Organic chemistry a modern approach 1

1.1 Hybridization, Bond Lengths, Bond Strengths or Bond Dissociation Enthalpies, Bond Angles and VSEPR Theory 1.1.1 Hybridization 1.1.2 Bond Length 1.1.3 Bond Dissociation Enthalpy or

ORGANIC CHEMISTRY

A Modern Approach

Volume-I

A BOUT THE A UTHOR

Nimai Tewari is a retired associate Professor, Department of Chemistry, Katwa College (affi liated to The University of Burdwan), West Bengal

A PhD in Organic Chemistry from Calcutta University, he has taught the subject for a period of more than three decades He has published various research papers in national and international journals Apart from Organic Chemistry—A Modern Approach, Dr Tewari has authored three more books on Organic Chemistry for undergraduate and postgraduate students His research interest includes Organic Synthesis and Heterocyclic Chemistry

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Organic Chemistry—A Modern Approach (Volume-I)

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1.4 Acids and Bases 1.46

1.4.1 Brönsted-Lowry Theory of Acids and Bases 1.46

1.4.2 Lewis Acid-Base Theory 1.52

1.6 Resonance and Resonance Effect or Mesomeric Effect 1.75

1.6.1 Resonance Energy 1.77

1.6.2 Rules for Writing Meaningful Resonance Structures 1.78

1.6.3 Relative Contribution of Resonance Structures towards Resonance Hybrid 1.78

1.6.4 Resonance or Mesomeric Effect 1.80

1.6.5 Isovalent and Heterovalent Resonance 1.81

1.6.6 Effect of Resonance on the Properties of Molecules 1.81

Solved Problems 1.98

Study Problems 1.112

1.7 Hyperconjugation 1.117

1.7.1 Sacrifi cial and Isovalent Hyperconjugation 1.119

1.7.2 Effect of Hyperconjugation on the Physical and Chemical Properties of Molecules and on the Stabilities of Intermediates 1.119

1.8.3 Face Strain or F-Strain 1.133

1.8.4 Steric Acceleration and Steric Retardation 1.134

1.9.2 van der Waals Forces 1.149

1.9.3 Effect of Intermolecular Forces on Different Properties of

Compounds 1.150Solved Problems 1.161

1.11.1 Mechanism of Keto-enol Tautomerism 1.221

1.11.2 Difference between Resonance and Tautomerism 1.222

1.11.3 Position of the Tautomeric Equilibrium 1.223

1.11.4 Ring-chain Tautomerism 1.228

1.11.5 Valence Tautomerism 1.228

Solved Problems 1.228

Study Problems 1.236

1.12 Aromaticity 1.240

1.12.1 Criteria for Aromaticity 1.240

1.12.2 Antiaromatic Compounds 1.242

1.12.3 Nonaromatic Compounds 1.242

1.12.4 Classifi cation of Compounds as Aromatic, Antiaromatic and Nonaromatic

by Comparing their Stabilities with that of the Corresponding Open Chain Compounds 1.243

1.12.5 Modern Defi nition of Aromaticity 1.243

1.12.6 Molecular Orbital Energy Diagram of Some Ions and Molecules 1.2441.12.7 Use of Inscribed Polygon Method to Determine the Relative Energies

of p Molecular Orbitals for Cyclic Planar and Completely Conjugated Compounds and to Classify Them as Aromatic and Antiaromatic 1.2451.12.8 Classifi cation of Some Molecules and Ions as Aromatic, Antiaromatic and

Nonaromatic 1.2471.12.9 Homoaromatic Compounds 1.251

1.12.10 Some Chemical and Physical Consequences of Aromaticity 1.251

1.14 Methods of Determining Mechanisms of Reactions 1.298

1.14.1 Kinetic Isotope Effects 1.305

Solved Problems 1.307

Study Problems 1.312

Introduction 2.2

2.1 Projection Formulas of Stereoisomers 2.3

2.1.1 Flying-Wedge Projection Formula 2.3

2.1.2 Fischer Projection Formula 2.4

2.1.3 Sawhorse Projection Formula 2.10

2.1.4 Newman Projection Formula 2.11

2.1.5 Interconversion of Projection Formulas 2.12

2.2.3 Centre of Symmetry (i) 2.36

2.2.4 Alternating Axis of Symmetry (Sn) 2.37

2.2.5 Symmetric, Asymmetric and Dissymmetric Molecules 2.38

2.5 Confi guration and Confi gurational Nomenclature 2.108

2.5.1 D, L-System of Confi gurational Designation 2.109

2.5.2 Specifi cation of Confi guration: The R, S-System 2.111

2.5.3 Erythro and Threo Nomenclature of Compounds with Two Adjacent Chiral Centres 2.124

2.5.4 The E-Z System of Designating Alkene Diastereoisomers 2.125

2.5.5 R, S and E, Z Assignment in the Same Molecule (Geometric

Enantiomerism) 2.1272.5.6 Syn-anti Nomenclature for Aldols 2.128

2.5.7 Number of Stereoisomers for Compounds with Chiral Centres 2.1292.5.8 Chirotopic and Achirotopic Atom in a Molecule 2.131

