Solutions Manual for Organic Chemistry, 4th Edition by David Klein INCLUDES STUDY GUIDE by welldoneassistant

Student Study Guide and Solutions Manual, 4e

for

Organic Chemistry, 4e

David Klein Johns Hopkins University

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ISBN: 978‐1‐119‐65958‐7 Printed in the United States of America 10 9 8 7 6 5 4 3 2 1 The inside back cover will contain printing identification and country of origin if omitted from this page. In addition, if the ISBN on the back cover differs from the ISBN on this page, the one on the back cover is correct.

CONTENTS Chapter 1 – Electrons, Bonds, and Molecular Properties

Chapter 2 – Molecular Representations

Chapter 3 – Acids and Bases

Chapter 4 – Alkanes and Cycloalkanes

110

Chapter 5 – Stereoisomerism

141

Chapter 6 – Chemical Reactivity and Mechanisms

173

Chapter 7 – Alkyl Halides: Nucleophilic Substitution and Elimination Reactions

195

Chapter 8 – Addition Reactions of Alkenes

262

Chapter 9 – Alkynes

317

Chapter 10 – Radical Reactions

368

Chapter 11 – Synthesis

409

Chapter 12 – Alcohols and Phenols

449

Chapter 13 – Ethers and Epoxides; Thiols and Sulfides

510

Chapter 14 – Infrared Spectroscopy and Mass Spectrometry

563

Chapter 15 – Nuclear Magnetic Resonance Spectroscopy

593

Chapter 16 – Conjugated Pi Systems and Pericyclic Reactions

639

Chapter 17 – Aromatic Compounds

682

Chapter 18 – Aromatic Substitution Reactions

717

Chapter 19 – Aldehydes and Ketones

790

Chapter 20 – Carboxylic Acids and Their Derivatives

865

Chapter 21 – Alpha Carbon Chemistry: Enols and Enolates

929

Chapter 22 – Amines

1010

Chapter 23 – Introduction to Organometallic Compounds

1071

Chapter 24 – Carbohydrates

1126

Chapter 25 – Amino Acids, Peptides, and Proteins

1154

Chapter 26 – Lipids

1183

Chapter 27 – Synthetic Polymers

1202

HOW TO USE THIS BOOK Organic chemistry is much like bicycle riding. You cannot learn how to ride a bike by watching other people ride bikes. Some people might fool themselves into believing that it’s possible to become an expert bike rider without ever getting on a bike. But you know that to be incorrect (and very naïve). In order to learn how to ride a bike, you must be willing to get on the bike, and you must be willing to fall. With time (and dedication), you can quickly train yourself to avoid falling, and to ride the bike with ease and confidence. The same is true of organic chemistry. In order to become proficient at solving problems, you must “ride the bike”. You must try to solve the problems yourself (without the solutions manual open in front of you). Once you have solved the problems, this book will allow you to check your solutions. If, however, you don’t attempt to solve each problem on your own, and instead, you read the problem statement and then immediately read the solution, you are only hurting yourself. You are not learning how to avoid falling. Many students make this mistake every year. They use the solutions manual as a crutch, and then they never really attempt to solve the problems on their own. It really is like believing that you can become an expert bike rider by watching hundreds of people riding bikes. The world doesn’t work that way! The textbook has thousands of problems to solve. Each of these problems should be viewed as an opportunity to develop your problem-solving skills. By reading a problem statement and then reading the solution immediately (without trying to solve the problem yourself), you are robbing yourself of the opportunity provided by the problem. If you repeat that poor study habit too many times, you will not learn how to solve problems on your own, and you will not get the grade that you want. Why do so many students adopt this bad habit (of using the solutions manual too liberally)? The answer is simple. Students often wait until a day or two before the exam, and then they spend all night cramming. Sound familiar? Unfortunately, organic chemistry is the type of course where cramming is insufficient, because you need time in order to ride the bike yourself. You need time to think about each problem until you have developed a solution on your own. For some problems, it might take days before you think of a solution. This process is critical for learning this subject. Make sure to allot time every day for studying organic chemistry, and use this book to check your solutions. This book has also been designed to serve as a study guide, as described below.

WHAT’S IN THIS BOOK This book contains more than just solutions to all of the problems in the textbook. Each chapter of this book also contains a series of exercises that will help you review the concepts, skills and reactions presented in the corresponding chapter of the textbook. These exercises are designed to serve as study tools that can help you identify your weak areas. Each chapter of this solutions manual/study guide has the following parts:

•

• •

•

Review of Concepts. These exercises are designed to help you identify which concepts are the least familiar to you. Each section contains sentences with missing words (blanks). Your job is to fill in the blanks, demonstrating mastery of the concepts. To verify that your answers are correct, you can open your textbook to the end of the corresponding chapter, where you will find a section entitled Review of Concepts and Vocabulary. In that section, you will find each of the sentences, verbatim. Review of Skills. These exercises are designed to help you identify which skills are the least familiar to you. Each section contains exercises in which you must demonstrate mastery of the skills developed in the SkillBuilders of the corresponding textbook chapter. To verify that your answers are correct, you can open your textbook to the end of the corresponding chapter, where you will find a section entitled SkillBuilder Review. In that section, you will find the answers to each of these exercises. Review of Reactions. These exercises are designed to help you identify which reagents are not at your fingertips. Each section contains exercises in which you must demonstrate familiarity with the reactions covered in the textbook. Your job is to fill in the reagents necessary to achieve each reaction. To verify that your answers are correct, you can open your textbook to the end of the corresponding chapter, where you will find a section entitled Review of Reactions. In that section, you will find the answers to each of these exercises. Review of Mechanisms. These exercises are designed to help you practice drawing the mechanisms. To verify that you have drawn the mechanism correctly, you can open your textbook to the corresponding chapter, where you will find the mechanisms appearing in numbered boxes throughout the chapter. In those numbered boxes, you will find the answers to each of these exercises. Common Mistakes to Avoid. This is a new feature to this edition. The most common student mistakes are described, so that you can avoid them when solving problems. A List of Useful Reagents. This is a new feature to this edition. This list provides a review of the reagents that appear in each chapter, as well as a description of how each reagent is used. Solutions. At the end of each chapter, you’ll find detailed solutions to all problems in the textbook, including all SkillBuilders, conceptual checkpoints, additional problems, integrated problems, and challenge problems.

The sections described above have been designed to serve as useful tools as you study and learn organic chemistry. Good luck! David Klein Johns Hopkins University

Chapter 1 A Review of General Chemistry: Electrons, Bonds and Molecular Properties Review of Concepts Fill in the blanks below. To verify that your answers are correct, look in your textbook at the end of Chapter 1. Each of the sentences below appears verbatim in the section entitled Review of Concepts and Vocabulary.           

_____________ isomers share the same molecular formula but have different connectivity of atoms and different physical properties. Second-row elements generally obey the _______ rule, bonding to achieve noble gas electron configuration. A pair of unshared electrons is called a ______________. A formal charge occurs when atoms do not exhibit the appropriate number of ___________________________. An atomic orbital is a region of space associated with ____________________, while a molecular orbital is associated with _______________________. Methane’s tetrahedral geometry can be explained using four degenerate _____-hybridized orbitals to achieve its four single bonds. Ethylene’s planar geometry can be explained using three degenerate _____-hybridized orbitals. Acetylene’s linear geometry is achieved via _____-hybridized carbon atoms. The geometry of small compounds can be predicted using valence shell electron pair repulsion (VSEPR) theory, which focuses on the number of  bonds and _______________ exhibited by each atom. The physical properties of compounds are determined by __________________ forces, the attractive forces between molecules. London dispersion forces result from the interaction between transient __________________ and are stronger for larger alkanes due to their larger surface area and ability to accommodate more interactions.

Review of Skills Fill in the blanks and empty boxes below. To verify that your answers are correct, look in your textbook at the end of Chapter 1. The answers appear in the section entitled SkillBuilder Review. SkillBuilder 1.1 Drawing Constitutional Isomers of Small Molecules

SkillBuilder 1.2 Drawing the Lewis Structure of a Small Molecule

CHAPTER 1

SkillBuilder 1.3 Calculating Formal Charge

SkillBuilder 1.4 Locating Partial Charges Resulting from Induction

SkillBuilder 1.5 Reading Bond-Line Structures

SkillBuilder 1.6 Identifying Electron Configurations Step 1 In the energy diagram shown here, draw the electron configuration of nitrogen (using arrows to represent electrons).

2p 2s 1s

Step 2 Fill in the boxes below with the numbers that correctly describe the electron configuration of nitrogen.

Nitrogen SkillBuilder 1.7 Identifying Hybridization States

CHAPTER 1 SkillBuilder 1.8 Predicting Geometry

SkillBuilder 1.9 Identifying the Presence of Molecular Dipole Moments

SkillBuilder 1.10 Predicting Physical Properties Dipole-Dipole Interactions Circle the compound below that is expected to have the higher boiling point.

H3 C

CH2

CH3 H3C

Hydrogen-Bonding Interactions Circle the compound below that is expected to have the higher boiling point.

H CH3

C H

H O

H H

Carbon Skeleton Circle the compound below that is expected to have the higher boiling point.

A Common Mistake to Avoid When drawing a structure, don’t forget to draw formal charges, as forgetting to do so is a common error. If a formal charge is present, it MUST be drawn. For example, in the following case, the nitrogen atom bears a positive charge, so the charge must be drawn:

As we progress though the course, we will see structures of increasing complexity. If formal charges are present, failure to draw them constitutes an error, and must be scrupulously avoided. If you have trouble drawing formal charges, go back and master that skill. You can’t go on without it. Don’t make the mistake of underestimating the importance of being able to draw formal charges with confidence.

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Solutions 1.1. (a) Begin by determining the valency of each atom that appears in the molecular formula. The carbon atoms are tetravalent, while the chlorine atom and hydrogen atoms are all monovalent. The atoms with more than one bond (in this case, the three carbon atoms) should be drawn in the center of the compound. Then, the chlorine atom can be placed in either of two locations: i) connected to the central carbon atom, or ii) connected to one of the other two (equivalent) carbon atoms. The hydrogen atoms are then placed at the periphery (ensuring that each carbon atom has a total of four bonds). The formula C3H7Cl has two constitutional isomers.

(b) Begin by determining the valency of each atom that appears in the molecular formula. The carbon atoms are tetravalent, while the hydrogen atoms are all monovalent. The atoms with more than one bond (in this case, the four carbon atoms) should be drawn in the center of the compound. There are two different ways to connect four carbon atoms. They can either be arranged in a linear fashion or in a branched fashion:

We then place the hydrogen atoms at the periphery (ensuring that each carbon atom has a total of four bonds). The formula C4H10 has two constitutional isomers:

(c) Begin by determining the valency of each atom that appears in the molecular formula. The carbon atoms are tetravalent, while the hydrogen atoms are all monovalent. The atoms with more than one bond (in this case, the five carbon atoms) should be drawn in the center of the compound. So we must explore all of the different ways to connect five carbon atoms. First, we can connect all five carbon atoms in a linear fashion:

Alternatively, we can draw four carbon atoms in a linear fashion, and then draw the fifth carbon atom on a branch. There are many ways to draw this possibility:

Finally, we can draw three carbon atoms in a linear fashion, and then draw the remaining two carbon atoms on separate branches.

Note that we cannot draw a unique carbon skeleton (a unique arrangement of carbon atoms) simply by placing the last two carbon atoms together as one branch, because that possibility has already been drawn earlier (a linear chain of four carbon atoms with a single branch):

In summary, there are three different ways to connect five carbon atoms:

We then place the hydrogen atoms at the periphery (ensuring that each carbon atom has a total of four bonds). The formula C5H12 has three constitutional isomers:

(d) Begin by determining the valency of each atom that appears in the molecular formula. The carbon atoms are tetravalent, the oxygen atom is divalent, and the hydrogen atoms are all monovalent. Any atoms with more than one bond (in this case, the four carbon atoms and the one oxygen atom) should be drawn in the center of the compound, with the hydrogen atoms at the periphery. There are several different ways to connect four carbon atoms and one oxygen atom. Let’s begin with the four carbon atoms. There are two different ways to connect four carbon atoms. They can either be arranged in a linear fashion or in a branched fashion.

CHAPTER 1

the center of the compound. There is only way to connect three carbon atoms:

Next, the oxygen atom must be inserted. For each of the two skeletons above (linear or branched), there are several different locations to insert the oxygen atom. The linear skeleton has four possibilities, shown here: O

C 1

Next, we must determine all of the different possible ways of connecting two chlorine atoms to the chain of three carbon atoms. If we place one chlorine atom at C1, then the second chlorine atom can be placed at C1, at C2 or at C3:

Furthermore, we can place both chlorine atoms at C2, giving a new possibility not shown above:

and the branched skeleton has three possibilities shown here:

Finally, we complete all of the structures by drawing the bonds to hydrogen atoms (ensuring that each carbon atom has four bonds, and each oxygen atoms has two bonds). The formula C4H10O has seven constitutional isomers:

There are no other possibilities. For example, placing the two chlorine atoms at C2 and C3 is equivalent to placing them at C1 and C2:

Finally, the hydrogen atoms are placed at the periphery (ensuring that each carbon atom has a total of four bonds). The formula C3H6Cl2 has four constitutional isomers:

(e) Begin by determining the valency of each atom that appears in the molecular formula. The carbon atoms are tetravalent, while the chlorine atom and hydrogen atoms are all monovalent. The atoms with more than one bond (in this case, the three carbon atoms) should be drawn in

1.2. The carbon atoms are tetravalent, while the chlorine atoms and fluorine atoms are all monovalent. The atoms with more than one bond (in this case, the two carbon atoms) should be drawn in the center of the compound. The chlorine atoms and fluorine atoms are then placed at the periphery, as shown. There are only two possible constitutional isomers: one with the three chlorine atoms all connected to the same carbon, and one in which they are distributed over both carbon atoms. Any other representations that one may draw must be one of these structures drawn in a different orientation.

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1.3. (a) Each carbon atom has four valence electrons, and each hydrogen atom has one valence electron. Only the carbon atoms can form more than one bond, so we begin by connecting the carbon atoms to each other. Then, we connect all of the hydrogen atoms, as shown.

(b) Each carbon atom has four valence electrons, and each hydrogen atom has one valence electron. Only the carbon atoms can form more than one bond, so we begin by connecting the carbon atoms to each other. Then, we connect all of the hydrogen atoms, and the unpaired electrons are shared to give a double bond. In this way, each of the carbon atoms achieves an octet.

1.5. Each of the carbon atoms has four valence electrons; the nitrogen atom has five valence electrons; and each of the hydrogen atoms has one valence electron. We begin by connecting the atoms that have more than one bond (in this case, the three carbon atoms and the nitrogen atom). There are four different ways that these four atoms can be connected to each other, shown here.

For each of these possible arrangements, we connect the hydrogen atoms, giving the following four constitutional isomers.

(c) Each carbon atom has four valence electrons, and each hydrogen atom has one valence electron. Only the carbon atoms can form more than one bond, so we begin by connecting the carbon atoms to each other. Then, we connect all of the hydrogen atoms, and the unpaired electrons are shared to give a triple bond. In this way, each of the carbon atoms achieves an octet.

(d) Each carbon atom has four valence electrons, and each hydrogen atom has one valence electron. Only the carbon atoms can form more than one bond, so we begin by connecting the carbon atoms to each other. Then, we connect all of the hydrogen atoms, as shown.

(e) The carbon atom has four valence electrons, the oxygen atom has six valence electrons, and each hydrogen atom has one valence electron. Only the carbon atom and the oxygen atom can form more than one bond, so we begin by connecting them to each other. Then, we connect all of the hydrogen atoms, as shown.

1.4. Boron is in column 3A of the periodic table, so it has three valence electrons. Each of these valence electrons is shared with a hydrogen atom, shown below. The central boron atom lacks an octet of electrons, and it is therefore very unstable and reactive.

In each of these four structures, the nitrogen atom has one lone pair. 1.6. (a) The carbon atom has four valence electrons, the nitrogen atom has five valence electrons and the hydrogen atom has one valence electron. Only the carbon atom and the nitrogen atom can form more than one bond, so we begin by connecting them to each other. Then, we connect the hydrogen atom to the carbon, as shown. The unpaired electrons are shared to give a triple bond. In this way, both the carbon atom and the nitrogen atom achieve an octet.

(b) Each carbon atom has four valence electrons, and each hydrogen atom has one valence electron. Only the carbon atoms can form more than one bond, so we begin by connecting the carbon atoms to each other. Then, we connect all of the hydrogen atoms as indicated in the given condensed formula (CH2CHCHCH2), and the unpaired electrons are shared to give two double bonds on the outermost carbons. In this way, each of the carbon atoms achieves an octet.

1.7. (a) Aluminum is in group 3A of the periodic table, and it should therefore have three valence electrons. In this

CHAPTER 1

case, the aluminum atom exhibits four valence electrons (one for each bond). With one extra electron, this aluminum atom will bear a negative charge.

the oxygen atom exhibits only five valence electrons (one for each bond, and two for the lone pair). This oxygen atom is missing an electron, and it therefore bears a positive charge.

(b) Oxygen is in group 6A of the periodic table, and it should therefore have six valence electrons. In this case, the oxygen atom exhibits only five valence electrons (one for each bond, and two for the lone pair). This oxygen atom is missing an electron, and it therefore bears a positive charge.

(h) Two of the atoms in this structure exhibit a formal charge because each of these atoms does not exhibit the appropriate number of valence electrons. The aluminum atom (group 3A) should have three valence electrons, but it exhibits four (one for each bond). With one extra electron, this aluminum atom will bear a negative charge. The neighboring chlorine atom (to the right) should have seven valence electrons, but it exhibits only six (one for each bond and two for each lone pair). It is missing one electron, so this chlorine atom will bear a positive charge.

(c) Nitrogen is in group 5A of the periodic table, and it should therefore have five valence electrons. In this case, the nitrogen atom exhibits six valence electrons (one for each bond and two for each lone pair). With one extra electron, this nitrogen atom will bear a negative charge.

(d) Oxygen is in group 6A of the periodic table, and it should therefore have six valence electrons. In this case, the oxygen atom exhibits only five valence electrons (one for each bond, and two for the lone pair). This oxygen atom is missing an electron, and it therefore bears a positive charge.

(e) Carbon is in group 4A of the periodic table, and it should therefore have four valence electrons. In this case, the carbon atom exhibits five valence electrons (one for each bond and two for the lone pair). With one extra electron, this carbon atom will bear a negative charge.

(f) Carbon is in group 4A of the periodic table, and it should therefore have four valence electrons. In this case, the carbon atom exhibits only three valence electrons (one for each bond). This carbon atom is missing an electron, and it therefore bears a positive charge.

(g) Oxygen is in group 6A of the periodic table, and it should therefore have six valence electrons. In this case,

(i) Two of the atoms in this structure exhibit a formal charge because each of these atoms does not exhibit the appropriate number of valence electrons. The nitrogen atom (group 5A) should have five valence electrons, but it exhibits four (one for each bond). It is missing one electron, so this nitrogen atom will bear a positive charge. One of the two oxygen atoms (the one on the right) exhibits seven valence electrons (one for the bond, and two for each lone pair), although it should have only six. With one extra electron, this oxygen atom will bear a negative charge.

1.8. (a) The boron atom in this case exhibits four valence electrons (one for each bond), although boron (group 3A) should only have three valence electrons. With one extra electron, this boron atom bears a negative charge.

(b) Nitrogen is in group 5A of the periodic table, so a nitrogen atom should have five valence electrons. A negative charge indicates one extra electron, so this nitrogen atom must exhibit six valence electrons (one for each bond and two for each lone pair).

CHAPTER 1

(c) One of the carbon atoms (below right) exhibits three valence electrons (one for each bond), but carbon (group 4A) is supposed to have four valence electrons. It is missing one electron, so this carbon atom therefore bears a positive charge.

1.9. Carbon is in group 4A of the periodic table, and it should therefore have four valence electrons. Every carbon atom in acetylcholine has four bonds, thus exhibiting the correct number of valence electrons (four) and having no formal charge.

Oxygen is in group 6A of the periodic table, and it should therefore have six valence electrons. Each oxygen atom in acetylcholine has two bonds and two lone pairs of electrons, so each oxygen atom exhibits six valence electrons (one for each bond, and two for each lone pair). With the correct number of valence electrons, each oxygen atom will lack a formal charge.

The nitrogen atom (group 5A) should have five valence electrons, but it exhibits four (one for each bond). It is missing one electron, so this nitrogen atom will bear a positive charge.

(b) Fluorine is more electronegative than carbon, and a C–F bond is polar covalent. For a C–F bond, the F will be electron-rich (‒), and the C will be electron-poor (+). Chlorine is also more electronegative than carbon, so a C–Cl bond is also polar covalent. For a C–Cl bond, the Cl will be electron-rich (‒), and the C will be electronpoor (+), as shown below.

(c) Carbon is more electronegative than magnesium, so the C will be electron-rich (‒) in a C–Mg bond, and the Mg will be electron-poor (+). Also, bromine is more electronegative than magnesium. So in a Mg–Br bond, the Br will be electron-rich (‒), and the Mg will be electron-poor (+), as shown below.

(d) Oxygen is more electronegative than carbon or hydrogen, so all C–O bonds and all O–H bond are polar covalent. For each C–O bond and each O–H bond, the O will be electron-rich (‒), and the C or H will be electronpoor (+), as shown below.

(e) Oxygen is more electronegative than carbon. As such, the O will be electron-rich (‒) and the C will be electronpoor (+) in a C=O bond, as shown below.

(f) Chlorine is more electronegative than carbon. As such, for each C–Cl bond, the Cl will be electron-rich (‒) and the C will be electron-poor (+), as shown below. 1.10. (a) Oxygen is more electronegative than carbon, and a C–O bond is polar covalent. For each C–O bond, the O will be electron-rich (‒), and the C will be electron-poor (+), as shown below. 1.11. Oxygen is more electronegative than carbon. As such, the O will be electron-rich (‒) and the C will be electron-poor (+) in a C=O bond. In addition, chlorine is more electronegative than carbon. So for a C–Cl bond,

CHAPTER 1 the Cl will be electron-rich (‒) and the C will be electronpoor (+), as shown below.

(d) Each corner represents a carbon atom (highlighted below), so this compound has seven carbon atoms. Each carbon atom has enough attached hydrogen atoms to have exactly four bonds, as shown:

Notice that two carbon atoms are electron-poor (+). These are the positions that are most likely to be attacked by an electron-rich anion, such as hydroxide. 1.12. Oxygen is more electronegative than carbon. As such, the O will be electron-rich (δ−) and the C will be electron-poor (δ+) in a C─O bond. In addition, chlorine is more electronegative than carbon. So for a C─Cl bond, the Cl will be electron-rich (δ−) and the C will be electronpoor (δ+), as shown below. As you might imagine, epichlorohydrin is a very reactive molecule!

1.13. (a) Each corner and each endpoint represents a carbon atom (highlighted below), so this compound has six carbon atoms. Each carbon atom has enough attached hydrogen atoms to have exactly four bonds, as shown:

(e) Each corner and each endpoint represents a carbon atom (highlighted below), so this compound has seven carbon atoms. Each carbon atom has enough attached hydrogen atoms to have exactly four bonds, as shown:

(f) Each corner represents a carbon atom (highlighted below), so this compound has seven carbon atoms. Each carbon atom has enough attached hydrogen atoms to have exactly four bonds, as shown:

1.14. Remember that each corner and each endpoint represents a carbon atom. This compound therefore has 16 carbon atoms, highlighted below: N

(b) Each corner and each endpoint represents a carbon atom (highlighted below), so this compound has twelve carbon atoms. Each carbon atom has enough attached hydrogen atoms to have exactly four bonds, as shown:

N NH

Each carbon atom should have four bonds. We therefore draw enough hydrogen atoms in order to give each carbon atom a total of four bonds. Any carbon atoms that already have four bonds will not have any hydrogen atoms: (c) Each corner represents a carbon atom (highlighted below), so this compound has six carbon atoms. Each carbon atom has enough attached hydrogen atoms to have exactly four bonds, as shown:

H H H H H C H C H C C N N H N C O C C N C C C N H C C C H H C C H H HH H H H H

1.15. (a) As indicated in Figure 1.10, carbon has two 1s electrons, two 2s electrons, and two 2p electrons. This

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information is represented by the following electron configuration: 1s22s22p2 (b) As indicated in Figure 1.10, oxygen has two 1s electrons, two 2s electrons, and four 2p electrons. This information is represented by the following electron configuration: 1s22s22p4 (c) As indicated in Figure 1.10, boron has two 1s electrons, two 2s electrons, and one 2p electron. This information is represented by the following electron configuration: 1s22s22p1 (d) As indicated in Figure 1.10, fluorine has two 1s electrons, two 2s electrons, and five 2p electrons. This information is represented by the following electron configuration: 1s22s22p5 (e) Sodium has two 1s electrons, two 2s electrons, six 2p electrons, and one 3s electron. This information is represented by the following electron configuration: 1s22s22p63s1 (f) Aluminum has two 1s electrons, two 2s electrons, six 2p electrons, two 3s electrons, and one 3p electron. This information is represented by the following electron configuration: 1s22s22p63s23p1 1.16. (a) The electron configuration of a carbon atom is 1s22s22p2 (see the solution to Problem 1.15a). However, if a carbon atom bears a negative charge, then it must have one extra electron, so the electron configuration should be as follows: 1s22s22p3 (b) The electron configuration of a carbon atom is 1s22s22p2 (see the solution to Problem 1.15a). However, if a carbon atom bears a positive charge, then it must be missing an electron, so the electron configuration should be as follows: 1s22s22p1 (c) As seen in SkillBuilder 1.6, the electron configuration of a nitrogen atom is 1s22s22p3. However, if a nitrogen atom bears a positive charge, then it must be missing an electron, so the electron configuration should be as follows: 1s22s22p2 (d) The electron configuration of an oxygen atom is 1s22s22p4 (see the solution to Problem 1.15b). However, if an oxygen atom bears a negative charge, then it must have one extra electron, so the electron configuration should be as follows: 1s22s22p5 1.17. Silicon is in the third row, or period, of the periodic table. Therefore, it has a filled second shell, like neon, and then the additional electrons are added to the third shell. As indicated in Figure 1.10, neon has two 1s electrons, two 2s electrons, and six 2p electrons. Silicon has an additional two 3s electrons and two 3p electrons to give a total of 14 electrons and an electron configuration of 1s22s22p63s23p2.

