ORGANIC CHEMISTRY E-BOOK W/SMARTWORK5
ORGANIC CHEMISTRY E-BOOK W/SMARTWORK5
2nd Edition
ISBN: 9780393664034
Author: KARTY
Publisher: NORTON
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Question
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Chapter 14, Problem 14.11P
Interpretation Introduction

(a)

Interpretation:

The given molecule is to be identified as aromatic, antiaromatic, or nonaromatic.

Concept introduction:

Huckel’s rule for aromaticity states that if a species is planar and possesses a π-system of molecular orbitals constructed from p-orbitals that are fully conjugated around a ring (sp2 hybridized), then the species is aromatic if the number of electrons in that π- system fits the formula 4n+2 (n = 0, 1, 2, ...). It is antiaromatic if the number of electrons in that π- system fits the formula 4n (n = 0, 1, 2, ...). All other species are considered nonaromatic.

Expert Solution
Check Mark

Answer to Problem 14.11P

The given molecule is nonaromatic.

Explanation of Solution

The structure of the given molecule is

ORGANIC CHEMISTRY E-BOOK W/SMARTWORK5, Chapter 14, Problem 14.11P , additional homework tip  1

In the given compound, six carbon atoms are sp2 hybridized, and the remaining four carbons are sp3 hybridized.

ORGANIC CHEMISTRY E-BOOK W/SMARTWORK5, Chapter 14, Problem 14.11P , additional homework tip  2

Because of the sp3 hybridized C5, C6, C9, and C10, the molecule will not be planar. These atoms also interrupt the conjugation.

Therefore, this molecule is nonaromatic.

Conclusion

The presence of four sp3 hybridized carbon atoms in ring indicates that the molecule is nonaromatic.

Interpretation Introduction

(b)

Interpretation:

The given molecule is to be identified as aromatic, antiaromatic, or nonaromatic.

Concept introduction:

Huckel’s rule for aromaticity states that if a species is planar and possesses a π-system of molecular orbitals constructed from p-orbitals that are fully conjugated around a ring (sp2 hybridized), then the species is aromatic if the number of electrons in that π- system fits the formula 4n+2 (n = 0, 1, 2, ...). These numbers are called Huckel numbers. It is antiaromatic if the number of electrons in that π- system fits the formula 4n (n = 0, 1, 2, ...). All other species are considered nonaromatic.

Expert Solution
Check Mark

Answer to Problem 14.11P

The given molecule is aromatic.

Explanation of Solution

Structure of the given molecule is

ORGANIC CHEMISTRY E-BOOK W/SMARTWORK5, Chapter 14, Problem 14.11P , additional homework tip  3

In this molecule, six carbon atoms are sp2 hybridized, and the remaining four carbons are sp3 hybridized. The six π-electrons in the ring on the left are completely conjugated, making it aromatic. The four sp3 hybridized carbons are not a part of this ring.

Therefore, the given molecule is aromatic.

Conclusion

The presence of six π electrons (a Huckel number) in completely conjugated cyclic π-system indicates that the molecule is aromatic.

Interpretation Introduction

(c)

Interpretation:

The given molecule is to be identified as aromatic, antiaromatic, or nonaromatic.

Concept introduction:

Huckel’s rule for aromaticity states that if a species is planar and possesses a π-system of molecular orbitals constructed from p-orbitals that are fully conjugated around a ring (sp2 hybridized), then the species is aromatic if the number of electrons in that π- system fits the formula 4n+2 (n = 0, 1, 2, ...). It is antiaromatic if the number of electrons in that π- system fits the formula 4n (n = 0, 1, 2, ...). All other species are considered nonaromatic.

Expert Solution
Check Mark

Answer to Problem 14.11P

The given molecule is nonaromatic.

Explanation of Solution

Structure of the given molecule is

ORGANIC CHEMISTRY E-BOOK W/SMARTWORK5, Chapter 14, Problem 14.11P , additional homework tip  4

In the given compound, four carbon atoms are sp2 hybridized, and the remaining carbon, C4, is sp3 hybridized. There are four electrons in the π-system, but it is not completely cyclic because of the presence of the sp3 carbon.

Therefore, the given molecule is nonaromatic.

Conclusion

The presence of one sp3 hybridized carbon atom in the ring indicates that the molecule is nonaromatic.

Interpretation Introduction

(d)

Interpretation:

The given molecule is to be identified as aromatic, antiaromatic, or nonaromatic.

