Post on 23-Sep-2019
Answers for Student Exercise 3.1 to 3.3
1
3.1(a) 1,2,3,4 are a spin system. Pople notation is A3M2S2X2. 1,2, and 3 are enantiotopic. 3.2 (a)
Spin Appr. δ Table Coupled with Multiplet (n+1) Integration 1 δ 3.40 Appendix A, Chart A.1 2 Triplet 2 2 δ 1.85 Appendix A, Chart A.2 1,3 Quintet 2 3 δ 1.20 Appendix A, Chart A.1 3, 4 Sextet 2 4 δ 0.85 Appendix A, Chart A.1 3 Triplet 3
3.3 (a)
Br
(a) 1-bromobutane
12
34
1 2 3 4
Answers for Student Exercise 3.1 to 3.3
2
3.1 (b) 1,2,3,4,5,6,7 are one spin system, and 9 is the second. 4,4’ ; 5,5’ ; 6,6’and 7,7’ are diastereotopic. 3.2 (b)
Spin Appr. δ Table or Calculation Coupled with Multiplet (n+1) Integration 1 δ 2.80 Appendix A, Chart A.2 6,2,7 Sextet* 1 2 δ 5.59 Appendix D, Chart D2 3,1 Triplet* 1 3 δ 5.59 Appendix D, Chart D2 Quartet* 1 4 δ 1.96 Appendix D, Chart D2 Quartet** 2 5 δ 1.65 Appendix D, Chart D2 4,6 Quartet** 2 6 δ 1.80 Appendix A, Chart A.2 1,5 Triplet** 2 7 δ 4.05 Appendix A, Chart A.2 1 Doublet** 2 9 δ 2.10 Appendix A, Chart A.1 none Singlet 3
* Coupling constants may not be the same, but make the assumption that they are. **These are diastereotopic, and are coupled to diastereotopic protons, they would not be first order. We will assume they are first order for the drawn spectrum. 3.3 (b)
2 and 3 7 1 9 4 6 5
O
1
23
4
5 6
7O
8 9
(b) (cyclohex-2-enyl)methyl acetate
2
Answers for Student Exercise 3.1 to 3.3
3
3.1 (c) 2,3,4,5,6 are a spin system, and 8,9,10 another spin system. 3,3’ ; 4,4’ ; 9,9’ are diastereotopic. 3.2 (c)
Spin Appr. δ Table or Calculation Coupled with Multiplet (n+1) Integration 2 δ 6.68 Appendix D, Chart D.1 3 Triplet 1 3 δ 2.05 Appendix A, Chart A1 2,4 Quartet*** 2 4 δ 2.05 Appendix A, Chart A1 3,5 Quartet*** 2 5 δ 5.68 Appendix D, Chart D1 4,6 Quartet* 1 6 δ 6.22 Appendix D, Chart D1 5 Doublet 1 8 δ 2.65 Appendix A, Chart A.1 9,10 Sextet 1 9 δ 3.20 Appendix A, Chart A.1 8 Doublet** 2 10 δ 1.05 Appendix A, Chart A.2 8 Doublet 3 ΟΗ δ 0.5-4.0 Appendix E 1
* Coupling constants may not be the same, but make the assumption that they are. ** These are diastereotopic, and are coupled to diastereotopic protons; they would not be first order. We will assume they are first order for the drawn spectrum. 3.3 (c)
OH
O
12
3
45
6
78
(c) 1-(cyclohexa-1,5-dienyl)-3-hydroxy-2-methylpropan-1-one
9
10
2 6 5 9 8 3,4 OH 10
Answers for Student Exercise 3.1 to 3.3
4
3.1 (d) 5,4,3 are a spin system and 1 is second. 5,4,3 is Pople notation is A3M2X2; may have long rang coupling. 3+4 are enantiotopic. 3.2 (d)
Spin Appr. δ Table or Calculation Coupled with Multiplet (n+1) Integration 5 δ 0.85 Appendix A, Chart A.2 4 Triplet 3 4 δ 1.50 Appendix A, Chart A2 3,5 Sextet 2 3 δ 2.20 Appendix A, Chart A1 4 Triplet 2 1 δ 1.80 Appendix D, Chart D3 none Singlet 1
3.3 (d)
H
(d) pent-1-yne
13 24
5
3 1 4 5
Answers for Student Exercise 3.1 to 3.3
5
3.1 (e) 3,4,5,6 are a spin system, and 1 is another spin system. Pople notation is A3G2MX. 5 is enantiotopic. 3.2 (e)
