5.27 heats of combustion z 5.27 a b c d química#orgânica# ... · 2ª#lista#de#exercícios#...
TRANSCRIPT
2ª Lista de Exercícios Química Orgânica I-‐ Prof. Marco Antonio Barbosa Ferreira
(1)
2)
3)
Write the structure of this substance or build a molecular model in a way that clearlyshows its stereochemistry.
(e) The sex pheromone of the honeybee is the E stereoisomer of the compound shown.Write a structural formula or build a molecular model for this compound.
(f) A growth hormone from the cecropia moth has the structure shown. Express thestereochemistry of the double bonds according to the E–Z system.
5.26 Which one of the following has the largest dipole moment (is the most polar)? Compareyour answer with the calculated dipole moments on the Learning By Modeling CD.
5.27 Match each alkene with the appropriate heat of combustion:
Heats of combustion (kJ/mol): 5293; 4658; 4650; 4638; 4632
Heats of combustion (kcal/mol): 1264.9; 1113.4; 1111.4; 1108.6; 1107.1
(a) 1-Heptene (d) (Z)-4,4-Dimethyl-2-pentene
(b) 2,4-Dimethyl-1-pentene (e) 2,4,4-Trimethyl-2-pentene
(c) 2,4-Dimethyl-2-pentene
5.28 Choose the more stable alkene in each of the following pairs. Explain your reasoning.
(a) 1-Methylcyclohexene or 3-methylcyclohexene
(b) Isopropenylcyclopentane or allylcyclopentane
(c)
(d) (Z)-Cyclononene or (E)-cyclononene
(e) (Z)-Cyclooctadecene or (E)-cyclooctadecene
5.29 (a) Suggest an explanation for the fact that 1-methylcyclopropene is some 42 kJ/mol (10 kcal/mol) less stable than methylenecyclopropane.
(b) On the basis of your answer to part (a), compare the expected stability of 3-methylcy-clopropene with that of 1-methylcyclopropene and that of methylenecyclopropane.
CH3
1-Methylcyclopropene
is less stable than CH2
Methylenecyclopropane
Bicyclo[4.2.0]oct-7-ene
or
Bicyclo[4.2.0]oct-3-ene
H3C
H3C CH3
CH3
C C
A
H3C
Cl Cl
CH3
C C
B
H3C
Cl
Cl
CH3
C C
C
Cl
Cl Cl
Cl
C C
D
O
CH3CH2
H3C
CH3CH2 CH3
COCH3
O
2637
CH3C(CH2)4CH2CHœCHCO2H
OX
Problems 203
Write the structure of this substance or build a molecular model in a way that clearlyshows its stereochemistry.
(e) The sex pheromone of the honeybee is the E stereoisomer of the compound shown.Write a structural formula or build a molecular model for this compound.
(f) A growth hormone from the cecropia moth has the structure shown. Express thestereochemistry of the double bonds according to the E–Z system.
5.26 Which one of the following has the largest dipole moment (is the most polar)? Compareyour answer with the calculated dipole moments on the Learning By Modeling CD.
5.27 Match each alkene with the appropriate heat of combustion:
Heats of combustion (kJ/mol): 5293; 4658; 4650; 4638; 4632
Heats of combustion (kcal/mol): 1264.9; 1113.4; 1111.4; 1108.6; 1107.1
(a) 1-Heptene (d) (Z)-4,4-Dimethyl-2-pentene
(b) 2,4-Dimethyl-1-pentene (e) 2,4,4-Trimethyl-2-pentene
(c) 2,4-Dimethyl-2-pentene
5.28 Choose the more stable alkene in each of the following pairs. Explain your reasoning.
(a) 1-Methylcyclohexene or 3-methylcyclohexene
(b) Isopropenylcyclopentane or allylcyclopentane
(c)
(d) (Z )-Cyclononene or (E)-cyclononene
(e) (Z )-Cyclooctadecene or (E)-cyclooctadecene
5.29 (a) Suggest an explanation for the fact that 1-methylcyclopropene is some 42 kJ/mol (10 kcal/mol) less stable than methylenecyclopropane.
(b) On the basis of your answer to part (a), compare the expected stability of 3-methylcy-clopropene with that of 1-methylcyclopropene and that of methylenecyclopropane.
