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 23, Problem 25P
Interpretation Introduction

Interpretation:

The structure of the given compound and the mechanism of the given reaction are to be determined.

Concept introduction:

1 NMR (hydrogen nuclear magnetic resonance) is an analytical technique in chemistry to determine the purity and molecular structure of a compound.

13C NMR, or the carbon-13 nuclear magnetic resonance, is an analytical technique in chemistry to identify the carbon atoms in the structure of a compound. It is an important tool in structure elucidation of organic compounds. However, it can only detect the C13 isotope, and not C12, as the latter has zero net spins.

To determine the actual structure of organic compounds, IR spectroscopy (Infrared spectroscopy) is an invaluable tool. The use of electromagnetic radiation is involved in IR spectroscopy with frequencies ranging from 4000400 cm1.

Transesterification is a process in which the R group of an ester is exchanged with an R group of an alcohol. This process takes place in the presence of a base catalyst or an acid catalyst.

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Please provide the IR, 1H NMR and UV Spectra values of Benzoic Acid, 3-bromo-, methyl ester. And discuss how the results were obtain.  For example the result for 3-(3-methoxyphenyl) propanoic acid would be written like this: 1 H NMR spectrumd (ppm in CDCI3 at 60 MHZ: 11.75 (Singlet, 1H, #7), 7.40-7.23 (Multiplet, 1H, #2), 6.85 (quartet, 3H, #1-3-6), 3.86 (Singlet, 3H, #5), 3.15-2.53 (Multiplet, 4H, #9-8). The discussion would be like this:  the unknown molecule contained 12H. Revealing the presence of a peak of medium intensity at 11.75 ppm that we can associate with a carbocyclic acid. It also showed a high intensity peak at 3.86 ppm that we can associate with an ether, an alcohol or an amino group. Since the signal also shows that 3 protons are attached to it, it is therefore considered an ether (OCH3) in the unknown molecule. We also see two average peaks around 7.23 ppm and 6.85 ppm which show the presence of an aromatic.
Acid-catalyzed hydrolysis of HOCH2CH2C(CH3)2CN forms compound A (C6H10O2). A shows a strong peak in its IR spectrum at 1770 cm-1 and the following signals in its 1H NMR spectrum: 1.27 (singlet, 6 H), 2.12 (triplet, 2 H), and 4.26 (triplet, 2 H) ppm. Draw the structure for A and give a stepwise mechanism that accounts for its formation.
2. Draw the structure of the products that will be formed when aniline reacts with nitrous acid at 5 °C, followed by: (a) CUCN (b) CuBr (c) KI (d) (1) HBF4, (2) heat (е) НзРО4, Н-0 (f) Cu2O, Cu2*, H2O
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