Organic Chemistry
Organic Chemistry
9th Edition
ISBN: 9781305080485
Author: John E. McMurry
Publisher: Cengage Learning
bartleby

Concept explainers

bartleby

Videos

Textbook Question
Book Icon
Chapter 13.SE, Problem 55GP

Propose structures for compounds that fit the following 1H NMR data:

(a) C4H6Cl2

2.18 δ (3 H, singlet)

4.16 δ (2 H, doublet, J=7 Hz)

5.71 δ (1 H, triplet, J=7 Hz)

(b) C10H14

1.30 δ (9 H, singlet)

7.30 δ (5 H, singlet)

(c) C4H7BrO

2.11 δ (3 H, singlet)

3.52 δ (2 H, triplet, J=6 Hz)

4.40 δ (2 H, triplet, J=6 Hz)

(d) C9H11Br

2.15 δ (2 H, quintet, J=7 Hz)

2.75 δ (2 H, triplet, J=7 Hz)

3.38 δ (2 H, triplet, J=7 Hz)

7.22 δ (5 H, singlet)

Expert Solution
Check Mark
Interpretation Introduction

a)

Organic Chemistry, Chapter 13.SE, Problem 55GP , additional homework tip  1

Interpretation:

The proposed structure of the compound to be identified for the given 1HNMR spectrum.

Concept introduction:

HDI calculation:

Organic Chemistry, Chapter 13.SE, Problem 55GP , additional homework tip  2

Where

C represent number of carbons.

N represent number of nitrogens.

H represent number of hydrogens.

X represent number of halogens.

Chemical shift: The frequency of the proton signal in the spectrum with reference to the standard compound which may be TMS(Tetramethylsilane) shows signal at 0 ppm(parts per million).

Multiplicity: The number of peaks on the each signal in NMR spectrum is defined as multiplicity; the multiplicity of each signal indicates the neighboring protons. It is generated by coupling of the subjected protons with the neighboring protons (both subjected and neighbor protons are to be chemically not equivalent) separated by either two or three sigma bonds.

Rule: Multiplicity of each signal is calculated using (n+1) rule only when the neighboring protons are chemically equivalent to each other.

(n+1)

where

n indicates number of neighboring protons

Integration value (I): The integration value at the bottom of the 1HNMR spectrum represents the number of protons giving rise to the signal.

To find:

The structure of the compound to be identified for the given molecular formula and 1HNMR spectrum.

Answer to Problem 55GP

Organic Chemistry, Chapter 13.SE, Problem 55GP , additional homework tip  3

Explanation of Solution

Calculate HDI value:

Organic Chemistry, Chapter 13.SE, Problem 55GP , additional homework tip  4

The HDI calculation confirms the presence of an aliphatic ring and double bond

Adjust the relative integration with the number of protons from the molecular formula.

The total number of protons in the molecular formula C10H14 is 6.

Interpret the given information.

Given information:

Three signals with multiplicity and integration values.

2.18ppm(3H, singlet)

4.16ppm(2H, doublet, J=7HZ)

5.71ppm(1H , triplet, J=7HZ)

The HDI value confirms the compound has either a ring or a double bond (one level of unsaturation). The total integration value (3+2+1=6 protons) is also an exact value with the protons of the molecular formula.

A signal at 2.18ppm with integration of 3H’s represents methyl groups which are chemically equivalent having one neighboring proton indicates the characteristic pattern of isopropyl group.

Organic Chemistry, Chapter 13.SE, Problem 55GP , additional homework tip  5

A signal with integration of 2H’s represent a methylene group appears at 4.16ppm rather 5.17ppm, consistent with the value of protons which present at alpha position to vinyl group (C=O) and accounts for the one degree of unsaturation.

The overall predicted structure is:

Organic Chemistry, Chapter 13.SE, Problem 55GP , additional homework tip  6

The methyl groups can be interchanged via no reflectional symmetry and the compound gives rise to totally three signals in spectrum.

Conclusion

The structure of the compound is identified using the details of spectrum and DHI calculation.

Expert Solution
Check Mark
Interpretation Introduction

b)

Organic Chemistry, Chapter 13.SE, Problem 55GP , additional homework tip  7

Interpretation:

The proposed structure of the compound to be identified for the given 1HNMR spectrum.

Concept introduction:

HDI calculation:

Organic Chemistry, Chapter 13.SE, Problem 55GP , additional homework tip  8

Where

C represent number of carbons.

N represent number of nitrogens.

H represent number of hydrogens.

X represent number of halogens.

