How would you use NMR (either 13 C or 1 H) to distinguish between the following pairs of isomers?

Question

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Answer 1

Textbook Question

Chapter 20.SE, Problem 55AP

How would you use NMR (either ¹³C or ¹H) to distinguish between the following pairs of isomers?

Chapter 20.SE, Problem 55AP, How would you use NMR (either 13C or 1H) to distinguish between the following pairs of isomers?

Expert Solution

Interpretation Introduction

a)

Organic Chemistry, Chapter 20.SE, Problem 55AP , additional homework tip 1

Interpretation:

How NMR (either ¹³C or ¹H) could be used to distinguish between benzene-1,2-dicarboxylic acid and benzene-1,3-dicarboxylic acid is to be stated.

Concept introduction:

In ¹HNMR, the acidic COOH proton normally absorbs as singlet, broadened in few cases, near 12 δ.

In ¹³CNMR, the carboxyl carbon absorb in the range 165-185 δ. The aromatic and α, β-unsaturated acids absorb near 165 δ and saturated aliphatic acids near 185 δ. The nitrile carbons absorb in the range 115-130 δ.

Compounds with symmetrical structures (like p-disubstituted phenyl) will have a simpler spectrum.

To state:

How NMR (either ¹³C or ¹H) could be used to distinguish between benzene-1,3-dicarboxylic acid and benzene-1,4-dicarboxylic acid.

Answer to Problem 55AP

1,4-Benzenedicarboxylic acid will have a simpler spectrum. In ¹H NMR, there will be a doublet of doublet in the aromatic region and a singlet for carboxyl proton around 12 δ. In ¹³C NMR there will be three signals, two in aromatic region and one for carboxyl carbon around 165 δ.

The ¹H NMR of 1,3-Benzenedicarboxylic acid will have a multiplet in the aromatic region due to the ring protons and a singlet for carboxyl proton around 12 δ. In ¹³C NMR there will be five signals, four in aromatic region and one for carboxyl carbon around 165 δ.

Explanation of Solution

1,4-Benzenedicarboxylic acid has a symmetrical structure. Hence it gives minimum peaks in both ¹H NMR and ¹³C NMR spectrum.

In ¹H NMR, the four aromatic hydrogens belong to two groups and thus give a doublet of doublet in the aromatic region. The carboxyl proton gives another signal. Thus there will be three signals.

In ¹³C NMR, the six ring carbons classify themselves into two groups and give two signals while the carboxyl carbons give a signal. Thus there will be three signals.

1,3-Benzenedicarboxylic acid does not have a symmetrical structure. Hence it gives more peaks in both ¹H NMR and ¹³C NMR spectrum than 1,4-Benzenedicarboxylic acid.

In ¹H NMR, the four aromatic hydrogens give complex multiplet signals in the aromatic region. The carboxyl protons yield another signal. Thus there will be two signals.

In ¹³C NMR, the six ring carbons classify themselves into four groups and give four signals while the carboxyl carbons give a signal. Thus there will be five signals.

Conclusion

1,4-Benzenedicarboxylic acid will have a simpler spectrum. In ¹H NMR, there will be a doublet of doublet in the aromatic region and a singlet for carboxyl proton around 12 δ. In ¹³C NMR there will be three signals, two in aromatic region and one for carboxyl carbon around 165 δ.

The ¹H NMR of 1,3-Benzenedicarboxylic acid will have a multiplet in the aromatic region due to the ring protons and a singlet for carboxyl proton around 12 δ. In ¹³C NMR there will be five signals, four in aromatic region and one for carboxyl carbon around 165 δ.

Expert Solution

Interpretation Introduction

b) HO₂CCH₂CH₂CO₂H and CH₃CH(CO₂H)₂

Interpretation:

How NMR (either ¹³C or ¹H) could be used to distinguish between succinic acid and ethane-1,1-dicarboxylic acid is to be stated.

Concept introduction:

In ¹HNMR, the acidic COOH proton normally absorbs as singlet, broadened in few cases, near 12 δ.

In ¹³CNMR, the carboxyl carbon absorb in the range 165-185 δ. The aromatic and α, β-unsaturated acids absorb near 165 δ and saturated aliphatic acids near 185 δ. The nitrile carbons absorb in the range 115-130 δ.

Compounds with symmetrical structures (like p-disubstituted phenyl) will have a simpler spectrum.

To state:

How NMR (either ¹³C or ¹H) could be used to distinguish between succinic acid and ethane-1,1-dicarboxylic acid.

Answer to Problem 55AP

Succinic acid will have a simpler spectrum. In ¹H NMR, there will be two signals, a triplet for the methylene groups and a singlet for carboxyl protons around 12 δ. In ¹³C NMR there will be two signals, one for the methylene carbon and one for carboxyl carbons around 165 δ.

The ¹H NMR, 2-methylmalonic acid will have three signals, a doublet for the methyl protons, a quartet for the methane proton and a singlet for carboxyl proton around 12 δ. In ¹³C NMR there will be three signals, one for methyl carbon another for the methine carbon and a third one for carboxyl carbons around 165 δ.

Explanation of Solution

Succinic acid has a simpler spectrum since it is symmetrical. In ¹H NMR, the two methylene protons split each other giving a triplet integrating to four protons. The two carboxyl protons yield another singlet signal. Thus there will be two signals a triplet and a singlet.

In ¹³C NMR, the four carbons in succinic acid classify themselves into two groups and give two signals. Thus there will be two signals, one for methylene carbons and other for carboxyl carbons.

2-methylmalonicacid does not have a symmetrical structure. Hence it gives more peaks in both ¹H NMR and ¹³C NMR spectrum than succinic acid.

In ¹H NMR, there will be a doublet for the methyl protons, a quartet integrating to one proton for the methine proton signal and a singlet for the carboxyl protons. Thus there will be three signals.

In ¹³C NMR, there will be separate signals for methyl carbon, methine carbon and carboxyl carbons. Thus there will be three signals.

Conclusion

Succinic acid will have a simpler spectrum. In ¹H NMR, there will be two signals, a triplet for the methylene groups and a singlet for carboxyl protons around 12 δ. In ¹³C NMR there will be two signals, one for the methylene carbons and one for carboxyl carbons around 165 δ.

The ¹H NMR, 2-methylmalonic acid will have three signals, a doublet for the methyl protons, a quartet for the methine proton and a singlet for carboxyl proton around 12 δ. In ¹³C NMR there will be three signals, one for methyl carbon another for the methine carbon and a third one for carboxyl carbons around 165 δ.

Expert Solution

Interpretation Introduction

c) CH₃CH₂CH₂CO₂H and HOCH₂CH₂CH₂CHO

Interpretation:

How NMR (either ¹³C or ¹H) could be used to distinguish between succinic acid and 4-hydroxybutanal is to be stated.

Concept introduction:

In ¹HNMR, the acidic COOH proton normally absorbs as singlet, broadened in few cases, near 12 δ.

In ¹³CNMR, the carboxyl carbon absorb in the range 165-185 δ. The aromatic and α, β-unsaturated acids absorb near 165 δ and saturated aliphatic acids near 185 δ. The nitrile carbons absorb in the range 115-130 δ.

Compounds with symmetrical structures will have a simpler spectrum.

In ¹HNMR the aldehyde protons absorb near 10 δ with a coupling constant, J=3Hz. Hydrogens on the carbon next to aldehyde group absorb near 2.0-2.3 δ.

In ¹³CNMR, saturated aldehydes and ketones usually absorb in the region from 200 to 215 δ while the aromatic and unsaturated carbonyl compounds absorb in the 190 to 200 δ region.

To state:

How NMR (either ¹³C or ¹H) could be used to distinguish between butanoic acid and 4-hydroxybutanal.

Answer to Problem 55AP

In ¹H NMR of butanoic acid there will be a signal around 12 δ while ¹H NMR of 3-hydroxybutanal there will be a signal for the aldehydic proton around 9-10 δ and for the hydroxyl proton in the range 3.0-4.5 δ. There will be no signal around 12 δ. In ¹³C NMR the carboxyl carbon absorbs around 165 δ while the absorption due to aldehydic carbon is observed around 200-215 δ.

Explanation of Solution

The two compounds can be easily distinguished. For butanoic acid there will be signals for the absorption of carboxyl proton in ¹H NMR (12 δ) and carboxyl carbon in ¹³C NMR (165 δ). In 3-hydroxybutanal, there will be signals for the absorption of hydroxyl proton in ¹H NMR (3.0-4.5 δ) and aldehydic proton around 9-10 δ. The aldehydic carbon will resonate in ¹³C NMR in the range 200-215 δ.

Conclusion

In ¹H NMR of butanoic acid there will be a signal around 12 δ while ¹H NMR of 3-hydroxybutanal there will be a signal for the aldehydic proton around 9-10 δ and for the hydroxyl proton in the range 3.0-4.5 δ. There will be no signal around 12 δ. In ¹³C NMR the carboxyl carbon absorbs around 165 δ while the absorptrion due to aldehydic carbon is observed around 200-215 δ.

Expert Solution

Interpretation Introduction

d)

Organic Chemistry, Chapter 20.SE, Problem 55AP , additional homework tip 2

Interpretation:

How NMR (either ¹³C or ¹H) could be used to distinguish between 4-methylpent-3-eneoic acid and cyclopentylcarboxylic acid is to be stated.

Concept introduction:

In ¹HNMR, the acidic COOH proton normally absorbs as singlet, broadened in few cases, near 12 δ.

In ¹³CNMR, the carboxyl carbon absorb in the range 165-185 δ. The aromatic and α, β-unsaturated acids absorb near 165 δ and saturated aliphatic acids near 185 δ. The nitrile carbons absorb in the range 115-130 δ.