1.18. The angles of an equilateral triangle are 60º, but each bond angle of cyclopropane is supposed to be 109.5º. Therefore, each bond angle is severely strained, causing an increase in energy. This form of strain, called ring strain, will be discussed in Chapter 4. The ring strain associated with a three-membered ring is greater than the ring strain of larger rings, because larger rings do not require bond angles of 60º. 1.19. (a) The C=O bond of formaldehyde is comprised of one  bond and one  bond. (b) Each C‒H bond is formed from the interaction between an sp2-hybridized orbital from carbon and an s orbital from hydrogen. (c) The oxygen atom is sp2 hybridized, so the lone pairs occupy sp2-hybridized orbitals. 1.20. Rotation of a single bond does not cause a reduction in the extent of orbital overlap, because the orbital overlap occurs on the bond axis. In contrast, rotation of a  bond results in a reduction in the extent of orbital overlap between the two p orbitals, because the orbital overlap is NOT on the bond axis. 1.21. (a) The highlighted carbon atom (below) has four bonds, and is therefore sp3 hybridized. The other carbon atoms in this structure are all sp2 hybridized, because each of them has three bonds and one  bond.

(b) Each of the highlighted carbon atoms has four bonds, and is therefore sp3 hybridized. The other two carbon atoms in this structure are sp hybridized, because each has two bonds and two  bonds.

(c) Each of the highlighted carbon atoms (below) has four bonds, and is therefore sp3 hybridized. The other two carbon atoms in this structure are sp2 hybridized, because each has three bonds and one  bond.

CHAPTER 1 (d) Each of the two central carbon atoms has two bonds and two  bonds, and as such, each of these carbon atoms is sp hybridized. The other two carbon atoms (the outer ones) are sp2 hybridized because each has three bonds and one  bond.

(e) One of the carbon atoms (the one connected to oxygen) has two bonds and two  bonds, and as such, it is sp hybridized. The other carbon atom is sp2 hybridized because it has three bonds and one  bond.

1.22. Each of the following three highlighted carbon atoms has four  bonds, and is therefore sp3 hybridized:

And each of the following three highlighted carbon atoms has three  bonds and one  bond, and is therefore sp2 hybridized:

Finally, each of the following five highlighted carbon atoms has two  bonds and two  bonds, and is therefore sp hybridized.

1.23. Carbon-carbon triple bonds generally have a shorter bond length than carbon-carbon double bonds, which are generally shorter than carbon-carbon single bonds (see Table 1.2).

1.24. (a) In this structure, the boron atom has four  bonds and no lone pairs, giving a total of four electron pairs (steric number = 4). VSEPR theory therefore predicts a tetrahedral arrangement of electron pairs. Since all of the electron pairs are bonds, the structure is expected to have tetrahedral geometry. (b) In this structure, the boron atom has three  bonds and no lone pairs, giving a total of three electron pairs (steric number = 3). VSEPR theory therefore predicts a trigonal planar geometry. (c) In this structure, the nitrogen atom has four  sigma bonds and no lone pairs, giving a total of four electron pairs (steric number = 4). VSEPR theory therefore predicts a tetrahedral arrangement of electron pairs. Since all of the electron pairs are bonds, the structure is expected to have tetrahedral geometry. (d) The carbon atom has four  bonds and no lone pairs, giving a total of four electron pairs (steric number = 4). VSEPR theory therefore predicts a tetrahedral arrangement of electron pairs. Since all of the electron pairs are bonds, the structure is expected to have tetrahedral geometry. 1.25. In the carbocation, the carbon atom has three bonds and no lone pairs. Since there are a total of three electron pairs (steric number = 3), and all three are bonds, VSEPR theory predicts trigonal planar geometry, with bond angles of 120°. In contrast, the carbon atom of the carbanion has three bonds and one lone pair, giving a total of four electron pairs (steric number = 4). For this ion, VSEPR theory predicts a tetrahedral arrangement of electron pairs, with a lone pair positioned at one corner of the tetrahedron, giving rise to trigonal pyramidal geometry with bond angles approximately 107°. 1.26. In ammonia, the nitrogen atom has three bonds and one lone pair. Therefore, VSEPR theory predicts trigonal pyramidal geometry, with bond angles of approximately 107°. In the ammonium ion, the nitrogen atom has four bonds and no lone pairs, so VSEPR theory predicts tetrahedral geometry, with bond angles of 109.5°. Therefore, we predict that the bond angles will increase (by approximately 2.5°) as a result of the reaction. 1.27. The silicon atom has four  bonds and no lone pairs, so the steric number is 4 (sp3 hybridization), which means

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that the arrangement of electron pairs will be tetrahedral. With no lone pairs, the arrangement of the atoms (geometry) is the same as the electronic arrangement. It is tetrahedral.

1.28. (a) This compound has three C–Cl bonds, each of which exhibits a dipole moment. To determine if these dipole moments cancel each other, we must identify the molecular geometry. The central carbon atom has four bonds so we expect tetrahedral geometry. As such, the three polar C–Cl bonds do not lie in the same plane, and they do not completely cancel each other out. There is a net molecular dipole moment, as shown:

(b) The oxygen atom has two bonds and two lone pairs (steric number = 4), and VSEPR theory predicts bent geometry. As such, the dipole moments associated with the polar C–O bonds do not fully cancel each other, and the dipole moments associated with the lone pairs also do not fully cancel each other. As a result, there is a net molecular dipole moment, as shown:

(c) The nitrogen atom has three bonds and one lone pair (steric number = 4), and VSEPR theory predicts trigonal pyramidal geometry (because one corner of the tetrahedron is occupied by a lone pair). As such, the dipole moments associated with the polar N–H bonds do not fully cancel each other, and there is also a dipole moment associated with the lone pair (pointing up). As a result, there is a net molecular dipole moment, as shown:

C–Br bonds, and as such, there is a net molecular dipole moment, shown here:

(e) The oxygen atom has two bonds and two lone pairs (steric number = 4), and VSEPR theory predicts bent geometry. As such, the dipole moments associated with the polar C–O bonds do not fully cancel each other, and the dipole moments associated with the lone pairs also do not fully cancel each other. As a result, there is a net molecular dipole moment, as shown:

(f) There are individual dipole moments associated with each polar C–O bond and the lone pairs (as in the previous solution), but due to the symmetrical shape of the molecule in this case, they fully cancel each other to give no net molecular dipole moment. (g) Each C=O bond has a strong dipole moment, and they do not fully cancel each other because they are not pointing in opposite directions. As such, there will be a net molecular dipole moment, as shown here:

(h) Each C=O bond has a strong dipole moment, and in this case, they are pointing in opposite directions. As such, they fully cancel each other, giving no net molecular dipole moment. (i) Each C–Cl bond has a dipole moment, and they do not fully cancel each other because the polar bonds are not pointing in opposite directions. As such, there will be a net molecular dipole moment, as shown here:

(d) The central carbon atom has four bonds (steric number = 4), and VSEPR theory predicts tetrahedral geometry. There are individual dipole moments associated with each of the C–Cl bonds and each of the C–Br bonds. If all four dipole moments had the same magnitude, then we would expect them to completely cancel each other to give no molecular dipole moment (as in the case of CCl4). However, because Cl is more electronegative than Br, each C–Cl bond is more polar than each C–Br bond. Therefore, the dipole moments for the C–Cl bonds are larger than the dipole moments of the

(j) Each C–Cl bond has a dipole moment, and in this case, they are pointing in opposite directions. As such, they fully cancel each other, giving no net molecular dipole moment. (k) Each C–Cl bond has a dipole moment, and they do not fully cancel each other because they are not pointing

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in opposite directions. As such, there will be a net molecular dipole moment, as shown here:

(c) The second compound is expected to have a higher boiling point, because it has an O–H bond, which will lead to hydrogen-bonding interactions between molecules.

(l) Each C–Cl bond has a dipole moment, but in this case, they fully cancel each other to give no net molecular dipole moment.

(d) Both compounds have the same molecular weight because they are isomers, and they are both capable of forming hydrogen bonds. The first compound is expected to have a higher boiling point, however, because it is less branched. The greater surface area in the first compound results in greater London dispersion forces and a higher boiling point (b.p.):

1.29. Each of the C–O bonds has an individual dipole moment, shown here:

To determine if these individual dipole moments fully cancel each other, we must determine the geometry around the oxygen atom. The oxygen atom has two  bonds and two lone pairs, giving rise to a bent geometry. As such, the dipole moments associated with the polar C–O bonds do NOT fully cancel each other,

and the dipole moments associated with the lone pairs also do not fully cancel each other. As a result, there is a net molecular dipole moment, as shown:

1.30. (a) Both compounds have the same molecular weight because they are isomers. The second compound is expected to have a higher boiling point, because it is less branched. The greater surface area in the second compound results in greater London dispersion forces and a higher boiling point (b.p.):

1.31. Compound 3 is expected to have a higher boiling point than compound 4, because only compound 3 has an O-H group. Compound 4 does not form hydrogen-bonds, so it will have a lower boiling point. When this mixture is heated, the lower boiling compound (4) can be collected first, leaving behind compound 3.

1.32. (a) The carbon atoms are tetravalent, while the chlorine atom and hydrogen atoms are all monovalent. The atoms with more than one bond (in this case, the two carbon atoms) should be drawn in the center of the compound. The chlorine atom and hydrogen atoms are then placed at the periphery (ensuring that each carbon atom has a total of four bonds), as shown:

The chlorine atom can be placed in any one of the six available positions. The following six drawings all represent the same compound, in which the two carbon atoms are connected to each other, and the chlorine atom is connected to one of the carbon atoms. (b) The second compound is expected to have a higher boiling point, because more carbon atoms results in a higher molecular weight. Larger compounds have greater London dispersion forces and a higher boiling point (b.p.):

higher b.p.

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(b) The carbon atoms are tetravalent, while the chlorine atoms and hydrogen atoms are all monovalent. The atoms with more than one bond (in this case, the two carbon atoms) should be drawn in the center of the compound. The chlorine atoms and hydrogen atoms are then placed at the periphery, and there are two different ways to do this. The two chlorine atoms can either be connected to the same carbon atom or to different carbon atoms, as shown.

(c) The carbon atoms are tetravalent, while the chlorine atoms and hydrogen atoms are all monovalent. The atoms with more than one bond (in this case, the two carbon atoms) should be drawn in the center of the compound. The chlorine atoms and hydrogen atoms are then placed at the periphery, and there are two different ways to do this. One way is to connect all three chlorine atoms to the same carbon atom. Alternatively, we can connect two chlorine atoms to one carbon atom, and then connect the third chlorine atom to the other carbon atom, as shown here:

(d) The carbon atoms are tetravalent, and the hydrogen atoms are all monovalent. Any atoms with more than one bond (in this case, the six carbon atoms) should be drawn in the center of the compound, with the hydrogen atoms at the periphery. There are five different ways to connect six carbon atoms, which we will organize based on the length of the longest chain.

1.33. (a) According to Table 1.1, the difference in electronegativity between Br and H is 2.8 – 2.1 = 0.7, so an H–Br bond is expected to be polar covalent. Since bromine is more electronegative than hydrogen, the Br will be electron-rich (‒), and the H will be electron-poor (+), as shown below:

(b) According to Table 1.1, the difference in electronegativity between Cl and H is 3.0 – 2.1 = 0.9, so an H–Cl bond is expected to be polar covalent. Since chlorine is more electronegative than hydrogen, the Cl will be electron-rich (‒), and the H will be electron-poor (+), as shown below:

(c) According to Table 1.1, the difference in electronegativity between O and H is 3.5 – 2.1 = 1.4, so an O–H bond is expected to be polar covalent. Oxygen is more electronegative than hydrogen, so for each O–H bond, the O will be electron-rich (‒) and the H will be electron-poor (+), as shown below:

Finally, we complete all of the structures by drawing the bonds to hydrogen atoms (ensuring that each carbon atom has a total of four bonds). There are a total of five isomers:

(d) According to Table 1.1, oxygen (3.5) is more electronegative than carbon (2.5) or hydrogen (2.1), and a C–O or H–O bond is polar covalent. For each C–O or H–O bond, the O will be electron-rich (‒), and the C or H will be electron-poor (+), as shown below:

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1.34. (a) The difference in electronegativity between Na (0.9) and Br (2.8) is greater than the difference in electronegativity between H (2.1) and Br (2.8). Therefore, NaBr is expected to have more ionic character than HBr. (b) The difference in electronegativity between F (4.0) and Cl (3.0) is greater than the difference in electronegativity between Br (2.8) and Cl (3.0). Therefore, FCl is expected to have more ionic character than BrCl. 1.35. (a) Each carbon atom has four valence electrons, the oxygen atom has six valence electrons, and each hydrogen atom has one valence electron. In this case, the information provided in the problem statement (CH3CH2OH) indicates how the atoms are connected to each other:

(b) Each carbon atom has four valence electrons, the nitrogen atom has five valence electrons, and each hydrogen atom has one valence electron. In this case, the information provided in the problem statement (CH3CN) indicates how the atoms are connected to each other:

The nitrogen atom has three bonds and one lone pair, so the steric number is 4, which means that the arrangement of electron pairs is expected to be tetrahedral. One corner of the tetrahedron is occupied by a lone pair, so the geometry of the nitrogen atom (the arrangement of atoms around that nitrogen atom) is trigonal pyramidal. As such, the individual dipole moments associated with the C–N bonds do not fully cancel each other, and there is also a dipole moment associated with the lone pair (pointing up). As a result, there is a net molecular dipole moment, as shown:

1.37. Bromine is in group 7A of the periodic table, so each bromine atom has seven valence electrons. Aluminum is in group 3A of the periodic table, so aluminum is supposed to have three valence electrons, but the structure bears a negative charge, which means that there is one extra electron. That is, the aluminum atom has four valence electrons, rather than three, which is why it has a formal negative charge. This gives the following Lewis structure:

The unpaired electrons are then paired up to give a triple bond. In this way, each of the atoms achieves an octet. The aluminum atom has four bonds and no lone pairs, so the steric number is 4, which means that this aluminum atom will have tetrahedral geometry. 1.36. Each of the carbon atoms has four valence electrons; the nitrogen atom has five valence electrons, and each of the hydrogen atoms has one valence electron. We begin by connecting the atoms that have more than one bond (in this case, the four carbon atoms and the nitrogen atom). The problem statement indicates how we should connect them:

Then, we connect all of the hydrogen atoms (ensuring that each carbon atom has four bonds), as shown.

1.38. The molecular formula of cyclopropane is C3H6, so we are looking for a different compound that has the same molecular formula, C3H6. That is, we need to find another way to connect the carbon atoms, other than in a ring (there is only one way to connect three carbon atoms in a ring, so we must be looking for something other than a ring). If we connect the three carbon atoms in a linear fashion and then complete the drawing by placing hydrogen atoms at the periphery, we notice that the molecular formula (C3H8) is not correct:

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We are looking for a structure with the molecular formula C3H6. If we remove two hydrogen atoms from our drawing, we are left with two unpaired electrons, indicating that we should consider drawing a double bond:

The structure of this compound (called propylene) is different from the structure of cyclopropane, but both compounds share the same molecular formula, so they are constitutional isomers. 1.39. (a) C–H bonds are considered to be nonpolar, although they do have a very small dipole moment, because there is a small difference in electronegativity between carbon (2.5) and hydrogen (2.1). With no polar bonds present, the molecule does not have a molecular dipole moment. (b) The nitrogen atom has trigonal pyramidal geometry. As such, the dipole moments associated with the polar N– H bonds do not fully cancel each other, and there is also a dipole moment associated with the lone pair (pointing up). As a result, there is a net molecular dipole moment, as shown:

(c) The oxygen atom has two bonds and two lone pairs (steric number = 4), and VSEPR predicts bent geometry. As such, the dipole moments associated with the polar O– H bonds do not cancel each other, and the dipole moments associated with the lone pairs also do not fully cancel each other. As a result, there is a net molecular dipole moment, as shown:

(d) The central carbon atom of carbon dioxide (CO2) has two bonds and no lone pairs, so it is sp hybridized and is expected to have linear geometry. Each C=O bond has a strong dipole moment, but in this case, they are pointing in opposite directions. As such, they fully cancel each other, giving no net molecular dipole moment. (e) Carbon tetrachloride (CCl4) has four C–Cl bonds, each of which exhibits a dipole moment. However, the central carbon atom has four bonds so it is expected to have tetrahedral geometry. As such, the four dipole moments

completely cancel each other out, and there is no net molecular dipole moment. (f) This compound has two C–Br bonds, each of which exhibits a dipole moment. To determine if these dipole moments cancel each other, we must identify the molecular geometry. The central carbon atom has four bonds so it is expected to have tetrahedral geometry. As such, the polar C–Br bonds do not completely cancel each other out. There is a net molecular dipole moment, as shown:

1.40. (a) As indicated in Figure 1.10, a neutral oxygen atom has two 1s electrons, two 2s electrons, and four 2p electrons. (b) As indicated in Figure 1.10, a neutral fluorine atom has two 1s electrons, two 2s electrons, and five 2p electrons. (c) As indicated in Figure 1.10, a neutral carbon atom has two 1s electrons, two 2s electrons, and two 2p electrons. (d) As seen in SkillBuilder 1.6, the electron configuration of a neutral nitrogen atom is 1s22s22p3 (e) This is the electron configuration of a neutral chlorine atom. 1.41. (a) The difference in electronegativity between sodium (0.9) and bromine (2.8) is 2.8 – 0.9 = 1.9. Since this difference is greater than 1.7, the bond is classified as ionic. (b) The difference in electronegativity between sodium (0.9) and oxygen (3.5) is 3.5 – 0.9 = 2.6. Since this difference is greater than 1.7, the Na–O bond is classified as ionic. In contrast, the O–H bond is polar covalent, because the difference in electronegativity between oxygen (3.5) and hydrogen (2.1) is less than 1.7 but more than 0.5. (c) Each C–H bond is considered to be covalent, because the difference in electronegativity between carbon (2.5) and hydrogen (2.1) is less than 0.5. The C–O bond is polar covalent, because the difference in electronegativity between oxygen (3.5) and carbon (2.5) is less than 1.7 but more than 0.5. The Na–O bond is classified as ionic, because the difference in electronegativity between oxygen (3.5) and sodium (0.9) is greater than 1.7. (d) Each C–H bond is considered to be covalent, because the difference in electronegativity between carbon (2.5) and hydrogen (2.1) is less than 0.5. The C–O bond is polar covalent, because the difference in electronegativity between oxygen (3.5) and carbon (2.5) is less than 1.7 but more than 0.5.

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The O–H bond is polar covalent, because the difference in electronegativity between oxygen (3.5) and hydrogen (2.1) is less than 1.7 but more than 0.5. (e) Each C–H bond is considered to be covalent, because the difference in electronegativity between carbon (2.5) and hydrogen (2.1) is less than 0.5. The C=O bond is polar covalent, because the difference in electronegativity between oxygen (3.5) and carbon (2.5) is less than 1.7 but more than 0.5.

1.42. (a) Begin by determining the valency of each atom in the compound. The carbon atoms are tetravalent, the oxygen atom is divalent, and the hydrogen atoms are all monovalent. Any atoms with more than one bond (in this case, the two carbon atoms and the oxygen atom) should be drawn in the center of the compound, with the hydrogen atoms at the periphery. There are two different ways to connect two carbon atoms and an oxygen atom, shown here:

We then complete both structures by drawing the remaining bonds to hydrogen atoms (ensuring that each carbon atom has four bonds, and each oxygen atom has two bonds):

(b) Begin by determining the valency of each atom in the compound. The carbon atoms are tetravalent, the oxygen atoms are divalent, and the hydrogen atoms are all monovalent. Any atoms with more than one bond (in this case, the two carbon atoms and the two oxygen atoms) should be drawn in the center of the compound, with the hydrogen atoms at the periphery. There are several different ways to connect two carbon atoms and two oxygen atoms (highlighted, for clarity of comparison), shown here:

(c) The carbon atoms are tetravalent, while the bromine atoms and hydrogen atoms are all monovalent. The atoms with more than one bond (in this case, the two carbon atoms) should be drawn in the center of the compound. The bromine atoms and hydrogen atoms are then placed at the periphery, and there are two different ways to do this. The two bromine atoms can either be connected to the same carbon atom or to different carbon atoms, as shown.

1.43. (a) Oxygen is more electronegative than carbon, and the withdrawal of electron density toward oxygen can be indicated with the following arrow:

(b) Carbon is more electronegative than magnesium, and the withdrawal of electron density toward carbon can be indicated with the following arrow:

(c) Nitrogen is more electronegative than carbon, and the withdrawal of electron density toward nitrogen can be indicated with the following arrow:

(d) Carbon is more electronegative than lithium, and the withdrawal of electron density toward carbon can be indicated with the following arrow:

We then complete all of these structures by drawing the remaining bonds to hydrogen atoms:

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(e) Chlorine is more electronegative than carbon, and the withdrawal of electron density toward chlorine can be indicated with the following arrow:

(e) The oxygen atom has two bonds and two lone pairs (steric number = 4), and VSEPR theory predicts bent geometry. Therefore, the C-O-C bond angle is expected to be around 105º. The remaining bond angles are all expected to be approximately 109.5º (because each carbon atom has four bonds and tetrahedral geometry).

(f) Carbon is more electronegative than silicon, and the withdrawal of electron density toward carbon can be indicated with the following arrow:

(f) The nitrogen atom has three bonds and one lone pair (steric number = 4), and VSEPR theory predicts trigonal pyramidal geometry, with bond angles of 107º. The carbon atom is also tetrahedral (because it has four bonds), although the bond angles around the carbon atom are expected to be approximately 109.5º.

(g) Oxygen is more electronegative than hydrogen, and the withdrawal of electron density toward oxygen can be indicated with the following arrow:

(h) Nitrogen is more electronegative than hydrogen, and the withdrawal of electron density toward nitrogen can be indicated with the following arrow:

1.44. (a) The oxygen atom has two bonds and two lone pairs (steric number = 4), and VSEPR theory predicts bent geometry. The C-O-H bond angle is expected to be approximately 105º, and all other bonds angles are expected to be approximately 109.5º (because each carbon atom has four bonds and tetrahedral geometry). (b) The central carbon atom has three bonds and no lone pairs (steric number = 3), and VSEPR theory predicts trigonal planar geometry. As such, all bond angles are approximately 120º.

(c) Each of the carbon atoms has three bonds and no lone pairs (steric number = 3), and VSEPR theory predicts trigonal planar geometry. As such, all bond angles are approximately 120º.

(g) Each of the carbon atoms has four bonds (steric number = 4), so each of these carbon atoms has tetrahedral geometry. Therefore, all bond angles are expected to be approximately 109.5º.

(h) The structure of acetonitrile (CH3CN) is shown below (see the solution to Problem 1.35b).

One of the carbon atoms has four bonds (steric number = 4), and is expected to have tetrahedral geometry. The other carbon atom (connected to nitrogen) has two bonds and no lone pairs (steric number = 2), so we expect linear geometry. As such, the C–C≡N bond angle is 180º, and all other bond angles are approximately 109.5º. 1.45. (a) The nitrogen atom has three bonds and one lone pair (steric number = 4). It is sp3 hybridized (electronically tetrahedral), with trigonal pyramidal geometry (because one corner of the tetrahedron is occupied by a lone pair). (b) The boron atom has three bonds and no lone pairs (steric number = 3). It is sp2 hybridized, with trigonal planar geometry.

(d) Each of the carbon atoms has two bonds and no lone pairs (steric number = 2), and VSEPR theory predicts linear geometry. As such, all bond angles are 180º.

(c) This carbon atom has three bonds and no lone pairs (steric number = 3). It is sp2 hybridized, with trigonal planar geometry. (d) This carbon atom has three bonds and one lone pair (steric number = 4). It is sp3 hybridized (electronically

CHAPTER 1 tetrahedral), with trigonal pyramidal geometry (because one corner of the tetrahedron is occupied by a lone pair). 1.46. (a) Each corner and each endpoint represents a carbon atom (highlighted), so this compound has nine carbon atoms. Each carbon atom will have enough hydrogen atoms to have exactly four bonds, as shown.

1.48. The double bond represents one bond and one  bond, while the triple bond represents one bond and two  bonds. All single bonds are bonds. Therefore, this compound has sixteen bonds and three  bonds.

1.49. (a) The second compound is expected to have a higher boiling point, because it has an O–H bond, which will lead to hydrogen bonding interactions.

(b) Each corner and each endpoint represents a carbon atom (highlighted below), so this compound has eight carbon atoms. Each carbon atom will have enough hydrogen atoms to have exactly four bonds, as shown.

(b) The second compound is expected to have a higher boiling point, because it has more carbon atoms, and thus a higher molecular weight and more opportunity for London dispersion forces. (c) The first compound has a C=O bond, which has a strong dipole moment, while the second compound is nonpolar. The first compound is therefore expected to exhibit strong dipole-dipole interactions and to have a higher boiling point than the second compound. 1.50. (a) This compound possesses an O–H bond, so it is expected to exhibit hydrogen bonding interactions.

(c) Each corner and each endpoint represents a carbon atom (highlighted below), so this compound has eight carbon atoms. Each carbon atom will have enough hydrogen atoms to have exactly four bonds, as shown. (b) This compound lacks a hydrogen atom that is connected to an oxygen or nitrogen atom. Therefore, this compound cannot serve as a hydrogen-bond donor (although the lone pairs can serve as hydrogen-bond acceptors, in the presence of a hydrogen-bond donor). As a pure compound, we do not expect there to be any hydrogen bonding interactions.

1.47. Each corner and each endpoint represents a carbon atom, so this compound has fifteen carbon atoms. Each carbon atom will have enough hydrogen atoms to have exactly four bonds, giving a total of eighteen hydrogen atoms, as shown here:

(c) This compound lacks a hydrogen atom that is connected to an oxygen or nitrogen atom. Therefore, this compound will not exhibit hydrogen bonding interactions.

(d) This compound lacks a hydrogen atom that is connected to an oxygen or nitrogen atom . Therefore, this compound will not exhibit hydrogen bonding interactions.

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(e) This compound lacks a hydrogen atom that is connected to an oxygen or nitrogen atom. Therefore, this compound cannot serve as a hydrogen-bond donor (although the lone pairs can serve as hydrogen-bond acceptors, in the presence of a hydrogen-bond donor). As a pure compound, we do not expect there to be any hydrogen bonding interactions.

(f) This compound possesses two N–H bonds, so it is expected to exhibit hydrogen bonding interactions.

1.52. (a) Each of the highlighted carbon atoms has three bonds and no lone pairs (steric number = 3). Each of these carbon atoms is sp2 hybridized, with trigonal planar geometry. Each of the other four carbon atoms has two bonds and no lone pairs (steric number = 2). Those four carbon atoms are all sp hybridized, with linear geometry.

(b) The highlighted carbon atom has three bonds and no lone pairs (steric number = 3). This carbon atom is sp2 hybridized, with trigonal planar geometry. Each of the other three carbon atoms has four bonds (steric number = 4). Those three carbon atoms are all sp3 hybridized, with tetrahedral geometry.

(g) This compound lacks a hydrogen atom that is connected to an oxygen or nitrogen atom. Therefore, this compound will not exhibit hydrogen bonding interactions.

