1000800600400200CPS500400300200100-Ó CPSH NMR60 MHz2502001501005010050806020Integral = 34030CAH&O2Integral = 3Integral = 28.07.06.05.04.03.02.01.0O PPM

Double bond equivalence value is given by, DBE= (no of Carbon +1) - monovalent atom/2 + trivalent…

Answered: 1000 500 250 400 300 200 100 Ó CPS 1H…

1000 500 250 400 300 200 100 Ó CPS 1H NMR 60 ΜHz 200 50 100 80 Integral = 3 50 Integral = 3 Integral = 2 8.0 7.0 6.0 5.0 4.0 3.0 2.0 1.0 O PPM

Principles of Instrumental Analysis

7th Edition

ISBN:9781305577213

Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch

Publisher:Douglas A. Skoog, F. James Holler, Stanley R. Crouch

Chapter19: Nuclear Magnetic Resonance Spectroscopy

Section: Chapter Questions

Problem 19.13QAP

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Question

1000
800
600
400
200
CPS
500
400
300
200
100
-Ó CPS
H NMR
60 MHz
250
200
150
100
50
100
50
80
60
20
Integral = 3
40
30
CAH&O2
Integral = 3
Integral = 2
8.0
7.0
6.0
5.0
4.0
3.0
2.0
1.0
O PPM

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