Organic Chemistry - Owlv2 Access (4 Term)
Organic Chemistry - Owlv2 Access (4 Term)
9th Edition
ISBN: 9781305867659
Author: McMurry
Publisher: Cengage
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Chapter 15.SE, Problem 23AP

Propose structures for aromatic hydrocarbons that meet the following descriptions:

(a) C9H12; gives only one C9H11Br product on substitution of a hydrogen on the aromatic ring with bromine

(b) C10H14; gives only one C10H13Cl product on substitution of a hydrogen on the aromatic ring with chlorine

(c) C8H10; gives three C8H9Br products on substitution of a hydrogen on the aromatic ring with bromine

(d) C10H14; gives two C10H13Cl products on substitution of a hydrogen on the aromatic ring with chlorine

Expert Solution
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Interpretation Introduction

a) C9H12; gives only one C9H11Br product on substitution of a hydrogen on the aromatic ring with bromine

Interpretation:

A possible structure for the hydrocarbon with molecular formula C9H12 that gives only one monobromination product C9H11Br on substitution of a hydrogen on the aromatic ring with bromine is to be given.

Concept introduction:

All aromatic compounds are derivatives of benzene. The benzene ring accounts for six carbon atoms. The remaining carbons can be attached as alkyl group like methyl, ethyl etc. to the benzene ring.

To propose:

A possible structure for the hydrocarbon with molecular formula C9H12 that gives only one monobromination product C9H11Br on substitution of a hydrogen on the aromatic ring with bromine.

Answer to Problem 23AP

A possible structure for the hydrocarbon with molecular formula C9H12 that gives only one monobromination product C9H11Br on substitution of a hydrogen atom on the aromatic ring with bromine is

Organic Chemistry - Owlv2 Access (4 Term), Chapter 15.SE, Problem 23AP , additional homework tip  1

Explanation of Solution

The molecular formula of the compound is C9H12. Benzene ring accounts for six carbons. When the three carbons remaining are arranged as three methyl groups alternatively on the benzene ring the three hydrogens on the other three carbons become equivalent. Hence upon bromination, the hydrocarbon gives only one monosubstituted product.

Conclusion

A possible structure for the hydrocarbon with molecular formula C9H12 that gives only one monobromination product C9H11Br on substitution of a hydrogen atom on the aromatic ring with bromine is

Organic Chemistry - Owlv2 Access (4 Term), Chapter 15.SE, Problem 23AP , additional homework tip  2

Expert Solution
Check Mark
Interpretation Introduction

b) C10H14; gives only one C10H13Cl product on substitution of a hydrogen on the aromatic ring with chlorine

Interpretation:

A possible structure for the hydrocarbon with molecular formula C10H14 that gives only one monochlorination product C10H13Cl on substitution of a hydrogen on the aromatic ring with chlorine is to be given.

Concept introduction:

All aromatic compounds are derivatives of benzene. The benzene ring accounts for six carbon atoms. The remaining carbons can be attached as alkyl group like methyl, ethyl etc. to the benzene ring. Only one monosubstituted product will be obtained only if all the hydrogen atoms in the aromatic ring are equivalent.

To propose:

A possible structure for the hydrocarbon with molecular formula C10H14 that gives only one monochlorination product C10H13Cl on substitution of a hydrogen on the aromatic ring with chlorine is to be given.

Answer to Problem 23AP

A possible structure for the hydrocarbon with molecular formula C10H14 that gives only one monochlorination product, C10H13Cl, on substitution of a hydrogen atom on the aromatic ring with chlorine is

Organic Chemistry - Owlv2 Access (4 Term), Chapter 15.SE, Problem 23AP , additional homework tip  3

Explanation of Solution

The molecular formula of the hydrocarbon is C10H14. Benzene ring accounts for six carbons. The remaining four carbons can be arranged as four methyl groups or as two ethyl groups on the benzene ring in four different ways as shown such that remaining hydrogen atoms become equivalent. Hence upon chlorination, the hydrocarbon will give only one monosubstituted product.

