The labeled bonds in the given compound are to be arranged in the order of increasing bond lengths. Concept introduction: In hybridization, one 2 s orbital is always used, but the number of 2 p orbitals varies with the type of hybridization. The percent of s-character gives the fraction of 2 s orbital which is present in a hybrid orbital. As the percent of s − character increases, the bond becomes shorter and stronger. This is due to the fact that 2 s orbital keeps electron density closer to a nucleus as compared to 2 p orbital.

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Answer 1

Question

Chapter 4, Problem 4.1P

Interpretation Introduction

Interpretation:

The labeled bonds in the given compound are to be arranged in the order of increasing bond lengths.

Concept introduction:

In hybridization, one 2s orbital is always used, but the number of 2p orbitals varies with the type of hybridization. The percent of s-character gives the fraction of 2s orbital which is present in a hybrid orbital. As the percent of s− character increases, the bond becomes shorter and stronger. This is due to the fact that 2s orbital keeps electron density closer to a nucleus as compared to 2p orbital.

Expert Solution & Answer

Answer to Problem 4.1P

The increasing order of bond-lengths in the given compound is d<a<c<e<b<f.

Explanation of Solution

The given compound is,

Organic Chemistry, Chapter 4, Problem 4.1P

Figure 1

The bond with higher percentage of s-character has electron density closer to the nucleus and thus, has shorter bond length. The order of percentage s− character in orbitals is sp3<sp2<sp.

• Bond ‘a’ is present between the sp3 hybridized carbon and hydrogen atom and bond angle denoted by “d” is present between the sp2 hybridized carbon and hydrogen atom. Bond angle denoted by “d” has higher percentage of s− character than bond ‘a’ and thus, bond angle denoted by “d” is shorter than the bond ‘a’.

• Bond ‘c’ is present between the two sp2 hybridized carbon atoms, while bond ‘f’ is present between the two sp3 hybridized carbon atoms. Bond ‘c’ has higher percentage of s− character than bond ‘f’ and thus, bond ‘c’ is shorter than the bond ‘f’.

• Due to the conjugation between two pi bonds, the bond length of ‘e’ is in between the carbon-carbon single bond and carbon-carbon double bond.

• Bond ‘b’ is present between one sp2 hybridized carbon atom and one sp3 hybridized carbon atom, while bond ‘f’ is present between the two sp3 hybridized carbon atoms. Bond ‘b’ has higher percentage of s− character than bond ‘f’ and thus, bond ‘b’ is shorter than the bond ‘f’.

• The C(sp2)−H(s) is shortest bond as this bond has the highest percentage s− character.

From the above points, it is concluded that the order of bond-lengths in the given compound is d<a<c<e<b<f.

Conclusion

The increasing order of bond-lengths in the given compound is d<a<c<e<b<f.

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Answer 2

Question

Chapter 4, Problem 4.1P

Interpretation Introduction

Interpretation:

The labeled bonds in the given compound are to be arranged in the order of increasing bond lengths.

Concept introduction:

In hybridization, one 2s orbital is always used, but the number of 2p orbitals varies with the type of hybridization. The percent of s-character gives the fraction of 2s orbital which is present in a hybrid orbital. As the percent of s− character increases, the bond becomes shorter and stronger. This is due to the fact that 2s orbital keeps electron density closer to a nucleus as compared to 2p orbital.

Expert Solution & Answer

Answer to Problem 4.1P

The increasing order of bond-lengths in the given compound is d<a<c<e<b<f.

Explanation of Solution

The given compound is,

Organic Chemistry, Chapter 4, Problem 4.1P

Figure 1

The bond with higher percentage of s-character has electron density closer to the nucleus and thus, has shorter bond length. The order of percentage s− character in orbitals is sp3<sp2<sp.

• Bond ‘a’ is present between the sp3 hybridized carbon and hydrogen atom and bond angle denoted by “d” is present between the sp2 hybridized carbon and hydrogen atom. Bond angle denoted by “d” has higher percentage of s− character than bond ‘a’ and thus, bond angle denoted by “d” is shorter than the bond ‘a’.

• Bond ‘c’ is present between the two sp2 hybridized carbon atoms, while bond ‘f’ is present between the two sp3 hybridized carbon atoms. Bond ‘c’ has higher percentage of s− character than bond ‘f’ and thus, bond ‘c’ is shorter than the bond ‘f’.