Spin Appr. δ Table or Calculation Coupled with Multiplet (n+1) Integration 1 δ 1.86 Appendix D, Chart D.2 None Singlet 3 3 δ 6.09 Appendix D, Chart D.1 4 Doublet 1 4 δ 6.82 Appendix D, Chart D.1 3, 5 Quartet* 1 5 δ 2.05 Appendix A, Chart A1 4, 6 Quintet* 2 6 δ 1.00 Appendix A, Chart A2 5 Triplet 3
* Coupling constants may not be the same, but make the assumption that they are. 3.3 (e)
4 3 5 1 6
O
(e) hex-3-en-2-one
1 34
52 6
Answers for Student Exercise 3.1 to 3.3
6
O
(f) 1-methoxybut-1-ene
1 23
45
3.1 (f) 3,4,5,6 are a spin system, and 1 is another spin system. Pople notation is A3G2MX. 4 is enantiotopic. 3.2 (f)
Spin Appr. δ Table or Calculation Coupled with Multiplet (n+1) Integration 1 δ 3.20 Appendix A, Chart A.1 None Singlet 3 2 δ 6.14 Appendix D, Chart D.1 3 Doublet 1 3 δ 4.63 Appendix D, Chart D.1 2, 4 Quartet* 1 4 δ 2.05 Appendix A, Chart A1 3, 5 Quintet* 2 5 δ 1.00 Appendix A, Chart A2 4 Triplet 3
* Coupling constants may not be the same, but make the assumption that they are. 3.3 (f)
2 3 1 4 5
Answers for Student Exercise 3.1 to 3.3
7
NH
O
O
(g) propyl methylcarbamate
1
2 34
5
3.1 (g) 3,4,5 are a spin system, and 1 is another spin system. Pople notation is A3M2X2. 3 and 4 are enantiotopic. 3.2 (g)
Spin Appr. δ Table or Calculation Coupled with Multiplet (n+1) Integration 1 δ 2.95 Appendix A, Chart A.1 None Singlet 3 3 δ 4.10 Appendix D, Chart D.1 3 Triplet 2 4 δ 1.60 Appendix A, Chart A.2 2, 4 Sextet 2 5 δ 0.85 Appendix A, Chart A1 3, 5 Triplet 3
NH δ 4.5-7.5 Appendix E * Singlet 1 *See Section 3.6.2 for coupling between 1 and N-H. 3.3 (g)
NH 3 1 4 5
Answers for Student Exercise 3.1 to 3.3
8
O O
(h) diethoxymethane
12 3
3.1 (h) 1 and 2 are a spin system, and 3 is another spin system. Pople notation is A3X2. 2 are enantiotopic. 3.2 (h)
Spin Appr. δ Table or Calculation Coupled with Multiplet (n+1) Integration 1 δ 1.20 Appendix A, Chart A.2 2 Triplet 3 2 δ 3.40 Appendix A, Chart A.1 1 Quartet 2 3 δ 4.95 Appendix B, Table B.1 none Singlet 2
3.3 (h)
3 2 1
Answers for Student Exercise 3.1 to 3.3
9
HO
(i) 2-methylpentan-2-ol
1
23
45
6
1 and 6 Large singlet of 6 H
3.1 (i) 3,4 and 5 is a spin system. 1 and 6 are enantiotopic to each other and 3 and 4 are also enantiotopic. Pople notation is A3M2X2. 3.2 (i)
Spin Appr. δ Table or Calculation Coupled with Multiplet (n+1) Integration 1 and 6 δ 1.20 Appendix A, Chart A2 none Singlet 6
3 δ 1.50 Appendix A, Chart A2 4 Triplet 2 4 δ 1.20 Appendix A, Chart A1 3,5 Sextet 2 5 δ 0.85 Appendix A, Chart A2 4 Triplet 3
ΟΗ δ 0.5-4.0 Appendix E none Singlet 1 3.3 (i)
OH 3 1,6,4
Answers for Student Exercise 3.1 to 3.3
10
SS
(j) 1,4-bis(methylthio)butane
1 23
3.1 (j) Not First order (no Pople). See Section 3.11.2. Protons 2 and 3 are enantiotopic. Protons on 1,2,3 are Chemical Shift Equivalent. 3.2 (j)
Spin Appr. δ Table or Calculation Coupled with Multiplet (n+1) Integration 1 δ 2.10 Appendix A, Chart A1 none Singlet 6 or 3 2 δ 2.60 Appendix A, Chart A1 3 Triplet* 4 or 2 3 δ 1.60 Appendix A, Chart A2 2,3 Quintet* 4 or 2
* Would not be first order. We will assume they are first order for the drawn spectrum. 3.3 (j) Actual spectrum would look like this!
