Chapter 14

Organic Chemistry - Some Basic

Principles and Techniques
Chapter Contents
 Introduction Introduction
 Classification of Organic
The development of organic chemistry is about 200-year-old. Around
Compounds
the year 1780, the chemists began to distinguish between organic
 Nomenclature System
compounds obtained from plants and animals and inorganic compounds
 Nomenclature of Substituted
Benzene Compounds prepared from mineral sources.
 Isomerism
In the earlier period of development of chemistry, Berzelius, a Swedish
 General Concepts in Organic
Reaction Mechanism chemist proposed that the synthesis of organic compounds within the
 Reaction Intermediates plants and animals required some mysterious force. This force was
 Classification of Organic called vital force and the theory was referred to as vital force theory
theory.
Reactions
heat
 Methods of Purification of NH4 CNO 
 NH2CONH2
Organic Compounds Ammonium Cyanate Urea

 Qualitative Analysis of This synthesis gave a death blow to vital force theory and clearly
Organic Compounds
demonstrated that no mysterious force was required in the formation of
 Quantitative Analysis
organic compounds in the lab. The elegant synthesis of acetic acid by
Kolbe in 1845 and that of methane by Berthelot in 1856 from purely

inorganic sources. The synthesis of organic compounds altogether

changed the concept of organic chemistry. With the downfall of vital

force theory, the term organic lost its original significance.

Thus, the modern definition of the organic chemistry is the chemistry

of the hydrocarbons and their derivatives. The development of electronic

theory of covalent bonding gave organic chemistry a modern shape.

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CLASSIFICATION OF ORGANIC COMPOUNDS

Organic compounds can be classified as per chart given below.

Organic Compounds

Acyclic compounds Cyclic compounds

(open chain compounds) (closed chain compounds)
e.g., hydrocarbons, alkylhalides,
alkyl amines, cyanides etc
Homocyclic Heterocyclic
(presence of heteroatoms
like N, O, S in ring) i.e.,
Aromatic Alicyclic
(Non aromatic)
Benzenoid Non-benzenoid e.g. Aromatic Alicyclic
, , etc.
e.g. , ,
Benzene , N O S etc. O
Cyclobutane Cyclopropane Cyclo Furan Thiophene
Cyclopropylium hexene H
Pyrrole
Naphthalene cation

Hydrocarbons and their Classifications

Compounds of C and H only are called hydrocarbons. They are parent organic compound and all other organic
compounds have been derived by replacing one or more H atoms from hydrocarbon.

Hydrocarbons

Acyclic Cyclic
(open chain hydrocarbon)

Alicyclic Aromatic CH3

Saturated hydrocarbon/Alkanes/Paraffins Unsaturated (cycloalkanes)
(GF CnH 2n + 2) hydrocarbons i.e., i.e., ,
, etc.
(having= and ) Toluene

 Concept of n, iso and neo alkane

n-alkane alkanes having C–C straight chain having only 1º and 2º C atoms.

CH3 – CH2 – CH2 – CH2 – CH3 n-pentane

iso-alkane alkanes having one time branching of methyl group at second last C only.

CH3
i.e., CH3 CH CH2 CH3 Isopentane

Neoalkane alkanes having two times branching of methylgroup at second last C only.
CH3
i.e., CH3 C CH3 neo pentan e

CH3

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 Concept of primary, secondary and tertiary C and their H atoms.
In a compound like

Tertiary C

H H CH3 CH3 Primary C

H C C C C CH3

H H H CH3
Quaternary C
Primary C Secondary C

H of 1º C is called Primary H
H of 2º C is called Secondary H
H of 3º C is called Tertiary H
Some Alkyl Groups :

CH3 CH2 (Ethyl )

CH3

CH3 CH (Isopropyl )

CH3 CH2 CH 2 CH2 (n-Butyl )

C4 H9
CH3 CH2 CH CH 3 (Sec-Butyl )
CH3
CH3 CH 3 CH CH2 (Isobutyl)
CH3 C (Tert-butyl)

CH3
CH3

CH3 C CH2 (Neo pentyl )

CH3
CH2 CH CH2 (Allyl)
CH2 CH (Vinyl)

or Ph (Phenyl)
or 
CH 2

(Benzyl)

CH

(Benzal)

(Benzo)

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Note : Bond line notation of organic compounds can be as

CH3 CH 3

CH3 CH2 CH3

CH3

CH3 CH CH3

CH CH

C H3 C C CH 3

CH3 CH CH CH2 C CH

CH3

Example 1 : Write the condensed formula and bond line formula for

(a) CH3 — CH — CH2 — CH2 — O H

CH3

(b) H2N — CH2 — CH2 — CH2 — CH2 — COOH

(c) CH3 — CH2 — CH — CH2 — CH2 — CH3

Br

Solution : (a) (CH3)2CH(CH2)2OH or

OH

(b) NH2(CH2)4COOH or NH2 OH

(c) CH3CH2 CH(Br) CH2 CH2 CH3 or

Br

Example 2 : Expand the structure

(a) (b) HO CH2 CH2 NH2

CH3 CH3 H H H
Solution : (a) (b) H—O—C—C—N
H C C H

CH3 H H H
CH3

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NOMENCLATURE SYSTEM
Since organic compounds formation enhanced after mid eighteenth century so as per requirements three types
of nomenclature system has been developed.

1. Common/Trivial name system: System is based on origin of a compound from which it was obtained
or depends upon scientist who discovered it.. e.g,

HCOOH

Formic acid obtained by formica called red ant

CH3COOH

acetic acid obtained by acitum called vinegar

similarly malic acid from malus called apple etc.

Table : Common or Trivial Names of Some organic Compound

Compound Common Name

CH4 Methane
CH3CH2 CH2CH3 n-Butane
(CH3)2CHCH3 Isobutane
(CH3)4C Neopentane
CH3CH2 CH2OH n-Propyl alcohol
HCHO Formaldehyde
(CH3)2CO Acetone
CHCl3 Chloroform
CH3COOH Acetic acid
C6H6 Benzene
C6H5OCH3 Anisole
C6H5NH2 Aniline
C6H5COCH3 Acetophenone
CH3OCH2CH3 Ethyl methyl ether

2. Derived name system: System is based on first member of homologous series.

e.g., CH CH acetylene

CH3 C CH methylacetylene

CH2 CH C CH vinyl acetylene

since organic compounds became more so this system was failed and new system called IUPAC was
developed.

3. IUPAC nomenclature system: Organic chemistry deals with millions of compounds in order to clearly
identify, then a systematic method of naming them has been developed and called IUPAC (International
Union of Pure and Applied Chemistry).

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z Some basic terminology used in IUPAC.

Word root  number of C atoms present in parent chain.

C1 meth -
C2 eth -
C3 prop -
C4 but -
C5 pent -
C6 hex -
C7 hept -
C8 oct -
C9 non -
C10 dec -
C11 undec -
C12 dodec -
––––––
––––––
––––––
C19 nondec
C20 icos

z Prefix (methyl) CH3 O

Functional group/secondary suffix
CH3 CH CH C CH C OH
(oic acid)
Cl
6C in word root Primary suffix(ene)
Prefix (chloro)

IUPAC Name of organic compounds can be written as

no. of C in parent chain ane/ene/yne Functional group

-----Prefix+-----word root +-----+Primary suffix+----Secondary suffix

written in alphabetical order

Note : Position/location in parent chain
A. Rules for writing IUPAC names of the branched alkanes
1. Select the longest C chain containing maximum number functional groups, =,  branching. It should be
straight line or zig zag or cyclic but not complex type.
ex. CH3 CH2 CH2 CH2 CH2 CH3

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Number of C in parent C chain = 6,

Number of C in parent chain = 6
CH3

CH2 C CH2 C
4
so if we open it, as

CH2

C CH3

CH3 CH2 CH3

CH2 C CH2 C CH2 C CH2

CH2 Parent chain

Number of C = 7
HC CH3

CH2
2. (a) For alkanes follow lowest substituent rule/lowest locant rule.for example.

1 2 3 4 5 6 7 8 9()
CH3 CH CH2 CH2 CH2 CH CH2 CH2 CH3 6-Ethyl-2-methyl nonane
9 8 7 6 5 4 3 2 1()
CH3 CH2 CH3
(b) If two substituents are present at same position then we follow alphabatical order

word root
1 2 3 4 5( )
CH3 CH CH2 CH CH3 2-Bromo-4-chloro pentane
5 4 3 2 1( )
Br Cl Primary
(Subs. in alphabetical order) suffix
(c) The name of complex alkyl radical is considered to begin with the first letter of its complete name.

2 CH3

1 CH CH 3
1 2 3 4 5 6 7 8 9
CH3 CH2 CH CH2 CH CH CH2 CH2 CH3

C 2H5 1 CH CH 3

2 CH CH 3

3 CH3

5 – (1, 2 – Dimethylpropyl) – 3 – ethyl – 6 – (1 – methylethyl) nonane.

Note: In alphabetical order for comparison iso-and neo-are considered to be part of fundamental
name of alkyl groups but not sec- and tert-.

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2 CH3

1 CH CH 2
1 2 3 4 5 6 7 8 9
e.g., CH3 CH2 CH2 CH CH CH2 CH2 CH2 CH3

CH CH 2 CH3
1 2 3
CH3

5-sec butyl -4 - isopropyl nonane.

here b is compared by i but not s.
B. Nomenclature of purely unsaturated hydrocarbon
(i) is preferred than if both has the same position.
1 2 3 4 5()
CH2 CH CH2 C CH ; Pent-1-en- 4-yne ()
5 4 3 2 1() Pent -1-yne - 4- ene()

When ene/yne comes in middle remove e from spelling of ene/yne in pure hydrocarbons.

1 2 3 4 5
(ii) HC C CH2 C CH

Pent –1, 4 – diyne (

Penta –1, 4 – diyne ( * When two or more than two times or
 is present then add a in spelling of root word.
1 2 3 4 5 6
(iii) CH 
C CH CH C CH
Hex a 3 ene-1, 5 diyne
1 2 3
(iv) H2C CH CH C CH

4 CH2

5 CH CH2
6

3-Ethynyl hexa – 1, 5 – diene

C. Nomenclature of compounds containing functional groups :
The order of seniority while selecting the principal functional group is as under –

O O O O O

SO3 C OH C OR C Cl C NH2 C N C H

O
C OH NH2 C C C C

The seniority order is applied in numbering when two or more than two functional group are present in
compounds.
4 3 2 1( )
CH 2 CH CH COOH
1 2 3 4( )
z OH
Junior functional group

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Table : Some Functional Groups and Classes of Organic compounds :

Functional IUPAC IUPAC Example

Class of group
group structure group
Compounds suffix
prefix
Alkanes – – –ane Butane,
CH3(CH2)2CH3
Alkenes C=C – –ene But-1-ene
CH2  CHCH2CH3
Alkynes –CC– – –yne But-1-yne
CH  CCH2CH3
Arenes – – –
Benzene,

Halides –X halo– – 1-Bromobutane,

(X = F, Cl, Br, I) CH3(CH2)2CH2Br

Alcohols – OH hydroxy– –ol Butan-2-ol,

CH3CH2CHOHCH3
Aldehydes – CHO formyl, –al Butanal,
or oxo* CH3(CH2)2CHO
Ketones C=O oxo– –one Butan-2-one
CH3CH2COCH3

Nitriles –CN cyano nitrile Pentanenitrile,

CH3CH2CH2CH2CN
Ethers –R–O–R– alkoxy– – Ethoxyethane,
CH3CH2OCH2CH3
Carboxylic –COOH carboxy –oic acid Butanoic acid,
acids CH3(CH2)2COOH

Carboxylate – COO– – –oate Sodium butanoate,

ions CH3(CH2)2COO– Na+
Esters – COOR alkoxycarbonyl –oate Methyl propanoate
CH3CH2COOCH3

Acyl halides – COX halocarbonyl –oyl halide Butanoyl chloride

(X = F, Cl, Br, I) CH3(CH2)2COCl

Amines – NH2, amino– –amine Butan-2-amine,

NH, N– CH3CHNH2CH2CH3

Amides –CONH2, –carbamoyl –amide Butanamide,

–CONHR, CH3(CH2)2CONH2
– CONR2

Nitros – NO2 nitro – 1-Nitrobutane,

CH3(CH2)3NO2

* If C of aldehyde functional group is counted in main chain of carbon then use oxo prefix otherwise formyl.

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O O
6 5 4 3 2 1
e.g. CH3 CH2 C CH2 CH CH 2
CH3 C O CH3 methyl ethanoate
Hex -1-en-4-one
O
O
6 5 4 3 2 1 CH3 CH CH C O C2 H5
CH3 CH CH C CH2 COOH 4 3 2 1
3-oxo hex-4-en-1- oic acid Ethyl but - 2 - en - 1- oate.
O
2 1
CH C OH
4
3 CH C OH
O
2 1
But – 2 – ene –1, 4 –dioic acid CH3 CN Ethanenitrile
O

CH C O
O
CH C CH3 C NH2 Ethanamide
2 1
O
O
CH3
But - 2 - ene- 1, 4 - dioic anhydride
(Succinic anhydride) CH3 C N N-methyl
H ethanamide
5 CH 2 OH
4 CH O
CH3
3 CH Pent- 3-ene -1,2, 5 -triol CH3 CH CH C N
4 3 2 1
2 CH OH CH3
N, N-Dimethyl but-2-en-1-amide
1 CH 2 OH
4 3 2
CH2 CH CH 2 3-Carboxy pentane -1, 5- dioic acid (1979 recommendation)
or
COOH COOH COOH
5 1 Propane –1, 2, 3, - tri carboxylic acid (1993 recommendation)
Note: When functional group starts from vowels as a, e, i, o, u directly, e is removed from spelling of
ane/ene/yne but if it starts from dioic, trioic etc., e is kept.
D. IUPAC Name of Cyclic compounds:
Type - I :

Cyclopropane

Cyclobut-1ene

Cyclopenta-1, 3, -diene

COOH

C N

1 2
3
CH3

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Type II.

CH 3 CH2 Ethylcyclopropane
(large C containing
will be parent chain)

CH3 CH2 CH2 CH2 1- Cyclopropylbutane

1 2 3 4
CH CH CH CH2

1-Cyclohexylbuta-1,3-diene

Type III : Bicyclo compounds having two condensed ring system having 2 common C–atoms.
Bridge head
2 9
2 1 9 1 8
3 8 3

4 7 CH 3
4 7 6
6 5
5
Bicyclo[4. 3. 0]nonane 7-Methylbicyclo[4. 3. 0]non-7-ene
written in descending order

z for numbering largest bridge is preferred first.

1 C2 H5
6 2
7
8
5 3 CH3
4
2-Ethyl-3-methylbicyclo[2. 2. 2]octane
Type IV. Spiro compounds have one common C atom.
1 2

3 bridge head
4
7
5 6
Spiro[2. 4] heptane

in ascending orde
z for numbering smallest cycle is preferred

Example 3 : Write its IUPAC name

CH3
CH3 — CH — CH2 — C — CH3
CH3 CH3
CH3
5 4 3 2 1
Solution : CH3 — CH — CH2 — C — CH3
CH3 CH3
2, 2, 4, trimethylpentane

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Example 4 : Give its IUPAC name.

Et
3 2
Solution : 1-Butyl-2-ethylcyclobutane.
1
4

Bu

Example 5 : Give its IUPAC name.

Solution : 1, 1, 2, 2, 3, 3, -Hexamethylcyclopropane.

Example 6 : Give IUPAC name of

CH3 — CH — CH2 — CH — CH3

C2H5 C2H5

5 4 3
Solution : CH3 — CH — CH2 — CH — CH3

6 CH2 2 CH2

7 CH3 1 CH3

3, 5-Dimetyl heptane.

Example 7 : Write the IUPAC names of the following compounds.

(i) CH3 — CH2 — CH — CH2 — CH2 — CH — CH2 — CH3
OH CH3
(ii) CH  C — CH = CH — CH = CH2
1 2 3 4 5 6 7 8
Solution : (i) CH3 — CH2 — CH — CH2 — CH2 — CH — CH2 — CH3

OH CH3
 The functional group present is an alcohol (OH). Hence, the suffix is ‘–ol’,
 The longest chain containing –OH has eight carbon atoms. Hence, the corresponding
saturated hydrocarbon is octane.
 The –OH is on carbon atom 3. In addition, a methyl group is attached at 6th carbon.
Hence, the IUPAC name is 6-Methyloctan-3-ol
6 5 4 3 2 1
(ii) CH  C — CH = CH — CH = CH2
The two C = C functional groups are present at carbon atoms 1 and 3, while the C  C
functional group is present at carbon 5. These groups are indicated by suffixes ‘diene’ and
‘yne’ respectively. The longest chain containing the functional groups has 6 carbon atoms;
hence, the parent hydrocarbon is hexane. The systematic name is,
Hexa-1, 3-dien-5-yne.

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Example 8 : Derive the structure of (i) Pent-4-en-2-ol, (ii) Cyclohex-2-en-1-ol.

Solution : (i) Pent-4-en-2-ol
‘Pent’ indicates that parent hydrocarbon contains 5 carbon atoms in the chain ‘en’ and ‘ol’
correspond to the functional groups C = C and –OH at carbon atoms 4 and 2 respectively.
Thus, the structure is
CH2 = CHCH2CH(OH)CH3
(ii) Cyclohex-2-en-1-ol
‘1–ol’ means that a –OH group is present at C – 1, OH is suffixed functional group and gets
preference over C = C bond. Thus, the structure is
OH
1
2

NOMENCLATURE OF SUBSTITUTED BENZENE COMPOUNDS

Monosubstituted Benzene Compounds :
(1) For naming the substituted benzene compounds, the prefix used for the substituent is prefixed to the word
‘benzene’ simply.
Example :

CH3 OMe NH2 NO2 Br

Methylbenzene Methoxybenzene Aminobenzene Nitrobenzene Bromobenzene

(IUPAC name)

(2) Many substituted benzene compounds are universally known by their common names.

CH3 OMe NH2

Toluene Anisole Aniline

Disubstituted Benzene Compounds

If two atoms of hydrogen of benzene are replaced by two monovalent atoms or group of atoms, the resulting
product is called disubstituted benzene derivate. For such compounds the position of substituents is defined
by numbering the carbon atoms of the ring such that the substituents get the lowest numbers. There can be
three forms of disubstituted benzene compounds.
(i) 1, 2 (ortho)-form : When the two substituents are on the adjacent carbon atoms.
Br 1, 2-Dibromobenzene
1 (IUPAC system of nomenclature)
2 Br Or
o-Dibromo benzene (ortho is abbreviated as o)
(Trivial system of nomenclature)

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(ii) 1, 3 (meta)-form : When the two substituents are on the alternate carbon atoms.

Br
1, 3-Dibromobenzene
1
2 (IUPAC system of nomenclature)
Or
m-Dibromo benzene (meta is abbreviated as m)
3
Br (Trivial system of nomenclature)

(iii) 1, 4 (para)-form : When the two substituents are on diagonally situated carbon atoms.

Br
1
2
1, 4-Dibromobenzene
(IUPAC system of nomenclature)
3 Or
4 p-Dibromobenzene (para is abbreviated as p)
Br (Trivial system of nomenclature)

Tri- or higher substituted benzene derivatives

For naming the tri- or higher substituted benzene derivatives, the position of the substituents on the ring is
identified following the lowest locant rule.
(i) Compound with three functional groups is named as a derivative of the compound with the principal
functional group at position 1.
COOH

Cl
Example : (i)
Br
O2N NO2
OH
1-Chloro-2, 4-dinitrobenzene 3-Bromo-4-hydroxybenzoic acid
(not 4-chloro-1, 3-dinitrobenzene) (– COOH is the principal functional group)

(ii) Suffix of the base compound is assigned number 1 and then the direction of numbering is chosen such
that the next substituent gets the lowest number.

Example : (i) OH
1 CH3
2

3
4
I
4-Iodo-2-methylphenol
(– OH is the principal functional group)

(ii) When a substituent is such which when taken together with the benzene ring gives a
special name to the molecule then it is named as derivative of that molecule with the
substituent.
OMe NH2 OH
1 Cl 1 CH3
2 2

3 3
4 4 CH3
CH3 C2H5 CH3
2-Chloro-4-methylanisole 4-Ethyl-2-methylaniline 3, 4-Dimethylphenol

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(iii) When the benzene ring is attached to an aliphatic chain having a functional group, it is
named as phenyl (C6H5–) derivative of that aliphatic compound.

Br
OH
4 3 2 1
CH2 — CH — CH2 — CHO
Br
2, 3-Dibromo-l-phenylpentane 3-hydroxy-4-phenylbutanal

ISOMERISM
Isomerism means (Greek, isos – same, meros – parts) compounds having same molecular formula but
differing in their physical and chemical properties; phenomenon is called isomerism and compounds which
show it are called isomers.
Isomerism

Structural isomerism Stereo isomerism

(arises due to difference in (arises due to difference in
arrangements of atoms with in the arrangements of atoms/group
molecules having different structural of atoms in space)
formula)

Geometrical Conformation/ Optical

isomerism rotational isomerism isomerism

Chain Position Functional Metamerism Tautomerism

isomerism isomerism isomerism
Structural isomerism/Isomers
(a) Chain isomers: Compounds having same molecular formula but differ in C – C chain length.
In alkanes for showing chain isomerism minimum 4C atoms are required.

CH 3CH 2 CH2 CH3

C4H 10

CH3 CH CH 3

CH3

CH3 CH2 CH2 CH2 CH3

CH3
C5 H12 CH3 CH CH2 CH3
CH3

CH3 C CH3

CH3

Similarly
C6H15 5 chain isomers
C7H16 9 chain isomers

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(b) Position isomers : Compounds having same molecular formula, same carbon chain, same functional
group but differ in position occupied by an atom or functional group in C chain.

CH3 CH 2 CH 2 CH 2 OH Butan- 1- ol 

C4H 9OH 
CH 3 CH 2 CH CH 3 Butan- 2- ol 

OH
e.g. 3 2 1
CH 3 CH 2 CH 2 O CH 3 1-Methoxypropane 
CH3 

CH 3 CH O CH 3 2-Methoxypropane 


CH3 CH2 CH CH2

C4 H8

CH3 CH = CHCH3

(c) Functional isomers : Compounds having same molecular formula but they differ in functional groups in
which compound can be seperated out.
generally

Aldehyde Acid

z Cn H2nO z Cn H2nO 2
Ester
Ketone

Alcohol

z Cn H2n + 2O z RCN and RNC

Ethers

z Alkyne and alkadiene

(d) Metamers : Compounds having same molecular formula and functional groups but differing from each other
due to unequal distribution of C atoms on either side of Polyvalent atom/Functional groups. It generally
exists in ethers, thio ether, ketone, esters, secondary amines etc.

C2 H5 O C2H5
I
e.g. (i) C4 H10O CH3 O CH2 CH2 CH3
II
CH3

CH3 O CH CH3
III

CH3 CH2 C CH2 CH3

I
(ii) C5 H10O O

CH3 C CH2 CH2 CH3

II
here I and II are also position isomers.

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(e) Ring chain isomers : Compounds differ in node of linking of C atoms i.e.,isomers differs in either closed
chain or open chain. They are indeed functional isomers.

CH3 CH2 C H2 CH2 CH CH 2

e.g. C6 H12 Heptene

Cyclohexane

(f) Tautomerism/Desmotropism. Tautomers are special type of functional isomers which exists
simultaneously in dynamic equilibrium.
 In majority cases it is due to shifting of H atom between two polyvalent atoms like O, N with
rearrangement of single or double bond.
 Tautomers are interconvertible by two ways.
(i) 1, 2 – Diad system

1 2
H C N C N H
cyano Isocyano

1 2 O
H O N O H N
O
Nitrite Nitroform

OH O

H O P OH H P OH

(ii) 1, 3 -Triad system : Here H atom shifts from first to 3rd polyvalent atom in a chain.
e.g. ketoenol, Nitro Aciform.

O3 OH

e.g. CH3 C H CH2 C H
1 2
Keto enol

at room temperature 100% Traces of keto form.

3O O OH O

CH3C CH 2 O C 2H 5
C CH3 C CH C O C 2H5
2 1 ketoform enolform
(Ethyl aceto acetate)

O OH

CH3 N CH2 N
1 2 O O
Aci-form
CH3
 O
Ph C N No tautomerism
O because no – H is present.
CH3

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Example 9 : Draw all the possible isomers of monochloro Butane (C4H9Cl)

Solution : (i) CH3 — CH2 — CH2 — CH2 — Cl (ii) CH3 — CH2 — CH — CH3
Cl

CH3 CH3
(iii) CH3 — CH — CH2 — Cl (iv) CH2 — C — CH3

Cl

Example 10 : Draw the polygon formulae for all the possible structural isomers having the molecular formula
C5H10.
Solution : Polygon formulae = Cyclic formulae

Cyclopentane Methyl cyclobutane Ethyl cyclopropane

1, 1-Dimethyl 1, 2-Dimethyl cyclopropane

cyclopropane

Stereoisomerism
Isomers which have the same structural formula but have different relative arrangement of atoms or groups in
space are called stereoisomers and the phenomenon is called stereoisomerism.
It has two types:
(i) Geometrical Isomerism
(ii) Optical Isomerism
We will learn more about Geometrical isomerism in next chapter and optical isomerism in class XII.

EXERCISE
1. The IUPAC name of CH2 = CH – CH2 – OH is
(1) Propynal (2) Propanol
(3) Prop-2-enol (4) Propenyl alcohol
2. Which of the following alkane cannot show isomerism?
(1) C4H10 (2) C3H8
(3) C5H12 (4) C6H14
3. The number of structural isomers possible for C4H8 is
(1) 4 (2) 3
(3) 5 (4) 6

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NEET Organic Chemistry - Some Basic Principles and Techniques 19
O
OH
4. IUPAC name of the compound is

(1) 5-Hydroxycyclopent-2-enone (2) 2-Hydroxycyclopent-4-enone

(3) 5-Hydroxycyclopent-4-enone (4) 2-Hydroxycyclopent-2-enone
5. In which of the following cases functional isomerism is not possible?
(1) C2H6O (2) C3H6O
(3) CH2O (4) C4H8O2
6. Which of the following is a heterocyclic compound?

(1) (2)
O S

(3) N (4) All of these

H
7. Which of the following will not show tautomerism?
(1) HCN (2) CH3CH2NO2
O
(3) (4) CH3OH

8. Which of the following is not the isomer of CH3COOH?

O O O
(1) H – C – C – H (2) H – C – OCH3
OH O
(3) H2C – C – H (4) All of these
9. IUPAC name of CH3 – C – CH2 – C  N is
O
(1) 1-Cyanopropan-2-one (2) 3-Cyanopropan-2-one
(3) 3-Oxobutanenitrile (4) 2-Oxopropanenitrile

GENERAL CONCEPTS IN ORGANIC REACTION MECHANISM

In organic reactions reactant which contains C and gets attacked are regarded as SUBSTRATE and which
attacks is called ATTACKING REAGENT. The new substances formed as a result of reaction are called
PRODUCTS.
For reaction there may be any path as given.
 Path I Substrate + reagent Activated complex Products
 Path II Substrate + reagent Activated complex Intermediate Products

Note :
 Concerted mechanism (without formation of intermediate)
 Non concerted mechanism (involves formation of intermediates)

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Thus mechanism is hypothesis which explains various facts regarding a chemical change.
In the complete discussion of reaction mechanism. There is involvement of following concepts.
1. Type of bond fission : We know that organic compounds have covalent bond generally. These covalent
bond can be broken by two ways

Homolysis / Homolytic fission A B A+ B

or
Covalent A B Free radicals
bond
Hetrolysis / Hetrolytic fission A B A+ B

 Homolysis is usually carried out at high temperature or under influence of UV radiations or in

presence of radical initiators such as peroxides.

UV light
Cl Cl Cl + Cl Free radica l

O O O
D
C6 H5 C O O C C6 H5 2C6 H5 C O

 Heterolytic bond fission requires more energy than Homolytic bond fission. The energy required to
cause hetrolytic bond fission is obtained by dipole-dipole interaction.

C H3 C H3
 
CH3 C Br C + Br

CH3 H3 C CH3
Carbocation

2. Types of reagents : In general more reactive reactant is called reagents less reactive reactant containing
C atom is called substrate.

Electrophiles / Electrophilic reagents

Reagent Nucleophile / Nucleophilic reagents
Radical

(i) Electrophile/Electron loving : Species which can act as electron pair acceptor are called as
electrophiles.

(a) Species having positive charge B
    
NO 2, R, H, Cl, Br, etc.

(b) Neutral species having incomplete octet of central atom

BF3, AlCl3, :CX2(Carbenes), SO3 etc.
(c) Neutral species having vacant d- orbital of central atom.
SnCl4, SiF4, etc.

¨ 3, H 2O¨ ,:Br :etc.

: :

(ii) Nucleophiles : Species having at least one lone pair of electrons. i.e., :CN : , NH
¨

(iii) Radical : Species having unpaired electron. i.e., ·CH3, :Br ·etc.
: :

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NEET Organic Chemistry - Some Basic Principles and Techniques 21

Example 11 : Classify the reagents shown in bold in the following equations as nucleophiles or electrophiles.
(a) CH3COOH + OH – –
CH3COO + H2O
(b) CH3CH2Br + SH – CH3CH2SH + Br
–

(c) C6H6 + CH 3CO+ C6H5COCH3 + H

+

Solution : (a) OH– is a nucleophile.

(b) SH– is a nucleophile.
(c) CH3CO+ is an electrophile.

3. Electronic displacement effects :

(i) Inductive effects
For a system like
    
     
C4 C3 C2 C1 X (X is more electronegatitive than C) and

   
    
C4 C3 C2 C1 X (X is less electronegative than C-atom)

The permanent displacement of electrons in a bond towards, the more electronegative element
is called inductive effect. The effects provides polarity to the molecules. It is measured with respect
to H.
+ I effect electron donating effect
e.g., (CH 3)3 C (CH3) 2 CH CH3 CH2 CH3 H
Inductive effect

– I effect electron with drawing effect.

Order of electron with drawing tendancy ie (– I effect).
O

e.g., R3N NO 2 CN SO3 H C OH F Cl Br I

OH OR NH2 NHR NR2

(ii) Electromeric Effect

The temporary displacement of  electrons towards the more electronegative element in the presence
of a reagent is electromeric effect. If reagent is an electrophile then the effect is (+E) and if a
nucleophile then (–E) effect.

Reagent 
C=O C– O

(iii) Resonance Effect

The effect involves permanent delocalization of conjugated  electrons in a conjugated system.
Intermediate structures formed are called resonating structure. The structure that collectively
represents all resonating structures is believed to be most stable and is called the resonance hybrid.
During resonance in a benzene molecule all C–C bond lengths happen to be the same and lie in
between those of a single and double bond.
Resonating structures must have the same position of nuclei, same number of total electrons. The
one having the more number of covalent bond, negative charge on electronegative and positive charge
on electropositive element with more charge separation found to be more stable then others.

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The difference between energies of most stable resonating structure and the resonance hybrid is
called the resonance energy of the molecule.

+ R or + M Effect : If a conjugated system has an electron-donating group is said to have + R or

+ M effect. Such as – OH, – OR, – SH, – NH2, – NHR, – X (halogen) etc.
For example,

  
O— H O— H O— H O— H O— H OH+
– –

–  – +
CH2 = CH — O — CH3 CH2 — CH = O — CH3 CH2 — CH — O — CH3

– R or – M Effect : If a conjugated system has an electron-withdrawing group is said to have – R

or
– M effect. Such as – CHO, – COOH, – COOR, – CN, – NO2, C = O etc.
For example,

O – – – –
N NO2 NO2 NO2 NO2 NO2
O
 

–
O O O –
 +
CH2 = CH — C — CH3 CH2 — CH = C — CH3 CH2 — CH — C — CH3

Example 12 : Draw resonance structures for the C6H5NH2.

  
NH2 NH2 NH2 NH2 NH2
– –
Solution :
–

(iv) Hyperconjugation effect : Hyperconjugation effect is an extension of the resonance effect or Baker-
Nathan effect. It describes the orbital interactions between the  systems and adjacent  bonds
of substituent group(s) in organic compound. It is also known as - conjugation.
 This effect arises due to partial overlap of sp3-s (a C –H bond) with adjacent empty p-orbital.
It is represented by .

H H H H

H C CH = CH2 H C = CH – CH2 H C = CH – CH2 H C = CH – CH2

H H H H


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NEET Organic Chemistry - Some Basic Principles and Techniques 23

H H H H
 
H C CH2 H C = CH2 H C = CH2 H C = CH2

H H H H


H H H H
H C CH 2 H C = CH2 H C = CH 2 H C = CH2
H H H H

There is no bond between carbon and hydrogen atoms in these structure. Therefore hyper
conjugation is also known as no-bond resonance.
In the above structure, although a free proton has been shown, it is still bound quite firmly to
the -cloud and hence is not free to move.
It is clear from the above structures that hyperconjugation occurs through the H-atoms present
on the carbon atom next to the double bond i.e., -hydrogen atom.

Number of  -Hydrogen atom  Number of Hyperconjugation structure

Number of Hyperconjugative structure  Stability

Note : Above effects are used to explain the stability of carbonium ion, Carbanion, Free radicals etc.

EXERCISE
10. Which of the following has maximum –I effect?
(1) – F (2) –NO2
(3) –CN (4) –OH
11. Carbon-carbon bond-order in benzene is 1.5 due to
(1) Inductive effect (2) Electromeric effect
(3) Resonance (4) H-bonding
12. Which of the following is the most acidic?
(1) HCOOH (2) CH3COOH
(3) CH3CH2COOH (4) CH3CH2CH2COOH
13. In which of the following, the group attached to the benzene ring shows +R effect?
OH NH2

(1) (2)

OCH3

(3) (4) All of these

14. Which of the following compound can show resonance?

(1) CH2 = CH – CH = CH2 (2) CH2 = CH – CHO
(3) CH2 = CH – NH2 (4) All of these

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15. Which of the following is the most stable carbocation?

 
(1) CH3 – CH – CN (2) CH3 – CH – OCH3

 
(3) CH3 – CH – NO2 (4) CH3 – CH – CH3

16. Electromeric effect involves the complete transfer of

(1) -electron (2) -electron
(3) Proton (4) Both  and  electrons
17. The most deactivating group for electrophilic substitution reaction in benzene ring is
(1) –NH2 (2) –NO2
(3) –OCH3 (4) –CHO
18. Hyperconjugation involves delocalisation of
(1) -electron (2) -electron
(3) Both  and  electron (4) Proton
19. Which of the following alkene is the most unstable?

H H
(1) C=C (2) CH3 – CH = CH2
H H
H3C CH3
(3) H3C – CH = CH – CH3 (4) C=C
H3C CH3
20. The maximum number of no bond resonating structure is possible in
 
(1) CH3 – CH2 (2) H3C – CH – CH3
CH3 CH3

(3) C2H5 – C – CH3 (4) H3C – C – CH3

 

REACTION INTERMEDIATES
The species produced during cleavage of bonds are called reaction intermediates, these are generally
short-lived and highly reactive and hence cannot be isolated. The typical intermediates are
(i) Carbocation
(ii) Carbanions
(iii) Free radicals
Carbocation
The stability of different carbocation by resonance.
   
(C6 H5 )3 C > (C6 H5 )2 C H > C6 H 5 C H2 > C H2 – CH = CH2
The stability of different carbocation by inductive and hyper conjugation.
CH3 CH3
 
CH3 C  > H C  > CH3 CH2 > CH3
CH3 CH3

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NEET Organic Chemistry - Some Basic Principles and Techniques 25
Carbanion
The stability of different carbanions by resonance

(C6 H5)3C > (C 6H5)2 CH > C6H5 – CH2 > CH2 – CH = CH2

The stability of different carbanion by inductive effect.

CH3
CH3 > CH3– CH2> CH3– CH > CH3 – C
CH3 CH3

Free Radicals
Stability of different free radicals by resonance

(C6 H5)3 C > (C 6H5 )2 CH > C6H5 – CH 2 > CH 2 – CH = CH2

Stability of different free radicals by inductive effect and hyperconjugation.

(CH3 )3 C > (CH 3)2 CH > CH3 – CH2 > CH3

CLASSIFICATION OF ORGANIC REACTIONS

Organic reactions are mainly of four types.
1. Substitution reactions
2. Addition reactions
3. Elimination reactions
4. Rearrangement reactions
1. Substitution reactions : In substitution reactions any atom/group of atoms is replaced by similar type
of atom/group of atom. It may be Nucleophilic /Electrophilic/Free radical substitution reaction.
+ –
e.g., Nu R X Nu R :X —Nucleophilic substitution reaction

H E

E+ H Electrophilic substitution reaction

UV ligh t
CH4 + Cl2 CH3Cl + HCl –– Free radical substitution reaction
2. Addition reactions : Those organic reactions in which the reactant combine to form a single product having
all the atoms of combining units are termed as addition reactions.
Depending upon the nature of the attacking species (electrophiles, nucleophiles free radicals), addition
reactions are of the following three types :
(a) Nucleophilic addition reactions
(b) Electrophilic addition reactions
(c) Free radical addition reactions
(a) Nucleophilic Addition Reactions : Addition reactions brought about by nucleophiles are called
nucleophilic addition reactions. These reactions are typical of aldehydes and ketones.

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Example : Base catalysed addition of HCN to aldehydes or ketones.

HO– + H — CN H2O + CN
–

Nucleophile

CN CN
R
H — CN
C = O + CN– R—C—O
–
R — C — OH
H
Aldehyde
H H
Cyanohydrin
(Addition Product)

(b) Electrophilic Addition Reactions : Addition reactions brought about by electrophiles are called
electrophilic addition reactions. These reactions are typical of alkenes and alkynes. For example,
addition of HBr to alkene in absence of peroxide.
Slow  Br–
CH3 — CH = CH2 + H+ CH3 — CH — CH2 CH3 — CH — CH3
Carbocation
Br
(Addition product)

(c) Free Radical Addition Reactions : Addition reactions brought about by free radicals are called free-
radical addition reactions.
Ex : Addition of HBr to alkenes in presence of peroxide

R—O—O—R Homolytic
2RO
Peroxide fission Free radicals

RO + H — Br ROH + Br
Free radical

CH3 — CH = CH2 + Br CH3 — CH — CH2Br

Free radical

H — Br

CH3 — CH2 — CH2 — Br + Br

Propylbromide
(Addition product)

3. Elimination reactions : In an elimination reactions two atoms/ group of atoms are eliminated to give the
product. In elimination reaction bond order always increases.
 Depending on relative position of atoms eliminated, these are classified as // elimination
reactions.
e.g.,  - elimination
R R R
:
a H Cl
R C Cl R C Cl R C: (Carbene, reactive intermediate)
:
H
 - elimination

H
 Alc.KOH
CH3 CH2 CH CH2 Br CH3 CH2 CH = CH2 + KBr + H2O
CH3 CH3
Conc. H 2SO 4
CH3 CH CH2 OH CH3 C = CH2+ H 2O

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 - elimination


C  C
 Zn dust
C C C C + ZnBr2

Cl Cl

 In  - elimination always 3 membered cyclic product is obtained.

4. Re-arrangement reactions : In these reactions, the atoms or group of atoms changes their positions
in the same molecule. e.g.,
CH3 1,2 CH3 CH3
shift
CH3 C CH2 CH3 C CH2 CH3 (1, 2- methyl shift)
3º
1º
(more stable)
CH3
(less stable)

C H3 H O H
C H C
(Beckmann-rearrangement)
N N
OH H 3C H

CH3
anhydrous AlCl3/HCl
CH3CH2CH2CH3 CH3 CH CH3 (Isomerisation)
D

CH3 H h CH3 CH3

C=C C=C
(Trans) H CH3 H H (Cis)

Note :
Pericyclic reactions : Those chemical reactions which do not involve ionic or free radical
intermediates. These reactions occur in single step via cyclic transition state. The breaking and
formation of bonds takes place simultaneously; such reactions are called pericyclic reactions.
CH 2

e.g. + (Diel's Alder Reaction)
CH 2
4 e 2 e
H


C
N
N
4 e
 Hydroboration of alkene is also a type of pericyclic reaction.

Note : Compounds having multiple bond are called Ambiphiles like.

–
O

C , CN etc.
+

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EXERCISE
21. The most stable carbocation is

 
(1) CH  CH (2) CH2  CH


(3) CH3 – CH2 (4) All are equally stable

22. The hybridisation of carbon in CH2 = CH – CH2 is

(1) sp3 (2) sp2
(3) sp (4) sp3d
23. The most stable carbocation amongst the following is


(1) (2) CH2  CH – CH2


 
(3) CH3 – CH2 – CH2 (4) CH2  C  CH

24. The carbanion which is most stable is

(–) (–)
(1) HC  C (2) CH2  CH – C H2

CH3
(–)
(–)
(3) CH2 (4) CH3 – C
CH3

25. The carbocation which is most stable is

 
CH2 CH2

(1) (2)

NO2 OCH3


CH2 
CH2

(3) (4)

CN

26. Towards catalytic hydrogenation, the least reactive compound is

(1) CH2 = CH2 (2) CH3 – CH = CH2
(3) H3C – CH = CH – CH3 (4) (CH3)2C = C(CH3)2

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27. Which of the following reaction is an example of -elimination?

Cu Base
(1) CH3 – CH2 – OH 

 CH3CHO (2) CH3 – CH2Br 

 CH2  CH2

Zn
(3) Br – CH2 – CH2 – CH2 – Br   (4) CH2  CH2 
H2
Pt
CH3 – CH3

28. Which of the following is aromatic in nature?

(1) (2)
 

(3) (4) All of these

29. Which of the following is not a heterocyclic aromatic compound?

(1) (2) N
O H

(3) (4)
O N

30. CH3 CH2 CH2 Cl 

aq. KOH
 CH3 CH2 CH2 OH .

The reaction is
(1) Free radical substitution (2) Addition reaction
(3) Elimination reaction (4) Nucleophilic substitution
31. Which is not an aromatic species?
(1) Graphite (2) Benzene

(3) Fullerene (4)

32. Which of the following is aromatic carbanion?

(1) (2)
( ) ( )

( )

(3) (4) All of these

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METHODS OF PURIFICATION OF ORGANIC COMPOUNDS

The organic compounds whether isolated from a natural source or prepared in the laboratory are mostly impure.
These are generally contaminated with some other substances. A number of methods are available for the
purification. The choice of a particular technique or method depends upon the nature of the compound whether
solid or liquid and also upon the nature of the impurities associated with it.
The common techniques used for purification are as follows :
(i) Sublimation
(ii) Crystallisation
(iii) Distillation
(iv) Differential extraction
(v) Chromatography
Once the compounds have been properly purified and the purity checked by finding out the m.p. or b.p.
depending upon their nature. Most of the pure compounds have sharp melting and boiling points. Nowadays,
the purity of an organic compound is determined by new methods based on different types of chromatographic
and spectroscopic techniques.
Let us briefly learn about different methods for purifying the organic compounds.
(i) Sublimation : Certain organic solids on heating directly change from solid to vapour state without passing
through a liquid state. Such substances are called sublimable. This process is called sublimation. The
vapours on cooling change back to the solid form.
Heat

Solid   Vapours
Cool

The sublimation process is used for the separation of sublimable volatile compounds (such as camphor,
naphthalene, anthracene, benzoic acid etc.) from non-sublimable impurities.
(ii) Crystallisation : This is the most common method for purifying organic solids. This method is based on
the differences in the solubility of the organic compound and its impurities in a suitable solvent. This
method involves the following steps :
(a) Preparation of the Solution : The organic compound is dissolved in a suitable solvent by heating.
The amount of solvent should be just sufficient to dissolve the whole of the solid on heating and the
solvent should not dissolve the impurities and chemically unreactive with the compound.
(b) Filtration of the Solution : The hot solution obtained above is then filtered immediately.
(c) Crystallisation : The hot filtrate is allowed to cool slowly and undisturbed in a beaker. After some
time the crystals of the pure compound are formed. When the crystallisation is complete, the crystals
are separated from the mother liquor by filtration. The crystals left in the funnel are washed once or
twice with a little of the cold solvent to remove adhering impurities, if any.
(d) Drying of Crystals : The crystals are dried by pressing between the folds of filter paper and then
placed in a steam or air oven for some time. The crystals are finally dried over sulphuric acid or
calcium chloride in a desicator.
(e) Removal of Colour : Sometimes, the crystals obtained are slightly coloured due to the presence
of certain coloured impurities. In such cases, crystals are redissolved in the same solvent and a small
amount of activated charcoal is added to it. The mixture is boiled for 15–20 minutes. During this
treatment, charcoal adsorbs all the coloured impurities. The charcoal is then filtered out and the filtrate
is allowed to cool when colourless crystals of the pure substance are obtained.
(iii) Distillation : “Distillation is the process of converting a liquid into vapours upon heating and then cooling
the vapours back to the liquid state”.
 Cool
Liquid Vapour Liquid

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The process of simple distillation is used to purify those organic liquids which are quite stable at their
boiling points and the impurities present are non-volatile. Liquids such as benzene, toluene, ethanol,
acetone, chloroform, carbon tetrachloride can be purified by simple distillation.
(a) Fractional Distillation : This method is used to separate two liquids whose boiling points are very
close to each other (say differ by 10 K to 30 K) and the separation cannot be achieved by simple
distillation. This is achieved by the use of fractionating column.
Mixtures whose components can be separated by fractional distillation process are
(i) Fractional distillation of crude oil (Petroleum)
(ii) Methyl alcohol (CH3OH) and acetone (CH3COCH3)
(iii) A mixture of C6H6 (Benzene) and CH3C6H5 (Toluene)
(b) Distillation Under Reduced Pressure or Vacuum Distillation
This method is used for the purification of high boiling points liquids and liquids which decompose
at or below their boiling points.
A liquid boils when its vapour pressure becomes equal to the external pressure. Obviously, the same
liquid would boil at a lower temperature if the pressure acting on it is reduced. Since the liquid now
boils at a lower temperature, its decomposition does not occur.
Pressure can be reduced up to 10 - 20 mm of Hg with the help of water section pumps in the
laboratory. However with vacuum pumps, pressure of the order of 0.1 mm of Hg can be easily obtained.
Some of the examples where vacuum distillation has been used to purify liquids.
(i) Glycerol which decomposes at its boiling point (563 K) can be distilled without decomposition
at 453 K under 12 mm Hg pressure.
(ii) Concentration of sugarcane in sugar industry.
(c) Steam Distillation : The process is employed for the distillation of those substances (solid or liquid)
which
(i) are insoluble in water
(ii) have high molecular mass
(iii) have fairly high vapour pressure at about 373 K and
(iv) are volatile in steam but the impurities present in them are non-volatile.
In this process, a mixture of two immiscible liquids. i.e., water and an organic liquid is heated. Each
would exert its own vapour pressure independently of the other and the mixture will begin to boil at
a temperature when the sum of the vapour pressure of the organic liquid (p1) and that of water (p2)
becomes equal to the atmospheric pressure (P).

P  p1  p2

Evidently, the temperature at which the mixture boils must be lower than the normal boiling point
of both the components. In other words, the organic liquid boils at a temperature lower than its normal
boiling point and hence, the decomposition is avoided.
The process of steam distillation can be applied for the separation of a mixture of o-nitrophenol and
p-nitrophenol.
In this process, water vapours carry along with them vapours of o-nitrophenol which is more volatile
and they get condensed in the receiver. p-nitrophenol with higher boiling point remains in the distillation
flask.
This method can also be used for the purification of impure sample of aniline.

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(iv) Differential Extraction : This technique is normally used to separate certain organic solids dissolved
in water by shaking with a suitable organic solvent. The process called extraction is done in a separating
funnel. The organic solvent selected should be such that
(a) The given solid must be more soluble in the organic solvent than in water.
(b) Water and organic solvent should not be miscible with each other.
The most common organic solvent is ether. Benzene and acetone can also be used. However alcohol is
not the proper solvent because water and alcohol are highly miscible with each other.
(v) Chromatography : Chromatography is a modern and sensitive techniques used for rapid and efficient
separation or analysis of components of a mixture and purification of the compounds.
The name chromatography is based on the Greek word Chroma meaning colour and Graphy for writing
because the method was Ist used for the separation of coloured substances found in plants.
Chromatography is essentially a physical method of separation. It is defined as follows :
“The technique of separating the components of a mixture in which separation is achieved by the differential
movement of individual components through a stationary phase under the influence of a mobile phase”.
Based on the principle involved chromatography is classified as

Column chromatography
(a) Adsorption
chromatography Thin layer chromatography (TLC)

(b) Partition chromatography – Paper chromatography

(a) Adsorption Chromatography : Principle : This method is based upon the differential adsorption of
the various components of a mixture on a suitable adsorbent such as silica gel or alumina. Since
some compounds are more strongly absorbed than the other, they will travel through the column at
different rates and thus get separated.
Column Chromatorgraphy : Column chromatography involves separation of a mixture over a column
of absorbent (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.

Solvent
Mixture of
Compounds a a
(a + b + c) b+c b
Adsorbent
(staitionary c
phase)
Glass wool

An appropriate element 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 absorbed substances are retained near the top and others come down to various distances
in the column as shown in the above figure.
Thin Layer Chromatography (TLC) : This is also a type of adsorption chromatography. In this
chromatography separation of components of a mixture is done over a thin layer of an adsorbent. Thin
layer of adsorbent (silica gel or alumina) is spread over a glass sheet. The plate is called as thin layer
chromatography plate (TLC-plate). The solution containing different component is applied as a small spot
at one end of plate. Now the plate is kept in solvent. The solvent moves up along with components of
mixture. The components depending on their degree of adsorption moves up to different distance resulting

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in separation. The relative adsorption of components represented in terms of retention factor i.e., (Rf) value
or retardation factor.

Jar Solvent
Adsorbent coated Front
on glass plate
Spot
Base line Sample dot
y
Solvent x Base line

Distance moved by the substance from base line (x)

Rf =
Distance moved by the solvent from base line  y 

The spots of coloured compounds are visible on TLC plate due to their original colour. The spot of colourless
compounds which are invisible to the eye but fluorescence can be detected by putting the plate under
UV light. Another detection technique is to place the plate in a covered jar containing a few crystals of
iodine (I2). Spots of compounds which absorbe 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.
(b) Partition Chromatography — Paper Chromatorgraphy : This is a special type of chromatography in
which stationary and mobile both phases are liquid. Its common example is paper chromatographly in
which chromato graphic paper is used as adsorbent surface as TLC plate in thin layer chromotography.
The water trapped in paper acts as stationary phase. In the paper chromatography, the chromatography
paper is spotted by a mixture containing different components and then placed in a solvent which acts
as mobile phase and rises up due to capillary action. The paper selectively retains different components
named as chromatogram. These chromatogram may be observed in ultraviolet light or by spraying suitable
reagents.

Card board Card board

Chromatography Chromatography
Base line paper
Jar Jar
Spot
Spot
Base line
Solvent Solvent

QUALITATIVE ANALYSIS OF ORGANIC COMPOUNDS

In the study of any organic compound, it is an important step to know the elements present in it. In addition
to carbon and hydrogen, organic compounds contain some other elements, e.g., nitrogen, sulphur, halogens,
etc. These are detected as follows :
(a) Detection of Carbon and Hydrogen : Carbon and hydrogen are detected by heating the compound with
copper (II) oxide. Carbon present in the compound is oxidised to carbon dioxide and hydrogen to water
vapours.
C + 2CuO 2 Cu + CO2
(From Organic Compound)

2H + CuO Cu + H2O
(From Organic Compound)

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The gaseous products of reaction are passed. Ist over anhydrous copper sulphate powder (white) taken
in a bulb and then through lime water. If the reaction products contain water vapour, copper sulphate will
turn blue and if only carbon dioxide is present, it will not have any effect on copper sulphate but it will
turn lime water milky.

CuSO4  5H2O  CuSO4 .5H2O

Colourless
 White 

Ca  OH2  CO2  CaCO3  H2O

Lime
water

Organic
Compound +
Cupric Oxide
Bulb with
a Trap

Anhydrous
CuSO4
Lime water

This test is also known as copper oxide test. Detection of other elements nitrogen, sulphur, halogens and
phosphorous present in an organic compound are detected by Lassaigne’s test. For this test we have
to prepare the sodium extract or Lassaigne’s extract. For the preparation of sodium extract we take a
small piece of a dry sodium metal in a fusion tube and heated gently till it melts to a shining globule.
Then, a small amount of organic compound is added and the tube is heated strongly till it becomes red
hot. The red hot tube is plunged into distilled water contained in a china dish and boiled for some time.
It is then cooled and filtered. The filterate liquid is known as sodium extract.
During fusion, the elements present in organic compound combine with sodium to form ionic compound
of sodium as

Na + C + N  NaCN

2Na + S   Na2S

Na + X 
 NaX [X = Cl, Br, I]
C, N, S and X come from organic compound.
(b) Test for Nitrogen : About 2 ml of sodium extract is taken in a test tube and made alkaline by adding
NaOH solution. To this reaction mixture is added freshly prepared FeSO4 solution and boiled for
3–4 minutes. The formation of Prussian blue colour or precipitate shows the presence of nitrogen. The
reactions involved are the following:
Na + C + N NaCN
Sodium cyanide
(sodium extract)

FeSO 4 + 2NaOH Na2SO 4 + Fe(OH)2

+2
6 NaCN + Fe(OH)2 Na4[Fe(CN)6] + 2NaOH
Sodium ferrocyanide
(sodium hexacyanoferrate II)

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On heating with conc. H2SO4, some iron (II) ions are oxidised to iron (III) ions.
3 2
Na4 [Fe  CN6 ]  Fe3 
 Fe4 [Fe  CN6 ]3
Sod. ferrocyanide Iron hexacyanoferrate II
 ferric ferrocyanide 
Prussian blue in colour

(c) Test for Sulphur

(i) Lead Acetate Test : Acetic acid and lead acetate are added to a portion of sodium extract, black
precipitate confirms sulphur.

Na2S +  CH3 COO 2 Pb  PbS  2CH3 COONa

From sodium extract Black ppt. Sodium acetate
Lead acetate

(ii) Sodium Nitroprusside Test : To another portion of sodium extract, add a few drops of sodium
nitroprusside. Violet colour confirms sulphur.

Na2S  Na2 [Fe  CN5 NO]  Na4 [Fe  CN5 NOS]
From sodium extract Sodium Nitroprusside Sodium thionitroprusside

(iii) Lassaigne’s Test : If organic compound contains both nitrogen and sulphur, sodium thiocyanate is
formed.

Na + C + N + S 
 NaSCN
From organic
compound

 Fe  SCN3
3NaSCN + Fe3 
Ferric sulphocyanide
Blood red colour 

The formation of blood red colour in Lassaigne’s test for nitrogen indicates the presence of both
nitrogen and sulphur in organic compound.
(d) Test for Halogens : A portion of the sodium extract is Boiled with 2–3 ml concentrated HNO3 followed
by cooling and addition of AgNO3 solution when
(i) A white precipitate, soluble in ammonia solution (NH4OH) and insoluble in dil HNO3 indicates the
presence of chlorine.
NaCl + AgNO3 AgCl + NaNO3
White ppt

NH4OH

Soluble
(ii) A pale yellow precipitate partially soluble in ammonia solution indicates the presence of bromine.

NaBr + AgNO3 
 AgBr + NaNO3
Pale yellow ppt

(iii) A yellow precipitate insoluble in ammonia solution indicates the presence of iodine.

