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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
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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.
1s
Nitrogen SkillBuilder 1.7 Identifying Hybridization States
2s
2p
3
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
O
C
C
CH3 H3C
Hydrogen-Bonding Interactions Circle the compound below that is expected to have the higher boiling point.
H CH3
H
C H
H O
C
H H
H
H
H
C
C
H
H
O
H
Carbon Skeleton Circle the compound below that is expected to have the higher boiling point.
H
H
H
H
C
C
C
H
H
H
H
H
H
H
H
H
H
C
C
C
C
C
H
H
H
H
H
H
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.
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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
O
C
C
C
C
C
C
C
C
1
2
3
4
1
2
3
4
O
C
C
C
C
C
2
3
4
1
2
O
C
C
3
4
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
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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).
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(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.
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(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:
O
N
N
N NH
N
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
N
H
C
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.
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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.
19
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.
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(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 CCl bond in chloroform partially cancels the dipole moments of the other two CCl 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 CCl 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.
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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 CCl 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
C
C
O
H
H
23
three bonds and one bond). There are certainly many other possible compounds for which all of the carbon atoms are sp2 hybridized.
H
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.
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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.
25
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
26
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
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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
O
H
C
NH2
NH2
CH3
C
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.
NH
H
C
NH
O
O CH3
C
CH2
CH3
C
CH2
H NH2
CH3 H3C
O
C
H H
C
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.
O
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.
O
SkillBuilder 2.9 Identifying Localized and Delocalized Lone Pairs
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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:
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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
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
C
C
H
H
O
H
H H
(c) 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, 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:
31
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:
Cl
Cl
O
C
C
O
H
Cl
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
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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:
(c) We begin by drawing the carbon chain and any atoms attached to the carbon chain:
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
O
H
C
C
C
H
H
H H H
H C C
C
H O
H
H
H
H
H
C
C
C
H
H
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
CHAPTER 2
33
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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CHAPTER 2
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
O
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
O
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.
36
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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37
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:
39
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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CHAPTER 2
(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.
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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
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(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)
O
O
Equivalent
CHAPTER 2
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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:
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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.
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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
O
HO
C C
H
H
C
N
C
C C
H C
H C
H
O H
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:
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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
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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
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(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:
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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
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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:
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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
O
O
O
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:
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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.
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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 15. The answers appear in the section entitled SkillBuilder Review. 15.1 Determining the Relationship between Two Protons in a Compound
15.2 Identifying the Number of Expected Signals in a 1H NMR Spectrum
15.3 Predicting Chemical Shifts
15.4 Determining the Number of Protons Giving Rise to a Signal Step 1 Compare the relative _____________ values, and choose the lowest number.
Step 2 Divide all integration values by the number from step 1, which gives the ratio of __________.
Step 3 Identify the number of protons in the compound (from the molecular formula) and then adjust the relative integration values so that the sum total equals the number of __________________.
15.5 Predicting the Multiplicity of a Signal
15.6 Drawing the Expected 1H NMR Spectrum of a Compound Step 1 Identify the number of __________.
Step 2 Predict the __________ ________ of each signal.
Step 3 Determine the __________ of each signal by counting the number of __________ giving rise to each signal.
Step 4 Predict the __________ of each signal.
Step 5 Draw each signal.
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15.7 Using 1H NMR Spectroscopy to Distinguish Between Compounds
15.8 Analyzing a 1H NMR Spectrum and Proposing the Structure of a Compound
15.9 Predicting the Number of Signals and Approximate Location of Each Signal in a 13C NMR Spectrum
15.10 Determining Molecular Structure using DEPT 13C NMR Spectroscopy
Mistakes to Avoid In 1H NMR spectroscopy, two neighboring CH2 groups will only split each other if they are not chemically equivalent. For example compare the structures of butane and pentane, and in particular, compare the CH2 group at the C2 position in each structure, highlighted below:
In the case of butane, the signal for the highlighted CH2 group is expected to be a quartet as a result of the neighboring methyl group. Notice that the signal is not further split by the neighboring CH2 group (C3), because the CH2 groups at C2 and C3 are chemically equivalent and therefore, they don’t split each other. In contrast, the signal for the
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highlighted CH2 group does experience splitting. In the case of pentane, the C2 and C3 positions are not chemically equivalent. They occupy different electronic environments, and they are not interchangeable by symmetry. As a result, the signal for the highlighted CH2 group (C2) is expected to be split into a sextet because it has 5 neighbors (n + 1 = 6). When analyzing a 1H NMR spectrum, make sure to take this into account, as students will often misinterpret a spectrum as a result of failing to take this into account.
Solutions 15.1. (a) When looking for symmetry, don’t be confused by the position of the double bonds in the aromatic ring. Recall that we can draw the following two resonance structures:
(b) These protons cannot be interchanged by rotational symmetry, so they are not homotopic. They can be interchanged by reflectional symmetry (the plane of symmetry is the plane of the page). Therefore, the protons are enantiotopic. Neither resonance structure is more correct than the other. For purposes of looking for symmetry, it will be less confusing to draw the compound like this:
When drawn in this way, we can see that the two highlighted protons can be interchanged by rotational symmetry (the axis of symmetry is shown below). Therefore, the protons are homotopic.
This conclusion can be verified by the replacement test. Specifically, each proton is replaced with deuterium, and the resulting compounds are found to be enantiomers. Therefore, the protons are enantiotopic:
(c) These protons cannot be interchanged by rotational symmetry, so they are not homotopic. They also cannot be interchanged by reflectional symmetry so they are not enantiotopic either. To determine if they are diastereotopic, we use the replacement test. Specifically, each proton is replaced with deuterium, and the resulting compounds are found to be diastereomers. Therefore, the protons are diastereotopic:
This conclusion can be verified by the replacement test. Specifically, each proton is replaced with deuterium, and the resulting compounds are found to be the same. Therefore, the protons are homotopic:
CHAPTER 15 (d) These protons cannot be interchanged by rotational symmetry, so they are not homotopic. They can be interchanged by reflectional symmetry (the plane of symmetry is the plane of the page). Therefore, the protons are enantiotopic.
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(c) Pentane has three different kinds of protons, shown here:
This conclusion can be verified by the replacement test. Specifically, each proton is replaced with deuterium, and the resulting compounds are found to be enantiomers. Therefore, the protons are enantiotopic: (d) Hexane has three different kinds of protons, shown here:
(e) These protons can be interchanged by rotational symmetry (the axis of symmetry is shown below). Therefore, the protons are homotopic.
This conclusion can be verified by the replacement test. Specifically, each proton is replaced with deuterium, and the resulting compounds are found to be the same. Therefore, the protons are homotopic:
15.2. (a) All four protons shown in red can be interchanged either via rotation or reflection, so they are all chemically equivalent. (b) The three protons of a methyl group are always equivalent (as we will soon see, immediately after the SkillBuilder), and in this case, the two methyl groups are equivalent to each other because they can be interchanged by rotation. Therefore, all six protons shown in blue are equivalent.
(e) The presence of a chlorine atom creates six different environments (in terms of proximity to the Cl), so there are six different kinds of protons, highlighted here:
15.3. The compound must have a high degree of symmetry in order to have only one kind of proton. The molecular formula indicates that there are twelve protons. The equivalence of twelve protons can be achieved by having four methyl groups in identical environments. Four methyl groups account for four of the five carbon atoms in the compound. So, we can draw a structure in which the fifth carbon atom is connected to each of the methyl groups (shown below), providing the necessary symmetry:
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15.4. In BZP, the two protons on each CH2 group are equivalent, because they can be interchanged by reflectional symmetry (the plane of symmetry is the plane of the page). In addition, a second plane of symmetry passing through both nitrogen atoms in the ring makes the CH2 groups on one half of the molecule equivalent to the corresponding CH2 groups on the other half of the molecule. Overall, BZP has two signals for the piperazine CH2 protons, as shown below:
(b) This compound has four different kinds of protons (highlighted below), giving rise to four signals.