012PPM
Answers for Student Exercise 3.1 to 3.3
11
N
O
(k) N,N,4-trimethylbenzamide
1
2
3
4
5
6
78
2'3'
3.1 (k) 2,2’; 3,3’ are CSE but not ME. 6 and7 are diastereotopic. Pople notation for ring is AA’ XX’ 3.2 (k)
Spin Appr. δ Table or Calculation Coupled with Multiplet (n+1) Integration 2, 2’ δ 7.80 Appendix D, Chart D1 none Doublet** 2 3, 3’ δ 7.25 Appendix D, Chart D1 3 Doublet** 2 6,7 δ 2.95 Appendix A, Chart A1 none Singlet* 6* 8 δ 2.25 Appendix A, Chart A1 none Singlet 3
* There will be two singlets because of the restricted rotation of the NR2 ** These can not be first order. Not magnetically equivalent. 3.3 (k)
2,2’ 3,3’ 6,7 8
Answers for Student Exercise 3.1 to 3.3
12
O
OH(l) 1-(2-hydroxyphenyl)ethanone
1
2
34
5
6
78
3.1 (i) N.A. 3.2 (i)
Spin Appr. δ Table or Calculation Coupled with Multiplet (n+1) Integration 3 δ 6.85 Appendix D, Chart D.1 4 Doublet* 1 4 δ 7.05 Appendix D, Chart D.1 3,5 Triplet* 1 5 δ 7.25 Appendix D, Chart D.1 4,6 Triplet* 1 6 δ 7.70 Appendix D, Chart D.1 5 Doublet* 1 8 δ 2.40 Appendix A, Chart A1 none Singlet 3
OH δ 5.5-12.5 Appendix E none Singlet 1 * Will have long range coupling. 3.3 (i) OH 6 5 4 3 8
Answers for Student Exercise 3.1 to 3.3
13
ClH
O(m) 2-chloroacetaldehyde
12
3.1(m) 1 and 2 is a spin system. Pople notation is A2X. 2 protons are enantiotopic. 3.2 (m)
Spin Appr. δ Table Coupled with Multiplet (n+1) Integration 1 δ 4.95 Appendix A, Chart A.1
Plus Appendix B, Table B.1* 2 Triplet 1
2 δ 9.80 Appendix D, Chart D.6 extrapolation
1 Doublet 2
Calc is based on Chart A1 3.45 for ClCH2 plus Table B.1 value of 1.50 for HC=O 3.3 (m)
2 1
Answers for Student Exercise 3.1 to 3.3
14
HH
H
F(n) fluoroethene
1
23
Protocol of the H-1 NMR Prediction:
Node Shift Base + Inc. Comment (ppm rel. to TMS)
H 4.85 5.25 1-ethylene -0.40 1 -F cis H 6.79 5.25 1-ethylene 1.54 1 -F gem H 4.23 5.25 1-ethylene -1.02 1 -F trans
01234567PPM
3.1(n) Charts are not available in text for this problem. Pople notation is AGMX, where X is F. 3.2 (n) 3.3 (n)
Answers for Student Exercise 3.1 to 3.3
15
O
(o) cyclohexyl acetate
O1
2
34
5
6
2'3'
3.1 (o) 1,2,2’,3,3’, and 4 is a spin system, and 6 is another spin system. 2,2,;2’,2’;3,3;3’,3’;4,4 protons are diastereotopic. 2 and 2’ are CSE, also 3 and 3’ are CSE. 3.2 (o)
Spin Appr. δ Table or Calculation Coupled with Multiplet (n+1) Integration 1 δ 4.95 Appendix A, Chart A.1 2,2’ Quintet 1