Alkenes show a signal in the range 4.5-6.5 δ in ¹HNMR and around 110-150 δ in ¹³CNMR.

Compounds with symmetrical structures will have a simpler spectrum.

To state:

How NMR (either ¹³C or ¹H) could be used to distinguish between 4-methylpent-3-eneoic acid and cyclopentylcarboxylic acid.

Answer to Problem 55AP

In ¹HNMR, 4-methylpent-3-eneoic acid will show the peak due to olefinic proton absorption in the range 4.5-6.5 δ. In ¹³CNMR, the two carbons in the double bond each will give a signal in the range 110-150 δ. There will be no signals in these ranges for cyclopentanecarboxylic acid in both the spectra.

Explanation of Solution

4-Methylpent-3-eneoic acid has a hydrogen atom attached to olefinic double bond. Since it is a vinylic hydrogen its absorption peak is observed in the range 4.5-6.5 δ. The two olefinic carbons are not equivalent. Hence they give two signals in the range 110-150 δ. In the ¹HNMR and ¹³CNMR spectra of cyclopentanecarboxylic acid no signals will be observed in the ranges 4.5-6.5 δ and 110-150 δ respectively.

Conclusion

In ¹HNMR, 4-methylpent-3-eneoic acid will show the peak due to olefinic proton absorption in the range 4.5-6.5 δ. In ¹³CNMR, the two carbons in the double bond each will give a signal in the range 110-150 δ. There will be no signals in these ranges for cyclopentanecarboxylic acid in both the spectra.

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Answer 2

Textbook Question

Chapter 20.SE, Problem 55AP

How would you use NMR (either ¹³C or ¹H) to distinguish between the following pairs of isomers?

Chapter 20.SE, Problem 55AP, How would you use NMR (either 13C or 1H) to distinguish between the following pairs of isomers?

Expert Solution

Interpretation Introduction

a)

Organic Chemistry, Chapter 20.SE, Problem 55AP , additional homework tip 1

Interpretation:

How NMR (either ¹³C or ¹H) could be used to distinguish between benzene-1,2-dicarboxylic acid and benzene-1,3-dicarboxylic acid is to be stated.

Concept introduction:

In ¹HNMR, the acidic COOH proton normally absorbs as singlet, broadened in few cases, near 12 δ.

In ¹³CNMR, the carboxyl carbon absorb in the range 165-185 δ. The aromatic and α, β-unsaturated acids absorb near 165 δ and saturated aliphatic acids near 185 δ. The nitrile carbons absorb in the range 115-130 δ.

Compounds with symmetrical structures (like p-disubstituted phenyl) will have a simpler spectrum.

To state:

How NMR (either ¹³C or ¹H) could be used to distinguish between benzene-1,3-dicarboxylic acid and benzene-1,4-dicarboxylic acid.

Answer to Problem 55AP

1,4-Benzenedicarboxylic acid will have a simpler spectrum. In ¹H NMR, there will be a doublet of doublet in the aromatic region and a singlet for carboxyl proton around 12 δ. In ¹³C NMR there will be three signals, two in aromatic region and one for carboxyl carbon around 165 δ.

The ¹H NMR of 1,3-Benzenedicarboxylic acid will have a multiplet in the aromatic region due to the ring protons and a singlet for carboxyl proton around 12 δ. In ¹³C NMR there will be five signals, four in aromatic region and one for carboxyl carbon around 165 δ.

Explanation of Solution

1,4-Benzenedicarboxylic acid has a symmetrical structure. Hence it gives minimum peaks in both ¹H NMR and ¹³C NMR spectrum.

In ¹H NMR, the four aromatic hydrogens belong to two groups and thus give a doublet of doublet in the aromatic region. The carboxyl proton gives another signal. Thus there will be three signals.

In ¹³C NMR, the six ring carbons classify themselves into two groups and give two signals while the carboxyl carbons give a signal. Thus there will be three signals.

1,3-Benzenedicarboxylic acid does not have a symmetrical structure. Hence it gives more peaks in both ¹H NMR and ¹³C NMR spectrum than 1,4-Benzenedicarboxylic acid.

In ¹H NMR, the four aromatic hydrogens give complex multiplet signals in the aromatic region. The carboxyl protons yield another signal. Thus there will be two signals.

In ¹³C NMR, the six ring carbons classify themselves into four groups and give four signals while the carboxyl carbons give a signal. Thus there will be five signals.

Conclusion

1,4-Benzenedicarboxylic acid will have a simpler spectrum. In ¹H NMR, there will be a doublet of doublet in the aromatic region and a singlet for carboxyl proton around 12 δ. In ¹³C NMR there will be three signals, two in aromatic region and one for carboxyl carbon around 165 δ.

The ¹H NMR of 1,3-Benzenedicarboxylic acid will have a multiplet in the aromatic region due to the ring protons and a singlet for carboxyl proton around 12 δ. In ¹³C NMR there will be five signals, four in aromatic region and one for carboxyl carbon around 165 δ.

Expert Solution

Interpretation Introduction

b) HO₂CCH₂CH₂CO₂H and CH₃CH(CO₂H)₂

Interpretation:

How NMR (either ¹³C or ¹H) could be used to distinguish between succinic acid and ethane-1,1-dicarboxylic acid is to be stated.

Concept introduction:

In ¹HNMR, the acidic COOH proton normally absorbs as singlet, broadened in few cases, near 12 δ.

In ¹³CNMR, the carboxyl carbon absorb in the range 165-185 δ. The aromatic and α, β-unsaturated acids absorb near 165 δ and saturated aliphatic acids near 185 δ. The nitrile carbons absorb in the range 115-130 δ.

Compounds with symmetrical structures (like p-disubstituted phenyl) will have a simpler spectrum.

To state:

How NMR (either ¹³C or ¹H) could be used to distinguish between succinic acid and ethane-1,1-dicarboxylic acid.

Answer to Problem 55AP

Succinic acid will have a simpler spectrum. In ¹H NMR, there will be two signals, a triplet for the methylene groups and a singlet for carboxyl protons around 12 δ. In ¹³C NMR there will be two signals, one for the methylene carbon and one for carboxyl carbons around 165 δ.

The ¹H NMR, 2-methylmalonic acid will have three signals, a doublet for the methyl protons, a quartet for the methane proton and a singlet for carboxyl proton around 12 δ. In ¹³C NMR there will be three signals, one for methyl carbon another for the methine carbon and a third one for carboxyl carbons around 165 δ.

Explanation of Solution

Succinic acid has a simpler spectrum since it is symmetrical. In ¹H NMR, the two methylene protons split each other giving a triplet integrating to four protons. The two carboxyl protons yield another singlet signal. Thus there will be two signals a triplet and a singlet.

In ¹³C NMR, the four carbons in succinic acid classify themselves into two groups and give two signals. Thus there will be two signals, one for methylene carbons and other for carboxyl carbons.

2-methylmalonicacid does not have a symmetrical structure. Hence it gives more peaks in both ¹H NMR and ¹³C NMR spectrum than succinic acid.

In ¹H NMR, there will be a doublet for the methyl protons, a quartet integrating to one proton for the methine proton signal and a singlet for the carboxyl protons. Thus there will be three signals.

In ¹³C NMR, there will be separate signals for methyl carbon, methine carbon and carboxyl carbons. Thus there will be three signals.

Conclusion

Succinic acid will have a simpler spectrum. In ¹H NMR, there will be two signals, a triplet for the methylene groups and a singlet for carboxyl protons around 12 δ. In ¹³C NMR there will be two signals, one for the methylene carbons and one for carboxyl carbons around 165 δ.

The ¹H NMR, 2-methylmalonic acid will have three signals, a doublet for the methyl protons, a quartet for the methine proton and a singlet for carboxyl proton around 12 δ. In ¹³C NMR there will be three signals, one for methyl carbon another for the methine carbon and a third one for carboxyl carbons around 165 δ.

Expert Solution

Interpretation Introduction

c) CH₃CH₂CH₂CO₂H and HOCH₂CH₂CH₂CHO

Interpretation:

How NMR (either ¹³C or ¹H) could be used to distinguish between succinic acid and 4-hydroxybutanal is to be stated.

Concept introduction:

In ¹HNMR, the acidic COOH proton normally absorbs as singlet, broadened in few cases, near 12 δ.

In ¹³CNMR, the carboxyl carbon absorb in the range 165-185 δ. The aromatic and α, β-unsaturated acids absorb near 165 δ and saturated aliphatic acids near 185 δ. The nitrile carbons absorb in the range 115-130 δ.

Compounds with symmetrical structures will have a simpler spectrum.

In ¹HNMR the aldehyde protons absorb near 10 δ with a coupling constant, J=3Hz. Hydrogens on the carbon next to aldehyde group absorb near 2.0-2.3 δ.

In ¹³CNMR, saturated aldehydes and ketones usually absorb in the region from 200 to 215 δ while the aromatic and unsaturated carbonyl compounds absorb in the 190 to 200 δ region.

To state:

How NMR (either ¹³C or ¹H) could be used to distinguish between butanoic acid and 4-hydroxybutanal.

Answer to Problem 55AP

In ¹H NMR of butanoic acid there will be a signal around 12 δ while ¹H NMR of 3-hydroxybutanal there will be a signal for the aldehydic proton around 9-10 δ and for the hydroxyl proton in the range 3.0-4.5 δ. There will be no signal around 12 δ. In ¹³C NMR the carboxyl carbon absorbs around 165 δ while the absorption due to aldehydic carbon is observed around 200-215 δ.