(h) This compound possesses N–H bonds, so it is expected to exhibit hydrogen bonding interactions.

1.53. Each of the highlighted carbon atoms has four bonds (steric number = 4), and is sp3 hybridized, with tetrahedral geometry. Each of the other fourteen carbon atoms in this structure has three bonds and no lone pairs (steric number = 3). Each of these fourteen carbon atoms is sp2 hybridized, with trigonal planar geometry.

1.51. (a) Boron is in group 3A of the periodic table, and therefore has three valence electrons. It can use each of its valence electrons to form a bond, so we expect the molecular formula to be BH3 (x=3). (b) Carbon is in group 4A of the periodic table, and therefore has four valence electrons. It can use each of its valence electrons to form a bond, so we expect the molecular formula to be CH4 (x=4). (c) Nitrogen is in group 5A of the periodic table, and therefore has five valence electrons. But it cannot form five bonds, because it only has four orbitals with which to form bonds. One of those orbitals must be occupied by a lone pair (two electrons), and each of the remaining three electrons is available to form a bond. Nitrogen is therefore trivalent, and we expect the molecular formula to be NH3 (x=3). (d) Carbon is in group 4A of the periodic table, and therefore has four valence electrons. It can use each of its valence electrons to form a bond, and indeed, we expect the carbon atom to have four bonds. Two of the bonds are with hydrogen atoms, so the other two bonds must be with chlorine atoms. The molecular formula is CH2Cl2 (x=2).

1.54. (a) Oxygen is the most electronegative atom in this compound. See Table 1.1 for electronegativity values. (b) Fluorine is the most electronegative atom. See Table 1.1 for electronegativity values. (c) Carbon is the most electronegative atom in this compound. See Table 1.1 for electronegativity values. 1.55. The highlighted nitrogen atom has two bonds and one lone pair (steric number = 3). This nitrogen atom is sp2 hybridized. It is electronically trigonal planar, but one of the sp2 hybridized orbitals is occupied by a lone pair, so the geometry (arrangement of atoms) is bent. The other nitrogen atom (not highlighted) has three bonds and a lone pair (steric number = 4). That nitrogen atom is sp3

CHAPTER 1 hybridized and electronically tetrahedral. One corner of the tetrahedron is occupied by a lone pair, so the geometry (arrangement of atoms) is trigonal pyramidal.

(c) Each C–Cl bond has a dipole moment, and in this case, the two dipole moments are pointing in opposite directions. As such, they fully cancel each other, giving no net molecular dipole moment. (d) The C–Cl bond has a dipole moment, and the C–Br bond also has a dipole moment. These two dipole moments are in opposite directions, but they do not have the same magnitude. The C–Cl bond has a larger dipole moment than the C–Br bond, because chlorine is more electronegative than bromine. Therefore, there will be a net molecular dipole moment, as shown here:

1.56. Each of the nitrogen atoms in caffeine achieves an octet with three bonds and one lone pair, while each oxygen atom in this structure achieves an octet with two bonds and two lone pairs, as shown:

1.59. The CCl bond in chloroform partially cancels the dipole moments of the other two CCl bonds, thereby reducing the molecular dipole moment relative to methylene chloride. 1.57. As seen in Section 1.2, the following two compounds have the molecular formula C2H6O.

The second compound will have a higher boiling point because it possesses an OH group which can exhibit hydrogen bonding interactions. 1.58. (a) Each C–Cl bond has a dipole moment, and the two dipole moments do not fully cancel each other because the polar bonds are not pointing in opposite directions. As such, there will be a net molecular dipole moment, as shown here:

(b) Each C–Cl bond has a dipole moment, and the two dipole moments do not fully cancel each other because the polar bonds are not pointing in opposite directions. As such, there will be a net molecular dipole moment, as shown here:

1.60. CHCl3 is expected to have a larger molecular dipole moment than CBrCl3, because the bromine atom in CBrCl3 serves to partially cancel out the dipole moments of the three CCl bonds (as is the case for CCl4). 1.61. The carbon atom of O=C=O has two bonds and no lone pairs (steric number = 2) and VSEPR theory predicts linear geometry. As a result, the individual dipole moments of each C=O bond cancel each other completely to give no overall molecular dipole moment. In contrast, the sulfur atom in SO2 has a steric number of three (because it also has a lone pair, in addition to the two S=O bonds), which means that it has bent geometry. As a result, the individual dipole moments of each S=O bond do NOT cancel each other completely, and the molecule does have a molecular dipole moment. 1.62. Two compounds possess OH groups. These two compounds will have the highest boiling points, because they can form hydrogen bonds. Among these two compounds, the one with more carbon atoms (six) will be higher boiling than the one with fewer carbon atoms (four), because the higher molecular weight results in greater London dispersion forces. The remaining three compounds all have equal molecular weights (five carbon atoms) and lack an OH group. The difference between these three compounds is the extent of branching. Among these three compounds, the compound with the greatest extent of branching has smallest surface area and therefore the lowest boiling point, and the one with the least branching has the highest boiling point.

CHAPTER 1 bonds, so it is sp3-hybridized, and the carbon on the right has three  bonds, so it is sp2-hybridized. The indicated  bond results from the overlap of one sp3 hybridized orbital (from the carbon atom on the left), and one sp2 hybridized orbital (from the carbon atom on the right).

1.68. The correct answer is (c). An atom with a trigonal planar arrangement of bonds has a bond angle of 120°, the largest bond angle among the choices listed. A tetrahedral arrangement of bonds corresponds to an angle of 109.5°, while trigonal pyramidal and bent geometries typically have bond angles that are slightly smaller than 109.5°. 1.63. The correct answer is (a). We must first draw the structure of HCN. To draw a Lewis structure, we begin by counting the valence electrons (hydrogen has one valence electron, carbon has four valence electrons, and nitrogen has five valence electrons, for a total of ten valence electrons). The structure must have ten valence electrons (no more and no less). Carbon should have four bonds, and it can only form a single bond with the hydrogen atom, so there must be a triple bond between carbon and nitrogen:

The single bond accounts for two electrons, and the triple bonds accounts for another six electrons. The remaining two electrons must be a lone pair on nitrogen. This accounts for all ten valence electrons, and it gives all atoms an octet. Since the carbon atom has a triple bond, it must be sp hybridized, with linear geometry. 1.64. The molecular formula of cyclobutane is C4H8. Of the four structures shown, only structure (c) has the same molecular formula (C4H8). 1.65. The correct answer is (b). Each of the structures has two carbon atoms and one oxygen atom. However, only the second structure has an OH group. This compound will have an elevated boiling point, relative to the other three structures, because of hydrogen bonding.

1.69. The correct answer is (c). There are 13  bonds in the given compound (highlighted in bold below). All of the single bonds are  bonds, and the double bond is comprised of one  bond and one  bond.

1.70. The correct answer is (b). Each C–Cl bond in Y has a dipole moment, and compound Y has a molecular dipole moment. Each CCl bond in X also has a dipole moment, but in this case, the two polar bonds are pointing in opposite directions. As such, the dipole moments fully cancel each other, giving no net molecular dipole moment for compound X. The polar isomer (Y) has more significant dipole-dipole attractions between molecules and, therefore, has a higher boiling point than the nonpolar isomer (X).

1.66. The first statement (a) is the correct answer, because an oxygen atom has a negative charge, and the nitrogen atom has a positive charge, as shown here:

1.67. The correct answer is (d). Carbon-carbon  bonds are formed by overlapping hybridized orbitals, so we must determine the hybridization of both carbon atoms (highlighted below). The carbon on the left has four 

1.71. The correct answer is (b). In the bond-line drawing, each corner and each endpoint represents a carbon atom, so this compound has three carbon atoms (highlighted below). Each carbon atom will have enough hydrogen atoms to have exactly four bonds. Together with the hydrogen atom that is connected to the oxygen atom, there are a total of six hydrogen atoms, as shown below:

CHAPTER 1 H

H OH

C H

three bonds and one  bond). There are certainly many other possible compounds for which all of the carbon atoms are sp2 hybridized.

1.72. The correct answer is (d). In the bond-line drawing, each corner represents a carbon atom (highlighted below), so this compound has nine carbon atoms. Each double bond is comprised of one  bond and one  bond, so there are five  bonds (in bold below).

1.73. (a) Compounds A and B share the same molecular formula (C4H9N) but differ in their constitution (connectivity of atoms), and they are therefore constitutional isomers. (b) The nitrogen atom in compound B has three bonds and one lone pair (steric number = 4). It is sp3 hybridized (electronically tetrahedral), with trigonal pyramidal geometry (because one corner of the tetrahedron is occupied by a lone pair). (c) A double bond represents one bond and one  bond, while a triple bond represents one bond and two  bonds. A single bond represents a bond. With this in mind, compound B has 14 bonds, as compared with compounds A and C, which have 13 and 11 bonds, respectively.

(b) In each of the following two compounds, all of the carbon atoms are sp3 hybridized (because each carbon atom has four bonds) with the exception of the carbon atom connected to the nitrogen atom. That carbon atom has two bonds and is therefore sp hybridized. There are certainly many other acceptable answers.

(c) In each of the following two compounds, there is a ring, and all of the carbon atoms are sp3 hybridized (because each carbon atom has four bonds). There are certainly many other acceptable answers.

(d) As explained in the solution to part (c), compound C has the fewest bonds. (e) A double bond represents one bond and one  bond, while a triple bond represents one bond and two  bonds. As such, compound C exhibits two  bonds. (f) Compound A has a C=N bond, in which the carbon atom has three bonds and no lone pairs (steric number = 3). It is sp2 hybridized. (g) Each of the carbon atoms in compound B is sp3 hybridized with four bonds (steric number = 4). Similarly, the nitrogen atom in compound B has three bonds and one lone pair (steric number = 4). This nitrogen atom is also sp3 hybridized. (h) Compound A has an N–H bond, and is therefore expected to form hydrogen bonding interactions. Compounds B and C do not contain an N–H bond, so compound A is expected to have the highest boiling point. 1.74. (a) In each of the following two compounds, all of the carbon atoms are sp2 hybridized (each carbon atom has

(d) In each of the following two compounds, all of the carbon atoms are sp hybridized (because each carbon atom has two bonds). There are certainly many other acceptable answers.

1.75. In each of the following two compounds, the molecular formula is C4H10N2, there is a ring (as suggested in the hint given in the problem statement), there are no  bonds, there is no net dipole moment, and there is an N-H bond, which enables hydrogen bonding interactions.

CHAPTER 1

1.76. If we try to draw a linear skeleton with five carbon atoms and one nitrogen atom, we find that the number of hydrogen atoms is not correct (there are thirteen, rather than eleven):

This will be the case even if try to draw a branched skeleton:

In fact, regardless of how the skeleton is branched, it will still have thirteen hydrogen atoms. But we need to draw a structure with only eleven hydrogen atoms (C5H11N). So we must remove two hydrogen atoms, which gives two unpaired electrons:

Now we have the correct number of hydrogen atoms (eleven), which means that our structure must indeed contain a ring. But this particular cyclic structure (cyclic = containing a ring) does not meet all of the criteria described in the problem statement. Specifically, each carbon atom must be connected to exactly two hydrogen atoms. This is not the case in the structure above. This issue can be remedied in the following structure, which has a ring, and each of the carbon atoms is connected to exactly two hydrogen atoms, as required by the problem statement.

1.77. (a) In compound A, the nitrogen atom has two bonds and no lone pairs (steric number = 2). It is sp hybridized. The highlighted carbon atom has one bond and one lone pair (steric number = 2), so that carbon atom is also sp hybridized. (b) The highlighted carbon atom is sp hybridized, so the lone pair occupies an sp-hybridized orbital. (c) The nitrogen atom is sp hybridized and therefore has linear geometry. As such, the C-N-C bond angle in A is expected to be 180°.

This indicates that we should consider pairing these electrons as a double bond. However, the problem statement specifically indicates that the structure cannot contain a double bond. So, we must find another way to pair the unpaired electrons. Let’s consider forming a ring instead of a double bond:

(d) The nitrogen atom in B has two bonds and one lone pair (steric number = 3). It is sp2 hybridized. The highlighted carbon atom has three bonds and no lone pairs (steric number = 3), and that carbon atom is sp2 hybridized. Each of the chlorine atoms has three lone pairs and one bond (steric number = 4), and the chlorine atoms are sp3 hybridized. (e) The nitrogen atom is sp2 hybridized, so the lone pair occupies an sp2-hybridized orbital. (f) The nitrogen atom is sp2 hybridized so the C-N-C bond angle in B is expected to be approximately 120°. 1.78. By analyzing the data, we can see that C(sp2)–Cl must be shorter than 1.79Å [compare with C(sp3)–Cl], while C(sp)–I must be longer than 1.79Å [compare with C(sp)–Br]. Therefore, C(sp)–I must be longer than C(sp2)–Cl. 1.79. (a) In the first compound, the fluorine isotope (18F) has no formal charge. Therefore, it must have three lone pairs (see Section 1.4 for a review of how formal charges are calculated). Since it has one bond and three lone pairs, it must have a steric number of 4, and is sp3 hybridized. The bromine atom also has no formal charge. So, it too, like the fluorine isotope, must have three lone pairs. Once again, one bond and three lone pairs give a steric number of 4, so the bromine atom is sp3 hybridized.

CHAPTER 1 In the second compound, the nitrogen atom has no formal charge. Therefore, it must have one lone pair. Since the nitrogen atom has three bonds and one lone pair, it must have a steric number of 4, and is sp3 hybridized. In the product, the fluorine isotope (18F) has no formal charge. Therefore, it must have three lone pairs. Since it has one bond and three lone pairs, it must have a steric number of 4, and is sp3 hybridized. The nitrogen atom does have a positive formal charge. Therefore, it must have no lone pairs. Since it has four bonds and no lone pairs, it must have a steric number of 4, and is sp3 hybridized. Finally, the bromine atom has a negative charge and no bonds. So it must have four lone pairs. With four lone pairs and no bonds, it will have a steric number of 4, and is expected to be sp3 hybridized. In summary, all of the atoms that we analyzed are sp3 hybridized. (b) The nitrogen atom is sp3 hybridized. With four bonds, we expect the geometry around the nitrogen atom to be tetrahedral. So, the bond angle for each C-N-C bond is expected to be approximately 109.5°. 1.80. (a) Boron is in group 3A of the periodic table and is therefore expected to be trivalent. That is, it has three valence electrons, and it uses each one of those valence electrons to form a bond, giving rise to three bonds. It does not have any electrons left over for a lone pair (as in the case of nitrogen). With three bonds and no lone pairs, the boron atom has a steric number of three, and is sp2 hybridized. (b) Since the boron atom is sp2 hybridized with three  bonds, we expect the geometry to be trigonal planar and the bond angles to be approximately 120°. However, in this case, the O-B-O system is part of a five-membered ring. That is, there are five different bond angles (of which the O-B-O angle is one of them) that together must form a closed loop. That requirement could conceivably force some of the bond angles (including the O-B-O bond angle) to deviate from the predicted value. In fact, we will explore this very phenomenon, called ring strain, in Chapter 4, and we will see that five-membered rings actually possess very little ring strain compared with smaller rings. (c) Each of the oxygen atoms has no formal charge, and must therefore have two bonds and two lone pairs. The boron atom has no lone pairs, as explained in the solution to part (a) of this problem.

1.81. (a) If we analyze each atom (in both 1 and 2) using the procedure outlined in Section 1.4, we find that none of the atoms in compound 1 have a formal charge, while compound 2 possesses two formal charges:

The nitrogen atom has a positive charge (it should have five valence electrons, but it is exhibiting only four), and the oxygen atom has a negative charge (it should have six valence electrons, but it is exhibiting seven). (b) Compound 1 possesses polar bonds, as a result of the presence of partial charges (+ and -). The associated dipole moments can form favorable interactions with the dipole moments present in the polar solvent molecules (dipole-dipole interactions). However, compound 2 has formal charges (negative on O and positive on N), so the dipole moment of the N-O bond is expected to be much more significant than the dipole moments in compound 1. The dipole moment of the N-O bond in compound 2 is the result of full charges, rather than partial charges. As such, compound 2 is expected to experience much stronger interactions with the solvent molecules, and therefore, 2 should be more soluble than 1 in a polar solvent. (c) In compound 1, the carbon atom (attached to nitrogen) has three bonds and no lone pairs (steric number = 3). That carbon atom is sp2 hybridized, with trigonal planar geometry. As such, the C-C-N bond angle in compound 1 is expected to be approximately 120°. However, in compound 2, the same carbon atom has two bonds and no lone pairs (steric number = 2). This carbon atom is sp hybridized, with linear geometry. As such, the C-C-N bond angle in 2 is expected to be 180°. The conversion of 1 to 2 therefore involves an increase in the C-C-N bond angle of approximately 60°.

1.82. (a) Ca has three bonds and no lone pairs, so it has a steric number of 3, and is sp2 hybridized. The same is true for Cc. In contrast, Cb has two bonds and no lone pairs, so it has a steric number of 2, and is therefore sp hybridized. (b) Since Ca is sp2 hybridized, we expect its geometry to be trigonal planar, so the bond angle should be approximately 120°. (c) Since Cb is sp hybridized, we expect its geometry to be linear, so the bond angle should be approximately 180°. (d) The central carbon atom (Cb) is sp hybridized, so it is using two sp hybridized orbitals to form its two bonds, which will be arranged in a linear fashion. The remaining two p orbitals of Cb used for  bonding will be 90° apart

CHAPTER 1

from one another (just as we saw for the carbon atoms of a triple bond; see Figure 1.33).

The other N-C-N unit (highlighted below) exhibits a central carbon atom that is sp2 hybridized and is therefore expected to have trigonal planar geometry. Accordingly, the bond angles about that carbon atom are expected to be approximately 120°.

As a result, the two  systems are orthogonal (or 90°) to each other. Therefore, the p orbitals on Ca and Cc are orthogonal. The following is another drawing from a different perspective (looking down the axis of the linear Ca-Cb-Cc system. (b) The interaction is an intramolecular hydrogen bond that forms between the + H (connected to the highlighted nitrogen atom) and the lone pair of the – oxygen atom:

1.83. (a) The N-C-N unit highlighted below exhibits a central carbon atom that is sp3 hybridized and is therefore expected to have tetrahedral geometry. Accordingly, the bond angles about that carbon atom are expected to be approximately 109.5°.

Chapter 2 Molecular Representations Review of Concepts Fill in the blanks below. To verify that your answers are correct, look in your textbook at the end of Chapter 2. Each of the sentences below appears verbatim in the section entitled Review of Concepts and Vocabulary.       

 

In bond-line structures, _______atoms and most ________ atoms are not drawn. A ________________ is a characteristic group of atoms/bonds that show a predictable behavior. When a carbon atom bears either a positive charge or a negative charge, it will have ___________, rather than four, bonds. In bond-line structures, a wedge represents a group coming ______ the page, while a dash represents a group _________ the page. ___________ arrows are tools for drawing resonance structures. When drawing curved arrows for resonance structures, avoid breaking a _______ bond and never exceed _____________ for second-row elements. The following rules can be used to identify the significance of resonance structures: 1. The most significant resonance forms have the greatest number of filled ___________. 2. The structure with fewer _________________ is more significant. 3. Other things being equal, a structure with a negative charge on the more _____________ element will be more significant. Similarly, a positive charge will be more stable on the less _____________ element. 4. Resonance forms that have equally good Lewis structures are described as ___________ and contribute equally to the resonance hybrid. A ______________ lone pair participates in resonance and is said to occupy a ____ orbital. A _____________ lone pair does not participate in resonance.

Review of Skills Fill in the blanks and empty boxes below. To verify that your answers are correct, look in your textbook at the end of Chapter 2. The answers appear in the section entitled SkillBuilder Review. SkillBuilder 2.1 Converting a Condensed Structure into a Lewis Structure

SkillBuilder 2.2 Drawing Bond-Line Structures

CHAPTER 2

SkillBuilder 2.3 Identifying Lone Pairs on Oxygen Atoms

SkillBuilder 2.4 Identifying Lone Pairs on Nitrogen Atoms

SkillBuilder 2.5 Identifying Valid Resonance Arrows

SkillBuilder 2.6 Assigning Formal Charges in Resonance Structures

SkillBuilder 2.7 Ranking the Significance of Resonance Structures Step 1 The most significant resonance forms have the greatest number of filled octets. Identify which form below is the most significant.

Step 2 The structure with the fewer formal charges is more significant. Based on steps one and two, identify which resonance form below is the most significant and which is the least significant.

H3C

NH2

CH3

Step 3 Other things being equal, the most significant resonance structure has the negative charge on the more electronegative atom. Consider the location of the formal charge(s), and identify the more significant contributor below.

O CH3

CH2

CH3

CH2

H NH2

CH3 H3C

H H

NH2

Step 4 Rank resonance forms. As a general rule, if one or more of the resonance forms has all filled octets, then any resonance form missing an octet will be a minor contributor.

SkillBuilder 2.8 Drawing a Resonance Hybrid Steps 1 and 2 After drawing all resonance structures, identify which one is more significant.

Step 3 Redraw the structure, showing partial bonds and partial charges.

Step 4 Revise the size of the partial charges to indicate distribution of electron density.

SkillBuilder 2.9 Identifying Localized and Delocalized Lone Pairs

CHAPTER 2

Common Mistakes to Avoid When drawing a structure, make sure to avoid drawing a pentavalent, hexavalent, or heptavalent carbon atom:

Carbon cannot have more than four bonds. Never draw a carbon atom with more than four bonds! While this type of mistake is not uncommon among new students of organic chemistry, it is important to correct it right away, so that it does not lead to further confusion. Also, when drawing a structure, either draw all carbon atom labels (C) and all hydrogen atom labels (H), like this Lewis structure:

or don’t draw any labels (except H attached to a heteroatom), like this bond-line structure:

That is, if you draw all C labels, then you should draw all H labels also. Avoid drawings in which the C labels are drawn and the H labels are not, as shown here:

These types of drawings (where C labels are shown and H labels are not shown) should only be used when you are working on a scratch piece of paper and trying to draw constitutional isomers. For example, if you are considering all constitutional isomers with the molecular formula C3H8O, you might find it helpful to use drawings like these as a form of “short-hand” so that you can identify all of the different ways of connecting three carbon atoms and one oxygen atom:

But your final structures should either show all C and H labels, or no labels at all. The latter is the more commonly used method:

CHAPTER 2

Solutions 2.1. (a) We begin by drawing the carbon chain and any atoms attached to the carbon chain, and then we look for atoms that do not have the correct number of bonds (highlighted):

To resolve this issue, we draw a pi bond between the two carbon atoms. Finally, we draw the lone pairs on the oxygen atom, giving the following Lewis structure, in which all atoms have filled octets:

(b) We begin by drawing the carbon chain and any atoms attached to the carbon chain:

In the structure drawn above, all atoms have the correct number of bonds. So, we complete the Lewis structure by drawing the lone pairs on the oxygen atom, so that the oxygen atom has a filled octet of electrons: H H H

To resolve this issue, we draw a pi bond between the two carbon atoms. Finally, we draw the lone pairs on the oxygen atom, giving the following Lewis structure, in which all atoms have filled octets:

(e) We begin by drawing the carbon chain and any atoms attached to the carbon chain:

H C

H H

C C

C H

(d) We begin by drawing the carbon chain and any atoms attached to the carbon chain, and then we look for atoms that do not have the correct number of bonds (highlighted):

H C

H H

In the structure drawn above, all atoms have the correct number of bonds, and there are no atoms with lone pairs, so this is the correct Lewis structure. (f) We begin by drawing the carbon chain and any atoms attached to the carbon chain:

CHAPTER 2 In the structure drawn above, all atoms have the correct number of bonds. So, we complete the Lewis structure by drawing the lone pairs on the bromine atom, so that the bromine atom has a filled octet of electrons:

2.2. (a) We begin by drawing the carbon chain and any atoms attached to the carbon chain:

be resolved by drawing a pi bond between the carbon and oxygen atoms:

Finally, we complete the octets on each oxygen atom by drawing the lone pairs (an uncharged oxygen atom has two bonds and two lone pairs), giving the following Lewis structure:

(b) We begin by drawing the carbon chain and any atoms attached to the carbon chain: Before we can deal with missing octets, we must correct the sigma bond framework shown, because it cannot be correct. Notice that there is a hydrogen atom on the right side of the structure that appears to have two bonds: and then we look for atoms that do not have the correct number of bonds (highlighted):

This is not possible. Hydrogen is monovalent – it can only form one bond, so let’s look more closely at the condensed structure, and see where we went wrong. The condensed structure ends with -CHO. In this grouping of atoms, notice that H is listed immediately after C, indicating that the H is connected directly to the C. Then, O is listed next, but it cannot be connected to the H (because hydrogen cannot have more than one bond), so it is understood that the O must also be connected directly to the C, like this:

The highlighted carbon atom has only two bonds (it should have four), and we cannot introduce pi bonds, because both neighboring atoms already have the correct number of bonds. So we must consider whether we have incorrectly drawn the framework of sigma bonds. Let’s look more closely at the condensed structure, which ends with -CO2H. In this grouping of atoms, there are two oxygen atoms listed. Certainly the first one is connected to C, but let’s consider the possibility that the second oxygen atom is also connected to C, like this:

In this grouping of atoms, the carbon atom is still missing a bond (it now has three and still needs to have four), but now the neighboring oxygen atom is also missing a bond, Now that we have drawn a framework of sigma bonds, the next step is to look for any atoms that do not have the correct number of bonds:

so we can draw a pi bond between the carbon and oxygen atoms:

The highlighted carbon has only three bonds (it should have four bonds), and the highlighted oxygen atom has only one bond (it should have two bonds). This issue can

CHAPTER 2

Finally, we complete the octets on each oxygen atom and each chlorine atom by drawing the lone pairs (oxygen typically has two bonds and two lone pairs, while chlorine typically has one bond and three lone pairs), giving the following Lewis structure:

Finally, we complete the octets on each oxygen atom by drawing the lone pairs (oxygen typically has two bonds and two lone pairs), giving the following Lewis structure:

2.3. Begin by drawing a Lewis structure for each isomer, so that the bonding of the carbon atoms is shown more clearly. Notice that in two of the isomers, a carbon atom is sharing a double bond with oxygen. Each of these carbon atoms is sp2 hybridized. All of the other carbon atoms exhibit four single bonds and are sp3 hybridized (highlighted below). The number of sp3-hybridized carbon atoms in the structures are two, three, and two, respectively:

and then we look for atoms that do not have the correct number of bonds (highlighted):

H2C

(CH3)2CO

H2C

CHOH

CH3CH2CHO

condensed / partially condensed structures

The highlighted carbon atom has only two bonds (it should have four), and we cannot introduce pi bonds, because both neighboring atoms already have the correct number of bonds. So we must consider whether we have incorrectly drawn the framework of sigma bonds. Let’s look more closely at the condensed structure, which has a -CO2- group. In this grouping of atoms, there are two oxygen atoms listed. Certainly, the first one is connected to C, but let’s consider the possibility that the second oxygen atom is also connected to C, like this:

H H H

H C C

H O

Lewis structures

2.4. (a) The carbon skeleton is drawn in a zig-zag format, in which each corner and each endpoint represents a carbon atom:

In this grouping of atoms, the carbon atom is still missing a bond (it now has three and needs to have four), but now the neighboring oxygen atom is also missing a bond,

(b) The carbon skeleton is drawn in a zig-zag format, in which each corner and each endpoint represents a carbon atom: so we can draw a pi bond between the carbon and oxygen atoms:

(c) The carbon skeleton is drawn in a zig-zag format, in which each corner and each endpoint represents a carbon atom. Hydrogen atoms generally don’t have to be drawn, but make sure to draw any hydrogen atoms that are connected to a heteroatom (such as oxygen). In this

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structure, there are three such hydrogen atoms, and all three must be drawn:

2.5. (a) The carbon skeleton is drawn in a zig-zag format, in which each corner and each endpoint represents a carbon atom:

(g) The carbon skeleton is drawn in a zig-zag format, in which each corner and each endpoint represents a carbon atom. In this case, a pi bond must be drawn between two carbon atoms, so that all atoms have filled octets:

(b) The carbon skeleton is drawn in a zig-zag format, in which each corner and each endpoint represents a carbon atom:

(h) The carbon skeleton is drawn in a zig-zag format, in which each corner and each endpoint represents a carbon atom. The hydrogen atom connected to oxygen must be drawn:

(c) The carbon skeleton is drawn in a zig-zag format, in which each corner and each endpoint represents a carbon atom. The hydrogen atom connected to oxygen must be drawn:

(i) The carbon skeleton is drawn in a zig-zag format, in which each corner and each endpoint represents a carbon atom. In this case, a pi bond must be drawn between two carbon atoms, so that all atoms have filled octets:

(d) The carbon skeleton is drawn in a zig-zag format, in which each corner and each endpoint represents a carbon atom. The hydrogen atom connected to oxygen must be drawn: (j) The carbon skeleton is drawn in a zig-zag format, in which each corner and each endpoint represents a carbon atom: (e) The carbon skeleton is drawn in a zig-zag format, in which each corner and each endpoint represents a carbon atom:

(f) The carbon skeleton is drawn in a zig-zag format, in which each corner and each endpoint represents a carbon atom. In this case, a pi bond must be drawn between two carbon atoms, so that all atoms have filled octets:

(k) The carbon skeleton is drawn in a zig-zag format, in which each corner and each endpoint represents a carbon atom. Hydrogen atoms generally don’t have to be drawn, but make sure to draw any hydrogen atoms that are connected to a heteroatom (such as nitrogen). In this

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structure, there are many hydrogen atoms that must be drawn:

ester

(l) The carbon skeleton is drawn in a zig-zag format, in which each corner and each endpoint represents a carbon atom. In this case, a pi bond must be drawn between two carbon atoms, so that all atoms have filled octets:

O N HO

carboxylic acid

(m) The carbon skeleton is drawn in a zig-zag format, in which each corner and each endpoint represents a carbon atom. Notice that there is a -CO2- group in the condensed structure. If you try to draw this group in a linear fashion (-C-O-O-), you will find that the carbon atom does not have the correct number of bonds. This is resolved by drawing both oxygen atoms being connected to the same carbon atom, and by also drawing a pi bond between the carbon atom and one of the oxygen atoms, as shown below. In this way, all atoms have filled octets:

O amide

N H

amine aromatic

2.8. (a) In this case, the oxygen atom has two bonds and no formal charge, so it must have two lone pairs in order to complete its octet (see Table 2.2).

