Solutions Manual NMR from spectra to structures: an experimental approach

Solutions Manual

NMR - From the Spectra to the Structure Problem number

Formula

Problem number

1

Cl(CH2)3CN

2

H3C CH CH CHO

3

O H3C C COOCH3

4

11

H3C CH2 CH CH2OH NH2

Formula

H2N

CH3 Cl

12

COOH COOH

O

C CH2Br O

13

O2N 5 N

H2N

15

H3CO

O NH C CH3

Cl

6 N

14

CH2 OH

NH2

CN 7

HO

F

16 H3C

8

17 H3CO

N

H3CO

NH2

Cl H3CO

18

O C CH 3

19

CH CH CHO

H3C 9

HO NH2 NO2

10

HOOC NO2

NMR - From the Spectra to the Structure Problem number 20

Formula H3CO

Problem Formula number CH2OH

28

H3C H3C

H3CO

H3C

COOH

21 H3C

C CH CH2CH2 C CH CH2 CH3 O C O CH3

H3C CH3 CH CH C CH3 29 O CH3

CH3

(CH3)2N

O 22

N(CH3)2

30 H3C

H3C O

C

CH2 C2H5OOC CH COOC2H5 CH2

CH3 O 31

23 H3C

CH3 CHO

24

H3C OH OH CH3 CH O C CH CH C O CH H3C O O CH3

32

CHO OH 25

33

H3C

OH CH3

H3C

26

34 C2H5OOC

27

COOC2H5

C C COOC2H5 NC OH

35

H3C

CH3

C CH CH2CH2

CH3

CH3 OH C CH CH2 C CH CH2 CH3

O P C C C4H9

Problems 36-50 (2nd Edition) Problem number

Formula and data

36 Cl 5

Cl

4

6

Cl

O 3 H C 2 1 P(OCH2CH3)2 O

Cl

O

 [ppm] 16.77 64.98 107.75 144.33 113.40 126.10

C atom 1 2 3 4 5 6

J (31P-13C) [Hz] 5.2 6.8 196.7 2.3 -

J (13C-1H) [Hz] 1 JCH = 127.5 ; 2JCH = 2.5 1 JCH = 149.9 ; 2JCH = 4.3 1 JCH = 176.6 3 JCH = 2.3 -

37 T1= 4.7 sec

110.3 102.6 T1= 3.4 sec Cl

O

Cl

O

CH2

124.5 147.3 T1= 61.8 sec T1= 70.4 sec

38a

38b O PB(Oi-C3H7)2

O PB(Oi-C3H7)2 T1= 6.5sec

T1= 4.2 sec

PA(C6H5)2

PA(C6H5)2 S

T1= 4.7 sec

39

T1= 8.5 sec

40 S NH N H

Cl

Cl

S N

OH

Problem number

Formula and data

41

protons of the CH2 group not equivalent : ( AB system) CH3O C CHAHB CHX C OCH3 O OH O

(HA) = 2.78 ppm 2J(HAHB) = 16.5 Hz ; 3J(HAHX) = 4.4 Hz (HB) = 2.71 ppm 2J(HAHB) = 16.5 Hz ; 3J(HBHX) = 6.2 Hz (HX) = 4.44 ppm 3J(HAHX) = 4.4 Hz ; 3J(HBHX) = 6.2 Hz

42

protons of the CH2 group not equivalent : ( AB system) O O CH3O C CHAHB CHX C OCH3 2 NH3 Cl J(H

= 18.3 Hz ; 3J(HAHX) = 5.1 Hz ; J(HBHX) = 5.1 Hz ; 3J(HX-NH) = 5.3 Hz AHB)

3

43 O CH

4 5

3 2

6

N 1

(CHO) (H4) (H5) (H6) (OCH3)

= 10.30 = 8.08 = 6.98 = 8.36 = 4.08

OCH3

ppm d ( 5JHH = 0.83Hz ) ppm dd ( 3JHH = 7.3 Hz ; 4JHH = 2.08 Hz ) ppm ddd ( 3JHH = 7.3 Hz ; 3JHH = 4.83 Hz ; 5JHH = 0.83Hz ) ppm dd ( 3JHH = 7.3 Hz ; 4JHH = 2.08 Hz ) ppm s

44 4

F3C 5

CF3 3

6

2 1

Br

characteristic coupling constants : (CF3) = 122.6 ppm q ( 1J(FC) = 273.0 Hz ) (C3) = 133.5 ppm q ( 2J(FC) = 34.2 Hz )

Problem number

Formula and data

45 Br

46 H3C CH2 PBr2 H3C

characteristic coupling constants : 2 J(PH) = 16.8 Hz ; 1J(PC) = 49.6 Hz 47

F Ti F 3

J(FH) = 1.7 Hz ;

48 Br

Br

49 O

O

50 S S

O CH P

2

J(FC) = 2.5 Hz