Chapter 9, Problem 9.65P

Draw the products of each reaction.

a. c.

b. d.

Interpretation Introduction

(a)

Interpretation: The product of the given reaction is to be drawn.

Concept introduction: The opening of an epoxide/ethylene oxide ring is regioselective either it takes place with a strong nucleophile $\left({\text{:Nu}}^{-}\right)$ or an acid $\left(\text{HZ}\right)$. The mechanism of the reaction with ${\text{:Nu}}^{-}$ is ${\text{S}}_{\text{N}}\text{2}$, and the mechanism of the reaction with $\text{HZ}$ is in between ${\text{S}}_{\text{N}}\text{2}\text{and}{\text{S}}_{\text{N}}1$.

Answer to Problem 9.65P

The product of the given reaction is,

Explanation of Solution

The given reaction involves treatment of unsymmetrical epoxide with ${\text{CH}}_3{\text{CH}}_{\text{2}}\text{OH;}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$, an acid.

The opening of an epoxide/ethylene oxide ring is regioselective either it takes place with a strong nucleophile $\left({\text{:Nu}}^{-}\right)$ or an acid $\left(\text{HZ}\right)$. The mechanism of the reaction with ${\text{:Nu}}^{-}$ is ${\text{S}}_{\text{N}}\text{2}$, and nucleophilic attack takes place at less substituted carbon atom. The mechanism of the reaction with $\text{HZ}$ is in between ${\text{S}}_{\text{N}}\text{2}\text{and}{\text{S}}_{\text{N}}1$, and attack of ${\text{Z}}^{-}$ takes place at highly substituted carbon atom.

Thus, the product of the given reaction is,

Figure 1

Conclusion

The product of the given reaction is drawn in Figure 1.

Interpretation Introduction

(b)

Interpretation: The product of the given reaction is to be drawn.

Concept introduction: The opening of an epoxide/ethylene oxide ring is regioselective either it takes place with a strong nucleophile $\left({\text{:Nu}}^{-}\right)$ or an acid $\left(\text{HZ}\right)$. The mechanism of the reaction with ${\text{:Nu}}^{-}$ is ${\text{S}}_{\text{N}}\text{2}$, and the mechanism of the reaction with $\text{HZ}$ is in between ${\text{S}}_{\text{N}}\text{2}\text{and}{\text{S}}_{\text{N}}1$.

Answer to Problem 9.65P

The product of the given reaction is,

Explanation of Solution

The given reaction involves treatment of unsymmetrical epoxide with $\left[1\right]{\text{CH}}_3{\text{CH}}_{\text{2}}{\text{O}}^{-}{\text{Na}}^{+};\left[2\right]{\text{H}}_{\text{2}}\text{O}$. Ethoxide ion $\left({\text{CH}}_3{\text{CH}}_{\text{2}}{\text{O}}^{-}\right)$ is a strong nucleophile.

The opening of an epoxide/ethylene oxide ring is regioselective either it takes place with a strong nucleophile $\left({\text{:Nu}}^{-}\right)$ or an acid $\left(\text{HZ}\right)$. The mechanism of the reaction with ${\text{:Nu}}^{-}$ is ${\text{S}}_{\text{N}}\text{2}$, and nucleophilic attack takes place at less substituted carbon atom. The mechanism of the reaction with $\text{HZ}$ is in between ${\text{S}}_{\text{N}}\text{2}\text{and}{\text{S}}_{\text{N}}1$, and attack of ${\text{Z}}^{-}$ takes place at highly substituted carbon atom.

Thus, the product of the given reaction is,

Figure 2

Conclusion

The product of the given reaction is drawn in Figure 2.

Interpretation Introduction

(c)

Interpretation: The product of the given reaction is to be drawn.

Concept introduction: The opening of an epoxide/ethylene oxide ring is regioselective either it takes place with a strong nucleophile $\left({\text{:Nu}}^{-}\right)$ or an acid $\left(\text{HZ}\right)$. The mechanism of the reaction with ${\text{:Nu}}^{-}$ is ${\text{S}}_{\text{N}}\text{2}$, and the mechanism of the reaction with $\text{HZ}$ is in between ${\text{S}}_{\text{N}}\text{2}\text{and}{\text{S}}_{\text{N}}1$.

Answer to Problem 9.65P

The product of the given reaction is,

Explanation of Solution

The given reaction involves treatment of unsymmetrical epoxide with $\text{HBr}$, an acid.

The opening of an epoxide/ethylene oxide ring is regioselective either it takes place with a strong nucleophile $\left({\text{:Nu}}^{-}\right)$ or an acid $\left(\text{HZ}\right)$. The mechanism of the reaction with ${\text{:Nu}}^{-}$ is ${\text{S}}_{\text{N}}\text{2}$, and nucleophilic attack takes place at less substituted carbon atom. The mechanism of the reaction with $\text{HZ}$ is in between ${\text{S}}_{\text{N}}\text{2}\text{and}{\text{S}}_{\text{N}}1$, and attack of ${\text{Z}}^{-}$ takes place at highly substituted carbon atom.

Thus, the product of the given reaction is,

Figure 3

Conclusion

The product of the given reaction is drawn in Figure 3.

Interpretation Introduction

(d)

Interpretation: The product of the given reaction is to be drawn.

Concept introduction: The opening of an epoxide/ethylene oxide ring is regioselective either it takes place with a strong nucleophile $\left({\text{:Nu}}^{-}\right)$ or an acid $\left(\text{HZ}\right)$. The mechanism of the reaction with ${\text{:Nu}}^{-}$ is ${\text{S}}_{\text{N}}\text{2}$, and the mechanism of the reaction with $\text{HZ}$ is in between ${\text{S}}_{\text{N}}\text{2}\text{and}{\text{S}}_{\text{N}}1$.

Answer to Problem 9.65P

The product of the given reaction is,

Explanation of Solution

The given reaction involves treatment of unsymmetrical epoxide with $\left[1\right]\text{NaCN};\left[2\right]{\text{H}}_{\text{2}}\text{O}$. Cyanide ion $\left({}^{-}\text{CN}\right)$ is a strong nucleophile.

The opening of an epoxide/ethylene oxide ring is regioselective either it takes place with a strong nucleophile $\left({\text{:Nu}}^{-}\right)$ or an acid $\left(\text{HZ}\right)$. The mechanism of the reaction with ${\text{:Nu}}^{-}$ is ${\text{S}}_{\text{N}}\text{2}$, and nucleophilic attack takes place at less substituted carbon atom. The mechanism of the reaction with $\text{HZ}$ is in between ${\text{S}}_{\text{N}}\text{2}\text{and}{\text{S}}_{\text{N}}1$, and attack of ${\text{Z}}^{-}$ takes place at highly substituted carbon atom.

Thus, the product of the given reaction is,

Figure 4

Conclusion

The product of the given reaction is drawn in Figure 4.