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