Chapter 8, Problem 8.56P

Draw all products, including stereoisomers, in each reaction.

a. c.

b. d.

Interpretation Introduction

(a)

Interpretation: All products formed in the given reaction are to be drawn.

Concept introduction: The one-step bimolecular elimination reaction that favors the removal of a proton by a base from carbon adjacent to the leaving group that results in the formation of a carbocation is termed as $\text{E2}$ elimination reaction. The formation of a double bond takes place simultaneously through the carbocation to form an alkene as the desired product.

The nucleophilic reaction that consists of bimolecular as well as bond-making and bond-breaking steps is termed as ${\text{S}}_{\text{N}}\text{2}$ reactions. This is a one-step reaction in which both nucleophile and substrate participate in the rate determining step.

Stereoisomers have the same molecular formula but they differ in the three-dimensional arrangement of their bonds.

Answer to Problem 8.56P

The products that are formed in the given reaction are $\left(S\right)-\text{hexan}-2-\text{ol}$, $\left(E\right)-\text{hex}-2-\text{ene}$, $\left(Z\right)-\text{hex}-2-\text{ene}$ and $\text{hex}-1-\text{ene}$.

Explanation of Solution

In the given reaction, the secondary alkyl halide is treated with a strong base due to which both substitution and elimination reactions takes place with the given secondary alkyl halide. This reaction results in the formation of three alkenes via $\text{E2}$ elimination reaction and ${\text{S}}_{\text{N}}\text{2}$ substitution reaction gives the product that has inversion of configuration. The reaction of the given alkyl halide is shown as,

Figure 1

Thus, the products formed in the given reaction are $\left(S\right)-\text{hexan}-2-\text{ol}$, $\left(E\right)-\text{hex}-2-\text{ene}$, $\left(Z\right)-\text{hex}-2-\text{ene}$ and $\text{hex}-1-\text{ene}$.

Conclusion

The products that are formed in the given reaction are $\left(S\right)-\text{hexan}-2-\text{ol}$, $\left(E\right)\text{-hex-2-ene}$, $\left(Z\right)\text{-hex-2-ene}$ and $\text{hex-1-ene}$.

Interpretation Introduction

(b)

Interpretation: All products formed in the given reaction are to be drawn.

Concept introduction: The one-step bimolecular elimination reaction that favors the removal of a proton by a base from carbon adjacent to the leaving group that results in the formation of a carbocation is termed as $\text{E2}$ elimination reaction. The formation of a double bond takes place simultaneously through the carbocation to form an alkene as the desired product.

The nucleophilic reaction that consists of bimolecular as well as bond-making and bond-breaking steps is termed as ${\text{S}}_{\text{N}}\text{2}$ reactions. This is a one-step reaction in which both nucleophile and substrate participate in the rate determining step.

Stereoisomers have the same molecular formula but they differ in the three-dimensional arrangement of their bonds.

Answer to Problem 8.56P

The products that are formed in the given reaction are s $\left(S\right)-5-\text{methylhexan}-2-\text{ol}$, $\left(R\right)-5-\text{methylhexan}-2-\text{ol}$, $\left(Z\right)-5-\text{methylhex}-2-\text{ene}$, $5-\text{methylhex}-1-\text{ene}$ and $\left(E\right)-5-\text{methylhex}-2-\text{ene}$.

Explanation of Solution

In the given reaction, the secondary alkyl halide is treated with a strong base due to which both substitution and elimination reactions takes place with the given secondary alkyl halide. This reaction results in the formation of three alkenes via $\text{E2}$ elimination reaction and ${\text{S}}_{\text{N}}\text{2}$ substitution reaction gives two alcoholic products that have inversion of configuration. The reaction of the given alkyl halide is shown as,

Figure 2

Thus, the products formed in the given reaction are s $\left(S\right)-5-\text{methylhexan}-2-\text{ol}$, $\left(R\right)-5-\text{methylhexan}-2-\text{ol}$, $\left(Z\right)-5-\text{methylhex}-2-\text{ene}$, $5-\text{methylhex}-1-\text{ene}$ and $\left(E\right)-5-\text{methylhex}-2-\text{ene}$.

