ORGANIC CHEMISTRY E-BOOK W/SMARTWORK5
ORGANIC CHEMISTRY E-BOOK W/SMARTWORK5
2nd Edition
ISBN: 9780393664034
Author: KARTY
Publisher: NORTON
Question
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Chapter 16, Problem 16.80P
Interpretation Introduction

Interpretation:

The structure for the compound having molecular formula C19H16 is to be drawn based on the distinct signals in its 1H NMR and 13C NMR spectra.

Concept introduction:

In 1H NMR spectroscopy, protons in different environments within a molecule have different chemical shifts, that is, they experience different degrees of shielding.

In addition to chemical shift, a 1H NMR spectrum provides structural information based on Number of signals, which tells how many different kinds of protons are there; Integrated areas, which tells the ratios of the various kinds of protons; and Splitting pattern, which gives information about the number of protons that are within two or three bonds of the one giving the signal. Spin-spin splitting of NMR signals results from the coupling of the nuclear spins that are separated by two bonds (geminal coupling) or three bonds (vicinal coupling). In these cases, the number of peaks into which a signal is split is equal to (n+1), where n is the number of protons to which the proton in question is coupled. Protons that have the same chemical shift do not split each other’s signals.

Complicated splitting patterns can result when a proton is unequally coupled to two or more protons that are different from one another.

The ideal range for alkane protons is δ0.9- δ1.4; for alcohol, it is δ2- δ5.0, and for aromatic protons, it is δ6.5- δ8.5. The chemical shift of a signal prompts about the aromatic rings, double bonds, or nearby electronegative atoms. The integration of each signal suggests the number of protons responsible for that signal. The splitting pattern of a signal indicates the number of neighboring protons that are distinct from the protons responsible for that signal. To deduce the structure of an unknown compound, the first step is to find the index of hydrogen deficiency if the molecular formula is given. Based on the data given in the 1H NMR, one can build molecular fragments with multiple carbon atoms.

13C NMR provides valuable information about the carbon skeleton. Each signal in the 13C NMR equals the number of distinct carbon atoms in the given unknown. In most of the 13C NMR spectra, almost every time, all the signals would appear as singlets. The saturated carbon atoms appear in the range δ0-35 ppm if it is a simple alkane fragment. If it is attached to any electronegative element such as halogens or nitrogen, the range is δ25-70 ppm. Alkene carbons range from δ105-150 ppm. Carbonyl carbon atoms in acids, esters, amides, and anhydrides range from δ120-185 ppm. Carbonyl carbons in aldehydes and ketones range from δ190-220 ppm.

Expert Solution & Answer
Check Mark

Answer to Problem 16.80P

The structure for the compound having molecular formula C19H16 based on the distinct signals in its 1H NMR and 13C NMR spectra is drawn as

ORGANIC CHEMISTRY E-BOOK W/SMARTWORK5, Chapter 16, Problem 16.80P , additional homework tip  1

Explanation of Solution

The molecular formula for the given unknown compound is C19H16.

The molecular formula shows an index of hydrogen deficiency of 12.

The 1H NMR of the compound has three distinct signals; therefore, there must be four distinct protons in the structure, and the data is as follows:

C19H16: δ 6.9 ppm (15 H), unresolved multiplet            δ 7.44 ppm (1 H), singlet             δ 5.5 ppm (1 H), singlet

In 1H NMR spectrum, there are two signals; one is an unresolved multiplet, so it is not possible to interpret the number of distinct H atoms using 1H NMR data.

In 1H NMR spectrum, the relative integration is the same as the number of protons because the sum of the relative integrations equals the total number of hydrogen atoms in the molecular formula. The structural feature that gives rise to the symmetry, with several C signals and several H signals, appears in the aromatic region, suggesting one or more benzene rings.

In the 13C NMR spectrum, there are five signals

C19H16 : δ 57 ppm              δ 126 - 144 ppm (Four)

The given molecular formula with 19 carbons, which gives only four signals in 13C NMR spectrum, represents that a high degree of symmetry must be present in the compound, that is, many carbons must be chemically equivalent. One different signal at 57 ppm for sp3 C atom, along with the remaining signals between 126 - 144 ppm, appears owing to the aromatic C atoms.

One benzene ring accounts for IHD 4; thus the given compound has three benzene rings, IHD 12 in the given compound. Only one C is distinct at 57 ppm with one chemically distinct proton, which gives the proton NMR signal at 5.5 ppm. Three equivalent benzene rings (C18H15) and one CH give C19H16. Thus, the compound having molecular formula C19H16 based on the distinct signals in its 1H NMR and 13C NMR spectra is

ORGANIC CHEMISTRY E-BOOK W/SMARTWORK5, Chapter 16, Problem 16.80P , additional homework tip  2

Conclusion

The structure for the given compound with molecular formula C19H16 is drawn on the basis of the given data.