NaI + AgNO3 
 AgI + NaNO3
Yellow ppt

In case, nitrogen and sulphur are present along with the halogens in the organic compound, the
Lassaigne’s extract contains sodium sulphide (Na2S) and sodium cyanide (NaCN) along with sodium
halide. These will form precipitates with silver nitrate solution. Therefore, nitric acid is added to
decompose sodium cyanide and sodium sulphide.
NaCN + HNO3 NaNO3 + HCN

Na2S + HNO3 2NaNO3 + H2S

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If cyanide and sulphide ions are not decomposed, they will react with silver nitrate and hence will
interfere with the rest.

NaCN + AgNO3 
 AgCN + NaNO3
Silver cyanide
 White ppt 

Na2S + 2AgNO3 
 Ag2S + 2NaNO3
Silver sulphide
Black ppt 

(e) Test for Phosphorus : Phosphorus is detected by fusing the organic compound with an oxidising agent
i.e., sodium peroxide when phosphorus is oxidised to sodium phosphate.


5Na2O2  2P   2Na3PO4  2Na2O
 from organic compoud

The fused mass is extracted with water. The aqueous solution is boiled with conc. HNO3 and then
ammonium molybdate solution is added. The appearance of a yellow precipitate or colouration due to the
formation of ammonium phosphomolybdate indicates the presence of phosphorus.


Na3PO4 + 3HNO3  H3PO4 + 3NaNO3

6 6
H3PO4 + 12 NH4 2 MoO 4  21HNO3  NH4 3 PO 4 .12 MoO3 + 21NH4NO3  12H2O
Ammonium Ammonium
Molybdate Phosphomolybdate
 Yellow colour 

QUANTITATIVE ANALYSIS
The estimation of percentage composition of the elements present in organic compounds is called quantitative
analysis. The following methods are commonly used for the estimation of various elements.
(i) Estimation of Carbon and Hydrogen (Leibig’s Method) : In this method, carbon and hydrogen are
estimated simultaneously.
Principle : A known mass of the given dry organic compounds is heated strongly with dry cupric oxide
in an atmosphere of air or oxygen free from CO2. The carbon and hydrogen of the organic compound are
oxidised to CO2 and water vapour as :

C + 2CuO 
 CO2  2Cu
From organic
compound

2H + CuO 
 H2O  Cu
From organic
compound

 y y
In general : CxHy +  x +  O2 
 xCO2 + H2 O
 4 2

The mass of water produced is absorbed in a U-tube containing anhydrous CaCl2 (Calcium chloride) or
anhydrous magnesium perchlorate while CO2 produced is absorbed in another U-tube containing a strong
solution of KOH. The tubes are weighed before and after the combustion. The increase in the mass of
CaCl2 in U-tube gives the mass of water produced while increase in the mass of KOH in U-tube gives
the mass of CO2 produced.

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The apparatus used for estimation of carbon and hydrogen is shown in the following figure.
Sample in
Platinum Boat
CuO pallets
Combustion
tube

Excess
O2
Pure dry
oxygen
Anhydrous CaCl2 KOH solution
Calculations : Suppose the mass of the organic compound taken = w g
Increase in the mass of calcium chloride tube = a g
= Mass of water formed
Increase in the mass of potash bulbs = b g
= mass of CO2 gas liberated
(a) Calculation of % of H : One mole of water contains two mole of hydrogen atom
 18 g of water contains 2 g hydrogen

 2 
 a g of water contains   a g hydrogen
 18 
2
a
Thus, The % of H  18  100
w
(b) Calculation of % of C : One mole of CO2 contains one mole of carbon atom.
 44 g of CO2 contains carbon = 12 g
 12 
 b g of CO2 contains carbon =   bg
 44 
12
b
Hence , The % of carbon = 44  100
w

Example 13 : 0.2475 g of an organic compound gave on combustion 0.4950 g of CO2 and 0.2025 g of H2O.
Calculate the % of C & H in it.
Solution : Wt. of organic compound (w) = 0.2475 g
Wt. of CO2 (b) = 0.4950 g
Wt. of H2O(a) = 0.2025 g
12
×b
% of C = 44  100
w
12
 0.4950
= 44  100 = 54.54
0.2475
2
×a
% of H = 18  100
w
2
× 0.2025
18  100 = 9.09
=
0.2475

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Example 14 : An organic compound contains 69% carbon and 4.8% hydrogen, the remainder being oxygen
calculate the masses of CO2 & H2O produced when, 0.2 g of this substance is subjected to
complete conbusion.
69
Solution : Amount of C in 0.2 g of compound = 0.2 × = 0.138 g
100
Now, C  CO2
12 g of carbon gives 44 g CO2 on combustion.
44
 0.138 g of carbon gives  0.138 g = 0.506 g
12
4.8
Amount of H in 0.2 g of compound = 0.2 × = 0.0096 g
100
Now, 2H  H2O
2 g of hydrogen gives 18 g water on combustion.
18
 0.0096 g of hydrogen gives  0.0096 g = 0.0864 g
2

(ii) Estimation of Nitrogen : There are two methods for estimation of nitrogen.
(a) Duma’s Method : The method is applicable to all organic compounds containing nitrogen.
Principle : A known weight of the organic compound is heated strongly with an excess of copper
oxide in an atmosphere of carbon dioxide. Carbon is oxidised to carbon dioxide, hydrogen to water
and nitrogen is set free, as nitrogen gas. If any oxide of nitrogen is set free, it is reduced back to
nitrogen by passing it over hot bright copper spiral.
C + 2CuO  2Cu + CO2 
2H + CuO  Cu + H2O 
Steam

Nitrogen + CuO  N2 + a little of oxide of nitrogen

Oxides of nitrogen + Cu  CuO + N2

 y y z  y
In general CxHyNz   2x +  CuO   xCO2 + H2O + N2   2x   Cu
 2  2 2  2 
The whole of nitrogen set free is collected over conc. KOH solution which absorbs CO2 and water
vapours. The volume of nitrogen is measured and from that the percentage of nitrogen is calculated.
Reduced copper
CuO + organic

Coarse CuO
CuO gauze

compound

Nitrogen
gauze

CO2

Furnace

Mercury seal

Nitrometer

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Calculation : Let the mass of organic compound = w g.
The volume of nitrogen collected = v ml. Atmospheric pressure (from Barometer) = P mm of Hg
Room temperature = T1K
If temperature is in °C, then T1K = t°C + 273
Aqueous tension at T1K = a mm of Hg.
Pressure of dry nitrogen = (P – a) mm of Hg.
Let us Ist convert the volume of nitrogen to volume at STP.
Experimental Condition STP Conditions
P1 = (P – a) mm of Hg P2 = 760 mm of Hg
T1 = T1K T2 = 273 K
V1 = v ml V2 = ?

P1V1 P2 V2
According to gas equation, =
T1 T2

P1V1 × T2 P – a × V × 273
 V2 =  = x ml
P2 T1 760  T1

Now, 22,400 ml of N2 at STP will weight = 28 g

28
x ml of N2 at STP will weigh =  xg
22400

Mass of Nitrogen
 % of Nitrogen =  100
Mass of organic compound

28x
% of N = × 100
22400 × w

Example 15 : In Duma’s method for estimation of nitrogen 0.30 g of an organic compound gave 50 ml of nitrogen
collected at 300 K and 715 mm pressure. Calculate the % composition of nitrogen in the
compound (aq. tension at 300 K is 15 mm)
Solution : w = 0.30 g V1 = 50 ml
P1 = (P – a) = 715 – 15 = 700 mm of Hg
T1 = 300 K

P1V1 P2 V2
Now, =
T1 T2

P1V1 × T2 700  50  273

 V2 = T1 × P2 = 300  760

V2 = 41.9 ml
28
Weight of nitrogen =  41.9 g
22400
28  41.9
Percentage of Nitrogen =  100 = 17.46%
22400  0.3

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(b) Kjeldahl’s Method : The method is simple and convenient and is largely used for the estimation of
nitrogen in food, fertilisers and drugs. This method is, however, not applicable to (i) compounds containing
nitrogen in the ring (e.g., pyridine or quinoline etc) and (ii) Compounds containing nitro (– NO2) and azo
(– N = N –) groups.
The apparatus used for the estimation of nitrogen by Kjeldahl’s method is in the following figure.

Kjeldahl's trap

Water
outlet
Contents of Kjeldahl's
Organic compound flask after digestion
+ conc. H2SO4 + NaOH
+ CuSO4

Kjeldahl's Water
flask inlet

Known volume of
standard acid

Kjeldahl's distillation
Digestion (heating in Kjeldahl's flask)
A known mass of the organic compound is digested with conc. H2SO4 in presence of potassium sulphate
and a little copper sulphate in a long necked flask. The nitrogen in the compound is quantitatively converted
into ammonium sulphate.
N  NH4  2 SO4
+ Conc.H2SO4 
From organic compound
The resulting liquid is then distilled with excess of sodium hydroxide solution and ammonia evolved is
passed into a known but excess volume of the standard acid (HCl or H2SO4).
NH4 2 SO4 + 2NaOH  Na2SO 4 + 2H2O + 2NH3
  NH4  2 SO 4
2NH3 + H2SO 4 
The acid left unused is estimated by titration with some standard alkali
2NaOH + H2SO4 
 Na2SO4 + 2H2O
The amount of acid used against ammonia can be known and from this, the percentage of nitrogen in
the compound can be calculated.
Calculations : Let the mass of organic compound = w g
Volume of standard H2SO4 of molality M = V ml.
Now, excess acid left after absorption of NH3 is titrated with NaOH of molarity, M. Let volume of NaOH
of molarity M used for titration of excess of H2SO4 = V1 ml
V1
Now, V1 ml of NaOH of molarity M  ml of H2SO4 of molarity M
2
V1
 Volume of H2SO4 of molarity M left unreacted = ml
2
 V
Volume of H2SO4 of molarity M used for neutralisation of NH3 =  V – 1  ml
 2

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 V1   V
Now,  V –  ml of H2SO4 of molarity M  2  V – 1  ml of NH3 solution of molarity M
 2  2
Now, 1000 ml of 1 M NH3 solution contain 17 g NH3 or 14 g nitrogen.

 V
14 × M × 2  V – 1 
 V  2
 2  V – 1  ml of NH3 of molarity M contains nitrogen = g
 2 1000

 V
14 × M × 2  V – 1 
 2  100
Thus % of N = ×
1000 w

 V
1.4M × 2  V – 1 
 2
% of N 
w

Example 16 : During nitrogen estimation of an organic compound by Kjeldahl’s method, the ammonia evolved
by 0.5 g of the compound neutralised 10 ml of 1 M H2SO4. Calculate the percentage of nitrogen
in the compound.
Solution : 1 M of 10 ml H2SO4 = 1M of 20 ml of NH3
1000 ml of 1 M ammonia contains = 14 g N
14  20
 20 ml of 1 M ammonia contains = gN
1000
14  20
 % of N =  100 = 56.0%
1000  0.5

Example 17 : 0.35 g of an organic substance was Kjeldahlised and the ammonia obtained was passed into
100 ml of M/10H2SO4. The excess acid required 154 ml of M/10 NaOH for neutralisation, calculate
the % of nitrogen in the compound.
M
Solution : Volume of H SO taken = 100 ml
10 2 4
M
Excess volume of H SO is
10 2 4
M 154 M
154 ml of NaOH = ml of H SO
10 2 10 2 4
M
 Volume of H SO left unused = 77 ml
10 2 4
M
Volume of H SO used for neutralisation of NH3 = 100 – 77 = 23 ml
10 2 4
M M
Now, 23 ml of H SO  2 × 23 ml of NH3
10 2 4 10
M
 46 ml of 10 NH3 now, 1000 ml of 1M NH3 contains 14 g N

M 14 46  1
 46 ml of NH3 contain 
10 1000 10
14  46  100
% of N =  18.4%
1000  10  0.35

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(iii) Estimation of Halogens : A known mass of the organic substance containing halogen is heated with
fuming nitric acid and a few crystals of silver nitrate in a sealed tube. The silver halide is formed which
is separated, washed, dried and weighed. From the mass of silver halide obtained, the percentage of
halogen can be calculated. The apparatus used has been shown in the following figure.

Sealed
capillary

50 cm
2 cm

Fuming HNO3
AgNO3

Organic
compound
Let the mass of the substance = x gram and
Mass of Ag × (Silver halide) = x gram
Now, 1 mole of AgX  1 mole atom of X where X = Cl, Br, I

% of Halogen  X  
 At. wt. of X    wt. of AgX   100
Mol. wt. of AgX   w
For chlorine,
AgCl = 108 + 35.5 = 143.5
35.5  wt. of AgCl
% of Cl   100
143.5  w
For Bromine,
AgBr = 108 + 80 = 188
80  wt. of AgBr
% of Br   100
188  w
For Iodine,
AgI = 108 + 127 = 235
127  wt. of AgI
% of I   100
235  w
Here At. wt. of Ag = 108 & I = 127
Cl = 35.5
Br = 80

Example 18 : 0.185 g of an organic substance when treated with conc HNO3 gave 0.328 g of silver bromide.
Calculate the % of bromine in the compound.
Solution : w = 0.185 g
wt. of AgBr = 0.320 g
80  0.320  100
% of Br = = 73.60
188  0.185

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NEET Organic Chemistry - Some Basic Principles and Techniques 43

Example 19 : 0.189 g of an organic substance gave in a carius determination 0.287 g of silver chloride. What
is the % of Cl in the given compound?
Solution : w = 0.189 g
wt. of AgCl = 0.287 g
35.5  0.287  100
% of Cl = = 37.57%
143.5  0.189

Example 20 : 0.2585 g of an organic compound containing iodine was heated with excess of strong nitric acid
and silver nitrate in a carius tube. The precipitate of silver iodide was filtered, washed and dried.
Its weight was found to be 0.3894 g. Calculate the % of iodine in the compound.
Solution : w = 0.2585
wt. of AgI = 0.3894
127  0.3894
% of I =  100 = 81.4%
235  0.2585

(iv) Estimation of Sulphur : A known mass of the substance is heated with sodium peroxide or fuming nitric
acid in a carius tube. Sulphur present in the compound is oxidised to sulphuric acid which is then
precipitated as barium sulphate by adding excess of barium chloride solution.

S + H2O + 3O   H2SO 4
From organic compound from HNO 3

 BaSO 4 + 2 HCl
H2SO4 + BaCl2 
White ppt

The precipitate of BaSO4 is filtered, washed, dried and weighed, knowing the mass of the substance taken
and the mass of BaSO4 precipitate formed, the percentage of sulphur can be calculated.
Calculation : Let the mass of the substance taken = w g
Mass of BaSO4 ppt formed = x g
Now 233 g or 1 mole of BaSO4 contains 1 gram atom or 32 g of S.
32
 x g mole of BaSO4 contains × x g of S
233
32  x
% of S   100
233  w

(v) Estimation of Phosphorus : Carius method can also be used to estimate phosphorus in the given
compound. A known mass of the organic compound is heated with fuming nitric acid when phosphorus
present in the compound gets oxidised to phosphoric acid. This is then treated with magnesia mixture
(a mixture containing MgCl2, NH4Cl and NH3 when the precipitate of magnesium ammonium phosphate
is formed (MgNH4PO4). The precipitate is now ignited to get magnesium pyrophosphate (Mg2P2O7) and
weighed.
Calculation : Let the mass of organic compound = w g.
The mass of Mg2P2O7 formed = x g
1 mole of Mg2P2O7(222 g) contains 2 mole of phosphorus atoms 62 g

62
 x g of Mg2P2O7 contains  x g of P
222
62 x
Now % of P =   100
222 w

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44 Organic Chemistry - Some Basic Principles and Techniques NEET

(vi) Estimation of Oxygen : The percentage of oxygen in an organic compound is usually calculated by the
difference method. For this, the percentages of all other elements present in the organic compound are
added and the sum is subtracted from 100.

% of oxygen = 100 – sum of the % of all other element

EXERCISE
33. Sulphur cannot be detected by
(1) Beilstein test (2) Lassaigne test
(3) Lead acetate test (4) Sodium nitroprusside test
34. If an organic compound contains both N and S ; the appearance of blood red colour takes place in Lassaigne-
test due to formation of
(1) NaCNS (2) Fe(CNS)3
(3) NaSCN (4) Fe(CN)2
35. Kjeldahl’s method for detection of nitrogen in organic compound, cannot be used in case of

(1) CH3 – NO2 (2)

N

(3) N=N (4) All of these

36. Carius method is used for the estimation of

(1) Halogens (2) Sulphur
(3) Phosphorus (4) All of these
37. During estimation by Duma’s method 2.36 g of an organic compound gave 448 mL of N2 (g) at NTP. The %
of N in the compound is
(1) 23.7% (2) 40%
(3) 47.4% (4) 12%
38. Naphthalene and benzoic acid can be separated from each other best by the method of
(1) Crystallisation (2) Sublimation
(3) Distillation (4) Chromatography
39. Which of the following compound will not give Lassaigne’s test for nitrogen?
(1) Azobenzene (2) Hydrazine
(3) Phenylhydrazine (4) Urea



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 t
en
nm nment
sig ssig
As A Assignment

Assignment
5. Which among the following is not a heterocyclic
SECTION - A compound? [NCERT Pg. 339]
NCERT Based MCQs OH
1. Among the following which element has highest (1) (2)
catenation property? [NCERT Pg. 334]
S
(1) Carbon (2) Silicon
(3) Phosphorous (4) Sulphur (3) (4)
2. Number of resonating structures possible for N
O
benzene are [NCERT Pg. 353]
H
(1) 1 (2) 2
(3) 3 (4) 4 6. The correct decreasing priority order of given
functional groups while writing the IUPAC
3. The sum of  and -bonds present in the given nomenclature is [NCERT Pg. 344]
molecule is [NCERT Pg. 335]
(1) – COOH, – SO3H, – COOR, – COCl
O
CH3 (2) – COOH, – SO3H, – COCl, – COOR
HC
(3) – COOH, – COOR, – SO3H, – COCl
(4) – COOH, – COCl, – SO3H, – COOR
CH3
7. Consider the given species [NCERT Pg. 342]
(1) 26 (2) 28
(CH3)3N BF3
(3) 29 (4) 30
4. Select the correct expanded form of following bond- r
C2H5 Os Cl
line structure [NCERT Pg. 336]
The total number of nucleophiles are
(1) 0
OH
(2) 2
H H H H
(3) 4
(1) H — C C—C—C—C—C—H
(4) 3
H OH H H
8. Correct IUPAC name of the given compound is
H H H H H [NCERT Pg. 344]
(2) H — C C—C—C—C—H
OH
H OH H
H
(3) H — C C—C C—C—H
(1) 2-methylbutan-2-ol
OH
(2) 2-methylbutan-4-ol
H H H
(3) 2-methylbutan-1-ol
(4) H — C C—C—C—C—H
(4) 2-Ethylbutan-1-ol
H OH H
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46 Organic Chemistry - Some Basic Principles & Techniques NEET

 15. CH2 = CH2 + HBr CH3 – CH2 – Br.

9. For carbocation CH3 , which of the following is not
The given reaction is an example of
true? [NCERT Pg. 350]
[NCERT Pg. 356]
(1) It has trigonal planar shape
(1) Elimination reaction
(2) Its shape is derived by the overlap of each of (2) Substitution reaction
the three sp2 hybridised orbitals of carbon with
corresponding 1 s orbital of hydrogen atoms (3) Electrophilic addition reaction

(3) It has one vacant orbital on carbon which is (4) Free radical addition reaction
perpendicular to molecular plane
(4) It has two vacant p orbital on carbon which are SECTION - B
in molecular plane Objective Type Questions
10. Electromeric effect can be observed in 1. The element which shows highest catenation
[NCERT Pg. 355] property is
(1) (CH3)2O (2) C2 H5OH
(1) Hydrogen (2) Carbon
(3) C2 H6 (4) C2H4
(3) Oxygen (4) Sulphur
11. The violet colour developed during the Lassaigne’s
2. Which of the following is heterocyclic compound?
test for sulphur is due to [NCERT Pg. 363]
(1) Pyrrole (2) Furan
(1) Na4[Fe(CN)5NOS] (2) Fe4[Fe(CN)6]3
(3) Thiophene (4) All of these
(3) Fe(SCN)3 (4) Fe3[Fe(CN)6]2
3. Which of the following shows structure of
12. On complete combustion of 0.25 g of an organic neohexane?
compound 0.22 g of CO2 and 0.18 g of H2O are
obtained. The percentage of C and H respectively CH3
in the organic compound are [NCERT Pg. 364] (1) CH3 – CH – CH 2 – CH2 – CH 2 – CH3
(1) 8 and 20 (2) 24 and 12
C H3 CH3
(3) 24 and 8 (4) 24 and 10
(2) CH3 – CH – C – CH3
13. Kjeldahl’s method for detection of nitrogen in CH3
organic compounds can be used for
CH3
[NCERT Pg. 365]
(3) CH3 – C – CH2 – CH3
(1) C2H5NO2
CH3

(2) CH3 CH3

N
(4) CH3 – CH – CH 2 – CH
CH3

(3) N 4. The incorrect match is

H 1. CnH2n – 2 – Alkyne
(4) NH2CONH2 2. CnH2n – Alkene
14. Which of the following is purified by sublimation? 3. CnH2n + 2 – Bicyclo alkane
[NCERT Pg. 357] 4. CnH2n – 2 – Alkadiene
(1) Naphthalene 5. Number of primary hydrogen atoms and tertiary ‘C’
(2) Benzoic acid present in neopentane is
(3) Camphor (1) 9 and 1 (2) 9 and 0
(4) All of these (3) 6 and 1 (4) 12 and 0
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NEET Organic Chemistry - Some Basic Principles & Techniques 47
6. The functional group is mainly responsible for 13. The IUPAC name of the compound
(1) Physical properties (2) Chemical reactivity CH2 = CHCOCH = CH2 is
(3) Biological properties (4) Both (2) & (3) (1) Divinyl ketone
7. The number of C atoms present in allyl vinyl (2) 3-Keto penta 1,4-diene
acetylene is _____ and it has _____ degree of
unsaturation (3) Penta-1,4-dien-3-one

(1) 7 and 4 (2) 7 and 3 (4) 3-Ethenyl prop-1-ene-3-one

(3) 5 and 4 (4) 5 and 3 14. The IUPAC name of the compound
8. IUPAC name for the compound CH2 CH2

O
is
(1) Ethene oxide

(1) 3-Methyl-4-ethyl hexane (2) Ethylene oxide

(2) 3-Ethyl-4-methyl hexane (3) Epoxy ethane

(3) 4-Ethyl-3-methyl hexane (4) Ethylene epoxide

(4) Nonane 15. The compound which has one isopropyl group is
9. IUPAC name of the compund (1) 2,2,3,3,-Tetramethyl pentane
ClCH2–CH2–CH2 –CH2 –CH2Br is (2) 2,2-Dimethyl pentane
(1) 1–Chloro–5–bromopentane (3) 2, 2, 3-Trimethyl pentane
(2) 5–Bromo–1–chloropentane (4) 2-Methyl pentane
(3) 1–Bromo–5–chloropentane 16. The IUPAC name of Cl3C.CHO is
(4) 5–Chloro–1–bromopentane
(1) Trichloroacetaldehyde
10. IUPAC name of the compound
(2) 1, 1, 1-Trichlorethanal
CH3 – CH=CH – C  C – C  CH is
(3) 2, 2, 2-Trichloroethanal
(1) Hepta–2–ene-4, 6–diyne
(4) Chloral
(2) Hepta–5–ene-1,3–diyne
17. The IUPAC name of CH3OC2H5 is
(3) Hepta-1,3-diyne-5-ene
(1) Ethoxymethane
(4) 5-Ene hepta-1, 3-diyne
(2) Methoxyethane
11. IUPAC name of Succinic acid is
(3) Methyl ethyl ether
(1) Ethane dioic acid
(4) Ethyl methyl ether
(2) Propane-1, 3–dioic acid
18. The IUPAC name of the compound
(3) Propane-1, 2–dioic acid
(4) Butane–1, 4–dioic acid CH2 — CH — CH 2
is
12. IUPAC name of Glucose is CN CN CN

(1) 2,3, 4, 5, 6–Pentahydroxy hexanal (1) 1, 2, 3-Cyanopropene

(2) 1, 2, 3, 4, 5–Pentahydroxy hexanal (2) 1, 2, 3-Cyanopropane

(3) Hexane–6–al–1,2,3,4,5 pentanol (3) 3-Cyanopentan-1, 5,-dinitrile

(4) Aldohexose (4) Propane-1,2,3-Tricarbonitrile
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48 Organic Chemistry - Some Basic Principles & Techniques NEET

19. The IUPAC name of the compound

OC H3
CHO CHO CHO Cl
is 25. The IUPAC name of ; is
CH2 CH CH2
CH 3
(1) 3-Formyl-1, 5-pentanedial
(2) 2–Aldo Pentane-1,5-dialdehyde (1) 4-methoxy-1-methyl chloro benzene
(2) 2-chloro-4-methyl anisole
(3) Propanetrialdehyde
(4) Propane-1, 2, 3-tricarbaldehyde (3) 1-methoxy-2 chloro-4-methyl benzene

20. The IUPAC name of (4) 4-methyl-2 chloro anisole

CH2 = CH — CH2 — C  CH is NO2

(1) Pent-2-en-5-yne (2) Pent-1-en-4-yne F
26. Compound is
(3) Pent-4-en-1-yne (4) Pent-4-en-2-yne C2H5

(1) 4-Ethyl-2-nitro-1-fluoro benzene

21. The IUPAC name of the compound is (2) 4-Ethyl-1-fluoro-2-nitro benzene
(1) Cyclo hexane (3) 1-Nitro-2-fluoro ethyl benzene
(2) Bicyclo [2. 2. 0] hexane (4) 1-Ethyl-3-nitro-4-fluoro benzene
(3) Bicyclo [0. 2. 2] hexane 27. The number of chain isomers possible in C5H12 is
(4) Bicyclohexane (1) 2 (2) 3
COOH (3) 4 (4) 5
22. IUPAC name of is 28. Propanol-1 and Propanol-2 are
OCH 3
(1) Chain isomers
(1) 2-methoxybenzoic acid
(2) Position isomers
(2) 2-carboxymethoxide
(3) Functional isomers
(3) 2-carboxyphenyl methoxide
(4) Tautomers
(4) 2-methoxy carboxybenzene
29. How many primary alcohols are possible of
molecular formula C4H10O?
23. The IUPACname of is (1) 3

(1) Bicyclo [2. 2. 2] octane (2) 2

(2) Cyclo octane (3) 5

(3) Bicyclo octane (4) 7

(4) Bicyclo [2. 2. 2] hexane 30. The tautomerism is not shown by

24. The IUPAC name of ; is O

O
|| ||
(1) (2)
||
(1) Spiro [2, 4] heptane O

(2) Spiro [4, 2] heptane

(3) Bicyclo [4, 2] heptane O
(3) (4) CH3CHO
(4) Both (2) & (3) O

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NEET Organic Chemistry - Some Basic Principles & Techniques 49
31. How many alkenes (excluding stereoisomers) are 39. The nucleophile is not
possible of molecular formula C4H8? (1) Lewis base (2) Lewis acid
(1) 3 (2) 4 
(3) H2 O (4) Carbanion
(3) 5 (4) 6 

32. CH3 – CH2 – CH2 – O – CH3 and 40. The most stable carbocation is
 +
CH3 (1) CH3  CH2 (2) CH3 – CH
CH3 – CH – O – CH3 are CH3

(1) Chain isomers  

(3) CH2 = CH – CH2 (4) CH3
(2) Position isomers
41. The most stable carbanion is
(3) Metamers
(4) Functional isomers (1) CH3 CH2 (2) (CH3 )3C
33. Which of the following is not an example of
stereoisomers? (3) CH (4) (CH3 )2CH
3

(1) Conformers and geometrical isomers 42. Which resonating structure of formic acid is most
stable?
(2) Metamers and optical isomers
–
(3) Geometrical and optical isomers O
| +
(1) H–C=O–H
(4) Conformers and optical isomers
–
O
34. The most stable alkene is |
(2) H – C – O – H
(1) CH3CH=CHCH3 +
+
(2) CH2=CHCH2CH3 O
|
(3) (CH3)2C=C(CH3)2 (3) H–C–O–H
–
(4) (CH3)3C–CH=CH2 (4) All have equal stability

35. The strongest deactivating group for benzene 43. Which is the weakest nucleophile?
nucleus is
(1) CH3 O (2) CH3 CH2 O
(1) – COOH (2) –NO2
(3) (CH3 )2 CHO (4) (CH3 )3CO
(3) –CN (4) –CHO
44. The electromeric effect is
36. Which of the following is not an electrophile?
 (1) Permanent effect
(1) CH3 (2) AlCl3
(2) Temporary effect
+ 
(3) SiF4 (4) [( CH3)4 N ] OH (3) Pie electron transfer effect
(4) Both (2) & (3)
37. :CH2 is an
– +
(1) Electrophile (2) Nucleophile O O
+ –
45. In CH2 –HC = C–H and CH2–CH = CH–C–H most
(3) Free radical (4) Ambiphiles
(I) (II)
38. In – NO2 , – NH2 , – SO 3 H ; The decreasing order of stable is
(I) (II) (III)
(1) (I)
–I effect is
(2) (II)
(1) (I) > (II) > (III) (2) (I) > (III) > (II)
(3) Sometimes (I) and sometimes (II)
(3) (III) > (II) > (I) (4) (III) > (I) > (II)
(4) Both are equal stable
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50 Organic Chemistry - Some Basic Principles & Techniques NEET

46. Which one is incorrect match of reactions? 51. Number of C atoms which present linearly in vinyl
acetylene is
CH3 Ph O Ph
C 
H C (1) 2 (2) 3
(1) ;
N N (3) 4 (4) 1
OH CH3 H 52. In the Lassaigne’s test for nitrogen in an organic
 Rearrangement compound, the prussian blue colour is obtained
aq.KOH due to formation of
(2) CH3 – CH2 – Br 
 CH3 – CH2 OH ;
(1) Fe4[Fe(CN)6]3 (2) [Fe(CNS)3]
 Nucleophilic substitution reaction
(3) [Fe(CN)5NOS]2  (4) [Fe(CN)5NO]2 
OH
53. 0.756 g of chloroderivative of organic compound
(3) H3PO 4

+ H2 O ; gave 1.148 g of AgCl. What is the percentage of
chlorine in the compound?
 Elimination reaction (1) 66.85% (2) 34.14%
(3) 49.5% (4) 37.59%
OH
(4) (CH3 )3 C–Cl (CH3 )2 C = CH 2 + H2O + Cl
54. On complete combustion of 0.492 g of an organic
 Nucleophilic substitution reaction compound gave 0.396 g of CO2 and 2.028 g of
H2O. What is the percentage of oxygen in the
47. The most stable free radical is
compound?
(1) 22% (2) 32.25%
(1) (2) (3) 51.5% (4) 78%
55. Match the column-I to column-II
Column-I Column-II
(3) (4)
a. Carius method (i) (NH4)2SO4
48. Which carboxylic acid has highest pKa value? b. Duma’s method (ii) Sodium fusion
extract
(1) CH2 –COOH
|
NO2
c. Kjeldahl’s method (iii) N2-gas

(2) HCOOH d. Lassaigne’s test (iv) AgNO3

(3) C6H5COOH (1) a(ii), b(i), c(iii), d(iv) (2) a(iv), b(i), c(iii), d(ii)
(3) a(ii), b(iii), c(i), d(iv) (4) a(iv), b(iii), c(i), d(ii)
(4) CCl3–COOH
56. The mixture of chloroform and aniline can be
49. The correct statement for -elimination is
separated into its component best by the method
(1) It forms cyclic compounds (1) Distillation (2) Solvent extraction
(2) It forms carbene or substituted carbene (3) Sublimation (4) Crystallisation
(3) Two atoms are removed from  and  positions 57. The percentage composition of an organic
Cl compound is as C = 40%, N = 46.66%, H = 13.3%.
The empirical formula of the compound is
(4) In CH3 – C – Cl ;  elimination is not possible
(1) C2NH5 (2) CHN
C2H5
(3) C2NH2 (4) CNH4
50. Pericyclic reaction occurs
58. Ammonia released by 1.0 g of an organic
(1) Via cyclic transition state compound containing N in Kjeldahl’s method is
(2) With formation of carbocation and carbanion neutralised by 20 mL of 1M H 2 SO 4 . The
percentage of nitrogen in the compound is
(3) Breaking and formation of bonds takes place
with time interval (1) 65% (2) 56%

(4) Both (1) & (3) (3) 46% (4) 24%

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NEET Organic Chemistry - Some Basic Principles & Techniques 51
(3) Electrophiles are generally neutral species and
SECTION - C
can form a bond by accepting a pair of
Previous Years Questions electrons from a nucleophile
1. The most stable carbocation, among the following, (4) Electrophile can be either neutral or positively
is [NEET-2019 (Odisha)] charged species and can form a bond by
 accepting a pair of electrons from a nucleophile
(1) CH3 – CH2 – CH2
6. The most suitable method of separation of 1 : 1
 mixture of ortho and para-nitrophenols is
(2)  CH3 3 C – CH – CH3
[NEET-2017]

(3) CH3 – CH2 – CH – CH2 – CH3 (1) Sublimation

 (2) Chromatography
(4) CH3 – CH – CH2 – CH2 – CH3
(3) Crystallisation
2. The number of sigma () and pi () bonds in (4) Steam distillation
pent-2-en-4-yne is [NEET-2019]
7. The IUPAC name of the compound [NEET-2017]
(1) 10 bonds and 3 bonds
O O
(2) 8 bonds and 5 bonds
C
(3) 11 bonds and 2 bonds H is ________.
(4) 13 bonds and no  bonds
3. Which of the following carbocations is expected to (1) 3-keto-2-methylhex-4-enal
be most stable? (2) 5-formylhex-2-en-3-one
[NEET-2018] (3) 5-methyl-4-oxohex-2-en-5-al
NO2 NO2 (4) 3-keto-2-methylhex-5-enal

 8. In which of the following molecules, all atoms are

coplanar? [NEET(Phase-2)-2016]
(1) (2)

Y H Y H
NO2 NO2
(1) (2)
H 
(3) Y (4) H
Y CH3 CN
 (3) C C (4)
4. Which of the following is correct with respect to CH3 CN
– I effect of the substituents? (R = alkyl) 9. In pyrrole
[NEET-2018] 4 3
(1) – NH2 < – OR < – F (2) – NR2 < – OR < – F 5 2
N1
(3) – NR2 > – OR > – F (4) – NH2 > – OR > – F
5. The correct statement regarding electrophile is H
the electron density is maximum on
[NEET-2017]
[NEET(Phase-2)-2016]
(1) Electrophile is a negatively charged species
and can form a bond by accepting a pair of (1) 2 and 3
electrons from a nucleophile (2) 3 and 4
(2) Electrophile is a negatively charged species (3) 2 and 4
and can form a bond by accepting a pair of
(4) 2 and 5
electrons from another electrophile
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52 Organic Chemistry - Some Basic Principles & Techniques NEET

10. Which among the given molecules can exhibit (3) A carbonyl compound with a hydrogen atom on
tautomerism? [NEET(Phase-2)-2016] its alpha-carbon rapidly equilibrates with its
O O corresponding enol and this process is known
as aldehyde-ketone equilibration
O
Ph
Ph (4) A carbonyl compound with a hydrogen atom on
I II III its alpha-carbon rapidly equilibrates with its
(1) III only corresponding enol and this process is known
(2) Both I and III as carbonylation

(3) Both I and II 14. The number of structural isomers possible from the
molecular formula C3H9N is [Re-AIPMT-2015]
(4) Both II and III
(1) 2 (2) 3
11. For the following reactions :
(3) 4 (4) 5
(a) CH3CH2CH2Br + KOH 
15. Which of the following statements is not correct for
CH3CH = CH2 + KBr + H2O a nucleophile? [Re-AIPMT-2015]
H3C CH3 H3C CH3 (1) Nucleophiles attack low e– density sites
(b) + KOH + KBr
Br OH (2) Nucleophiles are not electron seeking
Br (3) Nucleophile is a Lewis acid
(c) + Br2
Br (4) Ammonia is a nucleophile
Which of the following statements is correct? 16. Consider the following compounds
[NEET-2016]
CH3
(1) (a) is substitution, (b) and (c) are addition
reactions I. CH3 C CH

(2) (a) and (b) are elimination reactions and (c) is CH3
addition reaction
Ph
(3) (a) is elimination, (b) is substitution and (c) is
addition reaction II. Ph C Ph
(4) (a) is elimination, (b) and (c) are substitution
reactions
III.
12. The pair of electron in the given carbanion, CH3
CH3C  C is present in which of the following
Hyperconjugation occurs in [AIPMT-2015]
orbitals? [NEET-2016]
(1) I and III (2) I only
(1) sp
(3) II only (4) III only
(2) 2p
17. Which of the following is the most correct electron
(3) sp3 displacement for a nucleophilic reaction to take
(4) sp2 place? [AIPMT-2015]
13. The correct statement regarding a carbonyl H H2
compound with a hydrogen atom on its alpha- (1) H3C C = C — C — Cl
H
carbon, is [NEET-2016]
H H2
(1) A carbonyl compound with a hydrogen atom on (2) H3C C = C — C — Cl
its alpha-carbon rapidly equilibrates with its H
corresponding enol and this process is known
H H2
as keto-enol tautomerism (3) H3C C = C — C — Cl
(2) A carbonyl compound with a hydrogen atom on H
its alpha-carbon never equilibrates with its H H2
corresponding enol (4) H3C C = C — C — Cl
H
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NEET Organic Chemistry - Some Basic Principles & Techniques 53
18. Given : 21. In the Kjeldahl's method for estimation of nitrogen
H3C CH3 present in a soil sample, ammonia evolved from
0.75 g of sample neutralized 10 mL of 1 M H2SO4.
The percentage of nitrogen in the soil is
I.
CH3 [AIPMT-2014]

H3C CH2 (1) 37.33 (2) 45.33

(3) 35.33 (4) 43.33
II. 22. The structure of isobutyl group in an organic
CH3 compound is [NEET-2013]
H2C CH2 (1) CH 3–CH–CH 2–CH 3 (2) CH3–CH2–CH2–CH2–
CH3
III. CH
(3) CH3–C–
3

CH2 (4) CH CH–CH2–

CH3 3

The enthalpy of hydrogenation of these compounds

will be in the order as [AIPMT-2015] 23. The radical, CH2 . is aromatic because it
(1) II > I > III (2) I > II > III has [NEET-2013]
(3) III > II > I (4) II > III > I (1) 7 p-orbitals and 6 unpaired electrons
19. In Duma's method for estimation of nitrogen, 0.25 g (2) 7 p-orbitals and 7 unpaired electrons
of an organic compound gave 40 mL of nitrogen (3) 6 p-orbitals and 7 unpaired electrons
collected at 300 K temperature and 725 mm
(4) 6 p-orbitals and 1 unpaired electron
pressure. If the aqueous tension at 300 K is
25 mm, the percentage of nitrogen in the compound 24. The order of stability of the following tautomeric
is [AIPMT-2015] compounds is

(1) 15.76 (2) 17.36 OH O

| ||
(3) 18.20 (4) 16.76 

CH2  C — CH2 — C— CH3 

I
20. Given
O O
CH3 CH3 || ||


CH3 — C— CH2 — C— CH3 

II
I. OH O
| ||
O CH3 — C  CH — C— CH3
CH3 III
[NEET-2013]
CH3
II. (1) III > II > I (2) II > I > III
O (3) II > III > I (4) I > II > III
25. Structure of the compound whose IUPAC name is
3-ethyl-2-hydroxy-4-methylhex-3-en-5-ynoic acid is
CH3
III. [NEET-2013]
CH3
O OH OH
(1) COOH (2) COOH
Which of the given compounds can exhibit
tautomerism? [AIPMT-2015]
OH
(1) I, II and III (2) I and II COOH
(3) (4) COOH
(3) I and III (4) II and III OH

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54 Organic Chemistry - Some Basic Principles & Techniques NEET

26. Which nomenclature is not according to IUPAC 30. The correct IUPAC name of the compound
system ? [AIPMT (Prelims)-2012]

(1) CH3–CH–CH–CH2CH3,
CH3
2–Methyl–3–phenylpentane

(2) CH3–C–CH2–CH2COOH, is [AIPMT (Prelims)-2011]

(1) 3-(1-ethyl propyl) hex-1-ene
O
4–oxopentanoic acid (2) 4-Ethyl-3-propyl hex-1-ene

Br–CH2–CH = CH2, (3) 3-Ethyl-4-ethenyl heptane

(3)
1–Bromo–prop–2–ene (4) 3-Ethyl-4-propyl hex-5-ene

CH3 31. The IUPAC name of the following compound is

(4) CH3–CH2–C–CH2–CHCH3, Cl CH2CH3

C C

||
Br CH3 H 3C I
4–Bromo, 2, 4–di–methylhexane
[AIPMT (Mains)-2011]
27. In Dumas' method of estimation of nitrogen 0.35 g (1) trans-3-iodo-4-chloro-3-pentene
of an organic compound gave 55 mL of nitrogen
(2) cis-3-chloro-3-iodo-2-pentene
collected at 300 K temperature and 715 mm
pressure. The percentage composition of nitrogen (3) trans-2-chloro-3-iodo-2-pentene
in the compound would be: (Aqueous tension at (4) cis-3-iodo-4-chloro-3-pentene
300 K = 15 mm) [AIPMT (Prelims)-2011]
32. Which of the following species is not electrophilic
(1) 14.45 in nature? [AIPMT (Mains)-2010]
(2) 15.45

(3) 16.45 (1) Cl (2) BH3

(4) 17.45  
(3) H O (4) NO2
28. Considering the state of hybridization of carbon 3

atoms, find out the molecule among the following 33. The IUPAC name of the compound
which is linear? [AIPMT (Prelims)-2011] CH3CH =CHC  CH is [AIPMT (Mains)-2010]
(1) CH3 – CH2 – CH2 – CH3 (1) Pent-4-yn-2-ene
(2) CH3 – CH = CH – CH3 (2) Pent-3-en-1-yne
(3) CH3 – C  C – CH3 (3) Pent-2-en-4-yne
(4) CH2 = CH – CH2 – C  CH (4) Pent-1-yn-3-ene
29. The Lassaigne's extract is boiled with conc. HNO3 34. The state of hybridization of C2, C3, C5 and C6 of the
while testing for halogens. By doing so it
CH 3 CH3
[AIPMT (Prelims)-2011] hydrocarbon, CH3–C–CH = CH–CH–C  CH is in the
7 6 5 4 3 2 1
(1) Increases the concentration of NO3– ions CH3

(2) Decomposes Na2S and NaCN, if formed following sequence [AIPMT (Prelims)-2009]
(3) Helps in the precipitation of AgCl (1) sp3, sp2, sp2 and sp (2) sp, sp2, sp2 and sp3
(4) Increases the solubility product of AgCl (3) sp, sp2, sp3 and sp2 (4) sp, sp3, sp2 and sp3
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NEET Organic Chemistry - Some Basic Principles & Techniques 55
35. The IUPAC name of the compound having the formula 41. The correct order regarding the electronegativity of
CH C – CH = CH2 is [AIPMT (Prelims)-2009] hybrid orbitals of carbon is [AIPMT (Prelims)-2006]
(1) 1-butyn-3-ene (2) but-1-yne-3-ene (1) sp > sp2 < sp3 (2) sp > sp2 > sp3
(3) 1-butene-3-yne (4) 3-butene-1-yne (3) sp < sp2 > sp3 (4) sp < sp2 < sp3
36. Which of the following compounds will exhibit cis-
trans (geometrical) isomerism? Cl
42. The IUPAC name of is
[AIPMT (Prelims)-2009] O
(1) Butanol (2) 2-Butyne
[AIPMT (Prelims)-2006]
(3) 2-Butenol (4) 2-Butene
(1) 3, 4-dimethylpentanoyl chloride
37. In the hydrocarbon CH3–CH = CH–CH2–C  CH
6 4 3 2 1 (2) 1-chloro-1-oxo-2, 3-dimethylpentane
5
(3) 2-ethyl-3-methylbutanoyl chloride
The state of hybridization of carbons 1,3 and 5 are in
the following sequence [AIPMT (Prelims)-2008] (4) 2, 3-dimethylpentanoyl chloride
(1) sp3, sp2, sp (2) sp2, sp, sp3 43. Which amongst the following is the most stable
(3) sp, sp3, sp2 (4) sp, sp2, sp3 carbocation? [AIPMT (Prelims)-2005]
CH3
38. The stability of carbanions in the following
+
(1) CH3–CH (2) CH3C +
a. R—C C CH3 CH3
 

b. (3) C H 3 (4) CH3 C H 2

44. Which one of the following pairs represents
stereoisomerism? [AIPMT (Prelims)-2005]
c. R 2C CH
(1) Chain isomerism and rotational isomerism
d. R 3C CH2
(2) Structural isomerism and geometric isomersm
is in the order of [AIPMT (Prelims)-2008]
(3) Linkage isomerism and geometric isomerism
(1) a > c > b > d (2) a > b > c > d
(4) Optical isomerism and geometric isomerism
(3) b > c > d > a (4) d > b > c > a
45. Names of some compounds are given. Which one is
39. Base strength of not correct in IUPAC system?
a. H3CCH2 [AIPMT (Prelims)-2005]
(1) CH3–CH–CH–CH3
b. H 2C CH

c. OH CH3
H C C
3-Methylbutan-2-ol
is in the order of [AIPMT (Prelims)-2008] (2) CH3–CC–CH(CH3)2
4-Methylpent-2-yne
(1) a > b > c (2) b > a > c
(3) CH3 – CH2 – C — CH – CH3
(3) c > b > a (4) a > c > b
CH2 CH3
40. The general molecular formula, which represents the 2-Ethyl-3-methyl-but-1-ene
homologous series of alkanols is
CH3
[AIPMT (Prelims)-2006]
(4) CH3–CH2–CH2–CH–CH–CH2CH3
(1) CnH2nO2 (2) CnH2nO
CH2CH3
(3) CnH2n+1O (4) CnH2n+2O 3-methyl-4-ethylheptane

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56 Organic Chemistry - Some Basic Principles & Techniques NEET

Questions asked Prior to Medical Ent. Exams. 2005 50. For the given alkane
46. The correct order of increasing bond length of
C – H, C – O, C – C and C = C is

(1) C  H  C  O  C  C  C  C
The IUPAC name is
(2) C  H  C  C  C  O  C  C
(1) 1, 1 dimethyl-5-ethyl octane
(3) C  C  C  C  C  O  C  H (2) 6-ethyl-2-methyl nonane

(4) C  O  C  H  C  C  C  C (3) 4-ethyl-8-methyl nonane

47. Homolytic fission of the following alkanes forms (4) 2-methyl, -6-propyl octane
free radicals CH3 – CH3, CH3 – CH2 – CH3, (CH3)2 51. Most stable carbocation is
CH – CH3, CH3 – CH2 – CH (CH3)2. Increasing
CH3 +CH2
order of stability of the radicals is
+
(1) (CH3)3 C < (CH3)2 C – CH2CH3 < (1) (2)
CH3 – CH – CH3 < CH3 – CH2
CH3 CH3
(2) (CH3)2 C – CH2CH3 < CH3 – CH – CH3 +
(3) (4)
< CH3 – CH2 < (CH3)3 C +

(3) CH3 – CH2 < CH3 – CH – CH3 < (CH3)2 C 52. Arrange the following in increasing order of stability.
– CH2 – CH3 < (CH3)3 C  
(a) (CH3)2 C – CH2CH3 (b) (CH3)3 C
(4) CH3 – CH2 < CH3 – CH – CH3 <
 
(CH3)3 C < (CH3)2 C – CH2CH3 (c) (CH3)2 CH (d) CH3 CH2

48. Nitrogen detection in an organic compound is 

(e) CH3
carried out by Lassaigne's test. The blue colour
formed corresponds to which of the following (1) e < d < c < b < a (2) e < d < c < a < b
formulae?
(3) d < e < c < a < b (4) a < e < d < c < b
(1) Fe3[Fe(CN)6]3 (2) Fe3[Fe(CN)6]2
53. An organic compound contains carbon, hydrogen
(3) Fe4[Fe(CN)6]3 (4) Fe4[Fe(CN)6]2 and oxygen. Its elemental analysis gave C, 38.71%
49. Which one is a nucleophilic substitution reaction and H, 9.67%. The empirical formula of the
among the following? compound would be
(1) CH4O (2) CH3O
(1) CH3CHO + HCN  CH3CH  OH CN
(3) CH2O (4) CHO
(2) CH3 – CH = CH2  H2O H+ 54. The IUPAC name for
 CH3 — CH — CH3

OH CH3CH = CHCH2CHCH2COOH
(3) RCHO + R'MgX  R – CH – R'
NH 2
OH is
CH3
(1) 3-amino hept-5-enoic acid
(4) CH3 — CH2 — CH — CH2Br + NH3
(2) -amino –heptenoic acid
CH3
(3) 5-amino 2-heptenoic acid
CH3 — CH2 — CH — CH2NH2
(4) 5-amino hex-2-ene-carboxylic acid
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NEET Organic Chemistry - Some Basic Principles & Techniques 57
55. The number of structural isomers in C4H10O will be 62. Name of the compound given below is
(1) 7 (2) 8 CH3
(3) 5 (4) 6 H3C

56. The IUPAC name of (CH3)2 CH–CH2–CH2Br is CH3

(1) 1-bromo-3-methyl butane
(2) 2-methyl-3-bromo propane CH3
(1) 4-ethyl-3-methyloctane
(3) 1-bromo pentane
(2) 3-methyl-4-ethyloctane
(4) 2-methyl and 4-bromo butane
(3) 2, 3-diethylheptane
57. Tautomerism is exhibited by
(4) 5-ethyl-6-methyloctane
(1) R3CNO2 (2) RCH2NO2 63. The molecular formula of diphenyl methane,
(3) (CH3)3CNO (4) (CH3)2NH
CH2 , is C13H12. How many structural
58. Which one of the following orders is correct
isomers are possible when one of the hydrogen is
regarding the –I effect of the substituents?
replaced by a chlorine atom?
(1)  NR2 <  OR <  F (1) 6 (2) 4
(2)  NR2 >  OR >  F (3) 8 (4) 7
(3)  NR2 <  OR >  F 64. What is the hybridisation state of benzyl
+
(4)  NR2 >  OR <  F carbonium ion CH2 ?
59. The incorrect IUPAC name is (1) sp3 (2) sp2
(1) CH3 – C – CH – CH3 - 2 methylbutan-3-one (3) spd2 (4) sp2d
|| |
O CH3
SECTION - D
(2) CH3 – CH – CH – CH3 - 2, 3-dimethylpentane NEET Booster Questions
| |
CH3 CH2CH3 1. Which of the following is correct about the following
isomers?
(3) CH3 – C  CCH(CH3)2 – 4-methylpent-2-yne
|
(1) & are chain isomers
(4) CH 3 – CH – CH – CH 3 -2-bromo-3-chlorobutane
| |
Cl Br (2) | & are position
|
60. IUPAC name of the following is isomers
CH2 = CH – CH2 – CH2 – C  CH (3) HCN and HNC are tautomers
(1) 1, 5-hexenyne (2) 1-hexene-5-yne (4) All of these
(3) 1-hexyne-5-ene (4) 1, 5-hexynene 2. Which is paramagnetic?
(1) Singlet carbene (2) Nitrene
·· (3) Free radical (4) Carbonium ion
61. CH 2 – C – CH 3 and CH2 C – CH3 are
||

|| |
O O 3. Diphenyl methane is CH2 .
(1) Resonating structures
How many structural isomers are possible when
(2) Tautomers one of the hydrogen is replaced by chlorine atom?
(3) Geometrical isomers (1) 8 (2) 7
(4) Optical isomers (3) 4 (4) 2

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58 Organic Chemistry - Some Basic Principles & Techniques NEET

4. In which compound/compounds nitrogen cannot be (1) (I) & (II) are tautomers to each other
estimated by Kjeldahl's method?
(2) (II) is more stable than (I)
(1) CH3NO2
(3) (I) is keto form whereas (II) is enol-form
(2) CH3–N=N–CH3
(4) All of these
9. The maximum number of  bonds possible with
molecular formula C8H12 is
(3)
N (1) 2 (2) 3
(3) 4 (4) 5
(4) All of these
10. How many primary amines are possible with
5. Which of the following is bicyclo [4.4.0] decane?
molecular formula C4H11N?
(1) 3
(1)
(2) 2
(3) 4
(2) (4) 5
11. Lassaigne’s test for the detection of nitrogen fails
in case of
(3)
(1) NH2 –NH2

O
||
(4) (2) CH3–C–NH2

6. Choose the incorrect IUPAC name of an organic O

compound ||
(3) H2N–C–NH2
(1) Pent-3-ene
(4) All of these
(2) 2-Methylcyclopropene
12. 70 mg of an organic compound on vapourisation in
(3) 3-Hydroxypropene
Victor Meyer test displaces 22.4 mL of air
(4) All of these measured at NTP. The compound is
(1) C5H10
7. IUPAC name of is (2) C4H10
(3) C2H4
(1) Cyclohexa-1,2,3-triene (4) CH4
(2) Benzene 13. The number of structural isomers having molecular
formula C5H10 cannot decolourise bromine water is
(3) Cyclohexa-2,4,6-triene
(1) 3 (2) 4
(4) Cyclohexane
(3) 8 (4) 5
8. Consider the following conversion
14. Which of the following compound will not show
O O
|| || geometrical isomerism?
O OH
||

(1) But-2-ene

O O (2) 1, 2-dichloroethene
(I) (II) (3) Pent-1-ene
Choose the correct statement(s) (4) 1, 2-dimethylcyclopropane

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NEET Organic Chemistry - Some Basic Principles & Techniques 59
15. Choose the correct statement

  21. Molecular formula of the naphthalene

(1) CH3–CH2 is more stable than CH 3–O–CH 2
is
 
(2) CH2=CH is more stable than CH3–CH2 (1) C12H12
(2) C10H10
(3)  is aromatic in nature (3) C10H8
(4) C10H12

(4) is anti-aromatic 22. Estimation of halogens except F is done by

(1) Duma’s method
16. In which of the following molecular formula,
(2) Kjeldahl’s method
functional group isomerism is not possible?
(3) Carius method
(1) C4H10O
(4) All of these
(2) C3H6O
(3) C4H9X 23. During the test of S, it is precipitated as

(4) C3H6O2 (1) PbS

17. The principle involved in chromatography is (2) H2S

(1) Osmosis (3) SO2

(2) Adsorption (4) H2SO4

(3) Solubility 24. Which of the following is a temporary effect?