• Due to the conjugation between two pi bonds, the bond length of ‘e’ is in between the carbon-carbon single bond and carbon-carbon double bond.

• Bond ‘b’ is present between one sp2 hybridized carbon atom and one sp3 hybridized carbon atom, while bond ‘f’ is present between the two sp3 hybridized carbon atoms. Bond ‘b’ has higher percentage of s− character than bond ‘f’ and thus, bond ‘b’ is shorter than the bond ‘f’.

• The C(sp2)−H(s) is shortest bond as this bond has the highest percentage s− character.

From the above points, it is concluded that the order of bond-lengths in the given compound is d<a<c<e<b<f.

Conclusion

The increasing order of bond-lengths in the given compound is d<a<c<e<b<f.

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Answer 3

Question

Chapter 4, Problem 4.1P

Interpretation Introduction

Interpretation:

The labeled bonds in the given compound are to be arranged in the order of increasing bond lengths.

Concept introduction:

In hybridization, one 2s orbital is always used, but the number of 2p orbitals varies with the type of hybridization. The percent of s-character gives the fraction of 2s orbital which is present in a hybrid orbital. As the percent of s− character increases, the bond becomes shorter and stronger. This is due to the fact that 2s orbital keeps electron density closer to a nucleus as compared to 2p orbital.

Expert Solution & Answer

Answer to Problem 4.1P

The increasing order of bond-lengths in the given compound is d<a<c<e<b<f.

Explanation of Solution

The given compound is,

Organic Chemistry, Chapter 4, Problem 4.1P

Figure 1

The bond with higher percentage of s-character has electron density closer to the nucleus and thus, has shorter bond length. The order of percentage s− character in orbitals is sp3<sp2<sp.

• Bond ‘a’ is present between the sp3 hybridized carbon and hydrogen atom and bond angle denoted by “d” is present between the sp2 hybridized carbon and hydrogen atom. Bond angle denoted by “d” has higher percentage of s− character than bond ‘a’ and thus, bond angle denoted by “d” is shorter than the bond ‘a’.

• Bond ‘c’ is present between the two sp2 hybridized carbon atoms, while bond ‘f’ is present between the two sp3 hybridized carbon atoms. Bond ‘c’ has higher percentage of s− character than bond ‘f’ and thus, bond ‘c’ is shorter than the bond ‘f’.

• Due to the conjugation between two pi bonds, the bond length of ‘e’ is in between the carbon-carbon single bond and carbon-carbon double bond.

• Bond ‘b’ is present between one sp2 hybridized carbon atom and one sp3 hybridized carbon atom, while bond ‘f’ is present between the two sp3 hybridized carbon atoms. Bond ‘b’ has higher percentage of s− character than bond ‘f’ and thus, bond ‘b’ is shorter than the bond ‘f’.

• The C(sp2)−H(s) is shortest bond as this bond has the highest percentage s− character.

From the above points, it is concluded that the order of bond-lengths in the given compound is d<a<c<e<b<f.

Conclusion

The increasing order of bond-lengths in the given compound is d<a<c<e<b<f.

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Answer 4

Question

Chapter 4, Problem 4.1P

Interpretation Introduction

Interpretation:

The labeled bonds in the given compound are to be arranged in the order of increasing bond lengths.

Concept introduction:

In hybridization, one 2s orbital is always used, but the number of 2p orbitals varies with the type of hybridization. The percent of s-character gives the fraction of 2s orbital which is present in a hybrid orbital. As the percent of s− character increases, the bond becomes shorter and stronger. This is due to the fact that 2s orbital keeps electron density closer to a nucleus as compared to 2p orbital.

Expert Solution & Answer

Answer to Problem 4.1P

The increasing order of bond-lengths in the given compound is d<a<c<e<b<f.

Explanation of Solution

The given compound is,

Organic Chemistry, Chapter 4, Problem 4.1P

Figure 1

The bond with higher percentage of s-character has electron density closer to the nucleus and thus, has shorter bond length. The order of percentage s− character in orbitals is sp3<sp2<sp.

• Bond ‘a’ is present between the sp3 hybridized carbon and hydrogen atom and bond angle denoted by “d” is present between the sp2 hybridized carbon and hydrogen atom. Bond angle denoted by “d” has higher percentage of s− character than bond ‘a’ and thus, bond angle denoted by “d” is shorter than the bond ‘a’.

