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A covalent bond is formed when two atoms share a pair of electrons. A polar covalent bond is a covalent bond between atoms with different electronegativities. The greater the difference in electronegativity between the atoms forming the bond, the closer the bond is to the ionic end of the continuum. A polar covalent bond has a dipole (a positive end and a negative end), measured by a dipole moment. The dipole moment of a bond is equal to the size of the charge × the distance between the charges. The dipole moment of a molecule depends on the magnitude and direction of all the bond dipole moments. Core electrons are electrons in inner shells. Valence electrons are electrons in the outermost shell. Lone-pair electrons are valence electrons that do not form bonds. formal charge = # of valence electrons – # of electrons the atom has to itself (all the lone-pair electrons and one-half the bonding electrons) Lewis structures indicate which atoms are bonded together and show lone pairs and formal charges. When the atom is neutral: C forms 4 bonds, N forms 3 bonds, O forms 2 bonds, and H or a halogen forms 1 bond. When the atom is neutral: N has 1 lone pair, O has 2 lone pairs, and a halogen has 3 lone pairs. A carbocation has a positively charged carbon, a carbanion has a negatively charged carbon, and a radical has an unpaired electron. According to molecular orbital (MO) theory, covalent bonds result when atomic orbitals combine to form molecular orbitals.

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Atomic orbitals combine to give a bonding MO and a higher energy antibonding MO. Electrons in a bonding MO assist in bonding. Electrons in an antibonding MO detract from bonding. There is zero probability of finding an electron at a node. Cylindrically symmetrical bonds are called sigma (s) bonds; side-to-side overlap of parallel p orbitals forms a pi (P) bond. Bond strength is measured by the bond dissociation energy; a s bond is stronger than a p bond. To be able to form four bonds, carbon has to promote an electron from an s orbital to an empty p orbital. C, N, O, and the halogens form bonds using hybrid orbitals. The hybridization of C, N, or O depends on the number of p bonds the atom forms: no p bonds = sp3, one p bond = sp2, and two p bonds = sp. Exceptions are carbocations and carbon radicals, which are sp2. All single bonds in organic compounds are s bonds. A double bond consists of one s bond and one p bond; a triple bond consists of one s bond and two p bonds. The greater the electron density in the region of orbital overlap, the shorter and stronger the bond. Triple bonds are shorter and stronger than double bonds, which are shorter and stronger than single bonds. The shorter the bond, the stronger it is. Molecular geometry is determined by hybridization: sp3 is tetrahedral, sp2 is trigonal planar, and sp is linear. Bonding pairs and lone-pair electrons around an atom stay as far apart as possible. The more s character in the orbital used to form a bond, the shorter and stronger the bond and the larger the bond angle.

GLOSSARY The definitions of the key words used in each chapter can be found at the beginning of each pertinent chapter in the Study Guide/Solutions Manual. The definitions of all the key words used in this book can be found in the Glossary on page G-1.

PROBLEMS 46. Draw a Lewis structure for each of the following species: b. CO32 c. CH2O d. CO2 a. H2CO3 47. a. Which of the following has a nonpolar covalent bond? b. Which of the following has a bond closest to the ionic end of the bond spectrum? CH3NH2

CH3CH3

CH3F

CH3OH

48. What is the hybridization of all the atoms (other than hydrogen) in each of the following species? What are the bond angles around each atom? c. - CH3 e. + NH4 g. HCN i. H3O + a. NH3 b. BH3 d. . CH3 f. + CH3 h. C(CH3)4 j. H2C “ O

50

CHAPTER 1

Remembering General Chemistry: Electronic Structure and Bonding

49. Draw the condensed structure of a compound that contains only carbon and hydrogen atoms and that has a. three sp3 hybridized carbons. b. one sp3 hybridized carbon and two sp2 hybridized carbons. c. two sp3 hybridized carbons and two sp hybridized carbons. 50. Predict the approximate bond angles: + a. the C i N i C bond angle in (CH3)2NH2 b. the C i O i H bond angle in CH3OH

c. the C i N i H bond angle in (CH3)2NH d. the C i N i C bond angle in (CH3)2NH

51. Draw the ground-state electronic configuration for the following: a. Mg b. Ca2 + c. Ar d. Mg2 + 52. Draw a Lewis structure for each of the following species: a. CH3NH2 b. HNO2 c. N2H4 d. NH2O 53. What is the hybridization of each of the carbon and oxygen atoms in vitamin C? CH2OH HO

CH O HC C C C

HO

O OH

vitamin C

54. List the bonds in order from most polar to least polar. a. C i O, C i F, C i N b. C i Cl, C i I, C i Br

c. H i O, H i N, H i C

d. C i H, C i C, C i N

55. Draw the Lewis structure for each of the following compounds: a. CH3CHO b. CH3OCH3 c. CH3COOH 56. What is the hybridization of the indicated atom in each of the following molecules? O CH2

a. CH3CH

b. CH3CCH3

CH3CH2OH

c.

