Chapter 14
14.1 Allene, H2C=C=CH2, has a heat of hydrogenation of -298KJ/mol (-71.3 kcal/mol). Rank a
conjugated diene, a nonconjugated diene, and an allene in order of stability.
Solution:
Stability
H2C
H2C
C
H
C
H2
C
-298KJ/mol
CH2
C
H
CH2
least stable
-253KJ/mol
C
C
CH2
most stable
-236KJ/mol
H
H
14.2 Give the structures of both 1,2 and 1,4 adducts resulting from reaction of 1 equivalent of HCl with
H2C
1,3-petadiene.
Solution:
1st
H
C
H2C
C
H
C
H
CH
CH3
H
Cl
H3C
H2C
C
H
C
H
CH
CH3
H
Cl
H2C
C
H
CH
CH3
H3C
C
H
C
H
CH
CH3
H
C
CH2
CH3
H2C
C
H
C
H
CH2
CH3
C
H
C
H
H
C
CH
CH3
CH2
CH3
2nd
H3C
H2C
H
C
C
H
Therefore, the adducts are:
H3C
H
C
C
H
C
H
CH3
4-Chloro-pent-2-ene
CH3
1-Chloro-pent-2-ene
Cl
H2C
C
H
C
H
H2
C
Cl
H2C
C
H
H
C
H2
C
CH3
3-Chloro-pent-1-ene
Cl
14.3 Look at the possible carbocation intermediates produced during addition of HCl to 1,3-pentadiene
(Problem 14.2),and predict which 1,2 adduct predominates. Which 1,4 addition adduct
predominates?
Solutions
Cl
is the predominate adduct
14.4 Give the structure of both 1,2 and 1,4 adduct resulting from reaction of 1 equivalent of HBr with
the following substance?
Solutions:
The reaction carries out as follows:
CH3
CH3
CH3
CH3
CH3
CH3
CH3
Br
CH3
CH3
CH3
Br
Red stands for Br atom, green stands for hydrogen atoms and black stands for carbon atoms.
is
CH3
Br
CH3
the
structure
of
is
the
structure
of
CH3
CH3
Br
14.5 The 1,2 adduct and the 1,4 adduct formed by reaction of HBr with 1,3-butadiene are in
equilibrium at 40℃. Propose a mechanism by which the interconversion of products takes place.
(See Section 11.6)
Solution:
HBr
H2C
C
H
C
H
CH2
H
C
H3C
C
H
+
CH2
H2C
1,4 adduct
1,2 adduct
H
C
C
H
CH2
H3C
C
H
CH3
+H2C
H+
C
C
H
Br
Br
H3C
C
H
C
H
CH2 +
C
H
CH3
BrH
Br
H2C
C
H
C
H
+
Br
Br
14.6 Why do you suppose 1,4 adducts of 1,3-butadiene are generally more stable than 1,2 adducts?
Solution:
According to Zaitsev’s rule, the intermediate of the 1,4 adduct is more stable than the 1,2 adduct, so 1,4
adducts of 1,3-butadiene are generally more stable than 1,2 adducts.
14.7 Predict the product of the following Diels-Alder reaction:
CH3
O
C
H
+
C
OCH3
C
H3C
H
Solution:
O
O
C
H
+
H
C
C
OCH3
OCH3
C
H3C
H
H
CH3
14.8 Which of the following alkenes would you expect to be good Diels-Alder dienophiles?
O
O
(a)H2C
CHCCl
(b)H2C
CHCH2CH2COCH3
O
(c)
O
(e)
(d)
Solution:
(a),(d)
14.9 Which of the following dienes have an s-cis conformation, and which have an s-trans
conformation? Of the s-trans dienes, which can readily rotate to s-cis?
(a)
(b)
(c)
Solution:
(a) has an s-cis conformation, (b) and (c) have an s-trans conformation, and (c) rotate to s-cis.
14.10 Predict the product of the following Diels-Alder reaction:
+
→?
Solution:
The product is:
CH3
H
H
H
14.11
CO2CH3
CO2CH3
Draw a segment of the polymer that might be prepared from 2-phenyl-1, 3-butadiene.
C6H5
*
*
n
14.12
Show the mechanism of the acid-catalyzed polymerization of 1, 3-butadiene.