CH3
1-Methylcyclopropene
is less stable than CH2
Methylenecyclopropane
Bicyclo[4.2.0]oct-7-ene
or
Bicyclo[4.2.0]oct-3-ene
H3C
H3C CH3
CH3
C C
A
H3C
Cl Cl
CH3
C C
B
H3C
Cl
Cl
CH3
C C
C
Cl
Cl Cl
Cl
C C
D
O
CH3CH2
H3C
CH3CH2 CH3
COCH3
O
2637
CH3C(CH2)4CH2CHœCHCO2H
OX
Problems 203
666 Chapter 14 Aromatic Compounds
AROMATICITY
14.18 Which of the following molecules would you expect to be aromatic?
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
(k)
(l) S
!
"
"
!
!
N
"
N
N
N
"
O
"
"
14.19 Use the polygon-and-circle method to draw an orbital diagram for each of the following compounds.
(a) (b)
14.20 Write the structure of the product formed when each of the following compounds reacts with one molar equiva-lent of HCl.
(a) (b)
14.21 Which of the hydrogen atoms shown below is more acidic? Explain your answer.
A B
14.22 The rings below are joined by a double bond that undergoes cis–trans isomerization much more readily than thebond of a typical alkene. Provide an explanation.
!
HH
N
NH
HCl (1 equiv.)N
N
CH3
HCl (1 equiv.)
!"
solom_c14_632-675hr.qxd 6-10-2009 12:52 Page 666
4)
5)
6)
7)
Give the structres of C and D. 8)
666 Chapter 14 Aromatic Compounds
AROMATICITY
14.18 Which of the following molecules would you expect to be aromatic?
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
(k)
(l) S
!
"
"
!
!
N
"
N
N
N
"
O
"
"
14.19 Use the polygon-and-circle method to draw an orbital diagram for each of the following compounds.
(a) (b)
14.20 Write the structure of the product formed when each of the following compounds reacts with one molar equiva-lent of HCl.
(a) (b)
14.21 Which of the hydrogen atoms shown below is more acidic? Explain your answer.
A B
14.22 The rings below are joined by a double bond that undergoes cis–trans isomerization much more readily than thebond of a typical alkene. Provide an explanation.
!
HH
N
NH
HCl (1 equiv.)N
N
CH3
HCl (1 equiv.)
!"
solom_c14_632-675hr.qxd 6-10-2009 12:52 Page 666
NH
OHCl
(c)
666 Chapter 14 Aromatic Compounds
AROMATICITY
14.18 Which of the following molecules would you expect to be aromatic?
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
(k)
(l) S
!
"
"
!
!
N
"
N
N
N
"
O
"
"
14.19 Use the polygon-and-circle method to draw an orbital diagram for each of the following compounds.
(a) (b)
14.20 Write the structure of the product formed when each of the following compounds reacts with one molar equiva-lent of HCl.
(a) (b)
14.21 Which of the hydrogen atoms shown below is more acidic? Explain your answer.
A B
14.22 The rings below are joined by a double bond that undergoes cis–trans isomerization much more readily than thebond of a typical alkene. Provide an explanation.
!
HH
N
NH
HCl (1 equiv.)N
N
CH3
HCl (1 equiv.)
!"
solom_c14_632-675hr.qxd 6-10-2009 12:52 Page 666
666 Chapter 14 Aromatic Compounds
AROMATICITY
14.18 Which of the following molecules would you expect to be aromatic?
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
(k)
(l) S
!
"
"
!
!
N
"
N
N
N
"
O
"
"
14.19 Use the polygon-and-circle method to draw an orbital diagram for each of the following compounds.
(a) (b)
14.20 Write the structure of the product formed when each of the following compounds reacts with one molar equiva-lent of HCl.
(a) (b)
14.21 Which of the hydrogen atoms shown below is more acidic? Explain your answer.
A B
14.22 The rings below are joined by a double bond that undergoes cis–trans isomerization much more readily than thebond of a typical alkene. Provide an explanation.
!
HH
N
NH
HCl (1 equiv.)N
N
CH3
HCl (1 equiv.)