d. CH3C

N

e. CH3CH

NCH3

f. CH3OCH2CH3

57. Predict the approximate bond angles for the following: c. the C i C i N bond angle in CH3C ‚ N a. the H i C i H bond angle in H2C “ O d. the C i C i N bond angle in CH3CH2NH2 b. the F i B i F bond angle in - BF4 58. Show the direction of the dipole moment in each of the following bonds (use the electronegativities given in Table 1.3): a. H3C i Br b. H3C i Li c. HO i NH2 d. I i Br f. (CH3)2 N i H e. H3C i OH 59. Draw the missing lone-pair electrons and assign the missing formal charges for the following. H a. H

C

H O

b. H

H

H

H

C

O

H

H

c. H

H

C

H d. H

O

H

C

N

H

H

H

60. a. Which of the indicated bonds in each molecule is shorter? b. Indicate the hybridization of the C, O, and N atoms in each of the molecules. CH3

H

Solo b

1.

CH3CH

CHC

CH

3. CH3NH

O 2.

CH3CCH2

CH2CH2N

CHCH3

5.

C

CHC

C

C

H

CH3

6. Br

CH2CH2CH2 Cl

H OH

C

4. H

CHC

C

H

H

Problems

51

61. For each of the following molecules, indicate the hybridization of each carbon and give the approximate values of all the bond angles: a. CH3C ‚ CH

b. CH3CH “ CH2

d. CH2

c. CH3CH2CH3

CH

CH

CH2

62. Draw the Lewis structure for each of the following compounds: b. CH3CH(OH)CH2CN c. (CH3)2CHCH(CH3)CH2C(CH3)3 a. (CH3)3COH 63. Rank the following compounds from highest dipole moment to lowest dipole moment: CH HC C Cl

Cl

CH

C

HC

CH

HC

CH

CH

Cl C

HC C

C Cl

CH

Br

Cl

CH

C

HC

CH

HC

CH

Cl C CH

CH

64. In which orbitals are the lone pairs in nicotine? H2C CH HC

C

HC

CH

CH2

CH CH2 N CH3

N nicotine nicotine increases the concentration of dopamine in the brain; the release of dopamine makes a person feel good—the reason why nicotine is addictive

65. Indicate the formal charge on each carbon that has one. All lone pairs are shown. H H

H

C

H

H

C

H

H

H

C

H

C H

CH2

CH

CH2

CH

CH2

CH

H

66. Do the sp2 carbons and the indicated sp3 carbons lie in the same plane? CH3 CH3

CH3

67. a. Which of the species have bond angles of 109.5°? b. Which of the species have bond angles of 120°? H2O

H3O+

+

CH3

BF3

NH3

+

NH4

−

CH3

68. Which compound has a longer C—Cl bond? CH3CH2Cl

CH2

at one time it was used as a refrigerant, an anesthetic, and a propellant for aerosol sprays

CHCl

used as the starting material for the synthesis of a plastic that is used to make bottles, flooring, and clear packaging for food

69. Which compound has a larger dipole moment, CHCl3 or CH2Cl2? 70. The following compound has two isomers. One isomer has a dipole moment of 0 D, whereas the other has a dipole moment of 2.95 D. Propose structures for the two isomers that are consistent with these data. ClCH “ CHCl

52

CHAPTER 1

Remembering General Chemistry: Electronic Structure and Bonding

71. Explain why the following compound is not stable: H H

C

H

C H

H

H

H

C C

C C H

72. Explain why CH3Cl has a greater dipole moment than CH3F even though F is more electronegative than Cl. 73. Draw a Lewis structure for each of the following species: a. CH3N2+ b. CH2N2 c. N3− d. N2O (arranged NNO)

140

CHAPTER 3

An Introduction to Organic Compounds

PROBLEMS 52. Draw a condensed structure and a skeletal structure for each of the following compounds: a. sec-butyl tert-butyl ether e. 5-(1,2-dimethylpropyl)nonane i. 3-ethoxy-2-methylhexane b. isoheptyl alcohol f. triethylamine j. 5-(1,2-dimethylpropyl)nonane c. sec-butylamine g. 4-(1,1-dimethylethyl)heptane k. 3,4-dimethyloctane d. isopentyl bromide h. 5,5-dibromo-2-methyloctane l. 4-(1-methylethyl)nonane 53. List the following compounds from highest boiling to lowest boiling: OH

N H

NH2

N

O

O

OH

54. a. What is each compound’s systematic name? b. Draw a skeletal structure for each condensed structure given, and draw a condensed structure for each skeletal structure given in the following list:

Solo b 1.

CH3

(CH3)3CCH2CH2CH2CH(CH3)2

5. BrCH2CH2CH2CH2CH2NHCH2CH3

9.

6.

10.

