Answers for Student Exercise 3.1 to 3.3 1 - NTNUfolk.ntnu.no/audunfor/6....
Transcript of Answers for Student Exercise 3.1 to 3.3 1 - NTNUfolk.ntnu.no/audunfor/6....
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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)
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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
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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
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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
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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
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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
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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
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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
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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
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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.
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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.
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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.
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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.
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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
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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:
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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
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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:
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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
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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)
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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
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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