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CS – 1
SOME BASIC PRNICIPLES
OF ORGANIC CHEMISTRY
Einstein Classes,
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CS – 2
C O N C E P T S (Isomerism)
C1A Organic compounds having same molecular formulae but differing from each other at least in some
physical or chemical properties or both are known as isomers.
C1B
There are two mains types of isomerism :
(i)
Structural isomerism
(ii)
Space or stereoisomerism
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Chain Isomerism : This type of isomerism is due to difference in the arrangement of carbon atom constituting the chain i.e., straight or branched chain of carbon atoms.
For e.g., n-Butane and Iso-butane are chain isomer.
i.e.,
CH3CH2CH2CH3 and
Position Isomerism : This arises due to the difference in the positions occupied by particular atoms or
groups (substituents) in the same carbon chain or due to different position of double or triple bonds in the
alkenes and alkynes.
For e.g., C3H8O
Functional Isomerism : Compounds having the same molecular formulae but different functional groups
show functional isomerism.
As the functional group largely determines the properties of compound, such isomers differ in their
physical and chemical properties.
For e.g.,
(1)
Alcohols and Ethers (CnH2n+2O)
(2)
Aldehyde, Ketones & Unsaturated Alcohols (CnH2nO)
(3)
Carboxylic Acids and Esters (CnH2nO2)
(4)
Oximes and Amides (CnH2n+1NO)
(5)
10, 20 and 30 amines (CnH2n+3N)
(6)
Alkynes and Alkadiene (CnH2n–2)
(7)
Cyanides and Isocyanides
(8)
Nitroalkanes and Alkylnitrites
Ring and Chain Isomer : This type of isomerism is due to different mode of linking of carbon atoms i.e.,
the isomers possess either open or closed chain isomers.
For e.g.,
(1)
Alkenes and Cyclo Alkanes (CnH2n)
(2)
Alkynes and Cyclo Alkenes (CnH2n–2)
Tautomerism : This is a special type of functional isomerism where the functional isomers exist in
equilibrium with each other. The two forms which are in equilibrium in each other are called tautomers.
For e.g.,
(1)
Keto & Enol are tautomeric forms.
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CS – 3
(2)
Nitro and Aci
(3)
Oximes and Nitroso
Enolisation is in order as follows :
CH3CHO < CH3COCH3 < CH3COCH2CHO < CH3COCH2COCH3
Metamerism : It is the isomerism in the same homologous series. It is due to presence of different allkyl
groups attahced to the same polyvalent functional group or atom.
For e.g.,
(1)
C4H10O
C2H5OC2H5
Diethylether
C3H7OCH3
Methylpropyl ether
(2)
C5H10O
C2H5COC2H5
Diethyl ketone
C3H7COCH3
Methylpropyl ketone
(3)
C4H11N
C2H5NHC2H5
Diethylamine
C3H7NHCH3
Methylpropyl amine
1.
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Practice Problems :
The number of isomers of C6H14 is
(a)
2.
4
(b)
5
(c)
6
(d)
methanol
(b)
acetone
diethyl ether
(d)
dimethyl ether
metamerism
(b)
position isomerism
functional isomerism
(d)
all the three
7
An isomer of ethanol is
(a)
(c)
3.
The compound C4H10O can show
(a)
(c)
4.
C7H8O shows how many isomers (having onebenzene ring).
(a)
5.
2
(b)
3
(c)
4
(d)
5
Keto-enol tautomerism is observed in :
(a)
(c)
6.
(b)
(d)
(CH3)3CCHO
CH3COCH3
(b)
CH3COCH2CHO
CH3COCH2COCH3
(d)
=O
Maximum enolisation takes place of
(a)
(c)
[Answers : (1) b (2) d (3) d (4) d (5) b (6) d]
C2
Stereoisomerism :
They are of two types :
(1)
Geometrical isomerism : Such isomers, which possess the same molecular and structural
formula but differ int he arrangement of atoms or groups in space due to hindered rotation
around the double bond i.e. (
)
The isomers which has similar group on same side are called -cis isomer and similar groups on
opposite sides are called as -trans isomer.
However in case of all the four groups are different, for e.g., (
Einstein Classes,
), it is not possible
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CS – 4
to decide -cis and -trans isomers. In such cases E and Z system of nomenclature is used.
(2)
Optical isomerism : The compounds having similar physical and chemical properties but
differing only in the behaviour towards the plane polarised light are called as optical isomers.
On the basis of study of optical activity, the various organic compounds are divided into three
types :
(a)
The optical isomer with rotate the plane of polarised light to the right are called as
dextrorotatory or d-form or indicated by + sign.
(b)
The optical isomer which rotate the plane or polarised right to the left are called as
leavoratotory or l-form on indicated by –ve sign.
(c)
The optical powers of the above two isomers are same i.e. of same magnitude but of
positive sign.
(d)
The equimolar mixture of two forms, therefore will be optically inactive. This mixture
is termed is Racemic Mixture.
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Cause of Optical Activity or Optical Isomerism :
A plane which divides an object into two symmetrical halves is said to have a plane of symmetry. An object
lacking a plane of symmetry is called dissymmetric or chiral.
A symmetric object is called as achiral.
A dissymmetrical or chiral object can be defined as the one which is not super imposable on its mirror
image. In order to exhibit optical activity an object must be chiral.
The compounds which are mirror images of each other but not super imposable are called as enantiomers.
Generally compunds having carbon atom to which four different groups are when attached show
optical isomerism. Molecule having a symmetric carbon atom lacks in plane of symmetry. Thus compound
which consist of at least one asymmetric carbon atom is capable of showing the phenomenon of optical
isomerism.
Exception : Following compound with a cumulative system of double bond and having no chiral carbon
atom shows optical isomerims.
but not the following compound
i.e.,
.
Number of possible stereoisomers in compounds containing different numbers of asymmetric
carbons :
(i)
When the molecule has no symmetry : n ® number of asymmetric carbon atoms.
The number of d- and l-forms a = 2n
The number of meso forms m = 0
The total number of optical isomer = a + m = 2n
(ii)
When the molecule has symmetry :
n  number of asymmetric carbon atoms which is even.
The number of d- and l-form, a = 2n – 1
n
The number of meso forms, m = 2 2
1
n
The total number of optical isomer = a + m = 2 n 1  2 2
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1
.
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CS – 5
(iii)
When the molecule has symmetry and n  no. of asymmetric carbon atom which is odd :
 n 1 