NCH3

CH2CH3 2. Br

3.

CH2CH2CH2CH3

OH

CH3

11.

7. CH3CH2CHOCH2CH3

CH3CHCH2CHCH2CH3 CH3

4.

OH

Br

8. CH3OCH2CH2CH2OCH3

(CH3CH2)4C

55. Which of the following represents a cis isomer? CH3

CH3 CH3

CH3 CH3

CH3 CH3

CH3 A

B

C

D

56. a. How many primary carbons does each of the following compounds have? b. How many secondary carbons does each one have? c. How many tertiary carbons does each one have? 1.

CH2CH3

2.

CH2CHCH3 CH3 57. Which of the following conformers of isobutyl chloride is the most stable? CH3

CH3 H H

H

H

CH3

H

Cl Cl CH3

CH3

H3C H

H

Cl

H

H

A

B

C

Problems

141

58. Draw the structural formula for an alkane that has a. six carbons, all secondary. b. eight carbons and only primary hydrogens. c. seven carbons with two isopropyl groups. 59. What is each compound’s systematic name? a. CH3CH2CHCH3

e. CH3CHCH2CH2CH3

h. CH3CHCH2CH2CH2OH

NH2 b. CH3CH2CHCH3

CH3 f. CH3CHNH2

i.

Cl c. CH3CH2CHNHCH2CH3

CH3 Br

CH3 CH3 g. CH3C Br

CH3

OH

j.

CH2CH3

d. CH3CH2CH2OCH2CH3 60. Which has a. the higher boiling point: 1-bromopentane or 1-bromohexane? b. the higher boiling point: pentyl chloride or isopentyl chloride? c. the greater solubility in water: 1-butanol or 1-pentanol? d. the higher boiling point: 1-hexanol or 1-methoxypentane? e. the higher melting point: hexane or isohexane? f. the higher boiling point: 1-chloropentane or 1-pentanol? g. the higher boiling point: 1-bromopentane or 1-chloropentane? h. the higher boiling point: diethyl ether or butyl alcohol? i. the greater density: heptane or octane? j. the higher boiling point: isopentyl alcohol or isopentylamine? k. the higher boiling point: hexylamine or dipropylamine? 61. a. b. c. d.

Draw Newman projections of the two conformers of cis-1,3-dimethylcyclohexane. Which of the conformers would predominate at equilibrium? Draw Newman projections of the two conformers of the trans isomer. Which of the conformers would predominate at equilibrium?

62. Ansaid and Motrin belong to the group of drugs known as nonsteroidal anti-inflammatory drugs (NSAIDs). Both are only slightly soluble in water, but one is a little more soluble than the other. Which of the drugs has the greater solubility in water?

F

CH3

CH3

CH3

CH3CHCH2

CHCOOH Ansaid®

CHCOOH

Motrin®

63. Draw a picture of the hydrogen bonding in methanol. 64. A student was given the structural formulas of several compounds and was asked to give them systematic names. How many did the student name correctly? Correct those that are misnamed. a. 4-bromo-3-pentanol e. 5-(2,2-dimethylethyl)nonane i. 1-bromo-4-pentanol b. 2,2-dimethyl-4-ethylheptane f. isopentyl bromide j. 3-isopropyloctane c. 5-methylcyclohexanol g. 3,3-dichlorooctane k. 2-methyl-2-isopropylheptane d. 1,1-dimethyl-2-cyclohexanol h. 5-ethyl-2-methylhexane l. 2-methyl-N,N-dimethyl-4-hexanamine 65. Which of the following conformers has the highest energy (is the least stable)? CH3

Cl A

CH3

Cl

CH3 B

Cl C

142

CHAPTER 3

An Introduction to Organic Compounds

66. Give the systematic names for all alkanes with molecular formula C7H16 that do not have any secondary hydrogens. 67. Draw skeletal structures for the following: a. 5-ethyl-2-methyloctane c. 2,3,3,4-tetramethylheptane b. 1,3-dimethylcyclohexane d. propylcyclopentane

e. 2-methyl-4-(1-methylethyl)octane f. 2,6-dimethyl-4-(2-methylpropyl)decane

68. For rotation about the C@3 i C@4 bond of 2-methylhexane, do the following: a. Draw the Newman projection of the most stable conformer. b. Draw the Newman projection of the least stable conformer. c. About which other carbon–carbon bonds may rotation occur? d. How many of the carbon–carbon bonds in the compound have staggered conformers that are all equally stable? 69. Draw all the isomers that have molecular formula C5H11Br. (Hint: There are eight.) a. Give the systematic name for each of the isomers. b. Give a common name for each isomer that has a common name. c. How many of the isomers are primary alkyl halides? d. How many of the isomers are secondary alkyl halides? solo c-e e. How many of the isomers are tertiary alkyl halides? 70. What is each compound’s systematic name? a.

e.

i. NH2

b.

f.

