence of
(e) 3-Chlorophenol the reagent only
(1) a < b < c < d < e (2) c < d < e < b < a (4) Hydrogen ion (H+) shows negative
(3) d < b < e < c < a (4) e < c < d < b < a electromeric effect

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 CHEMISTRY General Organic Chemistry
Exercise 3
1. Statement I: Tropylium cation is aromatic 4. Statement I: The pKa value of acetic acid is
⊕ lower than that of phenol.
in nature Statement II: Phenoxide ion is more
resonance stabilized than acetate ion.
Statement II: The only property that (1) Both Statement I and Statement II are
determines its aromatic behaviour is its correct.
planar structure. (2) Both Statement I and Statement II are
(1) Both Statement I and Statement II are incorrect.
correct. (3) Statement I is correct but Statement II is
(2) Both Statement I and Statement II are incorrect.
incorrect. (4) Statement I is incorrect but Statement II is
(3) Statement I is correct but Statement II correct.
is incorrect.
⊕
(4) Statement I is incorrect but Statement II is 5. Statement I: CH 3 – C H2 is more stable
correct. ⊕
than C H3 due to hyperconjugation.
⊕
2. Statement I: Me3 C is more stable than Me2 Statement II : Hyperconjugation in
⊕
⊕ ⊕ CH 3 – C H2 involves following overlapping:
C H and Me2 C H is more stable than the Me
⊕ H ⊕
C H2 . H C CH2
H
Statement II: Greater the number of
hyperconjugative structures, more is the (1) Both Statement I and Statement II are
stability of carbocation. correct.
(1) Both Statement I and Statement II are (2) Both Statement I and Statement II are
correct. incorrect.
(2) Both Statement I and Statement II are (3) Statement I is correct but Statement II is
incorrect. incorrect.
(3) Statement I is correct but Statement II (4) Statement I is incorrect but Statement II is
correct.
is incorrect.
(4) Statement I is incorrect but Statement II is 6. Given below are two statements:
correct. Statement I : Aniline is less basic than
acetamide.
3. Statement I: Cyclopentadienyl anion is Statement II : In aniline, the lone pair of
much more stable than allyl anion. electrons on nitrogen atom is delocalised
Statement II: Cyclopentadienyl anion is over benzene ring due to resonance and
aromatic in nature. hence less available to a proton.
(1) Both Statement I and Statement II are Choose the most appropriate option;
correct. (1) Both statement I and statement II are
(2) Both Statement I and Statement II are true.
incorrect. (2) Both statement I and statement II are
false.
(3) Statement I is correct but Statement II
(3) Statement I is true but statement II is
is incorrect.
false.
(4) Statement I is incorrect but Statement II is
(4) Statement I is false but statement II is
correct.
true.
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General Organic Chemistry CHEMISTRY
7. Given below are two statements: 10. Match the terms mentioned in Column I with
Statement I: Hyperconjugation is a the terms in Column II
permanent effect. Column I Column II
Statement II: Hyper conjugation in ethyl (i) Carbocation (a) cyclohexane
 +
 and 1- hexene
cation  CH3 − CH2  involves the
(ii) Nucleophile (b) Conjugation of
 
overlapping of Csp2 − H1s bond with empty electrons of C–
H σ bond with
2p orbital of other carbon. empty p-
Choose the correct option: orbital present
(1) Both statement I and statement II are at adjacent
true positively
(2) Statement I is incorrect but statement charged
II is true carbon.
(3) Statement I is correct but statement II (iii) Hyperconjugation (c) sp2 hybridised
is false carbon with
(4) Both Statement I and statement II are empty p-
false. orbital
(iv) Isomers (d) Ethyne
8. Statement-I: Resonance affects bond (v) sp (e) Species that
length. hybridization can receive a
Statement-II: Resonance stabilises pair of
electrons
molecule.
(vi) Electrophile (f) Species that
(1) Both Statement-I and Statement-II are
can supply a
incorrect.
pair of
(2) Statement-I is correct but Statement-II electrons
is incorrect. (1) (i) → (c), (ii) → (f ), (iii) → (b),
(3) Statement-I is incorrect but Statement- (iv) → (a), (v) → (d), (vi) → (e)
II is correct. (2) (i) → (f), (ii) → (c ), (iii) → (b),
(4) Both Statement-I and Statement-II are (iv) → (a), (v) → (d), (vi) → (e)
correct. (3) (i) → (c), (ii) → (f ), (iii) → (a),
9. Statement I : In an organic compound (iv) → (b), (v) → (d), (vi) → (e)
when inductive and electrometric effect (4) (i) → (c)), (ii) → (f), (iii) → (b)
operate in opposite direction, the (iv) → (d), (v) → (a), (vi) → (e)
electrometric effect Predominates.
11. Match the intermediates given in Column I
Statement II : Polarisation of σ bond
caused by the polarisation of adjacent with their probable structure in Column II.
σ bond is referred to as inductive effect. Column I Column II
(1) Both statement I and statement II are (i) Free radical (a) Trigonal planar
true. (ii) Carbocation (b) Pyramidal
(2) Both statement I and statement II are (iii) Carbanion (c) Linear
false. (1) (i) → (a), (ii) → (a), (iii) → (b)
(3) Statement I is true but statement II is (2) (i) → (b), (ii) → (c), (iii) → (a)
false. (3) (i) → (a), (ii) → (b), (iii) → (c)
(4) Statement I is false but statement II is
(4) (i) → (a), (ii) → (b), (iii) → (b)
true.
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12. Match the ions given in Column I with their 14. Match the column
nature given in Column II.
Column-I Column-II
Column I Column II
i .. ⊕ (a) Stable due to (Compound) (Pkb value)
CH3 − O− CH2
.. resonance (i) Methan amine (p) 4.75
ii ⊕ (b) Destabilised
F3C – C N-Methyl
due to inductive (ii) (q) 3.38
effect methanamine
iii CH3 (c) Stabilised by
N, N-Dimethyl
CH3–C⊕ hyperconjugation (iii) (r) 3.27
methanamine
CH3
iv ⊕ (d) A secondary (iv) Ammonia (s) 4.22
CH3 − CH − CH3
carbocation (1) (i)–(p), (ii)–(q), (iii)–(r), (iv)–(s)
(1) (i) → (a), (b), (ii) → (b),
(iii) → (c), (iv) → (c), (d) (2) (i)–(p), (ii)–(r), (iii)–(s), (iv)–(q)
(2) (i) → (a), (b), (d), (ii) → (a), (3) (i)–(s), (ii)–(r), (iii)–(q), (iv)–(p)
(iii) → (b), (iv) → (c), (d)
(4) (i)–(q), (ii)–(r), (iii)–(s), (iv)–(p)
(3) (i) → (a), (b), (d), (ii) → (b),
(iii) → (b), (iv) → (b), (d)
(4) (i) → (a), (c), (d), (ii) → (b), 15. Match the column-I and column-II and
(iii) → (b), (iv) → (c), (d) select the answer:
Column-I Column-II
13. Match List -I with List II:
O
List-I Mechanism steps List–II Effect
(A) •• ⊕ (I) –E effect (a) C–OH + NaHCO3 (p) NH3
NH2 NH2 14

Θ
O
14 14
(b) C–OH + NaHCO3 (q) C O 2
(B) H (II) –R effect
⊕
+H
⊕ O
(C) (III) +E effect (c) C–OH + Na (r) CO2

Θ Θ
+CN O
CN (d) S–OH + NaNH2 (s) H2
(D) Θ
•• (IV) +R effect O
O ←N= O O ←N– O
⊕
(1) (a-r), (b-q), (c-s), (d-p)

Choose the correct answer from the (2) (a-q), (b-r), (c-s), (d-p)
options given below :
(1) (A)- (IV), (B)- (III), (C)- (I), (D)- (II) (3) (a-s), (b-q), (c-r), (d-p)
(2) (A)- (III), (B)- (I), (C)- (II), (D)- (IV)
(4) (a-p), (b-s), (c-q), (d-r)
(3) (A)- (II), (B)- (IV), (C)- (III), (D)- (I)
(4) (A)- (I), (B)- (II), (C)- (IV), (D)- (III)

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16. Assertion A : Enol form of acetone 18. Assertion : The order of acidic strength for
[CH3COCH3] exists in <0.1% quantity.
However, the enol form of the acetyl the following compound is
acetone [CH3COCH2COCH3] exists in
CH3–CH2–OH > CH3–CH–OH >
approximately 15% quantity.
Reason R : Enol form of acetyl acetone is CH3
stabilized by intramolecular hydrogen
CH3
bonding, which is not possible in enol form
of acetone. CH3–C–OH
Choose the correct statement: CH3
(1) If both Assertion and Reason are
correct and the Reason is a correct Reason : Electron releasing group (–CH3)
explanation of the Assertion.
increases electron density on oxygen
(2) If both Assertion and Reason are
correct but Reason is not a correct tending to decrease the polarity of O-H
explanation of the Assertion.
(3) If the Assertion is correct but Reason is bond
incorrect.
(4) If the Assertion is incorrect and Reason (1) If both Assertion and Reason are
is correct. correct and the Reason is a correct

17. Assertion: P-nitro benzoic acid is more explanation of the Assertion.

acidic than p-hydroxy benzoic acid.
Reason: EDG increases acidity of benzoic (2) If both Assertion and Reason are
acid while EWG decreases acidity of
correct but Reason is not a correct
benzoic acid.
(1) If both Assertion and Reason are explanation of the Assertion.
correct and the Reason is a correct
explanation of the Assertion. (3) If the Assertion is correct but Reason is
(2) If both Assertion and Reason are
correct but Reason is not a correct incorrect.
explanation of the Assertion. (4) If the Assertion is incorrect and Reason
(3) If the Assertion is correct but Reason is
incorrect. is correct.
(4) If the Assertion is incorrect and Reason
is correct.

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Exercise 4 (Previous Year's Questions)

1. In which of the following compounds, the 4. The pair of electron in the given carbanion,
C – Cl bond ionisation shall give most
CH3C ≡ CΘ , is present in which of the
stable carbonium ion? [AIPMT – 2015]
H following orbitals? [NEET-I – 2016]
H3C CH–Cl
(1) C – Cl (2) (1) 2p (2) sp3
H3C
CH3 (3) sp2 (4) sp
H H3C
(3) CH – Cl (4) CH – Cl
O2N–H2C H3C 5. In pyrrole: [NEET-II - 2016]
4 3
2. Consider the following compounds
5
.
CH3 Ph •• 2
N1
CH3–C–CH .
Ph–C–Ph
H
CH3
I II The electron density is maximum on:

. (1) 2 and 4 (2) 2 and 5

(3) 2 and 3 (4) 3 and 4
CH3

III
6. Which among the given molecules can
Hyperconjugation occurs in:
[AIPMT – 2015] exhibit tautomerism? [NEET-II - 2016]
(1) II only (2) III only O
(3) I and III (4) I only
O Ph
Ph
3. The correct statement regarding the I II
basicity of arylamines is: O
[NEET-I – 2016]
(1) Arylamines are generally less basic
than alkylamines because the nitrogen III
lone-pair electrons are delocalized by (1) Both I an II (2) Both II and III
interaction with the aromatic ring π (3) III only (4) Both I and III
electron system
(2) Arylamines are generally more basic
than alkylamines because the nitrogen 7. The correct order of strengths of the
lone-pair electrons are not delocalized carboxylic acid: [NEET-II - 2016]
by interaction with the aromatic ring π
COOH COOH COOH
electron system.
(3) Arylamines are generally more basic O
O
than alkylamines because of aryl I II III
group. is:
(4) Arylamines are generally more basic
(1) III > II > I (2) II > I > III
than alkylamines, because the
nitrogen atom in arylamines is sp- (3) I > II > III (4) II > III > I
hybridized.

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8. The correct increasing order of basic 14. Which of the following substituted
strength for the following compounds is: phenols is the strongest acid?
[NEET – 2017] [NEET(UG) 2020(COVID-19)]
NH2 NH2 NH2 OH OH

(1) (2)

NO2 CH3 NO2 OCH3

(I) (II) (III)
OH OH
(1) II < III < I (2) III < I < II
(3) III < II < I (4) II < I < III
(3) (4)
9. Which of the following is correct with
respect to -I effect of the substituents ? (R C2H5 CH3
= alkyl) [NEET(UG) 2018]
(1) –NH2 < –OR < –F (2) –NR2 < –F < –OR
15. Which compound amongst the following is
(3) –NR2 > –OR > –F (4) –NR2 < –OR > –F
not an aromatic compound? [NEET - 2022]
10. The correct order of the basic strength of
methyl substituted amines in aqueous (1) (2)
solution is [NEET(UG) 2019] Θ ⊕
(1) (CH3)2NH > CH3NH2 > (CH3)3N
(2) (CH3)3N > CH3NH2 > (CH3)2NH (3) (4)
(3) (CH3)3N > (CH3)2NH > CH3NH2
(4) CH3NH2 > (CH3)2NH > (CH3)3N ⊕

11. The compound that is most difficult to 16. Given below are two statements:
protonate is [NEET(UG) 2019] [NEET(UG) 2022]
O O Statement I : The acidic strength of
(1) H H (2) H3C H
O monosubstituted nitrophenol is higher
O
(3) H3C CH3 (4) Ph H than phenol because of electron
withdrawing nitro group.
12. The most stable carbocation among the Statement II : o-nitrophenol, m-
following is [NEET(UG) 2019(ODISHA)] nitrophenol and p-nitrophenol will have
⊕
(1) (CH3)3C– CH –CH3 same acidic strength as they have one nitro
⊕ group attached to the phenolic ring.
(2) CH3–CH2– CH –CH2–CH3
In the light of the above statements,
⊕
(3) CH3– CH –CH2–CH2–CH3 choose the most appropriate answer from
⊕ the options given below:
(4) CH3–CH2– C H 2
(1) Both Statement I and Statement II are
13. A tertiary butyl carbocation is more stable correct.
than a secondary butyl carbocation (2) Both Statement I and Statement II are
because of which of the following ? incorrect.
[NEET(UG) 2020] (3) Statement I is correct but Statement II
(1) Hyperconjugation is incorrect.
(2) –I effect of –CH3 groups
(4) Statement I is incorrect but Statement II is
(3) +R effect of –CH3 groups
correct.
(4) –R effect of –CH3 groups
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 CHEMISTRY General Organic Chemistry
17. Consider the following compounds/ 21. The correct order of decreasing acidity of
species: [NEET(UG) 2023] the following aliphatic acids is :
[NEET(UG)-2025]
(1) HCOOH > CH3COOH > (CH3)2
(i) (ii)
CHCOOH> (CH3)3CCOOH
 (2) HCOOH > (CH3)3CCOOH >
 (CH3)2CCOOH CH3COOH
(iii) (iv) (3) (CH3)3CCOOH > (CH3)2 CHCOOH>
CH3COOH > HCOOH
(4) CH3COOH > (CH3)2 CHCOOH >
⊕ (CH3)3CCOOH > HCOOH
(v) (vi)
22. Among the given compounds I-III, the
correct order of bond dissociation energy
(vii)
of C–H bond marked with * is):
The number of compounds/species which [NEET(UG)-2025]
obey Huckel’s rule is ______.
H
(1) 4 (2) 6 * C≡C—H
(3) 2 (4) 5 * * H
H
18. Which amongst the following Compounds Ι ΙΙ ΙΙΙ
/species is least basic ? [NEET(UG) 2023] (1) III > II > I (2) II > III > I
H2N H2N ⊕
(1) C=O (2) C – OH (3) II > I > III (4) I > II > III
H2N H2N
H2N H2N ⊕
(3) C = NH (4) C = NH2 23. The correct order of decreasing basic
H2N H2N
strength of the given amines is :
19. The most stable carbocation among the [NEET(UG)-2025]
following is: [NEET(UG)- 2024] (1) N-ethylethanamine > ethanamine
> N-methylanilne > benzenamine
⊕
(1) CH2 (2) benzenamine > ethanamine >
N-methylanilne > N-ethylethanamine
(2) ⊕ (3) N-methylanilne > benzenamine >
CH3
ethanamine > N-ethylethanamine
CH3 (4) N-ethylethanamine > ethanamine >
(3) H C benzenamine > N-methylanilne
3 ⊕ CH3
CH3
⊕
(4) H C
3 CH3

20. Methyl group attached to a positively

charged carbon atom stabilizes the
carbocation due to : [RE–NEET(UG) 2024]
(1) –I inductive effect
(2) electromeric effect
(3) hyperconjugation
(4) mesomeric effect

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General Organic Chemistry CHEMISTRY

ANSWER KEYS
Exercise 1.1
Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
Ans. 4 4 3 1 1 3 1 2 4 2 1 3 1 2 3 1 1 3

Exercise 1.2
Que. 1 2 3 4 5 6 7 8 9 10 11
Ans. 3 3 3 1 4 3 1 3 3 4 2

Exercise 1.3
Q ue. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
Ans. 4 4 2 2 4 4 1 1 2 3 1 1 2 3 2 1 1

Exercise 1.4
Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans. 4 1 4 2 2 1 4 3 3 1 4 2 1 2 1 1 2 3 2 1

Exercise 1.5
Q ue. 1 2 3 4 5 6 7 8 9 10 11
Ans. 1 3 4 2 3 3 1 3 1 4 3

Exercise 1.6
Que. 1 2 3 4 5 6 7 8 9 10 11
Ans. 1 2 3 4 1 2 4 3 4 3 3

Exercise 2
Q ue. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans. 4 2 1 4 2 1 3 4 2 2 4 1 1 4 3 1 2 3 4 4
Q ue. 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
Ans. 1 3 3 3 4 1 2 4 3 4 4 4 2 4 3 1 3 4 4 3
Q ue. 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60
Ans. 2 3 3 3 2 3 3 1 4 1 3 1 3 4 2 2 1 4 4 4
Q ue. 61 62 63 64 65 66 67 68 69 70
Ans. 2 4 1 2 2 3 1 1 4 4

Exercise 3
Q ue. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
Ans. 3 1 1 3 3 4 3 4 1 1 1 1 1 4 1 1 3 1

Exercise 4 (Previous Year's Questions)

Q ue. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans. 2 2 1 4 2 3 4 4 1 1 4 3 1 1 4 3 1 2 2 3
Q ue. 21 22 23
Ans. 1 1 1

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 Notes

Sarvam Career Institute

Hydrocarbons CHEMISTRY

Chapter HYDROCARBONS

4
Chapter Summary Alkanes
• Alkanes Introduction of Alkanes
• General Methods of • Alkanes are saturated hydrocarbons.
Preparation of Alkanes • Alkanes are inert under normal condition as they do not react
• Physical Properties of with acids, bases and other reagents, so these are called
Alkanes paraffins i.e parum (little), affinis (affinity).
• Chemical Reactions of
• General formula is CnH2n+2.
Alkanes
• Hybridisation state of carbon is sp3.
• Alkenes
• General Methods of
• Geometry is tetrahedral.
Preparation of Alkenes • Bond angle is 109o 28'.
• Physical Properties of • Petroleum and natural gas are the main sources of alkanes.
Alkenes
• Chemical Reactions of General Methods of Preparation of Alkanes
Alkenes
(1) From Unsaturated Hydrocarbons:
• Alkynes
• General Methods of • Addition of dihydrogen gas on alkene, alkyne in presence of
Preparation of Alkynes metal catalyst is called hydrogenation.
• Physical Properties of • Finely divided Pt, Pd or Ni is used as catalyst.
Alkynes • Catalyst adsorbs dihydrogen gas and activates hydrogen –
• Chemical Properties of hydrogen bond,
Alkynes
• Pt or Pd catalyse the reaction at room temperature but relatively
• Aromatic Hydrocarbon
high temperature and pressure is required with Ni.
• General Methods of
• Hydrogenation in presence of Ni is known as Sabatier-
Preparation of Aromatic
Hydrocarbon Senderen’s reaction.
Ni or Pt or Pd
• Physical Properties of H=
2C CH2 + H2  →H3C − CH3
Aromatic Hydrocarbon Ethene(Ethylene) Ethane

• Chemical Properties of Ni or Pt or Pd
HC ≡ CH + 2H2  →H3C − CH3
Aromatic Hydrocarbon Ethyne( Acetylene ) Ethane

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 CHEMISTRY Hydrocarbons
(2) From alkyl Halides
(i) By Reduction
Alkyl Halides can be reduced into alkanes by using metal and dilute HCl.
Zn
R − X 
HCl
→R − H
( aq )
2eΘ
δ⊕ δΘ Zn Θ H⊕
Mechanism: R—X R R–H
H 2O

Note: alkyl fluorides are not used due to their less reactivity
(ii) By Wurtz Reaction:
Alkyl halide reacts with sodium metal in dry ether free from moisture solution to form symmetrical
alkanes containing double the number of C atoms present in the alkyl halide.
Dry
R —X + 2Na + X—R  ether
→ R–R+2NaX
Dry
Ex. CH3—Br + 2Na + Br —CH3 
ether
→ CH3–CH3 + 2NaBr
a. Limitations of the Reaction : The reaction is not an appropriate method to prepare unsymmetrical
alkane (like R1 –R2) since other hydrocarbons (R1 – R1 & R2 – R2) are also formed along with R1– R2.
Dry
R1— X + 2Na + X—R2  Ether
→ R1–R2 + R1–R1 + R2–R2
Dry
Ex. CH3 —Br + 2Na + Br—C2H5 
Ether
→ CH3 − CH3 + CH3 − C2H5 + C2H5 − C2H5
Ethane Pr opane Butane

b. The reaction may complete through both free radical and ionic mechanism.
Free radical mechanism Ionic mechanism
2Na → 2Na+ + 2e– 2Na → 2Na+ + 2e–
δ⊕ δΘ 2eΘ Θ Θ
• Θ
R –X + e– → R + X ••
R–X → R + X
••

• Θ Θ ⊕
R –X + e– → R + X R+Na → RNa (organometallic compound)
• •
Θ⊕ Θ
R + R 
→R − R RNa+ R—X → R–R+ X
2Na⊕ + 2x Θ 
→ 2NaX 2Na⊕ + 2x Θ  → 2NaX
c. Reactivity order of different alkyl halide for the reaction is R–I > R – Br > R –Cl > R– F
d. In case of di halides, intramolecular reaction is observed to give unsaturated or cyclic compound
CH3–CH2–CH–CH2 Na CH3–CH2–CH=CH2
dry ether
x x
Na CH2–CH2
CH3–CH2–CH–CH2
dry ether
x x CH2–CH2

e. Side products of the reaction : When two alkyl free radicals approach each other two processes are
observed between them
Ex. Radical combination Radical Disproportionation
R⇒CH3–CH2 H
CH3–CH2 +CH2–CH3 H3C–CH2 +H–C–CH2
H
CH3–CH2 –CH2–CH3 CH3–CH3 +CH2=CH2
In case of 1º & 2 º radical combination dominates but in 3º radical disproportionation dominates.

