Organic Chemistry
Organic Chemistry
11th Edition
ISBN: 9781118133576
Author: T. W. Graham Solomons, Craig Fryhle
Publisher: Wiley, John & Sons, Incorporated
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Chapter 14, Problem 38P
Interpretation Introduction

Interpretation:

The structure of compound Y is to be proposed on the basis of H1NMR and IR.

Concept introduction:

Nuclear Magnetic Resonance (NMR) is one of the most capable analytical techniques used for determining the functional groups and how the atoms are structured and arranged in a molecule.

Few elements, such as 13C and 1H, have nuclei behaving as magnets about an axis. These elements are placed in magnetic field irradiated with electromagnetic energy of specific frequency and the nuclei tend to absorb energy via magnetic resonance. There is this graph that shows energy absorption frequencies and intensities of a sample kept in the magnetic field called nuclear magnetic resonance (NMR).

In NMR

spectroscopy, the proton nuclear magnetic resonance (1HNMR) is used to find out the structure of molecules with the help of 1H

atom within the molecules.

Induced magnetic field consists of electricity generated from movement in a magnetic field.

Infrared spectroscopy is a simple, instrumental technique, which helps to determine the presence of various functional groups.

It depends on the interactions of atoms or molecules with the electromagnetic radiation.

Infrared spectroscopy is most commonly used in the identification of the structure of the compound.

Infrared spectroscopy is the examination of the infrared light interacting with a molecule. The examination can be done in three ways, that is, by measuring absorption, emission, and reflection, and it can also measure the vibration of atoms.

13C

NMR is only used in the observation of isotopes of carbon atoms.

The H1NMR

signal for protons bonded to substituted benzene rings is between δ6 and 8.5.

The IR (infrared) spectra of benzene show the absorption occurring at 3030 cm1 for the stretching of C-H

bond. Further, the monosubstituted benzene results in the generation of peaks in the range of 690 and 710 cm1, and in the range of 730 and 770 cm1. The O-H

group gives a broad peak at 3550-3200 cm1.

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Chapter 14 Solutions

Organic Chemistry

Ch. 14 - Practice Problem 14.11 In addition to a signal...Ch. 14 - PRACTICE PROBLEM 14.12 Azulene has an appreciable...Ch. 14 - Practice Problem 14.13 (a) The -Sh group is...Ch. 14 - Practice Problem 14.14 Explain how NMR...Ch. 14 - PRACTICE PROBLEM 14.15 Four benzenoid compounds,...Ch. 14 - Prob. 16PCh. 14 - Write structural formulas and give acceptable...Ch. 14 - Prob. 18PCh. 14 - Prob. 19PCh. 14 - Prob. 20PCh. 14 - Which of the hydrogen atoms shown below is more...Ch. 14 - 14.22 The rings below are joined by a double bond...Ch. 14 - Prob. 23PCh. 14 - 14.24 (a) In 1960 T. Katz (Columbia University)...Ch. 14 - Prob. 25PCh. 14 - Prob. 26PCh. 14 - 14.27 5-Chloro-1,3-cyclopentadiene (below)...Ch. 14 - Prob. 28PCh. 14 - Furan possesses less aromatic character than...Ch. 14 - 14.30 For each of the pairs below, predict...Ch. 14 - Assign structures to each of the compounds A, B,...Ch. 14 - Prob. 32PCh. 14 - Give a structure for compound F that is consistent...Ch. 14 - Prob. 34PCh. 14 - Prob. 35PCh. 14 - A compound (N) with the molecular formula C9H10O...Ch. 14 - The IR and 1H NMR spectra for compound X(C8H10)...Ch. 14 - Prob. 38PCh. 14 - Prob. 39PCh. 14 - 14.39 Given the following information, predict the...Ch. 14 - Consider these reactions: The intermediate A is a...Ch. 14 - Prob. 42PCh. 14 - Compound E has the spectral features given below....Ch. 14 - Draw all of the molecular orbitals for...Ch. 14 - Prob. 1LGPCh. 14 - Prob. 2LGPCh. 14 - 3. The NMR signals for the aromatic hydrogens of...Ch. 14 - Prob. 4LGPCh. 14 - Prob. 5LGPCh. 14 - Prob. 1QCh. 14 - Which is the correct name of the compound shown?...Ch. 14 - Prob. 3QCh. 14 - Prob. 4QCh. 14 - Give the structure of a compound with the formula...Ch. 14 - Prob. 6Q
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