ORGANIC CHEMISTRY-OWL V2 ACCESS
ORGANIC CHEMISTRY-OWL V2 ACCESS
8th Edition
ISBN: 9781305582422
Author: Brown
Publisher: CENGAGE L
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Chapter 13, Problem 13.17P

Following are 1H-NMR spectra for compounds G, H, and I, each with the molecular formula C5H12O. Each is a liquid at room temperature, is slightly soluble in water, and reacts with sodium metal with the evolution of a gas.

(a) Propose structural formulas of compounds G, H, and I.

(b) Explain why there are four lines between δ 0.86 and 0.90 for compound G.

(c) Explain why the 2H multiplets at δ 1.5 and 3.5 for compound H are so complex.

Chapter 13, Problem 13.17P, Following are 1H-NMR spectra for compounds G, H, and I, each with the molecular formula C5H12O. Each , example  1

Chapter 13, Problem 13.17P, Following are 1H-NMR spectra for compounds G, H, and I, each with the molecular formula C5H12O. Each , example  2

Chapter 13, Problem 13.17P, Following are 1H-NMR spectra for compounds G, H, and I, each with the molecular formula C5H12O. Each , example  3

(a)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

Following compounds G,H and I stractural formula has to be proposed with the help of given molecular formula C5H12O and corrsponding 1H NMR spectrums.

Concept introduction:

The 1H NMR spectrum of a compound provides some vital information that is required to predict the structure of the compound.  The chemical shift values can predict the groups that are present in the molecule.  The splitting of signals by the (N+1) rule predicts the number of hydrogens or protons attached to the adjacent carbon in a carbon chain of a compound.  With this information, the structure of the compound can be predicted.

Chemical shift: The NMR spectrum of any compound is taken with reference to a standard compound called reference compound.  Generally, tetramethylsilane (TMS) is taken as the reference compound.  The methyl protons of TMS are equivalent and produces only one sharp peak at the rightmost end of the scale.

The distance between the TMS signal and the signals produced by the compound is called the chemical shift.  Chemical shift basically measures the shift in the signal position of the compound with respect to the reference signal.

Chemical shift in delta scale is given as,

chemical shift,δ(ppm)=distance from the TMS signal(Hz)operating frequency of the spectrometer (MHz)

Explanation of Solution

Index of Hydrogen Deficiency (IHD) calculation,

Given molecular formula is C5H12O

We calculate the IHD as follows:

HDI( 2C+2+N-H-X)2=2×5+2+0-12-02=0_

From the molecular formula, there is an index of hydrogen deficiency is zero for these molecules, so there are no rings (or) double bonds.

The fact that the compounds are slightly soluble in water and react with sodium metal indicates that each molecule has an hydroxyl (OH) group. This chemical shifts associated with each set of hydrogens are indicated on the corresponding structures.

The 1H NMR analysis for given compound-G:

There are 5 peaks observed in the given 1H NMR spectrum thus 5 sets of non-equivalent protons are present in the molecule.

A multiplet is observed for 1 hydrogen at around 3.5 ppm that indicates that has the adjacent groups (CH_OH) containing a total of one hydrogen.

The one doublet is observed for 1 hydrogen at around δ1.85ppm that indicate that has the hydroxyl group (-OH) containing a total of one hydrogen.

Another one multiplet is observed for one hydrogen at around 1.6 ppm that indicate that has the other adjacent groups (-CH_-(CH3)2)

One doublet is observed for 3 hydrogens at around δ1.15ppm that indicate that has the one methyl groups (-C(OH)-CH3) containing a total of three hydrogens.

The one doublet is observed for 6 hydrogens at around δ0.9ppm that indicate that has the two methyl groups (-CH-(CH3)2) containing a total of six hydrogens. This doublets overlapping, this is more complex than expected because it is adjacent to a chiral center.

Based on the above 1H NMR spectral details, the structural formula for compound-G is:

ORGANIC CHEMISTRY-OWL V2 ACCESS, Chapter 13, Problem 13.17P , additional homework tip  1

The 1H NMR analysis for given compound-H:

There are 5 peaks observed in the given 1H NMR spectrum thus 5 sets of non-equivalent protons are present in the molecule.