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(3) From Carboxylic acid:
(a) Sodalime decarboxylation:
When a sodium salt of carboxylic acid is heated with soda lime [NaOH + CaO (3 : 1)] an alkane with one
C atom less than carboxylic acid is formed.
RCOOH + NaOH → RCOONa + H2O
O
r.d.s. Θ⊕
R–C–ONa R Na+CO2
(Carbanion)

CO2 + 2NaOH 
→Na2CO3 + H2O
Θ⊕ ⊕ Θ
RNa + H2O 
→R − H + NaOH
{Reactively of R-COOH for decarboxylation ∝ Stability of RΘ}
CaO in soda lime is used to keep NaOH dry because it is hygroscopic in nature.
Note: Those acid which have multiple bond at beta position show high reactivity towards decarboxylation
and the process is possible even in absence of Soda line, only heat will be required to complete the
process
Ex. (1) β-keto acid R–C–CH2–C–OH ∆ R–C–CH3
–CO2
O O O
(2) Gem dicarboxylic acid

R–CH COOH ∆ R–CH2–COOH

COOH –CO2
Note: Vicinal dicarboxylic acid on heating forms cyclic anhydride by dehydration
O
R–CH–COOH ∆ R–CH–C
O
R–CH–COOH –H2O R–CH–C
O
(b) Kolbe's Electrolysis:
• Electrolysis of an aqueous solution of sodium or potassium salt of a carboxylic acid gives higher alkane.
• Mixture of different acid salts gives mixture of hydrocarbons
electrolysis
Ex. CH3 – COONa(aq) + C2H5–COONa(aq). → CH3–CH3+ CH3–CH2–CH3+CH3–CH2–CH2–CH3
• Electrolysis of an aqueous solution of sodium or potassium salt of dibasic acid gives alkene
CH2–COOK electrolysis CH2
Ex. CH –COOk
2 (aq) CH2
(Potassium succinate)
• Electrolysis of aqueous solution of sodium or potassium salt of dibasic unsaturated fatty acid gives
alkyne.
CH–COOK electrolysis CH
Ex.
CH–COOk(aq) CH
(Potassium maleate/potassium fumarate
Mechanism:
Electrolysis
2RCOONa(aq) → R–R+2CO2+2NaOH+H2
2RCOONa(aq) ⇋ 2RCOO + 2Na⊕Θ

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At Anode (Oxidation):
• •
2RCOOΘ 
−2eΘ
→ 2RCOO 
−2CO
→ 2R 
→R − R
2 Alkane

At Cathode (Reduction): Reduction of H2O takes place since reduction potential of H2O is greater than
Na⊕.
2H2O + 2e– → 2OHΘ+H2↑
2Na⊕ + 2OHΘ → 2NaOH
electrolysis
Ex. 2CH3 − COONa( aq ) → CH3 − CH3 + 2CO2 + 2NaOH + H2

(4) From Grignard Reagent:

Grignard’s reagents react with the compound having active hydrogen to form alkane by acid base
reaction.
Ex. δΘ δ⊕
R-Mgx + H–OH→R-H
δΘ δ⊕
R-Mgx + H–OR'→R-H
H−O −H
Ex. CH3MgBr  → CH4
R −O −H
CH3MgBr  → CH4
CH3CO2 −H
CH3MgBr  → CH4

(5) Reduction using (P + HI):

It is a powerful reducing agent which converts aldehyde, ketone, alcohol, carboxylic acid, alkane
iodide, ether and carbohydrates into alkanes with same number of carbon (except ether).
O O
HI HI HI HI
(i) R–C–OH R–C–H R–CH2–OH R–CH2–I R–CH3+I2

HI
(ii) R–CH2–O–CH2–R HI R–CH2–I+R–CH2–OH R–CH3+I2

(iii) Glucose HI n-Hexane + I2

Note: P is used to remove I2 from the end product.

(6) From Carbonyl compounds :

Zn-Hg/Hcl
CH2 (Clemmensen’s Reduction)
C=0
(i) N2H4
Θ CH2+ N2 (Wolf kishner Reduction)
(ii) OH/∆

(7) Hydrolysis of Metal carbide:

1. Al4C3 + 12H2O → 3CH4 ↑ + 4Al(OH)3
2. Be2C + 4H2O → CH4 ↑ + 2Be(OH)2

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Physical Properties of Alkanes

Hydrocarbons are non-polar molecules. Therefore, the intermolecular forces of attraction are of van der Waals
type, which mainly depends on the molecular weight and surface area of the contact between adjacent
molecules. Greater the molecular weight and surface area, stronger are the vander Waals forces of attraction.
a. Boiling Point:
Due to weak vander Waals forces first four members, C1–C4 are gases, C5 –C17 are liquids, and C18
onwards are colourless, waxy solids.
1
B. P ∝ Molecular weight ∝ (Branching↑, surface area ↓ vander Waals forces of attraction↓)
Branching

b. Melting Point:
Melting point of alkanes increases with increase in carbon content. Melting point of alkane increases
sharply when moving toward higher members specially from alkane with odd number of carbon to even
number of carbon, but flat increment is observed when moving from lower alkane with even number of
carbon to higher alkane with odd number of carbon.
This property is called alternation effect.
In n-alkane, due to sp3 hybridisation, C atoms are arranged in a zig-zig chain. So, n-alkanes with even
number of C atoms are more symmetrical than those containing odd number of C atoms and, hence,
can pack closely in the crystal lattice. For example.

n-Propane n-Butane n-Pentane n-Hexane n-Propane

M.P. 85.3K 134.6K 143.3K 178.5K 182.4K
Alternation effect is not observed in boiling points because in the liquid state the conformations of the
molecules are fixed but keep on changing as a result of collisions.
c. Solubility:
'Like dissolves like' is the general rule of solubility. Hydrocarbons (non polar molecules) are insoluble
in polar solvents such as H2O and alcohol, but are highly soluble in non-polar solvents such as
petroleum, benzene, and CCl4.
Chemical Reactions of Alkanes or Chemical Properties
(A) Free Radical Substitution Reactions: Alkanes are non polar and have strongly bonded sigma electrons
so reaction of alkanes are generally carried out in presence of light or at high temperature so,
substitution reaction of alkanes follow free radical mechanism.
(a) Halogenation : substation of H-atom by halogen atom
hv or
R–H + X2 
250 − 400ºC in dark
→ R–X + HX
• The reactivity order for halogens towards the reaction:
•
F2 > Cl2 > Br2 > I2 (based on reactivity of X )
• Reactivity order of hydrogen atom present in alkane is :
•
3° > 2° > 1° H (based on stability of R )
(i) Fluorination :
• Reaction is highly exothermic, so it is explosive reaction
• Reaction is possible even in dark.
• Direct halogenation gives carbon black, alkyl fluoride is formed if reaction is carried out in presence of
N2.

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(ii) Chlorination : CH4+ Cl2 hν CH3Cl+ CH2Cl2 + CHCl3+CCl4

If CH4(excess) → CH3Cl (major)

Cl2(excess) → CCl4 (major)
Mechanism:
hν 2X
•
Step I Chain initiation step: X—X
Or ∆ 
• •
Step II Chain propogation step: R–H+X →R +HX(r.d.s)
• •
R + X– X → R–X+X
• •
Step III Chain termination step : X+ X→X2
• •
R+R→R–R
• •
R+X→R–X

(iii) Bromination : Br2 reacts with alkanes in a similar manner like Cl2.
(iv) Iodination : R–H +I2 R–I + HI
(Low yield)
• The reaction is reversible since HI is a good reducing agent.
• The reaction is very slow since I2 is least reactive in halogenation
• R-I can be formed with better yield if reaction is carried out in presence of HIO3or HNO3.
HIO3 + HI → I2 + H2O
HNO3 + HI → NO2 +I2+H2O

(b) Nitration : The reaction is carried out in vapour phase at high temperature and pressure, it involved
breaking of C-C bond along with C–H, So mixture of all possible nitro alkanes is formed as product.
400 −500°C
R–H + HO–NO2 
pressure
→ R–NO2 + H2O
450ºC
Ex. CH3–CH3 + HNO3 
pressure
→ CH3 CH2NO2 + CH3NO2 + H2O

C–H CH3–CH2-CH2-NO2

CH3–CH-CH3
∆ & Pr
CH3CH2CH3 + Conc. HNO3 NO2
Vapour
Phase CH3–NO2
C–C

CH3–CH2–NO2

(c) Sulphonation : Replacement of H atom of alkane by –SO3H is known as sulphonation.

Alkane react with fuming H2SO4 or oleum (H2S2O7).
CH3 CH3
SO3
CH3 – C – H + HO – SO3H ∆ & Pr CH3 – C – SO3H + H2O
Ex.
CH3 CH3
2-Methyl propane
Note: The reaction is observed in higher alkanes ( ≥ 6 C) or the alkanes having 3°H.

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(d) Allylic or Benzylic Substitution by Br2 (Low Concentration)/hv or NBS/hv
Br2 /hν CH3 CH3-Br
CH3–CH = CH2 or NBS/hν
→ CH2 − CH = CH2 + HBr
| N.B.S./hν
Br Or Br2/ hν
Note: The reaction is preferred at the site where free radical formed is most stable
Ex.1 In the following reaction, the major product is :-
CH3

Br2
Major
hv
CH3
CH2Br CH3 CH3
Br Br
(1) (2) (3) (4)
Br
Ans. (3)
Br2(Low conc)
Ex.2 Major product
hν
Br
Br Br Br
(1) (2) (3) (4)
Br
Ans. (2)
(B) Combustion
Alkanes on heating in excess of air or oxygen are completely oxidized (Combustion) to CO2 and H2O,
with the evolution of large amount of heat.
13
i. CH4 + 2O2 → CO2 + 2H2O + Heat ii. C4H10 + O2 → 4CO2 + 5H2O + Heat
2
The general combustion equation for an alkane is-
 3n + 1  ∆
CnH2n+2 +   O2 → nCO2 + (n + 1) H2O + Heat
 2 
Note: Due to the evolution of large amount of heat during combustion, alkanes are used as fuels.
• During incomplete combustion of alkanes with insufficient amount of air or oxygen, carbon black
is formed which is used in the manufacturing of ink, black pigments and as filters
incomplete
CH4( g ) + O2( g ) 
compbustion
→ C( s ) + 2H2O

(C) Catalytic Oxidation

(Controlled oxidation)
Alkanes on heating with a regulated supply of oxygen (air) at high pressure and in the presence of
suitable catalysts give a variety of oxidation products
Cu/523K /100atm
i. 2CH4 + O2  → 2CH3OH
MO2O3
ii. CH4 + O2 ∆
→ HCHO + H2O
( CH COO ) Mn
iii. 2CH3CH3 + 3O2 
3
∆
2
→ 2CH3COOH + 2H2O
iv. With KMnO4, alkanes containing tertiary H atom are oxidized to alcohol.
Ex. (CH3)3CH KMnO4 (CH3)3C–OH

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(D) Isomerisation:- Unbranched alkanes are converted into branched alkane on heating in presence of
anhydrous AlCl3 and HCl gas
Me
Me anhy.AlCl3
Me +
Me Me Me
HCl,∆ Me Me
Hexane 2-Methylpentane 3-Methylpentane

(E) Aromatisation
Alkanes with six or more C atoms, when heated under pressure in the presence of a catalyst (e.g., Al2O3
+ Cr2O3 or V2O5 or Mo2O3 at 773 K and 10-20 atm) first forms cyclo alkane and then aromatise to give
benzene or its homologues by dehydrogenation.
(i) n-Hexane Cr2O3–Al2O3
∆
CH3
(ii) n-Heptane Cr2O3–Al2O3
∆
CH3 C2H5
CH3
Cr2O3–Al2O3
(iii) n-Octane +
∆
O–Xylene Ethyl benzene

(F) Cracking (Pyrolysis or Thermal Decomposition)

• Cracking may be defined as a process in which higher alkanes are converted into a mixture of lower
alkanes and alkenes by the action of heat alone or heat
• Cracking involves breaking of (C–C) and (C–H) bonds resulting in the formation of a mixture of lower
hydrocarbons.
• pyrolysis of alkanes involves free radical reaction. Preparation of oil gas or petrol gas from kerosene
oil or petrol involves the principle of pyrolysis.
C6H12+H2
773K
C6H14 C4H8+C2H6
C3H6+C2H4+CH4
770K
CH3(CH2)8CH3  → CH3(CH2)6CH3 + CH2 = CH2

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REACTION CHART FOR ALKANES

GNP GR
H2 , Ni X ,h ν or UV light or
(1) R – C ≡ CH 
200 −300°C

→ (1) →
2

250 − 400ºCin dark

R − X (Halogenations )
Or Sabatier senderen's Reaction
R – CH = CH2 (2) R–NO2 (Nitration)
H2 S2O7 ,∆
(2) R–X
Zn+HCl
→ (3) 
or(H SO + SO )
→ R–SO3H (Sulphonation)
2 4 3

Na,dryether
(3) R–X →
Wurtzreaction (4) O
2
→ CO2+H2O (Combustion)
∆

R −Zn −R R–H
(4) R–X 
Frankland'sreaction
→ Or
R–R
R Culi or Controlled
(5) R–X → 2
( Corey −Housesynthesis ) (5)
CnH2n+2 Oxidation

HOHorROH
(6) R–Mg–X 
orNH orRNH
→
3 2

(Alkane should
Red P/HI
(7) R–OH, R–CHO, R-I 
→ have 3ºH)
R − C − R, RCOOH
|| AlCl3 /HCl
(6) 
Isomerisation
→Branched alkanes
O
Zn−Hg/Conc.HCl
(8) R − C −R →
Clemmensen's reduction
|| Cr or Mo or V oxide
(7)  → Aromatic compound
O + Al O ,500ºC
2 3

R − C −R −
H2N NH2 /OH, ∆ Pyrolysis Lower Lower
(9) 
Wolf kishner reduction
→ (8)  → +
500 −700ºC
|| alkenes alkanes
O

NaOH+ CaO
(10) RCOONa 
∆
→

Kolbe's electrolysis
(11) RCOONa(aq) 
→

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Exercise 1.1
1. An alkane of molecular weight 86 g/mol on 6. Arrange the correct sequence for mechanism
monochlorination gives two products. The of chlorination of methane
alkane is •
I CH4 + Cl• 
hv
→ C H3 + H − Cl
(1) 2-Methylbutane
•
(2) n-butane I hv
Cl − Cl  → 2Cl
(3) 2, 2-Dimethyl propane •
III CH3 + Cl2 → CH3 − Cl + Cl•
(4) 2, 3-Dimethyl butane
• •
IV Cl + Cl → Cl − Cl
2. Total number of products including • •

stereoisomers formed during monochlorination CH3 + CH3 → CH3 − CH3

of 2-methyl butane? Choose the correct option.

(1) 2 (2) 4 (1) II, I, III and IV (2) I, II, III and IV
(3) 6 (4) 8 (3) IV, III, II and I (4) II, III, I and IV

Na V2O5 − Al2O3
3. Which of the following is a termination step in 7. CH3 − CH2 − CH2 − Cl 
dryether
→×  ∆
→
the free radical chlorination of methane? Y : Y is
•
(1) Cl2 → 2Cl (1)
• •
(2) Cl2 + Cl → Cl + Cl2
(2)
• •
(3) CH3 + Cl → CH3Cl (3) CH3– (CH2)4–CH3
• • (4)(CH3)2CH– CH(CH3)2
(4) CH4 + Cl → HCl + CH3
AlCl3, HCl
8. 25°C Product is :
4. Which of the following is a chain
propagation step in the free radical (1) (2)
chlorination of methane?
• •
(1) CH4 + Cl → CH3 + HCl (3) (4)
• •
(2) CH3 + CH4 → CH4 + CH3
• • 9 1-bromo-3-chlorocyclobutane is treated
(3) Cl + CH3 → CH3Cl with two equivalents of Na in the presence of
• dry ether. Which of the following will be
(4) Cl2 → 2Cl
formed?
Br Cl
5. Nitration of ethane gives
(1) Nitro methane
(1) (2)
(2) Nitroethane
(3) Ethylnitrite (3) (4)
(4) 1 & 2 both

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A 16. Kolbe’s electrolysis is convenient for the
10. Ph–C–CH3 Ph–CH2–CH3
O preparation of :
A could be : (1) Methane
(1) NH2NH2, glycol/OH– (2) Symmetrical alkanes containing even
number of carbon atoms
(2) Zn(Hg)/conc. HCl
(3) Alkanes containing even as well as odd
(3) Red P/HI
number of carbon atoms
(4) All (4) Alkanes containing odd number of
carbon atoms
11. Which is correct product?
Cl
O2 ,Cu
(1) CH3 − CH3 
∆ , pr
→ CH3 − CH2 − OH |
Na /ether
17. CH3 – CH – CH3  → (A),Major
O2 ,MO2O3
(2) CH3 − CH3 ∆ , pr
→ CH3 − CHO product
(A) is:
(3) (CH3 )3 CH → ( CH3 )3 C − OH
KMnO4
(1) CH3–CH2–CH2–CH2–CH2–CH3
(4) All (2) CH3— CH — CH—CH3
CH3 CH3
12. When aqueous solution of sodium ethanoate (3) CH3CH2–CH2CH3
is electrolysed, the product(s) at anode (4) CH3–CH=CH2
is/are :
(1) Ethane (2) CO2 18. The order of reactivity of alkyl halides in
(3) H2 (4) Both (1) & (2) Wurtz reaction is :
(1) R–I > R–Br > R–Cl
13. The no. of isomeric carboxylic acid that will (2) R–I < R–Br < R–Cl
be required to obtain neopentane by soda (3) R–Br >> R–I < R–Cl
lime: (4) R–I >> R–Cl > R–Br
(1) 3 (2) 1
19. Which of the following alkane is synthesized
(3) 4 (4) 6
from single alkyl halide by wurtz reaction :
(1) (2)
14. Which product is formed at the anode by the
electrolysis of sodium butyrate : (3) (4)
(1) Butane + CO2
(2) Pentane + CO2
20. Which of the following reaction of alkanes
(3) Hexane + CO2
involves free radical intermediates.
(4) Hexane + H2
(1) halogenation (2) Pyrolysis
15. The appropriate reagent for the (3) Nitration (4) All of these
transformation is: H3C
21. NH2–NH2/∆ product is
O C O KOH/CH2–CH2
H3C
OH OH
CH3
(1) CH3–C=CH2
HO
HO CH3
(1) Zn(Hg)/conc. HCl (2) CH3–CH2–CH3, N2
Θ (3) CH3–CH2–CH3, NH3
(2) NH2NH2/ OH¯ / Ethylene glycol
(4) H3C–CHCH3,N2
(3) H2/Ni
OH
(4) NaBH4
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22. The reaction, C2H5Br (i)Li

(ii)Cu Ι
→ [ X ] 
CH3Br
→[Y ] : 26. Which of the following compound can not be
obtained from single alkyl halide by wurtz
is called- reaction.
(1) Wurtz synthesis (1) ethane (2) butane
(2) Wolf-Kishner reduction (3) isopentane (4) hexane
(3) Corey-House sysnthesis
(4) Kolbe’s sysnthesis O
Zn–Hg
27. HCl
Products is
23. What is the chief product obtained when n-
butane is treated with bromine in the (1) (2)
OH Cl
presence of light at 130°C?
(3) (4)
(1) CH3 − CH2 − CH2 − CH2 − Br
(2) CH3 − CH2 − CH − Br
| COONa electrolysis H2+Ni
CH3 28. X Y; Y is
COONa(aq) ∆
(3) CH3 − CH − CH2 − Br
| (1) CH3–CH2–CH2–CH3
CH3 (2) (CH3)2CH–CH(CH3)2
CH3 (3) (CH3)2CH–CH2–CH3
| (4) CH3–CH = CH – CH3
(4) CH3 − C − CH2 − Br
|
CH3 29. Give reactivity order for decarboxylation?
(I) CH3–CH2–COOH
24. If 1-chloropropane reacts with Na in (II) CH2=CH –COOH
presence of dry ether. Which alkane is (III) CH ≡ C–COOH
obtained. (1) I > II > III
(1) Hexane (2) III > II > I
(2) 2,3-Dimethyl butane
(3) III > I > II
(3) Isopentane
(4) None of these
(4) Neopentane
25. If isopropyl chloride and methyl chloride 30. Which of the following reaction can not be
both react with Na in presence of dry ether used to obtained propane in good yield.
which alkanes are obtained. (1) Wurtz reaction
(1) Ethane (2) Kolbes elastolysis
(2) Isobutane (3) Sodalime decarboxylation
(3) 2,3-Dimethyl butane (4) (1) & (2) both
(4) All of them

Alkenes
Introduction of Alkenes
• Alkenes are unsaturated hydrocarbons.
• General formula : CnH2n.
• Alkenes are also called olefins (oil forming) since the first member, ethene forms an oily liquid on
reaction with chlorine.
• The doubly bonded carbon atoms are sp2 hybridized.
• Geometry is trigonal planar.
• Bond angle is 120°.
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General Methods of Preparation of Alkenes

(1) By the Partial Hydrogenation of Alkynes:
Alkynes can be converted into alkenes by controlled or partial hydrogenation in two ways :
(i) by using Lindlar's Catalyst :
• Lindlar's catalyst is palladium deposited on barium sulphate or calcium carbonate. The catalytic is
slightly poisoned with lead acetate, quinoline or sulphur.
• Alkynes are partially reduced with calculated amount of dihydrogen in presence of lindlar’s catalyst
into cis alkenes.
Lindlar’s catalyst CH3 CH3
Ex. CH3—C ≡ C–CH3 + H2 C=C
Pd/CaCO3 H H
But-2-yne cis-2-butene
In place of Lindlar's catalyst Nickel-boride (Ni-B also called P-2 catalyst) can also be used.
(ii) By using Sodium in Liquid Ammonia :
This is known as Birch reduction and the major product is a trans alkene.
CH3 H
Ex. CH3—C ≡ C— CH3 Na/Liquid NH3 C=C
H CH3
But-2-yne
trans-2-butene
Note: Birch reduction fails in case of terminal alkynes due of acid base reaction

(2) From Alkyl Halide (By Dehydrohalogenation):

Alcoholic KOH or NaNH2 is used as dehydrohalogenating agent
CH3–CH2 Cl + KOH(alc.) → CH2 = CH2 + KBr + H2O
Mechanism : The reaction involves E2 mechanism.
δΘ
H H HO H H
Θ
HO + H–C – C– H H–C C–H H2C = CH2
δΘ
H Cl H Cl
Transition state
• Rate of reaction ∝ [substrate] [base]
• Order of reaction = 2
• Carbocation rearrangement is not observed during the reaction
• Since reaction is endothermic (∆H = + Ve), so high temperature is a favourable condition to get product with good
yield
• It is an example of β-elimination
• The rate of dehydrohalogenation shows the order :
For alkyl group 
→ tertiary > secondary > primary
For halogen in alkyl halide 
→ Iodide > Bromide > Chloride > fluoride
KOH( alc. ), ∆
Ex. CH3 − CH2 − CH − CH3  → CH3CH = CH − CH3 + CH3CH2CH = CH2
−HX
|
X ( Saytzeff product ) (Hoffmann'sproduct )

Note: (i) If – X ⇒ –Cl |–Br| – I, saytzeff’s product is major (Based on stability of alkene)
(ii) If – X ⇒ –F, Hoffman’s product is major (Based on stability of carbanion)
(iii) If attacking base is bulky (like t-butoxide ion) Hoffnann’s product is major.

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(3) From Alkyl Dihalide (By Dehalogenation of Vicinal or Gem Dihalide):
• Removal of X2 from a substrate by Zn dust
(i) From Vicinal Dihalide : alkene with same number of carbon is obtained.
X X
| |
Zn
Ex. H − C − C − H 
∆
→ CH2= CH2 + ZnX 2
| |
H H
∆
Ex. H3C − CHBr − CH2Br + Zn → CH3 =
− CH CH2 + ZnBr2

(ii) From gem Dihalide : Higher alkene obtained

∆
CH3CH X 2 + 2Zn + X 2 CHCH3 → CH3 −=
CH CH − CH3 + 2ZnX 2

(4) From Alcohols: (By acidic dehydration)

Alkenes can be prepared from monohydric alcohols or alkanols by the loss of H2O in presence of an
acid, the reaction is known as acidic dehydration of alcohols.

β α
—C—C— —C = C— + H2O
H OH Alkene
Alcohol
The dehydration can be carried out in the presence of heat with Conc. H2SO4. Or Conc. H3PO4
conc.H2SO4
EX. CH3 –CH2 –OH →
443K
=
CH2 CH2 + H2O
Ethanol Ethene
Mechanism of Reaction : The reaction follows E1 mechanism
δ⊕ δΘ
•• H–OSO3H ⊕
Step-I : CH3–CH2 –O–H+
•• CH3–CH2 –O–H
Θ
–HSO4 H
Ethanol
Protonated ethanol
⊕
Step-II : CH3–CH2 –O–H
··⊕ Slow
CH3–CH2 +H2O
RDS
H Ethyl carbocation
This is slow step and is regarded as rate determining step in E1 reaction.
Base
Θ
⊕ HSO4
Step-III : Hα–CH2 –CH2 CH2=CH2
Ethene
• The reaction involves carbocation intermediate.
• Rate of reaction ∝ [substrate]
• The reaction is an example of β-elimination.
Saytzeff Rule : When more then one alkenes are obtained by the elimination reaction then that alkene
containing maximum number of alkyl group on double bonded C-atoms is called Saytzeff’s product and
formed as major product.

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Note: The alkene having less number of alkyl groups on double bonded C-atoms is called Hofmann’s
product.
OH
H2SO4
Ex.(i) CH3–CH2–CH–CH3 CH3–CH=CH–CH3 + CH3–CH2–CH=CH2
∆
2-butanol main product 1-butene 20%
2-butene 80% (Hoffmann’s product)
(Saytzeff’s product)
H2SO4
(ii) CH3 − CH2 − CH2 − CH2 − OH ∆
→ CH3 − CH = CH − CH3 + CH3CH2CH = CH2
1−butanol 2−butene 80% 1−butene 20%
Main product

The order of acidic dehydration in different alcohols is: Tertiary > Secondary > Primary
Note: Al2O3 can also be used at high temperature for dehydration of alcohols .