Explanation of Solution

The two compounds can be easily distinguished. For butanoic acid there will be signals for the absorption of carboxyl proton in ¹H NMR (12 δ) and carboxyl carbon in ¹³C NMR (165 δ). In 3-hydroxybutanal, there will be signals for the absorption of hydroxyl proton in ¹H NMR (3.0-4.5 δ) and aldehydic proton around 9-10 δ. The aldehydic carbon will resonate in ¹³C NMR in the range 200-215 δ.

Conclusion

In ¹H NMR of butanoic acid there will be a signal around 12 δ while ¹H NMR of 3-hydroxybutanal there will be a signal for the aldehydic proton around 9-10 δ and for the hydroxyl proton in the range 3.0-4.5 δ. There will be no signal around 12 δ. In ¹³C NMR the carboxyl carbon absorbs around 165 δ while the absorptrion due to aldehydic carbon is observed around 200-215 δ.

Expert Solution

Interpretation Introduction

d)

Organic Chemistry, Chapter 20.SE, Problem 55AP , additional homework tip 2

Interpretation:

How NMR (either ¹³C or ¹H) could be used to distinguish between 4-methylpent-3-eneoic acid and cyclopentylcarboxylic acid is to be stated.

Concept introduction:

In ¹HNMR, the acidic COOH proton normally absorbs as singlet, broadened in few cases, near 12 δ.

In ¹³CNMR, the carboxyl carbon absorb in the range 165-185 δ. The aromatic and α, β-unsaturated acids absorb near 165 δ and saturated aliphatic acids near 185 δ. The nitrile carbons absorb in the range 115-130 δ.

Alkenes show a signal in the range 4.5-6.5 δ in ¹HNMR and around 110-150 δ in ¹³CNMR.

Compounds with symmetrical structures will have a simpler spectrum.

To state:

How NMR (either ¹³C or ¹H) could be used to distinguish between 4-methylpent-3-eneoic acid and cyclopentylcarboxylic acid.

Answer to Problem 55AP

In ¹HNMR, 4-methylpent-3-eneoic acid will show the peak due to olefinic proton absorption in the range 4.5-6.5 δ. In ¹³CNMR, the two carbons in the double bond each will give a signal in the range 110-150 δ. There will be no signals in these ranges for cyclopentanecarboxylic acid in both the spectra.

Explanation of Solution

4-Methylpent-3-eneoic acid has a hydrogen atom attached to olefinic double bond. Since it is a vinylic hydrogen its absorption peak is observed in the range 4.5-6.5 δ. The two olefinic carbons are not equivalent. Hence they give two signals in the range 110-150 δ. In the ¹HNMR and ¹³CNMR spectra of cyclopentanecarboxylic acid no signals will be observed in the ranges 4.5-6.5 δ and 110-150 δ respectively.

Conclusion

In ¹HNMR, 4-methylpent-3-eneoic acid will show the peak due to olefinic proton absorption in the range 4.5-6.5 δ. In ¹³CNMR, the two carbons in the double bond each will give a signal in the range 110-150 δ. There will be no signals in these ranges for cyclopentanecarboxylic acid in both the spectra.

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Answer 3

Textbook Question

Chapter 20.SE, Problem 55AP

How would you use NMR (either ¹³C or ¹H) to distinguish between the following pairs of isomers?

Chapter 20.SE, Problem 55AP, How would you use NMR (either 13C or 1H) to distinguish between the following pairs of isomers?

Expert Solution

Interpretation Introduction

a)

Organic Chemistry, Chapter 20.SE, Problem 55AP , additional homework tip 1

Interpretation:

How NMR (either ¹³C or ¹H) could be used to distinguish between benzene-1,2-dicarboxylic acid and benzene-1,3-dicarboxylic acid is to be stated.

Concept introduction:

In ¹HNMR, the acidic COOH proton normally absorbs as singlet, broadened in few cases, near 12 δ.

In ¹³CNMR, the carboxyl carbon absorb in the range 165-185 δ. The aromatic and α, β-unsaturated acids absorb near 165 δ and saturated aliphatic acids near 185 δ. The nitrile carbons absorb in the range 115-130 δ.

Compounds with symmetrical structures (like p-disubstituted phenyl) will have a simpler spectrum.

To state:

How NMR (either ¹³C or ¹H) could be used to distinguish between benzene-1,3-dicarboxylic acid and benzene-1,4-dicarboxylic acid.

Answer to Problem 55AP

1,4-Benzenedicarboxylic acid will have a simpler spectrum. In ¹H NMR, there will be a doublet of doublet in the aromatic region and a singlet for carboxyl proton around 12 δ. In ¹³C NMR there will be three signals, two in aromatic region and one for carboxyl carbon around 165 δ.

The ¹H NMR of 1,3-Benzenedicarboxylic acid will have a multiplet in the aromatic region due to the ring protons and a singlet for carboxyl proton around 12 δ. In ¹³C NMR there will be five signals, four in aromatic region and one for carboxyl carbon around 165 δ.

Explanation of Solution

1,4-Benzenedicarboxylic acid has a symmetrical structure. Hence it gives minimum peaks in both ¹H NMR and ¹³C NMR spectrum.

In ¹H NMR, the four aromatic hydrogens belong to two groups and thus give a doublet of doublet in the aromatic region. The carboxyl proton gives another signal. Thus there will be three signals.

In ¹³C NMR, the six ring carbons classify themselves into two groups and give two signals while the carboxyl carbons give a signal. Thus there will be three signals.

1,3-Benzenedicarboxylic acid does not have a symmetrical structure. Hence it gives more peaks in both ¹H NMR and ¹³C NMR spectrum than 1,4-Benzenedicarboxylic acid.

In ¹H NMR, the four aromatic hydrogens give complex multiplet signals in the aromatic region. The carboxyl protons yield another signal. Thus there will be two signals.

In ¹³C NMR, the six ring carbons classify themselves into four groups and give four signals while the carboxyl carbons give a signal. Thus there will be five signals.

Conclusion

1,4-Benzenedicarboxylic acid will have a simpler spectrum. In ¹H NMR, there will be a doublet of doublet in the aromatic region and a singlet for carboxyl proton around 12 δ. In ¹³C NMR there will be three signals, two in aromatic region and one for carboxyl carbon around 165 δ.

The ¹H NMR of 1,3-Benzenedicarboxylic acid will have a multiplet in the aromatic region due to the ring protons and a singlet for carboxyl proton around 12 δ. In ¹³C NMR there will be five signals, four in aromatic region and one for carboxyl carbon around 165 δ.

Expert Solution

Interpretation Introduction

b) HO₂CCH₂CH₂CO₂H and CH₃CH(CO₂H)₂

Interpretation:

How NMR (either ¹³C or ¹H) could be used to distinguish between succinic acid and ethane-1,1-dicarboxylic acid is to be stated.

Concept introduction:

In ¹HNMR, the acidic COOH proton normally absorbs as singlet, broadened in few cases, near 12 δ.

In ¹³CNMR, the carboxyl carbon absorb in the range 165-185 δ. The aromatic and α, β-unsaturated acids absorb near 165 δ and saturated aliphatic acids near 185 δ. The nitrile carbons absorb in the range 115-130 δ.

Compounds with symmetrical structures (like p-disubstituted phenyl) will have a simpler spectrum.

To state:

How NMR (either ¹³C or ¹H) could be used to distinguish between succinic acid and ethane-1,1-dicarboxylic acid.

Answer to Problem 55AP

Succinic acid will have a simpler spectrum. In ¹H NMR, there will be two signals, a triplet for the methylene groups and a singlet for carboxyl protons around 12 δ. In ¹³C NMR there will be two signals, one for the methylene carbon and one for carboxyl carbons around 165 δ.

The ¹H NMR, 2-methylmalonic acid will have three signals, a doublet for the methyl protons, a quartet for the methane proton and a singlet for carboxyl proton around 12 δ. In ¹³C NMR there will be three signals, one for methyl carbon another for the methine carbon and a third one for carboxyl carbons around 165 δ.

Explanation of Solution

Succinic acid has a simpler spectrum since it is symmetrical. In ¹H NMR, the two methylene protons split each other giving a triplet integrating to four protons. The two carboxyl protons yield another singlet signal. Thus there will be two signals a triplet and a singlet.

In ¹³C NMR, the four carbons in succinic acid classify themselves into two groups and give two signals. Thus there will be two signals, one for methylene carbons and other for carboxyl carbons.

2-methylmalonicacid does not have a symmetrical structure. Hence it gives more peaks in both ¹H NMR and ¹³C NMR spectrum than succinic acid.

In ¹H NMR, there will be a doublet for the methyl protons, a quartet integrating to one proton for the methine proton signal and a singlet for the carboxyl protons. Thus there will be three signals.

In ¹³C NMR, there will be separate signals for methyl carbon, methine carbon and carboxyl carbons. Thus there will be three signals.

Conclusion

Succinic acid will have a simpler spectrum. In ¹H NMR, there will be two signals, a triplet for the methylene groups and a singlet for carboxyl protons around 12 δ. In ¹³C NMR there will be two signals, one for the methylene carbons and one for carboxyl carbons around 165 δ.

The ¹H NMR, 2-methylmalonic acid will have three signals, a doublet for the methyl protons, a quartet for the methine proton and a singlet for carboxyl proton around 12 δ. In ¹³C NMR there will be three signals, one for methyl carbon another for the methine carbon and a third one for carboxyl carbons around 165 δ.

Expert Solution

Interpretation Introduction

c) CH₃CH₂CH₂CO₂H and HOCH₂CH₂CH₂CHO

Interpretation:

How NMR (either ¹³C or ¹H) could be used to distinguish between succinic acid and 4-hydroxybutanal is to be stated.