(4) Evaporation (1) Inductive effect (2) Electromeric effect

18. Which of the following groups contain only (3) Resonance effect (4) All of these
electrophiles? 25. Identify the most stable carbanion
(1) NH3, SO2, Cl

(2) H2O, NH3, SO24 (1) H–C C (2) H2C CH


(3) H3C CH2 (4) H C
: :

(3) CH3–C=O, SO3, :OH 3

26. Identify the group showing –I and +R effect

(4) NO
2 ,
:CH2 , BF3 simultaneously
19. The number of metamers possible for molecular (1) –F (2) –NO2
formula C4H10O is
O O
(1) 2 (2) 3 || ||
(3) –C–H (4) –C–OH
(3) 7 (4) 4 27. Hybridisation of carbon bearing negative charge in
20. Choose the correct order of priority of the functional the following structure is
group in IUPAC naming
(1) –COOH > –CHO > –CN > –OH
(2) –CHO > –COOH > –OH > –CN
(1) sp (2) sp2
(3) –CN > –COOH > –CHO > –OH
(3) sp3 (4) sp3d
(4) –COOH > –CN > –CHO > –OH

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60 Organic Chemistry - Some Basic Principles & Techniques NEET

28. Which of the following is an example of elimination 33. Which among the following species is an ambident
reaction? nucleophile?
(1) Chlorination of CH4
(1) CH 3 CH 2
(2) Dehydration of alcohols
(2) CH2 CH2
(3) Nitration of benzene
(4) Hydroxylation of ethene (3) CN
29. Number of isomeric forms of C7H9N having benzene
ring will be (4) NH3

(1) 7 (2) 6 34. Given reaction, C2H5Br + NaOH  C2H5OH +

NaBr is called
(3) 5 (4) 4
(1) Electrophilic substitution
30. The most stable canonical structure of is (2) Nucleophilic substitution
(3) Both (1) & (2)
N
(4) Nucleophilic addition
N
35. For which of the following molecules (Dipole
moment)  0
(1) (2)

N N Cl Cl NC CN
(A) (B)
N N

(3) (4) HO OH HS SH
(C) (D)
N N
(1) Only (A)
N N
(2) (A) & (B)
31. The correct order of resonance energy is
(3) Only (C)
CH2 CH F CH2 CH OCH3
(4) (C) & (D)
I II
36. How many structural isomers of C5H11OH will be
CH3
CH2 CH N primary alcohol?
CH3
III (1) 2 (2) 3
(1) I > II > III (2) II > I > III (3) 4 (4) 5
(3) III > II > I (4) III > I > II OH
OH OH
  O2N
O O
32. CH 3 C CH3 C 37.
 OH  O
NO2
NO2
The correct order of bond lengths is (C)
(A) (B)
(1)  >  >  > 
The order of the C–O bond lengths among these
(2)  >  >  >  compounds is
(3)  > = >  (1) (A) > (B) > (C) (2) (C) > (B) > (A)
(4)  = >  >  (3) (B) > (A) > (C) (4) (C) > (A) > (B)

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NEET Organic Chemistry - Some Basic Principles & Techniques 61
38. Consider the following statements about 41. Which of the following is incorrect statement?
(1) An antiaromatic compound is destabilized by
O
a closed loop of electrons.

(2) ( ) 3 Cis more stable than (Ph)3C

(3) If there is conjugation between triple bond and
(i) The IUPAC nomenclature of the compound is
double bond then both -bonds of triple bond
cyclohexa-2,5-dienone.
are in conjugation with double bond
(ii) The tautomer of it is phenol.
(iii) Degree of unsaturation in it is 4. (4) CH 2 CH—CH2 is more stable than

(iv) Hybridisation of each carbon is sp2. CH 3—CH 2

Choose the correct statement(s) 42. The total number of structural isomers of C4H10O is

(1) Only (i) (1) 7

(2) (ii) & (iii) (2) 8

(3) 4
(3) (i) & (iv)
(4) 6
(4) (i), (ii) & (iii)
43. The most stable carbocation is
39. Match the terms mentioned in column I with the
terms in column II (1) +
Column I Column II
N
a. Hyperconjugation (i) Cyclohexane and
1-hexene
b. Isomers (ii) – conjugation (2) +
c. Nucleophile (iii) Species that can
N N
receive a pair of
+
electrons C

d. Electrophile (iv) Species that can (3)

supply a pair of
+
electrons C
(4)
(1) a(ii), b(i), c(iv), d(iii)
(2) a(i), b(ii), c(iii), d(iv) 44. 0.40 g of an organic compound containing
phosphorus gave 0.444 g of Mg2P2O7 by usual
(3) a(ii), b(i), c(iii), d(iv)
analysis. Calculate the percentage of phosphorus
(4) a(i), b(ii), c(iv), d(iii) in the organic compound
40. In which of the following species resonance is (1) 15.5% (2) 38%
possible? (3) 62% (4) 31%
(1) CH3—CH—NH2 45. The best and latest technique for purification and
separation of organic compound is
(2) CH 2 CH—NH 3 (1) Chromatography
(2) Sublimation
(3) O O
(3) Crystallization

(4) CH 3—CH—BH 2 (4) Steam distillation

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62 Organic Chemistry - Some Basic Principles & Techniques NEET

49. The correct order of +I effect is

H O—Hb
– – – –
46. Hd C—C—C—O—Ha (1) COO > O> NH > CH2

H Hc O – – –
(2) CH2 > O> COO > C(CH3)3

In this formula which of the hydrogen is the most – – –

(3) C(CH3)3 > O> COO > NH
acidic?
– – –
(1) b (4) O> COO > C(CH 3) 3 > NH

(2) c 50. The stability of alkyl free radicals is due to

(3) d (1) Hyperconjugation

(4) a (2) + I effect

CN CHO (3) – I effect

47. , (4) Both (1) & (2)

CHO CN
51. In the given compound
Above two molecules are

(1) Chain isomers CH2 H (1)

(6) H H(2)H (3)
(2) Position isomers

(3) Metamers (5) H

H
(4) All of these (4)

48. Aromatic enol is not possible in

The correct order of reactivity of H atoms towards
O free radical substitution is

(1) 2 > 4 > 6 > 5 > 3 > 1

(1)
O (2) 2 > 4 > 5 > 6 > 1 > 3

(3) 1 > 4 > 2 > 5 > 6 > 3

O
(4) 3 > 1 > 5 > 6 > 4 > 2

52. Which of the following is a correct order of

(2)
O reactivity of chlorides for given compounds?
O
Cl
Cl Cl Cl
O Cl
O
(3)
I II III IV V
O
(1) I > II > III > IV > V
O
(2) I > III > II > IV > V
(4) (3) III > I > IV > V > II
O (4) I > III > IV > V > II

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NEET Organic Chemistry - Some Basic Principles & Techniques 63
53. In the given reaction 55. Which of the following should be a sp3 hybridised
OH free radical?

+
Cl
A (major) CH2
NH2 (1)

NO2
+
(2) CF3
Cl
B (major)
OH
A and B are respectively (3)

OH NO2 (4) Both (2) & (3)

Cl 56. The charge on given atom cannot be dispersed on
(1) and other atom in
NH2 Cl OH
CH3
.. + –
(1) NH2 CH CH3 (2) CH2 B
OH NO2 CH3
Cl +
and NH3
(2)
NH2 OH
Cl (3) (4) All of these

57. The correct order of stability of given carbocations

OH NO2 is
Cl Cl
(3) and O
.. .. + ..
NH2 OH O
.. + O
..
+ + O
..

OH NO2
(1) I > IV > III > II (2) I > IV > II > III

(4) and (3) IV > I > III > II (4) IV > I > II > III
NH2 OH 58. The correct order of stability of given carbocations
Cl Cl is
+ + +
54. The least reactive alkene towards dil. HCl is NO2 CH2 CH2 NH2 CH2 CH2 CH3 O CH2
(I) (II) (III)
CH3 CH3 + + +
(1) C C HO CH2 NH2 CH2 NO2 CH2
CH3 CH3 (IV) (V) (VI)
+
CH3O CH2 CH2
(2) CH3—CH==CH2
(VII)
CH3 (1) VII > III > II > IV > V > I > VI
(3) C CH2 (2) V > III > IV > II > VI > I > VII
CH3
(3) V > IV > III > II > VII > I > VI
(4) CH2==CH2 (4) VI > I > VII > III > II > V > IV

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64 Organic Chemistry - Some Basic Principles & Techniques NEET

60. In which of the following rearranged 3° carbocation

CH2
will not be formed?
59. +
H
Intermediate (Major) CH3

The most stable carbocation is (1)

+
CH2 CH3
+
CH CH3
(1) + (2) + (2)

CH2 +
CH2
(3) +
(3) (4)
+ (4) Both (1) & (3)



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Chapter 15

Hydrocarbons
Chapter Contents
 Introduction Introduction
 Classification The term ‘hydrocarbon’ is self-explanatory meaning compounds of
 Alkanes carbon and hydrogen only. Industrial organic chemistry today can be
roughly divided into four major areas. In order of their current economic
 Alkenes
importance they are polymers, petrochemicals, synthetic materials
 Alkynes (other than polymers) and miscellaneous organic materials. Natural
 Aromatic Hydrocarbon gas and petroleum are chief sources of aliphatic hydrocarbons at the
present time, and coal is one of the major sources of aromatic
 Carcinogenicity and toxicity
hydrocarbons. Petroleum is a dark, viscous mixture of many organic
compounds, most of them being hydrocarbons, mainly alkanes,
cycloalkanes and aromatic hydrocarbons. Petroleum or crude oil also
known as ‘black gold’ when put to fractional distillation yields petrol,
diesel and kerosene oil. ‘LPG’ used in cooking is the abbreviated form
of Liquefied Petroleum Gas where ‘CNG’ stands for Compressed Natural
Gas. Another term ‘LNG’ (Liquefied Natural Gas) produced by cooking
natural gas below its boiling point, –162°C (–259°F) is also in news
these days.

CLASSIFICATION
As we are quite aware that there are different types of hydrocarbons.
Depending upon the types of carbon-carbon bonds present, they can be
classified into three main categories – (i) saturated (ii) unsaturated and
(iii) aromatic hydrocarbons. Saturated hydrocarbons contain carbon-carbon and
carbon-hydrogen single bonds. If different carbon atoms are joined together
to form open chain of carbon atoms with single bonds, they are termed as
alkanes. On the other hand, if carbon atoms form a closed chain or ring, they
are termed as cycloalkanes. Unsaturated hydrocarbons contain carbon-carbon
multiple bonds – double bonds, triple bonds or both. Aromatic hydrocarbons
are a special type of cyclic compounds.

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ALKANES
These are the saturated chains of hydrocarbons containing carbon-carbon single bonds. Methane (CH4), is
the first member of this family containing single carbon atom. Since it is found in coal mines and marshy
areas, is also known as ‘marsh gas’. The next member of the alkane family is ethane (C2H6) containing two
carbon atoms, both joined together by single bond.
These hydrocarbons exhibited low reactivity or no reactivity under normal conditions with acids, bases and
other reagents, they were earlier known as paraffins (Latin : parum, little; affinis, affinity). The general formula
for alkane is CnH2n + 2, where n stands for number of hydrogen atoms in the molecule. All the C – C bonds
in alkanes are sp3 – sp3 hybridized, (sigma) bond which is quite strong. They have three-dimensional
structures where all the four bonds are tetrahedrally arranged making a normal angle of 109°28. For example,
in methane, the carbon at the centre is bonded by four hydrogen atoms, where all the bonds lie in different
planes giving it a three-dimensional structure.

H H
3
sp hybridized
109°28
C or C
H H H H

H H
Structure of methane molecule
If we take an example of ethane (C2H6) which contains one C – C and six C – H bonds, each carbon atom
is sp3 hybridized where carbon–carbon bond is formed by sharing of sp3-orbitals of carbon and carbon-
hydrogen bond is formed by the head-on overlapping of sp3 hybridized carbon and 1s orbitals of hydrogen
atoms.
3 3
sp -s H H sp -s H
H

109°28 109°28
C C or C
C
H H H H

H H
H H
Structure of ethane molecule
The tetrahedra joined together in alkanes have the C–C and C–H bond lengths of 154 pm (1.54 Å) and
112 pm (1.12 Å) respectively.

Example 1 : What would be the formula of the next alkane if one hydrogen from butane is replaced by a methyl
group?
Solution : Formula of butane is C4H10. Replacement of one hydrogen by a methyl group forms the alkane
with formula C5H12.

Nomenclature and Isomerism

The first three members of alkanes have same common as well as IUPAC name. The higher members of this
family exists in different forms or isomers where isomers are those compounds which have the same molecular
formula but different structures.
Nomenclature and isomerism in alkanes can be understood with the help of few examples. Common names
are written in parenthesis. If alkanes with four or more than four carbon atoms are there, it can exist in more

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NEET Hydrocarbons 67
than one structure. For example, butane can have two structures, one having a straight chain and the other
one a branched structure with n-butane and iso-butane common names respectively.
H
H–C–H
H H H H H H
1 2 3 4 1 2 3
H–C–C–C–C–H H–C–C–C–H
H H H H H H H
(I) Butane (n-butane) (II) 2-Methylpropane (isobutane)
(b.p. 273 K) (b.p. 261 K)

As the number of atoms goes on increasing in the alkane its isomeric forms also increases. For eg.,: The next
consecutive higher alkane exists in greater number of structure forms. Pentane (C5H12) exists in three forms.
H
H–C–H
H H H H H H H H H H H
1 2 3 4 5 1 2 3 4 1 2 3
H–C–C–C–C–C–H H–C–C–C–C–H H–C–C–C–H
H H H H H H H H H H
(I) Pentane (n-pentane) H–C–H H–C–H
(b.p. 309 K)
H H
(II) 2-Methylbutane (isopentane) (III) 2, 2-Dimethylpropane (neopentane)
(b.p. 301 K) (b.p. 282.5 K)

So from the given examples we find that even though a given alkane possesses the same molecular formula
it differs in other properties, as evident from the boiling points. Since, the differences in their properties is due
to the difference in their structures, they are known as structural isomers. Those structural isomers which
differ in chain of carbon atoms are known as chain isomers. Based upon the number of carbon atoms
attached to the carbon atom, the carbon atom is termed as primary (1°), secondary (2°), tertiary (3°) or
quarternary (4°). Carbon atom attached to no other carbon atom as in methane or to only one carbon atom
as in ethane is called primary carbon atom. Terminal carbon atoms are always primary while carbon atom
attached to two carbon atoms is known as secondary. Tertiary carbon atom is attached to three carbon atoms
and quarternary carbon atom is attached to four carbon atoms.
1°
CH3
1° 2° 4° 3° 2° 1°
H3C – CH2 – C – CH – CH2 – CH3
1°
CH3 1°CH3
Type of carbon atoms shown in 3, 3, 4-Trimethylhexane

Example 2 : Write structures of different chain isomers of alkanes corresponding to the molecular formula C6H14.
Also write their IUPAC names.
Solution : (I) CH3 – CH2 – CH2 – CH2 – CH2 – CH3 (II) CH3 – CH – CH2 – CH2 – CH3
Hexane
CH3
2-Methylpentane
(III) CH3 – CH2 – CH – CH2 – CH3 (IV) CH3 – CH – CH – CH3
CH3 CH3 CH3
3-Methylpentane 2, 3-Dimethylbutane
CH3

(V) CH3 – CH2 – C – CH3

CH3
2, 2-Dimethylbutane

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We know that as the number of carbon atoms in the chain increases, number of structural isomers also
increases. The table given below gives an insight to it.

Molecular formula Number of structural isomer

C4H10 2
C5H12 3
C6H14 5
C7H16 9
C8H18 18
C9H20 35
C10H22 75

Alkyl Group
When one H-atom is replaced from alkane, the group obtained is called alkyl group. The general formula for
the alkyl group is CnH2n + 1. The alkyl groups being derivatives of the alkanes, are named accordingly.

Structural formula IUPAC accepted common name

CH3 – Methyl
CH3 – CH2 – or C2H5 – Ethyl

CH3 – CH2 – CH2 – or C3H7 – Propyl

CH3 – CH2 – CH2 – CH2 – or C4H9 – Butyl
CH3 – CH – CH3 Isopropyl

CH3
Isobutyl group
CH3 – CH – CH2 –
CH3 – CH2 – CH –
Secondary butyl group
CH3
CH3
CH3 – C – Tertiary butyl group
CH3
CH3 – CH – CH2 – CH2 –
Isopentyl group
CH3
CH3
CH3 – C – CH2 – Neopentyl group
CH3
CH3
CH3 – CH2 – C – Tertiary pentyl group
CH3
CH3
Isohexyl group
CH3 – CH – CH2 – CH2 – CH2 –

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Let us recall the general rules for nomenclature.

Remarks

CH3 CH2 – CH3 Lowest sum and

(i)
1 2 3
CH3 – CH – CH2 – CH – CH2 – CH3 4 5 6 alphabetical
(4-Ethyl-2-methylhexane)
arrangement

CH2 – CH3
8 2
(ii) CH3 – CH2 – CH2 – CH – CH – C – CH2 –1CH3
7 6 5 4 3
Lowest sum and
CH CH3 CH2 – CH3 alphabetical
arrangement
CH3 CH3
(3, 3-Diethyl-5-isopropyl-4-methyloctane)

CH(CH3)2 sec is not considered

(iii)
1 2 6
CH3 – CH2 – CH2 – CH –5CH – CH2 –7CH2 –8CH2 –9CH2 –10CH3
3 4 while arranging
alphabetically;
H3C – CH – CH2 – CH3 isopropyl is taken as
5-sec -Butyl-4-isopropyldecane one word

1 2 6
(iv) CH3 – CH2 –3CH2 –4CH2 –5CH – CH2 – 7CH2 –8CH2 –9CH3
CH2 Further numbering
2
CH3 – C – CH3 to the substituents
of the side chain
3
CH3
5-(2, 2-Dimethylpropyl)nonane

1 2 6
(v) CH3 – CH2 –3CH –4CH2 –5CH – CH2 –7CH3 Alphabetical
CH2 – CH3 CH3 priority order
3-Ethyl-5-methylheptane

It is equally important to write the correct structure from the given IUPAC name if it is important to write the
correct IUPAC name for a given structure. For this a certain set of rules need to be followed for the convenient
writing of the structure. Let the rules for writing the structure be applied for 3-Ethyl-2, 2-dimethylpentane.
Steps :
(i) Draw the chain of five carbon atoms since the root word ‘pent’ in the given name denotes the chain to
have five carbon atoms.
C–C–C–C–C
(ii) Give number to the carbon atoms in the chain so branches can be assigned their respective positions.
1 2 3 4 5
C–C–C–C–C
(iii) Attach ethyl group at carbon 3 and two methyl groups at carbon 2.

CH3
C1 – C2 – C3– C4 – C5
CH3 C2H5

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(iv) The valence of each carbon atom is satisfied by putting required number of hydrogen atoms.

CH3
CH3 – C – CH – CH2 – CH3
CH3 C2H5
So, following the likewise set of rules the correct structures of the compound can be obtained.

Example 3 : Write the structure of the compound 3, 4-Diethyl-3, 4-dimethyl heptane.

CH3 CH3
1 2 3 4 5
Solution : H3C – CH2 – C – C – CH2 –6CH2 –7CH3
C2H5 C2H5

Preparation of Alkanes
Though petroleum and natural gas are the main sources of alkanes, it can be prepared by several other
methods as well.
1. From unsaturated hydrocarbons
The addition of dihydrogen to unsaturated hydrocarbons like alkenes and alkynes in the presence of a
suitable catalyst under a given set of conditions produces saturated hydrocarbons or alkanes. This process
of addition of dihydrogen is known as hydrogenation process. The metals like nickel, platinum or
palladium are used as a catalysts in finely divided form to increase the surface area. Platinum and
palladium catalyse the reaction at room temperature but relatively higher temperature and pressure are
required with nickel catalysts.

CH 2 = CH 2 + H 2 Pt/Pd/Ni CH3 – CH3

Ethene Ethane
Pt/Pd/Ni
CH3 – CH = CH2 + H2 CH3 – CH2 – CH3
Propene Propane

CH3 – C  C – H + 2H2 Pt/Pd/Ni CH3 – CH2 – CH3

Propyne Propane

If the above catalytic hydrogenation reaction takes place in presence of nickel, it is also known as
Sabatier-Senderen’s reaction. Methane cannot be prepared by this method.
2. From alkyl halides
(a) Reduction : Alkyl halides undergo reduction with zinc and dilute hydrochloric acid to give alkanes.
In general the reaction can be represented as
+
Zn, H
CnH2n + 1X + H2 CnH2n + 2 + HX
+
Zn, H
CH3 – Cl + H2 CH4 + HCl
Chloromethane Methane
+
Zn, H
C2H5 – Cl + H2 C2H6 + HCl
Chloroethane Ethane
+
Zn, H
CH3CH2CH2Cl + H2 CH3CH2CH3 + HCl
1-Chloroethane Propane

The alkyl fluorides does not give this reaction due to strong bond showing low dissociation. Among
the halogens, the order of reactivity is iodides > bromides > chlorides.

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(b) Wurtz reaction : Alkyl halides on treatment with sodium metal in dry ether (free from moisture)
give higher alkanes. This reaction is known as Wurtz reaction. In this reaction, the product obtained
is an alkanes with new carbon-carbon bond. The reaction is used in general to produce alkanes
with even number of carbon atoms.
dry ether
CH3Br + 2Na + BrCH3 CH3 – CH3 + 2NaBr
Ethane

When two different types of alkyl halides are taken, a mixture of three alkanes with odd and even
number of carbon atoms are obtained.
dry
ether
2CH3 – CH2 – Br + 4Na + 2Br – CH3 CH3CH2CH2CH3 + CH3CH2CH3 + CH3CH3 + 4NaBr

3. From carboxylic acids

(a) By decarboxylation of carboxylic acids : Sodium salts of carboxylic acids on heating with soda
lime (mixture of sodium hydroxide and calcium oxide) give alkanes containing one carbon atom less
than the carboxylic acid. A molecule of carbon dioxide is eliminated which dissolves in NaOH to
form sodium carbonate.
– + CaO
CH3COO Na + NaOH  CH 4 + Na2CO 3
Sodium ethanoate Methane

Note : CaO in sodalime is used to keep NaOH dry due to its hygroscopic nature.

(b) Kolbe’s electrolytic method : An aqueous solution of sodium or potassium salt of a carboxylic
acid on electrolysis gives alkane containing even number of carbon atoms at anode.
– + Electrolysis
2CH3COO Na + 2H2O CH3 – CH3 + 2CO2 + H2 + 2NaOH
Sodium acetate Ethane

Mechanism :

The reaction follows a free-radical mechanism :

O O
– + – +
2CH3 – C – O Na 2CH3 – C – O + 2Na

At anode :

O O
–
– –2e
(i) 2CH3 – C – O 2CH3 – C – O 2CH3 + 2CO2
Acetate ion Acetate free Methyl free
radical radical

(ii) CH 3 + CH3 CH3 – CH3

At cathode :

(i) 2H2O + 2e–  2OH– + 2H

2H H2

(ii) 2Na+ + 2OH–  2NaOH

This method is applicable for the preparation of symmetrical alkanes only and not suitable for the
preparation of methane. In Kolbe’s electrolytic methods the alkane obtained is always a higher
alkane due to the formation of carbon-carbon (C – C) bond restricting the formation of methane by
this process.

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Example 4 : Which salt of carboxylic acid will be required to prepare ethane by sodalime decarboxylation? Give
equation for the reaction.
Solution : Sodium propanoate

– + CaO
CH3CH2COO Na + NaOH  CH3 – CH 3 + Na2CO3
Sodium propanoate Ethane

Some other important methods used for the preparation of higher alkanes are
1. Frankland’s reaction : This reaction is similar to Wurtz reaction. To obtain a higher alkane, the alkyl
halide is heated with zinc in an inert solvent like ether.
ether
R – X + Zn + X – R R – R + ZnX2
ether
CH3 – I + Zn + I – CH3 CH3 – CH3 + ZnI2
Methyl iodide Ethane

Note : Methane cannot be prepared by this method since higher alkane is obtained by the formation
of C – C bond.
2. Clemmensen reduction : Aldehydes and ketones undergo reduction with amalgamated zinc and conc.
HCl yielding alkanes.
Zn – Hg
– C = O Conc. HCl – CH2
(Ketone)
H
Zn – Hg
CH3 – C = O Conc. HCl
CH3 – CH3 + H2O
Acetaldehyde Ethane
CH3
Zn – Hg
CH3 – C = O Conc. HCl
CH3 – CH2 – CH3 + H2O
Acetone Propane
3. Wolff-Kishner reduction : Hydrazones of aldehydes and ketones can be converted in to alkane in an alkaline
medium (C2H5ONa or NaOH) at high temperature
R R C2H5ONa
–H2O or
C O + H2NNH2 C NNH2 NaOH, 
R – CH2 – R + N2
R Hydrazine R Hydrazone

CH3 CH3 C2H5ONa CH3

–H2O or
C O + H2 NNH2 C NNH2 NaOH, 
CH2 + N2
CH3 CH3 CH3
Propane

4. From Grignard reagents : Grignard reagents are organometallic halides discovered by the French chemist
Victor Grignard in 1900. These reagents have the general formula R – Mg – X. These undergo hydrolysis in the
presence of an acid, or reacts with ammonia, alcohol or amines having active hydrogen (H) atom to give
alkane corresponding to alkyl group of Grignard reagent.

H
+ OH
CH3 – Mg – Br + HOH CH3 – H + Mg
Methyl magnesium Methane Br
bromide

Br
Br Mg CH3 + H O C2H5 CH3 – H + Mg
Methyl magnesium Ethyl alcohol Methane OC2H5
bromide

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CH3 H NH2 NH2

Mg + CH3 – H + Mg
Cl Methane Cl
Methyl Ammonia
magnesium
chloride
H H
C2H5 H – N – CH3 N – CH3
Mg + C2H5 – H + Mg
Br Ethane Br
Ethyl magnesium Methyl amine
bromide

Properties of Alkanes
Physical Properties
Alkanes are considered to be almost non-polar molecules because of the covalent nature of C – C and C –
H bonds and due to very little difference of electronegativity between carbon and hydrogen atoms. They
possess very weak intermolecular forces of attraction i.e., van der Waal’s forces.
(i) State : Due to the weak van der Waals forces, the first four members C1 to C4 i.e., methane, ethane,
propane and butane are gases. From C5 to C17 are liquids and those containing 18 carbon atoms or
more are solids at 298 K. They all are colourless and odourless.
(ii) Solubility : Alkanes are generally insoluble in water or in polar solvents but they are soluble in non-
polar solvents like, ether, benzene, carbontetrachloride etc. The solubility of alkanes follow the property
“Like Dissolves like”. Being non-polar in nature, these are readily soluble in organic or non-polar solvents.
(iii) Boiling point : The boiling points of straight chain alkanes increase regularly with the increase of number
of carbon atoms. This is due to the fact that the intermolecular van der Waals forces increase with
increase in the molecular size or the surface area of the molecule.
Among the isomeric alkanes straight chain alkanes have a higher boiling point than the branched chain
alkanes. As the branching, increases the molecule becomes more symmetric due to which the surface area
decreases which in turn decreases the van der Wall interactions.
Table : Variation of Melting Point and Boiling Point in Alkanes

Molecular Name Molecular b.p./(K) m.p./(K)

formula mass/u
CH4 Methane 16 111.0 90.5
C2H6 Ethane 30 184.4 101.0
C3H8 Propane 44 230.9 85.3
C4H10 Butane 58 272.4 134.6
C4H10 2-Methylpropane 58 261.0 114.7
C5H12 Pentane 72 309.1 143.3
C5H12 2-Methylbutane 72 300.9 113.1
C5H12 2,2-Dimethylpropane 72 282.5 256.4
C6H14 Hexane 86 341.9 178.5
C7H16 Heptane 100 371.4 182.4
C8H18 Octane 114 398.7 216.2
C9H20 Nonane 128 423.8 222.0
C10H22 Decane 142 447.1 243.3
C20H42 Eicosane 282 615.0 236.2

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Chemical Properties
Generally alkanes show inertness or low reactivity towards acids, bases, oxidizing and reducing agents at
ordinary conditions because of their non-polar nature and absence of  bond. The C – C and C – H bonds
are strong sigma bonds which do not break under ordinary conditions but they undergo certain reactions under
given suitable conditions.
(1) Substitution Reactions
One or more hydrogen atoms of alkanes can be replaced by halogens, nitro group and sulphonic acid
group. These reactions in which hydrogen atoms of alkanes are substituted are known as substitution
reactions.
Halogenation reaction : When hydrogen atom of an alkane is replaced by a halogen, it is known as
halogenation reaction. Halogenation takes place either at high temperature (300–500°C) or in the
presence of diffused sunlight or ultraviolet light. The extent of halogenation depends upon the amount
of halogen used. For example, methane undergoes chlorination reaction in excess of chlorine in
presence of UV light to produce a mixture of chloromethane, dichloromethane (or methylene chloride),
trichloromethane (or chloroform) and tetrachloromethane (or carbon tetrachloride). The reaction of methane
with chlorine is given below :
h
CH4  Cl2  CH3 Cl  HCl
Chloromethane
(methylchloride)

h
CH3 Cl  Cl2  CH2 Cl2  HCl
Dichloromethane
(methylene chloride)

h
CH2Cl2  Cl2   CHCl3  HCl
Trichloromethane
(chloroform)

h
CHCl3  Cl2  CCl4  HCl
Tetrachloromethane
(carbon tetrachloride)

Ethane and higher alkanes react in a similar way and all types of possible substituted products are
obtained. For example, ethane undergoes chlorination to give chloroethane.

h
CH3 – CH3  Cl2   CH3 – CH2 – Cl  HCl
Chloroethane

However, as we ascend the order of higher alkane, the rate of halogenation differs due to their reactivity
since these contain hydrogen of two or three types i.e., primary, secondary or tertiary. The order of
reactivity is,
Tertiary hydrogen (3°) > Secondary hydrogen (2°) > Primary hydrogen (1°)
The rate of reaction of alkanes with halogens is F2 > Cl2 > Br2 > I2. Fluorination reaction of alkanes is
violent in nature while chlorination and bromination proceed smoothly. Iodination is very slow and a
reversible reaction, so the process is carried out in the presence of an oxidizing agents like HIO3 (iodic
acid) or HNO3 which decomposes HI into iodine.

CH4 + I2 CH3I + HI

HIO3 + 5HI  3I2 + H2O

2HNO3 + 2HI  2NO2 + 2H2O + I2
Halogenation is supposed to proceed via free radical chain mechanism involving three steps namely
initiation, propagation and termination as given below.

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Mechanism
(i) Initiation : The chlorine molecule undergoes homolytic cleavage producing chlorine free radical in the
presence of heat or light which initiates the reaction. The Cl – Cl bond is weaker than the C – C and
C – H bond and hence easier to break.

h
Cl – Cl homolytic
2 Cl or Cl + Cl
fission Chlorine free radicals

(ii) Propagation : Chlorine free radical attacks the methane molecule and takes the reaction in the forward
direction by breaking the C – H bond to generate methyl free radical with the formation of H – Cl.
h
(a) CH4 + Cl CH3 + H – Cl
The methyl radical thus obtained attacks the second molecule of chlorine to form CH3 – Cl with the
liberation of another chlorine free radical by homolytic fission of chlorine molecule.

h
(b) CH3 + Cl – Cl CH3 – Cl + Cl
Chlorine
free radical

The chlorine and methyl free radical generated in the reaction set-up a chain of reactions. The propagation
steps (a) and (b) are those which directly give principal products, but many other propagation steps are
possible and may occur. Two such steps given below explains the formation of more halogenated products
from main product, chloromethane.

 
CH3Cl  Cl 
 CH2Cl  HCl

 
CH2Cl  Cl – Cl 
 CH2Cl2  Cl

(iii) Termination : The reaction stops after some time due to consumption of reactants and/or due to the
following side reactions :
The possible chain-terminating steps are

(a) Cl + Cl Cl – Cl

(b) H3C + CH3 CH3 – CH 3

(c) CH 3 + Cl CH3 – Cl
Though in (c), CH3 – Cl, the one of the products is formed but free radicals are consumed and the chain
is terminated. The above mechanism helps us to understand the reason for the formation of ethane as
a by-product during chlorination of methane.
Like in halogenation process, the hydrogen in alkane is replaced by a halogen. When the hydrogen atom
of alkane is replaced by a nitro (–NO2) group or sulphonic (–SO3H) group, it is known as nitration or
sulphonation respectively. Lower alkanes do not undergo nitration and sulphonation reactions.
(2) Combustion
Alkanes on heating in presence of air gets completely oxidized to carbon dioxide and water. It burns
with a non-luminous flame. The combustion of alkanes is an exothermic process i.e., it produces a large
amount of heat.

CH4(g) + 2O2(g)  CO2(g) + 2H2O(l);  cH = –890 kJ mol–1

13
C4H10(g) + O (g)  4CO2(g) + 5H2O(l);  cH = –2875.84 kJ mol
–1
2 2
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The heat generated in the reaction shows that as the molecular mass of alkane increases, the generation
of heat also increases. The general combustion equation for any alkane is

 3n  1
CnH2n  2   O 
 nCO2  (n  1)H2O
 2  2

Due to large evolution of heat during combustion, alkanes are used as fuels.
During incomplete combustion of alkanes with insufficient amount of air or dioxygen, carbon black is
formed which is used in the manufacture of ink, printer ink, black pigments and filters.

Incomplete
CH 4(g) + O 2(g) combustion C(s) + 2H 2O(l)
Carbon black

Example 5 : Why iodination of alkanes is carried out in the presence of oxidizing agents?
Solution : Iodination is a slow and reversible process. Hence, the HI formed in the reaction again decomposes
to produces alkane. The oxidizing agents prevent the formation of alkane back by decomposing
HI into iodine.

CH 4 + I2 CH 3I + HI

HIO3 + 5HI 3I2 + 3H2O

Iodic acid
(oxidizing agent)

(3) Controlled oxidation

Alkanes undergo oxidation in the presence of regulated supply of dioxygen or air at high pressure and
a suitable catalyst. Bringing change in the conditions like that of catalyst, pressure etc. yields a variety
of oxidation products.

(i) When methane and dioxygen compressed at 100 atm are passed through heated copper tube at
523 K yield methanol.

Cu tube
2CH4 + O2 2CH3OH
523 K/100 atm
Methanol

(ii) When the mixture of methane and dioxygen is passed over molybdenum oxide at 543 K at 100
atm pressure, methanal (formaldehyde) is obtained.

Mo2O3
CH 4 + O 2 HCHO + H2O
543 K/100 atm
Formaldehyde

(iii) Higher alkanes are oxidized to carboxylic acid in presence of manganese acetate.

(CH3COO)2 Mn
2CH3 – CH3 + O2 2CH3COOH + 2H2O
Ethanoic acid

(iv) Ordinarily alkanes resist oxidation but alkanes having tertiary H atom can be oxidized to
corresponding alcohols by potassium permanganate.
CH3 CH3
KMnO4
CH3 – C – H (oxidation)
CH3 – C – OH
CH3 CH3
2-Methylpropane 2-Methylpropan-2-ol

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(4) Isomerisation
n-Alkanes on heating in presence of anhydrous aluminium chloride and hydrogen chloride gas isomerise
to branched chain alkanes. The molecular rearrangement of one isomer to another is known as
isomerisation.
Major products in the reaction is mentioned while possibilities of formation of minor products are also
there which are generally not mentioned in organic reactions.
CH3
Anhyd. AlCl3/HCl
CH3 – CH2 – CH2 – CH3 CH3 – C – CH3
n-Butane
H
2-Methylpropane
Anhyd. AlCl3/HCl
CH3 – CH2 – CH2 – CH2 – CH2 – CH3 CH3 – CH2 – CH2 – CH – CH3
n-Hexane
CH3
2-Methylpentane
+
CH3 – CH2 – CH – CH2 – CH3
CH3
3-Methylpentane

(5) Aromatization
The conversion of aliphatic compounds into aromatic compounds is known as aromatisation.
n-Alkanes having six or more carbon atoms on heating to 773 K at 10–20 atmospheric pressure in the
presence of oxides of vanadium, molybdenum or chromium supported over alumina get dehydrogenated
and cyclised to benzene and its homologues. This reaction is also known as reforming.
CH3 Cr2O3 or V2O3
CH2 CH3 or Mo2O3
+ 4H2
773 K,
CH2 CH2 10–20 atm
CH2
n-Hexane
n-Heptane isomerises to give toluene.
CH3 CH3
CH2 Cr2O3 or V2O3
C
CH2 CH3 or Mo2O3 CH CH
+ 4H2
773 K,
CH2 CH2 10–20 atm
CH CH
CH2 CH

(6) Reaction with steam

Methane reacts with steam at 1273 K in the presence of nickel catalyst to form carbon monoxide and
dihydrogen. This method is used for industrial preparation of dihydrogen gas.
Ni, 
CH4  H2O  CO  3H2
(7) Pyrolysis
The breakdown of a compound into smaller compounds is known as pyrolysis. Higher alkanes, on heating
to higher temperature decomposes into lower alkanes, alkenes etc. Such a decomposition of alkanes
into smaller fragments by the application of heat is called cracking.
C6H12 + H2
Hexene

773 K
C6H14 C4H8 + C2H6
Butene Ethane

C3H6 + C2H4 + CH4

Propene Ethene Methane

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Pyrolysis of alkanes is believed to be a free radical mechanism due to rupturing of either carbon-carbon
bond or carbon-hydrogen bond. Preparation of oil gas or petrol involves the principle of pyrolysis. For
example, dodecane, a constituent of kerosene oil on heating to 973 K in presence of platinum, palladium
or nickel gives a mixture heptane and pentene.

Pt/Pd/Ni
C12H26 
973 K
 C7H16  C5H10  Other products
Dodecane Heptane Pentene

Example 6 : Name few catalysts used in aromatization reaction.

Solution : Cr2O3, V2O3 or MoO3 are the oxides of metals used as catalyst in aromatization reaction.

EXERCISE

1. Which of the following reaction will not give methane?

NaOH  CaO 
(1) CH3COONa 


 (2) BeC2  H2O  


(3) Al4C3  H2O   (4) All of these

2. Which of the following isomer having molecular formula C6H14, will give minimum number of mono-chloro
derivatives?
(1) Hexane (2) 2-Methylpentane
(3) 3-Methylpentane (4) 2, 3-Dimethyl butane
3. Methane cannot be prepared by
(1) Corey-house synthesis (2) Wurtz reaction
(3) Fittig reaction (4) All of these
4. Which of the following alkane is not liquid at room temperature?
(1) C5H12 (2) C17H36
(3) C10H22 (4) C4H10
5. Which of the following compound can form during the free-radical chlorination of methane?
(1) CH3Cl (2) C2H6
(3) CCl4 (4) All of these

CH3 h
6. + Br2 A (major) . Identify A
298 K

Br
CH3
CH3
(1) (2) Br

CH2Br Br CH3
(3) (4)

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7. Which of the following reaction cannot be used for the preparation of alkane?
(1) Corey-House synthesis (2) Frankland reaction
(3) Clemmenson’s reduction (4) Aromatization
8. Which of the following has maximum boiling point?

(1) (2)

(3) (4)

9. Which of the following halogen is the most reactive towards alkane?

(1) F2 (2) Cl2
(3) Br2 (4) I2
10. Which of the following alkane upon dichlorination can give only two products?
(1) CH3 – CH2 – CH3 (2) CH4

(3) C(CH3 )4 (4)

Conformations
Alkanes contain carbon-carbon sigma () bonds. Electrons are symmetrically distributed around the
internucleus axis of the carbon-carbon bond which is not disturbed by the free rotation of the carbon-carbon
single bond around its axis. This rotation results into different spatial arrangements (arrangement in space)
of atoms in space which can change into one another. Such two different three-dimensional arrangements
in space of the atoms in a molecule which can be converted into one another just by free rotation about the
axis (carbon-carbon single bond) are known as conformations. Conformations represent conformers,
which are rapidly interconvertible and nonseparable, are also known as “Conformational isomers” or
“rotamers”.

Alkanes due to rotation around the C – C single bonds can have infinite number of conformations. However,
rotation around the C – C single bond is not completely free, it is hindered by a small energy barrier of
1–20 kJ mol–1 due to weak repulsive interaction between the adjacent bonds. Such a type of repulsive
interaction is called torsional strain.

Conformations of ethane : Ethane molecule (C2H6) is composed of two methyl groups bonded together by
a carbon-carbon single bond with each carbon atom attached to three hydrogen atoms. Each carbon atom
is sp3 hybridized with overlapping of orbitals forming the single bond. The bond maintains its linear bonding
overlap as the carbon atoms turn, giving rise to different conformations. Pure conformers cannot be isolated
in most cases because the energy required for rotation is very small and gains it sufficiently even from the
room temperature. So, molecules constantly rotate forming all possible conformations.

H H H H
H C sp3 sp3 C H H C sp3 sp3 C H

H H H H
Sigma bond formed by Overlaping is maintained
the linear overlaping even by rotation

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Considering the ball and stick model of ethane, one carbon atom is kept stationary and other carbon atom
is rotated around the carbon-carbon axis.

H H
Stationary
carbon atom
C C
H Rotate H
H H
H
H H
C C
H H
H
Change in spatial arrangement shown by a ball and stick model
The rotation results into infinite number of spatial arrangements of hydrogen atom w.r.t. the hydrogen atoms
attached to the carbon atom. Thus there are infinite number of conformations of ethane and isomers produced
are known as conformational isomers. Out of the conformations produced there are two extreme cases. One
such conformation in which hydrogen atoms attached to two carbons are as close together as possible is
called eclipsed conformation and the other in which hydrogens are as far apart as possible is known as the
staggered conformation. Any other intermediate conformation is called as skew conformation. In all the
conformations, the bond angles and bond lengths remain the same. Eclipsed and the staggered conformations
can be represented by Sawhorse and Newman projections.
1. Sawhorse projections : In this projection, the molecule is viewed along the molecular axis i.e., looking
down at an angle towards the carbon-carbon bond. Care should be taken while drawing molecular
conformation in sawhorse projections because it could be misleading, depending on how eye sees them.
So, it is projected on paper by drawing 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 of the line. 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. Sawhorse projections of eclipsed and staggered conformations of ethane are depicted as below.
H
H H
120°

H H H H
H

H H H H
(i) Eclipsed (ii) Staggered
Sawhorse projections of ethane
2. Newman projections : This is the way of drawing a molecule looking straight down the bond connecting
two carbon atoms such that carbon-carbon bond appears to be 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° each other, coming together in a Y shape. The rear carbon atom
(the carbon atom away from eye) is represented by a circle with three bonds pointing out from it shorter
in length at an angle of 120° to each other.
H H
H C C H

Viewed from H H
the end
Front carbon Rear
atom carbon atom
Perspective drawing

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The Newman’s projections are depicted below.
Angle of rotation
or angle of torsion
H or dihedral angle

H H H
H
Rear carbon atom

Front carbon atom

H H H
H H H H
(i) Eclipsed conformation (ii) Staggered conformation
Newman’s projection of ethane
Relative stability of conformations : In staggered form of ethane, the electron clouds in C–H bonds are
separated as far as possible. Thus, there are minimum repulsive forces, minimum energy and maximum
stability of the molecule. On the other hand, when the staggered form changes into the eclipsed form, the
electron clouds of carbon-hydrogen bonds come closer resulting in the increase in electron cloud repulsion.
To oppose the increased repulsive forces, molecule should possess more energy and thus lesser stability.
As the ethane molecule rotates towards the eclipsed conformation, its potential energy increases restricting
its rotation. This resistance to twisting which affects the stability of a conformation is called torsional strain
and the energy required in restriction is known as torsional energy. The magnitude of torsional strain depends
on the angle of rotation about C–C bond. This angle is also called as dihedral angle or torsional angle
denoted by .

Of all the conformations of ethane, the staggered form has the least torsional strain and the eclipsed form
has the maximum torsional strain. Thus, an inference can be drawn that rotation around the C–C bond is
not completely free. The energy difference between the two forms is very small of the order 12.5 kJ mol–1.
This small energy difference is easily overcome by the ethane molecules even at ordinary temperatures, gaining
thermal or kinetic energy through intermolecular collisions. Thus, it can be said that rotation around the
C–C bond is practically free. This is the reason it has not been possible to separate or isolate different
conformational isomers of ethane.


H  = 0° H
 = 60°
H
H H
Newman H H
Projections
H
H
H HH H H H H
H H
H
H H H
H
Sawhorse
Projections H H H H H
H H

H H H H H H

Conformations Eclipsed,  = 0° Staggered,  = 60° Skew,  = anything else

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Ethane conformations : The eclipsed conformation has a dihedral angle  = 0°, the staggered conformation
has  = 60° and any other angle between the eclipsed and staggered position is the skew conformation.

12.5 kJ mol–1
Potential
Energy

0° 60° 120° 180°

Dihedral angle ()

H  = 0° H  = 60° H H
H H
H H H H
Newman 120°
Projections 60° 180°
H H
H HH H H H HH H H
H H

Conformations Eclipsed,  = 0° Staggered,  = 60° Eclipsed,  = 0° Staggered,  = 60°

The torsional strain is lowest in staggered conformation and highest in eclipsed conformation shown by the
potential energy diagram. Low energy barrier in change of conformations keep the molecule to rotate
constantly. A rotation of the C-C bond by 60° produces the conformations.

Example 7 : On change from the staggered form to the eclipsed form in the ethane molecule conformation,
what happens to the electron cloud of carbon-hydrogen bonds?
Solution : The carbon-hydrogen bonded electron cloud comes closer to each other resulting in increase in
electron cloud repulsions consequently increasing the torsional strain.

EXERCISE
11. Which of the following has maximum angle strain?

(1) (2)

(3) (4)

12. Total number of conformations possible in cyclohexane is

(1) Zero (2) Infinite
(3) Four (4) Two
13. Conformations arise due to rotation around
(1) Carbon-carbon double bond (2) Carbon-carbon triple bond
(3) Carbon-carbon single bond (4) All of these
14. Which of the following is the most stable cycloalkane?

(1) (2)

(3) (4)

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15. Bond angle in chair form of cyclohexane is
(1) 109°28 (2) 120°
(3) 60° (4) 108°
16. The most stable conformation of Butane is
(1) Gauche-form
(2) Partially eclipsed form
(3) Anti-form
(4) Eclipsed form
17. Torsion strain is the repulsive interaction between
(1) Electron cloud of two bonds
(2) Electron cloud of two -bonds
(3) Electron cloud of two -bonds
(4) Electron cloud of two -bonds on adjacent atoms
18. Which form of cyclohexane is free from angle strain?
(1) Boat-form (2) Chair form
(3) Twist boat (4) All of these
19. The number of axial hydrogen atoms in chair form of cyclohexane is
(1) 3 (2) 6
(3) 12 (4) 2
20. Select the correct statement
(1) Deviation from normal tetrahedral angle in cycloalkane is called angle strain
(2) Due to torsional strain eclipsed form has higher energy than the staggered form of a compound
(3) Chair form of cyclohexane is the most stable conformation of cyclohexane
(4) All of these

ALKENES
Alkenes are unsaturated hydrocarbons containing at least one carbon-carbon double bond with general formula
CnH2n. Alkenes are also known as olefins (oil forming) since the first member, ethylene or ethene (C2H4) was
found to form an oily liquid on reaction with chlorine.

Structure of Double Bond

Carbon-carbon double bond in alkenes consists of one strong sigma () bond formed by the head-on
overlapping of the sp2 hybrid orbitals and one weak pi () bond formed by the sideways overlapping of the
two 2p-orbitals of the two carbon atoms. The bond enthalpy of the sigma bond is about 397 kJ mol–1
representing a strong bond, whereas -bond enthalpy is about 284 kJ mol–1. The low bond enthalpy of pi bond
shows the weak nature in comparison to sigma bond. The double bond is shorter in bond length (134 pm)
than the C–C single bond (154 pm). Since pi () bond is a weak bond due to poor lateral overlapping between
the two 2p-orbitals, thus making -bond vulnerable to the attack of electrophilic (electron searching) reagents
or compounds. This is the reason alkenes can be easily attacked by the electrophiles due to the loose
electrons making it as unstable molecules in comparison to alkanes and therefore can easily change to
saturated or single bond compounds by reaction with electrophilic compounds. Strength of the double bond

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(bond enthalpy, 681 kJ mol–1) is greater than that of a carbon-carbon single bond in ethane (bond enthalpy,
348 kJ mol–1). The orbital diagram of ethene molecule is shown in the given figure below :

H   H
sp2 – s
sp2 sp2


sp2 – sp2()
H   H
Orbital picture of ethene depicting  bonds only
p-p(-bond) -cloud
H H H H H 121.7° H
C C 116.6° C C

H H H H H H
134 pm 110 pm
(a) (b) (c)
Orbital picture of ethene showing formation of (a) -bond, (b) -cloud and
(c) bond angles and bond lengths

Nomenclature
For nomenclature of alkenes in IUPAC system, the longest chain of carbon atoms containing the double bond
is selected. Numbering of the chain is done from the end which is nearer to the double bond. The suffix ‘ene’
replaces ‘ane’ of alkanes. The first member of the alkene series is C2H4 known as ethylene (common name)
or ethene (IUPAC name). IUPAC names of a few members of alkenes are given below :
Structure IUPAC name
CH3 —CH==CH2 Propene
CH3 —CH2 —CH==CH2 But-1-ene
CH3 —CH==CH—CH3 But-2-ene
CH2 ==CH—CH==CH2 Buta-1, 3-diene
CH 2 C CH3 2-Methylprop-1-ene

CH 3
CH 2 CH CH CH3 3-Methylbut-1-ene

CH3

Example 8 : Write IUPAC name of the following

Solution : 4-Ethyl-2, 5, 7-decatriene.

Isomerism
Alkenes show both structural isomerism and geometrical isomerism.
Structural isomerism
Those isomers which differ in the bonding arrangement of atoms or groups within the molecule i.e., differ in
structure, are structural isomers and this phenomenon is known as structural isomerism.
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Alkenes possessing molecular formula C4H8 and higher ones can have different structures. For example,
butene (C4H8) can be written in different ways.
(i) H2C==CH—CH2 —CH3
But-1-ene
(ii) CH3 —CH==CH—CH3
But-2-ene
1 2 3
(iii) CH2 C CH3

CH3
2-Methylprop-1-ene
Structure (I) and (II) are the examples of chain isomerism whereas structures (I) and (II) are position isomers.
Geometrical isomerism
Those isomers which have different arrangements around carbon-carbon double bond are known as geometrical
isomers and the phenomenon exhibited by them is known as geometrical isomerism.
Doubly bonded carbon atoms have to satisfy the remaining two valences by joining with two atoms or groups.
If two atoms or groups attached to each carbon atom are different, they can be represented
in space by the following two ways :

X Y X Y
C C

C C
X Y Y X
(a) (b)

In (a) two identical atoms i.e., both the X or both the Y lie on the same side of the double bond but in (b)
two X or two Y lie across the double bond or on the opposite sides of the double bond. This difference in
the attachment of atoms or groups in space gives rise to difference in arrangements. Therefore, they are known
as stereoisomers.
The isomer of the type (a), in which two identical atoms or groups lie on the same side of the double bond
is called cis-isomer and the other isomer of the type (b), in which identical atoms or groups lie on the opposite
sides of the double bond is called trans-isomer.

X Y X Y
C C

C C
X Y Y X
Cis-isomer Trans-isomer
Thus, cis and trans isomers have the same structure but have different configuration (due to arrangement
difference of atoms or groups in space). Due to different arrangement of atoms or groups in space, these
isomers differ in their properties like melting point, boiling point, dipole moment, solubility etc. Geometrical
or cis-trans isomers of But-2-ene are represented below.

H3C CH3 H3C H

C C C C
H H H CH3
cis-But-2-ene trans-But-2-ene
(b.p. 277 K) (b.p. 274 K)

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Cis form of alkene is found to be more polar than the trans form. For example, dipole moment of
cis-but-2-ene is 0.33 Debye, whereas, dipole moment of the trans form is almost zero or it can be said that
trans-but-2-ene is non-polar. This can be understood by drawing geometries of the two forms as given below
from which it is clear that in the trans-but-2-ene, the two methyl groups are in opposite directions. Therefore,
dipole moments of C—CH3 bonds cancel, thus making the trans form non-polar.
+ + +
  
H3C + –
CH3 H3C – –
H
   
C C C C +

H H H CH3
cis-But-2-ene trans-But-2-ene
( = 0.33 D) ( = 0)
In the case of solids, it is observed that the trans isomer has higher melting point than the cis form. This is
due to the fact that trans isomers pack more closely in the crystal due to their higher symmetry. This is the
reason trans but-2-ene has a higher melting point than cis but-2-ene.
Geometrical or cis-trans isomerism is also shown by alkenes of the type XYC = CXZ and XYC = CZW.

X Y X Y
C C
C C
X Z Z X
(a) (b)

Such type of compounds where different atoms or group of atoms are attached to the double-bonded carbon
atom like in the above case, it becomes difficult to tell whether it is a cis-isomer or trans-isomer. In such
case a different notation based on the atomic number of the atom directly attached to the carbon atom with
double bond is taken into consideration. Such type of notation is known as E–Z notation. It was given by
Cahn, Ingold and Prelog. Here E stands for Entgegen meaning opposite and Z stands for Zusammen meaning
together.
For example :
Higher priority Br CH3
than Cl due to
high atomic mass C C
Cl I Higher atomic mass
than carbon of CH3,
so higher priority

Since groups of higher priorities are on opposite side hence, it is (E)-1-Bromo-1-chloro-2-iodopropene.

Example 9 : Give the E–Z designation of the following compound.

HOOC OH
C C
H Cl
Solution : Higher priority HOOC OH
C C
H Cl Higher atomic mass
hence higher priority
Since, higher priority groups lie on opposite sides, it is E-Isomer.

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Preparation
1. From alkynes : Alkynes undergo partial reduction with calculated amount of dihydrogen producing
alkenes.

(i) Alkynes on partial reduction with calculated amount of dihydrogen in presence of palladised charcoal
partially deactivated with poisons like sulphur compounds or quinoline give alkenes. Partially
deactivated palladised charcoal is known as Lindlar’s catalyst. Alkenes produced in the reaction
is of cis geometry.

R R
Pd/C
R C C R + H2 (Lindlar’s catalyst)
C C
Alkyne H H
cis-alkene

CH3 CH2 CH3

Pd/C
CH3 C C CH2 CH3 + H2 (Lindlar’s catalyst)
C C
H H
cis-pent-2-ene

(ii) Alkynes on reduction with sodium and liquid ammonia yields trans alkenes.
R H
Na/liq. NH3
R C C R C C
H R
trans-alkene

CH3 H
Na/liq. NH3
CH3 C C CH2 CH3 C C
H CH2 CH3
trans-pent-2-ene

This reaction is also known as Birch reduction.

2. From alkyl halides : Alkyl halides (R–X) on heating with alcoholic potash (potassium hydroxide dissolved
in alcohol, say, ethanol) eliminates one molecule of halogen acid to form alkenes. This reaction is known
as dehydrohalogenation i.e., removal of halogen acid. This is an example of -elimination reaction,
since hydrogen atom is eliminated from the -carbon atom (carbon next to the carbon to which halogen
is attached).
H H H H
 alc. KOH
H C C H 
C C
H X H H
(X = Cl, Br, I)
Nature of halogen atom and the alkyl group determine the ease of dehydrohalogenation i.e., rate of
reaction. The observed rate is in the order :
Iodine > Bromine > Chlorine for different halogens
and for alkyl groups it is
Tertiary > Secondary > Primary

H H H H
  alc. KOH
H C C H 
H C C H + KCl + H2O
H Cl

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Note : According to Saytzeff rule, when two alkenes are formed, the alkene which is more substituted
is major product. This is because the alkene which is the more substituted has a greater stability in
comparison to the alkene which is less substituted. The stability is related to the hyperconjugating
structures, greater is the number of hyperconjugative structures (greater conjugation) greater is the
stability of alkene.
For example :

  
alc. KOH
CH3 CH 2 CH CH3 
CH3 CH CH CH3 + CH3 CH2 CH CH2
But-2-ene (Major) But-1-ene (Minor)
Br
As we have seen the preferred alkene is the one which is more alkylated i.e., which carries more
number of alkyl groups attached to the double-bonded carbon atoms.

3. From vicinal dihalides : Dihalides in which two halogen atoms are attached to two adjacent carbon
atoms are known as vicinal dihalides. Vicinal dihalides on treatment with zinc metal in an alcoholic
solution, lose a molecule of ZnX2 to form an alkene. This reaction is known as dehalogenation reaction.
A pure alkene is produced by this reaction.