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Chapter 16 Solutions

ORGANIC CHEMISTRY E-BOOK W/SMARTWORK5

Ch. 16 - Prob. 16.11PCh. 16 - Prob. 16.12PCh. 16 - Prob. 16.13PCh. 16 - Prob. 16.14PCh. 16 - Prob. 16.15PCh. 16 - Prob. 16.16PCh. 16 - Prob. 16.17PCh. 16 - Prob. 16.18PCh. 16 - Prob. 16.19PCh. 16 - Prob. 16.20PCh. 16 - Prob. 16.21PCh. 16 - Prob. 16.22PCh. 16 - Prob. 16.23PCh. 16 - Prob. 16.24PCh. 16 - Prob. 16.25PCh. 16 - Prob. 16.26PCh. 16 - Prob. 16.27PCh. 16 - Prob. 16.28PCh. 16 - Prob. 16.29PCh. 16 - Prob. 16.30PCh. 16 - Prob. 16.31PCh. 16 - Prob. 16.32PCh. 16 - Prob. 16.33PCh. 16 - Prob. 16.34PCh. 16 - Prob. 16.35PCh. 16 - Prob. 16.36PCh. 16 - Prob. 16.37PCh. 16 - Prob. 16.38PCh. 16 - Prob. 16.39PCh. 16 - Prob. 16.40PCh. 16 - Prob. 16.41PCh. 16 - Prob. 16.42PCh. 16 - Prob. 16.43PCh. 16 - Prob. 16.44PCh. 16 - Prob. 16.45PCh. 16 - Prob. 16.46PCh. 16 - Prob. 16.47PCh. 16 - Prob. 16.48PCh. 16 - Prob. 16.49PCh. 16 - Prob. 16.50PCh. 16 - Prob. 16.51PCh. 16 - Prob. 16.52PCh. 16 - Prob. 16.53PCh. 16 - Prob. 16.54PCh. 16 - Prob. 16.55PCh. 16 - Prob. 16.56PCh. 16 - Prob. 16.57PCh. 16 - Prob. 16.58PCh. 16 - Prob. 16.59PCh. 16 - Prob. 16.60PCh. 16 - Prob. 16.61PCh. 16 - Prob. 16.62PCh. 16 - Prob. 16.63PCh. 16 - Prob. 16.64PCh. 16 - Prob. 16.65PCh. 16 - Prob. 16.66PCh. 16 - Prob. 16.67PCh. 16 - Prob. 16.68PCh. 16 - Prob. 16.69PCh. 16 - Prob. 16.70PCh. 16 - Prob. 16.71PCh. 16 - Prob. 16.72PCh. 16 - Prob. 16.73PCh. 16 - Prob. 16.74PCh. 16 - Prob. 16.75PCh. 16 - Prob. 16.76PCh. 16 - Prob. 16.77PCh. 16 - Prob. 16.78PCh. 16 - Prob. 16.79PCh. 16 - Prob. 16.80PCh. 16 - Prob. 16.81PCh. 16 - Prob. 16.82PCh. 16 - Prob. 16.83PCh. 16 - Prob. 16.84PCh. 16 - Prob. 16.85PCh. 16 - Prob. 16.86PCh. 16 - Prob. 16.87PCh. 16 - Prob. 16.88PCh. 16 - Prob. 16.89PCh. 16 - Prob. 16.1YTCh. 16 - Prob. 16.2YTCh. 16 - Prob. 16.3YTCh. 16 - Prob. 16.4YTCh. 16 - Prob. 16.5YTCh. 16 - Prob. 16.6YTCh. 16 - Prob. 16.7YTCh. 16 - Prob. 16.8YTCh. 16 - Prob. 16.9YTCh. 16 - Prob. 16.10YTCh. 16 - Prob. 16.11YTCh. 16 - Prob. 16.12YTCh. 16 - Prob. 16.13YTCh. 16 - Prob. 16.14YTCh. 16 - Prob. 16.15YTCh. 16 - Prob. 16.16YTCh. 16 - Prob. 16.17YTCh. 16 - Prob. 16.18YTCh. 16 - Prob. 16.19YTCh. 16 - Prob. 16.20YTCh. 16 - Prob. 16.21YTCh. 16 - Prob. 16.22YTCh. 16 - Prob. 16.23YTCh. 16 - Prob. 16.24YTCh. 16 - Prob. 16.25YTCh. 16 - Prob. 16.26YTCh. 16 - Prob. 16.27YTCh. 16 - Prob. 16.28YTCh. 16 - Prob. 16.29YTCh. 16 - Prob. 16.30YTCh. 16 - Prob. 16.31YTCh. 16 - Prob. 16.32YTCh. 16 - Prob. 16.33YTCh. 16 - Prob. 16.34YT
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