DBZP has an additional plane of symmetry (slicing top to bottom in the structure shown), so all eight CH2 protons are equivalent, and only one signal is expected for these eight protons.
H H3CO
H3CO H
The three protons of a methyl group are always chemically equivalent
These two protons can be interchanged by rotational symmetry, so they are chemically equivalent
15.5. (a) This compound has eight different kinds of protons (highlighted below), giving rise to eight signals.
(c) This compound has two different kinds of protons (highlighted below), giving rise to two signals.
(d) This compound has three different kinds of protons (highlighted below), giving rise to three signals.
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(i) Due to the location of the two bromine atoms, each CH2 group occupies a unique electron environment, giving rise to two separate signals (one for each CH2 group). In addition, the two methyl groups are also in different electronic environments, giving two separate signals. In total, we expect four signals.
(e) This compound has five different kinds of protons (highlighted below), giving rise to five signals.
(j) Each of the protons in each CH2 group is in a unique electronic environment, as a result of the presence of a chiral center. That is, each CH2 group gives rise to two separate signals, so the two CH2 groups collectively give rise to four different signals. The two methyl groups are also different from each other (because of their unequal proximity to the bromine atom), giving two more signals. In addition, there is one signal from the proton attached to the carbon bearing the bromine atom. In total, we expect seven signals. (k) Heptane has four different kinds of protons, shown here: H
H
These two protons can be interchanged by rotational symmetry, so they are chemically equivalent
(f) This compound has three different kinds of protons (highlighted below), giving rise to three signals.
H H
H H
(g) This compound has four protons and none of them can be interchanged by rotational or reflectional symmetry. Each of the four protons occupies a unique electronic environment, giving rise to four signals. (h) This compound has two different kinds of protons (highlighted below), giving rise to two signals.
H
H
H
All four of these protons can be interchanged by either rotational or reflectional symmetry, so they are chemically equivalent
H 3C
H
All four of these protons can be interchanged by either rotational or reflectional symmetry, so they are chemically equivalent
CH3
These six protons can be interchanged by either rotational or reflectional symmetry, so they are chemically equivalent
(l) Each of the three vinylic protons occupies a unique electronic environment, giving rise to three separate signals:
The two vinylic protons at the very end are different from each other because one is cis to the main chain and the other is trans to the main chain, as shown. Each of the CH2 groups provides one signal (because each CH2 group occupies a unique electronic environment), and the
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CH3 provides one more signal, giving a total of seven signals. 15.6. The presence of the bromine atom does not render C3 a chiral center because there are two ethyl groups connected to C3. Nevertheless, the presence of the bromine atom does prevent the two protons at C2 from being interchangeable by reflection. The replacement test gives a pair of diastereomers, so the protons are diastereotopic (which means that they are not chemically equivalent).
In addition, the following highlighted protons are also diastereotopic:
Furthermore, there are four more signals arising from the two methyl groups and the two CH groups, as shown: 15.7. Each of the protons in the following highlighted CH2 groups is in a unique electronic environment, as a result of the presence of the chiral center. The protons in each CH2 group are diastereotopic and not chemically equivalent. That is, each CH2 group will give rise to two separate signals, because one H is on the same face as (cis to) the propenyl substituent, while the other H is further away from it, on the opposite face as (trans to) the propenyl substituent. Therefore, these two CH2 groups collectively give rise to four different signals:
Therefore, we have seen that (S)-carvone has a total of ten different kinds of protons, giving rise to ten signals.
15.8. (a) The 1H NMR spectrum of this compound is expected to exhibit five signals. The calculation for the estimated chemical shift of each signal is shown here:
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(b) The 1H NMR spectrum of this compound is expected to exhibit three signals. The calculation for the estimated chemical shift of each signal is shown here: methylene protons (CH2) = 1.2 ppm alpha to oxygen = + 2.5 ppm alpha to carbonyl = + 1.0 ppm 4.7 ppm
methyl protons (CH3) = 0.9 ppm beta to oxygen = + 0.6 ppm 1.5 ppm
O H H
O O
methylene protons (CH2) = 1.2 ppm alpha to oxygen = + 2.5 ppm beta to oxygen = + 0.6 ppm 4.3 ppm
H
CH3 H
CH3
(c) The 1H NMR spectrum of this compound is expected to exhibit four signals. The calculation for the estimated chemical shift of each signal is shown here:
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(d) The 1H NMR spectrum of this compound is expected to exhibit four signals. The calculation for the estimated chemical shift of each signal is shown here:
(e) All four methylene groups are equivalent, so the compound will have only one signal in its 1H NMR spectrum. That signal is expected to appear at approximately (1.2 + 2.5 + 0.5) = 4.2 ppm.
15.9. First determine the number of expected signals. In this compound there are six different kinds of protons giving rise to six distinct signals. For each type of signal, identify whether it represents a methyl group (0.9 ppm), a methylene group (1.2 ppm), or a methine group (1.7 ppm). Finally, modify each of these values based on proximity to the oxygen atoms and the carbonyl group, as shown. These values are only estimates and the actual chemical shifts might differ slightly from the predicted values. The actual values are also shown, and they are fairly close to the estimated values.
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15.10. (a) Using the values provided in Tables 15.1 and 15.2, we expect the following chemical shifts: ~ 4.5 - 6.5 ppm
~ 2 ppm
~ 10 ppm O
H
H
H
H
H
H
H
H
H
~ 3 ppm
H
~ 4.5 - 6.5 ppm
~ 2.5 ppm
~ 1.2 ppm
(b) Using the values provided in Tables 15.1 and 15.2, we expect the following chemical shifts:
15.11. Among the integration values provided, the lowest number is 33.2, so we divide all integration values by 33.2, giving the following ratio: 1 : 1.5 : 1 : 1.5 Since there is no such thing as a half of a proton, these numbers must represent 2, 3, 2, and 3 protons, respectively. This is confirmed by the molecular formula, which indicates that the compound has ten hydrogen atoms. Therefore, The signal at 4.0 ppm represents two protons. The signal at 2.0 ppm represents three protons. The signal at 1.6 ppm represents two protons. The signal at 0.9 ppm represents three protons. 15.12. Among the integration values provided, the lowest number is 17.1, so we divide all integration values by 17.1, giving the following ratio: 1 : 5 : 1 : 3
(c) Using the values provided in Tables 15.1 and 15.2, we expect the following chemical shifts:
The molecular formula indicates that the compound has ten hydrogen atoms, so the numbers above are not only relative values, but they are also exact values. Therefore, The signal at 9.6 ppm represents one proton. The signal at 7.5 ppm represents five protons. The signal at 7.3 ppm represents one proton. The signal at 2.1 ppm represents three protons. 15.13. Among the integration values provided, the lowest number is 18.92, so we divide all integration values by 18.92, giving the following ratio: 1 : 1 : 1 The molecular formula indicates that the compound has six hydrogen atoms (not just three), so the numbers above are only relative values. Each signal must actually represent two protons.
(d) Using the values provided in Tables 15.1 and 15.2, we expect the following chemical shifts:
15.14. We expect a total of eight signals in the 1H NMR spectrum of phenylalanine; note the symmetry in the benzene ring and the nonequivalent diastereotopic protons next to the chiral center bearing the NH2 group.
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2H
1H 1H H
H
H
O
H
NH2
2H H
1H OH
1H H
H H
1H
(d) This compound has six different kinds of protons, highlighted here. In each case, we apply the n + 1 rule, giving the multiplicities shown:
2H
Cinnamic acid has a simpler 1H NMR spectrum with only six signals.
