1. Trang chủ
  2. » Giáo án - Bài giảng

Sách giải cuốn Organic Structures from Spectra 4th Wiley (2007)

39 367 2

Đang tải... (xem toàn văn)

Tài liệu hạn chế xem trước, để xem đầy đủ mời bạn chọn Tải xuống

THÔNG TIN TÀI LIỆU

Thông tin cơ bản

Định dạng
Số trang 39
Dung lượng 454,15 KB

Nội dung

Problem 327 5 Spin System AMX3 JAM = 15.3 Hz is typical of a coupling between vinylic protons which are trans to each other see Section 5.7... Problem 328 5 Spin System ADMX3 The criti

Trang 1

Solutions Manual

Trang 2

1,2-butanedione (biacetyl)

C4H6O2

CH2C

N N

2-propanol

C3H8O

Br H

Trang 3

+ -

1-phenylethanol

C8H10O

29

CH2 C CH3O

benzyl methyl ketone

Trang 4

33

O CH3

O O

t-butyl acetoacetate

C8H14O3

34 CH3CH2 O C

O H

ethyl formate

C3H6O2

35

O O

Odiethyl oxalate

C6H10O4

43

CH3 O O

O CH3O

ethylene glycol diacetate

C6H10O4

44

CH3O

OCH3O

O O C

diethyl succinate

C8H14O4

Trang 5

OCH31,2,3-trimethoxybenzene

C12H18

Trang 6

65 CH3 CH3

H CH3

pentamethylcyclopentadiene

3-hydroxybutanone (acetoin)

C4H8O2

O

pentanone

4-hydroxy-4-methyl-2-C6H12O2

71

O CH3O

Trang 7

18-crown-6

C12H24O6

83

H 2 C C CH2ClCl

4-chlorobutyl acetate

C6H11O2Cl

85

OH O Br

Trang 9

H O

C O

H3C

O OH

4-acetoxybenzoic acid

C9H8O4

121

NH2C

CH3O

H C O

O O

Trang 10

O O

O O

O O O

diethyl o-phthalate

C12H14O4

136

C O O C

O O

diethyl isophthalate

C12H14O4

C O

O

1,3-dihydroxyphenyl dipropionate

C12H14O4

138

C C

CH3O

O O

cyclopropane carboxylic acid

C4H6O2

C Obenzoylcylopropane

C10H10O

C O OCH2CH3

ethyl cyclobutanecarboxylate

C7H12O2

Trang 11

145 C(CH

3 )3O

4-t-butylcyclohexanone

C10H18O

146

H 3 C CH3N

benzyl benzoate

C14H12O2

150

O O

p-cresyl phenylacetate

C15H14O2

1,3-bis(trichloromethyl)- benzene

C8H4Cl6

C O NEt 2

C9H8O4

Br Br

3,5-dibromocumene

C9H10Br2

159

COOH Br

H O

piperonal

C8H6O3

Trang 12

2-hydroxycyclohex- 1-en-3-one

C13H8O

Trang 13

177

O O O

CH3

CH3

CH32,4,6-trimethyl- 1,3,5-trioxane

C6H12O3

178

O O

H4-ethyl-4-methyl-2,6-piperidinedione

2,5-diol

C8H14O2

C C H

H O H

4-amino-3,5-C12H9NO2

189

CH3 CH2Br Br

H

CH3H

H

2-butanol

C4H10O

Trang 14

193

S O

dibenzyl sulfoxide

C14H14OS

194

S COOH

O O

4

tetraethylene glycol ditosylate

phenylacetaldeyde ethylene glycol acetal

C10H12O2

204

O CH(CH 3 ) 2

(Z)-β-bromostyrene

C8H7Br

Trang 15

209

NO 2

Br

H H

homophthallic acid

C9H8O4

212

O O

CH 3 O

CH 3 O

5,6-dimethoxy-2- coumaranone

C10H10O4

213

C H

Cl

COOH Cl

2,4,5-trichlorophenoxy- acetic acid (2,4,5-T)