(b) In this case, each of the oxygen atoms has two bonds and no formal charge, so each oxygen atom must have two lone pairs in order to complete its octet (see Table 2.2).

(c) In this case, each of the oxygen atoms has two bonds and no formal charge, so each oxygen atom must have two lone pairs in order to complete its octet (see Table 2.2). (n) The carbon skeleton is drawn in a zig-zag format, in which each corner and each endpoint represents a carbon atom. The hydrogen atom connected to oxygen must be drawn:

2.6. The carbon skeleton is drawn in a zig-zag format, in which each corner and endpoint represents a carbon atom. The condensed formula for the highlighted substituent is provided below:

2.7. The functional groups in the following compounds are highlighted and identified, using the terminology found in Table 2.1.

(d) One of the oxygen atoms has two bonds and no formal charge, so that oxygen atom must have two lone pairs in order to complete its octet (see Table 2.2). The other oxygen atom has one bond and a negative charge, so that oxygen atom must have three lone pairs. Three lone pairs complete the octet on this oxygen atom, and it has a negative charge because it exhibits seven electrons.

(e) In this case, the oxygen atom has one bond and a negative charge, so it must have three lone pairs (see Table 2.2). Three lone pairs complete the octet on the oxygen atom, and it has a negative charge because it exhibits seven electrons.

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(f) In this case, the oxygen atom has two bonds and no formal charge, so it must have two lone pairs in order to fill its octet (see Table 2.2).

(g) In this case, the oxygen atom has three bonds and a positive charge, so it must have one lone pair (see Table 2.2). One lone pair completes the octet on the oxygen atom, and it has a positive charge because it exhibits only five electrons.

2.10. (a) In this case, the nitrogen atom has three bonds and no formal charge, so it must have one lone pair to complete its octet (see Table 2.3).

(b) In this case, the nitrogen atom has three bonds and no formal charge, so it must have one lone pair to complete its octet (see Table 2.3). (h) In this case, the oxygen atom has three bonds and a positive charge, so it must have one lone pair (see Table 2.2). One lone pair completes the octet on the oxygen atom, and it has a positive charge because it exhibits only five electrons.

(i) From left to right, the first oxygen atom has two bonds and no formal charge, so it must have two lone pairs to complete its octet (see Table 2.2). The second oxygen atom has three bonds and a positive charge, so it must have one lone pair. One lone pair completes the octet on the second oxygen atom, and it has a positive charge because it exhibits only five electrons. Finally, the third oxygen atom has one bond and a negative charge, so it has three lone pairs. Three lone pairs complete the octet on the third oxygen atom, and it has a negative charge because it exhibits seven electrons.

(j) One of the oxygen atoms has two bonds and no formal charge, so that oxygen atom must have two lone pairs to complete its octet (see Table 2.2). The other oxygen atom has three bonds and a positive charge, so that oxygen atom must have one lone pair. One lone pair completes the octet on the oxygen atom, and it has a positive charge because it exhibits only five electrons.

2.9. Each oxygen atom in hydroxymethylfurfural lacks a charge and has two bonds, so each oxygen atom must have two lone pairs to complete its octet.

(c) In this case, the nitrogen atom has three bonds and no formal charge, so it must have one lone pair to complete its octet (see Table 2.3).

(d) In this case, the nitrogen atom has four bonds and a positive charge, so it must have no lone pairs (see Table 2.3). Four bonds complete the octet on the nitrogen atom, and it has a positive charge because it exhibits only four electrons.

(e) In this case, the nitrogen atom has two bonds and a negative charge, so it must have two lone pairs (see Table 2.3). Two lone pairs complete the octet on the nitrogen atom, and it has a negative charge because it exhibits six electrons.

(f) In this case, the nitrogen atom has three bonds and no formal charge, so it must have one lone pair in order to complete its octet (see Table 2.3).

(g) In this case, the nitrogen atom has four bonds and a positive charge, so it must have no lone pairs (see Table 2.3). Four bonds complete the octet on the nitrogen atom, and it has a positive charge because it exhibits only four electrons.

CHAPTER 2 no formal charge, so it must have one lone pair to complete its octet.

(h) One of the nitrogen atoms has four bonds and a positive charge, so it must have no lone pairs (see Table 2.3). Four bonds complete the octet on the nitrogen atom, and it has a positive charge because it exhibits only four electrons. The other nitrogen atom has three bonds and 2.11. Every uncharged nitrogen atom in this compound has three bonds and needs one lone pair of electrons to fill its octet. Every positively charged nitrogen atom has four bonds and no lone pairs. Four bonds complete the octet on a nitrogen atom, and it is positively charged because it exhibits only four electrons. The missing lone pairs that have been added to nitrogen atoms are highlighted below:

2.12. (a) This curved arrow violates the second rule by giving a fifth bond to a nitrogen atom (thus exceeding its octet). (b) This curved arrow does not violate either rule. (c) This curved arrow violates the second rule by giving five bonds to a carbon atom (thus exceeding its octet). (d) This curved arrow violates the second rule by giving three bonds and two lone pairs to an oxygen atom (thus exceeding its octet). (e) This curved arrow violates the second rule by giving five bonds to a carbon atom (thus exceeding its octet). (f) This curved arrow violates the second rule by giving five bonds to a carbon atom (thus exceeding its octet). (g) This curved arrow violates the first rule by breaking a single bond, and violates the second rule by giving five bonds to a carbon atom (thus exceeding its octet). (h) This curved arrow violates the first rule by breaking a single bond, and violates the second rule by giving five bonds to a carbon atom (thus exceeding its octet). (i) This curved arrow does not violate either rule. (j) This curved arrow does not violate either rule. (k) This curved arrow violates the second rule by giving five bonds to a carbon atom (thus exceeding its octet).

(l) This curved arrow violates the second rule by giving five bonds to a carbon atom (thus exceeding its octet). 2.13. The tail of the curved arrow must be placed on the double bond in order to avoid violating the first rule (avoid breaking a single bond).

2.14. (a) This curved arrow violates the first rule (avoid breaking a single bond). (b) This curved arrow does not violate either rule. (c) This curved arrow violates the second rule (never exceed an octet for second-row elements) by giving five bonds to a carbon atom. (d) This curved arrow violates the second rule by giving five bonds to a carbon atom (thus exceeding its octet). 2.15. (a) The curved arrow indicates that we should draw a resonance structure in which the  bond has been pushed

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over. We then complete the resonance structure by assigning any formal charges. Notice that both resonance structures show a positive charge, but in different locations:

(b) The curved arrows indicate that we should draw a resonance structure in which the lone pair has been pushed to become a  bond, and the  bond has been pushed to become a lone pair. We then complete the resonance structure by assigning any formal charges. Notice that both resonance structures show a negative charge, but in different locations:

(c) The curved arrows indicate that we should draw a resonance structure in which a lone pair has been pushed to become a  bond, and the  bond has been pushed to become a lone pair. We then complete the resonance structure by assigning any formal charges. Notice that both resonance structures have zero net charge:

(d) The curved arrows indicate that we should draw a resonance structure in which a lone pair has been pushed to become a  bond, and the  bond has been pushed to become a lone pair. We then complete the resonance structure by assigning any formal charges. Notice that both resonance structures show a negative charge, but in different locations:

(e) The curved arrows indicate that we should draw the following resonance structure. Notice that both resonance structures have zero net charge:

(f) The curved arrows indicate that we should draw the following resonance structure. Notice that both resonance structures have zero net charge:

(g) The curved arrows indicate that we should draw a resonance structure in which a lone pair has been pushed to become a  bond, and a  bond has been pushed to become a lone pair. We then complete the resonance structure by assigning any formal charges. Notice that both resonance structures have zero net charge, but they differ in the location of the negative charge:

(h) The curved arrows indicate that we should draw the following resonance structure. Notice that both resonance structures have zero net charge:

2.16. (a) One curved arrow is required, showing the  bond being pushed to become a lone pair:

(b) Two curved arrows are required. One curved arrow shows the carbon-carbon  bond being pushed up, and the other curved arrow shows the carbon-oxygen  bond becoming a lone pair:

(c) Two curved arrows are required. One curved arrow shows a lone pair from the nitrogen atom becoming a  bond, and the other curved arrow shows the carbonoxygen  bond becoming a lone pair:

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(d) One curved arrow is required, showing the  bond being pushed over:

2.17. (a) Notice that there are two formal charges (positive and negative), but the positive charge is in the same location in all three resonance structures. Only the negative charge is spread out (over three locations). The negative charge in 2a is on a carbon atom. Such a carbon atom (bearing a negative charge) will have three bonds, and in order to exhibit five electrons, it must also have one lone pair of electrons. Two curved arrows are required to convert from resonance structure 2a to resonance structure 2b. One arrow shows that the lone pair on carbon can become a new carbon-nitrogen  bond while the other arrow shows that the electrons in the original carbon-nitrogen  bond can become a lone pair on a different carbon atom.

(b) Two curved arrows are required to convert resonance structure 2a to resonance structure 2c. One arrow shows that the lone pair on carbon can become a carbon-carbon  bond while the other arrow shows that the electrons in the carbon-oxygen  bond can become a third lone pair on oxygen.

2.18. (a) This pattern (lone pair next to a  bond) has two curved arrows. The first curved arrow is drawn showing a lone pair becoming a  bond, while the second curved arrow shows a  bond becoming a lone pair. Notice that both resonance structures have zero net charge:

(b) This pattern (lone pair next to a  bond) has two curved arrows. The first curved arrow is drawn showing a lone pair becoming a  bond, while the second curved arrow shows a  bond becoming a lone pair. Notice that both resonance structures show a negative charge, but in different locations:

(c) This pattern (lone pair next to a  bond) has two curved arrows. The first curved arrow is drawn showing a lone pair becoming a  bond, while the second curved arrow shows a  bond becoming a lone pair. Notice that both resonance structures have zero net charge:

(d) This pattern (lone pair next to a  bond) has two curved arrows. The first curved arrow is drawn showing a lone pair becoming a  bond, while the second curved arrow shows a  bond becoming a lone pair. Notice that both resonance structures have zero net charge:

(e) This pattern (lone pair next to a  bond) has two curved arrows. The first curved arrow is drawn showing a lone pair becoming a  bond, while the second curved arrow shows a  bond becoming a lone pair. Notice that both resonance structures show a negative charge, but in different locations:

(f) This pattern (lone pair next to a  bond) has two curved arrows. The first curved arrow is drawn showing a lone pair becoming a  bond, while the second curved arrow shows a  bond becoming a lone pair. Notice that both resonance structures show a negative charge, but in different locations:

CHAPTER 2 (g) This pattern (lone pair next to a  bond) has two curved arrows. The first curved arrow is drawn showing a lone pair becoming a  bond, while the second curved arrow shows a  bond becoming a lone pair. Notice that both resonance structures have an overall +1 net charge:

same pattern can be applied again, giving another resonance structure, as shown:

Notice that each resonance structure has an overall +1 net charge, and that each structure shows the positive charge in a different location.

(h) This pattern (lone pair next to a  bond) has two curved arrows. The first curved arrow is drawn showing a lone pair becoming a  bond, while the second curved arrow shows a  bond becoming a lone pair. Notice that both resonance structures have zero net charge:

2.19. (a) This pattern (an allylic C+) has just one curved arrow, showing the  bond being pushed over. Notice that both resonance structures have an overall +1 net charge, but they show the positive charge in different locations:

(d) This pattern (an allylic C+) has just one curved arrow, showing the  bond being pushed over. But when we draw the resulting resonance structure, we find that the same pattern can be applied again, giving another resonance structure. This process continues several more times:

Notice that each resonance structure has an overall +1 net charge, and that each structure shows the positive charge in a different location. We can see that the positive charge is spread (via resonance) over all seven carbon atoms of the ring

(b) This pattern (an allylic C+) has just one curved arrow, showing the  bond being pushed over. Notice that both resonance structures have an overall +1 net charge, but they show the positive charge in different locations:

2.20. (a) This pattern (a lone pair adjacent to C+) has just one curved arrow, showing the lone pair becoming a  bond. Notice that both resonance structures have an overall +1 net charge, but they show the positive charge in different locations:

(c) This pattern (an allylic C+) has just one curved arrow, showing the  bond being pushed over. But when we draw the resulting resonance structure, we find that the

(b) This pattern (a lone pair adjacent to C+) has just one curved arrow, showing the lone pair becoming a  bond. Notice that both resonance structures have zero net charge:

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(c) This pattern (a lone pair adjacent to C+) has just one curved arrow, showing the lone pair becoming a  bond. Notice that both resonance structures have an overall +1 net charge, but they show the positive charge in different locations:

2.21. (a) This pattern (a pi bond between two atoms of differing electronegativity) has just one curved arrow, showing the  bond becoming a lone pair on the more electronegative nitrogen atom. Notice that both resonance structures have zero net charge:

2.23. This pattern (a pi bond between two atoms of differing electronegativity) has just one curved arrow, showing the  bond becoming a lone pair on the more electronegative oxygen atom. Notice that both resonance structures have zero net charge:

2.24. The pattern that is present in fingolimod (conjugated  bonds enclosed in a ring) has three curved arrows that push the  bonds in either a clockwise or counter-clockwise direction: (b) This pattern (a pi bond between two atoms of differing electronegativity) has just one curved arrow, showing the  bond becoming a lone pair on the more electronegative oxygen atom. Notice that both resonance structures have zero net charge:

(c) This pattern (a pi bond between two atoms of differing electronegativity) has just one curved arrow, showing the  bond becoming a lone pair on the more electronegative oxygen atom. Notice that both resonance structures have zero net charge:

2.22. This pattern (a pi bond between two atoms of differing electronegativity) has just one curved arrow, showing the  bond becoming a lone pair on the more electronegative oxygen atom. Notice that both resonance structures have zero net charge:

2.25. (a) We begin by looking for the five patterns. In this case, there is a C=O bond (a  bond between two atoms of differing electronegativity), so we draw one curved arrow showing the  bond becoming a lone pair on the more electronegative oxygen atom. We then draw the resulting resonance structure and assess whether it exhibits one of the five patterns. In this case, there is a C+ (carbocation) that is allylic, so we draw the curved arrow associated with that pattern (pushing over the  bond), shown here. Notice that each of the three resonance structures has a zero net charge:

CHAPTER 2 (b) The C+ (a carbocation) occupies an allylic position, so we draw the one curved arrow associated with that pattern (pushing over the  bond). The C+ in the resulting resonance structure is again next to another  bond, so we draw one curved arrow and another resonance structure, as shown here:

Notice that each resonance structure has an overall +1 net charge, and that each structure shows the positive charge in a different location.

resonance structures have an overall +1 net charge, but they show the positive charge in different locations:

(f) This compound exhibits a C=O bond (a  bond between two atoms of differing electronegativity), so we draw one curved arrow showing the  bond becoming a lone pair on the more electronegative oxygen atom. We then draw the resulting resonance structure and assess whether it exhibits one of the five patterns. In this case, there is a C+ (carbocation) that is allylic, so we draw the curved arrow associated with that pattern (pushing over the  bond). The C+ in the resulting resonance structure is next to another  bond, so we draw one more resonance structure. Notice that all four resonance structures have zero net charge:

(c) The lone pair (associated with the negative charge) occupies an allylic position, so we draw the two curved arrows associated with that pattern. The first curved arrow is drawn showing a lone pair becoming a  bond, while the second curved arrow shows a  bond becoming a lone pair. Notice that each resonance structure has an overall –1 net charge, and that each structure shows the negative charge in a different location:

(d) We begin by looking for the five patterns, focusing first on any patterns that employ just one curved arrow (in this case, there is another pattern that requires two curved arrows, but we will start with the pattern using just one curved arrow). There is a C=O bond (a  bond between two atoms of differing electronegativity), so we draw one curved arrow showing the  bond becoming a lone pair on the more electronegative oxygen atom. We then draw the resulting resonance structure and assess whether it exhibits one of the five patterns. In this case, there is a lone pair adjacent to a C+ (a carbocation), so we draw the curved arrow associated with that pattern (showing the lone pair becoming a  bond), shown here. Notice that each of the three resonance structures has a zero net charge:

(g) This structure exhibits a lone pair that is adjacent to C+, so we draw one curved arrow, showing a lone pair becoming a  bond. Notice that both resonance structures have an overall +1 net charge, but they show the positive charge in different locations:

(h) This compound exhibits a C=N bond (a  bond between two atoms of differing electronegativity), so we draw one curved arrow showing the  bond becoming a lone pair on the more electronegative nitrogen atom. Notice that both resonance structures have zero net charge: (e) This structure exhibits a lone pair that is adjacent to a C+ (a carbocation), so we draw one curved arrow, showing a lone pair becoming a  bond. Notice that both

CHAPTER 2

(i) We begin by looking for the five patterns, focusing first on any patterns that employ just one curved arrow (in this case, there is another pattern that requires two curved arrows, but we will start with the pattern using just one curved arrow). There is a C=O bond (a  bond between two atoms of differing electronegativity), so we draw one curved arrow showing the  bond becoming a lone pair on the more electronegative oxygen atom. We then draw the resulting resonance structure and assess whether it exhibits one of the five patterns. In this case, there is a lone pair adjacent to C+, so we draw the curved arrow associated with that pattern (showing the lone pair becoming a  bond). Notice that each of the three resonance structures has a zero net charge:

(j) We begin by looking for the five patterns, focusing first on any patterns that employ just one curved arrow (in this case, there is another pattern that requires two curved arrows, but we will start with the pattern using just one curved arrow). There is a C=O bond (a  bond between two atoms of differing electronegativity), so we draw one curved arrow showing the  bond becoming a lone pair on the more electronegative oxygen atom. We then draw the resulting resonance structure and assess whether it exhibits one of the five patterns. In this case, there is a lone pair adjacent to C+, so we draw the curved arrow associated with that pattern (showing the lone pair becoming a  bond). Notice that each of the three resonance structures has a zero net charge:

2.26. (a) Using the skills developed in the previous SkillBuilders, we begin by drawing all significant resonance structures, shown below. None of the four structures have an atom with an incomplete octet. The first resonance structure is the most significant contributor because it has filled octets and the negative charge is on the more electronegative nitrogen atom. The other three resonance structures are approximately equivalent and they are less significant than the first structure because the negative charge is on a less electronegative carbon atom.

(b) Using the skills developed in the previous SkillBuilders, we begin by drawing all significant resonance structures, shown below. Both resonance structures have filled octets and a negative charge on an oxygen atom. Both resonance structures are equivalent and thus equally significant.

(c) Using the skills developed in the previous SkillBuilders, we begin by drawing all significant resonance structures, shown below. None of the three structures have an atom with an incomplete octet. The first resonance structure is the most significant because it has filled octets and no formal charges. The other two resonance structures are equivalent and less significant contributors because they contain formal charges.

O Most significant contributor (#1)

Equivalent

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(d) Using the skills developed in the previous SkillBuilders, we begin by drawing all significant resonance structures, shown below. The first resonance structure is the most significant because it is the only one with every second-row element having a filled octet. Recall that a structure with filled octets and no formal charges is an ideal Lewis structure. The other three resonance structures are approximately equivalent and they are less significant than the first structure because each one is missing an octet, highlighted below (they also have formal charges, but that is a less significant feature to consider when ranking resonance forms):

(e) Using the skills developed in the previous SkillBuilders, we begin by drawing all significant resonance structures, shown below. Neither resonance structure has an incomplete octet. The second resonance structure is the more significant contributor because it has the negative charge on the more electronegative nitrogen atom. The first resonance structure is less significant because the negative charge is on the less electronegative carbon atom.

(f) This cation has two different resonance patterns that can be employed, using the lone pair or the  bond to fill the vacancy on carbon, giving a total of three resonance structures. The middle resonance structure is the most significant contributor because it is the only one with filled octets. The other two structures are approximately equivalent, each having one missing octet (highlighted below) and a positive charge on a carbon atom. They are less significant and contribute equally to the hybrid. Note that the location of the charge (C+ vs. N+) is not relevant in this problem, because filled octets are more important.

2.27. When looking for resonance in the first structure, we can begin with the carbonyl (C=O) group by relocating the  bond electrons to the more electronegative oxygen atom. This provides a resonance structure with a C+ (carbocation) that is allylic. Allylic resonance throughout the ring provides for three more resonance structures. Be careful to use just one  bond at a time so you don’t accidentally “jump” over a possible resonance structure. Notice that each of the five resonance structures has a zero net charge:

For the second compound, the allylic lone pair can be delocalized using one of the  bonds in the benzene ring, and this pattern can continue to use the remaining  bonds in the ring (again, one at a time!). Notice that each of the four resonance structures has a zero net charge:

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Overall, when we consider the contributions made by all resonance structures, we find that the ring in the first compound is electron-poor, with several electron-deficient sites on the ring, and the ring in the second compound is electron-rich, with several  sites on the ring.

2.28. The benzene ring on the left has two oxygen atoms attached. Both oxygen atoms can donate electron density via allylic lone pair resonance, making this benzene ring electron-rich:

Note that the other oxygen atom directly attached to the benzene ring has a similar effect, resulting in additional resonance structures (not shown) with negative charges on the atoms of the ring. The benzene ring on the right is “connected” to the two carbonyl (C=O) groups by a series of conjugated  bonds, so the resonance of each carbonyl group extends into the benzene ring. The carbonyl groups withdraw electron density via allylic carbocation (C+) resonance, making this ring electron-deficient:

Note that the other carbonyl (C=O) group has a similar effect, resulting in resonance structures (not shown) with positive charges on the atoms of the ring. In summary, the benzene ring on the left is electron-rich due to resonance involving the lone pairs of electrons on both attached oxygen atoms. The benzene ring on the left is electron-poor due to resonance involving the carbonyl groups.

CHAPTER 2 2.29. (a) Begin by drawing all significant resonance structures. In this case, there are two:

Both resonance structures are equally significant, so the resonance hybrid is the simple average of these two resonance structures. There are no formal charges, so only partial bonds need to be drawn.

(b) Begin by drawing all significant resonance structures. In this case, there are two:

Both are equally significant, so the resonance hybrid is the simple average of these two resonance structures. Only the negative charge is delocalized, so partial charges are used for the negative charge but not for the positive charge.

(e) Begin by drawing all significant resonance structures. In this case, there are two:

The structure on the right is more significant because every atom has an octet. The resonance hybrid is a weighted average of these two resonance structures in which the nitrogen atom has more of the charge than the carbon atom.

Both are equally significant, so the resonance hybrid is the simple average of these two resonance structures. Both partial bonds and partial charges are required, as shown: (f) Begin by drawing all significant resonance structures. In this case, there are two: (c) Begin by drawing all significant resonance structures. In this case, there are two:

The left-hand structure is more significant because every atom has an octet. The resonance hybrid is a weighted average of these two resonance structures in which the oxygen atom has more of the charge than the carbon atom.

The structure on the left is more significant because every atom has an octet and it has no formal charges. The resonance hybrid is a weighted average of these two resonance structures, although we do not denote that by making the partial charges different sizes. In this example, the charges are opposite in sign, but they must be equal in magnitude so that the overall charge will be zero.