15.15. (a) This compound has four different kinds of protons, highlighted here. In each case, we apply the n + 1 rule, giving the multiplicities shown:
15.16. Considering the n + 1 rule, we recognize that a doublet must have only one neighboring proton. This compound has eight different kinds of protons, but only the two indicated below are each expected to give rise to a doublet signal:
(b) This compound has four different kinds of protons, highlighted here. In each case, we apply the n + 1 rule, giving the multiplicities shown:
(c) This compound has four different kinds of protons, highlighted here. In each case, we apply the n + 1 rule, resulting in all singlets:
15.17. (a) The spectrum exhibits the characteristic pattern of an isopropyl group (a septet with an integration of one, and a doublet with an integration of six). (b) The spectrum exhibits the characteristic pattern of an isopropyl group (a septet with an integration of one, and a doublet with an integration of six) as well as the characteristic pattern of an ethyl group (a quartet with an integration of two, and a triplet with an integration of three). (c) The spectrum exhibits a singlet with a relative integration of 9 (because 55.0 / 6.0 ≈ 9), which is the characteristic pattern of a tert-butyl group. (d) The spectrum does not exhibit the characteristic pattern of an ethyl group, an isopropyl group, or a tertbutyl group.
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15.18. To determine the expected splitting pattern for the signal corresponding to Ha, we must consider the effects of the two non-equivalent neighbors, Hb and Hc, The former is coupled to Ha with coupling constant Jab (11 Hz), and the latter is coupled to Ha with coupling constant Jac (17 Hz). We begin with the larger coupling constant (Jac in this case), which splits the signal into a doublet. Then, each peak of this doublet is then further split into a doublet because of the effect of Hb. The result is a doublet of doublets. It can be distinguished from a quartet because all four peaks should be approximately equal in height, as opposed to a quartet, in which the individual peaks have relative heights of 1 : 3 : 3 : 1.
To determine the expected splitting pattern for the signal corresponding to Hb, we must consider the effects of the two non-equivalent neighbors, Ha and Hc, The former is coupled to Hb with a coupling constant Jab (11 Hz), and the latter is coupled to Hb with a coupling constant Jbc (1 Hz). We begin with the larger coupling constant (Jab in this case), which splits the signal into a doublet. Then, each peak of this doublet is then further split into a doublet because of the effect of Hc. The result is a doublet of doublets.
To determine the expected splitting pattern for the signal corresponding to Hc, we must consider the effects of the two non-equivalent neighbors, Ha and Hb, The former is coupled to Hc with a coupling constant Jac (17 Hz), and the latter is coupled to Hc with a coupling constant Jbc (1 Hz). We begin with the larger coupling constant (Jac in this case), which splits the signal into a doublet. Then, each peak of this doublet is then further split into a doublet because of the effect of Hb. The result is a doublet of doublets.
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15.19. (a) This compound is expected to produce four signals in its 1H NMR spectrum. For each signal, its expected chemical shift, multiplicity, and integration are shown. = 1.2 +1 = 2.2 ppm m = triplet I = 2H
O
H H = 1.2 + 0.2 = 1.4 ppm m = quintet I = 2H
= 1.2 + 0.6 = 1.8 ppm m = quintet I = 2H
So we expect its 1H NMR spectrum to exhibit only five signals, corresponding with the following highlighted protons. For each signal, its expected chemical shift, multiplicity, and integration are shown.
O
H
H H
H
H
H
= 1.2 + 0.5 + 0.2 = 1.9 ppm m = triplet I = 2H
= 1.2 + 2.5 = 3.7 ppm m = triplet I = 2H
O = 1.2 + 3 = 4.2 ppm m = triplet I = 2H
(b) This compound has rotational and reflectional symmetry:
H = 10 ppm m = singlet I = 1H = 0.9 + 0.2 = 1.1 ppm m = singlet I = 6H
O O
= 1.2 + 2.5 + 1.0 = 4.7 ppm m = singlet I = 2H
15.20. GHB is expected to produce five signals in its 1H NMR spectrum. For each signal, its expected chemical shift, multiplicity, and integration are shown below. Note that OH protons typically do not couple with neighboring protons, and as a result, no splitting occurs. Furthermore, OH signals generally appear as broad singlets.
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15.21. (a) The first compound exhibits symmetry so it will have only three signals in its 1H NMR spectrum, while the second compound will have six signals.
is the easiest way to differentiate these compounds. The second compound is an aldehyde so it is expected to produce a signal near 10 ppm. The first compound will lack such a signal.
(b) Both compounds will exhibit 1H NMR spectra with only two singlets. In each spectrum, the relative integration of the two singlets is 1:3. In the first compound, the singlet with the smaller integration value will be at approximately 2 ppm (alpha to a carbonyl). In the second compound, the singlet with the smaller integration value will be at approximately 4 ppm (alpha to the oxygen of an ester).
15.23. (a) The molecular formula (C8H10O) indicates four degrees of unsaturation (see Section 14.16), which is highly suggestive of an aromatic ring. To determine the relative integration values, we divide each of the integration values by 9.1, giving a ratio of approximately 5 : 2 : 2 : 1. Since the compound has ten protons, the numbers above are not only relative values, but they are also exact values. Now let’s analyze each of the signals individually. The signal just above 7 ppm confirms our suspicion of an aromatic ring. This signal has an integration of 5H, indicating a monosubstituted aromatic ring:
(c) The first compound will have only two signals in its NMR spectrum, while the second compound will have three signals.
1H
(d) The first compound will have five signals in its 1H NMR spectrum, while the second compound exhibits symmetry so it will have only three signals. (e) The first compound exhibits symmetry so it will have only two signals in its 1H NMR spectrum, while the second compound will have four signals. (f) The first compound will have only one signal in its 1H NMR spectrum (a singlet), while the second compound will have two signals (one signal will be a doublet and the other signal will have ten peaks). 15.22. (a) Both compounds will have very similar 1H NMR spectra, with the same number of signals, splitting patterns and integration values. The major difference will be the location of the 2H quartet signal. For the first structure, this signal is expected to appear near 2.2 ppm, because it is a methylene group (benchmark value = 1.2 ppm) next to a carbonyl group (+1). For the second structure, this signal will appear at approximately 4.2 ppm, because it is a methylene group next to an oxygen atom of an ester group (+3):
(b) These compounds are indeed expected to exhibit a different number of signals in their 1H NMR spectra. However, most of the signals will appear in the region 12 ppm (in each spectrum), and may overlap, making it difficult to count the number of signals. The only distinctive peaks will be the vinyl and aldehyde signals. In fact, the presence or absence of a signal near 10 ppm
The spectrum also exhibits two triplets (just below 3 ppm and just below 4 ppm), indicating two methylene groups connected to each other:
Each of these signals appears more downfield than we might expect for a methylene group (1.2 ppm), so each of these methylene groups must be connected to a group that causes a deshielding effect. This must be taken into account in our final structure. If we inspect the two fragments that we have determined thus far (the monosubstituted aromatic ring and the methylene groups that neighbor each other), we will find that these two fragments account for nearly all of the atoms in the molecular formula (C8H10O). We only need to account for one more proton and one oxygen atom. The singlet at 2 ppm has an integration of 1, so this signal corresponds with only one proton (with no neighbors), so we conclude that the compound has an OH group. There is only one way to assemble the three fragments:
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This structure is consistent with all of the signals, including the chemical shifts of the two triplets, which can be explained by the electron withdrawing effect of the oxygen atom as well as the local magnetic field established by the aromatic ring. (b) The molecular formula (C7H14O) indicates one degree of unsaturation (see Section 14.16), which means that the compound must possess either a double bond or a ring. To determine the relative integration values, we divide each of the integration values by 10.8, giving a ratio of approximately 1 : 6. This spectrum has the characteristic pattern of an isopropyl group (a doublet with a relative integration of 6 and a septet with a relative integration of 1).
