Preview Organic Chemistry Study Guide and Solutions, 6th Edition by Jim Parise, Marc Loudon (2015)

Preview Organic Chemistry Study Guide and Solutions, 6th Edition by Jim Parise, Marc Loudon (2015) Preview Organic Chemistry Study Guide and Solutions, 6th Edition by Jim Parise, Marc Loudon (2015) Preview Organic Chemistry Study Guide and Solutions, 6th Edition by Jim Parise, Marc Loudon (2015) Preview Organic Chemistry Study Guide and Solutions, 6th Edition by Jim Parise, Marc Loudon (2015) Preview Organic Chemistry Study Guide and Solutions, 6th Edition by Jim Parise, Marc Loudon (2015)

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Study Guide and Solutions Manual

to Accompany

Organic Chemistry

Sixth Edition

Jim Parise Department of Chemistry and Biochemistry

University of Notre Dame

Marc Loudon

Department of Medicinal Chemistry and Molecular Pharmacology

Purdue University

ROBERTS & COMPANY PUBLISHERS

Greenwood Village, Colorado

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Cover art: Quade Paul

Reproduction or translation of any part of this work beyond that permitted by Section 107 or 108 of the

1976 United States Copyright Act without permission of the copyright owner is unlawful Requests for permission or further information should be addressed to the Permissions Department at Roberts and

Company Publishers

ISBN: 978-1-936221-86-8

10987654321

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1.2 Electron Density Distribution in Orbitals 3

Solutions to In-Text Problems 6 Solutions to Additional Problems 12

21 Nomenclature of Simple Branched Compounds 25

21 Atomic Radii and van der Waals Repulsion 26

ACIDS AND BASES THE CURVED-ARROW NOTATION 43

3.1 The Curved-Arrow Notation 43 3.2 Rules for Use of the Curved-Arrow Notation 44

33 Identification of Acids and Bases 45

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(f Study Guide Links val

41 ferent Ways to Draw the Same Structure n

42 Drawing Structures from Names 72

43 Solving Structure Problems 73

41 Relationship between Free Energy and Enthalpy 74

42 Sources of Heats of Formation 75

43 Molecular Orbital Description of Hyperconjugation 76

44 Stepwise View of Rearrangements 7

5.1 Transition Elements and the Electron-Counting Rules 107 5.2 How to Study Organic Reactions 108

53 Solving Structure Problems 112

5 Mechanism of Organoborane Oxidation 4 5.2 Mechanism of Ozonide Conversion into

Carbonyl Compounds us

53 Bond Dissociation Energies and Heats of Reaction 116

G9 Solutions to Problems 122 Solutions to In-Text Problems 122

Solutions to Additional Problems 131

61 Finding Asymmetric Carbons in Rings 147

62 Stereocenters and Asymmetric Atoms 148

63 Using Perspective Structures 148

6.2 Isolation of Conformational Enantiomers 150

G9 Solutions to Problems 152 Solutions to In-Text Problems 152

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(Uf Study Guide Links 167

7.1 Relating Cyclohexane Conformations 167

72 Reactions of Chiral Molecules 168

73 Analysis of Reaction Stereochemistry 169

74 Stereoselective and Stereospe 169

75 When Stereoselectivity Matters 169

7 Other Ways of Designating Relative Configuration 71

7.2 Alkenelike Behavior of Cyclopropanes 172

81 Common Nomenclature and the n-Prefix 207

8.1 Trouton’s Rule 207 {Gg Solutions to Problems 209

THE CHEMISTRY OF ALKYL HALIDES 225

91 Deducing Mechanisms from Rate Laws 225

9.2 Ring Carbons as Alkyl Substituents 226 9.3 Diagnosing Reactivity Patterns in Substitution

and Elimination Reactions 226

9.1 Reaction Rates 228

92 Absolute Rate Theory 228

93 Mechanism of Formation of Grignard Reagents 229

{9 Solutions to Problems 235

10.1 More on Half-Reactions 265

® Further Explorations 266

10.1 Solvation of Tertiary Alkoxides 266

10.2 Mechanism of Sulfonate Ester Formation 267

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(7 Study Guide Links 303

11.1 Learning New Reactions from Earlier Reactions 303

112 ‘Common Intermediates from Different

(2 Study Guide Links 367

13.1 Approaches to Problem Solving 367 13.2 More NMR Problem-Solving Hints 367

13.1 Quantitative Estimation of Chemical Shifts 369 13.2 Fourier-Transform NMR 370

Gy solutions to Problems 375 Solutions to In-Text Problems 375

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(2 study Guide Links 401

14.1 Functional Group Preparations 401

14.2 Ammonia, Solvated Electrons, and Amide lon 401

Solutions to In-Text Problems 408

CO study Guide Links 425

15.1 ATerminology Review 425

15.1 More on UV Spectroscopy 426

15.2 The Molecular Orbitals of Benzene 426

16 THE CHEMISTRY OF BENZENE AND ITS DERIVATIVES 469

16.1 NMR of Para-Substituted Benzene Derivatives 469

16.2 Different Sources of the Same Reactive Intermediate 470

16.3 Reaction Conditions and Reaction Rate 470

G9 solutions to Problems 476

(% Study Guide Links 507

17.1 Synthetic Equivalence 507

171 Addition versus Substitution with Bromine 508

13? solutions to Problems 513

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18 THE CHEMISTRY OF ARYL HALIDES, VINYLIC HALIDES,

U2 study Guide Links 547

18.1 Contrast of Aromatic Substitution Reactions 547

18.2 The Cumene Hydroperoxide Rearrangement 548

19.1 Lewis Acid Catalysis 609 19.2 Reactions That Form Carbon-Carbon Bonds 609 19.3 Alcohol Syntheses 610 19.4 Hemiacetal Protonation 610 19.5 Mechanism of Carbinolamine Formation 610 19.6 Dehydration of Carbinolamines 612 19.7 Mechanism of the Wolff-Kishner Reaction 612

19.1 IR Absorptions of Cyclic Ketones 613

20.1 Reactions of Bases with Carboxylic Acids 659

20.2 Resonance Effect on Carboxylic Acid Acidity 660

20.3 Mechanism of Acid Chloride Formation: 660

204 More on Synthetic Equivalents 661 20.5 Mechanism of Anhydride Formation 661

20.1 ‘Chemical Shifts of Carbonyl! Carbons, 663 20.2 More on Surfactants 663 20.3 Orthoesters 664 20.4 — Mechanism of the LiAIH4 Reduction of

Carboxylic Acids 665

9 Solutions to Problems 672 Solutions to In-Text Problems 672

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21.1 Solving Structure Problems Involving

Nitrogen-Containing Compounds 699

21.3 Mechanism of Ester Hydrolysis 700

21.4 Another Look at the Friedel-Crafts Reaction 701

215 Esters and Nucleophiles 701

211 NMR Evidence for Internal Rotation in Amides 702

21.2 Cleavage of Tertiary Esters and Carbonless

21.3 Reaction of Tertiary Amines with Acid Chlorides 705

9 Solutions to Problems a Solutions to In-Text Problems 715

22.1 lonization versus Nucleophilic Reaction at the

Carbonyl Carbon 749

22.2 Kinetic versus Thermodynamic Stability of Enols 750 22.3 Dehydration of B-Hydroxy Carbonyl Compounds 751 22.4 Understanding Condensation Reactions 751 22.5 Variants of the Aldol and Claisen Condensations 752 22.6 Further Analysis of the Claisen Condensation 752

22.7 Synthetic Equivalents in Conjugate Addition 753

OP study Guide Links 823

23.1 Nitration of Aniline 823

23.2 Mechanism of the Curtius Rearrangement 823

23.3 Formation and Decarboxylation of Carbamic Acids 824

23.1 Alkyl Group Polarization in lonization Reactions 825

23.2 Structures of Amide Bases 826 23.3 Mechanism of Diazotization 826

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24.1 Acid Catalysis of Carbohydrate Reactions 877 24.2 Configurations of Aldaric Acids 877

24.1 Nomenclature of Anomers 878 24.2 More on the Fischer Proof 878

'Y solutions to Problems 884

‘Solutions to In-Text Problems 885 Solutions to Additional Problems 895

25.1 Pentavalent Intermediates in Phosphate Ester Hydrolysis 915

G9 Solutions to Problems 920

27.1 Solid-Phase Peptide Synthesis 977

272 Reactions of a-Amino Acids with Ninhydrin 977

G9 Solutions to Problems 986

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sections are found within each chapter, each headed by its own icon

STUDY GUIDE LINKS

Study Guide Links are short supplements to text material that are called out in the margins of the text by the icon depicted above Study Guide Links provide hints for study, shortcuts, approaches

to problem-solving, or additional explanations of topics that typically cause difficulty For example, Study Guide Link 5.2, “How to Study Organic Reactions,” on p 108 of this manual, is called out at a point in the text (p 200) at which some students begin to have difficulty learning reactions It provides detailed hints on how to study and learn organic reactions efficiently

FURTHER EXPLORATIONS

Further Explorations are short discussions that provide additional depth on certain topics; these are called out in the margins of the text by the icon above For example, Further Exploration 1 called out with a margin icon on text p 666, gives a more in-depth discussion of Fourier- Transform NMR, a method used for obtaining NMR spectra

iil REACTION REVIEW

Reaction Reviews present the reactions in the chapter in detailed outline form This summary consolidates the reactions of the chapter in one place and presents not only the reactions themselves but also the essential features of their mechanisms and stereochemistry for cases in which these issues are discussed in the text Studying and learning reactions is a skill that should

be cultivated, and Study Guide Link 5.2 on page 108 of this manual will help you to use these reaction summaries to best advantage

The Reaction Review sections are found only in chapters in which specific reactions are covered in the text For example, Chapters 12 and 13, which cover spectroscopy, do not have

reaction reviews.

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? SOLUTIONS TO PROBLEMS

Undoubtedly, your instructor will stress the importance of problem solving for the mastery of organic chemistry This text provides over 1600 problems of varying difficulty levels, many with multiple parts The Solutions sections provide solutions to all problems in the text The solutions

to many of the problems are worked-out examples in which not only the answer is given, but also

the detailed approach to the solution is discussed

It is important not to rely too heavily on the solutions presented in this manual You learn organic chemistry by solving problems on your own, Many students develop a false sense of

understanding by trying a problem, giving up too soon, and then reading the detailed solution The

solution seems to make sense to them, they feel like they understand the concept, and move on What they've actually done is short-circuited the learning process, and they’re simply agreeing what the solution says; they haven't developed the capacity to solve problems on their own

Here is a better approach if you get stuck on a problem First, attempt to solve it by reviewing the text or your class notes, or by talking to a classmate, TA, or your instructor Be sure to ask for

guidance only, not the answer If you're still stuck, glance quickly at the solution in this manual without reading for detail Focus on the references within the solution—pages, equations, sections,

etc.—where the answer might be found Jot those down next to your work, and close the manual Refer back to the text and see if you can solve the problem after referring to relevant sources It

may take longer to solve problems this way, but your time will be much better spent if you take this approach

Solutions Icon Comments

Within the solutions, you will find two types of comments marked with icons

[>| Comments marked with this icon provide additional information or depth about the solution

T77_ Comments marked with this icon provide cautionary wannings —things tobe particularly

careful about or common traps

Acknowledgments

We would like to acknowledge Dr Joseph Stowell for his contributions to previous editions of this, manual We are also indebted to our proofreader, Dr Animesh Aditya of Purdue University

We have tried very hard to eliminate as many errors as possible Nevertheless, if errors are

found, we would like to know about them so that we can correct them on reprint if possible You

can send these or any comments to us by email at jparise@nd.edu A list of both textbook errors and study guide and solutions manual errors can be found at the link below

http://people.pharmacy.purdue.edu/~loudonm/teaching/

We also enjoy hearing from students with comments and suggestions about the text or the solutions manual We hope that you find this manual very useful in your study of organic chemistry

Jim Parise and Mare Loudon

May 2015

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About the Authors

Jim Parise

Jim Parise received his BS in chemistry from the State University of New York at Oswego in 2000 and his PhD in organic chemistry in 2007 from Duke University, where he worked with Professor Eric Toone After a postdoctoral fellowship with Professor David Lawrence at the University of North Carolina at Chapel Hill, Dr Parise joined the Department of Chemistry at Duke University, where he taught organic chemistry to both pre-professional students and science majors, and coordinated the organic chemistry laboratory courses In 2011, he joined the Department of Chemistry and Biochemistry at the University of Notre Dame He teaches primarily to pre- professional students and also oversees the pedagogy of the accompanying laboratory program He has received a number of teaching awards His research focuses on pedagogical techniques and the integration of classroom technology He has also developed a teaching mentorship program for new instructors Dr Parise has authored and co-authored laboratory manuals and a peer-reviewed book chapter on writing in the laboratory, and he is co-author of the accompanying textbook, Organic Chemistry, Sixth Edition,

in the School of Pharmacy and Pharmaceutical Sciences, where he teaches organic chemistry 10 pre-pharmacy students Dr Loudon served as Associate Dean for Research and Graduate Programs for the College of Pharmacy, Nursing, and Health Sciences from 1988 to 2007 He has received numerous teaching awards Dr Loudon was named the Gustav E Cwalina Distinguished Professor in 1996, one of the first three faculty members to be recognized by Purdue as distinguished professors for teaching and teaching scholarship Dr Loudon participated in the HHML-sponsored Nexus project from 2010-2014, one outcome of which was to completely redesign the first organic chemistry course to make is more relevant to the interests of students in pre-health profession programs Dr Loudon was named Cwalina Distinguished Professor emeritus following his retirement in 2015 In collaboration with Professor George Bodner, Dr Loudon has developed and practices collaborative-learning techniques for teaching organic chemistry to large classes Dr Loudon is the author of numerous research articles, a co-author of an in-house laboratory manual, and he is the author of the accompanying textbook, Organic Chemistry, Sixth Edition, The first edition was published in 1984

xv

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Some other common patterns are shown below

carbon nitrogen oxygen

~g—

(Verify these patterns by calculating the formal charge in each one.) Recognition of these patterns

can be a great time-saver in applying formal-charge rules Conversely, when you calculate a formal charge, always double-check your calculation if you seem to be violating one of these

common patterns.

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(7? 1.2 Vector Addition Review

Vectors can be added together to produce a new vector, called a resultant, There are two ways to add two vectors The first way is to line up the two vectors with their tails touching and construct a parallelogram Form the resultant by

connecting the vertex of the two vectors to be added to the opposite side of the

parallelogram, as shown on the right

Alternatively, let the head of one vector touch the tail of the second, and form the resultant by connecting the tail of the first to the head of the second to forma \R

triangle, as shown on the right

resultant = R

In either case, trigonometry can be used to calculate the magnitude of \

the resultant as well as its angle (consider the law of cosines) \

IN

To line up the vectors, you are allowed to move a vector in any we direction parallel to, or along, its path In other words, you can move a

vector anywhere so long as its length (magnitude) doesn't change, and \ Rh \

its angle (direction) relative to any arbitrary axis doesn't change

You now know that methane is tetrahedral But there is a good reason for using simpler but less

accurate structures anyway When geometrically accurate structures are not needed, there is no point in going to the additional trouble (and in the case of typesetting, the expense) of using them

In other words, we use the simplest structures that accomplish the purpose at hand, and so should you! When the shape of a molecule is an important issue, use line-and-wedge formulas or other types of structures that convey spatial information Otherwise, use the simpler structures

where q is the magnitude of the separated charge and r is a vector from the site of positive charge

to the site of negative charge When pis a bond dipole, as in H—Cl, the magnitude of r is simply

the bond length; in the case of H—Cl, it is the length of the H—CI bond In the HCI molecule the dipole moment vector pt lies along the H—CI bond, and it points from the H (the site of partial positive charge) to the Cl (the site of partial negative charge), as shown in the text That is, pe and

1 have the same direction

The dimensions of the dipole moment, as suggested by Eq SGI.1 above, correspond to charge x length As the text indicates, the units of the dipole moment are called debyes A dipole

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FURTHER EXPLORATIONS + CHAPTER 1 3

moment of one debye results when opposite charges, each with a magnitude of 1 x 10"

electrostatic unit (esu), are separated by one Angstrom That is,

I debye = 1 x 107! esu A

These units were established prior to the current trend toward the use of standard international

units To put these units on more familiar ground, the charge on an electron is 4.8 x 107° esu, and

one Angstrom = 10? meter = 100 picometers (In standard international (SI) units, in which the unit of length is the meter and the unit of charge is the coulomb (C), 1 D = 3.34 x 10° C m, or coulomb meter.)

An actual calculation of the charge separation in H—Cl can provide a more quantitative sense

of the meaning of the dipole moment A table of experimentally measured dipole moments gives =1.08 D for HCl A table of bond lengths gives the length of the H—CI bond as 1.274 A; this is rin Eq, SG1.1 The only unknown remaining in Eq SG1.1 is g Because 2 and r have the same direction, q can be calculated by dividing the magnitude of px by that of r

0.848 x 107!” esu

q=|pl + Irl = 1.08 x 107! esu A + 1.274

As noted above, the charge on an electron is 4.8 + 10~'” esu Consequently, the charge separation

in HCI in electronic charge units is

q(in electrons) = (0.848 x 10°! esu) = (4.8 x 10°" esu per electron)

= 0.18 electron

In other words, the dipole moment of HCI, 1.08 D, means that a partial negative charge of 0.18 units (“18% of an electron”) is on the chlorine of HCI, and a partial positive charge of 0.18 units is

on the hydrogen

Notice that the definition of dipole moment in Eq SG1.1 contains two elements: the amount

of charge separated (q) and the distance between the separated charges (r) A smaller amount of charge separated by a long bond can result in as significant a dipole moment as a larger amount of charge separated by a shorter bond, (See Problem 1.36, text p 43, in Chapter 1.)

1.2 Electron Density Distribution in Orbitals

The orbitals in the text (Figs 1.7 and 1.8, p 25, and Figs 1.10-1.11, p 26-27) are drawn as closed geometrical figures designed to encompass about 90% of the electron density within the orbital These figures do not show the distribution of the density within the orbital The purpose of

this Further Exploration is to provide a better idea of this distribution

In the text you learned that the wave motion of an electron in an orbital is described by a function yf called a wavefunction It turns out that the electron density at any point in space of an electron in an orbital z is proportional to the square of the wavefunction, y? Consequently, by

plotting the square of the wavefunction as a function of spatial coordinates we can obtain a description of the electron density at any point in space Because we are dealing with a function of

three spatial dimensions, such a plot would require four dimensions, or at least a contour plot in three dimensions A simpler plot that has all the information we need is obtained by plotting the electron density y? as a function of the radial distance r along any straight line originating at the

nucleus Consider, for example, the Is orbital Because the orbital is a sphere, we would get the

same plot in any direction The result for a 1s orbital is the plot in Fig SG1.la (dark line) From this plot you can see that the electron density is greatest at the nucleus and then falls off exponentially There are no nodes—no regions of zero electron density—in this plot

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CHAPTER 1 + CHEMICAL BONDING AND CHEMICAL STRUCTURE

Figure SG1.1 Plots of wavefunctions (2) and their probability distribution functions (rp?) for (a) a 15 orbital; (b) a 2s orbital The

square of the wavefunction (¥”) is scaled so that this quantity can be shown on the same plot The ‘sizes* of the orbitals can be

approximated by the distances at which 74? is a maximum Thus, the 1s orbital has a maximum at 0.529 A; the 2s orbital has a

maximum around 3 A

The plot just described gives the electron density at a point A different view of electron density is obtained if we plot the probability of finding the electron in a spherical shell at some distance r from the nucleus The appropriate function for this probability is :°y7, The reason this makes sense is that the further away we are from the nucleus, the more surface area there is in a spherical shell Another way of thinking about this is to imagine the dimples on two basketballs, one large and one small Assume that the dimples on both are the same size The large basketball, then, has more dimples Then think of each dimple as one probability unit for finding an electron, The larger the sphere (that is, the greater is r), the larger is the probability of finding an electron,

Since the surface area of a sphere increases as the square of the radius, that is, as 7°, then

multiplying the probability at one point Y7 by r? gives the probability of finding the electron in a spherical shell of radius r The gray curve in Fig SG1.1a is a plot of this probability Notice that this product has a maximum at 0.529 A This is called the Bohr radius It is exactly the same as the radial dimension of the electron orbit calculated in an earlier, planetary model of the atom This model said that the electron was in a fixed orbit with a radius of 0.529 A; the quantum model says that the maximum likelihood of finding an electron is on a spherical shell 0.529 A from the nucleus; but there is a finite probability at any distance from the nucleus

A similar plot of electron density for the 2s orbital is given in Fig SG1.1b In the plot of y?

you can clearly see the node—the region of zero electron density—at 1.1 A There is a small

maximum in the spherical-shell electron probability at the Bohr radius, but the largest maximum is

in the “outer ball” of electron density is at about 3 A If you contrast the value of r at which this maximum occurs with that for the Ls orbital you can get a very graphic sense of the different sizes

of the two orbitals

Plots of electron density for the 2p and 3p orbitals are given in Fig SG1.2a and Fig SG1.2b, respectively These plots can be imagined to be taken on a line running through the nucleus and along the axis of the orbital Both positive and negative values of r are shown to stress the lobe character of the orbitals Both orbitals have planar nodes at the nucleus that separate the orbital into two lobes The 3p orbital has, in addition, a spherical node separates the inner and outer regions of each lobe This node is shown in Fig SG1.2b as two dashed lines, which represent the +r and -r values, ie., the radius, of the spherical node The maximum in the spherical-shell electron probability :y7 for the 2p orbital (grey line, Fig SG1.2a) is at about 4 Bohr radii (2.1 A), whereas the largest maximum in the 3p orbital occurs at about 12 Bohr radii (6.3 A) Again, notice

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FURTHER EXPLORATIONS + CHAPTER 1 5

the larger size of the orbital with the larger quantum number

These orbital pictures are derived from the hydrogen atom Orbitals of atoms with higher

atomic number have basically the same shape and nodal characteristics, but their sizes differ

node at 2p orbital the nucleus

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a, SOLUTIONS TO PROBLEMS

Solutions to In-Text Problems

1.1 (a) Because sodium (Na) is in Group 1A, it has one valence electron

(b) The neutral calcium atom has a number of valence electrons equal to its group number, that is, 2

(©) Oxygen has 6 valence electrons; hence, O* has 8

(d) Neutral Br, being in Group 7A, has 7 valence electrons; therefore, Br* has 6

1.2 (a) Because neon has 10 electrons, the negative ion requested in the problem also has 10 The singly charged

negative ion with ten electrons is F-

(b) The positive ion isoelectronic with neon must have 10 electrons and 11 protons, and therefore must have

an atomic number = 11 This is the sodium ion, Na*

(©) The atomic number of Ar is 18; so, the dipositive ion with the same number of electrons must have 20

protons and atomic number = 20 This is the Ca** ion

(d) Because Ne has atomic number = 10 and F has atomic number =

(trichloromethane) ammonia ammonium ion hydronium ion

14 One structure is that of ethanol, and the other is that of dimethyl ether

ic tal

Hope eee

In all but the simplest cases there are many structures that have the same atomic composition,

‘Compounds that have the same atomic composition, but different atomic connectivities, are called constitutional isomers Thus, dimethyl ether and ethanol are constitutional isomers For molecules of moderate size hundreds or even thousands of constitutional isomers are possible

You'll learn about isomers in Chapter 2

15 (a) The overall charge is 0

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SOLUTIONS TO PROBLEMS - CHAPTER 1 7

1.10

‘The C—H bonds are the least polar, because carbon and hydrogen differ very little in electronegativity The

carbon that has the most partial positive character is the one indicated by the asterisk (*), because it is bound to two very electronegative atoms, O and Cl

(a) Formal charge does not give an accurate picture, because N is more electronegative than H; most of the

positive charge is actually on the hydrogens (b) Formal charge does give an accurate picture, because N is more electronegative than H; most of the

negative charge is on the nitrogen

(©) Ananalysis of relative electronegativities would suggest that, because C is slightly more electronegative

than H, a significant amount of the positive charge resides on the hydrogens However, carbon does not have its full complement of valence electrons—that is, it is short of the octet by 2 electrons In fact, both

C and H share the positive charge about equally

In addition to the structure shown on text p 17, five other line-and-wedge depictions of dichloromethane exist All are depictions of the same compound; think of them as the molecule tumbling in space, or as the view of the molecule from different perspectives

(a) The tetrafluoroborate anion is tetrahedral—that is, it has F—B—F bond angles of 109.5°

(b) Water has bent geometry; that is, the HO—H bond angle is approximately tetrahedral Repulsion

between the lone pairs and the bonds reduces this bond angle somewhat (The actual bond angle is 104.5°.)

(©) The formaldehyde molecule has trigonal planar geometry Thus, both the H—C—H bond angle and the

H—C=0 bond angle are about 120°

(@) Inacetonitrile, each of the H—C—H bond angles, as well as the H—C—C bond angle, is tetrahedral

(about 109.5° ) The C—C=N bond angle is linear (180°)

Bond angles: aa, ab, be, bd, ed, de, df, and ef are all about 120°, because all are centered on atoms with trigonal planar geometry; fg is predicted to have the tetrahedral value of 109.5 The bond lengths increase in the order

axg<e<b<d=f<c

(In Chapter 4, you'll learn that C—H bonds attached to carbons of double bonds are shorter than C—H bonds

attached to carbons of single bonds For this reason, a < g.)

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Lil

1.12

1.13

‘A molecule contains dihedral angles only if it contains at least four atoms connected consecutively Compound

A fits this requirement (H, C, N and H atoms bonded consecutively), and thus has dihedral angles Imagine two planes, one containing a C—H bond and another containing a N—H bond The dihedral angle is the angle

between the two planes The model with dihedral angle = 0° is as follow:

angle between the planes = the dihedral angle = 0°

The model with dihedral angle = 60° is as follows

The model with dihedral angle = 180° is identical to the one with dihedral angle = 60°

‘The resonance structures of benzene:

resonance structures of benzene hybrid structure of benzene

Each bond is a single bond in one structure and a double bond in the other On average, each bond has a bond

order of 1.5

(a) Structures of the allyl anion:

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SOLUTIONS TO PROBLEMS - CHAPTER 1 9

[H,c—cH—CH, <—> H,6—CH=CH, |

resonance structures of the allyl anion

Although the structure is shown in a linear representation, the central carbon is trigonal planar, and the carbon skeleton is bent, as application of the VSEPR rules will show

(b) Each CH; carbon contains 0.5 of a negative charge

(©) Ahybrid structure:

& é H,C=CH==CH,

hybrid structure of the allyl anion

Lid Fig SGI

domain is shows a plot of = sin nx for the different values of n Evidently, the number of nodes in this ne less than the “quantum number” n; that is, nodes = n — 1

Orbitals are described by mathematical functions of three dimensions, whereas w= sin nx is a

function of one dimension However, the quantum numbers of orbitals have the same effect as

in this simple equation: they control the number of nodes

1.15 (a) A 3s orbital is three concentric spheres of electron density, each separated by a node, as shown in the

“cutaway” diagram in Fig SG1.4a

(b) A 4s orbital is four concentric spheres of electron density, each separated by a node, as shown in

“cutaway” diagram (b) in Fig SG1.4b

Figure $G1.3 A plot of the function w = sin nx forn = 1, 2, and 3 for the solution to Problem 1.14 The

nodes are given by a gray filled circle (for w = sin 2x) and black filled circles (for w = sin 3x)

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wave trough

wave

nodes

cross-section of a 3s orbital Figure SG1.4a Cross section of a 3s orbital in the solution to Problem 1.15a

wave peaks

wave troughs

nodes cross-section of a 4s orbital Figure SG1.4b Cross section of a 4s orbital in the solution to Problem 1.15b,

1.16

1.17

(a) The oxygen atom (atomic number = 8, therefore 8 electrons): (1s)°(2s)°(2p,)°(2p,)'(2p.)' Two of the 2p

electrons are unpaired; each of these is in a different 2p orbital The valence orbitals are the 2s and 2p orbitals, and the valence electrons are the six electrons that occupy these orbitals

(b) The chloride ion, CI- (atomic number = 17 and one negative charge, therefore 18 electrons): This ion has

the same electronic configuration as argon: (1s)*(2s)*(2p)°(3s)°(3p)° The valence orbitals are the 3s and

3p orbitals, and the valence electrons are the eight electrons that occupy these orbitals

(c) The potassium ion, K*, has the same configuration as Ar, and therefore the same configuration as CI:

(1s)°(2s°(2p)°Bsy°Bp)* The valence orbitals are the 3s and 3p orbitals, and the valence electrons are the

cight electrons that occupy these orbitals

(d) The sodium atom (atomic number = 11) has 11 electrons Therefore its electronic configuration is,

(1s°2s)°(2p)%3s)! The valence orbital is the 3s orbital, and the valence electron is the one electron that occupies this orbital

The pictures and energy levels of the molecular orbitals for parts (a), (b), (c), and (d) are essentially the same as they are for the dihydrogen molecular orbitals in Fig 1.14 of the text

(a) The He,* ion contains three electrons This can be conceived as the combination of a He* ion with a He

atom, By the aufbau principle, two electrons occupy the bonding molecular orbital, and one occupies the antibonding molecular orbital This is shown in the electron-occupancy diagram (a) of Fig SG1.5

Because the bonding molecular orbital contains a greater number of electrons than the antibonding

molecular orbital, this species is stable

(b) The H>” ion contains three electrons This can be conceived as the combination of a hydrogen atom with

a hydride ion (H’) By the aufbau principle, two occupy the bonding molecular orbital, and one occupies the antibonding molecular orbital This is shown in the electron-occupancy diagram (b) of Fig SG1.5 Because the bonding molecular orbital contains a greater number of electrons than the antibonding molecular orbital, this species is stable Notice that in terms of electron occupancy, Hand He * are identical

(©) The H,?" ion can be conceived to result from the combination of two hydride ions (H”) This species

contains four electrons; two occupy the bonding molecular orbital, and two occupy the antibonding

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SOLUTIONS TO PROBLEMS + CHAPTER 111

molecular orbital This is shown in the electron-occupancy diagram (c) of Fig SG1.5 In this species, the

energetic advantage of the electrons in the bonding molecular orbital is cancelled by the energetic

disadvantage of the same number of electrons in the antibonding molecular orbital This species consequently has no energetic advantage over two dissociated hydride (H-) ions, and therefore it readily dissociates

‘The Hy" ion is conceptually derived from the combination of a hydrogen atom with a proton It contains one electron in a bonding molecular orbital This is shown in the electron-occupancy diagram (d) of Figure SG1.5, Because the bonding molecular orbital contains a greater number of electrons than the antibonding molecular orbital, this species ble

Because the H,* ion has one electron in a bonding molecular orbital, it should have about half the stability of Hy

itself relative to dissociated fragments, Thus, 217 kJ mol”! (52 kcal mol”') is an estimate of the bond

dis: ciation energy of this species

Referring to Figure SG1.5d, the H,* ion has | electron in the bonding MO and no electrons in the antibonding

MO Using Equation 1.8 in the text, the H)* ion has a bond order of (1 — 0)/2 = % This number corresponds with the answer to Problem 1.18, which states that the Hy" bond should be half as strong as the H2 bond

(a) If the oxygen of water is sp*-hybridized, then two of the sp’ hybrid orbitals contain unshared electron

pairs The other two sp’ hybrid orbitals contain one electron each; each of these overlaps with the 1s orbital of a hydrogen atom (which contains one electron) to give the two sp*-Ls @ bonds (the O—H bonds) of water

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requirement Either way, the H—O—H angle in water is less than 109.5° (In fact, it is 104.5°.)

Solutions to Additional Problems

Carbon has a sextet and a formal charge of +1

Nitrogen has a complete octet and a formal charge of 0

Carbon has a complete octet and a formal charge of —1

Boron has a sextet and a formal charge of 0

Iodine has a sextet and a formal charge of +1

Boron has an octet and a formal charge of 1

overall charge: ~1 overall charge = 0 overall charge = 0

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SOLUTIONS TO PROBLEMS + CHAPTER 1_13

(a) — Chlorine atom (atomic number = 17; therefore 17 electrons): (1s)*(2s)*(2p)%(3s)°3p,7°3p,)°3p)!

(b) Silicon atom (atomic number = 14; therefore 14 electrons): (1)°(2s)*(2p)°(3s)"(3p,)'(3p,)! or

(18)°(2s)°(2p)°3sY°3p) The valence electrons in silicon are the two 3s and the two 3p electrons, and

the valence orbitals are the 3s and the three 3p orbitals

(c) Argon atom (atomic number 18; therefore 18 electrons This has the same configuration as chloride

ion—that is, (1s)°(2s)*(2p)°(3s)°3p) (d) Magnesium atom (atomic number = 12; therefore 12 electrons): (1s)°(2s)*(2p)°3:

The 2d orbital is not permitted The reason: d means that the / quantum number = 2; the value of the / quantum number cannot exceed one less than the value of the principal quantum number n, which in this case is 2 Thus,

I cannot exceed | in principal quantum level 2

(a) The carbon is bound to three “groups,” the two Hs and an electron pair Hence, this carbon is trigonal

planar (if you count the electron pair) and the molecule is “bent” with an H—C—H angle of about 120° (or perhaps somewhat less because of the VSEPR rule that unshared pairs reduce bond angles)

(b) The beryllium (Be) bears two groups, and therefore has linear geometry The H—Be—H bond angle is

(f) The terminal carbons are bound to three groups, and are therefore trigonal planar The central carbon is

bound to two groups, and is therefore linear Hence, the H—C—H and H—C=C bond angles are 120°; the C=C=C bond angle is 180°

(g) The nitrogen is trigonal planar; the ON—O bond angle is 120°, as are both of the C—N—O bond

angles, The carbon is bound to four groups, and is tetrahedral; all bond angles centered on carbon are about 109.5°

c<a<d<b=fch<e<g

Notice that the rules on text pp 14-15 are in order of importance Thus, the row of the periodic table is

paramount in determining bond length Thus, bond b is between two period-2 elements (C and C), whereas

bond fis between a period-1 and a period-3 element The “average period” for both is 2 In fact, the two bonds

have about the same length

(a) The oxygen of the hydronium ion has approximately tetrahedral geometry and is therefore sp’-hybrid-

ized, Two of the sp* hybrid orbitals contain an unshared electron pair One of these becomes the lone pair

in H,O*; the other overlaps with a proton (H*) to give one of the O—H bonds The other two sp* hybrid

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bond (bond order = 1) in one resonance structure and a double bond (bond order = 2) in the other, the

average bond order is (1 + 2)/2 = 1.5 In other words, each carbon-carbon bond is equivalent to a single bond plus “half” of a double bond

By analogous reasoning, each of the terminal carbons bears +1 charge in one structure and 0 charge in

the other Hence, the charge on each carbon of the ion is the average of these two, or (+1 + 0)/2 = +0.5

The charge on the central carbon is zero in all resonance structures and therefore zero overall

‘The resonance structure on the right violates the octet rule for nitrogen

an octet This point will be considered further in Chapter 3

VW Notice from this example that a positive charge does not necessarily mean that an atom lacks

1.32 We use reasoning analogous to that used in the previous problem This ion is a hybrid of three equally important

structures Hence, each property will be the sum of that property in each structure divided by 3

Negative charge on each oxygen: (-1 + 0 +-1)/3 =-2/3

Bond order of each carbon-oxygen bond: (1 +2 + 1)/

4/3 or 1.33

Figure 1.8 on text p 25 shows that a 2s orbital has a single spherical node: and Figure 1.10 on text p 26

shows that the 2p orbital has one planar node Figure 1.11 shows that the 3p orbital has two nodes—one

planar node and one spherical node In both cases, the total number of nodes is one less than the principal quantum number In the 2s orbital, the value of /is 0, and the number of planar nodes is zero In the 2p and 3p orbitals, the value of / is 1, and each orbital has one planar node

The total number of nodes is n — 1, or 4, The value of / (0, which implies an s orbital) gives the number

of planar nodes, which is 0 Hence, the Ss orbital has four spherical nodes By analogous reasoning, a 3d

orbital has two nodes, both of which are planar; hence, it has no spherical nodes (Notice this property of the 3d orbital in the solution to the following problem.)

the x-z plane is anode

wave trough | wave peak

1.35 A 4p orbital is somewhat like a 3p orbital (Fig 1.11 on text p 27), except that it has one additional spherical

node and one additional “mushroom cap” region of electron density A cross section—that is, a planar slice along the axis of the orbital— is shown in Fig SG1.6

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Figure $G1.6 A cross section of a 4p orbital showing the nodes to accompany the solution to Problem 1.35

The actual shape of the orbital is the volume of revolution about the x-axis—that is, the volume "swept out”

when the figure is rotated around the x-axis

According to Eq 1.4 (p 11) in the text, the dipole moment is determined by two factors The first is the

separated charge, which should be greater for the compound containing the more electronegative atom, that is, for H;C—F However, the second factor is the bond length r, which is greater for the compound containing the

larger atom, that is, for H}C—Cl Evidently, these two opposing factors just about compensate in their effects

on the dipole moments of the two compounds

(a) The bond dipole for dimethylmagnesium should indicate that C is at the negative end of the C—Mg

bond, because carbon is more electronegative than magnesium,

++

H,C —Mg—CH;

(b) The two CH; groups should be 180° from each other VSEPR theory states that bonds should be oriented

‘as far apart as possible

(c) _ The two vectors indicating the bond dipoles of the C—Mg bonds are pointing in equal and opposite

directions, cancelling each other, for an overall dipole moment of 0 for dimethylmagnesium

No matter how any CH, group is turned, the resultant bond dipole is the same:

Hence, ethylenes with the two different dihedral angles should both have zero dipole moment because the

resultants of the two CH2 groups cancel each other in either arrangement The observation of zero dipole

moment does not permit a choice between these two dihedral angles Note: Actually, ethylene is “locked” in the planar orientation under normal conditions, as will be discussed in Chapter 4

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1.39

1.40

141

1.42

(a) Because the carbon has trigonal-planar geometry, the HCO bond angle is 120°

(b) The two structures are as follows:

fe}

~o 0O-C—0/C—-0—-C by,

dihedral angle = 0°

0—C—0/C—0—C dihedral angle = 180°

A linear water molecule would have zero dipole moment, because the O—H bond dipoles would be oriented in opposite directions and would cancel Hence, the nonzero dipole moment for water shows that it is a bent

molecule Out with Professor Szents!

To solve this problem, we first have to know the geometry of CClj This should be easy: it’s tetrahedral, because the carbon is bonded to four identical groups Now, as suggested in the hint, let’s do a vector addition for the bond dipoles of each C—C—CI pair (Use a dipole arrow of arbitrary length as long as it is the same for both We'll use an arrow with the length of the C—CI bonds.) Here’s the result of the vector addition, which

we've done graphically, for the first pair Notice that, because the C—Cl bonds are identical, the resultant must

bisect the angle between the two bonds

Now let’s do the resultant for the second pair Because the C—CI bonds are identical to those of the first pair, it

follows that the resultant is the same length, and it again bisects the C-—C—CI bond angle (We've rotated the

molecule about an axis along the dipole so we are viewing the other two chlorines from the perspective we used

to view the first two—in the plane of the page.)

(a) When the OH bonds lie at a dihedral angle of 180° their bond dipoles cancel; consequently, a

hydrogen peroxide molecule in this conformation would have a dipole moment of zero This conformation is ruled out by the observation of a significant dipole moment

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Note that 96.5° is not the dihedral angle, but the bond angle The idea is to calculate the dihedral

angle From the analysis in the previous paragraph, we know that the resultant dipole moment of the molecule (given as 2.13 D) is derived from vector addition of the y components of the two O—H bond dipoles, which we now know are 1.51 D All we have to do is determine the angle @ that will give 2.13 D

as the resultant from vector addition of two 1.51 D dipoles

angle of 93.8°!) Thus, the conformation of hydrogen peroxide is best represented as follows:

dihedral angle ~ 90°

1.43 This problem requires a construction similar to the one used in Problem 1.42 in which the resultant is known

and the individual O—H bond dipoles are unknown

angle = 180° - 6 1.84D

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Applying the law of cosines, with x as the O—H bond dipoles,

the calculated value of the O—H bond dipole, which is very similar to the value assumed in Problem

1.44 Dilithium, Lip, is derived from the combination of two lithium atoms It contains two bonding electrons derived

from two 2s valence electrons, one from each lithium atom, Each Li atom also has two Is electrons, but these are not valence electrons and therefore are not involved in bond formation and are not shown in the interaction diagram (Fig SG1.7) The interaction of the two 2s orbitals gives a bonding and an antibonding MO The nodes

present in the individual 2s orbitals are retained in the MOs This is a stable species because the bonding molecular orbital contains a greater number of electrons than the antibonding molecular orbital

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accompany Problem 1.45 The small spheres are nuclei, and the vertical gray lines are nodal planes viewed

end-on,

145 (a) The bonding molecular orbital is derived by the constructive overlap of wave peaks The antibonding

molecular orbital is derived by changing the peak to a trough and the trough to a peak in one of the two 2p orbitals The destructive overlap of a peak with a trough gives a node The resulting molecular orbitals are shown in Fig SG1.8

(b) The nodes are shown in Fig SG1.8 Both of the MOs retain the nodes of the original 2p orbitals The

antibonding MO has an additional node that results from the destructive overlap of a peak and a trough (©) Figure SG1.8 is the required interaction diagram

(d) The resulting bond is a o bond because it is cylindrically symmetrical about the orbital axis

1.46 (a) The bonding molecular orbital is derived by the constructive “side-to-side” overlap of peaks with peaks

and troughs with troughs The antibonding molecular orbital is derived by changing the peak to a trough and the trough to a peak in one of the two 2p orbitals Destructive overlap of troughs with peaks and peaks with troughs gives an additional node in the antibonding orbital These molecular orbitals are shown in Fig SGI

(b) The nodes are also shown in Fig SG1.9 Both of the MOs have the nodes of the original 2p orbitals,

which merge into a single node because the nodal plane is common to both orbitals The antibonding MO has in addition a node between the original orbitals that results from the destructive overlap of a peak and

a trough

(c) Fig SG1.9 is the required interaction diagram,

(d) The resulting bond is not ao bond because it is not cylindrically symmetrical about the internuclear axis

This type of bond, called a pi (77) bond, is important in the carbon-carbon double bond Pi bonds are discussed in Chapter 4

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2p, orbitals as shown in part (c), except that they are oriented on the z-axis, rotated 90° from the y-axis

The atomic orbitals of oxygen are combined according to parts (a-d) of this problem and are ordered in increasing energy as shown in Figure SG1.10e Each oxygen atom has six valence electrons, two in the 2s orbital, and four in the three 2p orbitals These electrons are placed in molecular orbitals according to

Hund’s rules

The oxygen molecule has a net (or, nonzero) spin, and is magnetic, An unpaired electron generates a magnetic field due to its spin The two unpaired electrons in the 2pz,** and 2px.* orbitals do not have electrons of opposing spin to cancel them out, so the oxygen molecule is “paramagnetic” and liquid

oxygen can be trapped between poles of a magnet

We eliminate structure A because it does not have unpaired electrons, We also eliminate structure D because it has more than 12 total electrons According to Eq 1.8 (text p 35), oxygen has a bond order of (8 ~4)/2 = 2 However, none of the remaining structures contains a double bond Yet we can use an

“average structure”, or resonance hybrid, of the molecule containing the single bond and the one

containing a triple bond This average would have a bond order = 2

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SOLUTIONS TO PROBLEMS - CHAPTER 1

O) is about the same as that of a conventional sing/e bond, but the bond length is about that of a

is dichotomy requires the use of resonance structures if we insist on

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Figure SG1.10b An orbital interaction diagram for the head-to-head interaction of two 2p, orbitals for the

solution to Problem 1.47b The small spheres are nuclei, and the vertical gray lines are nodal planes viewed

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SOLUTIONS TO PROBLEMS + CHAPTER 1 23

Figure SG1.10d An orbital interaction diagram for the side-to-side interaction of two 2p: orbitals for the

solution to Problem 1.47d The small spheres are nuclei, and nodal planes are the plane of the page for all

orbitals, with an additional plane perpendicular to the page (in the y/z plane) for the 2pm.* orbital

2s orbital 2s orbital

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