15.1 Show how loss of a proton can be represented using each of the three resonance structures for
the arenium ion, and show how each representation leads to the formation of a benzene ring with
three alternating double bonds (i.e., six fully delocalized πelectrons).
Answer:
E
H
E
A
E
E
H
A
E
H
E
A
15.2 Given that the pKa of H2SO4 is -9, and that of HNO3 is -1.4, explain why nitration occurs
more rapidly in a mixture of concentrated nitric and sulfuric acids, rather than in concentrated
nitric acid alone.
Answer:
Concentrated sulfuric acid increases the rate of the reaction by increasing the concentration of
the electrophile, the nitronium ion（NO2+）.
15.3 Outline all steps in a reasonable mechanism for the formation of isopropylbenzene from
propene and benzene in liquid HF. Your mechanism must account for the product being
isopropylbenzene, and not propylbenzene.
Mechanism:
H
H2C
CHCH3
H
CH(CH3)2
F-
CH(CH3)2
CH(CH3)2
F
15.4 Show how an acylium ion could be formed from acetic anhydride in the presence of AlCl3.
O
H3C
C
AlCl3
O
H3C
H3C
C
C
OAlCl3
+ H3C
C
O
H3C
C
O
O
O
15.5 When benzene reacts with neopentyl chloride, (CH3)3CCH2Cl, in the presence of aluminum
chloride, the major product is 2-methyl-2-penylbutane, not neopentylbenzene. Explain this result.
Cl
H
Cl
Cl
H
Al
Cl
H
15.6 When benzene reacts with propyl alcohol in the presence of boron trifluoride, both
propylbenzene and isopropylbeneze are obtained as products. Write a mechanism that account for
this.
H
OH
H
O
F
B
F
F
B
F
F
H
F
H
15.7 Starting with benzene and the appropriate acyl chloride or acid anhydride, outline a synthesis
of each of the following:
(a) Butylbenzene
O
Cl
AlCl3
+
Zn(Hg)
HCl
O
(b) (CH3)2CHCH2CH2C6H5
O
Cl
AlCl3
+
O
(c) Benzophenone (C6H5COC6H5)
Zn(Hg)
HCl
O
O
Cl
+
AlCl3
(d) 9,10-Dihydroanthracene
O
O
+
Zn(Hg)
AlCl3
HCl
O
HOOC
HO
O
O
H
H
AlCl3
SOCl2
Zn(Hg)
HCl
T. M.
Cl
O
O
15.8 Explain how the percentages just given show that the methyl group exerts an ortho-para
directive effect by considering the percentages that would be obtained if the methyl group had no
effect on the orientation of the incoming electrophile.
CH3
O
CH3
CH3
N
NO2
CH3
NO2
O
NO2
(a)
CH3
O
CH3
CH3
CH3
N
O
NO2
NO2
NO2
(b)
CH3
O
CH3
CH3
N
O2N
O2N
CH3
O2 N
O
(c)
Because the methyl group is a electron-donating group, in way (a) and (c), there is positive charge
on the carbon which the methyl group linked to, and the methyl group can donate electron to
stabilize the carbon with positive charge, and it is more stable than the intermediate of the way (b).
15.9 Use Table 15.2 to predict the major products formed when:
(a) Toluene is sulfonated.
(b) Benzoic acid is nitrated.
(c) Nitrobenzene is brominated.
(d) Phenol is subjected to Friedel-Crafts acetylation.
If the major products would be a mixture of ortho and para isomers you should so state.
Answer:
(a) Ortho-Para
(b) Meta
(c) Meta
(d) Ortho-Para
15.10 Use resonance theory to explain why the hydroxyl group of phenol is an activating group
and an ortho-para director. Illustrate your explanation by showing the arenium ions formed when
phenol reacts with a Br+ ion at the ortho, meta, and para positions.
Answer:
Ortho attack
H
H
O
H
O
Br+
H
H
O
O
O
Br
Br
Br
Br
H
H
H
H
Meta attack
O
O
O
O
H
H
H
H
Br+
Br
Br
Br
H
H
H
Para attack
H
Br+
H
H
O
O
Br
Br
H
H
H
O
O
O
H
Br
Br
H
H
15.11 Phenol reacts with acetic anhydride, in the presence of sodium acetate, to produce the ester,
(CH3CO)2O
phenyl acetate.