2,2’ δ 1.60 Appendix A, Chart A.2 1,(3or3’) Quartet* 4 3,3’ δ 1.44 Appendix C, Table C.1 (2 or 2’),4 Quintet* 4
4 δ 1.44 Appendix C, Table C.1 3,3’ Quintet* 2 6 δ 2.00 Appendix A, Chart A.1 none Singlet 3
* These are diastereotopic, and are coupled to diastereotopic protons; they would not be first order. We will assume they are first order for the drawn spectrum. 3.3 (o)
1 6 2,2’ 3,3’, and 4
4
Answers for Student Exercise 3.4 16
0.80.91.01.11.21.31.41.51.61.71.81.92.02.12.22.32.42.5 ppm
Problem 4.3 A C7H14O
Problem 4.3 B C7H16O
0.80.91.01.11.21.31.41.51.61.71.81.92.02.12.22.32.42.5 ppm
1.01.52.02.53.03.54.0 ppm
2.53.03.54.04.55.05.56.06.57.07.5 ppm
1.01.52.02.53.03.5 ppm
Problem 4.3 C C7H7Br
Problem 4.3 D C5H11Br
OH
Problem 3.4 A
Problem 3.4 D
Problem 3.4 C
Problem 3.4 B
O
4-Heptanone
1 23
4
OH
3-Heptanol
1 23 4 5 6
7
Br
4-Bromotoluene
1
2 3
45
Br
2-Bromopentane
1 23
45
1
2 3
1
7 6
2 4
5 3
2
5
3
1
2
4
5
3
Answers for Student Exercise 3.4 17
0.80.91.01.11.21.31.41.51.61.71.81.92.02.12.22.32.42.5 ppm
2.02.22.42.62.83.03.23.43.63.84.04.24.4 ppm
0.80.91.01.11.21.31.41.51.61.71.81.92.02.12.22.32.42.52.62.7 ppm
1.52.02.53.03.5 ppm
Problem 4.3 E C6H12O
Problem 4.3 F C3H6O2
Problem 4.3 G C4H11N
Problem 4.3 H C3H4O
11.512.0 ppm
x4
Problem 3.4 E
Problem 33.4 H
Problem 3.4 G
Problem 3.4 F
1 23
45 6
O
2-Hexanone
12
3
O
HOPropionic Acid
12
34
H2N
Butylamine
1 23
OH
Propargyl alcohol
1
6
4 5 3
2
3
OH
1 OH 3
1
2
4
3
NH2
Answers for Student Exercise 3.4 18
1.01.52.02.53.03.54.04.5 ppm
1.01.52.02.53.03.5 ppm
1.52.02.53.03.54.04.55.05.56.06.57.07.5 ppm
2.02.53.03.54.04.55.05.56.06.57.07.5 ppm
Problem 4.3 I C3H7NO2
Problem 4.3 J C8H10O2
Problem 4.3 K C8H10O
Problem 4.3 L C5H10O2
210021502200 Hz 115012001250 Hz
Problem 3.4 I
Problem 3.4 L
Problem 3.4 K
Problem 3.4 J
N+
O
-O1-Nitropropane
1 2 3
46
O
OH
2-Phenoxylethanol
1
23
5
45
6
O
Phenetole
12 3
5
OO
Methyl Butyrate
1 2 34
6
2 4
5
3
2 4
5 3
6 2
4
5
3
OH
1 2
3
Answers for Student Exercise 3.4 19
1.52.02.53.03.54.04.55.05.56.06.57.0 ppm
7.37.47.57.67.77.87.98.08.18.2 ppm
4.04.55.05.56.06.57.07.5 ppm
3.03.54.04.55.05.56.06.57.07.58.08.5 ppm
2490250025102520 Hz
Problem 4.3 M C6H11NO
Problem 4.3 N C5H6N2 A pyrazine
Problem 4.3 O C8H8O3
Problem 4.3 P C6H4ClNO2
2150220022502300 Hz
1213 ppm
1 proton (x32)