OH

j. Cl O

c.

g.

OH d.

h.

71. Draw the two chair conformers for each of the following, and indicate which conformer is more stable: a. cis-1-ethyl-3-methylcyclohexane d. cis-1,2-diethylcyclohexane b. trans-1-ethyl-2-isopropylcyclohexane e. cis-1-ethyl-3-isopropylcyclohexane c. trans-1-ethyl-2-methylcyclohexane f. cis-1-ethyl-4-isopropylcyclohexane 72. Why are lower molecular weight alcohols more soluble in water than higher molecular weight alcohols? 73. a. Draw a potential energy diagram for rotation about the C i C bond of 1,2-dichloroethane through 360⬚, starting with the least stable conformer. The anti conformer is 1.2 kcal/mol more stable than a gauche conformer. A gauche conformer has two energy barriers, 5.2 kcal/mol and 9.3 kcal/mole. b. Draw the conformer that would be present in greatest concentration. c. How much more stable is the most stable staggered conformer than the most stable eclipsed conformer? d. How much more stable is the most stable staggered conformer than the least stable eclipsed conformer? 74. For each of the following compounds, is the cis isomer or the trans isomer more stable? a.

b.

c.

75. How many ethers have molecular formula C5H12O? Draw their structures and give each a systematic name. What are their common names?

Problems

143

76. Draw the most stable conformer of the following molecule. (A solid wedge points out of the plane of the paper toward the viewer. A hatched wedge points back from the plane of the paper away from the viewer.) CH3

H3C

CH3

77. What is each compound’s systematic name? CH2CH3 a. CH3CH2CHCH2CH2CHCH3

c. CH3CHCHCH2Cl

e.

Cl

NHCH3 CH3 CH3 b. CH3CH2CHCH2CHCH2CH3

d. CH3CH2CHCH3

CH3CHCH3

f.

CH3CHCH3

78. Calculate the energy difference between the two chair conformers of trans-1,2-dimethylcyclohexane. 79. The most stable form of glucose (blood sugar) is a six-membered ring in a chair conformation with its five substituents all in equatorial positions. Draw the most stable conformer of glucose by putting the OH groups on the appropriate bonds in the structure on the right. CH2OH HO

CH2OH

O

HO

O OH

OH glucose

80. What is each compound’s systematic name? a. CH3CHCH2CHCH2CH3 CH3

d. CH3CHCH2CH2CHCH2CH2CH3 CH3

OH

Br

g. HO

Br

OH

b.

e. CH3

c. CH3CHCHCH2CH3

h. Br

OH OH

f.

Br

OH 81. Explain the following: a. 1-Hexanol has a higher boiling point than 3-hexanol. b. Diethyl ether has very limited solubility in water, but tetrahydrofuran is completely soluble.

O tetrahydrofuran

82. One of the chair conformers of cis-1,3-dimethylcyclohexane is 5.4 kcal/mol (23 kJ/mol) less stable than the other. How much steric strain does a 1,3-diaxial interaction between two methyl groups introduce into the conformer?

Problems

181

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Stereochemistry is the field of chemistry that deals with the structures of molecules in three dimensions. Isomers are compounds with the same molecular formula but different structures. Constitutional isomers differ in the way their atoms are connected. Stereoisomers differ in the way their atoms are arranged in space. There are two kinds of stereoisomers: cis–trans isomers and isomers that contain asymmetric centers. Because rotation about the bonds in a cyclic compound is restricted, disubstituted cyclic compounds exist as cis-trans isomers. The cis isomer has the substituents on the same side of the ring; the trans isomer has the substituents on opposite sides of the ring. Because rotation about a double bond is restricted, an alkene can exist as cis–trans isomers. The cis isomer has its hydrogens on the same side of the double bond; the trans isomer has its hydrogens on opposite sides of the double bond. A chiral molecule has a nonsuperimposable mirror image; an achiral molecule has a superimposable mirror image. An asymmetric center is an atom bonded to four different atoms or groups. Enantiomers are nonsuperimposable mirror images. Diastereomers are stereoisomers that are not enantiomers. Enantiomers have identical physical and chemical properties; diastereomers have different physical and chemical properties. An achiral reagent reacts identically with both enantiomers; a chiral reagent reacts differently with each enantiomer.

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A racemic mixture is a mixture of equal amounts of two enantiomers. The letters R and S indicate the configuration about an asymmetric center. If one member of a pair of stereoisomers has the R configuration and the other has the S configuration, they are enantiomers; if they both have the R configuration or both have the S configuration, they are identical. Chiral compounds are optically active; achiral compounds are optically inactive. If one enantiomer rotates the plane of polarization clockwise ( + ), its mirror image will rotate it the same amount counterclockwise ( - ). Each optically active compound has a characteristic specific rotation. A racemic mixture, indicated by ( { ), is optically inactive. In the case of compounds with two asymmetric centers, enantiomers have the opposite configuration at both asymmetric centers; diastereomers have the same configuration at one asymmetric center and the opposite configuration at the other asymmetric center. A meso compound has two or more asymmetric centers and a plane of symmetry; it is optically inactive. A compound with the same four groups bonded to two different asymmetric centers will have three stereoisomers—namely, a meso compound and a pair of enantiomers. Atoms other than carbons (such as N and P) can be asymmetric centers if they are bonded to four different atoms or groups.