CH3 CO2 Na
OH
O
O
C
CH3
The CH3COO– group of phenyl acetate, like the –OH group of phenol (Problem15.10), is an
ortho-para director. (a) What structural feature of the CH3COO– group explains this? (b) Phenyl
acetate, although undergoing reaction at the o and p positions, is less reactive toward electrophilic
aromatic substitution than phenol. Use resonance theory to explain why this is so. (c) Aniline is of
often so highly reactive toward electrophilic aromatic substitution that undesirable reactions take
place (see Section 15.14A). One way to avoid these undesirable reactions is to convert aniline to
acetanilide (below), by treating aniline with acetyl chloride or acetic anhydride.
(CH3CO)2O
O
H
N
C
CH3
NH2
Aniline
Acetanilide
What kind of directive effect would you expect the acetamido group (CH3CONH-) to have? (d)
Explain why it is much less activating than the amino group, –NH2.
(a)
Ortho attack
E
O
E
H
H
O
O
C
O
O
O
C
C
CH3
CH3
CH3
E
E
H
E
O
H
O
O
O
C
Relative stable contributor
C
CH3
CH3
para attack
O
O
O
C
C
CH3
E
O
O
O
C
H
CH3
CH3
E
E
H
C
C
CH3
CH3
H
O
O
O
O
Relative stable contributor
H
E
E
Meta attack
O
O
H
E
H
O
O
C
O
C
CH3
E
CH3
CH3
E
H
O
E
O
C
O
C
CH3
(b) Because the benzoyl group is a kind of electron withdrawing group.
(c) It is a ortho- and para- director.
(d) Because the acetyl group is a kind of electron withdrawing group.
O
N
O
C
N
CH3
C
CH3
H
H
15.12 Chloroethene adds hydrogen chloride more slowly than ethene, and the product is 1,
1-dichloroethane. How can you explain this using resonance and inductive effects?
Cl
C
H
CH2
HCl
Cl
Cl
C CH2
H
Resonance effect
H
C
CH2
Cl
H
Cl
C
H
CH2
Cl
C CH2
H
Inductive effect
15.13 Write resonance structures for the ortho and para arenium ions formed when ethylbenzene
reacts with a Br+ ion (as formed from Br2/FeBr3).
Answer:
Br
O.
Br
O.
Br
Br2
FeBr3
P.
Br
P.
Br
Br
15.14 When biphenyl (C6H5-C6H5) undergoes nitration, it reacts more rapidly than benzene, and
the major products are 1-nitro-2-phenylbenzene and 1-nitro-4-phenylbenzene. Explain these
results.
Answer: We know that the reaction have three intermediate:
OH
NO2
-O
N+
O
H2SO4
(I)
OH
OH
-O
-
H2SO4
N+
N+
O
H2SO4
O
NO2
O
(II)
NO2
(III)
The intermediates (I) and (III) formed during the process are more stable than intermediate (II).
Therefore, those are major products. Moreover, the intermediates (I) and (III) are stabilized by the
phenyl group, comparing to the normal arenium, biphenyl are more reactive than benzene.
15.15 When propylbenzene reacts with chlorine in the presence of UV radiation,
the major product is 1-cholro-1-phenylropane. Both 2-chloro-1-phenylpropane and
3-chloro-1-phenylpropane are minor products. Write the structure of the radical
leading to each product and account for the fact that
1-cholro-1-phenylropane
is the major product.
(1)
Cl2
radiation
2Cl
Cl
(2)
(3)
Cl
Cl2
Cl
Cl2
Cl
Cl2
Cl
(4)
is the most
1-cholro-1-phenylropane
is
stable
the
radical
major
15.16 Starting with phenylacetylene ,
(a) 1-phenyl-propyne
because of its
conjugated
form,
therefore,
product.
outline a synthesis of following
compounds:
CH3Br
NaNH2
NH3 (l)
T.M.
(b)1-phenyl-1-butyne
CH3CH2 Br
NaNH2
NH3 (l)
(c )
(Z)-1-phenylpropene, and
(d)
T.M.
(E)-1-phenylpropene
NaNH2
CH3Br
NH3 (l)
lindlar
Na
NH3 (l)
15.17 Write mechanism for the reactions whereby HBr adds to 1-phenylpropene. a) in the
presence of peroxides and
b) in the absence of peroxides. In each case account for the
regiochemistry of the addition (i.e., explain why the major product is 2-bromo-1-phenylpropane
when peroxides are present, and why it is 1-bromo-1-phenylpropane when peroxides are absent).
a) in the presence of peroxides
mechanism:
step 1:
RO
2OR
OR
step 2:
OR +
H
Br
R
OH
+
Br
step3:
Br
Br
+
step4:
Br
Br
+ H
OR
Br
is more stable.