Problem 3.4 M
Problem 3.4 P
Problem 3.4 O
Problem 3.4 N a pyrazine
56
O
HN
Caprolactam
1 23
4
6
7
N
N
2-methylpyrazine1
2
3 4
5
6
7
HOO
O
Methyl salicylate
1
2
3 4
5
8
N+O
-OCl
1-Chloro-4-Nitrobenzene
1
23
4
7
6
5
3
6
4
5 3 8
OH
2 3
6
2 4
5 3
NH
Answers for Student Exercise 3.4 20
1.52.02.53.03.54.04.5 ppm
1.01.21.41.61.82.02.22.42.62.8 ppm
1.52.02.53.03.54.04.55.0 ppm
1.61.82.02.22.42.62.83.03.23.43.6 ppm
150015101520 Hz
1112 ppm
1 proton (x32)
Problem 4.3 R C6H10O Unsaturated alcohol
Problem 4.3 Q C7H13BrO2 Acid
Problem 4.3 S C8H14O Unsaturated ketone
Problem 4.3 T C6H12O2
1213 ppm
Problem 3.4 Q
Problem 3.4 T
Problem 3.4 S Unsaturated ketone
Problem 3.4 R
7
O
HO Br7-Bromo Heptanoic Acid
12 3 4
5 6
HO5-Hexyn-1-ol
52
341
6
7
8O
6-Methyl-5-hepten-2-one2
341 5
6
O
HOHexanoic acid
342
1 5 6
6
2
4 5
3
OH
1
7
4
5
3
8
1
6
2 4
3 OH
7
6 2
5 4 3
OH
Answers for Student Exercise 3.4 21
5.75.85.96.06.16.26.36.46.56.66.76.86.97.07.17.27.37.47.5 ppm
2.02.53.03.54.04.55.05.56.06.57.0 ppm
2.53.03.54.04.55.05.56.06.57.0 ppm
Problem 4.3 W C6H8O Unsaturated ketone
Problem 4.3 V C8H10O
Problem 4.3 U C6H4OCl2
2060208021002120 Hz
Problem 3.4 U
Problem 3.4 W Unsaturated ketone
Problem 3.4 V
Cl
ClHO
2,6-Dichlorophenol
1
3
4
2
5HO
2,6-Dimethylphenol
3
4
1 2
O
2-Cyclohexen-1-one
34
12
5
6
6
2
4 5
3
4
3
5
OH
4
3 OH
22
3.5A. (1.58(300) – 0.90(300))/7 = about 29; (2.36(300) – 1.58(300))/7 = about 33 3.5E. (2.38(300) – 1.52(300))/7 = about 37; (1.52(300) – 1.38(300))/7 = about 6 (1.38(300) – 0.87(300))/7 = about 22 3.5F. (2.38(300) – 1.13(300))/7 = about 54 3.5G. (2.61(300) – 1.35(300))/7 = about 54; (1.35(300) – 1.29(300))/7 = about 2 to 3 (1.29(300) – 0.84(300))/7 = about 19 3.5H. Long range coupling; can’t measure coupling constant 3.5I. (4.33(300) – 2.01(300))/7 = about 99; (2.01(300) – 0.99(300))/7 = about 44 3.5K. (4.08(300) – 1.44(300))/7 = about 113 3.5L. (2.26(300) – 1.62(300))/7 = about 27; (1.62(300) – 0.92(300))/7 = about 30 3.5Q. (3.39(300) – 1.86(300))/7 = about 66; (2.36(300) – 1.65(300))/7 = about 30 (1.86(300) – 1.47(300))/7 = about 17; (1.65(300) – 1.37(300))/7 = about 12 3.5U. (7.25(300) – 6.81(300))/9 = about 15
Answers for Student Exercise 3.5
Answers for Student Exercises 3.6 23
O
4-Heptanone
A
1 23
4
Pople Notation -- A3M2X2; 2 and 3 (and 5 and 6) are enantiotopic. 3 and 5, 2 and 6 are CSE.
