Chapter 9, Problem 9.71P

Draw the products of each reaction, and indicate the stereochemistry where appropriate.

a. f

b g.

c. h.

d. i.

e. j.

Interpretation Introduction

(a)

Interpretation: The product of the given reaction is to be drawn including stereochemistry if appropriate.

Concept introduction: Alkyl tosylates undergo elimination reaction when they are allowed to react with strong nucleophilic base. The mechanism of the elimination reaction is $\text{E2}$.

Answer to Problem 9.71P

The product of the given reaction is,

Figure 2

Explanation of Solution

The given reaction involves treatment of alkyl tosylate with $\text{KOC}{\left({\text{CH}}_{\text{3}}\right)}_{\text{3}}$, a strong nucleophilic base.

Alkyl tosylates undergo elimination reaction when they are allowed to react with strong nucleophilic base. The mechanism of the reaction is $\text{E2}$, that is concerted and occurs in single step. The corresponding reaction is shown below.

Figure 1

Thus, the product of the given reaction is,

Figure 2

Conclusion

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

Interpretation Introduction

(b)

Interpretation: The product of the given reaction is to be drawn including stereochemistry if appropriate.

Concept introduction: The reaction of alcohols with halogen acids $\left(\text{HX}\right)$ is a general method to obtain ${\text{1}}^{\text{ο}}\text{,}{\text{2}}^{\text{ο}}\text{or}{\text{3}}^{\text{ο}}$ alkyl halides. This method is preferred due to the formation of ${\text{H}}_{\text{2}}\text{O}$ as leaving group. The more substituted alcohols generally react rapidly with halogen acids.

Answer to Problem 9.71P

The product of the given reaction is,

Figure 4

Explanation of Solution

The given reaction involves treatment of an alcohol with $\text{HBr}$.

The reaction of alcohols with halogen acids $\left(\text{HX}\right)$ is a general method to obtain ${\text{1}}^{\text{ο}}\text{,}{\text{2}}^{\text{ο}}\text{or}{\text{3}}^{\text{ο}}$ alkyl halides. This method is preferred due to the formation of ${\text{H}}_{\text{2}}\text{O}$ as leaving group. The more substituted alcohols generally react rapidly with halogen acids.

The corresponding reaction is shown below.

Figure 3

Thus, the product of the given reaction is,

Figure 4

Conclusion

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

Interpretation Introduction

(c)

Interpretation: The product of the given reaction is to be drawn including stereochemistry if appropriate.

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 9.71P

The product of the given reaction is,

Explanation of Solution

The given reaction involves treatment of epoxide ring with cyanide ion.

In the given reaction, $\left(\text{2S, 3S}\right)\text{-2, 3-diethyloxirane}$ reacts with $\text{CN}$ ion in the presence of water to form $\left(\text{2S, 3S}\right)\text{-2-ethyl-3-hydroxypentanenitrile}$ as the final product.

The corresponding reaction is shown below.

Figure 5

Thus, the product of the given reaction is,

Figure 6

Conclusion

The product of the given reaction is drawn in Figure 6.

Interpretation Introduction

(d)

Interpretation: The product of the given reaction is to be drawn including stereochemistry if appropriate.

Concept introduction: The ${}^{-}\text{SH}$ is a good nucleophile. Thiols are synthesized by the nucleophilic attack of ${}^{-}\text{SH}$ on alkyl halide or alkyl tosylates. The mechanism of the reaction is ${\text{S}}_{\text{N}}\text{2}$ that takes place in single step. Due to ${\text{S}}_{\text{N}}\text{2}$ mechanism the product of the reaction is obtained with inversion of configuration.

Answer to Problem 9.71P

The product of the given reaction is,

Figure 8

Explanation of Solution

The given reaction involves treatment of alkyl tosylate with $\text{KSH}$, which contains ${}^{-}\text{SH}$ nucleophile.

The ${}^{-}\text{SH}$ is a good nucleophile. Thiols are synthesized by the nucleophilic attack of ${}^{-}\text{SH}$ on alkyl halide or alkyl tosylates. The mechanism of the reaction is ${\text{S}}_{\text{N}}\text{2}$ that takes place in single step. Due to ${\text{S}}_{\text{N}}\text{2}$ mechanism the product of the reaction is obtained with inversion of configuration. The corresponding reaction is shown below.

