Chapter 7comsizo.com.br/resolucoes/OrganicChemistryChapter7.pdf · Chapter 7 7.1 One problem with...
Transcript of Chapter 7comsizo.com.br/resolucoes/OrganicChemistryChapter7.pdf · Chapter 7 7.1 One problem with...
Chapter 7 7.1 One problem with elimination reactions is that mixtures of products are often formed. For example,
treatment of 2-bromo-2-methylbutane with KOH in ethanol yields a mixture of two alkene products. What are their likely structures?
Solution:
Br
major product
7.2 How many alkene products, including E,Z isomers, might be obtained by dehydration of 3-methyl-3-hexanol (3-methyl-3-hydroxyhexane) with aqueous sulfuric acid?
Solution:
OH
dehydration
7.3 What product would you expect to obtain from addition of Cl2 to 1,2-dimethylcyclohexene? Show the stereochemistry of the product.
Solution:
H3C CH3
Cl Cl
H3C CH3
Cl
Cl
H3C
CH3Cl
Cl
H3C CH3
Cl
Cl
H3C
CH3Cl
Cl
7.4 Addition of HCl to 1,2-dimethylcyclohexene yields a mixture of two products. Show the
stereochemistry of each, and explain why a mixture is formed. Solution:
H3C CH3
H Cl
H3C
CH3
Cl
H3C CH3
Cl
H3C CH3
ClH3C Cl
CH3
7.5 What product would you expect from the reaction of cyclopentene with NBS and water? Show the
stereochemistry. Solution:
Br Br
Br
HO
H
OH
Br
H
H
H
H
OH
Br
Br
OH
H
H
H
H
Br
OH
7.6 When an unsymmetrical alkene such as propene is treated with NBS in aqueous DMSO, the major
product has the bromine atom bonded to the less highly substituted carbon atom. Is this Markovnikov or non-Markovnikov orientation? Explain.
H3C CH
CH2NBS,H2ODMSO H3C C
HCH2Br
OH
Solution:
CH2
BrH
H3CCH2
BrH
H3C
H2OH2O
vs
The product is Markovnikoff product. 7.7 What products would you expect from oxymercuration of the following alkenes? Solution: (a)
H3CH2C
H2C C
HCH2
Hg(OAc)2
H2OH3C
H2C
H2C
HC CH3
OH2-pentanol
(b)
H3C C CH
H2C CH3
CH3
Hg(OAc)2
H2OH3C C
H2C
H2C CH3
CH3
OH
2-methyl-2-pentanol
7.8 What alkanes might the following alcohols have been prepared from? Solution: (a)
H3C C CH
H2C
H2C CH3
CH3
H2C CH2C
H2C
H2C CH3
CH3
2-methyl-1-hexene
2-methyl-2-hexene
Hg(OAc)2
H2O
Hg(OAc)2
H2O
H3C CH2C
H2C
H2C CH3
OH
OH
(b)
Hg(OAc)2
H2O
1.BH3
OH-2.H2O2
OH
cyclohexylethylene
ethylidenecyclohexane
7.9 Show the structures of the products you would obtain by hydroboration/oxidation of the following alkenes:
(a)
CH3C CHCH2CH3
CH3 (b) CH3
Solution: (a)
CH3CH CHCH2CH3
CH3 OH
(b)
CH3
OH
7.10 What alkenes might be used to prepare the following alcohols by hydroboration/oxidation? (a) (CH3)2CHCH2CH2OH
(b)
(CH3)2CHCHCH3
OH
(c) CH2OH
Solution: (a) (CH3)2CHCH CH2
(b) (CH3)2C CHCH3
(c) CH2
7.11 The following cycloalkene gives a mixture of two alcohols on hydroborotion followed by
oxidation, Draw the structures of both, and explain the result.
H3C
HCH3
H
Solution: The structure of the two cycloalcohols:
H3C
HCH3
H
1. BH3 / THF
2. OH-, H2O
H3C
H
CH3
H
H3C
HCH3
H
H
OH
OH
H
7.12 What products would you expect from the following reactions?
(a)
CH2
C Cl
Cl
Cl
H
KOH?
(b) (H3C)2HCH2CHC CHCH3 CH2I2
Zn(Cu)?