2.5.9 Prostereoisomerism and Topicity 2.133

2.6.6 Racemic Modifi cation 2.187

2.6.7 Enantiomeric Excess (EE) or Optical Purity (EE) 2.187

2.6.8 Racemization 2.188

2.6.9 Resolution of Racemic Modifi cation 2.192

Solved Problems 2.199

Study Problems 2.208

2.7 Conformation of Acyclic Organic Molecules 2.211

3.1.5 Evidence in Favour of SN2 Mechanism 3.6

3.1.6 Factors Infl uencing SN2 Reaction Rate or SN2 Reactivity 3.7

3.2.5 Evidence in Favour of SN1 Mechanism 3.90

3.2.6 Factors Infl uencing SN1 Reaction Rate or SN1 Reactivity 3.91

3.2.7 Carbocation Rearrangements in SN1 Reactions 3.105

3.2.8 Comparison of the SN2 and SN1 Reactions 3.108

3.2.9 Summary of Reactivity of Alkyl Halide in Necleophilic Substitution

Reactions 3.1083.2.10 Factors Favouring SN1 and SN2 Reactions 3.108

3.2.11 SNi and SNi¢ Mechanisms 3.109

3.3.5 Evidence for Participation by a Neighbouring Group 3.148

3.3.6 Various Cases of Neighbouring Group Participation 3.148

4.1.5 Evidence in Favour of the E2 Mechanism 4.32

4.1.6 Factors Infl uencing E2 Reaction Rate or E2 Reactivity 4.33

4.1.7 Factors that Govern the Proportions of E2 and SN2 Reactions 4.34

4.1.8 Regioselectivity in b-elimination Reactions (Orientation of p Bond in the Product Alkene) 4.37

4.1.9 Hofmann Exhaustive Methylation or Hofmann Degradation 4.43

4.2.5 Evidence in Favour of the E1 Mechanism 4.90

4.2.6 Factors Infl uencing E1 Reaction Rate or E1 Reactivity 4.90

4.2.7 Regioselectivity of E1 Reactions 4.90

4.2.8 Rearrangement of the Carbocation Intermediate Involved in an E1

Reaction 4.934.2.9 Acid-Catalyzed Dehydration of Alcohols 4.95

4.2.10 Dehydration using POCl3 and Pyridine 4.99

4.2.11 Factors Infl uencing the Extent of E1 and E2 Reactions 4.100

4.2.12 Factors that Govern the Proportions of E1 and SN1 Reactions 4.101

Solved Problems 4.102

Study Problems 4.114

4.3 The E1CB Reaction 4.118

4.3.1 Example of E1cB Reaction 4.118

4.3.2 Kinetics of E1cB Reaction 4.118

4.3.3 Mechanism of E1cB Reaction 4.118

4.3.4 The Nature of the Substrate 4.119

4.3.5 To Distinguish between E1cB and E2 Mechanisms 4.119

Solved Problems 4.121

Study Problems 4.125

4.4 a- or 1,1-Elimination 4.126

4.4.1 Example of a- or 1,1-Elimination Reaction 4.126

4.4.2 Kinetics of a- or 1,1-Elimination Reaction 4.126

4.4.3 Mechanism of a- or 1,1-Elimination Reaction 4.126

4.4.4 Structure of the Substrate Involved in a-Elimination 4.127

Solved Problems 4.127

Study Problems 4.128

P REFACE

In the course of teaching Organic Chemistry to undergraduate students, I have been constantly feeling the need of a concise volume that deals with their important topics on the subject matter Students have often expressed their diffi culty caused by the absence of such a compact book My present effort is to meet this long-felt need and the book has been designed primarily for the students who have taken a basic course in Organic Chemistry

at the undergraduate level

The book covers some important topics on Organic Chemistry in four chapters It starts with a chapter on Structure, Bonding and Properties of Organic Molecules, highlighting concepts like hybridization, electronegativity and bond polarity, acids and bases, inductive effect, resonance, steric effect, intermolecular forces, etc

Stereochemistry is an essential part of the organic chemistry courses Chapter 2 deals with the Principles of Stereochemistry Chapter 3 covers Nucleophilic Substitution Reactions

at Saturated Carbon Atom and Chapter 4 covers the Elimination Reactions

By following a modern methodology of learning, the book presents a large number of reactions with discussions supported with mechanistic explanations and diagrams, wherever needed Organic chemistry is best learned by solving problems Each article of every chapter concludes with a number of Solved as well as Study Problems to provide an opportunity to the students for self-evaluation