(d) Begin by drawing all significant resonance structures. In this case, there are two:

(g) Begin by drawing all significant resonance structures. In this case, there are three:

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All three resonance structures are approximately equally significant, so the resonance hybrid is approximately the average of these three resonance structures, illustrating that the positive charge is delocalized over three carbon atoms.

The structure on the left is most significant because every atom has an octet and it has no formal charges. The resonance hybrid is a weighted average of these three resonance structures. Since the partial positive charge is delocalized over two carbon atoms and the partial negative charge is localized on only one oxygen atom, the partial negative charge is drawn larger than each of the individual partial positive charges.

(h) Begin by drawing all significant resonance structures. In this case, there are three:

2.30. Begin by drawing all significant resonance structures that show delocalization of the positive charge. In this case, there are four:

The first resonance structure and the last two resonance structures are the most significant, because all atoms have an octet in each of these three resonance structures. The second resonance structure (in which the carbon atom bears the positive charge) is the least significant because a carbon atom lacks an octet. If we compare the three most significant resonance structures, each has a positive charge on a nitrogen atom, so we expect these three resonance structures to contribute roughly equally to the resonance hybrid. To show this, we indicate + at all three positions, with a smaller + symbol at the central carbon atom (indicating the lower contribution of the second resonance structure). Also, if we compare the three significant resonance structures, we find that the  bond is spread over three locations, and these locations are indicated with dashed lines in the resonance hybrid:

2.31. (a) Let’s begin with the nitrogen atom on the left side of the structure. The lone pair on this nitrogen atom is delocalized by resonance (because it is next to a  bond). Therefore, this lone pair occupies a p orbital, which means that the nitrogen atom is sp2 hybridized. As a result, the geometry is trigonal planar. On the right side of the structure, there is a nitrogen atom with a localized lone pair (it does not participate in resonance). This nitrogen atom is therefore sp3 hybridized, with trigonal pyramidal geometry, just as

expected for a nitrogen atom with  sigma bonds and a localized lone pair.

CHAPTER 2 (b) As we saw with pyridine, the lone pair on this nitrogen atom is not participating in resonance, because the nitrogen atom is already using a p orbital for the  bond. As a result, the lone pair cannot join in the conduit of overlapping p orbitals, and therefore, it cannot participate in resonance. In this case, the lone pair occupies an sp2-hybridized orbital, which is in the plane of the ring. Since this lone pair is not participating in resonance, it is localized. The nitrogen atom is sp2 hybridized, and the geometry is bent.

2.34. (a) Each of the oxygen atoms has two bonds and no formal charge, so each oxygen atom will have two lone pairs to complete its octet.

(b) Each of the oxygen atoms has two bonds and no formal charge, so each oxygen atom will have two lone pairs to complete its octet. The nitrogen atom has three bonds and no formal charge, so it must have one lone pair of electrons to complete its octet. (c) The lone pair on this nitrogen atom is participating in resonance (it is next to a  bond), so it is delocalized via resonance. As such, the nitrogen atom is sp2 hybridized, with trigonal planar geometry.

H H

N HO

C C

H C

O H

2.32. As we saw with pyridine, the lone pair on the nitrogen atom in the 6-membered aromatic ring is not participating in resonance, because the nitrogen atom is already using a p orbital for the  bond. As a result, the lone pair cannot join in the conduit of overlapping p orbitals, and therefore, it cannot participate in resonance. In this case, the lone pair occupies an sp2-hybridized orbital, which is in the plane of the ring. Since this lone pair is not participating in resonance, it is localized. On the right side of the structure, the nitrogen atom in the 5membered ring also has a localized lone pair (it does not participate in resonance). Each of these lone pairs is localized, and, therefore, both lone pairs are expected to be reactive. 2.33. Lone pairs that participate in resonance are delocalized, while those that do not participate in resonance are localized: localized (not participating in resonance)

(c) Each of the oxygen atoms has two bonds and no formal charge, so each oxygen atom will have two lone pairs to complete its octet. Each nitrogen atom has three bonds and no formal charge, so each nitrogen atom must have one lone pair of electrons to complete its octet.

2.35. The molecular formula indicates that there are four carbon atoms. Recall that constitutional isomers are compounds that share the same molecular formula, but differ in constitution (the connectivity of atoms). So we are looking for different ways that four carbon atoms can be connected together. As described in the solution to Problem 1.1b, the carbon atoms can be connected in a linear fashion (below left), or they can be connected with a branch (below right).

localized (not participating in resonance) O N N

NH2

H delocalized (participating in resonance)

localized (not participating in resonance)

These two compounds are the only constitutional isomers that have the molecular formula C4H10, because there are no other ways to connect four carbon atoms without changing the number of hydrogen atoms. For example, if we try to connect the carbon atoms into a ring, we find that the number of hydrogen atoms is reduced:

CHAPTER 2 2.39. Carbon is in group 4A of the periodic table, and it therefore has four valence electrons. We are told that, in this case, the central carbon atom does not bear a formal charge.

2.36. As described in the solution to Problem 1.1c, there are only three constitutional isomers with the molecular formula C5H12, shown here again.

There are no other constitutional isomers with the molecular formula C5H12. The following two structures do NOT represent constitutional isomers, but are in fact two drawings of the same compound, as can be seen when the carbon skeletons are numbered, as shown:

Notice that in both drawings, the longest linear chain is four carbon atoms, and there is a CH3 group attached to the second carbon atom of the chain. As such, these two drawings represent the same compound. In contrast, we can see that all three constitutional isomers with the molecular formula C5H12 exhibit different connectivity of the carbon atoms:

Therefore, it must exhibit the appropriate number of valence electrons (four). This carbon atom already has two bonds (each of which requires one valence electron) and a lone pair (which represents two electrons), for a total of 1+1+2=4 valence electrons. This is the appropriate number of valence electrons, which means that this carbon atom does not have any bonds to hydrogen. Notice that the carbon atom lacks an octet, so it should not be surprising that this structure is highly reactive and very short-lived. 2.40. An oxygen atom will bear a negative charge if it has one bond and three lone pairs, and it will bear a positive charge if it has three bonds and one lone pair (see Table 2.2). A nitrogen atom will bear a negative charge if it has two bonds and two lone pairs, and it will bear a positive charge if it has four bonds and no lone pairs (see Table 2.3).

2.37. In each of the following structures, each corner and endpoint represents a carbon atom. Hydrogen atoms are only drawn if they are connected to heteroatoms (such as oxygen).

2.38. Each oxygen atom has two bonds and no formal charge. Therefore, each oxygen atom has two lone pairs, for a total of twelve lone pairs.

2.41. This compound exhibits a C=O bond (a  bond between two atoms of differing electronegativity), so we draw one curved arrow showing the  bond becoming a lone pair on the more electronegative oxygen atom. We then draw the resulting resonance structure and assess whether it exhibits one of the five patterns. In this case, there is a C+ (carbocation) that is allylic, so we draw the curved arrow associated with that pattern (pushing over the  bond). This pattern continues, several more times, spreading the positive charge over a total of four carbon atoms, as shown here:

CHAPTER 2

such, each of these six carbon atoms (highlighted) is sp3 hybridized.

2.42. Recall that constitutional isomers are compounds that share the same molecular formula, but differ in constitution (the connectivity of atoms). The problem statement shows a compound with the molecular formula C5H12 and the following structure:

So we are looking for other compounds that also have the molecular formula C5H12 but show a different connectivity of atoms. As seen in the solution to Problem 2.36, there are only two such compounds:

2.43. The following two compounds are constitutional isomers because they share the same molecular formula (C5H12O). The third compound (not shown here) has a different molecular formula (C4H10O).

2.44. Begin by drawing a Lewis structure, so that the bonding of each carbon atom is shown more clearly:

2.45. One of the oxygen atoms has two bonds and no formal charge, so that oxygen atom must have two lone pairs (see Table 2.2). The other oxygen atom has one bond and a negative charge, so that oxygen atom must have three lone pairs. The nitrogen atom has four bonds and a positive charge, so it does not have any lone pairs (see Table 2.3). Therefore, there are a total of five lone pairs in this structure.

2.46. (a) The lone pair on this nitrogen atom is not participating in resonance, because the nitrogen atom is already using a p orbital for the  bond. As a result, the lone pair cannot join in the conduit of overlapping p orbitals, and therefore, it cannot participate in resonance. In this case, the lone pair occupies an sp2-hybridized orbital, which is in the plane of the ring.

(b) There is a lone pair associated with the negative charge, and this lone pair is delocalized via resonance (the lone pair is allylic):

As such, the lone pair must occupy a p orbital. (c) The nitrogen atom has a lone pair, which is delocalized via resonance (there is an adjacent positive charge):

As such, the lone pair must occupy a p orbital. Notice that two of the carbon atoms are sharing a double bond. These two atoms are sp2 hybridized. Each of the other six carbon atoms exhibits four single bonds, and as

2.47. (a) This structure exhibits a lone pair that is adjacent to a C+ (carbocation). In fact, there are two such lone pairs

CHAPTER 2

(on the nitrogen and oxygen atoms). We will begin with a lone pair on the oxygen atom, although we would have arrived at the same solution either way (we will draw a total of three resonance structures, below, and it is just a matter of the order in which we draw them). We draw one curved arrow, showing a lone pair on the oxygen atom becoming a  bond. We then draw the resulting resonance structure and assess whether it exhibits one of the five patterns. In this case, there is a lone pair next to a  bond, so we draw the two curved arrows associated with that pattern. The first curved arrow is drawn showing a lone pair on the nitrogen atom becoming a  bond, while the second curved arrow shows a  bond becoming a lone pair on the oxygen atom:

2.48. (a) In a condensed structure, single bonds are not drawn. Instead, groups of atoms are clustered together, as shown here:

(b) In a condensed structure, single bonds are not drawn. Instead, groups of atoms are clustered together, as shown here: OH (CH3)2CHCH2CH2CH2OH or (CH3)2CH(CH2)3OH

(b) This structure exhibits a C+ (carbocation) that is allylic, so we draw one curved arrow showing the  bond being pushed over. We then draw the resulting resonance structure and assess whether it exhibits one of the five patterns. In this case, C+ is adjacent to a lone pair, so we draw the curved arrow associated with that pattern (the lone pair is shown becoming a  bond):

(c) This structure exhibits a C+ (carbocation) that is allylic, so we draw one curved arrow showing the  bond being pushed over. We then draw the resulting resonance structure and assess whether it exhibits one of the five patterns. In this case, C+ is again next to a  bond, so again, we draw the curved arrow associated with that pattern (pushing over the  bond again). The resulting resonance structure has C+ next to yet another  bond, so we draw a curved arrow showing the  bond being pushed over one more time to give our final resonance structure:

(c) In a condensed structure, single bonds are not drawn. Instead, groups of atoms are clustered together, as shown here:

2.49. (a) Each corner and each endpoint represents a carbon atom, so this compound has nine carbon atoms. Each carbon atom will have enough hydrogen atoms to have exactly four bonds, giving a total of twenty hydrogen atoms. So the molecular formula is C9H20.

(b) Each corner and each endpoint represents a carbon atom, so this compound has six carbon atoms. Each carbon atom will have enough hydrogen atoms to have exactly four bonds, giving a total of fourteen hydrogen atoms. So the molecular formula is C6H14O.

(c) Each corner and each endpoint represents a carbon atom, so this compound has eight carbon atoms. Each carbon atom will have enough hydrogen atoms to have

CHAPTER 2 exactly four bonds, giving a total of sixteen hydrogen atoms. So the molecular formula is C8H16.

2.50. As seen in the solution to Problem 2.35, there are only two ways to connect four carbon atoms in a compound with the molecular formula C4H10:

In our case, the molecular formula is C4H9Cl, which is similar to C4H10, but one H has been replaced with a chlorine atom. So, we must explore all of the different locations where a chlorine atom can be placed on each of the carbon skeletons above (the linear skeleton and the branched skeleton). Let’s begin with the linear skeleton. There are two distinctly different locations where a chlorine atom can be placed on this skeleton: either at position 1 or position 2, shown here:

Placing the chlorine atom at position 3 would be the same as placing it at position 2; and placing the chlorine atom at position 4 would be the same as it as position 1:

Next, we move on to the other carbon skeleton, containing a branch. Once again, there are two distinctly different locations where a chlorine atom can be placed: either at position 1 or position 2, shown here:

In summary, there are a total of four constitutional isomers with the molecular formula C4H9Cl:

2.51. (a) This compound exhibits a lone pair next to a  bond, so we draw the two curved arrows associated with that pattern. The first curved arrow is drawn showing a lone pair becoming a  bond, while the second curved arrow shows a  bond becoming a lone pair. Notice that both resonance structures have zero net charge:

(b) This structure exhibits a lone pair that is adjacent to C+ so we draw one curved arrow, showing a lone pair becoming a  bond. Notice that both resonance structures have an overall +1 net charge, but they show the positive charge in different locations:

(c) This structure exhibits a C+ (carbocation) that is allylic, so we draw one curved arrow showing the  bond being pushed over. Notice that both resonance structures have an overall +1 net charge, but they show the positive charge in different locations:

(d) This compound exhibits a C=N bond (a  bond between two atoms of differing electronegativity), so we draw one curved arrow showing the  bond becoming a lone pair on the more electronegative nitrogen atom. Notice that both resonance structures have zero net charge:

Placing the chlorine atom on any of the peripheral carbon atoms will lead to the same compound: 2.52. (a) This compound exhibits a lone pair next to a  bond, so we draw two curved arrows. The first curved arrow is drawn showing a lone pair becoming a  bond, while the second curved arrow shows a  bond becoming a lone

CHAPTER 2

pair. We then draw the resulting resonance structure and assess whether it exhibits one of the five patterns. In this case, the lone pair is now next to another  bond, so once again, we draw the two curved arrows associated with that pattern. The resulting resonance structure again exhibits a lone pair next to a  bond. This pattern continues again, thereby spreading a negative charge over three carbon atoms in the ring, as shown here:

curved arrow associated with that pattern (pushing over the  bond). This pattern continues, several more times, spreading the positive charge over four carbon atoms, as shown here:

Notice that all resonance structures have zero net charge. (b) This structure exhibits a lone pair next to a  bond, so we draw two curved arrows. The first curved arrow is drawn showing a lone pair becoming a  bond, while the second curved arrow shows a  bond becoming a lone pair. We then draw the resulting resonance structure and assess whether it exhibits one of the five patterns. In this case, the lone pair is now next to another  bond, so once again, we draw the two curved arrows associated with that pattern. The resulting resonance structure again exhibits a lone pair next to a  bond, so we draw one more resonance structure, as shown here:

Notice that all resonance structures have zero net charge. (d) We begin by looking for the five patterns, focusing first on any patterns that employ just one curved arrow (in this case, there is another pattern that requires two curved arrows, but we will start with the pattern using just one curved arrow). There is a C=O bond (a  bond between two atoms of differing electronegativity), so we draw one curved arrow showing the  bond becoming a lone pair on the more electronegative oxygen atom. We then draw the resulting resonance structure and assess whether it exhibits one of the five patterns. In this case, there is a lone pair adjacent to a C+, so we draw the curved arrow associated with that pattern (showing the lone pair becoming a  bond), shown here:

Notice that all resonance structures have zero net charge. Notice that all four resonance structures have an overall –1 net charge, but they show the negative charge in different locations: (c) This compound exhibits a C=O bond (a  bond between two atoms of differing electronegativity), so we draw one curved arrow showing the  bond becoming a lone pair on the more electronegative oxygen atom. We then draw the resulting resonance structure and assess whether it exhibits one of the five patterns. In this case, there is a C+ (carbocation) that is allylic, so we draw the

(e) This structure exhibits a C+ (carbocation) that is allylic, so we draw one curved arrow showing the  bond being pushed over. We then draw the resulting resonance structure and assess whether it exhibits one of the five patterns. In this case, C+ is again next to a  bond, so again, we draw the curved arrow associated with that pattern (pushing over the  bond). The resulting resonance structure has C+ adjacent to a lone pair, so we draw the one curved arrow associated with that pattern (showing the lone pair becoming a  bond), shown here:

CHAPTER 2

Notice that all resonance structures have an overall +1 net charge, but they show the positive charge in different locations. (f) This structure exhibits a C+ (carbocation) that is allylic, so we draw one curved arrow showing the  bond being pushed over. We then draw the resulting resonance structure and assess whether it exhibits one of the five patterns. In this case, C+ is again next to a  bond, so again, we draw the curved arrow associated with that pattern (pushing over the  bond). The resulting resonance structure again has C+ next to a  bond, so again, we draw the curved arrow associated with that pattern (pushing over the  bond). The resulting resonance structure has C+ adjacent to a lone pair, so we draw the one curved arrow associated with that pattern (showing the lone pair becoming a  bond), shown here: O

Notice that all resonance structures have an overall +1 net charge, but they show the positive charge in different locations. 2.53. These structures do not differ in their connectivity of atoms. They differ only in the placement of electrons that are in pi bonds and lone pairs. Therefore, these structures are resonance structures, as shown here:

2.54. (a) These compounds both have the same molecular formula (C7H12), but they differ in their connectivity of atoms, or constitution. Therefore, they are constitutional isomers. (b) These structures have the same molecular formula (C7H16), AND they have the same constitution (connectivity of atoms), so they represent the same compound. (c) The first compound has the molecular formula C5H10, while the second compound has the molecular formula C5H8. As such, they are different compounds that are not isomeric. (d) These compounds both have the same molecular formula (C5H8), but they differ in their connectivity of atoms, or constitution. Therefore, they are constitutional isomers. 2.55. (a) The condensed structure (shown in the problem statement) indicates the constitution (how the atoms are connected to each other). In the bond-line structure, hydrogen atoms are not drawn (they are implied). Each corner and each endpoint represents a carbon atom, so the carbon skeleton is shown more clearly.

(b) The condensed structure indicates how the atoms are connected to each other. In the bond-line structure, hydrogen atoms are not drawn (they are implied), except for the hydrogen atom attached to the oxygen atom (hydrogen atoms must be drawn if they are connected to a heteroatom, such as oxygen). Each corner and each endpoint represents a carbon atom, so the carbon skeleton is shown more clearly.

(c) The condensed structure indicates how the atoms are connected to each other. In the bond-line structure, hydrogen atoms are not drawn (they are implied). Each corner and each endpoint represents a carbon atom, so the carbon skeleton is shown more clearly.

(d) The condensed structure indicates how the atoms are connected to each other. In the bond-line structure, hydrogen atoms are not drawn (they are implied). Each corner and each endpoint represents a carbon atom, so the carbon skeleton is shown more clearly.

CHAPTER 2 nitrogen atom or 2) generate a nitrogen atom with less than an octet of electrons, or 3) generate a structure with three charges. The first of these would not be a valid resonance structure, and the latter two would not be significant resonance structures.

(e) The condensed structure indicates how the atoms are connected to each other. In the bond-line structure, hydrogen atoms are not drawn (they are implied). Each corner and each endpoint represents a carbon atom, so the carbon skeleton is shown more clearly.

2.57. The negatively charged oxygen atom has three lone pairs, the positively charged oxygen atom has one lone pair, and the uncharged oxygen atom has two lone pairs (see Table 2.2). Notice that this compound exhibits a lone pair that is next to a  bond, so we must draw two curved arrows associated with that pattern. The first curved arrow is drawn showing a lone pair becoming a  bond, while the second curved arrow shows a  bond becoming a lone pair:

(f) The condensed structure indicates how the atoms are connected to each other. In the bond-line structure, hydrogen atoms are not drawn (they are implied). Each corner and each endpoint represents a carbon atom, so the carbon skeleton is shown more clearly.

These two resonance forms have completely filled octets and are equivalent (they contribute equally to the resonance hybrid). There are no other valid resonance structures that are significant. 2.58. Each nitrogen atom has a lone pair that is adjacent to a  bond and is, therefore, delocalized via resonance. In order to be delocalized via resonance, the lone pair must occupy a p orbital, and therefore, each nitrogen atom must be sp2 hybridized. As such, each nitrogen atom is trigonal planar.

2.56. The nitronium ion does not have any significant resonance structures because any attempts to draw a resonance structure will either 1) exceed an octet for the

2.59. (a) This compound (estradiol) exhibits a lone pair next to a  bond, so we draw two curved arrows. The first curved arrow is drawn showing a lone pair becoming a  bond, while the second curved arrow shows a  bond becoming a lone pair. We then draw the resulting resonance structure and assess whether it exhibits one of the five patterns. In this case, the lone pair is next to another  bond, so once again, we draw the two curved arrows associated with that pattern. The resulting resonance structure again exhibits a lone pair next to a  bond, so again we draw two curved arrows and the resulting resonance structure. Once again, there is a lone pair next to a  bond, which requires that we draw one final resonance structure, shown below. This last resonance structure is not the same as the original resonance structure, because of the locations in which the  bonds are drawn (i.e., the resonance pattern of conjugated  bonds enclosed in a ring). Notice that all resonance structures have zero net charge: OH

OH H

H H

H H HO

H H

H HO

CHAPTER 2

(b) The following compound (testosterone) exhibits a C=O bond (a  bond between two atoms of differing electronegativity), so we draw one curved arrow showing the  bond becoming a lone pair on the more electronegative oxygen atom. We then draw the resulting resonance structure and assess whether it exhibits one of the five patterns. In this case, there is a C+ (carbocation) that is allylic, so we draw the curved arrow associated with that pattern (pushing over the  bond). Notice that all resonance structures have zero net charge:

2.60. This compound exhibits a C=O bond (a  bond between two atoms of differing electronegativity), so we draw one curved arrow showing the  bond becoming a lone pair on the more electronegative oxygen atom. We then draw the resulting resonance structure and assess whether it exhibits one of the five patterns. In this case, there is an allylic C+ (carbocation), so we draw the curved arrow associated with that pattern (pushing over the  bond), shown here. This pattern continues, several more times, spreading the positive charge over a total of six carbon atoms. Notice that all resonance structures have zero net charge:

By considering the significant resonance structures (drawn above), we can determine the positions that are electron deficient (+). There are six sites of low electron density:

2.61. This compound exhibits a lone pair next to a  bond, so we draw two curved arrows. The first curved arrow is drawn showing a lone pair becoming a  bond, while the second curved arrow shows a  bond becoming a lone pair. We then draw the resulting resonance structure and assess whether it exhibits one of the five patterns. In this case, there is a lone pair next to a  bond, so once again, we draw the two curved arrows associated with that pattern. The resulting resonance structure again exhibits a lone pair next to a  bond. This pattern continues, several more times, spreading a negative charge over a total of five carbon atoms. Notice that all resonance structures have zero net charge:

CHAPTER 2

By considering the significant resonance structures (drawn above), we can determine the positions that are electron rich (‒). There are five sites of high electron density:

2.62. Two patterns of resonance can be identified on the given structure: carbonyl resonance and allylic lone pair resonance (involving either the oxygen or nitrogen lone pairs). All three of these options will be used. Since it is possible to start with any one of the three, you may have developed the resonance forms in a different order than presented here, but you still should have found four reasonable resonance forms. Only one of the four structures has an atom with an incomplete octet (the last resonance structure shown), so that is identified as the least significant contributor to the hybrid. The first resonance form is the most significant contributor because it has filled octets and no formal charges. The middle two resonance structures both have filled octets and a negative charge on an oxygen atom, so they are ranked according to their only difference: the location of the positive charge. The third structure is the 2nd most significant resonance form because it has the positive charge on the less electronegative nitrogen atom. Note it is better to place a negative charge on a more electronegative atom, and it is better to place a positive charge on a less electronegative atom.

2.63. The only resonance pattern evident in the enamine is an allylic lone pair. After that pattern is applied, however, another allylic lone pair results so the resonance can ultimately involve both  bonds. There are a total of three major resonance forms that all have filled octets. Consideration of the hybrid of these resonance forms predicts two electronrich sites (‒).

Did you draw the following additional structure (or something similar, with C+ and C) and wonder why it was not shown in this solution?

This resonance form suffers from two major deficiencies: 1) it does not have filled octets, while the other resonance forms shown above all have filled octets, and 2) it has a negative charge on a carbon atom (which is not an electronegative atom). Either of these deficiencies alone would render the resonance form a minor contributor. But with both deficiencies together (C+ and C–), this resonance form is insignificant. The same is true for any resonance form that has both C+

CHAPTER 2

and C–. Such a resonance form will generally be insignificant (there are very few exceptions, one of which will be seen in Chapter 17). Also, note that the  bonds cannot be moved to other parts of the six-membered ring since the CH2 groups are sp3 hybridized. These carbon atoms cannot accommodate an additional bond without violating the octet rule.

2.64. The correct answer is (c). Structures (a), (b) and (d) are all significant resonance structures, as shown:

Specifically, this resonance structure has alternating single and double bonds in the ring and is considered to be aromatic (a stabilizing effect). 2.66. The correct answer is (d). The nitrogen atom in structure (a) is delocalized by resonance, and is therefore sp2 hybridized:

Structure (c) is not a resonance form at all. To see this more clearly, notice that the benzyl carbocation does not have any CH2 groups in the ring, but structure (c) does have a CH2 group in the ring:

Resonance structures differ only in the placement of electrons. Since structure (c) differs in the connectivity of atoms, it cannot be considered a resonance structure of the benzyl carbocation. 2.65. The correct answer is (a). The atoms in all four structures have complete octets. So we must consider the location of the negative charge. Structure (a) has a negative charge on an electronegative atom (oxygen). A negative charge is more stable on the more electronegative atom (oxygen) than it is on a nitrogen atom or a carbon atom. Therefore, structure (a) will contribute the most character to the resonance hybrid:

The nitrogen atom in structure (b) is also delocalized, so it is sp2 hybridized as well:

The nitrogen atom in structure (c) is also sp2 hybridized, because this nitrogen atom must be using a p orbital to participate in  bonding (C=N). The nitrogen atom in structure (d) has three  bonds, and its lone pair is localized. Therefore, this nitrogen atom is sp3 hybridized:

2.67. The correct answer is (c). We begin by considering all bonds in the compound, which is easier to do if we redraw the compound as shown:

This compound corresponds structure (c):

It turns out that this resonance structure is also significant for another reason that we will explore in Chapter 17.

CHAPTER 2

2.68. The correct answer is (c). For C+ that is allylic, only one curved arrow will be required to draw the resonance structures (pushing over the  bond):

2.69. The correct answer is (a). An anion will be resonance delocalized if the atom bearing the negative charge has a lone pair that is next to a π bond. Only compound I fits this pattern:

2.70. The correct answer is (a). There are only two resonance structures possible for this allylic carbocation. Drawing all the hydrogen atoms helps to demonstrate why the cation cannot be delocalized to any other position without breaking a single bond (and we cannot break a single bond when drawing resonance structures).