H
The progress goes on and product polymers.
14.13
Calculate the energy range of electromagnetic radiation in the UV region of the spectrum
from 200 to 400 nm. Recall the equation.
E=NAhc/λ=(1.20*10-4kJ/mol)/λ（m）
Solution: E1=1.20*10-4 /200(nm)=600kJ
E2=1.20*10-4 /400(nm)=300kJ
The energy range of UV is 300~600kJ/mol
14.14
How does the energy you calculated in Problem 14.13 for UV radiation compare with the
values calculated previously for IR and NMR spectroscopy?
Solution: The energy range of IR is 4.8~48kJ/mol;
NMR required about 8.0*10-5kJ/mol.
So the energy for UV is greater than IR and NMR.
14.15 A knowledge of molar absorptivities is particularly important in biochemistry where UV
spectroscopy can provide an extremely sensitive method of analysis. For example, imagine that
you wanted to determine the concentration of vitamin A in a sample. If pure vitamin A has
λmax=325(ε= 50,000), what is the vitamin A concentration in a sample whose absorbance at 325
nm is A=0.735 in a cell with a path length of 1.00 cm?
Solution: C=A/(ε*l)=1.47×10-5 mol/L
14.16 Which of the following compounds would you expect to show ultraviolet absorption in the 200
to 400 nm range?
(a) 1,4-cyclohexadiene
(b) 1,3-cyclohexadiene
(c)
H 2C
C
H
C
N
CH3
CO 2H
(d)
OCOCH3
(e) O
(f)
N
H
Solution: b, c, d, e, f
14.17 Write the structure of all possible adducts of the following diene with 1 equivalent of HCl.
Solution:
All possible adducts as follow:
Cl
Cl
H
H
H
Cl
H
Cl
14.18 Write the product of the Diels-Alder reaction of the following diene with 3-buten-2-one,
H2C=CHCOCH3. Make sure you show the full stereochemistry of the reaction product.
Solution:
H
H
O
H
H3C
14.19 The following diene does not undergo Diels-Alder reactions. Explain.
Solution: Because the ends of the diene partner are too far apart and steric strain between the two
methyl group prevents the molecule from adopting s-cis geometry.
CH3
CH3
14.20 Which of the following molecules are conjugated? Circle the conjugated part in each.
not conjugated
(a)
H2 C
conjugated
(b) H2 C
(c)
H2C
C
H
C
N
conjugated
O
O
C
O
C
CH3
(d)
O
CH3
conjugated
CH3
C
CH2
(e)
not conjugated
H
H
C
C
CH2
conjugated
(f)
14.21 Give IUPAC names for the following compounds:
CH3
(a) CH3CH
CCH
CH2
CHCH3
0
(b) H2 C
(c) CH3 CH
CH3 CH
(d)
CHCH
CHCH
C
CHCH
CCH
CHCH3
CHCH3
CH2
CH2CH2CH3
Solution:
(a) 3-methyl-2,4-hexadiene
(b) 1,3,5-heptatriene
(c) 2,3,5-heptatriene
(d) 3-propyl-1,3-pentadiene
14.22 What product(s) would you expect from 1,3-cyclohexadiene with each of the following?
(a) 1 mol Br2 in CH2Cl2
(b) O3 followed by Zn
(c) 1 mol HCl in ether
(d) 1 mol DCl in ether
(e) 3-Buten-2-one(H2C=CHCOCH3)
(f) Excess OsO4, followed by NaHSO3
Solution:
(a)
Br
Br
Br2
+
Br
CH2Cl2
+
Br
(b)
O
O
O3
HC
Zn
(c)
CH
+
CHOCH2CH2CHO
Cl
+ HCl
Ether
(d)
Cl
+ DCl
Cl
D
Ether
+
D
(e)
O
+
H2C
C
CHCOCH3
CH3
(f)
OH
HO
OsO4
NaHSO3
HO
OH
14.23 Draw and name the six possible diene isomers of formula C5H8.Which of the six are conjugated
dienes?
Solution:
1,3-Pentadiene
1,4-pentadiene
2-methyl-1,3-butadiene
2,3-pentadiene
2-methyl-1,3-butadiene
1,2-pentadiene
14.24. Treatment of 3,4-dibromohexane with strong base leads to loss of 2 equivalents of HBr and
formation of a product with formula C6H10. Three products are possible. Name each of the three, and
tell how you would use
13
C NMR spectroscopy to help identify them. How would you use UV
spectroscopy?