Odiethyl isopropylidene- malonate

C10H16O4

218

H

CN O

dimethylbutyraldehyde

4-cyano-2,2-C7H11NO

O H

2-chloroacetaldehyde diethylacetal

Trang 16

styrene epoxide

C8H8O

233

C C C

C O

O

CH3

Hcitraconic anhydride

2-furyl t-butyl ketone

C12H13N

Trang 17

2-methyl- tetrahydrofuran-3-one

Otetramethyl-1,3-cyclobutanedione

cyclopentanone oxime

C5H9NO

251

N OH

cyclohexanone oxime

C6H11NO

252

N H O

ε-caprolactam

C6H11NO

253

HO N(CH3)2OH

CH3

CH3H

pseudoephedrine

C10H15NO

255

N H

O

C(CH3)3H

H C O

CH 3

N-acetylcysteine

C5H9NO3S

Trang 18

2-methyl-2,4-pentanediol

C6H14O2

266

OH OCH 3

H

CH3O H

N

H

CH 3

O H O

CH 3

H N OCH 2 CH 3

N-acetyltyrosine ethyl ester

Trang 19

273 COOH

COOH

2,3-naphthalene- dicarboxylic acid

C12H8O4

NO21-methoxy-4-nitro-

naphthalene

C11H9NO3

CH31,5-dimethylnaphthalene

H

C O

dimethoxy-1-coumarinone

3-hydroxy-3-methyl-5,8-C13H16O4

Trang 20

Chapter 9.2 – The Analysis of MixturesProblem 284

Trang 23

Proton Chemical Shift ( δ)

5

Trang 24

a b c

Trang 25

1 2 3 4

Trang 28

Problem 310

environments

Number of 13C environments

Trang 29

Problem 313

Chemical Shifts δA = 150 Hz / 400 MHz = 0.375 ppm

δX = 300 Hz / 400 MHz = 0.750 ppm

Problem 314 Spin System A2X3

Chemical Shifts δA = 3.36 ppm = 3.36 x 60 = 202 Hz from TMS

δX = 1.11 ppm = 1.11 x 60 = 67 Hz from TMS

1st Order Analysis ΔνAX = νA - νX = 202 - 67 = 135 Hz

ΔνAX / JAX = 135 / 7 = 19.3 This ratio is much greater than 3 so a 1st order analysis is justified

Trang 30

Problem 315 Spin System AMX

Problem 316 Spin System AMX

Trang 31

Problem 317 Spin System AMX

Problem 318 Spin System AMX

Trang 32

Problem 319 Spin System AX3

Trang 33

Problem 321 4 Spin System AMX2

Trang 34

Problem 323 4 Spin System AM2X

Problem 324 4 Spin System A2MX

Trang 35

Problem 325 4 Spin System AMX2

Problem 326 5 Spin System A2M2X

Trang 36

Problem 327 5 Spin System AMX3

JAM = 15.3 Hz is typical of a coupling between vinylic protons which are

trans to each other (see Section 5.7)

Trang 37

Problem 328 5 Spin System ADMX3

The critical coupling constant is JDM = 15.8 Hz which is typical of a coupling

between vinylic protons which are trans to each other (see Section 5.7)

The compound is:

CH3

HC

H

OHA

XM

D

Trang 38

Problem 329 5 Spin System AMX3

Chemical Shifts δA = 80 Hz / 60 MHz = 1.33 ppm

δM = 220 Hz / 60 MHz = 3.67 ppm

δX = 320 Hz / 60 MHz = 5.33 ppm

Problem 330 3 Spin System A2X

Of the 6 isomeric anilines, only compounds 4 and 6 have the correct

symmetry to give a spectrum with only two chemical shifts in the aromatic region, in the ratio 2:1

Both 4 and 6 would give A2X spin systems The measured coupling constant is 7.7 Hz which is in the

range for protons which are ortho to

each other Compound 4 is the

correct answer

NH2BrBr

HX

Trang 39

Problem 331

The spectrum is obtained after D2O exchange so the carboxylic acid and

phenolic protons will not be present and the spectrum only contains the

aromatic and vinylic protons

The spectrum shows 6 distinct resonances therefore compounds 5 and 6

can be eliminated because they would each have only 4 resonances (on

symmetry grounds)

The proton at about δ 7.1 shows no large coupling (> 7 Hz), this means

that it has no protons ortho to it This eliminates compounds 1 and 2 since

all protons in these compounds will have at least one large ortho coupling

Compounds 3 and 4 differ by the stereochemistry at the double bond The

proton at δ 6.4 is clearly one of the vinylic protons and it is coupled to the

other vinylic proton at δ 7.6 The coupling constant is 16 Hz and this

characteristic of vinylic protons which are trans to each other

The correct answer is compound 3

OH

COOH

Problem 332

All of the protons in the 1H spectrum 1,5-dichoronaphthalene have protons

which are ortho to them This means that every proton must have at least one large (>7 Hz) ortho coupling The spectrum has one proton (at δ 7.1)

which has only a small coupling so this cannot be the spectrum of 1,5-dichloronaphthalene

The spectrum is an AMX spectrum with couplings between A and X of

about 8.3 Hz (typical of an ortho coupling) and coupling between M and X

of about 2.7 Hz (typical of a meta coupling) Two possible structure are

given below

Cl Cl

Ngày đăng: 05/08/2018, 10:10

TỪ KHÓA LIÊN QUAN

w