H H
H H
alcohol, 
H C C H + Zn C C + ZnX2
H H
X X

alcohol, 
CH2 CH2 + Zn CH2 CH2 + ZnBr2

Br Br

alcohol, 
CH2 CH3 CH2 + Zn CH3 CH CH2 + ZnBr2

Br Br

4. From alcohols by acidic dehydration : Alcohols are the hydroxy derivatives of alkanes, represented
by R–OH where, R is CnH2n + 1. Alcohols on heating with concentrated sulphuric acid form alkenes with
the elimination of one water molecule since a water molecule is eliminated from the alcohol molecule
in the presence of an acid, this reaction is known as acidic dehydration of alcohols. This reaction
is also the example of -elimination reaction since –OH group takes out one hydrogen atom from the
-carbon atom.
H H
  conc. H2SO4
H C C H 
CH2 CH2 + H2O
Ethene
H OH
Ethanol
The ease of dehydration of alcohols takes place in the order of
t-alcohol > s-alcohol > p-alcohol
Example :
OH
OH > OH >
H
3° 2° 1°

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Mechanism of dehydration of alcohols :
Primary alcohol


  
 HSO4  H
H2SO4 (conc.) 

H H H H H H
+
H H
  H
H C C H H C C H H C C H + H2O + C C
 –H
(Take a new molecule H H
H OH H O+ H H of alcohol) Ethene

Ethyl alcohol H
Protonated alcohol

Water molecule is eliminated to form ethyl carbocation.

Hydrogen from the -position is eliminated.

H
|
H2SO4 (conc.)
CH3CH2CH2CH2 OH 
 CH2 — CH=C — CH3
But-2-ene

The product obtained in the above case is but-2-ene instead of butene. This is according to saytzeff’s rule,
a more stable alkene is formed where the chances of two alkene formation are there. The mechanism can
be explain as such :


  
 H  HSO4
H2SO4 


(i) CH3 CH2 CH2 CH2 O H + CH3 CH2 CH2 CH2 O H
H (from acid)
H
Protonated alcohol

H H H H
 –H2O 
(ii) CH3 CH2 C C O H CH3 CH2 C C

H H H H H
1° carbocation formed

(iii) Rearrangement takes place by the hydride shift to form a more stable 2° carbocation.

H H H H

CH3 CH2 C C CH3 CH2 C C H
+
H H H
1, 2 hydride shift 2° carbocation formed

H H H H H
+
–H
(iv) CH3 C C C H CH3 C C CH3

H H
Unstable carbocation loses a proton But-2-ene
from -position to form alkene

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Example 10 : What is the major product obtained when 2-bromobutane is heated with alcoholic KOH? Write only
the major product expected to be obtained.

alc. KOH
Solution : CH3 CH2 CH CH3 
CH3 CH CH CH3 + KBr + H2O
2-Butene
Br

Properties
Physical properties : Alkenes have the physical properties similar to alkanes, except in types of isomerism
and difference in polar nature. The first three members of alkenes are gases, the next fourteen are liquids
and the higher ones are solids. Ethene is a colourless gas with a faint sweet smell. All other alkenes are
colourless and odourless, insoluble in water but fairly soluble in non-polar solvents like benzene, petroleum
ether. They show a regular increase in boiling point with increase in size i.e., every —CH2 group added
increase the boiling point by 20–30 K. Like alkanes, straight chain alkenes have higher boiling point than
isomeric branched chain compounds.

Chemical properties : Alkenes are rich source of loosely held pi() electrons due to the presence of the

carbon-carbon double bond C C by which it can undergo a wide variety of chemical reactions. They

show addition reactions due to the presence of -bonds, in which the electrophiles (electron-deficient species)
attack the carbon-carbon double bond to form the addition products. Some reagents also undergo addition
reaction through the free radical substitution reactions. Oxidation and ozonolysis reactions are also quite
prominent in alkenes. A brief description of different reactions of alkenes is given below :

1. Addition of dihydrogen : Alkenes adds one mole of dihydrogen gas in presence of catalysts such as
Ni (finely divided) at 200–250°C, or finely divided Pt or Pd at room temperature to give an alkane.

Catalyst
C C + H2 C C

Alkene Alkane
In this reaction one -bond is broken and two new sigma bonds are formed.

2. Addition of halogens : Halogens like bromine or chlorine add up to alkene to form vicinal dihalides in
presence of CCl4 as solvent. The order of reactivity of halogens is F > Cl > Br > I. Iodine does not show
addition reaction under normal conditions. The reddish orange colour of bromine solution in carbon
tetrachloride is discharged when bromine adds up to an unsaturation site. This reaction is used as a
test for unsaturation (test for presence of carbon-carbon double bond). Addition of halogens to alkenes
is an example of electrophilic addition reaction.
CCl4
(i) CH2 CH2 + Br Br CH2 CH2
Ethene Br Br
1, 2-Dibromoethane
CCl4
(ii) CH3 CH CH2 + Cl Cl CH3 CH CH2
Propene Cl Cl
1, 2-Dichloropropane
3. Addition of hydrogen halides : Hydrogen halides (HCl, HBr, HI) add upto alkenes to form alkyl halides.
The order of reactivity of hydrogen halides is HI > HBr > HCl. Like addition of halogens to alkenes,
addition of hydrogen halides is an example of electrophilic addition reaction. Let us illustrate this by taking
of HBr to symmetrical and unsymmetrical alkenes.

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(i) Addition reaction of HBr to symmetrical alkenes : Addition reactions of HBr to symmetrical
alkenes (similar groups attached to double bond) take place by electrophilic addition mechanism.
CH2 = CH2 + H — Br  CH3 —CH2 —Br

CH3 – CH = CH – CH3 + H – Br  CH 3 CH2 CH CH3

Br

(ii) Addition reaction of HBr to unsymmetrical alkenes : When HBr adds to an alkene which is
unsymmetrical, two types of products as given follows :

(i) CH3 CH CH3

Br
2-Bromopropane
CH3CH CH3 + HBr (Major product)

(ii) CH3 CH2 CH2 Br

1-Bromopropane
(Minor product)

Markovnikov, Russian chemist, after investigating several addition reactions, made a generalisation in 1869,
for such kind of addition reactions. These generalisations led to frame a rule known as Markovinkov rule.
According to the rule, the negative part of the addendum (adding molecule) adds to that carbon atom
of the unsymmetrical alkene which is maximum substituted or which possesses lesser number of
hydrogen atoms. So according to this rule, propene undergoes addition reaction forming 2-bromopropane in
majority. Let us try to understand in terms of mechanism of the reaction.

Mechanism
It is an electrophilic addition which is favoured by the formation of more stable carbonium ion intermediate.
(i) HBr undergoes ionisation in presence of propene molecule to give an electrophile (H) and a nucleophile
(Br – ).

H Br H + Br –
Electrophile Nucleophile

(ii) H attacks the double bond to produce a more stable secondary carbonium ion (carbocation).
H3C CH CH2 + H+

 
H3C CH2 CH2 H3C CH CH3
(a) 1° carbocation (b) 2° carbocation
(less stable) (more stable)
The secondary carbocation (b) is more stable than the primary carbocation (a), therefore, the former
predominates because it is formed at a faster rate.
(iii) In the following step the nucleophile (Br – ), attacks the secondary carbocation to form isopropyl bromide
(2-Bromopropane).
– +
Br + CH3 CH CH3 CH3 CH CH3

Br
2-Bromopropane
(Major product)

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Example 11 : What will be the major product obtained when isobutene undergoes reaction with HBr?

CH 3 C CH2 + HBr ?

CH3

Solution : CH3

CH3 C CH2 + HBr

+
H

CH3 CH3

(a) CH3 C CH3 (b) CH3 CH CH2

3° carbocation 1° carbocation
Since, tertiary (3°) carbocation is far more stable than the primary (1°) carbocation, the major
product obtained will be in the case (a).

Br

CH3 C CH2 + HBr CH3 C CH3

CH3 CH3

Anti Markovnikov addition or Peroxide effect or Kharash effect

In the presence of peroxide, addition of HBr to unsymmetrical alkenes like propene takes place contrary to
the Markovnikov rule. This happens only with HBr but not with HCl or HI. This reaction is known as peroxide
or Kharash effect or addition reaction anti to Markovnikov rule.

(C6H5CO)2O2
CH3 CH CH2 + HBr CH3 CH2 CH2 Br
1-Bromopropane

Mechanism : Peroxide effect proceeds via free radical chain mechanism.

(i) The peroxide breaks homolytically to yield free radicals

O O O
Homolysis
C6H5 C O O C C6H5 
2C6H5 C O
Benzoyl peroxide Benzoyl peroxide
free radical

O O
C6H5 C O + H — Br C6H5 — C — OH + Br
Homolysis
(ii) C6H5 + H Br C6H6 + Br
CH3 CH CH2 (a) Primary free radical
(Less stable)
Br
(iii) CH3 CH CH2 + Br

CH3 CH CH2 Br (b) Secondary free radical

(More stable)

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Primary free radical (a) and secondary free radical (b) are obtained when bromine free radical reacts with
propene. Since secondary free radical is more stable than the primary free radical, former will proceed
further to form the product.

Homolysis
(iv) CH3 CH CH2Br + H Br CH3 CH 2 CH2 Br + Br
(Major product)

Homolysis
(v) CH3 CH CH2 + H Br CH3 CH CH3 + Br
Br Br
(Minor product)

The peroxide effect is not noted in the case of HCl or HI. This is due to the fact that HCl bond is too
strong to be broken to produce free radicals (430.5 kJ mol–1) than the HBr bond (363.7 kJ mol–1). HI
has a lower bond energy (296.8 kJ mol–1) and in case of the addition of HI although H–I bond is easily
broken by homolysis, but the iodine atoms so produced immediately react to form iodine molecules rather
than attacking the double bond of carbon atom in alkene.
 
CH3  HI  CH4  I

 
I  I  I2

4. Addition of sulphuric acid : Cold conc. H2SO4 adds to alkenes in accordance to Markovnikov rule to
form alkyl hydrogen sulphate by the electrophilic addition reaction.

CnH2n  H2SO4  CnH2n  1HSO4

(Cold conc.) Alkyl hydrogen
sulphate

O O
CH2 CH2 + H O S O H CH3 CH2 O S O H or C2H5HSO4
O O Ethyl hydrogen
sulphate

O
CH3 CH CH2 + H O S O H CH3 CH CH3 or CH3 CH CH3
O O OSO2OH
O S O Propylhydrogen sulphate
O
H
5. Addition of water : Alkenes in the presence of few drops of conc. H2SO4 reacts with water to form
alcohols. In this reaction also, Markovnikov rule is followed

OH

H
CH3 C CH2 + H2O CH3 C CH3

CH3 CH3
2-Methylpropene 2-Methylpropane-2-ol

Mechanism :

H + HSO4
–
H 2SO 4 (conc.)

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(i) The first step involves the formation of a more stable carbonium ion by the addition of a proton
across the double bond.

CH3 C CH2 + H CH3 C CH3

CH3 CH3
(Carbonium ion)
(ii) The second step involves addition of water and loss of H+ (proton) to give alkanol.
CH3 CH3
CH3 C CH3 + H O H CH3 C CH3

+
O H

H
CH3 CH3

–H
CH3 C CH3 CH3 C CH3

O H O H

H
Loss of proton 2-Methylpropane-2-ol
6. Addition of S2Cl2 : Ethylene reacts with sulphur mono chloride (S2Cl2) to give mustard gas (a war gas)
CH2 CH2 – Cl
+ S2Cl2
CH2 CH2 – S – CH2 – CH2 – Cl
(Mustard gas)
7. Oxidation :
(a) Alkenes on reaction with cold, dilute, aqueous solution of potassium permanganate also known as
Baeyer’s reagent, produces vicinal glycols. Decolorisation of KMnO4 solution is used as a test for
unsaturation.
H H
dil. KMnO4
CH2 CH2 + H2O [O], 273 K
H C C H
OH OH
Ethane-1, 2-diol
(Glycol)
dil. KMnO4
CH3 CH CH2 + H2O 273 K, [O]
CH3 CH CH2 OH
OH
Propane-1, 2-diol
(b) Acid potassium permanganate or acidic potassium dichromate oxidises alkenes to Ketones and/
or acids.
+
KMnO4/H
CH3 C CH2 CH3 C O + CO2 + H2O

CH3 CH3
2-Methylpropene Propane-2-one
The terminal group is converted to HCOOH which is oxidized to CO2.
+
KMnO4/H
CH3 HC CH2 CH3 2CH3COOH

But-2-ene Ethanoic acid

Note : 1% cold and alkaline solution of KMnO4 is known as Baeyer’s reagent.

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Example 12 : What are the product obtained when but-1-ene undergoes addition reaction of HBr in different
conditions?
(i) In absence of peroxide
(ii) In presence of peroxide
(i)
Solution : CH3 CH CH2 CH3 CH2 CH CH2 CH3
No peroxide

Br
2-Bromobutane
(ii)
Br CH2 CH2 CH2 CH3 Peroxide
1-Bromobutane

8. Ozonolysis : Ozonolysis of alkenes involves the addition of ozone molecule to alkene to form ozonide.
This happens when ozone is passed through a solution of alkene dissolved in an inert solvent (CCl4 or
CHCl3 or ether), addition takes place across the double bond to form an ozonide.

Ozonide on warming with zinc and water (Zn–H2O) gets cleaved into smaller molecules. This reaction
is highly useful in detecting the position of the double bond in alkenes or other unsaturated compounds.

O
Zn + H2O
CH3 CH CH2 + O3 CH3 CH CH2 CH3CHO + HCHO
Ethanal Methanal
O O

Propene ozonide

H3C O
O O Zn + H2O
C CH2 + O3 H3C H3C C CH3 + HCHO
H3C C CH2 Propan-2-one Methanal
H3C O
Ozonide

9. Polymerisation : Polymerisation is the process where monomers combines together to form polymers.
For example, polythene is obtained by the combination of large number of ethene molecules at high
temperature, high pressure and in presence of a catalyst. The large molecules thus obtained are called
polymers. Other alkenes also undergo polymerisation.
High temperature/pressure
n(CH2 CH2) Catalyst
[ CH2 CH2 ]n
Polythene

High temperature/pressure
n(CH3 CH CH2) Catalyst
[ CH CH2 ]n

CH3
Polypropene

Polymers are used for the manufacture of plastic bags, squeeze bottles, refrigerator dishes, toys, pipes,
radio and TV cabinets etc. Polypropene is used for the manufacture of milk crates, plastic buckets and
other moulded articles. Though these materials have now become common, excessive use of polythene
and polypropylene is a matter of great concern for all of us.

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EXERCISE
21. In which of the following geometrical isomerism is possible?
(1) CH3CH = C(CH3)2 (2) C6H5N = NC6H5
(3) CH3CH = CH2 (4) All of these
22. Identify the product in the following reaction
alc. KOH
CH3 – CH2 – CH2Br 
Product
(major )

(1) CH3CH2OH (2) CH2 = CH2

(3) CH3CH2 – CH2 – OH (4) CH3CH = CH2

CH2Br
alc. KOH
23. Products .


Which of the following is not a possible product in the above reaction?

CH2 CH3
(1) (2)

CH3
(3) (4)

24. Consider the given reaction

CH3
conc. H2SO4
CH3 – C – CH2 – OH 
Alkene (major)
CH3

Identify alkene
(1) CH3 – CH2 – CH = CH2 (2) CH3 – CH = CH – CH3

(3) (CH3)2C = C(CH3)2 (4) CH 3 – CH = C – CH 3

CH3
CH3

40°C
25. + HBr (1 eq.) X (major) .
(1, 4-addition)

Identify X

CH3
CH3
Br
(1) (2)
Br
CH3 CH3
Br
Br
(3) (4)

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26. In which of the following reaction, the major product alkene is formed by E1cb mechanism?

alc. KOH
(1) CH3 – CH2 – CH2 – CH2 – Br 


alc. KOH
(2) CH3 – CH2 – CH – CH3 

Br

alc. KOH
(3) CH3 – CH2 – CH – CH3 

alc. KOH
(4) CH3F 


27. The alkene which is the most reactive towards catalytic hydrogenation is

(1) CH2 = CH2 (2)

(3) CH3 – CH = CH – CH3 (4) (CH3)2C = C(CH3)2

28. In which of the following alkenes, Markownikoff’s rule is not applicable?

CH3
(1) (2)

CH3
(3) (4) CH3 – CH = CH2

Benzoyl peroxide
29. CH3 – CH  CH2  HBr h
 Products

Which of the following is a possible product in the above reaction?

(1) CH3 – CH2 – CH2 – Br

(2) CH3 – CH – CH2 – Br

Br

(3) CH3 – CH —– CH – CH3

CH2Br CH2Br

(4) All of these

30. In E1, reaction, the intermediate formed is
(1) Carbanion
(2) Carbocation
(3) Carbon free radical
(4) Carbene

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ALKYNES
Like alkenes, alkynes are also unsaturated hydrocarbons with general formula CnH2n – 2. They contain at least
one triple bond between two carbon atoms. These have four H-atoms less compared to alkanes.
The first stable member of alkyne series is ethyne commonly known as acetylenes. Acetylene is used for
arc welding purposes in the form of oxyacetylene flame obtained by mixing acetylene with oxygen gas. Alkynes
are starting materials for a large number of organic compounds. Hence, it is interesting to study this class
of organic compounds.

Nomenclature and Isomerism

In common system, alkynes are named as derivatives of acetylene. In IUPAC system, they are named as
derivatives of the corresponding alkanes replacing ‘ane’ by the suffix ‘yne’. The position of the triple bond is
indicated by the first triply bonded carbon. Common and IUPAC names of a few members of alkyne series
are given in the table below :

Value of n Formula Structure Common name IUPAC name

2 C2H2 H–CCH Acetylene Ethyne
3 C3H4 CH3–CCH Methylacetylene Propyne
4 C4H6 CH3CH2–CCH Ethylacetylene But-1-yne
4 C4H6 CH3–CC–CH3 Dimethylacetylene But-2-yne

Alkynes exhibit isomerism. For example, ethyne and propyne have got only one structure but there are two
possible structures for butyne (i) but-1-yne and (ii) but-2-yne. Since these two compounds differ in their
structures due to the position of triple bond, they are known as position isomers. Similarly when the length
of alkyne chain differs keeping the formula same it gives rise to chain isomerism. Let us try to construct
the structure for the next homologue i.e., the next alkyne with molecular formula C5H8 and try to arrange five
carbon atoms with a continuous chain and with a side chain.
Structure IUPAC name
I. HC C CH2 CH2 CH3 Pent-1-yne

II. H3C C C CH2 CH3 Pent-2-yne

III. H3C CH C CH 3-Methylbut-1-yne

CH3
Structures I and II are position isomers and structures I and III or II and III are chain isomers.

Example 13 : What is the kind of isomerism exhibited by the compounds given below?

CH3 C C CH2 CH2 CH3 CH3 CH2 C C CH2 CH 3

(I) (II)
Solution : Position isomerism.

Structure of Triple Bond

Ethyne is the simplest molecule of alkyne series where each carbon atom of ethyne has two sp hybridised
orbitals. Carbon-carbon sigma () bond is obtained by the head-on overlapping of the two sp hybridised orbitals
of two carbon atoms. The remaining sp hybridised orbital of each carbon atom undergoes overlapping along
the internuclear axis with the 1s orbital of each of the two hydrogen atoms forming two C–H sigma bonds.
H–C–C bond angle is of 180°. Each carbon has two unhybridised p orbitals which are perpendicular to each
other as well as to the plane of the C–C sigma bond. The 2p orbitals of one carbon atom are parallel to the

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2p orbitals of the other carbon atom, which undergoes lateral or sideways overlapping to form two pi () bonds
between two carbon atoms. Thus, ethyne molecule consists of one C–C  bond, two C–H  bonds and two
C–C  bonds.

In the plane Below the plane

of paper of the paper

2px 2px

1s  1s 
H C C H 2py

 sp sp 
2py 
(a)

In the plane
Above the plane of the paper
of the paper
(b)
The strength of CC bond (bond enthalpy 823 kJ mol–1)
is more than those of C=C bond (bond enthalpy
681 kJ mol–1) and C–C bond (bond enthalpy 348 kJ mol–1). The CC bond length is shorter (120 pm) than
those of C=C bond (134 pm) and C–C (154 pm). Electron cloud between two carbon atoms is cylindrically
symmetrical about the internuclear axis. Thus, ethyne is a linear molecule.

H C C H H  bond C  bond C  bond H

Overlap of p orbitals Cylinder of electron density

154 pm 134 pm 120 pm

H H H H
H
C C C C H C C H
H H
H H
H 106 pm
109 pm 108 pm

Ethane Ethene Ethyne

Preparation
1. From calcium carbide : On industrial scale, ethyne is prepared by reacting calcium carbide with water.
Calcium carbide is prepared by heating quick lime with coke. Quick lime can be obtained by heating
limestone as shown in the following reactions :

CaCO3   CaO + CO2
CaO  3C  CaC2  CO
Calcium
carbide

CaC2 + 2H2O  Ca(OH)2 + C2H2

2. From vicinal dihalides : Vicinal dihalides on treatment with alcoholic potassium hydroxide undergo
dehydrohalogenation. One molecule of hydrogen halide is eliminated to form alkenyl halide which on
treatment with sodamide gives alkyne.
H H
H H + –
Alcohol Na NH2
H C C H + KOH –KBr
C C –NaBr
H C C H
–H2O H Br –NH3
Br Br
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Properties
Physical properties
Physical properties of alkynes are similar to those of alkanes and alkenes. The first three members (acetylene,
propyne and butynes) are gases, the next eight are liquids and higher ones are solids. All alkynes are
colourless. All alkynes except ethyne which have an offensive characteristic odour, are odourless. Alkynes
are weakly polar in nature and nearly insoluble in water. They are quite soluble in organic solvents like ethers,
carbon tetrachloride and benzene. Their melting point, boiling point and density increase with increase in molar
mass.
Chemical properties
Alkynes show acidic nature, addition reactions and polymerisation reactions as follows :
(i) Acidic character of alkyne : Sodium metal and sodamide (NaNH2) are strong bases. They react with
ethyne to form sodium acetylide with the liberation of dihydrogen gas. These reactions are not observed
in the case of ethene and ethane thus indicating that ethyne is acidic in nature in comparison to ethane
or ethene. The structure and hybridization has a role to play in this acidic character. In ethyne hydrogen
is attached to the sp hybridized carbon atoms whereas they are attached to sp2 hybridised carbon atoms
in ethene and sp3 hybridised carbon atoms in ethane. Due to sp hybridised carbon atom, it has the
maximum s-character (50%), they have highest electronegativity, hence, these attract the shared pair
of electrons of C–H bond of ethyne to a greater extent than that of the sp2 hybridised orbitals of carbon
in ethene and sp3 hybridised orbital of carbon in ethane. Thus, in ethyne, hydrogen atoms can be
liberated as protons more easily as compared to ethene and ethane. Hence, hydrogen atoms of ethyne
attached to triple bonded carbon atom are acidic nature. It should be noted that hydrogen atoms attached
to the triply bonded carbons are acidic but not all the hydrogen atoms of alkynes.

1
H C C H + Na H C C–Na+ + H
2 2
Monosodium
ethynide

CH3 C C H + Na+NH–2 CH3 C C–Na+ + NH3

Sodium
propynide

These reactions are not shown by alkenes and alkanes, hence used for distinction between alkynes,
alkenes and alkanes. Alkanes, alkenes and alkynes follow the following trend in their acidic behaviour:
(a) HCCH > H2C=CH2 > H3C–CH3
(b) HCCH > CH3–CCH >> CH3–CC–CH3
Increase in the alkyl group decreses the deficiency of electrons on the sp-hybridised carbon atoms, thus
decreasing the acidic character.
(ii) Addition reactions : Alkynes contain a triple bond, so they add up, two molecules of dihydrogen,
halogen, hydrogen halides etc.
Addition in unsymmetrical alkynes takes place according to Markovnikov rule. Majority of the reactions
of alkynes are the examples of electrophilic addition reactions. A few addition reactions are given below :
(a) Addition of dihydrogen :
Pt /Pd/Ni H
CnH2n  2  H2  CnH2n 
2
 CnH2n  2
Alkyne Alkene Alkane

Pt/Pd/Ni H
HC  CH  H2  H2 C  CH2 
2
 H3 C  CH3
Ethane
Pt/Pd/Ni H
CH3  C  CH  H2  CH3  CH  CH2 
2
 CH3  CH2  CH3
Propane

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(b) Addition of halogens :
X X X X
X2
C C + X2 C C C C
X X
Br Br Br Br
Br2
CH3 C CH + Br Br CH3 C C H CH3 C C H
1, 2-Dibromopropene
Br Br
1, 1, 2, 2-Tetrabromopropane

Reddish orange colour of the solution of bromine in carbon tetrachloride is decolourised. This is
used as a test for unsaturation.
(iii) Addition of hydrogen halides : Two molecules of hydrogen halides (HCl, HBr, HI) add to alkynes to
form gemdihalides (in which two halogens are attached to the same carbon atom). The initial product
is a vinyl halide when a hydrogen halide adds to a terminal alkyne, the product is governed by
Markovnikov’s rule. A second molecule of HX can add, usually with the same orientation as the first.
X H
HX
C C + HX C C H C C
X H
X H
H C C H+H Br [CH2 CH Br] CHBr2
Bromoethene
CH3
1, 1-Dibromoethane

Br
HBr
CH3 C CH + H Br [CH3 CH CH2] CH3 C CH3

Br Br
2-Bromopropene 2, 2-Dibromopropane

(iv) Addition of water :

Like alkanes and alkenes, alkynes are also immiscible and do not react with water. However, one
molecule of water adds to alkynes on warming with mercuric sulphate and dilute sulphuric acid at 333 K
to form carbonyl compounds.

H O
2+ +
Hg /H Isomerisation
HC CH + H OH 333 K
CH2 C OH CH3 C H
Ethyne Vinyl alcohol Ethanal

2+ +
Hg /H Isomerisation
CH3 C CH + H OH 333 K
CH3 C CH2 CH3 C CH3
Propyne
OH O
Propanone

All higher alkynes gives ketones under similar conditions.

(v) Polymerisation :
(a) Under suitable conditions, linear polymerisation of ethyne takes place to produce polyacetylene or
polyethyne which is a high molecular weight polyene containing repeating units of
( CH==CH—CH==CH ) and can be represented as ( CH CH CH CH )n . Under special
conditions, this polymer conducts electricity. This film of polyacetylene can be used as electrodes
in batteries. These thin films are good conductors, lighter and cheaper than the metal conductors.

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(b) Cyclic polymerisation : Ethyne on passing through red hot iron tube at 873 K undergoes cyclic
polymerization. Three molecules polymerise to form benzene, which is the starting molecule for the
preparation of derivatives of benzene, dyes, drugs and large number of organic compounds.

CH
CH
Red hot
CH CH iron tube CH CH
or
873 K
CH CH CH CH
CH
CH
Benzene

Cl
CuCl/NH4Cl HCl
(c) 2(CH  CH) or
CH2 = CH – C  CH CH2 = CH – C = CH2
Cu2Cl2/NH4Cl Vinyl acetylene Chloroprene

(d) Under high pressure and in presence Ni(CN)2, acetylene tetramerises,

Anhy. Ni(CN)2
 4 HC  CH
Acetylene THF
(solvent) Cyclooctatetraene
High pressure

Example 14 : What is the product obtained when two molecules of ethyne and one molecule of propyne
undergoes cyclic polymerisation, when the mixture is passed through red hot iron tube?

CH3 H

CH C C CH3
Red hot CH3
iron tube H C C
Solution : CH CH or
873 K
H C C H
CH CH C

H
Toluene

EXERCISE

31. How many structural isomers are possible for the molecular formula C4H8, which can undergo ozonolysis?

(1) 2 (2) 4

(3) 3 (4) 1
32. Which of the following reaction will give acetylene as a major product?

Ag alc.
(1) 2[CHI3 ] 

 (2) H3C – CH2 – Br 
KOH


Zn
(3) CH 2 – CH 2 
(4) All of these

Br Br

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33. Which of the following will give positive Tollen’s test?
(1) CH3 – CH3 (2) CH2 = CH2

(3) CH  CH (4)

34. Which of the following is an example of nucleophilic addition reaction?

(1) CH  CH  Cl2  CHCl  CHCl

HCl
(2) HC  CH 
CH COOH
 CH2  CHCl
3

O

H
(3) HC  CH + H2O 2 CH3 – C – H
Hg , 

NaNH2
(4) H – C  C – H   HC  C(–)Na

35. Which of the following compound will give only one type of carbonyl compound on reductive ozonolysis?

(1) (CH3)2C = C(CH3)2 (2)

CH3

(3) (4) All of these

H3C CH3
36. Ethylene reacts with S2Cl2 to give
(1) Mustard gas (2) Lewisite
(3) Thiophene (4) Ethanethiol
37. Actylene reacts with ammonical Cu2Cl2 to give precipitate of
(1) Red colour (2) Yellow colour
(3) White colour (4) Blue colour
38. Identify the product in the reaction

K 2Cr2O7  H2SO4
HC  CH  [O]
 Product

(1) CH3CHO (2) CH3CH2OH

(3) CH3COOH (4) CH3OH
39. In Wacker oxidation of ethene, the product formed is
(1) Ethanoic acid (2) Ethanal
(3) Ethanol (4) Ethanedial
40. Which of the following hydrocarbon on reductive ozonolysis, will give CO2 gas?
(1) CH2 = CH2 (2) CH2 = CH – CH = CH2
(3) CH3 – C = C = CH – CH3 (4) All of these
H

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AROMATIC HYDROCARBON
Aromatic hydrocarbons are also known as ‘arenes’. Since most of them possess pleasant odour (Greek; aroma
meaning pleasant smelling), the class of compounds are known as ‘aromatic compounds’. Most of the
compounds are found to have benzene ring. Benzene ring is highly unsaturated and in a majority of reactions
of aromatic compounds, the unsaturation of benzene ring is retained. However there are examples of aromatic
hydrocarbons which do not contain a benzene ring but contains highly unsaturated ring. Aromatic compounds
containing benzene ring are known as benzenoids and those, not containing a benzene ring are known as
non-benzenoids. Some examples of arenes are given below :
CH3 OH

Benzene Napthalene Anthracene Toluene Biphenyl Phenol

Since all these compounds contain benzene rings, these are known to be ‘benzenoids’.

Azulene
Though azulene is aromatic in nature but differs from benzene hence it is said to be non-benzoid aromatic
compound.

Nomenclature and Isomerism

Since all the six hydrogen atoms in benzene are equivalent; so it forms one and only one type of
monosubstituted product. When two hydrogen atoms in benzene are replaced by two similar or different
monovalent atoms or groups, three different position isomers are possible which differ in the position of
substituents. So we can say that disubstituted products of benzene show position isomerism. The three
isomers obtained are 1, 2 or 1, 6 which is known as the ortho (o-), the 1, 3 or 1, 5 as meta (m-) and 1, 4
as para (p-) disubstitued compounds.
H CH3
H C H H C H
C C C C
C C C C
H C H H C H
H H
Benzene (C6H6) with all the hydrogen atoms Methylbenzene (Toluene)
One hydrogen atom is substituted by the methyl group, forming a monosubstituted product.
The three position isomers of dimethylbenzene are
CH3 CH3 CH3
1 CH3 1 1
2 2 2
3
3 CH3 4
CH3
1, 2-Dimethyl benzene 1, 3-Dimethylbenzene 1, 4-Dimethyl benzene
or o-Dimethylbenzene or o-Xylene or m-Dimethylbenzene or m-Xylene or p-Dimethylbenzene or p-Xylene

Structure of Benzene
Benzene was isolated by Michael Faraday in 1825. The molecular formula of benzene, C6H6, indicates a high
degree of unsaturation as it has eight hydrogen atoms less then the normal alkane C6H14. This molecular
formula did not account for its relationship to corresponding alkanes, alkenes or alkynes. It was an uphill task
to come to its structure. The formation of a triozonide by adding three atoms of oxygen, on ozonolysis
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indicated the presence of three double bonds. The three double bonds presence was also indicated by the
addition of six hydrogen atoms, when benzene was subjected to reduction by dihydrogen forming cyclohexane.

Ni
C6H6  3H2 
200  250C
C6H12
Cyclohexane

Benzene was further found to produce one and only one monosubstituted derivative which indicated that all
the six carbon and six hydrogen atoms of benzene are identical. On the basis of this observation August
Kekule in 1865 proposed the following structure for benzene having cyclic arrangement of six carbon atoms
with alternate single bonds and double bonds where one hydrogen atom is attached to each carbon atom.
H
H H
or
H H
H
The Kekule structure indicates the possibility of two isomeric 1, 2-dibromobenzenes. In one of the isomers,
the bond between the substituted carbon atoms is a single bond while in the other it is a double bond.
Br Br
Br Br

However, benzene was found to form only one ortho disubstituted product. This problem was overcome by
suggesting the concept of oscillating nature of double bonds in benzene as given below :

Even with the modified theory, Kekule failed to explain the unusual stability of benzene and its preference
to the substitution reaction than addition reactions, which could later be explained by resonance.

Example 15 : What are the marked positions known in the disubstituted benzene compounds?
CH3
5 1

4 2
3

Solution : 1 = ortho position; 2 = meta position; 3 = para position

4 = meta position and 5 = ortho position

Resonance and Stability of Benzene

Two modern hypothesis had been given to explain the stability and structure of benzene. These two theories
are :
(i) Resonance or valence-bond theory : According to valence-bond theory, the oscillating nature of double
bonds in benzene (as proposed by Kekule) is now proposed by resonance. Even though the double bonds
keep on changing their positions. The structures produced is such that the position of nucleus remains
the same in each of the structure. The structural formula of such a compound is somewhat intermediate
(hybrid) between the various propose formulae. This state is known as Resonance.

or

(A) (B) (C)

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The structures proposed by Kekule are the two main contributing structures of benzene. The hybrid structure
is represented by inserting a circle or a dotted line in the hexagon as shown in figure (C). The circle
represents the six electrons which are delocalized between the six carbon atoms of the benzene ring.
X-ray diffraction data reveals that benzene is a planar molecule. Had any one of the above structures
of benzene (A and B) been correct. Two types of C– C bond lengths were expected. However, X-ray data
indicates that all the six C – C bond lengths are of the same order (139 pm) which is intermediate
between C – C single bond (154 pm) and C –C double bond (133 pm). The C – C bond length in benzene
(139 pm) indicates that each bond in benzene had acquired the characteristics differently due to
resonance. The single bond acquires the characteristic of double bond and the double bond acquires
the characteristic of single bond, thus changing the bond lengths. This is why the absence of pure double
bond in benzene accounts for the reluctance of benzene to show addition reactions under normal
conditions, thus explaining the unusual behaviour of benzene.
(ii) Molecular orbital theory to explain benzene structure : The orbital overlapping gives us better picture
about the structure of benzene. All the six carbon atoms in benzene are sp 2 hybridized. Two sp2 hybrid
orbitals of adjacent carbon atoms overlap with sp 2 hybrid orbitals of each carbon atom which also
overlaps with s-orbital of a hydrogen atom to form six C – H sigma bonds. Each carbon atom is now left
with one unhybridised p-orbital perpendicular to the plane of the ring as shown below :



 
 

 

 

The unhybridised p-orbital of carbon atoms are close enough to form a -bond by lateral overlap. There
are two equal possibilities of forming three -bonds by overlap of p-orbitals of C1–C2, C3–C4, C5–C6 or
C2–C3, C4–C5, C6–C1 respectively as shown in following figures.

1 2 1 2

6 3 6 3
or or

4
5 5 4

(a) (b)
Structures shown in the above figures correspond to two Kekule’s structures with localised -bonds. The
internuclear distance determined between all the carbon atoms in the ring by the X-ray diffraction is found
to be the same showing; there is equal probability for the p-orbital of each carbon atom to overlap with
the p-orbitals of adjacent carbon atoms (c). This can be represented in the form of two doughnuts (rings)
of electron clouds (d), one above the plane and one below the plane of the hexagonal ring as shown below:

(c)

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The six -electrons are thus delocalised and are free to move about the six carbon nuclei, instead of
any two as shown in (a) and (b). This spreading of -electrons in the form of ring of -electrons above
and below the plane of carbon atoms is called delocalisation of -electrons.
H H

C C

H C C H or

C C
(Electron cloud)
H H
The delocalized -electron cloud is attracted more strongly by the nuclei of the carbon atoms than the
electron cloud localised between two carbon atoms. Therefore, presence of delocalised -electrons in
benzene makes it more stable than the hypothetical cylohexatriene.

Example 16 : Why benzene is reluctant to show addition reaction?

Solution : Due to resonance benzene loses its pure double bond character rendering it partial single bond
character to double bond and double bond character to single bond. This causes for the reluctance
of -electrons to participate in addition reactions. Moreover addition reaction destroys the aromaticity
of the benzene molecule.

Aromaticity
Benzene was considered as parent ‘aromatic’ compound. Now, the name is applied to all the ring systems
whether or not having benzene ring, possessing the following characteristics.
(i) It should be planar.
(ii) It should have complete delocalisation of -electrons in the ring.
(iii) It should follow Huckel Rule i.e., it should have (4n + 2) electrons in the ring. Where n is an whole
number (n = 0, 1, 2, 3, ...).
Some examples of aromatic species are given below :

– +

Benzene Cyclopentadienyl Cycloheptatrienyl cation

anion

(n = 1, 6 electrons)

Napthalene Anthracene Phenanthrene

(n = 2, 10 electrons) (n = 3, 14 electrons)
Preparation of Benzene
Benzene is commercially isolated from coal tar. However, it may be prepared in the laboratory by the following
methods.
(i) Cyclic polymerisation of ethyne : Ethyne on passing through red hot iron tube at 873 K undergoes
cyclic polymerization.
CH
Red hot
CH CH iron tube
or
873 K
CH CH
CH

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(ii) Decarboxylation of aromatic acids : Sodium salt of benzoic acid i.e., sodium benzoate on heating
with sodalime gives benzene.

COONa
CaO
+ NaOH 
+ Na2CO3
Sodium Benzene
benzoate

This reaction is called decarboxylation.

(iii) Reduction of phenol : Phenol is reduced to benzene by passing its vapour over heated zinc dust.

OH


+ Zn + ZnO
(Dust)
Phenol

Example 17 : What is the product obtained when propyne is passed through hot iron tube at 873 K?
Solution :
CH3

C CH3
CH CH Fe tube, 873 K

C C CH3
CH3
CH3 CH3
CH Mesitylene

Physical Properties
Aromatic hydrocarbons are non-polar molecules and are usually colourless liquids or solids with a characteristic
aroma. You may be aware of the napthalene balls used in toilets and for preservation of clothes because of
unique smell of the compound and the moth repellent property, is an aromatic compound. Aromatic
compounds are insoluble in water but soluble in organic solvents such as alcohol and ether. They burn with
sooty flame.

Chemical Properties
Arenes are characterised by electrophilic substitution reactions. However, under special conditions they can
also undergo addition and oxidation reactions.
Electrophilic substitution reactions
The common electrophilic substitution reaction of arenes are nitration, halogenation, sulphonation, Friedel
Craft’s alkylation and acylation reactions in which attacking reagent is an electrophile (E+). The arenes
undergoes easy substitution reactions because of being electron-rich system due to presence of delocalized
-electrons. The reaction can be represented as :
H E
H H H H
+ –
+ E Nu + H Nu
H H H H
H H
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Mechanism of Electrophilic Substitution Reactions
According to experimental evidences, SE (S = Substitution; E = Electrophilic) reactions are supposed to
proceed via the following three steps :
(a) Generation of electrophile
E — Nu  E+ + Nu–
(b) Formation of arenium ion by the attack of electrophile : The attack of electrophile results in the formation
of -complex or arenium ion in which one of the carbon is sp3 hybridized.

sp3 hybridized carbon

H
+ E E
H
Sigma complex
(Arenium ion)

The arenium ion gets stabilised by resonance.

H H  H H
E E E  E

H H H H
Resonance forms of carbocation (arenium ion) Resonance
hybrid
Sigma complex or arenium ion loses its aromatic character because delocalisation of electrons stops
at sp 3 hybridised carbon.
(c) Removal of proton from the carbocation intermediate to give substitution product. To restore the aromatic
character, -complex releases proton from sp 3 hybridised carbon on attack by the nucleophile.

H E
E + Nu– + H – Nu
 H H

(i) Nitration : A nitro group is introduced into the benzene ring when benzene is heated with a mixture of
concentrated nitric acid and concentrated sulphuric acid
NO2

323–333 K
+ conc. HNO3 + conc. H2SO4 + H2O


Step (a) : Generation of electrophile : In case of nitration, the electrophile, nitronium ion, NO2 is
produced by transfer of a proton (from sulphuric acid) to nitric acid in the following manner :
O H
H O S O H + H O NO2 H O NO2 + HSO–4
O
H

H O NO2 H O H + NO2
Protonated Nitronium ion
nitric acid (Electrophile)

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It is interesting to note that in the process of generation of nitronium ion, sulphuric acid serves as an
acid and nitric acid acts as a base. Thus, it is a simple acid-base equilibrium.
Step (b) : Formation of carbocation : Nitronium ion attacks the benzene ring to give an intermediate
carbocation which is stabilized by resonance.

sp3 hybridized carbon

H

+ NO2  NO2
H
Sigma complex
(Carbocation)

H H  H H
 NO2 NO2 NO2  NO2

H H H H
Resonance forms of carbocation Resonance
hybrid

Step (c) : Removal of proton : Removal of proton from the carbocation gives the substitution product
nitrobenzene.

H NO2
 NO2 + HSO–4 + H2SO4
H
Nitrobenzene

(ii) Halogenation : Arenes undergo halogenation when it is treated with halogens in presence of Lewis
catalyst such as anhy. FeCl3, FeBr3 or AlCl3 to yield haloarenes.
Cl

anhyd. AlCl3
+ Cl2 + HCl

Chlorobenzene
Mechanism

Step (a) : Generation of electrophile

Cl – Cl + AlCl3  Cl+ + (AlCl4)–

Step (b) : Attack of electrophile to generate carbocation.

H

+ Cl  Cl
H

H H  H H
 Cl Cl Cl  Cl

H H H H
Carbocation stabilised by resonance Resonance
hybrid

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Step (c) : Removal of proton

H Cl
–
 Cl + [AlCl4] + HCl + AlCl3
H
Chlorobenzene
(iii) Sulphonation : The replacement of a hydrogen atom by a sulphonic acid group in a ring is called
sulphonation. It is carried out by heating benzene with fuming sulphuric acid or oleum (conc. H2SO4 + SO3)
SO3H

+ H2SO4(SO3) + H2O
Fuming Benzene
sulphuric acid sulphonic acid

Mechanism

Step (a) : Generation of electrophile

2H2SO4 H3O+ + SO3 + HSO4–

Here, SO3 acts as a neutral electrophile.

Step (b) : Attack of electrophile on the benzene ring to form a carbocation.

H
+ SO3  SO–3
H
Carbocation

or
– O–
O H
+ S
+ S  O O
O O H

H H  H H
SO–3 SO–3
 SO
–
3
 SO–3

H H H H
Resonance stabilised carbocation Resonance
hybrid

Step (c) : Loss of proton from the carbocation.

H SO–3
–
 SO3 + HSO–4 + H2SO 4
H
Step (d) : In the last steps an addition of proton takes place to benzene sulphonic acid anion to give
the final substitution product.

SO–3 +
SO3H
H O H
+ + H2O
H Benzene
sulphonic acid

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(iv) Friedel-Crafts alkylation reaction : When benzene is treated with an alkyl halide in the presence of
anhydrous aluminium chloride, alkylbenzene is formed.

CH3
C2H5
Anhyd. AlCl3 Anhyd. AlCl3
Ex. 1 : + CH3Cl + HCl ; + C2H5Cl + HCl
Toluene

Mechanism
Step (a) : Generation of electrophile

CH3Cl  AlCl3  CH3  [AlCl4 ]

Chloromethane Electrophile

Step (b) : Attack of electrophile on benzene ring and formation of stable carbocation.

H
+
+ CH3  CH3
H
Benzene Carbocation

H H  H H
 CH3 CH3 CH3  CH3

H H H H
Resonance stabilised carbocation Resonance
hybrid
Step (c) : Loss of proton from the carbocation

H CH3
–
 CH3 + [AlCl4] + HCl + AlCl3
H Toluene

Ex. 1 : When 1-chloropropane is put to Friedel Crafts alkylation reaction, isopropyl benzene instead
of n-propyl benzene is obtained.
CH3
Anhyd. AlCl3 CH
+ CH3CH2CH2Cl + HCl + AlCl3
CH3
Benzene 1-chloropropane Isopropyl
benzene

Mechanism :
Step (a) : Generation of electrophile
CH3 CH2 CH2 Cl + AlCl3 CH 3 CH2 CH+2 + [AlCl4] –
1° Carbocation

Primary carbocation being less stable undergoes rearrangement forming secondary carbocation which
is more stable.
H
+ Rearrangement 
CH3 C CH2 CH3 CH CH3

H
1,2 hybride shift 2° Carbocation
1° Carbocation

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NEET Hydrocarbons 113
Step (b) : Attack of electrophile on the benzene nucleus forming an arenium ion.

CH3
H
 CH3
+ CH3 CH CH3 
CH
H 3
Benzene Arenium ion

CH3 CH3 CH3

H H  H
CH CH CH H CH3
  CH

CH
H 3
CH
H 3
CH
H 3 H CH3
Resonance stabilised arenium ion Resonance hybrid

Step (c) : Loss of proton from the arenium ion

CH3
CH3
H
CH
CH

+ [AlCl4]– CH3 + HCl + AlCl3
CH3
H Isopropyl benzene

Addition Reaction: Under vigorous condition, i.e., at high temperature and/or pressure in presence of nickel
catalyst, hydrogenation of benzene gives cyclohexane.

Ni
+ 3H2 
Cyclohexane

Under ultraviolet light, three chlorine molecules add to benzene to produce benzene hexachloride, C6H6Cl6
which is also called gammaxane.
Cl
Cl Cl
UV
+ 3Cl2 500 K
Cl Cl
Cl
Benzene hexachloride (BHC)

Combustion: When heated in air, benzene burns with sooty flame producing CO2 and H2O.
15
C6H6  O2  6CO2  3H2O
2
General combustion reaction for any hydrocarbon may be given by the following chemical equation.
 y y
CxHy   x   O2  xCO2  H2O
 4 2

Dark
Example 18 : Complete the following reaction, + 3Cl2 ?

Solution : No reaction will take place. Since reaction takes place in presence of Lewis acid.

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CARCINOGENICITY AND TOXICITY

Benzene and polynuclear hydrocarbons containing more than two benzene rings fused together are toxic and
said to possess cancer-producing (carcinogenic) property. Such polynuclear hydrocarbons are formed on
incomplete combustion of organic materials like tobacco, coal and petroleum. They enter into human body
and undergo various biochemical reactions and finally damage DNA and cause cancer. Some of the
carcinogenic hydrocarbons are given below.

CH3

CH3
1,2-Benzanthracene 1,2-Benzpyrene 1,2,5,6-Dibenzanthracene 9,10-Dimethyl-1,2-benzanthracene

Note :
1. Rate of Hydrogenation of alkene follows the order :

H H H 3C H H 3C CH3 H3C H H3C H

C C > C C > C C > C C > C C >
H H H H H H H3C H H CH3

H3C CH3 H3C CH3

C C > C C
H CH3 H3C CH3

1
2. Rate of hydrogenation of alkene  Stability of alkene .

3. CH4 cannot be prepared by Wurtz reaction. Alkane with odd number carbon is difficult to prepare.
4. Eclispsed form of ethane is the least stable form due to torsional strain and van der Waals repulsion
between hydrogen atoms.
5. Skew form of ethane-1, 2-diol is the most stable conformation of it due to intramolecular H-bond.
6. Most stable conformation of cyclohexane is chair form.
7. Saytzeff rule: The preferred alkene is the one which is more alkylated during dehydrohalogenation.
alc. KOH
CH3 — CH2 — CH— CH3 
CH3 — CH  CH — CH3  CH3 — CH2 — CH  CH2
| (Major ) (Minor )
Br
8. In the presence of organic peroxide, addition of HBr to unsymmetrical alkenes takes place contrary to
the markovnikov rule. This is also known as Kharasch or peroxide effect.
9. Following are the aromatic systems :

, , , ,
( ) O N S
H
10. Following are the anti-aromatic systems :
, ,
( )

11. Cyclooctateraene is a conjugated system but it is non-aromatic.

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EXERCISE
41. Carbon-carbon bond length in benzene is
(1) 1.39 Å (2) 1.09 Å
(3) 1.54 Å (4) 1.34 Å
42. The resonance energy of benzene is
(1) 209 kJ/mol (2) 360 kJ/mol
(3) 151 kJ/mol (4) 109 kJ/mol

Sun light
43. The product formed in the reaction, + 3Cl2 Product , is

Cl Cl
Cl Cl Cl Cl
(1) (2)
Cl Cl Cl Cl
Cl Cl
Cl Cl
Cl
(3) (4)
Cl Cl Cl
44. Which of the following is the most reactive towards electrophilic substitution reaction?
NO2

(1) (2)

C2H5 CH3
(3) (4)

45. Which of the following is aromatic in nature?

(1) (2)


(3) (4) All of these

N
46. Which of the following is used for the preparation of benzene?
(1) Phenol (2) Ethyne
(3) Furan (4) Both (1) & (2)
47. Which of the following is an examples of Friedel Crafts reaction?

Anhydrous AlCl3
(1) + CH3Cl AlCl3 (2) + COCl2

AlCl3
(3) + (CH3CO)2O (4) All of these

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48. Which of the following is aromatic hydrocarbon?

(1) (2)
N O

(3) (4)

49. The number of  and -bonds in toluene is respectively

(1) 3 and 6 (2) 6 and 12
(3) 3 and 15 (4) 6 and 10
50. C – C – C bond angle in benzene is
(1) 120° (2) 60°
(3) 45° (4) 135°

  

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 nt
me nment
sigAnssig
Assignment
As

Assignment
(1) Trans-2-butene and cis-2-butene
SECTION - A
(2) Cis-2-butene and trans-2-butene
NCERT Based MCQs
(3) Both cis-2-butene
1. Consider the given reaction. [NCERT Pg. 382]
(4) Both trans-2-butene
Mn(CH COO)
2CH3  CH3  3O2 
3

2
 2X  2H2O.
6. The most stable conformation of n-butane is
Product X is [NCERT Pg. 384]
(1) CH3CHO
CH3 CH3
(2) CH3CH2OH H CH3
(3) CH3COOH (1) (2)
H H H H
(4) CH3 – CH2 – O – CH2 – CH3 H CH3 H H
2. The catalyst that can be used in aromatisation
reaction is/are [NCERT Pg. 382] CH3
CH3
(1) Cr2O3 (2) V2O3 H CH3 H H
(3) Mo2O3 (4) All of these (3) (4)
H H H H
3. The alkene X on reductive ozonolysis gives H CH3
CH3 C CH3 and HCHO. Alkene X is
O
sunlight
[NCERT Pg. 391] 7. +Cl2(Excess) P . Product P is
CH3
(1) CH 3 C CH2 [NCERT Pg. 402]
(1) C6H5Cl
(2) CH3 – CH = CH2
(2) C6H3Cl3
(3) CH3 – CH – CH = CH2
(3) C6Cl3
(4) CH3 – CH = CH – CH3
(4) C6H6Cl6
4. Alkyne X which when passes through hot iron tube
at 873 K forms Benzene. Alkyne X is 8. The number of carbon atoms in the alkane obtained
[NCERT Pg. 396] by the electrolysis of aqueous solution of sodium
salt of ethanoic acid is [NCERT Pg. 379]
(1) Butyne
(1) 8 (2) 2
(2) Propyne
(3) 6 (4) 4
(3) Acetylene
9. The alkane having high boiling point among the
(4) Pentyne
following is [NCERT Pg. 380]
Na/liq. NH3 (1) Methane
A
5. CH3 – CH C – CH 3 . Products A and (2) Ethane
Pd/CaCO3
+ H2
B
(3) Propane
B respectively are [NCERT Pg. 387] (4) Butane

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CCI4
10. CH2 CH2+ Br2 Product (A) . SECTION - B
[NCERT Pg. 389]
Objective Type Questions
Product A is
1. The difference in potential energy between eclipsed
(1) 1,2-Dibromoethane and staggered form of ethane is
(2) 1-bromoethane (1) Zero kJ/mol
(3) 1,1,2-Tribromoethane (2) 12.5 kJ/mol
(4) 1,1,2,2-Tetrabromoethane (3) 22 kJ/mol
HBr/R2O2 (4) 44 kJ/mol
A
11. CH3 CH2 CH CH2 . 2. Eclipsed form of ethane has higher energy due to
HBr
B
[NCERT Pg. 390] (1) Torsional strain

A and B are related as (2) Dipole moment

(1) Chain isomers (2) Position isomers (3) van der Waal’s strain

(3) Functional isomers (4) Tautomers (4) Hydrogen bonding

12. Dihedral angle in staggered conformation of ethane 3. The angle strain in cyclopentane is
is [NCERT Pg. 383]
(1) 72°
(1) 0° (2) 30°
(2) 1°28
(3) 60° (4) 90°
(3) 44
CH3
(4) 108°
V2O3, 773 K
13. A + 4 H2 [NCERT Pg. 382]
(10-20 atm) 4. Which one is not aromatic compound?