• Bond ‘c’ is present between the two sp2 hybridized carbon atoms, while bond ‘f’ is present between the two sp3 hybridized carbon atoms. Bond ‘c’ has higher percentage of s− character than bond ‘f’ and thus, bond ‘c’ is shorter than the bond ‘f’.

• Due to the conjugation between two pi bonds, the bond length of ‘e’ is in between the carbon-carbon single bond and carbon-carbon double bond.

• Bond ‘b’ is present between one sp2 hybridized carbon atom and one sp3 hybridized carbon atom, while bond ‘f’ is present between the two sp3 hybridized carbon atoms. Bond ‘b’ has higher percentage of s− character than bond ‘f’ and thus, bond ‘b’ is shorter than the bond ‘f’.

• The C(sp2)−H(s) is shortest bond as this bond has the highest percentage s− character.

From the above points, it is concluded that the order of bond-lengths in the given compound is d<a<c<e<b<f.

Conclusion

The increasing order of bond-lengths in the given compound is d<a<c<e<b<f.

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Answer 5

Chapter 4, Problem 4.1P

Interpretation Introduction

Interpretation:

The labeled bonds in the given compound are to be arranged in the order of increasing bond lengths.

Concept introduction:

In hybridization, one 2s orbital is always used, but the number of 2p orbitals varies with the type of hybridization. The percent of s-character gives the fraction of 2s orbital which is present in a hybrid orbital. As the percent of s− character increases, the bond becomes shorter and stronger. This is due to the fact that 2s orbital keeps electron density closer to a nucleus as compared to 2p orbital.

Expert Solution & Answer

Answer to Problem 4.1P

The increasing order of bond-lengths in the given compound is d<a<c<e<b<f.

Explanation of Solution

The given compound is,

Organic Chemistry, Chapter 4, Problem 4.1P

Figure 1

The bond with higher percentage of s-character has electron density closer to the nucleus and thus, has shorter bond length. The order of percentage s− character in orbitals is sp3<sp2<sp.

• Bond ‘a’ is present between the sp3 hybridized carbon and hydrogen atom and bond angle denoted by “d” is present between the sp2 hybridized carbon and hydrogen atom. Bond angle denoted by “d” has higher percentage of s− character than bond ‘a’ and thus, bond angle denoted by “d” is shorter than the bond ‘a’.

• Bond ‘c’ is present between the two sp2 hybridized carbon atoms, while bond ‘f’ is present between the two sp3 hybridized carbon atoms. Bond ‘c’ has higher percentage of s− character than bond ‘f’ and thus, bond ‘c’ is shorter than the bond ‘f’.

• Due to the conjugation between two pi bonds, the bond length of ‘e’ is in between the carbon-carbon single bond and carbon-carbon double bond.

• Bond ‘b’ is present between one sp2 hybridized carbon atom and one sp3 hybridized carbon atom, while bond ‘f’ is present between the two sp3 hybridized carbon atoms. Bond ‘b’ has higher percentage of s− character than bond ‘f’ and thus, bond ‘b’ is shorter than the bond ‘f’.

• The C(sp2)−H(s) is shortest bond as this bond has the highest percentage s− character.

From the above points, it is concluded that the order of bond-lengths in the given compound is d<a<c<e<b<f.

Conclusion

The increasing order of bond-lengths in the given compound is d<a<c<e<b<f.

Answer 6

Chapter 4, Problem 4.1P

Interpretation Introduction

Interpretation:

The labeled bonds in the given compound are to be arranged in the order of increasing bond lengths.

Concept introduction:

In hybridization, one 2s orbital is always used, but the number of 2p orbitals varies with the type of hybridization. The percent of s-character gives the fraction of 2s orbital which is present in a hybrid orbital. As the percent of s− character increases, the bond becomes shorter and stronger. This is due to the fact that 2s orbital keeps electron density closer to a nucleus as compared to 2p orbital.

Expert Solution & Answer

Answer to Problem 4.1P

The increasing order of bond-lengths in the given compound is d<a<c<e<b<f.