(5) By Kolbe's Method :

Electrolysis of potassium or sodium salt of saturated dicarboxylic acid gives alkene.
CH2COONa electrolysis CH2 + 2CO2 + 2NaOH +H2
CH2COONa(aq) CH2
CH3—CHCOONa CH3 —CH + 2CO2 +2NaOH + H2
electrolysis
CH3—CHCOONa(aq) CH3—CH

Physical Properties of Alkenes

• From C2 –C4 they are colourless, odourless gases, from C5 –C17 they are colourless liquids, and higher
alkenes are solids.
• Alkenes are insoluble in water but they are fairly soluble in organic solvents like benzene, chloroform,
CCl4, petroleum ether etc. (Like dissolves like)
• B.P increase with increase in size and every – CH2 group added increase B.P by 20–30K.
• The melting points of cis isomers are lower than trans isomers because cis isomer is less symmetrical
than trans. Thus trans packs more tightly in the crystal lattice and hence has a higher melting point.
• The boiling points of cis isomers are higher than trans isomers because cis–alkenes has greater polarity
(Dipole moment) than trans one.
• The increase in branching in carbon chain decreases the boiling point among isomeric alkenes.

Chemical Reactions of Alkenes

Alkenes are more reactive than Alkanes this is because -
(i) The π electrons of double bond are located much far from the carbon nuclei and are loosely bonded.
(ii) π bond is weaker than σ bond
Following Reactions are observed in alkenes:
(A) Addition reaction (B) Oxidation reaction (C) Free radical Substitution reaction
(D) Polymerization Reaction
(A) Addition reaction:
(a) Addition of H2 :
Ni,Pt or Pd
R–CH = CH2 + H2  → R–CH2–CH3 + Heat of Hydrogenation
• Reaction is exothermic, Heat released in reaction is called heat of hydrogenation.
• Heat of hydrogenation ∝ no. of π – bond
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if no of π–bond are same then.
1
Heat of hydrogenation ∝
stability of alkene
Note: Stability of alkene depends upon resonance, hyperconjugation and type of geometrical isomerism
• The process is used to obtain vegetable (saturated fats) ghee from hydrogenation of oil.

(b) Electrophilic Addition Reactions:

Characteristic reaction of alkenes are electrophilic addition reaction
Nu E
C=C + E—Nu —C — C—
Alkene Attacking
reagent Addition product
(Adduct)
From mechanism point of view, the addition in alkenes is generally electrophilic in nature which means
that attacking reagent which carries the initial attack is an electrophile (E+). This is quite expected also
as there is high electron density in the double bond. The mechanism proceeds in two steps.
Step-I: The reaction involves addition of E⊕ in first step to form carbocation as intermediate.
δ⊕ δΘ r.d.s ⊕ Θ
R—CH = CH2 + E—Nu R—CH – CH2 + Nu
E
Step-II: The nucleophile (:Nu–) released in the slow step combines with the carbocation to give the
addition product in the fast step.
Θ
⊕ Nu
R—CH — CH2 R—CH – CH2
E Nu E

Reactivity of alkenes / alkynes for electrophilic addition ∝ µ

Stability of carbocation formed in r.d.s
(i) Addition of Halogen : It is an electrophilic addition reaction.
X
|
R–CH=CH2+X2 → R − CH − CH2
|
X
( Vicinaldihalides )
• The addition of Br2 on alkenes is a useful test for unsaturation in molecule. The brown (or reddish
orange) colour of the bromine in CCl4 is discharged. Thus decolorization of Br2 in CCl4 by a compound
suggest unsaturation in it.
• I2 reacts with alkenes to form Vicinal di-iodides which are unstable and I2 gets eliminated to give back
alkene.
I

CH3–CH=CH2+I2 CH3–CH–CH2
I
Unstable

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Mechanism:
Br
⊕ Θ
Br—Br Br
R—CH= CH2 R—CH– CH2 R—CH– CH2 R—CH– CH2
CCl4 ⊕
Θ • • Br
–Br • •Br
••
Br
(Cyclic bromonium ion)
• Electrophile involved in the reaction is Br2 (due to vacant orbital)
• Cyclic bromonium ion is formed as intermediate
• Reaction involved anti addition of Br2
• Carbocation rearrangement is not possible

Stereochemistry of the reaction :

R Br Br
H
C Br2 R H H R
C CCl4 H R’ R’ H
R H Br Br
(Cis)
If R=R’ (d+l) Racemic mixture
R≠R’ (d+l) Racemic mixture

R Br Br
H
C Br2 R H H R
C CCl4 R’ H H R’
H R’ Br Br

If R=R’ (Meso)
R≠R’ (d+l) Racemic mixture
(ii) Addition of Halogen acid : (hydrogen halides)

X
|
R–CH = CH2 + HX → R − CH − CH3 + R − CH2 − CH2 − X
(major ) (minor )
• The order of reactivity of hydrogen halide is : HI > HBr > HCl
• Carbocation rearrangement is possible in the reaction.
• Addition on alkene proceeds via the formation of more stable carbonium ion.
• Addition of HX on unsymmetrical alkenes (R–CH = CH2) takes place according to Markovnikov’s rule.
X
R—CH– CH3 (Major)

• Addition of HX on unsymmetrical alkene : R—CH = CH2 + H— X

R—CH2 – CH2 —X (Minor)
Markovnikov’s Rule States :
First Rule : In addition of HX on unsymmetrical alkene, the electrophile (H+) mainly goes to the
unsaturated carbon atom having more number of hydrogen atoms, or negative part of the reagent
(addendum) mainly gets attached to the carbon with possesses less number of hydrogen atoms.

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Mechanism :
Θ X
⊕ X
δ⊕ δΘ
R—CH– CH3 R—CH— CH3(Major)
More stable
R—CH= CH2 + H–X Θ ⊕
Θ
X
–X R—CH2– CH2 R—CH2— CH2—X (Minor)
Less stable

• Markovnikov’s rule is based on stability of carbocation.

• The electrophilic addition to unsymmetrical alkenes always occurs through the formation of a more
stable carbocation intermediate.

(iii) Anti Markovnikov’s Rule / Kharasch Effect / Peroxide Effect –

Addition of HBr on unsymmetrical alkene in presence of peroxide follows free radical addition
mechanism and major product is formed according to anti Markovnikov’s rule.
H H H H
(C6H5–CO)2O2
R—CH= CH2 + HBr R—C — C —H + R—C — C —H
Or R2O2
Br H H Br
(Minor) (Major)
(Markovnikov’s (Anit Markovnikov’s
product) product)
Mechanism
(a) Chain initiation step:
O O O
• • • •
(i) R − O − O− R → R O+ OR or C6H5– C –O–O–C–C6H5 2C6H5—C — O →2C6 H5+2CO2
Peroxide

• • • •
(ii) RO+ H—Br → ROH + Br or C6H5 + H –Br → C6H6 + Br
(b) Chain propogation step:
H H H H H H
•
(iii) Br + R —C = C–H → R —C— C—H
•
+ R—C—C—H
•
Br Br
(Less stable) (More stable)

H H H H
•
(iv) R —C• — C— H + H–Br → R— C— C—H + Br
Br H Br (Majar)

• •
(v) R —CH — CH2 + H —Br → R— CH— CH3 + Br
Br Br (Minor)

(c) Chain termination step:

• •
(vi) Br + Br → Br2
• •
R − CH− CH2 + Br → R − CH − CH2
| | |
Br Br Br
• •
R —CH — CH2—Br + R—CH—CH2—Br → R—CH—CH2—Br
R—CH—CH2—Br
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Note: Anti Markovnikov addition in the presence of peroxide is not observed in HF, HCl and HI.
(i) In the case of H–F and H – Cl, one of the chain propagating steps is endothermic
Cl Cl H
| • | | •
H−Cl
Ex. CH2 − CH2  → CH2 − CH2 + CI is endothermic (as ∆H• = +12.6 KJ mol–1)
(ii) In the case of H – I, one of the chain propagating steps is endothrmic
Ι
• |•
CH=
2
CH2
+ I → CH2
− CH2
is endothermic (as ∆H = +46 KJ mol–1)
But in the case of HBr both of the chain propagating steps are exothermic, which results spontaneous
reaction.
(iv) Addition of cold conc. H2SO4:
Reaction follows electrophilic addition mechanism and forms major product according to
Markovnikov’s rule
O Θ
δ⊕ δΘ ⊕ O–SO3H
R —CH = CH2 + H —O —S—O—H Θ R —CH — CH3 R —CH — CH3 (Major)
–HSO4
O O–SO3H
(v) Addition of Water (Hydration of alkenes) : Alkenes react with water in the presence of a few drops of
concentrated sulphuric acid to form alcohol, this reaction is called acidic hydration.
OH
Ex. H+
CH3 – C= CH2 + H2O CH3 – C – CH3
CH3 CH3

2–Methylpropene 2–Methylpropan-2–ol
Mechanism:

⊕ (Slow) ⊕
CH3 – C =CH2 + H CH3 – C – CH3
CH3 CH3
Carbocation
CH3 CH3
⊕ ·· (Fast) –H+
CH3 – CH–CH3 + H–O–H
·· CH3 – C–CH3 CH3 – C–CH3
CH3 H—O—H OH
··
⊕ Propan-2-ol

• Carbocation rearrangement is possible

• Addition of water molecule proceed via formation of more stable carbocation

(vi) Addition of Hypohalous Acid (or X2/H2O, or HOX) :

••
OH
H2O ••
R —CH — CH2 (Major)
Br–Br ⊕ (Bulk)
R —CH = CH2 R —CH – CH2 R —CH – CH2 Br Br
H2 O BrΘ
–Br Θ • •
• Br • Br R —CH — CH2 (Minor)
•• ⊕
Br
• It is a electrophilic addition reaction.
• it follows Markovnikov’s rule
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• Anti addition takes place
• Cyclic bromonium ion is formed as intermediate
• carbocation rearrangement is not possible

(vii) Hydroboration Oxidation : Borane readily reacts with alkenes to give trialkyl boranes. The reaction is
called hydroboration.
T.H.F
R–CH=CH2 + BH3 (R–CH2–CH2)BH2
R–CH=CH2
R-CH=CH2
(R–CH2–CH2)2B (R–CH2–CH2)2BH
Trialkylborane
Θ
H2O2/OH(aq)
R–CH2–CH2–OH(Alcohol)
oxidation
(R–CH2–CH2)3B
Trialkyl borane CH3–COOH
R–CH2–CH3 (Alkane)
Reduction

Note: The overall process appears to be addition of water according to anti Markovnikov’s rule without
carbocation rearrangement and involves syn addition
(viii) Oxymercuration-Demercuration: overall process appears to be addition of water molecule according
to Markovnikov’s rule without carbocation rearrangement.
R–CH = CH2 → R − CH − CH3
|
OH
Reagents :
(i) (AcO)2Hg/H2O or (CH3COO)2Hg/H2O (oxymercuration)
(ii) NaBH4 (demercuration)

Mechanism :

CH3—COO
Hg(aq) CH3–COO– + CH3 –COOHg+ (Electrophile)
CH3—COO

+ +
R–CH=CH2+ HgOOCCH3 R–CH–CH2
• HgOOCCH
• 3

⊕
H–O–H OH OH
H2O NaBH4
R–CH–CH2 R-CH-CH2 R-CH-CH2 R–CH–CH3+Hg+CH3COOΘ
–H⊕
⊕
HgOOCCH3 HgOOCCH3 HgOOCCH3 (Demercuration)
(Cyclic cation) (Oxymercuration)

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Ex.
OH
H+/H2O
CH3—C–CH2–CH3 Markovnikov’s product (C⊕ rearrangement)
CH3

CH3—CH–CH=CH2 OH
(i) BH3/THF
(ii) H2O2/ΘOH
CH3—CH–CH–CH2 Anti Markovnikov’s (NO C⊕ rearrangement)
CH3
CH3 H product
Cl
HCl
CH3—C–CH2–CH3 Markovnikov’s product (C⊕ rearrangement)
CH3
OH
(i) (AcO)2 Hg/H2O
CH3—CH–CH–CH2 Markovnikov’s product (NO C⊕ rearrangement)
(ii) NaBH4
CH3 H

(B) Oxidation Reaction : Alkenes are easily oxidised by oxidising agents. Oxidising agents attack on double
bond and product formed during oxidation depends on oxidising agents and experimental conditions.
3n
(1) Combustion: CnH2n + O2 → nCO2 + nH2O
2
3n
One mole of alkene requires moles of O2 for complete combustion.
2
Hydroxylation
(2) Hydroxylation : C=C C C

OH OH
(a) Oxidation by Baeyer's Reagent (A test for unsaturation) : Alkenes on reaction with cold, dilute, 1%
alkaline KMnO4 (i.e., Baeyer's reagent) produces vicinal glycol and decolourise the pink colour of
KMnO4 and gives brown ppt of MnO2. The reaction involves syn addition.
C C < +H2O + [0] 
>= dil.KMnO4
273K
→ > C − C < (Syn− addition)
| |
OH OH
(Glycol)
Ex. CH3–CH=CH2 dil.KMnO4
CH3–CH–CH2
273K
OH OH
(b) reaction with OsO4 followed by hydrolysis also form vicinal glycol and involves syn addition.
Ex. CH3–CH=CH2 (i) OsO4 + CH3–CH–CH2(Glycol)
(ii) H2O/H
OH OH
(3) Oxidation by Strong Oxidising Agent (Oxidative Cleavage) :- Acidic potassium permanganate or acidic
potassium dichromate oxidises alkene to ketones and /or acids depending upon the nature of the
alkene and the experimental condition.
Formic acid further gets oxidised to give CO2 + H2O.
CH3
(i)
KMnO4 /H⊕
C= CH2  Or
→CH3 C = O+H–COOH→CO2+H2O
CH3 K 2Cr2O7 /H⊕ CH3

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⊕
KMnO4 /H
(ii) CH3CH = CHCH3 
Or
→ 2CH3COOH
K 2Cr2O7 /H⊕

(4) Ozonolysis :
• Ozonolysis of alkenes useful in detecting the position of double bond in alkenes or other unsaturated
compounds.
R
ii) H2O/Zn
R–C=O+O=CH-R
R
i) O3
R
Ex. R–C=CH-R ii) LiAlH4
R–CH–OH+HO-CH2–R
Or NaBH4
R O
H2O
R–C=O+HO-C-R
• Reaction of alkene with O3 involves formation of ozonide
O3 O
C=C C C
O–O
• It may be noted that reaction with bromine solution or Baeyer's reagent detects the presence of double
bond (or unsaturation) in an alkene while ozonolysis helps in detecting the position of the double bond.

(C) Substitution Reaction (Allylic Substitution) :

• When alkenes are treated with low concentration of Cl2 or Br2 at high temperature or with NBS/hv one
of their allylic hydrogen is replaced by halogen atom.
• Allylic position is the carbon adjacent to one of the doubly bonded carbon atoms.
• It is free radical substitution.
500°C
Ex. (i) CH3 − CH = CH2 + Cl2  → ClCH2 − CH = CH2 + HCl
Allyl chloride
(3-Chloro-1-propene)
Br
(ii) N.B.S
hν

(D) Polymerization :
  High temperature/pressure
Ex. n =CH2 CH2  catalyst
→ —(CH2 − CH2 — )n
 ethene  Polythene

n ( CH3 −=CH CH2 ) →

high themperature /pressure
catalyst
−(CH− CH2 )n
Propene |
CH3
Polypropene

• Polythene is used in the manufacture of polythene bags polythene sheets

• Polypropene is Used for manufacture of milk crates, plastic buckets and other moulded articles.

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Reaction Chart for Alkenes

GMP GR
O–SO3H
Pyrolysis
(1) R–CH2–CH3 H2SO4(conc.)
Cracking (1) Cold
R–CH–CH3
H2/Pd
(2) R–C≡ CH CaCO3 or BaSO4 dil. H2SO4
(2) R–CH(OH)–CH3
NaNH2 or alc. KOH
(3) R–CH2 – CH2–X R–CH=CH2 HX
–HX
Or (3) R–CHX–CH3
Zn/Na/Mg
(4) R–CH – CH2 CnH2n
dry ether HBr, Peroxide
(4) R–CH2–CH2Br
X X
X2
conc.H3PO4/∆ (5) R–CHX–CH2X (X2=Br2/Cl2)
(5) R–CH2 – CH2–OH CCl4
Or
conc.H2SO4/∆
HOBr
–H2O
(6) R–CH(OH)–CH2Br
or Br2/H2O

(6) R–CH CH2

Kolbe’s electrolysis OH
(i) (CH3COO)2Hg, H2O
NaOOC COONa(aq) (7) (ii) NaBH4
R–CH–CH3 (O.M.D.M)

∆ BH3 H 2O 2
(7) (RCH2CH2)4N+OH– (8) (RCH2CH2)3B Θ R–CH2–CH2–OH +
OH(aq)
H3BO3 (H.B.O)
Bayer reagent
(9) R–CH–CH2 (Syn addition)
(1% alkaline, Cold, KMnO4)
OH OH
(i) OSO4
(10) (ii) H2O
R–CH–CH2 (Syn addition)
OH OH
OH
(i) C6H5–COOOH
(11) R–CH–CH2 (Anti addition)
(ii) H3O⊕
OH
OH
(i) Ag2O
(12) R–C–H–CH2 (Anti addition)
(ii) H3O⊕
OH
(i) O3 R H
(13) (ii) H2O/Zn
C + C (Ozonolysis)
H H
O O
KMnO4
(14) R–C–OH+CO2+H2O
H⊕
O
O2
(15) ∆
CO2+H2O
Cl2
(16) 500ºC
Allylic halogenation

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Exercise 1.2

1. The compound X is 5. Propene on reaction with hypochlorous acid

(i) B2H6 gives?
(ii) H2O2.OH– X
Cl
CH3 CH3 Cl
OH
(1) H (2) H (1) OH (2)
OH H
H OH Cl OH
CH3
(3) HO (4) All of these (3) (4)
H
OH
6. The olefin, which on ozonolysis give
2. Hydroboration-oxidation and acidic
CH3CH2CHO and CH3CHO is
hydration will not give the same products in
(1) But-1-ene
case of (2) But-2-ene
(3) Pent-1-ene
(1)
(4) Pent-2-ene

(2) 7. 2-MethyIbutan-2-oI can be obtained by the

acid catalyzed hydration of
(3) CH2 = CH2
(1) CH3CH2CH = CH2
(4) CH3CH = CH – CH3
(2) CH3CH = CHCH2
(3) (CH3)2C = CHCH3
3. The major product(A) of the reaction given (4) Either of the three
below is
8. The final product obtained in the reaction
H3C CH=CH CH3

(1) Hg(OAc), H2O

HBr (2) NaBH4
A(major product) Peroxide
CH3 CH3
(1) Br–CH2 CH=CH OH
(1) (2) OH

(2) H3C CH=CH Br

CH3 CH3

(3) H3C CH2–CH

(3) H (4) HO
Br
OH H

(4) H3C CH–CH2 9. Which of the following will be the major

Br product when 3-phenylpropene reacts with
4. How many of the following compounds will HBr?
form acetic acid on reaction with acidic (1) C6H5CH2CHBrCH3
KMnO4 ? (2) C6H5–C=CH2–CH3
Prop-1-ene, 2-Methylbut-2-ene,
Br
2-Methylpropene, But-2-ene, Cyclohexene.
(1) 2 (2) 3 (3) C6H5CH2CH2CH2Br
(3) 4 (4) 5 (4) C6H5CHBrCH2CH3
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O 15. The compound formed when 2-butene is

10. O3 /H2O
Alkene (A) → CH3–C–CH3 + treated with hot acidic KMnO4 is :
O (1) CH3–CH=O
CH3COOH + CH3–C–COOH (2) CH3–C–OH
A can be : O
CHCH3
CH3 (3) CH2OH–CH2OH
(1) CH3–C–CH=C CH3 (4) CH3CH2–C–CH3
C(CH3)2 O

(2) CH3–C–CH = HC – CH3

(3) Both correct C4H9Br
alc.KOH
(B)
(i) O3
CH3–CHO
16. ∆ (ii) Zn/H2O
(4) None is correct (A)
(A) and (B) in the above reaction sequence
11. For the ionic reaction of hydrochloric acid
with the following alkenes, predict the are:
correct sequence of reactivity as measured (1) s-Butyl bromide, 1-Butene
by reaction rates : (2) t-Butyl bromide, 2-methyl propene
I. ClCH=CH2 II. (CH3)2C=CH2 (3) s-Butyl bromide , 2-Butene
III. OHC CH=CH2 IV. (NC)2C=C(CN)2
(4) n-Butyl bromide, 1-Butene
(1) IV > I > III > II (2) I > IV > II > III
(3) III > II > IV > I (4) II > I > III > IV
CH2OH
17. H2SO4 NBS
12. Which of the following will give different P(Major) Q (Major)
∆
product with HBr in presence or absence of
peroxide : The structure of Q is :
(1) Cyclohexene Br
(2) But-2-ene Br Br
(3) 1, 2-dimethylcyclohexene (1) (2)
(4) 1-butene
Br
13. Reactivity of alkenes towards HX increases
in the order : (3) Br (4)
(1) Isobutene > Propene > Ethene
(2) Isobutene > Ethene > Propene
(3) Ethene > Propene > Isobutene +

H
→ X (Major)
(4) Propene > Ethene > Isobutene 18. ∆
‘A’ is
CH3 OH
CH3
14. Give reactivity order towards EAR.
OH
(i) (ii)
(1) CH3 (2) CH3
CH2 CH2
Cl CH3
(iii) (iv)

(1) (i) > (ii) > (iii) > (iv) (3) CH3 (4) CH3
(2) (iv) > (iii) > (ii) > (i) CH3 CH3
(3) (ii) > (iv) > (i) > (iii)
(4) (iii) > (ii) > (iv) > (i)
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Br 23. CH3 − CH − CH = CH2 + HBr → ‘X’

|
CH3 Alc.KOH
19. Product. CH3
Major product X is-
The product can be (1) CH3—CH—CH2—CH2Br
CH2 CH3
(1) (2)
Br
CH3
CH3 (2) CH3—C—CH2—CH3
CH3
(3) (4)
CH3 CH3 (3) CH3—CH—CH—CH3
CH3 Br
20. The major product of the following reaction (4) None is correct
is
CH = CH—CH3
(i) Hg(OAC)2.H2O
24. Two jars A and B are filled with hydrocarbon.
(ii) NaBH4
Br2 in CCl4 is added to these jars. A does not
(1) decolourise the Br2 solutions, but B
CH2—CH2—CH2—OH
decolourises. What are A and B respectively.
(2) CH2—CH—CH3 (1) Alkane and alkene
(2) Alkene and alkane
OH
(3) Both 1 and 2
OH
(4) Alkene and Alkyne
(3)
CH—CH2—CH3
dil/H2SO4
25. CH3CH=CH2  →A
(4) HO CH=CH—CH3
B2 H6 H2O6
CH3–CH = CH2  → 
Θ →B
OH

OH Correct statement about the product is

A
21. (1) A and B have different functional groups
CH3 CH3 (2) A and B are position isomers.
Reagent ‘A’ is (3) A and B show metamerism
Θ (4) All of these
(1) BH3 /H2O2 / O H
Na H2 +Pd
⊕ 26. B ←
Liq.NH3
CH3 − C ≡ C − CH3 
CaCO3
→A ;
(2) H2O /H (Major )

(3) Hg(OCOCH3 )2 /NaBH4 A and B are

(1) Homomers
(4) Cl2 / aq.NaOH
(2) Position isomers
22. A hydrocarbon X adds on one mole of (3) Geomatrical isomers
hydrogen to give another hydrocarbon and (4) Homologues
also decolourises bromine water. X reacts
27. When 3,3-dimethyl-2-butanol is heated with
with KMnO4 in presenec of acid to give two
H2SO4 the major product obtained is
moles of the same carboxylic acid. The
structure of X is (1) cis and trans isomers of 2,3-dimethyl-2-
(1) CH3CH = CH. CH2CH2CH3 butene
(2) CH3CH2CH = CHCH2CH3 (2) 3,3-dimethyl-1-butene
(3) CH3CH2CH2—CH=CHCH3 (3) 2,3-dimethyl-2-butene
(4) CH2= CH CH2CH2CH3 (4) 2,3-dimethyl-1-butene

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28. But-2-ene can be obtained by electrolysis of CH3= BD3 /THF
CH CH2  →X ,
30. H O /OHΘ
2 2
an aqueous solution of
(1) 2,3-Dimethyl maleic acid Major product ‘X’ is
(2) 2,2-Dimethylbutanedioic acid
(3) 2-Methylbutanedioic acid (1) CH2CHCH2D
|
(4) 2,3-Dimethylbutanedioic acid OH

⊕ (2) CH3CHCH2OH
29. H2O/H |
CH3–CH–CH=CH2 P (major), P is:
Or D
dil.H2SO4
(3) CH3CHCH3
|
(1) CH3–CH–CH2–CH2–OH OD

(4) CH2 − CH − CH3

OH |
OH
(2) CH3–CH–CH–CH3

OH
(3) CH3–C–CH2–CH3

(4) CH3–CH–CH2 –CH2

OH

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Alkynes
Introduction of Alkynes
• Like alkenes, alkynes are also saturated hydrocarbons.
• Alkynes are hydrocarbons that contain carbon-carbon triple bonds
• First member is ethyne popularly known as Acetylene.
• Their general formula is CnH2n–2
• Triply bonded carbon atoms are sp hybridized
• Bond angle in alkyne is 180o.
• Shape is linear
• Bond length order: C–C (154PM) > C = C (133PM) > C ≡ C (120 PM)
• Bond strength order C≡C (823 KJ mol–1)> C = C (681K J mol–1)> C–C (348 KJ mol–1)

General Methods of Preparation of Alkynes

(1) Preparation of Ethyne or Acetylene :
(i) From Calcium carbide [Industrial method] : Acetylene is prepared on industrial scale by treating
calcium carbide with water.
CaC2 + 2H2O → CH ≡ CH + Ca(OH)2

(ii) From Haloform [CHI3, CHCI3] : Pure acetylene is obtained when iodoform or chloroform is heated with
Silver powder.
∆
CHI3 + 6Ag + I3CH → CH ≡ CH + 6AgI

(iii) By Kolbe's Electrolysis : By the electrolysis of aqueous Solution of sodium or potassium fumarate or
maleate, acetylene is formed at anode.
CH– COONa CH
Electrolysis +2CO2
CH– COONa(aq) CH

(2) From dihalides (by dehydrohalogenation):

H X H
⊕Θ Θ
KOH(alco.) NaNH2
(geminal) – C – C – Base –C=C– Base
–C≡C–
–HX –HX
H X X

H H H
⊕Θ Θ
KOH(alco.) NaNH2
(Vicinal) – C – C – –C=C– Base
–C≡C–
Base
–HX –HX
X X X

(3) From tetrahalides (by dehalogenation)

X X
R – C – C – H 2Zn R – C ≡ CH+2ZnX2
∆
X X

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(4) Preparation of Higher Alkynes by Grignard Reagent : By this method lower alkyne is converted into
higher alkyne

δ— δ+ δ— δ+ Br
CH ≡ C – H + CH3 – Mg–Br R–Ι
CH ≡ C–MgBr + CH4 CH ≡ C–R + Mg
Ι
Br
R — C ≡ CH + CH3Mg — Br R–C≡C–MgBr + CH4 R’–Ι
R’ — C ≡ C — R + Mg
Ι
Physical Properties of Alkynes
• First three members [C2 to C4] are gases, from C5 to C12 are liquid and higher ones are solid.
• Alkynes are immiscible with H2O but soluble in organic solvents like CCl4, benzene, and ether.
• B.P., M.P. and densities of alkynes are comparatively more than alkenes and alkanes due to more
polarisation, and increase with increase in molar mass.
1
B.P. ∝ mol.wt. ∝
number of sidechains
• Terminal alkynes are acidic in nature. It is due to greater electronegativity of sp hybridised 'C'.