PROBLEMS 53. Disregarding stereoisomers, draw the structures of all compounds with molecular formula C5H10. Which ones can exist as stereoisomers? 54. Draw all possible stereoisomers for each of the following. Indicate if no stereoisomers are possible. a. 1-bromo-2-chlorocyclohexane f. 1,2-dimethylcyclopropane b. 2-bromo-4-methylpentane g. 4-bromo-2-pentene c. 1,2-dichlorocyclohexane h. 3,3-dimethylpentane d. 2-bromo-4-chloropentane i. 1-bromo-2-chlorocyclobutane e. 1-bromo-4-chlorocyclohexane j. 1-bromo-3-chlorocyclobutane 55. Which of the following has an asymmetric center? CHBr2Cl

BHFCl

CH3CHCl2

CHFBrCl

BeHCl

182

CHAPTER 4

Isomers: The Arrangement of Atoms in Space

56. Name the following compounds using R,S designations: CH3 H a. HO H Cl CH2CH3

Br C CH3

CH3CH2

CH2OH b. HO CH3 CH2CH2CH2OH

C H HO

c.

CH2Br

57. Mevacor is used clinically to lower serum cholesterol levels. How many asymmetric centers does Mevacor have? HO

O O

O O

CH3

Mevacor®

58. Are the following pairs identical, enantiomers, diastereomers, or constitutional isomers?

Br

b.

e.

and

a. Br

Br

C H HO

Br

and Br

Br

Br

d.

CH3

f.

Br

CH3

and

CH3 C

C

Br h.

CH3

CH3

H

CH3

H and

C

Cl C

CH3

CH3 C

Br

Br C

and

CH3 C CH(CH3)2

Cl

H

Br

C

HO H

Cl

CH3

CH3

and

Cl

g.

CH3

CH3CH2

CH(CH3)2

H

CH3 and CH3 CH3 CH3 CH3 and

c.

Br C CH3

CH3CH2

C Br

CH3 C CH3

59. Which of the following are optically active? CH3

CH3 CH3

Br

Br

Br

Br

CH3

60. For many centuries, the Chinese have used extracts from a group of herbs known as ephedra to treat asthma. A compound named ephedrine has been isolated from these herbs and found to be a potent dilator of air passages in the lungs. a. How many stereoisomers does ephedrine have? b. The stereoisomer shown here is the one that is pharmacologically active. What is the configuration of each of the asymmetric centers? CH3

H

CHCHNHCH3 OH ephedrine

61. Name each of the following: a.

b. Cl

OH

Br

C

HO H

C NHCH3 CH3

Problems

62. Which of the following has an achiral stereoisomer? a. 2,3-dichlorobutane b. 2,3-dichloropentane c. 2,3-dichloro-2,3-dimethylbutane d. 1,3-dichlorocyclopentane e. 1,3-dibromocyclobutane

f. g. h. i. j.

183

2,4-dibromopentane 2,3-dibromopentane 1,4-dimethylcyclohexane 1,2-dimethylcyclopentane 1,2-dimethylcyclobutane

63. Are the following pairs identical, enantiomers, diastereomers, or constitutional isomers? Cl

a.

Cl

and

CH3 Cl

b.

Cl

c.

CH3

Cl

CH3 Cl

Cl

and

CH3

and

Cl

d.

CH3

and

CH3

CH3

CH3

64. Citrate synthase, one of the enzymes in the series of enzyme-catalyzed reactions known as the citric acid cycle (Section 25.10), catalyzes the synthesis of citric acid from oxaloacetic acid and acetyl-CoA. If the synthesis is carried out with acetyl-CoA that contains radioactive carbon (14C) in the indicated position (Section 1.1), the isomer shown here is obtained. (If two isotopes—atoms with the same atomic number, but different mass numbers—are being compared, the one with the greater mass number has the higher priority.) O HOOCCH2CCOOH oxaloacetic acid

14

CH2COOH

O +

14

CH3CSCoA

citrate synthase

C

HO

acetyl-CoA

COOH CH2COOH

citric acid

a. Which stereoisomer of citric acid is synthesized, R or S? b. Why is the other stereoisomer not obtained? c. If the acetyl-CoA used in the synthesis contains 12C instead of 14C, will the product of the reaction be chiral or achiral? 65. Are the following pairs identical, enantiomers, diastereomers, or constitutional isomers? a.