Concept introduction:

In ¹HNMR, the acidic COOH proton normally absorbs as singlet, broadened in few cases, near 12 δ.

In ¹³CNMR, the carboxyl carbon absorb in the range 165-185 δ. The aromatic and α, β-unsaturated acids absorb near 165 δ and saturated aliphatic acids near 185 δ. The nitrile carbons absorb in the range 115-130 δ.

Compounds with symmetrical structures will have a simpler spectrum.

In ¹HNMR the aldehyde protons absorb near 10 δ with a coupling constant, J=3Hz. Hydrogens on the carbon next to aldehyde group absorb near 2.0-2.3 δ.

In ¹³CNMR, saturated aldehydes and ketones usually absorb in the region from 200 to 215 δ while the aromatic and unsaturated carbonyl compounds absorb in the 190 to 200 δ region.

To state:

How NMR (either ¹³C or ¹H) could be used to distinguish between butanoic acid and 4-hydroxybutanal.

Answer to Problem 55AP

In ¹H NMR of butanoic acid there will be a signal around 12 δ while ¹H NMR of 3-hydroxybutanal there will be a signal for the aldehydic proton around 9-10 δ and for the hydroxyl proton in the range 3.0-4.5 δ. There will be no signal around 12 δ. In ¹³C NMR the carboxyl carbon absorbs around 165 δ while the absorption due to aldehydic carbon is observed around 200-215 δ.

Explanation of Solution

The two compounds can be easily distinguished. For butanoic acid there will be signals for the absorption of carboxyl proton in ¹H NMR (12 δ) and carboxyl carbon in ¹³C NMR (165 δ). In 3-hydroxybutanal, there will be signals for the absorption of hydroxyl proton in ¹H NMR (3.0-4.5 δ) and aldehydic proton around 9-10 δ. The aldehydic carbon will resonate in ¹³C NMR in the range 200-215 δ.

Conclusion

In ¹H NMR of butanoic acid there will be a signal around 12 δ while ¹H NMR of 3-hydroxybutanal there will be a signal for the aldehydic proton around 9-10 δ and for the hydroxyl proton in the range 3.0-4.5 δ. There will be no signal around 12 δ. In ¹³C NMR the carboxyl carbon absorbs around 165 δ while the absorptrion due to aldehydic carbon is observed around 200-215 δ.

Expert Solution

Interpretation Introduction

d)

Organic Chemistry, Chapter 20.SE, Problem 55AP , additional homework tip 2

Interpretation:

How NMR (either ¹³C or ¹H) could be used to distinguish between 4-methylpent-3-eneoic acid and cyclopentylcarboxylic acid is to be stated.

Concept introduction:

In ¹HNMR, the acidic COOH proton normally absorbs as singlet, broadened in few cases, near 12 δ.

In ¹³CNMR, the carboxyl carbon absorb in the range 165-185 δ. The aromatic and α, β-unsaturated acids absorb near 165 δ and saturated aliphatic acids near 185 δ. The nitrile carbons absorb in the range 115-130 δ.

Alkenes show a signal in the range 4.5-6.5 δ in ¹HNMR and around 110-150 δ in ¹³CNMR.

Compounds with symmetrical structures will have a simpler spectrum.

To state:

How NMR (either ¹³C or ¹H) could be used to distinguish between 4-methylpent-3-eneoic acid and cyclopentylcarboxylic acid.

Answer to Problem 55AP

In ¹HNMR, 4-methylpent-3-eneoic acid will show the peak due to olefinic proton absorption in the range 4.5-6.5 δ. In ¹³CNMR, the two carbons in the double bond each will give a signal in the range 110-150 δ. There will be no signals in these ranges for cyclopentanecarboxylic acid in both the spectra.

Explanation of Solution

4-Methylpent-3-eneoic acid has a hydrogen atom attached to olefinic double bond. Since it is a vinylic hydrogen its absorption peak is observed in the range 4.5-6.5 δ. The two olefinic carbons are not equivalent. Hence they give two signals in the range 110-150 δ. In the ¹HNMR and ¹³CNMR spectra of cyclopentanecarboxylic acid no signals will be observed in the ranges 4.5-6.5 δ and 110-150 δ respectively.

Conclusion

In ¹HNMR, 4-methylpent-3-eneoic acid will show the peak due to olefinic proton absorption in the range 4.5-6.5 δ. In ¹³CNMR, the two carbons in the double bond each will give a signal in the range 110-150 δ. There will be no signals in these ranges for cyclopentanecarboxylic acid in both the spectra.

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Answer 4

Textbook Question

Chapter 20.SE, Problem 55AP

How would you use NMR (either ¹³C or ¹H) to distinguish between the following pairs of isomers?

Chapter 20.SE, Problem 55AP, How would you use NMR (either 13C or 1H) to distinguish between the following pairs of isomers?

Expert Solution

Interpretation Introduction

a)

Organic Chemistry, Chapter 20.SE, Problem 55AP , additional homework tip 1

Interpretation:

How NMR (either ¹³C or ¹H) could be used to distinguish between benzene-1,2-dicarboxylic acid and benzene-1,3-dicarboxylic acid is to be stated.

Concept introduction:

In ¹HNMR, the acidic COOH proton normally absorbs as singlet, broadened in few cases, near 12 δ.

In ¹³CNMR, the carboxyl carbon absorb in the range 165-185 δ. The aromatic and α, β-unsaturated acids absorb near 165 δ and saturated aliphatic acids near 185 δ. The nitrile carbons absorb in the range 115-130 δ.

Compounds with symmetrical structures (like p-disubstituted phenyl) will have a simpler spectrum.

To state:

How NMR (either ¹³C or ¹H) could be used to distinguish between benzene-1,3-dicarboxylic acid and benzene-1,4-dicarboxylic acid.

Answer to Problem 55AP

1,4-Benzenedicarboxylic acid will have a simpler spectrum. In ¹H NMR, there will be a doublet of doublet in the aromatic region and a singlet for carboxyl proton around 12 δ. In ¹³C NMR there will be three signals, two in aromatic region and one for carboxyl carbon around 165 δ.

The ¹H NMR of 1,3-Benzenedicarboxylic acid will have a multiplet in the aromatic region due to the ring protons and a singlet for carboxyl proton around 12 δ. In ¹³C NMR there will be five signals, four in aromatic region and one for carboxyl carbon around 165 δ.

Explanation of Solution

1,4-Benzenedicarboxylic acid has a symmetrical structure. Hence it gives minimum peaks in both ¹H NMR and ¹³C NMR spectrum.

In ¹H NMR, the four aromatic hydrogens belong to two groups and thus give a doublet of doublet in the aromatic region. The carboxyl proton gives another signal. Thus there will be three signals.

In ¹³C NMR, the six ring carbons classify themselves into two groups and give two signals while the carboxyl carbons give a signal. Thus there will be three signals.

1,3-Benzenedicarboxylic acid does not have a symmetrical structure. Hence it gives more peaks in both ¹H NMR and ¹³C NMR spectrum than 1,4-Benzenedicarboxylic acid.

In ¹H NMR, the four aromatic hydrogens give complex multiplet signals in the aromatic region. The carboxyl protons yield another signal. Thus there will be two signals.

In ¹³C NMR, the six ring carbons classify themselves into four groups and give four signals while the carboxyl carbons give a signal. Thus there will be five signals.

Conclusion

1,4-Benzenedicarboxylic acid will have a simpler spectrum. In ¹H NMR, there will be a doublet of doublet in the aromatic region and a singlet for carboxyl proton around 12 δ. In ¹³C NMR there will be three signals, two in aromatic region and one for carboxyl carbon around 165 δ.

The ¹H NMR of 1,3-Benzenedicarboxylic acid will have a multiplet in the aromatic region due to the ring protons and a singlet for carboxyl proton around 12 δ. In ¹³C NMR there will be five signals, four in aromatic region and one for carboxyl carbon around 165 δ.

Expert Solution

Interpretation Introduction

b) HO₂CCH₂CH₂CO₂H and CH₃CH(CO₂H)₂

Interpretation:

How NMR (either ¹³C or ¹H) could be used to distinguish between succinic acid and ethane-1,1-dicarboxylic acid is to be stated.

Concept introduction:

In ¹HNMR, the acidic COOH proton normally absorbs as singlet, broadened in few cases, near 12 δ.

In ¹³CNMR, the carboxyl carbon absorb in the range 165-185 δ. The aromatic and α, β-unsaturated acids absorb near 165 δ and saturated aliphatic acids near 185 δ. The nitrile carbons absorb in the range 115-130 δ.

Compounds with symmetrical structures (like p-disubstituted phenyl) will have a simpler spectrum.

To state:

How NMR (either ¹³C or ¹H) could be used to distinguish between succinic acid and ethane-1,1-dicarboxylic acid.

Answer to Problem 55AP

Succinic acid will have a simpler spectrum. In ¹H NMR, there will be two signals, a triplet for the methylene groups and a singlet for carboxyl protons around 12 δ. In ¹³C NMR there will be two signals, one for the methylene carbon and one for carboxyl carbons around 165 δ.

The ¹H NMR, 2-methylmalonic acid will have three signals, a doublet for the methyl protons, a quartet for the methane proton and a singlet for carboxyl proton around 12 δ. In ¹³C NMR there will be three signals, one for methyl carbon another for the methine carbon and a third one for carboxyl carbons around 165 δ.

Explanation of Solution

Succinic acid has a simpler spectrum since it is symmetrical. In ¹H NMR, the two methylene protons split each other giving a triplet integrating to four protons. The two carboxyl protons yield another singlet signal. Thus there will be two signals a triplet and a singlet.

In ¹³C NMR, the four carbons in succinic acid classify themselves into two groups and give two signals. Thus there will be two signals, one for methylene carbons and other for carboxyl carbons.

