Chapter 8, Problem 8.44P

Interpretation Introduction

(a)

Interpretation:

The detailed mechanisms for the given reaction occurring via $\text{E2}$ mechanism and $\text{E1}$ mechanism are to be drawn. If more than one product is possible by the same mechanism, the complete mechanism for all possible products is to be drawn.

Concept introduction:

The $\text{E1}$ reaction proceeds with the formation of a carbocation intermediate by the elimination of the leaving group in the first step. In the second step, the hydrogen atom, adjacent to the carbocation, is removed by the base to form a double bond. As the carbocation is planar, the elimination product would be a mixture of diastereomers when the double bonded carbon atoms are asymmetrically substituted. The stereochemistry of the $\text{C=C}$ bond is indicated by the $\text{E/Z}$ notation.

In case of $\text{E2}$ elimination, the eliminating groups (leaving group and $\text{H}$) must be anti to each other which produce either $\text{E}$ or $\text{Z}$ isomer. If the substrate has two hydrogen atoms attached to the carbon adjacent to the leaving group, then elimination can occur in two ways with each hydrogen atom, and thus forms a mixture of diastereomer i.e. both $\text{E/Z}$ isomers.

Answer to Problem 8.44P

The $\text{E2}$ mechanism for the given reaction is:

The $\text{E1}$ mechanism for the given reaction is:

Explanation of Solution

The given reaction equation is:

$\text{E2}$ reaction:

In the given reaction, $\text{NaOH}$ is a base. The substrate has $\text{Cl}$ leaving group. There are two hydrogen atoms adjacent to $\text{Cl}$.

The hydrogen atom, indicated with the dash bond, is anti to $\text{Cl}$, and on elimination, gives $\text{E}$ isomer.

If the hydrogen atom, indicated with wedge bond, tends to eliminate, it must orient anti to $\text{Cl}$. Thus, a single bond rotation is needed, and this produces $\text{Z}$ isomer.

$\text{E1}$ reaction:

In $\text{E1}$ reaction, the first step is the formation of a carbocation by detachment of the leaving group $\text{Cl}$. In the second step, the base abstracts the proton from the carbon adjacent to the positively charged carbon and produces $\text{E/Z}$ isomers. The detailed mechanism is shown below:

Conclusion

The products formed for the given reaction from both $\text{E2}$ mechanism and $\text{E1}$ mechanism are determined on the basis of substrate stereochemistry.

Interpretation Introduction

(b)

Interpretation:

The detailed mechanisms for the given reaction occurring via $\text{E2}$ mechanism and $\text{E1}$ mechanism are to be drawn. If more than one product is possible by the same mechanism, the complete mechanism for all possible products is to be drawn.

Concept introduction:

The $\text{E1}$ reaction proceeds with the formation of a carbocation intermediate by elimination of the leaving group in the first step. In the second step, the hydrogen atom, adjacent to the carbocation, is removed by the base to form a double bond. As the carbocation is planar, the elimination product would be a mixture of diastereomers when the double bonded carbon atoms are asymmetrically substituted. The stereochemistry of the $\text{C=C}$ bond is indicated by $\text{E/Z}$ notation.

In case of $\text{E2}$ elimination, the eliminating groups (leaving group and $\text{H}$) must be anti to each other which produce either $\text{E}$ or $\text{Z}$ isomer. If the substrate has two hydrogen atoms attached to the carbon adjacent to the leaving group, then elimination can occur in two ways with each hydrogen atom, and thus forms a mixture of diastereomer i.e. both $\text{E/Z}$ isomers.

Answer to Problem 8.44P

The $\text{E2}$ mechanism for the given reaction is:

The $\text{E1}$ mechanism for the given reaction is:

Explanation of Solution

The given reaction equation is:

$\text{E2}$ reaction:

In the given reaction, $\text{NaOH}$ is a base. The substrate has $\text{Br}$ leaving group. As the $\text{C=C}$ double bond is formed in the cyclic system, only the $\text{Z}$ isomer is produced. The detailed mechanism is:

$\text{E1}$ reaction:

In $\text{E1}$ reaction, the first step is the formation of a carbocation by detachment of the leaving group $\text{Br}$. In the second step, the base abstracts the proton from the carbon adjacent to the positively charged carbon and produces only $\text{Z}$ isomers as the $\text{C=C}$ double bond is formed in the cyclic system. The detailed mechanism is shown below:

Conclusion

The products formed for the given reaction from both $\text{E2}$ mechanism and $\text{E1}$ mechanism are determined on the basis of substrate stereochemistry.

Interpretation Introduction

(c)

Interpretation:

The detailed mechanisms for the given reaction occurring via $\text{E2}$ mechanism and $\text{E1}$ mechanism are to be drawn. If more than one product is possible by the same mechanism, the complete mechanism for all possible products is to be drawn.