1H NMR δ3.43.5 (2H, multiplet, this is more complex than expected because it is adjacent to a chiral center CH2OH),  1.41.6 (2H, multiplet, this is more complex than expected because it is adjacent to a chiral center CH3CH2), 1.1 (1H, multiplet, CH), further 0.80.9 (6H, broad multiplet, both methyl groups). Based on the above 1H NMR spectral details, the structural formula for compound-H is:

ORGANIC CHEMISTRY-OWL V2 ACCESS, Chapter 13, Problem 13.17P , additional homework tip  2

The 1H NMR analysis for given compound-I:

There are 5 peaks observed in the given 1H NMR spectrum thus 5 sets of non-equivalent protons are present in the molecule.

1H NMR δ3.6 (2H, broad multiplet, CH2OH), 2.9(1H, broad peak, -OH), 1.55 (2H, multiplet, CH2_CH2OH) and 1.4 (4H, multiplet, CH3CH2CH2 and CH3CH2CH2), 0.9 (3H, triplet, corresponding to methyl group).  

Based on the above 1H NMR spectral details, the structural formula for compound-I is

ORGANIC CHEMISTRY-OWL V2 ACCESS, Chapter 13, Problem 13.17P , additional homework tip  3

(b)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The following compound-G has four lines observed in the range of δ0.86and0.90 the reason should be explained.

Concept introduction:

The 1H NMR spectrum of a compound provides some vital information that is required to predict the structure of the compound.  The chemical shift values can predict the groups that are present in the molecule.  The splitting of signals by the (N+1) rule predicts the number of hydrogens or protons attached to the adjacent carbon in a carbon chain of a compound.  With this information, the structure of the compound can be predicted.

Diastereotopic: If the protons are not interchangeable by either of the symmetry operations, then the protons are Diastereotopic; the protons are not chemically equivalent if a chiral center present in the molecule.

Explanation of Solution

Let us consider the given compound-G:

  ORGANIC CHEMISTRY-OWL V2 ACCESS, Chapter 13, Problem 13.17P , additional homework tip  4

Above the molecule, the carbon atom 2 attached with hydroxyl OH group, it is a chiral center.

That is makes the two methyl groups are diastereotopic, so they have different chemical shifts, the four lines are actually two doublets.

(c)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The Following compound-H is two hydrogens appread at multiplet, this compound is complex or not, the behaind reason should be explained.

 Concept introduction:

The 1H NMR spectrum of a compound provides some vital information that is required to predict the structure of the compound.  The chemical shift values can predict the groups that are present in the molecule.  The splitting of signals by the (N+1) rule predicts the number of hydrogens or protons attached to the adjacent carbon in a carbon chain of a compound.  With this information, the structure of the compound can be predicted.

Chemical shift: The NMR spectrum of any compound is taken with reference to a standard compound called reference compound.  Generally, tetramethylsilane (TMS) is taken as the reference compound.  The methyl protons of TMS are equivalent and produces only one sharp peak at the rightmost end of the scale.

The distance between the TMS signal and the signals produced by the compound is called the chemical shift.  Chemical shift basically measures the shift in the signal position of the compound with respect to the reference signal.

Chemical shift in delta scale is given as,

chemical shift,δ(ppm)=distance from the TMS signal(Hz)operating frequency of the spectrometer (MHz)

Diastereotopic: If the protons are not interchangeable by either of the symmetry operations, then the protons are Diastereotopic; the protons are not chemically equivalent if a chiral center present in the molecule.

Explanation of Solution

Let us consider the given compound-H:

  ORGANIC CHEMISTRY-OWL V2 ACCESS, Chapter 13, Problem 13.17P , additional homework tip  5

Above the molecule, the carbon atom 2 attached with hydroxyl OH group, it is a chiral center.

This chiral center makes the adjacent CH2 protons diastereotopic, so the corresponding signals are complex.

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Chapter 13 Solutions

ORGANIC CHEMISTRY-OWL V2 ACCESS

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