2 
The number of d and l-forms, a = 2 n 1  2 
The number of meso forms m  2
n 1
2
.
Total optical isomer = a + m = 2n – 1.
Diastereomers : Stereoisomers which are not mirror images of each other are called diastereomers.
Practice Problems :
1.
The molecular formula of a saturated compound is C2H4Cl2. This formula permits the existance of
two
2.
(a)
functional isomers
(b)
position isomers
(c)
optical isomers
(d)
cis-trans isomers
Number of optically active isomers of tartaric acid
(a)
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CH(OH)COOH
|
CH (OH)COOH is :
2
(b)
3
(c)
4
(d)
5
[Answers : (1) b (2) a]
C3A Reagents :
Following types of reagents are used
(i)
(ii)
C3B
Electrophiles : Electrophilic reagents (electron loving) are those reagents that are positively
charged or contain an electron deficient atom. i.e., whose octet is not complete. They are also
known as Lewis Acid (they are electron acceptor). For e.g., NO2+, Cl+, H+, SO3, BF3 etc.
Nucleophiles : Nucleophilic reagents (nucleus loving) or electron repelling reagents are
negatively charged or having neutral atom with lone pair. They are also called as Lewis base
(electron pair donor). For e.g., OH–, CN–, NH3, H2O etc.
Types of Intermediates Formed in the Organic Reaction :
(i)
Carbocation or Carbonium Ion : The carbon with a positive charge and six electron pair
in valence shell is known as carbocation. This carbon is sp2 hybridised
.
The carbocation can undergo rearrangemets reaction.
Order of Stability of Carbocation : Benzyl, Allyl > 30 > 20 > 10 > CH3+.
(ii)
Carbanion : The carbon with a negative charge having three bond pair and one lone pair is
called as carbanion. This carbon is sp3 hybridised.
Order of stability of carbanion is : Benzyl, Allyl > CH3– > 10 > 20 > 30.
(iii)
Free Radical : The carbon containing single unpaired electron is called as carbon free radical. It
is sp2 hybridised
Order of Stability of Free Radical : Benzyl, Allyl > 30 > 20 > 10.
(iv)
Carbene : There is another class of intermediate called carbene,
two forms :
(methylene). It exists in
(a)
Singlet Carbene : In which the unshared electrons are paired. i.e.,
. It is sp2 hybridised
(b)
Triplet Carbene : In which unshared electrons are not paired. i.e.,
. It is sp hybridised
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CS – 6
(v)
Nitrene
(vi)
Benzyne
Practice Problems :
1.
Most-stable carbonium ion is


(a)
p  NO 2  C 6 H 4  C H 2
(c)
p  Cl  C 6 H 4  C H 2
(b)
C6 H 5 C H 2
(d)
p  CH 3O  C 6 H 4  C H 2

Arrange the following in the decreasing order of the free radical
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2.

(a)
II > I > III
(b)
I > II > III
(c)
III > II > I
(d)
I > III > II
[Answers : (1) b (2) a]
C4
Different Effects for the Displacement of Shared Pair of Electron are as follows :
(i)
Inductive Effect
(ii)
Resonance Effect
(iii)
Electromeric Effect
(iv)
Hyperconjugation Effect
(i)
Inductiv Effect : Electrone withdrawing and electron releasing group is attached to the carbon
chain permanent polarity is induced on the carbon atom. This displacement of electrons towards
one of the atom because of difference in electronegativity is known as inductive effect.
This effect is of two types :
(a)
Positive Inductive Effect (+I) : This is due to electron releasing group. It develops a
negative charge on the carbon chain.
+I effect decreases as we move away from the electron releasing group.
Greater the size of alkyl group greater the electron releasing effect. Due to electron
releasing group electron density increases hence basic nature increased, thus acidic
decreases
(b)
Negative Inductive Effect (–I) : This is due to electron withdrawing groups. It developes
positive charge on the chain.
–I effect also decreases as we move away from the electron withdrawing group. After
third carbon, this effect is negligible. –I effect is in the order as follows :
NO2 > F > COOH >Cl > Br > I > OH > C6H5
Due to –I effect electron density decreases hence basic nature is decreased and acidic
nature is increased.
(ii)
Resonance Effect : The phenomena by which two or more structures involving identical
positions of atoms can be written for a particular compound is called Resonance. The various
structures are called as resonating structures.
Resonating structure can be written for the following :
(a)
Einstein Classes,
Presence of conjugated system (Alternate double-single bond or alternate
single-multiple bonds) for e.g.,
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CS – 7
(b)
Conjugated system attached to electron deficient atom with vacant p-orbital, for e.g.,
(c)
Conjugated system between electron rich atom i.e., containing lone pair and alternate
double bond for e.g.,
(In this case Cl is having –I as well as +R effect)
Resonance effect is of two types :
(a)
+R Effect : Groups which can released or donate the electron pairs through resonance
are called as +R effect. for e.g., OH–, OCH3–, NH2– etc.
(b)
–R Effect : Groups which attrach or withdraw the electron through resonance are of
–R effect for e.g., NO2–, CN–,
Electromeric Effect : It is a temperory effect, it comes into play instantaneously at the demand
of attacking reagent and as soon as the attaching reagent is removed the original conditions
restored.
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(iii)
, –COOH etc.
The effect involving the complete transfer of shared pair of -electrons to one of the atoms
joined by the multiple bond (double or triple) at the requirement of attacking reagent is known as
electromeric effect. It is of two types :
(a)
(iv)
+E Effect
(b)
–E Effect
Hyperconjugation Effect : It is the conjugation between  and  bond, for e.g.,
This type of electron released by alkyl group attached to the unsaturated group is called hyper
conjugation. There should be atleast one H atom at  carbon atom w.r.t. (C = C) bond.
1.
2.
3.
Practice Problems :
Delocalised molecular orbitals are found in
(a)
H2
(b)
HS–
(c)
CH4
(d)
CO32–
The correct order of the increasing C – O bond length of CO, CO32–, CO2 is
(a)
CO32– < CO2 < CO
(b)
CO2 < CO32– < CO
(c)
CO < CO32– < CO2
(d)
CO < CO2 < CO32–
Among the following which one is most basic
(a)
(c)
4.
(b)
CH3NH2
CH3CH2NH2
(d)
C6H5NH2
Most acidic compound is
(a)
5.
NH3
CH3COOH
(b)
C6H5COOH
(c)
O2NC6H4CO2H (d)
C6H5OH
In the following compounds phenol (I), p-cresol (II), m-nitrophenol (III) and p-nitrophenol (IV), the
order of acidity is
(a)
III > IV > I > II
(b)
I > IV > III > II
(c)
II > I > III > IV
(d)
IV > III > I > II
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CS – 8
6.
The most unlikely representation of resonance structures of p-nitrophenoxide ion is
(a)
(b)
(c)
(d)
Increasing values of dissociation constant K a of
8.
CH3COOH (I),
HCOOH (II),
(a)
(b)
I < II < III
– COOH (III) :
III < II < I
(c)
I < III < II
(d)
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7.
II < III < I
Give the order of the stabilities of the followings :
(a)
b>a>c>d>e
(b)
a>b>c>d>e
(c)
e>b>a>c>d
(d)
d>a>b>c>e