We know that this structure
b) in the absence of peroxides
mechanism:
+ H
Br
step1:
step 2:
+
Br
Br
We also know that the carbon cation is stable in this structure
,
so the product is reasonable.
15.18 a) What would you expect to be the major product when 1-phenylpropene reacts with HCl?
b) When it is subjected to oxymercuration-demercuration?
answer: a) I think the situation will be the same to that the HBr react with 1-phenylpropene,the
product is:
Cl
b ) When it is subjected to oxymercuration-demercuration, it obeys the Markovnikoff rule, and the
product is:
OH
15.19 Suppose you needed to synthesize m-chloroethylbenzene from benzene. You could begin by
chlorinating benzene and then follow with a Friedel-Crafts alkylation using CH3CH2Cl and AlCl3,
or you could begin with a Friedel-Crafts alkylation followed by chlorination. Neither method will
give the desired product,however.
(a) Why won’t either method give the desired product?
(b) There is a three-step method that will work if the steps are done in the right order. What is this
method?
The answer:
(a) If the chlorinate reaction first, for the chloride is o,p-director, the Friedel-Crafts alkylation
product would be the o- or p-ethylchlorobenzene. If it begins with Friedel-Crafts alkylation,
the chlorination product would be ortho- or para- product too.
(b)
O
O
Cl
CH3COCl
Cl2
FeCl3
Cl
Zn/Hg
HCl
15.20 Predict the major product (or products) that would be obtained when each of the following
compounds is nitrated.
OCH3
CN
OH
(b)
(a)
(c)
SO3H
CF3
The answer
NO2
OH
(a)
NO2
(b)
OCH3
OCH3
CN
(c)
O 2N
+
O2N
NO2
NO2
SO3H
CF3
NO2
15.21 Account for the following observations:
(a) When 1-chloro-2-butene is allowed to react with a relatively concentrated solution of sodium
ethoxide in ethanol, the reaction rate depends on the concentration of the allylic halide and on
the concentration of ethoxide ion. The product of reaction is almost exclusively
CH3CH=CHCH2OCH2CH3.
Cl
C2H5O
Cl
O
O
(b) When 1-chloro-2-butene is allowed to react with very dilute solution of sodium ethoxide in
ethanol (or with ethanol alone), the reaction rate is independent of the concentration of
ethoxide ion; it depends only on the concentration of the allylic halide. Under these
conditions the reaction produces a mixture of CH3=CHCH2OCH2CH3 and CH3CHCH=CH2.
|
OCH2CH3
C2 H5 OH
C2 H5
C2H5
OH
O
Cl
C2H5OH
C2H5
C2H5
OH
O
(c) In the presence of traces of water 1-chloro-2-butene is slowly converted to a mixture of
1-chloro-2-butene and 3-chloro-1-butene.
ClCl
Cl
Cl-
Cl
15.22 1-Chloro-3-methyl-2-butene undergoes hydrolysis in a mixture of water and dioxane at a
rate that is more than a thousand times that of 1-chloro-2-butene.
(a) What factor accounts for the difference in reactivity?
Cl
(I)
( II )
Cl
( III )
While (I) is more stable than (II) and (III).
(b) What products would you expect to obtain? [Dioxane is cyclic ether (below) that is miscible
with water in all proportions and is useful cosolvent for conducting reactions like these.
Dioxane is carcinogenic (i.e. cancer causing), however, and like most ethers, it tends to form
peroxides.]
O
O
Dioxane
OH
HO
+
15.23 Primary halides of the type ROCH2X apparently undergo SN1 type reaction, whereas most
primary halides do not. Can you propose a resonance explanation for the ability of halides of the
type ROCH2X to undergo SN1 reaction?
Solution:
When ROCH2X loses the halide atom, the intermediate RO
has two resonance structures as follows.
CH2 is formed. Since it
RO
CH2
RO
CH2
So the intermediate is more stable, and then it is easier to undergo an SN1 reaction.
15.24 The following chlorides undergo solvolysis in ethanol at the relative rates given in
parentheses. How can you explain these results?
C6H5CH2Cl
C6H5CHClCH3
(C6H5)2CHCl
(C6H5)3CCl
(0.08)
(1)
(300)
(3*106)
Solution:
The intermediate of the four compounds is as follows.
a.
b.