Solution:
The three products:
2,4-hexadiene
2,3-hexadiene
3-hexyne
The way to identify:
1.
Use of 13C NMR Spectroscopy:
2,4-Hexadiene has 1 sp3 carbon resonance in the 20 to 50 δ range and 2
sp2-carbon resonances in the 100 to 150 δrange.
2,3-hexadiene has 3 sp3-carbon resonances in the 20 to 50 δ range and 3
sp2-carbon resonances in the 100 to 150 δ range.
3-hexyne has 2 sp3-carbon resonances in the 20 to 50 δ range and one signal
between 65-85 ppm assigned to SP hybridized carbon.
2.
Use of UV spectroscopy:
2,3-hexadiene has a shorter λmax than 2,4-hexadiene and 3-hexyne hasn’t
absorption.
14.25 Electrophilic addition of Br2 to isoprene (2-methy-1.3-butadiene) yields the following product
mixture:
Br2
CH2Br
CH2Br
BrH2C
BrH2C
Br
Br
21%
3%
76%
Of the 1,2 and 1,4-additon products, explain why 3,4-dibromo-3-methyl-1-butene (21%)
predominates over 3,4-dibromo-2-methyl-1-butene (3%).
Solution:
1
4
2
3
For the methyl group is an electron pushing group. So the density of electron on C1
and C2 are bigger the electron density on C3 and C4. So, as an electrophile, the Br2 is more likely to
add to C1 and C2.
14.26 Propose a structure for a conjugated diene that gives the same product from both 1,2 and 1,4
addition of HBr.
Solution:
can be the conjugated diene the give the same product.
14.27 Draw the possible products resulting from addition of 1 equivalent of HCl to
1-phenyl-1,3-butadiene. Which would you expect to predominate, and why?
CH
CH
CH
CH2
1-Phenyl-1,3-butadiene
Solution:
Cl
Cl
Cl
Cl
10.28 2,3-Di-tert-butyl-1,3-butadiene does not undergo Diels-Alder reactions. Explain.
(H3C)3C
CH2
C
C
(H3C)3C
CH2
2,3-Di-tert-butyl-1,3-butadiene
Solution:
Because of the bulk of the tert-butyl, the C-C bond has the s-trans relationship.
This structure forbids the Diels-Alder reactions.
14.29 Diene polymers contain occasional vinyl branches along the chain. How do you think these
branches might arise?
H2C C C CH2
H H
CH2CH CHCH2CH2 CHCH2CH
CHCH2
C CH2
H
A vinyl branch
Solution: The presence of a vinyl branch is because of the occurring of 1,2-addition.
14.30 Tires whose sidewalls are made of natural rubber tend to crack and weather rapidly in areas
around cities where high levels of ozone and other industrial pollutants are found. Explain.
Solution: Because natural rubber is the polymer of isoprene containing carbon double bonds, which can
be oxidized by ozone, causing the polymer to be decomposed, as a result tires tend to crack and
weather rapidly in areas where high levels of ozone are found.
14.31 Would you expect allene, H2C
C
CH2, to show a UV absorption in the 200 to 400 nm
range? Explain.
Solution: No I wouldn’t expect it. Because the molecular is not conjugated π electron system.
14.32: Which of the following compounds would you expect to have a πÆπ* UV absorption in the 200
to 400 nm range?
(c)
(b)
(a)
(H3C)2C
CH2
N
C
O
A ketene
Pyridine
Solution: (b)
14.33 Predict the products of the following Diels-Alder reactions:
CHO
(a)
+
CHO
O
(b)
+
O
O
O
O
O
O
(c)
O
+
O
O
14.34 Show the structure, including stereochemistry, of the product from the following Diels-Alder
Reaction
O
H
C
OCH 3
COOCH3
+
H
H 3CO
H
COOCH3
C
O
14.35 How can you account for the fact that cis-1, 3-pentadiene is much less reactive than trans-1,
3-pentadiene in the Diels-Alder reaction?