2.71. The correct answer is (d). Begin by drawing all significant resonance structures. In this case, there are two:

Both are equally significant, so the resonance hybrid is the simple average of these two resonance structures. When drawing the resonance hybrid, partial bonds and partial charges are required, as shown below:

2.72. The correct answer is (b). The following do not represent a pair of resonance structures, because the first rule of resonance has been violated (by breaking a single bond):

2.73. (a) The molecular formula is C3H6N2O2. (b) Each of the highlighted carbon atoms (below) has four sigma bonds (the bonds to hydrogen are not shown). As such, these two carbon atoms are sp3 hybridized.

(c) There is one carbon atom that is using a p orbital to form a  bond. As such, this carbon atom (highlighted) is sp2 hybridized.

(d) There are no sp hybridized carbon atoms in this structure. (e) There are six lone pairs (each nitrogen atom has one lone pair and each oxygen atom has two lone pairs):

(f) Only the lone pair on one of the nitrogen atoms is delocalized via resonance (because it is next to a  bond). The other lone pairs are all localized.

(g) The geometry of each atom is shown below (see SkillBuilder 1.8):

CHAPTER 2 (h) We begin by looking for the five patterns, focusing first on any patterns that employ just one curved arrow (in this case, there is another pattern that requires two curved arrows, but we will start with the pattern using just one curved arrow). There is a C=O bond (a  bond between two atoms of differing electronegativity), so we draw one curved arrow showing the  bond becoming a lone pair on the more electronegative oxygen atom. We then draw the resulting resonance structure and assess whether it exhibits one of the five patterns. In this case, there is a lone pair adjacent to a C+, so we draw the curved arrow associated with that pattern (showing the lone pair becoming a  bond). Notice that all three resonance structures have zero net charge:

(f) Recall that a lone pair next to a  bond will be delocalized by resonance. The lone pairs on the oxygen of the C=O bond are localized. One of the lone pairs on the other oxygen atom (attached to the aromatic ring) is delocalized via resonance. The lone pair on the nitrogen atom is delocalized via resonance. (g) All sp2-hybridized carbon atoms are trigonal planar. All sp3 hybridized carbon atoms are tetrahedral. The nitrogen atom is trigonal planar. The oxygen atom of the C=O bond does not have a geometry because it is connected to only one other atom, and the other oxygen atom has bent geometry (see SkillBuilder 1.8).

2.74. (a) The molecular formula is C16H21NO2. (b) Each of the highlighted carbon atoms (below) has four sigma bonds (the bonds to hydrogen are not shown). As such, these nine carbon atoms are sp3 hybridized.

2.75. (a) In Section 1.5, we discussed inductive effects and we learned how to identify polar covalent bonds. In this case, there are two carbon atoms that participate in polar covalent bonds (the C‒Br bond and the C‒O bond). Each of these carbon atoms will be poor in electron density (+) because oxygen and bromine are each more electronegative than carbon:

(b) There are two carbon atoms that are adjacent to one or both oxygen atoms. These carbon atoms will be poor in electron density (+), because oxygen is more electronegative than carbon:

(c) There are seven carbon atoms that are each using a p orbital to form a  bond. As such, these seven carbon atoms (highlighted) are sp2 hybridized.

The carbon atom of the carbonyl (C=O) group is especially electron deficient, as a result of resonance. (c) There are two carbon atoms that are adjacent to electronegative atoms. These carbon atoms will be poor in electron density (+), because oxygen and chlorine are each more electronegative than carbon:

(d) There are no sp-hybridized carbon atoms in this structure. (e) There are five lone pairs (the nitrogen atom has one lone pair and each oxygen atom has two lone pairs):

The carbon atom of the carbonyl (C=O) group is especially electron deficient, as a result of resonance.

CHAPTER 2

2.76. We begin by drawing all significant resonance structures, and then considering the placement of the formal charges on carbon atoms in each of those resonance structures (highlighted below)

A position that bears a positive charge is expected to be electron deficient (+), while a position that bears a negative charge is expected to be electron rich (). The following is a summary of the electron-deficient positions and the electron-poor positions, as indicated by the resonance structures above.

2.77. (a) Compound B has one additional resonance structure that compound A lacks, because of the relative positions of the two groups on the aromatic ring. Specifically, compound B has a resonance structure in which one oxygen atom has a negative charge and the other oxygen atom has a positive charge: 2.78. (a) The following group is introduced, and it contains five carbon atoms:

Compound A does not have a significant resonance structure in which one oxygen atom has a negative charge and the other oxygen atom has a positive charge. That is, compound A has fewer resonance structures than compound B. Accordingly, compound B has greater resonance stabilization. (b) Compound C is expected to have resonance stabilization similar to that of compound B, because compound C also has a resonance structure in which one oxygen atom has a negative charge and the other oxygen atom has a positive charge:

(b) The following highlighted carbon atom is involved in the reaction:

CHAPTER 2

The structure on the left is the most significant, because every atom has an octet and it has no formal charges. The resonance hybrid is a weighted average of these three resonance structures. Since the partial positive charge is delocalized onto two carbon atoms and the partial negative charge is localized on only one oxygen atom, the

partial negative charge is drawn larger than each of the individual partial positive charges.

(d) The reactive site (highlighted above) has partial positive character, which means that it is electron deficient. This is what makes it reactive. In the actual synthesis, this compound is treated with a carbanion (a structure containing a carbon atom with a negative charge). The reaction causes formation of a bond between the electron-deficient carbon atom and the electron-rich carbon atom. We will learn about such reactions in Chapter 21.

2.79. We will need to draw two resonance hybrids, one for each of the highlighted carbon atoms. One highlighted position is part of an aromatic ring, and we will begin by focusing on that position. In doing so, we can save time by redrawing only the relevant portion of the molecule, like this:

This aromatic ring has eight significant resonance structures, shown here:

CHAPTER 2

The resonance hybrid is a weighted average of these eight resonance structures. Resonance structures 1 and 2 are equally most significant because all atoms have an octet AND there are no formal charges. Resonance structures 3-8 are less significant than 1 and 2 but approximately equally significant to each other because every atom has a full octet with a positive charge on an oxygen atom and a negative charge on a carbon atom. Since the partial negative charge is delocalized onto three carbon atoms and the partial positive charge is delocalized onto only two oxygen atoms, the partial positive charges are drawn slightly larger than the partial negative charges. The analysis allows us to draw the resonance hybrid for this portion of the molecule, and it demonstrates that the highlighted carbon atom is electron rich (-).

Now let’s focus on our attention on the other highlighted carbon atom (the one that is part of a carboxylic acid group). Just as we did before, we will draw only the portion of the molecule that is of interest:

This portion of the molecule has four significant resonance structures, shown below:

The resonance hybrid is a weighted average of these four resonance structures. Resonance structure 1 is most significant because all atoms have an octet AND there are no formal charges. Resonance structure 3 is the next most significant because there are formal charges, yet every atom has a full octet. Since the partial negative charge is localized on one oxygen atom, this partial charge is the largest. The partial positive charge is delocalized over three atoms, but it is larger on the oxygen (relative to either of the carbon atoms). The resonance hybrid demonstrates that this carbon atom is electron poor (+).

CHAPTER 2

2.80. (a) The molecular formula for CL-20 is C6H6N12O12. The molecular formula for HMX is C4H8N8O8. (b) The lone pair is next to a  bond, and therefore delocalized (see resonance structures below).

2.81. This intermediate is highly stabilized by resonance. The positive charge is spread over one carbon atom and three oxygen atoms.

2.82. (a) Both molecules have identical functional groups (alcohol + alkene), and both compounds have the same molecular formula (C27H46O):

The structure on the left exhibits two six-membered rings and two five-membered rings, while the structure on the right has three six-membered rings and only one five-membered ring. The long alkane group is apparently located in the wrong position on the five-membered ring of the incorrect structure. (b) Both structures contain an alkene group, an aromatic ring, an amide group, and two ether functional groups. But the incorrect structure has a third ether functional group (in the eight-membered ring), while the correct structure has an alcohol functional group. The incorrect structure has an eight-membered ring, while the correct structure has a fivemembered ring. The two carbon atoms and oxygen atom in the ring of the incorrect structure are not part of the ring for the correct structure.

CHAPTER 2

2.83. (a) The positive charge in basic green 4 is resonance-stabilized (delocalized) over twelve positions (two nitrogen atoms and ten carbon atoms), as seen in the following resonance structures.

(b) The positive charge in basic violet 4 is expected to be more stabilized than the positive charge in basic green 4, because the former is delocalized over thirteen positions, rather than twelve. Specifically, basic violet 4 has an additional resonance structure that basic green 4 lacks, shown here:

CHAPTER 2 In basic violet 4, the positive charge is spread over three nitrogen atoms and ten carbon atoms. 2.84. Polymer 2 contains only ester groups, so the IR spectrum of polymer 2 is expected to exhibit a signal near 1740 cm-1 (typical for esters), associated with vibrational excitation (stretching) of the C=O bond. Polymer 4 lacks any ester groups, so the signal near 1740 cm-1 is expected to be absent in the IR spectrum of polymer 4. Instead, polymer 4 has OH groups, which are expected to produce a broad signal in the range 3200-3600 cm-1. Polymer 3 has both functional groups (alcohol group and ester group), so an IR spectrum of polymer 3 is expected to exhibit both characteristic signals. When polymer 3 is converted to polymer 4, the signal near 1740 cm-1 is expected to vanish, which would indicate complete hydrolysis of polymer 3. In practice, the signal for the C=O stretch in polymer 2 appears at 1733 cm-1, which is very close to our estimated value of 1740 cm-1. 2.85. Compound 1 has an OH group, which is absent in compound 2. Therefore, the IR spectrum of 1 should

exhibit a broad signal in the range 3200-3600 cm-1 (associated with O-H stretching), while the IR spectrum of 2 would be expected to lack such a signal. The conversion of 1 to 2 could therefore be confirmed with the disappearance of the signal corresponding with excitation of the O-H bond. Another way to monitor the conversion of 1 to 2 is to focus on the C-H bond of the aldehyde group in compound 1, which is expected to produce a signal in the range 27502850 cm-1. Since the aldehyde group is not present in compound 2, we expect this signal to vanish when 1 is converted to 2. There is yet another way to monitor this reaction with IR spectroscopy. Compound 1 possesses only one C=O bond, while compound 2 has two C=O bonds. As such, the latter should exhibit two C=O signals. One signal is expected to be near 1680 cm-1 (for the conjugated ketone), and the other signal should be near 1700 cm-1 (corresponding to the conjugated ester). In contrast compound 1 has only one C=O bond, which is expected to produce a signal near 1680 cm-1 (for the conjugated aldehyde). Therefore, the conversion of 1 to 2 can be monitored by the appearance of a signal near 1700 cm-1.

2.86. (a) This compound contains the following functional groups:

(b) The nitrogen atom has a lone pair that is delocalized via resonance:

In the second resonance structure shown, the C-N bond is drawn as a double bond, indicating partial double-bond character. This bond is thus a hybrid between a single and double bond; the partial double bond character results in

CHAPTER 2

partially restricted rotation around this bond. In contrast, the C-N bond on the left experiences free rotation because that bond has only single-bond character.

2.87. (a) Each of the four amides can be represented as a resonance hybrid (one example shown below). The charge-separated resonance structure indicates that there is a δ+ on the amide nitrogen, which thus pulls the electrons in the N-H bond closer to the nitrogen atom, leaving the hydrogen atom with a greater δ+. This resonance effect is not present in the NH bond of the amines. Thus, the δ+ on an amide H is greater than that on an amine H, leading to a stronger hydrogen bond.

(b) The following intermolecular hydrogen bonds are formed during self-assembly:

CHAPTER 2

2.88. (a) Anion 2 is highly stabilized by resonance (the negative charge is delocalized over two oxygen atoms and three carbon atoms). The resonance structures for 2 are as follows:

Cation 3 is highly stabilized by resonance (the positive charge is delocalized over two oxygen atoms and four carbon atoms). The resonance structures of 3 are as follows:

(b) Double bonds are shorter in length than single bonds (see Table 1.2). As such, the C-C bonds in compound 1 will alternate in length (double, single, double, etc.):

The double bonds do have some single-bond character as a result of resonance, as can be seen in resonance structure 1c:

CHAPTER 2

Similarly, the single bonds have some double-bond character, also because of resonance. However, this effect is relatively small, because there is only one resonance structure (1a above) in which all atoms have an octet AND there are no formal charges. Therefore, it is the most significant contributor to the overall resonance hybrid. As such, the double bonds have only a small amount of single-bond character, and the single bonds have only a small amount of double-bond character. In contrast, anion 2 does not have a resonance structure that lacks charges. All resonance structures of 2 bear a negative charge. Among the resonance structures, two of them (2a and 2e) contribute the most character to the overall resonance hybrid, because the negative charge is on an electronegative oxygen atom (rather than carbon).

In fact, these two resonance contributors will contribute equally to the overall resonance hybrid. As such, the bonds of the ring will be very similar in length, because they have both single-bond character and double-bond character in equal amounts. A similar argument can be made for compound 3. (c) In compound 1, a hydrogen bonding interaction occurs between the proton of the OH group and the oxygen atom of the C=O bond:

This interaction is the result of the attraction between partial charges (+ and -). However, in cation 3, a similar type of interaction is less effective because the O of the C=O bond is now poor in electron density, and therefore less capable of forming a hydrogen bonding interaction, as can be seen in resonance structure 3a.

The other oxygen atom is also ineffective at forming an intramolecular hydrogen bond because it too is poor in electron density, as can be seen in resonance structure 3f:

2.89. In order for all four rings to participate in resonance stabilization of the positive charge, the p orbitals in the four rings must all lie in the same plane (to achieve effective overlap). In the following drawing, the four rings are labeled A-D. Notice that the D ring bears a large substituent (highlighted) which is trying to occupy the same space as a portion of the C ring:

This type of interaction, called a steric interaction, forces the D ring to twist out of plane with respect to the other three rings, like this:

In this way, the overlap between the p orbitals of the D ring and the p orbitals of the other three rings is expected to be less effective. As such, participation of the D ring in resonance stabilization is expected to be diminished with respect to the participation of the other three rings.

Chapter 3 Acids and Bases Review of Concepts Fill in the blanks below. To verify that your answers are correct, look in your textbook at the end of Chapter 3. Each of the sentences below appears verbatim in the section entitled Review of Concepts and Vocabulary.       

A Brønsted-Lowry acid is a proton ________, while a Brønsted-Lowry base is a proton _____________. A reaction mechanism utilizes curved arrows to show the flow of _____________ that account for a chemical reaction. The mechanism of proton transfer always involves at least _____ curved arrows. A strong acid has a ________ pKa, while a weak acid has a _________ pKa. There are four factors to consider when comparing the ___________ of anionic conjugate bases. The equilibrium of an acid-base reaction always favors the more ____________ negative charge. A Lewis acid is an electron-pair ___________, while a Lewis base is an electron-pair _________.

Review of Skills Fill in the blanks and empty boxes below. To verify that your answers are correct, look in your textbook at the end of Chapter 3. The answers appear in the section entitled SkillBuilder Review. SkillBuilder 3.1 Drawing the Mechanism of a Proton Transfer

SkillBuilder 3.2 Using pKa Values to Compare Acids Circle the compound below that is more acidic:

pKa = 19.2

pKa = 4.75

SkillBuilder 3.3 Using pKa Values to Compare Basicity

CHAPTER 3

SkillBuilder 3.4 Using pKa Values to Predict the Position of Equilibrium

SkillBuilder 3.5 Assessing Relative Stability. Factor 1 – Atom

SkillBuilder 3.6 Assessing Relative Stability. Factor 2 – Resonance

SkillBuilder 3.7 Assessing Relative Stability. Factor 3 – Induction

SkillBuilder 3.8 Assessing Relative Stability. Factor 4 – Orbital

SkillBuilder 3.9 Assessing Relative Stability. Using All Four Factors

SkillBuilder 3.10 Assessing the Relative Acidity of Cationic Acids

SkillBuilder 3.11 Predicting the Position of Equilibrium Without Using pKa Values

CHAPTER 3

SkillBuilder 3.12 Choosing the Appropriate Reagent for a Proton Transfer Reaction

SkillBuilder 3.13 Identifying Lewis Acids and Lewis Bases

Common Mistakes to Avoid When drawing the mechanism of a proton transfer, two curved arrows are required. The first curved arrow shows the base attacking the proton, and the second curved arrow shows the bond to H being broken.

It is a common mistake to draw only the first curved arrow and not the second, so make sure to draw both curved arrows. When drawing the second curved arrow, make sure that the tail is placed on the middle of the bond to the H, as shown:

When the acid is H3O+, you must draw at least one of the O–H bonds in order to draw the second curved arrow properly.

This is also the case for other acids:

CHAPTER 3

Useful reagents In Chapter 3, we explored the behavior of acids and bases. Throughout the remainder of the textbook, many acids and bases will be frequently encountered. It would be wise to become familiar with the following reagents (and their uses), as they will appear many times in the upcoming chapters:

Structure

Name

Use

Sulfuric acid

A very strong acid. Commonly used as a source of protons. Concentrated sulfuric acid is a mixture of H2SO4 and H2O, and the acid present in solution is actually H3O+, because of the leveling effect. That is, H3O+ is a weaker acid than H2SO4, so the protons are transferred from H2SO4 to water, giving a high concentration of H3O+.

Hydrochloric acid

A very strong acid. Similar in function to H2SO4. In an aqueous solution of HCl, the acid that is present is H3O+, because of the leveling effect, as described above for H2SO4.

Acetic acid

A weak acid. Mild source of protons.

Water

A weak acid and a weak base. It can function as either, depending on the conditions. When treated with a strong base, water will function as an acid (a source of protons). When treated with a strong acid, water will function as a base and remove a proton from the strong acid.

An alcohol

R represents the rest of the compound. Alcohols (the subject of Chapter 12) are compounds that possess an OH group connected to an sp3-hybridized carbon atom. Alcohols can function very much like water (either as weak acids or as weak bases).

Ammonia

A fairly strong base, despite the absence of a negative charge. It is a strong base, because its conjugate acid (NH4+), called an ammonium ion, is a weak acid (pKa = 9.2).

CH3CH2ONa

Sodium ethoxide

The ethoxide ion (CH3CH2Oˉ) is a strong base, and Na+ is the counterion. Other alkoxide ions (ROˉ, where R is an alkyl group), such as methoxide (CH3Oˉ), are also strong bases.

NaNH2

Sodium amide

H2Nˉ is a very strong base, and Na+ is the counterion.

n-Butyllithium

An extremely strong base. This is one of the strongest bases that you will encounter. It is often abbreviated as n-BuLi

O H

CHAPTER 3

Solutions 3.1. (a) Phenol (C6H5OH) loses a proton and is therefore functioning as an acid. Hydroxide (HOˉ) functions as the base that removes the proton. Two curved arrows must be drawn. The first curved arrow shows a lone pair of the base attacking the proton, and the second curved arrow comes from the O–H bond (being broken) and goes to the oxygen atom, as shown:

(b) H3O+ loses a proton and is therefore functioning as an acid. The ketone functions as the base that removes the proton. Two curved arrows must be drawn. The first curved arrow shows a lone pair of the base attacking the proton; the second curved arrow comes from the O–H bond (being broken) and goes to the oxygen atom, as shown:

(c) The ketone loses a proton and is therefore functioning as an acid. The other reagent functions as the base that removes the proton. Two curved arrows must be drawn. The first curved arrow shows a lone pair of the base attacking the proton, and the second curved arrow comes from the C–H bond (being broken) and goes to the carbon atom, as shown:

(d) Benzoic acid (C6H5CO2H) loses a proton and is therefore functioning as an acid. Hydroxide (HOˉ) functions as the base that removes the proton. Two curved

arrows must be drawn. The first curved arrow shows a lone pair of the base attacking the proton, and the second curved arrow comes from the O–H bond (being broken) and goes to the oxygen atom, as shown:

3.2. Two curved arrows must be drawn. The first curved arrow shows a lone pair of the base attacking the proton, and the second curved arrow comes from the O–H bond (being broken) and goes to the oxygen atom, as shown:

3.3. The nitrogen atom is accepting a proton, so dextromethorphan is the base in this proton-transfer reaction, and HBr is the acid (proton donor). As usual, the mechanism involves two curved arrows. The salt product contains both the conjugate acid (protonated dextromethorphan) and the conjugate base (Br—).

3.4. (a) According to Table 3.1, phenol (C6H5OH) has a pKa of 9.9, while water has a pKa of 15.7. Phenol is more acidic because it has a lower pKa value.

CHAPTER 3

(b) According to Table 3.1, (CH3)3COH has a pKa of 18.0, while water has a pKa of 15.7. Water is more acidic because it has a lower pKa value. (c) According to Table 3.1, ammonia (NH3) has a pKa of 38, while acetylene (H–C≡C–H) has a pKa of 25. As such, acetylene is more acidic because it has a lower pKa value.

The first compound is more acidic because it has a lower pKa value. We know that a stronger acid has a weaker conjugate base, so the conjugate base of the first compound will be a weaker base than the conjugate base of the second compound:

(d) According to Table 3.1, H3O+ has a pKa of -1.74, while HCl has a pKa of -7. As such, HCl is more acidic because it has a lower pKa value. (e) According to Table 3.1, ethane (C2H6) has a pKa of 50, while acetylene (H–C≡C–H) has a pKa of 25. As such, acetylene is more acidic because it has a lower pKa value.

(b) We first imagine protonating each base, and then we compare the pKa values of the resulting compounds (using Table 3.1):

(f) According to Table 3.1, a protonated ketone has a pKa of -7.3, while sulfuric acid (H2SO4) has a pKa of -9. As such, sulfuric acid is more acidic because it has a lower pKa value. 3.5. According to Table 3.1, the proton connected to the oxygen atom is expected to be the most acidic proton in the compound. That proton is expected to have a pKa value near 16 (similar to CH3CH2OH), though it may be somewhat different from this value, based on the effect of adjacent groups. The proton connected to the nitrogen atom is expected to have a pKa value near 38 (similar to NH3), so it is not the most acidic proton.

The second compound (CH3CH2OH) is more acidic because it has a lower pKa value. We know that a stronger acid has a weaker conjugate base, so the conjugate base of the second compound will be a weaker base than the conjugate base of the first compound:

3.6. According to Table 3.1, the proton of the carboxylic acid group (Ha below) is expected to be the most acidic (pKa ~ 5). The two protons labeled Hb and Hc are expected to have a pKa near 10 (like C6H5OH), and the protons labeled Hd are expected to have a pKa near 38 (like NH3). The order of acidity is shown below:

3.7. (a) We first imagine protonating each base, and then we compare the pKa values of the resulting compounds (using Table 3.1):

The first ion is more acidic because it has a lower pKa value. We know that a stronger acid has a weaker conjugate base, so the conjugate base of the first ion will be a weaker base than the conjugate base of the second ion:

(d) We first imagine protonating each base, and then we compare the pKa values of the resulting compounds (using Table 3.1):

The first compound (water) is more acidic because it has a lower pKa value. We know that a stronger acid has a weaker conjugate base, so the conjugate base of the

CHAPTER 3 second compound will be a stronger base than the conjugate base of water:

the lone pairs on the oxygen atom. When compound 1 is protonated, we expect the nitrogen atom to be protonated (rather than the oxygen atom) to give the following conjugate acid:

(e) We first imagine protonating each base, and then we compare the pKa values of the resulting compounds (using Table 3.1):

The second compound is more acidic because it has a lower pKa value. We know that a stronger acid has a weaker conjugate base, so the conjugate base of the second compound will be a weaker base than the conjugate base of the first compound:

(f) We first imagine protonating each base, and then we compare the pKa values of the resulting compounds (using Table 3.1). HCl has a pKa of -7, while H2O has a pKa of 15.7. Since HCl has a lower pKa value, it is more acidic than water. We know that a stronger acid has a weaker conjugate base, so the conjugate base of HCl will be a weaker base than the conjugate base of H2O.

3.8. The pKa values indicate that a proton is more acidic when it is connected to a positively charged oxygen atom (pKa ~ -2.2) than when it is connected to a positively charged nitrogen atom (pKa ~ 10.6). The weaker acid, (CH3)2NH2+, gives the stronger base upon deprotonation: H H3C

The second has a higher pKa value, and is therefore a weaker acid. The weaker acid gives the stronger base upon deprotonation, so we expect the lone pair on the nitrogen atom on the right in structure 1 to be more basic:

Therefore, when 1 is mono-protonated, the sp3 hybridized nitrogen atom is protonated, as shown:

CH3

H weaker acid

H3C

CH3

stronger base

H H3C

3.9. Inspection of structures 1 and 3 shows that the nitrogen atoms, and more specifically their lone pairs, serve as bases in the protonation of nicotine. In order to determine which nitrogen atom in structure 1 is more basic, we can compare the pKa values for each of the acidic protons in structure 3, using similar acids found in Table 3.1 for reference:

H stronger acid

H3C

weaker base

Therefore, in the structure of compound 1, we expect that the lone pair on the nitrogen atom will be more basic than

3.10. (a) We begin by identifying the acid on each side of the equilibrium. In this case, the acid on the left side is ethanol (CH3CH2OH) and the acid on the right side is water (H2O). We compare their pKa values (Table 3.1), and we find that ethanol (pKa = 16.0) is less acidic than water (pKa = 15.7). The equilibrium will favor the weaker acid (ethanol), so the left side of the reaction is favored.