Conclusion

The products that are formed in the given reaction are s $\left(S\right)\text{-5-methylhexan-2-ol}$, $\left(R\right)\text{-5-methylhexan-2-ol}$, $\left(Z\right)\text{-5-methylhex-2-ene}$, $5\text{-methylhex-1-ene}$ and $\left(E\right)\text{-5-methylhex-2-ene}$.

Interpretation Introduction

(c)

Interpretation: All products formed in the given reaction are to be drawn.

Concept introduction: The two-step unimolecular elimination reaction that favors the removal of a HX substituent and the formation of a carbocation intermediate takes place in its first step. In the second step of the reaction, the carbocation forms a double bond. This type of reaction is termed as $\text{E1}$ elimination reaction.

The two-step unimolecular reaction which favors the removal of a HX substituent and the formation of a carbocation intermediate takes place in its first step. Then, in the second step, the carbocation undergoes substitution. This type of reaction is termed as ${\text{S}}_{\text{N}}\text{1}$ reaction.

Stereoisomers have the same molecular formula but they differ in the three-dimensional arrangement of their bonds.

Answer to Problem 8.56P

The products that are formed in the given reaction are $\left(R\right)-\left(1-\text{methoxy}-2,2-\text{dimethylcyclohexyl}\right)\text{benzene}$, $\left(S\right)-\left(1-\text{methoxy}-2,2-\text{dimethylcyclohexyl}\right)\text{benzene}$ and $\left(6,6-\text{dimethylcyclohex}-1-\text{enyl}\right)\text{benzene}$.

Explanation of Solution

The given reaction takes place between the tertiary alkyl halide and methanol that acts as a weak base as well as a weak nucleophile. The given tertiary alkyl halide undergoes elimination reaction via $\text{E1}$ pathway that results in the formation of an alkene and substitution reaction takes place through S_N1 pathway forming two products. The reaction for the same is shown as,

Figure 3

Thus, the products formed in the given reaction are $\left(R\right)-\left(1-\text{methoxy}-2,2-\text{dimethylcyclohexyl}\right)\text{benzene}$, $\left(S\right)-\left(1-\text{methoxy}-2,2-\text{dimethylcyclohexyl}\right)\text{benzene}$ and $\left(6,6-\text{dimethylcyclohex}-1-\text{enyl}\right)\text{benzene}$

Conclusion

The products that are formed in the given reaction are $\left(R\right)\text{-}\left(\text{1-methoxy-2, 2-dimethylcyclohexyl}\right)\text{benzene}$, $\left(S\right)\text{-}\left(\text{1-methoxy-2, 2-dimethylcyclohexyl}\right)\text{benzene}$ and $\left(\text{6, 6-dimethylcyclohex-1-enyl}\right)\text{benzene}$.

Interpretation Introduction

(d)

Interpretation: All products formed in the given reaction are to be drawn.

Concept introduction: The nucleophilic reaction that consists of bimolecular as well as bond-making and bond-breaking steps is termed as ${\text{S}}_{\text{N}}\text{2}$ reactions. This is a one-step reaction in which both nucleophile and substrate participate in the rate determining step.

Stereoisomers have the same molecular formula but they differ in the three-dimensional arrangement of their bonds.

Answer to Problem 8.56P

The product that is formed in the given reaction is shown in Figure 4.

Explanation of Solution

In the given reaction, the secondary alkyl halide is treated with a strong base due to which both substitution and elimination reactions takes place with the given secondary alkyl halide. This reaction results in the formation of three alkenes via $\text{E2}$ elimination reaction and ${\text{S}}_{\text{N}}\text{2}$ substitution reaction gives the product that has inversion of configuration. Elimination reaction is not possible as the halogen atom and the proton are not anti-periplanar to each other. The reaction of the given alkyl halide is shown as,

Figure 4

Conclusion

The product that is formed in the given reaction is shown in Figure 4.