Compound A is 
(1) (2)
(1) Toluene
(2) Benzene
(3) Cyclohexane 
(4) Hexanol (3) (4)
14. Which of the following does not act as an oxidising
agent? [NCERT Pg. 382]
5. The incorrect match is
(1) Mo2O3
(2) (CH3COO)2Mn (1) – Antiaromatic
(3) KMnO4
(4) LiAIH4 (2) – Aromatic
15. Reductive ozonolysis of 2-methylpropene gives N
[NCERT Pg. 391] H
(1) Acetone and Ethanal
(2) Ethanal and Ethanol (3) – Non-aromatic

(3) Methanal and Acetone

(4) – Aromatic
(4) Methanal and Ethanal
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6. Which one of the following statements is not CH3 CH3
correct for sigma- and pi-bonds formed between | |
10. In CH3–C–OH, CH3–CH–OH, CH3–CH2–CH2OH
two carbon atoms? |
(1) Sigma-bond is stronger than a pi-bond CH3
(I) (II) (III)
(2) Bond energies of sigma- and pi-bonds are of
the order of 264 kJ/mol and 347 kJ/mol, The ease of rate of dehydration is
respectively (1) (I) > (II) > (III)
(3) Free rotation of atoms about a sigma-bond is (2) (II) > (I) > (III)
allowed but not in case of a pi-bond
(3) (III) > (II) > (I)
(4) Sigma-bond determines the direction between
carbon atoms but a pi-bond has no primary (4) (I) > (III) > (II)
effect in this regard
CH2CH3
CH3
11. can be prepared by
7. 
A + NaOH CH3 – C – CH2 – CH3 ;
(CaO)
(Sodalime) H (1) Wurtz reaction
(2) Fittig reaction
The possible compound "A" is
(3) Wurtz fittig reaction
CH3 (4) Frankland reaction
(1) CH3 – C – CH2COONa
Na CH3Br
H 12. Mg2C3 + H2O A B C ;

CH3 The incorrect statement for C is

(2) CH3 – C – CH2 – CH2 – COONa (1) Compound "C" is CH3 – C  C – CH3

H (2) "C" gives positive Tollens test

CH3 (3) In compound "C" all four C are linearly present

(3) CH3 – CH2 – CH – CH2 – COONa (4) Compound "C" on ozonolysis gives diketone

(4) Both (2) & (3) 13. In the following compounds, the decreasing order
of B.P. is
8. On electrolysis of sodium succinate the alkene
obtained is ______ and nature of solution after CH3 – CH2 – CH2 – CH2 – CH2 – CH3
electrolysis is ______ (I)

CH2 CH2 CH3

(1) and acidic (2) and basic CH3 – CH – CH2 – CH2 – CH3
CH2 CH2
(II)
CH CH3 CH3
(3) and acidic (4) and basic
CH CH3 CH3 – C – CH2 – CH3
9. Which one is not prepared by Wurtz reaction? CH3
(1) C2H6 (III)

(2) n-C4H10 (1) (I) > (II) > (III)

(3) CH4 (2) (I) > (III) > (II)

CH3 CH3 (3) (II) > (III) > (I)

(4) CH 3 – CH – CH – CH 3 (4) (III) > (II) > (I)

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CH3 20. In iodination, for preparation of iodomethane

Cl2 compound used is
14. CH3 – CH – CH2 – CH3 UV light (A) ;
Major (1) HIO3 (2) HgO
The compound "A" is (3) Both (1) & (2) (4) HI

CH3 21. In chlorination, the relative rate of abstraction of H

of 3°, 2° and 1° H atom respectively is
(1) CH3 – CH – CH – CH3
(1) 5 : 3.8 : 2 (2) 5 : 3.8 : 1
Cl
(3) 1600 : 82 : 1 (4) 1600 : 5 : 82
CH3
OH
(2) CH3 – C – CH2 – CH3 H3PO4
22. In "A" ; Compound "A" is
 Major
Cl
CH2
CH3
(3) CH3 – CH – CH2 – CH2Cl (1) (2)

O
CH2Cl
(4) CH3 – C – CH2 – CH3 (3) (4)

H 23. When butyne-2 is reacted with Na in liquid

ammonia in presence of C2H5OH then the major
15. Octane number is zero for
product formed is
(1) n-hexane (2) n-heptane
(1) Cis-butene-2
(3) Iso-octane (4) Gasoline
(2) Trans-butene-2
16. n-hexane on heating in presence of Cr2O3 gives
(3) Butene-1
(1) Cyclohexene (2) Benzene
(4) No reaction
(3) Cyclohexyne (4) None of these
(i) (CH COO) Hg
17. The incorrect match is 24. CH3  C  CH 
3 2

 (A)
(ii) NaBH4 /OH

Hydrocarbons Octane numbers Conc. H2SO4


 (B)
(1) 1-propanol – 118
The product (B) is
(2) 2,5-dimethylfuran – 101
(1) Mesitylene
(3) Ethanol – 109
(2) Toluene
(4) Isooctane – 0
(3) 1,3,5-trimethyl benzene
18. A mixture of ethyl iodide and n-propyl iodide is
(4) Both (1) & (3)
subjected to Wurtz reaction. The hydrocarbon
which will not be formed is Na/liq. NH3
;
25. CH3 – C  C – CH3 A
Pd/CaCO3
(1) Butane (2) Propane B ;
(CH3COO)2Pb
(3) Pentane (4) Hexane
The minimum heat of hydrogenation is in
19. Indicate the expected structure of the organic
product when ethyl magnesium bromide is treated (1) (A)
with heavy water (D2O) (2) (B)
(1) C2H5–C2H5 (2) C2H5OD (3) Both have equal
(3) Sodium benzoate (4) C2H5D (4) Cannot predict
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26. In which of the following reactions, Markownikov’s
rule is not observed ? (1) B2H6
33. "A" . "A" is
(1) CH3CH = CH2 + HCl  (2) H2O2/OH (Major)

Organic Peroxide
(2) CH3CH = CH2 + HBr 

(3) CH3CH = CH2 + HBr  (1) (2) OH

OH
(4) CH3CH = CH2 + H2SO4 
OH
27. CH 3 – CH 2 – CH 3 and CH 3 – C  CH can be OH
distinguished by (3) (4)
(1) Baeyer's reagent (2) Tollen's Test
CH3
(3) Br2/CCl4 (4) All of these
C KMnO4 + H2SO4 NaOH
28. C6H5CH2CH2CH3 is when oxidised in the presence 34. A B ; Compound B is
CH CaO
of alk. KMnO4, the product obtained is
(1) C2H6 (2) CH4
(1) C6H5CHO (2) C6H5COOH
(3) CH3COOH (4) CH3COONa
(3) C6H5CH2CH2CHO (4) None of these
CH2 Turns CuSO4
29. In an attempt to prepare propane by Wurtz reaction, crystal to blue
(1) KMnO4/
1 mole of methyl bromide and 1 mole of ethyl 35. A + B + C ;
(2) H+
bromide are treated with sodium. Assuming equal
probability for all possible reactions, how many grams Turns lime H2O milky
of propane will be obtained? Compound A, B and C are respectively
(1) 44 g (2) 22 g
O
44 (1) CH2O, H2O and
(3) 33 g (4) g
3

30. Addition of HBr to 2-methyl-1-propene in the O

presence of hydrogen-peroxide produces
(2) , CO2 and H2O
(1) 1-Bromobutane
(2) 2-Bromopropane O

(3) 2-Bromo-2-methylpropane (3) , SO2 and H2O

(4) 1-Bromo-2-methylpropane
CHO
K2Cr2O7
31. C6H5CH3
H2SO4
Y . Here Y is (4) , CO2 and H2O

(1) Benzaldehyde
CH3
(2) Toluene CH3 (1) O3
36. C = C – CH = CH2 A + B + C;
CH3 (2) Zn/H2O
(3) Benzoic acid
(4) Ethylbenzene Which one is not A, B, C?

Oxidation O
32. C 6H6     X . Here, X is
V2O 5 /  (1) CH 3 – C – CH 3 (2) CH2O

(1) Maleic anhydride (2) Acetic acid

O
(3) Propanoic acid (4) Succinic acid (3) CH 3 – C – CHO (4) CH3CHO
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44. In sulphonation of benzene; electrophilic reagent

CH3
used is
CH3 (1) O3
37. A + B + C ; 
(2) Zn/H2O (1) SO3H (2) SO3

A, B, C is/are O

(3) O = S = O (4) SO2OH

O O CHO
(1) CH3 – C – C – CH3 (2) Anhydrous
CHO 45. (CH3 )2 CHCH2Cl   P (Major) .
AlCl3
O
The product 'P' is
(3) CH 3 – C – CHO (4) All of these
CH2CH(CH3)2 C(CH3)3
38. Baeyer’s reagent is
(1) 1% Alkaline KMnO4 (2) Acidic KMnO4 (1) (2)
(3) Neutral KMnO4 (4) Aq. Br2 solution
CH2(CH2)2CH3
39. Acetylene when oxidized with chromic acid gives
(3) (4) All of these
(1) Ethylene glycol (2) Oxalic acid
(3) Formic acid (4) Acetic acid 46. Which one is o, p-directing group for electrophilic
substitution reaction?
40. Benzene undergoes substitution reaction more
easily than addition because O O
(1) It has a cyclic structure (1) – C – OH (2) – C – NH2
(2) It has three double bonds CH3
(3) – N (4) – NO2
(3) Of delocalization of electrons CH3

(4) It has six hydrogen atoms UV

47. Benzene  Cl2 
500 K
 A , A is
41. Ethylene + S2Cl2  "A". The compound "A" is
(1) Cyclohexane (2) Gammaxane
(1) Lewisite (2) Mustard oil
(3) Chlorobenzene (4) 1,4-dichlorobenzene
(3) Mustard gas (4) Insecticide
2+
Hg , H
+ HI
42. Ph – C  CH + H2O "A" . The reaction 48. CH3 – CH = CH – CH2 – CH3 A ;
Major
is an example of
Compound "A" is
(1) Electrophilic addition
(1) CH3 – CH2 – CH – CH2 – CH3
(2) Nucleophilic addition
I
(3) Free radical addition
(2) CH3 – CH – CH2 – CH2 – CH3
(4) Electrophilic substitution
I
+
H
43. + CH3 – CH = CH2 A . Compound "A" CH3
is (3) CH3 – C – CH2 – CH3
(1) Isopropyl benzene
I
(2) Cumene CH3
(3) An alkyl derivative of benzene (4) CH3 – C – CH2I
(4) All of these CH3

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49. The electrophile which attacks in Friedel-Craft 3. Among the following, the reaction that proceeds
acylation is through an electrophilic substitution, is
O [NEET-2019]
(1) R – C 
+ Cu 2Cl2
O
–
(1) N2CI CI + N2

(2) R – C
–
O AlCl 3
(2) + Cl2 CI + HCl
–
(3) R – C – O
+
CI CI
(4) Both (1) & (3)
50. Which of the following shows geometrical isomerism? UV light
(3) + Cl2 CI CI
(1) But-1-ene
(2) But-2-ene
CI CI
(3) Prop-1-ene
(4) Pent-1-ene heat
(4) CH 2OH + HCl

SECTION - C CH2Cl + H2O

Previous Years Questions 4. An alkene "A" on reaction with O3 and Zn–H2O

1. In the following reaction, gives propanone and ethanal in equimolar ratio.
Addition of HCl to alkene "A" gives "B" as the
red hot major product. The structure of product "B" is:
iron tube
H3C – C  CH 
873 K
A, the
[NEET-2019]
number of sigma () bonds present in the product
CH3 CH 2Cl
A, is [NEET-2019 (Odisha)]
(1) Cl–CH2–CH2–CH (2) H3C–CH2–CH–CH3
(1) 18
CH3
(2) 21
(3) 9
CH3 CH3
(4) 24
(3) H3C–CH2–C–CH3 (4) H3C–CH–CH
2. The most suitable reagent for the following
Cl Cl CH3
conversion is

H3C CH3 5. Hydrocarbon (A) reacts with bromine by

H3C C C CH3 substitution to form an alkyl bromide which by
H H Wurtz reaction is converted to gaseous
cis-2-butene
hydrocarbon containing less than four carbon
[NEET-2019] atoms. (A) is [NEET-2018]

(1) Na/liquid NH3 (1) CH  CH

(2) H2, Pd/C, quinoline (2) CH2  CH2

(3) Zn/HCl (3) CH4

(4) Hg2+/H+, H2O (4) CH3 – CH3

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124 Hydrocarbons NEET

6. The compound C 7H 8 undergoes the following 10. With respect to the conformers of ethane, which of
reactions [NEET-2018] the following statements is true? [NEET-2017]
3Cl / 
C7H8 
2
 A 
2 Br /Fe
Zn/HCl
 B  C (1) Bond angle remains same but bond length
changes
The product 'C' is
(2) Bond angle changes but bond length remains
(1) m-bromotoluene same
(2) o-bromotoluene
(3) Both bond angle and bond length change
(3) p-bromotoluene
(4) Both bond angles and bond length remains
(4) 3-bromo-2,4,6-trichlorotoluene same
7. Which of the following molecules represents the
11. Which of the following can be used as the halide
order of hybridisation sp2, sp2, sp, sp from left to
component for Friedel Crafts reaction?
right atoms? [NEET-2018]
(1) HC  C – C  CH [NEET(Phase-2)-2016]
(2) CH2 = CH – C  CH (1) Chlorobenzene
(3) CH3 – CH = CH – CH3 (2) Bromobenzene
(4) CH2 = CH – CH = CH2
(3) Chloroethene
8. Predict the correct intermediate and product in the
following reaction [NEET-2017] (4) Isopropyl chloride

12. Which of the following compounds shall not

H2O, H2SO4 produce propene by reaction with HBr followed by
H 3C C CH HgSO4
intermediate product elimination or direct only elimination reaction?
(A) (B)

[NEET(Phase-2)-2016]
(1) A : H3C C CH2 B : H3C C CH3
H2
SO4 O (1) H2C CH2 (2) H3C C CH2OH
C
(2) A : H3C C CH2 B : H3C C CH2 H2

OH SO4 H2
(3) H 2C C O (4) H3C C CH2Br
(3) A : H3C C CH3 B : H3C C CH
13. In the given reaction
O
HF
(4) A : H3C C CH2 B : H3C C CH3 + P
0°C

OH O
the product P is [NEET(Phase-2)-2016]
9. Which one is the correct order of acidity?
[NEET-2017] F F
(1) CH2 = CH2 > CH3 – CH = CH2 > CH3 – C  (1) (2)
CH > CH  CH
(2) CH  CH > CH3 – C  CH > CH2 = CH2 >
CH3 – CH3
(3) CH  CH > CH2 = CH2 > CH3 – C  CH >
CH3 – CH3 (3) (4)
(4) CH3 – CH3 > CH2 = CH2 > CH3 – C  CH >
CH  CH

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14. The compound that will react most readily with 19. In the reaction with HCl, an alkene reacts in
gaseous bromine has the formula accordance with the Markovnikov's rule, to give a
[NEET(Phase-2)-2016] product 1-chloro-1-methylcyclohexane. The
possible alkene is [Re-AIPMT-2015]
(1) C3H6 (2) C2H2
(3) C4H10 (4) C2H4 CH2 CH3
15. The correct statement regarding the comparison of
(1) (A) (2) (B)
staggered and eclipsed conformations of ethane is
[NEET-2016]
(1) The staggered conformation of ethane is more CH3
stable than eclipsed conformation, because
staggered conformation has no torsional strain (3) (A) and (B) (4)
(2) The staggered conformation of ethane is less
stable than eclipsed conformation, because 20. A single compound of the structure
staggered conformation has torsional strain
(3) The eclipsed conformation of ethane is more CH3 CH3
stable than staggered conformation, because
eclipsed conformation has no torsional strain OHC C C
H
C C O
(4) The eclipsed conformation of ethane is more H2 H2
stable than staggered conformation even
though the eclipsed conformation has torsional is obtainable from ozonolysis of which of the
strain following cyclic compounds? [AIPMT-2015]
16. In the reaction
H3C
(1) NaNH /liq. NH (1) NaNH /liq. NH
H – C  CH 2
(2) CH CH Br
3
X 2
(2) CH CH Br
3
Y (1) CH3 (2) CH3
3 2 3 2

X and Y are [NEET-2016] CH3

(1) X = 1-Butyne; Y = 2-Hexyne
(2) X = 1-Butyne; Y = 3-Hexyne H3C H3C CH3

(3) X = 2-Butyne; Y = 3-Hexyne (3) (4)

H3C
(4) X = 2-Butyne; Y = 2-Hexyne
17. Consider the nitration of benzene using mixed 21. The reaction of C 6 H 5 CH = CHCH 3 with HBr
conc. H 2SO 4 and HNO 3 . If a large amount of produces [AIPMT-2015]
KHSO 4 is added to the mixture, the rate of
CH = CHCH3
nitration will be [NEET-2016]
(1) Doubled (2) Faster
(1) (2) C6H5CHCH2CH3
(3) Slower (4) Unchanged
Br
18. 2,3-Dimethyl-2-butene can be prepared by heating Br
which of the following compounds with a strong (3) C 6H5CH2CHCH 3 (4) C6H5CH2CH2CH2Br
acid? [Re-AIPMT-2015]
Br
(1) (CH3)2C = CH – CH2 – CH3
22. Which of the following organic compounds has same
(2) (CH3)2CH – CH2 – CH = CH2
hybridization as its combustion product-(CO2)?
(3) (CH3)2CH – CH – CH = CH2
[AIPMT-2014]
CH3 (1) Ethane (2) Ethyne
(4) (CH3)3C – CH = CH2 (3) Ethene (4) Ethanol

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126 Hydrocarbons NEET

23. Some meta-directing substituents in aromatic 29. The reaction of toluene with Cl2 in presence of FeCl3
substitution are given. Which one is most gives 'X' and reaction in presence of light gives 'Y'.
deactivating? [NEET-2013] Thus, 'X' and 'Y' are [AIPMT (Prelims)-2010]
(1) –SO3H (1) X = Benzal chloride, Y = o-chlorotoluene

(2) –COOH (2) X = m-chlorotoluene, Y = p-chlorotoluene

(3) –NO2 (3) X = o-and p-chlorotoluene, Y = Trichloromethyl

benzene
(4) –CN
(4) X = Benzyl chloride, Y = m-chlorotoluene
24. Which of the following compounds will not undergo
Friedal-Craft's reaction easily? [NEET-2013] 30. In the following the most stable conformation of
n-butane is [AIPMT (Prelims)-2010]
(1) Xylene
(2) Nitrobenzene CH3 CH3
H CH3 H H
(3) Toluene
(4) Cumene (1) (2)
H H H H
25. Among the following compounds the one that is most H CH3
reactive towards electrophilic nitration is
[AIPMT (Prelims)-2012] CH3 CH 3
CH3 H
(1) Toluene (2) Benzene
(3) (4)
(3) Benzoic acid (4) Nitrobenzene H H
H H H H H3C H
26. Which of the following reagents will be able to
distinguish between 1-butyne and 2-butyne ? 31. Benzene reacts with CH3Cl in the presence of
[AIPMT (Mains)-2012] anhydrous AlCl3 to form [AIPMT (Prelims)-2009]

(1) NaNH2 (2) HCl (1) Chlorobenzene

(2) Benzylchloride
(3) O2 (4) Br2
(3) Xylene
27. Which one of the following is most reactive towards
electrophillic reagent? [AIPMT (Prelims)-2011] (4) Toluene
32. Nitrobenzene can be prepared from benzene by using
CH3 CH3
a mixture of conc. HNO3 and conc. H2SO4 in the
(1) (2)
mixture, nitric acid acts as a/an
CH2OH OCH3
[AIPMT (Prelims)-2009]

CH3 CH3 (1) Acid (2) Base

(3) (4) (3) Catalyst (4) Reducing agent
OH NHCOCH3
33. Which one of the following is most reactive towards
28. Liquid hydrocarbons can be converted to a mixture electrophilic attack? [AIPMT (Prelims)-2008]
of gaseous hydrocarbons by
[AIPMT (Prelims)-2010] Cl
CH2OH
(1) Oxidation (1) (2)
(2) Cracking
(3) Distillation under reduced pressure NO2 OH
(3) (4)
(4) Hydrolysis

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34. The order of decreasing reactivity towards an 38. Geometrical isomers differ in
electrophilic reagent, for the following would be
(1) Position of functional group
a. Benzene b. Toluene
(2) Position of atoms
c. Chlorobenzene d. Phenol
(3) Spatial arrangement of atoms
[AIPMT (Prelims)-2007]
(4) Length of carbon chain
(1) d > b > a > c
39. The correct order of reactivity towards the
(2) a > b > c > d electrophilic substitution of the compounds
aniline (I), benzene (II) and nitrobenzene (III) is
(3) b > d > a > c
(1) III > II > I (2) II > III I
(4) d > c > b > a
(3) I < II > III (4) I > II > III
35. Predict the product C obtained in the following
reaction 40. The reactive species in the nitration of benzene is

HΙ  C
CH3CH2 – C  CH + HCl  B  (1) NO3 (2) NO2

[AIPMT (Prelims)-2007] (3) NO2+ (4) NO2–

l 41. H3C CH CH CH2 + HBr  A

(1) CH3CH2 – C – CH3 CH3
Cl
A (predominantly) is
(2) CH3 – CH – CH2CH2l
(1) CH3 CH CH CH3
Cl
l CH3 Br
(2) CH3 CH CH2 CH2Br
(3) CH 3 – CH 2 – CH 2 – C – H
Cl CH3
l Br
(4) CH3 – CH2 – CH – CH2Cl (3) CH3 C CH2CH3
36. Which one of the following alkenes will react fastest CH3
with H2 under catalytic hydrogenation conditions ?
(4) CH3 CH CH CH3
[AIPMT (Prelims)-2005]
Br CH3
R H R R
42. The alkene R–CH = CH2 reacts readily with B2H6
(1) (2)
R R and the product on oxidation with alkaline hydrogen
H R
peroxides produces
R R R R
(1) R C O (2) R CH CH2
(3) (4)
H H R H CH3 OH OH
(R = Alkyl substituent)
(3) R–CH2–CHO (4) R–CH2–CH2–OH
Questions asked Prior to Medical Ent. Exams. 2005 43. Electrophile in the case of chlorination of benzene
37. Which is maximum stable? in the presence of FeCl3 is

(1) But-1-ene (2) cis-but-2-ene (1) Cl (2) FeCl3

(3) trans-but-2-ene (4) All have equal (3) Cl+ (4) Cl–

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128 Hydrocarbons NEET

44. The bond length between central carbon atom and 50. The most stable conformation of n-butane is
other carbon atom is minimum in
(1) Gauche (2) Staggered
(1) Propene (2) Propyne
(3) Skew boat (4) Eclipsed
(3) Propane (4) Pentane
51. Which one of these is not compatible with arenes?
45. Which of the following is used as an anti-knocking
(1) Electrophilic additions
material?
(2) Delocalisation of -electrons
(1) Glyoxal (2) Freon
(3) Greater stability
(3) T.E.L. (4) Ethyl alcohol
(4) Resonance
46. Which of the following reaction is expected to
readily give a hydrocarbon product in good yields? 52. When acetylene is passed through dil. H2SO4 in
the presence of HgSO4, the compound formed is
Cl
(1) CH3CH3 
2
 (1) Acetic acid (2) Ketone
h

C 2H5 OH (3) Ether (4) Acetaldehyde

(2) (CH3 )3 CCl  
53. In Friedel-Craft’s alkylation, besides AlCl3 the other
Electrolysis
(3) RCOOK    reactants are
Oxidation
(1) C6H6 + CH3Cl (2) C6H6 + CH4
I
(4) RCOOAg 
2
h
(3) C6H6 + NH2 – NH2 (4) C6H6 + CH3COCl

47. In a reaction 54. Gammexane is

CH2OH (1) Bromobenzene

Hypochlorous R
CH2  CH2    
 M  (2) Benzyl Chloride
acid CH2OH
(3) Chlorobenzene
where M = Molecule and R = Reagent. M and R
are (4) Benzene hexachloride
(1) CH3CH2OH and HCl 55. In Friedel-Crafts reaction, toluene can be prepared by
(2) CH3–CH3 and heat (1) C6H6 + CH3Cl
(3) CH3CH2Cl and NaOH (2) C6H5Cl + CH4
(4) CH2Cl–CH2OH and aq. NaHCO3 (3) C6H6 + CH2Cl2

48. The cylindrical shape of alkyne is due to (4) C6H6 + CH3COCl

(1) Two sigma C–C and one  C–C bonds 56. 2-butene shows geometrical isomerism due to

(2) One sigma C–C and two  C–C bonds (1) Restricted rotation about double bond

(3) Three sigma C–C bonds (2) Free rotation about double bond

(4) Three  C–C bonds (3) Free rotation about single bond

49. In the commercial gasolines, the type of (4) Chiral carbon

hydrocarbons which are more desirable is 57. Dihedral angle in staggered form of ethane is
(1) Linear unsaturated hydrocarbon (1) 0°
(2) Toluene (2) 120°
(3) Branched hydrocarbon (3) 60°
(4) Straight-chain hydrocarbon (4) 180°

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58. Which alkene on ozonolysis gives CH3CH2CHO 63. One of the following which does not observe the anti-
and CH3CCH3 ? Markownikoff’s addition of HBr, is
||
O (1) Pent-2-ene (2) Propene
CH3 (3) But-2-ene (4) But-1-ene
(1) CH3CH2CH C

||
CH3
64. Given
(2) CH3CH2CH = CHCH2CH3
(3) CH3CH2CH = CHCH3
Br Br H Br
H H
CH3
and
(4) CH3 – C CHCH3 CH3 CH3 CH3
||

| H
CH3 Br
I II
59. In preparation of alkene from alcohol using Al2O3
I and II are
which is the effective factor?
(1) A pair of optical isomers
(1) Porosity of Al2O3
(2) Identical
(2) Temperature
(3) A pair of conformers
(3) Concentration
(4) A pair of geometrical isomers
(4) Surface area of Al2O3 65. Which of the following conformers for ethylene
glycol is most stable?
CH3
OH OH
60. The compound CH3 C CH CH3 on reaction OH
H H
with NaIO4 in the presence of KMnO4 gives
(1) (2)
(1) CH3COCH3
H H H
OH H H
(2) CH3COCH3 + CH3COOH H
(3) CH3COCH3 + CH3CHO OH
H
(4) CH3CHO + CO2
(3) (4)
61. Products of the following reaction
HO
(i) O / ether H H
CH3 C  CCH2 CH3  
3
  are H
(ii) H2O / Zn
66. Reaction of HBr with propene in the presence of
O O peroxide gives

(1) CH3 – C – C – CH2 – CH3 (1) Isopropyl bromide (2) 3-bromo propane
(3) Allyl bromide (4) n-propyl bromide
(2) CH3COOH + HOOC.CH2CH3
67. Which one of the following is a free-radical
(3) CH3CHO + CH3CH2CHO
substitution reaction?
(4) CH3COOH + CH3COCH3
CH3 Boiling CH2Cl
62. Which of the compounds with molecular formula (1) + Cl2
C5H10 yields acetone on ozonolysis?
Anh. AlCl3 CH3
(1) 3-methylbut-1-ene (2) + CH3Cl

(2) Cyclopentane
CH2Cl CH2NO2
(3) + AgNO2
(3) 2-methylbut-1-ene
(4) 2-methylbut-2-ene (4) CH3CHO + HCN  CH3CH(OH)CN

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130 Hydrocarbons NEET

68. Using anhydrous AlCl3 as catalyst, which one of the 74. In the following reaction
following reactions produces ethylbenzene (PhEt)? 1. Mg, Ether
C6H5 CH2Br   X, the product ‘X’ is
(1) H3C – CH2OH + C6H6 2. H O 3

(2) CH3 – CH = CH2 + C6H6 (1) C6H5CH2OCH2C6H5 (2) C6H5CH2OH

(3) H2C = CH2 + C6H6 (3) C6H5CH3 (4) C6H5CH2CH2C6H5

75. When 3, 3-dimethyl 2-butanol is heated with
(4) H3C – CH3 + C6H6
H2SO4, the major product obtained is
69. Given reaction
(1) 2, 3-dimethyl 2-butene
ether HCl (2) cis and trans isomers of 2, 3-dimethyl 2-butene
Br Mg X H2O Y(main product)
(3) 2, 3-dimethyl 1-butene
Y in the reaction is (4) 3, 3-dimethyl 1-butene
(1) Hexane 76. Identify Z in the sequence of reaction
(2) Cyclohexane HBr
CH3 – CH2 – CH  CH2 
H O /h
Z
2 2
(3) Cyclohexylcyclohexane
(4) Cyclohexylether (1) CH3 CH2 CH CH3

70. Monobromination of 2-methylbutane gives how many Br

distinct isomers? (2) CH3 – CH2 – CH2 – CH2 – Br
(1) One (2) Two (3) CH3 CH CH CH2
(3) Three (4) Four Br
71. When CH3CH2CHCl2 is treated with NaNH2, the (4) None of these
product formed is
77. In the following reaction :
(1) CH3 – CH = CH2
CH3
(2) CH3 – C  CH
H2O/H
NH2 H3C–C–CH = CH2 A + B
(Major (Minor
(3) CH 3CH 2CH Product)
CH3 Product)
NH2
Cl The major product is
(4) CH3CH2CH
NH2 CH3

72. 2-Bromopentane is heated with potassium ethoxide (1) H 3C–C – CH–CH3

in ethanol. The major product obtained is CH3 OH
(1) trans-pent-2-ene
CH3
(2) Pent-1-ene
(2) H 3C–C – CH2–CH 2
(3) 2-ethoxypentane
CH3 OH
(4) cis-pent-2-ene
73. An organic compound A(C4H9Cl) on reaction with CH3
Na/diethyl ether gives a hydrocarbon which on
monochlorination gives only one chloro derivative (3) H 3C–C – CH–CH3
then, A is OH CH3
(1) t-butyl chloride
CH3
(2) Secondary butyl chloride
(4) CH 2–C–CH2–CH3
(3) Iso butyl chloride
OH CH3
(4) n-butyl chloride
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78. What products are formed when the following compound 81. Increasing order of electrophilic substitution for
is treated with Br2 in the presence of FeBr3? following compounds

CH3 CH3

I. II.

CH3 OCH3 CF3

CH3 CH3 III. IV.

Br
(1) and (1) IV < I < II < III
CH3 CH3 (2) III < II < I < IV
Br
(3) I < IV < III < II
CH3 CH3 (4) II < III < I < IV
Br Br 82. Among the following compounds (I-III) the correct
(2) and order of rate of electrophillic substitution is
CH3 CH3 OCH3 NO2

CH3 CH3
Br
(3) and
I II III
CH3 CH3 (1) I > II > III
Br
(2) I = II > III
CH3 CH3
(3) II > III > I
(4) and (4) III < I < II
CH3 Br CH3
Br
SECTION - D
79. Which of the following chemical system is
non aromatic? NEET Booster Questions

(1) (2) 1. Which of the following is most reactive towards

electrophilic substitution reaction?

(3) (4)
S (1)
80. Which one of the following compounds will be most
easily attacked by an electrophile?
NH2
CH3
(1)
(2)

(2)
NO2
Cl
(3) (3)

OH
(4) (4) All have same reactivity

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132 Hydrocarbons NEET

2. Which of the following are aromatic? HN O

+ C2H5 CH3

(1) (3)

Br
HC = CC6H5
HN O
C
(2) | C2H5 CH3
C6H5
(4)

(3) Br
5. Which of the following chemical species is/are
aromatic?
(4) All of these


:
N
Li/NH3(liq.)
3. C C C C X N:

: :
N O

:
‘X’ is H
(I) (II) (III)
Choose the correct option
(1)
(1) Only (I) (2) (I) & (II)
(3) (II) & (III) (4) (I), (II) & (III)
6. Which of the following conformer of cyclohexane
(2) has maximum stability?

(1) (2)
(3)

(3) (4)
(4)
7. Which of the following cycloalkane would have
maximum angle strain?
HN O

C2H5 CH3 (1) (2)

4. Br+
Product .
(3) (4)
Product is
8. Which of the following pair of compounds are
HN O geometrical isomers to each other?

C2H5 CH3 Cl I Cl Br
(1) (1) C C and C C
H Br I H
Br
I H Br Cl
(2) C C and C C
O Br Cl I H
HN
C2H5 CH3 Br Cl Br I
(2) (3) C C and C C
H I H Cl
Br
(4) All of these
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9. Which of the following has most acidic hydrogen? 14. Arrange the following in the order of the rate of
-elimination with alcoholic KOH
CH3 Br Br
| | |
(1) (2) CH3–CH2–Br H3C–C— C–CH3 H3C–C–CH3
H H
H H | | |
H CH3 CH3
H (I) (II) (III)
H (1) (I) > (II) > (III) (2) (III) > (II) > (I)
(3) (4)
H H (3) (I) > (III) > (II) (4) (II) > (III) > (I)
HBr H O
10. Which of the following group is metadirecting for 15. X   H3 C  CH  CH2 
2 2
HBr
Y
(Major ) (Major )
electrophilic substitution in benzene?
‘X’ and ‘Y’ are respectively
(1) – NO2 (2) – OCH3
(1) CH3–CH–CH3 & CH3–CH2–CH2–Br
(3) – OH (4) – Cl |
Br
11. Which of the following compound is most reactive
(2) CH 3–CH 2–CH 2–Br & CH 3–CH–CH 3
towards electrophilic substitution reaction? |
Br
CH2CH3 H3C–CH–CH3 (3) Both are CH3–CH–CH3
|
Br
(1) (2)
(4) Both are CH3–CH2–CH2–Br
16. Which of the following would undergo Wurtz
CH3 reaction least easily?
CH3 (1) CH3Br (2) CH3–CH2–Cl
H3C–C–CH3
(3) (H3C)3C–Cl (4) CH3CH–CH3
(3) (4) |
Cl
2 
Hg /H Tautomerisation
17. H3C–C CH + H2O X Y
12. A hydrocarbon on reductive ozonolysis gives only 333 K (major)
ethanal and CO2. The hydrocarbon is Identify ‘X’

(1) But-2-ene (1) Ethanal (2) Propanal

(3) Propanone (4) Propen-2-ol
(2) 2-methylpropene
CH3
(3) Penta-2,3-diene |
C–CH3 Conc. H2SO4
(4) Butene 18. | X . X is
OH  (major)
13. Consider the following hydrocarbons
CH3
CH3
| CH3
H3C–CH–CH3 H3C–C–CH3 CH3–CH2–CH2–CH3 CH3
| | CH3
CH3 CH3 CH3
(I) (II) (III) (1) (2)
CH3
The correct order of boiling point is CH3
(1) (III) > (I) > (II) CH3
CH3
(2) (I) > (II) > (III)
CH3
CH3
(3) (I) > (III) > (II) (3) (4)
(4) (II) > (III) > (I) CH3 CH3

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CCl4 alc. KOH NaNH 2 23. Propanal and pentan-3-one are the ozonolysis
19. H3C–C CH2 + Br2 X Y Z . products of an alkene.What is the IUPAC name of
|  
(1 eq.)
H the alkene?
Identify ‘Z’ (1) 4-Ethylhex-3-ene
(1) H3C–CH=CH2 (2) H3C–C CH (2) 3-Ethylhex-3-ene
(3) H3C–CH2–CH2–OH (4) CH3–CH2–CH3 (3) 4-Ethylpent-3-ene
20. Choose the aromatic species (4) 3-Ethylpent-2-ene
24. Which of the following has cancer producing
property?
(1) (2)

(1)

(3) (4)
 ( )

21.

Anhydrous Br2 (2)

+ CH3CH2CH2–Cl X Y.
AlCl3 (major) h (major)

Y is
CH3
CH2–CH2–CH2–Br
(1)
(3)

CH3
CH2–CH–CH3
|
(2) Br (4) All of these
25. Which of the following compounds on oxidation
with alk. KMnO4 give benzoic acid?
CH3
| (1) Toluene
C–Br (2) Cumene
(3) |
CH3 (3) Styrene
(4) All of these
CH3 26. Grignard reagent produces alkane if reacts with
|
CH–CH3 (1) Phenol
(4) (2) Carboxylic acid
Br (3) Amine

22. Which of the following is the strongest o, p- (4) All of these

directing group? 27. How many geometrical isomers are possible for
given compound?
O C6H5 CH CH CH CH COOH
||
(1) –NO2 (2) –C–OH (1) 3 (2) 4
(3) –SO3H (4) –NH2 (3) 2 (4) 1
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28. Which of the following is aromatic? 33. The trichloromethyl group is meta directing when
attached in benzene ring due to
O
(1) Hyperconjugation
(2) Reverse hyperconjugation
(1) (2)
(3) –I-effect
O
(4) +I-effect
C H NOCl
X
(3) (4) All of these 34. CH2 CH—CH3
HOCl
29. Which of the following compounds will exhibit Y
geometrical isomerism? The major product ‘X’ & ‘Y’ are respectively
(1) 2-phenyl-2-butene (1) CH 2—CH—CH 3 & CH 2—CH—CH 3
(2) 3-phenyl-1-butene Cl NO OH Cl
(3) 2-phenyl-1-butene
(2) CH2—CH—CH3 & CH 2—CH—CH 3
(4) 1,1-diphenyl-1-propene species
NO Cl Cl OH
30. Consider the following species
(3) CH 2—CH 2—CH 2 & CH2—CH2—CH2
NO Cl Cl OH
(i) (ii)
 (4) CH3—CH2—CH—Cl & CH3—CH2—CH—OH
NO Cl
(iii) (iv) O
C
The aromatic specie(s) is/are H2C anhydrous CaO/NaOH
35. + O (A) (B)
(1) (i) only (2) (i) & (ii) AlCl3
H2C
C
(3) (iii) & (iv) (4) (ii) & (iv)
O
31. In which of the following, the elements of the ring
The product ‘B’ is
do not have the same hybridisation?
O

(1) (2) (1) (2)

N
H O
O
C
O (3) (4) O
C
(3) (4) N
N O
H 36. Which of the following compound has maximum
32. Which of the following is correct representation for melting point?
Allene compound? (1)
H H H H
(1) C C C (2) C C C (2)
H H H H

H (3)
H H H
(3) C C C (4) C C C
H H H H (4) All have same melting point

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37. The reagent which is most appropriate for the 42. The following reaction is an example of
following conversion is/are
CH2
Cl Na Na Zn-dust
Br H2C CH2 (A)
(1) Alcoholic KOH Br Br
(2) Alcoholic KOH followed by 1-gram-equivalent Mg
dry ether
NaNH2
(3) Aqueous KOH followed by NaNH2 (1) Preparation of alkene only
(4) Excess NaNH2 (2) Preparation of cycloalkane only

H H (3) Preparation of both alkene and cycloalkane

H
Na (4) Preparation of both cycloalkene and
38. 2·CH3 C Cl dry ether
CH3—C—C CH3
cycloalkane
CH3 CH3 CH3
43. Which of the following compound has most acidic
The above reaction is known as hydrogen?

(1) Kolbe’s reaction (2) Wurtz reaction

(1) (2)
(3) Frankland reaction (4) Friedel Craft reaction

U.V.
39. C6H6  Cl2   Pr oduct
light (3) (4)
In the above reaction, product is
44. Kolbe’s electrolytic method can be applied on
(1) C6H6Cl6 (2) C6H5Cl aqueous solution of
(3) C6H8Cl2 (4) CCl3CH3
(1) CH2COOK
Red hot
(A) CH2COOK
'Fe' tube
40. CH CH 1
S8 (2) CH—COOK
8
(B)
CH—COOK
The product (A) & (B) are respectively
COOK
(1) Aromatic and non-aromatic
(3)
(2) Aromatic and antiaromatic COOK
(3) Non aromatic and antiaromatic
(4) All of these
(4) Aromatic and aromatic
41. The major product formed in the following O3/Zn
45. C3H4 (A) + (B)
reaction is H2O
Mo2O3/O2
CF3CH CH2 + HCl ? (C)
sp-hybridisation 543 K/100 atm
of central atom
(1) CF3CH—CH3 , by Markownikoff’s rule
Cl The product (A), (B) and (C) are respectively

(2) CCl3CH CH2 , by substitution reaction (1) CO2, HCHO and CH4

(3) CF2—CH—CH2 , by rearrangement reaction (2) CH3CHO, HCHO and C2H6

F Cl (3) CO2, CH3CHO and C2H6

(4) CF3CH2CH2Cl, by anti-Markownikoff’s addition (4) CO2, CH3COOH and C2H6

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O3/Zn — H2O 51. Natural rubber is a polymer of isoprene (2-methyl
46. C3H4 CO2 + HCHO buta-1, 3-diene). If natural rubber is treated with O3
followed by Zn/H2O, the final product will be
2 eqv. HCl (1) alc. KOH
A B 2+ + C (major) O
(2) NaNH2 Hg /H3O
+
(3) H H C
(1) C O
Product (C) is
H
(1) Propanal (2) Butanal O

(3) 2-pentanone (4) Propanone (2) CH3 C CH2 CH2 CHO

47. In the given reaction
O
O
COOH CH3 (3) CH3 CH2 C CH2 CHO
HI
O O
Red P
H
H C C H
OH (4) CH2 C
The number of moles of HI required for conversion is CH3

(1) 12 (2) 8 52. Dehydration by conc. H2SO4 is the most difficult in

(3) 14 (4) 10 OH

48. Which of the following can form methane gas with (1) (2) CH3
CH3 OH
methyl magnesium bromide?

OH
(1) (2)
(3) (4)
CH3 CH2OH
(3) (4) CH3

49. A
CH3 CH CH CHO CH3 CH3 CH2
53. C C A
CH 3 CH CH CH3 H H
The best suitable reagent 'A' is CH2
B
(1) C3H8S2/H2/Ni (2) N2H4/KOH
A & B are respectively
(3) Zn–Hg/conc. HCl (4) HI/P(red)
H H H CH3 CH3 CH3
50. In the given reaction (1) , mixture of &
CHO CH3 CH3 CH 3 H H H
N2H4/H2O2
Product
CH3 H CH3 H CH3 CH 3 H
(2) , mixture of &
The product will be
CH 3 H CH3 H H CH3
CHO CH3
H H CH3 H H
CH3 H H
(1) H (2) H (3) Mixture of ,
CH3 CH3 H CH3 CH3 H CH3 CH3
CHO
H CH3 H H
CH3 H CH3
(3) CH3 (4) Mixture of (1) & (2) (4) Mixture of ,
H CH3 CH3 H CH3 H
H
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138 Hydrocarbons NEET

54. In which of the following dehydration by conc. H2SO4 H2O Cu2Cl2 (1) 1 eqv. HCl
no rearrangement is favourable? 57. CaC2 A NH4Cl
B (2) Mg/ether
C
+
(3) CO2/H2O/H
CH2OH CH2OH
In above reaction product (C) is
(1) (2)
(1) CH2 ==CH—COOH
(2) HCC—COOH
CH2OH CH3 CH3
(3) HOOC—CC—COOH
OH
(3) (4)
(4) CH 2 CH C CH2

COOH
CHO CHO 58. Consider the following two structures

55.

C CH C CH (i) (ii)
Choose the correct statement
The most suitable reagent for given conversion is
(1) Both (i) & (ii) are conjugated system
(1) Diimide (2) H2/Ni2B
(2) (i) & (ii) both show resonance
(3) Zn/dil. HCl (4) LiAlH4
(3) (i) & (ii) both are aromatic
H (4) (i) is less stable than (ii)
56. CH3 C C C C
C C CH3 Na O3 Zn — Hg/conc. HCl
H 59. A B C
NH3(l) Zn — H2O
major
H2/Pd — BaSO4 Na/NH3(l)
(One equivalent)
A B Product (C) is
(1) Mixture of n-butane, ethane
In above reaction product (B) is
(2) Only propane
(1)
(3) Only ethane
(4) n-hexane
(2)
60. Which of the following compound is paramagnetic?
–
(3) (1) (2)


+
(4) (3) (4)

  

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Chapter 29

Environmental Chemistry
Chapter Contents
z Introduction Introduction
z Environmental Pollution Interrelation of biological, social, economical, physical and chemical
studies with our surrounding is called environmental studies.
z Atmospheric Pollution Environmental pollution is the greatest health hazard all over the world.

z Water Pollution An undesirable change in physical, chemical or biological

characteristics of air, water and land that is harmful to human life
z Soil Pollution and other living organisms, living conditions, cultural assets, industrial
progress and harms our resources is called pollution.
z Industrial Waste

z Strategies to Control
Environmental Pollution ENVIRONMENTAL POLLUTION
z Green Chemistry Undesirable changes that have harmful effects on plants, animals and human
beings in our surrounding is called environmental pollution.
Pollutant : The substance which causes pollution and is harmful for
environment. Pollutants are of two types :
Biodegradable and non-biodegradable : Those substances which are
degraded rapidly by natural process or artificial methods (radiation and
microbial action) are called biodegradable pollutants. While those substances
which degrade at very slow rate or does not degrade by natural biological
process, for example, DDT, arsenic salts of heavy metals, radioactive
materials and plastics are non-biodegradable pollutants.

ATMOSPHERIC POLLUTION

The addition of particulate matter, gases and other ingredients into air which
will have adverse effect on vegetation, animals human beings, human assets
and resources. Lowest layer of atmosphere is troposphere which have dust,
water vapour and clouds, it contains dust, water vapour and clouds while
stratosphere contains ozone. Atmospheric pollution includes both troposphere
and stratosphere pollution.

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124 Environmental Chemistry NEET

Tropospheric Pollution
(Extends upto height of ~10 km from sea level). It is due to pollutant. Pollutant is of two types : Primary
pollutant and Secondary pollutant. Primary pollutants are those pollutants which are persisting in the
environment in the form it is produced e.g., carbon monoxide. Secondary pollutants are formed from primary
pollutant by reaction or change. It is caused by gaseous or solid pollutants. Such as natural fog, pollen grains,
bacteria, volcanic eruptions, dust, smoke, mist fumes, gases and vapours.

1. Gaseous air Pollutants

(a) Oxides of Sulphur (SO3 and SO2) : 67% is by volcanic eruption and 33% by metallurgy of sulphide
ores combustion of sulphur-containing fuels. It, in low concentration, causes respiratory diseases e.g.,
asthma, bronchitis, emphysema, irritation of eyes. High concentration causes stiffness in flowers. In the
presence of pollutant, SO2 converts into SO3.
2SO2 + O2  2SO3
It can also take place by O3 and H2O2.
SO2 + O3  SO3 + O2
SO2 + H2O2  H2SO4
SO3 is more harmful than SO2 because it combines with water to form H2SO4 which causes acid rain
which causes yellowing of Taj Mahal and other monuments.
(b) Oxides of Nitrogen : Mainly produced by combustion of fossil fuels at high temperature in automobile
engines mainly NO and NO2.

1483 K
N2 + O2 
 2NO

1483 K
2NO + O2 
 2NO2

NO also reacts with ozone to give NO2.

NO + O3  NO2 + O2
It is also produced when lightning strikes then N2 and O2 combine to form (NOx). NO2 gives NO3– which
serves as fertilizer.
Effects : These produce reddish brown haze or brown air NO2 is more dangerous than NO. These oxides
can cause pulmonary oedema, dilation of arteries, eye irritation, heart problems, injury to liver and
kidneys and also causes acid rains.
Higher concentration of NO2 damage the leaves of plant and retard the rate of photosynthesis.
(c) Hydrocarbons : Produced naturally (e.g., marsh gas) as well as due to incomplete combustion. These
are carcinogenic and causes irritation of mucous membrane, eyes. They causes ageing, breakdown of
tissues, shedding of flower, leaves and twigs in plants.
(d) Oxides of carbon :
(i) Carbon monoxide : It is colourless, odourless gas. It is produced by incomplete combustion of
fuels, naturally it is produced by oceans or by decaying of organic matter by bacteria. It is
poisonous because it combines with haemoglobin to form 300 more times stable
carboxyhaemoglobin which reduces oxygen-carrying capacity of blood and results into giddiness,
headache, decreased vision, cardiovascular malfunction and asphyxia. Cigarette smoke also

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NEET Environmental Chemistry 125
contains a lot of CO which induces premature birth deformed babies and spontaneous abortions
in pregnant women.
(ii) Carbon dioxide : It is produced naturally by volcanic eruptions, respiration. It is also produced
by burning of fossil fuels. Increased level of CO 2 is controlled by green plants during
photosynthesis. It is a greenhouse gas and responsible for global warming. It causes headache
nausea and asphyxiation.

Note :
There are three methods of identifying air pollution :
1. By eye irritation, acid taste in mouth, unusual odour etc.
2. Physical measurement by standard methods of sampling and analysis.
3. Effects can be observed on growth of plants and health of animals.

Greenhouse Effect and Global Warming

This effect was discovered by Fourier and the term was coined by Arrhenius. 75% of solar radiation is absorbed
by earth surface and remaining is reflected back. Some of which is absorbed by greenhouse gases which
increases temperature of atmosphere is called greenhouse effect.
Radiation emitted
Incoming sunlight by the earth Some escapes
(UV-visible wavelengths) (infrared wavelengths) into space

Some is absorbed
and re-emitted by
CO 2 in the atmosphere
Greenhouse Gases :
(i) CO2 : Produced by burning fossil fuels, burning wood, respiration etc. Present growth rate of CO2 is
0.5%
(ii) Methane : Produced by incomplete combustion of fossil fuels. It is 25 times more effective than CO2.
(iii) CFCs : Thousand times more effective than CO2. Used in aerosols cans, jet fuels, refrigerant in air
conditioners, refrigerators, fire extinguishers. They are depleting ozone layer.
(iv) N2O : 320 times as powerful than CO2. Produced by combustion of livestock waste, breakdown of
nitrogenous fertilizers in soil etc.
Effects :
(i) Increase of global temperature results into extreme climatic conditions.
(ii) Global warming will push tropics into temperate areas and temperate areas towards pole called shifting
of climatic zones.
(iii) It increases infectious diseases like dengue, malaria, yellow fever, sleeping sickness etc.

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126 Environmental Chemistry NEET

Acid Rain

Sulphuric acid (H2SO4)

Nitric acid (HNO3)

Dry fallout
Acid snow
Water vapour
(cloud) Acid rain

Acid fog
Sulphur dioxide
and nitrogen
dioxide
Damage to emission
vegetation

Aquatic life in danger

Acid deposition
Acid rain term was given by Robert Augus. Acid rain has pH less than 5.6. Oxides of sulphur, nitrogen and
carbon are acidic in nature which dissolve in water and causes acid rain.
(i) H2O + CO2 H2CO3
1
(ii) SO2 + O + H2O  H2SO4
2 2
1 1
(iii) NO2 + O2 + H2O  HNO3
2 2
When these acids come to earth’s surface along with rain then it is called acid rain.
There are two types of acid deposition, wet and dry deposition. Wet deposition occurs through rain, snow
and fog. When acidic gases and particles blown with wind, settle over trees, articles and soil then it is called
dry deposition.
It causes chlorosis, necrosis, defoliation and die back of growing points.
It threatens human and aquatic life and destroy forests, reduces agricultural productivity. Other than it
damages railways, bridges, statues, monuments building etc.
Taj Mahal and Acid rain : High level of sulphur and nitrogen oxides caused by combustion of kerosene,
poor quality coal and firewood causes acid rain which reacts with marble and causes discolouring and
disfiguring.
CaCO3 + H2SO4  CaSO4 + H2O + CO2
Therefore, in order to save Taj, area including Mathura, Firozabad, Agra and Bharatpur called Taj trapezium
is switched over to use of natural gas or LPG in industries. People are also encouraged to use LPG as fuel.
Vehicles are encouraged to use low-sulphur content diesel.
2. Particulate pollutants
Minute solid particles or liquid droplets in air that cause pollution are called particulate pollutants.
Types : Viable and Non-viable
(a) Viable : These are living organism like bacteria, fungi, moulds, algae etc.
(b) Non-viable :
(i) Smoke from combustion of oil, fossil fuel, cigarette, dry leaves, garbage consist of solid or mixture
of solid and liquid particles.
(ii) Dust are particles from sand blasting wood works, pulverizing coal, ash of factories, dust storm,
size is over 1 m in diameter.

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(iii) Mist is obtained from condensation of vapours of spray liquids.
(iv) Fumes are obtained from condensation of vapours from boiling, distillation, sublimation etc.
Mist can be from sulphuric acid herbicide, insecticide etc., while fumes can be from organic
solvents, metallic oxide etc.
Particulate having size less than 5  enter in lungs while bigger than 5  lodge in nasal passage.
Lead pollution from leaded petrol interferes development and maturation of RBC.
Smog
Opaque or dark fog having smoke, dust, gases and water vapours.
Smoke + Fog  Smog
(i) Classical smog : It is also known as London smoke or reducing smog as it contains smoke fog and
sulphur dioxide. It occurs in cold humid climate.
(ii) Photochemical smog : It is also called Los Angles smog or oxidising smog as it contains O3, PAN
and NOx. It occurs in warm, dry and sunny climate.

Formation of Photochemical Smog

Photochemical
smog

Nitrogen oxides and volatile organic compounds

On burning fossils, hydrocarbons and nitric oxides are produced. NO converts into NO2 and in presence of sunlight
it gives NO and free oxygen. This free oxygen forms O3 with O2 and this O3 gives NO2 on reaction with NO.
2NO(g) + O2 (g)  2NO2(g)
Air
h
NO2(g)  NO(g) + O(g)
O(g) + O2(g) O3(g)
NO(g) + O3(g)  NO2(g) + O2(g)
Strong oxidising agent NO2 and O3 reacts with unburnt hydrocarbons to form formaldehyde, acrolein (CH2 =
CH – CHO) and peroxy acetyl nitrate (PAN).
3CH4 + 2O3  3CH2 = O + 3H2O
PAN causes reduced visibility, irritation of eyes and respiratory tract. Due to excess of NO2 which is brown
in coloure, photochemical smog is brown coloured.

Methods to Control Photochemical Smog

1. By controlling primary precursors of photochemical smog, such as NO2, hydrocarbons etc.
2. By controlling secondary precursors such as ozone, PAN etc.
3. By plants which metabolise nitrogen oxide e.g., Pinus, Juniparus, Quercus, Pyrus and Vitis.
4. By using catalytic converter which prevents release of nitrogen oxide and hydrocarbon.

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128 Environmental Chemistry NEET

Stratospheric Pollution (Between 10 to 50 km above sea level)

Ozone Formation and Its Breakdown
Molecular oxygen splits into free oxygen atom by UV radiations which combine with molecular oxygen to form
ozone.
 
UV
O2 (g)   O(g) + O(g)

UV


O(g) + O2 (g) 
 O3 (g)
As ozone is thermodynamically unstable hence, there exists dynamic equilibrium between its decomposition
and formation.
Ultraviolet radiations dissociate chlorofluorocarbon to give chlorine-free radical, which combines with ozone to
form chlorine monoxide radical which combines with atomic oxygen to form more chlorine-free radicals.
Reactions can be given as
 
UV
CF2Cl2 
 CF2Cl  Cl
 
Cl  O3 
 ClO  O2
 
ClO  O  Cl  O2
These chlorine radicals deplete ozone layer.
Ozone Hole
Depletion in the concentration of ozone over the South Pole in Antarctica is called Ozone Hole.

 
In summer season, there is reduction in ozone depleting free radicals ClO and Cl .



ClO  NO2  ClONO2
 
Cl  CH4  CH3  HCl
These are known as chlorine sinks.
In winter, however, conditions begin to appear, which lead to release of these ozone depleting free radicals

ClO and Cl in the spring.

 

In the winter season, polar stratospheric clouds appear over Antarctica. These clouds provide surface for the
hydrolysis of chlorine nitrate (ClONO2). It also reacts with HCl.
ClONO2 + H2O  HOCl + HNO3
ClONO2 + HCl  Cl2 + HNO3
In spring season, sunlight breaks HOCl and Cl2 to give chlorine radicals.
 
h
HOCl(g)  OH(g)  Cl(g)

h
Cl2 (g)  2Cl(g)
These chlorine radicals deplete ozone layer.
In spring season, sunlight breaks HOCl and Cl2 to give chlorine radicals.
 
h
HOCl(g)  OH(g)  Cl(g)

h
Cl2 (g)  2Cl(g)
These chlorine radicals deplete ozone layer.

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NEET Environmental Chemistry 129
Effects of Depletion of The Ozone Layer
Ozone is commonly called as chemical weed in troposphere. Bad ozone is formed in troposphere that harms
plants and animals while good ozone is formed in stratosphere which acts as shield. UV rays can enter in
earth’s atmosphere. It is harmful as can cause skin cancer. And also
(i) It increases transpiration hence decreases soil moisture.
(ii) It damages paints and fibres, causing them to fade faster.