CHAPTER 3

(b) Identify the acid on each side of the equilibrium. In this case, the acid on the left side is phenol (C6H5OH) and the acid on the right side is water (H2O). We compare their pKa values (Table 3.1), and we find that water (pKa = 15.7) is less acidic than phenol (pKa = 9.9). The equilibrium will favor the weaker acid (water), so the right side of the reaction is favored. (c) Identify the acid on each side of the equilibrium. In this case, the acid on the left side is HCl and the acid on the right side is H3O+. We compare their pKa values (Table 3.1), and we find that H3O+ (pKa = -1.74) is less acidic than HCl (pKa = -7). The equilibrium will favor the weaker acid (H3O+), so the right side of the reaction is favored. (d) Identify the acid on each side of the equilibrium. In this case, the acid on the left side of the equilibrium is acetylene (H–C≡C–H), and the acid on the right side is ammonia (NH3). We compare their pKa values (Table 3.1), and we find that ammonia (pKa = 38) is less acidic than acetylene (pKa = 25). The equilibrium will favor the weaker acid (ammonia), so the right side of the reaction is favored. 3.11. One way to answer this question is to look up (or at least estimate) the pKa of the acid on each side of the equilibrium. Because phenolic protons (pKa ~ 10) are more acidic than water (pKa = 15.7), we can determine that the equilibrium should favor deprotonation of hydroquinone by hydroxide. The equilibrium will favor the weaker acid (water), so the forward reaction is favored and hydroxide is a sufficiently strong base. 3.12. At a pH of 7.4, acids with a pKa below 7.4 are expected to be mostly deprotonated, while acids with a pKa above 7.4 are expected to be mostly protonated. According to Table 3.1, the pKa of a carboxylic acid group (RCO2H) is expected to be near 4.75 (which is below 7.4), while the pKa of an ammonium group (RNH3+) is expected to be near 9.3 (which is above 7.4) Therefore, at a pH of 7.4, the carboxylic acid group will exist primarily as its conjugate base (RCO2ˉ), called a carboxylate ion, while the ammonium group (RNH3+) will retain its proton, and will primarily exist in the charged form, as shown here:

3.13. (a) Carbon and oxygen are in the same row of the periodic table, so we must compare their electronegativity values. Oxygen is more electronegative than carbon and can better stabilize the negative charge that will be generated upon deprotonation. A more stable (weaker) conjugate base indicates a stronger parent acid. Therefore, a proton connected to an oxygen atom is expected to be more acidic than a proton connected to a carbon atom:

(b) Carbon and nitrogen are in the same row of the periodic table, so we must compare their electronegativity values. Nitrogen is more electronegative than carbon and can better stabilize the negative charge that will be generated upon deprotonation. A more stable (weaker) conjugate base indicates a stronger parent acid. Therefore, a proton connected to a nitrogen atom is expected to be more acidic than a proton connected to a carbon atom:

(c) Sulfur and oxygen are in the same column of the periodic table, so we must compare their size. Sulfur is larger than oxygen and can better stabilize the negative charge that will be generated upon deprotonation. A more stable (weaker) conjugate base indicates a stronger parent acid. Therefore, a proton connected to a sulfur atom is expected to be more acidic than a proton connected to an oxygen atom:

(d) Nitrogen and oxygen are in the same row of the periodic table, so we must compare their electronegativity values. Oxygen is more electronegative than nitrogen and can better stabilize the negative charge that will be generated upon deprotonation. A more stable (weaker) conjugate base indicates a stronger parent acid. Therefore, a proton connected to an oxygen atom is expected to be more acidic than a proton connected to a nitrogen atom:

3.14. We start by drawing the two possible conjugate bases:

The first conjugate base has the negative charge on a nitrogen atom, while the second conjugate base has the negative charge on a sulfur atom. Nitrogen and sulfur are neither in the same row nor in the same column of the periodic table. However, nitrogen and oxygen are in the same row of the periodic table and sulfur and oxygen are in the same column of the periodic table, so it makes sense to use oxygen as a means of comparing nitrogen and sulfur. Oxygen is better able to stabilize a negative charge than nitrogen due to its greater electronegativity (the dominant factor when making comparisons in a row). Sulfur is better able to stabilize a negative charge than oxygen due to its larger size (the dominant factor when making comparisons in a column). So, we can deduce that

CHAPTER 3

sulfur must be more capable of stabilizing the negative charge than nitrogen:

A more stable (weaker) conjugate base indicates a stronger parent acid. Therefore, a proton connected to a sulfur atom is expected to be more acidic than a proton connected to a nitrogen atom:

A more stable (weaker) conjugate base indicates a stronger parent acid. Removal of the blue proton would result in a carbanion that is less stable, because it is not resonance stabilized.

3.15. (a) The proton marked in blue is expected to be more acidic than the proton marked in red, because removal of the blue proton leads to a resonance-stabilized anion, shown here:

(d) The proton marked in red is expected to be more acidic than the proton marked in blue, because removal of the red proton leads to a resonance-stabilized anion, shown here:

A more stable (weaker) conjugate base indicates a stronger parent acid. Removal of the red proton would result in a carbanion that is less stable, because it is not resonance stabilized.

A more stable (weaker) conjugate base indicates a stronger parent acid. Removal of the blue proton would result in a carbanion that is less stable, because it is not resonance stabilized

(b) The proton marked in blue is expected to be more acidic than the proton marked in red, because removal of the blue proton leads to a resonance-stabilized anion, shown here:

(e) The proton marked in blue is expected to be more acidic than the proton marked in red, because removal of the blue proton leads to a resonance-stabilized anion in which the negative charge is spread over two oxygen atoms.

A more stable (weaker) conjugate base indicates a stronger parent acid. Removal of the red proton would result in a nitrogen anion that is less stable, because it is not resonance stabilized. (c) The proton marked in red is expected to be more acidic than the proton marked in blue, because removal of the red proton leads to a resonance-stabilized anion, shown here:

A more stable (weaker) conjugate base indicates a stronger parent acid. Removal of the red proton also leads to a resonance-stabilized anion, but the negative charge is spread over an oxygen atom and a carbon atom, which is less stable than spreading the charge over two oxygen atoms. (f) The proton marked in red is expected to be more acidic than the proton marked in blue, because removal of the red proton leads to a resonance-stabilized anion in which the negative charge is spread over one oxygen atom and three carbon atoms.

CHAPTER 3

A more stable (weaker) conjugate base indicates a stronger parent acid. Deprotonation at the location of the blue proton also leads to a resonance-stabilized anion, but the negative charge is spread over four carbon atoms, which is less stable than spreading the charge over one oxygen atom and three carbon atoms.

Deprotonation at the location marked by the blue proton leads to a resonance-stabilized anion in which the negative charge is spread over two oxygen atoms:

3.16. The proton highlighted is the most acidic proton in the structure:

This proton is the most acidic, because deprotonation at that location generates a resonance-stabilized anion, in which the negative charge is spread over two oxygen atoms and one carbon atom:

By comparing pKa values from Table 3.1, we conclude that the carboxylic acid proton is more acidic (lower pKa value):

HO O

OH HO O HO O

HO O HO O

The stronger acid has a more stable (weaker) conjugate base. This suggests that a negative charge will be more stabilized when spread over two oxygen atoms, rather than being spread over one oxygen atom and three carbon atoms (oxygen is more electronegative than carbon). 3.18. (a) The highlighted proton is the most acidic:

A more stable (weaker) conjugate base indicates a stronger parent acid. Deprotonation of any of the other oxygen atoms would result in an oxyanion (or oxygen anion) that is less stable, because it is either not resonance stabilized (for the two OH groups that are not part of the ring), or less resonance stabilized (for the other OH group attached to the ring). 3.17. Deprotonation at the location marked by the red proton leads to a resonance-stabilized anion in which the negative charge is spread over one oxygen atom and three carbon atoms:

When this proton is removed, the resulting anionic conjugate base is stabilized by the electron-withdrawing effects of the nearby electronegative fluorine atoms. A more stable (weaker) conjugate base indicates a stronger parent acid. (b) The highlighted proton is more acidic:

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capable of stabilizing the conjugate base that is formed when the proton is removed.

When this proton is removed, the resulting anionic conjugate base is stabilized by the electron-withdrawing effects of the electronegative chlorine atoms, which are closer to this proton than the other acidic proton (far left). 3.19. (a) The compound with two chlorine atoms is more acidic:

The electron-withdrawing effects of the additional chlorine atom help stabilize the conjugate base that is formed when the proton is removed. A more stable (weaker) conjugate base indicates a stronger parent acid. (b) The more acidic compound is the one in which the bromine atom is closer to the acidic proton:

The electron-withdrawing effects of the bromine atom stabilize the conjugate base that is formed when the proton is removed. A more stable (weaker) conjugate base indicates a stronger parent acid.

(c) There are many acceptable answers to this question, since there are many constitutional isomers that lack the carboxylic acid functional group. One example is shown below. This compound is not a carboxylic acid, so its conjugate base is not resonance-stabilized:

Without resonance stabilization, the conjugate base of this isomer is a fairly strong base, indicating a weaker parent acid. 3.21. The most acidic proton is highlighted in each of the following compounds. For each of the first two compounds, deprotonation leads to a conjugate base in which the negative charge is associated with an sphybridized orbital (which is more stable than being associated with an sp2- or sp3-hybridized orbital). In the final compound, deprotonation leads to a conjugate base in which the negative charge is associated with an sp2hybridized orbital (which is more stable than being associated with an sp3-hybridized orbital).

3.20. (a) In the following compound, one of the chlorine atoms has been moved closer to the acidic proton of the carboxylic acid group, which further stabilizes the conjugate base that is formed when the proton is removed.

Note that in the final compound, there are two equivalent sp2-hybridized protons (removal of either one of them would produce the same conjugate base). (b) In the following compound, one of the chlorine atoms has been moved farther away from the acidic proton of the carboxylic acid group, and the distant chlorine atom is less

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3.22. n-Butyllithium (n-BuLi) has a negative charge on an sp3-hybridized carbon atom, so it is a very strong base and will remove the most acidic proton in the compound. All protons have been drawn below, and the most acidic one is highlighted. Removal of this proton gives a conjugate base (shown below) in which the negative charge is stabilized because it is associated with an sp-hybridized orbital.

All of the other protons are connected to sp2-hybridized carbon atoms. So if any of those protons had been removed instead, the resulting conjugate base would have been less stable because the negative charge would have been associated with an sp2-hybridized orbital (rather than an sp-hybridized orbital). Note that the negative charge in the product is more stable than the negative charge in n-BuLi, so the forward reaction is favored.

3.23. (a) The proton marked in red is expected to be more acidic than the proton marked in blue, because removal of the red proton leads to a resonance-stabilized anion in which the negative charge is spread over one oxygen atom and three carbon atoms:

Formation of a more stable (weaker) conjugate base indicates the more acidic proton has been removed. Removal of the proton marked in blue results in an anion that is not resonance stabilized. (b) The proton marked in red is expected to be more acidic than the proton marked in blue. Both conjugate bases are resonance-stabilized, but removal of the red proton leads to a resonance-stabilized anion in which the negative charge is spread over one carbon atom and two oxygen atoms, rather than just being spread over one carbon atom and one oxygen atom:

Formation of a more stable (weaker) conjugate base indicates the more acidic proton has been removed. (c) The proton marked in red is expected to be more acidic than the proton marked in blue. Both conjugate bases are resonance stabilized, but removal of the red proton leads to a resonance-stabilized anion in which the negative charge is spread over two oxygen atoms, rather than being spread over two nitrogen atoms:

Formation of a more stable (weaker) conjugate base indicates the more acidic proton has been removed. (d) The proton marked in blue is expected to be more acidic than the proton marked in red, because removal of the blue proton leads to a resonance-stabilized anion in which the negative charge is spread over a nitrogen atom and a carbon atom:

Formation of a more stable (weaker) conjugate base indicates the more acidic proton has been removed. Removal of the red proton would result in a less stable carbanion, because it is not resonance stabilized. (e) The proton marked in red is expected to be more acidic than the proton marked in blue, because removal of the red proton leads to a resonance-stabilized anion in which

CHAPTER 3 the negative charge is spread over two nitrogen atoms, rather than being spread over two carbon atoms:

Formation of a more stable (weaker) conjugate base indicates the more acidic proton has been removed. Removal of the blue proton would result in a less stable anion because the negative charge would be spread over two carbon atoms (rather than two nitrogen atoms).

Formation of a more stable (weaker) conjugate base indicates the more acidic proton has been removed. (i) The proton marked in blue is expected to be more acidic than the proton marked in red. Both conjugate bases are resonance-stabilized, but removal of the blue proton leads to a conjugate base in which the negative charge is spread over three oxygen atoms, rather than being spread over one nitrogen atom and one oxygen atom:

(f) The proton marked in blue is expected to be more acidic than the proton marked in red. Both conjugate bases are resonance-stabilized, but removal of the blue proton leads to a resonance-stabilized anion in which the negative charge is spread over two sulfur atoms, rather than two oxygen atoms:

Formation of a more stable (weaker) conjugate base indicates the more acidic proton has been removed.

A negative charge on sulfur is more stable than a negative charge on oxygen, because sulfur is larger, and therefore more polarizable than oxygen. Formation of a more stable (weaker) conjugate base indicates the more acidic proton has been removed. (g) The proton marked in red is expected to be more acidic than the proton marked in blue, because removal of the red proton leads to a more stable conjugate base in which the negative charge is associated with an sp hybridized orbital. This case represents an exception to the ARIO priority scheme: factor 4 (orbital) trumps factor 1 (atom), and this can be confirmed by looking at Table 3.1 (the pKa of acetylene is 25, and the pKa of a proton on nitrogen is ~38).

(h) The proton marked in blue is expected to be more acidic than the proton marked in red. Both conjugate bases are resonance-stabilized, but removal of the blue proton leads to a conjugate base in which the negative charge is spread over two oxygen atoms, rather than being spread over one oxygen atom and three carbon atoms:

3.24. (a) Bromine and chlorine are in the same column of the periodic table (group 7A), so we must compare their size. Bromine is larger than chlorine and can better stabilize the negative charge that will be generated upon deprotonation. A more stable (weaker) conjugate base indicates a stronger parent acid, and therefore, HBr is expected to be more acidic than HCl. (b) Sulfur and oxygen are in the same column of the periodic table (group 6A), so we must compare their size. Sulfur is larger than oxygen and can better stabilize the negative charge that will be generated upon deprotonation. A more stable (weaker) conjugate base indicates a stronger parent acid, and therefore, H2S is expected to be more acidic than H2O. (c) Carbon and nitrogen are in the same row of the periodic table, so we must compare their electronegativity values. Nitrogen is more electronegative than carbon and can better stabilize the negative charge that will be generated upon deprotonation. A more stable (weaker) conjugate base indicates a stronger parent acid, and therefore, NH3 is expected to be more acidic than CH4. (d) Acetylene (H–C≡C–H) is more acidic. The conjugate base of acetylene has a negative charge associated with a lone pair in an sp-hybridized orbital, which is more stable than a negative charge associated with a lone pair in an sp2-hybridized orbital. A more stable (weaker) conjugate base indicates a stronger parent acid.

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3.25. (a) When the proton is removed, the resulting conjugate base is highly resonance-stabilized because the negative charge is spread over four nitrogen atoms and seven oxygen atoms. In addition, the inductive effects of the trifluoromethyl groups (-CF3) further stabilize the negative charge. This extraordinarily stable conjugate base is consistent with a particularly strong parent acid. (b) There are certainly many, many acceptable answers to this problem. A more stable (weaker) conjugate base indicates a stronger parent acid, so the goal is to further stabilize the conjugate base. Below are two separate modifications that would render the compound even more acidic: The OH group can be replaced with an SH group. Sulfur is larger than oxygen and more capable of stabilizing a negative charge:

Alternatively, the conjugate base could be further stabilized by spreading the charge over an even larger number of nitrogen and oxygen atoms. For example, consider the structural changes, highlighted here:

These additional structural units would enable the conjugate base to spread its negative charge over six nitrogen atoms and nine oxygen atoms, which should be even more stable than being spread over four nitrogen atoms and seven oxygen atoms. 3.26. The most acidic proton belongs to the carboxylic acid group (COOH). Deprotonation of this functional group gives a resonance-stabilized anion in which the negative charge is spread over two oxygen atoms. Formation of a more stable (weaker) conjugate base indicates the more acidic proton has been removed. Removal of any of the other protons would lead to a less stable conjugate base.

3.27. (a) To assess the relative acidity of cationic acids, we compare the stability of the positive charges. We are comparing a positive charge on an oxygen atom with a positive charge on a nitrogen atom. Nitrogen is better at stabilizing a positive charge (compared with oxygen), because nitrogen is less electronegative than oxygen.

With O+ being less stable than N+, we conclude that the following acid is the stronger acid (less stable = more acidic):

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(b) To assess the relative acidity of cationic acids, we compare the stability of the positive charges:

In this case, we cannot directly assess the relative stability of the two acids because the positive charge is on the same element (a nitrogen atom) in each structure. Therefore, we must draw the conjugate bases and compare the availability of the lone pairs in the conjugate bases:

The first conjugate base above has a lone pair that is delocalized by resonance:

In the other conjugate base, the nitrogen atom is connected to an sp3-hybridized atom, so the lone pair is localized:

In the other conjugate base, the lone pair is localized:

As a result, the first conjugate base is a weaker base (its lone pair is less available to function as a base). Therefore, we conclude that the first acid is the stronger acid (because the stronger acid has a weaker conjugate base):

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(d) To assess the relative acidity of cationic acids, we compare the stability of the positive charges. We are comparing a positive charge on an oxygen atom with a positive charge on a nitrogen atom. Nitrogen is better at stabilizing a positive charge (compared with oxygen), because nitrogen is less electronegative than oxygen. With O+ being less stable than N+, we conclude that the following acid is the stronger acid (less stable = more acidic):

while the second base has a lone pair that is localized:

As a result, the first base is weaker base (its lone pair is less available to function as a base). Therefore, the following base is the stronger base:

3.28. (a) For each base, we draw the conjugate acid, and we then compare the stability of the conjugate acids: (c) The first base has a nitrogen atom with a lone pair that is delocalized by resonance,

The second conjugate acid above is more stable (a weaker acid) than the first conjugate acid, because nitrogen is better than oxygen at stabilizing a positive charge. Therefore, we conclude that the following base is the stronger base (because the stronger base produces the weaker conjugate acid). while the second base has a nitrogen atom with a lone pair that is localized:

As a result, the first base is weaker base (its lone pair is less available to function as a base). Therefore, the following base is the stronger base:

Indeed, this follows the general trend that amines (RNH2) are stronger bases than alcohols (ROH). (b) The first base has a lone pair that is delocalized by resonance,

(d) For each base, we draw the conjugate acid, and we then compare the stability of the conjugate acids:

The second conjugate acid above is more stable (a weaker acid) than the first conjugate acid, because the less electronegative nitrogen atom is better than oxygen at stabilizing a positive charge. Therefore, we conclude that the second base is the stronger base (because the stronger base produces the weaker conjugate acid).

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The charge in the first conjugate acid is localized (not stabilized by resonance), while the charge in the second conjugate acid is delocalized via resonance, as shown here:

3.29. There are three nitrogen atoms in guanidine, although two of them (highlighted below) are identical because of symmetry:

So we only need to compare the availability of the lone pairs in the following two locations: NH H

Because of resonance stabilization, this conjugate acid is more stable and a weaker acid, so it must correspond with the most strongly basic nitrogen atom:

One of these lone pairs is localized,

while the other lone pair is delocalized by resonance:

In summary, the same conclusion can be reached by comparing the availability of the lone pairs in the bases or by comparing the stability of the positive charges in the conjugate acids. 3.30. (a) We compare the stability of the ions on either side of the equilibrium:

The localized lone pair is more available to function as a base, so the following position is the most basic nitrogen atom in guanidine: NH H

The same conclusion can be reached if we compare the structures of the two possible conjugate acids:

The first ion (left side) has a negative charge on a carbon atom, while the second ion (right side) has a negative charge on a nitrogen atom, so we turn to factor #1 (atom). Carbon and nitrogen are in the same row of the periodic table, so we compare their electronegativity values. Nitrogen is more electronegative than carbon, so a negative charge will be more stable on a nitrogen atom. The base on the product side is the more stable (weaker) base. As such, the reaction favors the products.

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(b) We compare the stability of the ions on either side of the equilibrium:

The first ion (left side) has a negative charge on an oxygen atom, while the second ion (right side) has a negative charge on a nitrogen atom, so we turn to factor #1 (atom). Oxygen and nitrogen are in the same row of the periodic table, so we compare their electronegativity values. Oxygen is more electronegative than nitrogen, so a negative charge will be more stable on an oxygen atom. The base on the reactant side is the more stable (weaker) base. As such, the reaction favors the reactants, and the equilibrium lies to the left. (c) We begin by comparing the stability of the ions on either side of the equilibrium. Both ions have a positive charge on nitrogen, so we must look at their conjugate bases (shown in the reaction) and compare the availability of the lone pairs:

The second conjugate base above has a localized lone pair, while the first conjugate base above has a lone pair that is delocalized by resonance:

(d) We compare the stability of the ions on either side of the equilibrium:

The first ion (left side) has a negative charge on a carbon atom, while the second ion (right side) has a negative charge on an oxygen atom, so we turn to factor #1 (atom). Carbon and oxygen are in the same row of the periodic table, so we compare their electronegativity values. Oxygen is more electronegative than carbon, so a negative charge will be more stable on an oxygen atom. The base on the product side is the more stable (weaker) base. As such, the reaction favors the products, and the equilibrium lies to the right. (e) We compare the stability of the ions on either side of the equilibrium:

In this case, we are comparing a positive charge on a nitrogen atom with a positive charge on an oxygen atom. Nitrogen is better at stabilizing a positive charge, because nitrogen is less electronegative than oxygen. The acid on the reactant side is the more stable (weaker) acid. Therefore, the equilibrium favors the reactants, not the products, and the equilibrium lies to the left. (f) We compare the stability of the ions on either side of the equilibrium:

As a result, this conjugate base is a weaker base (its lone pair is less available to function as a base). The base on the reactant side is the more stable (weaker) base. Therefore, the equilibrium favors the reactants, not the products, and the equilibrium lies to the left.

Both ions have a negative charge on a carbon atom, and neither ion is stabilized by resonance or induction. The difference between these ions is the type of orbital that is accommodating the lone pair (and its associated negative charge). The first ion above has the negative charge associated with an sp3-hybridized orbital, while the second ion has the negative charge associated with an sphybridized orbital, and the latter is more stable. The base on the product side is the more stable (weaker) base. Therefore, the reaction favors the products, and the equilibrium lies to the right.

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3.31. (a) First identify the acid and the base in the conversion of 2 into 3. Anion 2 has its negative charge on an oxygen atom, and this oxygen atom functions as a base and accepts a proton during the conversion of 2 to 3. The nearby hydroxyl group loses its proton during the reaction, so it is functioning as the acid. To draw a mechanism, remember to use two curved arrows. The tail of the first curved arrow is placed on a lone pair from the base. The head of the arrow is placed on the proton of the acid. The tail of the second curved arrow is placed on the bond between the acidic proton and the attached oxygen atom. The head of that arrow is placed on the oxygen atom that receives the negative charge as a result of the process.

(b) To determine the position of the equilibrium, we compare the stability of the negative charge in 2 with the negative charge in 3 using the four factors (ARIO). 1.

Atom. In each case, the negative charge is on an oxygen atom. Thus, the atom effect is not relevant.

Resonance. The anionic oxygen in 2 has no resonance stabilization, because it is attached to an sp3-hybridized carbon atom. In contrast, anion 3 is highly stabilized by resonance, as shown here. Notice that the negative charge is spread over many locations, including two oxygen atoms.

Induction. The anionic oxygen in 2 is stabilized by the inductive effect of the nitrile nitrogen, the hydroxyl oxygen, and the ester oxygen atoms. However, each of these electronegative atoms are fairly far removed from the negative charge and thus probably play a minimal role in stabilizing 2. Anion 3 is also stabilized by the nitrile nitrogen and the other oxygen atoms in the molecule, but here the electronegative atoms are even more distant, thus the effects are even less than in 2.

Orbitals. Orbitals are not a relevant factor in this case.

Recall that factors 1-4 are listed in order of priority. Thus, based on factor 2 (resonance), we would expect that anion 3 should be significantly more stable than anion 2. The inductive effects that stabilize 2 are significantly less important than the resonance effects that stabilize 3. Therefore, the equilibrium should favor 3 over 2.

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3.32. (a) Yes, because a negative charge on an oxygen atom will be more stable than a negative charge on a nitrogen atom. This proton transfer reaction is favorable because it produces a more stable (weaker) base. (b) Yes, because a negative charge on a nitrogen atom will be more stable than a negative charge on an sp3hybridized carbon atom. This proton transfer reaction is favorable because it produces a more stable (weaker) base. (c) No, because this base is resonance-stabilized, with the negative charge spread over two oxygen atoms and one carbon atom. Protonating this base with water would result in the formation of a hydroxide ion, which is less stable because the negative charge is localized on one oxygen atom. This proton transfer reaction is NOT favorable because it would produce a less stable, stronger base.

A proton transfer reaction between this anion (as a base) and water (as an acid) is NOT favorable because it would produce a less stable, stronger base (hydroxide). 3.34. The CH3CH2 group in ethanol provides a small amount of steric bulk that is absent in water. As a result, hydroxide is better solvated and is therefore more stable. This improved solvation makes hydroxide a weaker base than ethoxide (CH3CH2Oˉ). As such, water is more acidic than ethanol. Indeed, the pKa of water (15.7) is lower than the pKa of ethanol (16.0). 3.35. (a) A lone pair of the oxygen atom attacks the aluminum atom. AlCl3 functions as the Lewis acid by accepting the electrons, and the ketone functions as the Lewis base by serving as an electron-pair donor.

(d) Yes, because a negative charge on an oxygen atom will be more stable than a negative charge on a carbon atom. This proton transfer reaction is favorable because it produces a more stable (weaker) base. 3.33. (a) No, water will not be a suitable proton source to protonate this anion, because the anion is resonancestabilized and is more stable than hydroxide:

(b) A lone pair of the oxygen atom (in the starting ketone) attacks a proton, as shown below. H3O+ functions as the Lewis acid by accepting the electrons, and the ketone functions as the Lewis base by serving as an electron-pair donor. O

Lewis base

A proton transfer reaction between this anion (as a base) and water (as an acid) is NOT favorable because it would produce a less stable, stronger base (hydroxide). (b) No, water will not be a suitable proton source to protonate this anion, because the anion is resonancestabilized (with the negative charge being spread over two oxygen atoms) and is therefore more stable than hydroxide:

H +

H H Lewis acid

H +

(c) A lone pair of a bromine atom in molecular bromine (Br2) attacks the aluminum atom. AlBr3 functions as the Lewis acid by accepting the electrons, and molecular bromine (Br2) functions as the Lewis base by serving as an electron-pair donor.

(d) A lone pair on one of the oxygen atoms of the ester attacks a proton on H3O+, as shown below. H3O+ functions as the Lewis acid by accepting the electrons, and the ester functions as the Lewis base by serving as an electron-pair donor.

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(e) A lone pair of the oxygen atom attacks the boron atom. BF3 functions as the Lewis acid by accepting the electrons, and the ketone functions as the Lewis base by serving as an electron-pair donor.

3.36. In compound 1, there are two oxygen atoms that bear lone pairs, and either location could certainly function as a Lewis base to donate an electron pair. But the problem statement indicates that the carbonyl (C=O) group interacts with AlCl3. Let’s explore why the carbonyl groups is the stronger Lewis base, by considering the structure of the complex that is obtained in each scenario. If the oxygen atom of the ether group (left) interacts with AlCl3, the following complex is obtained:

Notice that the positive charge in this complex is not resonance stabilized. In contrast, if the oxygen atom of the C=O bond interacts with AlCl3, then the following resonance-stabilized complex is obtained:

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As such, we expect the oxygen atom of the C=O group to interact with the AlCl3 to form the lower-energy, resonancestabilized complex, shown above. The major contributor is the one that exhibits filled octets:

All other resonance structures exhibit a C+, which lacks an octet of electrons. Those resonance structures are less significant, and minor contributors to the resonance hybrid. 3.37. (a) A conjugate base is produced when an acid loses a proton. The most acidic proton in the compound is the proton of the O-H group, because deprotonation at that location gives a conjugate base with a negative charge on the electronegative oxygen atom:

(e) A conjugate base is produced when an acid loses a proton. The most acidic proton in the compound is the one attached to oxygen. Deprotonation at that location gives a resonance-stabilized conjugate base, in which the charge is spread over two electronegative oxygen atoms. Formation of the most stable (weakest) conjugate base indicates the most acidic proton has been removed.