Solution:
(a)
CH2
C ClCl
Cl
H
KOH CCl Cl
HCl
(b) (H3C)2HCH2CHC CHCH3 CH2I2
Zn(Cu)ZnI2
7.13 What product would you obtain from calalytic hydrogenation of the following alkenes?
(a) H3C C
CH3
CH
H2C CH3
(b)CH3
CH3 Solution:
(a) H3C CH
CH3
H2C
H2C CH3
(b)CH3
CH3 7.14 What alkene would you start with to prepare each of the following compounds?
(a)OH
OH
H
CH3
(b)H3CH2C C
H
OH
C
OH
CH3
CH3 (c) HOH2C C
H
OH
CH
OH
H2C OH
S
olution:
(a)
H
CH3
(b)H3CH2C C
HC
CH3
CH3 (c) H2C CH
CH
CH2
7.15 What products would you expect? Solution:
(a) Aqueous acidic KMnO4 Product CH3COCH2CH2CH2CH2COOH
(b) O3
Product CH3COCH2CH2CH2CH2CHO 7.16 Propose structure for alkenes. Solution:
(a): (CH3) 2C=O and H2C=O Structure (CH3) 2C =CH2 (b): 2 equiv CH3CH2CH=O Structure CH3CH2CH=CHCH2CH3
7.17 Show the monomer units you would use to prepare the following polymers:
(a)
H2C C
H
OCH3
H2C C
H
OCH3
H2C C
H
OCH3
Solution: The monomer unit is H2C CH
OCH3
.
(b)
CH
CH
CH
CH
CH
CH
Cl Cl Cl Cl Cl Cl
Solution: The monomer unit is ClHC CHCl. 7.18 One of the chain-termination steps that sometimes occurs to interrupt polymerization is the
following reaction between two radicals:
CH2CH22 CH2CH3 + CH
CH2
Propose a mechanism for this reaction, using fishhook arrows to indicate electron flow. Solution:
HC
CH2CH3+CH
CH2
H2C CH2
H
CH2
7.19 tert-Butyl vinyl ether is polymerized commercially for use in adhesives by a cationic process.
Draw a segment of poly(tert-butyl vinyl ether), and show the mechanism of the chain-carrying step.
O
tert-butyl vinyl ether
Solution:
OH
Acid catalyst
O
O
O O
Repeat many times O
CH
H2C* *
n 7.20 Name the following alkenes, and predict the products of their reaction with (i) KMnO4 in aqueous
acid and (ii) O3, following by Zn in acetic acid:
(a) (b) Solution: (a)2,5-dimethyl-2-heptene
(i)
O
OH
O
and
(ii)
O
H
O
and
(b)3,3-dimethylcyclopentene
(i)
OH
O
HO O
(ii)
H
O
H O
7.21 Draw the structures of alkenes that would yield the following alcohols on hydration (red=O). Tell
in each case whether you would use hydroboration/oxidation.
Solution: (a) CH2
I would use oxymercuration in this case.
(b)
Both oxymercuration / demercuration and hyboration / oxidation reactions could be used. 7.22 The following alkene undergoes hydroboration/oxidation to yiele a single product rather than a
mixture. Explain the result, and draw the product showing its stereochemistry. Solution: The reactant is symmetrical, so OH adding to either of the carbon that is double bonded
will yield the same product. The single product is: OH
7.23 Predict the products of the following reactions (the aromatic ring is unreactive in all cases).
Indicate regiochemistry when relevant.
CH CH2
(a) H2/Pd
(b) Br2
(c) HBr
(d)1.OsO42.NaHSO4
(e) D2/Pd
Solution: (a)
CH CH2H2/Pd
CH
H CH
HH
(b)
CH CH2Br2
C CH
BrH
HBr
(c)
CH CH2HBr
C CH
BrHH
H
(d)
CH CH21.OsO42.NaHSO4
C CH
HO OHH H
(e)
CH CH2D2/Pd
CD
HC
D
H
H
7.24 Suggest structures for alkenes that give the following reactions products. There may be more than one answer for some cases.
(a) ? CH3CHCH2CH2CH2CH3
CH3
(b)
? CH3
CH3
(c)
?Br2 CH3CHCHCH2CHCH3
CH3
Br
Br
(d)
?HCl CH3CHCHCH2CH2CH2CH3
CH3
Cl
(e) ?