I believe that this book will be of great utility for the students who have taken a basic course of Organic Chemistry in B.Sc (Chemistry) Hons., besides being equally effective for advance students of Chemistry because of the in-depth discussion in a reader-friendly language The book will also be useful for the students preparing for competitive examinations like NET, SLET, etc

Acknowledgements

I offer my sincere gratitude to Mr Kaushik Bellani, MD, McGraw Hill Education (India) Pvt Ltd and Mrs Vibha Mahajan, Director, Science & Engineering Portfolio for successful

publication of this book I also wish to thank Mr Sumen Sen, Mr Amit Chatterjee and

Mr P L Pandita for taking keen interest in publishing this book I am grateful to all of them

I also owe a debt of gratitude to my colleagues for constructive suggestions and to my students who encouraged me constantly I appreciate the interest and enthusiasm shown

by my wife Mrs Dali Tewari and my daughter Andrila Tewari (Mukherjee) during the long period of preparation of the manuscript

Valuable suggestions from the readers for the improvement of the book will be most welcome

Nimai Tewari

1.1 Hybridization, Bond Lengths, Bond

Strengths or Bond Dissociation

Enthalpies, Bond Angles and VSEPR

Theory

1.1.1 Hybridization

1.1.2 Bond Length

1.1.3 Bond Dissociation Enthalpy or

Bond Dissociation Energy

1.1.4 Bond Angle

1.1.5 VESPR Theory and Molecular

Geometry

1.2 Electronegativity and Bond Polarity

1.3 Molecular Formula as a clue to

structure: Double Bond Equivalent

(DBE) or Index of Hydrogen

Defi ciency (IHD)

1.4 Acids and Bases

1.4.1 Brönsted-Lowry Theory of Acids

and Bases

1.4.2 Lewis Acid-Base Theory

1.5 Inductive and Electrometric Effects

1.6.3 Relative Contribution of Resonance

Structures towards Resonance

Hybrid

1.6.4 Resonance or Mesomeric Effect

1.6.5 Isovalent and Heterovalent Resonance

1.6.6 Effect of Resonance on the Properties of Molecules 1.7 Hyperconjugation

1.7.1 Sacrifi cial and Isovalent Hyperconjugation 1.7.2 Effect of Hyperconjugation on the Physical and Chemical Properties

of Molecules and on the Stabilities

of Intermediates 1.8 Steric Effect

1.8.1 Properties of Molecules Infl uenced

by Steric Effect 1.8.2 Proton Sponges 1.8.3 Face Strain or F-Strain 1.8.4 Steric Acceleration and Steric Retardation

1.8.5 Bredt’s Rule 1.9 Intermolecular Forces 1.9.1 Dipole–Dipole Interactions 1.9.2 van der Waals Forces 1.9.3 Effect of Intermolecular Forces on Different Properties of Compounds 1.10 Reactive Intermediates

1.10.1 Nonclassical Carbocation 1.10.2 Carbonium Ion and Carbenium Ion

or Carbocation 1.11 Tautomerism 1.11.1 Mechanism of Keto-enol Tautomerism

1.11.2 Difference between Resonance and Tautomerism

1.1 HYBRIDIZATION, BOND LENGTHS, BOND STRENGTHS OR BOND

DISSOCIATION ENTHALPIES, BOND ANGLES AND VSEPR THEORY 1.1.1 Hybridization

The process of intermixing of atomic orbitals of the same atom having slightly different energies so as to redistribute their energies and give new orbitals of equal energies and identical shapes and sizes is called hybridization

The number of hybrid orbitals is equal to the number of pure atomic orbitals reshuffl ed The hybrid orbitals are more effective in forming stable bonds as compared to the pure atomic orbitals and this is because they can undergo more effective overlapping The relative overlap of orbitals decreases in the following order: sp > sp2 > sp3 >> p The hybrid orbitals are oriented in space in some preferred directions to have a stable arrangement in which there is minimum repulsion among themselves Therefore, the type of hybridization governs the geometrical shapes of the molecules

Hybrid orbitals Hybridization Geometry Bond angle

234

s + p = sp

s + p + p = sp2

s + p + p + p = sp3

LinearTrigonal planarTetrahedral

1.12.4 Classifi cation of Compounds

as Aromatic, Antiaromatic and

Nonaromatic by Comparing

their Stabilities with that of

the Corresponding Open Chain

Compounds

1.12.5 Modern Defi nition of Aromaticity

1.12.6 Molecular Orbital Energy Diagram

of Some Ions and Molecules

1.12.7 Use of Inscribed Polygon Method to

Determine the Relative Energies

of p Molecular Orbitals for Cyclic

Planar and Completely Conjugated

Compounds and to Classify Them

as Aromatic and Antiaromatic 1.12.8 Classifi cation of Some Molecules and Ions as Aromatic, Antiaromatic and Nonaromatic