Chemical Properties of Alkynes

Alkynes show acidic nature, addition reaction and polymerisation reaction.
(1) Addition reaction : Due to presence of loosely bonded π electrons, alkynes like alkenes, undergo
electrophilic addition reaction.
Alkynes are less reactive than alkenes towards electrophilic addition reactions, it is due to less stable
vinylic carbocation formed in r.d.s of electrophilic addition reaction of alkynes.

(2) Acidic character of alkynes : The hydrogen atom attached to the triply bonded carbon can be removed
by a strong base and hence acetylene and terminal alkynes show acidic behaviour

Explanation : The s-character in various types of C–H bonds is as –

Alkyne Alkene Alkane
H
–C–H
≡ C–H =C–H
H
H
S-Character 50% 33% 25%

Due to maximum % S character (50%), the sp hybridised orbitals of carbon atom in alkyne have highest
electronegativity. Hence these attract the shared pair of electron of C–H bond of alkyne to a greatest
extent, So H atom can be liberated more easily in alkynes as compared to alkenes and alkanes.
(1) Addition Reaction:
(a) Addition of Hydrogen:
(i) Alkynes reacts with hydrogen in presence of a catalyst like of Pt, Pd or Ni to give corresponding alkanes
with H2
Ni,H Ni,H
R–C ≡ CH 
∆
2
→ R–CH = CH2 
∆
2
→ R–CH2–CH3

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In presence of Lindlar's catalyst [Pd/CaCO3 + quinoline] alkynes give cis-alkene
Lindlar's catalyst R R’ (Cis)
R–C ≡ C–R'  H2
→ C=C
H H
In presence of Na/liq. NH3, alkynes give mainly trans-alkene. (Birch Reduction)
R H (Trans)
Na/NH3 C=C
R–C ≡ C–R'  → H
H2
R’
(Major)
(b) Electrophilic Addition:
E Nu E
δ⊕ δΘ | Θ | |
⊕
Nu
−C ≡ C − + E − Nu → − C = C −  → −C = C −
(Vinylic carbocation)
Addition product formed depends on stability of vinylic carbocation. Addition in unsymmetrical alkynes
takes place according to Markovnikov’s rule.
(i) Addition of Halogens : Alkynes react with CI2 or Br2 in presence of CCl4 to form di and tetra halo
derivatives.
Br Br Br
Br2
Ex. R—C≡CH  Ccl
→ R – C = CH 
Br2
→ R– C – C– H
4 Ccl4
Br Br Br
• Reddish orange colour of the solution of bromine in CCl4 is decolorised. This is used as a test for
unsaturation
(ii) Addition of Hydrogen Halides (H–X) : Addition takes place according to Markovnikov's Rule.
Reactivity order of H–X : HI > HBr > HCI > HF
Br Br
H−Br
Ex. R—C ≡ C—H  → R – C = CH2 
H−Br
→ R– C –CH3
Br
(Gem dihalide)
(iii) Addition of water (Hydration) : Addition of water to alkynes is carried out in the presence of dil.
sulphuric acid and mercuric sulphate. This is known as Kucherov’s reaction
O
OH Tautomerizes
H2O R – C – CH3
Ex. R – C ≡ CH 
HgSO ,H SO
→ R – C = CH2
4 2 4 (Carbonyl Compound)
H2O 
CH3— C ≡ CH 
HgSO4 ,H2SO4
→ CH3 — C = CH2  CH3—C—CH3
O—H O
(enol) (ketone)

C ≡ CH HO —C = CH2 O =C – CH3

H2O

HgSO ,H SO
→
4 2 4

Acetophenone
(2) Acidic Character of Terminal alkynes
In terminal alkyne or acetylene, the H which is attached with sp hybridised carbon is called acidic or
active H.
So Acetylene and Terminal alkynes show acidic behavior when reacted with strong base.
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Na
R — C ≡ CNa +1/2H2
δΘ δ⊕
R—C ≡ C—H (Sodium alkynide white PPt.)
Sp(acidic) NaH
R — C ≡ CNa + H2
(White PPt)
NaNH2
R — C ≡ CNa + NH3
(White PPt)
AgNO3+ NH4OH
Tollen’s reagent R — C ≡ CAg
(White PPt)
Cu2Cl2+NH4OH
R — C ≡ C Cu
(Red PPt)
Used of the reaction:
• The reaction can be used to differentiate terminal and Non terminal alkynes
• The reaction can be used to convert terminal alkynes into higher alkynes

(3) Oxidation Reactions:

(a) Combustion :
3n–1
CnH2n−2 + O2 
→ nCO2 + (n–1) H2O + Heat
2
2HC ≡ CH + 5O2 
→ 4CO2 +2H2O + 312 K.cal
The combustion of acetylene is used for welding and cutting of metals in which oxy-acetylene flame
having high temp (3000°C) is produced.

(b) Oxidation with Acidic KMnO4 /K2Cr2O7:

In presence of acidic KMnO4/ K2Cr2O7, alkynes are oxidised to monocarboxylic acids.
KMnO4/H+ [O]
R–C≡C–R'򨕍򨕍 R–C–C–R' RCOOH+R'COOH
O O
CH CHO
KMnO4/H+ [O] [O]
2HCOOH CO2 + H2O
CH CHO
glyoxal
KMnO4/H+ [O]
CH3–C≡CH+ CH3–C= O CH3COOH+HCOOH
[O]
CHO CO2+H2O
(c) Ozonolysis :
H2O/Zn
R—C — C—R
(i) O3
R—C ≡ C—R O O
LiAlH4
Or NaBH4 R—CH — CH—R
OH OH
+H2O2
H2O R—C — C—R  → R—C—OH + HO—C—R
O O O O

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 CHEMISTRY Hydrocarbons
(i) O3
(ii) H2O
CH3 — C—C—H + H2O2 → CH3—C—OH + HCOOH
Ex. CH3—C ≡ CH O O O [O]
(i) O3
CO2+H2O
(ii) H2O/Zn CH3 — C— C —H + ZnO
O O
(4) Polymerisation Reactions :
Alkyne mainly shows addition polymerisation reactions.
(a) Dimerisation :
Two mole acetylene reacts with Cu2Cl2 & NH4Cl and forms butenyne
Cu2Cl2 +NH4Cl
HC ≡ CH + H—C ≡ C — H → CH2= CH — C ≡ CH
(Butenyne)
(b) Trimerisation :
If acetylene is passed through red hot iron or Cu tube, then polymerization will occur in which three
molecules polymerise to form Benzene.
CH CH
CH CH CH
CH →
Red hot Fe/Cu
873K
CH CH CH
CH
CH CH
Benzene
Note: Propyne gives mesitylene
CH3

H3C CH3
(mesitylene)
Test for Alkynes :
• Decolourization of Br2 in CCl4 Solution.
• Decolourisation of 1% alkaline KMnO4 Solution.
• Terminal alkynes give white ppt. with ammoniacal AgNO3 and red ppt with ammoniacal cuprous
chloride Solution.

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Reaction Chart for Alkynes

GMP GR
(i) alc. KOH
(1) CH2Br–CH2Br H2 H2,Ni
(ii) NaNH2 (1) C2H4 C2H6
Ni,∆ ∆
(i) alc. KOH
(2) CH3—CHBr2
(ii) NaNH2 X2
(2) X2CH–CHX2
Ag powder
(3) CHCl3 ∆ CH ≡ CH HBr
(3) Peroxide
CH2BrCH2Br
Zn dust
(4) CHBr2 – CHBr2 HBr
∆ (4) CH2–CHBr2
Kolbe’s
(5) HC–COONa HOCl
electrolysis Cl2CHCHO
(5) or H2O/Cl2
HC–COONa(aq)
Hg+2, dil H2SO4
(6) CaC2 H 2O (6) (Kucherov’a reaction)
CH3CHO

(i) BH3/THF
(7) CH3–C ≡ CH
(i) Na(ii) R–X (7) Θ CH3–CHO
(ii) H2O2/OH(aq)
CH3–C ≡ C–R
(i) CH3MgI(ii) R–X NaNH2
(8) CH3–C ≡ CH (8) NaC ≡ CNa

AgNO3 + NH4OH
(9) (Tollen’s Reagent)
AgC ≡ CAg

Cu2Cl2 +NH4OH
(10) CuC ≡ CCu

Combustion O2
(11) CO2 + H2O

Bayer Reagent CHO

(12)
CHO
H2O/Zn HC—C–H
O O O
O3 H 2O H 2O 2
(13) H–C–C–H HC—C–H
Ozonolysis
O–O O O
H–COOH
H 2O 2

CO2+H2O
Trimerization
(14) (Red hot iron tube)

Cu2Cl2+NH4Cl
(15) CH2 = CH–C ≡ CH

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Exercise 1.3
1. The number of carbonyl carbon in the 7. Which of the following compounds will
product X is yield 2, 3-pentanedione on reductive
Propyne  i) O3
→X ozonolysis?
ii) Zn/H O
2
(1) 2, 3-Pentadiene (2) 2-Pentene
(1) 2 (2) 4
(3) Cyclopentene (4) 2-Pentyne
(3) 6 (4) 8
8. An alkyne which gives two moles of acetic
2. The product obtained on acidic hydration
acid on ozonolysis is :
(HgSO4 + H2SO4) of 1-butyne would be : (1) 1-Butyne
(1) CH3–CH2–C–CH3 (2) 2-Butyne
O (3) Methyl acetylene
(2) CH3–CH2–CH2–CHO (4) 3-methyl-1-butyne
(3) CH3–CH2–CHO + HCHO
(4) CH3–CH2–COOH + HCOOH Cl
NaNH2 CH3Br
9. CH3 – CH2 – CH  → X  →Y
Cl Excess
3. Propyne and propene can be distinguished H2 +Pd
 → Z ; Z is :-
by : CaCO3

(1) Conc. H2SO4 (2) Br2 in CCl4 (1) (2)

(3) Dil. KMnO4 (4) AgNO3 in NH4OH
(3) (4)
4. Hydrogenation of the above compound in
the presence of poisoned palladium 10. CH3–C ≡ CH + 2HBr → Product?
catalyst gives :
(1) CH3 – CH – CH2
Me Me
Br Br
H
H Br
Me (2) CH3 – CH – CH2
H
Br
(1) An optically active compound
(3) CH2 – CH2 – CH2
(2) An optically inactive compound
(3) A racemic mixture Br Br
(4) A diastereomeric mixture (4) CH3 – C – CH3
Br Br
5. What is the chief product of reaction
between 2-chloro but-2-ene and NaNH2 11. Acetylene can be obtained by the reaction?
Electrolysis
(1) 1,2-butadiene (1) HCOOK →
Δ
(2) 1,3-butadiene (2) CHI3 + Ag →
(3) 2-butyne Conc.H2SO4
(3) CH3CH2OH 
443K
→
(4) 1-butyne
(4) Be2C + H2O →
6 Acetylene can be prepared from
(1) Potassium fumarate 12. Acetylene may be prepared using Kolbe's
(2) Calcium carbide electrolytic method employing :
(3) Ethylene bromide (1) Pot. acetate (2) Pot. succinate
(4) All (3) Pot. fumarate (4) None of these

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13. Which is/are true reactions ? Reagent
17. R–CH2–CCl2–R  → R–C ≡ C–R. The
(1) Al4C3 + H2O → CH4 reagent is
(2) CaC2 + H2O → C2H2 (1) Na (2) KOH(aq)
(3) Me3C–H + KMnO4 → Me3C–OH (3) NaNH2 (4) Zn
(4) All

18.
BH3 THF
B ←
H O ,OH
 CH3 − C ≡ CH 
HgSO4 /H2SO4
dil.
→A ,
14. For preparing methyl acetylene, we take : 2 2

(1) CH3–C–COOK A and B are respectively

CHCOOK O
||
(2) CHCOOK (1) CH3CH2CHO,CH3 − C− CH3
CHCOOK O
(3) CH3–CH2–CCOOK ||
CHCOOK (2) CH3 − C− CH3 ,CH3CH2CHO
(4) All of these (3) CH3CH2CHO (both)
O
||
15. Propyne is formed by heating
(4) CH3 − C− CH3 (both)
1, 2-dibromopropane with :
(1) C2H5ONa  R − C ≡ C − R 
Lindlar
B ← Na/liq.NH3
→ A,
19.
(2) Alc. KOH and sodamide
A and B are
(3) Aqueous alkali
(1) A is trans, B is cis
(4) Sodalime
(2) A and B both are cis
(3) A and B both are trans
16. An alkyne C7H12 on reaction with Alk.
(4) A is cis, B is trans
KMnO4 and subsequent acidification with
HCl yields a mixture of (CH3)2—CH— 20. Which is the most suitable reagent among
COOH+CH3—CH2COOH. The alkyne is the following to distinguish compound 'C'
(1) 3—Hexyne from the rest of the compounds?
(2) 2—methyl—3—hexyne (A) CH3C ≡ CCH3 (B) CH3CH2–CH2CH3
(3) 3—methyl—2—hexyne (C) CH3CH2C ≡ CH (D) CH3CH = CH2
(4) 2—methyl—1—hexyne (1) Br2 in CCl4 (2) Br2 in H2O
(3) Cold KMnO4 (4) Tollen's reagent

Aromatic Hydrocarbon
• Aromatic hydrocarbon are also called arenes.
• Characteristic reaction of arenes is Electrophilic substitution reaction (ESR).
• Arenes are cyclic unsaturated compounds but do not give test of unsaturation with Br2/CCl4 or cold
KMnO4.
• Main source of Arenes is coal tar.
• They have higher precentage of carbon so burn with sooty flame.
BENZENE
[i] Discoverer : Michael Faraday (1825) [ii] C–hybridisation : sp2
[iii] Geometry – Hexagonal [iv] Bond angle : 120o
[v] C–C Bond length : 1.39Å [vi] C–H Bond length : 1.09 Å
[vii] Resonance in bezene
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≡ (Resonance hybrid)

Kekule
Structure
There are two equal possibilities of forming three π bond by overlapping of p orbitals
2 3 2 3
1 4
1 4

6
6 5
or
or

Overlapping Of p–orbital Overlapping Of p–orbital

C1—C2,C3—C4 and C5—C6 C2—C3,C4—C5 and C1—C6
Delocalized π-electrons : There is equal probability for the p-orbital of each carbon atom to overlap
with the p-orbital of adjacent carbon atoms. the six π-electron are thus delocalized and can move freely
about the six carbon nuclei instead of any two kekule structure. the delocalized π eΘ cloud is attracted
more strongly by the nuclei of the carbon atoms than the electron cloud localized between two carbons.
Therefore presence of delocalized π eΘ makes the benzene more stable than the hypothetical cyclo
hexatriene

• •
• •
• •

Aromatic hydrocarbons

CH3 CH3 CH3 CH3 CH=CH2 CH(CH3)2

CH3

CH3
Benzene Toluene O-xylene m-xylene CH3 Styrene Cumene
p-xylene

CH3

Biphenyl anthracene Phenanthrene H3 C CH3

naphthalene
(phenylbenzene)
1,3,5-trimethylbenzene
(mesitylene)

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General Methods of Preparation of Aromatic Hydrocarbon

(i) From Alkyne (By cyclic polymerisation) When acetylene is passed through red hot cu or Fe tube cyclic
polymerisation takes place and benzene is formed.

Redhot irontube
3CH ≡ CH  →

(ii) From Carboxylic Acids (By decarboxylation) :

COOH H
NaOH+CaO
+ CO2
∆

CH3 CH3
COOH
NaOH+CaO
+ CO2
∆

(iii) From Phenol (By reduction) :

OH H
∆
+ Zn → + ZnO

(iv) From Alkanes (By aromatization or reforming)

n-alkanes having six or more carbon atoms on heating to 773K at 10-20 atmospheric pressure in the
presence of oxides of V, Mo or Cr supported over alumina get dehydrogenated and cyclised to benzene
and its homologues. This is known as aromatization or reforming
Cr2O3 /Al2O3
n - Hexane 773K
→ Benzene + 4H2
Cr2O3 /Al2O3
n - Heptane 773K
→ Toluene + 4H2

(v) From Carbonyl Compounds (By reduction) :

CHO CH3
(1) NH2–NH2
[Wolf kishner reduction]
(2) HO/ ∆

O
C–CH3 CH2CH3
Zn–Hg
[Clemmensen’s reduction]
Conc.HCl

(vi) By Diazonium salts :

⊕
N2Cl H
H3PO2+H2O
Or C2H5OH

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 CHEMISTRY Hydrocarbons
(vii) By Grignard reagent :
MgCl H
H—OH

(viii) By Wurtz Fitting Reaction :

Cl CH3
Na/ether
+ CH3–Cl ∆

Physical Properties of Aromatic Hydrocarbon

• Benzene is non polar molecule
• Benzene is colourless liquid [B.P. is 80°C]
• Benzene is insoluble in water and density less than water
• Benzene is used as a solvent and it is soluble in organic solvents.
• It is highly inflammable and burns with sooty flame.

Chemical Properties of Aromatic Hydrocarbon

Benzene show following types of reaction -
(A) Addition reactions (under special condition)
(B) Electrophilic substitution reactions (Characteristic reaction )
(C) Oxidation reactions(under special condition)
(A) Addition Reaction :
(i) Addition of H2 :
Reaction completes under vigorous conditions i.e at high temperature and /or pressure in the presence
of Ni catalyst
Ni
+ 3H2 ∆

Benzene Cyclohexane
(ii) Addition of X2
Cl
Cl Cl
3Cl2/hv
500K
Cl Cl
Aromatic Compound Cl
Alicyclic compound (C6H6Cl6) or Benzene hexachloride (BHC)
or Gammaxane or lindane or 666
*BHC is used as insecticide
(iii) Addition of O3:
Addition of O3 followed by hydrolysis in presences of Zn (ozonolysis)
O
O3 O CH
H2O
∆
HC O CH O Zn 3 HC —CH
O O O
HC O CH O
O
O CH Benzenetriozonide

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Note: Addition of 3 mole Cl2 or 3 mole H2 or 3 mole O3 on benzene show presence of 3 double bonds in
benzene.
(B) Electrophilic Substitution Reaction [ESR] :
• Characteristic reaction of arenes is ESR.
• The reaction involves attack of E⊕ on benzene ring to replace H in the form of H⊕.
• ESR reactions occur via a two steps mechanism.
Step 1. Addition of the electrophile
H H ⊕ H
δ+ H
H E⊕ E E E E
Slow
⊕ ⊕ δ+ δ+
(RDS)
(σ-complex/arenium ion/
wheland intermediate)
Step 2. Loss of a proton to reform the aromatic ring

H Θ
Z E
E Fast
⊕ + H-Z

Note: In general 1st step of the reaction is r.d.s. but in case of Iodination and sulphonation 2nd step (removel
of H⊕) is r.d.s.
Example of ESR:
(i) Halogenation
(ii) Nitration
(iii) Sulphonation
(iv) Friedel craft reaction (F.C.R
(a) Alkylation (b) Acylation
(1) Halogenation: Source of electrophile (E⊕)→
•• ⊕ Θ
(i) X–X+ AlCl3 X+AlCl4
(Lewis Acid)
X2 ⊕ Θ
(ii) X2+Fe→ FeX3 X+FeX4
(Lewis Acid)

H Cl
Anhyd.Alcl3
+ Cl2 →

Cl
Cl Cl
Anhyd AlCl3
+ Cl2 →
(Excess)
Cl Cl
Cl
Hexachlorobenzene
(C6Cl6)
CH3 CH3 CH3
AlCl3
Cl
+ Cl2 
Anhyd
→ +

Cl
Note: CH3 group in toluene is o/p directing and activating group.

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 CHEMISTRY Hydrocarbons
(2) Nitration :
H Conc.HNO + Conc.H SO NO2
3 2 4

(Nitrating mixture)

⊕
Formation of E :
δ⊕ δΘ
· · — NO + H—HSO H ⊕ Θ ⊕
HO
·· 2 4 O — NO2 + HSO4 NO2 + H2O
(Bronsted base)(Bronsted acid) H (Attacking species)
(Nitrenium ion)

CH3 CH3 CH3

NO2
HNO3 /H2SO4
 → +

NO2
(3) Sulphonation :
H Conc.H2SO4
SO3H

Θ ⊕

Mechanism : 2H2SO4  SO3 + HSO4 + H3 O
⊕ SO3H
H HΘ
SO3
Attacking species
SO3

CH3 CH3 CH3

Conc.H2SO4
SO3H
Ex. +

SO3H
Note: Fuming sulphuric acid H2SO4(SO3) acts as a better source for sulphonation.

(4) Friedel Craft’s Reaction [FCR] : Alkylation or acylation of arenes in presence of lewis acid [FeCl3, AlCl3
or ZnCl2 .....] is called as FCR.
(a) Friedel craft alkylation
CH3
AlCl3
Ex. (i) + CH3–Cl

⊕
CH3
Intermediate carbocation is formed in Friedel craft alkylation so rearrangement is possible in presence
of strong Lewis acid like AlCl3.
CH3
(ii) AlCl3 CH CH3
+CH3–CH2CH2–Cl
⊕ Isopropyl benzene
CH3–CH2–CH2
(cumene)
1,2–H
Θ Shilt
⊕
CH3–CH–CH3(E⊕)
(b) Friedel craft acylation

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AlCl3 C–CH3
Ex. (i) + CH3–C–Cl
O
O

⊕
CH3–C (E⊕)
O
COCH3
Anhyd.Alcl3
(ii) + (CH3CO)2 → + CH3COOH
Acetic anhydride

Direction influence of a functional group in monosubstituted benzene:

(1) Ortho/Para Directing Group : Group which directs incoming electrophile to ortho and para position is
called ortho /directing group
• +M/+H effect of these group increases electron density at ortho and pare positions. hence the
substitution take place mainly at these position
• The groups which have net electron releasing effect increase electron density in benzene ring, activates
the benzene ring easily for the attack by an electrophile. So these groups are called activating groups.
These group are:
•• •• •• •• •• •• •• ••
NH2 , − NHR , − NR2 , − OH, − OR , − NHCOCH3 , − SH, − OCOR , −CH3 , −CH2CH3 , −CH ( CH3 )2 .....etc
••
•
•OR OR OR
E ⊕ E
+

E
Note: Although Halogens are deactivating groups (–Ι > + m effect) but they are ortho and para directing.