An isopropyl group only contains seven protons, but there are fourteen protons in the compound (C7H14O). We therefore conclude that the compound must contain two isopropyl groups, which are interchangeable by symmetry:
Notice that the two isopropyl groups account for all but one of the carbon atoms in the compound. Therefore, there can only be one carbon atom in between the two isopropyl groups. This carbon atom cannot have any protons, since we don’t see any other signals in the 1H NMR spectrum. Also, we must still account for one oxygen atom (C7H14O), and we said that the compound must contain one degree of unsaturation. This all points to a carbonyl group at the central position:
This structure is indeed consistent with the observed chemical shift at 2.7 ppm for the methine (CH) protons (1.7 + 1 = 2.7 ppm). (c) The molecular formula (C10H14O) indicates four degrees of unsaturation (see Section 14.16), which is highly suggestive of an aromatic ring. The signals near 7 ppm are likely a result of aromatic protons. Notice that the combined integration of these two signals is 4H. This, together with the distinctive splitting pattern (a pair of doublets), suggests a 1,4disubstituted aromatic ring:
The spectrum also exhibits a singlet with an integration of 9H (at approximately 1.4 ppm) which is characteristic of a tert-butyl group.
If we inspect the two fragments that we have determined thus far (the disubstituted aromatic ring and the tert-butyl group), we will find that these two fragments account for nearly all of the atoms in the molecular formula (C10H14O). We only need to account for one more proton and one oxygen atom. The peak just under 5 ppm has an integration of 1, so this signal corresponds with only one proton (with no neighbors), so we conclude that the compound has an OH group. This signal is broad, which is often (although not always) the case for signals arising from OH groups. There is only one way to assemble the three fragments:
(d) The molecular formula (C4H6O2) indicates two degrees of unsaturation (see Section 14.16), which means that the compound must possess either two double bonds, or two rings, or one ring and one double bond, or a triple bond. To determine the relative integration values, we divide each of the integration values by 18.0, giving a ratio of approximately 1 : 1 : 1. The molecular formula indicates six protons (rather than three), so the relative integration values must correspond with two protons for each signal. That is, the spectrum indicates the presence of three different methylene groups. From the splitting patterns (a triplet, a triplet, and a quintet), we can conclude that the three methylene groups are connected to each other:
Now let’s focus on the chemical shifts of the triplets (2.4 ppm and 4.3 ppm). Both signals are shifted downfield (relative to 1.2 ppm for a typical methylene group). One of these signals is significantly shifted downfield,
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perhaps because it is next to an oxygen atom (after all, the molecular formula indicates that there are two oxygen atoms in the compound):
The other triplet is also shifted downfield, but the effect is weaker. This seems consistent with the effect of a carbonyl group:
The central methylene group is beta to both the oxygen atom and the carbonyl group, and it feels the distant effects of both (explaining why that signal is shifted somewhat downfield itself). The fragment above accounts for ALL of the atoms in the molecular formula, yet we are still missing one degree of unsaturation (the product must contain two degrees of unsaturation, but the fragment above has only one degree of unsaturation). Therefore, we close the ends together, giving the following structure:
(e) The molecular formula (C9H10O) indicates five degrees of unsaturation (see Section 14.16), which is highly suggestive of an aromatic ring, in addition to either one double bond or one ring. To determine the relative integration values, we divide each of the integration values by 13.9, giving a ratio of approximately 1 : 1.5 : 1 : 1.5. Since the compound has ten protons, the numbers above must correspond with: 2:3:2:3 Now let’s analyze each of the signals individually. The signals near 7 ppm are likely a result of aromatic protons. Notice that the combined integration of these signals is 5H, indicating a monosubstituted aromatic ring:
The spectrum also exhibits the characteristic pattern of an ethyl group (a quartet with an integration of 2 and a triplet with an integration of 3):
If we inspect the two fragments that we have determined thus far (the monosubstituted aromatic ring and the ethyl group), we will find that these two fragments account for nearly all of the atoms in the molecular formula (C9H10O). We only need to account for one more carbon atom and one oxygen atom. And let’s not forget that our structure still needs one more degree of unsaturation, suggesting a carbonyl group:
There is only one way to connect these three fragments.
(f) The molecular formula (C5H12O) indicates no degrees of unsaturation (see Section 14.16), which means that the compound does not have a bond or a ring. To determine the relative integration values, we divide each of the integration values by 13.6, giving a ratio of approximately 1 : 2 : 6 : 3. Since the compound has twelve protons, the numbers above are not only relative values, but they are also exact values. Now let’s analyze each of the signals individually. Let’s begin with the two signals that represent the characteristic pattern for an ethyl group (a quartet with an integration of 2 and a triplet with an integration of 3).
The singlet at 1.2 ppm has an integration of 6, indicating two methyl groups that can be interchanged via symmetry (and they cannot have any neighboring protons).
The singlet at 2.2 ppm has an integration of 1, so this signal corresponds with only one proton (with no neighbors), so we conclude that this likely represents an OH group (the molecular formula indicates the presence of an oxygen atom).
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If we inspect the fragments that we have determined thus far (an ethyl group, two methyl groups and an OH group), we will find that these fragments account for all of the atoms in the molecular formula (C5H10O) except for one carbon atom. Indeed, this carbon atom is necessary to connect all of the fragments, as shown:
singlet, these methyl groups must not have any neighbors. The chemical shift is consistent with these methyl groups being connected to oxygen atoms:
If we inspect the fragments that we have determined thus far (a 1,4-disubstituted aromatic ring, and two methoxy groups), we will find that these fragments account for all of the atoms in the molecular formula (C10H10O4) except for two carbon atoms and two oxygen atoms. Recall that the compound must have six degrees of unsaturation, and the fragments above only account for four degrees of unsaturation. Therefore, the remaining two carbon atoms and two oxygen atoms are likely carbonyl groups:
15.24. The molecular formula (C10H10O4) indicates six degrees of unsaturation (see Section 14.16), which suggests an aromatic ring as well as two other degrees of unsaturation. To determine the relative integration values, we divide each of the integration values by 52, giving a ratio of approximately 1 : 1.5. Since the compound has ten protons (C10H10O4), the numbers above must correspond to four protons and six protons, respectively. Now let’s analyze each of the signals individually. The signal at 8.1 ppm is significantly downfield, and likely represents aromatic protons. Since it is a singlet with an integration of 4, it must correspond with a 1,4disubstituted aromatic ring in which both substituents are identical (therefore rendering all four aromatic protons equivalent).
There are certainly a few different ways to connect all of these fragments, but there is only one way to connect them without breaking the symmetry necessary to keep all four aromatic protons identical:
The other signal has an integration of 6, which likely represents two equivalent methyl groups (interchangeable by symmetry). Since the signal is a
15.25. Begin by calculating the HDI. The molecular formula indicates 9 carbon atoms and 1 nitrogen atom. These would require 21 hydrogen atoms in order to be fully saturated. There are only 13 hydrogen atoms, so 8 hydrogen atoms are missing. Therefore, the HDI is 4. This is a relatively large number, and it would be inefficient to think about all of the possible ways to have 4 degrees of unsaturation. Any time we encounter an HDI of 4 or more, we should be on the lookout for an aromatic ring. Keep this in mind when analyzing the spectrum, which we expect to exhibit aromatic protons (near 7 ppm). Next, consider the number of signals and the integration value for each signal. Be on the lookout for integration values that suggest the presence of symmetry. For example, a signal with an integration of 4 would suggest two equivalent CH2 groups.
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The spectral data indicates a total of 6 signals. Let’s begin with the pair of triplets just below 3 ppm, each of which has an integration of 2. This suggests that there are two adjacent methylene groups.
The singlet near 4 ppm has an integration of 3 which suggests an isolated methyl group, likely connected to an electronegative oxygen atom. A methoxy fragment, OCH3, seems likely.
Moving downfield, there are 2 doublets near 7 ppm, each of which has an integration of 2. This pattern is characteristic of a 1,4-disubstituted benzene ring, bearing two different substituents (with different electronic demands). In such a case, there are two types of aromatic protons, each of which has an integration of 2 and is split into a doublet by its one neighbor:
So far, we have the following three fragments:
These fragments collectively account for 9 carbon atoms, 11 hydrogen atoms, and 1 oxygen atom. If we inspect the molecular formula, we see that we have accounted for all of the carbon atoms and the oxygen atom, but we must still account for 2 more hydrogen atoms and 1 nitrogen atom. Therefore, we conclude that the broad singlet near 1 ppm is likely to be an NH2 group. We will see later (Chapter 22) that NH2 protons generally appear as broad signals between 0.5 and 5.0 ppm. Next, we assemble the fragments. There are two reasonable ways to connect the fragments in this case.