1.Hg(OAc)2,H2O2.NaBH4
CH3CH2CH2CHCH3
OH
Solution: (a)
CH3CHCH2CH2CH CH2
CH3 H3CHCH2CHC CHCH3
H3C
H3CHCHC CHCH2CH3
H3C
H3CC CHCH2CH2CH3
CH3
H2C CCH2CH2CH2CH3
CH3
(b)
CH3
CH3
CH3
CH3
(c) H3CHC CHCH2CHCH3
CH3
(d)
H2C CHCHCH2CH2CH2CH3
CH3 (e) H3CH2CH2CHC CH2 7.25 Predict the products of the following reactions, indicating both regiochemistry and stereiochemistry when appropriate:
(a)
CH3
H
1. O3
2.Zn, H3O+ ?
Solution:
CH3
H
1. O3
2.Zn, H3O+
O
O
CH3
H
(b)
KMnO4H3O+ ?
Solution:
KMnO4
H3O+
O
OH
O
OH
(c)
CH3
1. BH32. H2O2,- OH
?
Solution:
CH3
1. BH32. H2O2,- OH
H CH3
OH H
+
CH3 H
OHH
50% 50%
(d)
CH3
1. Hg(OAc)2,H2O2. NaBH4
?
Solution:
CH3
1. Hg(OAc)2,H2O2. NaBH4
CH3
OHH
H
H
H OH
CH3
50% 50% 7.26 How would you carry out the following transformations? Indicate the reagents you would use in each case.
(a)
?
H
H
OH
OH
Solution:
H
H
OH
OH
1.OsO4
2.NaHSO3,H2O
(b)
?
OH
Solution:
OH
1. Hg(Ac)22. NaBH4
(c)
?
H
H
Cl
Cl
Solution:
H
H
Cl
Cl
1. CHCl3
2. KOH
(d)
CH3
OH ?
CH3
Solution:
CH3
OH
CH3
1. H2SO4, H2O2. THF, 50 ℃
(e)
H3CHC CHCHCH3
CH3
? CH3CH
O
+ CH3CHCH
CH3
O
Solution:
H3CHC CHCHCH3
CH3
CH3CH
O
+ CH3CHCH
CH3
O
1. O32. Zn/H3O+
(f) H3CC CH2
CH3
?CH3CHCH2OH
CH3
Solution: H3CC CH2
CH3
CH3CHCH2OH
CH3
1. BH3, THF2. H2O2,OH-
7.27 What product will result from hydroboration/oxidation of 1-methylcyclopentene with deuterated
borane, BD3? Show both the stereochemistry (spatial arrangement) and the regiochemistry (orientation) of the product.
H CH3
BD3
H CH3
D2B D
H CH3
BD
D
D
δ
δ
-OHH2O2
H CH3
HO D
H CH3
D2B D
H CH3
BD
D
D
δδ
-OHH2O2
H CH3
HO D
7.28 Draw the structure of an alkene that yield only acetone, (CH3)2C=O, on ozonolysis followed by
treatment with Zn.
OO3
Zn/H3O+ O+
7.29 Draw the structure of a hydrocarbon that react with 1 molar equivalent of H2 on catalytic hydrogenation and gives only pentanal , CH3CH2CH2CH2CHO, on the ozonolysis followed by treatment with Zn. Write the reaction involved.
Solution: The structure of the hydrocarbon is showed in the following picture:
The reactions are showed as follow:
Catalyst
H2
Zn/H3O
O3
O
H2
7.30 Draw the structure of alkenes that give the following products on oxidative cleavage with KMnO4
in the acidic solution:
(a).
O
CO2+ (b).
O
+O
(c).
O +
O
Solution:
(a). (b).
(c).
7.31 Compound A has the formula C10H16. On catalytic hydrogenation over palladium, it reacts with
only 1 molar equivalent of H2. Compound A also undergoes reaction with ozone, followed by zinc treatment, to yield a symmetrical diketone, B (C10H16O2).
(a) How many rings does A have? A have two rings.
(b) What are the structures of A and B?
O
O
A
B
(c) Write the reactions.
+H2palladium
O
O
+O3Zn
7.32 An unknown hydrocarbon A, with formula C6H12, reacts with 1molar equivalent of H2 over a
palladium catalyst. Hydrocarbon A also reacts with OsO4 to give a diol, B. When oxidized with KMnO4 in acidic solution, A gives two fragments. One fragment is propanoic acid, CH3CH2CO2H, and the other fragment is a ketone. C. What are the structures of A, B, and C? Write all reactions, and show your reasoning.
