Propose structures for compounds that fit the following 1 H NMR data: (a) C 4 H 6 Cl 2 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) C 10 H 14 1.30 δ (9 H, singlet) 7.30 δ (5 H, singlet) (c) C 4 H 7 BrO 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) C 9 H 11 Br 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)

Question

Want to see more full solutions like this?

Answer 1

Textbook Question

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

Interpretation Introduction

a)

Organic Chemistry - With Access (Custom), 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 - With Access (Custom), 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 - With Access (Custom), Chapter 13.SE, Problem 55GP , additional homework tip 3

Explanation of Solution

Calculate HDI value:

Organic Chemistry - With Access (Custom), 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 - With Access (Custom), 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 - With Access (Custom), 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

Interpretation Introduction

b)

Organic Chemistry - With Access (Custom), 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 - With Access (Custom), 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 - With Access (Custom), Chapter 13.SE, Problem 55GP , additional homework tip 9

Explanation of Solution

Calculate HDI value:

Organic Chemistry - With Access (Custom), 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 - With Access (Custom), 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 - With Access (Custom), 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

Interpretation Introduction

c)

Organic Chemistry - With Access (Custom), 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 - With Access (Custom), 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 - With Access (Custom), Chapter 13.SE, Problem 55GP , additional homework tip 15

Explanation of Solution

Calculate HDI value:

Organic Chemistry - With Access (Custom), 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 - With Access (Custom), 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 - With Access (Custom), 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

Interpretation Introduction

d)

Organic Chemistry - With Access (Custom), 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 - With Access (Custom), 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 - With Access (Custom), Chapter 13.SE, Problem 55GP , additional homework tip 21

Explanation of Solution

Calculate HDI value:

Organic Chemistry - With Access (Custom), 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 - With Access (Custom), 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 - With Access (Custom), 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

1) a) Give the dominant Intermolecular Force (IMF) in a sample of each of the following compounds. Please show your work. (8) SF2, CH,OH, C₂H₂ b) Based on your answers given above, list the compounds in order of their Boiling Point from low to high. (8)

19.78 Write the products of the following sequences of reactions. Refer to your reaction road- maps to see how the combined reactions allow you to "navigate" between the different functional groups. Note that you will need your old Chapters 6-11 and Chapters 15-18 roadmaps along with your new Chapter 19 roadmap for these. (a) 1. BHS 2. H₂O₂ 3. H₂CrO4 4. SOCI₂ (b) 1. Cl₂/hv 2. KOLBU 3. H₂O, catalytic H₂SO4 4. H₂CrO4 Reaction Roadmap An alkene 5. EtOH 6.0.5 Equiv. NaOEt/EtOH 7. Mild H₂O An alkane 1.0 2. (CH3)₂S 3. H₂CrO (d) (c) 4. Excess EtOH, catalytic H₂SO OH 4. Mild H₂O* 5.0.5 Equiv. NaOEt/EtOH An alkene 6. Mild H₂O* A carboxylic acid 7. Mild H₂O* 1. SOC₁₂ 2. EtOH 3.0.5 Equiv. NaOEt/E:OH 5.1.0 Equiv. NaOEt 6. NH₂ (e) 1. 0.5 Equiv. NaOEt/EtOH 2. Mild H₂O* Br (f) i H An aldehyde 1. Catalytic NaOE/EtOH 2. H₂O*, heat 3. (CH,CH₂)₂Culi 4. Mild H₂O* 5.1.0 Equiv. LDA Br An ester 4. NaOH, H₂O 5. Mild H₂O* 6. Heat 7. MgBr 8. Mild H₂O* 7. Mild H₂O+

Li+ is a hard acid. With this in mind, which if the following compounds should be most soluble in water? Group of answer choices LiBr LiI LiF LiCl

Answer 2

Textbook Question

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

Interpretation Introduction

a)

Organic Chemistry - With Access (Custom), 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 - With Access (Custom), 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 - With Access (Custom), Chapter 13.SE, Problem 55GP , additional homework tip 3

Explanation of Solution

Calculate HDI value:

Organic Chemistry - With Access (Custom), 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 - With Access (Custom), 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 - With Access (Custom), 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

Interpretation Introduction

b)

Organic Chemistry - With Access (Custom), 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 - With Access (Custom), 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 - With Access (Custom), Chapter 13.SE, Problem 55GP , additional homework tip 9

Explanation of Solution

Calculate HDI value:

