Laboratory Techniques in Organic Chemistry
Laboratory Techniques in Organic Chemistry
4th Edition
ISBN: 9781464134227
Author: Jerry R. Mohrig, David Alberg, Gretchen Hofmeister, Paul F. Schatz, Christina Noring Hammond
Publisher: W. H. Freeman
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Chapter 22, Problem 3Q
Interpretation Introduction

Interpretation:

The structure and chemical shift values of C3H8O should be determined.

Concept introduction:

Nuclear magnetic resonance spectroscopy is applied for the identification of the structure of molecules. The energy in the radiofrequency region is suitable for NMR. Nuclear magnetic resonance results from the spin of the nucleus of an atom. The value of I is obtained using the atomic number and the mass number of an atom. The non-zero magnetic moment of an isotopic nucleus is detectable by the NMR technique.

Any nucleus with both an even atomic number and the mass number has 0 nuclear spins. There are a total of (2I+1) energy levels that are allowed for a nuclear spin (I). In the absence of a magnetic field, energy levels are equal in energy that is degenerate.

However, energy levels become non-degenerate in the presence of a magnetic field.

Deuterated chloroform (CDCl3) is a common NMR solvent generally used because it dissolutes a broad range of organic compounds and is inexpensive.

The total signal intensity of each set of proton is given by the height of each set of steps. The integration value defines the relative number of each kind of proton in the molecule.

In NMR spectrum, the intensity of signals is plotted against the magnetic field or frequency. Nuclei that are non-equivalent show only one peak in the NMR spectrum. However, protons absorb at different frequencies that are non-equivalent.

An increase in the electron density that surrounds the nucleus shields it from the applied field. This results in a net decrease in the field experienced by the nucleus. The value of the observed chemical shift of the signal therefore decreases, and, on a typical NMR spectrum, the signal moves to the right, which is called an upfield shift because, at a constant frequency, a slightly higher applied magnetic field is required for resonance to occur. De-shielding is the effect of a decline in the electron density around a nucleus which leads to shifting in the peaks of a chemical shift towards left in the NMR spectrum that results in an increase in delta values, hence downshift.

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The NMR spectra for compound 1 were acquired in a 7.5 mg / 0.6 mL solution ofCDCl3 . The 1H and 13C peaks are also listedbelow. Provide a full analysis of the NMR spectra for compound 1. correct assignment of NMR spectra of both 13C spectra.  correct rationalisation of 13C spectrum1H NMR (400 MHz, CDCl3) δ 7.73 (d, J = 9.5 Hz, 1H), 7.56 (ddd, J = 8.5, 7.5, 1.6 Hz, 1H),7.51 (dd, J = 7.5, 1.6 Hz, 1H), 7.36 (d, J = 8.5 Hz, 1H), 7.30 (dd, J = 8.5, 7.5 Hz, 1H), 6.45(d, J = 9.5 Hz, 1H).13C NMR (101 MHz, CDCl3) δ 160.79, 154.09, 143.43, 131.85, 127.87, 124.44, 118.86,116.94, 116.74.Note: There are two carbon peaks in the 13C spectrum that are so close together that they are not differentiable at the resolution in this experiment.  you should be able to assign these peaks to one of two carbon atoms in 1.
Compound 1, was found to consist of C,H,N, and Cl with an elemental analysis of 61.1% C, 2.9% H, and 10.2% N. IR Spectra displayed absorbances at 2215, 1605. Deduce the structure of this compound. (there is only one Cl)
Compound A is a hydrocarbon with a molar mass of 96g/mol, with the given C13 spectral data. When compound A reacts with BH3 followed by the treatment with basic H2O2 it is converted to compound B. Propose structures for A and B, explain your analysis.Compound A- Proton decoupled C NMR: 26.8, 28.7, 35.7, 106.9, 149.7 δ.DEPT-90: No peak.DEPT-135: No positive peaks; negative peaks at 26.8, 28.7, 35.7, 106.9 δ.Compound B- Proton decoupled C NMR: 26.1, 26.9, 29.9, 40.5, 68.2 δ.DEPT-90: 40.5 δ.DEPT-135: positive peak at 40.5 δ; negative peaks at 26.1, 26.9, 29.9, 68.2 δ

Chapter 22 Solutions

Laboratory Techniques in Organic Chemistry

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