Chapter 15, Problem 15.10P

Interpretation Introduction

(a)

Interpretation: The major product formed when given cycloalkane is heated with ${\text{Br}}_{\text{2}}$ is to be drawn.

Concept introduction: Free radicals are classified as $1°$, $2°$ or $3°$ depending upon the number of alkyl groups attached to it. The number of alkyl substituents increases, the stability of radical increases. This is due to the fact alkyl substituents donate their electrons to electron deficient radicals.

Halogens react with alkanes in presence of heat or light to form alkyl halides. This is known as halogenation reaction. This is a free radical substitution reaction. In this the halogen substitutes the hydrogen atom from $\text{C}-\text{H}$ sigma bond of the alkane.

Answer to Problem 15.10P

The major product formed when given cycloalkane is heated with ${\text{Br}}_{\text{2}}$ is shown below.

Explanation of Solution

The given species is,

Figure 1

Three types of radicals can be formed from cleavage of $\text{C}-\text{H}$ bond in the given molecule. Carbon atom highlighted produces $3°$ radical while the other carbon atoms of the cycloalkane produce $2°$ radical and methyl group produces $1°$ radical.

The number of alkyl substituents increases, the stability of radical increases. The order of stability is $1°<2°<3°$. Higher the stability of the radical formed, weaker is the $\text{C}-\text{H}$ bond. The most stable radical is $3°$ radical. Hence, the major product formed is by the cleavage of $3°\text{C}-\text{H}$ bond. The corresponding reaction is given below.

Figure 2

Conclusion

The major product formed when given cycloalkane is heated with ${\text{Br}}_{\text{2}}$ is shown in Figure 2.

Interpretation Introduction

(b)

Interpretation: The major product formed when given cycloalkane is heated with ${\text{Br}}_{\text{2}}$ is to be drawn.

Concept introduction: Free radicals are classified as $1°$, $2°$ or $3°$ depending upon the number of alkyl groups attached to it. The number of alkyl substituents increases, the stability of radical increases. This is due to the fact alkyl substituents donate their electrons to electron deficient radicals.

Halogens react with alkanes in presence of heat or light to form alkyl halides. This is known as halogenation reaction. This is a free radical substitution reaction. In this the halogen substitutes the hydrogen atom from $\text{C}-\text{H}$ sigma bond of the alkane.

Answer to Problem 15.10P

The major product formed when given cycloalkane is heated with ${\text{Br}}_{\text{2}}$ is shown below.

Explanation of Solution

The given species is,

Figure 3

Three types of radicals can be formed from cleavage of $\text{C}-\text{H}$ bond in the given molecule. Carbon atom highlighted produces $3°$ radical while the other carbon atoms of the cycloalkane produce $2°$ radical and methyl group produces $1°$ radical.

The number of alkyl substituents increases, the stability of radical increases. The order of stability is $1°<2°<3°$. Higher the stability of the radical formed, weaker is the $\text{C}-\text{H}$ bond. The most stable radical is $3°$ radical. Hence, the major product formed is by the cleavage of $3°\text{C}-\text{H}$ bond. The corresponding reaction is given below.

Figure 4

Conclusion

The major product formed when given cycloalkane is heated with ${\text{Br}}_{\text{2}}$ is shown in Figure 4.

Interpretation Introduction

(c)

Interpretation: The major product formed when given cycloalkane is heated with ${\text{Br}}_{\text{2}}$ is to be drawn.

Concept introduction: Free radicals are classified as $1°$, $2°$ or $3°$ depending upon the number of alkyl groups attached to it. The number of alkyl substituents increases, the stability of radical increases. This is due to the fact alkyl substituents donate their electrons to electron deficient radicals.

Halogens react with alkanes in presence of heat or light to form alkyl halides. This is known as halogenation reaction. This is a free radical substitution reaction. In this the halogen substitutes the hydrogen atom from $\text{C}-\text{H}$ sigma bond of the alkane.

Answer to Problem 15.10P

The major product formed when given cycloalkane is heated with ${\text{Br}}_{\text{2}}$ is shown below.

Explanation of Solution

The given species is,

Figure 5

Three types of radicals can be formed from cleavage of $\text{C}-\text{H}$ bond in the given molecule. Carbon atom highlighted produces $3°$ radical while the other carbon atoms of the cycloalkane produce $2°$ radical and methyl group produces $1°$ radical.

The number of alkyl substituents increases, the stability of radical increases. The order of stability is $1°<2°<3°$. Higher the stability of the radical formed, weaker is the $\text{C}-\text{H}$ bond. The most stable radical is $3°$ radical. Hence, the major product formed is by the cleavage of $3°\text{C}-\text{H}$ bond. The corresponding reaction is given below.

Figure 6

Conclusion

The major product formed when given cycloalkane is heated with ${\text{Br}}_{\text{2}}$ is shown in Figure 6.

Interpretation Introduction

(c)

Interpretation: The major product formed when given cycloalkane is heated with ${\text{Br}}_{\text{2}}$ is to be drawn.

Concept introduction: Free radicals are classified as $1°$, $2°$ or $3°$ depending upon the number of alkyl groups attached to it. The number of alkyl substituents increases, the stability of radical increases. This is due to the fact alkyl substituents donate their electrons to electron deficient radicals.

Halogens react with alkanes in presence of heat or light to form alkyl halides. This is known as halogenation reaction. This is a free radical substitution reaction. In this the halogen substitutes the hydrogen atom from $\text{C}-\text{H}$ sigma bond of the alkane.

Answer to Problem 15.10P

The major product formed when given cycloalkane is heated with ${\text{Br}}_{\text{2}}$ is shown below.

Explanation of Solution

The given species is,

Figure 7

Two types of radicals can be formed from cleavage of $\text{C}-\text{H}$ bond in the given molecule. Carbon atoms of the cycloalkane produce $2°$ radical and methyl group produces $1°$ radical.

The number of alkyl substituents increases, the stability of radical increases. The order of stability is $1°<2°<3°$. Higher the stability of the radical formed, weaker is the $\text{C}-\text{H}$ bond. The most stable radical is $3°$ radical. Hence, the major product formed is by the cleavage of $2°\text{C}-\text{H}$ bond. The corresponding reaction is given below.

Figure 8

Conclusion

The major product formed when given cycloalkane is heated with ${\text{Br}}_{\text{2}}$ is shown in Figure 8.