CH2
c.
d.
The relative stability of the carbocation: d>c>b>a.
Therefore, the relative reactivity:
C6H5CHCH3
(C6H5)3CCl > (C6H5)2CHCl >
Cl
> C6H5CH2Cl.
15.25 Birch reduction of toluene leads to a product with the molecular formula C7H10. On
ozonolysis followed by reduction with zinc and water, the product is transformed into
CH3COCH2CHO and OHCCH2CHO. What is the structure of the Birch reduction product?
Answer:
15.26 Give the major product (or products) that would be obtained when each of the following
compounds is subjected to ring chlorination with Cl2 and FeCl3.
(a) Ethylbenzene
(b) Anisole (C6H5OCH3)
(c) Fluorobenzene
(d) Benzoic acid
(e) Nitrobenzene
(f) Chlorobenzene
(g) Biphenyl (C6H5-C6H5)
(h) Ethyl phenyl ether
Answer:
(a)
Cl
(b)
Cl
O
O
Cl
(c)
Cl
F
F
Cl
(d)
Cl
O
OH
(e)
Cl
NO2
(f)
Cl
Cl
Cl
Cl
(g)
Cl
Cl
(h)
Cl
O
O
Cl
Cl
15.27 Predict the major product (or products) formed when each of the following compounds is
subjected to ring nitration.
(a) Acetanilide (C6H5NHCOCH3)
(b) Phenyl acetate (CH3COOPh)
(c) 4-Chlorobenzoic acid
(d) 3-Chlorobenzoic acid
(e) C6H5COC6H5
Answer:
(a)
NHCOCH3
NHCOCH3
NO2
+
(d)
COOH
O2N
COOH
+
Cl
Cl
NO2
NO2
OCOCH3
(b)
OCOCH3
NO2
+
(e)
O
NO2
NO2
(c)
COOH
O2N
Cl
15.28 Give the structures of the major products of the following reactions:
(a) Styrene + HCl
(b) 2-Bromo-1-phenylpropane + C2H5ONa
HA,heat
(c) C6 H5 CH2CHOHCH 2CH 3
(d) Product of (c) + HBr
peroxides
HA,heat
(e) Product of (c) + H2O
Pt
(f) Product of (c) + H2
25℃
(1) KMnO4,OH-,heat
(g) Product of (f)
(2)H 3O+
Answer:
Cl
(a)
(d)
(c)
(b)
Br
OH
(g)
(f)
(e)
COOH
15.29 Starting with benzene, outline a synthesis of each of the following:
(a) Isopropylbenzene
CH3
CH3CHClCH3
CH
CH3
AlCl3
(b) tert- Butylbenzene
H2SO4
(C) Propylbenzene
B r 2 , lig h t
M gB r
Mg
CH2CH2CH3
C H 3C H 2C H 2B r
THF
(d) Butylbenzene
CH3CH2CH2COCl
Zn / Hg, HCl
COCH2CH2CH3
CH2CH2 CH2CH3
(e) 1-tert-Butyl-4-chlorobenzene
Cl2
(CH3)3CCl
Cl
AlCl3
FeCl3
(f) 1-Phenylcyclopentene
Br
Cl
AlCl3
+
NaOEt
heat
(g) trans-2-Phenylcyclopentanol
NBS
OH
OH
Ph
H
1) THF, BH3
2) H2O2, OHfrom (f)
(h) m-Dinitrobenzene
NO2
NO 2
HNO3
HNO3
NO2
(i)
m-Bromonitrobenzene
NO2
HNO3
NO2
Br2
FeBr3
Br
(j) p- Bromonitrobenzene
Br
Br
Br2
HNO3
FeBr3
NO2
(k) p - Chlorobenzenesulfonic acid
Cl
Cl2
Cl
H2SO4
FeCl3
SO3H
(l) o - Chloronitrobenzene
Cl
Cl2
Cl
NO2
HNO3
FeCl3
(m) m - Nitrobenzenesulfonic acid
NO2
HNO3
NO2
H2SO4
SO3H
15.30 Starting with styrene, outline a synthesis of each of the following:
(a) C6H5CHClCH2Cl
Cl
H
Cl2
Cl
(b) C6H5 CH2 CH3
H2 ,Pt