2-methylmalonicacid does not have a symmetrical structure. Hence it gives more peaks in both ¹H NMR and ¹³C NMR spectrum than succinic acid.

In ¹H NMR, there will be a doublet for the methyl protons, a quartet integrating to one proton for the methine proton signal and a singlet for the carboxyl protons. Thus there will be three signals.

In ¹³C NMR, there will be separate signals for methyl carbon, methine carbon and carboxyl carbons. Thus there will be three signals.

Conclusion

Succinic acid will have a simpler spectrum. In ¹H NMR, there will be two signals, a triplet for the methylene groups and a singlet for carboxyl protons around 12 δ. In ¹³C NMR there will be two signals, one for the methylene carbons and one for carboxyl carbons around 165 δ.

The ¹H NMR, 2-methylmalonic acid will have three signals, a doublet for the methyl protons, a quartet for the methine proton and a singlet for carboxyl proton around 12 δ. In ¹³C NMR there will be three signals, one for methyl carbon another for the methine carbon and a third one for carboxyl carbons around 165 δ.

Expert Solution

Interpretation Introduction

c) CH₃CH₂CH₂CO₂H and HOCH₂CH₂CH₂CHO

Interpretation:

How NMR (either ¹³C or ¹H) could be used to distinguish between succinic acid and 4-hydroxybutanal is to be stated.

Concept introduction:

In ¹HNMR, the acidic COOH proton normally absorbs as singlet, broadened in few cases, near 12 δ.

In ¹³CNMR, the carboxyl carbon absorb in the range 165-185 δ. The aromatic and α, β-unsaturated acids absorb near 165 δ and saturated aliphatic acids near 185 δ. The nitrile carbons absorb in the range 115-130 δ.

Compounds with symmetrical structures will have a simpler spectrum.

In ¹HNMR the aldehyde protons absorb near 10 δ with a coupling constant, J=3Hz. Hydrogens on the carbon next to aldehyde group absorb near 2.0-2.3 δ.

In ¹³CNMR, saturated aldehydes and ketones usually absorb in the region from 200 to 215 δ while the aromatic and unsaturated carbonyl compounds absorb in the 190 to 200 δ region.

To state:

How NMR (either ¹³C or ¹H) could be used to distinguish between butanoic acid and 4-hydroxybutanal.

Answer to Problem 55AP

In ¹H NMR of butanoic acid there will be a signal around 12 δ while ¹H NMR of 3-hydroxybutanal there will be a signal for the aldehydic proton around 9-10 δ and for the hydroxyl proton in the range 3.0-4.5 δ. There will be no signal around 12 δ. In ¹³C NMR the carboxyl carbon absorbs around 165 δ while the absorption due to aldehydic carbon is observed around 200-215 δ.

Explanation of Solution

The two compounds can be easily distinguished. For butanoic acid there will be signals for the absorption of carboxyl proton in ¹H NMR (12 δ) and carboxyl carbon in ¹³C NMR (165 δ). In 3-hydroxybutanal, there will be signals for the absorption of hydroxyl proton in ¹H NMR (3.0-4.5 δ) and aldehydic proton around 9-10 δ. The aldehydic carbon will resonate in ¹³C NMR in the range 200-215 δ.

Conclusion

In ¹H NMR of butanoic acid there will be a signal around 12 δ while ¹H NMR of 3-hydroxybutanal there will be a signal for the aldehydic proton around 9-10 δ and for the hydroxyl proton in the range 3.0-4.5 δ. There will be no signal around 12 δ. In ¹³C NMR the carboxyl carbon absorbs around 165 δ while the absorptrion due to aldehydic carbon is observed around 200-215 δ.

Expert Solution

Interpretation Introduction

d)

Organic Chemistry, Chapter 20.SE, Problem 55AP , additional homework tip 2

Interpretation:

How NMR (either ¹³C or ¹H) could be used to distinguish between 4-methylpent-3-eneoic acid and cyclopentylcarboxylic acid is to be stated.

Concept introduction:

In ¹HNMR, the acidic COOH proton normally absorbs as singlet, broadened in few cases, near 12 δ.

In ¹³CNMR, the carboxyl carbon absorb in the range 165-185 δ. The aromatic and α, β-unsaturated acids absorb near 165 δ and saturated aliphatic acids near 185 δ. The nitrile carbons absorb in the range 115-130 δ.

Alkenes show a signal in the range 4.5-6.5 δ in ¹HNMR and around 110-150 δ in ¹³CNMR.

Compounds with symmetrical structures will have a simpler spectrum.

To state:

How NMR (either ¹³C or ¹H) could be used to distinguish between 4-methylpent-3-eneoic acid and cyclopentylcarboxylic acid.

Answer to Problem 55AP

In ¹HNMR, 4-methylpent-3-eneoic acid will show the peak due to olefinic proton absorption in the range 4.5-6.5 δ. In ¹³CNMR, the two carbons in the double bond each will give a signal in the range 110-150 δ. There will be no signals in these ranges for cyclopentanecarboxylic acid in both the spectra.

Explanation of Solution

4-Methylpent-3-eneoic acid has a hydrogen atom attached to olefinic double bond. Since it is a vinylic hydrogen its absorption peak is observed in the range 4.5-6.5 δ. The two olefinic carbons are not equivalent. Hence they give two signals in the range 110-150 δ. In the ¹HNMR and ¹³CNMR spectra of cyclopentanecarboxylic acid no signals will be observed in the ranges 4.5-6.5 δ and 110-150 δ respectively.

Conclusion

In ¹HNMR, 4-methylpent-3-eneoic acid will show the peak due to olefinic proton absorption in the range 4.5-6.5 δ. In ¹³CNMR, the two carbons in the double bond each will give a signal in the range 110-150 δ. There will be no signals in these ranges for cyclopentanecarboxylic acid in both the spectra.

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution

Interpretation Introduction

a)

Organic Chemistry, Chapter 20.SE, Problem 55AP , additional homework tip 1

Interpretation:

How NMR (either ¹³C or ¹H) could be used to distinguish between benzene-1,2-dicarboxylic acid and benzene-1,3-dicarboxylic acid is to be stated.

Concept introduction:

In ¹HNMR, the acidic COOH proton normally absorbs as singlet, broadened in few cases, near 12 δ.

In ¹³CNMR, the carboxyl carbon absorb in the range 165-185 δ. The aromatic and α, β-unsaturated acids absorb near 165 δ and saturated aliphatic acids near 185 δ. The nitrile carbons absorb in the range 115-130 δ.

Compounds with symmetrical structures (like p-disubstituted phenyl) will have a simpler spectrum.

To state:

How NMR (either ¹³C or ¹H) could be used to distinguish between benzene-1,3-dicarboxylic acid and benzene-1,4-dicarboxylic acid.

Answer to Problem 55AP

1,4-Benzenedicarboxylic acid will have a simpler spectrum. In ¹H NMR, there will be a doublet of doublet in the aromatic region and a singlet for carboxyl proton around 12 δ. In ¹³C NMR there will be three signals, two in aromatic region and one for carboxyl carbon around 165 δ.

The ¹H NMR of 1,3-Benzenedicarboxylic acid will have a multiplet in the aromatic region due to the ring protons and a singlet for carboxyl proton around 12 δ. In ¹³C NMR there will be five signals, four in aromatic region and one for carboxyl carbon around 165 δ.

Explanation of Solution

1,4-Benzenedicarboxylic acid has a symmetrical structure. Hence it gives minimum peaks in both ¹H NMR and ¹³C NMR spectrum.

In ¹H NMR, the four aromatic hydrogens belong to two groups and thus give a doublet of doublet in the aromatic region. The carboxyl proton gives another signal. Thus there will be three signals.

In ¹³C NMR, the six ring carbons classify themselves into two groups and give two signals while the carboxyl carbons give a signal. Thus there will be three signals.

1,3-Benzenedicarboxylic acid does not have a symmetrical structure. Hence it gives more peaks in both ¹H NMR and ¹³C NMR spectrum than 1,4-Benzenedicarboxylic acid.

In ¹H NMR, the four aromatic hydrogens give complex multiplet signals in the aromatic region. The carboxyl protons yield another signal. Thus there will be two signals.

In ¹³C NMR, the six ring carbons classify themselves into four groups and give four signals while the carboxyl carbons give a signal. Thus there will be five signals.

Conclusion

1,4-Benzenedicarboxylic acid will have a simpler spectrum. In ¹H NMR, there will be a doublet of doublet in the aromatic region and a singlet for carboxyl proton around 12 δ. In ¹³C NMR there will be three signals, two in aromatic region and one for carboxyl carbon around 165 δ.

The ¹H NMR of 1,3-Benzenedicarboxylic acid will have a multiplet in the aromatic region due to the ring protons and a singlet for carboxyl proton around 12 δ. In ¹³C NMR there will be five signals, four in aromatic region and one for carboxyl carbon around 165 δ.

Expert Solution

Interpretation Introduction

b) HO₂CCH₂CH₂CO₂H and CH₃CH(CO₂H)₂

Interpretation:

How NMR (either ¹³C or ¹H) could be used to distinguish between succinic acid and ethane-1,1-dicarboxylic acid is to be stated.

Concept introduction:

In ¹HNMR, the acidic COOH proton normally absorbs as singlet, broadened in few cases, near 12 δ.

In ¹³CNMR, the carboxyl carbon absorb in the range 165-185 δ. The aromatic and α, β-unsaturated acids absorb near 165 δ and saturated aliphatic acids near 185 δ. The nitrile carbons absorb in the range 115-130 δ.

