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:
EH
A
E
E
H
A
E
E
E
H
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:
H2C CHCH3
H F
CH(CH3)2
H
CH(CH3)2
F-CH(CH3)2
15.4 Show how an acylium ion could be formed from acetic anhydride in the presence of AlCl3.
C
O
C
H3C
H3C
O
O
AlCl3 H3C C
O
OAlCl3 + H3C C O H3C C 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
Al
ClCl
Cl
H
H
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.
OH
B
F
FF
HO
B
F
FF
H
H
H
15.7 Starting with benzene and the appropriate acyl chloride or acid anhydride, outline a synthesis of each of the following: (a) Butylbenzene
+Cl
O
O
Zn(Hg)
HCl
AlCl3
(b) (CH3)2CHCH2CH2C6H5
+Cl
O
O
Zn(Hg)
HCl
AlCl3
(c) Benzophenone (C6H5COC6H5)
O
+
O
Cl AlCl3
(d) 9,10-Dihydroanthracene
+
O
O
O
HOOC
SOCl2
O
HO
AlCl3
O
Zn(Hg)
HCl
H H
O
AlCl3
Cl
O
Zn(Hg)
HClT. M.
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
N
O
O
CH3
NO2
CH3
NO2
CH3
NO2
(a)CH3
N
O
O
CH3 CH3 CH3
NO2 NO2 NO2(b)
CH3
N
O
O
CH3 CH3 CH3
O2N O2N 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
Meta attack
Para attack
O
Br+
O
H
Br
O
H
Br
O
H
Br
O
H
Br
O
Br+
O
H
Br
O
H
Br
O
H
Br
O
Br+
O
Br H
O
Br H
O
Br H
O
Br H
H H H H H
H H H H
H H H H H
15.11 Phenol reacts with acetic anhydride, in the presence of sodium acetate, to produce the ester,
phenyl acetate. OH
(CH3CO)2O
CH3CO2Na O C CH3
O
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.
NH2
(CH3CO)2OHN C CH3
O
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)
OC
O
CH3
Ortho attack
OC
O
CH3
E H
E
OC
O
CH3
E H
OC
O
CH3
E H
OC
O
CH3
E H
Relative stable contributor
OC
O
CH3
para attack
OC
O
CH3
E
OC
O
CH3
Relative stable contributor
E
H
H
E
OC
O
CH3H
E
OC
O
CH3H
E
OC
O
CH3
Meta attack
OC
O
CH3
E
OC
O
CH3
EH H
E
OC
O
CH3
H
E
(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.
NC
CH3
O
NC
CH3
O
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 CH
CH2HCl
ClHC CH2
Cl H
Cl CH
CH2
Resonance effectCl C
HCH2
Cl CH
CH2
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:
Br2
FeBr3
Br
Br Br
Br Br
Br
O. O.
P. P.
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:
H2SO4
NO2
H2SO4
H2SO4
NO2
NO2
N+
OH
O
-O
N+
OH
O
-O
N+
OH
O
-O
(I)
(II)
(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.
Cl2(1) radiation 2Cl
Cl2(2)
Cl
Cl
Cl2
(3)Cl
Cl
Cl2(4)
Cl
is the most stable radical because of its conjugated form, therefore, 1-cholro-1-phenylropane is the major product. 15.16 Starting with phenylacetylene , outline a synthesis of following compounds: (a) 1-phenyl-propyne
NaNH2
NH3 (l)
CH3BrT.M.
(b)1-phenyl-1-butyne
CH3CH2BrT.M.