(c) C6H5CHOHCH2OH
OH
OH
cold KMnO4
OH
H
-
(d) C6H5COOH
H2,Pt
hot KMnO4
(e) C6 H5 CHOHCH3
OH
H2O
(f) C6 H5CHBrCH3
Br
HBr
(g) C6H5CH2CH2OH
OH
B2H6
H2O2/OH-
(h) C6 H5CH2CH2D
1) THF BH3
2) CH3CO2D
(i)
C6 H5 CH2CH2Br
D
COOH
Br
HBr
ROOR
(j) C6H5 CH2 CH2I
Br
the same as (i)
I
NaI
acetone
(k) C6H5CH2CH2CN
Br
the same as (i)
NaCN
CN
(l) C6H5CHDCH2D
D
D
H
D2 ,Pt
(m) Cyclohexylbenzene
H2,Pt
(n) C6H5CH2CH2OCH 3
Br
the same as (i)
OCH3
NaOCH3
15.31 Starting with toluene, outline a synthesis of each of the following:
(a) m-Chlorobenzoic acid
(f) p-Isopropyltoluene (p-cymene)
(b) p-Methylacetophenone
(g) 1-Cyclohexyl-4-methylbenzene
(c) 2-Bromo-4-nitrotoluene
(h) 2,4,6-Trinitrotoluene (TNT)
(d) p-Bromobenzoic acid
(i) 4-Chloro-2-nitrobenzoic acid
(e) 1-Chloro-3-trichloromethylbenzene
(j) 1-Butyl-4-methylbenzene
Answer:
Cl
(1)KMnO4,OH-,heat
(a)
Cl2
FeCl3
(2)H3O
CH3
COOH
COOH
CH3
CH3
CH3 COCl
(b)
AlCl3
COCH3
CH3
CH3
CH3
Br
(c)
Br2
HNO3
Fe
H2SO4
NO2
NO2
CH3
CH3
COOH
(1)KMnO4 ,OH-,heat
Br2
(d)
Fe
(2)H3O
Br
CH3
(f)
Br
CH3
HC(CH3)2 Cl
AlCl3
CH(CH3 )2
CH3
cyclohexene
(g)
H3C
HF
CH3
CH3
HNO3
(h)
CH3
fuming HNO3
H2 SO4
O2N
H2SO4
NO2
CH3
NO2
NO2
CH3
CH3
NO2
(i)
Cl2
HNO3
FeCl3
H2SO4
Cl
Cl
CH3
CH3
CH2=C(CH3)2
(j)
H2 SO4
C(CH3)3
15.32 Starting with aniline, outline a synthesis of each of the following:
(a) p-Bromoaniline
(d) 4-Bromo-2-nitroaniline
(b) o-Bromoaniline
(e) 2,4,6-Tribromoaniline
(c) 2-Bromo-4-nitroaniline
Answ
NH2
(a)
(1)H2O,H2SO4
Br2
CH3 COCl
base
NH2
NHCOCH3
NHCOCH3
(2) HO-
Fe
Br
Br
NH2
NHCOCH3
CH3COCl
(b)
NHCOCH3
H3COCHN
Br2
Fe
concd
H2SO4
base
HO3S
SO3H
Br
NH2
(1)H2O,H2SO4
(2) HOBr
NH2
NH2
H3COCHN
NHCOCH3
CH3 COCl
base
(c)
Br
HNO3
H2SO4
(1)H2O,H2SO4,heat
Fe
Br2
(2)O-H
NO2
NO2
NH2
NHCOCH3
CH3 COCl
base
(d)
H3COCHN
H3COCHN
HNO3
concd
H2SO4
O2N
SO3H
NHCOCH3
Br2
Fe
NH2
(1)H2O,H2SO4,heat
(2)O-H
Br
NO2
Br
NO2
NH2
Br
NH2
(e)
Br
Br2
H2O
Br
15.33 Both of the following syntheses will fail. Explain what is wrong with each one.
(a)
SO3H
NO2
(1)HNO 3/H 2SO4
(2)CH3COCl/AlCl3
(3)Zn(Hg),HCl
CH 2CH 3
Solution: In the first step -NO2 will attach to the ring, but it will deactivate the benzene
ring so that it can’t react with the reagent in step (2).
(b)
CH2CH3
(1) NBS,CCl4,light
(2)NaOEt,EtOH,heat
(3) Br2,FeBr3
Br
Solution: The product can’t be get, but react as following:
Br
NBS,CCl4,light
NaEt,EtOH
heat
BrH 2C
Br
Br2
FeBr3
Br
15.34 One ring of phenyl benzoate undergoes electrophilic aromatic substitution much more
readily than the other. (a) Which one is it? (b) Explain your answer.