HC

CH2

C CH3CH2

CH3 H

CH2CH3

CH2CH3 and

H

C

CH3 HC CH2

e.

and CH3

CH2OH b. H

CH2CH3 CH3 and CH3 H CH2CH3 CH2OH

and

f. Cl

CH3 CH3 HO H H OH c. and H Cl Cl H CH3 CH3

g.

CH3 CH2CH3 HO H HO H d. and H Cl H Cl CH2CH3 CH3

h.

Cl

Cl

Cl

and Cl Cl

Cl

Cl

and Cl

Cl

Cl

Cl

66. The specific rotation of (R)-( + )-glyceraldehyde is +8.7. If the observed specific rotation of a mixture of (R)-glyceraldehyde and (S)-glyceraldehyde is +1.4, what percent of glyceraldehyde is present as the R enantiomer?

184

CHAPTER 4

Isomers: The Arrangement of Atoms in Space

67. Indicate whether each of the following structures is (R)-2-chlorobutane or (S)-2-chlorobutane. Cl Cl H C H 3 Cl e. a. C c. CH 3 CH3CH2 H CH3 H H CH3

CH3 b. H

Cl

Cl

d.

f.

CH2CH3

H

H

H

Cl CH3

68. A solution of an unknown compound (3.0 g of the compound in 20 mL of solution), when placed in a polarimeter tube 2.0 dm long, was found to rotate the plane of polarized light 1.8⬚ in a counterclockwise direction. What is the specific rotation of the compound? 69. Butaclamol is a potent antipsychotic that has been used clinically in the treatment of schizophrenia. How many asymmetric centers does it have? OH C(CH3)3

N H

H

butaclamol

70. Explain how R and S are related to ( + ) and ( - ). 71. Are the following pairs identical, enantiomers, diastereomers, or constitutional isomers? CH3 OH CH3 CH3 H H H Br H H and a. and b. H OH CH3 H H CH2CH3 H H CH3 Cl

CH2CH3 Cl CH3 Br

72. a. Draw all possible stereoisomers of the following compound: HOCH2CH

CH

CHCH2OH

OH

OH

OH

b. Which stereoisomers are optically inactive? 73. What is the configuration of the asymmetric centers in the following structures? CH2CH2Br a.

Br

C

H CH2CH2CH3

BrCH2 b.

H

C Br

OH C H CH

H3C c. O

Br

C H

CH2CH3 C H Br

74. a. Draw all the isomers with molecular formula C6H12 that contain a cyclobutane ring. (Hint: There are seven.) b. Name the compounds without specifying the configuration of any asymmetric centers. c. Identify: 5. achiral compounds 1. constitutional isomers 6. meso compounds 2. stereoisomers solo 7. enantiomers 3. cis–trans isomers a y c 8. diastereomers 4. chiral compounds 75. A compound has a specific rotation of -39.0. A solution of the compound (0.187 g/mL) has an observed rotation of -6.52⬚ when placed in a polarimeter tube 10 cm long. What is the percent of each enantiomer in the solution?

Problems

185

76. Are the following pairs identical, enantiomers, diastereomers, or constitutional isomers? a.

b.

c.

and

and

d.

and

and

77. Draw structures for each of the following: a. (S)-1-bromo-1-chlorobutane b. (2R,3R)-2,3-dichloropentane c. an achiral isomer of 1,2-dimethylcyclohexane d. a chiral isomer of 1,2-dibromocyclobutane e. two achiral isomers of 3,4,5-trimethylheptane 78. Explain why the enantiomers of 1,2-dimethylaziridine can be separated, even though one of the “groups” attached to nitrogen is a lone pair. CH3

H 3C N

N CH3

H3C

enantiomers of 1,2-dimethylaziridine

79. A sample of (S)-( + )-lactic acid was found to have an enantiomeric excess of 72%. How much R isomer is present in the sample? 80. Indicate whether each of the structures in the second row is an enantiomer of, a diastereomer of, or is identical to the structure in the top row. CH3 H H

Br CH3 CH2CH3

Br H

Br

H CH3

Br CH3

CH2CH3 A

CH3

CH3

CH3

H CH2CH3

Br

H B

Br CH3CH2

H H CH3

C

D

E

81. a. Using the wedge-and-dash notation, draw the nine stereoisomers of 1,2,3,4,5,6-hexachlorocyclohexane. b. From the nine stereoisomers, identify one pair of enantiomers. c. Draw the most stable conformation of the most stable stereoisomer. 82. Tamiflu is used for the prevention and treatment of flu. What is the configuration of each of its asymmetric centers? (How Tamiflu Works is explained on page 1124.)