!"
solom_c14_632-675hr.qxd 6-10-2009 12:52 Page 666
Challenge Problems 673
TMS
Y, C9H12O
8 7 6 5 4 3 2 1 0dH (ppm)
4.6 1.8 0.9 0.81.64.4
Challenge Problems
14.40 Given the following information, predict the appearance of the 1H NMR spectrum aris-ing from the vinyl hydrogen atoms of p-chlorostyrene. Deshielding by the inducedmagnetic field of the ring is greatest at proton c (d 6.7) and is least at proton b (d 5.3).The chemical shift of a is about d 5.7. The coupling constants have the followingapproximate magnitudes: Jac 18 Hz, Jbc 11 Hz, and Jab 2 Hz. (These cou-pling constants are typical of those given by vinylic systems: Coupling constants fortrans hydrogen atoms are larger than those for cis hydrogen atoms, and coupling con-stants for geminal vinylic hydrogen atoms are very small.)
!!!
Figure 14.34 (Continued)
14.41 Consider these reactions:
The intermediate A is a covalently bonded compound that has typical 1H NMR signals for aromatic ring hydro-gens and only one additional signal at d 1.21, with an area ratio of 5:3, respectively. Final product B is ionic andhas only aromatic hydrogen signals.
What are the structures of A and B?
14.42 The final product of this sequence, D, is an orange, crystalline solid melting at 174°C and having molecular weight 186:
Cyclopentadiene ! Na 9: C ! H2
2 C ! FeCl2 9: D ! 2 NaCl
In its 1H and 13C NMR spectra, product D shows only one kind of hydrogen and only one kind of carbon,respectively.
Draw the structure of C and make a structural suggestion as to how the high degree of symmetry of D can beexplained. (D belongs to a group of compounds named after something you might get at a deli for lunch.)
A B!t-BuOK
"KCl
HBr
"t-BuOH
Cl
(a)
(b)
(c)
Cl
H
H
H
solom_c14_632-675hr.qxd 6-10-2009 12:52 Page 673
Openmirrors.com
(c) C6H5CHœCH2 ±£ C6H5CPCH
(d) C6H5CPCH ±£ C6H5CH2CH2CH2CH3
(e) C6H5CH2CH2OH ±£ C6H5CH2CH2CPCH
(f)
11.38 The relative rates of reaction of ethane, toluene, and ethylbenzene with bromine atoms havebeen measured. The most reactive hydrocarbon undergoes hydrogen atom abstraction a million timesfaster than does the least reactive one. Arrange these hydrocarbons in order of decreasing reactivity.
11.39 Write the principal resonance structures of o-methylbenzyl cation and m-methylbenzylcation. Which one has a tertiary carbocation as a contributing resonance form?
11.40 The same anion is formed by loss of the most acidic proton from 1-methyl-1,3-cyclopenta-diene as from 5-methyl-1,3-cyclopentadiene. Explain.
11.41 There are two different tetramethyl derivatives of cyclooctatetraene that have methyl groupson four adjacent carbon atoms. They are both completely conjugated and are not stereoisomers.Write their structures.
11.42 Evaluate each of the following processes applied to cyclooctatetraene, and decide whetherthe species formed is aromatic or not.
(a) Addition of one more ! electron, to give C8H8!
(b) Addition of two more ! electrons, to give C8H82!
(c) Removal of one ! electron, to give C8H8"
(d) Removal of two ! electrons, to give C8H82"
11.43 Evaluate each of the following processes applied to cyclononatetraene, and decide whetherthe species formed is aromatic or not:
(a) Addition of one more ! electron, to give C9H10!
(b) Addition of two more ! electrons, to give C9H102!
(c) Loss of H" from the sp3-hybridized carbon
(d) Loss of H" from one of the sp2-hybridized carbons
11.44 From among the molecules and ions shown, all of which are based on cycloundecapentaene,identify those which satisfy the criteria for aromaticity as prescribed by Hückel’s rule.
(a) (c)
(b) (d)
!
Cycloundecapentaenide anionCycloundecapentaenyl radical
"
Cycloundecapentaenyl cationCycloundecapentaene
Cyclononatetraene
C6H5CHCH2Br
OH
C6H5CH2CH2Br
440 CHAPTER ELEVEN Arenes and Aromaticity
9)
11.45 (a) Figure 11.16 is an electrostatic potential map of calicene, so named because its shaperesembles a chalice (calix is the Latin word for “cup”). Both the electrostatic potentialmap and its calculated dipole moment (! ! 4.3 D) indicate that calicene is an unusu-ally polar hydrocarbon. Which of the dipolar resonance forms, A or B, better correspondsto the electron distribution in the molecule? Why is this resonance form more importantthan the other?