O
O
C
Solution: The left ring is more readily. Because the group attach to it is RCO2, it is an electron
donating group, it can activate the benzene ring. But the group attached to the right ring is an
electron with-drawing group, so the reactivity of it is limited.
15.35 What product (or products) would you expect to obtain when the following compounds
undergo ring bromination with Br2 and FeBr3?
O
O
Br
(a)
O
O
N
H
(b)
Br
N
H
Br
O
O
C
(c)
O
O
C
15.36 Many polycyclic aromatic compounds have been synthesize by a cyclization reaction
known as the Bradsher reaction or aromatic cyclodehydration. This method can be illustrated by
following synthesis of 9-methylphenanthrene.
HBr
O
acetic acid
heat
An arenium ion is an intermediate in this reaction, and the last step involves the dehydration of an
alcohol. Propose a plausible mechanism for this example of the Bradsher reaction.
H
H
O
OH
OH
OH2
H+
15.37 Propose structures for compounds G－I.
OH
concd H2SO4
60-65
OH
Solution:
concd HNO3
G
concd H2SO4
(C6 H6S2O8)
H
(C6 H5 NS2O10)
H3 O,H2 O
I
(C6H5NO4 )
OH
OH
OH
HO3S
HO3S
NO2
OH
OH
G.
H.
SO3H
NO2
I.
SO3H
OH
15.38 2,6-Dichlorophenol has been isolated from the females of two species of ticks (Amblyomma
americanum and A.maculatum), where it apparently serves as a sex attractant. Each female tick
yields about 5 ng of 2,6-dichlorophenol. Assume that you need larger quantities than this, and
outline a synthesis of 2,6-dichlorophenol from phenol. (Hint: When phenol is sulfonated at 100℃,
the product is chiefly p-hydroxybenzenesulfonic acid.)
Answer:
OH
OH
OH
Cl
concd H2SO4
100℃
OH
Cl
Cl2
H3O,H2O
SO3H
Cl
Cl
SO3H
15.39 The addition of a hydrogen halide (hydrogen bromide or hydrogen chloride) to
1-phenyl-1,3-butadiene produces (only) 1-phenyl-3-halo-1-butene. (a) Write a mechanism that
accounts for the formation of this product. (b) Is this 1,4 addition or 1,2 addition to the butadiene
system? (c) Is the product of the reaction consistent with the formation of the most stable
intermediate carbocation? (d) Dose the reaction appear to be under kinetic control or equilibrium
control? Explain.
Answer:
(a)
HC
C
H
C
H
CH2
H+
HC
C
H
H
C
CH3
-
Br
HC
C
H
H
C
CH3
Br
(b) 1,2 addition
(c) Yes
(d) Since the reaction produces only the more stable isomer, that is, the one in which the double
bond is conjugated with the benzene ring, the reaction is likely to be under equilibrium control.
15.40 2-Methylnaphthalene can be synthesized from toluene through the following sequence of
reactions. Write the structure of each intermediate.
AlCl3
Toluene+succinic anhydride
SOCl2
H2SO4
heat
A
(C11H12 O3)
AlCl3
C
(C11H13ClO)
NBS
D
(C11H12 O)
G
CCl4 ,light (C11H11Br)
F
(C11H12)
answer:
A:
OH
O
O
B:
OH
O
C:
Cl
O
D:
O
E:
OH
F:
G:
Br
Zn(Hg)
HCl
NaBH4
NaOEt
EtOH
heat
B
(C11 H14 O2 )
E
(C11H14O)
2-Methylnaphthalene
15.41 Ring nitration of a dimethylbenzene (a xylene) results in the formation of only one
nitrodimethylbenzene. What is the structure of the dimethylbenzene?
Answer:
15.42 Write mechanisms that account for the products of the following reactions:
HA
(a)
phenanthrene
-H2O
CH2OH
H3C
(b)
2
H3C
C
C6H5
HA
CH2
C6H5
H3C
CH3
Answer:
(a)
H2C
CH2OH
H
OH2
CH2
A
H
H
(b)
H3C
C
CH3
HA
CH2
H3C
C6H5
C
C6H5
CH3
H3C
CH3
C6H5
H2
C
C6H5
A
H
H3C
H3C
H3C
C6H5
CH3
H3C
C6H5
CH3
15.43 Show how you might synthesize each of the following starting with α－tetralone.
(a)
(b)
(c)
CH3
H3C
OH
OH
(d)
C6H5
Answer:
(a): Zn(Hg)/HCl
(b): LiAlH4
(c): CH3MgBr, H3+O
(d)C6H5Li, H3+O; heat; Ni / H2
15.44 The compound phenylbenzene is called biphenyl, and the rings are numbered in the
following manner.