NaNH2
NH3 (l)
(c ) (Z)-1-phenylpropene, and (d) (E)-1-phenylpropene
NaNH2
NH3 (l)
CH3Br
Na
NH3 (l)
lindlar
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 OR 2OR
step 2:
OR + H Br R OH + Br
step3:
Br +
Br
step4: Br
+ H OR
Br
We know that this structure
Br
is more stable.
b) in the absence of peroxides mechanism:
step1:
+ H Br
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)
CH3COCl
O
Cl2
FeCl3
O
Cl
Zn/Hg
HCl
Cl
15.20 Predict the major product (or products) that would be obtained when each of the following compounds is nitrated.
(a)
OH
CF3
(b)
SO3H
CN
(c)
OCH3
NO2
The answer
(a)
OH
CF3
NO2 (b)
SO3H
CN
O2N
(c)
OCH3
NO2
NO2
OCH3
NO2
+
O2N
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.
ClC2H5O
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
Cl
C2H5OH
C2H5OH
OH
OH
C2H5
C2H5
O
C2H5
O
C2H5
(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.
Cl
Cl
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 )
Cl ( II ) ( 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 CH2RO is formed. Since it
has two resonance structures as follows.
CH2RO 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:
(C6H5)3CCl > (C6H5)2CHCl >
C6H5CHCH3
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)
(b)
Cl
Cl
(c)
O
Cl
O
Cl
(d)
F
Cl
F
Cl
(e)
OHO
Cl
(f) Cl
NO2
(g)
Cl
Cl
Cl
Cl
(h) Cl
Cl
O
Cl
O
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:
NHCOCH3
NO2
(a)
(b)OCOCH3
NO2
(c) COOH
Cl
O2N
(d)COOH
O
NO2
(e)
NHCOCH3NO2
+
+
OCOCH3
NO2
Cl
NO2
COOH
Cl
+
O2N
15.28 Give the structures of the major products of the following reactions: (a) Styrene + HCl (b) 2-Bromo-1-phenylpropane + C2H5ONa
(c) C6H5CH2CHOHCH2CH3 HA,heat
(d) Product of (c) + HBr peroxides
(e) Product of (c) + H2O HA,heat
(f) Product of (c) + H2
Pt
25℃
(g) Product of (f) (1) KMnO4,OH-,heat
(2)H3O+
Answer:
(a) (b)
Cl
(c) (d)
Br
(e)
OH
(f) (g) COOH
15.29 Starting with benzene, outline a synthesis of each of the following: (a) Isopropylbenzene
CH3CHClCH3
AlCl3
CH
CH3
CH3
(b) tert- Butylbenzene
H2SO4
(C) Propylbenzene
B r2, ligh t
M g
M gB rC H 3C H 2C H 2B r
T H F
CH2CH2CH3
(d) Butylbenzene
COCH2CH2CH3CH3CH2CH2COCl CH2CH2CH2CH3Zn / Hg, HCl
(e) 1-tert-Butyl-4-chlorobenzene
(CH3)3CCl
AlCl3
Cl2
FeCl3
Cl
(f) 1-Phenylcyclopentene
+Cl
AlCl3 NBS
Br
NaOEt
heat
(g) trans-2-Phenylcyclopentanol
1) THF, BH3
2) H2O2, OH-
OH OH
HPhfrom (f) (h) m-Dinitrobenzene
HNO3
NO2HNO3
NO2
NO2 (i) m-Bromonitrobenzene
HNO3 NO2 Br2
FeBr3
NO2
Br
(j) p- Bromonitrobenzene
Br2
FeBr3
BrHNO3
Br
NO2
(k) p - Chlorobenzenesulfonic acid
Cl2
FeCl3
Cl H2SO4
Cl
SO3H (l) o - Chloronitrobenzene
Cl2
FeCl3
ClHNO3
Cl
NO2
(m) m - Nitrobenzenesulfonic acid
