EBK EXPERIMENTAL ORGANIC CHEMISTRY: A M
EBK EXPERIMENTAL ORGANIC CHEMISTRY: A M
6th Edition
ISBN: 9781305687875
Author: Gilbert
Publisher: CENGAGE LEARNING - CONSIGNMENT
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Chapter 10.7, Problem 12E
Interpretation Introduction

Interpretation: Differences observed in the IR and NMR spectra of norborneneand exo-norborneol should be discussed.

Concept introduction:The IR spectrum of compound helps to detect the presence of the functional group in molecule. The functional groups that are present in compound absorb certain IR radiation at a characteristic wavelength or wavenumber. For example, the carbonyl group present in compound shows a characteristic peak around 1700 cm1 in the IR spectrum.

Infrared absorption frequencies values are given as follows:

  EBK EXPERIMENTAL ORGANIC CHEMISTRY: A M, Chapter 10.7, Problem 12E

The scale used for the NMR spectrum is delta scale. δ denotes parts per million units. The lower values of δ denote upfield while higher values of δ denote downfield region. Each peak in the NMR spectrum corresponds to distinct hydrogen.

The area within each peak corresponds to the number of equivalent protons found at that chemical shift values.

Spin-spin splitting is observed as a result of the interaction amongst non-equivalent NMR active nuclei. This is independent of the strength of the external magnetic field. The formula to calculate the number of peaks in the 1H NMR spectra is as follows:

  Number of peaks=N+1

Where,

  • N is the number of equivalent protons on the adjacent carbon atoms.

Thus if a fragment is CH2CH3 then it would show a three proton triplet due to CH3 and a two proton quartet in accordance with N+1 the rule.

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Chapter 10 Solutions

EBK EXPERIMENTAL ORGANIC CHEMISTRY: A M

Ch. 10.2 - Prob. 11ECh. 10.2 - Prob. 12ECh. 10.2 - Prob. 13ECh. 10.2 - Prob. 14ECh. 10.2 - Prob. 15ECh. 10.2 - Prob. 16ECh. 10.2 - Prob. 17ECh. 10.3 - Prob. 1ECh. 10.3 - Prob. 2ECh. 10.3 - Prob. 3ECh. 10.3 - Prob. 4ECh. 10.3 - Prob. 5ECh. 10.3 - Prob. 6ECh. 10.3 - Prob. 7ECh. 10.3 - Prob. 8ECh. 10.3 - Prob. 9ECh. 10.3 - Prob. 10ECh. 10.3 - Prob. 11ECh. 10.3 - Prob. 12ECh. 10.3 - Prob. 13ECh. 10.3 - Prob. 14ECh. 10.3 - Prob. 15ECh. 10.3 - Prob. 16ECh. 10.3 - Prob. 17ECh. 10.3 - Prob. 18ECh. 10.3 - Prob. 19ECh. 10.3 - Prob. 20ECh. 10.3 - Prob. 21ECh. 10.3 - Prob. 22ECh. 10.3 - Prob. 23ECh. 10.3 - Prob. 24ECh. 10.3 - Prob. 25ECh. 10.3 - Prob. 26ECh. 10.3 - Prob. 27ECh. 10.3 - Prob. 28ECh. 10.3 - Prob. 29ECh. 10.3 - Prob. 30ECh. 10.3 - Prob. 31ECh. 10.3 - Prob. 32ECh. 10.5 - Prob. 1ECh. 10.5 - Prob. 2ECh. 10.5 - Prob. 3ECh. 10.5 - Prob. 4ECh. 10.5 - Prob. 5ECh. 10.5 - Prob. 6ECh. 10.5 - Prob. 7ECh. 10.5 - Prob. 8ECh. 10.5 - Prob. 9ECh. 10.5 - Prob. 10ECh. 10.5 - Prob. 11ECh. 10.5 - Prob. 12ECh. 10.5 - Prob. 14ECh. 10.5 - Prob. 15ECh. 10.5 - Prob. 16ECh. 10.5 - Prob. 17ECh. 10.5 - Prob. 18ECh. 10.5 - Prob. 19ECh. 10.5 - Prob. 20ECh. 10.5 - Prob. 23ECh. 10.6 - Prob. 1ECh. 10.6 - Prob. 2ECh. 10.6 - Prob. 3ECh. 10.6 - Prob. 4ECh. 10.6 - Prob. 5ECh. 10.6 - Prob. 6ECh. 10.6 - Prob. 7ECh. 10.6 - Prob. 8ECh. 10.6 - Prob. 9ECh. 10.6 - Prob. 10ECh. 10.6 - Prob. 11ECh. 10.6 - Prob. 12ECh. 10.6 - Prob. 13ECh. 10.6 - Prob. 14ECh. 10.6 - Prob. 15ECh. 10.6 - Prob. 16ECh. 10.6 - Prob. 17ECh. 10.6 - Prob. 18ECh. 10.6 - Prob. 20ECh. 10.6 - Prob. 21ECh. 10.6 - Prob. 22ECh. 10.6 - Prob. 23ECh. 10.6 - Prob. 24ECh. 10.6 - Prob. 25ECh. 10.6 - Prob. 26ECh. 10.6 - Prob. 28ECh. 10.6 - Prob. 29ECh. 10.6 - Prob. 30ECh. 10.7 - Prob. 1ECh. 10.7 - Prob. 2ECh. 10.7 - Prob. 3ECh. 10.7 - Prob. 4ECh. 10.7 - Prob. 5ECh. 10.7 - Prob. 6ECh. 10.7 - Prob. 7ECh. 10.7 - Prob. 8ECh. 10.7 - Prob. 9ECh. 10.7 - Prob. 10ECh. 10.7 - Prob. 11ECh. 10.7 - Prob. 12ECh. 10.8 - Prob. 1ECh. 10.8 - Prob. 2ECh. 10.8 - Prob. 4ECh. 10.8 - Prob. 5ECh. 10.8 - Prob. 6ECh. 10.8 - Prob. 7ECh. 10.8 - Prob. 8ECh. 10.8 - Prob. 9ECh. 10.8 - Prob. 10ECh. 10.8 - Prob. 11ECh. 10.8 - Prob. 12ECh. 10.8 - Prob. 13ECh. 10.8 - Prob. 14ECh. 10.8 - Prob. 15E
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