(2) Meta Directing: Due to –M/–H of groups electron density at ortho and para position is decreased, so
electrophile mainly attack at meta position. So, groups which direct electrophile to meta position are
called meta directing groups.
• The groups which have net electron withdrawing effect decrease electron density in benzene ring and
deactivate the benzene ring for the attack by electrophile are called deactivating groups.
These groups are :
– CHO , –COOH ,–COOR ,–COR ,–CN ,–NO2 ,–SO3H ,–CX3
C–H ⊕
E
C–H
O O
E
Note: (i) M and H effect does not depend on distance so attack at ortho and meta position equally while I-
effect depends on distance .
(ii) M-effect does not affect meta position.

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Relative Reactivity of Substituted Benzene

Ortho para directors Mata directors

Strongly activating Strongly deactivating

·· ·· ·· ·· ⊕ ⊕
–OΘ–NH2,–NR2,–OH,–OR
·· ·· –NH3, –NR3,–NO2,– C ≡ N, – CF3, –CCl3
··
• Moderately activating Moderately deactivating
O O O O O O O O
·· ·· –C–H, –C–,–C–Cl,–C–NH2,–C–OH,–C–OR
–NH – C – CH3, –O–C–CH3
··
• Weakly activating
–CH3, –CH2CH3,
, – CH = CH2

··· ·· · ·· · ···
Note: Halogens (– F
··· CI
·· ·, –Br
·· · , – ··Ι· )are weekly deactivating but they are ortho para directing

Ex. Give reactivity order for electrophilic substitution reaction.

 1 
 Reactivityµ∝ +Ι / +H / +Mµ
µ∝ 
 −Ι / −H / −M 
NO2 C–H
(i) <
O
More –M of –NO2 & more –I less –M of –CHO & less –I

CH3
NH2 CHO
(ii)
+H +M –M
III > II > I > IV

NH2 OH
(iii) CH3 Cl
(More + M due to) (Less + M due to)
+H (–Ι > +M) less EN of N More EN of O

III > IV > I > II

(iv) Cl CHO NH2

(–Ι> + M) (–M) +M (reference)

III > IV > I > II

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(v) CCl3 CH3 NH-R NO2

–H +H +M –M
III > II > I > IV

⊕
(vi) CH3 CH2 CH3 NR3 OH

+H(more) +H(less) –Ι +M
IV > I > II > III

(C) Oxidation reactions :

(i) Combustion :
15
C6H6 + O2 → 6CO2 + 3H2O
2
(ii) Catalytic Oxidation :
OH
300ºC

V2O5
Phenol
O
450ºC CH–C
O
CH–C
O
Maleic anhydride
(iii) Side Chain Oxidation : At least one benzylic -H containing alkyl benzene gives benzoic acid in presence
of strong oxidising agent.

—C—H (Benzylic-H) COOH

⊕
Ex. 
KMnO4/H
∆
→

CH3
CH3 CH2CH3 CH
CH3

, ,

CH2Cl CH2OH CH2CH2CH3 CH2CH2……CH3 CH= CH2

etc
, , ,

Note: t-butyl benzene does not give benzoic acid in presence of H⊕/KMnO4, since t-butyl benzene does
not contain benzylic-H

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Exercise 1.4
1. O-xylene on ozonolysis will give : 5. Correct order of reactivity towards ESR is
(1) CHO O CH3
& CH3
CHO CH3–C–CHO
(i) (ii)
(2) CH3–C=O O OMe
&
CH3–C=O CH3–C–CHO CHO CHO
(3) CH3–C=O CHO
& (iii) (iv)
CH3–C=O CHO
(4) CH3–C=O O CHO NO2
, &
(1) i > iii > ii > iv
CH3–C=O CH3–C–CHO CHO
(2) ii > i > iii > iv
(3) i > iv > iii > ii
2. In the following sequence of reactions the
(4) iv > ii > iii > i
maximum number of atoms present in
product 'C' in sp3 hybridised state is
6. Consider the following statements :
Red hot CH3Cl(1eq.)
A 
Fe tube
→B →
AlCl3
C I. The —OH group present in phenol is
(A is a lowest molecular weight alkyne) ortho and para-directing.
(1) 1 (2) 3 II. Directive influence of a functional group
(3) 5 (4) 7 in monosubstituted benzene depends on
the nature of the substituent already
3. The correct increasing order of reactivity for present in the benzene ring.
following molecules towards electrophilic III. The —OH group activates the benzene
aromatic substitution. ring for the attack by an electrophile.
OH OH OH IV. Groups such as —
OH
−NH2 , −NHR, −NHCOCH3 ,
−OCH3 , −CH3 , −C2H5 etc, are examples of
OMe Cl activating groups.
NO2
Select the correct option
(I) (II) (III) (IV)
(1) I and II (2) II and III
(1) I < IV < II < III (2) I < IV < III < II (3) II, III and IV (4) All of these
(3) I < III < II < IV (4) I < III < IV < II

4. Increasing order of reactivity towards E.S.R. 7. MgBr + OH → X ,

for following compounds is ? X is:
OCH3 Cl NO2 CH3
(1) O

(2) OH
(I) (II) (III) (IV)
(1) III < II < IV < I (3)
(2) II < III < IV < I
(3) III < II < I < IV (4)
(4) III < I < II < IV

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8. In the sulphonation, acetylation and 13. Chlorobenzene is O, P-directing in
formylation [CO+HCl/AlCl3] of benzene the electrophilic substitution reaction. The
group of effective electrophiles would be directing influence is explained by
+ (1) +M of Ph (2)–Ι of Cl
(1) SO3+ , CH3C ≡ O+ , HC = O
(3)–Ι of Ph (4) + M of Cl
+
(2) SO3 , CH3 —C ≡ O+ , H—C = O
(3) SO3, CH3 —C ≡ O+ , CO + HCl 14. In which reaction, polysubstitution takes
place :
(4) HSO3, CH3 CO, HC=O
COCH3
9. The electrophile involved in chlorination of
benzene is (1) +CH3COCl
(1) Chloride ion (2) Chloronium ion
CH3
(3) Cl2 mlecule (4) Alcl3
(2) +CH3Cl
10. Which of the following is ortho and para
directing. SO3H
(1) –CCl3 (2) –NHCOCH3
(3) +H2SO4
(3) –SO3H (4) –NO2
NO2
11. In the nitration of benzene with
concentrated HNO3 and H2SO4, the attack on H2SO4
(4) +HNO3
ring is made by :
(1) NO2Θ (2) NO2⊕
(3) NO3 Θ (4) HNO3 15. Identify correct statement/s :
(A) -OCH3 and -NHCOCH3 are activating
AlCl3 group
12. +CH3–CH2–CH2– CH2–Cl ’X’
(B) -CN and -OH are meta directing group
Major product X is (C) -CN and -SO3H are meta directing group
(D) Activating groups act as ortho directing
(1) CH2CH–CH3
groups and para
CH3 (E) Halides are activating groups
CH3 Choose the correct answer from the options
(2) CH–CH2.CH3 given below:
(1) (A), (C) and (D) only
(3) CH2 CH2CH2CH3 (2) (A), (B) and (E) only
(3) (A) only
CH3
(4) (A) and (C) only
(4) C–CH3
CH3

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Exercise 2
Θ⊕ 6. Consider the reaction given below.
COONa
Anhyd. AlCl3
CaO + CH3Cl A, here A is-
1. + NaOH ∆
A+Na2CO3 , A is

(1) C6H6 (2) C6H5 — CH3 (1) methylbenzene (2) Halo benzene
(3) C6H5 — ONa (4) C6H5 — OH (3) benzyl chloride (4) P-Chloro toluene

2. On passing vapours of phenol over heated 7. In the reaction,

zinc, dust it gets reduced to
H2SO4/SO3,
(1) toluene (2) aniline X Cl2 Y
AlCl3 Fuming sulphuric acid
(3) benzene (4) phenol
X and Y respectively are
3. For an electrophilic substitution reaction, the (1) C6H5Cl and C6H5SO3H
presence of a halogen atom in the benzene (2) C6H5Cl and C6H5OH
ring (3) C6H4Cl2 and C6H5SO2
(1) deactivates the ring by inductive effect (4) C6H5Cl and C6H5CHO
(2) deactivates the ring by resonance
(3) increases the charge density at ortho 8. Acetophenone (C6H5COCH3) is prepared from
position relative to para-position by reaction of benzene (C6H6) with A in the
resonance presence of anhyd. AlCl3 . Here, A refers to
(4) directs the incoming electrophile to meta (1) CH3COCH3 (2) CH3CHO
-position by increasing the charge density (3) CH3CH2CHO (4) CH3COCl
relative to ortho and para -position
9. Which of the following carbocations is
4. Consider the reaction given below. expected to be most stable?
(323-333)K NO2 NO2
+ Conc. HNO3 + Conc. H2SO4 A+H2O,
⊕
Here, A refers to (1) (2)
H
NO2
NH2 Y ⊕ Y H
(1) (2) NO2
NO2
H ⊕
(3) ⊕ (4) Y
(3) (4) All of these
SO3H Y H

5. In an electrophilic substitution reaction of 10. Benzene hexachloride is obtained by addition

nitrobenzene, the presence of nitro group of X moles of Cl2 with C6H6 in the presence of
(1) activates the ring by resonance effect UV/500 K. The value of X is
(2) activates the ring by inductive effect (1) 2 (2) 4
(3) decreases the charge density at ortho and (3) 3 (4) 8
para -position of the ring relative to meta
-position by resonance 11. Which of the following deactivates the
(4) increases the charge density at meta - benzene ring towards electrophilic
position relative to the ortho and para - substitution and is o/p directing?
positions of the ring by resonance (1) –Cl (2) –OCH3
(3) –CHO (4) All of these
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12. Which of the following set of functional 18. Amongst the following hydrocarbons, the one
groups is meta-directing? having lowest boiling point is-
(1) −NO2 , −NH2 , −COOH, −COOR (1) n-Hexane (2) n-Pentane
(2) −NO2 , −CHO, −SO3H, −COR (3) Isopentane (4) Neopentane
(3) −CN, −CHO, −NHCOCH3 , −COOR
(4) −CN, −NH2 , −NHR, −OCH3 19. Which of the following on reaction with RMgX
does not give RH ?
13. Correct order of electrophilic substitution (1) Ethylamine (2) Dimethylamine
reaction is (3) Trimethylamine (4) Phenol
CH3 OCH3 Cl CHO
20. How many different alkene including
geometrical isomer on catalytic
(A) hydrogenation produces 2-methyl pentane.
(B) (C) (D)
(1) Three (2) Two
(1) A > B > C > D (2) D > B > A > C
(3) Five (4) One
(3) B > A > C > D (4) B > A > D > C

14. Which of the following statements are correct 21. BrCH2 –CH2 –CH2Br reacts with Na in the
regarding Kolbe’s electrolytic process? presence of ether at 100 °C to produce –
I. Alkanes containing even number of C- (1) BrCH2 − CH =CH2
atoms are prepared. (2) CH2= C= CH2
II. Sodium or potassium salts of carboxylic (3) CH2—CH2
acids are taken as substrate.
CH2
III. PH decreases with reaction progress
(4) All of these
IV. H2 gas is liberated at anode
(1) Both I and II (2) Both II and III
(3) Both I and III (4) All of these 22. What is the slow, rate-determining step, in the
acid-catalyzed dehydration of 2-methyl-2-
15. The reactivity of the halogens towards propanol?
methane decreases in the order - ( CH3 )3 COH 
H2SO4
→
(1) F2 > Cl2 > Br2 > I2 (1) Protonation of the alcohol to form an
(2) I2 > Br2 > Cl2 > F2 oxonium ion
(3) F2 > Cl2 > I2 > Br2 (2) Loss of water from the oxonium ion to
(4) Cl2 > F2 > Br2 > I2 form a carbocation.
(3) Loss of a β-hydrogen from the carbocation
16. An aqueous solution containing sodium to form an alkene.
acetate and sodium propionate is (4) The simultaneous loss of a β-hydrogen
electrolyzed. Which of the following alkanes and water from the oxonium ion.
is expected as product -
(1) Ethane (2) Propane 23. Propene when heated with chlorine at about
(3) Butane (4) All of these 500ºC forms majorly :
17. A mixture of ethyl iodide and n-propyl iodide (1) CH2Cl.CH = CH2
is subjected to Wurtz reaction. The (2) CH3CHCl.CH2Cl
hydrocarbon that will not be formed is- (3) CH2Cl.CHClCH2Cl
(1) n–Butane (2) Propyne (4) All the three
(3) n–Pentane (4) n–Hexane

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24. When two alkenes maybe formed by 29. +Br2→A ; A will have configuration
dehydrohalogenation of an alkyl halide, the alkene
which is more substituted is the major or preferred Br Br
product. This generalization is known as : (1) (2)
(1) Markownikoff’s rule
Br Br
(2) Anti-Markownikoff’s rule
(3) both true (4) none is true
(3) Saytzeff rule
(4) Hofmann’s rule
30. Which is most easily dehydrohalogenated –
25. Which is correct statement for P & Q
Cl Cl Cl
HCl (P)
RO–OR
I II III
Cl
HBr (1) I
(Q)
RO–OR Br (2) II
(1) P is product of ionic reaction & Q is (3) III
product of radical reaction. (4) all with same rate
(2) P & Q both are product of ionic reaction.
(3) P & Q both are product of free radical Br
reaction 31. CH3 alc. KOH
Product.
(4) P is free radical reaction & Q is ionic CH3
reaction.
The product can be—
26. The compound formed when 2-butene is CH3 CH3
treated with hot alk KMnO4 is - (1) CH3 (2) CH3
(1) Acetaldehyde (2) Acetic acid
(3) CH2OH ⋅ CH2OH (4) CH3CH2COCH3 CH3
CH3
CH3
27. Which of the following do not give Allyl (3) (4)
halide ? CH3
475K
(1) CH=
2 CH − CH3 + SO2Cl2 
traces of peroxide
→
32. Consider the following reaction
(2) CH2 =CH − CH3 + Cl2 400ºC −500ºC CH3
 → CH3 CH3
CH3–C–CH2CH3 base C=C
O  → CH3 H
Br (I)
(3) CH3–CH=CH2 + N–Br→ +

O CH2CH3
N–bromo Succinimide (NBS) CH2=C
CH3
(4) CH3 = Peroxide
− CH CH2 + HBr  → (II)
Which of the following base will give the best
28. The simplest branched chain alkene cannot be yield of the alkene II as the major product -
prepared by heating which of the following (1) CH3O−
with ethanolic potassium hydroxide -
(1) t-Butyl halide (2) C2H5O−
(2) s-Butyl halide (3) ( CH3 )3 CO−
(3) Isobutyl halide
(4) ( C2H5 )3 CO−
(4) 2-Halo-2-methylpropane
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33. HCl does not form Anti-Markownikoff’s 39. The catalyst required for the reaction :
product with propene, because - Catalyst
HC ≡ CH + dil.H2SO4 → CH3CHO is
(1) HCl is a polar covalent molecule
(1) HgSO4 (2) Pd
(2) Dipole-Dipole attraction exists between
(3) Pt (4) AlCl3
HCl molecules
(3) Formation of Cl• is energetically 40. Which of the following statements is correct?
unfavourable (1) Alkynes are more reactive than alkenes
(4) Chlorine has the highest electron affinity towards halogen additions
34. Choose the correct statement - (2) Alkynes are less reactive than alkenes
(1) Addition of HCl to vinyl chloride yield towards halogen additions.
vicinal dichloride (3) Both alkynes and alkenes are equally
reactive towards halogen additions
(2) There is difference between peroxide
(4) Primary vinylic cations (RC = CH+ ) is more
effect and Kharash effect
stable than secondary vinylic cation ( RC+
(3) Peroxide is a ready source of free radical
= CH2)
in the Markownikoff’s reaction
(4) Propene with HCl in presence of peroxide 41. A gas which reacts with cold aqueous alkaline
gives isopropyl chloride KMnO4 solution but does not give precipitate
with ammoniacal Cu2Cl2 solution is:
35. Diborane reacts with terminal alkenes to form
(1) ethylene (2) methane
trialkylboranes. These react with alkaline
(3) ethane (4) acetylene
hydrogen peroxide to form -
(1) Secondary alcohols 42. Addition of 2 mol of HCl to 1-butyne would
(2) Tertiary alcohols. yield :
(3) Isobutyl alcohol (1) CH3CH2CH2CHCl2
(4) Primary alcohols (2) CH3CH2CCl2CH3
(3) CH3CH2CHClCH2Cl
36. 2-Butyne and 1-Butyne show resemblance in (4) CH3CH2CH = CHCl
all except –
(1) Both decolourise alkaline KMnO4 43 Acetylene gives-
(2) Both turn bromine water colourless (1) White ppt with AgNO3 and red ppt with
(3) Both undergo addition reaction Cu2Cl2
(4) Both from white precipitate with Tollen’s (2) White ppt with Cu2Cl2 and red ppt with
reagent AgNO3
(3) White ppt with both
37. Consider the reaction : (4) Red ppt with both
CH ≡ CH(g) + 2HCl(g) → product. 44. Which of the following reductions of an alkyne
The product of this reaction is : is not correct :
2H2
(1) CH3CHCl2 (g) (2) CH2 = CHCl(g) (1) CH3CH2C ≡ CCH3 →Pt
CH3CH2CH2CH2CH3
H2
(3) CHCl = CHCl(g) (4) CH2ClCH2Cl(g) (2) CH3CH2C ≡ CCH3 
Pd −BaSO
→
4

CH3CH2 CH3
38. The end product of the following sequence is : C=C
H2O H2SO4 H H
CaC2 → A 
HgSO
→B Na
4
(3) CH3CH2C ≡ CCH3 
liq.NH
→
(1) ethanol 3

(2) ethyl hydrogen sulphate CH3CH2 CH3

C=C
(3) acetaldehyde H H
(4) ethylene glycol (4) All of the above are correct

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45. Which reagents would be best to perform the 51. Which one of these is incorrect for benzene?
following reaction ? (1) It forms only one type of monosubstituted
product
H (2) There are three C-C single bonds and
O
(1) 1) BH3, 2) NaOH, H2O2 three C–C double bonds
(2) HgSO4 , H2O, H2SO4 (3) The heat of hydrogenation of benzene is
(3) 1) NaNH2 , 2) CH3 I, 3) HgSO4 , H2SO4 less than the theoretical value
(4) 1) (CH3COO)2Hg, H2O 2) NaBH4 (4) The bond-angle between c–c–c bond is
120°
46. Indicate the expected structure of the organic
product when ethyl magnesium bromide is H2O Cl2 Na
52. A → B  → C  → C2H6
treated with heavy water (D2O) - Ether

(1) C2H5 − C2H5 (2) C2H5OD In the above reaction sequence A and B
respectively are-
(3) C2H6 (4) C2H5D
(1) CaC2, C2H2 (2) Al4C3, C2H6
47. C3H8 + Cl2 Light
 → C3H7Cl + HCl is an example (3) Al4C3, CH4 (4) CaC2, CH4
of which of the following types of reactions -
(1) Substitution (2) Elimination H2 /Ni
(3) Addition (4) Rearrangement 53.  →A
D D
48. Major product of the reaction is A is
D D
Conc. H2SO4
(1) CH3–(CH2)4–CH3 (2) H
Product
∆ H
D H
(1) (2)
(3) H (4)
D H H
(3) (4) NH3 /NaNH2 Et Br
54. MeCH2C ≡ CH  → A  →B
, A and B respectively are-
(1) MeCH2C ≡ C – Na, MeCH2 C ≡ C–Et
(1) O3 Zn−Hg (2) MeCH2CH=CH2, MeCH2 —CHEt –CH3
49.  → X  → Y; Y is
(2) H2O/Zn HCl (3) MeCH2CH=CHNH2, MeCH2CH=CH–NHBr
(4) MeCH2C≡C–NH2, MeC≡C–NH–Br
(1) OH (2) H
C ≡ CH
55. Hydration of in
(3) (4)
Presence of dil. H2SO4/HgSO4 gives-
COCH3
50. Consider the following reaction sequence. (1)
CH MgBr CH CH Br
CH3C ≡ CH 3
→ I  3 2 → II
CH2CHO
The final product (II) formed is : (2)
(1) CH3C ≡ CCH3
COCH3
(2) CH3C ≡ CCH2CH3 (3)
(3) CH3C ≡ CMgBr
CH3 CH2CHO
(4)
(4) CH3–C=CHCH2CH3

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56. In which of the following compound the 59. Which one of the following carbonyl
electrophile attack on O– and P– positions compounds cannot be prepared by addition of
NO2 Cl water on an alkyne in the presence of HgSO4
(1) (2)
and H2SO4 ?
COOH O
(1) CH –CH –C–H
CCl3 3 2
(3) (4) O
(2) C–CH3
57. Arrange the following alkanols A, B and C in O
order of their reactivelty towards acid (3) CH –C–H
3
catalysed dehydration:-
(A) CH3 − CH − CH2 − CH2 − OH O
(4) CH –C–CH CH
| 3 2 3
CH3
OH ‘A’
| KMnO4
(major product)
(B) CH3 − C − CH2 − CH3 60. H2SO4,∆
|
CH3 ‘B’
KMnO4
(major product)
(C) CH3 − CH − CH − CH3 H2O, 273K

| |
CH3 OH For above chemical reactions, identify the
(1) A > B > C (2) B > A > C correct statement from the following:
(3) B > C > A (4) C > B > A (1) Compound’A’ is dicarboxylic acid and

Br2 /hv KCN compound ‘B’ is diol

58. PhC2H5  → A → B; B is
(2) Compound ‘A’ is diol and compound ‘B’ is
CN
CH2–CH3 dicarboxylic acid
(1) PhCH2—CH2—CN (2)
(3) Both compound ‘A’ and compound ‘B’ are
CN diols
CH2–CH CN CH–CH3 (4) Both compound ‘A’ and compound ‘B’ are
CN
(3) (4) dicarboxylic acids

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Exercise 3
1. Match the column : 4. Match the column-I with column-II
Column I Column II Column I Column II
(Reactants) (Product)
(Reaction) (Product)
(A) CaC2 + H2O (p) CH3–CH3
(B) CH3–CH2MgBr+H2O (q) H2 (A) 2CH3CH2Br (p)
(C) HC ≡ CH + Na (r) CH4 2Na
 →
dryether
(D) CH3COOH + NaOH + (s) C2H2
CaO (B) CH3COONa+ (q) CH3CH2CH2CH3
(1) A → (q); B → (p); C → (r); D → (s) CaO+NaOH
 →
(2) A → (p); B → (q); C → (s); D → (r) ∆

(3) A → (s); B → (p); C → (q); D → (r) (C) CH3 ( CH2 )4 CH3 (r) 2CH3OH
(4) A → (p); B → (q); C → (r); D → (s)
V2O5

773K,
→
2. Match the column : 10–20 atm
Column I Column II
(Reaction) (Product) (D 2CH4 + O2 (s) CH4
(A) CH3 (p) CH2–Cl
Cu/523K
 →
100atm
AlCl3
+Cl2 (1) A → (q); B → (p); C → (r); D → (s)
(2) A → (q); B → (s); C → (p); D → (r)
(B) CH3 (q) CH3
(3) A → (q); B → (p); C → (s); D → (r)
hν
+Cl2 H3C (4) A → (p); B → (p); C → (r); D → (s)
(C) CH3 (r) CH3
5. Match the column-l with Column II
AlCl3
+CH3 –Cl Cl Column I Column II
(1) A → (q); B → (p); C → (r) (Reaction) (Product)
(2) A → (r); B → (p); C → (q) (A) CH4 + O2 (p) ( CH3 )3 C − OH
(3) A → (q); B → (r); C → (p) Mo2O3
 →
(4) A → (p); B → (q); C → (r) Δ

(B) 2CH3CH3 + 3O2 (q) HCHO + H2O

3. Match the column-I with column-II
( CH COO ) Mn
Column I Column II 
3
Δ
2
→
(A) Lindlar's (p Na / liq.NH3
catalyst
(C) ( CH3 )3 CH (r) CO + 3H2
KMnO4
(B) Birch catalyst (q H2 /Pd − CaCO3  →
Oxidation