Figure 7

Thus, the product of the given reaction is,

Figure 8

Conclusion

The product of the given reaction is drawn in Figure 8.

Interpretation Introduction

(e)

Interpretation: The product of the given reaction is to be drawn including stereochemistry if appropriate.

Concept introduction: Alkyl bromides are obtained by the reaction of ${\text{1}}^{\text{ο}}\text{or}{\text{2}}^{\text{ο}}$ alcohols with phosphorous tribromide $\left({\text{PBr}}_{\text{3}}\right)$. The reaction yields ${\text{HOPBr}}_{\text{2}}$ as by-product. The mechanism of the reaction is ${\text{S}}_{\text{N}}2$ that results an inversion of configuration product. However, the further strong nucleophilic attack results the inversion in configuration product through ${\text{S}}_{\text{N}}\text{2}$ mechanism.

Answer to Problem 9.71P

The product of the given reaction is,

Figure 10

Explanation of Solution

The given reaction involves treatment of secondary alcohol with ${\text{PBr}}_{\text{3}}$.

Alkyl bromides are obtained by the reaction of ${\text{1}}^{\text{ο}}\text{or}{\text{2}}^{\text{ο}}$ alcohols with phosphorous tribromide $\left({\text{PBr}}_{\text{3}}\right)$. The reaction yields ${\text{HOPBr}}_{\text{2}}$ as by-product. The mechanism of the reaction is ${\text{S}}_{\text{N}}2$ that results an inversion of configuration product. The corresponding reaction is shown below.

Figure 9

Thus, the product of the given reaction is,

Figure 10

Conclusion

The product of the given reaction is drawn in Figure 10.

Interpretation Introduction

(f)

Interpretation: The product of the given reaction is to be drawn including stereochemistry if appropriate.

Concept introduction: Alcohols are converted into alkyl tosylates by treatment with $p\text{-toluenesulfonyl}\text{chloride}\left(\text{TsCl}\right)$ in the presence of pyridine. The overall reaction converts ${}^{-}\text{OH}$ (a poor leaving group) into ${}^{-}\text{OTs}$ (a good leaving group), and stereochemistry of the reaction is retained. However, the further reaction with strong nucleophile leads to the formation of an ${\text{S}}_{\text{N}}\text{2}$ product with inversion of configuration.

Answer to Problem 9.71P

The product of the given reaction is,

Figure 12

Explanation of Solution

The given reaction involves treatment of an alcohol with $\text{TsCl}$, pyridine, followed by nucleophilic attack of ${\text{CH}}_{\text{3}}{\text{CO}}_{\text{2}}{}^{-}$.

Alcohols are converted into alkyl tosylates by treatment with $p\text{-toluenesulfonyl}\text{chloride}\left(\text{TsCl}\right)$ in the presence of pyridine. The overall reaction converts ${}^{-}\text{OH}$ (a poor leaving group) into ${}^{-}\text{OTs}$ (a good leaving group), and stereochemistry of the reaction is retained. However, the further reaction with strong nucleophile leads to the formation of an ${\text{S}}_{\text{N}}\text{2}$ product with inversion of configuration. The corresponding reaction is shown below.

Figure 11

Thus, the product of the given reaction is,

Figure 12

Conclusion

The product of the given reaction is drawn in Figure 12.

Interpretation Introduction

(g)

Interpretation: The product of the given reaction is to be drawn including stereochemistry if appropriate.

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 9.71P

The product of the given reaction is,

Figure 14

Explanation of Solution

The give reaction involves treatment of an epoxide with halogen acid $\left(\text{HBr}\right)$.

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 ${\text{S}}_{\text{N}}\text{2}\text{and}{\text{S}}_{\text{N}}1$. With halogen acids epoxides yield corresponding halohydrin.

The corresponding reaction is shown below.

Figure 13

Thus, the product of the given reaction is,

Figure 14

Conclusion

The product of the given reaction is drawn in Figure 14.

