1D TopSPIN Multiplet Table ID Shift [ppm] J[Hz] M 1 7.9901 7.0159 2 7.6243 6.9655 7.2951 6.9655 3 2 4 5 6 7.4201 6.9488 2 1.1999 5.9920 3 2.8059 5.9976 4 5.056 ppm/682.51 Hz Index 14527-14553 Value--1.909e-05 rel 11 00001 25660 8066 0 0.9917 Connection J(1, 0) J(2, 0) J(3, 0) J(4,0) J(5,0) J(6, 0) Unknown II C10H12O, Ketone R 1.9863 5267 Z [ppm]
1D TopSPIN Multiplet Table ID Shift [ppm] J[Hz] M 1 7.9901 7.0159 2 7.6243 6.9655 7.2951 6.9655 3 2 4 5 6 7.4201 6.9488 2 1.1999 5.9920 3 2.8059 5.9976 4 5.056 ppm/682.51 Hz Index 14527-14553 Value--1.909e-05 rel 11 00001 25660 8066 0 0.9917 Connection J(1, 0) J(2, 0) J(3, 0) J(4,0) J(5,0) J(6, 0) Unknown II C10H12O, Ketone R 1.9863 5267 Z [ppm]
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...
Related questions
Question
Find unknown and show ALL work and arrows etc
![1D TopSPIN Multiplet Table
ID Shift [ppm] J[Hz] M
7.9901 7.0159 2
7.6243 6.9655 3
7.2951 6.9655 3
7.4201 6.9488 2
1.1999 5.9920
2.8059 5.9976 4
4
5
6
5.056 ppm / 682.51 Hz
Index - 14527-14553
Value-1.909e-05 rel
Connection
J(1, 0)
J(2, 0)
J(3, 0)
J(4,0)
J(5,0)
J(6, 0)
HE
Unknown II
C10H12O, Ketone
1
-2.9) 45
98762
(ppm)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F09c533f0-773b-4ddb-abaf-f88b025c7ffb%2F24f1cb05-0e97-4601-9cc4-d28545fc38ba%2F4ivymvls_processed.jpeg&w=3840&q=75)
Transcribed Image Text:1D TopSPIN Multiplet Table
ID Shift [ppm] J[Hz] M
7.9901 7.0159 2
7.6243 6.9655 3
7.2951 6.9655 3
7.4201 6.9488 2
1.1999 5.9920
2.8059 5.9976 4
4
5
6
5.056 ppm / 682.51 Hz
Index - 14527-14553
Value-1.909e-05 rel
Connection
J(1, 0)
J(2, 0)
J(3, 0)
J(4,0)
J(5,0)
J(6, 0)
HE
Unknown II
C10H12O, Ketone
1
-2.9) 45
98762
(ppm)
Expert Solution
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Step 1: NMR spectroscopy
- 1H NMR spectroscopy is useful to determine structure of organic compounds.
- Proton NMR spectroscopy gives information about types of protons, their signals, splitting etc hence we can determine structure of organic compounds.
- Each different set of protons gives signal in NMR spectrum. Number of signals obtained as per number of sets of protons present in a compound.
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