ORGANIC CHEMISTRY SAPLING ACCESS + ETEX
ORGANIC CHEMISTRY SAPLING ACCESS + ETEX
6th Edition
ISBN: 9781319306977
Author: LOUDON
Publisher: INTER MAC
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Chapter 23, Problem 23.71AP
Interpretation Introduction

(a)

Interpretation:

The compound A (C9H13NO) is to be identified by the use of given spectroscopy data.

Concept introduction:

NMR spectroscopy is a technique used to determine the unique structure of the compounds. It identifies the carbon-hydrogen bonding of an organic compound. A hydrogen atom is known as a proton in the NMR spectroscopy. The number of NMR signal in a compound is equal to the number of chemically non-equivalent protons present in that compound. The more the shielded proton less will be its chemical shift value and the corresponding signal will be produced at the right-hand side or lower frequency region.

Interpretation Introduction

(b)

Interpretation:

The compound B (C6H16N2) is to be identified by the use of given spectroscopy data.

Concept introduction:

NMR spectroscopy is a technique used to determine the unique structure of the compounds. It identifies the carbon-hydrogen bonding of an organic compound. A hydrogen atom is known as as a proton in the NMR spectroscopy. The number of NMR signal in a compound is equal to the number of chemically non-equivalent protons present in that compound. The more the shielded proton less will be its chemical shift value and the corresponding signal will be produced at the right-hand side or lower frequency region.

Interpretation Introduction

(c)

Interpretation:

The compound C (C6H13N) is to be identified by the use of given spectroscopy data.

Concept introduction:

NMR spectroscopy is a technique used to determine the unique structure of the compounds. It identifies the carbon-hydrogen bonding of an organic compound. A hydrogen atom is known as a proton in the NMR spectroscopy. The number of NMR signal in a compound is equal to the number of chemically non-equivalent protons present in that compound. The more the shielded proton less will be its chemical shift value and the corresponding signal will be produced at the right-hand side or lower frequency region.

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Chapter 23 Solutions

ORGANIC CHEMISTRY SAPLING ACCESS + ETEX

Ch. 23 - Prob. 23.11PCh. 23 - Prob. 23.12PCh. 23 - Prob. 23.13PCh. 23 - Prob. 23.14PCh. 23 - Prob. 23.15PCh. 23 - Prob. 23.16PCh. 23 - Prob. 23.17PCh. 23 - Prob. 23.18PCh. 23 - Prob. 23.19PCh. 23 - Prob. 23.20PCh. 23 - Prob. 23.21PCh. 23 - Prob. 23.22PCh. 23 - Prob. 23.23PCh. 23 - Prob. 23.24PCh. 23 - Prob. 23.25PCh. 23 - Prob. 23.26PCh. 23 - Prob. 23.27PCh. 23 - Prob. 23.28PCh. 23 - Prob. 23.29PCh. 23 - Prob. 23.30PCh. 23 - Prob. 23.31PCh. 23 - Prob. 23.32PCh. 23 - Prob. 23.33PCh. 23 - Prob. 23.34PCh. 23 - Prob. 23.35PCh. 23 - Prob. 23.36PCh. 23 - Prob. 23.37PCh. 23 - Prob. 23.38PCh. 23 - Prob. 23.39PCh. 23 - Prob. 23.40PCh. 23 - Prob. 23.41PCh. 23 - Prob. 23.42PCh. 23 - Prob. 23.43PCh. 23 - Prob. 23.44APCh. 23 - Prob. 23.45APCh. 23 - Prob. 23.46APCh. 23 - Prob. 23.47APCh. 23 - Prob. 23.48APCh. 23 - Prob. 23.49APCh. 23 - Prob. 23.50APCh. 23 - Prob. 23.51APCh. 23 - Prob. 23.52APCh. 23 - Prob. 23.53APCh. 23 - Prob. 23.54APCh. 23 - Prob. 23.55APCh. 23 - Prob. 23.56APCh. 23 - Prob. 23.57APCh. 23 - Prob. 23.58APCh. 23 - Prob. 23.59APCh. 23 - Prob. 23.60APCh. 23 - Prob. 23.61APCh. 23 - Prob. 23.62APCh. 23 - Prob. 23.63APCh. 23 - Prob. 23.64APCh. 23 - Prob. 23.65APCh. 23 - Prob. 23.66APCh. 23 - Prob. 23.67APCh. 23 - Prob. 23.68APCh. 23 - Prob. 23.69APCh. 23 - Prob. 23.70APCh. 23 - Prob. 23.71APCh. 23 - Prob. 23.72APCh. 23 - Prob. 23.73APCh. 23 - Prob. 23.74APCh. 23 - Prob. 23.75APCh. 23 - Prob. 23.76APCh. 23 - Prob. 23.77APCh. 23 - Prob. 23.78APCh. 23 - Prob. 23.79APCh. 23 - Prob. 23.80APCh. 23 - Prob. 23.81AP
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NMR Spectroscopy; Author: Professor Dave Explains;https://www.youtube.com/watch?v=SBir5wUS3Bo;License: Standard YouTube License, CC-BY