Concept introduction:

Huckel’s rule for aromaticity states that if a species is planar and possesses a π-system of molecular orbitals constructed from p-orbitals that are fully conjugated around a ring (sp2 hybridized), then the species is aromatic if the number of electrons in that π- system fits the formula 4n+2 (n = 0, 1, 2, ...). It is antiaromatic if the number of electrons in that π- system fits the formula 4n (n = 0, 1, 2, ...). All other species are considered nonaromatic.

Expert Solution
Check Mark

Answer to Problem 14.11P

The given molecule is nonaromatic.

Explanation of Solution

Structure of the given molecule is

ORGANIC CHEMISTRY E-BOOK W/SMARTWORK5, Chapter 14, Problem 14.11P , additional homework tip  5

In the given compound, six carbon atoms are sp2 hybridized, and the remaining one, C6, is sp3 hybridized. The π-system is not completely cyclic because of the presence of the sp3 carbon.

Therefore, the given molecule is nonaromatic.

Conclusion

The presence of one sp3 hybridized carbon atom in ring indicates that the molecule is nonaromatic.

Interpretation Introduction

(e)

Interpretation:

The given molecule is to be identified as aromatic, antiaromatic, or nonaromatic.

Concept introduction:

Huckel’s rule for aromaticity states that if a species is planar and possesses a π-system of molecular orbitals constructed from p-orbitals that are fully conjugated around a ring (sp2 hybridized), then the species is aromatic if the number of electrons in that π- system fits the formula 4n+2 (n = 0, 1, 2, ...). It is antiaromatic if the number of electrons in that π- system fits the formula 4n (n = 0, 1, 2, ...). All other species are considered nonaromatic.

Expert Solution
Check Mark

Answer to Problem 14.11P

The given molecule is nonaromatic.

Explanation of Solution

Structure of the given molecule is

ORGANIC CHEMISTRY E-BOOK W/SMARTWORK5, Chapter 14, Problem 14.11P , additional homework tip  6

In the given compound, two carbon atoms are sp2 hybridized, and the remaining two, C3 and C4, are sp3 hybridized. The π system is not completely cyclic because of the presence of the two sp3 hybridized carbon atoms.

Therefore, the given molecule is nonaromatic.

Conclusion

The presence of two sp3 hybridized carbon atoms in the ring indicates that the molecule is nonaromatic.

Interpretation Introduction

(f)

Interpretation:

The given molecule is to be identified as aromatic, antiaromatic, or nonaromatic.

Concept introduction:

Huckel’s rule for aromaticity states that if a species is planar and possesses a π-system of molecular orbitals constructed from p-orbitals that are fully conjugated around a ring (sp2 hybridized), then the species is aromatic if the number of electrons in that π- system fits the formula 4n+2 (n = 0, 1, 2, ...). It is antiaromatic if the number of electrons in that π- system fits the formula 4n (n = 0, 1, 2, ...). All other species are considered nonaromatic.

Expert Solution
Check Mark

Answer to Problem 14.11P

The given molecule is nonaromatic.

Explanation of Solution

Structure of the given molecule is

ORGANIC CHEMISTRY E-BOOK W/SMARTWORK5, Chapter 14, Problem 14.11P , additional homework tip  7

The molecule is not a cyclic system.

Therefore, the given molecule is nonaromatic.

Conclusion

Non-cyclic form of the given molecule indicates that the molecule is nonaromatic.

Interpretation Introduction

(g)

Interpretation:

The given molecule is to be identified as aromatic, antiaromatic, or nonaromatic.

Concept introduction:

Huckel’s rule for aromaticity states that if a species is planar and possesses a π-system of molecular orbitals constructed from p-orbitals that are fully conjugated around a ring (sp2 hybridized), then the species is aromatic if the number of electrons in that π- system fits the formula 4n+2 (n = 0, 1, 2, ...). It is antiaromatic if the number of electrons in that π system fits the formula 4n (n = 0, 1, 2, ...). All other species are considered nonaromatic.

Expert Solution
Check Mark

Answer to Problem 14.11P

The given molecule is aromatic.

Explanation of Solution

The structure of the given molecule is

ORGANIC CHEMISTRY E-BOOK W/SMARTWORK5, Chapter 14, Problem 14.11P , additional homework tip  8

The molecule has a cyclic part with alternating single and double bonds. This means all ring atoms are sp2 hybridized. The ring has a completely cyclic, conjugated π system containing six electrons, a Huckel number.

Therefore, this molecule is aromatic.

Conclusion

The presence of six π electrons (a Huckel number) in a completely conjugated cyclic π-system indicates that the molecule is aromatic.

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Chapter 14 Solutions

ORGANIC CHEMISTRY E-BOOK W/SMARTWORK5

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