Chemical shift: The frequency of the proton signal in the spectrum with reference to the standard compound which may be TMS(Tetramethylsilane) shows signal at 0 ppm(parts per million).

Multiplicity: The number of peaks on the each signal in NMR spectrum is defined as multiplicity; the multiplicity of each signal indicates the neighboring protons. It is generated by coupling of the subjected protons with the neighboring protons (both subjected and neighbor protons are to be chemically not equivalent) separated by either two or three sigma bonds.

Rule: Multiplicity of each signal is calculated using (n+1) rule only when the neighboring protons are chemically equivalent to each other.

(n+1)

where

n indicates number of neighboring protons

Integration value (I): The integration value at the bottom of the 1HNMR spectrum represents the number of protons giving rise to the signal.

To find:

The structure of the compound to be identified for the given molecular formula and 1HNMR spectrum.

Answer to Problem 55GP

Organic Chemistry, Chapter 13.SE, Problem 55GP , additional homework tip  9

Explanation of Solution

Calculate HDI value:

Organic Chemistry, Chapter 13.SE, Problem 55GP , additional homework tip  10

The HDI calculation confirms the presence of an aliphatic ring and double bond

Adjust the relative integration with the number of protons from the molecular formula.

The total number of protons in the molecular formula C10H14 is 14.

Interpret the given information.

Given information:

Three signals with multiplicity and integration values.

1.30ppm(9H, singlet)

7.30ppm(5H, singlet)

The HDI value confirms the compound has either a ring or a double bond (four level of unsaturation). The total integration value (9+5=14 protons) is also an exact value with the protons of the molecular formula.

A signal at 1.30ppm with integration of 9H’s represents methyl groups which are chemically equivalent having one neighboring proton indicates the characteristic pattern of isopropyl group.

Organic Chemistry, Chapter 13.SE, Problem 55GP , additional homework tip  11

A signal with integration of 5H’s represent a benzene appears at 7.30ppm which present at aromatic group and accounts for the four degree of unsaturation.

The overall predicted structure is:

Organic Chemistry, Chapter 13.SE, Problem 55GP , additional homework tip  12

The methyl groups can be interchanged via reflectional symmetry and the compound gives rise to totally two signals in 1HNMR spectrum.

Conclusion

The structure of the compound is identified using the details of 1HNMR spectrum and DHI calculation.

Expert Solution
Check Mark
Interpretation Introduction

c)

Organic Chemistry, Chapter 13.SE, Problem 55GP , additional homework tip  13

Interpretation:

The proposed structure of the compound to be identified for the given 1HNMR spectrum.

Concept introduction:

HDI calculation:

Organic Chemistry, Chapter 13.SE, Problem 55GP , additional homework tip  14

Where

C represent number of carbons.

N represent number of nitrogens.

H represent number of hydrogens.

X represent number of halogens.

Chemical shift: The frequency of the proton signal in the spectrum with reference to the standard compound which may be TMS(Tetramethylsilane) shows signal at 0 ppm(parts per million).

Multiplicity: The number of peaks on the each signal in NMR spectrum is defined as multiplicity; the multiplicity of each signal indicates the neighboring protons. It is generated by coupling of the subjected protons with the neighboring protons (both subjected and neighbor protons are to be chemically not equivalent) separated by either two or three sigma bonds.

Rule: Multiplicity of each signal is calculated using (n+1) rule only when the neighboring protons are chemically equivalent to each other.

(n+1)

where

n indicates number of neighboring protons

Integration value (I): The integration value at the bottom of the spectrum represents the number of protons giving rise to the signal.

To find:

The structure of the compound to be identified for the given molecular formula and 1HNMR spectrum.

Answer to Problem 55GP

Organic Chemistry, Chapter 13.SE, Problem 55GP , additional homework tip  15

Explanation of Solution

Calculate HDI value:

Organic Chemistry, Chapter 13.SE, Problem 55GP , additional homework tip  16

The HDI calculation confirms the presence of an aliphatic ring and double bond

Adjust the relative integration with the number of protons from the molecular formula.

The total number of protons in the molecular formula C4H7Bro is 7.

Interpret the given information.

Given information:

Three signals with multiplicity and integration values.

2.11ppm(3H, singlet)

3.52ppm(2H, triplet, J=6HZ)

4.40ppm(2H, triplet, J=6HZ)

The HDI value confirms the compound has either a ring or a double bond (one level of unsaturation). The total integration value (3+2+2=7 protons) is also an exact value with the protons of the molecular formula.

A signal at 2.11ppm with integration of 3H’s represents methyl groups which are chemically equivalent having one neighboring proton indicates the characteristic pattern of isopropyl group.