..
The acylium ion, the structure R — C  O : is more stable than R  C  O :
9.

(a)
As in R — C  O : octet of every atom is complete.
(b)
In R — C  O : structure there are moree covalent bonds than R  C  O :
(c)
Both are correct
(d)
Given statement is wrong as R  C  O : is more stable than R — C  O :



..
..

[Answers : (1) d (2) d (3) c (4) c (5) d (6) c (7) c (8) a (9) c]
Chemical Bonding
C5
INTRODUCTION
A molecule will only be formed if it is more stable, and has a lower enrgy than the individual atoms.
Normally only electrons in the outermost shell of an atom are involved in forming bonds. We divide
elements into three classes :
(A)
Electropositive elements, whose atoms give up one or more electrons easily. They have low
ionisation potentials.
(B)
Electronegative elements, which can gain electrons. They have higher value of electronegativity.
(C)
Elements which have little tendency to lose or gain electrons.
C6A Ionic Bonding
An ionic bond is formed when a metal atom transfers one or more electrons to a nonmetal atom.
Na 
 Na   e 
F  e 
 F 
As a result of this transfer the metal atom becomes cation and the nonmetal the anion.
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CS – 9
Cations and anions attain noble gas configuration with outer shell octet of electrons. In some cases cations
may not have octet but different outershell.
Properties of Ionic solids
They are good conductor of electricity in fused state and in aq. solution. They are soluble in polar solvents
and insoluble in non polar solvents. They have high m.p. and b.p. They have strong force of attraction
2
between cation and anion (Coulombic force) F  z 1 z 2 e
Dr 2
2.
C6B
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Practice Problems :
Which of following elements has the strongest tendency to form electrovalent compound ?
(a)
Li
(b)
Na
(c)
Be
(d)
Mg
Which of the following is more ionic ?
(a)
NaCl
(b)
KCl
(c)
MgCl2
(d)
CaCl2
[Answers : (1) b (2) b]
1.
Solubility of Ionic Solids :
By the Coulombic force of attraction F 
z 1 z 2e 2
. We can study the solubility of ionic solids in different
Dr 2
solvents. Greater the force of attraction between ions, smaller the tendency of the ions to go into solution
and thus smaller the solubility.
For a given solute, greater the value of D then smaller the force of attraction between positive and negative
ions, hence greater the solubility.
D(H2O) > D(CH3CH2OH) > D(CH3OCH3)
hence solubility of ionic solid (say NaCl) in H2O > CH3CH2OH > CH3OCH3
Practice Problems :
MgSO4 is soluble while BaSO4 in insoluble in H2O. This is because
(a)
lattice energy of BaSO4 is greater than MgSO4
(b)
BaSO4 is more covalent than MgSO4
(c)
hydration energy of Mg2+ is greater than Ba2+
(d)
lattice energy of MgSO4 is greater than BaSO4
[Answers : (1) c]
1.
C7A Covalent Bonding : If duplet (2) or octet (8) is completed by sharing of electrons between two
electronegative elements, the bond formed is called covalent bond.
C7B
The Octet Rule : For many light atoms a stable arrangement is attained when the atom is surrounded by
eight electrons – the octet rule. (In case of H2, duplet is completed)
Exceptions to the Octet Rule : In so many cases, the octet rule is violated :
In BeF2, octet of Be is not complete, in BF3, octet of B is not complete. Other examples are PCl5, SF6, IF7
where centre atom is having more than eight electrons.
C7C Coordinate Bonding
A covalent bond results from the sharing of pair of electrons between two atoms where each atom
contributes one electron to the bond. It is also possible to have an electron pair bond where both electrons
originate from one atom and none from the other. Such bonds are called coordinate bonds or dative
bonds. Since in coordinate bonds, two electrons are shared by two atoms, they differ from normal covalent
bonds. It is represented as  Atom/ion/molecule donating electron pair is called DONOR or LEWIS
BASE. Atom/ion/molecule accepting electron pair is called ACCEPTOR or LEWIS ACID () points
donor to acceptor.
NH4+ : NH3 has three (N — H) bonds and one lone pair. In NH4+ formation this lone pair is donated to H+
(having no electron).
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CS – 10
C8A Valence Shell Electron Pair Repulsion (VSEPR) Theory
(Gillespie Theory)
The shape of the molecule is determined by repulsions between all of the electron pairs represent in the
valence shell.
Repulsion is in order : (lp – lp) > (lp – bp) > (bp – bp)
The magnitude of repulsion between pairs of electrons depends on the electronegativity difference between
the central atom and the other atom. Double bond cause more repulsion than single bonds, and triple bonds
cause more repulsion than a double bond.
The Effect of Bonding and Lone Pairs of Geometry and Bond Angles
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C8B
Example
Orbitals on
Central Atom
Theoretical
Shape
Bond Angle
Distorted Geometry
due to Repulsion
BeCl2
2
Linear
1800
Linear
BF3
SO2
CH4
NH3
NF3
H2 O
3
0
Plane triangle
0
Angular
Tetrahedral
0
109 28’
Tetrahedral
Tetrahedral
0
107 48’
Trigonal pyramidal
Tetrahedral
0
102 30’
Pyramidal
Tetrahedral
0
104 27’
Angular
0
Angular
Plane triangle
3
120
Plane triangle
4
4
4
4
119
F2 O
4
Tetrahedral
102
H2 S
4
Tetrahedral
900
Angular
0
0
PCl5
5
Trigonal bipyramidal
120 and 90
SF4
5
Trigonal bipyramidal
101 36’ and 86 33’ Irregular tetrahedral
CIF3
5
Trigonal bipyramidal
87040’
I3
–
SF6
BrF5
XeF4
5
Trigonal bipyramidal
6
6
180
Trigonal bipyramidal
0
0
T-shaped
Linear
0
Octahedral
Octahedral
0
84 30’
Square pyramidal
Octahedral
0
Square planar
Octahedral
6
0
90
90
Practice Problems :
Molecular shapes of SF4, CF4 and XeF4 are
(a)
the same, with 2, 0 and 1 lone pair of electrons respectively.
(b)
the same, with 1, 1 and 1 lone pair of electrons respectively.
(c)
different, with 0, 1 and 2 lone pairs of electrons respectively.
(d)
different, with 1, 0 and 2 lone pairs of electrons respectively.
[Answers : (1) d]
1.
C9