1.12.9 Homoaromatic Compounds 1.12.10 Some Chemical and Physical Consequences of Aromaticity 1.13 Thermodynamics, Energy Diagrams and Kinetics of Organic Reactions 1.13.1 Thermodynamics

1.13.2 Energy Diagram 1.13.3 Kinetics

1.13.4 Catalysis 1.13.5 Hammond Postulate 1.13.6 Kinetic Control versus Thermodynamic Control of a Chemical Reaction

1.14 Methods of Determining Mechanisms

of Reactions 1.14.1 Kinetic Isotope Effects

In fact, the number of groups surrounding a particular atom determines its geometry A group is either an atom or a lone pair of electrons Any atom surrounding by two, three and four groups are linear, trigonal planar and tetrahedral, respectively, and they have bond angles of 180°, 120° and 109.5°, respectively.

The hybridization of a C, O or N atom can be determined by the number of p bonds it forms If it forms no p bond, one p bond and two p bonds, it is sp3, sp2 and sp hybridized, respectively All single bonds are s bonds A double bond consists of one s bond and one p bond A triple bond consists of one s bond and two p bonds

“ Hybrid orbital number” method can be used to determine the hybridized state of an atom within a molecule

Hybrid orbital number = (number of s bonds) + (number of unshared pair of electrons)

If the hybrid orbital number is 2, the atom is sp hybridized, if it is 3, the atom is sp2hybridized and if it is 4, the atom is sp3 hybridized

The electronic confi guration of carbon atom in its ground state is 1s2 2s2 2px1 2py1 2pz°, i.e., one odd electron is present in each of 2px and 2py orbitals of carbon atom The number

of odd electrons present in the valence shell of an atom generally gives the measure of covalency of that atom So, the valency of carbon should be two However, the valency of carbon in almost all organic compounds is 4, except a few extremely unstable compounds, where its valency is 2, like methylene (:CH2), dichloromethylene (:CCl2), etc

During chemical reaction, the two electrons present in 2s orbital become unpaired by absorbing energy and one of them is promoted to 2pz orbital This is the excited state of carbon atom and the electronic confi guration of carbon atom in this state is 1s2 2s1 2px12py1 2pz1 Thus, in the excited state, four odd electrons are present in the outermost shell of carbon atom The presence of these four unpaired electrons accounts for the tetravalency

The four sp3 hybrid orbitals each containing one electron are directed towards the four corners of a regular tetrahedron, making an angle of 109°28′ with one another and the atom lies at the centre of the tetrahedron The hybrid orbitals are oriented in such a fashion

in space that there occurs minimum repulsion between them The formation of sp3 hybrid orbitals by the combination of s, px, py and pz atomic orbitals may be shown as follows:

Formation of methane (CH4) molecule: During the formation of methane molecule, one 2s orbital and three 2p orbitals of excited carbon atom undergo hybridization to form four equivalent sp3 hybrid orbitals The hybrid orbitals are directed towards the four corners of a regular tetrahedron Each hybrid orbitals containing an unpaired electron overlaps with the 1s orbital of a hydrogen atom resulting in the formation of four C—H s bonds Thus, methane molecule possesses a highly stable tetrahedral geometry with each H—C—H bond angle equal to 109°28′

It is to be noted that if the four atoms linked covalently to the carbon atom are not the same, the geometry of the molecule would still be tetrahedral but it may not be regular

in shape, e.g., methyl bromide (CH3Br), bromoform (CHBr3), etc In these cases, the bond angles differ slightly from the normal value of 109°28′

sp2-Hybridization: When one s orbital and two p orbitals of the valence shell of a carbon atom merge together and redistribute their energies to form three equivalent new orbitals of equal energy and identical shape, the type of hybridization occurs is called

sp2-hybridization The new orbitals formed as a result of this hybridization are called sp2hybrid orbitals

All three hybrid orbitals each containing one electron lie in one plane and make an angle

of 120° with each other, i.e., they are directed towards the three corners of an equilateral triangle with the carbon atom in the centre of the triangle The unhybridized 2pz orbital (containing one electron) remains perpendicular to the plane of the triangle with its two lobes above and below that plane Therefore, a molecule in which the central atom is

sp2-hybridized possesses triangular planar shape and the hybridization is called planar trigonal hybridization The formation of sp2 hybrid orbitals by combination of s, px and pyatomic orbitals is shown below

