CHз он нас но "CHз H2N CHз НаС .CH 3 + CH нс CHз н н CHз -CH3 Hас 2,4-Bis(a.,a-dimethylbe nzyl) phenol Ligand 6 formaldehyde N,N-dimethylethylenediamine 4.0096 g 0.9 mL 0.66 mL 1H- Ligand 7 12 11 10 8 7 6 5 3 2 1 ppm 1.94 _10.116 000 7.205 000 6.870 CO 3.93 3.557 2.613 2.598 2.583 1.98 2.01 2.468 2.456 5.92 2.444 2.363 1.98 1.794 1.782 1.771 17.93 18.02 90 1.278

Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Question

Using the reaction below and using Ligand 6 to determine the HNMR shown below.

Ligand 6 is a symmetrical Ligand.

Label all the H protons.

 

CHз
он нас
но
"CHз
H2N
CHз
НаС
.CH 3
+
CH
нс
CHз
н
н
CHз
-CH3
Hас
2,4-Bis(a.,a-dimethylbe nzyl)
phenol
Ligand 6
formaldehyde
N,N-dimethylethylenediamine
4.0096 g
0.9 mL
0.66 mL
Transcribed Image Text:CHз он нас но "CHз H2N CHз НаС .CH 3 + CH нс CHз н н CHз -CH3 Hас 2,4-Bis(a.,a-dimethylbe nzyl) phenol Ligand 6 formaldehyde N,N-dimethylethylenediamine 4.0096 g 0.9 mL 0.66 mL
1H- Ligand 7
12
11
10
8
7
6
5
3
2
1
ppm
1.94
_10.116
000
7.205
000
6.870
CO
3.93
3.557
2.613
2.598
2.583
1.98
2.01
2.468
2.456
5.92
2.444
2.363
1.98
1.794
1.782
1.771
17.93
18.02
90
1.278
Transcribed Image Text:1H- Ligand 7 12 11 10 8 7 6 5 3 2 1 ppm 1.94 _10.116 000 7.205 000 6.870 CO 3.93 3.557 2.613 2.598 2.583 1.98 2.01 2.468 2.456 5.92 2.444 2.363 1.98 1.794 1.782 1.771 17.93 18.02 90 1.278
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