Solution: When cis-1,3-pentadiene rotates to the s-cis conformation, a steric interaction occurs between
the methyl group protons and a hydrogen on C1. Since it’s more difficulty for
cis-1,3-pentadiene to assume the s-cis conformation, it is less reactive in D-A reaction.
H
H
H
H
H
H
H
H
CH3
H
cis-1,3-pentadiene
H
H3C
trans-1,3-pentadiene
14.36 Would you expect a conjugated diyne such as 1, 3-butadiyne to undergo Diels-Alder reaction
with a dienophile? Explain.
Solution: it is hard to take place this reaction. First, the diyne is linear structure, the two ends are too far
apart just like trans-1, 3-butadiene. Second, suppose it can react, the product will be
, they are so reactive that they are hardly exist.
14.37 Reaction of isoprene (2-mehtyl-1,3-butadiene ) with ethyl propenoate gives a mixture of two
Diels-Alder adducts. Show the structures of each, and explain why a mixture is formed.
COOCH2CH3
COOCH2CH3
+
COOCH2CH3
For isoprene both the 1,4 carbons are partial negative, and for ethyl propenoate the
terminal carbon connected with double bond are partial positive. So we get two products.
14.38 Rank the following dienophile in order of their expected reactivity in the Diels-Alder
reaction. Explain.
N
N
C
1
C
C
C
C
C
N
2
N
H2C
C
H
CHO
H2C
C
H
CH3
3
4
Reactivity 1>2>4>3.The Diels-Alder cycloaddition reaction occurs most rapidly and in
highest yield if the dienophile has an electron withdrawing substituent group. So the
dienophile with the stronger electron withdrawing will be more reactive.
14.39 Cyclopentadiene is very reactive in Diels-Alder cycloaddition reactions, but 1,3-cyclohexadiene
is less reactive and 1,3-cycloheptadiene is nearly inert. Explain. (Molecular models are helpful)
Solution:
The structures of the three cyclodiene are showed as follow:
In the reaction process, the dienophile need to overcome the repulsion of steric strain, as we know that
the more methylene groups the cyclodiene has, the more steric strain it contains. So the reactivity of
cyclodiene is in the given order.
14.40 How would you use Dies-Alder cycloaddition reactions to prepare the following product? Show
the starting diene and dienophile in each case.
O
O
(a).
H
CN
(b).
O
O
(c).
(d).
O
CO2 CH3
Solution:
O
CN
O
(a).
(b).
O
O
CO2CH3
(c).
(d).
O
14.41 Aldrin, a chlorinated insecticide now banned for use in the United States, can be made by
Diels-Alder reaction of hexachloro-1, 3-cyclopentadiene with norbornadiene. What is the
structure of aldrin?
Norbornadiene
Solution:
Cl
Cl
Cl
Cl
Cl
Cl
Cl
+
Cl
Cl
Cl
Cl
Cl
hexachloro-1, 3-cyclopentadiene
norbornadiene
aldrin
14.42 Norbornadiene (Problem 14.41) can be prepared by reaction of chloroethylene with
cyclopentadiene, followed by treatment of the product with sodium ethoxide. Write out the
overall scheme, and identify the two kinds of reactions.
Solution:
H
H
H
+
(1)
H
H
Cl
H
The Diels-Alder
cycloaddition reaction
Cl
H
(2)
sodium ethoxide
H
The E2 reaction
H
Cl
H
Na+
H
H
+
+
-O
OH
+
NaCl
Cl
14.43 We’ve seen that the Diels-Alder cycloaddition reaction is a one-step, pericyclic process that
occurs through a cyclic transition state. Propose a mechanism for the following reaction:
Heat
Solution:
+ H2C
CH2
+ H2C
CH2
14.44 Propose a mechanism to explain the following reaction (see Problem 14.43):
CO2CH3
CO2CH3
O
C
Heat
+
+ CO2
C
CO2CH3
O
CO2CH3
α-Pyrone
Solution:
O
CO2CH3
CO2CH3
C
C
C
C
CO2CH3
CO2CH3
CO2CH3
+ CO2
+ CO2
+
CO2CH3
O
14.45 The triene shown below reacts with two equivalents of maleic anhydride to yield C17H16O6 as
product. Predict a structure for the product.