Formation of ethylene (C2H4) molecule: During the formation of ethylene molecule, each of the two carbon atoms undergoes sp2-hybridization, leaving the 2pz orbitals unhybridized The three sp2 hybrid orbitals of each carbon atom are planar and oriented

at an angle of 120° to each other The unhybridized 2pz orbitals are perpendicular to the plane of sp2 hybrid orbitals One sp2 orbital of one carbon overlaps axially with one

sp2 orbital of the other carbon to form a C—C s bond The remaining two sp2 orbitals of each carbon overlap with the half-fi lled 1s orbitals of two hydrogen atoms resulting in the formation of a total of three C—H s bonds The unhybridized 2pz orbitals of one carbon overlap with that of the other carbon in a sideways fashion to from a p bond between the two carbon atoms The p bond consists of two equal electron clouds distributed above and

below the plane of carbon and hydrogen atoms Since all the six atoms in the molecule lie in one plane, ethylene is a planar molecule Each C—C—H or H—C—H bond angle is nearly equal to 120°.

sp-Hybridization: When one s and one p orbital of the valence shell of a carbon atom merge together and redistribute their energies to form two equivalent hybrid orbitals of equal energy and identical shape, it results in sp-hybridization or diagonal hybridization

Formation of acetylene (C2H2) molecule: In acetylene molecule, the two carbon atoms are sp-hybridized There are two unhybridized orbitals (2py and 2pz) on each C atom Two

sp hybrid orbitals are linear and directed at an angle of 180° The unhybridized p orbitals are perpendicular to the sp hybrid orbitals and also perpendicular to each other One sp hybrid orbital of one carbon overlaps axially with the similar orbital of the other carbon

to form a C—C s bond The remaining hybrid orbital of each C atom overlaps with

half-fi lled 1s orbital of H atom to form a total of two C—H s bond Thus, acetylene molecule

is linear The unhybridized py orbitals of two carbons and the unhybridized pz orbitals of two carbons overlap sideways separately to form two different p bonds Electron clouds

of one p bond lie above and below the internuclear axis representing the s bond while the electron clouds of the other p bond lie in front and backside of the internuclear axis These two sets of p electron cloud merge into one another to form a cylindrical cloud of electrons around the internuclear axis surrounding the C—C s bond Each C—C—H bond angle is equal to 180°

C

p

Csp

2py

2px2py

Linear acetylene molecule

1.1.2 Bond Length

Bond length is defi ned as the equilibrium distance between the centres of the nuclei of two bonded atoms in a covalent molecule Bond length depends on the factors such as(i) size of atoms, (ii) multiplicity of bonds, (iii) s-character of the orbitals and (iv) resonance, hyperconjugation, etc Bond length increases with increase in the size of atoms but it decreases with increase in bond multiplicity Thus, a triple bond is shorter than a double bond which in turn is shorter than a single bond An s orbital is closer to the nucleus than a p orbital So, electrons in the s orbital is more tightly held by the nucleus than the electrons in the p orbital For this reason, with increase in s-character of the hybrid orbital, the attractive force on the electron(s) increases and so, the size of the hybrid orbital decreases As a consequence, the length of the bond obtained by overlapping of the hybrid orbitals with the s orbitals of hydrogen, for example, decreases The s-character of sp3, sp2and sp hybrid orbitals are 25 percent, 33.33 percent and 50 percent respectively Thus, the lengths of the C—H bonds involving C atoms with different hybridization follow the order:

3

C

sp —H (1.093 Å) > C 2

sp —H (1.078 Å) > Csp — H (1.057 Å)

1.1.3 Bond Dissociation Enthalpy or Bond Dissociation Energy

The bond dissociation enthalpy or bond dissociation energy which is a measure of bond strength may be defi ned as the amount of energy required to break one mole of a particular type of bond between two atoms in the gaseous state so as to produce neutral gaseous atoms or free radicals Bond dissociation energies are usually abbreviated by the symbol DH° and are usually expressed in kJ mol–1 or kcal mol–1 The greater the bond dissociation energy, stronger the bond The factors affecting bond dissociation energy are: (i) size of the bonded atom, (ii) multiplicity of bonds, (iii) s-character of the hybrid orbital involved

in bond formation, etc

Larger the size of the bonded atoms, greater the bond length and lesser is the bond dissociation energy The bond dissociation energy increases with increase in bond multiplicity A C ∫∫ C bond is stronger than a C == C bond which in turn is stronger than

a C – C bond The bond dissociation energy also increases with increase in the s-character

of the hybrid orbital and this is because with increase in s-character of the hybrid orbital, the electron density in the region of overlap increases

Bond type Bond dissociation energy

kJ mol –1 (kcal mol –1 )

C(sp3) — C(sp3)C(sp3) — C(sp2)C(sp3) — C(sp)C(sp2) — C(sp2)C(sp2) — C(sp)C(sp) — C(sp)