Isoelectronic Principle : Isoelectronic species usually have the same structure. This may be extended to
species with the same number of valence electrons.
Species
Structures
+
4
CH4, NH , BF
2–
3
–
4
–
3
CO , NO , SO3
–
3
+
2
CO2, N , NO
Einstein Classes,
tetrahedral
planar triangle
linear
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CS – 11
C10A Valence Bond Theory : The covalent bond is a region of high electron charge density (high electron
probability) that results from the overlap of atomic orbitals between two atoms. In general, the greater the
amount of overlap between two orbitals, the stronger the bond. For each bond there is a condition of
maximum atomic orbital overlap leading to maximum bond strength at a particular internuclear distance
(bond length). This is called Valence Bond Approach.
–
This is a localised electron model of bonding.
–
Most of the electrons retain the same orbital locations as in a separated atoms, and the bonding
electrons are localised (fixed) in the region of atomic orbital overlap.
–
In H2 molecules, H – H  bond is by s – s overlapping
–
In F2 molecule : F – F  bond is by axial overlapping of two p-orbitals.
–
In O2 molecule : One  bond is by axial overlap of p-orbitals and  bond is by lateral overlap of
p-orbitals.
–
In N2 molecule : One  bond is by axial overlap of p-orbitals and two  bonds are by lateral
overlap of p-orbitals.
–
–
–
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C10B Hybridisation : Hybridisation is defined as the concept of intermixing of orbitals of same energy or of
slightly different energy to produce entirely new orbitals of equivalent energy, identical shapes and symmetrically disposed in plane. New orbitals formed are called hybrid orbitals.
Only the orbitals of an isolated single atom can undergo hybridisation.
The hybrid orbitals generated are equal in number to that of the pure atomic orbitals which mix
up.
A hybdrid orbitals, like the atomic orbitals, cannot have more than two electrons of opposite
signs.
C10C Sigma and PI Bond There can be following types of overlaping along the axes (end to end) :
–
–
–
–
(s - s) overlapping when s-orbital overlaping with another s-orbital
(s - p) overlapping, (p can be px or py or pz)
(px – px) overlapping
any of the hybrid orbitals overlaps with another hybrid orbitals or s or p orbital.
Bond formed in the manner is called sigma () bond in which electron density is concentrated in between
the two atoms, and on a line joining the two atoms.
Double or triple bonds occur by sideways overlap of orbitals (like (py – py) and (pz – pz) orbitals) giving pi
() bonds in which electron density also concentrates between the atoms, but on either side of the line
joining the atoms.
The shape of the molecule is determined by the  bonds (and lone pairs) but not by the  bonds. Pi bonds
merely shorten bond length. Thus
1.
Practice Problems :
In which of the following molecules would you expect the nitrogen to nitrogen bond to be the
shortest ?
(a)
N2 H 4
(b)
N2
(c)
N2 O 4
(d)
N2 O
Einstein Classes,
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CS – 12
2.
Allyl cyanide has
(a)
9  bonds and 4  bonds
(b)
9 -bonds, 3  bonds and one lone pair
(c)
8  bonds and 5  bonds
(d)
8 -bonds, 3  bonds
[Answers : (1) b (2) b]
C10D Hybrid Orbitals and their Geometric Orientation
Hybrid Orbitals Orientation
2.
3.
4.
5.
Predict Bond Angle
0
Actual Shape
sp (two)
linear
BeCl2, CO2
C2H 2
180
linear
sp2 (three)
trigonal planar
BF3,C2H4
SO2, SO3
1200
trigonal planar
sp3 (four)
tetrahedral
CH 4
1090 28’
tetrahedral
NH3
0
107 48’
H2 O
1040 27’
trigonal pyramidal (due
to lp–bp repulsion)
angular (V-shaped) due
to lp – lp and lp – bp
repulsion
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1.
Example
sp3d(five)
trigonalbi
pyramidal
PCl5
SF4
I –3
1200 and 900
1010, 36’ and 860 33’
1800
trigonal pyramidal
irregular tetrahedral
linear
sp3d2
octahedral
SF6
900
octahedral
Practice Problems :
The hybridisation of atomic orbitals of N in NO2+, NO3– and NH4+ are, respectively
(a)
sp, sp2, sp3
(b)
sp, sp3, sp2
(c)
sp2, sp, sp3
(d)
sp2, sp3, sp
If a molecule MX3 has zero dipole moment, the sigma bonding orbitals used by M are
(a)
pure p
(b)
sp hybrids
(c)
sp2 hybrids
(d)
sp3 hybrids
The shape of sulphate ion is
(a)
hexagonal
(b)
square planar
(c)
trigonal bipyramidal
(d)
tetrahedral
+
The hybridisation of atomic orbitals of nitrogen in NO2 , NO3— and NH4+ are
(a)
sp, sp3 and sp2 respectively
(b)
sp, sp2 and sp3 respectively
2
3
(c)
sp , sp and sp respectively
(d)
sp2, sp3 and sp respectively
Specify the coordination geometry around and hybridization of N and B atoms in a 1 : 1 complex of
BF3 and NH3
(a)
N : tetrahedral, sp3 ; B : tetrahedral, sp3 (b)
N : pyramidal, sp3 ; B : pyramidal, sp3
(c)
N : pyramidal, sp3 ; B : planar, sp2
(d)
N : pyramidal, sp3 ; B : tetrahedral, sp3
[Answers : (1) a (2) c (3) d (4) b (5) a]
C11
Dipole Moment : In H2, there is no displacement of the electric charge due to same electron affinity of both
H-atoms and the bond is non-polar. In HCl, the Cl atom has a more electron affinity than does the H atom.
Electronic charge distribution is shifted towards the Cl atom. The H-Cl bond is said to be polar.
The magnitude of the charge displacement in a polar covalent bond is measured through a quantity called
the dipole moment µ. It is the product of the magnitude of charges () and the distance separating them (d).
(Here the symbol () suggests a small magnitude of charge, less than the charge on an electron)
µ=×d
If
 = 4.8 × 10
Einstein Classes,
–10
esu and d = 1Å = 1 × 10–8 cm
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CS – 13
then
µ = 4.8 × 10–10 × 1 × 10–8 = 4.8 × 10–18 esu cm
In S.I. unit, 1 D = 3.33 × 10–30 coulomb meter (when charge  = 3.33 × 10–20 C and d = 1 × 10–10 m).
In diatomic molecule,µ =  × d
but in polyatomic molecule with angle , resultant dipole moment is the vector summation of the vector
moments. Also µresultant
1