Organic Chemistry - With Access (Custom), 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 - With Access (Custom), 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 - With Access (Custom), 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

Interpretation Introduction

c)

Organic Chemistry - With Access (Custom), 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 - With Access (Custom), 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 - With Access (Custom), Chapter 13.SE, Problem 55GP , additional homework tip 15

Explanation of Solution

Calculate HDI value:

Organic Chemistry - With Access (Custom), 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 - With Access (Custom), 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 - With Access (Custom), 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

Interpretation Introduction

d)

Organic Chemistry - With Access (Custom), 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 - With Access (Custom), 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 - With Access (Custom), Chapter 13.SE, Problem 55GP , additional homework tip 21

Explanation of Solution

Calculate HDI value:

Organic Chemistry - With Access (Custom), 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 - With Access (Custom), 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 - With Access (Custom), 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!

Answer 3

Textbook Question

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

Interpretation Introduction

a)

Organic Chemistry - With Access (Custom), 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 - With Access (Custom), 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 - With Access (Custom), Chapter 13.SE, Problem 55GP , additional homework tip 3

Explanation of Solution

Calculate HDI value:

Organic Chemistry - With Access (Custom), 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 - With Access (Custom), 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 - With Access (Custom), 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

Interpretation Introduction

b)

Organic Chemistry - With Access (Custom), 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 - With Access (Custom), 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 - With Access (Custom), Chapter 13.SE, Problem 55GP , additional homework tip 9

Explanation of Solution

Calculate HDI value:

Organic Chemistry - With Access (Custom), 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 - With Access (Custom), 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 - With Access (Custom), 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

Interpretation Introduction

c)

Organic Chemistry - With Access (Custom), 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 - With Access (Custom), 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 - With Access (Custom), Chapter 13.SE, Problem 55GP , additional homework tip 15

Explanation of Solution

Calculate HDI value:

Organic Chemistry - With Access (Custom), 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 - With Access (Custom), 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 - With Access (Custom), 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

Interpretation Introduction

d)

Organic Chemistry - With Access (Custom), 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 - With Access (Custom), 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 - With Access (Custom), Chapter 13.SE, Problem 55GP , additional homework tip 21

Explanation of Solution

Calculate HDI value:

Organic Chemistry - With Access (Custom), 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 - With Access (Custom), 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 - With Access (Custom), 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!

Answer 4

Textbook Question

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

Interpretation Introduction

a)

Organic Chemistry - With Access (Custom), 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 - With Access (Custom), 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 - With Access (Custom), Chapter 13.SE, Problem 55GP , additional homework tip 3

Explanation of Solution

Calculate HDI value:

Organic Chemistry - With Access (Custom), 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 - With Access (Custom), 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 - With Access (Custom), 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

Interpretation Introduction

b)

Organic Chemistry - With Access (Custom), 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 - With Access (Custom), 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 - With Access (Custom), Chapter 13.SE, Problem 55GP , additional homework tip 9

Explanation of Solution

Calculate HDI value:

Organic Chemistry - With Access (Custom), 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 - With Access (Custom), 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 - With Access (Custom), 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

Interpretation Introduction

c)

Organic Chemistry - With Access (Custom), 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 - With Access (Custom), 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 - With Access (Custom), Chapter 13.SE, Problem 55GP , additional homework tip 15

Explanation of Solution

Calculate HDI value:

Organic Chemistry - With Access (Custom), 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 - With Access (Custom), 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 - With Access (Custom), 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

Interpretation Introduction

d)

Organic Chemistry - With Access (Custom), 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 - With Access (Custom), 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 - With Access (Custom), Chapter 13.SE, Problem 55GP , additional homework tip 21

Explanation of Solution

Calculate HDI value:

Organic Chemistry - With Access (Custom), 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 - With Access (Custom), 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 - With Access (Custom), 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!

Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution

Interpretation Introduction

a)

Organic Chemistry - With Access (Custom), 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 - With Access (Custom), 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 - With Access (Custom), Chapter 13.SE, Problem 55GP , additional homework tip 3

Explanation of Solution

Calculate HDI value:

Organic Chemistry - With Access (Custom), 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 - With Access (Custom), 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 - With Access (Custom), 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

Interpretation Introduction

b)

Organic Chemistry - With Access (Custom), 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 - With Access (Custom), 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 - With Access (Custom), Chapter 13.SE, Problem 55GP , additional homework tip 9

Explanation of Solution

Calculate HDI value:

Organic Chemistry - With Access (Custom), 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 - With Access (Custom), 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 - With Access (Custom), 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