Conclusion

A possible structure for the hydrocarbon with molecular formula C10H14 that gives only one monochlorination product, C10H14Cl, on substitution of a hydrogen atom on the aromatic ring with chlorine is

Organic Chemistry - Owlv2 Access (4 Term), Chapter 15.SE, Problem 23AP , additional homework tip  4

Expert Solution
Check Mark
Interpretation Introduction

c) C8H10; gives three C8H9Br products on substitution of a hydrogen on the aromatic ring with bromine

Interpretation:

A possible structure for the hydrocarbon with molecular formula C8H10 that gives three monobromination products. C8H9Br, on substitution of a hydrogen on the aromatic ring with bromine is to be given.

Concept introduction:

All aromatic compounds are derivatives of benzene. The benzene ring accounts for six carbon atoms. The remaining carbons can be attached as alkyl group like methyl, ethyl etc. to the benzene ring.

To propose:

A possible structure for the hydrocarbon with molecular formula C8H10 that gives three products, C8H9Br, on substitution of a hydrogen on the aromatic ring with bromine.

Answer to Problem 23AP

A possible structure for the hydrocarbon with molecular formula C8H10 that gives three monobromination products C9H11Br on substitution of a hydrogen atom on the aromatic ring with bromine is

Organic Chemistry - Owlv2 Access (4 Term), Chapter 15.SE, Problem 23AP , additional homework tip  5

Explanation of Solution

The molecular formula of the hydrocarbon is C8H10. Benzene ring accounts for six carbons. When the two carbons remaining are arranged as two methyl groups meta to each other or as an ethyl group on the benzene ring hydrogens on the other carbon atoms in the ring classify themselves into three different groups. Hence upon bromination, the hydrocarbon gives three monosubstituted products.

Conclusion

A possible structure for the hydrocarbon with molecular formula C8H10 that gives three monobromination products, C9H11Br, on substitution of a hydrogen atom on the aromatic ring with bromine is

Organic Chemistry - Owlv2 Access (4 Term), Chapter 15.SE, Problem 23AP , additional homework tip  6

Expert Solution
Check Mark
Interpretation Introduction

d) C10H14; gives two C10H13Cl products on substitution of a hydrogen atom on the aromatic ring with chlorine

Interpretation:

A possible structure for the hydrocarbon with molecular formula C10H14 that gives two monochlorination products C10H13Cl on substitution of a hydrogen atom on the aromatic ring with chlorine is to be given.

Concept introduction:

All aromatic compounds are derivatives of benzene. The benzene ring accounts for six carbon atoms. The remaining carbons can be attached as alkyl group like methyl, ethyl etc. to the benzene ring.

To propose:

A possible structure for the hydrocarbon with molecular formula C10H14 that gives two products, C10H13Cl, on substitution of a hydrogen on the aromatic ring with chlorine.

Answer to Problem 23AP

Possible structure for the hydrocarbon with molecular formula C10H10 that gives two monobromination products, C10H13Cl, on substitution of a hydrogen atom on the aromatic ring with chlorine is

Organic Chemistry - Owlv2 Access (4 Term), Chapter 15.SE, Problem 23AP , additional homework tip  7

Explanation of Solution

The molecular formula of the compound is C10H14. Benzene ring accounts for six carbons. The four carbons remaining can be arranged as two ethyl groups or as an ethyl & two methyl groups or as propyl & methyl groups or as isopropyl & methyl groups on the benzene ring as shown so that the hydrogens on the other carbon atoms in the ring classify themselves into two different groups. Hence upon chlorination, the hydrocarbon gives two monosubstituted products.

Conclusion

A possible structure for the hydrocarbon with molecular formula C10H10 that gives two monobromination products, C10H13Cl, on substitution of a hydrogen atom on the aromatic ring with chlorine is

Organic Chemistry - Owlv2 Access (4 Term), Chapter 15.SE, Problem 23AP , additional homework tip  8

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Chapter 15 Solutions

Organic Chemistry - Owlv2 Access (4 Term)

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