Formation of the most stable (weakest) conjugate base indicates the most acidic proton has been removed. Deprotonation at any other location would lead to a conjugate base with a negative charge on carbon (which is MUCH less stable). (b) A conjugate base is produced when an acid loses a proton. The most acidic protons are the four protons attached to the carbon atoms adjacent to the C=O group. Deprotonation at one of those locations gives a resonancestabilized conjugate base:

Deprotonation of the sp3-hybridized carbon atom would lead to a resonance-stabilized conjugate base with a negative charge shared between carbon and oxygen. This is significantly less stable than a negative charge distributed between two oxygen atoms. (f) A conjugate base is produced when an acid loses a proton. The most acidic proton in the compound is the proton connected to nitrogen, because deprotonation at that location gives a conjugate base with a negative charge on the electronegative nitrogen atom:

Formation of the most stable (weakest) conjugate base indicates the most acidic proton has been removed. Deprotonation at any other location would lead to a conjugate base that is not resonance-stabilized, and therefore much less stable. (c) A conjugate base is produced when an acid loses a proton. Deprotonation of NH3 gives the following conjugate base:

(d) A conjugate base is produced when an acid loses a proton. Deprotonation of H3O+ gives the following conjugate base:

Formation of the most stable (weakest) conjugate base indicates the most acidic proton has been removed. Deprotonation at any other location would lead to a conjugate base with a negative charge on carbon (which is less stable). (g) A conjugate base is produced when an acid loses a proton. Deprotonation of NH4+ gives the following conjugate base:

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3.38. (a) A conjugate acid is produced when a base accepts a proton. Protonation occurs at the site bearing the negative charge, giving the following compound:

(h) A conjugate acid is produced when a base accepts a proton. In this case, the base is HOˉ (the sodium cation is simply a counterion, also called a spectator ion, and does not participate). Protonation of HOˉ gives H2O, as show below:

(b) A conjugate acid is produced when a base accepts a proton. Protonation occurs at the site bearing the negative charge, giving the following ketone:

3.39. The difference in acidity between compounds A and B is 10 – 7 = 3 pKa units. Each pKa unit represents one order of magnitude (a power of 10), so compound A is 103, or 1000 times more acidic than compound B.

(c) A conjugate acid is produced when a base accepts a proton. In this case, the base is H2Nˉ (the sodium cation is simply a counterion, also called a spectator ion, and does not participate). Protonation of H2Nˉ gives NH3, as shown below:

3.40. (a) A lone pair of the oxygen atom attacks the carbocation (C+), as shown below. The carbocation functions as the Lewis acid by accepting the electrons, and ethanol (CH3CH2OH) functions as the Lewis base by serving as an electron-pair donor. H O OH

(d) A conjugate acid is produced when a base accepts a proton. Protonation of H2O gives H3O+, as shown below:

Lewis base

+ Lewis acid

(b) A lone pair of the oxygen atom attacks the boron atom. BF3 functions as the Lewis acid by accepting the electrons, and ethanol (CH3CH2OH) functions as the Lewis base by serving as an electron-pair donor.

(e) A conjugate acid is produced when a base accepts a proton. One of the lone pairs of the oxygen atom can serve as a base. Protonation gives the following oxonium ion (a cation in which the positive charge is located on an oxygen atom):

(f) A conjugate acid is produced when a base accepts a proton. The lone pair of the nitrogen atom can serve as a base. Protonation gives the following ammonium ion (a cation in which the positive charge is located on a nitrogen atom):

(g) A conjugate acid is produced when a base accepts a proton. One of the lone pairs of the oxygen atom can serve as a base. Protonation gives the following resonancestabilized cation:

(c) A lone pair of a chlorine atom attacks the aluminum atom. AlCl3 functions as the Lewis acid by accepting the electrons, and ethyl chloride (CH3CH2Cl) functions as the Lewis base by serving as an electron-pair donor.

3.41. The amide ion (H2Nˉ) is a strong base, and in the presence of H2O, a proton transfer reaction will occur, generating hydroxide, which is a more stable (weaker) base than the amide ion (Factor #1: oxygen is more electronegative than nitrogen). The reaction will greatly favor products (ammonia and hydroxide).

3.42. No, the reaction cannot be performed in the presence of ethanol, because the leveling effect would

CHAPTER 3

cause deprotonation of ethanol to form more-stable ethoxide ions (CH3CH2Oˉ) (Factor #1: oxygen is more electronegative than nitrogen and carbon). The desired anion would not be formed under these conditions.

are required. The first curved arrow shows a lone pair of the base attacking the proton, and the second curved arrow comes from the O–H bond (being broken) and goes to the oxygen atom, as shown:

3.43. No, water would not be a suitable proton source in this case. Note that the sodium cation (Na+) serves as a counterion. The anion is the conjugate base of a carboxylic acid, and the negative charge is resonance stabilized. A proton transfer reaction between this anion (as a base) and water (as an acid) is NOT favorable because it would produce a less stable, stronger base (hydroxide). Hydroxide is less stable (and thus more strongly basic) than the compound shown, because the negative charge in hydroxide is not resonance stabilized.

(c) Water functions as an acid in this case, by giving a proton to the strong base, as shown below. Two curved arrows are required. The first curved arrow shows a lone pair of the base attacking the proton, and the second curved arrow comes from the O–H bond (being broken) and goes to the oxygen atom, as shown:

3.44. (a) Water functions as a base and deprotonates HBr. Two curved arrows are required. The first curved arrow shows a lone pair of the base attacking the proton, and the second curved arrow comes from the H–Br bond (being broken) and goes to the bromine atom, as shown:

(d) Water functions as an acid in this case, by giving a proton to the strong base, as shown below. Two curved arrows are required. The first curved arrow shows a lone pair of the base attacking the proton, and the second curved arrow comes from the O–H bond (being broken) and goes to the oxygen atom, as shown:

(b) Water functions as a base and deprotonates sulfuric acid (H2SO4). Two curved arrows are required. The first curved arrow shows a lone pair of the base attacking the proton. The second curved arrow comes from the O–H bond (being broken) and goes to the oxygen atom, as shown:

(c) Water functions as a base and deprotonates this strong acid (See Table 3.1). Two curved arrows are required. The first curved arrow shows a lone pair of the base attacking the proton, and the second curved arrow comes from the O–H bond (being broken) and goes to the oxygen atom, as shown:

3.45. (a) Water functions as an acid in this case, by giving a proton to the strong base, as shown below. Two curved arrows are required. The first curved arrow shows a lone pair of the base attacking the proton, and the second curved arrow comes from the O–H bond (being broken) and goes to the oxygen atom, as shown:

(b) Water functions as an acid in this case, by giving a proton to the base, as shown below. Two curved arrows

3.46. (a) The second anion is more stable because it has an allylic lone pair, and the negative charge is delocalized by resonance.

(b) The second anion is more stable because the negative charge is on a more electronegative nitrogen atom (factor #1 of ARIO), rather than an sp3 hybridized carbon atom.

(c) The second anion is more stable because the negative charge is on an sp-hybridized carbon atom, rather than an sp3-hybridized carbon atom.

CHAPTER 3 3.47. (a) The second compound is more acidic, because its conjugate base (shown below) has a negative charge on a sulfur atom, which is more stable than a negative charge on an oxygen atom (factor #1 of ARIO), A more stable (weaker) conjugate base indicates a stronger parent acid.

(b) The first compound (called phenol) is more acidic, because its conjugate base (shown below) is resonance stabilized (factor #2 of ARIO). A more stable (weaker) conjugate base indicates a stronger parent acid.

(e) The first compound is more acidic because its conjugate base (shown below) is resonance stabilized.

A more stable (weaker) conjugate base indicates a stronger parent acid. The conjugate base of the second compound is not resonance stabilized. (f) The first compound is more acidic because its conjugate base (shown below) is resonance stabilized, and one of the resonance structures has the negative charge on an electronegative oxygen atom.

(c) The conjugate base for each of these compounds is resonance-stabilized (as in part b of this problem). The difference between these compounds is the presence of electron-withdrawing chlorine atoms. The first compound is more acidic as a result of the combined inductive effects of the five chlorine atoms which stabilize the conjugate base shown below (factor #3 of ARIO). A more stable (weaker) conjugate base indicates a stronger parent acid.

A more stable (weaker) conjugate base indicates a stronger parent acid. The conjugate base of the second compound is not resonance stabilized (the negative charge would be localized on a carbon atom). (d) The second compound is more acidic, because its conjugate base (shown below) has a negative charge associated with an sp-hybridized orbital, while the first compound is less acidic because its conjugate base has a negative charge on a nitrogen atom. We learned that this example constitutes an exception to the order of priorities, ARIO. In this case, factor #4 (orbital) trumps factor #1 (atom). This exception applies whenever we compare a negative charge on an sp-hybridized carbon atom and a negative charge on an sp3-hybridized nitrogen atom.

(h) The second compound is more acidic because its conjugate base is resonance stabilized, with the negative charge being spread over two oxygen atoms (shown below). A more stable (weaker) conjugate base indicates a stronger parent acid.

The conjugate base of the first compound is also resonance stabilized, but the negative charge would be

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spread over an oxygen atom and a nitrogen atom, which is less stable than being spread over two oxygen atoms (because oxygen is more electronegative than nitrogen, as described in factor #1 of ARIO). 3.48. NaA represents an ionic compound, comprised of cations (Na+) and anions (Aˉ). When H–B is treated with Aˉ, a proton can be transferred from H–B to Aˉ, as shown in the following equilibrium:

The equilibrium will favor the weaker acid (the acid with the higher pKa value). In this case, the equilibrium favors formation of HA.

3.49. (a) Water (H2O) loses a proton and is therefore functioning as an acid. Two curved arrows must be drawn. The first curved arrow shows a lone pair of the base attacking the proton of water, and the second curved arrow comes from the O–H bond (being broken) and goes to the oxygen atom of water, as shown. To determine the position of the equilibrium, we compare the stability of the ions on either side of the equilibrium (the competing bases, in this case). Because of better solvation, hydroxide is the more stable anion. The base on the product side is the more stable (weaker) base. As such, the reaction favors the products, and the equilibrium lies to the right. Alternatively, we could compare the pKa values of the acids on either side of the equilibrium [H2O, 15.7 and (CH3)2CHOH, ~16] and we would arrive at the same conclusion. The acid on the product side is the more stable (weaker) acid. As such, the reaction favors the products, and the equilibrium lies to the right.

(b) Two curved arrows must be drawn. The first curved arrow shows a lone pair of the base attacking the proton, and the second curved arrow comes from the S–H bond (being broken) and goes to the sulfur atom, as shown below. To determine the position of the equilibrium, we compare the stability of the ions on either side of the equilibrium (the competing bases, in this case). A negative charge on a larger sulfur atom is expected to be more stable than the negative charge on an oxygen atom (factor #1 of ARIO). The base on the product side is the more stable (weaker) base. As such, the reaction favors the products, and the equilibrium lies to the right. Alternatively, we could compare the pKa values of the acids on either side of the equilibrium and we would arrive at the same conclusion. The acid on the product side is the more stable (weaker) acid. As such, the reaction favors the products, and the equilibrium lies to the right.

(c) Two curved arrows must be drawn. The first curved arrow shows a lone pair of the base (HSˉ in this case) attacking the proton, and the second curved arrow comes from the S–H bond (being broken) and goes to the sulfur atom, as shown. To determine the position of the equilibrium, we compare the stability of the ions on either side of the equilibrium (the competing bases, in this case). The base on the right side of the equilibrium is resonance stabilized, with the negative charge being spread over two sulfur atoms. The base on the left side of the equilibrium is not resonance stabilized, and the negative charge is localized on one sulfur atom. The equilibrium favors the more stable (weaker) base, and the product side is favored.

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(d) Two curved arrows must be drawn. The first curved arrow shows a lone pair of the base (the nitrogen atom) attacking the proton, and the second curved arrow comes from the O–H bond (being broken) and goes to the oxygen atom, as shown. The equilibrium favors the product, which can be determined by comparing the stability of the ions on either side of the equilibrium (the competing bases, in this case). Factor #1 of ARIO indicates that the product is favored in this case, because a negative charge is more stable on an oxygen atom than a nitrogen atom. The equilibrium favors the more stable (weaker) base, and the product side is favored.

3.50. One of the anions is resonance stabilized, with the negative charge spread over two oxygen atoms. That anion is the weakest (most stable) base. Among the remaining three anions, they do not differ from each other in any of the four factors (ARIO), but they are expected to differ from each other in terms of solvent effects. That is, an anion will be less stable (stronger base) if it has steric bulk in close proximity with the negative charge. The steric bulk reduces the stability of the anion by limiting its ability to interact with solvent molecules, as described in Section 3.8.

3.51. (a) The proton highlighted below is expected to be the most acidic, because deprotonation leads to a conjugate base in which the negative charge is associated with an sphybridized orbital. This case represents an exception to the ARIO priority scheme, because factor #4 (orbital) trumps factor #1 (atom). This exception applies whenever we compare a negative charge on an sp-hybridized carbon atom and a negative charge on an sp3-hybridized nitrogen atom.

(b) The proton highlighted below is expected to be the most acidic, because deprotonation leads to a conjugate base in which the negative charge is on a sulfur atom (more stable than being on an oxygen, nitrogen, or carbon atom, according to factor #1 of ARIO). Formation of the

most stable (weakest) conjugate base indicates that the most acidic proton has been removed.

(c) The proton highlighted below is expected to be the most acidic, because deprotonation leads to a resonancestabilized conjugate base. Formation of the most stable (weakest) conjugate base indicates that the most acidic proton has been removed. Removal of any of the other protons would lead to an anion that is not resonance stabilized.

(d) The proton highlighted below is expected to be the most acidic, because deprotonation leads to a resonancestabilized conjugate base in which the negative charge is spread over two carbon atoms and one oxygen atom. Formation of the most stable (weakest) conjugate base indicates that the most acidic proton has been removed. Removal of any of the other protons would lead to an anion that is either not resonance stabilized or less resonance stabilized.

(e) The proton highlighted below is expected to be the most acidic, because deprotonation leads to a resonance-

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stabilized conjugate base in which the negative charge is spread over two oxygen atoms (factor #2 of ARIO). Formation of the most stable (weakest) conjugate base indicates that the most acidic proton has been removed. Removal of any other proton would lead to a less stable anion.

(f) The proton highlighted below is expected to be the most acidic, because deprotonation leads to a resonancestabilized conjugate base in which the negative charge is spread over one carbon atom and one oxygen atom (factor #2 of ARIO). Formation of the most stable (weakest) conjugate base indicates that the most acidic proton has been removed. Deprotonation of either of the other carbon atoms would lead to an anion that is not resonance stabilized, and therefore less stable.

acidity of cationic acids, we compare the stability of the positive charges. We are comparing a positive charge on an oxygen atom with a positive charge on a nitrogen atom. As we have seen, nitrogen is better at stabilizing a positive charge (compared with oxygen), because nitrogen is less electronegative than oxygen. With O+ being less stable than N+, we conclude that ROH2+ is a stronger acid than RNH3+. So the most acidic proton in this structure is connected to the positively charged oxygen atom.

3.53. (a) Ammonia is protonated by HCl to give an ammonium ion, as shown:

(b) There are two nitrogen atoms in DBN. One of the lone pairs is delocalized via resonance:

(g) There are three carboxylic acid (CO2H) groups, each of which bears an acidic proton. Removing any one of these protons will result in a resonance-stabilized conjugate base. Among these three protons, the highlighted proton is the most acidic because of the electron-withdrawing effects of the nearby chlorine atoms. When this proton is removed, the conjugate base is stabilized not only by resonance, but also by induction (factor #3 of ARIO). Formation of the most stable (weakest) conjugate base indicates that the most acidic proton has been removed.

while the other lone pair is localized:

The localized lone pair is more available to function as a base, and protonation at that location leads to the following resonance-stabilized cation: (h) The highlighted proton is expected to be the most acidic, because deprotonation leads to a conjugate base in which the negative charge is on a large sulfur atom, which is more stable than being on an oxygen or carbon atom (factor #1 of ARIO). Formation of the most stable (weakest) conjugate base indicates that the most acidic proton has been removed.

3.52. In this case, we are assessing the relative acidity of two cationic acids: ROH2+ and RNH3+. To compare the

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Protonation of the other nitrogen atom would lead to a conjugate acid that is not resonance stabilized. (c) When we compare the two conjugate acids to each other, we find that the conjugate acid of DBN is more stable (because the charge is delocalized by resonance). A more stable (weaker) conjugate acid indicates a stronger parent base, and therefore, DBN is expected to be a stronger base than NH3.

3.54. In each case, the lone pair on the nitrogen atom is delocalized via resonance throughout the ring.

But compound 2 has additional delocalization of electrons that compound 1 lacks, because of the additional resonance structure shown below:

Therefore, the lone pair on the nitrogen atom in compound 2 is less available to function as a base, rendering compound 2 a weaker base than compound 1. 3.55. The correct answer is (c). We begin by drawing methyl magnesium bromide (CH3MgBr):

The difference in electronegativity between C and Mg is so great, that the C-Mg bond is sometimes drawn as ionic:

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A carbanion (negative charge on C) is highly unstable. That is, the carbanion is a very strong base. However, as a result of the leveling effect, a base stronger than hydroxide is not possible in the presence of water. When treated with water, the carbanion will be irreversibly protonated, giving the more stable (weaker) hydroxide base (highlighted below).

3.56. The correct answer is (a). We compare the ions on either side of the equilibrium (the competing bases, in this case). On the left side, the base is an amide ion (H2Nˉ), which has a negative charge localized on a nitrogen atom. On the right side, the base is a resonance-stabilized anion, with the negative charge being delocalized:

A negative charge is more stable when it is delocalized, and in the case, the charge is spread over two oxygen atoms. The base on the product side is the more stable (weaker) base. As such, the reaction favors the products, and the equilibrium lies to the right. 3.57. The correct answer is (c). A more stable (weaker) conjugate base indicates a stronger parent acid. The conjugate base of (a) is HSˉ, which is more stable than HOˉ, because of factor #1 (the negative charge is more stable on a larger atom). Therefore, H2S is expected to be more acidic than H2O. The conjugate base of (b) is resonance stabilized, with the negative charge being spread over two oxygen atoms:

This resonance-stabilized anion is more stable (and a weaker base) than hydroxide. Therefore, (b) is more acidic than water. Answer (d) is also more acidic than water, because its conjugate base is resonance-stabilized, with the negative charge delocalized over two oxygen atoms:

As a result of solvent effects, (c) tert-butanol is less acidic than water. 3.58. The correct answer is (d). Each of the first three structures has at least one lone pair and can therefore function as a Lewis base:

The final structure (CH4) does not possess a lone pair, and therefore cannot function as a Lewis base.

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3.59. The correct answer is (b). The highlighted proton in compound I (below) is the most acidic proton. When this proton is removed, the resulting conjugate base is stabilized by the electron-withdrawing effects of the electronegative fluorine atom. The conjugate base of acid III is also stabilized by an electron-withdrawing group, but chlorine is less electronegative than fluorine, so the effects are smaller. The least stable conjugate base comes from deprotonating acid II. In the conjugate base of acid II, the chlorine atom is farther away from the negative charge, so the electron-withdrawing effects are the smallest. The least stable conjugate base (the strongest conjugate base) has the weakest parent acid. When arranged from the weakest acid to the strongest acid, the answer is: II < III < I.

3.62. The correct answer is (c). Removing a proton from compound X results in an anion that is not resonancestabilized, while removing a proton from compound Y results in an anion that is resonance-stabilized. The more stable conjugate base (the weaker conjugate base) has the stronger parent acid, and therefore compound Y is the stronger acid. O OH

H+

OH H+

3.60. The correct answer is (d). In this example, we are comparing the stability of the conjugate bases, Cl− and Br−, which appear in the same column of the periodic table. In such a case, electronegativity is not the dominant effect. Instead, the dominant effect is size. Bromine is larger than chlorine and can therefore better stabilize a negative charge by spreading the charge over a larger volume of space. As such, Br− is the more stable (weaker) conjugate base, and it has the stronger parent acid. Therefore, HBr is a stronger acid than HCl. 3.61. The correct answer is (d). Proton D (highlighted below) is the most acidic proton. Deprotonation at this position leads to a conjugate base in which the negative charge is associated with an sp-hybridized orbital (which is more stable than being associated with an sp2- or sp3hybridized orbital). Formation of the most stable (weakest) conjugate base indicates that the most acidic proton has been removed.

conjugate base of X stronger base

O O

conjugate base of Y weaker base

3.63. The correct answer is (a). Carbon, nitrogen and oxygen are in the same row of the periodic table, so we must compare their electronegativity values. Oxygen is the most electronegative and can best stabilize the negative charge. Carbon is the least electronegative and is the least effective at stabilizing the negative charge. Nitrogen has an electronegativity value between that of carbon and oxygen, so it has an intermediate ability to stabilize the negative charge. As such, the carbanion is the least stable, strongest base.

3.64. (a) Acetic acid (CH3CO2H) loses a proton and is therefore functioning as an acid. Hydroxide (HOˉ) functions as the base that removes the proton. Two curved arrows must be drawn. The first curved arrow shows a lone pair of the base attacking the acidic proton, and the second curved arrow comes from the O–H bond (being broken) and goes to the oxygen atom, as shown. The equilibrium favors the products, because the base on the right side of the equilibrium is resonance stabilized and therefore more stable than a hydroxide ion. The equilibrium favors the more stable (weaker) base. Alternatively, we could compare the pKa values of the acids on either side of the equilibrium and we would arrive at the same conclusion (the equilibrium will favor the products in this case because the equilibrium favors the weaker acid).

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(b) Water (H2O) loses a proton and is therefore functioning as an acid. The carbanion (an anion in which the negative charge is on a carbon atom) functions as the base that removes the proton. Two curved arrows must be drawn. The first curved arrow shows a lone pair of the base attacking the proton, and the second curved arrow comes from the O–H bond (being broken) and goes to the oxygen atom, as shown. The reaction favors the products, because hydroxide is more stable than a carbanion (factor #1 of ARIO). Alternatively, we could compare the pKa values of the acids on either side of the equilibrium and we would arrive at the same conclusion (the reaction favors the products). In fact, the pKa values of H2O (15.7) and C4H10 (~50) are so vastly different that the reaction is essentially irreversible.

(c) Hydroxide (HOˉ) functions as a base and removes a proton from the acid. Two curved arrows must be drawn. The first curved arrow shows a lone pair of the base attacking the proton, and the second curved arrow comes from the C–H bond (being broken) and goes to the carbon atom, as shown.

Notice that the conjugate base is resonance stabilized, so its formation can alternatively be shown with more than two curved arrows (leading directly to the resonance structure that contributes the most character to the overall resonance hybrid).

The equilibrium favors the products, because the base on the right side of the equilibrium is resonance stabilized, with the negative charge spread over two oxygen atoms. This is more stable than the negative charge being localized on one oxygen atom, as in hydroxide. The equilibrium favors the more stable (weaker) base. Alternatively, we could compare the pKa values of the acids on either side of the equilibrium and we would arrive at the same conclusion. That is, the equilibrium will favor the products in this case because the equilibrium favors formation of the weaker acid (H2O).

3.65. Each of the carbon atoms is tetravalent; the sulfur atom is divalent; and each of the hydrogen atoms is monovalent. We begin by connecting the atoms that have more than one bond (in this case, the two carbon atoms and the sulfur atom). There are only two different ways that these three atoms can be connected to each other, shown below:

The second isomer is more acidic because deprotonation of that isomer gives a conjugate base with a negative charge on a sulfur atom. A more stable (weaker) conjugate base indicates a stronger parent acid. The first compound above is less acidic, because its conjugate base would have a negative charge on a carbon atom, which is much less stable (factor #1 of ARIO). 3.66. There are three constitutional isomers with the molecular formula C3H8O, shown here:

For each of these arrangements, we connect the hydrogen atoms, giving the following two constitutional isomers:

One of these compounds lacks an O-H group, so that compound will be the least acidic (its conjugate base will

CHAPTER 3 have an unstable negative charge on a carbon atom). Of the two remaining compounds, the compound with the least branching will be the most acidic, because its conjugate base is the most stable (due to better solvating effects, discussed in Section 3.8). A more stable (weaker) conjugate base indicates a stronger parent acid.

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(f) There are five hydrogen atoms in the conjugate base.

(g) As seen in the resonance structures (see the solution to part c of this problem), there is one lone pair in the conjugate base, and it is highly delocalized. 3.68. We begin by drawing the conjugate base of each compound and comparing them: 3.67. (a) A carbon atom must have four sigma bonds in order to be sp3 hybridized. There is only one such carbon atom in cyclopentadiene, highlighted below.

(b) The most acidic proton in cyclopentadiene is highlighted below:

The corresponding conjugate base is highly resonance stabilized, as shown in the solution to part (c), below. In addition, the conjugate base is further stabilized by yet another factor that we will discuss in Chapter 18. Formation of the most stable (weakest) conjugate base indicates that the most acidic proton has been removed.

The first conjugate base is stabilized by resonance, while the second conjugate base is not. However, the second conjugate base exhibits a negative charge on a sulfur atom, which is larger than an oxygen atom. Therefore, there is a competition between two factors. Using the ARIO order of priority that is generally applied (“atom” is more important than “resonance”), we would expect that the second conjugate base should be more stable than the first. Yet, when we compare the pKa values, we find that our prediction is not correct. Therefore, this is an exception, in which “resonance” is more important than “atom.” 3.69. In an intramolecular acid-base reaction, a proton is transferred from one region of the molecule to another region within the same molecule, because the acid and base are tethered together, as in this case.

The equilibrium favors the product shown, because the negative charge in the product is resonance-stabilized (factor 2), while the negative charge on the reactant is not. The reaction will therefore favor the formation of a weaker base from a stronger one.

(d) There are no sp3-hybridized carbon atoms in the conjugate base. All five carbon atoms are sp2 hybridized. (e) In order to be delocalized via resonance, the lone pair must occupy a p orbital, and therefore, all carbon atoms must be sp2 hybridized and trigonal planar. Therefore, the entire compound has planar geometry.

3.70. The stronger acid has a more stable, (weaker) conjugate base. Two possible explanations for the increased acidity of salicylic acid can be given: 1) In each compound, the carboxylic acid proton (CO2H) is the most acidic proton. In salicylic acid, the inductive effect (electron-withdrawal) of the OH group is expected to be more pronounced because of its proximity to the negative charge in the conjugate base. 2) When salicylic acid is deprotonated, the resulting conjugate base can be significantly stabilized by