Concept introduction:

The $\text{E1}$ reaction proceeds with the formation of a carbocation intermediate by elimination of the leaving group in the first step. In the second step, the hydrogen atom, adjacent to the carbocation, is removed by the base to form a double bond. As the carbocation is planar, the elimination product would be a mixture of diastereomers when the double bonded carbon atoms are asymmetrically substituted. The stereochemistry of the $\text{C=C}$ bond is indicated by $\text{E/Z}$ notation.

In case of $\text{E2}$ elimination, the eliminating groups (leaving group and $\text{H}$) must be anti to each other which produce either $\text{E}$ or $\text{Z}$ isomer.. If the substrate has two hydrogen atoms attached to the carbon adjacent to the leaving group, then elimination can occur in two ways with each hydrogen atom, and thus forms a mixture of diastereomer i.e. both $\text{E/Z}$ isomers. In case of a substituted cyclic structure, a proton can be removed from two different carbon atoms.

Answer to Problem 8.44P

The $\text{E2}$ mechanism for the given reaction is:

The $\text{E1}$ mechanism for the given reaction is:

Explanation of Solution

The given reaction equation is:

$\text{E2}$ reaction:

In the given reaction, ${\text{KOC(CH}}_{\text{3}}{\text{)}}_{\text{3}}$ is a base. The substrate has $\text{I}$ leaving group. There are two products formed because the proton is eliminated from two different carbon atoms. The detailed mechanisms are:

$\text{E1}$ reaction:

In $\text{E1}$ reaction, the first step is the formation of the carbocation by detachment of the leaving group $\text{I}$.

In the second step, the base abstracts the proton from the carbon adjacent to the positively charged carbon. Two products are possible because the proton gets eliminated from two different carbon atoms. The detailed mechanism is shown below:

Conclusion

The products formed for the given reaction from both $\text{E2}$ mechanism and $\text{E1}$ mechanism are determined on the basis of substrate stereochemistry.

Interpretation Introduction

(d)

Interpretation:

The detailed mechanisms for the given reaction occurring via $\text{E2}$ mechanism and $\text{E1}$ mechanism are to be drawn. If more than one product is possible by the same mechanism, the complete mechanism for all possible products is to be drawn.

Concept introduction:

The $\text{E1}$ reaction proceeds with the formation of a carbocation intermediate by elimination of the leaving group in the first step. In the second step, the hydrogen atom, adjacent to the carbocation, is removed by the base to form a double bond. As the carbocation is planar, the elimination product would be a mixture of diastereomers when the double bonded carbon atoms are asymmetrically substituted. The stereochemistry of $\text{C=C}$ bond is indicated by $\text{E/Z}$ notation. There is a possibility of carbocation rearrangement by $\text{1, 2-}$ hydride shift.

In case of $\text{E2}$ elimination, the eliminating groups (leaving group and $\text{H}$) must be anti to each other which produce either $\text{E}$ or $\text{Z}$ isomer.. If the substrate has two hydrogen atoms attached to the carbon adjacent to the leaving group, then elimination can occur in two ways with each hydrogen atom, and thus forms a mixture of diastereomer i.e. both $\text{E/Z}$ isomers. If both the carbon atoms adjacent to the leaving group have hydrogen atoms attached, then a proton can be removed by the base from two different carbon atoms.

Answer to Problem 8.44P

The $\text{E2}$ mechanism for the given reaction is:

The $\text{E1}$ mechanism for the given reaction is:

Explanation of Solution

The given reaction equation is:

$\text{E2}$ reaction:

In the given reaction, ${\text{NaOCH}}_{\text{3}}$ is a base. The substrate has $\text{Cl}$ leaving group. There are two products formed because the proton gets eliminated from two different carbon atoms. The detailed mechanisms are:

$\text{E1}$ reaction:

In $\text{E1}$ reaction, the first step is the formation of a carbocation by detachment of the leaving group $\text{Cl}$.

In the second step, the base abstracts the proton from the carbon adjacent to the positively charged carbon. Two products are possible from the secondary carbocation because the proton is eliminated from two different carbon atoms.

The carbocation formed is a secondary carbocation, which can be rearranged to a more stable tertiary carbocation by $\text{1, 2-}$ hydride shift.

Two products are possible from the tertiary carbocation because the proton is eliminated from two different carbon atoms.

Thus, in all the reactions above, the products are formed by $\text{E1}$ mechanism.

Conclusion

The products formed for the given reaction from both $\text{E2}$ mechanism and $\text{E1}$ mechanism are determined on the basis of substrate stereochemistry.