To distinguish between these two options, we consider chemical shifts. Note that the adjacent methylene groups appear fairly close to one another, near 3 ppm, indicating that neither is attached to the highly electronegative and deshielding oxygen atom. The first compound above shows a CH2 group connected to oxygen, and we expect that compound to produce a triplet somewhere near 4 ppm, which is absent from the spectral data. The second compound above has the two CH2 groups connected to an aromatic ring and an NH2 group, respectively, which is consistent with the observed chemical shifts of these triplets. The structure is redrawn here in conventional bond-line notation.
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15.26. (a) The two methyl groups occupy identical environments (there is a conformation of this molecule in which the two methyl groups are interchangeable by reflectional symmetry), so the methyl groups are chemically equivalent. All of the other carbon atoms are unique, giving rise to a total of four signals. The expected chemical shift for each of these signals (listed below) can be found in Table 15.4.
The two, equivalent methyl (CH3) groups will produce one signal in the region 10 – 30 ppm. The methylene (CH2) group will produce one signal in the region 15 – 55 ppm. The methine (CH) group will produce one signal in the region 20 – 60 ppm. The carbonyl group will produce one weak signal in the region 185 – 220 ppm.
(b) This compound exhibits symmetry, rendering the two methyl groups equivalent. Similarly, the ring has only four unique signals, because of symmetry. In total, the 13C NMR spectrum of this compound should exhibit five signals. The expected chemical shift for each of these signals (listed below) can be found in Table 15.4.
The two, equivalent methyl (CH3) groups will produce one signal in the region 10 – 30 ppm. The five methylene (CH2) groups will produce three signals in the region 15 – 55 ppm. The quaternary carbon (C) will produce one signal in the region 20 – 60 ppm.
(c) Monosubstituted aromatic rings exhibit symmetry, giving only four unique carbon atoms in the aromatic ring. Each of the carbon atoms of the ethyl group is unique, giving a total of six signals. The expected chemical shift for each of these signals (listed below) can be found in Table 15.4.
The methyl (CH3) group will produce one signal in the region 10 – 30 ppm. The methylene (CH2) group will produce one signal in the region 15 – 55 ppm. The aromatic ring will produce four signals in the region 110 – 170 ppm.
(d) This compound has no symmetry. Each of the carbon atoms is unique, giving a total of nine signals. The expected chemical shift for each of these signals (listed below) can be found in Table 15.4.
One of the methyl (CH3) groups (the methyl group that is NOT connected to oxygen) will produce one signal in the region 10 – 30 ppm. The methyl (CH3) group connected to oxygen will produce one signal in the region 40 – 80 ppm. The methylene (CH2) group will produce one signal in the region 15 – 55 ppm.
The aromatic ring will produce six signals in the region 110 – 170 ppm.
(e) The aromatic ring is 1,4-disubstituted so it exhibits symmetry, giving four unique carbon atoms in the aromatic ring (four signals). In addition, there will be a signal for the carbon atom of the methoxy group, and finally, there will be two signals for the carbon atoms of the ethyl group. In total, there are seven signals. The expected chemical shift for each of these signals (listed below) can be found in Table 15.4.
One of the methyl (CH3) groups (the methyl group NOT connected to oxygen) will produce one signal in the region 10 – 30 ppm. The methyl group connected to oxygen will produce one signal in the region 40 – 80 ppm. The methylene (CH2) group will produce one signal in the region 15 – 55 ppm. The aromatic ring will produce four signals in the region 110 – 170 ppm.
(f) This compound has symmetry, giving only five unique positions (highlighted):
So the 13C NMR spectrum should have five signals. The expected chemical shift for each of these signals (listed below) can be found in Table 15.4.
The four methyl (CH3) groups will produce two signals in the region 10 – 30 ppm. The two methylene (CH2) groups will produce one signal in the region 15 – 55 ppm. The two vinylic carbon atoms will produce two signals in the region 100 – 150 ppm.
(g) None of the carbon atoms in this compound can be interchanged with any of the other carbon atoms in this compound via either reflection or rotation. All of the carbon atoms are unique, giving rise to seven signals. The expected chemical shift for each of these signals (listed below) can be found in Table 15.4.
The four methyl (CH3) groups will produce four signals in the region 10 – 30 ppm. The methylene (CH2) group will produce one signal in the region 15 – 55 ppm. The two vinylic carbon atoms will produce two signals in the region 100 – 150 ppm.
(h) This compound exhibits a high degree of symmetry. All four methyl groups are interchangeable by either rotation or reflection, and therefore, all four methyl groups give rise to one signal. The two vinylic carbon
CHAPTER 15 atoms are also chemically equivalent, giving rise to one signal. In total, there are only two signals. The expected chemical shift for each of these signals (listed below) can be found in Table 15.4.
613
its 13C NMR spectrum. This would be an easy way to distinguish between these compounds.
The four methyl (CH3) groups will produce one signal in the region 10 – 30 ppm. The two vinylic carbon atoms will produce one signal in the region 100 – 150 ppm.
(i) This compound exhibits a high degree of symmetry. All four carbon atoms are interchangeable by either rotation or reflection, and therefore, all four carbon atoms give rise to one signal, appearing in the region 40 – 80 ppm (as expected for an sp3 hybridized carbon atom attached to an oxygen atom). (j) None of the carbon atoms in this compound can be interchanged with any of the other carbon atoms in this compound via either reflection or rotation. All of the carbon atoms are unique, giving rise to five signals. The expected chemical shift for each of these signals (listed below) can be found in Table 15.4.
The methyl (CH3) group will produce one signal in the region 10 – 30 ppm. The methylene (CH2) group next to oxygen will produce one signal in the region 40 – 80 ppm. The two vinylic carbon atoms will produce two signals in the region 100 – 150 ppm. The carbonyl group will produce one weak signal in the region 165 – 185 ppm.
15.27. Let’s first determine which diastereomer has the R configuration and which has the S configuration. When we place the hydrogen on a dash, the three substituents will be arranged such that the sequence of priorities (1– 2–3) is clockwise; therefore, compound 3 (having the R configuration) will have the H on a dash. Since compound 4 has the S configuration, we would simply need to invert the configuration at that center, meaning the H should be placed on a wedge.
15.28. The molecular formula (C5H10O) indicates one degree of unsaturation (see Section 14.16), which means that the compound must either have a double bond or a ring. The signal above 200 ppm (in the broadbanddecoupled spectrum) indicates the source of the degree of unsaturation (C=O). The four signals in the range 10 – 60 ppm represent four unique sp3 hybridized carbon atoms. Two of them are upside-down in the DEPT-135 spectrum, indicating that they are methylene groups. Based on the number of protons in the compound (C5H10O), the other two signals must be methyl groups (to give a total of ten protons). Now we must connect a carbonyl group, two methylene groups and two methyl groups. There are only two ways to do that: O O
The first possibility cannot be correct because it has symmetry and would have only three signals in its broadband-decoupled 13C spectrum. The latter structure is the only structure that is consistent with the data.
