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
E
H
A
E E
H
A
E H
E
A
15.2 Given that the pK a of H2SO4 is -9, and that of HNO 3 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 by increasing the concentration 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. 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 AlCl 3.
O H3C
C O
H3C
AlCl3
H3C
C
C
OAlCl3
+ H 3C
C
O
H 3C
C
O
O
O
15.5 When benzene reacts with neopentyl chloride, (CH 3)3CCH2Cl, in the presence of aluminum chloride, the major product is i s 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
F
B 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
AlCl 3
+ O
(c) Benzophenone (C 6H5COC6H5)
Zn(Hg) HCl
O H3C
C O
H3C
AlCl3
H3C
C
C
OAlCl3
+ H 3C
C
O
H 3C
C
O
O
O
15.5 When benzene reacts with neopentyl chloride, (CH 3)3CCH2Cl, in the presence of aluminum chloride, the major product is i s 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
F
B 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
AlCl 3
+ O
(c) Benzophenone (C 6H5COC6H5)
Zn(Hg) HCl
O
O
Cl
+
AlCl3
(d) 9,10-Dihydroanthracene 9,10-Dihydroanthracene O
O
Zn(Hg)
AlCl3 +
O
HCl HOOC
HO O
O H
H
AlCl3
SOCl2
Zn(Hg)
T. M.
HCl 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
NO 2
(a) CH3
O
CH3
CH3
CH3
N O
NO2
NO2
NO2
(b) CH3
O
CH3
CH3
N O
O2N
O2N
CH3 O2N
(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 O
Br +
H
H O
O
Br
Br
H
H
H O
O
H
Br
Br
H
H
15.11 Phenol reacts with acetic anhydride, in the presence of sodium acetate, to produce the ester, O
(CH3CO)2O
phenyl acetate.
CH3 CO2 Na OH
O
C
CH 3
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 CH 3COO– 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. O
(CH 3CO) 2O H N
C
CH3
NH2
Aniline
Acetanilide
What kind of directive effect would you expect the acetamido group (CH 3CONH-) 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
C
O C
CH3
Relative stable contributor CH3
para attack O
O O
C
O
O
O C
C
CH3
H
CH3
CH3
E
E
E H O
O
O C
CH3
CH3
H
O C
H
Relative stable contributor E
E
Meta attack O
O
H
E O
C
H
O
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 Br 2/FeBr 3). Answer:
Br
Br
O.
Br
O.
Br 2 FeBr 3
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
-
N+
H2SO4
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
the
1-cholro-1-phenylropane. Both 2-chloro-1-phenylpropane
major
product
is
3-chloro-1-phenylpropane
are
leading
to each
is
major
product.
radiation
(1)
the
Cl2
product
and
minor
chlorine in
the
products. Write the
account
for
the
fact
presence of structure that
of
UV radiation, the
and radical
1-cholro-1-phenylropane
2Cl Cl
(2)
(3)
Cl
Cl2
Cl
Cl2
Cl
is the most 1-cholro-1-phenylropane 15.16 Starting
Cl2
Cl
(4)
with
(a) 1-phenyl-propyne
is
stable the
radical
major
phenylacetylene ,
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
We know that this structure
is more stable.
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
OCH3
OCH3
CN
O2N
(c)
(b)
+ NO2
NO2
SO3H
O2N
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 CH 3=CHCH2OCH2CH3 and CH3CHCH=CH2. | OCH2CH3 C2 H5 OH
C2 H5
C2H5
OH
O
Cl
C2H5OH
(c)
C2H5
C2H5
OH
O
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 S N1 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 S N1 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 S N1 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
(0.08)
(C6H5)2CHCl
(1)
(300)
(C6H5)3CCl 6
(3*10 )
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 C 7H10. 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 Cl 2 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 (C6H5 NHCOCH3) (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 CH2CHOHCH2CH 3 (d) Product of (c) + HBr
pero xides
HA,heat
(e) Product of (c) + H 2O Pt
(f) Product of (c) + H 2
25 ℃
(1) KMnO 4,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 , light
MgBr
Mg
CH2CH2CH3
C H 3 C H 2 C H 2 B r THF
(d) Butylbenzene
CH3CH2CH2COCl
COCH2CH2CH3
Zn / Hg, HCl
CH2CH2 CH2CH3
(e) 1-tert-Butyl-4-chlorobenzene
Cl2
(CH3)3CCl AlCl3
Cl
FeCl 3
(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 NO2
HNO3
HNO3
NO2
(i)
m-Bromonitrobenzene NO2
HNO3
NO2
Br 2 FeBr 3 Br
( j) p- Bromonitrobenzene Br Br
Br 2
HNO3
FeBr3
NO2
(k) p - Chlorobenzenesulfonic acid Cl
Cl
Cl2
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
H
OH-
(d) C6H5COOH H2,Pt
hot KMnO4
(e) C6 H5 CHOHCH 3 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)