 cos   , µ  . Symmetrical molecules without lone pair will have µ = 0.

 2
% ionic character in a molecule =
2.
Practice Problems :
A molecule possessing dipole moment is
(a)
CH4
(b)
H2 O
(c)
BF3
Correct order of dipole moment for the following molecule is
(I)
3.
(II)
CO2
(III)
(a)
I = II = III
(b)
I < II < III
(c)
I > II > III
Which of the following hydrocarbons has the lowest dipole moment ?
(a)
4.
(d)
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1.
observed value of µ
 100
theoretical value of µ
(c)
CH3CH2C = CH
Dipole moment is exhibited by
(a)
1, 4-dichlorobenzene
(c)
trans-1, 2 dichloroethene
[Answers : (1) b (2) d (3) b (4) b]
(d)
II < III < I
(b)
CH3C = CCH3
(d)
CH2 – CH – C = CH
(b)
(d)
1, 2-dichlorobenzene
trans 1, 2-dichloro-2-butene
C12A Hydrogen Bonding : Hydrogen bonding is said to be formed when slightly acidic hydrogen attached to a
strongly electronegative atom such as F, N and O, is held with weak electrostatic forces by the non-bonded
pair of electrons of another atom. The coordination number of hydrogen in such cases is two. It acts as a
bridge between two atoms, to one of which it is covalently bonded, and to the other attached through
electrostatic forces, also called Hydrogen Bond.
Of all the electronegative donor atoms, only F, N and O atoms enter into stable hydrogen bond formation.
The weak electrostatic interaction leading to the hydrogen bond formation is shown by dott (....) lines. Thus
X – H....Y represent hydrogen bonding between hydrogen and Y atom.
C12B Intramolecular H-Bonding :
This type of H-bonding occurs when polar H and electronegative atom are present in the same molecule.
(a)
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(b)
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CS – 14
C12C Intermolecular H-Bonding :
This type of H-bonding takes place between H and electronegative element present in the different
molecules of the same substance (as in between H2O and H2O) or different substances (as in between H2O
and NH3)
e.g.
In water molecules :
Due to polar nature of H2O, there is association of water molecules giving a liquid state of abnormally high
b.p.
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C13A Molecular Orbital Theory (MOT) : According to Valence Bond Theory only the half-filled orbitals of
one atom overlaps with the half filled orbitals of other atom to form the covalent bond. According to MOT,
on the other hand, the all atomic orbitals of one atom overlap with the all the atomic orbitals of the other
atom provided the overlapping orbitals are of the same symmetry and of similar energy. The resulting
polynuclear molecular orbitals contain the all electrons of the molecule.
C13B The basic principles of the MOT :
1.
When nuclei of two atoms come close to each other, their atomic orbitals interact leading to the
formation of molecular orbital. The atomic orbitals of the atoms in a molecule completely lose
their identity after the formation of molecular orbital.
2.
Each molecular orbital is described by a wave function  , known as molecular orbital wave
function.
3.
The molecular orbital wave function  is such that  2 represents the probability density or
electron charge density.
4.
Each molecular orbital wave function (  ) is associated with a set of quantum numbers which
determine the energy and shape of the molecular orbital.
5.
Each  is associated with definite energy value.
6.
Electrons fill the molecular orbitals in the same way as they fill the atomic orbital following the
aufbau principle, Pauli’s exclusion principle, and the Hund’s rule of maximum multiplicity.
7.
Each electron in a molecular orbital belongs to all the nuclei present in the molecule.
8.
Each electron moving in a molecular orbital has a spin of 
1
1
or  .
2
2
The basic difference between an atomic orbital and molecular orbital is that while an electron in an atomic
orbital belongs to or influenced by one positive nucleus only, an electron in a molecular orbital is
influenced by all nuclei of atoms contained in a molecule.
Linear combination of Atomic orbitals (LCAO) in case of H2+
1.
A linear combination of two atomic orbitals  A and  B leads to the formation of two
molecular orbitals   and  
2.
The energy E+ of molecular orbital   is lowe than either of EA and EB (energies of isolated
atoms). It is therefore designated as bonding molecular orbital (BMO).
3.
The energy E– of molecular orbital   is higher than either of EA and EB. It is therefore
designated as antibonding molecular orbital (ABMO)
