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A Q&A Approach to
Organic Chemistry

A Q&A Approach to
Organic Chemistry
Michael B. Smith

First edition published 2020
by CRC Press
6000 Broken Sound Parkway NW, Suite 300, Boca Raton, FL 33487-2742
and by CRC Press
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© 2020 Taylor & Francis Group, LLC
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Typeset in Times
by Deanta Global Publishing Services, Chennai, India

Contents
Preface ......................................................................................................................................................ix
Common Abbreviations............................................................................................................................xi
Author .....................................................................................................................................................xiii
Part A A Q&A Approach to Organic Chemistry
1 Orbitals and Bonding....................................................................................................................... 3
1.1 ORBITALS............................................................................................................................. 3
1.1.1 Atomic Orbitals......................................................................................................... 3
1.1.2 Electron Confguration.............................................................................................. 5
1.1.3 Molecular Orbitals.................................................................................................... 5
1.2 BONDING.............................................................................................................................. 6
1.2.1 Ionic Bonding............................................................................................................ 6
1.2.2 Covalent Bonding...................................................................................................... 7
1.3 HYBRIDIZATION............................................................................................................... 12
1.4 RESONANCE ...................................................................................................................... 15
END OF CHAPTER PROBLEMS ...................................................................................................18
2 Structure of Molecules................................................................................................................... 19
2.1 BASIC STRUCTURE OF ORGANIC MOLECULES........................................................ 19
2.1.1 Fundamental Structures.......................................................................................... 19
2.1.2 Structures with Other Atoms Bonded to Carbon ................................................... 22
2.2 THE VSEPR MODEL AND MOLECULAR GEOMETRY............................................... 23
2.3 DIPOLE MOMENT ............................................................................................................. 25
2.4 FUNCTIONAL GROUPS.................................................................................................... 26
2.5 FORMAL CHARGE............................................................................................................ 28
2.6 PHYSICAL PROPERTIES................................................................................................... 28
END OF CHAPTER PROBLEMS .................................................................................................. 32
3 Acids and Bases .............................................................................................................................. 33
3.1 ACIDS AND BASES ........................................................................................................... 33
3.2 ENERGETICS...................................................................................................................... 35
3.3 THE ACIDITY CONSTANT, Ka ......................................................................................... 38
3.4 STRUCTURAL FEATURES THAT INFLUENCE ACIDITY.......................................... 40
3.5 FACTORS THAT CONTRIBUTE TO MAKING THE ACID MORE ACIDIC ............... 45
4 Alkanes, Isomers, and Nomenclature .......................................................................................... 49
4.1 DEFINITION AND BASIC NOMENCLATURE............................................................... 49
4.2 STRUCTURAL ISOMERS.................................................................................................. 50
4.3 IUPAC NOMENCLATURE ................................................................................................ 52
4.4 CYCLIC ALKANES............................................................................................................ 57
v

vi Contents
5 Conformations.................................................................................................................................61
5.1 ACYCLIC CONFORMATIONS...........................................................................................61
5.2 CONFORMATIONS OF CYCLIC MOLECULES ............................................................. 67
6 Stereochemistry.............................................................................................................................. 77
6.1 CHIRALITY ........................................................................................................................ 77
6.2 SPECIFIC ROTATION ........................................................................................................ 81
6.3 SEQUENCE RULES............................................................................................................ 83
6.4 DIASTEREOMERS............................................................................................................. 87
6.5 OPTICAL RESOLUTION ................................................................................................... 89
7 Alkenes and Alkynes: Structure, Nomenclature, and Reactions .............................................. 93
7.1 STRUCTURE OF ALKENES ............................................................................................. 93
7.2 NOMENCLATURE OF ALKENES ................................................................................... 95
7.3 REACTIONS OF ALKENES .............................................................................................. 98
7.4 REACTION OF ALKENES WITH LEWIS ACID-TYPE REAGENTS ......................... 107
7.4.1 Hydroxylation........................................................................................................ 107
7.4.2 Epoxidation............................................................................................................111
7.4.3 Dihydroxylation .....................................................................................................113
7.4.4 Halogenation ..........................................................................................................114
7.4.5 Hydroboration ........................................................................................................117
7.5 STRUCTURE AND NOMENCLATURE OF ALKYNES............................................... 122
7.6 REACTIONS OF ALKYNES............................................................................................ 124
END OF CHAPTER PROBLEMS ................................................................................................ 129
8 Alkyl Halides and Substitution Reactions ..................................................................................133
8.1 STRUCTURE, PROPERTIES, AND NOMENCLATURE OF ALKYL HALIDES........133
8.2 SECOND-ORDER NUCLEOPHILIC SUBSTITUTION (SN2) REACTIONS ................ 134
8.3 OTHER NUCLEOPHILES IN SN2 REACTIONS.............................................................143
8.4 FIRST-ORDER SUBSTITUTION (SN1) REACTIONS.....................................................151
8.5 COMPETITION BETWEEN SN2 vs. SN1 REACTIONS.................................................. 156
8.6 RADICAL HALOGENATION OF ALKANES ............................................................... 158
END OF CHAPTER PROBLEMS .................................................................................................162
9 Elimination Reactions...................................................................................................................165
9.1 THE E2 REACTION...........................................................................................................165
9.2 THE E1 REACTION...........................................................................................................172
9.3 PREPARATION OF ALKYNES........................................................................................176
9.4 SYN ELIMINATION..........................................................................................................178
10 Organometallic Compounds ........................................................................................................183
10.1 ORGANOMETALLICS ......................................................................................................183
10.2 ORGANOMAGNESIUM COMPOUNDS .........................................................................183
10.3 ORGANOLITHIUM COMPOUNDS.................................................................................185
10.4 BASICITY ...........................................................................................................................187
10.5 REACTION WITH EPOXIDES .........................................................................................188
10.6 OTHER METALS...............................................................................................................188

Contents vii
11 Spectroscopy ..................................................................................................................................191
11.1 THE ELECTROMAGNETIC SPECTRUM .......................................................................191
11.2 MASS SPECTROMETRY ................................................................................................. 192
11.3 INFRARED SPECTROSCOPY (IR)................................................................................. 196
11.4 NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY (nmr)................................. 201
12 Aldehydes and Ketones. Acyl Addition Reactions .....................................................................219
12.1 STRUCTURE AND NOMENCLATURE OF ALDEHYDES AND KETONES.............219
12.2 REACTION OF ALDEHYDES AND KETONES WITH WEAK NUCLEOPHILES...... 221
12.3 REACTIONS OF ALDEHYDES AND KETONES. STRONG NUCLEOPHILES ........ 230
12.4 THE WITTIG REACTION................................................................................................ 233
Part B A Q&A Approach to Organic Chemistry
13 Oxidation Reactions..................................................................................................................... 239
13.1 OXIDATION REACTIONS OF ALKENES ..................................................................... 239
13.2 OXIDATION OF ALKENES: EPOXIDATION................................................................244
13.3 OXIDATIVE CLEAVAGE: OZONOLYSIS....................................................................... 247
13.4 OXIDATIVE CLEAVAGE. PERIODIC ACID CLEAVAGE OF 1,2-DIOLS ................... 250
13.5 OXIDATION OF ALCOHOLS TO ALDEHYDES OR KETONES .................................251
14 Reduction Reactions..................................................................................................................... 257
14.1 CATALYTIC HYDROGENATION................................................................................... 258
14.2 DISSOLVING METAL REDUCTION: ALKYNES......................................................... 264
14.3 HYDRIDE REDUCTION OF ALDEHYDES AND KETONES ..................................... 265
14.4 CATALYTIC HYDROGENATION AND DISSOLVING METAL REDUCTIONS.
ALDEHYDES AND KETONES....................................................................................... 269
15 Carboxylic Acids, Carboxylic Acid Derivatives, and Acyl Substitution Reactions ............... 275
15.1 STRUCTURE OF CARBOXYLIC ACIDS....................................................................... 275
15.2 PREPARATION OF CARBOXYLIC ACIDS................................................................... 280
15.3 CARBOXYLIC ACID DERIVATIVES ............................................................................ 283
15.4 PREPARATION OF ACID DERIVATIVES..................................................................... 290
15.5 HYDROLYSIS OF CARBOXYLIC ACID DERIVATIVES ............................................ 301
15.6 REACTIONS OF CARBOXYLIC ACIDS AND ACID DERIVATIVES........................ 305
15.7 DIBASIC CARBOXYLIC ACIDS......................................................................................310
16 Benzene, Aromaticity, and Benzene Derivatives........................................................................315
16.1 BENZENE AND NOMENCLATURE OF AROMATIC COMPOUNDS ........................315
16.2 ELECTROPHILIC AROMATIC SUBSTITUTION ..........................................................319
16.3 SYNTHESIS VIA AROMATIC SUBSTITUTION ...........................................................335
16.4 NUCLEOPHILIC AROMATIC SUBSTITUTION........................................................... 337
16.5 REDUCTION OF BENZENE AND BENZENE DERIVATIVES................................... 344
16.6 POLYCYCLIC AROMATIC COMPOUNDS AND HETEROAROMATIC
COMPOUNDS ................................................................................................................... 347

viii Contents
17 Enolate Anions and Condensation Reactions............................................................................ 357
17.1 ALDEHYDES, KETONES, ENOLS, AND ENOLATE ANIONS .................................. 357
17.2 ENOLATE ALKYLATION................................................................................................361
17.3 CONDENSATION REACTIONS OF ENOLATE ANIONS AND ALDEHYDES
OR KETONES.................................................................................................................... 366
17.4 ENOLATE ANIONS FROM CARBOXYLIC ACIDS AND DERIVATIVES ................ 372
18 Conjugation and Reactions of Conjugated Compounds........................................................... 385
18.1 CONJUGATED MOLECULES......................................................................................... 385
18.2 STRUCTURE AND NOMENCLATURE OF CONJUGATED SYSTEMS .................... 387
18.3 REACTIONS OF CONJUGATED MOLECULES ............................................................391
18.4 THE DIELS–ALDER REACTION ................................................................................... 393
18.5 [3+2]-CYCLOADDITION REACTIONS .......................................................................... 401
18.6 SIGMATROPIC REARRANGEMENTS.......................................................................... 403
18.7 ULTRAVIOLET SPECTROSCOPY.................................................................................. 406
19 Amines............................................................................................................................................413
19.1 STRUCTURE AND PROPERTIES....................................................................................413
19.2 PREPARATION OF AMINES ...........................................................................................416
19.3 REACTIONS OF AMINES ............................................................................................... 420
19.4 HETEROCYCLIC AMINES ............................................................................................. 424
20 Amino Acids, Peptides, and Proteins ......................................................................................... 429
20.1 AMINO ACIDS.................................................................................................................. 429
20.2 SYNTHESIS OF AMINO ACIDS......................................................................................435
20.3 REACTIONS OF AMINO ACIDS .................................................................................... 437
20.4 PROTEINS ......................................................................................................................... 441
21 Carbohydrates and Nucleic Acids............................................................................................... 449
21.1 CARBOHYDRATES ......................................................................................................... 449
21.2 DISACCHARIDES AND POLYSACCHARIDES............................................................ 457
21.3 SYNTHESIS OF CARBOHYDRATES ............................................................................ 459
21.4 REACTIONS OF CARBOHYDRATES............................................................................ 461
21.5 NUCLEIC ACIDS, NUCLEOTIDES, AND NUCLEOSIDES......................................... 464
Appendix: Answers to End of Chapter Problems ............................................................................ 473
Index...................................................................................................................................................... 505

Preface
What is organic chemistry?
Organic chemistry is the science that studies molecules containing the element carbon. Carbon can
form bonds to other carbon atoms or to a variety of atoms in the periodic table. The most common
bonds observed in an organic chemistry course are C—C, C—H, C—O, C—N, C—halogen (Cl, Br, I),
C—Mg, C—B, C—Li, C—S and C—P.
This book is presented in the hope that it will provide extra practice to students taking organic chem-
istry for the frst time and also serve as a cogent review to those who need to refresh their knowledge
of organic chemistry. This book of questions began life as Organic Chemistry in 1993 to assist those
students taking undergraduate organic chemistry and was part of a HarperCollins Outline series that was
never completed. My book, along with those other books in the series that were completed, was sold as
a reference book rather than a textbook. In 2006, a second edition of Organic Chemistry was published
and marketed more or less the same way. The book laid fallow for several years until this version became
possible. With this book, published by CRC Press/Taylor & Francis Group, I continue the idea of teach-
ing organic chemistry by asking leading questions.
A Q&A Approach to Organic Chemistry is intended as a supplement to virtually any organic chem-
istry textbook rather than a stand-alone text and it will allow a “self-guided tour” of organic chemistry.
Teaching organic chemistry with a Q&A format uses leading questions along with the answers and is
presented in a manner that allows a student to refresh and renew their working knowledge of organic
chemistry. Such an approach will also be of value to those reviewing organic chemistry for MCATs
(Medical College Admission Test); graduate record exams (a standardized test), which is an admissions
requirement for many graduate schools); the PCAT (Pharmacy College Admission Test), which identi-
fes qualifed applicants to pharmacy colleges before commencement of pharmaceutical education; and
so on.
This Q&A format was classroom-tested here at the University of Connecticut for many years where
one of the earlier versions of this book was used as a supplement. Indeed, the book was not required for
purchase and used only on a voluntary basis by students. According to their end-of-semester evaluations,
students who wanted or needed additional homework found the book very useful and helpful. Classroom
experience and comments from students have been used for the preparation of this new student-friendly
book.
This book is organized into 21 chapters and will supplement most of the organic textbooks on the
market. In all chapters, there are leading questions to focus attention on a principle or reaction and the
answer is immediately provided. The organization of the book provides an initial review of fundamental
principles followed by reactions based on manipulation of functional groups. The intent in all cases is
to provide a focused question about a specifc principle or reaction and the answer immediately follows.
There is also a chapter on spectroscopy as well as chapters on amino acid and peptide chemistry and
carbohydrate and nucleoside chemistry. Each chapter ends with several homework questions for that
chapter, and the answers are provided in an Appendix at the end of the book.
I thank all of the organic chemistry students I taught over the years. They provided the inspira-
tion for the book as well as innumerable suggestions that were invaluable. I thank Ms. Hilary Lafoe
and Ms. Jessica Poile, the Taylor & Francis editors for this book, and also Dr. Fiona Macdonald, the
publisher. This book would not have been possible without their interest in chemistry and their help
as the book was written. I thank Professor John D’Angelo of Alfred University who provided a very
useful and helpful review of the manuscript. I thank PerkinElmer who provided a gift of ChemDraw
Professional (Version 18.0.0.231[4318]). All the reactions and fgures were done with ChemDraw except
for those images that use molecular models and the artist-rendered drawings. All molecular models
were rendered with Spartan’18 software and I thank Warren Hehre and Sean Ohlinger of Wavefunction,
ix

x Preface
Inc., who provided a gift of Spartan’18 software, version 1.2.0 (181121). I thank Ms. Christine Elder
(https://christineelder.com), graphics design artist, for her graphic arts expertise to render the drawings
on pages 14 (C1), 66 (C5), 93 and 122 (C7), 208 and 209 (C11), 280 (C15), 315 and 342 (C16). Finally, I
thank my wife, Sarah, for her patience and understanding while I was putting this book together.
Where there are errors, I take complete responsibility. Please contact me at michael.smith@uconn.edu
if there are questions, problems, or errors.
Michael B. Smith
Professor Emeritus
December 2019

Common Abbreviations
Other, less common abbreviations are given in the text when the term is used.
O
Ac Acetyl Me
AIBN azobisisobutyronitrile
aq aqueous
AIBN Azobisisobutyronitrile
AMP Adenosine monophosphate
ATP Adenosine triphosphate
Ax axial
9-BBN 9-Borabicyclo[3.3.1]nonane
Bn Benzyl -CH2Ph
Boc tert-Butoxycarbonyl Ot-B
Bu
Bu n-Butyl -CH2CH2CH2CH3
Bz Benzoyl
°C Temperature in Degrees Celsius
13C NMR Carbon Nuclear Magnetic Resonance
cat Catalytic
OC
CH2Ph
Cbz Carbobenzyloxy
CIP Cahn–Ingold–Prelog
mCPBA 3-Chloroperoxybenzoic acid
DCC 1,3-Dicyclohexylcarbodiimide c-C6H11-N=C=N-c-C6H11
DEA Diethylamine HN(CH2CH3)2
DMAP 4-Dimethylaminopyridine
O
H NM
Me2
DMF N,N'-Dimethylformamide
DMSO Dimethyl sulfoxide
EDTA Ethylenediaminetetraacetic acid
ee or % ee % Enantiomeric excess
Equiv Equivalent(s)
Et Ethyl -CH2CH3
Ether diethyl ether CH3CH2OCH2C3
Eq equatorial
FDNB Sanger’s reagent, 1-fuoro-2,4-dinitrobenzene
FMO Frontier molecular orbitals
FVP Flash Vacuum Pyrolysis
GC Gas chromatography
h Hour (hours)
1H NMR Proton Nuclear Magnetic Resonance
HMPA Hexamethylphosphoramide
HOMO Highest occupied molecular orbital
hν Irradiation with light
IP Ionization potential
O
O
xi

xii Common Abbreviations
iPr Isopropyl -CH(Me)2
IR Infrared
IUPAC International Union of Pure and Applied Chemistry
K Temperature in kelvin
LCAO Linear combination of atomic orbitals
LDA Lithium diisopropylamide LiN(i-Pr)2
LUMO Lowest unoccupied molecular orbital
mcpba meta-Chloroperoxybenzoic acid
Me Methyl -CH3 or Me
min minutes
MO Molecular orbital
mRNA Messenger ribonucleic acid
MS Mass spectrometry
NMR nuclear magnetic resonance
N.R. No reaction
NAD+ Nicotinamide adenine dinucleotide
NBS N-Bromosuccinimide
NCS N-Chlorosuccinimide
Ni(R) Raney nickel
NMO N-Methylmorpholine N-oxide
Nu (Nuc) Nucleophile
PCC Pyridinium chlorochromate
PDC Pyridinium dichromate
PES Photoelectron spectroscopy
Ph Phenyl
PhMe Toluene
PPA Polyphosphoric acid
Ppm Parts per million
Pr n-Propyl -CH2CH2CH3
N
Py Pyridine
RNA Ribonucleic acid
rt Room temperature
s seconds
(Sia)2BH Disiamylborane (Siamyl is sec-Isoamyl)
sBuLi sec-Butyllithium CH3CH2CH(Li)CH3
SEAr Electrophilic aromatic substitution
SET Single electron transfer
SNAr Nucleophilic aromatic substitution
SOMO singly occupied molecular orbital
T Temperature
t-Bu tert-Butyl -CMe3
TBHP (t-BuOOH) t-Butylhydroperoxide Me3COOH
TFA Trifuoroacetic acid CF3COOH
ThexBH2 Thexylborane (tert-hexylborane)
THF Tetrahydrofuran
TMEDA Tetramethylethylenediamine Me2NCH2CH2NMe2
Tol Tolyl 4-(Me)C6H4
Ts(Tos) Tosyl = p-Toluenesulfonyl 4-(Me)C6H4SO2
UV Ultraviolet spectroscopy
VIS visible
VDW van der Waals

Author
Professor Michael B. Smith was born in Detroit, Michigan, and moved to Madison
Heights, Virginia, in 1957. He graduated from Amherst County High School in
1964. He worked at Old Dominion Box Factory for a year and then began studies
at Ferrum Junior College in 1965. He graduated in 1967 with an AA and began
studies at Virginia Tech later that year, graduating with a BS in Chemistry in
1969. He worked as a chemist at the Newport News Shipbuilding Dry Dock
Co., Newport News, Virginia, from 1969 until 1972. In 1972, he began studies in graduate school at
Purdue University in West Lafayette, Indiana, working with Professor Joseph Wolinsky, graduating in
1977 with a PhD in Organic Chemistry. He took a postdoctoral position at Arizona State University in
Tempe, Arizona, working on the isolation of anti-cancer agents from marine animals with Prof. Bob
Pettit. After one year, he took another postdoctoral position at MIT in Cambridge, Massachusetts, work-
ing on the synthesis of the anti-cancer drug bleomycin with Prof. Sidney Hecht.
Professor Smith began his independent career as an assistant professor in the Chemistry department
at the University of Connecticut, Storrs, Connecticut, in 1979. He received tenure in 1986, and spent six
months on sabbatical in Belgium with Professor Leon Ghosez at the Université Catholique de Louvain
in Louvain la Neuve, Belgium. He was promoted to full professor in 1994 and spent his entire career at
UConn. Professor Smith’s research involved the synthesis of biologically interesting molecules. His most
recent work involved the preparation of functionalized indocyanine dyes for the detection of hypoxic
cancerous tumors (breast cancer), and also the synthesis of infammatory lipids derived from the den-
tal pathogen, Porphyromonas gingivalis. He has published 26 books including Organic Chemistry: An
Acid-Base Approach, 2nd edition (Taylor Francis), the 5th–8th edition of March’s Advanced Organic
Chemistry (Wiley), and Organic Synthesis, 4th edition (Elsevier), which is the winner of a 2018 Texty
Award. Professor Smith has published 96 peer-reviewed research papers and retired from UCONN in
January of 2017.
xiii

Part A
A QA Approach
to Organic Chemistry
What is organic chemistry?
Organic chemistry is the science that studies molecules containing the element carbon. Carbon can
form bonds to other carbon atoms or to a variety of atoms in the periodic table. The most common
bonds observed in an organic chemistry course are C—C, C—H, C—O, C—N, C—halogen (Cl, Br, I),
C—Mg, C—B, C—Li, C—S, and C—P.

1
Orbitals and Bonding
This chapter will introduce the carbon atom and the covalent bonds that join carbon atoms together in
organic molecules. The most fundamental properties of atoms and of covalent bonds will be introduced,
including hybridization, electronic structure, and a brief introduction to using molecular orbital theory
for bonding.
1.1 ORBITALS
1.1.1 Atomic Orbitals
What is the electronic structure of an atom?
A given atom has a fxed number of protons, neutrons, and electrons, and the protons and neutrons are
found in the nucleus. The electrons are located at discreet energy levels (quanta) away from the nucleus.
The nucleus is electrically positive, and electrons are negatively charged.
What is the Schrödinger wave equation?
The Schrödinger equation, Hψ = Eψ, is a linear partial differential equation that describes the wavefunc-
tion or state function of a quantum-mechanical system. The motion of an electron is expressed by a wave
equation, which has a series of solutions and each solution is called a wavefunction. Each electron may
be described by a wavefunction whose magnitude varies from point to point in space. The equation is a
partial differential equation that describes how the wavefunction of a physical system changes over time.
What are atomic orbitals?
An atomic orbital is a mathematical function that describes the wave-like behavior of either one electron
or a pair of electrons in an atom. If certain simplifying assumptions are made, it is possible to use the
Schrödinger equation to generate a different wavefunction for electrons with differing energies relative
to the nucleus. A particular solution to the so-called Schrödinger wave equation, for a given type of elec-
tron, is determined from the Schrödinger equation, and a solution for various values of ψ that correspond
to different energies shows the relationship between orbitals and the energy of an electron. The wave-
function is described by spatial coordinates ψ(x,y,z), and using Cartesian coordinates a point is defned
that describes the position of the electron in space.
What is a node?
A node is derived from a solution to the Schrödinger equation where the wavefunction changes phase,
and it is taken to be a point of zero electron density.
What is the Heisenberg uncertainty principle?
The Heisenberg uncertainty principle states that the position and momentum of an electron cannot be
simultaneously specifed so it is only possible to determine the probability that an electron will be found
at a particular point relative to the nucleus. The probability of fnding the electron in a unit volume of
three-dimensional space is given by |ψ(x,y,z)|2, or |ψ|2dτ, which is the probability of an electron being in
a small element of the volume dτ. This small volume can be viewed as a charge cloud if it contains an
3

4 A QA Approach to Organic Chemistry
electron, and the charge cloud represents the region of space where we are most likely to fnd the electron
in terms of the (x,y,z) coordinates. These charge clouds are orbitals.
What is a s-orbital?
Different orbitals are described by their distance from the nucleus, which formally corresponds to the
energy required to “hold” the electron. One solution to the Schrödinger equation is symmetrical in that
the wave does not change phase (zero nodes; a node is the point at which the wave changes its phase).
This corresponds to the frst quantum level and known as a s-atomic orbital. The 1s-orbital represents
the frst energetically favorable level where electrons can be held by the nucleus. The space in which
the electron may be found is spherically symmetrical in three-dimensional space. All spherically sym-
metrical orbitals are referred to as s-orbitals. The nucleus is represented by the “dot” in the middle of
the sphere.
Nucleus
s-Orbital
What is a p-orbital?
When the solution for the Schrödinger equation has one node (the wave changes phase once), electron
density is found in two regions relative to the node. When the space occupied by this electron is shown
in three-dimensional space, it is a p-orbital with a “dumbbell” shape. In the (x,y,z) coordinate system,
the single node could be in the x, the y, or the z plane and all three are equally likely. Therefore, three
identical p-orbitals must be described: px, py, and pz relative to the nucleus, as shown. Identifcation of
three identical p-orbitals means that there are three p-wavefunctions that describe three electrons that
are found at the same energy.
Nucleus
p-orbital pxpypz-Orbital
What is a degenerate orbital?
Orbitals with identical energies are said to be degenerate, and the three 2p-orbitals shown in the preced-
ing question are degenerate orbitals. The three electrons in different orbital lobes have the same energy
and have the same charge. Due to the presence of like charges, the orbital lobes repel and will assume
positions as far apart as possible in a tri-coordinate system. In other words, the three orbitals will be
directed to the x-, y-, and z-directions in a three-dimensional coordinate system as shown.
Do the electrons in a p-orbital migrate from one lobe to the other?
No! The picture of the p-orbital represents the uncertainty of where to fnd the electrons. The diagram
shows an equal probability of fnding the electrons in each of the three dumbbell-shaped orbitals, above
and below a node, which is taken as a point of zero electron density and corresponds to the position of
the nucleus in the diagram. Therefore, the electrons are found in the entire p-orbital (both lobes), and the
diagram simply indicates the uncertainty of their exact location.
How many orbitals are there in each valence shell?
Each orbital can hold two electrons. For the frst valence shell containing H and He, there is one s-orbital.
For the next valence shell (containing B, C, N, O, F), there is one 2s orbital, but three 2p-orbitals. The
2p-orbitals have different spatial orientations, correlated with the x, y and z axes of a three-dimensional
coordinate system. In other words, the three p-orbitals are px, py, and pz.

5
1.1.2 Electron Configuration
What is electron confguration?
The electron confguration is the distribution of electrons of an atom or molecule in atomic or molecular
orbitals. Electrons are distributed in shells, each of which has different types of electrons: s, p, d, f. Each
orbital (energy level) occurs further from the nucleus; the electrons are held less tightly. Each orbital can
hold a maximum of two electrons and each energy level will contain different numbers of electrons (one
electron for the 1s1 and two electrons for the 1s2 orbital, as shown. There are six electrons for p-orbitals;
two each is possible for each of the three-degenerate p-orbitals. There are ten electrons for d-orbitals;
two each for the fve d-orbitals. Orbitals will fll from lowest energy to highest energy orbital, according
to the order shown in the mnemonic for the electronic flling order of orbitals.
H He
1s1 1s2
1s 2s 2p 3s 3p
4s 3d 4p 5s 4d
5p 6s 4f 5d 6p
Filling Order
What is the Aufbau principle?
Orbitals “fll” according to the Aufbau principle. The principle states that each orbital in a sublevel s,
p, or d will contain one electron before any contains two. Orbitals containing two electrons will have
opposite spin quantum numbers (they are said to be spin paired, ↑↓). An example is helium in the pre-
ceding question.
What is the order in which the three degenerate p-orbitals fll with electrons through the 2p
level? Ignore the 1s and 2s levels.
The order for the 2p-orbitals will be 2p →2p →2p →2p →2p →2p :
x y z x y z
↑ _ _ →↑↑_ → ↑↑↑ → ↑↓↑↑ → ↑↓↑↓↑ → ↑↓↑↓↑↓
1.1.3 Molecular Orbitals
What is the difference between a molecular orbital and an atomic orbital?
An atomic orbital is a mathematical function that describes the wave-like behavior of either one electron
or a pair of electrons in an atom. A molecular orbital (MO) is a mathematical function describing the
wave-like behavior of an electron in a molecule.
An atomic orbital is associated with a specifc atom. The electrons found on an individual atom of an
element are in atomic orbitals whereas the electrons found in an atom that is part of a covalent bond are
in molecular orbitals. A molecular orbital is formed once two atoms are joined in a covalent bond. Much
of the electron density is shared between the two nuclei of the two atoms rather than being exclusively on
the nuclei of the two atoms. This energy level for the electrons found in the molecular orbital is different
from electrons that are on an individual atom such as that found in an element.
What is the Linear Combination of Atomic Orbitals (LCAO) model?
The LCAO model is the superposition of atomic orbitals that constitutes a technique for calculating
molecular orbitals in quantum chemistry. The LCAO model is a mathematical model that is used to mix
the atomic orbitals of two atoms to get new orbitals for the resulting bond between those two orbitals.
In the LCAO method, the atomic orbitals of each “free” atom are mixed to form molecular orbitals. The
model requires that there can be no more or no less orbitals and no more or no less electrons in the orbit-
als for the new bond than are found in the atomic orbitals for the two atoms. These new orbitals must be

of a different energy than the atomic orbitals in a non-degenerate system. In other words, when mixing
two atomic orbitals, one new orbital is formed that must be lower in energy and one is formed that is
higher in energy relative to the atomic orbitals.
How does the LCAO model apply to covalent bonds in simple diatomic molecules such as
hydrogen?
Two atoms are combined to form a covalent bond, and the atomic orbitals of each atom are combined to
form a molecular orbital. Assume that the electrons in each atomic orbital are transferred from energy
levels near the atom to different energy levels that correspond to electron density between the nuclei of
the bonded atoms. A molecular orbital is an orbital associated with the molecule rather than the indi-
vidual atoms, as shown below for H2. For molecules containing more electrons than hydrogen or helium,
and for those containing electrons in orbitals other than s-orbitals, the diagram is more complex and the
LCAO approach usually fails.
How are molecular orbitals formed from atomic orbitals?
Using the LCAO model, the orbitals for the molecule diatomic hydrogen (H2) can be formed by mix-
ing the atomic orbitals of two hydrogen atoms. The orbitals formed are not atomic orbitals, but they are
associated with a molecule, in this case H2, and are called molecular orbitals. Each of the two hydrogen
1s atomic orbitals (H atomic orbital) contains one electron, and these orbitals have the same energy.
When mixed to form the molecular orbital, the molecular orbital electrons have a different energy, and
those orbitals are in a different position relative to atomic orbitals, as shown for the molecule H—H.
Therefore, if the two atomic orbitals mix, two molecular orbitals are generated, one higher in energy and
one lower than the original atomic orbitals. It is noted that this model does not work well for atoms that
have degenerate p-orbitals.
H
1s1
H
1s1
Bonding molecular orbital
Anti-bonding molecular orbital
Increasing Energy
1.2 BONDING
1.2.1 Ionic Bonding
What is a Lewis dot structure?
A Lewis electron dot formula generates a bond between two atoms by simply using dots for electrons for
the two electrons that comprise a bond. In other words, each bond is represented by two dots between the
appropriate atoms, and unshared electrons are indicated by dots (one or two) on the appropriate atom.
What is the Lewis dot structure of lithium fuoride? Add the charges!
Li F

7
What is an ionic bond?
An ionic bond occurs when two atoms are held together by electrostatic forces, where one atom or group
assumes a positive charge and the other atom or group assumes a negative charge. Sodium chloride
(NaCl), for example, exists in the solid state as Na+Cl–.
Na Cl
Why does sodium assume a positive charge in NaCl?
If the valence electrons associated with each atom are represented as dots (one dot for each electron), the
structure for NaCl will be that shown above. Sodium chloride has an ionic bond, and in an ionic bond
all of the electrons are on chlorine and none are on sodium. Sodium (Na) is in Group 1 and has the elec-
tronic confguration 1s22s22p63s1. If one assumes that the sodium atom can react, it can either lose one
electron (ionization potential) or gain seven electrons (the ability to gain one electron is called electron
affnity) in order to achieve a “flled” shell. The loss of one electron gives the electronic confguration
1s22s22p6, which is a flled shell and very stable, and requires much less energy than gaining seven to give
another flled shell. After transfer of one electron, sodium has no electrons around it. In other words, it
is Na+, which is missing one electron relative to atomic sodium; this means it is electron defcient and so
assumes a positive charge (see formal charge in Section 2.5).
Why does chlorine assume a negative change in NaCl?
In the ionic bond for NaCl, the chlorine atom has eight electrons around it. The chlorine (Cl) atom
has the electronic confguration 1s22s22p63s23p5. If one imagines that the Cl atom reacts, and since
chlorine is in Group 17 with seven electrons in the outmost shell, it can either gain one electron or lose
seven electrons. Clearly, the loss of seven electrons will require a great deal of energy. Energetically,
it is far easier for Cl to gain an electron, leading to formation of a negatively charged atom. Therefore,
Cl gains an electron in contrast to Na, which loses an electron. With an excess of electrons, given that
electrons carry a negative charge, the Cl will take a negative charge. The strong electrostatic attraction
between the positive sodium and the negatively charged chlorine binds the two atoms together into an
ionic bond.
1.2.2 Covalent Bonding
What is a covalent bond?
A covalent bond has two electrons that are shared between two atoms. In the case of hydrogen (H2), the
covalent bond can be represented as H:H or H—H, where the (:) or the (—) indicates the presence of two
electrons. In a covalent bond, the bulk of the electron density is localized between the hydrogen nuclei.
This type of bond usually occurs when the atom cannot easily gain or lose electrons. Another way to
view this is that there is a small electronegativity difference between atoms. A model of fuorine (F—F
or F2) shows the electron density around both atoms, but signifcant electron density is clearly between
the two fuorine nuclei that represent the covalent F—F bond.

What is the Lewis dot structure of diatomic hydrogen?
H : H
What is the octet rule?
The octet rule states that every atom wants to have eight valence electrons in its outermost electron shell.
What is valence?
Valence is the number of bonds an atom can form to satisfy the octet rule and remain electrically neutral.
Valence is not to be confused with valence electrons, which are the number of electrons in the outermost
shell. In the second row from C to F, the valence is (8 – the last digit of the group number): C: 8 – 4, or
4; N: 8 – 5, or 3; O: 8 – 6, or 2; F: 8 – 7, or 1. Boron is an exception. There are only three electrons and,
therefore, boron can form no more than three covalent bonds and remain neutral. In other words, an atom
can form only as many bonds as there are electrons available to share. Note that the valence of boron is
three and it is electron defcient.
What is the Lewis dot structure of a carbon–carbon bond when drawn as a covalent bond but
ignoring all other electrons and the other valences of each carbon?
C : C
What is a Lewis acid?
A Lewis acid is any substance that can accept a pair of nonbonding electrons, so it is an electron-pair
acceptor. An example is boron, which has only three electrons in the outermost shell, can only form
three covalent bonds but is electron defcient because it requires two more electrons to satisfy the octet
rule. These two electrons are gained by reaction with an electron-rich molecule and trivalent boron com-
pounds are Lewis acids.
What is a Lewis base?
A Lewis base is any substance that can donate a pair of nonbonding electrons, so it is an electron-pair
donor. Electron donation to a hydrogen atom is not included in the Lewis base defnition, however. In other
words, a base that donates two electrons to a hydrogen atom is a Brønsted–Lowry base not a Lewis base.
Why does carbon have a valence of four?
Carbon is in Group 14 so there are four valence electrons. Therefore, carbon forms a total of four cova-
lent bonds by sharing the electrons with many other atoms, including another carbon atom. With carbon
(C: 1s22s22p2), there are four electrons in the highest valence shell. The gain of four electrons or the loss
of four electrons would require a prohibitively high amount of energy. In a thought experiment, assume
that a carbon atom can form bonds directly with up to four other atoms. In such an experiment, carbon,
because it is lower in energy, will form covalent bonds to share electrons with another atom rather than
“donate” or “accept” four electrons to form an ionic bond. A carbon atom with appropriate functionality
attached can undergo chemical reactions with other molecules to form covalent bonds to other carbon
atoms, to hydrogen atoms, as well as to many atoms in the periodic table.
What is covalent bond?
In a covalent bond, the electrons are mutually shared between two nuclei in that bond, so each nucleus has
a flled shell (eight in the case of carbon and two in the case of hydrogen). The most common way to show

9
mutual sharing of electrons for two carbon atoms is to draw a single line between the two atoms (C—C) rather
than using the Lewis dot structure, C:C. The two electrons are equally distributed between the two carbon
atoms and the resultant bond has a symmetrical distribution of electron density between the two atoms.
What does a covalent bond between two hydrogen atoms look like in the molecule H2?
The two hydrogen atoms are identical, the mutual sharing of electrons leads to a symmetrical distribution
of the electron density between the two hydrogen nuclei, as shown in the accompanying molecular model
(an electronic potential map)
What does a covalent bond between two carbon atoms with identical atoms attached?
If the two carbon atoms are identical, the mutual sharing of electrons leads to a symmetrical distribution
of the electron density between the two carbon nuclei.
What is electronegativity?
Electronegativity is a measure of the attraction that an atom has for the bonding pair of electrons in a
covalent bond. A more electronegative atom will attract more electron density toward itself than a less
electronegative atom.
What is a polar covalent bond?
If two atoms are part of a covalent bond, and one atom is more electronegative than the other, the shared
electron density is distorted toward the more electronegative atom, as shown in the molecular model of
H—F, rather than the symmetrical distribution found in H—H. The shaded area on the far right indicates
higher electron density, which is clearly on the more electronegative fuorine atom.
When a bond is formed between two atoms that are not identical, the electrons do not have to be equally
shared. If one atom is more electronegative (electronegativity is the ability of an atom to attract electrons
to itself; the electronegativity is higher), it will pull a greater share of electrons from the covalent bond
toward itself. The larger the difference in electronegativity, the greater the distortion of electron density
in the covalent bond toward the more electronegative atom.

What is an example of a molecule with a polar covalent bond?
An example is H—F (see the molecular model in the preceding question), where the fuorine is signifcantly
more electronegative than the hydrogen atom. This difference in electronegativity will lead to electron dis-
tortion in the covalent bond toward the fuorine, away from hydrogen, and an unsymmetrical covalent bond
will form that has less electron density between the nuclei. In other words, it is a weaker bond. The model
shown indicates this electron distortion, with the shaded area on the far right (higher electron density)
toward the fuorine and the shaded area on the far left (lower electron density) toward the hydrogen atom.
How can HF be drawn to represent the polarized covalent bond?
In a polarized bond such as that found in HF, fuorine is more electronegative, and the bond density is
distorted such that there is more electron density on fuorine relative to the hydrogen. The more electro-
negative atom will be “more negative” and the other will be “more positive.” The molecule is neutral so
there are no ionic charges, but the distortion of electron density in the polarized covalent bond is repre-
sented by a “partial charge” (δ+) at the atom with the least electron density and (δ-) at the atom with the
most electron density. Therefore, the polarization of HF is represented as δ+H—Fδ–.
Can the dipole of a polarized covalent bond be represented by an appropriate symbol?
A common way to represent this distortion of electrons is with the symbol +⟶, with the + representing
the positive atom and ⟶ representing the direction of electron fow. As noted in the previous question,
it is perhaps more common to use a δ+ at the atom with the least electron density and δ– at the atom with
the most electron density, as shown for H—F. Such a covalent bond is polarized, and this disparity in
electron density leads to a dipole moment. Any covalent bond between two atoms where one is less elec-
tronegative, and the other is more electronegative, will be a polar covalent bond.
What is dipole moment?
Bond dipole moment is a measure of the polarity of a chemical bond, generally induced by differences
in electronegativity of the two atoms in that bond. The bond dipole symbol is μ and the unit of measure-
ment is the Debye (D).
Is the C—H bond considered to be polarized?
No! Although C and H have different electronegativities (H = 2.1 and C = 2.5 on the Pauling electronega-
tivity scale), the C—H bond is not considered to be polarized. This assumption is based on the polarity
of molecules containing only C—H bonds, but the chemical reactivity of molecules that contain only
C—H bond will support this view.
What is a heteroatom?
A heteroatom is defned as any atom other than carbon or hydrogen. Examples are O, N, S, P, Cl, Br, F,
Mg, Na, etc.
Which of the following are polar covalent bonds? For the polarized bonds, identify the negative
and positive poles: (a) C—O (b) C—C (c) O—H (d) H—H (e) Br—Br (f) C—N (g) N—N (h)
O—O (i) H—Br (j) NaCl
Only those bonds between dissimilar atoms will be polarized, therefore (a), (c), (f), (i), and (j) are polar-
ized covalent. NaCl (j) is an ionic bond. The negative poles will be oxygen in (a) and (c), nitrogen in (f),
and bromine in (i). The positive poles are carbon in (a) and (f), hydrogen in (c) and (i).
What is van der Waals attraction?
When there are no polarizing atoms in the molecules, the only attraction between molecules results from
the electrons of one molecule being attracted to the positive nuclei of atoms in another molecule. This
interaction is known as van der Waals attraction (sometimes called London forces).

11
What are dipole–dipole interactions?
The intermolecular electrostatic interaction between polarized atoms in one molecule with polar-
ized atoms in a nearby molecule is referred to as a dipole–dipole interaction. The net result of this
δ+ ⟵ δ–interaction is that these molecules will be associated together, and some energy will be required
to disrupt this association.
How does the physical size of the group that is attached to the heteroatom infuence dipole–
dipole interactions?
The electrostatic interaction of the two dipolar molecules is diminished by the physical size imposed by
the carbon groups since the molecules cannot come as close together. As the molecules approach each
other, larger groups compete for the same space (this is called steric hindrance) and repel each other,
counteracting the electrostatic attraction to some extent.
What causes two molecules with C—O or C=O bonds to associate together in the liquid phase?
When two molecules, each with a polarized covalent bond, come into close proximity, the charges for
one bond will be infuenced by the charge on the adjacent molecule. In the example shown with two
molecules of propane-2-one (acetone), the negative oxygen of one molecule is attracted to the positive
carbon of the second molecule. Likewise, the positive carbon of that molecule is attracted to the negative
oxygen of the other. In general, the greater the dipole moment, the stronger the interaction and the greater
the energy will be required to disrupt that interaction between the molecules.
H3C
C
O
CH3
H3C
C
O
CH3
+
+
What are hydrogen bonds?
When hydrogen forms a polar covalent bond with heteroatoms (atoms other than carbon or hydrogen,
the most common are O—H, N—H, S—H), the hydrogen takes the δ+ charge of the dipole. There is a
strong interaction with a negative heteroatom that is brought into close proximity to a positive polarized
hydrogen of another molecule. The interaction shown in the fgure for the O—H units of methanol is
referred to as a hydrogen bond.
CH3
O
+ H
H +
O
H3C
Which is stronger, a dipole–dipole interaction or a hydrogen bond?
The attraction between a polarized hydrogen atom and a heteroatom in a hydrogen bond is generally
signifcantly stronger than the dipole–dipole attraction between the two atoms in a polarized covalent
bond (not involving H). The bond polarization of an X—H bond is generally greater, so there is more
attraction with a negative dipole, and the hydrogen atom is small. Due to the small size of hydrogen, there
is minimal steric hindrance due to intramolecular interactions with another nearby atoms.
Why is the attraction between two molecules of methanol, each with an O—H unit, stronger than
the attraction between two molecules of acetone (see above), each of which bears a C=O group?
The electronegative oxygen atom makes the O—H bond in methanol more polarized than the C=O bond
in acetone. The hydrogen atom is also smaller, so it is easier for the oxygen atom on a methanol molecule
to approach the hydrogen atom in another molecule of methanol. As the carbon atom of one acetone

molecule approaches the oxygen atom of another acetone molecule, there is a signifcant intermolecular
steric interaction of the methyl groups between the different molecules. Therefore, it is anticipated that
the intermolecular hydrogen bond interaction of one methanol molecule for another is stronger than the
intermolecular dipole–dipole interactions of one acetone molecule for another.
1.3 HYBRIDIZATION
What are “core” electrons?
Electrons that are close to the nucleus that they are too tightly bound to be shared with another atom to
form a covalent bond. Such electrons are referred to as core electrons; they are not the valence electrons.
Core electrons are not involved in chemical reactions and so they are not involved in covalent bonding.
What are valence electrons?
Valence electrons are electrons in the outermost electronic shells of an atom, and they are available for
chemical reactions and thus for sharing with another atom to form a covalent bond.
What is a molecular orbital and what is a molecular orbital diagram?
A molecular orbital is a mathematical function that describes the wave-like behavior of an electron
in a molecule. Another way to look at a molecular orbital is as a function that describes the electronic
behavior of electrons in a covalent bond. A molecular orbital diagram describes the chemical bonding in
molecules and the linear combination of atomic orbitals is used to generate the molecular orbital diagram
for a particular bond.
How is the LCAO method used to combine two carbon atoms to give the molecular orbital dia-
gram for C—C?
2p 2p
2s
1s 1s
2s
Valence
Core
Increasing
Energy
Atomic carbon has an electronic confguration of 1s22s22p2. The molecular orbital diagram is shown.
Using the LCAO method to form a carbon–carbon single bond, the two 2s orbitals and the three, degen-
erate p-orbitals of two carbon atoms are combined to give six molecular orbitals in a C—C molecule.
The total number of orbitals remains constant, but the molecular orbitals must be of a different energy
relative to the atomic orbitals and they are split into high energy and low energy components. The
orbitals are symmetrically split, as shown. The total number of electrons cannot change, and they are
added and spin-paired, beginning with the lowest energy orbitals. The three, degenerate p-orbitals on
the carbon atoms lead to three lower energy degenerate molecular orbitals as well as three degenerate
higher energy orbitals, as shown in the diagram. Note that when the four “p” electrons are added to the
molecular orbitals, the LCAO method suggests two unshared electrons, which is not the case in real
molecules.

13
Why does the molecular orbital diagram for diatomic carbon give an incorrect result?
The attempt to mix s- and p-atomic orbitals leads to the discrepancy. The LCAO-generated molecular
orbital diagram shown for a C—C bond suggests more than one type of bond. The 2s molecular orbit-
als combine to form one type of bond; the p-orbitals combine to form another type of bond, and there
appear to be unshared electrons in the orbitals. The prediction of two different kinds of bonds is a result
of trying to mix s- and p-orbitals, and this prediction is not correct. In fact, it is known from many years
of experiments that one type of bond is formed by each carbon atom. Indeed, CH4 (methane) forms four
identical bonds to four hydrogen atoms, for example.
A new model that generates a better description of a C—C bond is required that recognizes all four
bonds as the same type in the fnal molecule.
How is the LCAO model used to give the molecular orbital diagram for C—C by mixing two
sp3-hybrid orbitals?
When these identical orbitals (they are the same energy, therefore, degenerate) from two “hybridized”
carbon atoms are mixed to form a molecular orbital, the core electrons remain the same, but the covalent
orbitals now show four identical bonds (two electrons per bond). Since the correct answer was known in
advance, the modifed model must give the correct answer.
sp3
1s 1s
Core
Increasing
Energy
sp3
Can one gain or lose orbitals during hybridization?
No! The number of hybrid orbitals that are formed cannot be more or less than the total number of atomic
orbitals.
Can one gain or lose the total number of electrons during hybridization?
No! The number of electrons in the molecular orbitals cannot be more or less than the total number of
electrons in the atomic orbitals.
What is a sp3-hybrid orbital?
In sp3-hybridization, all three p-orbitals are mathematically mixed with the s-orbital to generate four
new hybrids that can form four covalent bonds of the same type.
What is a sigma bond?
A sigma (σ) bond is the usual covalent bond between two atoms such as C—C, C—H, C—O, or C—N,
where the electron density is concentrated between the two nuclei (essentially on a “line” between the
two nuclei). The σ-bond is associated with what is known as a “single covalent bond.”
How is a σ-bond normally drawn between two sp3-hybridized carbon atoms?
A carbon–carbon σ-bond is drawn as C—C. When drawn with a single line to represent the covalent
bond between two carbon atoms, that bond is assumed to be a covalent σ-bond.

What is the classifcation for the bond in a C—C unit?
When drawn with a single line to represent the covalent bond between two carbon atoms, that bond is
assumed to be a covalent σ-bond: e.g., C—C.
How is hybridization applied to a covalently bound atom?
When the s- and p-atomic orbitals of an atom are mathematically “mixed,” four identical hybrid orbitals
are formed. Since one 2s and three 2p are mixed, the resulting hybrid orbitals are called sp3-hybrid orbit-
als. When these sp3-hybrid orbitals for two carbon atoms are used in the Linear Combination of Molecular
Orbitals model, four identical bonding molecular orbitals are formed. When the atomic orbitals are rear-
ranged, mixtures of them (hybrids) are formed and used for bonding. For carbon and other elements of
the second row, the hybridization is limited to mixing one 2s and one or more of the three 2p-orbitals.
How are hybrid orbitals different when using different numbers of 2p-orbitals?
There are three basic types of hybridization: sp3, sp2, and sp1 (or just sp). In each case, the sp refers to the
hybridization of the atom where the superscript indicates the number of p-orbitals used to form hybrids
in combination with the 2s-orbital.
What is a sp2-hybrid orbital?
In sp2-hybridization, two p-orbitals are mixed with the s-orbital to generate three new hybrids that can
form three covalent σ-bonds. The “unused” p-orbital will participate in π-type bonding (see the chemis-
try of alkenes in Section 7.1).
What is a sp-hybrid orbital?
In sp-hybridization, one p-orbital is mixed with the s-orbital to generate two new hybrids that can form
two covalent σ-bonds. The two “unused” p-orbital will participate in two, mutually perpendicular π-type
bonds (see the discussion of alkynes in Section 7.5).
What is a π-bond?
Sideways overlap of p-orbitals
-Bond
-Bond
-Bond
-Bond
A pi- (π) bond occurs when two sp2-hybridized are connected by a covalent bond, and each atom has
an “unused” p-orbital, as described above. Sigma bonds using sp2-hybrid orbitals connect the two car-
bon atoms and all four C—H bonds. Each sp2-hybridized carbon has a “unused” p-orbital. When these
p-orbitals are parallel and on adjacent atoms, they can share electron density via “sideways” overlap to
form a new bond (called a π-bond) that is much weaker than the σ-bond. Effectively, there are two bonds
between the carbon atoms, a strong σ-bond, and a weak π-bond, as shown for ethene.
How is the double bond between two carbon atoms represented, ignoring the other bonds to
carbon?
A carbon–carbon double bond is represented as C=C. Note that in the C=C representation it is not pos-
sible to indicate which is the σ-bond and which is the π-bond, and it is not necessary to do so. One line
represents the σ-bond and the other line represents the π-bond.

15
1.4 RESONANCE
What is a carbon–carbon single bond?
A C—C single bond is a normal covalent σ-bond between two carbon atoms, involving mutual sharing
of two electrons between the two carbon nuclei.
What is a carbon–carbon double bond?
A C=C unit contains one σ-bond and one π-bond between the two carbon atoms (see the fgure of ethene
in Section 1.3 and see Section 7.1)
Is it possible to form a double bond between other atoms?
Yes! A double bond can form between carbon and other atoms, or between atoms other than carbon.
Examples are C=O, C=S, C=N, as well as N=N, O=O, N=O, S=O.
What is the structure of a molecule with an oxygen–oxygen double bond, one with an N=O
bond, and one that contains a S=O bond?
An example of a molecule with an oxygen–oxygen double bond is diatomic oxygen, O2. An example of
a molecule that contains a nitrogen–oxygen double bond (N=O) is nitric acid. An example of a molecule
that contains a sulfur–oxygen double bond (S=O) is sulfuric acid.
O O
O
N H
O O
O
O H
S O
O
Oxygen Nitric acid Sulfuric acid
What is a point charge?
A point charge is a charge that is completely localized on a single atom. In the chloride ion (Cl-), for exam-
ple, the two electrons that comprise the negative charge are in an orbital localized on the chlorine atom.
Can bonds be formed that are “in-between” single and double bonds?
In some cases, it is possible for electron density to be delocalized in such a way that the actual bond is in
between a single and a double bond. In such a bond, electron density is delocalized over several atoms
rather than localized on a single atom. This phenomenon occurs most commonly when a charge is pres-
ent in an atom and is also commonly associated with the presence of π-bonds.
What is resonance?
Resonance occurs when the electron density of bonds is not localized between two atoms but rather delo-
calized over three or more atoms. In other words, bonds between two atoms can be intermediate between
single and double bonds, due to the sharing of electrons by several atoms, which is described as delocaliz-
ing the electron density over several atoms. In effect, the electron density is “smeared” over several atoms.
Which is more stable, a point charge or a delocalized charge?
If a charge is delocalized over several atoms rather than localized on a single atom, the charge density on
each atom is diminished. Lower charge density is usually associated with lower energy, so a delocalized
charge should be more stable (less reactive) than a localized charge.
How is resonance in a molecule represented?
Resonance delocalization for a molecule is represented by drawing two or more resonance contribu-
tors, with different localized bonds and charges as required. The double-headed arrow shows that the

resonance contributors are linked to represent a single, delocalized structure. An example is the oxocar-
benium ion where the positive charge is delocalized over the O and the C, and this delocalization is rep-
resented by the two structures shown with the double-headed arrow. The different structures represent
the extent of electron delocalization in the actual structure.
H H
O O
C C
R R R R
What are the resonance contributors for the allylic carbocation?
There are two resonance contributors as shown. The positive charge is not localized on the two structures
but delocalized over three atoms. The double-headed arrow represented links the two structures with the
positive charge on the frst and third carbon atoms to represent the delocalization. Likewise, the π-bond
is not localized on the two structures, but the resonance contributors linked by the double-headed arrow
are used to show that the π-electrons are delocalized.
What structural features are required for resonance?
A molecule of three or more atoms that contains a π-bond with a third atom that has a (+) charge, a (−)
charge, or a single electron. When the attached atom has a (+) charge it is a cation. When the third atom
has a (–) charge it is an anion, and when it has a single electron it is a radical. When there are two atoms
with a (+) charge on one atom and an attached heteroatom such as O, S, or N has at least one unshared
electron, resonance can occur.
X Y Z X Y Z
Cation-1 Anion
X Y
Radical
Z Y Z
Cation-2
What is the energetic consequence of resonance?
The delocalization of the point charge to dispersal of charge over several atoms (a greater surface area)
leads to greater stability of the molecule. In general, the greater the point charge, the less stable (more
reactive), and the lesser the point charge, the more stable (less reactive) the molecule. Resonance enhances
the stability of a molecule and the term “resonance stability” is common. A resonance-stabilized struc-
ture is more stable and therefore less reactive.
What is a resonance contributor?
The two cation structures (cation-1 and cation-2) are used as an example. Cation-1 type structures are
characterized by a X=Y unit (C=C, C=O, etc.) attached to an atom with a p-orbital (a cation, an anion,
or a radical). A cation is essentially an empty p-orbital, an anion is essentially a flled p-orbital, and a
radical is a p-orbital with a single electron.
Note the use of a double-headed arrow to indicate the involvement of two electrons to delocalize
electron density (resonance) over the three atoms in X—Y—Z. Indeed, the curved arrow in cation-1
is used to show electron dispersal over the X—Y—Z unit. The arrow in cation-1 indicates dispersal
of the electron density from X=Y to the Z+ orbital, leaving behind a positive charge on X. The actual
structure of cation-1 is not X=Y—Z+ or is it +X—Y=Z. In the actual structure, the electron density is
delocalized over all three atoms, so both structures are used to represent the actual structure. This
phenomenon is called resonance and the two structures are resonance contributors to the actual

17
structure. Similar structures can be drawn as resonance contributors for the anion and the radical
shown in a preceding question.
X Y Z Y Z
X Y Z Y Z
Cation-1 Cation-2
Cation-2 type structures have a positive charged carbon attached to an atom that has lone electron pairs,
such as O, S, Cl, Br, etc. In cation-2, the lone electron pair on Y is shared with the orbital of Z+ (see the
use of the curved arrow to indicate the dispersal of two electrons). The resulting resonance contributor
has a double bond between Y=Z with a positive charge on Y. If the two electrons in the double bond are
donated back to the positively charged Y+, the original resonance contributor with Z+ is regenerated.
Therefore, molecules with this structural motif have the two resonance contributors shown.
Are the two structures shown as resonance contributors in equilibrium with each other?
No! They are not equilibrating structures, but rather two structures in resonance with one another that,
taken together, represent the bonding between the atoms. Such a molecule has bonds in-between single
and double bonds due to delocalization of the charge. The two structures are not in equilibrium.
What are the resonance contributors for the nitrate anion, NO3
–? For the perchlorate anion
(ClO4
–)?
The nitrate anion has three resonance contributors and the perchlorate anion has four resonance con-
tributors, as shown in the fgure.
N
O
O
O
O
Cl
O
O
O
Nitrate anion
Perchlorate anion
N
O
O
O
N
O
O
O
O
Cl
O
O
O
O
Cl
O
O
O
Cl
O
O
O
O
What are the resonance contributors for (a) and for (b)?
O
(a) (b)
CH CH2
H2C C O
H
The so-called allylic cation (a) has the two resonance contributors shown, which the positive charge delo-
calized over the three carbon atoms. The formate anion (the anion of formic acid) has the two resonance
contributors shown with the negative charge delocalized over the two oxygen atoms and the carbon.
O O
(a) (b)
CH CH2 H2C CH CH2
H2C C O C O
H H
In example (b) in the immediately preceding question, is the negative charge equally distributed
over all three atoms, or is the charge density higher on some atoms?
In the formate anion, the oxygen atoms are more electronegative and attract electron density more than
carbon, so the negative charge (the two electrons) are more concentrated on the oxygen atoms than on
the carbon atom.
What are two types of systems with a charge or a radical that exhibit resonance?
The resonance structures shown for cation-1 in the preceding question represent one type of resonance
involving a three-atom array with one π-bond between two atoms and the third atom attached with a (+),

a (−) charge, or a single electron (a radical). The second type of structure is cation-2, with one atom that
has a positive charge and the attached atom (usually O, S, or N) has a lone electron pair.
END OF CHAPTER PROBLEMS
1. Describe a 3s-orbital and a 3p-orbital.
2. What is the difference between a 2px and a 2py orbital?
3. Give the electronic confguration for each of the following: O, F, Cl, S, Si.
4. Identify each of the following molecules as having an ionic bond, a covalent bond or both:
(a) K–Cl (b) Na–C≡N (c) H3C—Br (d) H2N—H (e) N–Na (f) HO–Na (g) HO—H.
5. Give the number of covalent bonds each atom can form and remain electrically neutral: (a) C
(b) N (c) F (d) B (e) O.
6. Why is BF3 considered to be a Lewis acid?
7. Which of the following bonds is most polarized: C—C, C—N, C—O, C—F?
8. Identify the valence electrons in each of the following (also identify the atom involved):
(a) 1s22s22p3 (b) 1s22s22p63s1 (c) 1s2 (d) 1s22s22p63s23p3
9. Indicate which bonds are consistent with hydrogen bonding, which are consistent with dipole–
dipole interactions, and which would be consistent with only van der Waals interactions:
(a) C—O—H (b) C—F (c) C—C (d) N—H
10. Indicate which structures are likely to have resonance contributors. In those cases where there
is resonance, draw the resonance contributors.
C
(a) C C C (b) Cl C (c) C C C C (d) O C O
H

2
Structure of Molecules
This chapter will describe the fundamental structures of organic molecules, the properties associated
with those structures, and the structural features that lead to important functional groups.
2.1 BASIC STRUCTURE OF ORGANIC MOLECULES
2.1.1 Fundamental Structures
What is valence?
Valence is a whole number that represents the ability of an atom or a group of atoms to combine with
other atoms or groups of atoms. The valence is determined by the number of electrons that an atom can
lose, add, or share. Therefore, valence is the number of electrons available for chemical bond formation
for each atom. Valence also relates to the number of bonds an atom can form and remain neutral. Valence
is 8-#, where # is the last digit of the group number for an atom: # is 4 for Group 14, 5 for Group 15, 6
for Group 16, and 7 for Group 17.
What is the valence of each of the following: C, H, N, O, F?
The valence of each atom is determined by subtracting the last digit of group number from 8 for C, N,
O, F since these atoms are in the second row. Therefore, C = 4, N = 3, O = 2, and F = 1. The valence is
the number of bonds that can be formed to each of these atoms and the resulting atom will be neutral (no
charge). In other words, carbon can form four bonds, nitrogen can form two, oxygen can form two, and
fuorine can form one. In all cases, the resulting molecule is neutral.
What is a simple molecule that satisfes the valence of oxygen and a simple molecule that satis-
fes the valence of nitrogen?
Oxygen has a valence of two, and H—O—H (water) satisfes the valence of oxygen. Nitrogen has a
valence of three, and NH3 (ammonia) satisfes the valence of nitrogen.
H
O Water Ammonia
H H N
H H
What are valence electrons?
Valence electrons are outer shell electrons that are associated with an atom and participate in the forma-
tion of a chemical bond. Valence electrons can participate in the formation of a chemical bond if the
outer shell is not full. As practical matter, valence electrons are the number of electrons in the outermost
shell of an atom that are available to form a covalent bond with another atom. Boron has three valence
electrons, carbon has four valence electrons, nitrogen three, oxygen two, and fuorine has one.
How many valence electrons does N have? O?
Since the valence electrons for N and for O are those in the second electronic shell, nitrogen has fve
valence electrons while oxygen has six.
19

What is the structure of the molecule with one carbon connected to four hydrogen atoms by
σ-covalent bonds?
H
H C H Methane
H
What is the experimentally determined geometry of methane (CH4)?
Experiments have determined that CH4 has the structure with the four hydrogen atoms in a tetrahedral
array about a central carbon atom, and the H—C—H bond angles are 104°28′ (104.47°). All four of the
bond angles are identical, consistent with the tetrahedral shape.
The use of a solid wedge and a dashed line represents the three-dimensional shape of methane. The
solid wedge indicates the bond and the attached atom is projected out of the page, and the dashed line
indicates that the bond and the attached atom are projected behind the page. The normal lines are used
to indicate that the bonds and the attached atoms are in the plane of the paper. This convention is used to
indicate the three-dimensional shape of molecules about a specifc atom.
H
C
H
H
H
Based on the geometry of methane, what can be inferred about the geometry of organic mol-
ecules based on carbon?
With a valence of four, and four other atoms attached, each carbon atom in an organic molecule should
be tetrahedral, as they are in methane. The bond angles will vary with the attached atom, but each carbon
should have a tetrahedral-type geometry.
Can carbon form bonds to itself?
Yes! Carbon can form linear chains of carbon atoms or chains of carbon with carbon atoms or carbon
chains branched from a carbon chain. This property of carbon, to form covalent bonds to other carbon
atoms, leads to a huge number of different molecules. Carbon can also form covalent bonds to many
other atoms in the periodic table, leading to a huge variety of molecules that contain carbon.
Can carbon form bonds to atoms other than hydrogen, or another carbon?
Yes! Carbon can form bonds to oxygen, nitrogen, halogens, sulfur, phosphorus, and other atoms, and to
various metals.
How can fve carbon atoms be connected together in as many different ways as possible?
Complete all of the remaining valences on each carbon with a hydrogen atom.
There are three different possibilities. There is one possibility with fve carbon atoms in a linear chain,
one four-carbon liner chain with a one-carbon branch, and one three-carbon linear chain with two
branched carbon atoms.
CH3
CH3
CH3 CH2 CH3 C CH3
CH2 CH2 CH3
CH3 CH CH2 CH3
CH3

21
Note that the linear chain of carbon atoms is drawn in such a way that each carbon has four attached
atoms. However, the attachments are not drawn in a way that shows all the bonds. The CH3–C unit means
that the frst carbon is connected to three hydrogen atoms and one carbon. The three hydrogen atoms are
shown to the right of the carbon, indicating that the three hydrogen atoms are attached to that carbon,
and the carbon that is also attached is shown to the right. The C—CH2—C unit means that the central
carbon has four bonds, two to the carbon atoms, and two to the hydrogen atoms shown to the right of the
carbon. The same protocol is used for every carbon atom.
What is the empirical formula for all of the molecules drawn in the previous question?
In all three cases, the empirical formula is C5H12.
Two structures have a total of six carbon atoms. One appears to have a linear chain of fve car-
bon atoms and a one-carbon branch at the frst carbon and the second structure has a linear
chain of six carbon atoms. Are these two structures the same or different?
CH3
CH3 CH2 CH2 CH2 CH2 CH3 CH3 CH2 CH2 CH2 CH2
Do not be fooled by appearances, the connectivity of these molecules is identical, with six carbons
directly connected in a chain. Since both structures have the same connectivity, the two structures shown
are the same molecule. In other words, the structure on the left is identical to the one on the right; they are
the same. It is not important if the atoms are twisted up or down as long as the connectivity is identical.
What is line notation, which is used for drawing organic molecules?
A carbon atom attached to two other carbon atoms
A carbon atom attached to another carbon (on the
right) and the three other valences are understood A carbon atom attached to another carbon (on the
to have hydrgoen atoms. This is a CH3 unit. left) and the three other valences are understood
to have hydrgoen atoms. This is a CH3 unit.
and the other two valences are assumed to be hydrogen
atoms. 2- unit.
A carbon atom attached to two other carbon atoms
and the other two valences are assumed to be hydrogen
atoms. 2- unit.
The molecule shown is a linear chain of six carbon atoms, and all other valences for each carbon are
understood to be attached hydrogen atoms. Line notation is a shorthand method of drawing organic
molecules. Each carbon atom is represented by a “dot,” and a line is used to connect each carbon. Only
carbon and hydrogen are present, and the hydrogen atoms do not have to be shown; they are understood
to be there. In other words, each line represents a bond, and the remaining valences are understood to
be hydrogen atoms.
If the connectivity of molecules is different, what can be said about the molecules?
If the connectivity is different, they are different molecules.
What is a fve-carbon linear chain with a one-carbon branch on carbon 2 and a four-carbon
linear chain with two one-carbon branches on C2 using line notation, both with the empirical
formula C6H14?

Both molecules shown in the preceding question have the same empirical formula, C6H14, but
different connectivity of the atoms, so they are different molecules. What is their relationship?
These two molecules with the same empirical formula but different connectivity are said to be isomers.
What is the term for molecules that have the same empirical formula but different connectivity
for the carbon atoms?
Isomers. Specifcally, constitutional isomers.
Which are of the following are isomers?
(a) (b) (c) (d)
The empirical formula of (a), (b), and (d) are the same, C7H16, but they have different connectivity and
are different molecules. Therefore, (a), (b), and (d) are constitutional isomers, or just isomers. Note that
(c) has a different empirical formula, C9H20, so it is a different molecule, but it is not an isomer of (a),
(b), or (d).
(a) (b) (c) (d)
C7H16 C7H16 C9H20 C7H16
2.1.2 Structures with Other Atoms Bonded to Carbon
Can carbon form covalent bonds to carbon?
Yes!
What class name is used for molecules that have carbon or hydrogen atoms attached to a nitro-
gen so that three units are attached to the nitrogen atom?
Amine. Such molecules are called amines.
What is the molecule that has three carbon atoms attached to a nitrogen atom, and each carbon
has three hydrogen atoms?
CH3
N
CH3 CH3
This molecule is known as an amine, specifcally trimethylamine (see Sections 2.4 and 19.1). Note that
there is an electron pair on nitrogen that is not used to form covalent bonds in this structure. The lone
electron pair is not included in the structure shown, but it is understood to be there.
What is the molecule that has two carbon atoms attached to a nitrogen atom, a hydrogen on the
nitrogen, and three hydrogen atoms of each of the carbon atoms?
CH3
N
CH3 H

23
Nitrogen can form covalent bonds to C or H. The molecule shown is an amine, specifcally dimethyl-
amine (see Sections 2.4 and 19.1). As with the preceding question, the lone electron pair is not shown in
the structure but is understood to be there.
Can carbon form covalent bonds to oxygen?
Yes!
What is the molecule that has one carbon attached to an oxygen atom, one hydrogen atom
attached to oxygen, and three hydrogen atoms attached to the carbon?
O
CH3 H
There are two lone electron pairs on the oxygen, which has a valence of two and forms two covalent
bonds, but the lone electron pairs are not shown. This molecule has an O—H unit and it is classifed as
an alcohol. Specifcally, this molecule is methanol (see Sections 2.4, 7.4, and 8.3). In alcohols, oxygen
forms bonds to both H and to C.
Another way to draw the structure given as an answer to the previous question is CH3OH. How
is this possible?
The three hydrogen atoms on each carbon are “condensed” to CH3. Drawing CH3O means that the
carbon atom is bonded to the three H atoms to its right, as well as to the oxygen to complete the fourth
valence. Likewise, the oxygen is understood to be bonded to C, not to H.
What is the molecule that has two carbon atoms connected to an oxygen atom, and each of the
carbon atoms has three hydrogen atoms attached?
This molecule with a C—O—C structure is called an ether, specifcally dimethyl ether (Sections 2.4 and
8.3). There are also two lone electron pairs on the oxygen, which has a valence of two, and forms two
covalent bonds, but the lone electron pairs are not shown.
Another way to draw the molecule shown in the previous question is CH3OCH3. How is this
possible?
In CH3OCH3, the oxygen is bonded to the carbon on its left, and the one on its right. Remember, the
valence of O is two. The carbon to the right of O is bonded to the O, as well as to the three hydrogen
atoms to its right. The valence of carbon is four.
What is the class of molecule with a carbon attached to oxygen and also a hydrogen atom?
Alcohol. Compounds such as this are known as alcohols, ROH, where R is any carbon group. Alcohols
will be discussed in Sections 2.4 and 13.5.
What is the class of molecule with two carbon groups attached to oxygen?
Ether. Compounds such as this are known as ethers, ROR, where R is any carbon group.
2.2 THE VSEPR MODEL AND MOLECULAR GEOMETRY
What model is used determine the shape of a molecule when the atoms are found in the frst row
of the periodic table?
All molecules have a three-dimensional structure, of course. A simple model that will predict the
approximate shape of a given molecule formed from atoms in the second row of the periodic table is the

Valence Shell Electron Pair Repulsion (VSEPR) model. In this model there are two key components: (1)
carbon, oxygen, and nitrogen are assumed to be at the center of a tetrahedron, (2) atoms bonded to car-
bon, oxygen, and nitrogen form a tetrahedral array of atoms around each central atom, and (3) electron
pairs occupy space and are counted as “groups” attached to the central atom. The shape of the molecule
is determined by the relative position of the atoms, assuming that the lone electron pairs are not seen.
What is meant by the shape of a molecule?
The shape is the three-dimensional shape of a molecule, which is determined by the bond angles and
bond distances of the atoms directly connected to a given atom. Shapes can be determined in some cases
using various experimental techniques or inferred in other cases by indirect methods. Using carbon as
an example, the tetrahedral array of atoms attached to a given carbon leads to the tetrahedral shape or
tetrahedral geometry that is associated with organic molecules.
How can the shape of a molecule be probed?
Organic chemists use a simple model, the VSEPR model, for a preliminary prediction of the three-
dimensional structure of molecules that contain atoms in the second row of the periodic table. This
model is used to estimate the structure and properties of that molecule. The model is overly simplistic,
and the bond angles and bond distances are not taken into account, as the nature of the atoms or groups
attached to carbon changes. Nonetheless, it is a useful tool to estimate the three-dimensional shapes of
organic molecules.
How is the VSEPR model used?
The electrons in each bond around the central atom (C, N, O) will repel (like charges repel), but the bonds
are connected to a central locus and the attached atoms or groups attached cannot dissociate from the
central atom. Lone electron pairs are similarly “connected” to the central atom. The most effcient spatial
arrangement that minimizes electronic repulsion is to put each atom or electron pair at the corner of a reg-
ular tetrahedron (bond angles are 109°28′). Using this observation, the model assumes a tetrahedral array
of atoms and electrons around the atoms in the second row, specifcally carbon, oxygen, and nitrogen.
Can the VSEPR model be used for boron compounds?
Boron has only three groups around it since there are only three valence electrons. When the electrons in
the bonds repel, they distribute to the corners of a planar triangle. Therefore, molecules of boron tend to
be planar. In other words, the tetrahedral VSEPR model is not used for boron compounds.
How can the VSEPR model be used to predict the three-dimensional shape of CH4, CH2Cl2,
H2O, and NH3?
For methane (Section 4.1) the four hydrogen atoms are distributed to the corners of a regular tetrahedron
with carbon as the central locus. There are no unshared electron pairs on the carbon and the “shape” of
the entire molecule is dictated by the covalent bonds and is tetrahedral. When two of the hydrogens are
replaced with chlorine (dichloromethane), there are no unshared electron pairs on carbon and the over-
all shape remains tetrahedral. When water is analyzed, there are two hydrogens and two lone electron
pairs, distributed to the corners of a tetrahedron. When viewing the molecule, however, only the atoms are
observed (H—O—H), and these atoms assume an angular or bent shape for the water molecule. When
ammonia is analyzed, the three hydrogens and one lone pair distribute to the corners of the tetrahedron. On
viewing the atoms, however, the atoms assume a pyramidal shape, with nitrogen at the apex of the pyramid.
H Cl
H
C
H
H
H
C
Cl
H
O
H
H
H
N
H
H
Methane Dichloromethane Water Ammonia

25
What are the main shortcomings of the VSEPR model?
This VSEPR model is fawed since it does not predict differences in shape due to the size of the various
atoms and ignores attractive and repulsive forces that are present in some molecules. It does a reason-
able job for simple molecules, however, and it usually used for a “frst guess” of the shape of a molecule.
2.3 DIPOLE MOMENT
What is a dipole?
A dipole is the separation of charges within a molecule between two covalently bonded atoms or atoms
that share an ionic bond. A dipole is the unsymmetrical dispersion of electron density in a covalent bond
toward a concentration of higher electron density in one atom of that bond, in a molecule that has at
least one polarized covalent bond. A polarized covalent bond has a partial negative charge on one atom
and a partial positive charge on the other. The term dipole can also be used in connection with an entire
molecule, where the electron density is not symmetrically dispersed.
What is dipole moment?
Dipole moment is the quantity that describes two opposite charges separated by a distance. Dipole moment
for a bond is determined by discovering the size of the partial charges on the molecule and the bond length.
The dipole moment for a bond (the bond dipole moment) is defned as the product of the total separation of
positive or negative charge and the distance between the atoms. If there is a difference in electronegativity
of the elements involved in the bond (3.0 – 2.1 = 0.9, for example), there is a shift in electron density toward
the more electronegative atom, leading to polarization of the bonding electrons, and of the whole molecule.
The dipole moment can therefore be associated with an individual bond, and also with the entire molecule.
Dipole moment, μ, is calculated by the equation μ = δ d, and measured in units of Debye, where 1
Debye = 3.34 × 10–30 coulomb/meter. The charge difference is measured in coulombs and the bond dis-
tance is measured in meters. In this equation, μ is the dipole moment in Debye, d is the bond distance,
and δ is the charge difference between the atoms.
The dipole moment of C—F is 1.847 Debye and that of C—Cl is 1.860 Debye. Why does the C—
Cl bond have a larger dipole moment relative to C—F?
The bond distance between the C and Cl atoms is 174 pm whereas the bond distance between the C and
F atoms is 134 pm, due to the larger size of the chlorine atom relative to the fuorine atom. The atomic
radius of Cl is 175 pm and the atomic radius of F is 147 pm.
What is the dipole moment for the C—F bond using both the arrow notation and the partial
charge notation?
+
C F
The dipole moment is indicated by the +-arrow and also by δ+ and δ– are used to indicate the dipole. The
point of the arrow is pointed toward the more electronegative fuorine atom, which is labeled with the δ–.
The less electronegative carbon atom is positioned by the + end of the arrow and it is labeled with the δ+.
How can the VSEPR model be used to predict the dipole moment for a molecule rather than a
single bond?
A molecule will have several bonds and each carbon atom, for example, may have up to four single
covalent bonds. Each bond will have a dipole moment called a bond moment. Dipole moments are addi-
tive, and the dipole moment for a molecule is the additive value of all individual bond moments and
the sum of the individual bond moment direction. The VSEPR model can be used to approximate the

three-dimensional shape of molecules. The individual dipole moment of each bond in that molecule is
superimposed on the VSEPR model, the direction and the magnitude of each bond moment is deter-
mined, and the dipole moment for the molecule can be estimated. Using the dipole moment for the mol-
ecule, the relative polarity (polar vs. non-polar) of the molecule can be estimated.
How can the VSEPR model be used to predict the direction of the dipole moment of methane
(CH4) and of dibromomethane (CH2Br2), if any?
Methane Dibromomethane
In a molecule such as methane, the C—H bond is not considered to be polarized (the electronegativity of
C and H are assumed to be the same, although they are not the same). Since there is no dipole moment
for any bond and the dipole moment for the molecule is the sum of all bond dipole moments, the bond
moment is zero. In other words, there is no dipole moment for methane.
For dibromomethane, there are two bond moments, along the C—Br bonds. The dipole moment for the
molecule is the sum of all bond dipole moments. The negative pole of each of those bonds is the bromine.
Since dipole moments are directional, the dipole moment for the molecule is the vector sum of all the indi-
vidual bond moments. The vector sum of the two C—Br bond moments is in the direction that bisects the
Br—C—Br bond as shown in the model. The dipole moment for dibromomethane will be equal in magni-
tude to the vector sum of the two individual bond moments. The experimental dipole moment is calculated
to be 1.32 Debye. Note that the unit for dipole moment is the Debye, and has both magnitude and direction
What is the direction of the dipole moment for the molecule CHCl3 (known as chloroform)?
The dipole moment for each C—Cl bond is shown. There are three dipole moments for the bonds (bond
moments are adjacent to each bond) and with the tetrahedral shape of the molecule, the vector sum (bond
distance and direction; see the simple model) of the three C—Cl bond moments bisects the base of the
“tetrahedron” in the direction shown in the model. Note that the unit for dipole moment is the Debye and
has both magnitude and direction.
Cl
C
H
Cl
Cl
Chloroform
2.4 FUNCTIONAL GROUPS
What is a functional group?
When atoms other than carbon or hydrogen are incorporated into an organic molecule, certain col-
lections of heteroatoms taken as a unit have unique chemical and physical properties and are called

27
functional groups. Functional groups include the hydroxyl unit (OH), an amine unit (a nitrogen atom
with three attached atoms or groups), and the thiol unit (SH). Other functional groups include collections
of atoms that contain π-bonds, including C=C, C≡C, and C=O, C=N, and C≡N.
Why are functional groups important?
Certain collections of atoms have chemical and physical properties that are unique. Further, when such collec-
tions of atoms appear in different organic molecules, those molecules often have similar chemical and physi-
cal properties. For this reason, it is convenient to categorize organic molecules by such collections of atoms,
which are called functional groups. The functional group is the basis used to name most organic molecules.
These functional groups are also part of various classes of compounds, such as alcohols, ethers, amines, etc.
What functional groups involve only carbon and hydrogen?
The C=C and C≡C units in a molecule are functional groups that defne the alkene and alkyne classes
of compounds, respectively.
What functional groups involve oxygen and carbon and/or hydrogen?
These functional groups include OH (hydroxyl), C–O–C (ether), C=O (carbonyl), CO2H (carboxyl). The class
of compounds with a hydroxyl group are the alcohols; with an ether unit are called ethers; the carbonyl is
found in aldehydes and ketones; a carboxyl group defnes the class of compounds known as carboxylic acids.
What functional groups involve nitrogen, carbon, and/or hydrogen?
Amines are the class of compounds that have one, two, or three carbons attached to a nitrogen atom, with
hydrogen attached to nitrogen when there are only one or two carbon groups. In addition, C≡N (cyano)
is another nitrogen-containing functional group that is attached to carbon molecules. Molecules with a
cyano functional group are called nitriles.
What functional groups involve sulfur, carbon, and/or hydrogen?
These include SH (thioxy) is found in the class of compounds known as thiols (the older name is mercap-
tan) and the C—S—C unit is found in sulfdes.
What functional group correlates with the following names: hydroxyl, thiol, carbonyl, carboxyl,
amino, cyano, alkene, alkyne, and ether?
When these functional groups are incorporated into a molecule, they usually form a class of molecules
that tend to have unique chemical and physical properties. Examples of each type of functional group are
obtained by attaching one or more R group to each functional group, where R is a generic carbon group.
The key functional groups are:
O—H hydroxyl S—H thiol C=O carbonyl
H—O—C=O carboxyl C—N amino C≡N cyano
C=C alkene C≡C alkyne C—O—C ether
What functional group is found in each of the following classes of molecule: alcohol, thiol, alde-
hyde, ketone, carboxylic acid, amine, nitrile, alkene, alkyne, and ether?
When the functional group OH (hydroxyl) is in a molecule, the class is called alcohols. When the car-
bonyl (C=O) is incorporated, there are two structural variations. The carbonyl can be connected to both a
hydrogen atom and a carbon group (use R as a generic abbreviation) to give an aldehyde or the carbonyl
can be connected to two carbon groups (R) to give a ketone. If the carbonyl also contains a hydroxyl
(H—O—C=O, or it is abbreviated –COOH), the molecule is called a carboxylic acid. If a trisubstituted
nitrogen is in the molecule, the class is called an amine. If there is one R group and two hydrogens, it is a
primary amine (1° = primary). If there are two R groups and one H, it is a secondary (2°) amine, and three

R groups give a tertiary (3°) amine. The class of compounds that contains a cyano group (C≡N) is called
a nitrile. If the molecule contains a carbon–carbon double bond (C=C), it is an alkene and if it contains
a carbon–carbon triple bond (C≡C), it is an alkyne. A compound characterized by a C—O—C unit is an
ether. These are the major functional groups that will be discussed, and a few others will be added later.
2.5 FORMAL CHARGE
What is formal charge?
In a given structure, an atom that has one less bond than its valence it will have a (–) charge, whereas if it
has one more bond it usually will have a (+) charge. Formal charge can reside on individual atoms, and the
sum of all the formal charges on the individual atoms leads to the formal charge for the overall molecule.
What is the formula used to determine formal charge?
The formula is:
1
Formal Charge = W = (Valence Number)- (Number Unshared Electron
ns)- (Number Shared Electrons)
2
Determine the formal charge of each atom and also for the entire molecule of dimethylamine,
(CH3)2NH.
N
H3C
CH3
1
2
H
Dimethylamine
Before determining the formal charge for the entire molecule, frst determine the formal charge for each
atom. The sum of all atomic formal charges will give the formal charge for the molecule. For both carbon
atoms, the valence number is the group number (four for C). There are no unshared electrons and eight
covalently shared electrons. For C1, Ω = 4 – 0 – ½(8) = 0. For C2, Ω also = 0. For the nitrogen, the valence
(group) number is 5 (since N is in Group 15) and Ω is 5 – (1) – ½(8)) = 0. The formal charge for each
hydrogen is 0 [1 – 0 – ½ (2)]. To determine the formal charge for the molecule Ωmol, the sum of the formal
charge for all atoms is used: Ωmol = Σ (Ωatom). For this example, Ωmol = 0 + 0 + [8 × 0] + 0 = 0. Therefore,
dimethylamine exists as a neutral species.
What is the criterion for a “real” molecule based on formal charge?
For the molecules encountered in the frst organic course, it is very unusual for a molecule to exist as
anything but a neutral compound (formal charge = 0), a mono-cation (formal charge of +1), or a mono-
anion (formal charge of –1). For the purposes of structures encountered in a frst-year organic chemistry
course, if a structure is drawn and has a charge of greater than +2 or less than –2, that structure is likely
to be very high in energy and should be questioned.
2.6 PHYSICAL PROPERTIES
How are functional groups related to the physical properties of a molecule?
The presence or absence of a functional group in an organic molecule has a profound effect on the boil-
ing point, melting point, adsorption characteristics, etc. These parameters are physical properties, com-
monly measured to help identify a unique molecule.

29
What is a physical property?
A physical property is a parameter associated with a pure molecule that is unique and assists in the iden-
tifcation of that molecule. Examples are dipole moment, polarity, boiling point, melting point, adsorp-
tivity, refractive index, and solubility. While some of these physical properties may overlap with other
similar or related molecules, it is highly unusual for all of the physical properties to overlap. This set of
physical properties allows each molecule to be identifed as unique.
What is the defnition of boiling point?
Boiling point is formally defned as the temperature at which the molecules in the gas phase (over a liquid
phase) and the molecules in the liquid phase are at equilibrium.
What are the factors that infuence boiling point?
To get molecules into the gas phase, the intermolecular forces holding them together in the liquid phase
must be disrupted, which requires heat. In general, the greater the mass of the molecule, the higher the
boiling point, and larger molecules (more atoms) should have a higher boiling point than smaller mol-
ecules (less atoms). Intermolecular interactions of the molecule are very important. Three examples will
illustrate this statement: (a) CH3CH2CH3 (propane; molecular weight = 44), (b) H3C–O–CH3 (dimethyl
ether; molecular weight = 46), and (c) CH3OH (methanol; molecular weight = 32). The molecular weights
are not signifcantly different, although methanol has the lowest molecular weight. However, methanol
has the highest boiling point.
In case (a) there are no dipole interactions and the only intermolecular forces are van der Waals forces
(sometimes called London forces). These are quite weak, and the boiling point of propane is –42°C for a
molecular weight of 44.10. For case (b), the polarized C–O bonds lead to a stronger intermolecular inter-
action relative to London forces, and the dipole-dipole interactions help keep the molecules associated.
It takes more energy to disrupt these interactions and the boiling point is –24°C for a molecular weight
of 46.07. The modest increase in mass cannot account for the 17.3°C increase in temperature required to
boil dimethyl ether. When methanol in case (c) is examined, the boiling point of 64.7°C for a molecular
weight of 32.04 is far higher than either propane or dimethyl ether. This large increase is due to the
hydrogen bonding interactions between the oxygen and the hydrogen atom attached to oxygen, which are
much stronger than the dipole interactions found in dimethyl ether.
What are van der Waals interactions?
The van der Waals force is a distance-dependent interaction between atoms or molecules. This force
arises when adjacent atoms come close enough that their outer electron clouds just barely touch, induc-
ing charge fuctuations that result in a nonspecifc, nondirectional attraction. Part of this interaction is
the London dispersion force, which is a temporary attractive force that occurs when the electrons in two
adjacent atoms occupy positions that make the atoms form temporary dipoles. An example is the asso-
ciative interaction of one nonpolar molecule, such as butane, with a second nonpolar butane molecule.
What are dipole–dipole interactions?
Dipole–dipole interactions result from the through-space interaction of two molecules and can have
polarized covalent bonds. In this interaction, the partially negative portion of one of the polar molecules
is attracted to the partially positive portion of the second polar molecule. An example is the through-
space interaction of the carbonyl on one ketone with the carbonyl of a second ketone molecule.
What is hydrogen bonding?
A hydrogen bond results from the attraction of a hydrogen atom that is part of a polarized bond such as
O—H, S—H, N—H, etc. In all cases, the hydrogen will have a δ+ dipole. The through-space interaction
of such a hydrogen atom with any electronegative atom (O, N, S, Cl, Br, etc.) is not a covalent bond, and
is weak and easily broken. A hydrogen bond is one of the strongest associative interactions and hydrogen

bonding is strong enough to hold molecules together by a strong associative between atoms within a mol-
ecule or between two separate molecules. The strength of a hydrogen bond is dependent on the strength
of the dipole of the X—H bond.
Why are intermolecular associative forces important for the boiling point of a molecule?
Boiling point is the temperature at which enough energy is added for an equilibrium to be established
between molecules in the liquids phase and molecules in the gas phase. For this to occur, the intermolec-
ular associative forces such as van der Waals, dipole–dipole, or hydrogen bonding must be disrupted. In
other words, suffcient energy must be added to overcome to attractive energy of these associative forces.
Which of the following is likely to have the highest boiling point: CH3CH3, CH3CH2CH3, or
CH3CH2CH2CH2CH3?
Since all three compounds have only carbon and hydrogen atoms, and no functional groups that are
capable of dipole–dipole interactions or hydrogen bonding, all will have only van der Waals interac-
tions; the mass of the compounds must be considered as more important to boiling point. The molecule
with fve carbon atoms (molecular weight = 72) has a higher mass than the compound with three carbon
atoms (molecular weight = 44), which is higher than the molecule with only two carbon atoms (molecular
weight = 30). The fve-carbon compound has the highest boiling point: 2C = –89°C, 3C = –42°C, 5C =
+36.1°C.
Which of the following is likely to have the highest boiling point: CH3(C=O)CH3, CH3OH, or CH3CH3?
There is not a large difference in molecular weight for each of the molecules, so differences in inter-
molecular interactions are expected to dominate. Since CH3OH (molecular weight = 32; boiling point
= 64.7°C) is the only molecule with an OH unit, which is capable of hydrogen bonding, it is likely to
have the highest boiling point. Acetone (molecular weight = 58; boiling point = 56°C), with the carbonyl
group, has dipole–dipole interactions, which are weaker intermolecular interactions relative to hydrogen
bonding, and has the next highest boiling point. Ethane (molecular weight = 30; boiling point = –89°C),
with no functional groups, has only van der Waals interactions and has the lowest boiling point.
What is the defnition of melting point?
Melting point is defned as the temperature at which the molecules in the solid phase are in equilibrium
with those in the liquid phase.
What factors infuence melting point?
An increase in molecular weight often leads to an increase in melting point. The most important feature
of a molecule that infuences melting point is, however, more diffcult to describe. The more symmetri-
cal a molecule, the higher the melting point. Conversely, the more amorphous a molecule, the lower the
melting point.
Which has the higher melting point, A or B?
A B
Both A (hexadecane) and B (pentane) have only C and H and no functional groups. Neither A nor B are
“compact,” and mass is probably the dominant factor. The molecular weight of A is 226 and the molecu-
lar weight of B is 72, and A is expected to have the higher melting point (+18°C for A vs. –129.8°C for B).

31
Which has the higher melting point, A or B?
A B
Comparing A (pentane) with B (2,2-dimethylpropane), the melting points are –129.8°C for pentane and
–16.6°C for dimethylpropane. Both molecules have the same empirical formula (same number and kind
of atoms), so they clearly have the same mass. Note that 2,2-dimethylpropane is more compact, however,
and will “ft” into a crystal structure better than the “foppy” linear molecule, pentane. Such enhanced
“packing” leads to a lower melting point for pentane, A.
What is solubility?
Solubility is the property of one molecule (a solid, a liquid, or a gas called the solute) to dissolve into
another molecule that is in the liquid state, called the solvent. Solubility is measured by the maximum
amount of solute dissolved in a solvent at equilibrium, and the result is a saturated solution.
Is a ten-carbon molecule with only C and H likely to be soluble in water?
No! Water is highly polar, and a ten-carbon hydrocarbon is very nonpolar. Therefore, they are not
expected to be mutually soluble.
Is a two-carbon alcohol, CH3CH2OH, likely to be soluble in water?
Yes! The OH unit can hydrogen bond so there will be strong associative interactions between water and
the alcohol. The result of these interactions is an increased solubility for alcohols that have only a few
carbon atoms. In general, alcohols of less than fve carbons are soluble in water. Molecules of fve to
eight carbon atoms may be partially soluble in water and molecules of more than eight carbon atoms are
generally insoluble.
How does solubility give information about the structure of a molecule?
The old axiom “like dissolves like” can be used with remarkable accuracy. In general, non-polar
molecules will dissolve well in non-polar liquids such as hydrocarbons but not in polar liquids such as
water. Conversely, a polar molecule will not dissolve in a non-polar liquid but will dissolve in a polar
liquid. If a molecule contains one or more C-heteroatom bonds (a heteroatom is an atom other than C
or H) it is considered to be polar. If it contains no polarized bonds, it is considered to be non-polar.
There are exceptions to this latter statement that use the qualifying statement for the heteroatom
case. A molecule such as CCl4 has a net dipole moment of zero for the molecule and is non-polar.
The actual test of polarity (and usually solubility) is, therefore, the dipole moment for the molecule.
There are degrees of polarity and degrees of solubility. Molecules that undergo extensive hydrogen
bonding are very polar and very soluble in polar liquids such as water. A molecule with a small dipole
(CH3Cl) will be less polar than one with more dipole interactions (CHCl3) and this will be refected
in partial solubilities in polar solvents but some solubility in non-polar solvents. Solubility in water
is suggested when a molecule can hydrogen bond to water. If there are more than fve to eight car-
bon atoms, however, the one hydrogen bond is counteracted by the carbon atoms (with the attached
hydrogen atoms).

END OF CHAPTER PROBLEMS
Line notation will be used for these and subsequent problems. When atoms other than C and H are pres-
ent, the atom or group will be shown, as in problem 1. A line is drawn to the Br, OH, etc. Where there is a
multiple bond, two or three lines are used to indicate the double or triple bond, respectively. An example
is in problem 1b. In some cases, the functional group is condensed, as in the COOH unit shown in prob-
lem 1b and in 6c, drawn in two different ways.
1. For each series of compounds, identify the one of the following that has the highest boiling
point? Explain.
(a) Br OH
O O
(b) OH H
CH4
(c)
2. Predict the shape of each of the following: (a) the C–O–C unit of CH3OCH3 (b) the four atoms
about carbon in Cl3CH (c) the C–O–H unit of CH3OH (d) the N and three carbon atoms in
NH(CH3)2.
3. Indicate the direction of the dipole moment on the VSEPR model for each of the following:
(a) CH3OCH3 (b) Cl3CH (c) CH3OH (d) NH3 (e) CHBrCl.
4. What is the functional group identifed with each of the following: (a) alcohol (b) ketone
(c) alkyne (d) aldehyde (e) thiol (f) nitrile?
5. Determine the formal charge for all atoms in each molecule. Calculate the fnal charge for each
molecule.
O2
1 O H3 H4
2 C
C O1 H11 C1
H11
N C4
H2 N H3
C2 C3
H2 H5 H6
H5 H4
(a) (b)
6. Indicate which molecule has the higher boiling point in each of the following pairs. In each case
explain your answer.
OH
(a) (b) (c)
OH
OH CO2H HO
OH
7. Which molecule has the higher melting point? Explain.
(a) (b) (c)

3
Acids and Bases
Acid–base reactions are perhaps the most important category of chemical reactions in all of organic
chemistry. It will be seen that many, if not most, organic reactions have an acid–base component. Seeing
this relationship usually requires modifying the traditional concept of what is an acid and what is a base.
3.1 ACIDS AND BASES
There are two fundamental defnitions of acids and bases, the Lewis defnition and the Brønsted–Lowry
defnition. Both defnitions will be viewed through the lens of electron donation and electron acceptor
ability.
What is the generic symbol for an acid?
The proton, which is usually represented by H+.
What is a Brønsted–Lowry acid?
A Brønsted–Lowry acid is defned as a hydrogen ion donor. As a practical matter, a Brønsted–Lowry
acid must have a hydrogen atom with a δ+ dipole.
What structural feature makes a molecule able to donate a hydrogen ion?
The hydrogen atom must be polarized at H+ or Hδ+. In the neutral HCl molecule, not in water, the H has a
positive dipole and can be donated to a base by the Brønsted–Lowry acid defnition. Another way to say
this, and arguably more correct, is that the base donates two electrons to the hydrogen atom and “pulls it
off.” An alcohol, which has a C—OH unit, is a Brønsted–Lowry acid since the H has a positive dipole.
What are four examples of mineral acids that qualify as Brønsted–Lowry acids?
Four examples are hydrochloric acid (HCl), sulfuric acid (H2SO4), nitric acid (HNO3), and perchloric
acid (HClO4).
Do different molecules differ in Brønsted–Lowry acid strength?
Yes! Assuming that a structure such as C—X—H, where X is O, S, N, or another atom other than C or
H, the atoms or groups attached to C will “push” or “pull” electron density from the X—H bond, which
makes it more diffcult or less diffcult for the H to be removed. In other words, C—X—H is a weaker
acid or a stronger acid, respectively. This analysis is overly simplistic, and it ignores the products formed
after the hydrogen is donated and, of course, does not mention the base to be used. A stronger base will
remove the hydrogen more easily than a weaker base. This analysis is a simple illustration that different
molecules can be stronger or weaker Brønsted–Lowry acids.
What is a Lewis acid?
The classical defnition of a Lewis acid is an electron pair acceptor.
33

What is the structural difference between a Lewis acid and a Brønsted–Lowry acid?
A Brønsted–Lowry acid has a polarized hydrogen atom, such as X—H, whereas a Lewis acid does not
have a polarized hydrogen atom but rather has an electron defcient atom such as boron that can accept
electrons from another molecule.
What are two common examples of a Lewis acid?
Both BCl3 (boron trichloride) and AlCl3 (aluminum trichloride) are Lewis acids. Both boron and aluminum
are in Group 13 of the periodic table. They can form three covalent bonds, using the three valence electrons,
to generate a neutral molecule. These atoms can attain the octet only by accepting an electron pair from
another molecule (a Lewis base). Therefore, both B and Al are electron defcient in these compounds.
What is a Brønsted–Lowry base?
The classical defnition of a Brønsted–Lowry base is a hydrogen atom acceptor.
What is a Lewis base?
A Lewis base is an electron pair donor. A Lewis base is a molecule that can donate two electrons to an
atom other than hydrogen or carbon.
What is an “ate” complex?
An “ate” complex is the product of a Lewis acid–Lewis base reaction, which is the Lewis acid–Lewis
base adduct.
What is the product when the Lewis acid AlCl3 reacts with ammonia (NH3)?
The product of a Lewis acid–Lewis base reaction is a Lewis acid–Lewis base adduct. More commonly,
this product is called an “ate” complex. The reaction of ammonia and AlCl3 is the ate complex shown. The
arrow between N and Al in the fgure represents the dative bond of the ate complex. This bond can also be
shown as a line with the nitrogen assuming a positive charge and the aluminum assuming a negative charge.
Cl H Cl H
Cl Al N H Cl Al N H
Cl H Cl H
What is a dative bond?
A dative bond is a two-center, two-electron covalent bond in which both electrons come from the same
atom. A dative bond is also called a coordinate covalent bond or simply a coordinate bond.
Is it possible to view a Brønsted–Lowry acid in terms of the Lewis acid defnition?
Although the defnition of a Brønsted–Lowry acid is a hydrogen atom donor, to “donate” a hydrogen atom to
another molecule, the elecropositive hydrogen atom of an acid must react with an atom that has an electron
pair. In other words, a Brønsted–Lowry acid has a hydrogen atom that is “pulled off” by an atom that donates
an electron pair (a base). A Brønsted–Lowry acid reacts with a molecule that can donate an electron pair to
H whereas a Lewis acid reacts with a molecule that can donate an electron pair to an atom other than C or H.
How is a Brønsted–Lowry base like a Lewis base?
To accept a hydrogen atom, a Brønsted–Lowry base must donate an electron pair to the hydrogen atom.
Therefore, a Brønsted–Lowry base is as a molecule that donates an electron pair to a hydrogen atom. A
Brønsted–Lowry base donates an electron pair to a hydrogen whereas a Lewis base donates an electron
pair to atoms other than C or H.

35
Acids and Bases
What structural features are required for a molecule to be considered an acid?
For a Brønsted–Lowry acid, the molecule must have a hydrogen atom, generally attached to a hetero-
atom, that has a positive dipole. For a Lewis acid, the atom must be electron defcient, not C or H, and
able to accept a pair of electrons from another molecule.
What structural features are required for a molecule to be considered a base?
For a Brønsted–Lowry base, the molecule must be able to form a bond to a hydrogen atom that has a
positive dipole by donating a pair of electrons. For a Lewis base, the atom must be electron rich, and able
to donate a pair of electrons to another atom other than C or H.
3.2 ENERGETICS
What is the “driving force” for a chemical reaction?
Changes in energy accompany the transformation of starting materials to products in what is known as
a chemical reaction.
What is the bond dissociation enthalpy?
The term is the inherent energy (bond dissociation enthalpy), and that energy is released when a bond is
broken or is required to form that bond.
What is H°?
The term H° is the enthalpy of a given bond or an entire molecule. Formally, enthalpy is a thermody-
namic quantity equivalent to the total heat content of a system.
What is the parameter to describe the energy of a chemical transformation?
The parameter is ΔG°, the change in free energy during the reaction.
What is the Gibbs free energy equation?
The Gibbs free energy equation is ΔG° = ΔH° − TΔS°.
What is T in the equation in the preceding question?
The term T is the temperature in degrees centigrade.
What is ΔH?
The term ΔH° is the change in enthalpy.
What is ΔS°?
The term ΔS° is the change in entropy. The entropy term measures the “disorder” of a given system.
Formally, entropy is the lack of order or predictability: gradual decline into disorder. For most of the
chemical reactions in this book, the change in disorder is small as a starting material is transformed into
a product and the ΔS° term is very small compared to the ΔH° term.
What is ΔG°?
This term is the change in free energy for a reaction. If the ΔG° is positive, the reaction is endergonic
(endothermic) and proceeds to the products spontaneously. If the ΔG° is negative, the reaction is exer-
gonic (exothermic) and proceeds to the products spontaneously.

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train was as yet invisible, saw its smoke growing larger and larger in
volume over the hills. At last the train itself came into view. Jack saw
with surprise that the engine was at the other end of it; could the
goods train, he wondered, have been stopped in some inexplicable
way and started back after him? In two minutes it would be upon
him. He waited for one minute; then, seeing that a gap of some
fifteen or twenty yards had been made in the track, he summoned
his men back to the train and pressed the regulator handle. To his
eager impatience it seemed that the engine would never get under
way. The wheels slipped on the rails; he had pushed the regulator
too far; he drew it back, the wheels held, and, gathering speed
every moment, the locomotive raced on once more.
The thunder of the pursuing train was roaring in Jack's ears. It
seemed to him, looking back, that the foremost carriage was
charging at the gap. He hoped the work of destruction had not been
perceived; but in this he was disappointed, for when the rear of his
own train was barely two hundred yards from the break, steam was
shut off on the engine of the pursuer, and, helped by the rising
gradient, it succeeded in coming to a stand-still just as the buffers of
the foremost carriage were within half a dozen yards of the gap.
CHAPTER XXIV
Lieutenant Potugin in Pursuit
From a Hilltop—Mystified—In Full Chase—A Runaway—In Sight—A Railway Duel

Those Cossacks are taking their time, Akim Akimitch.
Yes, little father; 'tis to be hoped Ah Lum has not swallowed
them.
Lieutenant Potugin smiled.
Ah Lum has been a bogey to them, truly, ever since Captain
Kargopol walked into his trap. But I think we'll run the fox to earth
this time. General Bekovitch will soon start the rounding up; and 'tis
high time.
A half-company of Siberian infantry, including a few engineers,
were seated on the rocks in the hills above the Ma-en-ho, engaged
in a meagre luncheon of black bread and vodka. They had arrived
early that morning by special troop train, in company with a sotnia
of Cossacks, from Harbin. Their errand was to establish a temporary
signal-station on a convenient hilltop. The hole for the signal-pole
had been dug, not without difficulty, in the hard and frozen soil, and
before the completion of the job was taken in hand, Lieutenant
Potugin, in command of the working party, was allowing his men a
short respite for rest and food. The Cossacks meanwhile were
scouting in the hills beyond—a task they were by no means fond of,
—and seeking a suitable place for the erection of a corresponding
signal some miles distant, whence communication could be
established with the height now occupied by the infantry.
Lieutenant Potugin was very popular with his men, largely
because he never overworked them and was quite content when on
duty to share their humble rations. He was seated now beside the
sergeant, in the midst of the circle, munching his bread, and every
now and then raising his field-glass to scan the surrounding heights.
It was a fine morning; a breath of spring was already in the air, even

in these heights; the atmosphere was clear, and the outlines of the
country were sharply defined against the unclouded sky.
Over the shoulder of a low hill beneath him he could just see a
stretch of the main railway line, some three miles away. The little
branch line along which his train had come that morning was out of
sight immediately below; but he expected every moment to see the
empty train reappear on the main line. It was to return to Harbin;
rolling stock was urgently needed on all parts of the system; and
when his work was done Lieutenant Potugin was to report himself to
General Bekovitch and join that officer's carefully-planned expedition
against the Chunchuses. The branch line ended at a disused quarry
which had been largely drawn upon when the main railway was
under construction; and there was no telegraphic communication
between the main line and the terminus of the branch—if, indeed,
the latter could be said to have a terminus: it simply left off. The
empty troop train would doubtless remain at the junction until it was
signalled by trolley-car from Imien-po to proceed.
The sergeant, a famous raconteur, was telling a story, long-
winded, not at all humorous, yet received by the men with shouts of
laughter. Lieutenant Potugin smiled good-humouredly at the naïve
amusement of the honest fellows, and once more idly scanned the
panorama beneath him. In the far distance he saw a dense line of
smoke lying flat in the still air, betokening a train travelling eastward
at a high speed. He watched it with languid curiosity as it appeared
in the open and vanished into cuttings in the winding valley of the
river. It passed the junction, slackening speed, and then, to his
surprise, pulled up. Distant though it was, he could distinctly see
through his powerful glass a little knot of men hurrying from the

train up the line. They disappeared for a time, apparently beneath a
culvert. The circumstance awakened Lieutenant Potugin's curiosity;
he watched with a certain eagerness for the men to reappear; one
or two small groups could be seen against the snow, but a
considerable time elapsed before the most of the men joined them
and the whole party ran back to the train. Scarcely had they reached
it when a cloud of dust rose high into the air above the bridge, and a
few seconds later the sound of two dull explosions reached the
lieutenant's ear, followed by miniature echoes from the rocks.
The lieutenant sprang up and gazed intently through his glass.
The sounds had been heard by the men also; they turned their
heads for a moment, but, seeing nothing, resumed their
conversation. But Potugin stood as if stupefied. An attempt had been
made to wreck the culvert; that was clear. But who were the
wreckers? Were they Russians, cutting the railway to check pursuit
by the Japanese? Surely the enemy was not already at Harbin?
Accustomed as he was in this terrible war to sudden and startling
movements, the lieutenant could not believe that the Japanese had
made such strides. No, he thought; it was more likely to be a party
of Japanese who had captured the train and were engaged on a
wrecking foray. Such things had happened south of Moukden; a
flying squadron might have evaded the Cossacks and made a daring
attack on some inadequately protected train.
The train was moving forward. But what is that? It has stopped
again; it is running back towards the stream. The madmen! Are they
going to hurl themselves to destruction on the ruins of the culvert?
Potugin's gaze is fascinated. Ah! he sees through it now; three
carriages have left the rest of the train, which is again at a standstill;

they are rushing down the gradient, faster, faster. Good heavens!
they have crashed into the culvert, piling themselves one above
another, and the sound comes to him like the breaking of some
giant's crockery afar.
Then Potugin found his wits. Nothing in the whole course of the
war had given the Russians so much anxiety as their railway.
Depending on it for the rapid transit of reinforcements and munitions
of war, they were constantly in nervous dread of this their sole
communication with St. Petersburg being cut by Japanese or
Chunchuses. The dreaded thing had happened. Fully realizing the
situation, Lieutenant Potugin was prompt to act.
Fall in! he shouted.
The men sprang from their seats and were aligned in a
twinkling.
Sergeant, signal the Cossacks that a train is in the hands of the
enemy, and going eastward. Men, follow me.
He led the way at a breakneck pace down the hill towards the
spot where they had left the empty troop train. Three minutes
brought them within sight of the train; at that moment the engine
whistled and began to puff along. The officer shouted, waving his
hand; the engine-driver saw his urgent gesture, and shut off steam.
In another ten minutes sixty breathless men, heated with their
headlong scamper, were on board the train; the lieutenant was
beside the driver; and the engine was steaming as rapidly as the
crazy irregular track permitted towards the main line.
Arrived at the junction, Lieutenant Potugin himself leapt down
and switched the points close. The pointsman had apparently been
startled by the crash and run off to inform the guardsmen at the

nearest block-house. The troop in was just moving forward to cross
the points when a tremendous rumbling was heard from the
direction of Imien-po, moment by moment increasing. The engine of
the troop train was already on the main line. But the lieutenant,
standing with his hand on the switch and looking down the track,
was horrified at what he saw rapidly approaching.
Reverse the engine! he shouted; for God's sake reverse the
engine!
The driver with frenzied haste threw over his reversing lever
and put on more steam; the engine stopped, moved slowly
backward; it had reached safety by only a few inches when a goods
train came thundering past at furious speed, and disappeared in the
direction of the bridge. As it flashed by, Lieutenant Potugin was
almost sure that the engine had neither driver nor fireman. Startled
though he was by the hair's-breadth escape from destruction, he
immediately recovered his presence of mind. Setting the points, he
ran to his retreating train, clambered into the cab, and before the
driver had pulled himself together the lieutenant seized the lever,
reversed the engine, and drove the train on to the main line, then
sprang down, unlocked the points, and in two minutes was running
the train backward towards Imien-po.
The engine was a powerful Baldwin; the train though long was
nearly empty; it gathered way, and with the regulator fully open had
soon attained a high speed. But the engine was at the wrong end; it
was difficult to see ahead. The lieutenant was now outside the
engine, hanging on to the rail, and bending outwards in order to get
a clear view down the line. Half-way to Imien-po he caught sight of
a trolley approaching. He called to the driver to shut off steam and

apply the brakes. The man working the trolley stopped the moment
he caught sight of the train, and seemed in doubt whether to go
back or to remain. The train had almost come to rest; the officer
bellowed a few words to the trolley-man; he sprang to the ground,
promptly tipped the trolley off the track and over the embankment,
and, running to the engine, climbed up beside Potugin, the train still
moving. Again the brakes were released and the regulator opened,
and as the train forged ahead the trolley-man explained in a few
words to the lieutenant what had occurred.
At Imien-po a few minutes' stop was made while appliances for
repairing the line were hastily brought on board and a number of
skilled platelayers taken up. The opportunity was taken to shunt
several of the carriages on to a siding. The engine could not be
transferred to the front of the train without a serious waste of time,
and every second was precious. A fresh start was made; greatly
lightened, the train made fine running for some miles. Then the
lieutenant, using his glass, saw the smoke of a train about five miles
down the line. As he watched it, the smoke ceased; the train must
have stopped, for the gradient was rising. A few minutes more and
the runaway came in sight. But the fireman, stooping from his side
of the engine, observed with his trained eyes that a portion of the
track had been torn up, and steam was shut off and the brakes
applied only just in time to avert a disaster. Jumping from the train,
half a dozen platelayers hurried with their tools behind the engine,
and, spurred by the voice of the officer and helped by his men, in an
incredibly short space of time they had wrenched up some rails from
the track already covered, and bridged the gap at the other end.

Slowly and carefully the train was run over the shaky metals
only half-secured to the sleepers. When the danger point was
passed, the driver opened the valve and the engine pushed along at
full speed. It was to be a trial, not only of speed between the two
magnificent engines, but of wits between the two leaders: between
the ingenuity of the pursued in obstructing the progress of the
pursuer, and of the pursuer in overcoming the obstacles raised by
the pursued. It was more; it was a competition in daring and the
readiness to take risks. The track was hilly, winding, roughly laid; not
intended for, wholly unsuited to, great speed; with steep gradients
and sharp curves never rounded by the regular drivers of the line
but with caution. Over this track the two trains were leaping at a
pace unknown on the Siberian railway—a pace that would have
turned the chief engineer's hair white with dismay. On the one train
Jack Brown, on the other Lieutenant Potugin, had to think out their
decisions, or rather to flash them unthought, clinging to the outer
rail of a rattling, swaying, jolting, throbbing engine threatening at
any moment to jump the rails, with the noise of escaping steam, the
roaring of the furnace heaped to the mouth with fuel, the whistle
constantly sounding to warn off any obstruction ahead, small though
the chances were that the signal, if needed, could be heard and
acted on in time. Accident apart, the race would be to the coolest
head and the quickest wit. On the one side the stake was life or
death. Into whose hand would fortune give it?

CHAPTER XXV
The Pressure-Gauge
Timber on the Track—Fuel and Water—The Station House—A Trap—Neck or
Nothing—Screwing down the Valve—A Slip Carriage—Nearing the End—Kao-ling-
tzü—Indiscreet Zeal—A Lady Passenger—Traffic Suspended
Jack glanced anxiously back along the line; his engine was jolting,
bumping, up the incline at the rate of forty miles an hour; steam was
escaping from the safety-valves; the gauge registered over 10
atmospheres, considerably above working pressure; yet to his
impatience it seemed to be moving with exasperating slowness. Dust
was whirling behind; through the cloud, five minutes after he
started, he saw a puff of steam in the distance; the pursuing train
was again under way. Turning to see if he could put on more steam,
he was dismayed to find that the water was just disappearing in the
gauge glass. In a few minutes—he could not tell how few—the water
would be below the level of his fire-box crown, the fusible plug
would drop, and the fire would be put out by the escaping steam.
This was ominous indeed.
There were, he saw, two conditions in his favour: he had a start
of nearly five minutes; and he could choose his own place to
obstruct the pursuer. But the other conditions were all against him.
He must needs stop for water, and at the present rate of
consumption for fuel also; and whenever he passed a station it

would be necessary to cut the telegraph wires. Moreover, on board
the pursuing train there must be men skilled in repairing the line, or
the chase could not have been resumed so promptly; and Jack could
not expect to do more damage in a given time than could be
remedied by expert hands in the same period. Worst of all, the
pursuing engine was evidently more powerful than his; and though it
was somewhat handicapped by its position at the wrong end of the
train, yet an experienced driver can always get more work out of his
engine than a tyro,—and Jack was making his trial trip!
He cudgelled his brains for some means of checking the pursuit
without bringing his own train to a stand-still. He wished that he had
thought to instruct his men when tearing up the rails to lift some of
the sleepers into the train; these placed on the line would prove
serious obstacles. It was too late to repine; he made up his mind not
to lose the chance if it should occur again. While his thoughts were
still on the matter, his eye caught the balks of timber used for fuel
on this part of the line. The stock in the tender was much
diminished; more fuel must soon be obtained; but surely one or two
might be spared for the experiment. Without delay he sent Hi Lo to
the back of the tender with an order to Wang Shih to carry two of
the balks through the train and to drop them on the line from the
communication door at the rear of the last carriage. In a few
moments the command was carried out, but Wang Shih reported
that owing to the high speed he had found it difficult to see what
happened to the logs when they reached the ground. One, he
thought, had remained on the inside rail; the other appeared to
jump off. Narrowly watching the riband of steam from the pursuing
train, Jack believed he detected a momentary diminution about the

time when it should have reached the spot where the logs had been
thrown out; but if there was a delay it was very brief, and a few
minutes later the tail of the advancing train came into full view, the
growing size of the carriage-end showing that it was making up on
him.
Looking ahead with greater anxiety, Jack saw a station within a
mile. This must be Pei-su-ho. He had already decided that to stop
there would be absolutely necessary, and in a short colloquy with
Wang Shih when he returned from throwing the logs on the track he
had arranged what should be done. Immediately on the stoppage of
the train twelve men were to engage the station staff and destroy
the telegraphic instruments; ten were to tear up the rails behind the
train, and, if possible, bring some sleepers on board; four were to
cut the telegraph wire, and twenty to load wood from the station
stock on to the nearest carriage. In the meanwhile he himself, with
the assistance of the man acting as fireman and others riding on the
engine, would take in a supply of water from the tank.
The train rattled into the station. In his anxiety Jack found that
he had shut off steam too late; the engine ran some yards beyond
the water-tower. As he had already found at Imien-po, it was not
easy to the amateur to bring a train to a stand-still at a given spot.
But although the greater part of the train had run beyond the
platform, the Chunchuses, who were standing ready with the doors
open, swung themselves out, and before the gaping officials were
aware of what was happening they were disarmed and helpless. Not
for the first time had Jack reason to be glad that his men were the
pick of Ah Lum's band, and a standing proof of the efficacy of
discipline with the Chinese.

While Jack was backing the engine to the tank the work of
ripping up the track and demolishing the wire had already been
begun, and a string of men were hauling timber into the nearest
carriage. But before the supply of water was fully replenished Jack
had to blow his whistle to recall the various parties; the pursuer was
drawing perilously near. The train moved off before all the men were
in their places; the last of them running along the platform and
being helped in by his comrades. Up came the second train; again it
had to halt before the gap, and the driver, being at the other end,
was compelled for safety's sake to reduce speed earlier than he
would have done had he been able to judge the distance more
exactly. But this time the gap was shorter; the time required to
restore the line would be correspondingly less. Yet Jack had gained
one advantage; knowing that the enemy's water supply, like his own,
must have run low, he had brought the station hose away with him,
and he looked at it with grim satisfaction, lying coiled at the rear of
the tender.
As Jack's engine, Alexander the Second, gained impetus and
charged up the gradient towards the hills looming in the distance, it
was followed by a dropping fire from the pursuing train: some of
Lieutenant Potugin's men had climbed to the roof of the stationary
carriages. Whether any of the bullets struck the train was doubtful;
no harm was done; and in the excitement of the moment the idea of
firing rifles seemed almost as childish as shooting at the moon.
Nothing less than a siege-gun would have appeared formidable in
the circumstances.
The brigands' last cutting of the line and the removal of the
hose had evidently gained several minutes for the fugitive, for many

miles had been covered before the smoke of the pursuer was again
seen. With so considerable a start Jack felt it safe to pull up once
more and try a device that had occurred to him. His engine was at
the summit of a long descent where the line curved. Hitherto his
track-breakers had forced up both the rails, but the curve was here
so sharp that he thought he might save time by having only one rail
lifted, hoping that the partial gap might not be seen by the enemy
until it was too late to do more than check the train, which would in
all probability be derailed. An alternative plan suggested itself, only
to be dismissed. It was to remove the rail, and then replace it
without the bolts. The pursuer would then rush on at full speed
expecting no danger; the train would be hurled from the track, and
probably all on board would be killed or injured. But even in the heat
of the moment, and with the knowledge that if he were caught he
could expect no mercy from the Russians, Jack could not bring
himself to compass such wholesale destruction. Play the game: the
phrase of the school song stuck to him. His purpose would be amply
served by the mere derailment of the train, the speed of which
would no doubt be sufficiently checked, when the gap was descried,
to avert fatal consequences.
So confident was he of the success of his scheme that when,
after the single rail was removed and flung over the embankment,
he again crowded on steam, his mind was occupied rather with the
question of what should be done at the next station than with the
prospect of further difficulties with his dogged pursuer. He was now
approaching the place in the hills to which Ah Lum was to advance
by forced marches, and whence he was to be prepared to dash
across the line on receiving a message that the scheme had

succeeded. Jack had already selected his messenger; the man was
clinging to the rail of the engine, and only awaited the word to
spring during a temporary slackening of speed and plunge into the
hills.
The chosen spot lay between Pei-su-ho and Kao-ling-tzü, and
had been minutely described by Ah Lum. Jack was glad that his
anxieties appeared to be over, for the country flashed by so rapidly
that he ran the risk of over-shooting the mark unless he could keep
a good look-out. He was narrowly watching for the opening on his
right when Hi Lo suddenly drew his attention westward. With greater
alarm than he had yet felt, even when he first caught sight of the
pursuer, he saw, scarcely a mile and a half behind him, the relentless
enemy leaping along in his wake. He was half-way up a steep
incline; the second train was rushing with wholly reckless speed
down a steep straight gradient on which Jack, no longer fearing
pursuit, had thought it desirable to clap on the brakes. All notion of
going cautiously must now be abandoned. Amazed at the failure of
his last effort to delay the pursuer, Jack set his men with desperate
energy to pile up the furnace to its utmost capacity; and when he
topped the hill, and the enemy was just beginning the ascent, he let
the engine go at its own pace down the opposite side. He and his
men had to hold on with both hands as they rounded another sharp
curve; the wheels on the inside seemed to be raised from the track,
the train keeping the rails only by the grip of the outside wheels.
Jack held his breath as the panting engine plunged along; would it
come safely on to the straight? Even in the excitement of the
moment he solved as in a flash the mystery of the pursuers' escape
from derailment, and he could have beaten his head for his

thoughtlessness. The rail that had been lifted was an inside rail;
rounding a curve the weight of a train going at speed is always
thrown on the outer rail, which is raised above the level of the other.
Either designedly or by accident the pursuing train had passed at full
speed over the gap, its very speed proving its salvation.
Although there were many ups and downs, the general trend of
the line was still chiefly on the up grade, and Jack found that while
the enemy made as good timing as himself down the slopes, their
more powerful engine gained rapidly wherever the track began to
rise. As mile after mile was passed, the huts of the line guards at
intervals of ten versts seeming like the milestones on an ordinary
journey, the space between the two trains steadily diminished. Every
now and again the pursuer was lost to view; but whenever it next
came in sight it was always perceptibly nearer. The noble Alexander
the Second rattled and groaned like a creature in pain; the working
parts were smoking; some of the bearings were melting, and Jack
dared not risk the perils of oiling. He knew that he was getting out
of it every ounce of which it was capable, unless indeed he adopted
the desperate expedient of screwing down the safety-valve, from
which a dense cloud of steam was escaping. He glanced at the
gauge—13 atmospheres; then his eye went backwards along the
track—the pursuer was still gaining; he turned to look ahead, there
was a long steep ascent to be climbed. The pace lessened to an
alarming extent: puffing, panting, creaking, the engine toiled up a
hillside on which the track could be seen rising for at least two miles.
He must risk it.
Three minutes later, the valve now screwed down, he again
glanced at the gauge—14 atmospheres. Bursting pressure, Jack

knew, was calculated at five or six times the working pressure; but
the Alexander the Second was an old engine, he doubted whether
her boilers would stand anything like this strain.
For a time Jack's train drew away; but the gain was only
temporary; the pursuers, he guessed, must have adopted the same
desperate expedient. Gradually they crept up, while Jack alternately
watched them and the track ahead, and the gauge, which now
registered 15 atmospheres—the limit which it was constructed to
indicate. Beyond this point he had no means of knowing how the
pressure was increasing. The rapidity of his thoughts seemed to
keep pace with the tremendous speed at which he was travelling.
His mind worked with marvellous clearness; the minutes seemed like
hours; he even found himself speculating which of the three risks
was the greatest—derailment, capture by the Russians, or the
imminent explosion of the boiler.
To look for the spot chosen for the despatch of his messenger
was out of the question; it had probably been already passed. Jack
felt that he had no longer any alternative; he must play what
seemed his last card. The pursuing train was only half a mile behind
on the steep upward track when at his order Wang Shih, at the risk
of his life, uncoupled the rearmost of the three carriages. For a short
distance it followed the rest; then it stopped, and began to run back
at a pace that threatened to telescope at least one carriage of the
oncoming train. A turn in the track hid both the detached carriage
and the pursuer from sight; Jack listened with a beating heart for
the sound of the collision, which he felt would be audible even above
the thundering roar of his own train.

Lightened of part of its load, his engine was forging its way
uphill at considerably higher speed. At one moment he thought he
heard the expected crash, and it seemed that the move had been
successful, for when next he obtained a fair view of the line behind,
the enemy was not in sight. Alternating between compunction and
elation, he ventured, the line being more level, to reduce speed until
it was safe to drop his messenger, who must perforce find his way to
Ah Lum. But the man had barely left the track when, to Jack's
amazement, the indomitable pursuer reappeared. A glance showed
him that it was pushing the discarded carriage before it. His move
had been detected, probably before the cast-off carriage began its
backward journey; the pursuing engine had been able to reverse in
time; chased and overtaken by the runaway carriage, the train had
no doubt been badly bumped, but not with force enough to cause
any serious damage. Now, to all appearance, it was following the
quarry at the same breakneck pace as before. Jack felt a glow of
admiration for the wary Russians, who showed themselves so intent
to mark his every move, so quick to take measures to defeat it.
His mouth hardened as he watched the pursuer gaining upon
him yard by yard. He knew that the pressure must now be
enormous; would the boilers stand the strain? Yet in spite of all he
was steadily being overhauled. Yard by yard the gap lessened.
Nothing but an accident could now prevent him from being
overtaken; his only course seemed to be to stop before the enemy
was too close, reverse his engine, and with his men take to the hills.
But then he reflected with a kind of agony that the task he had set
himself was even yet only half done. There was no longer, indeed,
any chance of Ah Lum's retreat being cut from the west; but the

Russians could still despatch a force from Ninguta in ample time to
check the Chunchuses before they got across the railway; and if they
were once checked, the forces behind would at once close in and
crush them. While, therefore, the slightest hope remained, Jack
resolved to cling to his train; but he gave his men orders to jump
clear at a moment's notice. They must now be very near to Kao-ling-
tzü: if they failed to cut the line there the race was clearly run, for a
warning would certainly be flashed over the wire to the next station
at Han-ta-ho-tzü, giving ample time for preparations to be made to
meet him. He was in a bath of sweat; his throat was parched; his
limbs were trembling; but collecting all his forces, he watched the
gauge and grasped the lever.
There remained, he clearly saw, one small chance, and only
one. If there happened to be a train at Kao-ling-tzü side-tracked in
obedience to his instructions, it might be possible—how long would
it take?—to interpose it between himself and his pursuers. There
would be a minute, nay, less than a minute, to gain possession of it
and set it in motion. Could he increase the margin? Yes; by
detaching the saloon, now the rearmost carriage, and crowding the
whole of his men and the two prisoners into the single carriage in
front. The enemy had all along shown himself so alert that he would
doubtless be on the look-out for such a move; there was no longer
any likelihood that it would end the chase; but at least it would
check the pursuer's progress, forcing him to stop or reverse. Even if
it caused the delay of only a few seconds, it was worth attempting; a
few seconds might make all the difference.
The station was already in sight when, the transference of men
having been quickly effected, Wang Shih broke the couplings and left

the saloon solitary upon the line. Looking with blood-shot eyes
ahead, Jack saw—and his labouring heart leapt at the sight—not
one, as he had hoped, but two trains, one behind the other,
completely filling a siding, where they were halted to allow General
Bekovitch's expected train to pass.
But the same glance that gave Jack such elation showed him
that he had to deal with perhaps the greatest danger he had yet
encountered. He had intended to follow the same plan that had
proved successful at the other stations: dispose of the officials, cut
the wires, and block the line. But he saw almost with dismay that
the platform here was thronged. Drawn, no doubt, by curiosity to
see the train of General Bekovitch, and excited by the urgent
messages received along the wire, not only the station officials were
waiting, but a considerable number of workers on the railway,
Russian riflemen, and Chinese passengers. These, together with the
attendants of the standing passenger train, were massed upon the
platform. They formed so numerous a crowd that it would tax all the
energies of the Chunchuses to deal with them; there might be a
prolonged fight, and, even if it ended in a victory for the brigands,
so much time would have been consumed that the pursuers must
arrive before anything could be done to stop their progress. It was a
moment when many a man might have despaired. But Jack was not
made of the stuff that yields. As his engine plunged along towards
the station he conceived an alternative plan; it would test his nerve
and self-command to the uttermost; but it might succeed by its very
audacity.
Passing the word to his men that they were to remain in the
carriage and hold their revolvers ready in case an attack was made,

he halted the engine with a jerk a yard beyond the spot where the
station-master was standing. He sprang to the platform, clutched the
astonished official by the arm, and dragged him along, speaking in
low, rapid, urgent tones.
Come with me. There is not a moment to lose. We are pursued
by a train in the possession of the enemy. General Bekovitch is laid
up. We have done our best to check the pursuit, but they'll be upon
us in a few minutes. Only one thing can be done: uncouple the
engine on the siding, and start it up the line. Quick! our lives depend
on it. I will take the responsibility.
As Jack had hoped, the suddenness and unexpectedness of the
news, and the urgency of his manner, bereft the station-master of all
power of independent thought. He hurried along the platform,
shoving aside all who stood in his path, every man in the crowd
looking on with wonderment. He sprang on to the line, with his own
hands uncoupled the engine, signalled for the points to be closed,
and ordered the driver to send it ahead at full speed.
Two minutes saved! thought Jack, as the engine started. But
he could not afford to let the flurried official regain his self-
command.
That is not enough, he said. They will see the engine,
reverse, couple it on, and come at greater speed. I've tried it
already. You must empty the passenger train, and then push it along
with the goods engine. It would be well to throw a carriage or two
off the rails at the points. Anything to block the line.
Certainly, your nobility, said the station-master. It is the only
way.

They were now on the track between the waiting train and
Jack's. Many of the passengers had their heads out of the windows,
wondering what was going on. Waving his arms, the station-master
summoned them in urgent tones to alight.
I'll now push on, said Jack. Do your best, nichalnik;
remember how much depends on you.
He walked rapidly along between the trains to reach his engine.
Passengers, anxious, wonder-struck, were already leaving the train.
One of them, a Russian army doctor, stopped Jack and asked what
was the matter.
Train behind in possession of the enemy, returned Jack
laconically.
Bozhe moï! ejaculated the doctor, drawing his revolver and
making for the platform.
Jack passed on, not venturing to delay even long enough to
assist a lady, for whom the jump from carriage to track was
somewhat difficult. She sprang down unassisted.
Monsieur Brown, Monsieur Brown!
Jack shivered from top to toe, and never in his life felt so much
inclined to take to his heels as then. He could hardly believe he had
heard aright; yet amid the bustle now filling the station he had
caught the whisper of his name. On a sudden impulse he swung
round.
Monsieur Brown, said Gabriele Walewska, running up to him,
I have news for you: I have something to show you.
Come with me, Mademoiselle, said Jack instantly. I haven't a
minute to lose.
But Masha is here; I cannot leave her.

For heaven's sake, Mademoiselle, climb up into this carriage. I
will fetch Masha.
With anxiety tearing at his heart Jack hurried back down the
train. He saw Gabriele's old nurse at the door of a carriage; she was
almost the only passenger who had not yet alighted.
Spring into my arms, he said, forgetting that she knew no
tongue but Polish. But his outstretched arms spoke for him. The
woman jumped clumsily; but Jack kept his feet, and, straining his
muscles, he carried the burden, as rapidly as he could stagger, to his
own train. Gabriele's hands were ready to help the woman; with an
unceremonious heave Jack pushed her into the carriage. Then he
ran to his engine, swung himself up, and pressed the lever just as
the empty passenger train moved off in the other direction. Before
he had run a hundred yards he heard a crash behind. Glancing back,
he saw that the first carriage had jumped the points, ploughed up
the permanent way, and overturned. One after another the other
carriages followed; and in a brief minute there was a pile of wrecked
trucks and coaches in inextricable confusion across the rails.
Jack had not time to give a second thought to Gabriele. He was
again urging Alexander the Second along at full speed. He must run
to within a few miles of the next station, and lift enough rails to
delay for some hours any train despatched from the direction of
Ninguta. Twenty minutes brought him to a likely spot—a high culvert
over a brawling hill stream. Employing the whole strength of his
detachment in the work, he lifted fifty yards of the track and flung
the rails and sleepers into the stream's rocky bed.
At last! he exclaimed. The load of anxiety he had borne for
over two hours was gone. From the place where he had wrecked the

bridge nearly a hundred miles westward to the spot where he now
stood, traffic on the Siberian railway was hopelessly blocked.
CHAPTER XXVI
A Double Quest
Gabriele's Story—A Hasty Word—Lex Talionis—Bribery and Corruption—Cause and
Effect—The Natural Man—The Filial Obligation—The Choice of Routes—A Fair
Pleader—In the Circumstances—Improving the Occasion
Jack's part was done. The way had been cleared for the passage of
the Chunchuses across the railway, and knowing Ah Lum's rapidity of
movement he felt tolerably sure that the crossing might easily be
made. He could now afford to think of his own safety. He determined
to run the train back as near as he dared to Pei-su-ho, then to leave
it standing on the line and make off in a northerly or north-westerly
direction, trusting to join hands with Ah Lum at some distance north
of the line. The railway guards were amazed to see the train running
swiftly backwards; but, whatever their suspicions, they were
powerless. Jack came to a stop between two of the block-houses; in
a few minutes his men alighted with Bekovitch and Sowinski,
Gabriele, and her nurse; and then Jack abandoned the noble
Alexander the Second that had served him so well, and started on
his northward march. Some distance above the line he instinctively
turned for a last look. There was the short train, motionless on the
rails, a derelict in a vast solitude. But it represented activities that

had disorganized the whole traffic of the line for a hundred miles,
nullified a military scheme, and saved hundreds of lives. It was not
without a certain grim amusement Jack remembered that the final
card in that game had been played by the Russians themselves. I
only hope the station-master won't be cashiered, he thought, as he
turned his back upon the scene.
Not till now had he an opportunity of learning what strange fate
had entrusted Gabriele to his care. Some time after he had left the
missionary's house the girl, unable to endure the separation from
her father, again ventured into Vladivostok. Acting on the knowledge
that Jack had bribed a Russian official, she succeeded in persuading
a colonist about to re-embark for Sakhalin to carry a letter from her
to Count Walewski. She told him of her intentions, assuring him that
in spite of her failure to gain permission to enter the island, she still
meant to persevere. Several weeks later she received a reply,
brought by the same man, who had crossed the sea in probably the
last boat before the ports became ice-bound. It was addressed in a
strange handwriting, and as she tore it open she was oppressed by
the fear that her father was dead. But the first line of the letter,
written in French, dispelled her anxiety. The count was ill in hospital,
unable to write; but he had availed himself of the ready help of a
fellow-prisoner—a political prisoner who had recently arrived in the
island. He thanked his daughter for her affectionate solicitude, but
pled with her to abandon her purpose: Sakhalin was no place for a
woman; she would only suffer without alleviating his lot. As for
himself, until the arrival of his new friend he had despaired of ever
regaining his liberty. But the surprising news that the Japanese were
winning victory after victory had sown a seed of hope. The prisoners

on the island had been fed with lies by the officials, who reported
constant victories for Russia. But the new-comer had thrown a fresh
light on the war; he could not foresee its end: the Russians had still
enormous powers of resistance; it was possible that the great fleet
on its way eastward might break through to Vladivostok and change
the aspect of things. Yet, if it should be defeated, the Japanese
might capture Sakhalin; possibly the political prisoners would then
be released if they had not been previously removed to the
mainland. It was only a possibility, but sufficient to give new courage
to a sorely-tried man.
Jack read all this himself, for Gabriele, immediately after
explaining how the letter came into her possession, handed it to
him. The writing was his father's. At the first moment he felt
unutterable relief in finding that his father was alive; then rage
burned within him as he saw before him, marching at some distance
apart, each manacled to a Chunchuse, the two men whose villainy
had sent Mr. Brown to the bleak island of the dead. Gabriele
noticed his look.
I understand, she said. But if your anger is great, how much
greater is mine! Your father's persecutor is a Russian, a foreigner;
my father was betrayed by one of his own countrymen,—one of his
own house. The traitor there recognized me as I entered the saloon
carriage; bound as he was, he shrank from me as though expecting
that I would kill him.
But he did not recognize you when he saw you at Father
Mayenobe's?
No. But something must have put him on my track, for it is
through him that I was a passenger on the train. I was arrested in

Vladivostok and ordered to go back to Europe. He was with the
soldiers who arrested me: in fact, he pointed me out to them. I do
not know how he came to recognize me after all.
At the moment no explanation occurred to Jack, who indeed did
not give a thought to it. But later he remembered that, on the well-
remembered evening in Moukden when he had got the better of
Sowinski, he had mentioned the man's true name, Streleszki. This
had no doubt set the Pole wondering how Jack could have learnt his
name; and the chain of incidents had led him to connect the
disclosure with the European girl he had met at the missionary's. So
that Jack's almost inadvertent explanation had ultimately led to this
meeting with Gabriele at the station, and to the end of his long
search for his father's whereabouts.
The party marched as rapidly as possible, rising gradually
towards the barren hills. After two hours they stopped for a brief
rest, and for the first time since his capture at Mao-shan General
Bekovitch was within arm's-length of the Chunchuse leader. Jack
wondered whether he would be recognized; but the change of
costume, the hardening of his features and the development of his
physique due to his active rigorous life, made him a different being
from the lad whom Bekovitch had seen for five minutes at the
Moukden railway-station. And the general was certainly not in such a
calm and collected mood as might quicken his memory. He was
indeed in a condition of boiling rage and indignation.
Here, you— he cried, seeing Jack so near to him. Do you
understand Russian?
Moderately well, sir.

His very voice had become more manly; its deeper tones did not
awaken recollection.
Then what do you mean, confound you! by treating a Russian
general officer thus? What do you mean, I say? Do you know what
you are doing? Made to tramp over these hills—fettered to a filthy
Chinaman—why—why——
The general could find no further words to express his
indignation.
Is it not the Russian custom to manacle prisoners? asked Jack
quietly.
The Russian's cheeks took a purple hue.
An officer—a general! Do you know who I am, you—you——
You are General Bekovitch.
Well—well—loose me at once, then; I insist on this indignity
being removed; it is monstrous!
Possibly; but quite Russian. You are no worse treated than you
treat your prisoners. If a Chunchuse, myself for instance, had fallen
into your hands, what would have been his fate?
The mild reasonableness of the Chunchuse's reply, together with
his firm attitude, seemed to suggest to the general that he should
try another tack.
Come, he said, with sudden suavity, I know you gentlemen; I
suppose it is a matter of dollars. How much will you take to let me
go?
Jack looked at him.
Say a thousand dollars—that's a very fair sum, more than you'd
get in the ordinary way of your—business. Eh?

Yes: our business, as you call it, is certainly not profitable, but
we do make a haul at times.
The general looked furious. Jack quietly continued:
But you are making a mistake—you are treating me as you
would a Russian and an official. I am merely a brigand—but we
Chunchuses have our code. Dirty though he is, General Bekovitch,
the man you are bound to has cleaner hands than you: he at least is
an honest man according to his lights. It is he who should complain
of contamination.
Bekovitch quivered with rage, but gulping down the indiscreet
words his anger prompted he returned to the point.
I could make you a rich man. I said a thousand dollars; come, I
will make it two thousand. It will buy you a pardon, and an official
post as well. Batiushki! no brigand ever had such a chance.
Jack laughed.
We have our code, General Bekovitch, I repeat. There are
some things bribery cannot effect. Your release just now is one of
them. But for bribery you would not be here.
The general stared.
What do you mean?
It is all very simple. If the Pole Sowinski yonder had not bribed
you, General Bekovitch, you would not have conspired against Mr.
Brown at Moukden, and you would not have needed to deport his
son. If you had not deported his son, his son would not still be in
Manchuria; and if he had not been in Manchuria he could not have
captured you, General Bekovitch, and you need not have attempted
to bribe him.

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