3
2
2'
3'
4'
4
5
6
6'
5'
Use method to answer the following questions about substituted biphenyls. (a) When certain large
groups occupy three or four of the ortho positions, the substituted biphenyl may exists in
enantiomeric forms. An example of biphenyl that exists in enantiomeric forms in the compound in
which the following substitutents are present: 2-NO2; 6-CO2H; 2’-NO2 ; 6’-CO2H what factors
account for this? (b) Would you except a biphenyl with 2-Br; 6-CO2H; 2’-CO2H 6’-H to exist in
‘
enantiomeric forms? (c)The biphenyl with 2-NO2; 6-NO2 2’-CO2H ，6 -Br can’t be resolved into
enantiomeric forms .Explain.
Answer:
(a) Two phenyl groups are perpendicular
(b) Yes. I would.
(c) It will have a symmetrical planar, so this molecule is achiral.
15.45 Give structure (including stereochemistry where appropriate) for compounds A-G.
O
(a) Benzene + CH3CH2CCl
C(C9 H8 )
H2 ,Ni2B(P-2)
AlCl3
D(C9H10)
A
PCl5
2NaNH2
B(C9 H10Cl2)
o
mineral oil,
0C
heat
Hint: The 1H NMR spectrum of compound C consists of a multiplet at δ7.20 (5H) and a
singletδ2.0 (3H).
(1)Li,liq,NH3
E(C9H10)
(2)H2 O
Br2,CCl4
(c) D
F + enantiomer(major products)
2-5oC
Br2,CCl4
G + enantiomer(major products)
(d) E
2-5oC
(b) C
Solution:
The structure of the compounds A-G:
A:
O
C
CH2CH3
B:
Cl
C
CH2CH3
Cl
C:
C
C
CH3
D:
H
H
C
C
CH3
E:
H
CH3
C
C
H
F:
H
Br
CH3
C
H
Br
H
C
C
C
Br
CH3
Br
H
G:
CH3
H
Br
C
H
Br
H
C
C
Br
C
Br
H
CH3
15.46 Treating cyclohexene with acetyl chloride and AlCl3 leads to the formation of a product with
the molecular formula C8H13ClO. Treating this product with a base leads to the formation of
1-acetylcyclohexene. Propose mechanism for both steps of this sequence of reactions.
Solution:
Mechanism:
O
O
Setp1: AlCl3 + CH3 CCl
H3C
C + [AlCl4 ]
O
O
Setp2:
+
C
C
CH3
CH3
O
C
Setp3:
O
CH3
C
Cl
Cl
OH
H
O
O
C
Setp4:
CH3
C
CH3
CH3
Cl
15.47 The tert-butyl group can be used as a blocking group in certain syntheses of aromatic
compounds. (a) How would you introduce a tert-butyl group, and (b) how would you remove it?
(c) What advantage might a tert-butyl group have over a –SO3H group as a blocking group?
Answer:
CH3
+ CH3CCH3
(a)
AlCl3
+
HCl
Cl
(b) Because the Friedel-Craft alkylation reaction is reversible, it is easily removed by the acidic
condition.
(c) Alkyl group can activate the benzene and it is a o,p-director. But –SO3H group is a deactivating
group and m-director.
15.48 When toluene is sulfonated (concentrated H2SO4) at room temperature, predominantly
(about 95% of the total) ortho and para substitution occurs. If elevated temperatures (150-200oC)
and longer reaction times are employed, meta (chiefly) and para substitution account for some
95% of the products. Account for these differences. (Hint: m-Toluenesulfonic acid is the most
stable isomer.)
Answer:
At low temperature, the reaction is kinetically controlled, and the usual o/p directive effects of the
methyl group are observed. At the high temperature, the reaction is thermodynamically controlled.
At the reaction times long enough for equilibrium to be reached, the most stable isomer,
m-toluenesulfonic acid, is the principal product.
15.49 A C-D bond is harder to break than a C-H bond, and, consequently, reactions in which C-H
bonds are broken. What mechanistic information comes from the observation that perdeuterated
benzene, C6D6, is nitrated at the same rate as normal benzene, C6D6?