A
CH3
CH3HHO OH
H2C
B H3C
C
O
O
+ H2palladium
CH3
CH3HHO OH
H2C
H3COsO4
KMnO4 + H3CCH2
CO
H
O
7.33 Using an oxidative cleavage reaction explain how you would distinguish between the following
two isomeric dienes:
and Solution: When they react with KMnO4 in acid, the first one will yield two products and the second one
will only yield one product. 7.34 Compound A, C10H18O, undergoes reaction with dilute H2SO4 at 2500C to yield a mixture of two
alkenes, C10H16. The major alkene product, B, gives only cyclopentanone after ozone treatment followed by reduction with zine in acetic acid. Identify A and B, and write the reaction.
Solution: A: B:
HO
Reaction:
HOH2SO4
2500C
major
O3
Zn/H3OO
7.35 Which reaction would you expect to be faster, addition of HBr to cyclohexene or to
1-methylcyclohexene? Explain. Solution: The intermediates of the two reactions are not equal in stability. Judge by Hammond postulate,
the more stable intermediate forms faster, so addition of HBr to 1-methylcyclohexene may be faster.
7.36 Predict the products of the following reactions, and indicate regiochemistry if relevant:
(a) HBrBr
(c)CH I , Zn-Cu2 2
HH
(d)CH I , Zn-Cu2 2
H
H
7.37 Iodine azide, IN3, adds to alkenes by an electrophilic mechanism similar to that of bromine. If a
monosubstituted alkene such as 1-butene is used, only one product results:
H3CH2CHC CH2 + I N N N CH3CH2CHCH2I
N N N
(a) Add lone-pair electrons to the structure shown for IN3, and draw a second resonance form for the
molecule.
I N N I NN N N
(b) Calculate formal charges for the atoms in both resonance structures you drew for IN3 in part
For the iodine: Iodine valence electrons = 7 Iodine bonding electrons =2 Iodine nonbonding electrons =6 Formal charge = 7 - 2/2 – 6 = 0
For the left nitrogen: Nitrogen valence electrons =5 Nitrogen bonding electrons =6 Nitrogen nonbonding electrons =2 Formal charge = 5 – 6/2 – 2 = 0 For the middle nitrogen: Nitrogen valence electrons =5 Nitrogen bonding electrons =8
Nitrogen nonbonding electrons =0 Formal charge = 5 – 8/2 – 0= 1 For the right nitrogen: Nitrogen valence electrons =5 Nitrogen bonding electrons =4 Nitrogen nonbonding electrons =4 Formal charge = 5 – 4/2 – 4= - 1 (c) In light of the result observed when IN3 adds to 1-butebe, what is the polarity of the I-N3bomd?
Propose a mechanism for the reaction using curved arrows to show the electron flow in each step.
H3CH2CHC CH2 + I N N N
H3CH2CHC CH2I N N NCH3CH2CHCH2I
N N N
7.38 Draw the structure of a hydrocarbon that absorbs 2 molar equivalents of H2 on catalytic hydrogenation and gives only butanedial on ozonolysis. O
HCH2C
H2C CH
O
Butanedial
Solution:
7.39 Simmons–Smith reaction of cyclohexene with diiodomethane gives a single cycloproprane
product, but the analogous reaction of cyclohexene with 1,1-diiodoethane gives (in low yield) a mixture of two isomeric methylcyclopropane products. What are the two products, and how do they differ?
Solution: The reaction occurs as below:
+ I2HC CH3Zn(Cu)Ether
H CH3H3C H
+
7.40 In planning the synthesis of one compound from another, it’s just as important to know what not to
do as to know what to do. The reactions all have serious drawbacks to them. Explain the potential problems of each.
Solution: (a)
H3C C
CH3CHCH3
HI H3C CH
CH3CHCH3I
According to Markovnikov’s rule the reaction should be:
H3C C
CH3CHCH3
HI H3C CCH3
CH2CH3I
(b)
1.OsO42.NaHSO3
OH
H
H
OH
Actually it should be syn addition as below:
1.OsO42.NaHSO3
OH
OH
H
H
(c)
1.O32.Zn CHO
CHO
Actually the reaction should be:
1.O32.Zn
H2C CHOOHC2
(d)
CH31.BH32.H2O2, OH
CH3
OHH
H
Actually the reaction should be:
CH31.BH32.H2O2, OH
CH3
HOH
H
7.41 Which of the following alcohols could be made selectively by hydroboration/oxidation of an alkene? Explain.
(a) CH3CH2CH2CHCH3
OH
(b) (CH3)2CHC(CH3)2
OH
(c)
CH3
OH
H
H (d)
CH3
H
OH
H
Solution: (a) No selectivity (b) Selectively synthesized (c)Can’t be formed
(d) Can’t be formed 7.42 What alkenes might be used to prepare the following cyclopropane?