O
+ 2
C17H16O6
O
O
O
O
O
O
+ 2
O
O
O
O
Solution:
O
14.46 The following ultraviolet absorption maxima have measured:
Λmax(nm)
1,3-Butadiene
217
2-Methyl-1,3-butadiene
220
1,3-pentadiene
223
2,3-Dimethyl-1,3-butadiene
226
2,4-Hexadiene
227
2,4-Dimethyl-1,3-pentadiene
232
2,5-Dimethyl-2,4-hexadiene
240
What conclusion can you draw about the effect of alkyl substitution on UV absorption maxima?
Approximately what effect does each added alkyl group have?
Solution: 1)When alkyl substitution is added to dienes, the UV absorption wavelength maxima will be
longer.
2)The alkyl group added to the edge of the dienes make the wavelength maxima increase
more than the alkyl group added to the middle of the dienes.
The more alkyl group added to dienes, the longer the wave length maxima will be.
14.47 1,3,5-Hexatriene has λmax=258nm. In light of your answer to Problem 14.46, approximately
where would you expect 2, 3-dimethyl-1, 3, 5-hexatriene to absorb? Explain.
Solution: According to the figures of Problem 14.46: 1,3-Butadiene hasλmax=217nm, while
2,3-Dimethyl-1,3-butadiene hasλmax=226nm, so I would expect 2, 3-dimethyl-1,3,5-hexatriene
to absorb atλmax=267nm.
14.48 ß-Ocimene is a pleasant-smelling hydrocarbon found in the leaves of certain herbs. It has the
molecular formula C10H16 and exhibits a UV absorption maximum at 232 nm. On
hydrogenation with a palladium catalyst, 2,6-dimethyl-octane is obtained. Ozonolysis of
ß-Ocimene, followed by treatment with zinc and acetic acid, produces four fragments: acetone,
formaldehyde, pyruvaldehyde, and malonaldehyde:
O O
O
O
HC H
H3C C CH3
H3C C
CH
Pyruvaldehyde
Formaldehyde
Acetone
(a) How many double bonds does ß-Ocimene have?
O
H C
O
H2
C CH
Malonaldehyde
(b) Is ß-Ocimene conjugated or nonconjugated?
(c) Propose a structure for ß-Ocimene.
(d) Formulate the reactions, showing starting material and products.
Solution: (a) ß-Ocimene has three double bonds.
(b) ß-Ocimene is conjugated.
H3C C
(c)
C
H
H2
C C
H
C
H
CH3
C
CH2
CH3
(d)
H3C C
C
H
H2
C C
H
C
H
CH3
H3C C
CH3
C
CH2 + 3H2
palladium
catalyst
H
H3C C
CH3
CH3
C
H
H2
C C
H
C
H
C
H2
C
CH2
1.O3
2.Zn/H3O+
O
O
H C H + H3C C
O
H2 H2 H2 H
C C C C
CH3
CH3
O
CH + H C
O
O
H2
C CH + H3C C CH3
CH3
14.49 Myrcene, C10H16, is found in oil of bay leaves and is isomeric with B-ocimene (see Problem
14.48). It show an ultraviolet absorption at 226 nm and can be catalytically hydrogenated to yield
2,6-dimethyloctane. On ozonolysis followed by zinc/acetic acid treatment, myrcene yields
O
O
formaldehyde, acetone, and 2-oxopentanedial:
O
2-oxopentanedial
Propose a structure for myrcene, and formulate the reaction, showing starting material and products.
Solution:
mycrcene
O
O
O
O
O
O
O
O
O
O
O
O
O
O
two same steps
O
O
O
O
O
O
O
O
Zn/H2O
O
O
O
O
O
+
O
O
O
O
+
O
14.50 Addition of HCl to 1-methyloxycyclohexene yields 1-chloro-1-methoxycyclohexane as the sole
product. Why is none of the other regioisomer formed?
Cl
O
O
HCl
O
Solution: Because the interaction intermediate
is more stable than the other one:
Cl
O
O
, so there is only one product
, and no other regioisomer formed.
14.51 Hydrocarbon A, C10H14, has a UV absorption at λmax = 236 nm and gives hydrocarbon B, C10H18,
on catalytic hydrogenation. Ozonolysis of A followed by zinc/acetic acid treatment yields the
following diketo dialdehyde:
O
HC
CH2CH2CH2
O
O
C
C
O
CH2CH2CH2
CH
(a) Propose two possible structures for A.
and the other is
Solution: One is
.