4.
The extent of lowering of energy of the bonding molecular orbital is equal to the extent of
increase of energy of antibonding molecular orbital.
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CS – 15
Energies of bonding and antibonding molecular orbitals.
The order of energy of molecular orbital for lighter elements like boron, carbon and nitrogen are as
follows : (1s) < *(1s) < (2s) < *(2s) < 2py = 2pz < 2px < *2py = *2pz < *px
The order of energy of molecular orbitals for heavier elements after nitrogen are :
(1s) < *(1s) < (2s) < *(2s) < 2px < 2py = 2pz < *2py = *2pz < *2px
Practice Problems :
Which of the following molecular orbital in N2 has least energy ?
1.
(a)
(b)
 2p y
 2p z
(c)
 2s
(d)
 *2p z
C14
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[Answers : (1) c]
Heteronuclear Diatomic Molecule : The same principle apply when combining atomic orbitals from two
different atoms as applied when atoms are identical that is
(a)
(b)
(c)
only atomic orbitals of about same energy can combine effectively
They should have maximum overlap.
They must have the same symmetry
Since the two atoms are different, the energies of their atomic orbitals are slightly different.
C15
Electronic Configuration of Molecules : While discussing the electronic configuration of molecules, we
shall frequently make use of a term called bond order.
B.O. =
1
[Nb – Na]
2
Nb  number of bonding electrons
Na  number of anti-bonding electrons
If B.O. = 0, 1, 2, 3 so on it means that no bond is formed, one bond is formed, two bonds or three bonds are
formed between the atoms respectively.
1.
2.
B.O,  Bond dissociation energy and B.O. 
1
Bond Length
Electronic configuration helps to predict the magnetic character of the molecule.
If all the electrons in a molecule are paired they are diomagnetic and if unpaired electron is present they are
paramagnetic.
1.
2.
3.
4.
Practice Problems :
Among KO2, AlO2–, BaO2 and NO2+, unpaired electron in present in
(a)
KO2 only
(b)
NO2+ and BaO2
(c)
KO2 and AlO2–
(d)
BaO2 only
+
When N2 is formed from N2, bond order ...... and when O2+ is formed from O2, bond order ..........
(a)
increases
(b)
decreases
(c)
increases, decreases
(d)
decreases, increases
Which of the following has longest bond length ?
(a)
O2
(b)
O 2+
(c)
O 2–
(d)
O22–
Which of the following has identical bond order ?
(I)
CN–
(II)
O 2–
(III)
NO+
(IV)
CN+
(a)
I, III
(b)
II, IV
(c)
I, II, III
(d)
I, IV
[Answers : (1) a (2) d (3) d (4) a]
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CS – 16
INITIAL STEP EXERCISE
2.
3.
The IUPAC name of picric acid is
(a)
2, 4, 6-trinitrophenol
Markownikoff’s rule is applicable to which of the
following reactions
(b)
2, 4, 6-trinitro-1-hydroxy hexane
(a)
C2H4 + HBr
(b)
C3H6 + Cl2
(c)
2, 4, 6-trinitro-1-hydroxy benzene
(c)
C3H6 + HBr
(d)
C3H6 + Br2
(d)
1, 3, 5-trinitro-6-hydroxy benzene
(a)
geometrical isomers
(b)
keto-enol tautomers
(c)
chain isomers
(d)
position isomers
Number of isomers of molecular formula C2H2Br2
is
(c)
(b)
(c)
(a)
it reacts with nucleophile
(b)
it can undergo rearrangement
(c)
it can eliminate an H+ to form an olefin
(d)
all are correct
What is the decreasing order of strength of the bases
OH — (I), NH 2 — (II), H — C  C — (III),
CH3CH2— (IV) ?
(b)
2
3
(d)
4
11.
2-methylbutene-1
(a)
IV > II > III > I (b)
III > IV > II > I
(c)
I > II > III > IV (d)
II > III > I > IV
The intermediate during the addition of HCl to
propene in presence of peroxide is

3-methylbutyne-1
(a)
CH3 — C H — CH2Cl
(b)
CH3 — C H — CH3
(c)
CH3 — CH2 — C H 2
(d)
CH3 — CH2 — C H 2

2-methylbutanoic acid
Which of the following hydrocarbons has the
lowest dipole moment ?
(a)
(b)
(c)
(d)
12.
(b)
In the following compounds, anisole (I), benzene
(II) and nitro benzene (III), the case of reaction with
electrophiles is
(a)
II > III > I
(b)
III > II > I
(c)
II > I > III
(d)
I > II > III
CH2 = CH – C  CH
13.


C H 2  CH  CH  O

14.
Among the given compounds, the most susceptible
to nucleophilic attack at the carbonyl group is
(a)
MeCOCl
(b)
MeCHO
(c)
MeCOOMe
(d)
MeCOOCOMe
The IUPAC name of the compound having formula

C H 2  CH  C H  O

C H2  C H  C H  O
(d)

CH3CH2C  CH

(c)

CH3C  CCH3
Polarisation of electrons in acrolein may be written as
(a)
7.
10.
Which of the following statements is correct about
a carbonium ion ?
3-methylbutanoic acid
(d)
6.
1
Which of the following compounds exhibits stereoisomerism ?
(a)
5.
9.
Vinyl alcohol and acetaldehyde are
(a)
4.
8.
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1.