Compounds with symmetrical structures (like p-disubstituted phenyl) will have a simpler spectrum.

To state:

How NMR (either ¹³C or ¹H) could be used to distinguish between succinic acid and ethane-1,1-dicarboxylic acid.

Answer to Problem 55AP

Succinic acid will have a simpler spectrum. In ¹H NMR, there will be two signals, a triplet for the methylene groups and a singlet for carboxyl protons around 12 δ. In ¹³C NMR there will be two signals, one for the methylene carbon and one for carboxyl carbons around 165 δ.

The ¹H NMR, 2-methylmalonic acid will have three signals, a doublet for the methyl protons, a quartet for the methane proton and a singlet for carboxyl proton around 12 δ. In ¹³C NMR there will be three signals, one for methyl carbon another for the methine carbon and a third one for carboxyl carbons around 165 δ.

Explanation of Solution

Succinic acid has a simpler spectrum since it is symmetrical. In ¹H NMR, the two methylene protons split each other giving a triplet integrating to four protons. The two carboxyl protons yield another singlet signal. Thus there will be two signals a triplet and a singlet.

In ¹³C NMR, the four carbons in succinic acid classify themselves into two groups and give two signals. Thus there will be two signals, one for methylene carbons and other for carboxyl carbons.

2-methylmalonicacid does not have a symmetrical structure. Hence it gives more peaks in both ¹H NMR and ¹³C NMR spectrum than succinic acid.

In ¹H NMR, there will be a doublet for the methyl protons, a quartet integrating to one proton for the methine proton signal and a singlet for the carboxyl protons. Thus there will be three signals.

In ¹³C NMR, there will be separate signals for methyl carbon, methine carbon and carboxyl carbons. Thus there will be three signals.

Conclusion

Succinic acid will have a simpler spectrum. In ¹H NMR, there will be two signals, a triplet for the methylene groups and a singlet for carboxyl protons around 12 δ. In ¹³C NMR there will be two signals, one for the methylene carbons and one for carboxyl carbons around 165 δ.

The ¹H NMR, 2-methylmalonic acid will have three signals, a doublet for the methyl protons, a quartet for the methine proton and a singlet for carboxyl proton around 12 δ. In ¹³C NMR there will be three signals, one for methyl carbon another for the methine carbon and a third one for carboxyl carbons around 165 δ.

Expert Solution

Interpretation Introduction

c) CH₃CH₂CH₂CO₂H and HOCH₂CH₂CH₂CHO

Interpretation:

How NMR (either ¹³C or ¹H) could be used to distinguish between succinic acid and 4-hydroxybutanal is to be stated.

Concept introduction:

In ¹HNMR, the acidic COOH proton normally absorbs as singlet, broadened in few cases, near 12 δ.

In ¹³CNMR, the carboxyl carbon absorb in the range 165-185 δ. The aromatic and α, β-unsaturated acids absorb near 165 δ and saturated aliphatic acids near 185 δ. The nitrile carbons absorb in the range 115-130 δ.

Compounds with symmetrical structures will have a simpler spectrum.

In ¹HNMR the aldehyde protons absorb near 10 δ with a coupling constant, J=3Hz. Hydrogens on the carbon next to aldehyde group absorb near 2.0-2.3 δ.

In ¹³CNMR, saturated aldehydes and ketones usually absorb in the region from 200 to 215 δ while the aromatic and unsaturated carbonyl compounds absorb in the 190 to 200 δ region.

To state:

How NMR (either ¹³C or ¹H) could be used to distinguish between butanoic acid and 4-hydroxybutanal.

Answer to Problem 55AP

In ¹H NMR of butanoic acid there will be a signal around 12 δ while ¹H NMR of 3-hydroxybutanal there will be a signal for the aldehydic proton around 9-10 δ and for the hydroxyl proton in the range 3.0-4.5 δ. There will be no signal around 12 δ. In ¹³C NMR the carboxyl carbon absorbs around 165 δ while the absorption due to aldehydic carbon is observed around 200-215 δ.

Explanation of Solution

The two compounds can be easily distinguished. For butanoic acid there will be signals for the absorption of carboxyl proton in ¹H NMR (12 δ) and carboxyl carbon in ¹³C NMR (165 δ). In 3-hydroxybutanal, there will be signals for the absorption of hydroxyl proton in ¹H NMR (3.0-4.5 δ) and aldehydic proton around 9-10 δ. The aldehydic carbon will resonate in ¹³C NMR in the range 200-215 δ.

Conclusion

In ¹H NMR of butanoic acid there will be a signal around 12 δ while ¹H NMR of 3-hydroxybutanal there will be a signal for the aldehydic proton around 9-10 δ and for the hydroxyl proton in the range 3.0-4.5 δ. There will be no signal around 12 δ. In ¹³C NMR the carboxyl carbon absorbs around 165 δ while the absorptrion due to aldehydic carbon is observed around 200-215 δ.

Expert Solution

Interpretation Introduction

d)

Organic Chemistry, Chapter 20.SE, Problem 55AP , additional homework tip 2

Interpretation:

How NMR (either ¹³C or ¹H) could be used to distinguish between 4-methylpent-3-eneoic acid and cyclopentylcarboxylic acid is to be stated.

Concept introduction:

In ¹HNMR, the acidic COOH proton normally absorbs as singlet, broadened in few cases, near 12 δ.

In ¹³CNMR, the carboxyl carbon absorb in the range 165-185 δ. The aromatic and α, β-unsaturated acids absorb near 165 δ and saturated aliphatic acids near 185 δ. The nitrile carbons absorb in the range 115-130 δ.

Alkenes show a signal in the range 4.5-6.5 δ in ¹HNMR and around 110-150 δ in ¹³CNMR.

Compounds with symmetrical structures will have a simpler spectrum.

To state:

How NMR (either ¹³C or ¹H) could be used to distinguish between 4-methylpent-3-eneoic acid and cyclopentylcarboxylic acid.

Answer to Problem 55AP

In ¹HNMR, 4-methylpent-3-eneoic acid will show the peak due to olefinic proton absorption in the range 4.5-6.5 δ. In ¹³CNMR, the two carbons in the double bond each will give a signal in the range 110-150 δ. There will be no signals in these ranges for cyclopentanecarboxylic acid in both the spectra.

Explanation of Solution

4-Methylpent-3-eneoic acid has a hydrogen atom attached to olefinic double bond. Since it is a vinylic hydrogen its absorption peak is observed in the range 4.5-6.5 δ. The two olefinic carbons are not equivalent. Hence they give two signals in the range 110-150 δ. In the ¹HNMR and ¹³CNMR spectra of cyclopentanecarboxylic acid no signals will be observed in the ranges 4.5-6.5 δ and 110-150 δ respectively.

Conclusion

In ¹HNMR, 4-methylpent-3-eneoic acid will show the peak due to olefinic proton absorption in the range 4.5-6.5 δ. In ¹³CNMR, the two carbons in the double bond each will give a signal in the range 110-150 δ. There will be no signals in these ranges for cyclopentanecarboxylic acid in both the spectra.

Answer 8

Expert Solution

Interpretation Introduction

a)

Organic Chemistry, Chapter 20.SE, Problem 55AP , additional homework tip 1

Interpretation:

How NMR (either ¹³C or ¹H) could be used to distinguish between benzene-1,2-dicarboxylic acid and benzene-1,3-dicarboxylic acid is to be stated.

Concept introduction:

In ¹HNMR, the acidic COOH proton normally absorbs as singlet, broadened in few cases, near 12 δ.

In ¹³CNMR, the carboxyl carbon absorb in the range 165-185 δ. The aromatic and α, β-unsaturated acids absorb near 165 δ and saturated aliphatic acids near 185 δ. The nitrile carbons absorb in the range 115-130 δ.

Compounds with symmetrical structures (like p-disubstituted phenyl) will have a simpler spectrum.

To state:

How NMR (either ¹³C or ¹H) could be used to distinguish between benzene-1,3-dicarboxylic acid and benzene-1,4-dicarboxylic acid.

Answer to Problem 55AP

1,4-Benzenedicarboxylic acid will have a simpler spectrum. In ¹H NMR, there will be a doublet of doublet in the aromatic region and a singlet for carboxyl proton around 12 δ. In ¹³C NMR there will be three signals, two in aromatic region and one for carboxyl carbon around 165 δ.

The ¹H NMR of 1,3-Benzenedicarboxylic acid will have a multiplet in the aromatic region due to the ring protons and a singlet for carboxyl proton around 12 δ. In ¹³C NMR there will be five signals, four in aromatic region and one for carboxyl carbon around 165 δ.

Explanation of Solution

1,4-Benzenedicarboxylic acid has a symmetrical structure. Hence it gives minimum peaks in both ¹H NMR and ¹³C NMR spectrum.

In ¹H NMR, the four aromatic hydrogens belong to two groups and thus give a doublet of doublet in the aromatic region. The carboxyl proton gives another signal. Thus there will be three signals.

In ¹³C NMR, the six ring carbons classify themselves into two groups and give two signals while the carboxyl carbons give a signal. Thus there will be three signals.

1,3-Benzenedicarboxylic acid does not have a symmetrical structure. Hence it gives more peaks in both ¹H NMR and ¹³C NMR spectrum than 1,4-Benzenedicarboxylic acid.

In ¹H NMR, the four aromatic hydrogens give complex multiplet signals in the aromatic region. The carboxyl protons yield another signal. Thus there will be two signals.

In ¹³C NMR, the six ring carbons classify themselves into four groups and give four signals while the carboxyl carbons give a signal. Thus there will be five signals.