Organic Chemistry, Chapter 13.SE, Problem 55GP , additional homework tip  17

A two signal with integration of 2H’s represent a methylene group appears at 3.52ppm rather 4.40ppm, consistent with the value of protons which present at alpha position to carbonyl group(C=O) and accounts for the one degree of unsaturation.

The overall predicted structure is:

Organic Chemistry, Chapter 13.SE, Problem 55GP , additional homework tip  18

The methyl groups can be interchanged via no reflectional symmetry and the compound gives rise to totally three signals in spectrum.

Conclusion

The structure of the compound is identified using the details of 1HNMR spectrum and DHI calculation.

Expert Solution
Check Mark
Interpretation Introduction

d)

Organic Chemistry, Chapter 13.SE, Problem 55GP , additional homework tip  19

Interpretation:

The proposed structure of the compound to be identified for the given 1HNMR spectrum.

Concept introduction:

HDI calculation:

Organic Chemistry, Chapter 13.SE, Problem 55GP , additional homework tip  20

Where

C represent number of carbons.

N represent number of nitrogens.

H represent number of hydrogens.

X represent number of halogens.

Chemical shift: The frequency of the proton signal in the spectrum with reference to the standard compound which may be TMS(Tetramethylsilane) shows signal at 0 ppm(parts per million).

Multiplicity: The number of peaks on the each signal in NMR spectrum is defined as multiplicity; the multiplicity of each signal indicates the neighboring protons. It is generated by coupling of the subjected protons with the neighboring protons (both subjected and neighbor protons are to be chemically not equivalent) separated by either two or three sigma bonds.

Rule: Multiplicity of each signal is calculated using (n+1) rule only when the neighboring protons are chemically equivalent to each other.

(n+1)

where

n indicates number of neighboring protons

Integration value (I): The integration value at the bottom of the spectrum represents the number of protons giving rise to the signal.

To find:

The structure of the compound to be identified for the given molecular formula and 1HNMR spectrum.

Answer to Problem 55GP

Organic Chemistry, Chapter 13.SE, Problem 55GP , additional homework tip  21

Explanation of Solution

Calculate HDI value:

Organic Chemistry, Chapter 13.SE, Problem 55GP , additional homework tip  22

The HDI calculation confirms the presence of an aliphatic ring and double bond

Adjust the relative integration with the number of protons from the molecular formula.

The total number of protons in the molecular formula C9H11Br is 11.

Interpret the given information.

Given information:

Three signals with multiplicity and integration values.

2.15ppm(2H, quintet, J=7HZ)

2.75ppm(2H, triplet, J=7HZ)

3.38ppm(2H, triplet, J=7HZ)

7.22ppm(5H, singlet)

The HDI value confirms the compound has either a ring or a double bond (four level of unsaturation). The total integration value (2+2+2+5=11 protons) is also an exact value with the protons of the molecular formula.

A signal at 2.15ppm with integration of 2H’s represents three methyl groups which are chemically equivalent having one neighboring proton indicates the characteristic pattern of alkyl group.

A signal with integration of 2H’s represent benzylic appears at 2.75ppm which present at aromatic group

Organic Chemistry, Chapter 13.SE, Problem 55GP , additional homework tip  23

A signal with integration of 5H’s represent benzylic appears at 7.30ppm which present at aromatic group and accounts for the four degree of unsaturation.

The overall predicted structure is:

Organic Chemistry, Chapter 13.SE, Problem 55GP , additional homework tip  24

The methyl groups can be interchanged via reflectional symmetry and the compound gives rise to totally four signals in spectrum.

Conclusion

The structure of the compound is identified using the details of 1HNMR spectrum and DHI calculation.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
Propose a structure for each of the following two isomers with formula C6H14 given their 1H-NMR spectra. Isomer A:  δ = 0.84 (d, 12 H), 1.39 (septet, 2H) ppm Isomer B:  δ = 0.84 (t, 3 H), 0.86 (t, 9H), 1.22 (q, 2H) ppm
What is the structure of the compound with the formula C5H12O, if it has a strong broad IR signal centered near 3330 cm-2, and the 1H-NMR spectrum is: 0.91 ppm (3H triplet) 1.19 ppm (6H singlet) 1.50 ppm (2H quartet) 2.24 ppm (1H singlet)
An unknown compound C3H2NCl shows moderately strong IR absorptions around 1650 cm-1 and 2200 cm-1. Its NMR spectrum consists of two doublets (J = 14 Hz) at δ 5.9 and δ 7.1. Propose a structure consistent with this data?