15.29. The molecular formula (C5H12O) indicates no degrees of unsaturation (see Section 14.16), which means that the compound cannot have either a double bond or a ring. The signal at 73.8 ppm must be produced by the carbon atom that is connected to the oxygen atom. Notice that only one signal can be found in the range 40 – 80 ppm, which is consistent with the compound being an alcohol, as indicated in the problem statement. There are only two other signals, indicating symmetry. Both of those signals represent sp3 hybridized carbon atoms. The following structure meets these requirements: Upon close inspection, compound 3 has a plane of symmetry that compound 4 lacks. In fact, compound 3 is a meso compound. As a result, compound 3 should exhibit only thirteen signals in its 13C NMR spectrum, while compound 4 should have twenty-three signals in
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This structure is consistent with the information presented in the DEPT spectra. The DEPT-90 spectrum indicates there is only one CH group, and the chemical shift of that signal (73.8 ppm) is fairly downfield for a CH group, indicating that it is likely connected to oxygen (consistent with the structure above). And the DEPT135 spectrum confirms the presence of a methyl (CH3) group and a methylene (CH2) group, also consistent with the symmetrical structure above (the two methyl groups are equivalent and produce one signal, while the two methylene groups are equivalent and also produce one signal)
a CH group, while the last two signals must be methyl groups. There are several possible structures that are consistent with the information above, but only one of them is consistent with the 1H NMR spectrum. Specifically, the singlet at 1.9 ppm (with an integration of 3) indicates a methyl ketone.
15.30. The molecular formula (C7H14O) indicates one degree of unsaturation (see Section 14.16), which means that the compound must possess either a double bond or a ring. The signal above 200 ppm (in the broadbanddecoupled spectrum) indicates the source of the degree of unsaturation (C=O). The five signals in the range 20 – 40 ppm represent five unique sp3 hybridized carbon atoms. As such, the spectrum has a total of six signals for a compound with seven carbon atoms. That means that one of the signals in the range 20 – 40 ppm must represent two carbon atoms (for example, two equivalent methyl groups – the 1H NMR data confirms the two equivalent methyl groups). Also notice that two of the signals are upside-down in the DEPT-135 spectrum, indicating that they are methylene groups. Based on the DEPT-90 spectrum, we can see that one of the signals is
15.31. Diisobutyl phthalate has an internal plane of symmetry, so the two isobutyl groups are chemically equivalent (since they can be interchanged by rotational symmetry); the symmetry also bisects the aromatic ring. Although there are 16 carbon atoms in DIBP, only 7 signals are thus expected in its broadband-decoupled 13C NMR spectrum. Using the values provided in Table 15.4, we expect the following approximate chemical shifts:
The DEPT-90 spectrum exhibits three signals for the CH groups: two in the sp2 hybridized region 110-170 (C1 and C2), and one in the 20-60 ppm region (C6). The DEPT-135 spectrum exhibits five signals (only the quaternary carbon atoms, C3 and C4, are missing); there is one signal in the 40-80 ppm region that is a negative signal, indicating the presence of a methylene group (CH2) attached to an oxygen atom, C5.
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15.32. (a) The molecular formula (C5H10) indicates one degree of unsaturation (see Section 14.16), which means that the compound must possess either a double bond or a ring. With only one signal in the spectrum, the structure must have a high degree of symmetry, such that all ten protons are equivalent. This is indeed the case for a fivemembered ring (cyclopentane). There is no alkene with the molecular formula C5H10 in which all ten protons occupy identical electronic environments. So cyclopentane is the only structure consistent with the spectral data:
(b) The molecular formula (C5H8Cl4) indicates no degrees of unsaturation (see Section 14.16), which means that the compound cannot have any double bonds, triple bonds, or rings. That is, the structure must be acyclic and cannot have any bonds. With only one signal in the spectrum, the structure must have a high degree of symmetry, such that all eight protons are equivalent. This can be achieved with four equivalent methylene (CH2) groups. Four CH2 groups will be chemically equivalent if they are all attached to the same carbon atom (provided that all four CH2 groups are connected to identical groups):
The molecular formula indicates five carbon atoms (which are now all accounted for) and four chlorine
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atoms, which can serve to cap the four loose ends shown above, like this:
(c) The molecular formula (C12H18) indicates four degrees of unsaturation (see Section 14.16), which is highly suggestive of an aromatic ring. With only one signal in the spectrum, the structure must have a high degree of symmetry, such that all eighteen protons are equivalent. This can be achieved with six equivalent methyl groups, as seen in the following structure:
15.33. The molecular formula (C12H24) indicates one degree of unsaturation (see Section 14.16), which means that the compound must possess either a double bond or a ring. With only one signal in the 1H NMR spectrum, the structure must have a high degree of symmetry, such that all twenty-four protons are equivalent. This can be accomplished with either twelve equivalent methylene (CH2) groups or eight equivalent methyl (CH3) groups. Since the former would use up all of the carbon atoms in the structure (all twelve), it is tempting to explore that possibility first. Indeed, a twelve-membered ring is comprised of twelve equivalent methylene groups, which should give rise to one signal in the 1H NMR spectrum. And this compound (cyclododecane) also exhibits the correct degree of unsaturation (HDI = 1). However, this structure would give only one signal in its 13C NMR spectrum, and the problem statement indicates that the
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13C NMR spectrum has two signals. So, we consider our other alternative (eight equivalent methyl groups). If the structure has eight equivalent methyl groups, then we still need to account for four more carbon atoms, as well as one degree of unsaturation. We can account for all four carbon atoms in a four-membered ring (HDI = 1). In the following structure, all eight methyl groups are equivalent.
This structure has a high degree of symmetry, and it will indeed give rise to two signals in its 13C NMR spectrum (one signal for all of the methyl groups, and another signal for all four carbon atoms of the ring). 15.34. The molecular formula (C17H36) indicates no degrees of unsaturation (see Section 14.16), which means that the compound cannot have any double bonds, triple bonds, or rings. That is, the structure must be acyclic and cannot have any bonds. With only one signal in the 1H NMR spectrum, the structure must have a high degree of symmetry, such that all thirty-six protons are equivalent. If all of these protons were methylene groups, then we would need at least eighteen carbon atoms, but the molecular formula indicates fewer than eighteen carbon atoms. So the thirty-six protons must be twelve equivalent methyl groups. Twelve methyl groups cannot all be attached to the same carbon atom, because the central carbon atom cannot have twelve bonds. However, twelve methyl groups will be equivalent if they comprise four equivalent tert-butyl groups, attached to one carbon atom.
(c) This molecule has no symmetry, so all four aromatic protons are in unique environments. Therefore, we expect four signals. (d) This compound has two different kinds of protons (highlighted below), giving rise to two signals.
(e) The methine (CH) proton gives one signal, while the two methyl groups collectively give one signal (with an integration of 6). In total, we expect two signals. (f) The replacement test indicates that the two protons of the methylene group are diastereotopic. Therefore, each of these protons will produce its own signal. That is, the methylene group will give rise to two signals (because of the presence of the chiral center). The structure also has two methyl groups which are not equivalent to each other (because of their proximity to the Cl) so they also produce two different signals. Finally, the methine (CH) proton gives a signal, for a total of five signals. 15.36. (a) This compound has four different kinds of carbon atoms (highlighted below), giving rise to four signals.
This structure is consistent with all of the available data, and it is expected to give rise to three signals in its 13C NMR spectrum (one signal for the central carbon atom, another signal for the four carbon atoms connected to the central carbon atom, and then one last signal for all twelve, equivalent methyl groups). 15.35. (a) All six methyl groups are equivalent (giving rise to one signal), while all four aromatic protons are also equivalent (giving rise to another signal). In total, we expect two signals.
(b) This molecule has no symmetry, so each of the six carbon atoms is in a unique environment. Therefore, we expect six signals.
(b) This molecule has no symmetry, so all four aromatic protons are in unique environments. Therefore, we expect four signals.
(c) This molecule has no symmetry, so each of the six carbon atoms is in a unique environment. Therefore, we expect six signals.