OH
3-Heptanol
B
1 23 4 5 6
7
One spin system. 2,2’; 4,4’; 5,5’, and 6,6’ are diastereotopic.
Br
4-Bromotoluene
C
1
2 3
45
Two spin systems. Pople Notation -- AA’XX’. 2 and 2’ are CSE but not magnetically equivalent as are 3 and 3’.
Br
2-Bromopentane
D
1 23
45
One spin system. 3 and 3’; 4 and 4’ are diastereotopic.
1 23
45 6
O
2-Hexanone
E Two spin systems. 3, 4, and 5 are enantiotopic.
12
3
O
HOPropionic Acid
F A3X2. Protons on two are enantiotopic.
Answers for Student Exercises 3.6 24
12
34
H2N
Butylamine
G Pople Notation – A3G2M2X2One spin system. 1, 2, and 3 are enantiotopic.
1 23
OH
Propargyl alcohol
H One spin system with long range coupling. Protons on 2 enantiotopic.
N+
O
-O1-Nitropropane
I
1 2 3
Pople Notation -- A3M2X2. 1 and 2 are enantiotopic.
46
O
OH
2-Phenoxylethanol
J
1
23
5
2'3'
Two spins systems; Pople Notation -- AA’MM’X and A2X2. 2 and 2’are CSE but not magnetically equivalent as are 3 and 3’. 5 and 6 are enantiotopic.
45
6
O
Phenetole
K
12 3
2' 3'
Two spins systems; Pople Notation -- AA’MM’X and A3X2. 2 and 2’are CSE but not magnetically equivalent as are 3 and 3’. 5 and 6 are enantiotopic.
5
OO
Methyl Butyrate
L
1 2 34
Two spin systems; Pople Notation --A3M2X2. 2 and 3 are enantiotopic.
Answers for Student Exercises 3.6 25
56
O
HN
Caprolactam
M
1 23
4
One spin system. 2,2’; 3,3’; 4,4’; 5,5’ and 6,6’ are diastereotopic.
6
7
N
N
2-methylpyrazine
N
1
2
3 4
5
Pople Notation –AB for 5 and 6. Three spins systems.
6
7
HOO
O
Methyl salicylate
O
1
2
3 4
5
8
Two spin systems. Pople Notation – AGMX.
N+O
-OCl
1-Chloro-4-Nitrobenzene
P
1
23
4
3' 2'
One spin system. 2 and 2’ are CSE but not magnetically equivalent as are 3 and 3’.
7
O
HO Br7-Bromo Heptanoic Acid
Q
12 3 4
5 6
One spin system. 2, 3, 4, 5, 6, and 7 are enantiotopic.
Answers for Student Exercises 3.6 26
HO5-Hexyn-1-ol
R
52
341
6
One spin system with long range coupling. 1, 2, 3, and 4 are enantiotopic.
7
8O
6-Methyl-5-hepten-2-one
S
23
41 56
At 2 spins systems (7 and 8 probably have long range coupling). 3 and 4 are enantiotopic.
O
HOHexanoic acid
T
342
1 5 6
One spin system. 2, 3, 4, and 5 are enantiotopic.
Cl
ClHO
2,6-Dichlorophenol
U
1
3
4
2
3'
One spin system. 3 and 3’ are CSE but not magnetically equivalent.