(a) CH(CH3)2
(b)
Cl
Cl
Solution:
(a) CH(CH3)2+ (CH3)2CHCHI2
Zn(Cu)
Ether
(b)
Cl
Cl
+ CHCl3OH
7.43 Predict the products of the following reactions. Don’t worry about the size of the molecule;
concentrate on the functional groups.
HO
CH3
CH3
Br2
HBr
1. OsO4
2. NaHSO3
1. BH3, THF2. H2O2, OH-
CH2I2, Zn (Cu)
Solution:
a) HO
CH3
CH3
Br2
HO
CH3
CH3
BrBr
b) HO
CH3
CH3
HBr
HO
CH3
CH3
Br
c) HO
CH3
CH3
OsO4
NaHSO3
HO
CH3
CH3
OHOH
d) HO
CH3
CH3
BH3,THFH2O2,OH
HO
CH3
CH3
OH
e) HO
CH3
CH3
CH2I2Zn(Cu)
HO
CH3
CH3
7.44 The sex attractant of the common housefly is a hydrocarbon with the formula C23H46. On
treatment with aqueous acidic KMnO4, two products are obtained, CH3(CH2)12CO2H and CH3(CH2)7CO2H.Propose a structure.
Solution:
H3CH2C
HC
12CH
H2C CH3
7 should be the right structure.
7.45 Compound A has the formula C8H8. It reacts rapidly with KMnO4 to give CO2 and carboxylic
acid, B (C7H6O2), but reacts with only 1molar equivalent of H2 on catalytic hydrogenation over a palladium catalyst. On hydrogenation under conditions that reduce aromatic rings, 4 equivalents of H2 are taken up and hydrocarbon C (C8H16) is produced. What are the structure of A, B, and C? Write the reactions.
SOLUTION:
A CC
H
H
H
B COH
O
CH2C
CH3
The reactions:
CC
H
H
H
H2C
CH3
KMnO4C
OH
O
CO2
CC
H
H
H
H2
palladium catalyst
H2C
CH3
CC
H
H
H
4mol H2
1mol
7.46. Plexiglas, a clear plastic used to make many molded articles, is made by polymerization of
methyl methacrylate. Draw a representative segment of Plexiglas.
H2C C
CH3
C OCH3
O
Methyl methacrylate
SOLUTION:
The segment of Plexiglas:
H2C C
C
CH3
O
OCH3
7.47 Poly (vinyl pyrrolidone), prepared from N-vinylpyrrolidone, is used both in cosmetics and as a synthetic blood substitute. Draw a representative segment of the polymer.
N
O
N-vinylpyrrolidone Solution:
HC
H2C **
N
O
7.48 Reaction of 2-methylpropene with CH3OH in the presence of H2SO4 catalyst yields methyl
tert-butyl ether, CH3OC(CH3)3, by a mechanism analogous to that of acid-catalyst alkene hydration. Write the mechanism, using curved arrows for each step.
Solution:
CH2CH3C
CH3
+ CH3OHH HSO4
CH3CH3C
CH3
+ CH3OH
CH3CH3C
CH3
OH3C H
HSO4
CH3CH3C
CH3
OH3C
+ H2SO4
7.49 When 4-penten-1-ol is treated with aqueous Br2, a cyclic bromo ether is formed, rather than the
expected bromohydrin. Propose a mechanism, using curved arrows to show electron movement. Solution:
O
O CH2Br
H HO
O
Br
H
Br
7.50 How would you distinguish between the following pairs of compounds using simple chemical
tests? Tell what you would do and what you would see. (a) cyclopentene and cyclopentane (b) 2-hexene and benzene Solution: Bromine in carbon tetrachloride solution is reddish-brown color. Using this one to identify the double bond by electrophilic addition reaction. 7.51
Cl
CCl
Cl
C O-Na+
O Cl
CCl
Cl
C O
O- Cl
C-Cl
Cl
+O C O
Cl
C
Cl
Cl-+ +CO2+Na+
They have the same intermediate which is CCl3-.