O
H
NO2
slow
N
O
This step determined the rate of the reaction.
15.50 Show how you might synthesize each of the following compounds starting with either
benzyl bromide or allyl bromide.
(a) C6H5CH2CN
Br
+
CN
NaCN
NaBr
+
(b) C6H5CH2OCH3
Br
+
O
CH3ONa
NaBr
+
(c) C6H5CH2O2CCH3
O
Br
O
O
+
+
ONa
NaBr
(d) C6H5CH2I
Br
I
+
NaI
+
NaBr
N3
(e)
Br
+
NaN3
N3
+
NaBr
O
(f)
Br
+
O
O
+
Na
NaBr
15.51-Provide structures for compounds A,B and C.
Na
A(C6H8)
Benzene
liq,NH3 .EtOH
An:
NBS
CCl4
B(C6H7Br)
(CH3 )2CuLi
C(C7H10)
Br
A
CH3
B
C
15.52 Heating 1,1,1-triphenylmethanol with ethanol containing a trace of a strong acid causes the
formation of 1-ethoxy-1,1,1-triphenylmethane. Write a plausible mechanism that accounts for the
formation of this product
An:
Ph
Ph
Ph
Ph
Ph
Ph
Ph
Ph
Ph
Ph
Ph
Ph
Ph
Ph
OH
OH2
Ph
OCH2CH3
CH3CH2OH
H
OCH2CH3
H
15.53 Which of the following halides would you expect to be most reactive in an SN2 reaction? (b)
In an SN1 reaction? Explain your answer.
H3CH2CHC
CHCH2Br
H3CHC
(A)
CHCHBrCH3
H2C
(B)
Answer: (a) A>B>C
CHCBr(CH3)2
(C)
(b) A<B<C
15.54 Acetanilide was subjected to the following sequence of reaction: (1) concd H2SO4; (2)
HNO3,heat; (3) H2O, H2SO4, heat, then OH-.The 13C NMR spectrum of the final product gives six
signals. Write the structure of the final product.
Answer:
O
O
O
NHCCH3
NHCCH3
NHCCH3
H2SO4
HO 3S
HNO3
heat
HO 3S
NH 2
NO2
(1)H2O H2SO heat
(2)OH -
NO 2
15.55 The lignins are macromolecules that are major components of the many types of wood,
where they bind cellulose fibers together in these natural composites. The lignins are built up out
of a variety of small molecules (most having phenylpropane skeletons). These precursor
molecules are covalently connected in varying ways, and this gives the lignins great complexity.
To expain the formation of compound B below as one of many products obtained when lignins are
ozonized. Lignin model compound A was treated as shown. What is the structure of B?
CH3
1)NaBH4
2)O3
3)H2 O
B
H3CO
O
H
O
OH
To make B volatile enough for GC/MS(gas chromatography-mass spectroscopy, Section9.17), it
was first converted to its tris (O-thimethylsilyl) derivative, which had M+ 308m/z.[“Tris” means
that three of the indicated complex groups named (e.g..trimethylsily groups here) are present. The
capital, italicized O means these are attached to oxygen atoms of the parent compound, taking the
place of hydrogen atoms. Similarly, the prefix “bis” indicates the presence of two complex groups
subsequently named, and “tetrakis” (used in the problem below), means four.] The IR spectrum of
B had a broad absorption at 3400cm-1 ,and its 1H NMR spectrum shoued a single multiplet at δ3.6.
Answer:
H2C
OH
HC
OH
H2C
OH
15.56 When compound C, which is often used to model a more frequently occurring unit in lignins,
was ozonized, product D was obtained. In a variety of ways it has been established that the
stereochemistry of the three-carbon side chain of such lignin units remains largely if not
completely unchanged during oxidations like this.
OCH3
H3CO
O3
HO
H
H
O
H 2O
D
CH2OH
H3CO
For GC/MS, D was converted to its tetrakis (O-trimethylsilyl) derivative, which had M+ 424m/z.
The IR spectrum of D has bands at 3000cm-1 (broad, strong ) and 1710cm-1 (strong). Its 1H NMR
spectrum had peaks at δ4.2 (doublet, 1H) after treatment with D2O. Its DEPT 13C NMR spectra
had peaks at δ64 (CH2), δ75 (CH), δ82 (CH), and δ177 (C).
What is the structure of D, including its stereochemistry?
COOH
HO
H
H
OH
CH2OH