CHAPTER 15 (d) This compound has four different kinds of carbon atoms (highlighted below), giving rise to four signals. Cl
Cl
Br
Br This carbon atom gives one signal
Cl
This carbon atom gives one signal
Cl Br
These two carbon atoms are equivalent (giving one signal)
Br These two carbon atoms are equivalent (giving one signal)
(e) The carbon atom of the methine (CH) group gives one signal, and the two methyl groups give one signal, for a total of two signals. (f) Each of the four carbon atoms is in a unique environment, because of its substitution and its proximity to the chlorine atom. Therefore, we expect four signals. 15.37. The first compound exhibits symmetry that causes some of the carbon atoms to be equivalent (similar to the symmetry present in 15.36d). As such, the first compound will have five signals in its 13C NMR spectrum. In contrast, the second compound lacks this symmetry. Each carbon atom occupies a unique environment, so the second compound is expected to produce seven signals in its 13C NMR spectrum. 15.38. This compound has six different kinds of protons, highlighted here. In each case, we apply the n + 1 rule, giving the multiplicities shown:
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(b) The first compound is a meso compound. Two of the protons are enantiotopic (the protons that are alpha to the chlorine atoms) and are therefore chemically equivalent. As such, the first compound will only have two signals in its 1H NMR spectrum, while the second compound will have three signals. For a similar reason, the first compound will only have two signals in its 13C NMR spectrum, while the second compound will have three signals. (c) The 13C NMR spectrum of the second compound will have one more signal than the 13C NMR spectrum of the first compound. The 1H NMR spectra will differ in the following way: the OH group in the first compound will produce a singlet somewhere between 2 and 5 ppm with an integration of 1, while the methoxy group in the second compound will produce a singlet at approximately 3.4 ppm with an integration of 3. (d) The first compound has symmetry that is not present in the second compound. As such, the first compound will have three signals in its 13C NMR spectrum, while the second compound will have five signals. For similar reasons, the first compound will have two signals in its 1H NMR spectrum, while the second compound will have four signals. 15.40. The molecular formula (C8H18) indicates no degrees of unsaturation (see Section 14.16), which means that the compound does not have a bond or a ring. With only one signal in its 1H NMR spectrum, the structure must have a high degree of symmetry, such that all eighteen protons are equivalent. This can be achieved with six equivalent methyl groups, which account for six of the eight carbon atoms. The remaining two carbon atoms can be placed at the center of the structure, rendering all six methyl groups equivalent, like this:
This compound will exhibit two signals in its 13C NMR spectrum (one signal for the two central carbon atoms, and another signal for the six methyl groups). 15.41. (a) The replacement test gives the same compound, so the protons are homotopic:
15.39. (a) The first compound has a very high degree of symmetry, and will produce only four only signals in its 13C NMR spectrum, while the second compound will produce twelve signals. Also, the first compound will produce only two signals in its 1H NMR spectrum, while the second compound will produce eight signals.
(b) The replacement test gives enantiomers, so the protons are enantiotopic:
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(c) The replacement test gives enantiomers, so the protons are enantiotopic:
(h) The replacement test gives diastereomers, so the protons are diastereotopic:
D D
Enantiomers
(d) The replacement test gives the same compound, so the protons are homotopic:
(i) The replacement test gives the same compound, so the protons are homotopic:
(e) The replacement test gives diastereomers, so the protons are diastereotopic: (j) The replacement test gives the same compound, so the protons are homotopic:
These compounds are diastereomers because they are stereoisomers that are not mirror images of each other. If it seems to you like they should be enantiomers, keep in mind that each of these compounds has three chiral centers, and these compounds differ only in the configuration of one of the three chiral centers (each of the bridgehead positions is a chiral center).
(k) The replacement test gives the same compound, so the protons are homotopic:
(f) The replacement test gives the same compound, so the protons are homotopic:
(l) The replacement test gives diastereomers, so the protons are diastereotopic: (g) The replacement test gives diastereomers, so the protons are diastereotopic:
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CHAPTER 15 D
HO
H
H
HO
H
D
H
OH
OH
H
H H H
O
H Cl
H H H
H H
O
H
Diastereomers
(m) The replacement test gives enantiomers, so the protons are enantiotopic: MeO MeO
D
H
MeO MeO
H
D
The signal for the CH2 of the ethyl group is expected to appear at 1.2 + 1 = 2.2 ppm, while the signal for the CH3 of the ethyl group is expected to appear at 0.9 + 0.2 = 1.1 ppm. Finally, there are two neighboring methylene groups, shown here:
Enantiomers
(n) The replacement test gives diastereomers, so the protons are diastereotopic: Cl
Cl
H
H
D
Cl
Cl
D
Diastereomers
(o) The replacement test gives the same compound, so the protons are homotopic: Cl
Cl
D
Cl
Cl
H
H
Each of these signals is expected to be a triplet with an integration of 2. And each of these signals is expected to be shifted downfield, because of the electronwithdrawing effects of the chlorine atom and oxygen atom. As seen in Section 16.5, each Cl adds approximately +2, and oxygen adds approximately +2.5 ppm. The CH2 next to the oxygen atom should produce a signal near 1.2 + 2.5 + 0.4 = 4.1 ppm. The last term (+0.4) was for the effect of the distant Cl (one-fifth of 2.0). The CH2 next to the Cl should produce a signal near 1.2 + 2.0 + 0.5 = 3.7 ppm. The following hand-drawn spectrum shows all of the signals described above.
D
Same compound
15.42. This compound has four aromatic protons. Because of their relationship with the respect to the ring (symmetry), we expect two types of protons, giving rise to a pair of doublets between 7 and 8 ppm: H
H
H
H H
H O H H
H H
O
H Cl
H
H
The structure also has an ethyl group, so we expect the characteristic pattern of signals for an ethyl group. Specifically, we expect a triplet with an integration of 3 (corresponding to the CH3 of the ethyl group) and a quartet with an integration of 2 (corresponding to the CH2 of the ethyl group).
15.43. (a) Each of the protons occupies a unique environment, and therefore, we expect four signals in the 1H NMR spectrum of this compound. (b) All of the halogens withdraw electron density from the neighboring proton, causing a downfield shift. But the effect will be strongest for fluorine (the most electronegative) and weakest for iodine (the least electronegative of the halogens in this compound). F
Br
Ha Hd
Hc
I
Hb Cl
Increasing chemical shift Ha > Hb > H c > Hd
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(c) Each of the carbon atoms occupies a unique environment, and therefore, we expect four signals in the 13C NMR spectrum of this compound. (d) The carbon atoms follow the same trend exhibited by the protons (the carbon atom connected to fluorine will produce the signal that is farthest downfield). 15.44. The molecular formula (C9H18) indicates one degree of unsaturation (see Section 14.16), which means that the compound must possess either a double bond or a ring. With only one signal in the 1H NMR spectrum, the structure must have a high degree of symmetry, such that all eighteen protons are equivalent. This can be accomplished with either nine equivalent methylene (CH2) groups or six equivalent methyl (CH3) groups. Since the former would use up all of the carbon atoms in the structure (all nine), it is tempting to explore that possibility first. Indeed, a nine-membered ring is comprised of nine equivalent methylene groups, which should give rise to one signal in the 1H NMR spectrum:
And this compound (cyclononane) also exhibits the correct degree of unsaturation (HDI = 1). However, this structure would give only one signal in its 13C NMR spectrum, and the problem statement indicates that the 13C NMR spectrum has two signals. So, we consider our other alternative (six equivalent methyl groups). If the structure has six equivalent methyl groups, then we still need to account for three more carbon atoms, as well as one degree of unsaturation. We can account for all three carbon atoms in a three-membered ring (HDI = 1). In the following structure, all six methyl groups are equivalent.
environments, because one is trans to the main chain, while the other is cis to the main chain):
Each of the vinylic CH groups are different, giving rise to two more signals:
And finally, the proton of the OH group gives one last signal, for a total of 3 + 3 + 2 + 1 = 9 signals. (b) The methyl group gives one signal, and then each of the remaining protons gives rise to its own signal, for a total of six signals:
Note that each of the vinylic protons is unique. For example, the following two vinylic protons are different from each other even though they are connected to the same carbon atom:
15.46. Four CH2 groups can be chemically equivalent if they are all attached to the same carbon atom (provided that the four CH2 groups are all connected to identical groups:
This structure has a high degree of symmetry, and it will indeed give rise to two signals in its 13C NMR spectrum (one signal for all of the methyl groups, and another signal for all three carbon atoms of the ring). 15.45. (a) This compound has three different methylene (CH2) groups, giving rise to three separate signals.
