ORGANIC CHEMISTRY-PACKAGE >CUSTOM<
ORGANIC CHEMISTRY-PACKAGE >CUSTOM<
10th Edition
ISBN: 9781260028355
Author: Carey
Publisher: MCG CUSTOM
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Chapter 14, Problem 31P

Each of the following compounds is characterized by a 1H NMR spectrum that consists of

only a single peak having the chemical shift indicated. Identify each compound.

C 8 H 18 ;δ 0 .9

C 5 H 10 ;δ 1 .5

C 8 H 8 ;δ 5 .8

C 4 H 9 Br;δ 1 .8

C 2 H 4 Cl 2 ;δ 3 .7

C 2 H 3 Cl 3 ;δ 2 .7

C 5 H 8 Cl 4 ;δ 3 .7

C 12 H 18 ;δ 2 .2

C 3 H 6 Br 2 ;δ  2.6

Expert Solution & Answer
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Interpretation Introduction

Interpretation:

The compounds gives only single peak in 1H NMR spectrum as indicated by the chemical shift is to be identified.

Concept introduction:

The information related to the proton in the compound can be deduced by the number of signals in a spectrum.

The proton chemical shift in the compound is due to the environment of the proton.

The signal of a particular proton can be split by the presence of protons in vicinal position.

The NMR spectrum of a compound will consist of a single peak if all protons in the compound are structurally equivalent.

Index of hydrogen deficiency is calculated as

Indexofhydrogendeficiency=12(CnH2n+2-CnHx)

Here CnHx is the molecular formula of the compound.

Oxygen atoms do not affect the index of hydrogen deficiency.

Answer to Problem 31P

Solution:

C8H18; δ0.9

ORGANIC CHEMISTRY-PACKAGE >CUSTOM<, Chapter 14, Problem 31P , additional homework tip  1

C5H10; δ1.5

ORGANIC CHEMISTRY-PACKAGE >CUSTOM<, Chapter 14, Problem 31P , additional homework tip  2

C8H8; δ5.8

ORGANIC CHEMISTRY-PACKAGE >CUSTOM<, Chapter 14, Problem 31P , additional homework tip  3

C4H9Br; δ1.8

ORGANIC CHEMISTRY-PACKAGE >CUSTOM<, Chapter 14, Problem 31P , additional homework tip  4

C2H4Cl2; δ3.7

ORGANIC CHEMISTRY-PACKAGE >CUSTOM<, Chapter 14, Problem 31P , additional homework tip  5

C2H3Cl3; δ2.7

ORGANIC CHEMISTRY-PACKAGE >CUSTOM<, Chapter 14, Problem 31P , additional homework tip  6

C5H8Cl4; δ3.7

ORGANIC CHEMISTRY-PACKAGE >CUSTOM<, Chapter 14, Problem 31P , additional homework tip  7

C12H18; δ2.2

ORGANIC CHEMISTRY-PACKAGE >CUSTOM<, Chapter 14, Problem 31P , additional homework tip  8

C3H6Br2; δ2.6

ORGANIC CHEMISTRY-PACKAGE >CUSTOM<, Chapter 14, Problem 31P , additional homework tip  9

Explanation of Solution

Chemical formula: C8H18; Chemical shift: δ=0.9

A single peak at 0.9 indicates all protons in the compound are equivalent protons. Further, they must be symmetrical alkyl (methyl) protons.

Therefore, the structure of the compound is:

ORGANIC CHEMISTRY-PACKAGE >CUSTOM<, Chapter 14, Problem 31P , additional homework tip  10

Chemical formula: C5H10; Chemical shift: δ=1.5

A single peak at 1.5 indicates all protons are equivalent. Further, they must be alkyl protons. The chemical formula shows the index of hydrogen deficiency to be 1.

Therefore, it must be a symmetric cycloalkane:

ORGANIC CHEMISTRY-PACKAGE >CUSTOM<, Chapter 14, Problem 31P , additional homework tip  11

Chemical formula: C8H8; Chemical shift: δ=5.8

A single peak at 5.8 indicates all protons are equivalent. The chemical shift of 5.8 indicates that the protons are all vinylic protons. The chemical formula shows the index of hydrogen deficiency to be 5. Therefore, the structure of the compound is:

ORGANIC CHEMISTRY-PACKAGE >CUSTOM<, Chapter 14, Problem 31P , additional homework tip  12

Chemical formula: C4H9Br; Chemical shift: δ=1.8

The chemical formula shows it is a saturated alkyl halide. A single peak means that all protons are equivalent. The presence of the bromine atoms accounts for the higher chemical shift of the protons. Therefore, the structure of the compound is:

ORGANIC CHEMISTRY-PACKAGE >CUSTOM<, Chapter 14, Problem 31P , additional homework tip  13

Chemical formula: C2H4Cl2; Chemical shift: δ=3.7

A single peak shows all protons to be equivalent. The formula is of a saturated alkyl halide. The presence of two chlorine atoms on the same carbon as the protons will increase the shift to 3.7. Considering this, the structure of the compound is:

ORGANIC CHEMISTRY-PACKAGE >CUSTOM<, Chapter 14, Problem 31P , additional homework tip  14

Chemical formula: C2H3Cl3; Chemical shift: δ=2.7

A single peak means all protons must be equivalent. The chemical formula shows it to be a saturated alkyl halide. The higher shift of 2.7 indicates that the alkyl protons are all on one carbon (methyl group) with the three chlorines on the carbon next to it. Therefore, the structure of the compound is:

ORGANIC CHEMISTRY-PACKAGE >CUSTOM<, Chapter 14, Problem 31P , additional homework tip  15