Interpretation Introduction

(e)

Interpretation:

The detailed mechanisms for the given reaction occurring via $\text{E2}$ mechanism and $\text{E1}$ mechanism are to be drawn. If more than one product is possible by the same mechanism, the complete mechanism for all possible products is to be drawn.

Concept introduction:

The $\text{E1}$ reaction proceeds with the formation of a carbocation intermediate by elimination of the leaving group in the first step. In the second step, the hydrogen atom, adjacent to the carbocation, is removed by base to form a double bond. As the carbocation is planar, the elimination product would be a mixture of diastereomers when the double bonded carbon atoms are asymmetrically substituted. The stereochemistry of $\text{C=C}$ bond is indicated by $\text{E/Z}$ notation.

In case of $\text{E2}$ elimination, the eliminating groups (leaving group and $\text{H}$) must be anti to each other which produce either $\text{E}$ or $\text{Z}$ isomer.. If the substrate has two hydrogen atoms attached to the carbon adjacent to the leaving group, then elimination can occur in two ways with each hydrogen atom, and thus forms a mixture of diastereomer i.e. both $\text{E/Z}$ isomers. If both the carbon atoms adjacent to leaving group have hydrogen atoms attached, then a proton can be removed by the base from two different carbon atoms.

Answer to Problem 8.44P

The $\text{E2}$ mechanism for the given reaction is:

The $\text{E1}$ mechanism for the given reaction is:

Explanation of Solution

The given reaction equation is:

$\text{E2}$ reaction:

In the given reaction, $\text{KOH}$ is a base. The substrate has $\text{Cl}$ leaving group. Both carbon atoms, adjacent to leaving group, have hydrogen atoms anti to $\text{Cl}$ thus proton gets eliminated from two different carbon atoms forming two products. The detailed mechanisms are:

$\text{E1}$ reaction:

In $\text{E1}$ reaction, the first step is the formation of a carbocation by the detachment of the leaving group $\text{Cl}$.

In the second step, the base abstracts the proton from the carbon adjacent to the positively charged carbon. Two products are possible because the proton is eliminated from two different carbon atoms. The detailed mechanism is shown below:

Conclusion

The products formed for the given reaction from both $\text{E2}$ mechanism and $\text{E1}$ mechanism are determined on the basis of substrate stereochemistry.

Interpretation Introduction

(f)

Interpretation:

The detailed mechanisms for the given reaction occurring via $\text{E2}$ mechanism and $\text{E1}$ mechanism are to be drawn. If more than one product is possible by the same mechanism, the complete mechanism for all possible products is to be drawn.

Concept introduction:

The $\text{E1}$ reaction proceeds with the formation of a carbocation intermediate by elimination of the leaving group in the first step. In the second step, the hydrogen atom adjacent to the carbocation is removed by the base to form a double bond. As the carbocation is planar, the elimination product would be a mixture of diastereomers when the double bonded carbon atoms are asymmetrically substituted. The stereochemistry of $\text{C=C}$ bond is indicated by $\text{E/Z}$ notation.

In case of $\text{E2}$ elimination, the eliminating groups (leaving group and $\text{H}$) must be anti to each other which produce either $\text{E}$ or $\text{Z}$ isomer. If the substrate has two hydrogen atoms attached to the carbon adjacent to the leaving group, then elimination can occur in two ways with each hydrogen atom, and thus forms a mixture of diastereomer i.e. both $\text{E/Z}$ isomers. If both the carbon atoms adjacent to leaving group have hydrogen atoms attached, then a proton can be removed, by the base, from two different carbon atoms.

Answer to Problem 8.44P

The $\text{E2}$ mechanism for the given reaction is:

The $\text{E1}$ mechanism for the given reaction is:

Explanation of Solution

The given reaction equation is:

$\text{E2}$ reaction:

In the given reaction, $\text{KOH}$ is a base. The substrate has $\text{Cl}$ leaving group. In the given substrate, only one hydrogen atom is anti to the leaving group $\text{Cl}$, and there is no single bond rotation possible in case of the cyclic system. Thus, a single product is possible. The detailed mechanism is:

$\text{E1}$ reaction:

In $\text{E1}$ reaction, the first step is the formation of a carbocation by detachment of the leaving group $\text{Cl}$.

In the second step, the base abstracts the proton from the carbon adjacent to positively charged carbon. Two products are possible because the proton is eliminated from two different carbon atoms. The detailed mechanism is shown below:

The carbocation formed is a secondary carbocation, which can be rearranged to form more stable tertiary carbocation by $\text{1, 2-}$ hydride shift.

Two products are possible from the tertiary carbocation because the proton IS eliminated from two different carbon atoms.

Conclusion

The products formed for the given reaction from both $\text{E2}$ mechanism and $\text{E1}$ mechanism are determined on the basis of substrate stereochemistry.