O

O

O HN O H2N Tamiflu®

186

CHAPTER 4

Isomers: The Arrangement of Atoms in Space

83. A student decided that the configuration of the asymmetric centers in a sugar such as d-glucose could be determined rapidly by simply assigning the R configuration to an asymmetric center with an OH group on the right and the S configuration to an asymmetric center with an OH group on the left. Is he correct? (We will see in Chapter 21 that the “d” in d-glucose means the OH group on the bottommost asymmetric center is on the right.) HC O H OH HO H H OH H OH CH2OH D-glucose

84. What the configuration of each of the asymmetric centers in the following compounds? O CH3 b. c. a. CH2CH3 CH3 O

OH

Br

85. a. Draw the two chair conformers for each of the stereoisomers of trans-1-tert-butyl-3-methylcyclohexane. b. For each pair, indicate which conformer is more stable. 86. a. Do the following compounds have any asymmetric centers? 1. CH2 “ C “ CH2 2. CH3CH “ C “ CHCH3 b. Are the compounds chiral? (Hint: Make models.) 87. Is the following compound optically active? Cl

Br

Br

Cl

Problems ■

■

■

■

■

■

■

■

■

Approximate pKa values are as follows: protonated alcohols, protonated carboxylic acids, protonated water  6 0; carboxylic acids ~5; protonated amines ~10; alcohols and water ~15. The pH of a solution indicates the concentration of protons in the solution; the smaller the pH, the more acidic the solution. In acid–base reactions, the equilibrium favors formation of the weaker acid. Curved arrows indicate the bonds that are broken and formed as reactants are converted into products. The strength of an acid is determined by the stability of its conjugate base: the more stable (weaker) the base, the stronger its conjugate acid. When atoms are similar in size, the strongest acid will have its hydrogen attached to the more electronegative atom. When atoms are very different in size, the strongest acid will have its hydrogen attached to the larger atom. Hybridization affects acidity because an sp hybridized atom is more electronegative than an sp2 hybridized atom, which is more electronegative than an sp3 hybridized atom. Inductive electron withdrawal increases acidity: the more electronegative the electron-withdrawing group

■

■

■

■

■

■

and the closer it is to the acidic hydrogen, the stronger is the acid. Delocalized electrons (electrons that are shared by more than two atoms) stabilize a compound. A resonance hybrid is a composite of the resonance contributors, structures that differ only in the location of their p electrons and lone-pair electrons. The Henderson–Hasselbalch equation gives the relationship between pKa and pH: a compound exists primarily in its acidic form (with its proton) in solutions more acidic than its pKa value and primarily in its basic form (without its proton) in solutions more basic than its pKa value. A buffer solution contains both a weak acid and its conjugate base. In this text, the term acid is used to mean a protondonating acid, the term Lewis acid is used to refer to non-proton-donating acids such as AlCl3 or BF3, and the term electrophile refers to both proton-donating and non-proton-donating acids. In this text, the term base is used to mean a compound that shares its lone pair with a proton, and the term nucleophile is used for a compound that shares its lone pair with an atom other than a proton.

PROBLEMS 47. a. List the following alcohols in order from strongest acid to weakest acid: CCl3CH2OH

CH2ClCH2OH

Ka = 5.75 × 10−13

CHCl2CH2OH

Ka = 1.29 × 10−13

Ka = 4.90 × 10−13

b. Explain the relative acidities. 48. Which is a stronger base? a. HS− or HO− − b. CH3O− or CH3NH

c. CH3OH or CH3O− d. Cl− or Br−

e. CH3COO− or CF3COO−

f. CH3CHClCOO - or CH3CHBrCOO -

49. Draw curved arrows to show where the electrons start from and where they end up in the following reactions: +

OH

O a.

NH3 + H

Cl

+

NH4

+

Cl

−

c.

C H

b.

H2O

+ FeBr3

+ H OH

+

C

Cl H

Cl

−

OH

−

H2O+ FeBr3

50. a. List the following carboxylic acids in order from strongest acid to weakest acid: CH3CH2CH2COOH Ka = 1.52 × 10−5

CH3CH2CHCOOH Cl Ka = 1.39 × 10−3

79

ClCH2CH2CH2COOH Ka = 2.96 × 10−5

CH3CHCH2COOH Cl Ka = 8.9 × 10−5

b. How does the presence of an electronegative substituent such as Cl affect the acidity of a carboxylic acid? c. How does the location of the substituent affect the acidity of the carboxylic acid?

80

CHAPTER 2

Acids and Bases: Central to Understanding Organic Chemistry

51. Draw the products of the following reactions: a. CH3OCH3 + BF3

b. CH3OCH3 + H

52. For the following compound, a. draw its conjugate acid.

c. CH3NH2

Cl

+ AlCl3

b. draw its conjugate base.