Chemical formula: C5H8Cl4; Chemical shift: δ=3.7

A single peak shows all protons to be equivalent. The chemical formula shows it to be a saturated alkyl halide. The high chemical shift of 3.7 indicates that the chlorine atoms are on the same carbon atom as the protons. Therefore, the structure of the compound is:

ORGANIC CHEMISTRY-PACKAGE >CUSTOM<, Chapter 14, Problem 31P , additional homework tip  16

Chemical formula: C12H18; Chemical shift: δ=2.2

A single peak shows all protons are equivalent. The chemical formula shows an index of hydrogen deficiency of four. This, taken together with a chemical shift of 2.2, indicates that the compound has a benzene ring, with six methyl groups attached to it. Therefore, the structure of the compound is:

ORGANIC CHEMISTRY-PACKAGE >CUSTOM<, Chapter 14, Problem 31P , additional homework tip  17

Chemical formula: C3H6Br2; Chemical shift: δ=2.6

A single peak shows all protons are equivalent. The higher chemical shift of 2.6 indicates that the protons are all methyl protons, with the two bromine atoms bonded to the adjacent carbon.

Therefore, the structure of the compound is:

ORGANIC CHEMISTRY-PACKAGE >CUSTOM<, Chapter 14, Problem 31P , additional homework tip  18

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1Compound 1 has molecular formula C7H16. It shows three signals in the 1H-NMR spectrum, one at 0.85 ppm, one at 1.02 ppm, and one at 1.62 ppm. The relative integrals of these three signals are 6, 1, and 1, respectively. Compound 2 has molecular formula C7H14. It shows three signals in the 1H-NMR spectrum, one at 0.98 ppm, one at 1.36 ppm, and one at 1.55 ppm. The relative integrals of these three signals are 3, 2, and 2, respectively. Propose structures for compounds 1 and 2, explaining how you reach your conclusion.
Compound 1 has molecular formula C7H16. It shows three signals in the 1H-NMR spectrum, one at 0.85 ppm, one at 1.02 ppm, and one at 1.62 ppm. The relative integrals of these three signals are 6, 1, and 1, respectively. Compound 2 has molecular formula C7H14. It shows three signals in the 1H-NMR spectrum, one at 0.98 ppm, one at 1.36 ppm, and one at 1.55 ppm. The relative integrals of these three signals are 3, 2, and 2, respectively. Propose structures for compounds 1 and 2, explaining how you reach your conclusion.
14. Compound B has molecular formula C9H12. It shows five signals in the 1H-NMR spectrum - a doublet of integral 6 at 1.22 ppm, a septet of integral 1 at 2.86 ppm, a singlet of integral 1 at 5.34 ppm, a doublet of integral 2 at 6.70 ppm, and a doublet of integral 2 at 7.03 ppm. The 13C-NMR spectrum of B shows six unique signals (23.9, 34.0, 115.7, 128.7, 148.9, and 157.4). Identify B and explain your reasoning.

Chapter 14 Solutions

ORGANIC CHEMISTRY-PACKAGE >CUSTOM<

Ch. 14.7 - Describe the appearance of the 1H NMR spectrum of...Ch. 14.8 - Describe the appearance of the 1H NMR spectrum of...Ch. 14.11 - Prob. 13PCh. 14.11 - Prob. 14PCh. 14.12 - Hydrogen bonding between the oxygen of dimethyl...Ch. 14.14 - Prob. 16PCh. 14.15 - The 13C NMR spectrum of 1-bromo-3-chloropropane...Ch. 14.15 - Consider carbons x, y, and z in p-methylanisole....Ch. 14.15 - Prob. 19PCh. 14.16 - To which of the compounds of Problem 14.16 does...Ch. 14.18 - DEPT spectra for a compound with the formula...Ch. 14.20 - Vibrational frequencies are sensitive to isotopic...Ch. 14.21 - Prob. 23PCh. 14.22 - Prob. 24PCh. 14.23 - Prob. 25PCh. 14.23 - Which one of the C5H8 isomers shown has its max at...Ch. 14.24 - Knowing what to look for with respect to isotopic...Ch. 14.24 - The base peak appears at m/z105 for one of the...Ch. 14.24 - Mass spectra of 1-bromo-4-propylbenzene and...Ch. 14.25 - Prob. 30PCh. 14 - Each of the following compounds is characterized...Ch. 14 - Deduce the structure of each of the following...Ch. 14 - From among the isomeric compounds of molecular...Ch. 14 - The H1NMR spectrum of fluorene has signals at 3.8...Ch. 14 - Prob. 35PCh. 14 - H1NMR spectra of four isomeric alcohols with...Ch. 14 - Prob. 37PCh. 14 - We noted in Section 14.13 that an NMR spectrum is...Ch. 14 - Identify each of the C4H10O isomers on the basis...Ch. 14 - A compound (C3H7ClO2) exhibited three peaks in its...Ch. 14 - Label nonequivalent carbons in the following...Ch. 14 - Compounds A and B are isomers of molecular formula...Ch. 14 - C13 NMR spectra for four isomeric alkyl bromides...Ch. 14 - Prob. 44PCh. 14 - Prob. 45PCh. 14 - Identify the C3H5Br isomers on the basis of the...Ch. 14 - Prob. 47PCh. 14 - A compound (C8H10O) has the IR and H1NMR spectra...Ch. 14 - Deduce the structure of a compound having the...Ch. 14 - Figure 14.53 presents IR, H1NMR, C13NMR and mass...Ch. 14 - H1NMR, C13NMR, IR, and mass spectra are shown for...Ch. 14 - 1H NMR and IR spectra for a compound with the...Ch. 14 - FriedelCraftsalkylation of benzene with...Ch. 14 - Prob. 54DSPCh. 14 - Prob. 55DSPCh. 14 - Prob. 56DSPCh. 14 - Prob. 57DSPCh. 14 - Prob. 58DSP
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