Chapter 25, Problem 25.37P

Interpretation Introduction

(a)

Interpretation:

Which H would most likely be abstracted by bromine radical ${\text{Br}}^{\text{•}}$, is to be identified.

Concept introduction:

The H atom with the weakest bond and leads to the formation of stable ion is the most reactive H atom i.e. likely is abstracted by radical. The H atom attached to the carbon with more number of alkyl group form weaker bond $\text{(i}\text{.e}{\text{. 1}}^{\text{o}}{\text{ >2}}^{\text{o}}{\text{ > 3}}^{\text{o}}\text{)}$ because the resulting ${\text{C}}^{\text{•}}$ is stabilized by addition electron-donating alkyl group. Also bonds of H with allylic or benzylic C atoms tend to be weaker because of the resonance stabilization in the resulting alkyl radical.

Answer to Problem 25.37P

The H atom which is likely to be abstracted by bromine radical ${\text{Br}}^{\text{•}}$ is marked in box:

Explanation of Solution

The given compound is:

In the above compound there are three types of C atom; methyl, secondary and tertiary carbon atoms. Hydrogen atoms attached to respective carbons are also methyl H, secondary H and tertiary H atoms.

The H atom attached to the carbon with more number of alkyl group form weaker bond $\text{(i}\text{.e}{\text{. 1}}^{\text{o}}{\text{ >2}}^{\text{o}}{\text{ > 3}}^{\text{o}}\text{)}$ because the resulting ${\text{C}}^{\text{•}}$ is stabilized by addition electron-donating alkyl group. Thus the H atom attached to tertiary C atom is likely abstracted by the radical to form a most stable radical, marked in the box:

Conclusion

Which H would most likely be abstracted by bromine radical ${\text{Br}}^{\text{•}}$ is identified on the basis of the types of H atoms present in the given compound to form most stable ${\text{C}}^{\text{•}}$ radical.

Interpretation Introduction

(b)

Interpretation:

Which H would most likely be abstracted by bromine radical ${\text{Br}}^{\text{•}}$, is to be identified.

Concept introduction:

The H atom with the weakest bond and leads to the formation of stable ion is the most reactive H atom i.e. likely is abstracted by radical. The H atom attached to the carbon with more number of alkyl group form weaker bond $\text{(i}\text{.e}{\text{. 1}}^{\text{o}}{\text{ >2}}^{\text{o}}{\text{ > 3}}^{\text{o}}\text{)}$ because the resulting ${\text{C}}^{\text{•}}$ is stabilized by addition electron-donating alkyl group. Also bonds of H with allylic or benzylic C atoms tend to be weaker because of the resonance stabilization in the resulting alkyl radical.

Answer to Problem 25.37P

The H atom which is likely to be abstracted by bromine radical ${\text{Br}}^{\text{•}}$ is marked in box:

Explanation of Solution

The given compound is:

In the above compound there are three types of C atom; methyl, secondary and tertiary carbon atoms. Hydrogen atoms attached to respective carbons are also methyl H, secondary H and tertiary H atoms.

The H atom attached to the carbon with more number of alkyl group form weaker bond $\text{(i}\text{.e}{\text{. 1}}^{\text{o}}{\text{ >2}}^{\text{o}}{\text{ > 3}}^{\text{o}}\text{)}$ because the resulting ${\text{C}}^{\text{•}}$ is stabilized by addition electron-donating alkyl group. Thus the H atom attached to tertiary C atom is likely abstracted by the radical to form a most stable radical, marked in the box:

Conclusion

Which H would most likely be abstracted by bromine radical ${\text{Br}}^{\text{•}}$ is identified on the basis of the types of H atoms present in the given compound to form most stable ${\text{C}}^{\text{•}}$ radical.

Interpretation Introduction

(c)

Interpretation:

Which H would most likely be abstracted by bromine radical ${\text{Br}}^{\text{•}}$, is to be identified.