(C) Baeyer's (r) HBr / (PhCO)2 O2 (D) CH4 + H2O (s) 2CH3COOH+2H2O
Reagent Ni
 →
∆
(D Kharasch (s) Dil.KMnO4 / 273K
effect (1) A → (q); B → (p); C → (r); D → (s)
(1) A → (q); B → (p); C → (r); D → (s) (2) A → (r); B → (s); C → (p); D → (r)
(2) A → (p); B → (q); C → (s); D → (r) (3) A → (q); B → (s); C → (p); D → (r)
(3) A → (q); B → (p); C → (s); D → (r)
(4) A → (p); B → (q); C → (r); D → (s)
(4) A → (p); B → (q); C → (r); D → (s)

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6. Match the column-I with column-II 9. Match the column-I with column-II
Column I Column II Column I Column II
(Reaction) (Type of reaction) (A) CH3 − CH =CH2 (p) Catalytic
(A HC ≡ CH  H2
→ (p Pyrolysis Hydrogenation
Pd H⊕
H2 O
(B Anhydrou (q Catalytic
AlCl3+HCl CH3 – CH–CH3
hydrogenat
ion OH
(C 773K (r) Free radical (B) CH3 − CH =
CH2 (q) Acidic
substitution HBr
 → Hydration
R2O2
(D Cl 2
CH3CH3  → (s Isomerization
hv
CH3 − CH2
(1) A → (q); B → (s); C → (p); D → (r) −CH2 − Br
(2) A → (p); B → (q); C → (s); D → (r) (C) CH3 − CH =
CH2 (r) Halogenation
(3) A → (q); B → (p); C → (s); D → (r)
(4) A → (p); B → (q); C → (r); D → (s)
Br2
CCl4
7. Match the column-I with column-II Br
Column I Column II
CH3 – CH–CH2–Br
(A) CH3 − C ≡ C − CH3 (p)
H2 /Pd
CH3 CH3 (D) CH2 = CH2 (s) Hydrohalogenation
 → C=C
H2
H H 
Pd
→ CH3 − CH3
(B) CH3 − C ≡ C − CH3 (q)  ⊕
CH3 − C ≡ CNa (1) A → (s); B → (r); C → (q); D → (p)
Na/liq⋅NH3
 → (2) A → (q); B → (s); C → (r); D → (p)
(C) CH3 − C ≡ C − CH3 (r) CH3CH2CH2CH3 (3) A → (r); B → (s); C → (p); D → (q)
H2 /Pd −C

Quinoline
→ (4) A → (p); B → (q); C → (r); D → (s)
(D) CH3 − C ≡ CH (s)
CH3 H
Na/liq⋅NH3
→ C=C 10. Match List-I with List-II and select the
H CH3
correct answer.
(1) A → (s); B → (r); C → (q); D → (p)
List-I (Reactions)
(2) A → (p); B → (s); C → (r); D → (q)
(3) A → (r); B → (s); C → (p); D → (q) (A) CH3–CH=CH2 → CH3–CHBr–CH3
(4) A → (p); B → (q); C → (r); D → (s) (B) CH3–CH=CH2 → CH3–CH2–CH2Br
(C) CH3–CH=CH2 → BrCH2–CH=CH2
8. Match the column-I with column-II
(D) CH3–CH=CH2 → CH3–CHBr–CH2Br
Column I Column II
(Compound) (Boiling
List-II (Reagents)
point)
(A) 2-Methyl propane (p) 309.1 K (P) HBr
(B) Pentane (q) 282.5 K (Q) Br2/CCl4
(C) 2-Methyl butane (r) 261 K (R) HBr/Peroxide
(D) 2,2- (s) 300.9 K (S) NBS
Dimethylpropane
(1) A-P, B-R, C-S, D-Q
(1) A → (r); B → (p); C → (s); D → (q)
(2) A → (q); B → (r); C → (s); D → (p) (2) A-R, B-P, C-Q, D-S
(3) A → (r); B → (s); C → (q); D → (p) (3) A-P, B-S, C-R, D-Q
(4) A → (s); B → (r); C → (q); D → (p) (4) A-Q, B-P, C-S, D-R

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11. Statement-I : Alkenes decolourise KMnO4 14. Statement-I : Dihydrogen gas adds to
solution alkenes and alkynes in the presence of finely
Statement-II: Alkenes are unsaturated divided catalyst like platinum, palladium or
hydrocarbons. Nickel to form alkanes.
In the light of the above statements, choose Statement-II : Metals adsorb dihydrogen gas
the most appropriate answer from the
on their surfaces. and activate the hydrogen
options given below:
hydrogen bond. In the light of the above
(1) Statement-I and Statement-II are
correct, statements, choose the most appropriate
(2) Both Statement-I and II are incorrect. answer from the options given below:
(3) Statement-I is correct but Statement-II is (1) Both statements-I and II are correct.
incorrect. (2) Both Statement-I and II are incorrect
(4) Statement-I is incorrect but Statement-II is (3) Statement-I is correct but Statement-II
correct. is incorrect.
(4) Statement-I is incorrect but Statement-II
12. Statement-I : Ozonolysis of alkenes involves is correct.
the addition of ozone molecule to alkene to
form ozonide and then cleavage of the
ozonide by Zn- H2O to smaller molecules. 15. Statement-I : Sodamide reacts with Ethyne
Statement-II : Ozonolysis reaction is highly to form sodium acetylide with liberation of
useful in detecting the position of double hydrogen gas.
bond in alkenes or in other unsaturated Statement-II : Acetylene is more acidic than
compounds. In the light of the above ethene . In the light of the above statements,
statements, choose the most appropriate choose the most appropriate answer from
answer from the options given below: the options given below:
(1) Both Statement-I and Statement-II both (1) Statements-I and II both are incorrect.
are correct. (2) Statement-I is correct but Statement-II
(2) Statement-I and II both are incorrect.
is incorrect.
(3) Statement-I is wrong but Statement-II is
correct. (3) Statement-I is incorrect but Statement-II
(4) Statement-I is correct Statement-II is is correct.
wrong. (4) Statement-I and II both are correct

13. Statement-I: Cis-But-2-ene has higher 16. Statement-I: Ethyne is formed from 1,2-
dipole moment than trans-but-2-ene. dibromoethane by using alcoholic KOH
Statement-II: R − C ≡ C − R′ reacts with H2 in followed by NaNH2.
presence of Na/liq. NH3 to form mainly cis- Statement-II: 1,2- dibromo ethane on
alkene. reaction with Na in dry ether gives ethene.
In the light of the above statements, choose In the light of the above statements, choose
the most appropriate answer from the
the most appropriate answer from the
options given below:
options given below:
(1) Statement-I is correct but Statement-II
is incorrect. (1) Statement-I and Statement-II both are
(2) Statement-I is incorrect but Statement-II correct.
is correct (2) Statement-I and II both are incorrect.
(3) Statement-I and Statement-II are (3) Statement-I is correct but Statement-II is
correct. incorrect.
(4) Both Statement-I and II are incorrect. (4) Statement-I is incorrect but Statement-II is
correct.

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Hydrocarbons CHEMISTRY
17. Statement-I : Alkyl halides on treatment 19. Assertion (A): 1-butene on reaction with HBr
with sodium metal in moist ether gives in the presence of a peroxide, produces 1-
higher alkane. bromobutane as main product
Statement-II : Wurtz reaction is used for the Reason (R): It involves the formation of more
preparation of higher alkanes containing stable free radical.
even number of carbon atoms. (1) If both Assertion and Reason are correct
In the light of the above statements, choose and the Reason is a correct explanation
the most appropriate answer from the of the Assertion.
options given below: (2) If both Assertion and Reason are correct
(1) Statement-I and Statement-II both are
but Reason is not a correct explanation
correct.
of the Assertion.
(2) Statement-I and II both are incorrect.
(3) If the Assertion is correct but Reason is
(3) Statement-I is correct but Statement-II
incorrect.
is incorrect.
(4) If the Assertion is incorrect and Reason
(4) Statement-I is incorrect but Statement-II
is correct.
is correct.

20. Assertion (A): Propene reacts with HI in the

18. Given below are two statements
presence of peroxide to give
Statement-I : Ozonolysis of o-xylene (1,2-
1-iodopropane.
dimethyl benzene) gives three products. The
Reason (R): 1° free radical is less stable than
result supports kekule structure of benzene.
2° free radical.
Statement-II : Benzene have 6πe¯ in cycle
(1) If both Assertion and Reason are correct
and have cyclic delocalisation. In the light of
and the Reason is a correct explanation
the above statements, choose the most
appropriate answer from the options given of the Assertion.

below: (2) If both Assertion and Reason are correct

(1) Statement-I is correct but Statement-II but Reason is not a correct explanation

is incorrect. of the Assertion.

(2) Statement-I is incorrect but Statement-II (3) If the Assertion is correct but Reason is
is correct. incorrect.
(3) Statement-I and Statement-II are (4) If the Assertion is incorrect and Reason
incorrect. is correct.
(4) Both Statement-I and Statement-II are
correct.

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 CHEMISTRY Hydrocarbons
21. Assertion (A): Nitration of benzene requires 24. Assertion: benzene does not decolourise Br2
the use of nitrating mixture.
–water.
Reason (R): The mixture of concentrated
sulphuric acid and concentrated nitric acid Reason: Benzene is stabilized by aromaticity
produces the NO2+ . and normally does not under addition
(1) If both Assertion and Reason are correct
reaction.
and the Reason is a correct explanation
of the Assertion. (1) If both Assertion & Reason are True & the
(2) If both Assertion and Reason are correct Reason is a correct explanation of the
but Reason is not a correct explanation
of the Assertion. Assertion.
(3) If the Assertion is correct but Reason is (2) If both Assertion & Reason are True but
incorrect.
(4) If the Assertion is incorrect and Reason Reason is not a correct explanation of
is correct. the Assertion.
(3) If Assertion is True but the Reason is
22. Assertion: 2, 3-dimethyl but-2-ene
decolorizes Br2 water False.
Reason: 2, 3-dimethyl but-2-ene is an
(4) If both Assertion & Reason are False.
unsaturated compound
(1) If both Assertion & Reason are True & the
Reason is a correct explanation of the 25. Assertion: Phenol is more reactive than
Assertion.
(2) If both Assertion & Reason are True but benzene towards electrophilic substitution.
Reason is not a correct explanation of Reason: In case of phenol, the intermediate
the Assertion.
(3) If Assertion is True but the Reason is carbocation is more resonance stabilised.
False. (1) If both Assertion & Reason are True & the
(4) If both Assertion & Reason are False.
Reason is a correct explanation of the
23. Assertion: Iodination of alkane is carried out Assertion.
in presence of iodic acid.
(2) If both Assertion & Reason are True but
Reason: Iodic acid is an oxidizing agent.
(1) If both Assertion & Reason are True & the Reason is not a correct explanation of
Reason is a correct explanation of the
the Assertion.
Assertion.
(2) If both Assertion & Reason are True but (3) If Assertion is True but the Reason is
Reason is not a correct explanation of False.
the Assertion.
(3) If Assertion is True but the Reason is (4) If both Assertion & Reason are False.
False.
(4) If both Assertion & Reason are False.

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Exercise 4 (Previous Year's Questions)

1. Which of the following compounds will not 3. A single compound of the structure :-
CH3 CH3
undergo Friedel-raft's reaction:
OHC CH C
CH2 CH2 O
[NEET-UG - 2013]
is obtainable from ozonolysis of which of
(1) Toluene
the following cyclic compounds?
(2) Cumene [AIPMT - 2015]
(3) Xylene H3 C H3 C CH3
(4) Nitrobenzene (1) (2)
H3 C
2. What products are formed when the CH3 H3 C
following compound is treated with Br2 in (3) (4) CH3
CH3
the presence of FeBr3? [AIPMT - 2014]

CH3 4. The reaction of C6H5CH = CHCH3 with HBr

produces: [AIPMT - 2015]
CH3 (1) C6H6 – CH2 – CHCH3

CH3 CH3 Br
Br (2) C6H5CH2CH2CH2Br
(1) and CH3
CH=CHCH3
CH3
Br (3)

CH3 CH3
Br
(2) Br and Br
(4) C6H6–CHCH2CH3
CH3 CH3
Br

CH3
CH3 5. 2,3-Dimethyl-2-butene can be prepared by
(3) Br and heating which of the following compounds
CH3
with a strong acid? [Re-AIPMT - 2015]
CH3 Br
(1) (CH3)2C=CH–CH2–CH3
CH3 (2) (CH3)2 CH–CH2–CH = CH2
CH3
(3) (CH3)2CH–CH–CH = CH2
(4) and
CH3 CH3
Br Br CH3 (4) (CH3)3 C–CH = CH2

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 CHEMISTRY Hydrocarbons

6. The oxidation of benzene by V2O5 in the 11. In the given reaction

presence of air produces : HF
[Re-AIPMT - 2015] + 
0° C
→P
(1) benzoic acid
(2) benzaldehyde the product P is : [NEET - I - 2016]
(3) benzoic anhydride
(4) maleic anhydride (1)

7. Which of the following is not the product of

dehydration of OH ?
(2)
[Re-AIPMT - 2015]

(1) (2)
F

(3) (4) (3)

8. In the reaction with HCl, an alkene reacts in F

accordance with the Markovnikov's rule, to give
a product 1-chloro-1- methylcyclohexane. The (4)
possible alkene is: [Re-AIPMT - 2015]
CH2 CH3
12. The compound that will react most readily
(1) (2)
with gaseous bromine has the formula
[NEET - I - 2016]
CH3 (1) C4H10 (2) C2H4
(3) (1) and (2) Both (4) (3) C3H6 (4) C2H2

13. Which of the following can be used as the

9. Consider the nitration of benzene using mixed halide component for Friedel-Crafts
conc. H2SO4 and HNO3. If a large amount of reaction? [NEET - I - 2016]
KHSO4 is added to the mixture, the rate of (1) Chloroethene
nitration will be: [NEET-I -2016] (2) Isopropyl chloride
(1) Faster (2) Slower
(3) Chlorobenzene
(3) Unchanged (4) Doubled
(4) Bromobenzene
10. Which of the following compounds shall
not produce propene by reaction with HBr 14. Hydrocarbon (A) reacts with bromine by
followed by elimination or direct only by substitution to form an alkyl bromide which
elimination reaction? [NEET-I-2016] by Wurtz reaction is converted to gaseous
(1) H2C = C = O
hydrocarbon containing less than four
(2) H3C–CH2 –CH2Br
carbon atoms. (A) is [NEET-2018]
(3) H2C – CH2
(1) CH ≡ CH (2) CH2 = CH2
CH2
(4) H3C–CH2 –CH2OH (3) CH3–CH3 (4) CH4
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Hydrocarbons CHEMISTRY
15. The compound C7H8 undergoes the 18. An alkene “A” on reaction with O3 and
following reaction: [NEET-2018] Zn–H2O gives propanone and ethanal in
3Cl2 /∆ Br2 /Fe Zn/HCl
equimolar ratio. Addition of HCl to alkene
C7H8  → A  → B  →C “A” gives “B” as the major product. The
The product ‘C’ is structure of product “B” is: [NEET-2019]
(1) m-bromotoluene CH3
(1) Cl–CH2–CH2–CH
(2) o-bromotoluene
CH3
(3) 3-bromo-2, 4, 6-trichlorotoluene CH2Cl
(4) p-bromotoluene (2) H3C–CH2–CH–CH3
CH3
16. Identify the major products P, Q and R in (3) H3C–CH2–C–CH3
this following sequence of reactions: Cl
[NEET-2018] CH3
Anhydrous
(4) H3C–CH–CH
AlCl3
+ CH3CH2CH2Cl P Cl CH3

 (i) O2
→Q +R 19. Among the following the reaction that
(ii)H O+ / ∆
2 proceeds through an electrophilic
P Q R substitution is [NEET-2019]
⊕ Θ Cu2Cl2
(1) CH(CH3)2 OH CH3CH(OH)CH3 (1) N2Cl Cl + N2

AlCl3
(2) + Cl2 Cl + HCl
(2) CH2CH2CH3 CHO COOH
Cl Cl
UV light
(3) + Cl2 Cl Cl
(3) CH2CH2CH3 CHO CH3CH2–OH
Cl Cl

Heat
(4) CH2OH+HCl CH2Cl+H2O
(4) CH(CH3)2 OH CH3CO–CH3
20. The alkane that gives only one mono-chloro
product on chlorination with Cl2 in
presence of diffused sunlight is-
17. The most suitable regent for the following [NEET-(Odisha)2019]
conversion is: [NEET-2019] (1) 2, 2-dimethylbutane
(2) neopentane
H3C CH3
(3) n-pentane
H3C–C≡C–CH3 →
(4) Isopentane
H H
21. Which of the following alkane cannot be
(1) Na/liquid NH3
made in good yield by Wurtz reaction ?
(2) H2, Pd/C, quinoline [NEET-2020]
(3) Zn/HCl (1) n-Butane
(2) n-Hexane
(4) Hg2+/H+, H2O
(3) 2,3-Dimethylbutane
(4) n-Heptane
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22. Identify compound X In the following 26. NaOH,+ ?
CH3CH2COO¯Na+  →
Heat
sequence of reactions: [NEET-2020] CH3CH3 + Na2CO3 .
CH3 CHO Consider the above reaction and identify
the missing reagent/chemical.
H2 O
Cl2/hν
X 373K
[NEET-2021]
(1) B2H6 (2) Red Phosphorus
CCl3 Cl (3) CaO (4) DIBAL-H

(1) (2) 27. The major product of the following

chemical reaction is: [NEET-2021]
CH3 (C6H5CO)2O2
CH2Cl CHCl2 CH–CH=CH2+HBr
CH3
(3) (4) CH3
(1)
CBr–CH2–CH3
CH3
CH3
23. An alkene on ozonolysis gives methanal as (2)
CH–CH2–CH2–Br
one of the product. Its structure is CH3
[NEET-2020] CH3
(3) CH–CH2–CH2–O–COC6H5
CH2CH2CH3 CH=CH–CH3 CH3
CH3
(1) (2) CH–CH–CH3
(4)
CH3
Br
CH2–CH2–CH3 CH2–CH=CH2
28. Compound X on reaction with O3 followed
(3) (4) by Zn/H2O gives formaldehyde and 2-
methyl propanal as products. The
compound X is: [NEET-2022]
(1) 2-Methylbut-1-ene
(i) B2H6
24. CH3CH2CH = CH2 →
(ii) H O /OH¯
Z (2) 2-Methylbut-2-ene
2 2
(3) Pent-2-ene
What is Z? [NEET-2020(Covid-19)]
(4) 3-Methylbut-1-ene
(1) CH3CH2CH2CH2OH
(2) (CH3)2CH–CH–CH = CH2 29. Identify the correct reagents that would
CH3 bring about the following transformation.
(3) CH3CH2CH2CHO [NEET-2024]
(4) CH3CH2CH2CH3 CH2–CH=CH2→ CH2–CH2–CHO

25. The major product formed in

(1) (i) BH3 (2) (i) H2O/H+
dehydrohalogenation reaction of 2-Bromo Θ
(ii) H2O2/OH (ii) PCC
pentane is Pent-2-ene. This product (iii) alk. KMnO4
formation is based on ? [NEET-2021] (iv) H3OΘ
(1) Saytzeff’s Rule
(2) Hund’s Rule (3) (i) H2O/H
+
(4) (i) BH3 Θ
(3) Hofmann Rule (ii) CrO3 (ii) H2O2/OH
(4) Huckel’s Rule (iii) PCC

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Hydrocarbons CHEMISTRY
30. For the given reaction : [NEET-2024] 32. Match List I With List II. [NEET-2024]
KMnO4/H+ ‘P’ List-I (Reaction)
C=CH
H (major product) A. 2 O

‘P’ is : O

OH OH B.
(1)
CH–CH
OH O
O O C.
(2) C—C
CH2CH3 COOK
(3) D.
CHO

(4) COOH List-II reagents/Condition

O

31. Given below are two statements: I. Cl/Anhdy.AlCl3

[NEET-2024]
Statement I : The boiling point of three II. CrO3
isomeric pentanes follows the order n- III. KMnO4/ KOH, ∆
pentane > isopentane > neopentane IV. (i) O3
Statement II : When branching increases, (ii) Zn—H2O
the molecule attains a shape of sphere. This Choose the correct answer from the options
results in smaller surface area for contact,
given below.
due to which the intermolecular forces
(1) A-IV, B-I, C-II, D-III
between the spherical molecules are weak,
(2) A-I, B-IV, C-II, D-III
thereby lowering the boiling point.
(3) A-IV, B-I, C-III, D-II
In the light of the above statements, choose
(4) A-III, B-I, C-II, D-IV
the most appropriate answer from the
options given below:
(1) Statement I is correct but Statement II 33. Baeyer's reagent is: [RE-NEET-2024]
is incorrect. (1) acidic potassium permanganate
(2) Statement I is incorrect but Statement solution
II is correct. (2) acidic potassium dichromate solution
(3) Both Statement I and Statement II are (3) cold, dilute, aqueous solution of
correct. potassium permanganate
(4) Both Statement I and Statement II are (4) hot, concentrated solution of potassium
incorrect.
permanganate

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 CHEMISTRY Hydrocarbons
34. The following reaction method : 36. Which one of the following reactions does
[RE-NEET-2024] NOT give benzene as the product?
CH3 Fe [NEET-2025]
+X2 dark (1) H–C≡ C–H
red hot Iron Tube
at 873 K

CH3 CH3
+
(2) ⊕ Θ H 2O
N≡N Cl
X warm
X
Is not suitable for the preparation of the O
(3) sodalime
corresponding haloarene products, due to C–O Na
∆
high reactivity of halogen, when X is :
(4) Mo2O3
(1) F (2) I
(3) Cl (4) Br n-hexane 773 K, 10–20 atm.

35. The alkane that can be oxidized to the 37. How many products (including
corresponding alcohol by KMnO4 as per the stereoisomers) are expected from
equation : [RE-NEET-2024] monochlorination of the following
compound? [NEET-2025]
R1 R1
H3C
KMnO4 CH – CH2–CH3
R2–C–H R2–C–OH
H3C
R3 R3 (1) 5 (2) 6
is, when : (3) 2 (4) 3
(1) R1 = H; R2 = H; R3 = H
(2) R1 = CH3; R2 = CH3; R3 = CH3
(3) R1 = CH3; R2 = H; R3 = H
(4) R1 = CH3; R2 = CH3; R3 = H

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ANSWER KEYS
Exercise 1.1
Que . 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans . 4 3 3 1 4 1 2 3 4 4 4 4 2 3 2 2 2 1 4 4
Que . 21 22 23 24 25 26 27 28 29 30
Ans . 2 3 2 1 4 3 3 1 2 4

Exercise 1.2
Que . 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans . 2 1 3 2 1 4 3 1 4 3 4 4 1 3 2 3 3 4 3 3
Que . 21 22 23 24 25 26 27 28 29 30
Ans . 1 2 2 1 2 3 3 4 3 2

Exercise 1.3
Que . 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans . 1 1 4 2 3 4 4 2 4 4 2 3 4 1 2 2 3 2 1 4

Exercise 1.4
Que . 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Ans . 4 1 2 1 2 4 4 2 2 2 2 2 4 2 1

Exercise 2
Que . 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans . 1 3 1 1 3 1 1 4 1 3 1 2 3 1 1 4 2 4 3 3
Que . 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
Ans . 3 2 1 3 1 2 4 2 2 1 1 4 3 4 4 4 1 3 1 2
Que . 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60
Ans . 1 2 1 3 3 4 1 1 2 2 2 3 2 1 3 2 3 4 1 1

Exercise 3
Que . 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans . 3 2 3 2 3 1 3 1 2 1 1 1 1 1 3 1 4 4 1 4
Que . 21 22 23 24 25
Ans . 1 1 1 1 1

Exercise 4 (Previous Year's Questions)

Que . 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans . 4 3 4 4 4 4 4 3 2 1 1 3 2 4 1 4 2 3 2 2
Que . 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37
Ans . 4 4 4 1 1 3 2 4 4 4 3 1 3 1 2 2 2

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 Notes

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Purification & Analysis of Organic Compounds CHEMISTRY

Chapter PURIFICATION
AND ANALYSIS
5 OF ORGANIC
COMPOUNDS

Chapter Summary Methods of Purification of Organic Compounds

The common techniques used for purification are as follows:
• Methods of Purification of
(i) Sublimation
Organic Compounds
(ii) Crystallisation
• Qualitative analysis of (iii) Distillation
organic compounds (iv) Differential extraction and
(v) Chromatography
• Quantitative analysis of
organic compounds Sublimation
Some solid substance changes from solid to vapour state
without passing through liquid state
The purification technique based on the above principle is
known as sublimation and is used to
separate sublimable compounds from non-sublimable
impurities.
Naphthalene and anthraquinone can be separated from non
volatile impurities using sublimation method. Camphor and
NaCl can also be separated by this method.
Crystallisation
This is one of the most commonly used techniques for the
purification of solid organic compounds. It is based on the
difference in the solubilities of the compound and the impurities
in a suitable solvent.
The impure compound is dissolved in a solvent in which it is
sparingly soluble at room temperature but appreciably soluble
at higher temperature. The solution is concentrated to get a
nearly saturated solution. On cooling the solution, pure
compound crystallises out and is removed by filtration. The
filtrate (mother liquor) contains impurities and small quantity
of the compound. If the compound is highly soluble in one
solvent and very little soluble in another solvent, crystallisation

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 CHEMISTRY Purification & Analysis of Organic Compounds
can be satisfactorily carried out in a mixture of these solvents. Impurities, which impart colour to the
solution are removed by adsorbing over activated charcoal. Repeated crystallisation becomes
necessary for the purification of compounds containing impurities of comparable solubilities.