Interpretation Introduction

c)

Organic Chemistry - With Access (Custom), 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 - With Access (Custom), 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 - With Access (Custom), Chapter 13.SE, Problem 55GP , additional homework tip 15

Explanation of Solution

Calculate HDI value:

Organic Chemistry - With Access (Custom), 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 - With Access (Custom), 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 - With Access (Custom), 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

Interpretation Introduction

d)

Organic Chemistry - With Access (Custom), 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 - With Access (Custom), 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 - With Access (Custom), Chapter 13.SE, Problem 55GP , additional homework tip 21

Explanation of Solution

Calculate HDI value:

Organic Chemistry - With Access (Custom), 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 - With Access (Custom), 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 - With Access (Custom), 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.

Answer 8

Expert Solution

Interpretation Introduction

a)

Organic Chemistry - With Access (Custom), 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 - With Access (Custom), 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 - With Access (Custom), Chapter 13.SE, Problem 55GP , additional homework tip 3

Explanation of Solution

Calculate HDI value:

Organic Chemistry - With Access (Custom), 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 - With Access (Custom), 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 - With Access (Custom), 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.

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

Interpretation Introduction

a)

Organic Chemistry - With Access (Custom), 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 - With Access (Custom), 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 13

Answer to Problem 55GP

Organic Chemistry - With Access (Custom), Chapter 13.SE, Problem 55GP , additional homework tip 3

Answer 14

Explanation of Solution

Calculate HDI value:

Organic Chemistry - With Access (Custom), 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 - With Access (Custom), 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 - With Access (Custom), 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.

Answer 15

Conclusion

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

Answer 16

Expert Solution

Interpretation Introduction

b)

Organic Chemistry - With Access (Custom), 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 - With Access (Custom), 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 - With Access (Custom), Chapter 13.SE, Problem 55GP , additional homework tip 9

Explanation of Solution

Calculate HDI value:

Organic Chemistry - With Access (Custom), 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 - With Access (Custom), 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 - With Access (Custom), 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.

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

Expert Solution

Answer 20

Interpretation Introduction

b)

Organic Chemistry - With Access (Custom), 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 - With Access (Custom), 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 21

Answer to Problem 55GP

Organic Chemistry - With Access (Custom), Chapter 13.SE, Problem 55GP , additional homework tip 9

Answer 22

Explanation of Solution

Calculate HDI value:

Organic Chemistry - With Access (Custom), 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 - With Access (Custom), 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 - With Access (Custom), 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.

Answer 23

Conclusion

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

Answer 24

Expert Solution

Interpretation Introduction

c)

Organic Chemistry - With Access (Custom), 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 - With Access (Custom), 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 - With Access (Custom), Chapter 13.SE, Problem 55GP , additional homework tip 15

Explanation of Solution

Calculate HDI value:

Organic Chemistry - With Access (Custom), 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 - With Access (Custom), 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 - With Access (Custom), 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.

Answer 25

Expert Solution

Answer 26

Expert Solution

Answer 27

Expert Solution

Answer 28

Interpretation Introduction

c)

Organic Chemistry - With Access (Custom), 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 - With Access (Custom), 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 29

Answer to Problem 55GP

Organic Chemistry - With Access (Custom), Chapter 13.SE, Problem 55GP , additional homework tip 15

Answer 30

Explanation of Solution

Calculate HDI value:

Organic Chemistry - With Access (Custom), 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 - With Access (Custom), 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 - With Access (Custom), 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.

Answer 31

Conclusion

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

Answer 32

Expert Solution

Interpretation Introduction

d)

Organic Chemistry - With Access (Custom), 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 - With Access (Custom), 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 - With Access (Custom), Chapter 13.SE, Problem 55GP , additional homework tip 21

Explanation of Solution

Calculate HDI value:

Organic Chemistry - With Access (Custom), 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 - With Access (Custom), 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 - With Access (Custom), 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.

Answer 33

Expert Solution

Answer 34

Expert Solution

Answer 35

Expert Solution

Answer 36

Interpretation Introduction

d)

Organic Chemistry - With Access (Custom), 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 - With Access (Custom), 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 37

Answer to Problem 55GP

Organic Chemistry - With Access (Custom), Chapter 13.SE, Problem 55GP , additional homework tip 21

Answer 38

Explanation of Solution

Calculate HDI value:

Organic Chemistry - With Access (Custom), 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 - With Access (Custom), 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 - With Access (Custom), 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.