5HO
2,6-DimethylphenolV
3
4
1 26
3'
One spin system. 3 and 3’ are CSE but not magnetically equivalent. 5 and 6 are CSE.
O
2-Cyclohexen-1-one
W
34
12
5
6
One spin system. 4, 5, and 6 are enantiotopic.
27Answers for Student Exercise 3.7
JAX
AX
A2X
A3X
11
11
1 2 1
11
1 3 3 1
1 2 1
JXA
11
JAX
11
JXA
XA
JXA
JAX
11
A2X2
11
1 2 1
11
1 2 1
A3X2
11
1 2 1
JAX
JAX
11
1 3 3 1
1 2 1
JXA
JXA
A X
A2 X
A3
X
A2 X2
A3
X2
28Answers for Student Exercise 3.8
JAM
JAM
JMX
AMX
A2MX
A3MX
11
11
JMX
11
1 2 1
JAM
JMX
11
1 3 3 1
1 2 1
JAM
JMX
11
1 3 3 1
1 2 1
1 4 6 4 1
A M X
A2 M X
A3 M X
With the AMX systems A and X do not change; they are represented as follows:
29Answers for Student Exercise 3.8 Continued
A2MX2
JAM
JMX
11
1 3 3 1
1 2 1
1 4 6 4 1
A2MX3
JAM
JMX
11
1 3 3 1
1 2 1
1 4 6 4 1
1 5 10 10 5 1
JAM
JMX
11
1 3 3 1
1 2 1
1 4 6 4 1
1 5 10 10 5 1
A3MX3
1 6 15 20 15 6 1
A2 M X2
A2
M
X3
A3 M X3
30Answers for Student Exercise 3.9
JAM
JAM
JMX
JMX
AMX
A2MXJAM
JAM
1 1 1
11 11
11
1
2 2
A3MX
1 3 3 1
1 1
1 12
JMX1 1 1 13 3 3 3
1 12
1 1
11
11
JMX
With the AMX systems A and X do not change; they are represented as follows:
The following are for M:
31Answers for Student Exercise 3.9 Continued
A2MX3 JAM
11 11 2 2
1 12
1 1
JMX
1 2 3 4 3 2 1
3 5 7 7 5 3 1
JAMA3MX3
1
1 3 3 1
1 1
1 12
JMX1 1 1 13 3 3 3
2 4 6 6 6 4 2 1
1 3 6 10 12 12 10 6 3 1
A2MX2 JAM
11 11 2 2
1 1
1 2 3 4 3 2 1
1 12
JMX
1
32Answers for Student Exercise 3.10A
Jca
11
1 2 1 Jcb
1 1 2 2 1 1
(c)
Spin System ab2c3Couplinga-b 6 Hzb-c 12Hza-c 3 Hz
11
1 2 1
Jbc
Jba
1 3 3 1
1 1 3 3 3 3 1 1(b)
Jab
1 1
1 2 1
1 1 2 2 1 1
1
1
2 3 4 3 2 1
3 5 7 7 5 3 1
Jac(a)
33
Jcb
Jca
(c)
Spin System ab2c3Couplinga-b 6 Hzb-c 4Hza-c 9 Hz
Jbc
Jba
(b)
Jab
Jac(a)
1 1
1 1 1 1
1 2 1 1 2 1
1 1
1 1 1 1
1
1 1
12 11 2
3 1 3 3 1 3
1 1
1
1
1
1
2 1
3 3 1
1 3 3 3 3 1 1
2 3 1 6 3 3 6 1 3 2 1
Answers for Student Exercise 3.10B
34Answers for Student Exercise 3.10C
Jcb
Jca
(c)
Spin System a2b2c2Couplinga-b 6 Hzb-c 4Hza-c 9 Hz
Jbc
Jba
(b)
Jab
Jac
(a)
1
1
1
1 1
1
1
21 41
1
2
1
4
11
1 2
2
22
22
1
1
1
1
1
1
1
11
1 2
2
2
2
22 241 1
1
1
1
1
1
1
1
1
1
1
2
2
2
2
2 2