WATER POLLUTION
Any unwanted change which deteriorates quality of water and make it unfit for drinking. Pollution of water
originates from human activities. Pollution sources are of two types.
1. Point sources : Involve discharge from identifiable points e.g., discharge of waste water from factory.
2. Non-point sources : Involve discharge from unidentifiable points. It includes discharge from land run off,
atmospheric washout etc.
Causes of Water Pollution
(i) Organic matter such as leaves, grass, trash etc. as well as excessive phytoplankton growth in water
causes water pollution as this matter is decomposed through microbial activity is known as putrescibility
which requires oxygen. Degree of impurity of water due to organic matter is measured in terms of
Biochemical Oxygen Demand (BOD).
For clean water, BOD should be less than 5 ppm whereas in highly polluted water more than 17 ppm.
BOD is amount of oxygen required by bacteria to breakdown organic matter present in a certain volume
of a sample of water.
(ii) Pathogens : Disease-causing agents are called pathogens e.g., viruses, bacteria, protozoa, helminthes,
algae etc. Human excreta contains E.coli and Streptococcus faecalis bacteria which cause
gastrointestinal diseases.
(iii) Chemical pollutants : These are of two types, inorganic and organic.
Inorganic pollutants constitute acids, salts, heavy metals such as Cd, Hg, Ni etc. Heavy metals can
damage central nervous system, liver and kidneys.
Organic pollutants constitute, pesticides, petroleum pollutants, PCBs, detergents, fertilizers etc. PCBs
(Polychlorinated Biphenyls) are carcinogenic and phosphatic fertilizers increase algae growth. Acidic water
is harmful for aquatic life as well as for drinking.
The process in which nutrient-enriched water bodies support a dense plant population which results in
depleting oxygen level hence, killing animal life hence, results in loss of biodiversity is called as Eutrophication.
International standards for drinking water
S. No. Chemical Tolerable limit Effects
1. Nitrates 50 ppm In excess, it causes blue-baby syndrome.
2. Fluoride 1 ppm to 2 ppm In excess, it is poisonous and harmful for
bones and teeth.
3. Sulphates 500 ppm In excess, it causes laxative effect.
4. Lead 50 ppb In excess, it damages kidney, brain and liver.
5. pH Between 5.5 to 9.5 Buffer system of body gets disturbed.

Measures to Control Water Pollution

1. Treatment of sewage : First it is churned properly by machines then passed into a tank where heavier
particles settle down then flowing pure water is sterilised with chlorine then it is treated with alum, lime etc.
2. Treatment of industrial waste : First its pH is determined then it is neutralised with acid or alkali.
Dissolved chemical substances can be precipitated by chemicals. Nowadays we used ion exchangers
and photocatalysis.

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130 Environmental Chemistry NEET

SOIL POLLUTION
It is unfavourable alteration of soil by addition or removal of substances and factors which decrease soil
productivity, quality of plants and ground water is called soil pollution. Mainly caused by chemicals added
into soil as pesticides, herbicides and fertilizers for better productivity. These chemicals reached in human
beings through food chain and causes harmful effects.

Causes of Soil Pollution

1. Pesticides : These are actually synthetic toxic chemicals with ecological repercussions. These are used
in killing pathogens, pests and in inhibiting unwanted growth in agriculture, horticulture, forestry and water.
Organochlorines are persistent insecticides which show biological magnification e.g., DDT, BHC, Aldrin,
Endrin, Dieldrin etc. These are water insoluble, non-biodegradable, hence get transferred from lower level
to higher level in food chain. Pollutant gets 10 times concentrated at each trophic level. Carbamates
and organophosphates are less persistent and more biodegradable are nerve toxins. Insects also
becomes resistant to particular pesticides after successive use. Now a days we use herbicides in place
of pesticides. Herbicides are generally metabolic inhibitors e.g., sodium chlorate, sodium arsinite and
many others. These decompose in few months and not very much persistent. Arsenic-containing
herbicides are harmful to mammals. Some herbicides cause birth defect also.
2. Fertilizers : Excessive use of fertilizers decreases natural microflora hence detiorate soil. Therefore, now
a days organic farming is encouraged which involves organic pesticides, biofertilizers and disease
resistant varieties.
Note :
World Environment Day – 5th June
Ozone Day – 16th September
Bhopal Gas Disaster – 2nd December 1984
Hiroshima Day – 6th August
Nagasaki Day – 9th August
Earth Day – 22nd April

INDUSTRIAL WASTE
Industrial wastes are both solid and liquid and are dumped over the soil. These contain toxic chemicals like
mercury, copper, zinc, lead, cadmium, cyanides, acid, alkalies etc.
(i) Biodegradable : Includes wastes by cotton mills, food processing units, paper mills, textile factories etc.
(ii) Non-biodegradable : Gypsum from fertilizer industries, mud and tailings from metal industries, slag from
steel industries, fly ash from thermal power plants etc.

Prevention
(i) Cement industry can utilize fly ash and slag.
(ii) Small quantity toxic wastes are removed by burning in open bins and large quantity by controlled
incineration. These pollutants can be removed by various processes specified by pollution control board.

STRATEGIES TO CONTROL ENVIRONMENTAL POLLUTION

Two sources of environment pollutant are household waste and industrial waste. Following method can be used
to control them.
1. Recycling : Waste are recycled into manufacturing of new material. For example, scrap iron, broken glass,
clothes can be made from recycled plastic waste and soon becomes available in market.
We can also recover energy from burning combustible waste.
2. Digestion : Waste material can be degraded by anaerobic micro-organisms in absence of air. It can be
used to produce electricity. First biodegradable and non-biodegradable waste are separated then
biodegradable wastes are mixed with water and cultured by bacterial species which produce methane.

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NEET Environmental Chemistry 131
3. Dumping : Sewage sludge acts as fertilizer because it contains nitrogen and phosphorus hence, it is
dumped in land areas which increases soil fertility.
It is very important to sort out garbage into biodegradable and non-biodegradable and so it should be
properly collected. Otherwise it can cost lives of cattle or can choke the sewers.

GREEN CHEMISTRY
Green chemistry is focused on processes and products that minimise the generation and use of hazardous
waste. It increases using of those addition reactions which do not produce hazardous by-products. It emphasizes
to choose those starting materials which can be converted into end products with 100% yield (approx.).
Green chemistry is a way of thinking and is about utilizing the existing knowledge and principles of chemistry
and other sciences to reduce the adverse impact on environment. Utilization of existing knowledge base for
reducing the chemical hazards along with the development activities are the foundation of green chemistry.
Note : Previously water polluting, suspected carcinogen substance tetrachloroethene was also used as
solvent for dry-cleaning and chlorine gas for bleaching of paper but now in place of both these H2O2 is
used having batter result without harming to ground water. Also, liquid CO2 is used instead of halogenated
solvent.

Achievements of Green Chemistry

1. Catalytic dehydrogenation of diethanolamine produces a herbicide which is environment friendly.
2. Preparation of ibuprofen without producing waste and using large amount of solvent with 99% yield.
3. Preparation of ethanal by one step oxidation of ethene in presence of ionic catalyst in aqueous medium
with 90% yield.
Catalyst
CH2 = CH2 + O2 Pd(II)/Cu(II) (in water) CH 3 CHO(90%)

Yves Chauvin, Robert H. Grubbs and Richard R. Schrock won the 2005 Nobel prize in chemistry for their
development of metathesis method in organic synthesis, a way to arrange groups of atom within the
molecules that has huge commercial potential in the pharmaceuticals, biotechnology and food stuffs
production industries. This is a great contribution towards green chemistry.

EXERCISE
1. Which of the following is a greenhouse gas?
(1) CH4 (2) H2
(3) N2 (4) All of these
2. Which of the following is most responsible for global warming?
(1) CFC (2) CH4
(3) CO2 (4) N2O
3. Which of the following gases can cause ACID rain?
(1) SO2 (2) NO2
(3) H2 (4) Both (1) and (2)
4. The formation of photochemical smog depends upon
(1) CO (2) NO
(3) CO2 (4) SO2
5. The gas responsible for Bhopal Gas Tragedy is
(1) Methyl isocyanide (2) Propanenitrile
(3) Methyl isocyanate (4) Peroxy acetyl nitrate

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132 Environmental Chemistry NEET

6. The pH of acid rain is

(1) 7.0 (2) Less than 5.6
(3) Between 5.6-7.0 (4) Between 7.0-9.0
7. High concentration of Pb in human blood damages
(1) Kidneys (2) Liver
(3) Reproductive system (4) All of these
8. The culprit for the ozone layer depletion is
(1) Nitric oxide (2) CFCs
(3) UV-rays (4) Both (1) & (2)
9. BOD value less than 5 ppm indicates a water sample, which is
(1) Clean (2) Poor in dissolved oxygen
(3) Not suitable for aquatic life (4) Highly polluted
10. Which of the following is correct match?
(1) Photochemical smog : NO2
(2) Nitrate in drinking water : Blue baby syndrome
(3) Phosphate fertilisers in water : BOD level of water increases
(4) All are correct

Note :
1. Environmental pollution is the effect of undesirable changes in the surroundings that have harmful effects
on living things.
2. Tropospheric pollution is basically due to various oxides of sulphur, nitrogen, carbon, halogens etc.
3. pH of drinking water should lie between 5.5 – 9.5.
4. Clean water would have a BOD value of less than 5 ppm whereas highly polluted water could have a
BOD value of 17 ppm or more.
5. NaClO3 is used as herbicide.
6. Freons which cause damage to ozone layer are chemically unreactive, non-toxic and odourless.
7. The utilization of the existing knowledge and practices so as to bring about reduction in the production
of pollutant is called green chemistry.
8. Environmental pollution : Contamination of the environment with harmful wastes arises mainly from
certain human activities.
9. Primary pollutant : A pollutant which exists in the form in which it is produced in environment.
10. Secondary pollutant : A pollutant which is obtained on reaction of primary pollutant.
11. Ozone hole : Hole created due to depletion in the concentration of ozone over a restricted area as
over Antarctica is called ozone hole.
12. Smog : Opaque or dark fog having smoke, dust, water vapours and gases is called smog.
13. Eutrophication : Presence of extra nutrients in water bodies brings about dense growth of plants and
animals life is called eutrophication.
14. Water pollution : Degradation of quality of water due to addition of substances, leads to deprivation
that makes it health hazard, unfit for human use and growth of aquatic biota is called water pollution.
15. Greenhouse gases : The gases which absorbs some of the solar radiations and reflect back long wave
heat radiations which are trapped in atmosphere are called greenhouse gases and causes global
warming.
16. Soil pollution : It is alteration in soil caused by removal or addition of substances and factors which
decreases its productivity, quality of plants and ground water.
17. Green chemistry : A new branch of chemistry, which utilizes the existing knowledge and practices
so as to bring about reduction in the production of pollutant is called green chemistry.

‰ ‰ ‰
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 t
en
nm nment
s s igAs s ig
A ssignment

Assignment
A

8. The upper limit of lead in drinking water is

SECTION - A
[NCERT Pg. 416]
NCERT Based MCQs
(1) 50 ppb (2) 100 ppb
1. Which of the following is not a gaseous pollutant?
(3) 200 ppb (4) 500 ppb
[NCERT Pg. 407] catalyst
9. CH2  CH2  O2 
Pd(III)/CuII(in water )
 A (Major)
(1) Hydrocarbon (2) Ozone
A is [NCERT Pg. 420]
(3) Oxide of carbon (4) Smog
2. The irritant red haze in the traffic is due to the oxides (1) C2H6 (2) CH3CH2OH
of [NCERT Pg. 407] (3) CH3CHO (4) CH3COOH
(1) Carbon (2) Sulphur 10. Green chemistry involves reduction in
(3) Nitrogen (4) Lead [NCERT Pg. 420]
3. Which of the following is/are correct regarding CO? (1) Material (2) Energy consumption
[NCERT Pg. 408] (3) Waste generation (4) All of these
(1) Mainly released into air by automobile exhaust 11. Which of the following is used for bleaching of paper
(2) It is poisonous nowdays? [NCERT Pg. 420]
(3) It binds to haemoglobin (1) Cl2 (2) SO2
(4) All of these (3) H2O2 (4) CO2
4. Which of the following is not true? 12. Which of the following is/are true regarding “green
fuel”? [NCERT Pg. 418]
[NCERT Pg. 411]
(1) Obtained from plastic waste
(1) Classical smog occurs in cool humid climate
(2) Contains no lead
(2) Classical smog is reducing smog
(3) High octane rating
(3) Classical smog is due to hydrocarbon and
nitrogen dioxide (4) All of these
(4) Classical smog is a mixture of smoke, fog and 13. Which of the following is transporting agents for
sulphur dioxide chlorine radicals into stratosphere?
5. Which of the following is of brown colour? [NCERT Pg. 413]
[NCERT Pg. 412] (1) CFCs (2) O3
(1) NO (2) NO2 (3) O2 (4) CO2
(3) O3 (4) CH4 14. Marble of Taj Mahal is getting discoloured and
6. Upper stratosphere consists of considerable amount lusterless due to formation of [NCERT Pg. 410]
of [NCERT Pg. 413] (1) HCl (2) H2SO4
(1) O3 (2) CO2 (3) CO2 (4) CH4
(3) CO (4) All of these 15. Which of the following pollutants are difficult to
7. Eutrophication is caused by [NCERT Pg. 415] remove? [NCERT Pg. 406]
(1) Toxic heavy metals (2) Sediments (1) DDT (2) Plastic waste
(3) Pesticides (4) Fertilizers (3) Nuclear waste (4) All of these

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134 Environmental Chemistry NEET

10. Clean water has BOD value

SECTION - B
(1) Less than 5 ppm
Objective Type Questions (2) More than 10 ppm
(3) Between 20 and 30 ppm
1. Which of the following component of
photochemical smog is a powerful eye irritant? (4) Equal to 22 ppm
11. Normally rain water has a pH value of
(1) PAN (2) NO
(1) 4.5 (2) 5.6
(3) HCHO (4) CO
(3) 6.2 (4) 8
2. Which of the following species if present in drinking
12. The reason of ozone layer depletion is due to
water at high concentration causes brown mottling
of teeth? (1) Freons (2) Perhydrol
(3) CFC (4) Both (1) & (3)
(1) NO3– ion (2) F– ion
13. Which of the following ions in excessive amount in
(3) Pb (4) SO42– ion
water causes laxative effect?
3. Excess nitrate (more than 50 ppm) in drinking
water can (1) F– (2) SO24
(1) Cause methemoglobinemia
(3) NO3 (4) Pb+2
(2) Damage kidney
14. Fuel obtained from plastic waste is called
(3) Cause laxative effect
(1) Biodegradable fuel
(4) Cause eutrophication
(2) Polymeric fuel
4. Which of the following are the most dangerous
(3) Green fuel
water pollutants?
(4) Lead fuel
(1) Organic wastes (2) Pathogens
15. Range of stratosphere is
(3) Toxic heavy metals (4) Plant nutrients
(1) 0 – 10 km (2) 50 – 90 km
5. For which of the following pH of rain water is called (3) 10 – 50 km (4) 10 – 100 km
acid rain?
16. If BOD of a river is high it means that the river is
(1) 4.5 (2) 7 (1) Not polluted
(3) 6 (4) 6.5 (2) Very much polluted with inorganic chemicals
6. CO2 exists in which part of the atmosphere? (3) Very much polluted with organic chemical
(1) Troposphere (2) Stratosphere which are decomposed by microorganism
(4) Polluted with pesticides
(3) Ionosphere (4) Mesosphere
17. The primary precursors of photochemical smog are
7. Which gas is responsible for global warming?
(1) SO2 and NO2
(1) CO (2) NO
(2) SO2 and hydrocarbons
(3) CO2 (4) N2 (3) NO2 and hydrocarbons
8. Solar energy reaching the earth is absorbed by the (4) NO2 and PAN
earth surface is 18. Which one of the following statements regarding
(1) 50% (2) 45% photochemical smog is not correct?
(3) 90% (4) 75% (1) CO does not play any role in photochemical
smog formation
O (2) Photochemical smog is oxidising in character
9. The compound CH3 – C – O – O – NO2 is called (3) Photochemical smog is formed through
photochemical reaction involving solar energy
(1) PAN (2) Acrolein
(4) Photochemical smog does not cause irritation
(3) NAP (4) Picric acid in eye and throat

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NEET Environmental Chemistry 135
19. Ozone layer is present in 28. Which of the following is a secondary pollutant?
(1) Troposphere (2) Stratosphere (1) CO2 (2) N2O
(3) Mesosphere (4) Thermosphere (3) SO2 (4) PAN
20. Which of the following is not an air pollutant? 29. Which one of the following is herbicide?
(1) CO (2) SO2 (1) Organochlorides (2) Sodium arsenite
(3) NO (4) N2 (3) Organophosphates (4) Carbamates
21. Maximum permissible limit of lead in drinking water 30. The gas leaked from a storage tank of the union
is carbide plant in Bhopal gas tragedy was
(1) 50 ppb (2) 50 ppm (1) Ammonia (2) Phosgene
(3) 500 ppm (4) 500 ppb (3) Methyl isocyanate (4) Methylamine
22. Green chemistry means such reaction which
SECTION - C
(1) Are related to the depletion of ozone layer
(2) Produce colour during reactions Previous Years Questions
(3) Study the reaction in plant 1. The liquified gas that is used in dry cleaning along
(4) Reduce the use and production of hazardous with a suitable detergent is [NEET-2019 (Odisha)]
chemicals
(1) CO2
23. High concentration of fluoride is poisonous and
harmful to bones and teeth at level over (2) Water gas
(1) 1 ppm (2) 3 ppm (3) Petroleum gas
(3) 5 ppm (4) 10 ppm (4) NO2
24. London smog is found in 2. Among the following, the one that is not a green
(1) Summer during day time house gas is [NEET-2019]
(2) Summer during morning time (1) Nitrous oxide (2) Methane
(3) Winter during morning time
(3) Ozone (4) Sulphur dioxide
(4) Winter during day time
3. Which oxide of nitrogen is not a common pollutant
25. Which of the following is not a greenhouse gas? introduced into the atmosphere both due to natural
(1) Methane (2) Ozone and human activity? [NEET-2018]
(3) Carbon dioxide (4) Nitrogen (1) N2O5 (2) NO2
26. The process of ‘eutrophication’ is due to (3) NO (4) N2O
(1) Increase in concentration of insecticide in water
4. Which of the following is a sink for CO?
(2) Increase in concentration of fluoride ion in water [NEET-2017]
(3) The reduction in concentration of the dissolved
(1) Haemoglobin
oxygen in water due to phosphate pollution in
water (2) Micro-organisms present in the soil
(4) Increase in concentration of radioactive (3) Oceans
substance in water
(4) Plants
27. Identify the wrong statement in the following
5. Which one of the following is not a common
(1) Ozone layer does not permit infrared radiation
component of Photochemical Smog?
from the sun to reach the earth
(2) Acid rain is mostly because of oxides of [AIPMT-2014]
nitrogen and sulphur (1) Ozone
(3) Chlorofluorocarbons are responsible for ozone
(2) Acrolein
layer depletion
(4) Greenhouse effect is responsible for global (3) Peroxyacetyl nitrate
warming (4) Chlorofluorocarbons
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136 Environmental Chemistry NEET

6. Which one of the following statements regarding 11. BOD5 is

photochemical smog is not correct ?
(1) Waste decomposed in 5 days
[AIPMT (Prelims)-2012]
(2) Oxygen used in 5 days
(1) Photochemical smog is formed through
(3) Microorganisms killed in 5 days
photochemical reaction involving solar energy
(4) Dissolved oxygen left after 5 days
(2) Photochemical smog does not cause irritation
in eyes and throat 12. Ozone in the stratosphere is deleted by

(3) Carbon monoxide does not play any role in (1) CF2Cl2
photochemical smog formation (2) C7F16
(4) Photochemical smog is an oxidising agent in (3) C6H6Cl6
character
(4) C6F6
7. Which one of the following statement is not true ?
13. Which one of the following is responsible for depletion
[AIPMT (Prelims)-2011] of the ozone layer in the upper strata of the
(1) Oxides of sulphur, nitrogen and carbon are the atmosphere?
most widespread air pollutant (1) Polyhalogens
(2) pH of drinking water should be between (2) Ferrocene
5.5 – 9.5
(3) Fullerenes
(3) Concentration of DO below 6 ppm is good for
the growth of fish (4) Freons

(4) Clean water would have a BOD value of less 14. The smog is essentially caused by the presence of
than 5 ppm (1) O2 and O3
8. Green chemistry means such reactions which (2) O2 and N2
[AIPMT (Prelims)-2008] (3) Oxides of sulphur and nitrogen
(1) Study the reactions in plants (4) O3 and N2
(2) Produce colour during reactions 15. Pick up the correct statement :
(3) Reduce the use and production of hazardous (1) CO which is a major pollutant resulting from
chemicals the combustion of fuels in automobiles plays a
major role in photochemical smog
(4) Are related to the depletion of ozone layer
(2) Classical smog has an oxidizing character
Questions asked Prior to Medical Ent. Exams. 2005 while the photochemical smog is reducing in
9. Which of the following is a secondary pollutant? character

(1) CO2 (3) Photochemical smog occurs in day time

whereas the classical smog occurs in the
(2) N2O morning hours
(3) SO2 (4) During formation of smog the level of ozone in
(4) PAN the atmosphere goes down

10. ‘White lung cancer’ is caused by 16. In Antartica, ozone depletion is due to the formation
of the following compound
(1) Asbestos
(1) Acrolein
(2) Silica
(2) Peroxy acetyl nitrate
(3) Textiles
(3) SO2 and SO3
(4) Paper (4) Chlorine nitrate
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NEET Environmental Chemistry 137
17. Which of the following is not a greenhouse gas? 6. Hydrogen sulphide is considered as
(1) CO2 (1) Water pollutant (2) Soil pollutant
(2) CH4 (3) Air pollutant (4) None of these
(3) O3 7. Which of the following statements is/are correct?
(4) N2 (1) Bhopal gas tragedy was caused by methyl
18. High concentration of fluoride is poisonous and isocyanate
harmful to bones and teeth at levels over (2) pH of drinking water should lie between 5.5 –
(1) 1 ppm (2) 3 ppm 9.5
(3) pH of acid rain is approximately 6
(3) 5 ppm (4) 10 ppm
(4) All of these
19. Nitrogen dioxide and sulphur dioxide have some
properties in common. Which property is shown by 8. Which of the following is not a property of
one of these compounds, but not by the other? photochemical smog?

(1) Is used as a food-preservative (1) Photochemical smog is an oxidising agent in

character
(2) Forms 'acid-rain'
(2) Photochemical smog does not cause irritation
(3) Is a reducing agent in eyes and throat
(4) Is soluble in water (3) It is also called Los Angeles smog
(4) It contains O3, PAN and NOx
SECTION - D
9. Which of the following is/are greenhouse gases?
NEET Booster Questions (1) CO2
1. An increase in CO 2 concentration in the (2) CFCs
atmosphere will result in (3) N2O
(1) Adverse effect of natural vegetation
(4) All of these
(2) Global warming
10. Which of the following is the incorrect match?
(3) Temperature decrease in global atmosphere
(1) Classical smog – Reducing smog
(4) Genetic disorder in plants and animals
(2) Ozone layer – Troposphere
2. Photochemical smog is related to the pollution of
(3) Fluoride – In excess, causes harmful effect to
(1) Soil (2) Water
bones
(3) Noise (4) Air
(4) Carbon monoxide– Carboxyhaemoglobin
3. When huge amount of sevage is dumped into a
river, the BOD will 11. Which of the following statements is incorrect?
(1) Increase (2) Remain unchanged (1) When huge amount of sewage is dumped into
a river, the BOD will increase
(3) Slightly decrease (4) Decrease
4. Green house effect is related to (2) Excess nitrate in water causes ‘blue baby
syndrome’
(1) Cultivation of green plants
(3) Clean water has maximum BOD value
(2) Cultivation of vegetables in houses
(3) Global warming (4) Methane, water vapour and CFCs are
greenhouse gases
(4) Global green algae
12. Bhopal gas tragedy was caused by
5. Indiscriminate use of DDT is undesirable because
(1) MIC
(1) It is harmful
(2) It is degradable (2) TEL

(3) It causes mutation (3) DDT

(4) It is accumulated in food chain (4) BHC

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138 Environmental Chemistry NEET

13. Which of the following is/are correct for ‘acid rain’? 21. Which of the following is not caused due to SO2?
(1) Acid rain has pH less than 5.6 (1) Bronchitis
(2) It threatens human and aquatic life and (2) Shortness of breath
reduces agricultural productivity
(3) Emphysema
(3) Oxide of sulphur and nitrogen dissolve in water
(4) Blindness
and causes acid rain
22. Which of the following has no adverse effect on
(4) All of these
human health?
14. The maximum limit of nitrate in drinking water is
(1) N2 (2) NO2
(1) 500 ppm (2) 5000 ppm
(3) SO2 (4) CO
(3) 50 ppm (4) 103 ppm
23. Which of the following is Freon-11?
15. Which of the following is a herbicide?
(1) CCl3F
(1) NaClO3 (2) DDT
(3) Aldrin (4) All of these (2) CCl2F2

16. Formation of photochemical smog does not involve (3) CClF3

(1) O3 (2) Hydrocarbons (4) CF4

(3) NO2 (4) SO2 24. Which of the following is the correct match?
17. Freon used as refrigerant is (1) Classical smog – NO2
(1) CF2 = CF2 (2) CH2F2 (2) Photochemical smog – SO2
(3) CCl2F2 (4) CF4 (3) Greenhouse gas – CH4
18. Which one of the following statements regarding (4) Both (1) & (3)
photochemical smog is not correct?
25. Which group correctly represent the photochemical
(1) Photochemical smog is formed through smog produced during reaction of hydrocarbon in
photochemical reaction involving solar energy the combustion chamber of an engine?
(2) Photochemical smog does not cause irritation
(1) Formaldehyde and PAN
in eyes and throat
(2) Acrolein and PAN
(3) Carbon monoxide does not play any role in
photochemical smog formation (3) Formaldehyde and acrolein
(4) Photochemical smog is an oxidising agent in (4) Formaldehyde, acrolein and PAN
character
26. Freons which cause damage to ozone layer are
19. About 20 km above the earth, there is an ozone
(1) Chemically unreactive
layer. Which one of the following statement about
ozone and ozone layer is true? (2) Non-toxic
(1) It is beneficial to us as it stops U.V. radiation (3) Odourless
(2) Conversion of O3 to O2 is an endothermic (4) All of these
reaction
27. Ozone depletion over Antarctica is maximum in the
(3) Ozone has a triatomic linear molecule month of
(4) It is harmful as it stops useful radiation (1) September and October
20. Which one of the following is responsible for (2) March and April
depletion of the ozone layer in the upper strata of
the atmosphere? (3) June and July
(1) Polystyrene (4) December and January
(2) Ferrocene 28. Which of the following is a particulate pollutant?
(3) Fullerenes (1) O3 (2) SO2
(4) Freons (3) Dust (4) NO2

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NEET Environmental Chemistry 139
29. Which one of the following statements is not true? 37. Greenhouse gases
(1) Oxides of sulphur, nitrogen and carbon are (1) Allow shorter wavelength to enter earth’s
some of the most widespread air pollutants atmosphere while doesn’t allow longer
wavelength to leave the earth’s atmosphere
(2) pH of drinking water should be between
5.5 – 9.5 (2) Allow longer wavelength to enter earth
atmosphere while doesn’t allow shorter
(3) Concentration of DO below 6 ppm is good for
wavelength to leave the surface
the growth of fish
(3) Don’t have wavelength specific character
(4) Clean water would have a BOD value of less
than 5 ppm (4) Near to earth show wavelength specific
behaviour while far from earth have wavelength
30. At which of the following pH, rain is called as acid
independent behaviour
rain?
38. Heat pollution affects mainly
(1) 4 (2) 6
(1) Vegetation
(3) 6.8 (4) All of these
(2) Aquatic creature
31. Measurement of dissolved oxygen that would be
needed by the microorganism to oxidise the (3) Rocks
organic matter is known as (4) Air
(1) Acid rain 39. The layer of atmosphere which extends upto 10 km
(2) BOD from sea level is called

(3) COD (1) Troposphere (2) Stratosphere

(4) Ozone layer depletion (3) Ionosphere (4) Mesosphere

32. At which of the following concentration of sulphate 40. Troposphere pollutant contains
it will have a laxative effect? (1) Dust (2) Water vapours
(1) Above 300 ppm (3) Mist (4) All of these
(2) Above 500 ppm 41. Which one of the following is a secondary
(3) Above 400 ppm pollutant?
(4) Above 600 ppm (1) CO (2) O3
33. Which of the following is herbicide? (3) Pb (4) CH4
(1) DDT (2) (CH3)2Hg 42. Which one of the following pollutants causes
(3) NaClO3 (4) C2H2 irritation in eye?

34. Bhopal gas tragedy was caused by (1) CO2 (2) SO2

(1) CH3CN (3) NO2 (4) All of these

(2) CH3NC 43. A primary pollutant is
(3) CH3NCO (1) CO (2) DDT
(4) CH3NO (3) CO2 (4) All of these
35. DDT stands for 44. Which one of the following is colourless and highly
(1) o, o– Dichlorodiiodotrichloroethane toxic gas which reduces oxygen carrying capacity
of blood?
(2) o, m – Dichlorodiphenyltrichloroethane
(1) SO2 (2) CO
(3) p, p – Dichlorodiphenyltrichloroethane
(3) NO (4) SO3
(4) o, o – Dichlorodibromotrichloroethane
45. Which one of the following leads to stiffness of
36. Which of the following is not a greenhouse gas? flower buds?
(1) CO2 (2) CH4 (1) Sulphur dioxide (2) Nitrogen oxides
(3) CCl2F2 (4) O2 (3) Carbon monoxide (4) Both (1) & (2)
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140 Environmental Chemistry NEET

46. Which one of the following oxides produces brown 57. Which one of the following gases is known as
air or reddish brown haze? chemical weed in troposphere?
(1) Oxides of sulphur (2) Oxides of nitrogen (1) C2H6 (2) SO2
(3) Oxides of carbon (4) All of these (3) O3 (4) NO2
47. Which one of the following gaseous pollutants is 58. Depletion of ozone layer causes
carcinogenic?
(1) Increased transpiration
(1) Oxides of sulphur (2) Oxides of nitrogen
(2) Damages paints and fibres
(3) Arsenic (4) Oxides of carbon
(3) Causes aging of skin
48. Photochemical smog involves
(1) O3 (2) NO (4) All of these

(3) NO2 (4) All of these 59. pH of acid rain is approximately

49. Select viable pollutant. (1) 5.6 (2) 6.2
(1) Smoke (2) Moulds (3) 6.7 (4) 7
(3) Mist (4) Fumes 60. Which is not a cause of acid rain?
50. Which one of the following is not a property of (1) NO2 (2) CH4
classical smog?
(3) SO2 (4) SO3
(1) Occurs at low temperature
61. Acid rain causes
(2) Secondary pollutant plays significant role
(1) Chlorosis
(3) Contains SO2
(2) Increasing acidity of soil
(4) Reducing smog
(3) Corroding bridges
51. Photochemical smog always contains
(1) SO2 (2) O3 (4) All of these

(3) HNO3 (4) All of these 62. Interference in development and maturation of RBC
is caused by
52. Photochemical smog is actually a cause of
pollution of (1) Hg pollution (2) Pb pollution

(1) Soil (2) Air (3) S pollution (4) All of these

(3) Noise (4) Water 63. Which causes water pollution?
53. Photochemical smog includes the reduction of (1) Jet planes
(1) SO2 (2) NO (2) Herbicides
(3) Cl2 (4) NO2 (3) Smoke
54. Earth is protected from UV rays by (4) Combustion of fossils
(1) N2 (2) O2 64. Most common and commercial water pollutant is
(3) O3 (4) SO3 (1) Industrial waste (2) Fertilizers
55. Months in which ozone hole begins to appear over
(3) Detergents (4) PCB’s
poles is
(1) June (2) September 65. For clean water BOD is less then

(3) March (4) January (1) 17 ppm (2) 12.2 ppm

56. Depletion of ozone layer can not cause (3) 7 ppm (4) 5 ppm

(1) Damage of DNA 66. Permissible level of sulphate ions in the drinking
water is
(2) Skin cancer
(3) Cataract (1) 40 ppm (2) 500 ppm

(4) Rickets (3) 60 ppm (4) 70 ppm

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NEET Environmental Chemistry 141
67. Tooth decay is caused by deficiency of 77. Non-biodegradable industrial waste that can be
(1) Fluorides (2) Lead utilized by cement industry is

(3) Sulphates (4) Nitrates (1) Fly ash

68. Cause of eutrophication is (2) Slag
(1) Nutrient enrichment of water bodies (3) Gypsum
(2) Increase of oxygen in water bodies (4) Both (1) & (2)
(3) Increase in number of aquatic organisms 78. Small quantity of industrial toxic waste is removed by
(4) All of these (1) Burning in open bins
69. Inorganic pollutant can damage (2) Controlled incineration
(1) Central nervous system (3) Burning in closed bins
(2) Liver (4) Both (1) & (2)
(3) Kidneys 79. Select non-biodegradable industrial waste
(4) All of these (1) Waste from cotton mills
70. Decomposition of organic matter through microbial (2) Waste from paper mills
activity is known as
(3) Fly ash from thermal power plants
(1) Eutrophication (2) Stratification
(4) Waste from food processing units
(3) Putrescibility (4) BOD
80. Which one of the following is not a non-
71. Laxative effect is caused by biodegradable industrial waste?
(1) Nitrates (2) Sulphates (1) Gypsum from fertilizer industries
(3) Lead (4) Fluoride (2) Mud and tailings from metal industries
72. Soil pollution can result in decreased (3) Slag from steel industries
(1) Soil productivity (4) Waste from textile industries
(2) Quality of plants
81. Biodegradable wastes are mixed with water and
(3) Purity of ground water cultured from bacterial species which produces a
(4) All of these gas called

73. Which one of the following is not an organochlorine? (1) Methane gas

(1) Parathion (2) DDT (2) Ethane gas

(3) Propane gas
(3) BHC (4) Aldrin
74. Which one of the following is least persistent (4) Butane gas
insecticide? 82. Sewage sludge contains least amount of
(1) Aldrin (2) Dieldrin (1) Nitrogen (2) Phosphorus
(3) BHC (4) Carbamates (3) Potassium (4) Sulphur
75. Which one of the following is a herbicide? 83. Greenhouse gases
(1) Organochlorines (2) Organophosphates (1) Trap UV rays
(3) Carbamates (4) Sodium arsinite (2) Increase temperature of atmosphere
76. Those pollutants which are readily degradable by (3) Prevent global warming
natural process are called
(4) All of these
(1) Biodegradable pollutant
84. Greenhouse gas other than CO2 is
(2) Non-biodegradable pollutant
(1) H2 (2) N2O
(3) Primary pollutant
(4) Secondary pollutant (3) O2 (4) SO2

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142 Environmental Chemistry NEET

85. One of the greenhouse gases is 87. In place of tetrachloroethene in dry-cleaning we use
(1) CO (1) Sodium lauryl sulphate
(2) O2 (2) Sodium stearate
(3) H2
(3) Hydrogen peroxide
(4) CFCs
(4) Sodium bisulphite
86. Green chemistry emphasizes on
88. Chauvin, Grubbs and Schrock won 2005 Nobel
(1) Reaction that takes place in non-aqueous
prize for the development of
medium
(1) Dumas method
(2) Utilization of sciences to reduce adverse impact
on environment (2) Kjeldahl’s method
(3) Use of condensation polymers (3) Metathesis method
(4) Minimise the use of non-hazardous waste (4) Carius method

‰ ‰ ‰

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Chapter 11

The Solid State

Chapter Contents
z General Characteristics of GENERAL CHARACTERISTICS OF SOLID STATE
Solid State
In solid state, the constituent particles i.e., atoms, molecules or ions
z Space Lattice or Crystal
are closely packed and are held together by strong intermolecular
Lattice
attractive forces.
z Closed Packed Structures
Solids have
z Density of Unit Cell 1. Definite shape and volume
z Calculations of Packing 2. High density and low compressibility
Fractions
3. Less vapour pressure
z Radius Ratio
Classification of Solids
z Structure of Ionic Crystals
(A) Classification on the basis of packing of constituents in
z Imperfection in Solids lattice.
z Properties of Solids
Properties Crystalline solids Amorphous solids
1. Crystal geometry These have a definite crystal These do not have a definite
shape due to orderly geometrical shape due to haphazard
arrangement of particles, in arrangement of constituent particles.
three dimensions.
2. Melting point These show sharp m.p. These do not give sharp m.p.
3. Physical state These are hard and rigid solids. These are comparatively soft and not
very rigid.
4. Symmetry Crystalline solids can have Amorphous solids do not have any
(a) plane of symmetry symmetry.
(b) centre of symmetry
(c) axis of symmetry.
5. Crystal system Crystalline solids have a These do not have any crystal
definite crystal system e.g., system i.e. these do not have a
cubic, tetragonal, hexagonal, regular repeating units.
etc.
6. Anisotropic or Crystalline solids show some of Their physical properties are same in
Isotropic nature the physical properties different all directions. Thus these are termed
in different directions of the as isotropic.
crystal e.g. refractive index and
electrical conductivity. So these
are termed as anisotropic.
7. Examples Crystals of NaCl, CsBr, CaF2 Rubber, plastic, glass, etc.
and ZnS.

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(B) Classification of crystalline solids on the basis of nature of intermolecular forces.

Property Ionic crystals Covalent crystals Metallic Crystals Molecular crystals

1. Nature of Positve and Neutral atoms Positive kernel in Molecules

constituent negative ions sea of electrons polar or nonpolar
particles
2. Binding force Electrostatic Shared pair of Electrostatic Van der-Waals' forces
between force of electrons attraction between of attraction
ions/molecules attraction positive ions and
electrons
3. Melting and High Very high Moderate to high Low
Boiling point
4. Electrical Bad conductor Bad conductors Good conductors Bad conductors
conductivity in solid state but
good conductors
in fused state
5. Hardness Hard and brittle Very hard Hard or soft Very soft
6. Examples KNO3, ZnS, Diamond, graphite, Ni, Mn, Cu, Na Ice, solid, CO2
NaCl, CaF2, KI carborundum(SiC) and Fe metals (Dry ice)

SPACE LATTICE OR CRYSTAL LATTICE

z All crystals have a regular repetition of some constituent particles (atoms, molecules or ions).
z To represent the relative arrangement of these particles in a crystal, each particle is considered as a point.
z The position taken by these points in the three dimensional crystal are called the lattice points or lattice
sites.
z The representation of a crystal in which the location of the ions or atoms etc. are shown by lattice points
is called crystal lattice or space lattice.
z Therefore, the space lattice or crystal lattice can be defined as an array of lattice points showing
the arrangement of constituent particles in different positions in three dimensional space.
Y
Lattice points

c
b
O
a X

Z
Space lattice and lattice points
z In a space lattice, a certain group of lattice points set the pattern for the whole lattice. The smallest group
of lattice points is called a unit cell. Therefore, the unit cell can be defined as the smallest portion of
the crystal lattice which defines completely the repeating pattern in the crystal in all directions.

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Crystal System : On basis of geometrical consideration, seven type of crystal systems are there which differ
three-dimensionally in axial edge lengths (a, b and c) and axial angles (, , ).

Name of Crystal Possible Edge Inter-axial Example

System Variations Lengths Angles

1. Cubic (a) Primitive a=b=c  =  =  = 90° Ag, Cu, Zinc blende

(b) Body-Centred NaCl, KCl, diamond,
(c) Face-Centred Pb, Ag, CaF2

2. Tetragonal (a) Primitive a=bc 90° CaSO4, white tin, TiO 2,

(b) Body-Centred SnO2

3. Orthorhombic (a) Primitive a bc 90° Rhombic Sulphur,

(b) Body-Centred KNO3, PbCO3, CaCO3,
BaSO4, K2SO4,
(c) Face-Centred
Epsom salt (MgSO4.7H2O)
(d) End-Centred

4. Monoclinic (a) Primitive a bc 90° Monoclinic Sulphur,

Na2SO4.10H2O,
(b) End-Centred 90°
PbCrO4 ,
Gypsum (CaSO4.2H2 O)

5. Triclinic Primitive a bc  90° K2 Cr2O7, CuSO4.5H2O

H3 BO3

6. Hexagonal Primitive a=bc  =  = 90° Graphite, ZnO, CdS,

 = 120°

7. Rhombohedral Primitive a=b=c 90° Calcite (CaCO3 ), Cinnabar

or Trigonal (HgS),
Quartz, Sb, KMnO4

Types of the cubic unit cell :

z Cubic units cells are of three types. This is because of the positions taken by lattice points other than
the corners are faces and also the body of the cube :

(a) (b) (c)

(a) Simple cubic, (b) Face centered, (c) Body centered.

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Number of Particles Per Unit Cell

Number of atoms Total number Relation between

Type of unit cell of atoms per atomic radius
at corners within body on faces unit cubic cell (r) and edge
length (a)
1 a
1. Simple cube ×8=1 0 0 1 r
8 2
1 3
2. Body centred cube (BCC) ×8=1 1 0 2 r a
8 4
1 1 a
3. Face centred cube (FCC) ×8=1 0 ×6=3 4 r
8 2 2 2
Contribution of atom / ion / molecule at
1
z Corner = per unit cell
8
1
z Edge centre = per unit cell
4
1
z Face centre = per unit cell
2
z Body centre = 1 per unit cell

Example 1 : A compound is made up of two elements A & B, having cubic structure. Atoms A occupy corners
as well as face centres whereas atoms B occupy edge centre as well as body centre. What is
formula of the compound ?
Solution : Number of atoms of 'A'
1
(a) From 8 corners = 8 1
8
1
(b) From 6 faces =  6  3 ,  Total atoms of A = 3 + 1 = 4
2
Number of atoms of ‘B’
(a) From body centre = 1
1
(b) From edge centres =  12  3
4
Total = 4
A : B = 4 : 4 or 1 : 1
Hence, formula of compound = AB

Example 2 : A unit cell consists of a cube in which X atoms are at the corners and Y atoms are at the face
centres. If two atoms are missing from two corners of the unit cell, what is the formula of the
compound?
1 3
Solution : Total contribution of 'X' atoms from 6 corners = 6
8 4
Number of atoms of Y from face centres = 3
3
x:y= : 3 = 3 : 12 or 1 : 4
4
Hence, formula is XY4.

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NEET The Solid State 133

EXERCISE
1. Which one of the following property is attributed to amorphous solid ?
(1) Isotropy (2) Anisotropy
(3) Sharp melting point (4) Long range order
2. Which of the following covalent solid is the conductor of electricity ?
(1) SiC (2) AlN
(3) Diamond (4) Graphite
3. Which of the following is the characteristic of a crystalline solid?
(1) Sharp melting point (2) Short-range order
(3) Irregular arrangement (4) Indefinite heat of fusion
4. Which of the following is an example of hydrogen bonded solid ?
(1) Water (2) CO2(s)
(3) H2O(s) (4) SO2(s)
5. ‘Glass’ is an example of
(1) Pseudo solid (2) True solid
(3) Super cooled liquid (4) Both (1) & (3)
6. Which of the following set of solids contains only ionic solids ?
(1) NaCl, MgO & ZnS (2) CaF2, ZnS, SiO2
(3) MgO, SiO2, KCl (4) SiO2, SiC, AlN
7. When constituent particles are present only at the corners of a cubic unit cell is called
(1) Primitive unit cell (2) FCC unit cell
(3) BCC unit cell (4) All of these

CLOSE PACKED STRUCTURES

Packing in Metallic Crystals :
z In the formation of crystals, the constituent particles may be either atoms, ions or molecules. These
constituent particles may be of different size and so it may give different mode of packing of particles
in the crystal.
z The actual mode of packing of these particles in the crystal is determined experimentally by X-ray
diffraction method.
Two Dimensional Packing :
z The spheres in a single plane can have two different modes of arrangement as shown in the figure.

Square packing Hexagonal packing

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z Square Packing : Each sphere in this mode of packing has four other spheres in its contact. Its packing
fraction is 78.5 %. In two dimensional square close packing C.N of every sphere is 4.
z Hexagonal Packing : In this mode of packing, the packing fraction is 90.6% (in 2-D). C.N of every sphere
is 6.
Close Packing in three Dimension :
When layers are arranged over each other they form three- dimensional packing.
(1) Three Dimensional Close packing from Two Dimensional Square Close Packed Layers : It is layer
packing in which second layer is placed over first layer in such a way that all the spheres are exactly
above each other and all the spheres align horizontally as well as vertically. This arrangement forms
AAA... type of lattice. It forms simple cubic lattice and its unit cell is primitive cubic unit cell.

(a) (b)
Simple cubic lattice formed by AAA... arrangement
(2) Three Dimensional Close Packing from Two Dimensional Hexagonal Close Packed Layers: When
layers containing hexagonal close packing are arranged over each other, two types of arrangements are
feasible.
(i) Arrangement of 2nd Layer over the First Layer : Start with a two-dimensional hexagonal close
packed layer 'A' and arrange another similar layer B on it in such a way that spheres of 2nd layer are
placed in the depressions of first layer. In this case two types of voids are formed.

A layer
a a a
a a a A layer b b
b b B layer
c c
a a a a
b b b b b b

(a)
(b)
Packing of second layer (B) on first layer (A)

(ii) Formation of 3rd layer over the 2nd layer : There are two types of voids which are to be covered
in the third layer. These are the octahedral voids (a) which remain unoccupied for two consecutive
layers and tetraheral voids (c) formed in the second layer.
If third layer is formed in such a way that tetrahedral voids (c) are covered. In this way the spheres
of the third layer lie directly above those in first layer. It means third layer becomes exactly identical
to the first layer. This type of packing is ABABAB ...... arrangement and it is known as hexagonal
close packing (hcp)
e.g., Mg, Zn, Cd, Be etc.

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A A

B B

A
A
(a) hcp arrangement (b)
ABAB... or hcp arrangement of spheres
If third layer is formed in such a way that spheres of third layer must cover octahedral
voids (a). It forms a new third layer C. It forms ABCABC ... type arrangement called cubic closed
packing (ccp). In ccp each unit cell is face-centred type. e.g. Ag, Cu, Fe, Ni, Pt etc.
A
A
B
C
C B

A A
ccp arrangement
(a) (b) (c)
ABCABC.. or ccp arrangement of spheres
In both the arrangements i.e., hcp and ccp, each lattice point is in contact with 12 more nearest spheres
which is called their coordination number (C.N.)

Number and Position of Tetrahedral and Octahedral Voids :

Number of voids formed depends on the number of close packed spheres.
Let number of close packed spheres = N
The numbers of octahedral voids formed = N
The number of tetrahedral voids formed = 2N
(a) Position of Tetrahedral Voids in ccp or fcc Lattice : Let a unit cell is divided into eight small cubes
in which each small cube has atoms at alternate corners. It means there are 4 atoms in each small cube,
which form a regular tetrahedron when joined together. In this tetrahedron there is a tetrahedral void. In
whole there are 8 tetrahedral voids in all 8 tetrahedrons. Hence, we conclude that number of tetrahedral
voids is double the number of lattice points.
Tetrahedral
void
t t
t t

t
t
t t

(a) Locating tetrahedral void in (b) Eight tetrahedral voids in ccp

ccp struture structure shown
Locating tetrahedral voids in ccp structure

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(b) Position of Octahedral Voids in ccp or fcc Lattice : In a ccp or fcc lattice there are lattice points
occupied at 8 corners and 6 face-centres. At the body centre of the cube there is a point 'O' which is
not occupied. It is surrounded by six atoms on all the 6 face centres. On joining these face-centers an
octahedron is formed. Hence, it is an octahedral void. Simultaneously there are twelve octahedral voids
at twelve edge centres. Hence, we conclude that there are octahedral voids at twelve edge centres and
one is at the body centre.
1
Hence, Octahedral void (O.V) from 12 edge centres =  12  3
4
Octahedral void (O.V) from body centre = 1

Total number of octahedral voids per unit cell = 4

(a) Octahedral void at body (b) 12 octahedral voids at the edges

centre of ccp unit cell in ccp unit cell
Locating octahedral voids in ccp structure

Example 3 : A solid is made of two elements X and Y. Atoms X are in fcc arrangement and Y atoms occupy
all the octahedral sites and alternate tetrahedral sites. What is the formula of compound?

Solution : Atoms of X forming fcc lattice means they occupy 8 corners and 6 face centres hence number
of atoms = 4

Tetrahedral sites = 8 Number of Y atoms = 4

Octahedral sites = 4 Number of Y atoms = 4

 Total number of Y atoms = 8

 X:Y

4 : 8 or 1 : 2. Hence formula of compound is XY2.

Example 4 : Ferric oxide crystallises in a hcp of oxide ions with two out of every three octahedral voids occupied
by ferric ions. Derive the formula of ferric oxide.
Solution : In a closed packed arrangement there is one octahedral site corresponding to each atom
constituting the lattice. Therefore number of oxide ions per unit cell in hcp arrangement is 6.
Number of octahedral voids = 6 × 1 = 6

2
Number of Fe+3 ions = 6  4
3

 Formula of the compound Fe3+ : O–2 or is Fe2O3

4:6
2:3

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DENSITY OF UNIT CELL

Mass of unit cell

Density =
Volume of unit cell
M Z
= g cm–3
a3  10 30  N0
where M is Molar mass, N0 is Avogadro’s number, Z is the number of rigid solid particle per unit cell and 'a'
is the edge length in pm.
Note : Density of substance is same as density of unit cell.

Example 5 : An element having bcc geometry has atomic mass 50. Calculate density of unit cell if its edge
length is 290 pm.
Solution : For bcc geometry, number of atoms per unit cell (Z) = 2
Atomic mass (A) = 50
Edge length (a) = 290 × 10–10 cm
Density () = ?

Z×A 2  50
  = 6.82 g/cm3
 
3
N0 × a 3 6.02  10 23  290  10 –10

Example 6 : An element x with an atomic mass of 60 g/mol has density of 6.23 g/cm3. If the edge length of
the unit cell is 400 pm, identify the type of cubic unit cell. Calculate the radius of an atom of this
element.
Solution : A = 60 g/mol  = 6.25 g/cm3
a = 400 pm or 400 × 10–10 cm
 V = a3 = (400 × 10–10)3
Z=?

 
3
 × N0 × a3 6.25  6.02  1023  400  10–10
Z= =
M 60
 Z = 4. Hence, it is face-centred cubic unit cell.
For fcc, 4r = 2a
2a
 r=
4
1.414  400
 r=  141.4 pm
4

Example 7 : Aluminium crystallizes in a cubic close packed structure. Its metallic radius is 125 pm.
(a) What is the length of side of unit cell ?
(b) How many unit cells are there in 1.00 cm3 of Aluminium ?
2a
Solution : For a. c.c.p length of the side of unit cell is related to radius r =
4
4r 4  125
a= =  353.5 pm
2 1.414
 Volume of unit cell = a3 = (353.3 × 10–10)3 cm3 = 4.42 × 10–23 cm3
1
Number of unit cells in 1cm3 = = 2.26 × 1022 unit cells
4.42  10 –23

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CALCULATIONS OF PACKING FRACTIONS

Volume occupied by spheres

Packing fraction =
Total volume of cube

% Volume occupied = PF × 100

Summary of the Packing of Spheres

Stacking Coordination Space Unit

Structure Pattern Number Used (%) Cell
Simple cubic a-a-a-a- 6 52 Primitive cubic
Body-centered cubic a-b-a-b- 8 68 Body-centered cubic
Hexagonal close-packed ABAB------ 12 74 End-centered hexagonal
Cubic closed-packed ABCABC------- 12 74 Face-centered cubic

a, b are the layers having two dimensional square close packing while A, B are the layers having two
dimensional hexagonal close packing.

RADIUS RATIO

z For the stability of ionic compounds, each cation should be surrounded by maximum number of anion and
vice versa.
z The maximum number of oppositely charged ions surrounding each ion is called coordination number.
Since ionic bonds are non directional the arrangement of ions in crystal is determined by sizes.
z The ratio of the radius of the cation to that of the anion is called radius ratio, i.e

Radius of the cation (r )

Radius Ratio = Radius of the anion (r )
–

z Evidently, greater is the radius ratio, the larger is the size of the cation and hence greater is its
co-ordination number.

z The relationships between the radius ratio and the co-ordination number and the structural arrangement
are called radius ratio rules and are given below :

z The most stable arrangement is one in which the anions are touching each other as well as the cation
simultaneously.

Radius Ratio Co-ordination Type of void Example

No. of cation

r
0.155   0.225 3 Triangular void B2O3
r
r
0.225   0.414 4 Tetrahedral void ZnS
r

r
0.414   0.732
r 6 Octahedral void NaCl

r
0.732  1
r 8 Cubic void CsCl

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STRUCTURE OF IONIC CRYSTALS

AB Type Crystal :

 r 
1. Rock Salt (NaCl) :  0.414   0.732 
 r 
 

rNa 95
  0.5
rCl¯ 181

z The anion (Cl¯) is present at the lattice points of a face centred cubic close structure.

z Na+ ions are occupying all the octahedral voids, so each Na+ ion is surrounded by six Cl¯ ion.

z Since there will be six octahedral voids (holes) around each chloride ion, so each chloride ions is
surrounded by six sodium ions.

z The co-ordination number of Cl¯ as well as of Na + ion is six. Therefore, it is termed as 6 : 6

co-ordination crystal..

z The unit cell of NaCl has contribution of 4 Na+ ions and 4 Cl¯ ions i.e. Z = 4 for NaCl. The sodium chloride
structure is also called rock-salt structure.

z Example having rock-salt structure i.e. NaCl type of structure are

NaCl, LiCl, KBr, RbI, AgCl, AgBr

MgO, CaO, TiO, FeO, NiO

–
Cl
+
Na

 r 
2. Cesium Chloride (CsCl) :  0.732    1
 r 

rCs
 0.93
rCl¯

z The Cl¯ ions are present at the corners of the cubic unit cell and the Cs+ ions are at the body centre
of the unit cell.

z One unit cell contains one Cs+ and one Cl– i.e., Z = 1 for CsCl type crystals.

z Each Cs+ ion in this mode of packing is touching 8 chloride ions and each Cl¯ are touching eight Cs+
ions. Therefore, this structure will have 8 : 8 co-ordination.

z Example: CsBr, CsI, TlBr, TlCl etc.

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 r 
3. Sphalerite - Zinc Blende (ZnS) :  0.225    0.414 
 r 

–2
S

Zn+2

rZn2
= 0.40
rS2
z The sulphide ions (S2-) occupy face centred cubic lattice points and the zinc ions (Zn2+) occupy half of
the total number of tetrahedral voids.
z From the crystal structure of ZnS, it is clear that each sulphide ion is surrounded by 4 Zn2+ ions and
each Zn2+ ions is surrounded by 4 S2- ions. Therefore, ZnS has 4 : 4 co-ordination.
z Thus the number of ZnS units per unit cell is 4.
z The following ionic solids are observed to form ZnS type of crystal structure
ZnS, CuCl, CuBr, CdS, AgI

AB2 and A2B Type of Crystal

1. Fluorite Structure (CaF2) :

z Fluorite structure; Ca++ ions form CCP while F— ions are in all the tetrahedral voids, coordination
ratio 8 : 4.

–
F

+2
Ca

z As shown in the above figure, the Ca2+ ions are present at the face centred cubic lattice and the flouride
ions (F¯) occupy all the tetrahedral voids.
z This shows that the coordination number of calcium ion is eight, i.e. each calcium cation is surrounded
by eight flouride anions in a body centred cubic arrangement. Each flouride ion is in contact with four
calcium ions. Thus CaF2 has 8 : 4 co-ordination.
z Therefore, each unit cell has 4 CaF2 molecules.
Solids forming flourite type of structure are CaF2, SrF2, BaF2, BaCl2, etc.

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2. Antifluorite structure (Na2O)
z O–2 ions are present at CCP, while Na+ lies in all tetrahedral voids.
z Number of ions per unit cell are in the ratio of 8 : 4 for Na+ : O–2.
z Co-ordination ration is in the ratio of 4 : 8 for Na+ : O–2.
z Thus, each O–2 ion is surrounded by 8 Na+ and is present at faces and corner. Similarly, each Na+ is surrounded
by 4O–2 and are present at all tetrahedral voids.