(b) Which one of the following should be stabilized by resonance to a greater extent? (Hint:Consider the reasonableness of dipolar resonance forms.)
11.46 Classify each of the following heterocyclic molecules as aromatic or not, according toHückel’s rule:
(a) (c)
(b) (d)
O
O
NH
BH
NH
NH
HN
HB
HBO
C
or
D
Calicene
"
#
A
#
"
B
Problems 441
FIGURE 11.16 Elec-trostatic potential map ofcalicene (problem 11.45).
11.45 (a) Figure 11.16 is an electrostatic potential map of calicene, so named because its shaperesembles a chalice (calix is the Latin word for “cup”). Both the electrostatic potentialmap and its calculated dipole moment (! ! 4.3 D) indicate that calicene is an unusu-ally polar hydrocarbon. Which of the dipolar resonance forms, A or B, better correspondsto the electron distribution in the molecule? Why is this resonance form more importantthan the other?
(b) Which one of the following should be stabilized by resonance to a greater extent? (Hint:Consider the reasonableness of dipolar resonance forms.)
11.46 Classify each of the following heterocyclic molecules as aromatic or not, according toHückel’s rule:
(a) (c)
(b) (d)
O
O
NH
BH
NH
NH
HN
HB
HBO
C
or
D
Calicene
"
#
A
#
"
B
Problems 441
FIGURE 11.16 Elec-trostatic potential map ofcalicene (problem 11.45).
10)
11)
12)
11.45 (a) Figure 11.16 is an electrostatic potential map of calicene, so named because its shaperesembles a chalice (calix is the Latin word for “cup”). Both the electrostatic potentialmap and its calculated dipole moment (! ! 4.3 D) indicate that calicene is an unusu-ally polar hydrocarbon. Which of the dipolar resonance forms, A or B, better correspondsto the electron distribution in the molecule? Why is this resonance form more importantthan the other?
(b) Which one of the following should be stabilized by resonance to a greater extent? (Hint:Consider the reasonableness of dipolar resonance forms.)
11.46 Classify each of the following heterocyclic molecules as aromatic or not, according toHückel’s rule:
(a) (c)
(b) (d)
O
O
NH
BH
NH
NH
HN
HB
HBO
C
or
D
Calicene
"
#
A
#
"
B
Problems 441
FIGURE 11.16 Elec-trostatic potential map ofcalicene (problem 11.45).
496 Chapter 10 Radical Reactions
Problems
Note to Instructors: Many of the homework problems are available for assignment via WileyPLUS, an onlineteaching and learning solution.
RADICAL MECHANISMS AND PROPERTIES
10.20 Write a mechanism for the following radical halogenation reaction.
10.21 Explain the relative distribution of products below using reaction energy diagrams for the hydrogen abstractionstep that leads to each product. (The rate-determining step in radical halogenation is the hydrogen abstraction step.)In energy diagrams for the two pathways, show the relative energies of the transition states and of the alkyl radi-cal intermediate that results in each case.
10.22 Which of the following compounds can be prepared by radical halogenation with little complication by formationof isomeric by-products?
10.23 The radical reaction of propane with chlorine yields (in addition to more highly halogenated compounds) 1-chloro-propane and 2-chloropropane.
Write chain-initiating and chain-propagating steps showing how each of the products above is formed.
10.24 In addition to more highly chlorinated products, chlorination of butane yields a mixture of compounds with theformula C4H9Cl.
C4H9ClCl2hn
(multipleisomers)
Cl
ClCl2hn
!
Cl
Br
Br Br
Cl
I
BrBr
(92%) (8%)
!Br2
hn
BrBr2
hn
Key Terms and ConceptsThe key terms and concepts that are highlighted in bold, blue text within the chapter aredefined in the glossary (at the back of the book) and have hyperlinked definitions in theaccompanying WileyPLUS course (www.wileyplus.com).
solom_c10_459-501hr.qxd 28-09-2009 15:09 Page 496
Problems 497
(a) Taking stereochemistry into account, how many different isomers with the formula C4H9Cl would you expect?
(b) If the mixture of C4H9Cl isomers were subjected to fractional distillation (or gas chromatography), how manyfractions (or peaks) would you expect?
(c) Which fractions would be optically inactive?
