Structure & Reactivity

Nuclear Magnetic Resonance Spectroscopy

NMR2D.7.  2D NMR Solutions

Problem NMR2D1.1.

ethyl butanoate

Problem NMR2D1.2.

Problem NMR2D1.3.

Problem NMR2D1.4.

Problem NMR2D1.5.

Problem NMR2D1.6.

Problem NMR2D1.7.

Problem NMR2D1.8.

Problem NMR2D1.9.

Problem NMR2D1.10.

Problem NMR2D1.11.

 

Problem NMR2D1.12.

Problem NMR2D2.1.

Problem NMR2D2.2.

Problem NMR2D2.3.

Problem NMR2D3.1.

Problem NMR2D3.2.

Problem NMR2D3.3.

Problem NMR2D3.4.

 

Problem NMR2D3.5.

Problem NMR2D3.6.

Problem NMR2D3.7.

Problem NMR2D3.8.

 

Problem NMR2D.4.1.

Problem NMR2D.4.2.

Problem NMR2D.4.3.

 a) Due to resonance, there is substantial pi character to the amide bond which restricts free rotation around that bond.

b)

c)

 

Problem NMR2D.4.4.

 

Problem NMR2D.4.5.

 

Problem NMR2D.4.6.

a)

b)

c)

 

Problem NMR2D.4.7.

Problem NMR2D5.1.

Problem NMR2D5.2.

Problem NMR2D5.3.

Problem NMR2D5.4.

Problem NMR2D5.5.

 

Problem NMR2D6.1.

1H NMR

Chemical shift (ppm) Integration Multiplicity Partial Structure
5.54 2H multiplet CH=C (x 2)
4.17 2H doublet O-CH2-CH
2.08 2H quintet CH-CH2-CH3
1.47 1H boad singlet OH
0.95 3H triplet CH2-CH3

* total # H: 10

 

13C NMR

Chemical shift (ppm)

Type of carbon
134 sp2
128

sp2

58 sp3-O
21 sp3
14 sp3

*total # C: 5

 

COSY

Assignment 1H COSY
A 0.95 2.08
B 2.08 0.95, 5.54
C 4.17 5.54
D 5.54 2.08
E 5.54 4.17

*HMQC indicates two hydrogens at 5.54 are in two different environments

 

HMQC

Assignment 13C 1H
A 14 0.95
B 21 2.08
C 58 4.17
D 128 5.54
E 134 5.54

 

Formula:

C5H10O (1 O indicated from shift in 13C, 1H NMR)

FW = (5 x 12) + (10 x 1) + 1 x 16) = 86

Compare C5H10 ratio to C5H12 in hydrocarbon

Degrees of unsaturation =  [(2 x 5) + 2 - 10] / 2 = 1 unit (1 double bond)

 

The data tables should be consitent with this structure:

pent-2-en-1-ol (could be cis or trans based on this analysis)

 

Problem NMR2D6.2.

1H NMR:

Chemical shift (ppm) Integration Multiplicity Partial structure
4.7 5H singlet solvent
3.93 1H triplet CH2-CH-N
2.40 2H multiplet CH2-CH2?
2.09 2H multiplet CH2-CH2?
1.4 9H singlet C(CH3)3

*Total number of H: 19 H

 

13C NMR:

Chemical shift (ppm) Type of carbon
170 sp2 (C=O)
80 sp3 (C-O)
52 sp3 (C-N)
32 sp3
28 sp3
26 sp3

*Total number of C: 6 apparent, but two more suggested by symmetry (3 methyl groups in 1H NMR) for 8 C; a third extra suggested by MW fit for 9 C

 

COSY:

Assignment 1H COSY
Solvent 4.7 --
B 3.93 2.40
D 2.40 2.09
C 2.09 2.40, 2.09
A 1.4 --

 

Formula:

C9H18O3N2 (extra O indicated from shift in 13C, 1H NMR; second O suggested by C=O in 13C NMR; additional CO needed to fit MW)

FW = (9 x 12) + (18 x 1) + (3 x 16) + (2 x 14) = 202

FW = (9 x 12) + (18 x 1) + (3 x 16) + (2 x 14) = 202

Compare C9H18 to C9H22 for the corresponding hydrocarbon corrected for two nitrogens (therefore two extra hydrogens)

Degrees of unsaturation = [(2 x 9) + 2 + 2 - 10] / 2 = 2 units (2 double bonds)

 

The data tables should be consistent with this structure:

  

Problem NMR2D6.3.