346.3 (82.76)357.6 (85.48)382.5 (91.42)383.2 (91.58)403.7 (96.48)433.5 (103.6)

Bond Dissociation Energies (kcal mol –1 ) of Some Chemical Bonds

Bond DH° Bond DH° Bond DH°

1 4

Therefore, EC—H = 397/4 = 99.25 kcal mol–1

However, bond dissociation energy (DH°) is more convenient than bond energy (E) for our purpose

1.1.5 VESPR Theory and Molecular Geometry

We can predict the arrangement of atom in molecules and ions on the basis of a relatively simple idea called the valence shell electron pair repulsion (VSEPR) theory

We can apply this theory on the basis of the following considerations:

1 In a molecule or ion, the central atom is covalently bonded to two or more atoms or

groups

2 Covalent bonds contain shared pair of electrons which are often called bond pairs

or bonding pairs The unshared electrons of the central atom are called nonbonding pairs or unshared pairs or lone pairs

3 Since electron pairs repel each other, the electron pairs of the valence shell tend to

stay as far apart as possible to avoid electronic repulsion

4 If the central atom is surrounded by bond pairs as well as lone pairs of electrons,

the repulsions among themselves are different As a result, the molecule possesses

an irregular or distorted geometry The repulsive interactions of various electron pairs decrease in the order: lone pair–lone pair (lp – lp)> lone pair–bond pair (lp – bp) > bond pair–bond pair (bp – bp)

5 The geometry of a molecule is to be settled by considering all of the electron pairs,

bonding and nonbonding However, the shape of the molecule is to be described

by referring to the positions of the atoms and not by the position of the electron pairs

Let us consider the following examples

Methane (CH4) molecule: In CH4 molecule, the total number of electrons surrounding the central carbon atom = 4 valence electrons of C atom + 4 electrons of four singly-bonded

H atom = 8 electron or 4 electrons pairs = 4 s bond pairs The four bond pairs experience minimum repulsion if they are tetrahedrally oriented, i.e., if all the H—C—H bond angles are of 109.5° Hence, the shape of CH4 molecule is tetrahedral

(109.5°) and is reduced to 107°, i.e., the tetrahedron is somewhat distorted Excluding the lone pair, the shape of the molecule is trigonal pyramidal.

Water (H2O) molecule: In water molecule, the total number of electrons surrounding the central O atom = 6 valence electron of oxygen atom + 2 electrons of two singly-bonded H atoms = 8 electrons or 4 electron pairs = 2 s bond pairs + 2 lone pairs

In order to minimise the extent of mutual repulsion, these four electron pairs are oriented towards the four corners of a tetrahedron However, the tetrahedron is somewhat distorted due to the strong repulsive forces exerted by the lone pairs on each bond pair of electrons

In fact, the H—O—H bond angle is reduced to 104.5° from the normal tetrahedral angle of 109.5° Excluding the lone pairs, the shape of the molecule is angular or V-shaped

Angular or V-shapedwater molecule

O104.5°

HH

B120°

F

Trigonal planarboron trifluoride moleculeAcetylene (HCCH) molecule: The number of electrons surrounding each carbon atom

of acetylene molecule = 4 valence electron of carbon + 3 electrons of one triply-bonded C atom + 1 electron of one singly-bonded H atom = 8 electrons = 4 electron pairs = 2 s bond

pairs + 2p bond pairs In order to minimise the repulsive forces between the bond pairs, the shape of acetylene molecule is linear The effect of electrons involved in the formation of a

p bond is not generally considered in determining the geometrical shape of a molecule

H——C∫∫C——H180° 180°

Linear acetylene molecule

1 Give the state of hybridization of the central atom of each of the following

species and predict their shapes

+

(d)

CHH

2 Draw the structure of a hydrocarbon which contains:

(a) Three sp3-hybridized carbon atoms;

(b) One sp3 and two sp2-hybridized carbon atoms;

(c) One sp and two sp2-hybridized carbon atoms;

(d) Two sp3 and two sp-hybridized carbon atoms

sp ) and each N—H bond is formed by the overlap of

an sp3 orbital of nitrogen with the s orbital of hydrogen (N 3—H1s)

4 Answer the following question for the acetaldehyde —|| —

(c) Mention the type of orbital in which the lone pairs reside

Solution

(a)

(b) The s bond is formed by the end-on overlap of an sp2 orbital of carbon with an sp2

orbital of oxygen and the p bond is formed by the side-by-side overlap of the 2p orbital of carbon with the 2p orbital of oxygen

(c) The two sp2 hybrid orbitals are occupied by the two lone pairs of oxygen

5 Mention the state of hybridization of the starred (*) carbon atoms in each

of the following compounds:

3

CH CN (c) HC∫∫C—CHO*(d) CH == C == CH 2 * 2 (e) * (f)

6 How many s and p bonds are present in each of the following molecules?

(a) CH3—C ∫∫ C—CH == CH—CH3

(b) –CH CH2 3

(c) CH3 CH == C == CH CH2 CH3

Solution The hybridized atoms within the molecule is designated as a, b and c according

to their hybridization status: a = sp3; b = sp2 and c = sp

8 Draw the orbital picture for each of the following molecules: (a) ethylene

(b) acetylene (c) ketene (CH2 == C == O) (d) acraldehyde (e) acrylonitrile(f) but-1, 2, 3-triene

Solution

(a) Ethylene:

(b) Acetylene:

(c) Ketene:

(d) Acraldehyde:

(e) Acrylonitrile:

(f) But-1, 2, 3-triene:

9 Which atoms in each of the following molecules always remain in the

same plane and why?

(a) CH3CH==CH CH3 (b) C6H5C ≡≡ CCH3

(c) CH3CH==C==C==CH2 (d) CH2==CH—C ≡≡ CH

Solution The two sp2-hybridized carbon atoms and the atom directly attached to them always remain in the same plane This is applicable also when the two sp2-hybridzed carbon atoms are linked through an even number of sp-hybridized carbon atoms Therefore, in (a), all atoms excluding the six methyl hydrogens remain in the same plane In (b), all atoms excluding the three methyl hydrogens remain in the same plane In (c), all atoms excluding the three methyl hydrogens remain in the same plane and in (d), all atoms remain in the same plane This is also because the two sp-hybridized carbon atoms and the atoms directly attached to them remain in the same line

10 Explain why a p bond is weaker than a s bond

Solution A p bond is weaker than a s bond because the end-on overlap that forms s bonds

is better than the side-to-side overlap that forms p bonds This is also because the electron density in a p bond is farther from the two nuclei as compared to that in a s bond

11 In CH3CH3, the C—H bond is shorter and stronger than the C—C bond –

explain

Solution The s orbital of hydrogen is closer to the nucleus than the sp3 orbital of carbon having less s character So, the carbon and hydrogen nuclei are closer together in sp3-s overlap than the carbon nuclei in sp3-sp3 overlap Because of this, the C—H bond in ethane (CH3—CH3) is shorter and stronger than the C—C bond Again, as the percentage

of s character of the overlapping orbitals increases, the electron density in the region

of overlap increases and as a result, the strength of the bond increases Since there is greater electron density in the region of sp3-s overlap than in the region of sp3-sp3 overlap, therefore, the C—H bond is stronger than the C—C bond

12 Arrange the indicated bonds in each of the following compounds in order

of increasing bond strength and increasing bond length:

bond 3bond 1

Solution Greater the bond multiplicity, shorter the bond and greater the bond strength Therefore, in compound (a), bond length increases in the order: bond 1 < bond 2 < bond

3 and bond strength increase in the order: bond 3 < bond 2 < bond 1 In compound (b), the bond length increases in the order: bond 3 < bond 2 < bond 1 and the bond strength increases in the order: bond 1 < bond 2 < bond 3

13 Which of the indicated bonds in each pair of compounds is shorter and

sp —H bond in the fi rst compound;(b) the N 2

sp —H bond in the second compound is shorter than the N 3

sp —H bond in the fi rst compound, and (c) the C 2

sp —H bond in the fi rst compound is shorter than the

3

C

sp —H bond in the second compound

14 Determine the state of hybridization of the indicated atom in each of the

Solution In any system of the type X = Y – Z:, Z is sp2-hybridized and the unshared electron pair on it occupies a p orbital to delocalize the electron pair and make the system conjugated Therefore, the indicated atoms in (a), (b), (c), (d) and (f) are sp2-hybridized However, due to violation of Bredt’s rule (introduction of a double bond is not possible at the bridgehead position in bridged bicyclic compounds with small rings) delocalization of the unshared electron pair is not possible in the compound (e) and therefore, the indicated carbon atom is sp3-hybridized.

15 “Bromination of methane is less exothermic than that of chlorination” —

explain with DH° calculation

[Bond dissociation energies for C—H = 104 kcal/mol; Br—Br = 46 kcal/mol; H—Br = 87.5 kcal/mol; Cl—Cl = 58 kcal/mol; H—Cl = 103 kcal/mol; C—Cl = 83.5 kcal/mol and C—Br =

16 The H—X bond in hydrogen halides becomes longer and weaker as the

atomic mass of the halogen increases—explain

Solution A p orbital of halogen overlaps with the s orbital of hydrogen to form the hydrogen–halogen (H—X) bond in hydrogen halides For bond formation, fl uorine uses the p orbital that belongs to the second shell of electrons while chlorine uses the p orbital that belongs to the third shell of electrons Since the average distance from the nucleus is greater for an electron in the third shell than for an electron in the second shell, therefore, the average electron density is less in a 3p orbital than in a 2p orbital Consequently, the electron density in the region where the s and p orbitals overlap decreases as the size of the halogen increases Thus, the hydrogen–halogen bond becomes longer and weaker as the atomic mass of the halogen increases

1 A carbon-carbon bond formed by sp2–sp2 overlap is stronger than the one formed

by sp3–sp3 overlap Explain

2 Which atom in the ammonium ion ≈

4

(NH ) has the least electron density and why?