(d) Which fractions could theoretically be resolved into enantiomers?
(e) Predict features in the 1H and 13C DEPT NMR spectra for each that would differentiate among the isomersseparated by distillation or GC.
(f) How could fragmentation in their mass spectra be used to differentiate the isomers?
10.25 Chlorination of (R)-2-chlorobutane yields a mixture of dichloro isomers.
(a) Taking into account stereochemistry, how many different isomers would you expect? Write their structures.(b) How many fractions would be obtained upon fractional distillation?(c) Which of these fractions would be optically active?
10.26 Peroxides are often used to initiate radical chain reactions such as in the following radical halogenation.
(a) Using bond dissociation energies in Table 10.1, explain why peroxides are especially effective as radicalinitiators.
(b) Write a mechanism for the reaction above showing how it could be initiated by di-tert-butyl peroxide.
10.27 List in order of decreasing stability all of the radicals that can be obtained by abstraction of a hydrogen atom from2-methylbutane.
10.28 The relative stability of a series of primary, secondary, and tertiary alkyl radicals can be compared using R CH3 car-bon–carbon bond dissociation energies instead of R H bond dissociation energies (the method used in Section 10.2B).Bond dissociation energies (DH°) needed to make such a comparison for various R CH3 species can be calculatedfrom values for the heat of formation (Hf) of radicals R·, CH3·, and the molecule R CH3 using the following equa-tion: DH°[R R ] Hf[R·] Hf[CH3·] Hf[R CH3]. Using the data below, calculate the R CH3 bond disso-ciation energies for the examples given, and from your results compare the relative stabilities of the respective primary,secondary, and tertiary radicals in this series.
Chemical Species Hf (Heat of Formation, kJ mol!1)
CH3CH2CH2CH2 CH3 !146.8CH3CH2CH(CH3) CH3 !153.7(CH3)3C CH3 !167.9CH3CH2CH2CH2· 80.9CH3CH2CH(CH3)· 69(CH3)3C· 48CH3· 147
9
9
9
99!"#$999
99
Br
O O
Br2
(Di-tert-butylperoxide)
" HBr
Cl
C4H8Cl2Cl2hn
(multipleisomers)
solom_c10_459-501hr.qxd 28-09-2009 15:09 Page 497
Openmirrors.com
13)
14) Dentre os compostos a seguir, preveja quais são quirais, aquirais, meso. Quando indicado, forneça a nomenclatura (R) e (S) de cada centro de quiralidade.
25) Qual a relação de cada um dos pares de compostos? Forneça configuração (R) e (S) quando for o caso.
Challenge Problems 499
Challenge Problems
10.34 The following reaction is the first step in the industrial synthesis of acetone and phenol (C6H5OH). AIBN (2,2!-azobisisobutyronitrile) initiates radical reactions by breaking down to form two isobutyronitrile radicals andnitrogen gas. Using an isobutyronitrile radical to initiate the reaction, write a mechanism for the following process.
10.35 In the radical chlorination of 2,2-dimethylhexane, chlorine substitution occurs much more rapidly at C5 than it doesat a typical secondary carbon (e.g., C2 in butane). Consider the mechanism of radical polymerization and then sug-gest an explanation for the enhanced rate of substitution at C5 in 2,2-dimethylhexane.
10.36 Write a mechanism for the following reaction.
10.37 Hydrogen peroxide and ferrous sulfate react to produce hydroxyl radical (HO·), as reported in 1894 by Englishchemist H. J. H. Fenton. When tert-butyl alcohol is treated with HO· generated this way, it affords a crystallinereaction product X, mp 92°, which has these spectral properties:
MS: heaviest mass peak is at m/z 131IR: 3620, 3350 (broad), 2980, 2940, 1385, 1370 cm"1
1H NMR: sharp singlets at d 1.22, 1.58, and 2.95 (6:2:1 area ratio)13C NMR: d 28 (CH3), 35 (CH2), 68 (C)
Draw the structure of X and write a mechanism for its formation.
10.38 The halogen atom of an alkyl halide can be replaced by the hydrogen atom bonded to tin in tributyltinhydride (Bu3SnH). The process, called dehalogenation, is a radical reaction, and it can be initiated by AIBN (2,2!-azobisisobutyronitrile). AIBN decomposes to form nitrogen gas and two isobutyronitrile radicals, which ini-tiate the reaction. Write a mechanism for the reaction.