Conclusion

1,4-Benzenedicarboxylic acid will have a simpler spectrum. In ¹H NMR, there will be a doublet of doublet in the aromatic region and a singlet for carboxyl proton around 12 δ. In ¹³C NMR there will be three signals, two in aromatic region and one for carboxyl carbon around 165 δ.

The ¹H NMR of 1,3-Benzenedicarboxylic acid will have a multiplet in the aromatic region due to the ring protons and a singlet for carboxyl proton around 12 δ. In ¹³C NMR there will be five signals, four in aromatic region and one for carboxyl carbon around 165 δ.

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

Interpretation Introduction

a)

Organic Chemistry, Chapter 20.SE, Problem 55AP , additional homework tip 1

Interpretation:

How NMR (either ¹³C or ¹H) could be used to distinguish between benzene-1,2-dicarboxylic acid and benzene-1,3-dicarboxylic acid is to be stated.

Concept introduction:

In ¹HNMR, the acidic COOH proton normally absorbs as singlet, broadened in few cases, near 12 δ.

In ¹³CNMR, the carboxyl carbon absorb in the range 165-185 δ. The aromatic and α, β-unsaturated acids absorb near 165 δ and saturated aliphatic acids near 185 δ. The nitrile carbons absorb in the range 115-130 δ.

Compounds with symmetrical structures (like p-disubstituted phenyl) will have a simpler spectrum.

To state:

How NMR (either ¹³C or ¹H) could be used to distinguish between benzene-1,3-dicarboxylic acid and benzene-1,4-dicarboxylic acid.

Answer 13

Answer to Problem 55AP

1,4-Benzenedicarboxylic acid will have a simpler spectrum. In ¹H NMR, there will be a doublet of doublet in the aromatic region and a singlet for carboxyl proton around 12 δ. In ¹³C NMR there will be three signals, two in aromatic region and one for carboxyl carbon around 165 δ.

The ¹H NMR of 1,3-Benzenedicarboxylic acid will have a multiplet in the aromatic region due to the ring protons and a singlet for carboxyl proton around 12 δ. In ¹³C NMR there will be five signals, four in aromatic region and one for carboxyl carbon around 165 δ.

Answer 14

Explanation of Solution

1,4-Benzenedicarboxylic acid has a symmetrical structure. Hence it gives minimum peaks in both ¹H NMR and ¹³C NMR spectrum.

In ¹H NMR, the four aromatic hydrogens belong to two groups and thus give a doublet of doublet in the aromatic region. The carboxyl proton gives another signal. Thus there will be three signals.

In ¹³C NMR, the six ring carbons classify themselves into two groups and give two signals while the carboxyl carbons give a signal. Thus there will be three signals.

1,3-Benzenedicarboxylic acid does not have a symmetrical structure. Hence it gives more peaks in both ¹H NMR and ¹³C NMR spectrum than 1,4-Benzenedicarboxylic acid.

In ¹H NMR, the four aromatic hydrogens give complex multiplet signals in the aromatic region. The carboxyl protons yield another signal. Thus there will be two signals.

In ¹³C NMR, the six ring carbons classify themselves into four groups and give four signals while the carboxyl carbons give a signal. Thus there will be five signals.

Answer 15

Conclusion

1,4-Benzenedicarboxylic acid will have a simpler spectrum. In ¹H NMR, there will be a doublet of doublet in the aromatic region and a singlet for carboxyl proton around 12 δ. In ¹³C NMR there will be three signals, two in aromatic region and one for carboxyl carbon around 165 δ.

The ¹H NMR of 1,3-Benzenedicarboxylic acid will have a multiplet in the aromatic region due to the ring protons and a singlet for carboxyl proton around 12 δ. In ¹³C NMR there will be five signals, four in aromatic region and one for carboxyl carbon around 165 δ.

Answer 16

Expert Solution

Interpretation Introduction

b) HO₂CCH₂CH₂CO₂H and CH₃CH(CO₂H)₂

Interpretation:

How NMR (either ¹³C or ¹H) could be used to distinguish between succinic acid and ethane-1,1-dicarboxylic acid is to be stated.

Concept introduction:

In ¹HNMR, the acidic COOH proton normally absorbs as singlet, broadened in few cases, near 12 δ.

In ¹³CNMR, the carboxyl carbon absorb in the range 165-185 δ. The aromatic and α, β-unsaturated acids absorb near 165 δ and saturated aliphatic acids near 185 δ. The nitrile carbons absorb in the range 115-130 δ.

Compounds with symmetrical structures (like p-disubstituted phenyl) will have a simpler spectrum.

To state:

How NMR (either ¹³C or ¹H) could be used to distinguish between succinic acid and ethane-1,1-dicarboxylic acid.

Answer to Problem 55AP

Succinic acid will have a simpler spectrum. In ¹H NMR, there will be two signals, a triplet for the methylene groups and a singlet for carboxyl protons around 12 δ. In ¹³C NMR there will be two signals, one for the methylene carbon and one for carboxyl carbons around 165 δ.

The ¹H NMR, 2-methylmalonic acid will have three signals, a doublet for the methyl protons, a quartet for the methane proton and a singlet for carboxyl proton around 12 δ. In ¹³C NMR there will be three signals, one for methyl carbon another for the methine carbon and a third one for carboxyl carbons around 165 δ.

Explanation of Solution

Succinic acid has a simpler spectrum since it is symmetrical. In ¹H NMR, the two methylene protons split each other giving a triplet integrating to four protons. The two carboxyl protons yield another singlet signal. Thus there will be two signals a triplet and a singlet.

In ¹³C NMR, the four carbons in succinic acid classify themselves into two groups and give two signals. Thus there will be two signals, one for methylene carbons and other for carboxyl carbons.

2-methylmalonicacid does not have a symmetrical structure. Hence it gives more peaks in both ¹H NMR and ¹³C NMR spectrum than succinic acid.

In ¹H NMR, there will be a doublet for the methyl protons, a quartet integrating to one proton for the methine proton signal and a singlet for the carboxyl protons. Thus there will be three signals.

In ¹³C NMR, there will be separate signals for methyl carbon, methine carbon and carboxyl carbons. Thus there will be three signals.

Conclusion

Succinic acid will have a simpler spectrum. In ¹H NMR, there will be two signals, a triplet for the methylene groups and a singlet for carboxyl protons around 12 δ. In ¹³C NMR there will be two signals, one for the methylene carbons and one for carboxyl carbons around 165 δ.

The ¹H NMR, 2-methylmalonic acid will have three signals, a doublet for the methyl protons, a quartet for the methine proton and a singlet for carboxyl proton around 12 δ. In ¹³C NMR there will be three signals, one for methyl carbon another for the methine carbon and a third one for carboxyl carbons around 165 δ.

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

Expert Solution

Answer 20

Interpretation Introduction

b) HO₂CCH₂CH₂CO₂H and CH₃CH(CO₂H)₂

Interpretation:

How NMR (either ¹³C or ¹H) could be used to distinguish between succinic acid and ethane-1,1-dicarboxylic acid is to be stated.

Concept introduction:

In ¹HNMR, the acidic COOH proton normally absorbs as singlet, broadened in few cases, near 12 δ.

In ¹³CNMR, the carboxyl carbon absorb in the range 165-185 δ. The aromatic and α, β-unsaturated acids absorb near 165 δ and saturated aliphatic acids near 185 δ. The nitrile carbons absorb in the range 115-130 δ.

Compounds with symmetrical structures (like p-disubstituted phenyl) will have a simpler spectrum.

To state:

How NMR (either ¹³C or ¹H) could be used to distinguish between succinic acid and ethane-1,1-dicarboxylic acid.

Answer 21

Answer to Problem 55AP

Succinic acid will have a simpler spectrum. In ¹H NMR, there will be two signals, a triplet for the methylene groups and a singlet for carboxyl protons around 12 δ. In ¹³C NMR there will be two signals, one for the methylene carbon and one for carboxyl carbons around 165 δ.

The ¹H NMR, 2-methylmalonic acid will have three signals, a doublet for the methyl protons, a quartet for the methane proton and a singlet for carboxyl proton around 12 δ. In ¹³C NMR there will be three signals, one for methyl carbon another for the methine carbon and a third one for carboxyl carbons around 165 δ.

Answer 22

Explanation of Solution

Succinic acid has a simpler spectrum since it is symmetrical. In ¹H NMR, the two methylene protons split each other giving a triplet integrating to four protons. The two carboxyl protons yield another singlet signal. Thus there will be two signals a triplet and a singlet.

In ¹³C NMR, the four carbons in succinic acid classify themselves into two groups and give two signals. Thus there will be two signals, one for methylene carbons and other for carboxyl carbons.

2-methylmalonicacid does not have a symmetrical structure. Hence it gives more peaks in both ¹H NMR and ¹³C NMR spectrum than succinic acid.

In ¹H NMR, there will be a doublet for the methyl protons, a quartet integrating to one proton for the methine proton signal and a singlet for the carboxyl protons. Thus there will be three signals.

In ¹³C NMR, there will be separate signals for methyl carbon, methine carbon and carboxyl carbons. Thus there will be three signals.

Answer 23

Conclusion

Succinic acid will have a simpler spectrum. In ¹H NMR, there will be two signals, a triplet for the methylene groups and a singlet for carboxyl protons around 12 δ. In ¹³C NMR there will be two signals, one for the methylene carbons and one for carboxyl carbons around 165 δ.

The ¹H NMR, 2-methylmalonic acid will have three signals, a doublet for the methyl protons, a quartet for the methine proton and a singlet for carboxyl proton around 12 δ. In ¹³C NMR there will be three signals, one for methyl carbon another for the methine carbon and a third one for carboxyl carbons around 165 δ.