C H2  C H  C H  O
Which of the following is not an electrophile
(a)
AlBr3
(b)
BF3
(c)
SnCl4
(d)
NH3
Einstein Classes,
(a)
3-amino-hydroxy propionic acid
(b)
2-amino-propan-3-oic acid
(c)
Amino hydroxy propanoic acid
(d)
2-amino-3-hydroxy propanoic acid
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CS – 17
15.
The
IUPAC
name
of
the
compound,
20.
Stability order of ...... in order :
is

C H2 ,
–
16.
(a)
2-Bromo-3-chloro-4-oxopentanoic acid
(b)
3-Chloro-2-bromo-4-oxopentanoic acid
(c)
4-Carboxybromo-3-chloro-2-butanone
(d)
None
I

IV
(a)
IV < III < II < I (b)
IV < II < I < III
(c)
I < II < III < IV (d)
IV < I < III < II
Which shows aromatic character ?
(a)
1, 3-Dibromo-3-methylbutane
(b)
3-Methyl-1, 2-bromobutane
(c)
3-Methyl-1, 3-bromopropane
(d)
None
(c)
22.
(d)
Degree of unsaturation in
6 g of the organic compound on heating with NaOH
gave NH3 which is neutralised by 200 mL of 1 N
HCl. Percentage of nitrogen is
(a)
12%
(b)
60%
(c)
46.67 %
(d)
26.67%

is
C H 3 , N H 2 , OH  , F in increasing pKb values
are :

19.
(b)
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(a)

18.

CH 2  C H
III
The IUPAC name of the compound,
C H2
II
(CH 3 )3 C
21.
17.

CH2 = CH –



(a)
C H 3  N H 2  OH  F
(b)
F   OH   NH 2  CH 3
(c)
OH   NH 2  CH 3  F 
(d)
none is correct
23.
24.
B.E. of C — H bonds designated by I, II, III, IV,
25.
26.
27.
V in increasing order is :
(a)
V < IV < III < II < I
(b)
III < II < IV < V < I
(c)
III < II < V < IV < I
(d)
all equal
Einstein Classes,
(a)
2
(b)
3
(c)
4
(d)
5
The electronegativity of cesium is 0.7 and that of
fluorine is 4.0. The bond formed between the two is
(a)
covalent
(b)
electrovalent
(c)
coordinate
(d)
metallic
In which of the following molecules the bond angle
is maximum ?
(a)
CH4
(b)
H2O
(c)
NH3
(d)
CO2
Total number of line pair of electrons in XeF4 is
(a)
0
(b)
1
(c)
2
(d)
3
CO2 is isostructural with
(a)
SO2
(b)
HgCl2
(c)
C2H 2
(d)
SnCl2
Which of the following is the most polar bond ?
(a)
Cl – Cl
(b)
N–F
(c)
C–F
(d)
O–F
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CS – 18
29.
30.
31.
(A)
Tetracyanomethane
(B)
Carbon dioxide
(C)
Benzene
(D)
1, 3 butadiene
Ratio of  and  bonds is in order
(a)
A = B < C < D (b)
A=B<D<C
(c)
A = B = C = D (d)
C <D<A<B
3
In sp d hybridisation, the d orbital that participates
in hyubridisation is
(a)
dx2 – y2
(b)
dz2
(c)
dxy
(d)
dxz
C2 – C3 sigma single bond in vinyl acetylene is due
to the overlapping of
(a)
sp – sp
(b)
sp2 – sp2
(c)
sp – sp2
(d)
sp2 – sp
In the compound of the type ECl3, where E = B, P,
As or Bi the angles Cl – E – Cl for different E are in
the order
(a)
B > P = As = Bi (b)
B > P > As > Bi
(c)
B < P = As = Bi (d)
B < P < As < Bi
32.
33.
34.
A -bonds is formed by the overlap of
(a)
s-s orbital
(b)
s-p orbital
(c)
p-p orbital in end to end fashion
(d)
p-p orbital in sidewise manner
Mg2 C3 reacts with water and forms propyne C34–
which has
(a)
two sigma and two pi - bond
(b)
three sigma and one pi - bond
(c)
two sigma and one pi - bond
(d)
two sigma and three pi - bonds
Element X is strongly electropositive and Y is
strongly electronegative. Both are univalent. The
compound formed would be
(a)
X+ Y –
(b)
X–Y
–
+
(c)
X Y
(d)
XY
The correct order of the O – O bond length in
O2, H2O2 and O3 is
(a)
O3 > H2O2 > O2 (b)
O2 > H2O2 > O3
(c)
O2 > O3 > H2O2 (d)
H2O2 > O3 > O2
The molecule which has highest dipole moment
amongts the following is
(a)
CH3Cl
(b)
CH2Cl2
(c)
CHCl3
(d)
CCl4
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28.
35.
36.
FINAL STEP EXERCISE
1.
Pairs of related chemical species are given below.
Which pair is not related through resonance ?
(a)
(c)
CH3 – CO – CH2 – COOC2H5, CH3 –
C(OH) = CH – COOC2H5
(d)
3.
(b)
(c)
(d)
2.
Which is maximum basic in nature ?
Give the decreasing order of acidic characters of
the following :
(i)
(ii)
(iii)
(iv)
(a)
(a)
I > II > III > IV (b)
III > IV > I > II
(c)
II > III > IV > I (d)
I > III > IV > II
(b)
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., New Delhi -18
Ph. : 9312629035, 8527112111, E-mail [email protected],
www.einsteinclasses.com
CS – 19
4.
Arrange the following according to their stability
9.
Acidic character of the following in the increasing
order is :
(i)
II > IV > I > III > V > VI
(b)
II > IV > I > III > V > VI
(c)
I > II > III > IV > V > VI
(d)
VI > V > IV >III > II > I
The 1, 2 addition product of free radical addition
of CBrCl3 to 1, 3-butadiene
(a)
(b)
10.
(c)
(d)
6.
None
I
C6H5NH2
iii < ii < i
(b)
ii < i < iii
(c)
i < ii < iii
(d)
ii < iii < i
Arrange the following compounds in order of increasing b.pts.
(i)
CH3COCl
(ii)
(CH3CO)2O
(iii)
CH3CONH2
(iv)
CH3COOH
(a)
CH3COCl < CH3COOH < (CH3CO)2O <
CH3CONH2
(b)
CH3COCl > CH3COOH = (CH3CO)2O =
CH3CONH2
(c)
CH3COCl = CH3COOH > (CH3CO)2O >
CH3CONH2
(d)
CH3COCl < CH3COOH = (CH3CO)2O >
CH3CONH2
II
p - NO2 - C6H4NH2
o - NO2 - C6H4NH2
III
8.
(a)
Arrange in decreasing order of basicity.
m - NO2 - C6H4 - NH2
7.
(iii)
w E
w IN
w S
.e T
in E
st IN
ei C
nc L
la AS
ss S
es ES
.c
om
5.
(a)
(ii)
IV
(a)
II > I > III > IV (b)
I > II > III > IV
(c)
IV > III > II > I (d)
IV > II > III > I
Among 3-Butenoic acid and 3-Butynoic acid the
greater acidic character is
(a)
3-Butenoic acid
(b)
3-Butynoic acid
(c)
both
(d)
none
Basic character of following in the decreasing order is :
(i)
(CH3)3N
(iii)
CH3 – C  N
(a)
iii > ii > i
(b)
i > ii > iii
(c)
ii > i > iii
(d)
i > iii > ii
Einstein Classes,
(ii)
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., New Delhi -18
Ph. : 9312629035, 8527112111, E-mail [email protected],
www.einsteinclasses.com
ANSWERS (INITIAL STEP
EXERCISE)
a
13.
a
25.
c
2.
b
14.
d
26.
c
3.
c
15.
a
27.
c
4.
d
16.
a
28.
a
5.
b
17.
c
29.
b
6.
a
18.
a
30.
d
7.
d
19.
c
31.
b
8.
9.
10.
11.
12.
1.
2.
3.
4.
5.
a
b
c
a
a
6.
7.
8.
9.
10.
a
b
b
c
a
w E
w IN
w S
.e T
in E
st IN
ei C
nc L
la AS
ss S
es ES
.c
om
1.
CS – 20
ANSWERS (FINAL STEP EXERCISE)
c
20.
a
32.
d
d
21.
b
33.
a
a
22.
c
34.
a
b
23.
b
35.
d
d
24.
d
36.
b
Einstein Classes,
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., New Delhi -18
Ph. : 9312629035, 8527112111, E-mail [email protected],
www.einsteinclasses.com
CS – 21
TEST YOURSELF
7.
1.
2.
The IUPAC name of
is
(a)
3-ethyl-4, 4-dimethylheptane
(b)
1, 1-diethyl-2, 2-dimethylpentane
(c)
4, 4-dimethyl-5, 5-diethylpentane
(d)
5,5-diethyl-4, 4-dimethylpentane
The increasing order of stability of the following free
radicals is
8.
Which of the following pair of compounds are not
matamers ?
(a)
CH3OCH2CH2CH3 and
CH3CH2OCH2CH3
(b)
CH3CH2OCH2CH3 and CH3OCH(CH3)2
(c)
CH3NHCH2CH2CH3 and
CH3CH2NHCH2CH3
(d)
CH3NHCH2CH2CH3 and
CH3NHCH(CH3)2
Which of the following molecule has the maximum
number of stereoisomers ?
(a)