10.39 Write a mechanism that accounts for the following reaction. Note that the hydrogen atom bonded to tin in trib-utyltin hydride is readily transferred in radical mechanisms.
10.40 Molecular orbital calculations can be used to model the location of electron density from unpaired electrons ina radical. Open the molecular models on the book’s website for the methyl, ethyl, and tert-butyl radicals. Thegray wire mesh surfaces in these models represent volumes enclosing electron density from unpaired electrons.What do you notice about the distribution of unpaired electron density in the ethyl radical and tert-butyl radi-cal, as compared to the methyl radical? What bearing does this have on the relative stabilities of the radicals inthis series?
(PhCO2)2, Bu3SnH+
Br
(Major)
AIBN+ Bu3SnHI
CN
CNN
N
AIBN
(PhCO2)2, heatH
O+ CO
O2, AIBN
OOH
CN
CNN
N
AIBN
solom_c10_459-501hr.qxd 28-09-2009 15:09 Page 499
Openmirrors.com
7.24 In each of the following pairs of compounds one is chiral and the other is achiral. Identifyeach compound as chiral or achiral, as appropriate.
(a)
(b)
(c)
(d)
7.25 Compare 2,3-pentanediol and 2,4-pentanediol with respect to the number of stereoisomerspossible for each constitution. Which stereoisomers are chiral? Which are achiral?
7.26 In 1996, it was determined that the absolute configuration of (!)-bromochlorofluoromethaneis R. Which of the following is (are) (!)-BrClFCH?
7.27 Specify the configuration at R or S in each of the following.
(a) (!)-2-Octanol
(b) Monosodium L-glutamate (only this stereoisomer is of any value as a flavor-enhancing agent)
H3N"
H
CO2!
CH2CH2CO2! Na"
Cl H
F
Br
C
Br
H
F
Cl C
Br
H
F ClH
ClF
Br
Cl
and
Cl
H
H2N
H
H
NH2
CH3
CH3
CH3
H
H NH2
NH2
CH3
and
CH3CH CHCH2Br and CH3CHCH CH2
Br
OH
ClCH2CHCH2OH and
Cl
HOCH2CHCH2OH
294 CHAPTER SEVEN Stereochemistry
7.28 A subrule of the Cahn–Ingold–Prelog system specifies that higher mass number takes prece-dence over lower when distinguishing between isotopes.
(a) Determine the absolute configurations of the reactant and product in the biologicaloxidation of isotopically labeled ethane described in Section 7.2.
(b) Because OH becomes bonded to carbon at the same side from which H is lost, theoxidation proceeds with retention of configuration (Section 6.13). Compare this factwith the R and S configurations you determined in part (a) and reconcile any appar-ent conflicts.
7.29 Identify the relationship in each of the following pairs. Do the drawings represent consti-tutional isomers or stereoisomers, or are they just different ways of drawing the same compound?If they are stereoisomers, are they enantiomers or diastereomers? (Molecular models may proveuseful in this problem.)
(a)
(b)
(c)
(e)
(f)
H3C H
H Cl
and
andH OH
CH2OH
CH2OH
HO H
CH2OH
CH2OH
Br H
CH3
CH2CH3
and
C
CH3
CH3CH2
H
Br and C
H
Br
H3C
CH2CH3
C
CH3
CH3CH2
H
Br C
Br
CH3CH2
H
CH3and
C
CH3
HO
H
CH2Br C
H
Br
H3C
CH2OHand
C
T
H
D
CH3 C
T
HO
D
CH3biological oxidation
Problems 295
(d)
(g)
(h)
(i)
( j)
(k)
(l)
(m)
(n)
(o)
(p) andH
H
Br
CO2H
CH3
Br
HBr
CH3
Br
CO2H
H
andBr H
CH3
CO2H
H BrH
H
Br
CO2H
CH3
Br
H
H
Br
CO2H
CH3
H Br
CH3
Brand
Br
CO2H
H
andCH3
CH3
H
HH
HH3C
CH3
CH3
H
OH
HO
HCH2OH
and
OH
CH2OH
HH3C
HO H
andCH2OHHO CH2OH
HO
andCH2OHHO CH2OH
HO
andHHO OHH
andHHO
OHH
andHHO OHH
296 CHAPTER SEVEN Stereochemistry