Explanation of Solution

The given compound is,

Organic Chemistry, Chapter 4, Problem 4.1P

Figure 1

The bond with higher percentage of s-character has electron density closer to the nucleus and thus, has shorter bond length. The order of percentage s− character in orbitals is sp3<sp2<sp.

• Bond ‘a’ is present between the sp3 hybridized carbon and hydrogen atom and bond angle denoted by “d” is present between the sp2 hybridized carbon and hydrogen atom. Bond angle denoted by “d” has higher percentage of s− character than bond ‘a’ and thus, bond angle denoted by “d” is shorter than the bond ‘a’.

• Bond ‘c’ is present between the two sp2 hybridized carbon atoms, while bond ‘f’ is present between the two sp3 hybridized carbon atoms. Bond ‘c’ has higher percentage of s− character than bond ‘f’ and thus, bond ‘c’ is shorter than the bond ‘f’.

• Due to the conjugation between two pi bonds, the bond length of ‘e’ is in between the carbon-carbon single bond and carbon-carbon double bond.

• Bond ‘b’ is present between one sp2 hybridized carbon atom and one sp3 hybridized carbon atom, while bond ‘f’ is present between the two sp3 hybridized carbon atoms. Bond ‘b’ has higher percentage of s− character than bond ‘f’ and thus, bond ‘b’ is shorter than the bond ‘f’.

• The C(sp2)−H(s) is shortest bond as this bond has the highest percentage s− character.

From the above points, it is concluded that the order of bond-lengths in the given compound is d<a<c<e<b<f.

Conclusion

The increasing order of bond-lengths in the given compound is d<a<c<e<b<f.

Answer 7

Chapter 4, Problem 4.1P

Interpretation Introduction

Interpretation:

The labeled bonds in the given compound are to be arranged in the order of increasing bond lengths.

Concept introduction:

In hybridization, one 2s orbital is always used, but the number of 2p orbitals varies with the type of hybridization. The percent of s-character gives the fraction of 2s orbital which is present in a hybrid orbital. As the percent of s− character increases, the bond becomes shorter and stronger. This is due to the fact that 2s orbital keeps electron density closer to a nucleus as compared to 2p orbital.

Answer 8

Expert Solution & Answer

Answer to Problem 4.1P

The increasing order of bond-lengths in the given compound is d<a<c<e<b<f.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

The given compound is,

Organic Chemistry, Chapter 4, Problem 4.1P

Figure 1

The bond with higher percentage of s-character has electron density closer to the nucleus and thus, has shorter bond length. The order of percentage s− character in orbitals is sp3<sp2<sp.

• Bond ‘a’ is present between the sp3 hybridized carbon and hydrogen atom and bond angle denoted by “d” is present between the sp2 hybridized carbon and hydrogen atom. Bond angle denoted by “d” has higher percentage of s− character than bond ‘a’ and thus, bond angle denoted by “d” is shorter than the bond ‘a’.

• Bond ‘c’ is present between the two sp2 hybridized carbon atoms, while bond ‘f’ is present between the two sp3 hybridized carbon atoms. Bond ‘c’ has higher percentage of s− character than bond ‘f’ and thus, bond ‘c’ is shorter than the bond ‘f’.

• Due to the conjugation between two pi bonds, the bond length of ‘e’ is in between the carbon-carbon single bond and carbon-carbon double bond.

• Bond ‘b’ is present between one sp2 hybridized carbon atom and one sp3 hybridized carbon atom, while bond ‘f’ is present between the two sp3 hybridized carbon atoms. Bond ‘b’ has higher percentage of s− character than bond ‘f’ and thus, bond ‘b’ is shorter than the bond ‘f’.

• The C(sp2)−H(s) is shortest bond as this bond has the highest percentage s− character.

From the above points, it is concluded that the order of bond-lengths in the given compound is d<a<c<e<b<f.

Answer 12

Conclusion

The increasing order of bond-lengths in the given compound is d<a<c<e<b<f.

Answer 13

Ch. 4 - Prob. 4.1P Ch. 4 - Prob. 4.2P Ch. 4 - Prob. 4.3P Ch. 4 - Prob. 4.4P Ch. 4 - Prob. 4.5P Ch. 4 - Prob. 4.6P Ch. 4 - Prob. 4.7P Ch. 4 - Prob. 4.8P Ch. 4 - Prob. 4.9P Ch. 4 - Prob. 4.10P

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