w E
w IN
w S
.e T
in E
st IN
ei C
nc L
la AS
ss S
es ES
.c
om

(CH 3 ) 2 C H  (CH 3 ) 3 C  (C6 H 5 ) 2 C H  (C6 H 5 ) 3 C
(b)



(a)

(C6 H 5 ) 3 C  (C6 H 5 ) 2 C H  (CH 3 ) 3 C  (CH 3 ) 2 C H
(c)



(b)

(C6 H 5 ) 2 C H  (C 6 H 5 ) 3 C  (CH 3 ) 3 C  (CH 3 ) 2 C H
(d)



(c)

(CH 3 ) 2 C H  (CH 3 ) C  (C6 H 5 ) 3 C  (C6 H 5 ) 2 C H
3.
4.
5.
6.
Increasing order of stability among the three main
conformations of 2-fluoroethanol is
(a)
eclipse, gauche, anti
(b)
gauche, eclipse, anti
(c)
eclipse, anti, gauche
(d)
anti, gauche, eclipse
(d)
9.
Which type of isomerism is shown by
2, 3-dichlorobutane ?
10.
“H3CO–” show which type of effect.
(a)
+ R and + I
(b)
+ R and – I
(c)
– R and + I
(d)
– R and – I
Triplet carbene is hybridised of
(a)
Geometrical
(b)
Structural
(a)
sp
(b)
sp2
(c)
Optical
(d)
Diastereo
(c)
sp3
(d)
dsp2
Which one of the following does not have sp 2
hybridized carbon ?
(a)
Acetone
(b)
Acetic acid
(c)
Acetonitrile
(d)
Acetamide
Which of the following will have a meso-isomer
also ?
(a)
2-chlorobutane
(b)
2,3-dichlorobuane
(c)
2,3-dichloropentane
(d)
2-hydroxypropanoic acid
Einstein Classes,
ANSWERS
1.
a
6.
b
2.
a
7.
d
3.
a
8.
d
4.
c
9.
b
5.
c
10.
a
Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., New Delhi -18
Ph. : 9312629035, 8527112111, E-mail [email protected],
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