I really need help analyzing and interpreting these spectra to find a chemical formula and compound structure!

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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I really need help analyzing and interpreting these spectra to find a chemical formula and compound structure!

210
200
190
84
-17094
170
160
150
140
130
120
110
100
f1 (ppm)
90
-77.16 CDCB
-64.05
13C NMR in CDCl3 (solvent is labeled)
-30.34
30
69 02-
2881
20
10
-10
-4500000
--4000000
-3500000
-3000000
-2500000
-2000000
-1500000
-1000000
-500000
Transcribed Image Text:210 200 190 84 -17094 170 160 150 140 130 120 110 100 f1 (ppm) 90 -77.16 CDCB -64.05 13C NMR in CDCl3 (solvent is labeled) -30.34 30 69 02- 2881 20 10 -10 -4500000 --4000000 -3500000 -3000000 -2500000 -2000000 -1500000 -1000000 -500000
10.0
2.5
4.0
9.0
3.5
8.5
T
3.0
8.0
7.26 CDC13
25
11 (ppm)
7.5
7.0
2.0
SS
6.5
15
B
10
5.5
-4.0x10²
--3.0x10²
-2.0x10²
10x10
00
5.0
11 (ppm)
4.5
4.07
---4.05
007
4.0
2.03
3.5
291
29 T
1.60
3.0
AST
AST
1.58
1.57
KT.
25
1.57
1.55
1.54
1.40
F000-2
20
-1.38
at
86 19
Forz
102
TTT T
15
1.36
F-2009
1.38
0.94
0.5
260
¹H NMR in CDC13 (solvent is labeled). Multiplicity for signals 1.25-1.75 is multiplet.
0.89
4.5×10²
40x10
-35x10
-3.0x10²
--25×10²
-20x10
15x10²
10x10
--50x10⁰
Loo
Transcribed Image Text:10.0 2.5 4.0 9.0 3.5 8.5 T 3.0 8.0 7.26 CDC13 25 11 (ppm) 7.5 7.0 2.0 SS 6.5 15 B 10 5.5 -4.0x10² --3.0x10² -2.0x10² 10x10 00 5.0 11 (ppm) 4.5 4.07 ---4.05 007 4.0 2.03 3.5 291 29 T 1.60 3.0 AST AST 1.58 1.57 KT. 25 1.57 1.55 1.54 1.40 F000-2 20 -1.38 at 86 19 Forz 102 TTT T 15 1.36 F-2009 1.38 0.94 0.5 260 ¹H NMR in CDC13 (solvent is labeled). Multiplicity for signals 1.25-1.75 is multiplet. 0.89 4.5×10² 40x10 -35x10 -3.0x10² --25×10² -20x10 15x10² 10x10 --50x10⁰ Loo
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hello, thanks so much for this!! i also have an IR and Mass Spec spectra with this assignment. I figured out the IR to match these same signals and come up  with this same compound, but I cant seem to figure out how to do the fragmentation on the Mass Spec. It is given then M+ = 116, but there's baRely a peak there

Relative Intensity
100
80
Ö
O
20
миниито
10
M+= 116
20
30 40
50
60
m/z
70
powqurubluquruqtangengungstag
100 110 120
80
90
Transcribed Image Text:Relative Intensity 100 80 Ö O 20 миниито 10 M+= 116 20 30 40 50 60 m/z 70 powqurubluquruqtangengungstag 100 110 120 80 90
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