In addition, each of the three methyl groups gives its own unique signal, as none of the methyl groups are in identical electronic environments (the two methyl groups on the left side of the structure are not in identical
The molecular formula indicates a total of nine carbon atoms and twenty hydrogen atoms, but the structure above only accounts for five carbon atoms and eight hydrogen atoms. We must still account for another four carbon atoms and twelve hydrogen atoms. This can be accomplished if we simply connect a methyl group to each of the CH2 groups, giving the following structure:
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The four methyl groups are chemically equivalent, giving rise to only signal. As such, the 1H NMR spectrum of this compound is expected to exhibit only two signals (one for the CH2 groups and the other for the CH3 groups). 15.47. Below are the expected chemical shifts for each of the seven signals in this compound:
The signal from the carbon atom of the methyl group (sp3 hybridized) will appear in the region 10 – 30 ppm. The carbon atom of the methylene (CH2) group is also sp3 hybridized, but it is next to an oxygen atom. So we expect that signal to appear in the region 40 – 80 ppm, together with the signal from the sp hybridized carbon. Finally, the signal from the ester carbonyl group is expected to appear in the region 165 – 185 ppm. 15.49. Let’s begin by drawing the reaction described in the problem statement:
15.48. (a) Symmetry in the ring gives four different signals for the carbon atoms of the ring 110 – 170, in addition to two signals for the vinylic carbon atoms. So in total, we expect six signals, all of which result from sp2 hybridized carbon atoms, and therefore, we expect all six signals to appear in the region 110 – 170 ppm.
The Markovnikov product has symmetry that the antiMarkovnikov product lacks. As such, a 1H NMR spectrum of the Markovnikov product should have fewer signals than a 1H NMR spectrum of the antiMarkovnikov product. 15.50. (a) The compound has a high degree of symmetry, and there are only two unique aromatic protons, highlighted below, giving rise to two signals.
(b) Each of the carbon atoms of the ring occupies a unique environment, giving six signals. The two methyl groups occupy identical environments (they are interchangeable by reflectional symmetry), so they produce one signal. This can be seen more clearly if we draw wedges and dashes to illustrate the 3D orientation of the methyl groups: O CH3 CH3
This gives a total of seven signals. The signal resulting from the carbon atom of the ketone carbonyl group is expected to appear in the region 185 – 220 ppm, while the remaining six signals should appear in the region 10 – 60 ppm. (c) The compound is symmetrical, so we only need to consider half of the structure. We expect a total of four signals, corresponding with the following unique positions:
Each of the remaining aromatic protons can be interchanged with one of these positions (via either rotational or reflectional symmetry). (b) The presence of the methyl group renders all of the aromatic protons different from each other (because of their proximity to the methyl group). As such, we expect eight signals:
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(c) The compound has symmetry that renders some positions identical to other positions. As such, there are only four unique types of protons, highlighted below, giving rise to four signals. H
H
H
CH3
H
(h) The methyl group will produce one signal.
CH3 H
H
(d) The compound has a high degree of symmetry that renders some positions identical to other positions. As such, there are only two unique types of protons, highlighted below, giving rise to two signals.
H
Now let’s consider the remaining four protons. The two protons on wedges (highlighted below) are interchangeable via reflectional symmetry, so they are enantiotopic and therefore chemically equivalent.
H
H3C
CH3
H3C
CH3 H
H
(e) Each of the three vinylic protons is in a unique environment, so we expect three signals.
(f) Each of the highlighted protons occupies a unique environment, giving rise to six signals:
Similarly, the two protons on dashes (highlighted below) are also interchangeable via reflectional symmetry, so they too are enantiotopic and therefore chemically equivalent.
In summary, we expect this compound to produce three signals in its 1H NMR spectrum. 15.51. Among these three compounds, the first one (benzene) has aromatic protons, which are expected to produce a signal the farthest downfield (between 6.5 and 8 ppm). Acetylenic protons give signals that are relatively upfield (near 2.5 ppm) while vinylic protons are expected to produce a signal in the range of 4.5 – 6.5 ppm.
Note that the following positions are different from each other:
(g) The compound has a high degree of symmetry. As such, the two methyl groups occupy identical environments and collectively give rise to one signal. Similarly, all four protons of the two methylene (CH2) can be interchanged by either rotational or reflection symmetry, so these four protons will collectively give rise to one signal. In total, we expect only two signals:
15.52. In Section 15.5, the term “chemical shift” was defined in the following way:
CHAPTER 15 The problem statement indicates that the chemical shift of the proton is 1.2 ppm and the operating frequency of the spectrometer is 300 MHz. We then plug these values into the equation above, as shown:
which gives the following observed shift from TMS (in Hz):
15.53. The molecular formula (C13H28) indicates no degrees of unsaturation (see Section 14.16), which means that the compound does not have a bond or a ring. The 1H NMR spectrum exhibits the characteristic pattern of an isopropyl group (a septet with an integration of 1 and a doublet with an integration of 6):
However, there are no other signals in this spectrum, and the molecular formula indicates that there are 28 protons (not just 7 protons, as we would expect for an isopropyl group). So the compound must be highly symmetrical, with four equivalent isopropyl groups (to account for all 28 protons). This also accounts for 12 of the 13 carbon atoms in this compound. The remaining carbon atom must be at the center, connected to all four isopropyl groups:
15.54. The molecular formula (C8H10) indicates four degrees of unsaturation (see Section 14.16), which is highly suggestive of an aromatic ring. This accounts for six of the eight carbon atoms in the structure. The other two carbon atoms must be connected to the ring, either as an ethyl group or as two methyl groups. Ethylbenzene would give an 1H NMR spectrum with five signals and a 13C NMR spectrum with six signals. The problem statement indicates fewer signals in each of these spectra, which means that the compound must have more symmetry than ethylbenzene. If we explore the three possible ways to connect two methyl groups to a ring (1,2- or 1,3- or 1,4-), we will find that only 1,4dimethylbenzene has the necessary symmetry to give only two signals in the 1H NMR spectrum and three signals in the 13C NMR spectrum.
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15.55. The molecular formula (C3H8O) indicates no degrees of unsaturation (see Section 14.16), which means that the compound does not have a bond or a ring. The broad signal between 3200 and 3600 cm-1 indicates the presence of an OH group. The molecular formula indicates that the structure has three carbon atoms, yet the 13C NMR spectrum exhibits only two signals (not three), indicating the presence of symmetry. This is only true for 2-propanol (not for 1-propanol):
15.56. The molecular formula (C4H6O4) indicates two degrees of unsaturation (see Section 14.16), which means that the compound must possess either two double bonds, or two rings, or one ring and one double bond, or a triple bond. The very broad signal (2500 – 3600 cm-1) in the IR spectrum indicates the presence of a carboxylic acid group. The 1H NMR spectrum has only two signals, with a total integration of 3, however the molecular formula indicates the presence of 6 protons. Therefore, the actual integration values for the signals are 2H and 4H, respectively. The broad singlet at 12.1 ppm is characteristic of a carboxylic acid group (COOH), as suggested by the IR spectrum, and since this signal has an integration value of 2H, we conclude that the compound must have two carboxylic acid groups. This accounts for both degrees of unsaturation, which means that the compound does not possess a ring. In order for the remaining four protons to be identical, they must be interchangeable by symmetry, which is indeed the case when we place two methylene (CH2) groups in between the two carboxylic acid groups, like this:
Note that the CH2 groups do not split each other (they appear as singlets) because they are chemically equivalent. The n + 1 rule refers to n as the number of nonequivalent neighboring protons. 15.57. (a) The molecular formula (C5H10O) indicates one degree of unsaturation (see Section 14.16), which means that the compound must possess either a double bond or a ring. The 1H NMR spectrum exhibits the characteristic pattern of an isopropyl group (a doublet with an integration of 6, and a septet with an integration of 1):