(b) Hydrocarbon A reacts with maleic anhydride to yield a Diels-Alder adduct. Which of your
structures for A is correct?
Solution: The left one is correct.
(c) Formulate the reactions showing starting material and products.
H2
catalyst
Solution:
O3
Zn/H3O
O
HC
CH2CH2CH2
O
O
C
C
O
CH2CH2CH2
O
+
CH
O
O
O
O
O
14.52 Adiponitrile, a starting material used in the manufacture of nylon, can be prepared in three steps
from 1,3-butadiene. How would you carry out this synthesis?
H 2C C
H
H
C CH2
3 STEPS
N CCH2 CH2CH2CH2C
N
Solution:
Br2
1,4 addition
Br
Br
H2
Br
Br
NaCN
CH3CN
CN
CN
14-53 Ergosterol, (C28H44O), λmax=282nm and molar absorption ε=11,900.What is the concentration of
it in a solution whose absorbance A=0.065 with a sample l=1.oocm.
Solution:
ε=A/C×l
so C=A/ε×l=5.462.
14.54 Cyclopentadiene polymerizes slowly at room temperature to yield a polymer that has no double
bonds. On heating, the polymer breaks down to regenerate cyclopentadiene. Propose a structure
for the product.
Solution:
n
14.55 Dimethyl butynediocte undergoes a Diels-Alder reaction with (2E, 4E)-hexadiene. Show the
structure and stereochemistry of the product.
O
O
H3C O C C C C O CH3
Dimethyl butynedioate
Solution:
The product of Dimethyl butynedioate undergoes a Diels-Alder reaction with (2E, 4E)-hexadiene.
O
C O CH3
C O CH3
O
14.56 Dimethyl butynedioate also undergoes a Diels-Alder reaction with (2E,4Z)-hexadiene, but the
stereochemistry of the product is different from that of the (2E,4E) isomer. Explain.
Solution: The reaction with (2E,4Z)-hexadiene can be written as follows:
O
O
O
O
+
O
O
O
O
and with the (2E,4E) isomer can be written as follows:
O
O
O
O
+
O
O
O
O
Because their s-Cis conformations are different, the products are different.
14.57 How would you carry out the following synthesis (more than one step is required)? What
stereochemical relationship between the CO2CH3 group attached to the cyclohexane ring and
the CHO groups would your synthesis produce?
Solution：
CHO
+
CH2
1.O3
Bnzene
Heat
HC
C
O
O
CH3
CHO
2+
2.Zn /H3O+
endo
CO2CH3
CO2CH3
flip
CHO
OHC
H3CO2C
Obviously the group CO2CH3 and the red CHO are cis, however , CO2CH3 and green CHO are
trans.
14.58 The double bond of an enamine (alkene+amine) is much more nucleophilic than a typical alkene
double bond. Assuming that the nitrogen atom in an enamine is sp2-hybridized, draw an orbital
picture of an enamine, and explain why the double bond is electron-rich.
C
C
R
N
An enamine
R
Solution:
N
R
the carbon atom and nitrogen atom are all sp2-hybridized, so the p
R
orbitals will form a π34 conjugated system, the electrons are delocalized, it will surely lower the
system’s energy and make the electron cloud between the carbon-carbon more rich, so it looks as if the
double bond of an enamine (alkene+amine) is much more nucleophilic than a typical alkene double
bond.
14.59 Benzene has an ultraviolet absorption at λmax = 204nm, and para-toluidine hasλmax = 235nm. How
do you account for this difference?
H3C
Benzene
NH 2
para-toluidine
Solution:
H3C
NH2
H3C
NH2
H3C
H3C
Because CH3― and NH2― are electro-donating groups, and
The NH2― group enlarges the conjugation system, so it’s ultraviolet absorption becomes lager than
benzene.
14.60 Phenol, a weak acid with pKa=10.0, has a UV absorption at λmax=210nm in the ethanol solution.
When dilute NaOH is added, the absorption increases toλmax=235nm. Explain.
NH2
NH2
OH
Phenol
Solution:
O
When NaOH is added, it reacts with phenol to yield phenol anion
. And the conjugation of O-C in
phenol anion is stronger than it in phenol, so theλmax becomes a little longer.