Distillation :
This important method is used to separate
(i) Volatile liquids from nonvolatile impurities and
(ii) The liquids having sufficient difference in their boiling points. Liquids having different boiling points
vaporise at different temperatures. The vapours are cooled and the liquids so formed are collected
separately. Chloroform (b.p 334 K) and aniline (b.p. 457 K) are easily separated by the technique of
distillation figure.

Fig. Simple distillation. The vapours of a substance formed are condensed and the liquid is collected
in conical flask.

Fractional Distillation:
If the difference in boiling points of two liquids is not much, simple distillation cannot be used to
separate them. The vapours of such liquids are formed within the same temperature range and are
condensed simultaneously. The technique of fractional distillation is used in such cases. In this
technique, vapours of a liquid mixture are passed through a fractionating column before condensation.
The fractionating column is fitted over the mouth of the round bottom flask.

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Fig. Fractional distillation. The vapours of lower boiling fraction reach the top of the column first
followed by vapours of higher boiling fractions.
Vapours of the liquid with higher boiling point condense before the vapours of the liquid with lower
boiling point. The vapours rising up in the fractionating column become richer in more volatile
component. By the time the vapours reach to the top of the fractionating column, these are rich in the
more volatile component. Fractionating columns are available in various sizes and designs as shown in
Fig. A fractionating column provides many surfaces for heat exchange between the ascending vapours
and the descending condensed liquid. Some of the condensing liquid in the fractionating column
obtains heat from the ascending vapours and revaporises. The vapours thus become richer in low
boiling component. The vapours of low boiling component ascend to the top of the column. On reaching
the top, the vapours become pure in low boiling component and pass through the condenser and the
pure liquid is collected in a receiver. After a series of successive distillations, the remaining liquid in
the distillation flask gets enriched in high boiling component. Each successive condensation and
vaporisation unit in the fractionating column is called a theoretical plate. Commercially, columns with
hundreds of plates are available. One of the technological applications of fractional distillation is to
separate different fractions of crude oil in petroleum industry.

Ex. Mixture of methanol (B.P. 65°C) and acetone (B.P. 56°C) can be separated using fractional distillation
method.
Distillation under reduced pressure: (Vaccum distillation)
This method is used to purify liquids having very high boiling points and those, which decompose at or
below their boiling points. Such liquids are made to boil at a temperature lower than their normal
boiling points by reducing the pressure on their surface. A liquid boils at a temperature at which its
vapour pressure is equal to the external pressure. The pressure is reduced with the help of a water
pump or vacuum pump. Glycerol can be separated from spent-lye in soap industry by using this
technique.

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 CHEMISTRY Purification & Analysis of Organic Compounds

Fig. Distillation under reduced pressure. A liquid boils at a temperature below its vapour pressure
by reducing the pressure.
Steam Distillation:
This technique is applied to separate substances which are steam volatile and are immiscible with
water. In steam distillation, steam from a steam generator is passed through a heated flask containing
the liquid to be distilled. The mixture of steam and the volatile organic compound is condensed and
collected. The compound is later separated from water using a separating funnel. In steam distillation,
the liquid boils when the sum of vapour pressures due to the organic liquid (p1) and that due to water
(p2) becomes equal to the atmospheric pressure (p), i.e. p = p1 + p2. Since p1 is lower than p2 the organic
liquid vaporises at lower temperature than its boiling point. Thus, if one of the substances in the
mixture is water and the other, a water insoluble substance, then the mixture will boil close to but
below, 373K. A mixture of water and the substance is obtained which can be separated by using a
separating funnel. Aniline is separated by this technique from aniline – water mixture.
Ex. Ortho nitro phenol and para nitro phenol can be separated using steam distillation method.

Fig. Steam distillation. Steam volatile component volatilizes, the vapours condense in the
condenser and the liquid collects in conical flask.

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 Purification & Analysis of Organic Compounds CHEMISTRY
Differential Extraction :
When an organic compound is present in an aqueous medium, it is separated by shaking it with an
organic solvent in which it is more soluble than in water. The organic solvent and the aqueous solution
should be immiscible with each other so that they form two distinct layers which can be separated by
separatory funnel. The organic solvent is later removed by distillation or by evaporation to get back the
compound. Differential extraction is carried out in a separatory funnel. If the organic compound is less
soluble in the organic solvent, a very large quantity of solvent would be required to extract even a very
small quantity of the compound. The technique of continuous extraction is employed in such cases. In
this technique same solvent is repeatedly used for extraction of the compound.

Fig. Differential extraction. Extraction of compound takes place based on difference in solubility.
Chromatography :
The name chromatography is based on the Greek word chroma, for colour since the method was first
used for the separation of coloured substances found in plants. In this technique, the mixture of
substances is applied on to a stationary phase, which may be a solid or a liquid. A pure solvent, a
mixture of solvents, or a gas is allowed to move slowly over the stationary phase. The components of
the mixture get gradually separated from one another. The moving phase is called the mobile phase.
Based on the principle involved, chromatography is classified into different categories. Two of these
are:
(a) Adsorption chromatography, and
(b) Partition chromatography.

(a) Adsorption Chromatography:

Adsorption chromatography is based on the fact that different compounds are adsorbed on an
adsorbent to different degrees. Commonly used adsorbents are silica gel and alumina. When a mobile
phase is allowed to move over a stationary phase (adsorbent), the components of the mixture move by
varying distances over the stationary phase. Following are two main types of chromatographic
techniques based on the principle of differential adsorption.
(i) Column chromatography, and
(ii) Thin layer chromatography.

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 CHEMISTRY Purification & Analysis of Organic Compounds

(i) Column Chromatography:

Column chromatography involves separation of a mixture over a column of adsorbent (stationary
phase) packed in a glass tube. The column is fitted with a stopcock at its lower end.
The mixture adsorbed on adsorbent is placed on the top of the adsorbent column packed in a glass tube.
An appropriate eluant which is a liquid or a mixture of liquids is allowed to flow down the column
slowly. Depending upon the degree to which the compounds are adsorbed, complete separation takes
place. The most readily adsorbed substances are retained near the top and others come down to various
distances in the column.

Fig. Column chromatography. Different stages of separation of components of a mixture.

(ii) Thin Layer Chromatography:

Thin layer chromatography (TLC) is another type of adsorption chromatography, which involves
separation of substances of a mixture over a thin layer of an adsorbent coated on glass plate. A thin
layer (about 0.2mm thick) of an adsorbent (silica gel or alumina) is spread over a glass plate of suitable
size. The plate is known as thin layer chromatography plate or chromaplate. The solution of the mixture
to be separated is applied as a small spot about 2 cm above one end of the TLC plate. The glass plate is
then placed in a closed jar containing the eluant. As the eluant rises up the plate, the components of
the mixture move up along with the eluant to different distances depending on their degree of
adsorption and separation takes place. The relative adsorption of each component of the mixture is
expressed in terms of its retardation factor i.e. Rf value
Distance moved by the substance from base line (x)
Rf = .
Distance moved by the solvent from base line (y)

Fig. (a) Thin layer chromatography. Chromatogram being developed

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Purification & Analysis of Organic Compounds CHEMISTRY
The spots of coloured compounds are visible on TLC plate due to their original colour. The spots of
colourless compounds, which are invisible to the eye but fluoresce in ultraviolet light, can be detected
by putting the plate under ultraviolet light. Another detection technique is to place the plate in a
covered jar containing a few crystals of iodine. Spots of compounds, which adsorb iodine, will show up
as brown spots. Sometimes an appropriate reagent may also be sprayed on the plate. For example,
amino acids may be detected by spraying the plate with ninhydrin solution.

Fig. (b) Developed chromatogram.

Partition Chromatography:
Partition chromatography is based on continuous differential partitioning of components of a mixture
between stationary and mobile phases. Paper chromatography is a type of partition chromatography.
In paper chromatography, a special quality paper known as chromatography paper is used.
Chromatography paper contains water trapped in it, which acts as the stationary phase. A strip of
chromatography paper spotted at the base with the solution of the mixture is suspended in a suitable
solvent or a mixture of solvents. This solvent acts as the mobile phase. The solvent rises up the paper
by capillary action and flows over the spot. The paper selectively retains different components
according to their differing partition in the two phases. The paper strip so developed is known as a
chromatogram. The spots of the separated coloured compounds are visible at different heights from
the position of initial spot on the chromatogram. The spots of the separated colourless compounds may
be observed either under ultraviolet light or by the use of an appropriate spray reagent as discussed
under thin layer chromatography.

Fig. Paper chromatography. Chromatography paper in two different shapes.

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 CHEMISTRY Purification & Analysis of Organic Compounds
Exercise 1.1
1. Benzoic acid is soluble in hot water but 6. In steamzdistillation, the steam-volatile
sparingly soluble in cold water. An impure compound and water distils together at :
(1) The normal boiling point of water
sample of benzoic acid can be purified by:
(2) The normal boiling point of the
(1) distillation compound
(2) steam distillation (3) A temperature below the normal boiling
(3) crystallisation point of the compound
(4) A temperature below the normal boiling
(4) solvent extraction
points of both the compound and water

2. Naphthalene is a highly volatile solid. If can 7. If two volatile liquids have almost identical
boiling points, they are best separated from
be separated from non-volatile impurities
a mixture by :
by : (1) simple distillation
(1) crystallisation (2) fractional distillation
(2) sublimation (3) column chromatography
(4) paper chromatography
(3) fractional crystallisation
(4) distillation 8. The separation of a mixture of organic
compounds into individual components by
column chromatography is based on the
3. From a mixture of methanol (b.p. 65°C) and
differences in their :
acetone (b.p. 56°C) the components can be (1) solubilities
separated by : (2) adsorptivities
(3) absorptivities
(1) simple distillation
(4) partition coefficients
(2) vacuum distillation
(3) steam distillation 9. A mixture of naphthalene and sodium
(4) fractional distillation chloride can be separated into pure
compounds by :
4. Which of the following compounds can be (1) crystallisation
purified by steam distillation? (2) sublimation
(1) Benzamide (3) distillation
(2) Benzoic acid (4) vacuum distillation
(3) Aniline
(4) Cinnamic acid 10. The most commonly used adsorbent in
column chromatography is :
(1) sand
5. Which of the following pairs of compounds
(2) Calcium carbonate
can be separated from their mixture by (3) Charcoal
steam distillation? (4) silica gel
(1) o-Nitrophenol and p-nitrophenol
11. Method which is based on the difference in
(2) Camphor and NaCl the solubilities of the compound-
(3) NaCl and KCl (1) Crystallisation (2) Distillation
(4) All the above (3) Sublimation (4) All

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 Purification & Analysis of Organic Compounds CHEMISTRY
12. Different fractions of crude oil are 14. Which purification technique is used for
separated by high boiling organic compound
(1) Simple distillation (decomposes near its boiling point)?
(2) Fractional distillation (1) Simple distillation
(3) Steam distillation (2) Steam distillation
(4) Vacuum distillation
(3) Fractional distillation
13. Which process is used for purification of liquids (4) Reduced pressure distillation
having high B.P. and decomposition below their
B.P.
(1) Distillation
(2) Steam distillation
(3) Distillation under reduced pressure
(4) Differential extraction

Qualitative Analysis of Organic Compounds

(A) Detection of carbon and hydrogen → By Liebig's method
Carbon and hydrogen are detected by heating the compound with copper (II) oxide. carbon present in
the compound is oxidised to carbon dioxide and Hydrogen to water
CuO
CxHy ∆ H2O + CO2
organic
Compound

‘C’ 2Cu
O∆
2Cu + CO2
(i) Compound Lime water
Ca(OH)2
CaCO3↓+ H2O
(Turbidity)
Cuo
(ii) 2'H' → cu+H2O
∆ 
|
|
AnhydrousCuSO4
→ CuSO ·5H O
White 4 2

(Blue color )

(B) Detection of other Elements:

• P, S, N and X Present in an organic compound are detected by "Lassaigne's Test"
• The elements present in the compound are converted from covalent form into the ionic form by fusing
the compound with sodium metal.
∆
Na + C + N → NaCN
∆
2Na + S  → Na2S
Na + X  ∆
→ ( X =Nax
Cl,Br orI)

↓
Part of organic compound
Process→

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 CHEMISTRY Purification & Analysis of Organic Compounds

Organic compound
(C. N. S and x) (‘Na’ Metal)

NaCN
Boiled
Na2S With distill
NaX Water
Filter

Extract

Sodium Fusion extract (SFE)

(Burner)
(1) Test For Nitrogen→
The sodium fusion extract is boiled with Iron (II) sulphate and then acidified with concentrated
sulphuric acid. The formation of prussian blue color confirm the Presence of Nitrogen.
∆
Na + C + N →NaCN ( SFE )
4−
Fe2+ + 6CNΘ → Fe ( CN)6 
Hexacyanido ferrate(II)ion

Fe2+ 
conc·H2SO4
→Fe3+ + e−
4−
4Fe3+ + 3 Fe ( CN)6  → Fe4 Fe ( CN)6 
3
Ferri ferro cyanide
↓
(Prussion blue color)

Note: Lassaigne Test of Nitrogen is not given if

(i) carbon is absent in the given compound
Ex. NH2 − NH2 ,NH4 OH etc.

They dont form NaCN

(ii) Diazonium compound


Ex. Alkyl diazoniumor Aryl diazoniumcompound  →
∆
N2 ↑
NaCN is not formed
(R −N2 cl )
+ −
( Ar −N2 cl )
+ − 


(2) Test For sulphur→

(i) On treating sodium fusion extract with sodium nitro prusside, appearance of a violet/purple color
further indicates the presence of sulphur
∆
2Na + S →Na2S ( SFE )
2− 4−
S2− + Fe ( CN)5 No  → Fe ( CN)5 NOS 

Violet/Purplecolor

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 Purification & Analysis of Organic Compounds CHEMISTRY
(ii) 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
∆
2Na + S →Na2S ( SFE )

Na2S + Pb(CH 3COO)2 → Pbs

↓ + 2CH3COONa
 (Black ppt )
Lead acetale

Note: (i) if 's' and 'N' both are present in the compound
∆
Na + S + C + N → NaSCN
( Sodium thiocyanate )
2+
Fe3+ + SCN− → Fe ( SCN)  or Fe ( SCN)3 
(Ferricthiocyanate )
↓
(Blood redcolor )

• Preussian blue color is not observed since there are no free cyanide ions (CN–)
Note: (ii) if sodium fusion is carried out with excess of Sodium, the thiocyanate decomposes to yield
cyanide and sulphide ion. These ions give their usual tests
∆ Na
Na + S + C + N →NaSCN 
( excess )
→NaCN + Na2S

(3) Test for Halogens→

Sodium fusion extract acidified with added AgNO3

HNO3 (Silver nitrate)
(Nacl/NaBr/NaI)

Agcl↓ + NaNO3 AgBr↓+NaNO3 AgI↓+NaNO3

(white PPT) (Pale yellow) (yellow)

NH4OH NH4OH NH4OH

Complete soluble Sparingly soluble Completely

Insoluble
ClΘ ion Present BrΘ ion Present
IΘ ion Present
Note: sodium fusion extract is acidified with conc. HNO3 before adding AgNO3 because if 'N' or 's' is
also Present in the compound then formed NaCN or Na2S during Lassaigne test will interfare with silver
nitrate Test for halogen. therefore to decompose NaCN or Na2S, sodium fusion extract is acidified with
conc. HNO3

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 CHEMISTRY Purification & Analysis of Organic Compounds
AgNO3
Na + ( C + N) ∆
→ NaCN  →
( Side exn) AgCN ↓ +NaNO3
(PPt )
or Or
∆
2Na + S  → Na2S
HCN + NaNO3 
decomposed by   → No side reaction with AgNO

conc.HNO3
→ Or  3
 H2S + NaNO3 

(4) Test For Phosphorus:

• The compound is heated with an oxidizing agent (sodium peroxide). The phosphorus present in the
compound is oxidized to phosphate. The solution is boiled with nitric acid and then treated with
ammonium molybdate. A yellow colored ppt indicate the presence of phosphorus
‘P’ containing Na2O2/∆ (Na3Po4) Boiled with
H3Po4
compound HNO3
added
(NH4)2MoO4
Canery Yellow (ammonium
or (NH4)3Po4·12MoO3 molybdate)
yellow Color ppt
(Ammonium Phosphomolybdate)
Or
(NH4)3P·Mo12O40
Or
(NH4)3P·(Mo3O10)4

Exercise 1.2
1. When a nitrogenous organic compound is 3. The sodium fusion extract of an organic
fused with metallic sodium, its nitrogen is compound on treatment with sodium
converted to : nitroprusside solution develops violet
colour. So, the compound contains.
(1) sodium cyanate
(1) nitrogen (2) sulphur
(2) sodium nitrate
(3) iodine (4) phosphorus
(3) sodium amide
(4) sodium cyanide 4. The sodium fusion extract of an organic
compound containing both nitrogen and
2. The sodium fusion extract of an organic sulphur when acidified with HCl and then
compound on acidification with dilute HCl treated with FeCl3 solution develops a :
(1) blue colour (2) green colour
and subsequent treatment with FeCl3
(3) red colour (4) yellow colour
develops a blood-red colour. The organic
compound contains. 5. Lassaigne's Test for the detection of
(1) bromine nitrogen will fail in the case of-
(2) nitrogen (1) NH2CONH2
(2) NH2—CO—NH—NH2.HCl
(3) sulphur
(3) NH2—NH2.HCl
(4) both nitrogen and sulphur
(4) C6H5NH—NH2. 2HCl

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6. The Prussian blue color obtained during the 11. Which of the following will give positive
test of nitrogen by lassaigne's test is due to lassaigne's test for nitrogen
the formation of- (1) NH2OH
(2) NH2—NH2
(1) Fe4[Fe(CN)6]3
(3) KCN
(2) Na3[Fe(CN)6]
(3) Fe(CN)3 (4) N=N—Cl
(4) Na4[Fe(CN)5NOS]
12. Which of the following will give blood red
7. In lassaigne's extract of an organic color in lassaigne's test for nitrogen
compound both nitrogen and sulphur are (1) PhNH2
present, which given blood red color with (2) PhNO2
Fe3+ due to the formation of- (3) O2N SO3H
(1) NaSCN
4− (4) PhSO3H
(2) Fe ( CN)5 NOS 
2+ 13. The violet color that develops on addition of
(3) Fe ( SCN) 
nitro prusside solution to sodium fusion
(4) Fe4[Fe(CN)6]3.H2O extract of a sulphur containing organic
compound is due to the formation of
8. Nitric acid added to sodium extract before (1) Na4[Fe(CN)5NOS]
(2) Na2[Fe(CN)5NO]
adding silver nitrate for testing halogen
(3) Na3[Fe(SCN)6]
because
(4) Na4[Fe(CN)5S]
(1) Nitric acid reduces sulphide
(2) Nitric acid decomposes "NaCN" and 14. The sodium fusion extract of an organic
"Na2S" compound on acidification with acetic acid
(3) Nitric acid oxidises the organic followed by addition of lead acetate
compound solution produced a black ppt. So the
compound contains-
(4) Nitric acid acts as a dehydrating agent
(1) Nitrogen (2) phosphorous
(3) Sulphur (4) Bromine
9. Which one of the following set of elements
can be detected using sodium fusion 15. The sodium fusion extract of an organic
extract? compound is boiled with nitric acid and
(1) Sulfur, Nitrogen, phosphorous, halogen then treated with silver nitrate solution. A
(2) Phosphorous, oxygen, nitrogen, halogen white precipitate is formed which dissolves
in aqueous Ammonia solution. The organic
(3) Nitrogen, phosphorous, carbon, sulphur
compound contain
(4) Halogen, nitrogen, oxygen, sulfur
(1) Sulphur (2) Chlorine
(3) Iodine (4) Bromine
10. Black precipitate in the detection of
sulphur with lead acetate and acetic acid is 16. The formula of sodium nitro prusside is
due to the formation of (1) Na4[Fe(CN)6]
(1) Pb2S (2) PbS (2) Na3[Fe(CN)5NO]
(3) PbS2 (4) PbSO4 (3) Na2[Fe(CN)5NO]
(4) Na4[Fe(CN)5NOS]

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Quantitative analysis for organic compound
(1) % Calculation For carbon & hydrogen → by Liebig method
• Both carbon and hydrogen are estimated in one experiment. A known mass of an organic compound is
burnt in the presence of excess of oxygen and copper (II) oxide. carbon and hydrogen in the compound
are oxidised to CO2 and H2O respectively
 y y
CxHy +  x +  O2 → xco2 + H2o
  4  2
'w' gm w gm 
1
w2gm

Let mass of organic compound is ‘w’ gm

44gmCO2 → 12gm'c'
12
1gmCO2 → gm'c'
44
 12 
w1 gmCO2 →  × w1  gm'c'
 44 
12 w1
% of 'C' = × × 100
44 w
Similarly: 18gm H2O → 2gm'H'
2
1gm H2O → gm'H'
18
 2 
w2 gm H2O →  × w 2  gm'H'
 18 
2 w2
% of 'H' = × × 100
18 w
Ex. (1) 0.3960 gm of an organic compound on combustion gives 0.792 gm CO2 and 0.324 gm of H2O. Calculate
the % of C and H
Ans. 9.09%
12 Wt. of CO2
Sol. % of C = × × 100
44 Wt. of compound
12 0.792
= × × 100 = 54.55%
44 0.396
2 Wt.of H2O
% of H = × × 100
18 Wt. of compound
2 0.324
= × × 100 = 9.09%
18 0.396

(2) % Calculation for Halogen → by Carius method

Case (a):- for ‘Cl’
Chlorine containing Heated withfuming HNO3 w1gm
Agcl
organic compound In the presence of AgNO3 mw = 108+35.5
=143.5gm/mol
(W gm)

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35.5 w1
% of cl= × × 100
143.5 w

Case (b):- for ‘Br’

Bromine containing Heated withfuming HNO3 w1gm
AgBr
organic compound In the presence of AgNO3 mw = 108+80
=188 gm mol–1
(W gm)

80 w1
% of Br = × × 100
188 w

Case (c):- for ‘I’

Iodine containing Heated withfuming HNO3 w1gm
Agl
organic compound In the presence of AgNO3 mw = 108+127
=235 gm mol–1
(W gm)

127 w1
% of I = × × 100
235 w
Ex. (2) In carius method of estimation of halogen, 0.15g of an organic compound gave 0.12g of AgBr. Find out
the percentage of bromine in the compound.
Ans. 34.04%
Sol. Molar mass of AgBr = 108 + 80 = 188 g mol–1
188 g AgBr contains 80g bromine.
0.12g AgBr contains 80 g bromine.
80 × 0.12
0.12g AgBr contains g bromine.
188
80 × 0.12 × 100
Percentage of bromine = = 34.04%
188 × 0.15

(3) % Calculation for Sulphur→ by Carius method.