Answer 24

Expert Solution

Interpretation Introduction

c) CH₃CH₂CH₂CO₂H and HOCH₂CH₂CH₂CHO

Interpretation:

How NMR (either ¹³C or ¹H) could be used to distinguish between succinic acid and 4-hydroxybutanal is to be stated.

Concept introduction:

In ¹HNMR, the acidic COOH proton normally absorbs as singlet, broadened in few cases, near 12 δ.

In ¹³CNMR, the carboxyl carbon absorb in the range 165-185 δ. The aromatic and α, β-unsaturated acids absorb near 165 δ and saturated aliphatic acids near 185 δ. The nitrile carbons absorb in the range 115-130 δ.

Compounds with symmetrical structures will have a simpler spectrum.

In ¹HNMR the aldehyde protons absorb near 10 δ with a coupling constant, J=3Hz. Hydrogens on the carbon next to aldehyde group absorb near 2.0-2.3 δ.

In ¹³CNMR, saturated aldehydes and ketones usually absorb in the region from 200 to 215 δ while the aromatic and unsaturated carbonyl compounds absorb in the 190 to 200 δ region.

To state:

How NMR (either ¹³C or ¹H) could be used to distinguish between butanoic acid and 4-hydroxybutanal.

Answer to Problem 55AP

In ¹H NMR of butanoic acid there will be a signal around 12 δ while ¹H NMR of 3-hydroxybutanal there will be a signal for the aldehydic proton around 9-10 δ and for the hydroxyl proton in the range 3.0-4.5 δ. There will be no signal around 12 δ. In ¹³C NMR the carboxyl carbon absorbs around 165 δ while the absorption due to aldehydic carbon is observed around 200-215 δ.

Explanation of Solution

The two compounds can be easily distinguished. For butanoic acid there will be signals for the absorption of carboxyl proton in ¹H NMR (12 δ) and carboxyl carbon in ¹³C NMR (165 δ). In 3-hydroxybutanal, there will be signals for the absorption of hydroxyl proton in ¹H NMR (3.0-4.5 δ) and aldehydic proton around 9-10 δ. The aldehydic carbon will resonate in ¹³C NMR in the range 200-215 δ.

Conclusion

In ¹H NMR of butanoic acid there will be a signal around 12 δ while ¹H NMR of 3-hydroxybutanal there will be a signal for the aldehydic proton around 9-10 δ and for the hydroxyl proton in the range 3.0-4.5 δ. There will be no signal around 12 δ. In ¹³C NMR the carboxyl carbon absorbs around 165 δ while the absorptrion due to aldehydic carbon is observed around 200-215 δ.

Answer 25

Expert Solution

Answer 26

Expert Solution

Answer 27

Expert Solution

Answer 28

Interpretation Introduction

c) CH₃CH₂CH₂CO₂H and HOCH₂CH₂CH₂CHO

Interpretation:

How NMR (either ¹³C or ¹H) could be used to distinguish between succinic acid and 4-hydroxybutanal is to be stated.

Concept introduction:

In ¹HNMR, the acidic COOH proton normally absorbs as singlet, broadened in few cases, near 12 δ.

In ¹³CNMR, the carboxyl carbon absorb in the range 165-185 δ. The aromatic and α, β-unsaturated acids absorb near 165 δ and saturated aliphatic acids near 185 δ. The nitrile carbons absorb in the range 115-130 δ.

Compounds with symmetrical structures will have a simpler spectrum.

In ¹HNMR the aldehyde protons absorb near 10 δ with a coupling constant, J=3Hz. Hydrogens on the carbon next to aldehyde group absorb near 2.0-2.3 δ.

In ¹³CNMR, saturated aldehydes and ketones usually absorb in the region from 200 to 215 δ while the aromatic and unsaturated carbonyl compounds absorb in the 190 to 200 δ region.

To state:

How NMR (either ¹³C or ¹H) could be used to distinguish between butanoic acid and 4-hydroxybutanal.

Answer 29

Answer to Problem 55AP

In ¹H NMR of butanoic acid there will be a signal around 12 δ while ¹H NMR of 3-hydroxybutanal there will be a signal for the aldehydic proton around 9-10 δ and for the hydroxyl proton in the range 3.0-4.5 δ. There will be no signal around 12 δ. In ¹³C NMR the carboxyl carbon absorbs around 165 δ while the absorption due to aldehydic carbon is observed around 200-215 δ.

Answer 30

Explanation of Solution

The two compounds can be easily distinguished. For butanoic acid there will be signals for the absorption of carboxyl proton in ¹H NMR (12 δ) and carboxyl carbon in ¹³C NMR (165 δ). In 3-hydroxybutanal, there will be signals for the absorption of hydroxyl proton in ¹H NMR (3.0-4.5 δ) and aldehydic proton around 9-10 δ. The aldehydic carbon will resonate in ¹³C NMR in the range 200-215 δ.

Answer 31

Conclusion

In ¹H NMR of butanoic acid there will be a signal around 12 δ while ¹H NMR of 3-hydroxybutanal there will be a signal for the aldehydic proton around 9-10 δ and for the hydroxyl proton in the range 3.0-4.5 δ. There will be no signal around 12 δ. In ¹³C NMR the carboxyl carbon absorbs around 165 δ while the absorptrion due to aldehydic carbon is observed around 200-215 δ.

Answer 32

Expert Solution

Interpretation Introduction

d)

Organic Chemistry, Chapter 20.SE, Problem 55AP , additional homework tip 2

Interpretation:

How NMR (either ¹³C or ¹H) could be used to distinguish between 4-methylpent-3-eneoic acid and cyclopentylcarboxylic acid is to be stated.

Concept introduction:

In ¹HNMR, the acidic COOH proton normally absorbs as singlet, broadened in few cases, near 12 δ.

In ¹³CNMR, the carboxyl carbon absorb in the range 165-185 δ. The aromatic and α, β-unsaturated acids absorb near 165 δ and saturated aliphatic acids near 185 δ. The nitrile carbons absorb in the range 115-130 δ.

Alkenes show a signal in the range 4.5-6.5 δ in ¹HNMR and around 110-150 δ in ¹³CNMR.

Compounds with symmetrical structures will have a simpler spectrum.

To state:

How NMR (either ¹³C or ¹H) could be used to distinguish between 4-methylpent-3-eneoic acid and cyclopentylcarboxylic acid.

Answer to Problem 55AP

In ¹HNMR, 4-methylpent-3-eneoic acid will show the peak due to olefinic proton absorption in the range 4.5-6.5 δ. In ¹³CNMR, the two carbons in the double bond each will give a signal in the range 110-150 δ. There will be no signals in these ranges for cyclopentanecarboxylic acid in both the spectra.

Explanation of Solution

4-Methylpent-3-eneoic acid has a hydrogen atom attached to olefinic double bond. Since it is a vinylic hydrogen its absorption peak is observed in the range 4.5-6.5 δ. The two olefinic carbons are not equivalent. Hence they give two signals in the range 110-150 δ. In the ¹HNMR and ¹³CNMR spectra of cyclopentanecarboxylic acid no signals will be observed in the ranges 4.5-6.5 δ and 110-150 δ respectively.

Conclusion

In ¹HNMR, 4-methylpent-3-eneoic acid will show the peak due to olefinic proton absorption in the range 4.5-6.5 δ. In ¹³CNMR, the two carbons in the double bond each will give a signal in the range 110-150 δ. There will be no signals in these ranges for cyclopentanecarboxylic acid in both the spectra.

Answer 33

Expert Solution

Answer 34

Expert Solution

Answer 35

Expert Solution

Answer 36

Interpretation Introduction

d)

Organic Chemistry, Chapter 20.SE, Problem 55AP , additional homework tip 2

Interpretation:

How NMR (either ¹³C or ¹H) could be used to distinguish between 4-methylpent-3-eneoic acid and cyclopentylcarboxylic acid is to be stated.

Concept introduction:

In ¹HNMR, the acidic COOH proton normally absorbs as singlet, broadened in few cases, near 12 δ.

In ¹³CNMR, the carboxyl carbon absorb in the range 165-185 δ. The aromatic and α, β-unsaturated acids absorb near 165 δ and saturated aliphatic acids near 185 δ. The nitrile carbons absorb in the range 115-130 δ.

Alkenes show a signal in the range 4.5-6.5 δ in ¹HNMR and around 110-150 δ in ¹³CNMR.

Compounds with symmetrical structures will have a simpler spectrum.

To state:

How NMR (either ¹³C or ¹H) could be used to distinguish between 4-methylpent-3-eneoic acid and cyclopentylcarboxylic acid.

Answer 37

Answer to Problem 55AP

In ¹HNMR, 4-methylpent-3-eneoic acid will show the peak due to olefinic proton absorption in the range 4.5-6.5 δ. In ¹³CNMR, the two carbons in the double bond each will give a signal in the range 110-150 δ. There will be no signals in these ranges for cyclopentanecarboxylic acid in both the spectra.

Answer 38

Explanation of Solution

4-Methylpent-3-eneoic acid has a hydrogen atom attached to olefinic double bond. Since it is a vinylic hydrogen its absorption peak is observed in the range 4.5-6.5 δ. The two olefinic carbons are not equivalent. Hence they give two signals in the range 110-150 δ. In the ¹HNMR and ¹³CNMR spectra of cyclopentanecarboxylic acid no signals will be observed in the ranges 4.5-6.5 δ and 110-150 δ respectively.

Answer 39

Conclusion

In ¹HNMR, 4-methylpent-3-eneoic acid will show the peak due to olefinic proton absorption in the range 4.5-6.5 δ. In ¹³CNMR, the two carbons in the double bond each will give a signal in the range 110-150 δ. There will be no signals in these ranges for cyclopentanecarboxylic acid in both the spectra.