HNO3NO2 H2SO4
NO2
SO3H
15.30 Starting with styrene, outline a synthesis of each of the following: (a) C6H5CHClCH2Cl
Cl2
Cl
Cl
H
(b) C6H5CH2CH3
H2,Pt
(c) C6H5CHOHCH2OH
cold KMnO4
OH-
OH OH
H
(d) C6H5COOH
H2,Pt hot KMnO4COOH
(e) C6H5CHOHCH3
H2O
OH
(f) C6H5CHBrCH3
HBr
Br
(g) C6H5CH2CH2OH
B2H6H2O2/OH-
OH
(h) C6H5CH2CH2D
D1) THF BH3
2) CH3CO2D
(i) C6H5CH2CH2Br
HBr
ROOR
Br
(j) C6H5CH2CH2I
the same as (i)Br
NaI
acetone
I
(k) C6H5CH2CH2CN
the same as (i)Br
NaCN CN
(l) C6H5CHDCH2D
D2,Pt
D D
H
(m) Cyclohexylbenzene
H2,Pt
(n) C6H5CH2CH2OCH3
the same as (i)Br
NaOCH3OCH3
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:
Cl2FeCl3
COOH
(a) (1)KMnO4,OH-,heat
(2)H3O
COOHCH3
Cl
AlCl3
CH3COCl
CH3
COCH3
(b)
CH3
H2SO4
HNO3
CH3
NO2
Br2
Fe
CH3
Br
NO2
(c)
CH3
Br2
Fe
CH3
Br
COOH
Br
(d)(1)KMnO4,OH-,heat
(2)H3O
CH3
AlCl3
HC(CH3)2Cl
CH3
(f)
CH3
CH(CH3)2 CH3
(g)cyclohexene
HFH3C
HNO3
H2SO4 H2SO4
CH3
NO2O2N
NO2
(h)fuming HNO3
CH3 CH3
NO2
HNO3
H2SO4
Cl2
FeCl3
CH3
NO2
Cl
(i)
CH3 CH3
Cl
H2SO4
CH2=C(CH3)2
CH3
C(CH3)3
(j)
CH3
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
(a)
NH2
CH3COClbase
NHCOCH3
Br2
FeBr
NHCOCH3
(1)H2O,H2SO4
NH2
BrHO-(2)
HO-(2)
(b)base
NHCOCH3
concd Br2Fe
NHCOCH3
BrHO3S
(1)H2O,H2SO4
NH2
Br
NH2
CH3COCl
H2SO4
H3COCHN
SO3H
(c)base
HNO3H2SO4
(1)H2O,H2SO4,heat(2)O-HBr2
FeCH3COCl
NH2 NHCOCH3H3COCHN
NO2
NH2
NO2
Br
(d) CH3COClbase H2SO4
H3COCHN
SO3H
H3COCHN
SO3H
(2)O-H
concd HNO3
O2N
(1)H2O,H2SO4,heatBr2Fe
NHCOCH3
NO2Br
NH2 NHCOCH3
NH2
NO2Br
(e) Br2H2O
NH2
Br Br
Br
NH2
15.33 Both of the following syntheses will fail. Explain what is wrong with each one. (a)
(2)CH3COCl/AlCl3
(1)HNO3/H2SO4
(3)Zn(Hg),HCl
NO2
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
(2)NaOEt,EtOH,heat(1) NBS,CCl4,light
(3) Br2,FeBr3
Br Solution: The product can’t be get, but react as following:
NBS,CCl4,light
Br
NaEt,EtOH
heat
BrH2C 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 C
O
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?
(a)
O
O
Br
(b) NH
O
NH
O
Br
(c) C
OO
C
OO
Br
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.
O
HBr
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.
O
H+OH OH
H
OH2
H
15.37 Propose structures for compounds G-I.
OH
OH
concd H2SO4
60-65G
(C6H6S2O8)
concd HNO3concd H2SO4
H(C6H5NS2O10)
H3O,H2O I(C6H5NO4)
Solution:
G.
OH
OH
SO3H
HO3S
H.
OH
OH
SO3H
HO3S NO2
I.
OH
OH
NO2
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
concd H2SO4
100℃
OH
SO3H
Cl2
OH
Cl Cl
SO3H
H3O,H2O
OH
Cl Cl
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 CH
CH CH2
H+
HC CH
HC CH3
Br-
HC CH
HC 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.