Characteristic properties of various types of ionic solids

Structure Ions forming Ions present Co-ordination Number of Example

the close packed in voids number units per
structure unit cell
2- 2+ 2+
ZnS type S ions form CCP Zn ions in Zn = 4 4 ZnS, AgI, CuCl
2–
(Sphalerite) structure tetrahedral S =4 CuBr, CdS, HgS
void
+ +
NaCl type Cl¯ ions form CCP Na ion in Na = 6 4 NaCl, LiCl, KBr,
structure octahedral Cl¯ = 6 RbI, AgBr, MgO,
holes CsI, CaO, FeO, NiO
+ +
CsCl type BCC structure Cl¯ Cs ion at the Cs = 8 1 CsCl, CsCN,
ions are at the centre of the Cl¯ = 8 CaS, CsI
corners of a cube cube
2+ 2+
CaF2 type Ca ions form F¯ ions in all Ca = 8 4 CaF2, SrF2, BaF2
(Fluorite) CCP structure tetrahedral F¯ = 4 BaCl2
holes
2– + +
Na2O type O ions form CCP Na ions are in Na = 4 4 K2O, Li2O,
2–
(Antifluorite structure all tetrahedral O =8 K2S
structure)

Note : Diamond has similar lattice structure of zinc blende. Its packing efficiency is 34%.
Factors Influencing Crystal Structure :
z Pressure : Increase in pressure increases the coordinaiton number e.g. applying high pressure, the NaCl
crystal structure having 6 : 6 coordination number changes to CsCl having coordination number 8 : 8.

High pressure
NaCl   CsCl type structure
6:6 8:8

z Temperature : Increase of temperature, however decrease the coordination number.

heating
e.g. CsCl   NaCl type structure
8:8 6:6

Example 8 : KF has NaCl structure. If distance between K+ & F– is 269 pm, find the density of KF. (At. mass
of K = 39 and F = 19)
Solution : For NaCl type structure of KF

a = 2  rk+ + rF–  or 2 × 269 = 538 pm

Z×M 4  58

 = N × a 3 = 6.02  1023  538  10 –10

3
0

 = 2.48 g/cm3

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Example 9 : Density of KBr is 2.75 g/cm3. The edge length of unit cell is 654 pm. Find the type of unit cell
of KBr. (At. mass of K = 39 and Br = 80).
Solution : For KBr, M = 39 + 80 = 119
a = (654 × 10–10) cm
d = 2.75 g/cm3

Z×M  × N0 × a3
  = N × a3 or Z =
0 M

 
3
2.75  6.02  1023  654  10 –10
Z=
119

On solving, Z = 3.89  4
Hence, it is a fcc lattice.

IMPERFECTION IN SOLIDS

The imperfection i.e., irregularities in crystal is known as crystal defects.

z When some ions are missing from ionic crystals from their theortical lattice points, then this type of defect
is known as point defect.
z At absolute zero temperature, ionic crystal may not have any defect, with increase in temperature, point
defect increases.
Defects are of two types
1. Stoichiometric Defect
2. Non-stoichiometric Defect
1. Stoichiometric Defect : In these defects stoichiometric ratio of constituent ions is not disturbed.
(a) Schottky Defect : When equal number of cation and anion are missing from its normal site Schottky
defect is created. This defect causes decrease in density of crystal.
e.g., In NaCl, there are approximately 106 Schottky pair per cm3 at room temperature. [In 1 cm3 there
are about 1022 ions and therefore there is one Schottky defect per 1016 ions].
(b) Frenkel Defect or Dislocation Defect : Found mostly in ionic solids. The ion leaves its position in
the lattice and occupies an interstitial void. In this defect although density of solid remains unaffected
but stability of solid decreases. Simultaneously electrical conductivity and dielectric constant of solid
increases. This is majorly shown by crystals having large size variation between cation and diaion
and low co-ordination numbers. e.q. : Zns, AgCl, AgBr, AgI etc.
2. Non Stoichiometric Defects : In these defects stoicheometric ratio of constituent ions is changed.
(a) Metal excess defect :
1. Due to Anionic Vacancies : In few cases, negative ions may be missing from their lattice sites
leaving holes in which electrons remain trapped to maintain the electrical neutrality. The holes
with entrapped electrons are called F-centres (F-means Farbe centre, means colour centre). They
impart colour to the crystal lattice.
Alkali halides like NaCl & KCl show this defect. On heating crystals of NaCl in vapours of Na,
a few Cl– ions diffuse to the surface of crystal to react with Na atoms to form NaCl. Na atoms
get oxidised to Na+ and free electrons go to the holes created by Cl– ions and make F-centres.

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NEET The Solid State 143
When these electrons absorb energy in visible region they get excited and are responsible to
give the yellow colour to crystal.
Similarly.
KCl crystals appear violet in the atmosphere of K.
LiCl crystals appear pink in the atmosphere of Li.
2. Presence of Extra Cations in Interstitial Sites : When some extra positive ions occupy
interstitial sites to maintain electrical neutrality some extra electrons occupy some other interstitial
sites.
Example : ZnO is white at room temperature but turns yellow on heating

 1
ZnO   Zn2  2e –  O2
2
It causes excess of Zn in the lattice causing metal excess.
+ +
A
–
B– A+
–
B A+ B A+ B
Cation and
+ electron in
– + + +
B A B
– +
A B
+
A B A
+

Electron Interstitial states

remains e–
–
A+ e A
+
B
–
trapped in A+ B+ A+ B+
anion vacancy
+ +
B+ A B– A B
+
A
+
B
+
A
+

Metal excess defects Metal excess defects

due to loss of anion due to extra cation
(b) Metal deficiency : Crystals with metal deficiency defect behaviour like p-type semiconductor. It is
a common defect in transition metal compounds in which metal shows variable valency. In fact a few
metal cations in lower oxidation state are missing from crystal lattice and their charge is balanced
by the other cations of same metal in higher oxidation state. It causes some vacant points or holes
and there is a net loss of metal causing metal dificiency Due to loss of metal ion stoichiometric formula
of compound is changed. e.g., Fe0.96O, Fe0.93O.
(c) Impurity Defects : Some impurity atoms may either come and occupy interstitial sites or in case of
some ionic solid, some cations from regular sites are replaced by some other cations and cause
impurity defects. For example, NaCl having some amount of SrCl2 is crystallised, some of the sites
of Na+ ions are occupied by Sr+2 ions. As charge on Sr is + 2, one Sr+2 ion will replace two Na+ ions.
This lattice has cationic vacancies.

Example 10 : Iron oxide has formula Fe0.94O. What fraction of Fe exists as Fe+3?
Solution : Let there are 100 oxide ions and 94 iron ions.
Fe+x : O–2 Total negative charge on 100 Oxide ions = – 200
= 94 : 100
It means as per the concept of charge neutrality, 94 iron ions should have total charge of + 200.
Let number of Fe+3 = x
 Number of Fe+2 = (94 – x)
Total positive charge = 3x + 2(94 – x) = 200.
On solving, x = 12
12
 %Fe+3 =  100 = 12.8%
94

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144 The Solid State NEET

Example 11 : If NaCl is doped with 10–3 mole % of SrCl2. Calculate the number of cationic vacancies.
Solution : Number of moles of cationic vacancies = 100.
Mole of NaCl = 10–3
10 –3
 Number of moles of cationic vacancies per mole of NaCl =  10 –5
100
 Total number of cationic vacancies = 10–5 × N0 = 6.02 × 1018

PROPERTIES OF SOLIDS
Electrical Properties of Solids
z Solids are classified as conductors, semiconductors and insulators, depending on the ease with which
they conduct electricity.
z In ionic solids conduction is by ions. The magnitude of conductivity depends on the number of electrons
available to participate in the conduction process.
z In metals, conductivity strongly depends on the number of valence electrons available per atom. The
atomic orbital form molecular orbital which are close in energy to form a band (conduction band and
valence band).
z In semiconductors, gap between valence band and conduction band is small.
Note : With increase in temperature, electrical conductivity increases.
z Doping means introducing small impurities like P, As, B into pure crystal e.g., conductivity of Si increases
by doping 15 group element forming n-type semiconductor or doping by 13 group element forming
p-type semiconductors.
z Combination of p and n-type semiconductors are used to make electronic component. e.g. diode.

Type Conductivity in ohm–1 m–1

4 7
1. Conductors 10 – 10
–20 –10
2. Insulators 10 – 10

3. Semiconductors 10– 6 – 10 4

Magnetic Properties of Solids

z Magnetic property is due to the magnetic moment of electron. Magnetic moment is due to
(i) Orbital motion of electron around the nucleus and
(ii) Spin of electron around its own axis.
z This moving electron is considered as small current loop, generating small magnetic field and having a
magnetic moment along its axis.
z Each electron is with permanent orbital and spin magnetic moment.
z Solids are classified in different categories of magnetic materials depending on its behaviour in front of
external magnetic field.
Diamagnetic : If weakly repelled by the external magnetic field, contain no unpaired electrons, e.g., NaCl,
TiO2, Benzene, H2O, N2 etc.
Paramagnetic : If attracted by the external magnetic field, contain one or more unpaired electrons,
e.g., O2, Cu2+, Fe3+, CuO, TiO, VO2 etc.
Ferromagnetic : These have large paramagnetism and show paramagnetism even in the absence of magnetic
field e.g., Fe, Ni, Co, CrO2, Fe3O4 and alnico. CrO2 is used in magnetic tapes of casette recorders.

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In ferromagnetism there is spontaneous alignment of magnetic moments.
In antiferromagnetism the magnetic moment oppose each other, net magnetic moment is zero (e.g. MnO)
In ferrimagnetism the magnetic moments are aligned in unequal numbers resulting in net magnetic moment
(e.g., Fe3O4, MgFe2O4, ZnFe2O4).

Ferromagnetic

Antiferromagnetic

Ferrimagnetic

Note : Ferromagnetic, ferrimagnetic and antiferromagnetic, substances show paramagnetic nature at higher
temperature. This arises due to randomisation of spins of high temperature.
e.g.: Fe3O4 is ferrimagnetic at room temperature but becomes paramagnetic at 850 K.

Dielectric properties
Piezoelectricity : In piezoelectric crystals electricity is produced due to displacement of ions from their orders
arrangement by the application of mechanical stress.
Pyroelectricity : Upon heating, if the orderly arrangements of ions or atoms in the crystal gas displaced and
produces electricity.

EXERCISE
8. Stoichiometric defect is also known as
(1) Intrinsic defect (2) Impurity defect
(3) Thermodynamic defect (4) Both (1) & (3)
9. Which one of the following compounds can show Frenkel defect ?
(1) ZnS (2) AgCl
(3) AgI (4) All of these
10. The solids which are good conductor of electricity should have conductivities in the order of
(1) 107–1m–1 (2) 10–20–1m–1
(3) 10–6–1m–1 (4) 104–1 m–1
11. Identify the antiferromagnetic substance
(1) MnO (2) Fe3O4
(3) CrO2 (4) NaCl
12. Which of the following substance is diamagnetic ?
(1) Fe3O4 (2) MgFe2O4
(3) ZnFe2O4 (4) NaCl
13. Metal excess defect arises due to
(1) Anionic vacancies (2) The presence of extra cations at interstitial sites
(3) Cationic vacancies (4) Both (1) & (2)

‰ ‰ ‰

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 t
en
nm nment
ssigAssig
Assignment

Assignment
A

8. The number of possible three dimensional lattices

SECTION - A
(Bravais lattices) is [NCERT Pg. 11]
NCERT Based MCQs (1) 7

1. I2 is [NCERT Pg. 6] (2) 4

(1) Molecular solid (2) Covalent solid (3) 14

(3) Amorphous solid (4) Metallic Solid (4) 5

2. a = b = c,  =  =   90º represents which of the 9. The formula of FeO generally ranges from Fe0.93O to
following crystal system? [NCERT Pg. 10] Fe0.96O due to [NCERT Pg. 26]

(1) Rhombohedral (2) Orthorhombic (1) Metal excess defect

(3) Tetragonal (4) Cubic (2) Metal deficiency defect

3. The lattice having AAA type pattern is (3) Impurity defect

[NCERT Pg. 14] (4) Frenkel defect

(1) Simple cubic 10. Zinc oxide, white in colour at room temperature,
acquires yellow colour on heating due to
(2) Face-centred cubic
[NCERT Pg. 26]
(3) Body-centred cubic
(1) Zn being a transition element
(4) End-centred
(2) Paramagnetic nature of the compoud
4. The formula of crystalline solid having atoms ‘B’ in
ccp arrangement, atoms ‘A’ occupying half of (3) Trapping of electrons at the site vacated by oxide
octahedral and half of tetrahedral voids is ions

[NCERT Pg. 18] (4) High coordination number

(1) A2B3 (2) A4B3 11. Select the correct statement(s) [NCERT Pg. 27]

(3) A3B2 (4) A3B4 (1) The conductivity of metals depends upon the
number of valence electrons available per atom
5. Number of moles of tetrahedral voids present in FCC
type structure having 3 moles of atoms is (2) The atomic orbitals of metal atoms form
molecular orbitals which are so close in energy
[NCERT Pg. 17] to each other as to form a band
(1) 3 (2) 6 (3) Silicon and germanium are called intrinsic
(3) 9 (4) 1/3 semiconductors
6. The coordination number in one dimensional close (4) All of these
packed arrangement is [NCERT Pg. 14] 12. Which of the following is one of the characteristics
(1) 2 (2) 3 of paramagnetic substance? [NCERT Pg. 29]
(3) 4 (4) 6 (1) Strongly attracted by a magnetic field
7. If one of the atoms is removed from simple cubic (2) Magnetized in a magnetic field in the same
unit cell, then the effective number of atoms remained direction
in a unit cell is [NCERT Pg. 12] (3) Do not lose their magnetism in the absence of
(1) 7 (2) 7/8 magnetic field
(3) 8/7 (4) 8 (4) Contains no unpaired electrons

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13. Which of the following compounds will show 4. Lithium metal has a body centred cubic structure.
Schottky defect? [NCERT Pg. 25] Its density is 0.53 g cm–3 and its molar mass is 6.94
(1) ZnS (2) AgCl g mol–1. Calculate the edge length of a unit cell of
Lithium metal
(3) CsCl (4) AgI
(1) 153.6 pm (2) 351.6 pm
14. An element forms ccp lattice with a cell edge length
of 400 pm. The density of the element is 10 g cm–3. (3) 527.4 pm (4) 263.7 pm
The atomic mass of the element will be [ Take NA = 5. What is the volume of a face centred cubic unit cell,
6.02  1023] [NCERT Pg. 22]
when its density is 2.0 g cm–3 and the molar mass of
(1) 65 u (2) 54.2 u the substance is 60.22 g mol–1?
(3) 96.3 u (4) 205 u (1) 4 × 10—22 cm3 (2) 2 × 10—22 cm3
15. Semiconductors have electrical conductivities in the (3) 44 × 10—22 cm3 (4) 22 × 10—22 cm3
range from [NCERT Pg. 27]
6. KF has NaCl type of structure. The edge length of its
(1) 10–6 to 104 ohm–1 m–1 unit cell has been found to be 537.6 pm. The distance
(2) 104 to 107 ohm–1 m–1 between K+ F– in KF is
(3) 10–20 to 1010 ohm–1 m–1 (1) 26.88 pm (2) 268.8 pm
(4) 10 to 103 ohm–1 m–1 (3) 2688 pm (4) 236.54
7. If NaCl is doped with 10–3 mole of BaCl2. What is the
concentration of cationic vacancies?
SECTION - B
(1) 10–3 mole–1
Objective Type Questions
(2) 6.022 × 1018 mole–1
1. Fraction of the total volume occupied by atoms in a
simple cube is (3) 1050 mole–1
(4) 6.022 × 1020 mole–1
 3
(1) (2) 8. In HCP arrangement, the co-ordination number is
2 8
(1) 6 (2) 12
2  (3) 8 (4) 10
(3) (4)
6 6
9. The radius of the Na is 95 pm and that of Cl– ion is
+

2. If a is the length of unit cell, then which one is correct 181 pm. The co-ordination number of Na+ ion will be
relationship?
(1) 4 (2) 6
(1) For simple cubic lattice,
(3) 8 (4) 2
a 10. Zinc oxide on heating changes to yellow. This is
Radius of metal atom =
2 because
(2) For bcc lattice,
(1) Zinc oxide is a stoichiometric compound
3a (2) Zinc oxide is a covalent compound
Radius of metal atom =
4
(3) Zinc oxide shows metal excessive defect
(3) For fcc lattice,
(4) Zinc oxide is a neutral oxide
a
Radius of metal atom = 11. When an element of Group 14 is doped with an
2 2
element of Group 15 then
(4) All of these
(1) p-type of semiconductors are formed
3. The number of octahedral sites in a cubical close
pack array of N spheres is (2) n-type of semiconductors are formed
(1) N/2 (2) 2 N (3) Zeolites are formed
(3) 4 N (4) N (4) Electrolytes are formed
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12. In Antiferromagnetism 17. In any ionic crystal A has formed cubical close
(1) Alignments of magnetic moments is additive packing and B atoms are present at every tetrahedral
voids. If any sample of crystal contain ‘N’ number of
(2) Alignments of magnetic moments in one B atoms then number of A atoms in that sample is
direction is compensated by alignments in the
opposite directions N
(1) N (2)
2
(3) Alignments of magnetic moments does not
take place (3) 2 N (4) 2N
(4) Alignments of magnetic moments in one 18. When a crystal having rock salt type geometry is
direction is not completly compensated by heated in the presence of it’s metal vapour then defect
alignments in opposite direction in it will be
13. The gold crystallizes in a cubic closest packed (1) Stoichiometric defect
structure. If its molar mass is M, number of spheres
in one unit cell is Z and edge length of unit cell is x (2) Metal excess defect
pm then its density in g/cm3 will be (3) Anion excess defect
ZM ZM (4) Metal defficiency defect
(1) d  (2) d 
xNA x3 19. Substances which are magnetic but having less
magnetic moment than theoretically calculated value
ZM ZM
(3) d  (4) d 
are called
x 3N A x 10 30 NA
3
(1) Ferromagnetic
14. For face centered cubic structure edge length ‘a’ can
(2) Ferrimagnetic
be related with radius ‘r’ as
(3) Antiferromagnetic
(1) a  r  2 (2) a = r
(4) Diamagnetic
4
(3) a  2 2 r (4) a  r 20. Which of the following statement is correct?
3
(1) On increasing temperature the coordination
15. The site labelled as ‘X’ in fcc arrangement is number of solid remains unchanged
X (2) On increasing pressure the coordination number
of solid increases
(3) On increasing temperature the coordination
number of solid increases

1 (4) On increasing pressure the coordination number

(1) Face centre with contribution of solid decreases
4
21. A compound formed by element A and B crystallizes
1 in the cubic structure, where A atoms are at the
(2) Edge centre with contribution
4 corners of a cube and B atoms are at the centre of
1 the body. The formula of the compounds is
(3) Corner with contribution
4 (1) AB
1 (2) AB2
(4) Tetrahedral void with contribution
8 (3) A2B3
16. Which of the following crystal is represented by (4) AB3
a  b  c and  90°? 22. Packing fraction of diamond is
(1) Orthorhombic (1) 0.74
(2) Monoclinic (2) 0.68
(3) Triclinic (3) 0.52
(4) Tetragonal (4) 0.34

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NEET The Solid State 149
23. In a solid A, B and C arranged as below. The formula 30. A Match box exhibits
of solid is (1) Cubic geometry
C
(2) Monoclinic geometry
 (3) Orthorhombic geometry
B
A (4) Hexagonal geometry


(1) ABC (2) AB2C2
SECTION - C

(3) A2BC (4) AB8C2 Previous Years Questions

24. In a cubic close packed structure of mixed oxides, 1. Formula of nickel oxide with metal deficiency
the lattice is made up of oxide ions, one eighth of defect in its crystal is Ni 0.98 O. The crystal
tetrahedral voids are occupied by divalent (X2+) ions, contains Ni2+ and Ni3+ ions. The fraction of nickel
while one-half of the octahedral voids are occupied existing as Ni2+ ions in the crystal is
by trivalent ions (Y3+), then the formula of the oxide [NEET-2019 (Odisha)]
is
(1) 0.31 (2) 0.96
(1) XY2O4 (2) X2YO4
(3) 0.04 (4) 0.50
(3) X4Y5O10 (4) X5Y4O10
2. A compound is formed by cation C and
25. In a face centred cubic arrangement of A and B atoms, anion A. The anions form hexagonal close packed
atoms of A are at the corner of the unit cell and atoms
(hcp) lattice and the cations occupy 75% of
of B are at the face centres. One of the A atom is
octahedral voids. The formula of the compound is:
missing from one corner in unit cell. The simplest
formula of compound is [NEET-2019]

(1) A7B3 (2) AB3 (1) C2A3 (2) C3A2

(3) A7B24 (4) A7B8 (3) C3A4 (4) C4A3

26. A mineral having the formula AB2 crystallizes in the 3. Iron exhibits bcc structure at room temperature.
cubic close-packed lattice, with the A atoms Above 900°C, it transforms to fcc structure. The
occupying lattice points. What is the co-ordination ratio of density of iron at room temperature to that
number of the A, B and what fraction of tetrahedral at 900°C (assuming molar mass and atomic radii
sites is occupied by B atoms? of iron remains constant with temperature) is
(1) 4, 8, 50% (2) 8, 4, 50% [NEET-2018]
(3) 8, 4, 100% (4) 4, 8, 100%
3 4 3
27. Number of unit cells in 0.1 g molecule NaCl is (1) (2)
2 3 2
(1) 0.1 NA (2) 0.025 NA
1 3 3
(3) 0.5 NA (4) 0.25 NA (3) (4)
2 4 2
28. An element occurs in BCC structure with a edge 4. Which is the incorrect statement? [NEET-2017]
length of 288 pm. The density of the element is 7.2
(1) FeO0.98 has non stoichiometric metal deficiency
gm cm–3. How many atoms of the element does 208
defect
g of the element contain?
(2) Density decreases in case of crystals with
(1) 24.16 × 1022 (2) 24.16 × 1023
Schottky's defect
(3) 24.16 × 1024 (4) 24.16 × 1025 (3) NaCl(s) is insulator, silicon is semiconductor,
29. Co-ordination number of first nearest neighbour of silver is conductor, quartz is piezo electric
Cl– in NaCl lattice is crystal

(1) 4 (2) 8 (4) Frenkel defect is favoured in those ionic

compounds in which sizes of cation and
(3) 6 (4) 12 anions are almost equal
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150 The Solid State NEET

5. In calcium fluoride, having the fluorite structure, the (1) 30 g mol–1 (2) 27 g mol–1
coordination numbers for calcium ion (Ca2+) and
(3) 20 g mol–1 (4) 40 g mol–1
fluoride ion (F–) are [NEET-Phase-2-2016]
(1) 4 and 2 (2) 6 and 6 13. The number of carbon atoms per unit cell of diamond
unit cell is [NEET-2013]
(3) 8 and 4 (4) 4 and 8
6. Lithium has a bcc structure. Its density is (1) 8 (2) 6
530 kg m–3 and its atomic mass is 6.94 g mol–1. (3) 1 (4) 4
Calculate the edge length of a unit cell of Lithium
metal (NA = 6.02 × 1023 mol–1) [NEET-2016] 14. A metal crystallizes with a face-centered cubic lattice.
The edge of the unit cell is 408 pm. The diameter of
(1) 264 pm (2) 154 pm
the metal atom is [AIPMT (Prelims)-2012]
(3) 352 pm (4) 527 pm
(1) 144 pm (2) 204 pm
7. The ionic radii of A+
and B– ions are 0.98 ×10–10
–10
m and 1.81 ×10 m. The coordination number of (3) 288 pm (4) 408 pm
each ion in AB is [NEET-2016]
15. The number of octahedral void(s) per atom present
(1) 2 (2) 6 in a cubic close-packed structure is
(3) 4 (4) 8
[AIPMT (Prelims)-2012]
8. The vacant space in bcc lattice unit cell is
(1) 2 (2) 4
[Re-AIPMT-2015]
(1) 23% (2) 32% (3) 1 (4) 3
(3) 26% (4) 48% 16. Structure of a mixed oxide is cubic close packed
9. The correct statement regarding defects in (ccp). The cubic unit cell of mixed oxide is composed
crystalline solids is [Re-AIPMT-2015] of oxide ions. One fourth of the tetrahedral voids are
(1) Frenkel defect is a dislocation defect occupied by divalent metal A and the octahedral voids
are occupied by a monovalent metal B. The formula
(2) Frenkel defect is found in halides of alkaline
of the oxide is [AIPMT (Mains)-2012]
metals
(3) Schottky defects have no effect on the density (1) ABO2
of crystalline solids (2) A2BO2
(4) Frenkel defects decrease the density of
(3) A2B3O4
crystalline solids
10. A given metal crystallizes out with a cubic structure (4) AB2O2
having edge length of 361 pm. If there are four 17. A solid compound XY has NaCl structure. If the radius
metal atoms in one unit cell, what is the radius of of the cation is 100 pm, the radius of the anion (Y–)
one atom? [AIPMT-2015] will be [AIPMT (Mains)-2011]
(1) 108 pm (2) 40 pm
(1) 241.5 pm (2) 165.7 pm
(3) 127 pm (4) 80 pm
(3) 275.1 pm (4) 322.5 pm
11. If a is the length of the side of a cube, the distance
between the body centered atom and one corner 18. AB crystallizes in a body centred cubic lattice with
atom in the cube will be [AIPMT-2014] edge length ‘a’ equal to 387 pm. The distance
2 4 between two oppositively charged ions in the
(1) a (2) a lattice is [AIPMT (Prelims)-2010]
3 3
(1) 335 pm (2) 250 pm
3 3
(3) a (4) a
4 2 (3) 200 pm (4) 300 pm
12. A metal has a fcc lattice. The edge length of the unit 19. Among the following which one has the highest cation
cell is 404 pm. The density of the metal is to anion size ratio? [AIPMT (Mains)-2010]
2.72 g cm–3.The molar mass of the metal is
(NA is Avogadro's constant = 6.02 × 1023 mol–1) (1) Cs (2) CsF

[NEET-2013] (3) LiF (4) NaF

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NEET The Solid State 151
20. Lithium metal crystallises in a body centred cubic 26. The fraction of total volume occupied by the atoms
crystal. If the length of the side of the unit cell of present in a simple cube is
lithium is 351 pm, the atomic radius of the lithium
[AIPMT (Prelims)-2007]
will nearly be [AIPMT (Prelims)-2009]
 
(1) 152 pm (2) 75 pm (1) (2)
4 6
(3) 300 pm (4) 240 pm
 
21. Copper crystallises in a face-centred cubic lattice (3) (4)
3 2 4 2
with a unit cell length of 361 pm. What is the radius
–4
of copper atom in pm ? [AIPMT (Prelims)-2009] 27. If NaCl is doped with 10 mol % SrCl 2, the
concentration of cation vacancies will be
(1) 157 (2) 181 (NA = 6.02 × 1023 mol–1) [AIPMT (Prelims)-2007]
(3) 108 (4) 128 (1) 6.02 × 1014 mol–1
22. If ‘a’ stands for the edge length of the cubic systems : (2) 6.02 × 1015 mol–1
simple cubic, body centred cubic and face centred (3) 6.02 × 1016 mol–1
cubic, then the ratio of radii of the spheres in these
systems will be respectively (4) 6.02 × 1017 mol–1
28. CsBr crystallises in a body centred cubic lattice.
[AIPMT (Prelims)-2008]
The unit cell length is 436.6 pm. Given that the atomic
(1) 1a : 3a : 2a mass of Cs = 133 and that of Br = 80 amu and
Avagadro number being 6.02 × 1023 mol–1, the density
1 3 1 of CsBr is [AIPMT (Prelims)-2006]
(2) a: a: a
2 4 2 2 (1) 42.5 g/cm3
1 1 (2) 0.425 g/cm3
(3) a : 3a : a
2 2 (3) 8.25 g/cm3
1 3 2 (4) 4.25 g/cm3
(4) a: a: a
2 2 2
29. The appearance of colour in solid alkali metal halides
23. Percentage of free space in a body centred cubic is generally due to [AIPMT (Prelims)-2006]
unit cell is [AIPMT (Prelims)-2008]
(1) F-centres
(1) 28% (2) 30%
(2) Schottky defect
(3) 32% (4) 34%
(3) Frenkel defect
24. With which one of the following elements silicon
(4) Interstitial positions
should be doped so as to give p-type of
semiconductor ? [AIPMT (Prelims)-2008] 30. In a face-centered cubic lattice, a unit cell is shared
equally by how many unit cells ?
(1) Boron (2) Germanium
[AIPMT (Prelims)-2005]
(3) Arsenic (4) Selenium
(1) 8 (2) 4
25. Which of the following statements is not correct ?
(3) 2 (4) 6
[AIPMT (Prelims)-2008]
Questions asked Prior to Medical Ent. Exams. 2005
(1) The number of Bravais lattices in which a crystal
31. Ionic solids, with Schottky defects, contain in their
can be categorized is 14
structure
(2) The fraction of the total volume occupied by the (1) Cation vacancies only
atoms in a primitive cell is 0.52
(2) Cation vacancies and interstitial cations
(3) Molecular solids are generally volatile
(3) Equal number of cation and anion vacancies
(4) The number of carbon atoms in a unit cell of
diamond is 4 (4) Anion vacancies and interstitial anions

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152 The Solid State NEET

32. The number of atoms in 100 g of an FCC crystal 39. In crystals of which one of the following ionic
with density d = 10 g/cm3 and cell edge equal to compounds would you expect maximum distance
100 pm, is equal to between centres of cations and anions?
(1) 2 × 1025 (2) 1 × 1025 (1) CsI (2) CsF
(3) 4 × 1025 (4) 3 × 1025 (3) LiF (4) LiI
33. An element (atomic mass = 100 gm/mol) having 40. In cube of any crystal A-atom placed at every
BCC structure has unit cell edge 400 pm. The corners and B-atom placed at every centre of face.
density of element is
The formula of compound
(1) 7.289 gm/cm3 (2) 2.144 gm/cm3
(1) AB (2) AB3
(3) 10.376 gm/cm3 (4) 5.188 gm/cm3
(3) A2B2 (4) A2B3
34. If we mix a pentavalent impurity in a crystal lattice
41. Coordination number in ABAB... type arrangement
of germanium, what type of semiconductor formation
will occur? is

(1) n-type semiconductor (1) 6 (2) 8

(2) p-type semiconductor (3) 12 (4) 4

(3) Both (1) & (2) 42. The pyknometric density of sodium chloride crystal
is 2.165 × 103 kg m–3 while its X-ray density is
(4) None of these
2.178 × 103 kgm–3. The fraction of unoccupied sites
35. The intermetallic compound LiAg crystallizes in in sodium chloride crystal is
cubic lattice in which both lithium and silver have
coordination number of eight. The crystal class is (1) 5.96 (2) 5.96 × 10–2

(1) Face-centred cube (3) 5.96 × 10–1 (4) 5.96 × 10–3

(2) Simple cube 43. A compound formed by elements X and Y crystallizes

in a cubic structure in which the X atoms are at the
(3) Body-centred cube
corners of a cube and the Y atoms are at the face-
(4) None of these centres. The formula of the compound is
36. Schottky defect in crystals is observed when (1) XY3 (2) X3Y
(1) Density of the crystal is increased (3) XY (4) XY2
(2) Unequal number of cations and anions are
missing from the lattice
SECTION - D
(3) An ion leaves its normal site and occupies an
interstitial site NEET Booster Questions
(4) Equal number of cations and anions are missing 1. Metallic radius of aluminium is 125 pm. If it
from the lattice crystallises in a cubic closed packed structure, the
37. The edge length of rock salt type unit cell is number of unit cells present in 1 cm3 of aluminium is
508 pm. If the radius of the cation is 110 pm, the (1) 1.13 × 1011
radius of the anion assuming NaCl type structure is (2) 2.26 × 1018
(1) 144 pm (2) 398 pm (3) 2.26 × 1022
(3) 288 pm (4) 618 pm (4) 1.13 × 1016

38. The second order Bragg diffraction of X-rays with 2. Which of the following will be a p-type
semiconductor?
 = 1.00 Å from a set of parallel planes in a metal
occurs at an angle 60°. The distance between the (1) Si doped with B
scattering planes in the crystal is (2) Ge doped with In
(1) 2.00 Å (2) 1.00 Å (3) Si doped with C
(4) Both (1) & (2)
(3) 0.575 Å (4) 1.15 Å
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NEET The Solid State 153
3. Iron oxide crystallises in a hexagonal close-packed 11. AgCl is crystallized from molten AgCl containing a
arrangement of oxide ion with two out of every three little CdCl2, the solid obtained will have
octahedral holes occupied by iron. What should be
(1) Cationic vacancies equal to number of Cd2+ ions
the formula of the oxide? incorporated.
(1) FeO (2) Fe2O3 (2) Cation vacancies equal to double the no.
(3) Fe3O4 (4) Fe3O2 of Cd2+ ions

4. Which of the following is ferrimagnetic substance? (3) Anionic vacancies

(1) MnO (2) CrO2 (4) Neither cationic nor anionic vacancies.

(3) C6H6 (4) MgFe2O4 12. Given an alloy of Cu, Ag and Au in which Cu atoms
constitute the ccp arrangement. If the hypothetical
5. ZnO is white in colour at room temperature on heating formula of the alloy is Cu4Ag3Au. What are the
it turns yellow due to probable locations of Ag & Au atoms.
(1) Metal excess defect due to the presence of extra (1) Ag – all tetrahedral voids, Au – all octahedral
cations at interstitial sites voids
(2) Metal deficiency defect due to the presence of 3 1
extra anion at interstitial sites (2) Ag  th tetrahedral voids, Au – th
8 4
(3) Metal excess defect due to extra charge on Zn- octahedral voids
atom 1 1
(3) Ag  octahedral voids, Au – tetrahedral
(4) All of these 2 2
voids
6. Which of the following is H-bonded molecular solid?
(4) Ag – All octahedral voids, Au = All tetrahedral
(1) CO2(s) (2) H2O(s) voids
(3) SiO2(s) (4) CCl4(s) 13. A FCC crystal of O2– (oxide) ion has Mn+ ions in all
7. Total number of voids per unit cell if a metal OHV, then value of n is
crystallises into hcp is (1) 4 (2) 1
(1) 4 (2) 12 (3) 2 (4) 3
(3) 18 (4) 8 14. Fraction of empty space in ABAB type arrangement
in 3D close packing
8. Packing fraction of diamond if it has arrangement of
carbon atom similar to the arrangement of ions in (1) 0.74 (2) 0.32
ZnS is
(3) 0.26 (4) 0.68
(1) 34% (2) 68% 15. BaO has rock-salt type structure. When subjected
(3) 52% (4) 90% to high pressure, the ratio of the coordination of Ba2+
ion to O2– changes to
9. Total number of Na+ at edge centers of the unit cell
of NaCl if Cl– has fcc arrangement is (1) 4 : 8 (2) 8 : 4

(1) 4 (2) 3 (3) 8 : 8 (4) 4 : 4

(3) 2 (4) 1 16. A certain sample of cuprus sulphide is found to have

composition Cu1.8 S, because of presence of some
10. Which of the following is not a crystalline form of Cu2+ ions is the lattice. What is the mole % of Cu2+
matter? in crystal.
(1) Snow (1) 99.8 %
(2) Wood (2) 11.11 %
(3) Glass (3) 88.88 %
(4) All of these (4) 52.37%

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154 The Solid State NEET

17. AxOy oxide crystallises in hcp array of oxide ions 25. In a unit cell, atoms A, B, C and D are present at half
with two out of every three octahedral void occupied of total corners, all face-centres, body-centre and
by A+3. The formula of oxide is : one third of all edge-centres respectively. Then formula
of unit cell is
(1) AO (2) AO2
(1) AB3CD3 (2) ABCD
(3) A2O3 (4) A3O2
(3) AB6C2D4 (4) AB6C2D2
18. In which of following FCC arrangement cation
occupies alternate tetrahedral void ? 26. In a unit cell, atoms A, B, C and D are present at
corners, face-centres, body-centre and edge-centres
(1) NaCl (2) Na2O respectively. If atoms touching one of the plane
(3) ZnS (4) CaF2 passing through two diagonally opposite edges are
removed, then formula of compound is
19. If radius ratio (r+/r–) is equal to 0.7 then coordination
number will be : (1) ABCD2 (2) ABD2

(1) 6 (2) 8 (3) AB2D2 (4) AB4D5

(3) Both 6 as well as 8 (4) 3 27. Number of atoms per unit cell, if atoms are present
at the corner of unit cell and 2 atoms at each body
20. The nearest distance between two tetrahedral voids diagonal
in FCC Lattice is
(1) 9 (2) 10
a 2a
(1) (2) (3) 6 (4) 4
2 2
28. Second nearest neighbours of Na+ ion in NaCl unit
3a cell is
(3) (4) All of these
2 (1) 4 (2) 6
21. In crystalline solids, which of the following element (3) 8 (4) 12
of symmetry is not present?
29. Glass is a
(1) Axis of symmetry
(1) Micro-crystalline solid
(2) Angle of symmetry
(2) Super cooled liquid
(3) Centre of symmetry
(3) Gel
(4) Plane of symmetry
(4) Polymeric mixture
22. Amorphous solids have
30. If radius of a metal atom (A) is 5 pm and radius of an
(1) Orderly arrangement of atoms electronegative atom (B) is 20 pm, then in the unit
(2) Repeating unit of unit cell cell

(3) Long range of melting point (1) A in octahedral voids, B in FCC unit

(4) Anisotropy (2) A in FCC unit, B in tetrahedral void

23. The type of crystal system having a = b  c and  = (3) A in BCC unit, B in cubic void
 =  = 90° is (4) A in tetrahedral void, B in FCC unit
(1) Cubic (2) Orthorhombic 31. In a unit cell containing X2+, Y3+ and Z2– where X2+
occupies 1/8th of tetrahedral voids, Y3+ occupies
(3) Monoclinic (4) Tetragonal
1/2 of octahedral voids and Z2– forms ccp structure.
24. In a unit cell, atoms A, B, C and D are present at Then formula of compound is
corners, face-centres, body-centre and edge-centre
(1) X2Y4Z
respectively in a cubic unit cell. The total number of
atoms present per unit cell is (2) XY2Z4
(1) 4 (2) 8 (3) XY3Z4

(3) 15 (4) 27 (4) X4YZ2

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NEET The Solid State 155
32. In a ccp type structure, if half of the face-centred 39. Second nearest neighbour of Cl– in CsCl solid is
atoms are removed, then percentage void in unit cell
is approximately (1) 8 (2) 6

(1) 54% (3) 16 (4) 10

(2) 46.25% 40. In a calcium fluoride structure, the co-ordination

number of cation and anion is respectively.
(3) 63%
(1) 6, 6 (2) 8, 4
(4) 37%
33. In a BCC unit cell, if half of the atoms per unit cell (3) 4, 4 (4) 4, 8
are removed, then percentage void is 41. The packing efficiency of the 2D square unit cell
(1) 68% (2) 32% shown below is
(3) 34% (4) 66%
34. In a CsCl structure, if edge length is a, then distance
between one Cs atom and one Cl atom is

a 3 a 3
(1) (2)
2 4 (1) 39.27% (2) 68.02%

a (3) 74.05% (4) 78.54%

(3) a 2 (4)
2 42. On rising temperature and decreasing pressure in
35. The correct statement about, CCP structure is CsCl solid

(1) Packing fraction = 26% (1) C.N. of metal ion increases from 6 to 8

(2) Coordination number = 6 (2) Number of formula unit per unit cell (Z) changes
from one to four
(3) Unit cell is face centred cubic
(3) Density of unit cell remains constant
(4) AB–AB type of packing
36. In a NaCl structure, if positions of Na atoms and Cl r
(4) (radius ratio) is increased
atoms are interchanged, then in the new unit cell r
(1) Na atom is present at body centre 43. Some of the molecular solids upon heating produces
(2) Cl atom is present at face centre small amount of electricity, hence solid is

(3) Na atom is present in tetrahedral voids (1) Piezoelectric

(4) Cl atom is present in octahedral voids (2) Pyroelectric

37. A metal can be crystallized in both BCC and FCC (3) Ferrielectric
unit cells whose edge lengths are 2 pm and 4 pm
(4) Ferroelectric
respectively. Then ratio of densities of FCC and BCC
unit cells is 44. NaCl becomes paramagnetic at high temperature due
1 to
(1) (2) 4
4 (1) Formation of F-centre
1 (2) Molten state
(3) (4) 16
16
(3) Change in oxidation state
38. Number of unit cells in 10 g NaCl
(4) Conversion of Na+ to Na
1.5 2.5
(1)  1024 (2)  1023 45. In which of the following defect density increases?
58.5 58.5
5.6 5.6 (1) Schottky defect (2) Frenkel defect
(3)  1020 (4)  1021
58.5 58.5 (3) F-centre (4) Impurity defect
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156 The Solid State NEET

46. For a Hexagonal crystal, correct set of 54. Number of unit cell in 58.5 g NaCl is
crystallographic parameter is
(1) 0.25 (2) 0.25 NA
(1) = = 90°, a = b = c
(3) 0.5 NA (4) 0.5
(2) = = = 90°, a = b c
55. Percentage tetrahedral void vacant in NaCl unit cell
(3) = = =90°, a  b  c
is
(4) = = 90°, = 120°, a = b  c
(1) 10 (2) 20
47. The unit cell present in ABAB ... closed packing of
atoms is (3) 30 (4) 100

(1) Hexagonal (2) Tetragonal 56. The fraction of the total volume occupied by the atoms
present in a simple cube
(3) Face centred cubic (4) Primitive cubic
48. When NaCl is doped with 10–5 mole % of SrCl2, what (1) /4 (2) /6
is the no. of cationic vacancies per mol of NaCl?
 
(1) 10–5 NA (2) 10–7 NA (3) (4)
3 2 4 2
(3) 2 × 10–7 NA (4) 8 × 10–7 NA 57. An element crystallises in a structure having a fcc unit
49. In fcc, octahedral voids are present at cell of an edge length 200 pm. If 200 g of this element
contains 24 × 1023 atoms then its density is :
(1) Edge centers (2) Face centers
(3) Body centers (4) Both 1 and 3 (1) 41.66 g cm–3 (2) 313.9 g cm–3

50. The hcp and ccp structure for a given element would (3) 8.117 g cm–3 (4) 400 g cm–3
be expected to have different
58. How many effective Na+ and Cl– ions are present
(1) Number of atoms is unit cell respectively in a unit cell of NaCl solid (Rock salt
(2) Packing fraction structure). If ions along one of the line connecting
opposite face centres are absent?
(3) Co-ordination number of atom
7 7 7
(4) Void fraction (1) , (2) 4,
2 2 2
51. Ferromagnetic substance is
7
(1) NaCl (2) H2O (3) 3, 3 (4) ,4
2
(3) CrO2 (4) MnO
59. In the solid compound Cu2Hgl4 cations occupy
52. Conductivity of conductors ranging from tetrahedral holes in a closed packed anion lattice.
(1) 104 to 107 ohm–1 m–1 What fraction of tetrahedral holes is filled?

(2) 10–20 to 10–10 ohm–1 m–1 1 3

(1) (2)
(3) 10–6 to 104 ohm–1 m–1 4 8

(4) 10–6 to 103 ohm–1 m–1 3 1

(3) (4)
53. Select the incorrect statement in the following 4 2

(1) Fe0.96O has metal deficiency defect 60. Tungsten has an atomic radius of 0.136 nm. The
density of gungsten is 19.4 g/cm3. What is the
(2) Intrinsic defects are also known as crystal structure of tungsten? (Atomic weight of
thermodynamic defects Tungsten = 184)
(3) Schottky defect also known as dislocation defect
(1) Simple cubic (2) Body centered cubic
(4) There is 1 Schottky defect per 1016 ions in NaCl
at room temperature (3) Face centered cubic (4) None of these

‰‰‰

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Chapter 10

Solutions
Chapter Contents
z Introduction Introduction
z Concentration Terms Solution is a homogenous mixture of two or more components. Solvent is
z Solubility of Gases the component whose physical state is the same as that of solution. In case
both the components are in same phase, solvent is the component which is
z Vapour Pressure of Solutions
of Solids in Liquids in larger proportions.
Types of solutions
z Azeotropic Mixture
A solution can be solid, liquid or a gas depending upon the physical state
z Colligative properties
of the components. The various types of solutions are tabulated below.
S. No. Solute Solvent Types of Solution Examples
1. Solid Solid Solid in Solid Alloys
2. Liquid Solid Liquid in Solid Hydrated salts
3. Gas Solid Gas in Solid Dissolved gases in minerals
4. Solid Liquid Solid in Liquid Salt solution in water
5. Liquid Liquid Liquid in Liquid Alcohol in water
6. Gas Liquid Gas in Liquid Aerated drinks
7. Solid Gas Solid in Gas Iodine vapours in air
8. Liquid Gas Liquid in Gas Humidity in air
9. Gas Gas Gas in Gas Air

CONCENTRATION TERMS
S.No. Concentration Term Formula Definition
1. Mole fraction nA The ratio of the number of moles of
A 
n A + nB one component to the total number
of moles of all the components
present in the solution.
2. Mass percent wA The number of parts by mass of
Mass % of A = × 100
wA + wB one component (solute or solvent)
per 100 parts by mass of solution.
3. Volume percent VA The number of parts by volume of
Volume % of A = × 100 one component (solute or solvent)
V A + VB per 100 parts by volume of the
solution.

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102 Solutions NEET

S.No. Concentration Term Formula Definition

4. Parts per million Mass of A 6 The number of parts by mass (or
(ppm) ppm = × 10 by volume) of one component per
Mass of solution
million parts by mass (or by
volume of solution).
5. Molarity (M) Gram moles of solute The number of gram moles of the
M= 3
Volume of solution in litres solute dissolved per litre (or dm )
of the solution.
6. Normality (N) Gram equivalents of solute The number of gram equivalents of
N= the solute dissolved per litre of the
Volume of solution in litres
solution.
7. Molality (m) Gram moles of the solute The number of gram moles of
m= solute dissolved in 1000 gm of
Mass of solution in kg
solvent.

Some Important Relations :

z Molarity on mixing solutions of same solute
M1V1 + M2V2 = M3 (V1 + V2)
z Normality on mixing solutions of same solute
N1V1 + N2V2 = N3 (V1 + V2)
z Relation between Molarity and density (g/cc)

Percentage strength (w/w)  Density (gm/ml)  10

Molarity (M) =
Molar mass of solute

z Relation between normality and density (g/cc)

Percentage strength (w/w)  Density (gm/ml)  10

Normality (N) =
Equivalent mass of solute

z Relation between Normality and Molarity : Normality = n factor × molarity

Molarity  100
z Relation between molarity and molality : Molality 
1000  density  Molarity  molar mass of solute

z PPM = mass fraction × 106

SOLUBILITY OF GASES
z Gases can diffuse with each other similarly they dissolve in liquids and solids.

z The solubility of gas in a liquid is determined by several factors like temperature and pressure.
z The solubility of a gas in a liquid is given by Henry's Law, according to which solubility of gas in a liquid
is directly proportional to the pressure of gas.
z The solubility of gases depends on the partial pressure of gas and mole fraction of the gas in the solution
is proportional to the partial pressure of the gas in vapour phase hence

Partial pressure of the gas  KH  mole fraction of the gas

i.e. p = KH × , where KH is Henry’s constant and is the function of nature of the gas.

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NEET Solutions 103

Note :
1. Higher the value of KH at a given pressure, lower is the solubility of the gas in the liquid.
2. With increase in temperature solubility decreases.
3. Plot of partial pressure of the gas versus mole fraction of gas in solution is a straight line and slope
of the straight line is equal to Henry’s Constant (KH).

5
Example 1 : Henry's law constant of CO2 in water at 298 K is k bar. If pressure of CO2 is 0.01 bar, find
3
its mole fraction.
Solution : p = KH  X

 pCO2( g)  K H  X(CO2 )

5
 0.01   1000  XCO2
3
 XCO2  6  10 6

Vapour Pressure

The pressure exerted by the vapours in equilibrium with its liquid at a given temperature.
Factors on which vapour pressure depends :
z Nature of liquid : When the intermolecular forces of attractions are stronger then the vapour pressure
will be low because less number of molecules can leave the liquid.
o o
e.g. pC2H5OC2H5  pC2H5OH
z Temperature : Higher the temperature, greater would be the vapour pressure. This is because when
temperature is raised, kinetic energy of the molecules increases and therefore more number of molecules
leave the surface of the liquid
Raoult's Law
z In a solution, the vapour pressure of a component at a given temperature is equal to the mole fraction of
that component in the solution multiplied by the vapour pressure of that component in the pure state.
z Let us consider a mixture of two completely miscible volatile liquids A and B, having the mole fraction
A and B. Suppose at a certain temperature, their partial vapour pressures are pA and pB and the vapour
pressures in the pure state are p°A and p°B. According to the Raoult's Law,
pA = Ap°A and pB = Bp°B ...(i)
z If P is the total pressure of the system at the same temperature, then
P = pA + pB = Ap°A + Bp°B ...(ii)
or P = (1 – B)p°A + Bp°B = (p°B - p°A)B + p°A ...(iii)
z Since p°A and p°B are constant at a particular temperature, it is evident from eqn. (iii) that the total vapour
pressure is a linear function of the mole fraction B(or A = 1 – B).
Examples : Benzene + Toluene, Ethanol + Methanol, and CH3CHO + C2H5CHO etc.

z Mole Fraction of a Component in the Vapour Phase: p0A  A  PT  YA and pB0 B  PT  YB .

 A p0A
Mole fraction in vapour phase, (YA )  .
p A  A  pB0 B
0

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104 Solutions NEET

Example 2 : Vapour pressure of CH3Cl and CH2Cl2 are 540 mm Hg and 402 mm Hg respectively. 101 g of
CH3Cl and 85 g of CH2Cl2 are mixed together. Determine
(i) The pressure at which the solution starts boiling.
(ii) Molar ratio of solute v/s solvent in vapour phase in equilibrium with solution.
Solution : (i) Boiling occurs when external pressure becomes equal to the vapour pressure. So, the boiling
pressure = V.P. of solution
 pA x A  pB xB
Let A = CH3Cl, B = CH2Cl2, then
2 1
Toal pressure  540   402 
3 3
= 360 + 134 = 494 mm Hg
(ii) Here the solute is CH2Cl2 as mass is less
1
402 
xCH2Cl2  3  134
494 494
2
540 
xCH3Cl  3  360
494 494
nCH2Cl2 nsolute(g) 134
Now,    0.372
nCHCl3 nsolvent(g) 360

VAPOUR PRESSURE OF SOLUTIONS OF SOLIDS IN LIQUIDS

If a non-volatile solute is added to a solvent to give a solution, the vapour pressure of solution is solely from
the solvent alone. This vapour pressure of the solution at a given temperature is found to be lower than the
vapour pressure of the pure solvent at the same temperature.
p°
According to Raoults's law if p is vapour pressure of the solvent, 1 be Vapour pressure
its mole fraction, p1 be its vapour pressure in the pure state,
Then,
p1  1
and p1 = 1 p1
The proportionality constant is equal to the vapour pressure of pure
0  solvent 1
solvent, p1 . A plot between the vapour pressure and the mole fraction
of the solvent is linear.
o
Ideal Solutions + pB pB
Vapour pressure

p = pA
o
z The solution in which solute-solute interaction in pure state and pA
solvent-solvent interaction in pure state are almost similar to the pB
solute-solvent interactions in solution.
z H(mix) = 0 pA
z V(mix) = 0
B
z S(mix) > 0 B = 0 B = 1
z G(mix) < 0 A = 1 A = 0
z pA = pºA × A and pB = pºB × B.

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NEET Solutions 105
Ideal and Non-Ideal Solutions

Non-Ideal Solution
Ideal Solution
Positive Deviation Negative Deviation

1. Obey’s Raoult’s law 1. Dis-obey’s Raoult’s law 1. Dis-obey’s Raoult’s law

2. pA = p°A.XA 2. pA > p°AXA 2. pA < p°AXA
pB = p°B XB pB > p°BXB pB < p°BXB
Ps(obs) = Ps(ideal) Ps(obs) > Ps(ideal) Ps(obs) < Ps(ideal)

3. Hmix = 0, Smix > 0 3. Hmix = +ve, Smix > 0 3. Hmix = –ve, Smix > 0
Vmix = 0, Gmix < 0 Vmix = +ve, Gmix < 0 Vmix = –ve, Gmix < 0
4. Interaction 4. Interaction 4. Interaction
A – B = A – A/B – B A – B < A – A/B – B A – B > A – A/B – B
e.g. Chlorobenzene + e.g. C2H5OH + H2O, e.g. CH3COCH3 + CHCl3,
Bromobenzene,
Benzene + Toluene, Acetone + Carbon Chloroform + Benzene,
n-pentane + n-hexane, disulphide, Acetone + Aniline,
C2H5Br + C2H5I, Acetone + Ethyl alcohol, HCl + Water,
CCl4 + SiCl4 Acetone + Benzene, HNO3 + Water
Ethyl alcohol + Water,
Carbon tetrachloride +
Chloroform

p°A p°A p°A

p°B
p°B p°B

A = 1 A = 0 A = 1 A = 0 A = 1 A = 0
B = 0 B = 1 B = 0 B = 1 B = 0 B = 1

 For these solutions, at a  For these solutions a

particular composition Vapour particular compositon Vapour
pressure in maximum, at this pressure in minimum, at this
point minimum boiling Azeotrope point maximum boiling
is formed. Azeotrope is formed.

AZEOTROPIC MIXTURE

This type of liquid mixture, having a definite composition and boiling like a pure liquid is called azeotropic
mixture or constant boiling mixture. The azeotropic mixture cannot be separated by fractional distillation, such
solutions are called azeotropic solutions and this phenomenon is known as azeotropy.
Maximum Boiling Azeotropes are the solutions formed by solutions with negative deviation under azeotropic
conditions. Examples of this category are aqueous 68% (w/w) solution of HNO3 and 20.3% (w/w) aq. HCl.
Minimum Boiling Azeotropes are the solutions formed by solution with positive deviation from the ideal
behaviour. Examples of this category are aq. 95.37% (w/w) ethanol solution and aq. 71.7% (w/w) propanol.
Note : Azeotropic composition of solution is dependent on external pressure.