HOCH2CH2CH2NH2 53. List the following compounds in order from strongest acid to weakest acid: CH3CH2NH2

CH3CH2OH

CH3CH2SH

CH3CH2CH3

54. For each of the following compounds, draw the form in +which it will predominate at pH = 3, pH = 6, pH = 10, and pH = 14: b. CH3CH2NH3 c. CF3CH2OH a. CH3COOH pKa ⴝ 4.8

pKa ⴝ 11.0

pKa ⴝ 12.4

55. Give the products of the following acid–base reactions, and indicate whether reactants or products are favored at equilibrium (use the pKa values that are given in Section 2.3): O

O −

a. CH3COH + CH3O b. CH3CH2OH +

c. CH3COH + CH3NH2

−

d. CH3CH2OH + HCl

NH2

56. a. List the following alcohols in order from strongest acid to weakest acid. b. Explain the relative acidities. CH2

CHCH2OH

CH3CH2CH2OH

HC

CCH2OH

57. For each compound, indicate the atom that is most apt to be protonated. CH3 CH3 a. CH3

CH

b. CH3

CH2NH2

OH

C

OH

c. CH3

NH2

C

CH2OH

NH2

58. a. Given the Ka values, estimate the pKa value of each of the following acids without using a calculator (that is, is it between 3 and 4, between 9 and 10, and so on?): 1. nitrous acid (HNO2), Ka = 4.0 * 10 - 4 3. bicarbonate 1HCO3- 2 , Ka = 6.3 * 10 - 11 4. hydrogen cyanide (HCN), Ka = 7.9 * 10 - 10 2. nitric acid (HNO3), Ka = 22 5. formic acid (HCOOH), Ka = 2.0 * 10 - 4 b. Determine the exact pKa values, using a calculator. c. Which is the strongest acid? 59. A single bond between two carbons with different hybridizations has a small dipole. What is the direction of the dipole in the indicated bonds? a. CH3

CH

CH2

b. CH3

C

CH

60. Tenormin, a member of the group of drugs known as beta-blockers, is used to treat high blood pressure and improve survival after a heart attack. It works by slowing down the heart in order to reduce its workload. Which hydrogen in Tenormin is the most acidic? OCH2CHCH2NHCH(CH3)2

O H2N

C

OH CH2

Tenormin® atenolol

61. From which acids can HO- remove a proton in a reaction that favors product formation? CH3COOH A

CH3CH2NH2 B

+

CH3CH2NH3 C

CH3C

CH D

Problems

81

62. a. For each of the following pairs of reactions, indicate which one has the more favorable equilibrium constant (that is, which one most favors products): + + 2. CH3CH2OH + NH3 L CH3CH2O - + NH4 1. CH3CH2OH + NH3 L CH3CH2O - + NH4 or or + + CH3OH + NH3 L CH3O - + NH4 CH3CH2OH + CH3NH2 L CH3CH2O - + CH3NH3 b. Which of the four reactions has the most favorable equilibrium constant? 63. You are planning to carry out a reaction that produces protons. The reaction will be buffered at pH = 10.5. Would it be better to use a protonated methylamine/methylamine buffer or a protonated ethylamine/ethylamine buffer? (pKa of protonated methylamine = 10.7; pKa of protonated ethylamine = 11.0) 64. Which is a stronger acid? a. CH2 CHCOOH or CH3CH2COOH

c. CH2

O

b.

or

+

N H

d.

+

H

or

+

N H

CHCOOH or HC

+

N H

H

CCOOH

N H

H

65. Citrus fruits are rich in citric acid, a compound with three COOH groups. Explain the following: a. The first pKa (for the COOH group in the center of the molecule) is lower than the pKa of acetic acid. b. The third pKa is greater than the pKa of acetic acid. O pKa = 4.5

OH

C HO

CH2 O

O C

C

CH2

C

OH pKa = 5.8

OH pKa = 3.1

66. Given that pH + pOH = 14 and that the concentration of water in a solution of water is 55.5 M, show that the pKa of water is 15.7. (Hint: pOH = -log 3 HO - 4 ) 67. How could you separate a mixture of the following compounds? The reagents available to you are water, ether, 1.0 M HCl, and 1.0 M NaOH. (Hint: See Problem 42.) COOH

pKa = 4.17

+

NH3Cl−

pKa = 4.60

OH

pKa = 9.95

Cl

+

NH3Cl−

pKa = 10.66

68. Carbonic acid has a pKa of 6.1 at physiological temperature. Is the carbonic acid/bicarbonate buffer system that maintains the pH of the blood at 7.4 better at neutralizing excess acid or excess base? 69. a. If an acid with a pKa of 5.3 is in an aqueous solution of pH 5.7, what percentage of the acid is present in its acidic form? b. At what pH will 80% of the acid exist in its acidic form? 70. Calculate the pH values of the following solutions. (Hint: See Special Topic I in the Study Guide and Solutions Manual.) a. a 1.0 M solution of acetic acid (pKa = 4.76) b. a 0.1 M solution of protonated methylamine (pKa = 10.7) c. a solution containing 0.3 M HCOOH and 0.1 M HCOO - (pKa of HCOOH = 3.76)