Chapter 9, Problem 9.64P

Interpretation Introduction

(a)

Interpretation: The product formed by the treatment of ethylene oxide with the given reagent 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 9.64P

The product formed by the treatment of ethylene oxide with the given reagent is,

Explanation of Solution

The given reagent is $\text{HBr}$. Hydrogen bromide $\left(\text{HBr}\right)$ is a strong 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 formed by the treatment of ethylene oxide with the given reagent is,

Figure 1

Conclusion

The product formed by the treatment of ethylene oxide with the given reagent is drawn in Figure 1.

Interpretation Introduction

(b)

Interpretation: The product formed by the treatment of ethylene oxide with the given reagent 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 9.64P

The product formed by the treatment of ethylene oxide with the given reagent is,

Explanation of Solution

The given reagent is ${\text{H}}_{\text{2}}\text{O}\left({\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\right)$. Sulfuric acid $\left({\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\right)$ is a strong 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 formed by the treatment of ethylene oxide with the given reagent is,

Figure 2

Conclusion

The product formed by the treatment of ethylene oxide with the given reagent is drawn in Figure 2.

Interpretation Introduction

(c)

Interpretation: The product formed by the treatment of ethylene oxide with the given reagent 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 9.64P

The product formed by the treatment of ethylene oxide with the given reagent is,

Explanation of Solution

The given reagent is $\left[1\right]{\text{CH}}_3{\text{CH}}_{\text{2}}{\text{O}}^{-};\left[2\right]{\text{H}}_{\text{2}}\text{O}$. 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 formed by the treatment of ethylene oxide with the given reagent is,

Figure 3

Conclusion

The product formed by the treatment of ethylene oxide with the given reagent is drawn in Figure 3.

Interpretation Introduction

(d)

Interpretation: The product formed by the treatment of ethylene oxide with the given reagent 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 9.64P

The product formed by the treatment of ethylene oxide with the given reagent is,

Explanation of Solution

The given reagent is $\left[1\right]\text{HC}\equiv {\text{C}}^{-};\left[2\right]{\text{H}}_{\text{2}}\text{O}$. Acetylide ion $\left(\text{HC}\equiv {\text{C}}^{-}\right)$ 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 formed by the treatment of ethylene oxide with the given reagent is,

Figure 4

Conclusion

The product formed by the treatment of ethylene oxide with the given reagent is drawn in Figure 4.

Interpretation Introduction

(e)

Interpretation: The product formed by the treatment of ethylene oxide with the given reagent 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 9.64P

The product formed by the treatment of ethylene oxide with the given reagent is,

Explanation of Solution

The given reagent is $\left[1\right]{}^{-}\text{OH};\left[2\right]{\text{H}}_{\text{2}}\text{O}$. Hydroxyl ion $\left({}^{-}\text{OH}\right)$ 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 formed by the treatment of ethylene oxide with the given reagent is,

Figure 5

Conclusion

The product formed by the treatment of ethylene oxide with the given reagent is drawn in Figure 5.

Interpretation Introduction

(f)

Interpretation: The product formed by the treatment of ethylene oxide with the given reagent 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 9.64P

The product formed by the treatment of ethylene oxide with the given reagent is,

Explanation of Solution

The given reagent is $\left[1\right]{\text{CH}}_3{\text{S}}^{-};\left[2\right]{\text{H}}_{\text{2}}\text{O}$. Methyl sulfide ion $\left({\text{CH}}_3{\text{S}}^{-}\right)$ 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 formed by the treatment of ethylene oxide with the given reagent is,

Figure 6

Conclusion

The product formed by the treatment of ethylene oxide with the given reagent is drawn in Figure 6.