Concept introduction:

The H atom with the weakest bond and leads to the formation of stable ion is the most reactive H atom i.e. likely is abstracted by radical. The H atom attached to the carbon with more number of alkyl group form weaker bond $\text{(i}\text{.e}{\text{. 1}}^{\text{o}}{\text{ >2}}^{\text{o}}{\text{ > 3}}^{\text{o}}\text{)}$ because the resulting ${\text{C}}^{\text{•}}$ is stabilized by addition electron-donating alkyl group. Also bonds of H with allylic or benzylic C atoms tend to be weaker because of the resonance stabilization in the resulting alkyl radical.

Answer to Problem 25.37P

The H atom which is likely to be abstracted by bromine radical ${\text{Br}}^{\text{•}}$ is marked in box:

Explanation of Solution

The given compound is:

In the above compound there are four types of C atom; methyl, secondary, secondary benzylic and aromatic carbon atoms. Hydrogen atoms attached to respective carbons are also methyl H, secondary, secondary benzylic and aromatic H atoms.

The H atom attached to the carbon with more number of alkyl group form weaker bond $\text{(i}\text{.e}{\text{. 1}}^{\text{o}}{\text{ >2}}^{\text{o}}{\text{ > 3}}^{\text{o}}\text{)}$ because the resulting ${\text{C}}^{\text{•}}$ is stabilized by addition electron-donating alkyl group. Also bonds of H with allylic or benzylic C atoms tend to be weaker because of the resonance stabilization in the resulting alkyl radical. Thus the H atom attached to secondary benzylic C atom is likely abstracted by the radical to form a most stable radical, marked in box:

Here, the another benzylic radical could form by abstracting a H atom from the benzylic ${\text{CH}}_{\text{3}}$, but the resulting radical would be less stable because the radical would be primary.

Conclusion

Which H would most likely be abstracted by bromine radical ${\text{Br}}^{\text{•}}$ is identified on the basis of the types of H atoms present in the given compound to form most stable ${\text{C}}^{\text{•}}$ radical.

Interpretation Introduction

(d)

Interpretation:

Which H would most likely be abstracted by bromine radical ${\text{Br}}^{\text{•}}$, is to be identified.

Concept introduction:

The H atom with the weakest bond and leads to the formation of stable ion is the most reactive H atom i.e. likely is abstracted by radical. The H atom attached to the carbon with more number of alkyl group form weaker bond $\text{(i}\text{.e}{\text{. 1}}^{\text{o}}{\text{ >2}}^{\text{o}}{\text{ > 3}}^{\text{o}}\text{)}$ because the resulting ${\text{C}}^{\text{•}}$ is stabilized by addition electron-donating alkyl group. Also bonds of H with allylic or benzylic C atoms tend to be weaker because of the resonance stabilization in the resulting alkyl radical.

Answer to Problem 25.37P

The H atom which is likely to be abstracted by bromine radical ${\text{Br}}^{\text{•}}$ is marked in box:

Explanation of Solution

The given compound is:

In the above compound there are three types of C atom; methyl, allylic, and aromatic carbon atoms. Hydrogen atoms attached to respective carbons are also methyl H, allylic, and aromatic H atoms.

The H atom attached to the carbon with more number of alkyl group form weaker bond $\text{(i}\text{.e}{\text{. 1}}^{\text{o}}{\text{ >2}}^{\text{o}}{\text{ > 3}}^{\text{o}}\text{)}$ because the resulting ${\text{C}}^{\text{•}}$ is stabilized by addition electron-donating alkyl group. Also bonds of H with allylic or benzylic C atoms tend to be weaker because of the resonance stabilization in the resulting alkyl radical. Thus the H atom attached to secondary allylic C atom is likely abstracted by the radical to form a most stable radical, marked in box:

Conclusion

Which H would most likely be abstracted by bromine radical ${\text{Br}}^{\text{•}}$ is identified on the basis of the types of H atoms present in the given compound to form most stable ${\text{C}}^{\text{•}}$ radical.