Chapter 13 Solutions

Organic Chemistry

Ch. 13.6 - The integrated 1H NMR spectrum of a compound of...Ch. 13.7 - Identify the indicated sets of protons as...Ch. 13.7 - How many kinds of electronically nonequivalent...Ch. 13.7 - How many absorptions would you expect (S)-malate,...Ch. 13.8 - 3-Bromo-1-phenyl-1-propene shows a complex NMR...Ch. 13.9 - How could you use 1H NMR to determine the...Ch. 13.11 - Prob. 17PCh. 13.11 - Propose structures for compounds that fit the...Ch. 13.11 - Prob. 19PCh. 13.12 - Prob. 20PCh. 13.12 - Prob. 21PCh. 13.12 - Prob. 22PCh. 13.13 - Prob. 23PCh. 13.SE - Into how many peaks would you expect the 1H NMR...Ch. 13.SE - How many absorptions would you expect the...Ch. 13.SE - Sketch what you might expect the 1H and 13C NMR...Ch. 13.SE - How many electronically nonequivalent kinds of...Ch. 13.SE - Identify the indicated protons in the following...Ch. 13.SE - Prob. 29APCh. 13.SE - Prob. 30APCh. 13.SE - When measured on a spectrometer operating at 200...Ch. 13.SE - Prob. 32APCh. 13.SE - Prob. 33APCh. 13.SE - How many types of nonequivalent protons are...Ch. 13.SE - The following compounds all show a single line in...Ch. 13.SE - Prob. 36APCh. 13.SE - Propose structures for compounds with the...Ch. 13.SE - Predict the splitting pattern for each kind of...Ch. 13.SE - Predict the splitting pattern for each kind of...Ch. 13.SE - Identify the indicated sets of protons as...Ch. 13.SE - Identify the indicated sets of protons as...Ch. 13.SE - The acid-catalyzed dehydration of...Ch. 13.SE - How could you use 1H NMR to distinguish between...Ch. 13.SE - Propose structures for compounds that fit the...Ch. 13.SE - Propose structures for the two compounds whose 1H...Ch. 13.SE - Prob. 46APCh. 13.SE - How many absorptions would you expect to observe...Ch. 13.SE - Prob. 48APCh. 13.SE - How could you use 1H and 13C NMR to help...Ch. 13.SE - How could you use 1H NMR, 13C NMR, and IR...Ch. 13.SE - Assign as many resonances as you can to specific...Ch. 13.SE - Assume that you have a compound with the formula...Ch. 13.SE - The compound whose 1H NMR spectrum is shown has...Ch. 13.SE - The compound whose 1H NMR spectrum is shown has...Ch. 13.SE - Propose structures for compounds that fit the...Ch. 13.SE - Long-range coupling between protons more than two...Ch. 13.SE - The 1H and 13C NMR spectra of compound A, C8H9Br,...Ch. 13.SE - Propose structures for the three compounds whose...Ch. 13.SE - The mass spectrum and 13C NMR spectrum of a...Ch. 13.SE - Compound A, a hydrocarbon with M+=96 in its mass...Ch. 13.SE - Propose a structure for compound C, which has...Ch. 13.SE - Prob. 62GPCh. 13.SE - Propose a structure for compound E, C7H12O2, which...Ch. 13.SE - Compound F, a hydrocarbon with M+=96 in its mass...Ch. 13.SE - 3-Methyl-2-butanol has five signals in its 13C NMR...Ch. 13.SE - A 13C NMR spectrum of commercially available...Ch. 13.SE - Carboxylic acids (RCO2H) react with alcohols (ROH)...Ch. 13.SE - Prob. 68GPCh. 13.SE - The proton NMR spectrum is shown for a compound...Ch. 13.SE - The proton NMR spectrum of a compound with the...Ch. 13.SE - The proton NMR spectrum is shown for a compound...
Knowledge Booster
Background pattern image
Chemistry
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Similar questions
Recommended textbooks for you
  • Text book image
    Organic Chemistry
    Chemistry
    ISBN:9781305080485
    Author:John E. McMurry
    Publisher:Cengage Learning
    Text book image
    Organic Chemistry
    Chemistry
    ISBN:9781305580350
    Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. Foote
    Publisher:Cengage Learning
Text book image
Organic Chemistry
Chemistry
ISBN:9781305080485
Author:John E. McMurry
Publisher:Cengage Learning
Text book image
Organic Chemistry
Chemistry
ISBN:9781305580350
Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. Foote
Publisher:Cengage Learning
NMR Spectroscopy; Author: Professor Dave Explains;https://www.youtube.com/watch?v=SBir5wUS3Bo;License: Standard YouTube License, CC-BY