Interpretation Introduction

(h)

Interpretation: The product of the given reaction is to be drawn including stereochemistry if appropriate.

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 9.71P

The product of the given reaction is,

Figure 16

Explanation of Solution

The give reaction involves treatment of an epoxide with ${\text{NaOCH}}_{\text{3}}\text{,}{\text{H}}_{\text{2}}\text{O}$. Sodium methoxide contains a good nucleophile that is ${}^{-}{\text{OCH}}_{\text{3}}$.

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 ${\text{S}}_{\text{N}}\text{2}\text{and}{\text{S}}_{\text{N}}1$.

The corresponding reaction is shown below.

Figure 15

Thus, the product of the given reaction is,

Figure 16

Conclusion

The product of the given reaction is drawn in Figure 16.

Interpretation Introduction

(i)

Interpretation: The product of the given reaction is to be drawn including stereochemistry if appropriate.

Concept introduction: Ethers are the most common organic products of nucleophilic substitution reaction. They are prepared from alkyl halides and strong nucleophiles. The reaction proceeds through ${\text{S}}_{\text{N}}\text{2}$ mechanism. The highest yields of products are obtained from methyl (unhindered) and primary alkyl halides. The synthesis of ether by this route is called Williamson ether synthesis.

Answer to Problem 9.71P

The product of the given reaction is,

Figure 18

Explanation of Solution

The given reaction involves treatment of a ${\text{2}}^{\text{ο}}$ alcohol with $\text{NaH}$, then with ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{I}$.

The alkoxide salts are prepared from alcohols through the Bronsted-Lowry acid-base reaction. In this reaction, $\text{NaH}$ or ${\text{NaNH}}_{\text{2}}$ are especially utilized, due to the by-product of the reaction $\left({\text{H}}_{\text{2}}\text{or}{\text{NH}}_{\text{3}}\right)$. Both ${\text{H}}_{\text{2}}$ and ${\text{NH}}_{\text{3}}$ are gases that bubbles out rapidly from reaction mixture.

The obtained alkoxide from this reaction contains ${\text{RO}}^{-}$ as a strong nucleophile. It reacts with ethyl chloride to yield corresponding ether. The mechanism of the reaction is ${\text{S}}_{\text{N}}\text{2}$, which causes inversion of configuration product.

The corresponding reaction is shown below.

Figure 17

Thus, the product of the given reaction is,

Figure 18

Conclusion

The product of the given reaction is drawn in Figure 18.

Interpretation Introduction

(j)

Interpretation: The product of the given reaction is to be drawn including stereochemistry if appropriate.

Concept introduction: Ethers react with strong acids, (only $\text{HBr}$ and $\text{HI}$) and yield two alkyl halides as products.

$\begin{array}{l}\text{R}-\text{O}-{\text{R}}^{\text{'}}\underset{2\text{equivalents}}{\overset{\text{HX}}{\to }}\text{R}-\text{X}+{\text{R}}^{\text{'}}-\text{X}+{\text{H}}_{\text{2}}\text{O}\\ \text{X}=\text{Br}\text{or}\text{I}\end{array}$

In this reaction, both $\text{C}-\text{O}$ bonds of ether are cleaved by two successive nucleophilic substitution reactions.

Answer to Problem 9.71P

The product of the given reaction is,

Figure 20

Explanation of Solution

The given reaction involves treatment of an ether with two equivalents of $\text{HI}$.

Ethers react with strong acids, (only $\text{HBr}$ and $\text{HI}$) and yield two alkyl halides as products.

$\begin{array}{l}\text{R}-\text{O}-{\text{R}}^{\text{'}}\underset{2\text{equivalents}}{\overset{\text{HX}}{\to }}\text{R}-\text{X}+{\text{R}}^{\text{'}}-\text{X}+{\text{H}}_{\text{2}}\text{O}\\ \text{X}=\text{Br}\text{or}\text{I}\end{array}$

In this reaction, both $\text{C}-\text{O}$ bonds of ether are cleaved by two successive nucleophilic substitution reactions.

The corresponding reaction is shown below.

Figure 19

Thus, the product of the given reaction is,

Figure 20

Conclusion

The product of the given reaction is drawn in Figure 20.