Toluene+succinic anhydrideAlCl3 A
(C11H12O3)
Zn(Hg)HCl
B(C11H14O2)
C(C11H13ClO)
D(C11H12O)
E(C11H14O)
F(C11H12)
G(C11H11Br)
2-Methylnaphthalene
SOCl2 AlCl3 NaBH4
H2SO4 NBS NaOEt
heat CCl4,light EtOHheat
answer: A:
OH
O O
B: OH
O
C: Cl
O
D:
O E:
OH F:
G:
Br
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:
(a)
CH2OH
HA
-H2Ophenanthrene
(b) 2 H3C C CH2
C6H5
HA
H3CC6H5
H3C CH3
Answer:
(a)
CH2OH
H
H2C OH2 CH2
H A
(b) H3C C CH2
C6H5
HA
H3CC6H5
H3C CH3
H
H3C C CH3
C6H5
CH3
C6H5
H3C
CH3
H2C
C6H5
CH3
H3CC6H5
CH3
H
A
H3C 15.43 Show how you might synthesize each of the following starting with α-tetralone. (a) (b) (c)
OH OHH3C (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
5 6 6' 5'
4'
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.
Benzene + AlCl3 PCl5 2NaNH2mineral oil, heat
H2,Ni2B(P-2)
B(C9H10Cl2)
C(C9H8) D(C9H10)
0oCCH3CH2CCl
O
A(a)
Hint: The 1H NMR spectrum of compound C consists of a multiplet at δ7.20 (5H) and a
singletδ2.0 (3H).
C(1)Li,liq,NH3
(2)H2O E(C9H10)(b)
(c) DBr2,CCl4 F enantiomer(major products)
2-5oC+
(d) 2-5oC+E G enantiomer(major products)Br2,CCl4
Solution: The structure of the compounds A-G: A:
C
O
CH2CH3
B:
C
Cl
CH2CH3
Cl
C:
C C CH3
D:
C C
CH3
HH
E:
C C
CH3H
H
F:
C C
H
HCH3
Br
Br
C C
BrCH3
H
H
Br
G:
C C
CH3
HH
Br
Br
C C
BrH
H
CH3
Br
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:
AlCl3 CH3CCl
O O
CH3C [AlCl4]+ +Setp1:
+
O
C CH3
C
O
CH3Setp2:
C
O
CH3
Cl
C
O
CH3
Cl
Setp3:
C
O
CH3
Cl
OH H
C
O
CH3
Setp4:
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:
(a)
+ CH3CCH3
Cl
CH3
AlCl3 + HCl
(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?
N
O
Oslow
H NO2
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
+ NaCNCN
+ NaBr
(b) C6H5CH2OCH3
Br
+O
+ NaBrCH3ONa
(c) C6H5CH2O2CCH3
Br
+O
+ NaBr
O
ONa
O
(d) C6H5CH2I
Br
+I
+ NaBrNaI
(e) N3
+ NaBrN3Br + NaN3
(f)
O
OBr +
Na
O
+ NaBr
15.51-Provide structures for compounds A,B and C.
BenzeneNa
liq,NH3.EtOHA(C6H8)
NBS
CCl4B(C6H7Br) (CH3)2CuLi C(C7H10)
An:
A
Br
B
CH3
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:
PhPhPh
H
PhPhPh
OH OH2CH3CH2OH
PhPhPh
OCH2CH3
Ph
Ph
Ph
PhPhPh
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 CHCHBrCH3 H2C CHCBr(CH3)2
(A) (B) (C) Answer: (a) A>B>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:
NHCCH3
O
H2SO4
NHCCH3
O
HO3S
HNO3
heat
NHCCH3
O
HO3S NO2
NH2
NO2(1)H2O H2SO heat
(2)OH-
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
H3CO
O H
O
OH
1)NaBH4 2)O3 3)H2OB
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
HC
H2C
OH
OH
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
HHO
OH
CH2OH
H3CO
O3 H2O D
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
CH2OH
HHO
OHH