3 What is the state of hybridization of each of the C, O and N atom in the following compound?

H C3CO

5 Predict the mentioned bond angles:

(a) C—O—C bond angle in CH3COCH3 (b) F—B—F bond angle in @

4

BF(c) C—C—N bond angle in C2H5CN (d) H—N—H bond angle in ≈

4

NH(e) C—O—H bond angle in CH3CH2OH (f) C—N—C bond angle in (CH )3 2 N H! 2

6 Which of the indicated bonds is shorter and why?

7 Give the state of hybridization of the indicated atoms:

1.2 ELECTRONEGATIVITY AND BOND POLARITY

Electronegativity is the tendency of an atom to pull the bonding electrons towards itself

It increases across a row of the periodic table (excluding the noble gases) and it decreases down a column of the periodic table The electronegativity values of some common elements are given in the following table

A bond with the electrons shared equally between the two atoms, i.e., a bond in which each electron spends as much time in the vicinity of one atom as in the other, is called

a nonpolar covalent bond For example, the H—H covalent bond, the Br—Br covalent bond and the C—C covalent bond in ethane etc are nonpolar covalent bonds When two atoms of different electronegativities form a covalent bond, the electrons are not shared equally between them The atom with greater electronegativity draws the electron pair closer to it As a result of this unequal distribution of the bonding electrons, the bond acquires a slight positive charge (indicated by the symbol d+) on the end that has the less electronegative atom and a slight negative charge (indicated by the symbol d–) on the other end that has the more electronegative atom, i.e., a polarity developes within a bond Such a bond is called a polar covalent bond An example of a polar covalent bond is

the one in hydrogen fl uoride (H—F) The fl uorine atom, with its greater electronegativity, pulls the bonding electrons towards itself As a consequence, the hydrogen atom becomes somewhat electron defi cient and acquires a partial positive charge (d+) while the fl uorine atom becomes somewhat electron rich and acquires a partial negative charge (d–) So, the hydrogen fl uoride molecule is a dipole (Hd+— Fd–) The direction of bond polarity is indicated by an arrow, the head of which indicates the negative end of the bond A short perpendicular line is drawn near the tail of the arrow to indicate the positive end of the bond.

The extent of polarity in a covalent bond is expressed quantitatively by the physical property known as ‘ dipole moment’ m, which is, in fact, the product of the magnitude of charge e on any polar end and the distance d between the centres of positive and negative charges, i.e., m = e × d

Dipole moment is expressed in Debye unit (D) The charge e and the distance d are of the order 10–10 esu and 10–8 cm, respectively Therefore, m is of the order of 10–10 × 10–8 = 10–18esu.cm (in C.G.S system) This value of dipole moment is equal to one Debye unit, i.e.,1D = 10–18 esu cm The SI unit of dipole moment is coulomb-meter (C.m.) and 1 C.m = 2.9962 × 1029 D

The value of m for a nonpolar molecule is zero For example, m = 0D for the molecules like

H2, N2, O2, etc For polar molecules, m has a defi nite value, e.g., for HF molecule m = 1.91

D Polarities of molecules increase with increase in the value of m

Dipole moment is a vector quantity The dipole moment of a molecule is the result of the vector sum of the individual bond moments present in the molecule When a molecule

is formed by two atoms of different electronegativities (i.e., by two different atoms), the molecule must possess a dipole moment because there is no question to cancellation of the moment of this only polar bond However, when a molecule contains two or more polar bonds, i.e., when a molecule contains more than two atoms, the molecule may or may not possess dipole moment and in that case the polarity of the molecule depends on the shape

of the molecule A symmetrical polyatomic molecule possesses no dipole moment because the individual dipole moments of the bonds cancel each other It is to be noted that the symmetry in calculating dipole moment is not the molecular symmetry H2O molecule, for example, has a symmetrical structure because it has two planes of symmetry (su planes) But, it possesses dipole moment (1.85 D) because of its angular structure The value of dipole moment of a molecule is equal to the product of the charge e and the distance, d, between the positive and negative charge centres, i.e., m = e × d, it can be well understood by the following examples: