Solutions Manual For Organic Chemistry 9th Edition By John McMurry by digitaldownload87

Preface What enters your mind when you hear the words “organic chemistry?” Some of you may think, “the chemistry of life,” or “the chemistry of carbon.” Other responses might include “premed, “pressure,” “difficult,” or “memorization.” Although formally the study of the compounds of carbon, the discipline of organic chemistry encompasses many skills that are common to other areas of study. Organic chemistry is as much a liberal art as a science, and mastery of the concepts and techniques of organic chemistry can lead to improved competence in other fields. As you work on the problems that accompany the text, you will bring to the task many problem-solving techniques. For example, planning an organic synthesis requires the skills of a chess player; you must plan your moves while looking several steps ahead, and you must keep your plan flexible. Structure-determination problems are like detective problems, in which many clues must be assembled to yield the most likely solution. Naming organic compounds is similar to the systematic naming of biological specimens; in both cases, a set of rules must be learned and then applied to the specimen or compound under study. The problems in the text fall into two categories: drill and complex. Drill problems, which appear throughout the text and at the end of each chapter, test your knowledge of one fact or technique at a time. You may need to rely on memorization to solve these problems, which you should work on first. More complicated problems require you to recall facts from several parts of the text and then use one or more of the problem-solving techniques mentioned above. As each major type of problem—synthesis, nomenclature, or structure determination—is introduced in the text, a solution is extensively worked out in this Solutions Manual. Here are several suggestions that may help you with problem solving: 1.

The text is organized into chapters that describe individual functional groups. As you study each functional group, make sure that you understand the structure and reactivity of that group. In case your memory of a specific reaction fails you, you can rely on your general knowledge of functional groups for help.

Use molecular models. It is difficult to visualize the three-dimensional structure of an organic molecule when looking at a two-dimensional drawing. Models will help you to appreciate the structural aspects of organic chemistry and are indispensable tools for understanding stereochemistry.

Every effort has been made to make this Solutions Manual as clear, attractive, and error-free as possible. Nevertheless, you should use the Solutions Manual in moderation. The principal use of this book should be to check answers to problems you have already worked out. The Solutions Manual should not be used as a substitute for effort; at times, struggling with a problem is the only way to teach yourself.

Look through the appendices at the end of the Solutions Manual. Some of these appendices contain tables that may help you in working problems; others present information related to the history of organic chemistry.

Preface

Although the Solutions Manual is written to accompany Organic Chemistry, it contains several unique features. Each chapter of the Solutions Manual begins with an outline of the text that can be used for a concise review of the text material and can also serve as a reference. After every few chapters a Review Unit has been inserted. In most cases, the chapters covered in the Review Units are related to each other, and the units are planned to appear at approximately the place in the textbook where a test might be given. Each unit lists the vocabulary for the chapters covered, the skills needed to solve problems, and several important points that might need reinforcing or that restate material in the text from a slightly different point of view. Finally, the small self-test that has been included allows you to test yourself on the material from more than one chapter. I have tried to include many types of study aids in this Solutions Manual. Nevertheless, this book can only serve as an adjunct to the larger and more complete textbook. If Organic Chemistry is the guidebook to your study of organic chemistry, then the Solutions Manual is the roadmap that shows you how to find what you need. Susan McMurry

Acknowledgments Cengage Learning would like to acknowledge KC Russell of Northern Kentucky University and James Vyvyan of Western Washington University for providing solutions to the new mechanism and spectroscopy problems. We would also like to thank Paul Adams of the University of Arkansas for his review work.

Chapter 1 – Structure and Bonding

Chapter Outline I. Atomic Structure (Sections 1.1–1.3). A. Introduction to atomic structure (Section 1.1). 1. An atom consists of a dense, positively charged nucleus surrounded by negatively charged electrons. a. The nucleus is made up of positively charged protons and uncharged neutrons. b. The nucleus contains most of the mass of the atom. c. Electrons move about the nucleus at a distance of about 2 u 10 10 m (200 pm). 2. The atomic number (Z) gives the number of protons in the nucleus. 3. The mass number (A) gives the total number of protons and neutrons. 4. All atoms of a given element have the same value of Z. a. Atoms of a given element can have different values of A. b. Atoms of the same element with different values of A are called isotopes. B. Orbitals (Section 1.2). 1. The distribution of electrons in an atom can be described by a wave equation. a. The solution to a wave equation is an orbital, represented by Ȍ. b. Ȍ 2 predicts the volume of space in which an electron is likely to be found. 2. There are four different kinds of orbitals (s, p, d, f). a. The s orbitals are spherical. b. The p orbitals are dumbbell-shaped. c. Four of the five d orbitals are cloverleaf-shaped. 3. An atom’s electrons are organized into electron shells. a. The shells differ in the numbers and kinds of orbitals they contain. b. Electrons in different orbitals have different energies. c. Each orbital can hold up to a maximum of two electrons. 4. The two lowest-energy electrons are in the 1s orbital. a. The 2s orbital is the next higher in energy. b. The next three orbitals are 2px, 2py and 2pz, which have the same energy. i. Each p orbital has a region of zero density, called a node. c. The lobes of a p orbital have opposite algebraic signs. C. Electron Configuration (Section 1.3). 1. The ground-state electron configuration of an atom is a listing of the orbitals occupied by the electrons of the atom in the lowest energy configuration. 2. Rules for predicting the ground-state electron configuration of an atom: a. Orbitals with the lowest energy levels are filled first. i. The order of filling is 1s, 2s, 2p, 3s, 3p, 4s, 3d. b. Only two electrons can occupy each orbital, and they must be of opposite spin. c. If two or more orbitals have the same energy, one electron occupies each until all are half-full (Hund’s rule). Only then does a second electron occupy one of the orbitals. i. All of the electrons in half-filled shells have the same spin.

Chapter 1

II. Chemical Bonding Theory (Sections 1.4–1.5). A. Development of chemical bonding theory (Section 1.4). 1. Kekulé and Couper proposed that carbon has four “affinity units”; carbon is tetravalent. 2. Kekulé suggested that carbon can form rings and chains. 3. Van’t Hoff and Le Bel proposed that the 4 atoms to which carbon forms bonds sit at the corners of a regular tetrahedron. 4. In a drawing of a tetrahedral carbon, a wedged line represents a bond pointing toward the viewer, a dashed line points behind the plane of the page, and a solid line lies in the plane of the page. B. Covalent bonds. 1. Atoms bond together because the resulting compound is more stable than the individual atoms. a. Atoms tend to achieve the electron configuration of the nearest noble gas. b. Atoms in groups 1A, 2A and 7A either lose electrons or gain electrons to form ionic compounds. c. Atoms in the middle of the periodic table share electrons by forming covalent bonds. d. The neutral collection of atoms held together by covalent bonds is a molecule. 2. Covalent bonds can be represented two ways. a. In electron-dot structures, bonds are represented as pairs of dots. b. In line-bond structures, bonds are represented as lines drawn between two bonded atoms. 3. The number of covalent bonds formed by an atom depends on the number of electrons it has and on the number it needs to achieve an octet. 4. Valence electrons not used for bonding are called lone-pair (nonbonding) electrons. a. Lone-pair electrons are often represented as dots. C. Valence bond theory (Section 1.5). 1. Covalent bonds are formed by the overlap of two atomic orbitals, each of which contains one electron. The two electrons have opposite spins. 2. Bonds formed by the head-on overlap of two atomic orbitals are cylindrically symmetrical and are called ı bonds. 3. Bond strength is the measure of the amount of energy needed to break a bond. 4. Bond length is the optimum distance between nuclei. 5. Every bond has a characteristic bond length and bond strength. III. Hybridization (Sections 1.6–1.10). A. sp3 Orbitals (Sections 1.6, 1.7). 1. Structure of methane (Section 1.6). a. When carbon forms 4 bonds with hydrogen, one 2s orbital and three 2p orbitals combine to form four equivalent atomic orbitals (sp3 hybrid orbitals). b. These orbitals are tetrahedrally oriented. c. Because these orbitals are unsymmetrical, they can form stronger bonds than unhybridized orbitals can.

Structure and Bonding

d. These bonds have a specific geometry and a bond angle of 109.5°. 2. Structure of ethane (Section 1.7). a. Ethane has the same type of hybridization as occurs in methane. b. The C–C bond is formed by overlap of two sp3 orbitals. c. Bond lengths, strengths and angles are very close to those of methane. 2 B. sp Orbitals (Section 1.8). 1. If one carbon 2s orbital combines with two carbon 2p orbitals, three hybrid sp2 orbitals are formed, and one p orbital remains unchanged. 2. The three sp2 orbitals lie in a plane at angles of 120°, and the unhybridized p orbital is perpendicular to them. 3. Two different types of bonds form between two carbons. a. A ı bond forms from the overlap of two sp2 orbitals. b. A ʌ bond forms by sideways overlap of two p orbitals. c. This combination is known as a carbon–carbon double bond. 4. Ethylene is composed of a carbon–carbon double bond and four ı bonds formed between the remaining four sp2 orbitals of carbon and the 1s orbitals of hydrogen. a. The double bond of ethylene is both shorter and stronger than the C–C bond of ethane. C. sp Orbitals (Section 1.10). 1. If one carbon 2s orbital combines with one carbon 2p orbital, two hybrid sp orbitals are formed, and two p orbitals are unchanged. 2. The two sp orbitals are 180° apart, and the two p orbitals are perpendicular to them and to each other. 3. Two different types of bonds form. a. A ı bond forms from the overlap of two sp orbitals. b. Two ʌ bonds form by sideways overlap of four unhybridized p orbitals. c. This combination is known as a carbon–carbon triple bond. 4. Acetylene is composed of a carbon–carbon triple bond and two ı bonds formed between the remaining two sp orbitals of carbon and the 1s orbitals of hydrogen. a. The triple bond of acetylene is the strongest carbon–carbon bond. D. Hybridization of nitrogen and oxygen (Section 1.10). 1. Covalent bonds between other elements can be described by using hybrid orbitals. 2. Both the nitrogen atom in ammonia and the oxygen atom in water form3 sp hybrid orbitals. a. The lone-pair electrons in these compounds occupy sp3 orbitals. 3. The bond angles between hydrogen and the central atom is often less than 109° because the lone-pair electrons take up more room than the ı bond . 4. Because of their positions in the third row, phosphorus and sulfur can form more than the typical number of covalent bonds. IV. Molecular orbital theory (Section 1.11). A. Molecular orbitals arise from a mathematical combination of atomic orbitals and belong to the entire molecule. 1. Two 1s orbitals can combine in two different ways.

Chapter 1

a. The additive combination is a bonding MO and is lower in energy than the two hydrogen 1s atomic orbitals. b. The subtractive combination is an antibonding MO and is higher in energy than the two hydrogen 1s atomic orbitals. 2. Two p orbitals in ethylene can combine to form two ʌ MOs. a. The bonding MO has no node; the antibonding MO has one node. 3. A node is a region between nuclei where electrons aren’t found. a. If a node occurs between two nuclei, the nuclei repel each other. V. Chemical structures (Section 1.12). A. Drawing chemical structures. 1. Condensed structures don’t show C–H bonds and don’t show the bonds between CH3, CH2 and CH units. 2. Skeletal structures are simpler still. a. Carbon atoms aren’t usually shown. b. Hydrogen atoms bonded to carbon aren’t usually shown. c. Other atoms (O, N, Cl, etc.) are shown.

Structure and Bonding

Solutions to Problems 1.1

(a)

To find the ground-state electron configuration of an element, first locate its atomic number. For oxygen, the atomic number is 8; oxygen thus has 8 protons and 8 electrons. Next, assign the electrons to the proper energy levels, starting with the lowest level. Fill each level completely before assigning electrons to a higher energy level.

Notice that the 2p electrons are in different orbitals. According to Hund’s rule, we must place one electron into each orbital of the same energy level until all orbitals are half-filled. Remember that only two electrons can occupy the same orbital, and that they must be of opposite spin. A different way to represent the ground-state electron configuration is to simply write down the occupied orbitals and to indicate the number of electrons in each orbital. For example, the electron configuration for oxygen is 1s2 2s2 2p4.

(b)

Nitrogen, with an atomic number of 7, has 7 electrons. Assigning these to energy levels: The more concise way to represent ground-state electron configuration for nitrogen: 1s2 2s2 2p3

(c)

Sulfur has 16 electrons.

1s2 2s2 2p6 3s6 3p4

1.2

Chapter 1

The elements of the periodic table are organized into groups that are based on the number of outer-shell electrons each element has. For example, an element in group 1A has one outershell electron, and an element in group 5A has five outer-shell electrons. To find the number of outer-shell electrons for a given element, use the periodic table to locate its group. (a)

Magnesium (group 2A) has two electrons in its outermost shell.

(b)

Cobalt is a transition metal, which has two electrons in the 4s subshell, plus seven electrons in its 3d subshell.

(c)

Selenium (group 6A) has six electrons in its outermost shell.

1.3 A solid line represents a bond lying in the plane of the page, a wedged bond represents a bond pointing out of the plane of the page toward the viewer, and a dashed bond represents a bond pointing behind the plane of the page.

1.4

1.5 Identify the group of the central element to predict the number of covalent bonds the element can form. (a)

Carbon (Group 4A) has four electrons in its valence shell and forms four bonds to achieve the noble-gas configuration of neon. A likely formula is CCl4.

(b) (c) (d) (e)

Element

Group

Likely Formula

Al C Si N

3A 4A 4A 5A

AlH3 CH2Cl2 SiF4 CH3NH2

Structure and Bonding

1.6

Start by drawing the electron-dot structure of the molecule. (1)

Determine the number of valence, or outer-shell electrons for each atom in the molecule. For chloroform, we know that carbon has four valence electrons, hydrogen has one valence electron, and each chlorine has seven valence electrons.

(2)

Next, use two electrons for each single bond.

(3)

Finally, use the remaining electrons to achieve a noble gas configuration for all atoms. For a line-bond structure, replace the electron dots between two atoms with a line. Molecule

Electron-dot structure

Line-bond structure

(a)

(b)

(c)

(d) 1.7

Each of the two carbons has 4 valence electrons. Two electrons are used to form the carbon–carbon bond, and the 6 electrons that remain can form bonds with a maximum of 6 hydrogens. Thus, the formula C2H7 is not possible.

1.8

Chapter 1

Connect the carbons and add hydrogens so that all carbons are bonded to four different atoms.

The geometry around all carbon atoms is tetrahedral, and all bond angles are approximately 109°. 1.9

1.10

The C3–H bonds are ı bonds formed by overlap of an sp3 orbital of carbon 3 with an s orbital of hydrogen. The C2–H and C1–H bonds are ı bonds formed by overlap of an sp2 orbital of carbon with an s orbital of hydrogen. The C2–C3 bond is a ı bond formed by overlap of an sp3 orbital of carbon 3 with an sp orbital of carbon 2. 2

There are two C1–C2 bonds. One is a ı bond formed by overlap of an sp3 orbital of carbon 1 with an sp2 orbital of carbon 2. The other is a ʌ bond formed by overlap of a p orbital of carbon 1 with a p orbital of carbon 2. All four atoms connected to the carbon– carbon double bond lie in the same plane, and all bond angles between these atoms are 120°. The bond angle between hydrogen and the sp3-hybridized carbon is 109°. 1.11 All atoms lie in the same plane, and all bond angles are approximately 120°.

Structure and Bonding

1.12

Aspirin. 2

All carbons are sp hybridized, with the exception of the indicated carbon. All oxygen atoms have two lone pairs of electrons.

1.13

The C3-H bonds are ı bonds formed by overlap of an sp3orbital of carbon 3 with an s orbital of hydrogen. The C1-H bond is a ı bond formed by overlap of an sp orbital of carbon 1 with an s orbital of hydrogen. The C2-C3 bond is a ı bond formed by overlap of an sp orbital of carbon 2 with an sp3 orbital of carbon 3. There are three C1-C2 bonds. One is a ı bond formed by overlap of an sp orbital of carbon 1 with an sp orbital of carbon 2. The other two bonds are ʌ bonds formed by overlap of two p orbitals of carbon 1 with two p orbitals of carbon 2. The three carbon atoms of propyne lie in a straight line: the bond angle is 180°. The H–C1ŁC2 bond angle is also 180°. The bond angle between hydrogen and the sp3hybridized carbon is 109°. 1.14

(a)

The sp3-hybridized oxygen atom has tetrahedral geometry.

(b)

Tetrahedral geometry at nitrogen and carbon.

(c)

Like nitrogen, phosphorus has five outer-shell electrons. PH3 has tetrahedral geometry.

Chapter 1

(d)

The sp3-hybridized sulfur atom has tetrahedral geometry.

1.15 Remember that the end of a line represents a carbon atom with 3 hydrogens, a two-way

intersection represents a carbon atom with 2 hydrogens, a three-way intersection represents a carbon with 1 hydrogen and a four-way intersection represents a carbon with no hydrogens. (a)

(b)

1.16 Several possible skeletal structures can satisfy each molecular formula. (a)

(b) (c)

(d)

Structure and Bonding

1.17

Visualizing Chemistry 1.18 (a)

(b)

1.19 Citric acid (C6H8O7) contains seven oxygen atoms, each of which has two electron lone pairs. Three of the oxygens form double bonds with carbon.

Chapter 1

1.20

All carbons are sp2 hybridized, except for the carbon indicated as sp3. The two oxygen atoms and the nitrogen atom have lone pair electrons, as shown. 1.21

Additional Problems Electron Configuration 1.22

(a) (b) (c) (d)

Element

Atomic Number

Number of valence electrons

Zinc Iodine Silicon Iron

30 53 14 26

2 7 4 2 (in 4s nutshell),6 (in 3d subshell)

Element

Atomic Number

Ground-state electron configuration

Potassium Arsenic Aluminum Germanium

19 33 13 32

1s2 2s2 2p6 3s2 3p6 4s1 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p3 1s2 2s2 2p6 3s2 3p1 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p2

1.23

(a) (b) (c) (d)

Structure and Bonding

Electron-Dot and Line-Bond Structures 1.24 (a)

NH2OH

1.25 (a)

The 4 valence electrons of carbon can form bonds with a maximum of 4 hydrogens. Thus, it is not possible for the compound CH5 to exist.

(b) AlCl3

(d) CH2O

(b)

If you try to draw a molecule with the formula C2H6N, you will see that it I impossible for both carbons and nitrogen to have a complete octet of electrons. Therefore, C2H6N is unlikely to exist.

(c)

A compound with the formula C3H5Br2 doesn’t have filled outer shells for all atoms and is thus unlikely to exist.

1.26 Acetonitrile

In the compound acetonitrile, nitrogen has eight electrons in its outer electron shell. Six are used in the carbon-nitrogen triple bond, and two are a nonbonding electron pair. 1.27

Vinyl chloride Vinyl chloride has 18 valence electrons. Eight electrons are used for 4 single bonds, 4 electrons are used in the carbon–carbon double bond, and 6 electrons are in the 3 lone pairs that surround chlorine. 1.28 (a)

(b)

(c)

1.29 In molecular formulas of organic molecules, carbon is listed first, followed by hydrogen. All other elements are listed in alphabetical order. Compound (a) Aspirin (b) Vitamin C (c) Nicotine (d) Glucose

Molecular Formula C9H8O4 C6H8O6 C10H14N2 C6H12O6

Chapter 1

1.30 To work a problem of this sort, you must draw all possible structures consistent with the rules of valence. You must systematically consider all possible attachments, including those that have branches, rings and multiple bonds. (a)

(b)

(c)

(d)

(e)

(f)

1.31

1.32

Structure and Bonding

1.33 (a)

(b)

(c)

(d)

1.34

Hybridization 1.35 The H3C– carbon is sp3 hybridized, and the –CN carbon is sp hybridized. 1.36 (a)

(b)

(c)

(d)

1.37

Benzene

All carbon atoms of benzene are sp2 hybridized, and all bond angles of benzene are 120°. Benzene is a planar molecule. 1.38 (a)

(b)

(c)

Chapter 1

1.39 Examples: (a)

CH3CH2CHőCH2

(b) H2CőCHņCHőCH2

1.40 (a)

(c)

H2CőCHņCŁCH

(b)

1.41

The bond angles formed by atoms having sp3 hybridization are approximately 109°. The bond angles formed by atoms having sp2 hybridization are approximately 120°. Skeletal Structures 1.42 (a)

(b)

(c)

(d)

Structure and Bonding

1.43 (a)

(b)

(c)

1.44

1.45

Chapter 1

General Problems 1.46 In a compound containing a carbon–carbon triple bond, atoms bonded to the sp-hybridized carbons must lie in a straight line. It is not possible to form a five-membered ring if four carbons must have a linear relationship. 1.47

The central carbon of allene forms two ı bonds and two ʌ bonds. The central carbon is sp-hybridized, and the two terminal carbons are sp2-hybridized. The bond angle formed by the three carbons is 180°, indicating linear geometry for the carbons of allene. 1.48

Carbon dioxide is a linear molecule. 1.49

All of the indicated atoms are sp2-hybridized.

Structure and Bonding

1.50 (a)

The positively charged carbon atom is surrounded by six valence electrons; carbon has three valence electrons, and each hydrogen brings three valence electrons.

(b)

The positively charged carbon is sp2-hybridized.

(c)

A carbocation is planar about the positively charged carbon.

(a)

A carbanion is isoelectronic with (has the same number of electrons as) a trivalent nitrogen compound.

(b)

The negatively charged carbanion carbon has eight valence electrons.

(c)

The carbon atom is sp3-hybridized.

(d)

A carbanion is tetrahedral.

1.51

1.52 According to the Pauli Exclusion Principle, two electrons in the same orbital must have opposite spins. Thus, the two electrons of triplet (spin-unpaired) methylene must occupy different orbitals. In triplet methylene, sp-hybridized carbon forms one bond to each of two hydrogens. Each of the two unpaired electrons occupies a p orbital. In singlet (spin-paired) methylene the two electrons can occupy the same orbital because they have opposite spins. Including the two C–H bonds, there are a total of three occupied orbitals. We predict sp2 hybridization and planar geometry for singlet methylene.

1.53

The two compounds differ in the way that the carbon atoms are connected.

Chapter 1

1.54

One compound has a double bond, and one has a ring. 1.55 CH3CH2OH

CH3OCH3

The two compounds differ in the location of the oxygen atom. 1.56

The compounds differ in the way that the carbon atoms are connected and in the location of the double bond. 1.57

(a), (b)

(c)

Compound

sp3-Hybridized carbons

sp2-Hybridized carbons

Ibuprofen Naproxen Acetaminophen

6 3 1

7 11 7

Each of the structures has a six-membered ring containing three double bonds, each has a methyl group, and each has a C=O group.

Chapter 2 – Polar Covalent Bonds; Acids and Bases

Chapter Outline I. Polar covalent bonds (Sections 2.1–2.3). A. Electronegativity (Section 2.1). 1. Although some bonds are totally ionic and some are totally covalent, most chemical bonds are polar covalent bonds. a. In these bonds, electrons are attracted to one atom more than to the other atom. 2. Bond polarity is due to differences in electronegativity (EN). a. Elements on the right side of the periodic table are more electronegative than elements on the left side. b. Carbon has an EN of 2.5. c. Elements with EN > 2.5 are more electronegative than carbon. d. Elements with EN < 2.5 are less electronegative than carbon. 3. The difference in EN between two elements can be used to predict the polarity of a bond. a. If ǻǼȃ < 0.4, a bond is nonpolar covalent. b. If ǻǼȃ is between 0.4 and 2.0, a bond is polar covalent. c. If ǻǼȃ > 2.0, a bond is ionic. d. The symbols ǻ+ and ǻ– are used to indicate partial charges. e. A crossed arrow is used to indicate bond polarity. i. The tail of the arrow is electron-poor, and the head of the arrow is electronrich. 4. Electrostatic potential maps are also used to show electron-rich (red) and electronpoor (blue) regions of molecules. 5. An inductive effect is an atom’s ability to polarize a bond. B. Dipole moment (Section 2.2). 1. Dipole moment is the measure of a molecule’s overall polarity. 2. Dipole moment (ȝ) = Q x r, where Q = charge and r = distance between charges. a. Dipole moment is measured in debyes (D). 3. Dipole moment can be used to measure charge separation. 4. Water and ammonia have large values of D; methane and ethane have D = 0. C. Formal charge (Section 2.3). 1. Formal charge (FC) indicates electron “ownership” in a molecule. ª # of valence º ª # of bonding electrons º ª # nonbonding º 2. FC « » « »¼ « electrons » 2 ¬ electrons ¼ ¬ ¬ ¼ II. Resonance (Sections 2.4–2.6). A. Chemical structures and resonance (Section 2.4). 1. Some molecules (acetate ion, for example) can be drawn as two (or more) different electron-dot structures. a. These structures are called resonance structures.

Chapter 2

b. The true structure of the molecule is intermediate between the resonance structures. c. The true structure is called a resonance hybrid. 2. Resonance structures differ only in the placement of ʌ and nonbonding electrons. a. All atoms occupy the same positions. 3. Resonance is an important concept in organic chemistry. B. Rules for resonance forms (Section 2.5). 1. Individual resonance forms are imaginary, not real. 2. Resonance forms differ only in the placement of their ʌ or nonbonding electrons. a. A curved arrow is used to indicate the movement of electrons, not atoms. 3. Different resonance forms of a molecule don’t have to be equivalent. a. If resonance forms are nonequivalent, the structure of the actual molecule resembles the more stable resonance form(s). 4. Resonance forms must obey normal rules of valency. 5. The resonance hybrid is more stable than any individual resonance form. C. A useful technique for drawing resonance forms (Section 2.6). 1. Any three-atom grouping with a multiple bond adjacent to a nonbonding p orbital has two resonance forms. 2. One atom in the grouping has a lone electron pair, a vacant orbital or a single electron. 3. By recognizing these three-atom pieces, resonance forms can be generated. III. Acids and bases (Sections 2.7–2.11). A. Brønsted–Lowry definition (Section 2.7). 1. A Brønsted–Lowry acid donates an H+ ion; a Brønsted–Lowry base accepts H+. 2. The product that results when a base gains H+ is the conjugate acid of the base; the product that results when an acid loses H+ is the conjugate base of the acid. 3. Water can act either as an acid or as a base. B. Acid and base strength (Section 2.8–2.10). 1. A strong acid reacts almost completely with water (Section 2.8). 2. The strength of an acid in water is indicated by Ka, the acidity constant. 3. Strong acids have large acidity constants, and weaker acids have smaller acidity constants. 4. The pKa is normally used to express acid strength. a. pKa = –log Ka b. A strong acid has a small pKa, and a weak acid has a large pKa. c. The conjugate base of a strong acid is a weak base, and the conjugate base of a weak acid is a strong base. 5. Predicting acid–base reactions from pKa (Section 2.9). a. An acid with a low pKa (stronger acid) reacts with the conjugate base of an acid with a high pKa (stronger base). b. In other words, the products of an acid–base reaction are more stable than the reactants. © 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Polar Covalent Bonds; Acids and Bases

6. Organic acids and organic bases (Section 2.10). a. There are two main types of organic acids: i. Acids that contain hydrogen bonded to oxygen. ii. Acids that have hydrogen bonded to the carbon next to a C=O group. b. The main type of organic base contains a nitrogen atom with a lone electron pair. C. Lewis acids and bases (Section 2.11). 1. A Lewis acid accepts an electron pair. a. A Lewis acid may have either a vacant low-energy orbital or a polar bond to hydrogen. b. Examples include metal cations, halogen acids, group 3 compounds and transition-metal compounds. 2. A Lewis base has a pair of nonbonding electrons. a. Most oxygen- and nitrogen-containing organic compounds are Lewis bases. b. Many organic Lewis bases have more than one basic site. 3. A curved arrow shows the movement of electrons from a Lewis base to a Lewis acid. IV. Noncovalent interactions in molecules (Section 2.12). A. Dipole–dipole interactions occur between polar molecules as a result of electrostatic interactions among dipoles. 1. These interactions may be either attractive or repulsive. 2. The attractive geometry is lower in energy and predominates. B. Dispersion forces result from the constantly changing electron distribution within molecules. 1. These forces are transient and weak, but their cumulative effect may be important. C. Hydrogen bonds. 1. Hydrogen bonds form between a hydrogen bonded to an electronegative atom and an unshared electron pair on another electronegative atom. 2. Hydrogen bonds are extremely important in living organisms. 3. Hydrophilic substances dissolve in water because they are capable of forming hydrogen bonds. 4. Hydrophobic substances don’t form hydrogen bonds and usually don’t dissolve in water.

Chapter 2

Solutions to Problems 2.1 After solving this problem, use Figure 2.2 to check your answers. The larger the number, the more electronegative the element. More electronegative (a) H (2.1) (b) Br (2.8) (c) Cl (3.0) (d) C (2.5)

Less electronegative Li (1.0) B (2.0) I (2.5) H (2.1)

Carbon is slightly more electronegative than hydrogen. 2.2 As in Problem 2.1, use Figure 2.2. The partial negative charge is placed on the more electronegative atom, and the partial positive charge is placed on the less electronegative atom. (a)

(b)

(c)

(d)

(e)

(f)

2.3 Use Figure 2.2 to find the electronegativities of each element. Calculate ǻEN and rank the answers in order of increasing ǻEN. Carbon: Lithium:

Carbon: Magnesium:

EN = 2.5 EN = 1.0 ǻEN = 1.5 EN = 2.5 EN = 1.2 ǻEN = 1.3

Carbon: Potassium:

Oxygen: Carbon:

EN = 2.5 EN = 0.8 ǻEN = 1.7

Fluorine: Carbon:

EN = 4.0 EN = 2.5 ǻEN = 1.5

EN = 3.5 EN = 2.5 ǻEN = 1.0

The most polar bond has the largest ǻEN. Thus, in order of increasing bond polarity: H3C — OH < H3C — MgBr < H3C — Li, H3C — F < H3C — K

Polar Covalent Bonds; Acids and Bases

2.4

In an electrostatic potential map, the color red indicates regions of a molecule that are electron-rich. The map shows that chlorine is the most electronegative atom in chloromethane, and the direction of polarity of the C–Cl bond is:

Chloromethane

2.5 Ethylene glycol

The dipole moment of ethylene glycol is zero because the bond polarities of the two carbon–oxygen bonds cancel. 2.6

2.7

For each bond, identify the more electronegative element, and draw an arrow that points from the less electronegative element to the more electronegative element. Estimate the sum of the individual dipole moments to arrive at the dipole moment for the entire molecule. (a)

(b)

(c)

(d)

To find the formal charge of an atom in a molecule, follow these two steps: (1)

Draw an electron-dot structure of the molecule.

(2)

Use the formula in Section 2.3 (shown below) to determine formal charge for each atom. The periodic table shows the number of valence electrons of the element, and the electron-dot structure shows the number of bonding and nonbonding electrons.

Formal charge FC

ª # of valence º ª # of bonding electrons º ª # nonbonding º « electrons » « »¼ « electrons » 2 ¬ ¼ ¬ ¬ ¼

Chapter 2

(a)

Remember: Valence electrons are the electrons characteristic of a specific element. Bonding electrons are those electrons involved in bonding to other atoms. Nonbonding electrons are those electrons in lone pairs. (b)

(c)

Polar Covalent Bonds; Acids and Bases

2.8

Oxygen atoms 3 and 4 each have a formal charge of –1, and oxygen atoms 1 and 2 have a formal charge of 0. 2.9

Try to locate the three-atom groupings that are present in resonance forms. (a)

These two structures represent resonance forms. The three-atom grouping (C–C double bond and an adjacent vacant p orbital) is pictured on the right.

(b)

These two structures represent different compounds, not resonance structures.

2.10 Look for three-atom groupings that contain a multiple bond next to an atom with a p orbital. Exchange the positions of the bond and the electrons in the p orbital to draw the resonance form of each grouping. (a)

Methyl phosphate anion has 3 three-atom groupings and thus has 3 resonance forms.

Recall from Chapter 1 that phosphorus, a third-row element, can form more than four covalent bonds

Chapter 2

(b)

(c)

(d)

2.11 When an acid loses a proton, the product is the conjugate base of the acid. When a base gains a proton, the product is the conjugate acid of the base.

2.12 Recall from Section 2.8 that a stronger acid has a smaller pKa and a weaker acid has a larger pKa. Accordingly, phenylalanine (pKa = 1.83) is a stronger acid than tryptophan (pKa = 2.83). 2.13 HO–H is a stronger acid than H2N–H. Since H2N– is a stronger base than HO–, the conjugate acid of H2N– (H2N–H) is a weaker acid than the conjugate acid of HO– (HO–H). 2.14 Use Table 2.3 to find the strength of each acid. A reaction takes place as written if the stronger acid is the reactant. (a)

Remember that the lower the pKa, the stronger the acid. Thus CH3CO2H, not HCN, is the stronger acid, and the above reaction will not take place to a significant extent in them direction written.

Polar Covalent Bonds; Acids and Bases

(b)

Using the same reasoning as in part (a), we can see that the above reaction will not occur to a significant extent. 2.15

As written, the above reaction will take place to virtual completion due to the large difference in pKa values. 2.16 Enter –9.31 into a calculator and use the INV LOG function to arrive at the answer Ka = 4.9 x 10–10. 2.17 Locate the electron pair(s) of the Lewis base and draw a curved arrow from the electron pair to the Lewis acid. The electron pair moves from the atom at the tail of the arrow (Lewis base) to the atom at the point of the arrow (Lewis acid).(Note: electron dots have been omitted from Cl– to reduce clutter.) (a)

(b)

Chapter 2

2.18 (a)

The nitrogen on the left is more electron-rich and more basic. The indicated hydrogen is most electron-poor (bluest) and is most acidic.

(b)

2.19

Vitamin C is water-soluble (hydrophilic) because it has several polar –OH groups that can form hydrogen bonds with water. Vitamin A is fat-soluble (hydrophobic) because most of its atoms can’t form hydrogen bonds with water.

Visualizing Chemistry 2.20 Naphthalene has three resonance forms.

Polar Covalent Bonds; Acids and Bases

2.21

2.22 Electrostatic potential maps show that the electron-rich regions of the cis isomer lie on the same side of the double bond, leading to a net dipole moment. Because the electron-rich regions of the trans isomer are symmetrical about the double bond, the individual bond dipole moments cancel, and the isomer has no overall dipole moment.

2.23 (a)

(b)

Mechanism Problems 2.24 (a)

BF3 O

BF3

Chapter 2

(b)

(c)

O O

H N

2.25 (a)

H H

(b)

N H

(c)

(d)

O H

Polar Covalent Bonds; Acids and Bases

2.26 (a)

NH2

(b)

(c)

2.27 (a)

H2C

Cl H2C

CH2

CH3

(b)

H C

C H

(c)

H C

H2C

Cl C H2

Additional Problems Electronegativity and Dipole Moments 2.28 Use Figure 2.2 if you need help. The most electronegative element is starred. (a)

CH2FC1

(b)

FCH2CH2CH2Br

(c)

* 2CH2NH2 HOCH

(d)

CH3OCH2Li

Chapter 2

2.29

More polar

Less polar

(a) (b) (c)

(d) 2.30 (a)

2.31 (a) (b)

(b)

(c)

(d)

In Section 2.2, we found that ȝ = Q x r. For a proton and an electron separated by 100 pm, ȝ = 4.80 D. If the two charges are separated by 136 pm, ȝ = 6.53 D. Since the observed dipole moment is 1.08 D, the H–Cl bond has (1.08 D / 6.53 D) x 100 % = 16.5 % ionic character.

2.32 In phosgene, the individual bond polarities tend to cancel, but in formaldehyde, the bond polarities add to each other. Thus, phosgene has a smaller dipole moment than form aldehyde.

2.33 The magnitude of a dipole moment depends on both charge and distance between atoms. Fluorine is more electronegative than chlorine, but a C–F bond is shorter than a C–Cl bond. Thus, the dipole moment of CH3F is smaller than that of CH3Cl. 2.34 The observed dipole moment is due to the lone pair electrons on sulfur.

Polar Covalent Bonds; Acids and Bases

Formal Charges 2.35 To save space, molecules are shown as line-bond structures with lone pairs, rather than as electron-dot structures. (a)

(b)

(c)

(d)

(e)

(f)

Chapter 2

2.36 As in Problem 2.31, molecules are shown as line-bond structures with lone-pair electrons indicated. Only calculations for atoms with non-zero formal charge are shown. (a)

(b)

(c)

Resonance 2.37 Resonance forms do not differ in the position of nuclei. The two structures in (a) are not resonance forms because the positions of the carbon and hydrogen atoms outside the ring are different in the two forms.

The pairs of structures in parts (b), (c), and (d) represent resonance forms. 2.38 (a)

(b)

Polar Covalent Bonds; Acids and Bases

(c)

The last resonance structure is a minor contributor because its carbon lacks a complete electron octet. (d)

(e)

2.39 The two structures are not resonance forms because the positions of the carbon atoms are different in the two forms. Acids and Bases 2.40

2.41

The O–H hydrogen of acetic acid is more acidic than the C–H hydrogens. The –OH oxygen is electronegative, and, consequently, the –O–H bond is more strongly polarized than the –C–H bonds. In addition, the acetate anion is stabilized by resonance. 2.42 (a)

(b)

(c)

Chapter 2

(d)

(e)

(f)

The Lewis acids shown below can accept an electron pair either because they have a vacant orbital or because they can donate H+. The Lewis bases have nonbonding electron pairs.

2.43 (a)

(b)

(c)

2.44 The substances with the largest values of pKa are the least acidic.

2.45 To react completely (> 99.9%) with NaOH, an acid must have a pKa at least 3 units smaller than the pKa of H2O. Thus, all substances in the previous problem except acetone react completely with NaOH. 2.46 The stronger the acid (smaller pKa), the weaker its conjugate base. Since NH4+ is a stronger acid than CH3NH3+, CH3NH2 is a stronger base than NH3.

Polar Covalent Bonds; Acids and Bases

2.47

The reaction takes place as written because water is a stronger acid than tert-butyl alcohol. Thus, a solution of potassium tert-butoxide in water can’t be prepared. 2.48

2.49 (a)

Acetone: K a

2.50 (a)

Nitromethane: pK a

5 x 10 20

10.30

(b)

Formic acid: K a

1.8 x 10 4

(b)

Acrylic acid: pK a

4.25

2.51

If you let 0.050 – x = 0.050, then x = 3.0 × 10–3 and pH = 2.52. If you calculate x exactly using the quadratic equation, then x = 2.9 × 10–3 and pH = 2.54. 2.52 Only acetic acid will react with sodium bicarbonate. Acetic acid is the only substance in Problem 2.40 that is a stronger acid than carbonic acid.

Chapter 2

General Problems 2.53 In maleic acid, the individual dipole moments add to produce a net dipole moment for the whole molecule. The individual dipole moments in fumaric acid cancel, resulting in a zero dipole moment.

2.54 Sodium bicarbonate reacts with acetic acid to produce carbonic acid, which breaks down to form CO2. Thus, bubbles of CO2 indicate the presence of an acid stronger than carbonic acid, in this case acetic acid, as the pKa values indicate. Phenol does not react with sodium bicarbonate. 2.55 Reactions (a) and (c) are reactions between Brønsted–Lowry acids and bases; the stronger acid and stronger base are identified. Reactions (b) and (d) occur between Lewis acids and bases. (a)

(b)

(c)

(d)

Polar Covalent Bonds; Acids and Bases

2.56 Pairs (a) and (d) represent resonance structures; pairs (b) and (c) do not. For two structures to be resonance forms, all atoms must be in the same positions in all resonance forms. 2.57 (a)

(b)

(c)

2.58 The cation pictured can be represented by two resonance forms. Reaction with water can occur at either positively charged carbon, resulting in two products.

2.59 (a)

(b)

(c)

(d)

2.60

Chapter 2

2.61

When phenol loses a proton, the resulting anion is stabilized by resonance. The methanol anion is not stabilized by resonance. 2.62

2.63 (a)

The central carbon of carbonate ion is sp2 and trigonal planar. The three resonance forms contribute equally to the overall resonance hybrid. Thus, the molecule has no dipole.

Polar Covalent Bonds; Acids and Bases

(b)

The oxygen is sp3 hybridized and tetrahedral. Therefore, there is a dipole pointing towards the lone pairs.

O (c)

H3C

CH3

The carbocation is sp2 hybridized and trigonal planar. Thus, there is no dipole.

CH3 C(CH3)3 H3C

CH3

2.64 The equilibrium always favors the weaker acid/base pair (acid with the higher pKa) (a) OH

CO2

CO2H +

pKa 9.9

pKa 4.2

(b) CH3CH2CH2OH

NH2

CH3CH2CH2O

NH3

pKa 16.1

pKa 36

The “a” in pKa here should be subscript. (c) CH3

CH3NO2

CH4

pKa 10.3

pKa 60

CH2NO2

The “a” in pKa here should be subscript. 2.65 (a)

London dispersion forces

(b)

Hydrogen bonding

(c)

Hydrogen bonding causes the carboxylic acids interact with one another and that force is stronger than the dispersion forces between acetic acid and oil.

Chapter 2

2.66 (a)

(b)

H2C (d)

(c)

O O

2.67 Being more electronegative than carbon, the three chlorine atoms inductively remove electron density from the carbon atom to which they are all attached. This effect propagates down the chain, ultimately reducing the electron density on the oxygen. This makes the oxygen-hydrogen bond weaker and the molecule more acidic.

Cl Cl

Cl C

C H

Review Unit 1: Bonds and Bond Polarity Major Topics Covered (with vocabulary:) Atomic Structure: atomic number mass number

wave equation orbital shell node electron configuration

Chemical Bonding Theory: covalent bond Lewis structure lone-pair electrons line-bond structure valence-bond theory sigma (ı) bond bond strength bond length molecular orbital theory bonding MO antibonding MO Hybridization: sp3 hybrid orbital bond angle sp2 hybrid orbital pi (ʌ) bond sp hybrid orbital Polar covalent bonds: polar covalent bond electronegativity (EN) electrostatic potential maps inductive effect dipole moment formal charge dipolar molecule Resonance: resonance form resonance hybrid Acids and Bases: Brønsted-Lowry acid Brønsted-Lowry base conjugate acid conjugate base acidity constant Ka pKa organic acid organic base Lewis acid Lewis base Chemical Structures: condensed structure skeletal structure

space-filling models

ball-and-stick models

Types of Problems: After studying these chapters you should be able to: – – – – – – – – – – – –

Predict the ground state electronic configuration of atoms. Draw Lewis electron-dot structures of simple compounds. Predict and describe the hybridization of bonds in simple compounds. Predict bond angles and shapes of molecules. Predict the direction of polarity of a chemical bond, and predict the dipole moment of a simple compound. Calculate formal charge for atoms in a molecule. Draw resonance forms of molecules. Predict the relative acid/base strengths of Brønsted acids and bases. Predict the direction of Brønsted acid/base reactions. Calculate: pKa from Ka, and vice versa. pH of a solution of a weak acid. Identify Lewis acids and bases. Draw chemical structures from molecular formulas, and vice versa.

Review Unit 1

Points to Remember: *

In order for carbon, with valence shell electron configuration of 2s22p2, to form four sp3 hybrid orbitals, it is necessary that one electron be promoted from the 2s subshell to the 2p subshell. Although this promotion requires energy, the resulting hybrid orbitals are able to form stronger bonds, and compounds containing these bonds are more stable.

Assigning formal charge to atoms in a molecule is helpful in showing where the electrons in a bond are located. Even if a bond is polar covalent, in some molecules the electrons “belong” more to one of the atoms than the other. This “ownership” is useful for predicting the outcomes of chemical reactions, as we will see in later chapters.

Resonance structures are representations of the distribution of ʌ and nonbonding electrons in a molecule. Electrons don’t move around in the molecule, and the molecule doesn’t change back and forth, from structure to structure. Rather, resonance structures are an attempt to show, by conventional line-bond drawings, the electron distribution of a molecule that can’t be represented by any one structure.

As in general chemistry, acid-base reactions are of fundamental importance in organic chemistry. Organic acids and bases, as well as inorganic acids and bases, occur frequently in reactions, and large numbers of reactions are catalyzed by Brønsted acids and bases and Lewis acids and bases.

Self-Test:

A Ricinine (a toxic component of castor beans)

B Oxaflozane (an antidepressant)

C 1,3,4-Oxadiazole

For A (ricinine) and B (oxaflozane): Add all missing electron lone pairs. Identify the hybridization of all carbons. Indicate the direction of bond polarity for all bonds with ǻEN 0.5. In each compound, which bond is the most polar? Convert A and B to molecular formulas. Draw a resonance structure for B. Which atom (or atoms) of B can act as a Lewis base? Add missing electron lone pairs to C. Is it possible to draw resonance forms for C? If so, draw at least one resonance form, and describe it.

Review Unit 1

Multiple Choice: 1.

Which element has 4s24p2 as its valence shell electronic configuration? (a) Ca (b) C (c) Al (d) Ge

Which compound (or group of atoms) has an oxygen with a +1 formal charge? (a) NO3– (b) O3 (c) acetone anion (d) acetate anion The following questions involve these acids: (i) HW (pKa = 2); (ii) HX (pKa = 6); (iii) HY (pKa = 10); (iv) HZ (pKa = 20).

Which of the above acids react almost completely with water to form hydroxide ion? (a) none of them (b) all of them (c) HY and HZ (d) HZ

The conjugate bases of which of the above acids react almost completely with water to form hydroxide ion? (a) none of them (b) all of them (c) HZ (d) HY and HZ

If you want to convert HX to X–, which bases can you use? (a) W– (b) Y– (c) Z– (d) Y– or Z–

If you add equimolar amounts of HW, X– and HY to a solution, what are the principal species in the resulting solution? (a) HW, HX, HY (b) W–, HX, HY (c) HW, X–, HY (d) HW, HX, Y–

What is the approximate pH difference between a solution of 1 M HX and a solution of 1 M HY? (a) 2 (b) 3 (c) 4 (d) 6

If you wanted to write the structure of a molecule that shows carbon and hydrogen atoms as groups, without indicating many of the carbon-hydrogen bonds, you would draw a: (a) molecular formula (b) Kekulé structure (c) skeletal structure (d) condensed structure

Which of the following molecules has zero net dipole moment? (a)

10.

(b)

(c)

(d)

In which of the following bonds is carbon the more electronegative element? (a) C — Br (b) C — I (c) C — P (d) C — S

Chapter 3 – Organic Compounds: Alkanes and Their Stereochemistry Chapter Outline I. Functional Groups (Section 3.1). A. Functional groups are groups of atoms within a molecule that have a characteristic chemical behavior. B. The chemistry of every organic molecule is determined by its functional groups. C. Functional groups described in this text can be grouped into three categories: 1. Functional groups with carbon–carbon multiple bonds. 2. Groups in which carbon forms a single bond to an electronegative atom. 3. Groups with a carbon–oxygen double bond. II. Alkanes (Sections 3.2–3.5). A. Alkanes and alkane isomers (Section 3.2). 1. Alkanes are formed by overlap of carbon sp3 orbitals. 2. Alkanes are described as saturated hydrocarbons. a. They are hydrocarbons because they contain only carbon and hydrogen. b. They are saturated because all bonds are single bonds. c. The general formula for alkanes is CnH2n+2. 3. For alkanes with four or more carbons, the carbons can be connected in more than one way. a. If the carbons are in a row, the alkane is a straight-chain alkane. b. If the carbon chain has a branch, the alkane is a branched-chain alkane. 4. Alkanes with the same molecular formula can exist in different forms known as isomers. a. Isomers whose atoms are connected differently are constitutional isomers. i. Constitutional isomers are always different compounds with different properties but with the same molecular formula. b. Most alkanes can be drawn in many ways. 5. Straight-chain alkanes are named according to the number of carbons in their chain. B. Alkyl groups (Section 3.3). 1. An alkyl group is the partial structure that results from the removal of a hydrogen atom from an alkane. a. Alkyl groups are named by replacing the -ane of an alkane name by -yl. b. n-Alkyl groups are formed by removal of an end hydrogen atom of a straight chain alkane. c. Branched-chain alkyl groups are formed by removal of a hydrogen atom from an internal carbon. i. The prefixes sec- and tert- refer to the degree of substitution at the branching carbon atom. 2. There are four possible degrees of alkyl substitution for carbon. a. A primary carbon is bonded to one other carbon. b. A secondary carbon is bonded to two other carbons.

Organic Compounds: Alkanes and Their Stereochemistry

c. A tertiary carbon is bonded to three other carbons. d. A quaternary carbon is bonded to four other carbons. e. The symbol R refers to the rest of the molecule. 3. Hydrogens are also described as primary, secondary and tertiary. a. Primary hydrogens are bonded to primary carbons (RCH3). b. Secondary hydrogens are bonded to secondary carbons (R2CH2). c. Tertiary hydrogens are bonded to tertiary carbons (R3CH). C. Naming alkanes (Section 3.4). 1. The system of nomenclature used in this book is the IUPAC system. In this system, a chemical name has a locant, a prefix, a parent and a suffix. i. The locant shows the location of substituents and functional groups. ii. The prefix indicates the type of substituent or functional group. iii. The parent shows the number of carbons in the principal chain. iv. The suffix identifies the functional group family. 2. Naming an alkane: a. Find the parent hydrocarbon. i. Find the longest continuous chain of carbons, and use its name as the parent name. ii. If two chains have the same number of carbons, choose the one with more branch points. b. Number the atoms in the parent chain. i. Start numbering at the end nearer the first branch point. ii. If branching occurs an equal distance from both ends, begin numbering at the end nearer the second branch point. c. Identify and number the substituents. i. Give each substituent a number that corresponds to its position on the parent chain. ii. Two substituents on the same carbon receive the same number. d. Write the name as a single word. i. Use hyphens to separate prefixes and commas to separate numbers. ii. Use the prefixes, di-, tri-, tetra- if necessary, but don’t use them for alphabetizing. e. Name a complex substituent as if it were a compound, and set it off within parentheses. i. Some simple branched-chain alkyl groups have common names. ii. The prefix iso is used for alphabetizing, but sec- and tert- are not. D. Properties of alkanes (Section 3.5). 1. Alkanes are chemically inert to most laboratory reagents. 2. Alkanes react with O2 (combustion) and Cl2 (substitution).

Chapter 3

3. The boiling points and melting points of alkanes increase with increasing molecular weight. a. This effect is due to weak dispersion forces. b. The strength of these forces increases with increasing molecular weight. 4. Increased branching lowers an alkane’s boiling point. III. Conformations of straight-chain alkanes (Sections 3.6–3.7). A. Conformations of ethane (Section 3.6). 1. Rotation about a single bond produces isomers that differ in conformation. a. These isomers (conformers) have the same connections of atoms and can’t be isolated. 2. These isomers can be represented in two ways: a. Sawhorse representations view the C–C bond from an oblique angle. b. Newman projections view the C–C bond end-on and represent the two carbons as a circle. 3. There is a barrier to rotation that makes some conformers of lower energy than others. a. The lowest energy conformer (staggered conformation) occurs when all C–H bonds are as far from each other as possible. b. The highest energy conformer (eclipsed conformation) occurs when all C–H bonds are as close to each other as possible. c. Between these two conformations lie an infinite number of other conformations. 4. The staggered conformation is 12 kJ/mol lower in energy than the eclipsed conformation. a. This energy difference is due to torsional strain from interactions between C–H bonding orbitals on one carbon and C–H antibonding orbitals on an adjacent carbon, which stabilize the staggered conformer. b. The torsional strain resulting from a single C–H interaction is 4.0 kJ/mol. c. The barrier to rotation can be represented on a graph of potential energy vs. angle of rotation (dihedral angle). B. Conformations of other alkanes (Section 3.7). 1. Conformations of propane. a. Propane also shows a barrier to rotation that is 14 kJ/mol. b. The eclipsing interaction between a C–C bond and a C–H bond is 6.0 kJ/mol. 2. Conformations of butane. a. Not all staggered conformations of butane have the same energy; not all eclipsed conformations have the same energy. i. In the lowest energy conformation (anti) the two large methyl groups are as far from each other as possible. ii. The eclipsed conformation that has two methyl–hydrogen interactions and a H–H interaction is 16 kJ/mol higher in energy than the anti conformation. iii. The conformation with two methyl groups 60° apart (gauche conformation) is 3.8 kJ/mol higher in energy than the anti conformation. © 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Organic Compounds: Alkanes and Their Stereochemistry

(a). This energy difference is due to steric strain – the repulsive interaction that results from forcing atoms to be closer together than their atomic radii allow. iv. The highest energy conformations occur when the two methyl groups are eclipsed. (a). This conformation is 19 kJ/mol less stable than the anti conformation. The value of a methyl–methyl eclipsing interaction is 11 kJ/mol. b. The most favored conformation for any straight-chain alkane has carbon–carbon bonds in staggered arrangements and large substituents anti to each other. c. At room temperature, bond rotation occurs rapidly, but a majority of molecules adopt the most stable conformation.

Chapter 3

Solutions to Problems 3.1 Notice that certain functional groups have different designations if other functional groups are also present in a molecule. For example, a molecule containing a carbon–carbon double bond and no other functional group is an alkene; if other groups are present, the group is referred to as a carbon–carbon double bond. Similarly, a compound containing a benzene ring, and only carbon- and hydrogen-containing substituents, is an arene; if other groups present, the ring is labeled an aromatic ring. (a)

(b)

(c)

3.2 (a)

(b)

(e)

(f)

(c)

(d)

3.3

Organic Compounds: Alkanes and Their Stereochemistry

3.4

We know that carbon forms four bonds and hydrogen forms one bond. Thus, draw all possible six-carbon skeletons and add hydrogens so that all carbons have four bonds. To draw all possible skeletons in this problem: (1) Draw the six-carbon straight-chain skeleton. (2) Draw a five-carbon chain, identify the different types of carbon atoms on the chain, and add a –CH3 group to each of the different types of carbons, generating two skeletons. (3) Repeat the process with the four-carbon chain to give rise to the last two skeletons. Add hydrogens to the remaining carbons to complete the structures.

3.5

(a)

Nine isomeric esters of formula C5H10O2 can be drawn. The procedure is described in Problem 3.4.

(b)

Two isomers can be drawn.

(c)

Three isomers can be drawn.

3.6

Chapter 3

(a)

Two alcohols have the formula C3H8O.

(b)

Four bromoalkanes have the formula C4H9Br.

(c)

Four thioesters have the formula C4H8OS.

3.7

3.8 (a)

3.9

(b)

(c)

The carbons and the attached hydrogens have the same classification. (a)

(b)

(c)

3.10 (a)

(b)

(c)

Organic Compounds: Alkanes and Their Stereochemistry

3.11 (a)

(b)

Step 1: Find the longest continuous carbon chain and use it as the parent name. In (b), the chain is a pentane. Step 2: Identify the substituents. In (b), both substituents are methyl groups. Step 3: Number the substituents. In (b), the methyl groups are in the 2- and 3positions. Step 4: Name the compound. Remember that the prefix di- must be used when two are the same. The IUPAC name is 2,3-dimethylpentane.

. (c)

(d)

3.12 When you are asked to draw the structure corresponding to a given name, draw the parent carbon chain, attach the specified groups to the proper carbons, and fill in the remaining hydrogens. (a)

(b)

(c)

(d)

Chapter 3

3.13

3.14

3.15 The graph shows the energy of a conformation as a function of angle of rotation.

Organic Compounds: Alkanes and Their Stereochemistry

3.16 (a)

(b)

(c), (d)

3.17 This conformation of 2,3-dimethylbutane is the most stable because it is staggered and has the fewest CH3ļCH3 gauche interactions.

3.18 The conformation is a staggered conformation in which the hydrogens on carbons 2 and 3 are 60° apart. Draw the Newman projection.

Chapter 3

The Newman projection shows three gauche interactions, each of which has an energy cost of 3.8 kJ/mol. The total strain energy is 11.4 kJ/mol (3 × 3.8 kJ/mol). Visualizing Chemistry 3.19 (a)

(b)

3.20 (a)

(b)

(c)

(d)

3.21 In this conformation, all groups are staggered and the two methyl groups are 180° apart.

Organic Compounds: Alkanes and Their Stereochemistry

Additional Problems Functional Groups 3.22 (a)

(b)

(c)

(d)

(e)

(f)

3.23 Different answers to this problem and to Problem 3.24 are acceptable. (a)

(b)

(c)

(d)

(e)

(f)

Chapter 3

3.24 For (a) and (h), only one structure is possible. (a)

(b)

(c)

(d)

(e)

(f)

(g)

(h)

3.25 (a)

3.26 (a)

(b)

(c)

Although it is stated that biacetyl contains no rings or carbon–carbon double bonds, it is obvious from the formula for biacetyl that some sort of multiple bond must be present. The structure for biacetyl contains two carbon–oxygen double bonds.

(b)

Ethyleneimine contains a three-membered ring.

(c)

Glycerol contains no multiple bonds or rings.

(a)

(b)

(c)

Organic Compounds: Alkanes and Their Stereochemistry

Isomers 3.27 (a)

(b)

Eighteen isomers have the formula C8H18. Three are pictured.

Structures with the formula C4H8O2 may represent esters, carboxylic acids or many other complicated molecules. Three possibilities:

3.28

3.29 (a)

Chapter 3

(b)

(c)

Give the number “1” to the carbon bonded to –OH, and count to find the longest chain containing the –OH group. 3.30 The isomers may be either alcohols or ethers.

3.31 First, draw all straight-chain isomers. Then proceed to the simplest branched structure. (a)

There are four alcohol isomers with the formula C4H10O.

Organic Compounds: Alkanes and Their Stereochemistry

(b)

There are 17 isomers of C5H13N. Nitrogen can be bonded to one, two or three alkyl groups.

(c)

There are 3 ketone isomers with the formula C5H10O.

(d)

There are 4 isomeric aldehydes with the formula C5H10O. Remember that the aldehyde functional group can occur only at the end of a chain.

(e)

There are 4 esters with the formula C4H8O2.

Chapter 3

(f)

There are 3 ethers with the formula C4H10O.

3.32 (a)

(b)

(e)

(f)

(c)

(d)

Naming Compounds 3.33

3.34

3.35 (a)

(b)

(c)

(d)

Organic Compounds: Alkanes and Their Stereochemistry

(e)

(f)

3.36 (a)

(b)

(a)

(b)

(c)

3.37

3.38 (a)

(b)

(c)

(d)

(e)

(f)

Chapter 3

3.39

3.40 Structure and Correct Name

Error

(a)

The longest chain is an octane and has only methyl branches.

(b)

The longest chain is a hexane. Numbering should start from the opposite end of the carbon chain, nearer the first branch.

(c)

Numbering should start from the opposite end of the carbon chain. See step 2(b) in Section 3.4.

(d)

Numbering should start from the opposite end of the carbon chain.

(e)

The longest chain is an octane.

Organic Compounds: Alkanes and Their Stereochemistry

3.41 (a)

(b)

Remember that you must choose an alkane whose principal chain is long enough so that the substituent does not become part of the principal chain.

Conformations 3.42 (a), (b)

The energy difference between the two conformations is (11.0 + 6.0 + 4.0) kJ/mol – 3.8 kJ/mol = 17.2 kJ/mol. (c)

Consider the least stable conformation to be at zero degrees. Keeping the front of the projection unchanged, rotate the back by 60° to obtain each conformation.

Chapter 3

Organic Compounds: Alkanes and Their Stereochemistry

3.43 Each CH3ļCH3 gauche interaction has a value of 3.8 kJ/mol.

3.44 Since we are not told the values of the interactions for 1,2-dibromoethane, the diagram can only be qualitative.

The anti conformation is at 180°. The gauche conformations are at 60°, 300°. 3.45 The eclipsed conformation at 0° rotation has the largest dipole moment but is a high energy that is present in low abundance. The anti conformation has no net dipole because the polarities of the individual bonds cancel. The gauche conformation, however, has a dipole moment. Because the observed dipole moment is 1.0 D at room temperature, a mixture of gauche and anti conformations must be present.

Chapter 3

3.46 The best way to draw pentane is to make a model and to copy it onto the page. A model shows the relationship among atoms, and its drawing shows how these relationships in two dimensions. From your model, you should be able to see that all atoms are staggered in the drawing.

3.47

General Problems 3.48 (a)

(b)

(c)

(d)

(e)

(f)

3.49 (a)

Because malic acid has two –CO2H groups, the formula for the rest of the molecule is C2H4O. Possible structures for malic acid are:

(b)

Because only one of these compounds (the second one) is also a secondary alcohol, it must be malic acid.

Organic Compounds: Alkanes and Their Stereochemistry

3.50 To solve this type of problem, read the problem carefully, word for word. Then try to interpret parts of the problem. For example:

Formaldehyde is an aldehyde,

It trimerizes – that is, 3 formaldehydes come together to form a compound C3H6O3. Because no atoms are eliminated, all of the original atoms are still present.

There are no carbonyl groups. This means that trioxane cannot contain any –C=O functional groups. If you look back to Table 3.1, you can see that the only oxygencontaining functional groups that can be present are either ethers or alcohols.

A monobromo derivative is a compound in which one of the –H’s has been replaced by a –Br. Because only one monobromo derivative is possible, we know that there can only be one type of hydrogen in trioxane. The only possibility for trioxane is:

3.51 The highest energy conformation of bromoethane has a strain energy of 15 kJ/mol. Because this includes two H–H eclipsing interactions of 4.0 kJ/mol each, the value of an H–Br eclipsing interaction is 15 kJ/mol – 2(4.0 kJ/mol) = 7.0 kJ/mol.

Chapter 3

3.52

Most stable

Strain enegy

Least stable

Strain energy

(a)

(b)

(c)

(d)

Organic Compounds: Alkanes and Their Stereochemistry

3.53

The carboxylic acid and alcohol groups in Pravachol are an ester (lactone) group in Zocor. The alcohol at the bottom left of Pravachol is a methyl group in Zocor. 3.54 A puckered ring allows all the bonds in the ring to have a nearly tetrahedral bond angle. (If the ring were flat, C–C–C bond angles would be 120°.) Also, a puckered conformation relieves strain due to eclipsed hydrogens.

3.55

In one of the 1, 2-dimethylcyclohexanes the two methyl groups are on the same side of the ring, and in the other isomer the methyl groups are on opposite sides.

Chapter 4 – Organic Compounds: Cycloalkanes and Their Stereochemistry Chapter Outline I. Cycloalkanes – alicyclic compounds – (Sections 4.1–4.2). A. Cycloalkanes have the general formula CnH2n, if they have one ring. B. Naming cycloalkanes (Section 4.1). 1. Find the parent. a. If the number of carbon atoms in the ring is larger than the number in the largest substituent, the compound is named as an alkyl-substituted cycloalkane. b. If the number of carbon atoms in the ring is smaller than the number in the largest substituent, the compound is named as a cycloalkyl-substituted alkane. 2. Number the substituents. a. Start at a point of attachment and number the substituents so that the second substituent has the lowest possible number. b. If necessary, proceed to the next substituent until a point of difference is found. c. If two or more substituents might potentially receive the same number, number them by alphabetical priority. d. Halogens are treated in the same way as alkyl groups. C. Cis–trans isomerism in cycloalkanes (Section 4.2). 1. Unlike open-chain alkanes, cycloalkanes have much less rotational freedom. a. Very small rings are rigid. b. Large rings have more rotational freedom. 2. Cycloalkanes have a “top” side and a “bottom” side. a. If two substituents are on the same side of a ring, the ring is cis-disubstituted. b. If two substituents are on opposite sides of a ring, the ring is trans-disubstituted. 3. Substituents in the two types of disubstituted cycloalkanes are connected in the same order but differ in spatial orientation. a. These cycloalkenes are stereoisomers that are known as cis–trans isomers. b. Cis–trans isomers are stable compounds that can’t be interconverted. II. Conformations of cycloalkanes (Sections 4.3–4.9). A. General principles (Section 4.3). 1. Ring strain. a. A. von Baeyer suggested that rings other than those of 5 or 6 carbons were too strained to exist. b. This concept of angle strain is true for smaller rings, but larger rings can be easily prepared. 2. Heats of combustion of cycloalkanes. a. To measure strain, it is necessary to measure the total energy of a compound and compare it to a strain-free reference compound. b. Heat of combustion measures the amount of heat released when a compound is completely burned in oxygen. i. The more strained the compound, the higher the heat of combustion. ii. Strain per CH2 unit can be calculated and plotted as a function of ring size.