The data should be consistent with this structure:

 

Problem NMR2D6.4.

1H NMR:

Chemical shift (ppm)

Integration Multiplicity Partial Structure
4.71 -- singlet solvent
4.17 1H doublet? CO-CH-N
3.75 3H singlet O-CH3
2.25 1H multiplet CH-CH-(CH3)2
0.92 6H triplet? 2 x CH3

 

13C NMR:

Chemical shift (ppm) Type of carbon
170 sp2 C=O
60 sp3 C-N
52 sp3 C-O
30 sp3 C
19 sp3 C

 

COSY:

Assignment 1H COSY
A 4.17 2.25
B 3.75 --
C 2.25 4.17
D 0.92 2.25

Formula:

C6H13O2N (1 O indicated from shift in 13C, 1H NMR)

FW = (6 x 12) + (13 x 1) + (2 x 16) + (1 x 14) = 131

Degrees of unsaturation = [(2 x 6) + 2 + 1 - 13] / 2 = 1 unit (1 double bond)

 

The data should be consistent with this structure:

 

Problem NMR2D6.5.

The data should be consistent with this structure:

 

Problem NMR2D6.6.

The data should be consistent with this structure:

 

Problem NMR2D6.7.

The data should be consistent with this structure:

 

Problem NMR2D6.8.

The data should be consistent with this structure:

 

Problem NMR2D6.9.

The data should be consistent with this structure:

 

Problem NMR2D6.10.

The data should be consistent with this structure:

 

Problem NMR2D6.11.

The data should be consistent with this structure:

 

Problem NMR2D6.12.

The data should be consistent with this structure:

 

Problem NMR2D6.13.

1H NMR:

Chemical shift (ppm) Integration Multiplicity Partial structure
5.7 1H singlet C=CH-CO
2.77 1H multiplet CH2-CH-CO
2.62 1H multiplet C-CH-C
2.39 1H multiplet C-CH-C
2.05 1H doublet? C-CH-CH?
1.96 3H singlet C-CH3
1.48 3H singlet C-CH3
0.98 3H singlet C-CH3

 

13C NMR:

Chemical shft (ppm) Type of carbon
204 sp2 C=O
170 sp2
121 sp2
59 sp3
55 sp3
50 sp3
41 sp3
28 sp3
24 sp3
22 sp3

 

COSY:

Assignment 1H COSY
1 2.39 2.77, 2.62?
3 5.7 2.05?
5 2.62 2.62?
7a 2.77 2.05
7b 2.05 2.77
8 1.48 --
9 0.98 --
10 1.96 --

 

Formula:

C10H14O (1 O indicated from shift in 13C, 1H NMR)

FW = (10 x 12) + (14 x 1) + (1 x 16)  = 150

Degrees of unsaturation = [(2 x 10) + 2  - 14] / 2 = 4 units (e.g. 2 rings, 2 double bonds)

 

The data should be consistent with this structure:

 

Problem NMR2D6.14.

The data should be consistent with this structure:

 

Problem NMR2D6.15.

The data should be consistent with this structure:

 

 

 

Contributions from Kate Graham, College of Saint Benedict | Saint John's University

 

This site is written and maintained by Chris P. Schaller, Ph.D., College of Saint Benedict / Saint John's University (with contributions from other authors as noted).  It is freely available for educational use.

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Structure & Reactivity in Organic, Biological and Inorganic Chemistry by Chris Schaller is licensed under a Creative Commons Attribution-NonCommercial 3.0 Unported License

Send corrections to cschaller@csbsju.edu

 

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