CN
NaCN
(l) C6H5CHDCH2D D
D
H
D2,Pt
(m) Cyclohexylbenzene H2,Pt
(n) C6H5CH 2CH 2OCH 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)
Br 2
HNO3
Fe
H2SO4
NO2
NO2
CH3
CH3
COOH
(1)KMnO4 ,OH-,heat
Br 2
(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
H2SO4
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)H 2O,H2SO4
Br 2
CH3COCl base
NH2
NHCOCH3
NHCOCH3
(2) HO-
Fe
Br
Br NH2
(b)
NHCOCH3
H3COCHN
NHCOCH3
CH3COCl
concd
base
H2SO4
Br 2 Fe HO3S
SO3H
Br
NH2
(1)H2O,H2SO4 (2) HOBr NH2 NH2
H3COCHN
NHCOCH 3
CH3 COCl base
(c)
Br
HNO3 H2SO4
(1)H2O,H2SO4,heat
Fe Br 2
(2)O-H
NO2 NO2 NH2
(d)
H3COCHN
NHCOCH3
CH3 COCl base
H3COCHN
HNO3
concd H2SO4
O2N
SO3H NHCOCH3
Br 2
NH2
(1)H2O,H2SO4,heat
Fe
(2)O-H Br
NO2
Br
NO2
NH2 Br
NH2
Br
Br 2 H2O
(e)
Br
15.33 Both of the following syntheses will fail. Explain what is wrong with each one. (a)
SO3H
NO2
(1)HNO3/H2SO4 (2)CH3COCl/AlCl3 (3)Zn(Hg),HCl CH2CH3
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) Br 2,FeBr 3
Br
Solution: The product can’t be get, but react as following: Br
NBS,C Cl4,light
NaEt,EtO H heat
BrH2C
Br
Br 2 FeBr 3
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 RCO 2, 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 Br 2 and FeBr 3? O
O
Br
(a) O
O
N H
(b)
Br
N H
Br O
O
O
O
C
C
(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)
H3O,H2 O
I (C6H5 NO4)
OH
OH
OH
HO 3S
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
Cl
concd H2SO4
OH
Cl
H3O, H2O
2
Cl
Cl
100℃
SO3H
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. accounts for the formation of this product. system?
(a) Write a mechanism that
(b) Is this 1,4 addition or 1,2 addition to the butadiene
(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
HC
C H
H C
CH3 -
Br +
H
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
H2SO 4 heat
A (C11H12 O3)
AlCl3
C (C11H13ClO)
NBS
D (C11H12 O)
G CCl4,light (C11H11Br)
F (C11 H12 )
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
CH2
CH3
HA 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 t he 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-NO 2; 6-CO2H; 2 -NO2 ; 6 -CO2H ’
what factors ’
account for this? (b) Would you except a biphenyl with 2-Br; 6-CO 2H; 2 -CO2H 6 -H to exist in ’
‘
enantiomeric forms? (c)The biphenyl with 2-NO 2; 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 0oC
B(C9 H10 Cl2)
2NaNH2 mineral oil, heat
1
Hint: The H NMR spectrum of compound C consists of a multiplet at δ7.20 (5H) and a
singletδ2.0 (3H).
(b) C (c) D
(d) E
(1)Li,liq,NH3 (2)H2 O Br 2,CCl 4 2-5oC Br 2,CCl 4
2-5oC
E(C9H10)
F + enantiomer(major products)
G + enantiomer(major products)
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 AlCl 3 leads to the formation of a product with the molecular formula C 8H13ClO. 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
O CH3
C
CH3
Setp3: Cl
Cl
OH
H
O
O C
C
CH3
CH3
Setp4: 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 –SO 3H group as a blocking group? Answer: CH3
+
(a)
CH 3CCH3
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 H 2SO4) at room temperature, predominantly o
(about 95% of the total) ortho and para substitution occurs. If elevated temperatures (150-200 C) 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 t emperature, 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, C 6D6?
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
+
+ NaBr
ONa
(d) C6H5CH2I Br
I
+
NaI
+ NaBr
N3
(e) Br
+
NaN3
N3
+
NaBr
O
(f) Br
+
O
O
+ NaBr
Na
15.51-Provide structures for compounds A,B and C.
Na
Benzene liq,NH3.EtOH An:
NBS A(C6H8)
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 S N2 reaction? (b) In an S N1 reaction? Explain your answer. H3CH2CHC
CHCH2Br
H3CHC
(A)
CHCHBrCH3
H2C
(B)
Answer: (a) A>B>C
CHCBr(CH3)2
(C)
(b) A
15.54 Acetanilide was subjected to the following sequence of reaction: (1) concd H 2SO4; (2) -
HNO3,heat; (3) H 2O, H2SO4, heat, then OH .The
13
C NMR spectrum of the final product gives six
signals. Write the structure of the final product. Answer: O
O NHCCH3
NHCCH3 H2 SO4
O HNO3
NHCCH3
heat
HO 3S
HO 3S
NH 2 NO2 (1)H2 O H2SO heat (2)OH -
NO2
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 -1
1
B had a broad absorption at 3400cm ,and its H NMR spectrum shoued a single multiplet at δ3.6. Answer: H 2C
OH
HC
OH
H 2C
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.