7.52 (a) 3 (b) double bonds:2
(c) 7.53 Evidence that cleavage of 1,2-diols by HIO4 occurs through a five membered cyclic periodate
intermediate is based on kinetic data – the measurement of reaction rates. When diols A and B were prepared and the rates of their reaction with HIO4 were measured, it was found that diol A cleaved approximately 1 million times faster than diol B. Make molecular models of A and B and of potential cyclic periodate intermediates, and then explain the kinetic results.
OH
H
H
HOH
H
OH
OH
A Bcis diol trans diol
Solution:
A:
OH
H
H
HOI OH
OO
HIO4
O
OO
O
In cis diol, two –OH are in one plane, so it is easier to form a cyclic periodate inetermediate, while in trans diol, the two –OH aren’t in one plane, there is larger steric strain in the five membered cyclic periodate intermediate, it is less stable, so the rates of A is more faster.
7.54 Reaction of HBr with 3-methylcyclohexene yields a mixture of four products: cis- and
trans-1-3-methylbromocyclohexane and cis- and trans-1-2-methylbromocyclohexane. The analogous reaction of HBr with 3-bromocyclohexane trans-1, 2-dibromocyclohe- xane as the sole product. Draw structures of the possible intermediates, and then Explain why only a single product is formed in the reaction of HBr with 3-bromocyclohexane.
Solution:
Br
H Br
BrBr
Br
Br
H
H
Br
7.55 The following reaction takes place in high yield:
CO2CH3
Hg(OAc)2
CO2CH3
HgAcO
Use your general knowledge of alkene chemistry to propose a mechanism, even though you’ve never seen this reaction before.
Solution:
CO2CH3
+AcO Hg
OAc
CO2CH3
Hg
AcO
CO2CH3
HgAcO
CO2CH3
HgAcO
OAc
OAcH
CO2CH3
HgAcO
7.56 Reaction of cyclohexene with mercury(II) acetate in CH3OH rather than H2O, followed by
treatment with NaBH4, yields cyclohexyl methyl ether rather than cyclohexanol. Suggest a mechanism.
1.Hg(OAc)2, CH3OH
2.NaBH4OCH3
Solution:
OCH3
Hg OAc
OAcHg
OAc
HO CH3
HgOAc
O
H
CH3
OAc
HgOAc
O CH3
NaBH4
7.57 Hydroboration of 2-methyl-2-pentene at 25℃ followed by oxidation with alkaline H2O2 yields 2-methyl-3-pentano, but hydroboration at 160℃ followed by oxidation yields 4-methyl-1-pentanol. Explain.
1.BH3,THF,25℃
2.H2O2,OH-CH3CHCHCH2CH3
CH3
OH
2-Methyl-3-pentanol
1.BH3,THF,160℃
2.H2O2,OH-CH3CHCH2CH2CH2OH
CH3
4-Methyl-1-pentanol
CH3C CHCH2CH3
CH3
Solution: Case 1:
CH3C CHCH2CH3
CH3
3 + BH3 THF,25℃CH3CH CHCH2CH3
CH3
BH2
3 H2O2,OH-
CH3CHCHCH2CH3
CH3
OH
3
Case 2:
BH2
BH2
BH2
H2O2 / OH-
OH
7.58 Alkynes undergo many of the same reactions that alkenes do. What products would you expect
from each of the following reactions?
H3C CH
H2C
H2C C CH
CH3
(1) with 1 equiv Br2; (2) with 2 equiv H2 ,Pd/C (3)with 1 equiv HBr Solutions: (1)
H3C CH
H2C
H2C C CH
CH3
H3C CH
H2C
H2C C CH
CH31 equiv Br2Br
Br (2)
H3C CH
H2C
H2C C CH
CH3
H3C CH
H2C
H2C
H2C CH3
CH32 equiv H2 ,Pd/C
(3)
H3C CH
H2C
H2C C CH
CH3
H3C CH
H2C
H2C C CH2
CH31 equiv HBr Br
7.59 Explain the observation that hydroxylation of cis-2-butene with OsO4 yields a different product
than hydroxylation of trans -2-butene. First draw the structure and show the stereochemistry of each product, and then make molecular models.
Solution:
H3C CH
CH
CH3
CH3
CHHC
H3C
cis-But-2-ene
H3CCH
HC
CH3
trans-But-2-ene
CH3
CC
H3C
OH OHH H
Butane-2,3-diolmeso
compound
H3CC
CCH3
OH
OH
H H
Butane-2,3-diolenantiomer