Sulphur containing Heated with
[H2SO4]
organic compound fuming HNO3
Added aqueous
solution of BaCl2
(Wgm) W1gm
{BaSO4↓}

Mw = 233gm mol–1
32 w1
% of 'S' = × × 100
233 w

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Ex. (3) In Sulphur estimation, 0.157g of an organic compound gave 0.4813 g of barium sulphate. What is the
percentage of Sulphur in the compound ?
Ans. 42.10%
Sol. Molecular mass of BaSO4 = 137 + 32 + 64 = 233g
32 × 0.4813
233g BaSO4 contains g Sulphur
233
32 × 0.4813 × 100
percentage of Sulphur = = 42.10%
233 × 0.157

(4) % Calculation for Phosphorus →

Phosphorus containing Heated with
[H3PO4]
organic compound fuming HNO3 (Phosphoric acid)

(Wgm)
W1gm
Case (a):- H3Po4 (NH4)2MoO4 (NH4)3 Po4·12MoO3
(ammonium molybdate) (ammonium phospho Molybdate) mw = 1877gm mol-1
31 w1
% of 'P'= × × 100
1877 w

Magnesia mixture
Case (b): - H3Po4 (MgNH4Po4)
(MgCl2 + NH4Cl+Aq.NH3) White ppt
Heat

Mg2P2O7 W1gm
(Magnesium pyro phosphate)
Mw= 222 gm
62 w1
% of P = × × 100
222 w
(5) % Calculation for Nitrogen
There are two methods for estimation of nitrogen
(i) Dumas method (ii) Kjeldahl's method

(i) Dumas method :

when nitrogen containing organic compound is heated with copper (II) oxide in an atmosphere of
carbondioxide, yields free nitrogen in addition to carbondioxide and water.
Balance
 
  y ∆ y z  y 
reaction CxHyNz +  2x +  CuO → xCO2 + H2O + N2 +  2x +  cu
   2 2 2  2 
 ( 'w'gm ) 
if traces amount of nitrogen oxides formed then they are reduced to N2 by Passing the gaseous mixture
over heated copper gauze.
• Produced gaseous mixture is collected over aqueous Solution of KOH which absorbs CO2
{2KOH + CO2 → K 2CO3 + H2O}

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Fig. Dumas method. The organic compound yields nitrogen gas on heating it with copper (II) oxide in
the presence of CO2 gas. The mixture of gases is collected over potassium hydroxide solution in
which CO2 is absorbed and volume of nitrogen gas is determined.

From ideal gas equation

PV = nRT
PVαT
(P1 − P ) V1 P V 
T1
= 2 2  at STP
 T2 

P1 = Pressure of moist gas P2 = 1 atm or 760 mm of Hg

P = Aqueous tension T2 = 273k
T1 = Temp.(k) at which V2 = Volume of N2 gas collected
N2 gas is collected at STP
V1 = Volume of N2 gas (m)
At STP; 22.4L or 22400 ml → 1 mole N2 or 28 gm’N2’
 28 
1ml  →  gm'N2 '
 22400 
 28 
( V2 )ml 
→ × V2  gm'N2 '
 22400 
28 V2(ml )
% of 'N' = × × 100
22400 w
Ex. (4) 0.25gm of an organic compound NTP gives 31 ml of N2 gas by Duma’s method. Find out of % of N.
Ans. 15.5%
28 V
Sol. % of N = × × 100
22400 W
28 31
= × × 100 = 15.5%
22400 0.25

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Ex. (5) 31.7 ml of moist N2 was obtained from 0.2033 gm of an organic compound in Duma’s method at 14°C
and 758 mm pressure. If aq. tension at 14°C = 14 mm then calculate of % of nitrogen in the compound
Ans. 18.15%
Sol. P1 = Pressure of N2–Aq. tension = 758 – 14 = 744 mm
P2 = 760 mm
T2 = 273 K
T1 = 273 + 14
T1 = 287 K
P V 273 744 × 31.7 × 273
V2 = 1 1 × = = 29.51
T1 760 287 × 760
28 V
%N = × × 100
22400 W
28 29.51
= × × 100 = 18.15%
22400 0.2033

(ii) Kjeldahl’s method→

Nitrogen containing Heated with conc·H2SO4 (NH4)2SO4
Organic compound and Ammonium sulphate
(added cuso4 catalyst) 2NaOH(excess)
‘W’ gm
2NH3↑+2H2O+Na2SO4)
(Liberated)
• Liberated ammonia is absorbed in excess of standard solution of sulphuric acid
• The amount of ammonia Produced is determined by estimating the amount of H2SO4 consumed in the
reaction it is done by estimating unreacted sulphuric acid Left after the absorption of ammonia by
titrating it with standard alkali solution (NaOH).
• The difference between the initial amount of acid taken and that Left after the reaction gives the
amount of acid reacted with ammonia.

Fig. Kjeldahl method. Nitrogen-containing compound is treated with concentrated H2SO4 to get
ammonium sulphate which liberates ammonia on treating with NaOH; ammonia is absorbed in
known volume of standard acid.
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From low of equivalent:
gm equivalent of H2SO4 = gm equivalent of NH3
gm eq. of solute (H2SO4) ×1000 W
gm eq. =
N= E
Volume of solution (ml)
Weight of NH3
N×Vml gm eq. of NH3 =
gm eq. of H2SO4 = 17/1
1000
N × Vml weight of NH3
=
1000 17 / 1
 N × ( Vml ) × 17 
Weight of NH3 =   gm
 1000 
17 gm NH3 
→ 14gm ‘N’
14
1 gm 
→ gm ‘N’
17
 N × Vml × 17   14 N × Vml × 17 
  gm NH3  → ×  gm'N'
 1000   17 1000 
 14 
⇒ × N × Vml  gm'N'
 1000 
14 N × Vml
% of ‘N’ = × × 100
1000 w
1.4 × N × Vml
% of 'N' =
Wgm
W → weight of organic compound
N × Vml = milli eq. of H2SO4 consumed in the reaction to neutralized ammonia.
= milli eq. of H2SO4 taken – milli eq. of H2SO4 remaining.
= milli eq. of H2SO4 taken – milli eq. of NaOH.
Note:- This method is not applicable in the following Compound.
(I) Nitro comp (–NO₂) (II) Azo compound (–N=N–)
(III) Nitrogen in the ring

Ex.
N
(Pyridine)
(reason:- 'N' of these compound does not change to ammonium sulphate under these condition)

Ex. (6) 30 ml 0.25 N H2SO4 is used in neutralizing NH3 obtained from 0.75 gm of an organic compound in
Kjeldahl’s method. Find out %of N in the compound.
Ans. 14%
1.4
Sol. N% = × NV
W
14
= × 30 × 0.25 = 14%
0.75

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Ex. (7) 0.35 gm of an organic compound was analysed by Kjeldahl’s method. Ammonia evolved was absolved in
N N
100 ml H2SO4. Unused acid required 30 ml NaOH for neutralization calculate % of N
10 10
Ans. 28%
Sol. W = 0.35 gm NH3
↓
N
100 ml H2SO4(10 ml H2SO4)
10
↓
N
30 ml NaOH (3 ml N NaOH)
10
Remaining acid 3 ml N H2SO4
Consumed acid (10 ml N – 3ml N) = 7 ml N H2SO4
1.4
% of N = × NV
W
1.4
= × 1 × 7 = 28%
0.35

Exercise 1.3
1. Carbon and hydrogen in an organic 6. Kjeldahl's method of estimation of nitrogen
compound are estimated by : is not applicable to :
(1) Dumas method (1) Amides
(2) Carius method (2) Amines
(3) Liebig's method
(3) Diazo compounds
(4) Kjeldahl's method
(4) Amino acids
2. In Dumas method of estimation of nitrogen, 7. 0.20 g of an organic compound on
the gas collected in the nitrometer is : combustion gave 0.50 g of CO2. The
(1) N2 (2) NH3 percentage of carbon in the compound is
(3) NO2 (4) N2O
(1) 48.25 (2) 68.18
(3) 58.28 (4) 78.81
3. Kjeldahl's method is used for the estimation
of : 8. In the estimation of nitrogen by Dumas
(1) Sulphur (2) Nitrogen method, 0.112 g of an organic compound
(3) Halogen (4) Phosphorus
gave 22.4 mL of nitrogen gas at stp. The
percentage of nitrogen in the compound is :
4. In Dumas method of estimation of nitrogen,
an organic compound containing nitrogen (1) 20.2 (2) 47.5
is heated with : (3) 25.0 (4) 15.5
(1) Oxygen (2) Carbon dioxide
9. Halogens present in organic compounds
(3) Copper (4) Copper oxide
are estimated by :
5. In Kjeldalhl's method, nitrogen present in a (1) Carius method
compound is estimated by converting it (2) Dumas method
finally into : (3) Liebig's method
(1) NH3 (2) (NH4)2SO4 (4) Kjeldahl's method
(3) N2 (4) NH2-NH2

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10. In the Carius method, halogens present in 12. 0.32 g of an organic compound containing
organic compounds are estimated in the
sulphur was heated with fuming nitric acid
form of :
(1) Barium halide to convert sulphur into sulphuric acid. The
(2) Silver halide
latter was then precipitated as BaSO4 by
(3) Hydrogen halide
(4) Gaseous halogen adding excess of BaCl2 solution. 0.233 g of
BaSO4 was obtained. The percentage of
11. Phosphorus present in an organic
compound is generally estimated sulphur in the compound is approximately :
gravimetrically in the form of :
(1) 15.0 (2) 10.0
(1) Magnesium pyrophosphate
(2) Ammonium phosphomolybdate (3) 30.5 (4) 20.8
(3) Either of the above
(4) None of the above

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Exercise 2
1. Match Column I with Column II 7. When 0.01 mol of an organic compound
Column I Column II containing 60% carbon was burnt
(i) Dumas method (a) AgNO3 completely. 4.4g of CO2 was produced. The
(ii) Kjeldahl's (b) Silica gel molar mass of comp. is------ g mol-1.
method (1) 300 (2) 250
(iii) Carius (c) Nitrogen (3) 200 (4) 150
Method Gas
(iv) chromatography (d) Free
8. A sample of 0.125 g of an organic compound
radicals
when analyzed by duma's method yields
(e) Ammonium
sulphate 22.78 ml of nitrogen gas collected over KOH
(1) i → c; ii → e; iii → a; iv → b solution at 280K and 759 mm Hg. The % of
(2) i → d; ii → a; iii → c; iv → e nitrogen in the given organic compound
(3) i → b; ii → c; iii → a; iv → d is……? (Consider aqueous tension zero)
(4) i → e; ii → b; iii → d; iv → c (1) 36 (2) 22
(3) 12 (4) 48
2. The principle involved in paper
chromatography is : 9. The transformation occurring in duma's
(1) Adsorption (2) Partition method is given below:-
(3) Solubility (4) Volatility  y y
C2H 7N +  2x +  CuO → xCO2 + H2O +
 2 2
3. In carius method of estimation of halogen
0.15 g of an organic compound gave 0.12g of z  y
N2 +  2x +  Cu
AgBr. Find out the percentage of bromine in 2  2
the compound. The value of 'y' is_
(1) 23.03% (2) 34.04% (1) 14 (2) 7
(3) 42.10% (4) 45.05% (3) 7/2 (4) None of these

4. In Sulphur estimation, 0.157g of an organic 10. Which of the following compounds will give
compound gave 0.4813g of barium sulphate.
Prussian blue color when it is converted into
What is the percentage of sulphur in the
lassaigne's extract and FeSO4 is added
compound ?
(1) 23.03% (2) 34.04% followed by FeCl3 (Fe3+ ion)
(3) 42.10% (4) 48.05% NH
(I)
S
5. In kjeldahl's method nitrogen present is
estimated as:
(II)
(1) N2 (2) NO2
(3) NH4SO4 (4) NH3 SO3H
(III) NH2OH
6. An organic compound give 0.220 gm of CO2
and 0.126gm of H2O on complete O
||
combustion if the % of carbon is 24 then % (IV) NH2 —NH—C—NH2
hydrogen is-
(1) I and IV (2) IV only
(1) 76 (2) 8.6
(3) 5.6 (4) 50 (3) I, III and IV (4) I, II, III and IV

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11. The number of moles of CuO, that will be 15. The fragnance of flower is due to the
utilised in Dumas method for estimation of presence of some steam volatile organic
nitrogen in a sample of 57.5gm of C7H17N is- compound called essential oils. These are
(1) 11.25 (2) 18.6 generally insoluble in water at room
45 temperature but are miscible with water
(3) (4) 30.5 vapour in the vapour phase.
2
A suitable method for the extraction of these
oils from the flowers is :
12. The kjeldahl method of nitrogen estimation
(1) Distillation
fails in which of the following reaction major (2) Crystallisation
product? (3) Distillation under reduced pressure
NO2 (4) steam distillation

(A) Sn/HCl 16. If a liquid compound decomposes at its

boiling point, which method (s) can you
CN choose for its purification it is known that
the compound is stable at low pressure,
(B) LiAlH4 steam volatile and insoluble in water
(1) Steam distillation
CH2CN (2) fractional distillation
(3) chromatography
(i) SnCl2+HCl (4) Crystallisation
(C) (ii) H2O
17. A sample of 0.50 g of an organic compound
NH2
was treated according to Kjeldahl's method.
NaNO2 The ammonia evolved was absorbed in 50 ml
(D) HCl/0°–5°C of 0.25 M H2SO4. The residual acid required
30 ml of 0.5 M solution of NaOH for
(1) (C) and (D)
neutralization. Find % composition of
(2) (A) and (D)
nitrogen in the compound.
(3) (A), (C) and (D) (1) 56% (2) 28%
(4) (B) and (C) (3) 36% (4) 63%

13. In the recrystallisation of solid organic 18. Nitrogen can be estimated by kjeldahl's
compounds, the coloured impurities are method for which of the following
removed by boiling the solution of the compound?
compound with :
(1) (2)
(1) Activated charcoal
(2) oxalic acid ⊕
N≡NCl CH2—NH2
(3) magnesium sulphate
(4) calcium chloride
(3) (4)
N
14. The best and latest technique for isolation
NO2
purification and separation of organic
compounds is 19. In carius method, halogen containing
(1) Crystallisation organic compound is heated with fuming
(2) Distillation nitric acid in the presence of
(3) Sublimation (1) HNO3 (2) AgNO3
(4) Chromatography
(3) CuSO4 (4) FeSO4
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20. Which of the following is a false statement? 23. An organic compound is fused with sodium
(1) Carius tube is used in the estimation of peroxide and the fused mass is extracted
with water. The extract is heated with nitric
sulphur in an organic compound acid and ammonium molybdate. A yellow
(2) Carius method is used for the estimation precipitate is obtained. The compound
contains
of nitrogen in an organic compound
(1) Sulphur (2) Chlorine
(3) Phosphoric acid produced on oxidation (3) nitrogen (4) Phosphorus
of phosphorus present in an organic
24. An organic compound with 51.6% sulphur is
compound is precipitated as Mg2P2O7 by heated in a carius tube. The amount of this
adding magnesia mixture compound which form 18.74gm of barium
sulphate is-
(4) Kjeldahl's method is used for the (1) 15.9 gm (2) 11.8 gm
estimation of nitrogen in an organic (3) 18.6 gm (4) 4.99 gm
compound 25. Distillation under reduced pressure is
usefull for?
(1) Crude oil (2) Glycerol
21. The compound formed upon subjecting an
(3) Benzene (4) Naphthalene
aliphatic amine to lassaigne's test is
(1) NaNH2 (2) NaNO2 26. The most suitable method of separation of
(3) NaCN (4) NaN3 1 : 1 mixture of ortho and para nitrophenols is
(1) Chromatography (2) Crystallisation
22. Which of the following compound is formed (3) Steam distillation (4) Sublimation
when an organic compound containing both 27. The distillation Technique most suited for
nitrogen and sulphur is fused with sodium separating glycerol from spent lye in the
(avoiding excess) soap industry is-
(1) Sodium sulphide (1) Distillation under reduced pressure
(2) sodium cyanide (2) Simple distillation
(3) sodium thiocyanate (3) Fractional distillation
(4) A mixture of (1) & (2) (4) Steam distillation

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Exercise 3
4. Assertion (A) Carbon and hydrogen are
1. Assertion (A) If one of the substances in the
detected by heating the compound with
mixture is water and the other a water
copper (II) oxide.
insoluble substance, then the mixture will boil
Reason (R) Carbon and hydrogen present in
close to but higher than 373 K.
the compound are oxidised to carbon dioxide
Reason (R) Aniline is separated by using
and water respectively.
steam distillation technique from aniline-
(1) Both A and R are correct ; R is the correct
water mixture.
explanation of A.
(1) Both A and R are correct ; R is the correct
(2) Both A and R are correct ; R is not the
explanation of A.
correct explanation of A.
(2) Both A and R are correct ; R is not the
(3) A is correct ; R is incorrect.
correct explanation of A.
(4) R is correct ; A is incorrect.
(3) A is correct ; R is incorrect.
(4) R is correct ; A is incorrect.
5. Assertion (A) Phosphorus present in the
organic compound is oxidised to phosphoric
2. Assertion (A) A liquid boils at a temperature at
acid. It is precipitated as ammonium
which its vapour pressure is lower than the
phosphomolybdate by adding ammonium
external pressure.
hydroxide
Reason (R) Glycerol can be separated from
Reason (R) Phosphoric acid may be
spent-lye in soap industry by using the
precipitated as MgNH4PO4 by adding
technique of distillation under reduced
magnesia mixture which on ignition yields
pressure. Mg2P2O7.
(1) Both A and R are correct ; R is the correct (1) Both A and R are correct ; R is the correct
explanation of A. explanation of A.
(2) Both A and R are correct ; R is not the (2) Both A and R are correct ; R is not the
correct explanation of A. correct explanation of A.
(3) A is correct ; R is incorrect. (3) A is correct ; R is incorrect.
(4) R is correct ; A is incorrect. (4) R is correct ; A is incorrect.
6. Assertion (A) In Dumas method, the nitrogen
3. Assertion (A) In steam distillation, the liquid containing organic compound, when heated
boils when the sum of vapour pressures due to with copper oxide in an atmosphere of carbon
the organic liquid (p1) and that due to water dioxide yields free nitrogen in addition to
(p2) becomes equal to the atmospheric carbon dioxide and water.
pressure (p), i.e. p = p1 + p2 Reason (R) The correct equation representing
Reason (R) p1 is lower than p2 the organic this method is
liquid vaporises at lower temperature than its C x H y N z + ( 2x + y / 2 ) CuO → xCO 2
boiling point. + y / 2H 2 O + z / 2N 2 + (2x + y / 2)Cu
(1) Both A and R are correct ; R is the correct (1) Both A and R are correct ; R is the correct
explanation of A. explanation of A.
(2) Both A and R are correct ; R is not the (2) Both A and R are correct ; R is not the
correct explanation of A. correct explanation of A.
(3) A is correct ; R is incorrect. (3) A is correct ; R is incorrect.
(4) R is correct ; A is incorrect. (4) R is correct ; A is incorrect.

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 CHEMISTRY Purification & Analysis of Organic Compounds
7. Assertion (A) Kjeldahl's method is not 9. Match list I with list II
applicable to compounds containing nitrogen List-I (Mixture) List -II
in the form of nitro and azo groups and (Separation
nitrogen present in the ring (e.g. pyridine). Technique)
Reason (R) Nitrogen of nitro and azo group (A) CHCl3+C6H5NH2 (i) Steam
containing compounds does not change to distillation
ammonium sulphate by Kjeldahl's method. (B) C6H14+C5H12 (ii) Differential
(1) Both A and R are correct ; R is the correct extraction
explanation of A. (C) C6H5NH2+H2O (iii) Distillation
(2) Both A and R are correct ; R is not the (Simple)
correct explanation of A. (D) Organic (iv) Fractional
(3) A is correct ; R is incorrect. compound in distillation
(4) R is correct ; A is incorrect. water
(1) A-(iv), B-(i), C-(iii), D-(ii)
8. (2) A-(iii), B-(iv), C-(i), D-(ii)
Mixture Purification (3) A-(ii), B-(i), C-(iii), D-(iv)
technique (4) A-(iii), B-(i), C-(iv), D-(ii)
(A) Chloroform (P) Distillation
& aniline 10. Match list I with list II
(B) Glycerol & (Q) Chromatography List-I List -II Reagent
spent lye Element used/ Product
(C) Aniline & (R) Steam detected formed
water distillation (A) Nitrogen (i) Na2[Fe(CN)5NO]
(D) Amino acid (S) Distillation (B) Sulphur (ii) AgNO3
under reduced (C) Phosphorous (iii) Fe4[Fe(CN)6]3
pressure (D) Halogen (iv) (NH4)2MoO4
Correct match (1) A-(ii), B-(iv), C-(i), D-(iii)
(1) A-P, B-Q, C-R, D-S (2) A-(iv), B-(ii), C-(i), D-(iii)
(2) A-P, B-S, C-R, D-Q (3) A-(ii), B-(i), C-(iv), D-(iii)
(3) A-P, B-R, C-S, D-Q (4) A-(iii), B-(i), C-(iv), D-(ii)
(4) A-R, B-S, C-P, D-Q

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Solutions CHEMISTRY

Exercise 4 (Previous Year's Questions)

1. In the Kjeldahl’s method for estimation of 6. In Lassaigne’s extract of an organic
nitrogen present in a soil sample, ammonia compound, both nitrogen and sulpur are
evolved from 0.75 g of sample neutralized 10 present, which gives blood red colour with
mL of 1 M H2SO4. The percentage of nitrogen Fe3+ due to the formation of – [NEET-2023]
in the soil is : [AIPMT-2014] (1) NaSCN (2) [Fe(CN)5NOS]4–
2+
(3) [Fe(SCN)] (4) Fe4[Fe(CN6)]3.xH2O
(1) 35.33 (2) 43.33
(3) 37.33 (4) 45.33 7. On heating, some solid substances change
from solid to vapour state without passing
2. In Duma’s method for estimation of nitrogen, through liquid state. The technique used for
0.25 g of an organic compound gave 40 mL of the purification of such solid substances
nitrogen collected at 300 K temperature and based on the above principle is known as
[NEET-2024]
725 mm pressure. If the aqueous tension at
(1) Distillation (2) hromatography
300 K is 25 mm, the percentage of nitrogen
(3) Crystallization (4) Sublimation
in the compound is - [AIPMT-2015]
(1) 16.76 (2) 15.76 8. A steam volatile organic compound which is
(3) 17.36 (4) 18.20 immiscible with water has a boiling point of
250ºC. During steam distillation, a mixture
of this organic compound and water will boil.
3. Paper chromatography is an example of:
[RE-NEET-2024]
[NEET 2020]
(1) above 100ºC but below 250ºC.
(1) Column chromatography (2) above 250ºC.
(2) Adsorption chromatography (3) at 250ºC.
(3) Partition chromatography (4) close to but below 100ºC.
(4) Thin layer chromatography
9. Which one of the following reactions does
NOT belong to "Lassaignes' test"?
4. A liquid compound (x) can be purified by ∆
(1) Na + X → NaX
steam distillation only if it is: ∆
[NEET(UG) 2020 (Covid-19)] (2) 2CuO + C → 2Cu + CO2
∆
(1) Steam volatile, immiscible with water (3) Na + C + N → NaCN
∆
(2) Not steam volatile, miscible with water (4) 2Na + S → Na2S
(3) Steam volatile, miscible with water
10. Match List I with List II
(4) Not steam volatile, immiscible with
List- I List- II
water (Mixture) (Method of
Separation)
5. The Kjeldahl’s method for the estimation of A. CHCl3 + I. Distillation under
nitrogen can be used to estimate the amount C6H5NH2 reduced pressure
of nitrogen in which one of the following B. Crude oil in II. Steam distillation
compounds - [AIPMT-2022] petroleum
NO2 industry
N=N C. Glycerol from III. Fractional
(1) (2) spent-lye distillation
D. Aniline-water IV. Simple distillation
NH2
Choose the correct answer from the options
(3) (4) given below :
N (1) A-III, B-IV, C-I, D-II (2) A-III, B-IV, C-II, D-I
(3) A-IV, B-III, C-I, D-II (4) A-IV, B-III, C-II, D-I
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 CHEMISTRY Purification & Analysis of Organic Compounds
ANSWER KEYS
Exercise 1.1
Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14
Ans. 3 2 4 3 1 4 2 2 2 4 1 2 3 4

Exercise 1.2
Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Ans. 4 4 2 3 3 1 3 2 1 2 3 3 1 3 2 3

Exercise 1.3
Que. 1 2 3 4 5 6 7 8 9 10 11 12
Ans. 3 1 2 4 1 3 2 3 1 2 3 2

Exercise 2
Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans. 1 2 2 3 4 3 3 2 2 2 1 1 1 4 4 1 2 2 2 2
Que. 21 22 23 24 25 26 27
Ans. 3 3 4 4 2 3 1

Exercise 3
Que. 1 2 3 4 5 6 7 8 9 10
Ans. 4 4 1 1 4 1 1 2 2 4

Exercise 4 (Previous Year's Questions)

Q ue. 1 2 3 4 5 6 7 8 9 10
Ans. 3 1 3 1 4 3 4 4 2 3

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 Notes

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 Notes

Sarvam Career Institute