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106 Solutions NEET

EXERCISE
1. The example of solid solution is

(1) Glucose in water

(2) Copper in gold

(3) Camphor in nitrogen

(4) Oxygen in nitrogen

2. The tanks used by scuba divers are filled with air diluted with 11.7% He,

(1) 56.2% N2 and 32.1% O2

(2) 56.2% O2 and 32.1% N2

(3) 50.2% N2 and 38.1% O2

(4) 50.2% O2 and 38.1% N2

3. The increase in the temperature of the aqueous solution placed in a closed vessel will result in its

(1) Molarity to increase

(2) Molarity to decrease

(3) Mole fraction to increase

(4) Mass % to increase

4. In a binary solution

(1) Solvent may be liquid (2) Solvent may be solid

(3) Solute may be gas (4) Any of these

5. The temperature at which the vapour pressure of a liquid equals external pressure is called

(1) Freezing point (2) Boiling point

(3) Melting point (4) Critical temperature

6. Vapour pressure is the pressure exerted by vapours

(1) In equilibrium with liquid (2) In any condition

(3) In an open system (4) In atmospheric conditions

7. A sample of toothpaste weighing 500 g, on analysis was found to contain 0.2 g of fluorine. The concentration
of fluorine in ppm is

(1) 4 × 103 (2) 4 × 102

(3) 4 × 10 (4) 2 × 102

8. 100 ml of liquid A and 25 ml of liquid B is mixed to give a solution which does not obey Raoult's law. The
volume of the solution

(1) Will be 125 ml (2) Can be > or < than 125 ml

(3) Can be >, = or < than 125 ml (4) Will be less than 125 ml

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NEET Solutions 107
COLLIGATIVE PROPERTIES

The properties of a solution which depend entirely upon the number of particles of the solute (ions, molecules
or associated molecules) in a given volume of solvent and not on the chemical nature of the solute molecules
are defined as colligative properties
(i) Relative lowering of vapour pressure or lowering of vapour pressure
(ii) Osmotic pressure
(iii) Elevation in boiling point
(iv) Depression in freezing point
z Relative Lowering of Vapour Pressure: The relative lowering in vapour pressure of an ideal solution
is equal to the mole fraction of solute at that temperature.

p  p n2
= 2 =
p n1  n2
p  p w M
For dilute solutions,  
p m W
w = wt of solute, m = Mol. wt. of solute
W = wt. of solvent, M = Mol. wt. of solvent

Example 3 : What weight of solute (mol. wt. 60) is required to dissolve in 180 g of water to reduce the vapour
pressure to 4/5 th of pure water?

p  pS w M
Solution : 
p m W
4
p  p
5 w  18

p 60  180
w = 120 g

z Osmotic Pressure: The excess pressure which must be applied on a solution to prevent the passage of
solvent through a semipermeable membrane.
 = CRT
z For isotonic solutions, 1 = 2
z Solution of higher osmotic pressure is hypertonic while solution of lower osmotic pressure is hypotonic.
z 0.9% (mass/volume) NaCl solution, called normal saline solution and it is safe to inject it intravenously.
z Reverse Osmosis : If external pressure is applied higher than Osmotic Pressure of solution then the solvent
will start flowing from solution to pure solvent. This is called Reverse Osmosis.

Example 4 : What will be the concentration of sucrose solution which develops an osmotic pressure of 2 atm
at 27°C?
Solution : Osmotic pressure
  = CRT
 2 = C × 0.0821 × 300
 C = 0.081 M

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108 Solutions NEET

Example 5 : What would be osmotic pressure of 0.05 N sucrose solution at 5°C? Find out the concentration
of glucose solution (in g/L) which is isotonic with this sucrose solution.
Solution :  = CRT
 = 0.05 × 0.082 × 278 atm
Concentration of sucrose solution is 0.05 N or 0.05 × 342 g = 17.1 g/L
Isotonic solutions have same osmotic pressure.
sugar = glucose
Csugar = Cglucose
17.1 x
=
342 180
x = 9 g/L

z Elevation in Boiling Point or Ebullioscopy: The temperature at which vapour pressure of the liquid
becomes equal to the atmospheric pressure is known as its boiling point.
We know that the vapour pressure of the solution is lower than that of the pure solvent and vapour pressure
increases with increase in temperature. Hence the solution has to be heated more to make the vapour
pressure equal to the atmospheric pressure.
Alternatively, the elevation in boiling point may be expained on the basis of plots of vapour pressure versus
temperature as follows :
B D

Atmospheric E
F
Pressure t
Vapour pressure

n
lve
So
on
A luti
So
C

TO T
Temperature
Elevation in boiling point
Vapour pressure of the solvent increases with increase in temperature as shown by the curve AB. As at
any temperature, vapour pressure of the solution is less than that of the solvent, the curve for the solution
lies below that of the solvent, as shown by the curve CD. The temperatures at which the vapour pressure
of the solvent and the solution become equal to the atmospheric pressure are T0 and T respectively.
Obviously T > T0. The difference, called the elevation in boiling point, Tb, is given by
Tb = T – T0
Molal Elevation Constant or Ebullioscopic Constant, Kb. It is the increase in boiling point when
the molality of the solution is unity. It is a property of solvent.
Tb = Kbm when m = 1, Tb = Kb
K b  WB  1000
Tb 
MB  WA
WB  1000
MB =  Kb
Tb  WA

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NEET Solutions 109

RTb2 MoRTb2
Kb   (K kg mol1)
1000 l v 1000 Hv
where, Tb = boiling point of the liquid (pure solvent)
lv = latent heat of vaporization per gram of the solvent,
Hv
lv = Mo , where Mo = molecular weight of solvent,
Hv = heat of vaporization per mole of solvent

Example 6 : The boiling point of pure acetone is 56.38°C. When 0.707 g of a compound is dissolved in
10 g of acetone there is elevation to 56.88°C in b.p. What is the mol. wt. of the compound?
(Kb of acetone = 1.72 K kg mole–1)
Solution : Tb = Kb × m

0.707 1000
 (56.88  56.38)  1.72  
MB 10
 MB = 243.2

z Depression in Freezing Point or Cryoscopy : The property of decrease in freezing point when some
non-volatile solute is dissolved. The depression in freezing point is given by Tf .
Freezing Point : Temperature at which the liquid and the solid forms of the same substance are in
equilibrium and hence have same vapour pressure.
We know that vapour pressure of the solution is less than that of the pure solvent. As freezing point is
the temperature at which the vapour pressure of the liquid and the solid phase are equal, therefore for the
solution, this will occur at lower temperature (lower the temperature lower the vapour pressure).
The graph explains this.
Liquid solvent

n
tio
Vapour pressure

lu
Solid So
solvent
Tf

Tf Tf°
Temperature
Tf = T°f – Tf
Molal Depression Constant or Cryoscopic Constant (Kf). It is the decrease in freezing point when the
molality of solution is unity. It is also a property of solvent.
Tf = Kf.m
when m = 1, Tf = Kf

K f  w 2 / M2
Tf =
w1 / 1000

K f  w 2  1000
M2 =
Tf  w1

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110 Solutions NEET

M0RTf2
Kf  (K  kg  mol1)
1000  freezH
freezH° = standard enthalpy of freezing.
Tb and Tf can be represented graphically as
Solvent
Solution
1
uid
liq

lid P
so
uid
VP liq
(atm)
lid Q
so Tb
Tf

Tf Tf Tb Tb
T(K)
At both these points P and Q there is a phase transfer equilibrium between liquid solvent and solid solvent
that clearly show that even in case of solutions, it is only the solvent that freezes.

Example 7 : 5 g of a substance when dissolved in 50 g water lowers the freezing by 1.2°C. Calculate molecular
wt. of the substance if molal depression constant of water is 1.86° K kg mole–1.

1000  K f  w1
Solution : M=
Tf  w 2

1000  1.86  5
M= = 155
1.2  50

Example 8 : An aqueous solution has 5% urea and 10% glucose by weight. What will be the freezing point
of this solution? (Kf water = 1.86 K kg mole–1)
Solution : T = Turea + Tglucose

1000  1.86  5 1000  1.86  10

T = 
60  85 180  85

= 1.824 + 1.216
= 3.04
Freezing point = 0 – 3.04 = – 3.04°C.

Abnormal Molecular Mass

z When the mol. mass determined by any of the colligative properties comes out to be different than the
theoretically expected value, the substance is said to show abnormal mol. mass. This is observed when
the solution is non ideal, when the solute undergoes association or dissociation.
z The various relations as derived above for the colligative properties are applicable only to the solutions of
non-electrolytes which do not undergo any dissociation or association in the solution. In case of the

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NEET Solutions 111
aqueous solutions of electrolytes i.e. acids (HCl, H2SO4, CH3COOH, etc), inorganic bases (NaOH, KOH,
Ca(OH)2 etc) and salts (NaCl, KCl, KNO3, BaCl2 etc) which dissociate completely or to small extent in
the solution, the number of particles in the solution increases.
z For example in case of KCl, in the aqueous solution, each molecule dissociates to give two ions
(KCl  K+ + Cl¯). Thus the number of particles become double, the observed value of colligative property
is double than expected value and molecular mass is half of the expected value.
z Similarly, in case of certain substances, association takes place in the solution. For example, acetic acid,
benzoic acid etc. when dissolved in benzene exist as doubly associated molecules in benzene due to
hydrogen bonding which may be represented as

O H O

R C C R

O H O

z To sum up, abnormal molecular masses are observed in any one of the following cases :
(1) When the solution is non-ideal i.e. the solution is not dilute.
(2) When the solute undergoes association in the solution.
(3) When the solute undergoes dissociation in the solution.

van’t Hoff’s Factor

z It is defined as the ratio of the experimental value of the colligative property to the calculated value of
the colligative property i.e.

Experimental value of the colligative property

z van't Hoff factor (i) =
Calculated value of the colligative property when the solution behaves ideally

Observed value of the colligative property Normal molecular mass

= 
Normal value of the same property Observe molecular mass

i = T(obs)/T(cal) = (obs)/(cal) = p(obs)/p(cal) = M(cal)/M(obs)

z When i = 1 Neither association nor dissociation
i >1 Dissociation
i < 1 Association

1
We know that colligative properties 
Molecular mass of solute

 Normal Molecular mass of solute

i
Observed Molecular mass of solute

Degree of Association :
z Degree of association is defined as the fraction of the total substance which exists in the form of
associated molecules i.e.

No. of moles associated

z Degree of association,  
Total number of moles taken
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112 Solutions NEET

z Suppose n simple molecules of the solute A associate to form the associated molecule An so that we have
the equilibrium.

nA An

z If  is the degree of association and we start with one mole of A, then at equilibrium
Number of moles of A = 1 – 
Number of moles of An = /n
 The total number of moles = 1 –  + /n.
z Since the colligative property is proportional to the number of moles of the solute present in solution,
therefore,


1  1 i
n 1 
van't Hoff factor i =  i = 1    1   
1 n  1
1
n

Mc M  Mc n
But i  ,  o  , where n is the no. of simple molecules to form an associated molecule.
Mo Mo (n  1)

Degree of Dissociation
z The degree of dissociation is defined as the fraction of the total substance that undergoes
dissociation i.e.,

z Degree of dissociation ( )  No. of moles dissociated

Total number of moles taken

z Suppose a molecule of an electrolyte dissociates to give n ions An  nA

1 0
1  n

1  (n  1) i1
 van't Hoff factor, i = or    i = 1 + (n - 1)
1 n 1

Normal (calculated) molecular mass Mc

But i = 
Observed molecular mass Mo

Mc  Mo

Mo (n  1)

Modified Equation of Colligative Property Involving Van't Hoff Factor :

po  p in
z 
po in  N

z Tf = iKfm
z TB = iKbm
z  = iCRT.

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NEET Solutions 113

Example 9 : In a solvent 50% of benzoic acid dimerises while rest ionises, determine molar mass of acid which
is observed and also its van’t Hoff factor.
Solution : Let we have 2 moles of benzoic acid
 
(1    n )   1    
 n
i
2
1
1 1 2  1 1
 2  1.25
2
Molar mass of solute is found to be less than its normal value.

EXERCISE
9. A solution with osmotic pressure 1 is separated from another solution of osmotic pressure 2 by SPM, solvent
flows from 1  2, then
(1) 1 > 2
(2) 1 < 2
(3) Solutions are isotonic
(4) Solutions are ideal
10. 1 mole glucose is added to 1 L of water. Kb(H2O) = 0.512 K kg mole–1, boiling point of solution will be
(1) 373.512°C (2) 100.512°C
(3) 99.488°C (4) 372.488°C
11. The value of ebullioscopic constant depends upon
(1) HSolution (2) Nature of solvent
(3) Nature of solute (4) Freezing point of solution
12. 3% solution of glucose is isotonic with 1% solution of a non-volatile non-electrolyte substance. The molecular
mass of the substance would be
(1) 180 (2) 360
(3) 420 (4) 60
13. van’t Hoff factor for SrCl2 at 0.01 M is 1.6. Percent dissociation of SrCl2 is
(1) 70 (2) 55
(3) 40 (4) 30
14. Correct increasing order of osmotic pressure for the following is
(1) Sucrose(0.1 M) < glucose(0.5 M) < urea(1M) < NaCl(2M)
(2) Glucose(0.5 M) < urea(1M) < NaCl(2M) < Sucrose(0.1 M)
(3) Urea(1M) < NaCl(2M) < glucose(0.5 M) < Sucrose(0.1 M)
(4) NaCl(2M) < sucrose(0.1 M) < glucose(0.5 M) < urea(1M)
15. Which among the following has highest boiling point?
(1) 1 M glucose (2) 1 M KCl
(3) 1 M Al(NO3)3 (4) 1 M Na2SO4

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114 Solutions NEET

16. The solution having minimum boiling point is

(1) 0.1 M C6H12O6 (2) 0.1 M CaCl2
(3) 0.1 M NaCl (4) 0.5 M AlCl3
17. What is the boiling point of 1 molal aqueous solution of NaCl [Kb = 0.52 K molal–1]
(1) 99.48°C (2) 98.96°C
(3) 100.52°C (4) 101.04°C
18. For a non electrolytic solution
(1) i = 0.5 (2) i = –1
(3) i = 0 (4) i = 1

‰ ‰ ‰

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 nt
me nment
sigAnssig
Assignment
As

Assignment
7. The average osmotic pressure of human blood is 7.8
SECTION - A
bar at 37ºC. The concentration of aqueous NaCl
NCERT Based MCQs solution that could be used in the blood stream is
[NCERT Pg. 54]
1. Which of the following liquid pairs show a positive
deviation from Raoult’s law? [NCERT Pg. 47] (1) 0.15 mol/L (2) 0.32 mol/L
(1) Water – Hydrochloric acid (3) 0.30 mol/L (4) 0.45 mol/L
(2) Ether – Chloroform 8. A 5 molar solution of H2SO4 is diluted from 1 litre to
(3) Water – Nitric acid 10 litres. The normality of the resulting solution is
[NCERT Pg. 38]
(4) Benzene – Acetone
(1) 0.25 N (2) 1.0 N
2. If the elevation in boiling point of a solution of
10 g of solute (mol wt. = 100) in 100 g of water is (3) 2.0 N (4) 0.5 N
Tb, then the ebullioscopic constant of water is 9. Which of the following with respect to negative
[NCERT Pg. 50] deviation is correct? [NCERT Pg. 47]
(1) 100 Tb (2) Tb (1) GSol < 0, HSol < 0, SSol < 0

Tb (2) GSol < 0, HSol < 0, SSol > 0

(3) (4) 10 Tb
10 (3) GSol > 0, HSol < 0, SSol < 0
3. The order of K H values of A, B and C gases is (4) GSol > 0, HSol > 0, SSol > 0
KHA > KHC > KHB . Then, the correct order of their
10. The total vapour pressure, Ptotal (in torr) for a mixture
solubility in a liquid at given temperature is
of two volatile components, A and B is given by:
[NCERT Pg. 41]
Ptotal = 220 – 110 XB
(1) A > C > B (2) B > A > C
where, XB is mole fraction of component, B is mixture
(3) B > C > A (4) A > B > C
Hence, pressure (in torr) of pure components, A and
4. Which of the following aqueous solution will have the
highest freezing point? [NCERT Pg. 58] B are respectively [NCERT Pg. 43]

(1) 0.1 m NaCl solution (2) 0.1 m Sugar solution (1) 110, 220 (2) 220, 110

(3) 0.1 m MgCl2 solution (4) 0.1 m AlCl3 solution (3) 220, 330 (4) 220, 150

5. The molarity of a solution of sodium chloride in 11. 2 m aqueous solution of H2SO4 contains
500 ml of water containing 5.85 g of sodium chloride [NCERT Pg. 39]
is [NCERT Pg. 38] (1) 49 g of H2SO4 per 250 g of solution
(1) 0.25 M (2) 1.0 M (2) 98 g of H2SO4 per 250 g of solvent
(3) 2.0 M (4) 0.2 M (3) 49 g of H2SO4 per 250 g of solvent
6. Which of the following is/are colligative property? (4) 98 g of H2SO4 per 250 g of solution
[NCERT Pg. 49]
12. The concentration of sucrose solution which is
(1) Lowering in vapour pressure isotonic with 0.02 M solution of Na2SO4 (assume
(2) Freezing point complete dissociation) at 30°C is [NCERT Pg. 56]
(3) Boiling point (1) 0.04 M (2) 0.02 M
(4) Both (2) & (3) (3) 0.06 M (4) 0.08 M

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116 Solutions NEET

13. The effect of increase in the temperature of an (1) 135 and 254 torr (2) 119 and 135 torr
aqueous solution of a substance is (3) 119 and 254 torr (4) 334 and 135 torr
[NCERT Pg. 36]
4. Calculate the mole fraction of toluene in the vapour
(1) Decrease in its molality phase which is in equilibrium with a solution of
(2) Increase in its molality benzene and toluene having a mole fraction of toluene
0.5. (The vapour pressure of pure Benzene is 119 torr
(3) Increase in its mole fraction
and that of toluene is 37 torr at the same
(4) Decrease in its molarity temperature)
14. When a solute associates in solution, then van’t Hoff (1) 0.5 (2) 0.75
factor (i) is [NCERT Pg. 58]
(3) 0.625 (4) 0.237
(1) Greater than one (2) Less than one
5. What is the molality of C2H5OH in water solution
(3) Equal to one (4) Zero which will freeze at – 10oC? [Mol. wt. of C2H5OH =
15. If degree of dissociation of Al2(SO4)3 is 25% in a 46, Kf for water = 1.86]
solvent, then [NCERT Pg. 58] (1) 6.315 m (2) 63.15 m
(1) Normal boiling point = experimental boiling point (3) 3.540 m (4) 5.3 m
(2) Normal osmotic pressure = 2 × experimental 6. Which of the following pairs of solutions can we
osmotic pressure expect to be isotonic at the same temperature?
(3) Normal molecular weight = 2 × experimental (1) 0.1 M NaCl and 0.1 M Na2SO4
molecular weight (2) 0.1 M urea and 0.1 M NaCl
1 (3) 0.1 M urea and 0.2 M MgCl2
(4) Normal freezing point = × experimental
4
freezing point (4) 0.1 M Ca(NO3)2 and 0.1 M Na2SO44
7. In the case of osmosis, solvent molecules move from
(1) Higher vapour pressure to lower vapour pressure
SECTION - B
(2) Higher concentration to lower concentration
Objective Type Questions
(3) Lower vapour pressure to higher vapour pressure
1. The boiling points of C6H6, CH3OH, C6H5 NH2 and (4) Higher osmotic pressure to lower osmotic
C 6 H 5 NO 2 are 80 o C, 65 o C, 184 o C and 212 o C pressure
respectively. Which of the following will have highest
8. Assuming salts to be 90% dissociated, which of the
vapour pressure at room temperature?
following will have highest osmotic pressure?
(1) C6H6 (2) CH3OH (1) Decimolar Al2(SO4)3
(3) C6H5NH2 (4) C6H5NO2 (2) Decimolar BaCl2
2. The vapour pressure of a solvent decreased by 10 mm (3) Decimolar Na2SO4
of Hg when a non-volatile solute was added to the (4) A solution obtained by mixing equal volumes of
solvent. The mole fraction of solute in solution is 0.2, (2) and (3) and filtering
what would be the mole fraction of solute if decrease
9. Which of the following aqueous solution has
in vapour pressure is 20 mm of Hg?
maximum freezing point?
(1) 0.8 (2) 0.6 (1) 0.01 M NaCl (2) 0.005 M C2H5OH
(3) 0.4 (4) 0.2 (3) 0.005 M MgI2 (4) 0.01 M MgSO4
3. At 40°C the vapour pressure (in torr) of mixture of 10. The vapour pressure of a dilute aqueous solution of
methyl alcohol and ethyl alcohol is represented by glucose is 750 mm Hg at 373 K. Calculate the mole
P = 199x + 135 fraction of solute. (The vapour pressure of pure water
is 760 mm Hg at 373 K)
(where x is the mole fraction of methyl alcohol.) What
are the vapour pressures of pure methyl alcohol and (1) 0.013 (2) 1.3
pure ethyl alcohol at 40°C? (3) 0.13 (4) None of these

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NEET Solutions 117
11. If Pº and P are the vapour pressure of solvent and 18. Which of the following equimolar solution have
solution n1 and n2 are the moles of solute and solvent highest vapour pressure?
then (1) Glucose (2) NaCl
 n2   n2  (3) K2SO4 (4) K4[Fe(CN)6]
(1) Pº  P   (2) P  Pº  
 n1  n2   n1  n2  19. If solute-solvent interactions are more than
solute-solute and solvent - solvent interactions then
 n1 
(3) Pº  P   (4) Pº = P × n1 (1) It is ideal solution
 n1  n2 
(2) It is non-ideal solution with positive deviation
12. Which of the following is correct about a solution
showing positive deviation? (3) It is non-ideal solution with negative deviation

(1) Vapour pressure observed will be the less than (4) None of these
that calculated from Raoult’s law 20. On mixing 10 ml of ethanol with 10 ml of benzene
(2) Minimum boiling azeotrope will be formed the total volume of the solution is

(3) Hmix < 0 (1) > 20 ml (2) < 20 ml

(4) Vmix < 0 (3) = 20 ml (4) Can't be predicted

13. If any solute ‘A’ dimerises in water at 1 atm pressure 21. W gm of non-volatile organic substance of molecular
and the boiling point of this solution is 100.52°C. If mass M is dissolved in 250 gm benzene. Molal
2 moles of A is added to 1 kg of water and Kb for water elevation constant of benzene is Kb. Elevation in its
is 0.52°C/molal, calculate the percentage association boiling point is given by
of A M 4K b W
(1) K W (2)
(1) 50% (2) 30% b M
(3) 25% (4) 100% Kb W Kb W
(3) (4)
14. An aqueous solution freezes at –0.36°C. Kf and Kb 4M M
for water are 1.8 and 0.52°C - kg mol–1 respectively 22. The molal elevation constant for water is
then value of boiling point of solution at 1 atm pressure 0.56° K kg mol–1. Calculate the boiling of the solution
is made by dissolving 6.0 gm of urea in
(1) 101.04°C (2) 100.1°C 200 gm of water

(3) 0.104°C (4) 100°C (1) 100.28° C

15. The relationship between osmotic pressure (P) at 273 (2) 100° C
K when 10 g glucose (P1), 10 g urea (P2) and 10 g (3) 0.28° C
sucrose (P3) are dissolved in 250 ml of water is (4) 200.28° C
(1) P1 > P2 > P3 (2) P2 > P1 > P3 23. The graph plotted vapour pressure vs mole fraction
(3) P2 > P3 > P1 (4) P3 > P2 > P1 is the graph for which of the following example
16. Arrange the following aqueous solutions in the order
of their increasing boiling points
(i) 10–4 M NaCl (ii) 10–3 M Urea PB
10–3 10–3
PB
(iii) M MgCl2 (iv) M NaCl V.P
(1) (i) < (ii) < (iv) < (iii) (2) (ii) < (i) = (iii) < (iv)
(3) (i) < (ii) < (iii) < (iv) (4) (iv) < (iii) < (i) = (ii)
17. K4Fe(CN)6 is supposed to be 40% dissociated when
Mole fraction
1M solution prepared. Its boiling point is equal to
another solution of 20% A. Considering molality = (1) Hexane and Heptane
molarity. The molecular weight of A is (2) Ethyl bromide and Ethyl chloride
(1) 77 (2) 67 (3) Ethanol and Water
(3) 57 (4) 47 (4) Chloroform and Acetone
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118 Solutions NEET

24. Minimum boiling azeotropes and Maximum boiling 32. The van’t Hoff factor (i) for a 0.3 molal aqueous solution
azeotropes respectively on the examples of of urea is
(1) Non-ideal solution following +ve deviation and (1) 0.3 (2) 0.1
ideal solution (3) 1.3 (4) 1.0
(2) Ideal solution and non-ideal solution following
33. Which of the following mixtures will show positive
negative deviation
deviation from ideal behaviour?
(3) Both are ideal solution (1) H2O and HNO3
(4) Non-ideal solution showing positive deviation and (2) H2O and C2H5OH
Non-ideal solution showing negative deviation
25. The molecular weight of NaCl determined by studying O
freezing point depression of its 0.5% aqueous (3) CHCl3 and CH 3 — C — CH 3
solution is 30. The apparent degree of dissociation (4) n-hexane and n-heptane
of NaCl is
34. What is the mole fraction of the solute in 2.5 m
(1) 0.95 (2) 0.45
aqueous solution?
(3) 0.60 (4) 0.35
(1) 0.043 (2) 0.012
26. Expression relating mole fraction and molality is
(3) 0.43 (4) 4.3
(A : solute and B ; solvent)
35. Consider the following statements.
  1000   1000 (I) Maximum boiling azeotrope is formed by negative
(1) m  A (2) m  B
(1   A )MB (1   A )MB deviation from Raoult’s law.
mMB (II) The value of Kb & Kf depends upon nature of
(3) B  (4) None of these
1  mMB solute.
27. For CrCl 3 .xNH 3 , elevation in boiling point of (III) Minimum boiling azeotrope is formed by positive
1 molal solution is double of 1 molal solution of deviation from Raoult’s law.
glucose; hence x is, if complex is 100% ionised
Choose the correct statements.
(1) 4 (2) 5
(1) (I) & (II) (2) (I), (II) & (III)
(3) 6 (4) 3
(3) (I) & (III) (4) (II) & (III)
28. pH of 1M HA (weak acid) is 3. Hence, van’t Hoff factor
is 36. Which of the following statement is/are correct?

(1) 1.01 (2) 1.001 (1) Higher the value of KH, higher is the solubility of
the gas in the liquid.
(3) 1.1 (4) 1.04
(2) Different gases have different KH values at the
29. Which of the following aqueous solutions has the
same temperature
highest boiling point?
(3) Mole fraction of gas in the solution is inversely
(1) 0.1 M KNO3 (2) 0.1 M Al2(SO4)3
proportional to the partial pressure of the gas
(3) 0.1 M BaCl2 (4) 0.1 M Na3PO4
(4) All of these
30. Which of the following is not a colligative property?
37. 0.04 M CaCl2 is isotonic with 0.1 M glucose solution
(1) Vapour pressure of pure liquid solvent at 300 K temperature, then calculate the percentage
(2) Elevation in boiling point of ionisation of CaCl2.
(3) Depression in freezing point (1) 55% (2) 70%
(4) Osmotic pressure (3) 60% (4) 75%
31. Which of the following is correct for a solution showing 38. If a reactant ‘A’ is tetramerised, then the van’t Hoff
positive deviation from Raoult’s law? factor (i) is equal to
(1)  Vmix  0,  Hmix  0 (2)  Vmix  0,  Hmix  0 (1) 0.75 (2) 0.20
(3)  Vmix  0,  Hmix  0 (4)  Vmix  0,  Hmix  0 (3) 0.15 (4) 0.25
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NEET Solutions 119
39. Which of the following will not obey Henry law? 3. For an ideal solution, the correct option is
(1) CO (2) H2 [NEET-2019]
(1) mix S = 0 at constant T and P
(3) O2 (4) NH3
(2) mix V  0 at constant T and P
40. Which one of the following gases will have minimum
value of KH (kbar) at same temperature in aqueous (3) mix H = 0 at constant T and P
medium? (4) mix G = 0 at constant T and P
(1) H2 (2) CO2 4. Which of the following is dependent on
temperature? [NEET-2017]
(3) N2 (4) O2
(1) Molality (2) Molarity
41. Match the following correctly.
(3) Mole fraction (4) Weight percentage
a. Positive deviation (i)  Hmix = – ve 5. If molality of the dilute solution is doubled, the
value of molal depression constant (Kf) will be
b. Ideal solution (ii)  Vmix = 0
[NEET-2017]
c. Cryoscopic constant (iii)  Vmix =  ve
(1) Doubled (2) Halved
2
RT0 (3) Tripled (4) Unchanged
d. Negative deviation (iv)
1000lf 6. The van't Hoff factor (i) for a dilute aqueous solution
of the strong electrolyte barium hydroxide is
Rlf 2
(v) [NEET-Phase-2-2016]
1000 To
(1) 0 (2) 1
(1) a(iii), b(ii), c(iv), d(i)
(3) 2 (4) 3
(2) a(i), b(ii), c(v), d(iii) 7. Which one of the following is incorrect for ideal
(3) a(iii), b(ii), c(v), d(i) solution? [NEET-Phase-2-2016]
(1) Hmix = 0
(4) a(i), b(ii), c(iv), d(iii)
(2) Umix = 0

SECTION - C (3) P = Pobs – Pcalculated by Raoult's law = 0

(4) Gmix = 0
Previous Years Questions 8. At 100°C the vapour pressure of a solution of 6.5 g
1. Which of the following statements is correct of a solute in 100 g water is 732 mm. If Kb = 0.52,
regarding a solution of two components A and B the boiling point of this solution will be
exhibiting positive deviation from ideal behaviour? [NEET-2016]
[NEET-2019 (Odisha)] (1) 103°C (2) 101°C
(1) Intermolecular attractive forces between A-A (3) 100°C (4) 102°C
and B-B are equal to those between A-B 9. Which of the following statements about the
composition of the vapour over an ideal 1 : 1 molar
(2) Intermolecular attractive forces between A-A
mixture of benzene and toluene is correct?
and B-B are stronger than those between A-B
Assume that the temperature is constant at 25°C.
(3) mixH = 0 at constant T and P (Given, Vapour Pressure Data at 25°C, benzene
= 12.8 kPa, toluene = 3.85 kPa) [NEET-2016]
(4) mixV = 0 at constant T and P
(1) Not enough information is given to make a
2. The mixture that forms maximum boiling azeotrope prediction
is [NEET-2019]
(2) The vapour will contain a higher percentage of
(1) Water + Nitric acid benzene
(2) Ethanol + Water (3) The vapour will contain a higher percentage of
toluene
(3) Acetone + Carbon disulphide
(4) The vapour will contain equal amounts of
(4) Heptane + Octane benzene and toluene

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120 Solutions NEET

10. What is the mole fraction of the solute in a (1) Greater than one and greater than one
1.00 m aqueous solution? [Re-AIPMT-2015] (2) Less than one and greater than one
(1) 0.0354 (2) 0.0177 (3) Less than one and less than one
(3) 0.177 (4) 1.770 (4) Greater than one and less than one
11. Which one is not equal to zero for an ideal 17. The freezing point depression constant for water is
solution? [AIPMT-2015] –1.86° Cm –1. If 5.00 g Na2SO4 is dissolved in 45.0
g H2O, the freezing point is changed by –3.82°C.
(1) P = Pobserved – PRaoult Calculate the van't Hoff factor for Na2SO4
(2) Hmix [AIPMT (Prelims)-2011]
(3) Smix (1) 0.381 (2) 2.05
(4) Vmix (3) 2.63 (4) 3.11
12. The boiling point of 0.2 mol kg–1 solution of X in 18. A 0.1 molal aqueous solution of a weak acid is 30%
water is greater than equimolal solution of Y in ionized. If Kf for water is 1.86C/m, the freezing point
water. Which one of the following statements is of the solution will be [AIPMT (Mains)-2011]
true in this case? [AIPMT-2015] (1) –0.36C (2) –0.24C
(1) Y is undergoing dissociation in water while X (3) –0.18C (4) –0.54C
undergoes no change
19. 200 mL of an aqueous solution of a protein contains
(2) X is undergoing dissociation in water its 1.26g. The osmotic pressure of this solution at
300 K is found to be 2.57 × 10–3 bar. The molar mass
(3) Molecular mass of X is greater than the of protein will be (R = 0.083 L bar mol–1 K–1)
molecular mass of Y
[AIPMT (Mains)-2011]
(4) Molecular mass of X is less than the
molecular mass of Y (1) 31011 g mol–1 (2) 61038 g mol–1

13. Which one of the following electrolytes has the (3) 51022 g mol–1 (4) 122044 g mol–1
same value of van't Hoff's factor (i) as that of 20. A solution of sucrose (molar mass = 342 g mol–1)
Al2(SO4)3 (if all are 100% ionised)? [AIPMT-2015] has been prepared by dissolving 68.5 g of sucrose
in 1000 g of water. The freezing point of the solution
(1) K4[Fe(CN)6] (2) K2SO4
obtained will be (Kf for water = 1.86 K kg mol–1)
(3) K3[Fe(CN)6] (4) Al(NO3)3
[AIPMT (Prelims)-2010]
14. Of the following 0.10 m aqueous solutions, which
(1) –0.372C (2) –0.520C
one will exhibit the largest freezing point depression?
(3) +0.372C (4) –0.570C
[AIPMT-2014]
21. An aqueous solution is 1.00 molal in K. Which
(1) KCl (2) C6H12O6 change will cause the vapour pressure of the
(3) Al2(SO4)3 (4) K2SO4 solution to increase? [AIPMT (Prelims)-2010]

15. pA and pB are the vapour pressure of pure liquid (1) Addition of NaCl
components, A and B, respectively of an ideal binary (2) Additon of Na2SO4
solution. If x A represents the mole fraction of
(3) Addition of 1.00 molal K
component A, the total pressure of the solution will
be [AIPMT (Prelims)-2012] (4) Addition of water
(1) pB + xA (pB – pA) (2) pB + xA (pA – pB) 22. A 0.0020 m aqueous solution of an ionic compound
Co(NH3)5 (NO2)Cl freezes at –0.00732C. Number
(3) pA + xA (pB – pA) (4) pA + xA (pA – pB) of moles of ions which 1 mol of ionic compound
16. The van't Hoff factor i for a compound which produces on being dissolved in water will be
undergoes dissociation in one solvent and (Kf = 1.86C /m) [AIPMT (Prelims)-2009]
association in other solvent is respectively (1) 3 (2) 4
[AIPMT (Prelims)-2011] (3) 1 (4) 2

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NEET Solutions 121
23. 0.5 molal aqueous solution of a weak acid (HX) is 29. A solution of urea (mol. mass 60g mol-1) boils at
20% ionized. If Kf for water is 1.86 K kg mol–1, the 100.18C at the atmospheric pressure. If kf and kb
lowering in freezing point of the solution is for water are 1.86 and 0.512K kg mol–1 respectively,
the above solution will freeze at
[AIPMT (Prelims)-2007]
(1) –0.56 K (2) –1.12 K [AIPMT (Prelims)-2005]

(3) 0.56 K (4) 1.12 K (1) –6.54C (2) 6.54C

24. A solution containing 10g per dm 3 of urea (3) 0.654C (4) –0.654C
(molecular mass = 60 g mol–1) is isotonic with a
30. A solution has a 1 : 4 mole ratio of pentane to hexane.
5% solution of a non-volatile solute. The molecular
The vapour pressure of the pure hydrocarbons at 20C
mass of this non-volatile solute is
are 440 mm of Hg for pentane and 120 mm of Hg for
[AIPMT (Prelims)-2006] hexane. The mole fraction of pentane in the vapour
phase would be [AIPMT (Prelims)-2005]
(1) 250 g mol–1 (2) 300 g mol–1
(1) 0.549 (2) 0.200
(3) 350 g mol–1 (4) 200 g mol–1
25. 1.00 g of a non-electrolyte solute (molar mass (3) 0.786 (4) 0.478
250g mol–1) was dissolved in 51.2 g of benzene. If 31. The mole fraction of the solute in one molal aqueous
the freezing point depression constant, K f of solution is [AIPMT (Prelims)-2005]
benzene is 5.12 K kg mol–1, the freezing point of
benzene will be lowered by (1) 0.027 (2) 0.036

[AIPMT (Prelims)-2006] (3) 0.018 (4) 0.009

(1) 0.4 K (2) 0.3 K Questions asked Prior to Medical Ent. Exams. 2005
(3) 0.5 K (4) 0.2 K 32. Which of the following compounds can be used as
26. A solution of acetone in ethanol antifreeze in automobile radiators?

[AIPMT (Prelims)-2006] (1) Methyl alcohol (2) Glycol

(1) Shows a negative deviation from Raoult's law (3) Nitrophenol (4) Ethyl alcohol

(2) Shows a positive deviation from Raoult's law 33. Mole fraction of the solute in a 1.00 molal aqueous
solution is
(3) Behaves like a near ideal solution
(4) Obeys Raoult's law (1) 1.7700 (2) 0.1770

27. During osmosis, flow of water through a semi- (3) 0.0177 (4) 0.0344
permeable membrane is [AIPMT (Prelims)-2006] 34. 1 × 10–3 m solution of Pt(NH3)4Cl4 in H2O shows
(1) From solution having higher concentration only depression in freezing point by 0.0054°C. The
structure of the compound will be (Given
(2) From both sides of semi-permeable membrane Kf = 1.860 km–1)
with equal flow rates
(1) [Pt(NH3)4]Cl4 (2) [Pt(NH3)3Cl]Cl3
(3) From both sides of semi-permeable membrane
with unequal flor rates (3) [Pt(NH3)4Cl2]Cl2 (4) Pt(NH3)Cl3]Cl
(4) From solution having lower concentration only 35. Which of the following salt has the same value of
28. The vapour pressure of two liquids P and Q are 80 van’t Hoff’s factor i as that of K3[Fe(CN)6]?
and 60 torr, respectively. The total vapour pressure (1) Na2SO4
of solution obtained by mixing 3 moles of P and 2
moles of Q would be [AIPMT (Prelims)-2005] (2) Al(NO3)3

(1) 140 torr (2) 20 torr (3) Al2(SO4)3

(3) 68 torr (4) 72 torr (4) NaCl

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122 Solutions NEET

36. At 25°C, the highest osmotic pressure is exhibited mole fraction of the solvent if the decrease in the
by 0.1 M solution of vapour pressure is to be 20 mm of mercury?
(1) Glucose (2) Urea (1) 0.4 (2) 0.6
(3) CaCl2 (4) KCl (3) 0.8 (4) 0.2
37. According to Raoult’s law, the relative lowering of 44. If 0.15 g of a solute, dissolved in 15 g of solvent,
vapour pressure for a solution is equal to is boiled at a temperature higher by 0.216°C, than
(1) Mole fraction of solute that of the pure solvent. The molecular weight of
the substance, (Molal elevation constant for the
(2) Mole fraction of solvent
solvent is 2.16°C) is
(3) Moles of solute
(1) 10.1 (2) 100
(4) Moles of solvent
(3) 1.01 (4) 1000
38. The concentration units, independent of
temperature, would be 45. The vapour pressure of benzene at a certain
temperature is 640 mm of Hg. A non-volatile and
(1) Normality
non-electrolyte solid, weighing 2.175 g is added to
(2) Weight volume percent 39.08 of benzene. The vapour pressure of the
(3) Molality solution is 600 mm of Hg. What is the molecular
weight of solid substance?
(4) Molarity
(1) 69.45 (2) 59.6
39. In liquid-gas equilibrium, the pressure of vapours
above the liquid is constant at (3) 49.50 (4) 79.8
(1) Constant temperature 46. From the colligative properties of solution which one
(2) Low temperature is the best method for the determination of
molecular weight of proteins and polymers?
(3) High temperature
(1) Osmotic pressure
(4) None of these
(2) Lowering in vapour pressure
40. The vapour pressure of CCI4 at 25ºC is 143 mm Hg.
If 0.5 gm of a non-volatile solute (mol. weight = 65) is (3) Lowering in freezing point
dissolved in 100 g CCI4, the vapour pressure of the
(4) Elevation in boiling point
solution will be
(1) 199.34 mm Hg (2) 143.99 mm Hg 47. Molarity of liquid HCl, if density of solution is
1.17 gm/cc is
(3) 141.43 mm Hg (4) 94.39 mm Hg
(1) 36.5
41. What is the molarity of H2SO4 solution, that has a
density 1.84 g/cc at 35ºC and contains 98% by (2) 18.25
weight of solute?
(3) 32.05
(1) 18.4 M (2) 18 M
(4) 42.10
(3) 4.18 M (4) 8.14 M
48. A solution contains non volatile solute of molecular
42. A 5% (w/v) solution of cane sugar (mol. wt. = 342) mass M2. Which of the following can be used to
is isotonic with 1% (w/v) solution of a substance X. calculate the molecular mass of solute in terms of
The molecular weight of X is osmotic pressure? (m2 - mass of solute, V - volume
(1) 68.4 (2) 171.2 of solution,  - osmotic pressure)
(3) 34.2 (4) 136.8  m2   m  RT
(1) M2    VRT (2) M2   2 
43. The vapour pressure of a solvent decreased by     V  
10 mm of mercury when a non-volatile solute was
added to the solvent. The mole fraction of the m  m  
(3) M2   2   RT (4) M2   2 
solute in the solution is 0.2. What should be the  V   V  RT
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NEET Solutions 123
49. A solution containing components A and B follows 2. A solution of A and B with 30 mole percent of A is
Raoult’s law in equilibrium with its vapour which contains 60 mole
percent of A. Assuming ideality of the solution and
(1) A - B attraction force is greater than A - A and B - B the vapour, the vapour pressure of pure A to that of
pure B is
(2) A - B attraction force is less than A - A and B - B
(1) 3 : 2 (2) 4 : 5
(3) A - B attraction force remains same as A - A
and B - B (3) 2 : 1 (4) 7 : 2
3. Consider the following statement
(4) Volume of solution is different from sum of
volume of solute and solvent I. If one component in a binary solution shows
positive deviation, the second component would
50. Formation of a solution from two components can also show positive deviation.
be considered as
II. The gases which are easily liquefied, are more
(i) Pure solvent  separated solvent molecules, H1 soluble in common solvents.
(ii) Pure solute  separated solute molecules, H2 III. Maximum boiling azeotrope is formed by positive
deviation.
(iii) Separated solvent and solute molecules 
solution, H3 Choose the correct statement
(1) I & III (2) II & III
Solution so formed will be ideal if
(3) I & II (4) I, II & III
(1) Hsoln= H1 + H2 + H3
(2) Hsoln= H1 + H2 – H3 4. Which of the following mixture will show positive
deviation from ideal behaviour?
(3) Hsoln= H1 – H2 – H3
(1) H2O + C2H5OH
(4) Hsoln= H3 – H1 – H2
(2) H2O + HNO3
51. Camphor is often used in molecular mass
determination because O
(1) It is readily available (3) CHCl3 + CH 3 C CH3
(2) It has very high cryoscopic constant (4) All of these
(3) It is volatile
5. The ratio of the value of colligative properties of
(4) It is solvent for organic substances K4[Fe(CN)6](aq) to that of Fe4[Fe(CN)6]3(aq) is
52. Which condition is not satisfied by an ideal [Assuming 100% dissociation]
solution?
(1) 4 : 3 (2) 3 : 4
(1) mix H = 0
(3) 5 : 7 (4) 7 : 5
(2) mix V = 0
(3) mix S = 0 6. The vapour pressures of two liquids A and B are 100
and 160 torr respectively. The total vapour pressure
(4) Obeyance to Raoult's Law obtained by mixing 4 mole of A and 5 mole of B would
be
SECTION - D
(1) 126.66 torr (2) 140.44 torr
NEET Booster Questions (3) 150.00 torr (4) 133.33 torr
1. What would be osmotic pressure of a solution
7. The freezing point of a solution containing 0.4 g of
containing 5.85 g NaCl and 3.42 g sugar in 500 mL
acetic acid in 40 g of benzene is lowered by 0.512°C.
at 27°C?
What is the degree of dimerisation of acetic acid?
(1) 103.4 atm
[Kf(benzene) = 5.12 K kg mol–1]
(2) 10.34 atm
(3) 1.034 atm (1) 80% (2) 90%

(4) 0.1034 atm (3) 100% (4) 60%

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124 Solutions NEET

8. Henry law is not applicable to the gas, when 2xy yz

dissolved in water (1) y (2)
x
(1) N2(g) (2) O2(g)
xz yz
(3) NH3(g) (4) Ar(g) (3) y (4)
2x
9. What should be the amount of CaCl2 (i = 2.5) dissolved 16. For 1 molal solution of each compound maximum
in 2.5 L water, so that its osmotic pressure become freezing point will be assuming complete ionisation
0.75 atm at 300 K? in each case
(1) 0.34 g (2) 6 g (1) [Fe(H2O)6]Cl3
(3) 3.4 g (4) 0.6 g (2) [Fe(H2O)5Cl]Cl2.H2O
10. In which of the aqueous solution the van’t Hoff factor (3) [Fe(H2O)4Cl2]Cl.2H2O
(i) is maximum?
(4) [Fe(H2O)3Cl3].3H2O
(1) 1.2 M CH3COOH (2) 0.001 M CH3COOH 17. When some NaCl was dissolved in water, the freezing
(3) 0.1 M urea (4) 1.2 M glucose point depression was numerically equal to twice the
molal depression constant. The relative lowering of
11. The boiling point of C6H6,CH 3OH,C6H5NH2 and
vapour pressure of the solution nearly is
C 6 H 5 NO 2 are 80°C, 65°C, 184°C and 212°C
respectively. Which will show highest vapour pressure (1) 0.036 (2) 0.018
at room temperature? (3) 0.0585 (4) 0.072
(1) C6H6 (2) CH3OH 18. The osmotic pressure of 1M solution of urea is at 27°C
(3) C6H5NH2 (4) C6H5NO2 is
12. An ideal solution was found to have a vapour pressure (1) 2.46 atm (2) 24.6 atm
of 80 torr when the mole fraction of a non-volatile (3) 1.21 atm (4) 12.1 atm
solute was 0.2. What would be the vapour pressure
19. A certain aqueous solution of FeCl 3 (formula
of the pure solvent at the same temperature?
mass = 162) has a density of 1.1g/mL and contains
(1) 64 torr (2) 80 torr 20.0% by mass FeCl3 solution. Molar concentration
(3) 100 torr (4) 400 torr of this solution is

13. At 25°C, the vapour pressure of pure liquid (1) 0.028 (2) 1.36
A (mol. wt. = 40) is 100 torr, while that of pure liquid (3) 1.27 (4) 1.47
B is 40 torr, (mol. wt. = 80). The vapour pressure at
20. The rise in the boiling point of a solution containing
25°C of a solution containing 20 g of each A and B
1.8g of glucose in 100 g of a solvent is 0.1°C. The
is
molal elevation constant of the liquid is
(1) 80 torr (2) 59.8 torr
(1) 0.01 K/m (2) 0.1 K/m
(3) 68 torr (4) 48 torr (3) 1 K/m (4) 10 K/m
14. The van't Hoff factor i for an electrolyte which 21. At 17°C osmotic pressure of sugar solution was 580
undergoes dissociation and association in solvent are torr. The solution is diluted and temperature is raised
respectively to 57°C, osmotic pressure become 165 torr. The
(1) Greater than one and less than one extent of dilution is
(2) Less than one and greater than one (1) 2 times (2) 3 times
(3) Less than one and less than one (3) 4 times (4) 5 times
(4) Greater than one and greater than one 22. 25 ml of an aqueous solution of KCl was found to
require 20 ml of 1M AgNO3 solution for titration. The
15. If the elevation in boiling point of a solution of
depression in freezing point of KCl solution with 100%
non-volatile, non-electrolytic and non-associating
ionisation will be
solute in solvent (Kb = x K. kg.mol–1) is yK, then the
depression in freezing point of solution of (Kf = 2.0°K mol–1 kg and molality = molarity)
same concentration would be (1) 5.0 (2) 3.2
(Kf of the solvent = z K. kg. mol–1) (3) 1.6 (4) 0.8

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NEET Solutions 125
23. Henry's constant at 298 K for solubility of nitrogen 29. Which of the following concentration terms is
gas is 1.0 × 105 atm. The mole fraction of nitrogen temperature independent?
in air is 0.8. The moles of nitrogen from air dissolved I. Molarity
in 10 mol of water at 298 K and 5 atm pressure is
II. Molality
(1) 4.0 × 10–6
III. Normality
(2) 4.0 × 10–5
IV. Mole fraction
(3) 4.0 × 10–4
(1) I & II
(4) 5.0 × 10–4
(2) I & III
24. What weight of H2C2O4.2H2O (mol. wt. = 126) should (3) II only
be dissolved in water to prepare 250 mL of
centinormal solution (4) II & IV

(1) 0.63 g 30. 1 molar aqueous solution is _____ concentrated than

1 m aqueous solution
(2) 0.157 g
(1) More (2) Less
(3) 0.126 g
(3) Equally (4) Very less
(4) 0.875 g
31. Which gas is most soluble in water?
25. The boiling point of an azeotropic mixture of water (1) He (2) H2
and ethyl alcohol is less than that of theoretical value
of water and alcohol mixture. Mixture shows (3) NH3 (4) CO2
32. What is the concentration of NO3– ions when equal
(1) Solution is highly staurated
volumes of 0.1 M AgNO3 and 0.1 M NaCl are mixed
(2) Positive deviation from Raoult's law together?
(3) Negative deviation from Raoult's law (1) 0.1 N (2) 0.25 M
(4) Nothing can be said (3) 0.05 M (4) 0.2 M

26. How much ethyl alcohol must be added to 1.0L of 33. 15 g urea and 20 g NaOH dissolved in water. Total
water so that solution will not freeze at –20°C mass of solution is 250 g. Mole fraction of NaOH in
the mixture
(1) < 20 g
(1) 0.04 (2) 0.62
(2) < 10.75 g
(3) 0.5 (4) 0.4
(3) < 494.6 g 34. Vapour phase diagram for a solution is given below
(4) > 494.6 g if dotted line represents deviation

27. 100 ml of 1 M NaOH is mixed with 50 ml of

1 N KOH solution. Normality of mixture is
(1) 1 N
V.P.
(2) 0.5 N
(3) 0.25 N
(4) 2 N A = 1 A = 0
B = 0 Mole fraction B = 1
28. Mass of NaCl required to prepare 0.01 m aqueous
solution in 1 kg water is Correct observation for this solution

(1) 0.01 g (1) Hmix : +ve

(2) 0.585 g (2) Smix : +ve

(3) 58.8 g (3)  Vmix : +ve

(4) 5.88 g (4) All of these

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126 Solutions NEET

35. A mixture of two liquids A and B having boiling point (1) Vapour pressure of solution I is lowest
of A is 70°C, and boiling point of B is 100°C, distills
(2) Relative lowering of vapour pressure is maximum
at 101.2°C as single liquid, hence this mixture is
in III
(1) Ideal solution
(3) Freezing point is maximum for III
(2) Non ideal solution showing +ve deviation
(4) Boiling point is minimum for II
(3) Non ideal solution showing –ve deviation
40. An aqueous solution of sugar is taken in a beaker.
(4) Immiscible solution At freezing point of solution
36. Vapour pressure diagram of some liquids plotted (1) Crystals of sugar separated
against temperature are shown below
(2) Crystals of glucose and fructose are separated
P A B C
D (3) Crystals of ice separated
(4) Mixture of ice and some sugar crystals separated

E 41. The phenomenon taking place

T
Egg lining
Most volatile liquid
(1) A (2) B
(3) C (4) D Water

37. During evaporation of liquid

(1) The temperature of liquid rises
(1) Exo-osmosis
(2) The temperature of liquid falls
(2) Endo-osmosis
(3) The temperature of liquid remains uneffected
(3) Reverse-osmosis
(4) The liquid molecules becomes inert
(4) All of these
38. Two solutions marked as A and B are separated
42. Osmotic pressure of solution containing 0.6 g urea
through semipermeable membrane as below. The
phenomenon undergoing is and 3.42 g sugar in 100 ml at 27°C
(1) 492 atm
0.01 M NaCl 0.1 M NaCl
solution solution (2) 4.92 atm
(3) 49.2 atm
(4) 28.1 atm
A B

SPM 1 atm
(1) Na+ moves from solution A to solution B V.P.
(2) Both Na+ and Cl– moves from solution (A) to 43. H2O
solution (B) I
II III
(3) Both Na+ and Cl– moves from solution (B) to (A) T
(4) Solvent molecules moves from solution (A) to (B)
Which is having highest elevation in boiling point?
39. Correct observation
(1) H2O
(2) Aqueous solution I
I II III
(3) Aqueous solution II
0.1 M Urea 0.1 M NaCl 0.1 M CaCl2 (4) Aqueous solution III

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NEET Solutions 127
44. H2O 49. If any solute ‘A’ dimerises in water at 1 atm pressure
NaCl and the boiling point of this solution is 100.52°C. If
V.P. C solution
2 moles of A is added to 1 kg of water and kb for water
B D is 0.52°C/molal, calculate the percentage association
A of A
T (1) 50%
Freezing point of solution is marked as (2) 30%
(1) A (3) 25%
(2) B (4) 100%
(3) C 50. Substance A tetramerises in water to the extent of
(4) D 80%. A solution of 2.5 g of A in 100 g of water lowers
the freezing point by 0.3°C. The molar mass of A is
45. van't Hoff factor for acetic acid in aqueous medium (Kf for water = 1.86 K kg mol–1)
at infinite dilution is
(1) 122
(1) 2
(2) 31
(2) 1
(3) 244
(3) 1/2
(4) 62
(4) 3 51. K4[Fe(CN)6] is supposed to be 40% dissociated when
46. Correct order of freezing point of given solution 1M solution prepared. Its boiling point is equal to
another 20% mass by volume of non-electrolytic
I. 0.1 M glucose
solution A. Considering molality = molarity. The
II. 0.2 M urea molecular weight of A is
III. 0.1 M NaCl (1) 77 (2) 67

IV. 0.05 M CaCl2 (3) 57 (4) 47

(1) I < II < III < IV 52. For a binary ideal liquid solution, the total vapour
pressure of solution is given as
(2) I > II > III > IV
(1) PT = (PAo + PBo) XB
(3) III = II < IV < I
(2) PT = PAo × PBo
(4) IV > II > III > I
(3) PT = PAo/PBo
47. Boiling point of 0.01 M AB2 which is 10% dissociated
(4) PT = PBo + (PAo – PBo) XA
in aqueous medium ( K b (H2O)  0.52 ) as A+2 and B–
53. van't Hoff factor (i) for 30% dissociation of K4[Fe(CN)6]
(1) 273.006 K is
(2) 373.006 K (1) 1.1

(3) 0.006 K (2) 1.2

(4) 272.006 (3) 2.5

48. At higher altitude, the boiling point of water is lowered (4) 2.2
because 54. Which solution has the highest vapour pressure?
(1) Atmosphere pressure is low (1) 0.02 M NaCl
(2) Temperature is low (2) 0.03 M sucrose
(3) Atmospheric pressure increases (3) 0.005 M NaCl
(4) Water solidifies to ice (4) 0.005 M sucrose

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128 Solutions NEET

55. When equal volume of 0.1 M urea and 0.1 M glucose 58. The cryoscopic constant depends on
are mixed then the mixture will have
(1) The molar mass of the solvent
(1) Lower osmotic pressure
(2) The enthalpy of vapourization of solvent
(2) Same osmotic pressure
(3) The freezing point of the solvent
(3) Higher osmotic pressure
(4) All of these
(4) None of these
59. Among the following which gas has maximum
56. 0.1 M KBr and 0.2 M AgNO3 are mixed in 3 : 1 volume solubility in water at constant temp (KH: Henrry
ratio. The depression of freezing point of the resulting constant of gas in kbar)
solution will be (Kf of H2O = 1.86 K kg mol–1)
(1) H2(KH = 69.16)
(1) 0.8 K (2) 0.28 K
(2) N2(KH = 88.84)
(3) 1.2 K (4) 1.53 K
(3) CH4(KH = 0.413)
57. If 0.2 M aq. solution of KCl is isotonic with 0.2 M
(4) Ar·(KH = 40.3)
K2SO4 at same temperature then van't Hoff factor of
K2SO4 is 60. Incorrect relation among the following is

(1) 0.2 (1) Tb = iKbm

(2) 0.5 (2) Tb = (Tb)Solvent – (Tb)solution

(3) 0.8 (3) Tf = (Tf)Solvent – (Tf)solution

(4) 0.3 (4) = iCRT

‰‰‰

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