Chapter 9, Problem 9.27P

Draw the product formed when ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{CHOH}$ is treated with each reagent.

a. ${\text{SOCl}}_2,\text{pyridine}$ c. ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ e. ${\text{PBr}}_3$, then $\text{NaCN}$

b. $\text{TsCl},\text{pyridine}$ d. $\text{HBr}$ f. ${\text{POCl}}_3,\text{pyridine}$

Interpretation Introduction

(a)

Interpretation: The product formed by the treatment of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{CHOH}$ with the given reagent is to be drawn.

Concept introduction: Alkyl chlorides are obtained by the reaction of ${\text{1}}^{\text{ο}}\text{or}{\text{2}}^{\text{ο}}$ alcohols with thionyl chloride $\left({\text{SOCl}}_2\right)$, and pyridine. The reaction yields ${\text{SO}}_{\text{2}}$, and $\text{HCl}$ as by-products. The mechanism of the reaction involves two steps. The first step is conversion of the $\text{OH}$ group into a good leaving group. The second step is nucleophilic attack by ${\text{Cl}}^{-}$ through an ${\text{S}}_{\text{N}}2$ reaction.

Answer to Problem 9.27P

The product formed by the treatment of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{CHOH}$ with the given reagent is,

Explanation of Solution

The given reagent is ${\text{SOCl}}_2,\text{pyridine}$.

Alkyl chlorides are obtained by the reaction of ${\text{1}}^{\text{ο}}\text{or}{\text{2}}^{\text{ο}}$ alcohols with thionyl chloride $\left({\text{SOCl}}_2\right)$, and pyridine. The reaction yields ${\text{SO}}_{\text{2}}$, and $\text{HCl}$ as by-products. The mechanism of the reaction involves two steps. The first step is conversion of the $\text{OH}$ group into a good leaving group. The second step is nucleophilic attack by ${\text{Cl}}^{-}$ through an ${\text{S}}_{\text{N}}2$ reaction.

Thus, the product formed by the treatment of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{CHOH}$ with the given reagent is,

Figure 1

Conclusion

(a) The product formed by the treatment of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{CHOH}$ with the given reagent is drawn in Figure 1.

Interpretation Introduction

(b)

Interpretation: The product formed by the treatment of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{CHOH}$ with the given reagent is to be drawn.

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).

Answer to Problem 9.27P

The product formed by the treatment of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{CHOH}$ with the given reagent is,

Explanation of Solution

The given reagent is $\text{TsCl},\text{pyridine}$.

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).

Thus, the product formed by the treatment of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{CHOH}$ with the given reagent is,

Figure 2

Conclusion

The product formed by the treatment of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{CHOH}$ with the given reagent is drawn in Figure 2.

Interpretation Introduction

(c)

Interpretation: The product formed by the treatment of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{CHOH}$ with the given reagent is to be drawn.

Concept introduction: Alcohols undergo dehydration reaction in the presence of strong acids like ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ or $\text{p-toluenesulfonic}\text{acid}\left(\text{TsOH}\right)$ and yield alkenes. The dehydration of primary alcohols occurs in two steps and it is $\text{E2}$ reaction.

Answer to Problem 9.27P

The product formed by the treatment of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{CHOH}$ with the given reagent is,

Explanation of Solution

The given reagent is ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$.

Alcohols undergo dehydration reaction in the presence of strong acids like ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ or $\text{p-toluenesulfonic}\text{acid}\left(\text{TsOH}\right)$ and yield alkenes. The dehydration of primary alcohols occurs in two steps and it is $\text{E2}$ reaction.

Thus, the product formed by the treatment of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{CHOH}$ with the given reagent is,

Figure 3

Conclusion

The product formed by the treatment of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{CHOH}$ with the given reagent is drawn in Figure 3.

Interpretation Introduction

(d)

Interpretation: The product formed by the treatment of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{CHOH}$ with the given reagent is to be drawn.

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 mechanism of the reaction is ${\text{S}}_{\text{N}}\text{1}$ for ${\text{2}}^{\text{ο}}\text{or}{\text{3}}^{\text{ο}}$ alcohols, and ${\text{S}}_{\text{N}}2$ for ${\text{1}}^{\text{ο}}$ alcohols.

Answer to Problem 9.27P

The product formed by the treatment of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{CHOH}$ with the given reagent is,

Explanation of Solution

The given reagent is $\text{HBr}$.

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 mechanism of the reaction is ${\text{S}}_{\text{N}}\text{1}$ for ${\text{2}}^{\text{ο}}\text{or}{\text{3}}^{\text{ο}}$ alcohols, and ${\text{S}}_{\text{N}}2$ for ${\text{1}}^{\text{ο}}$ alcohols.

Thus, the product formed by the treatment of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{CHOH}$ with the given reagent is,

Figure 4

Conclusion

The product formed by the treatment of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{CHOH}$ with the given reagent is drawn in Figure 4.

Interpretation Introduction

(e)

Interpretation: The product formed by the treatment of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{CHOH}$ with the given reagent is to be drawn.

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. Alkyl bromides react with $\text{NaCN}$, and yield alkyl nitriles. The mechanisms of both reactions are ${\text{S}}_{\text{N}}2$.

Answer to Problem 9.27P

The product formed by the treatment of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{CHOH}$ with the given reagent is,

Explanation of Solution

The given reagents are ${\text{PBr}}_3$, and $\text{NaCN}$.

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. Alkyl bromides react with $\text{NaCN}$, and yield alkyl nitriles. The mechanisms of both reactions are ${\text{S}}_{\text{N}}2$.

Thus, the product formed by the treatment of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{CHOH}$ with the given reagent is,

Figure 5

Conclusion

The product formed by the treatment of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{CHOH}$ with the given reagent is drawn in Figure 5.

Interpretation Introduction

(f)

Interpretation: The product formed by the treatment of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{CHOH}$ with the given reagent is to be drawn.

Concept introduction: Alcohols undergo dehydration reaction in the presence of ${\text{POCl}}_{\text{3}}\text{,}\text{pyridine}$ and yield alkenes. The mechanism of the reaction is $\text{E2}$ and it does not involve carbocation rearrangement.

Answer to Problem 9.27P

The product formed by the treatment of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{CHOH}$ with the given reagent is,

Explanation of Solution

The given reagent is ${\text{POCl}}_3,\text{pyridine}$.

Alcohols undergo dehydration reaction in the presence of ${\text{POCl}}_{\text{3}}\text{,}\text{pyridine}$ and yield alkenes. The mechanism of the reaction is $\text{E2}$ and it does not involve carbocation rearrangement.

Thus, the product formed by the treatment of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{CHOH}$ with the given reagent is,

Figure 6

Conclusion

The product formed by the treatment of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{CHOH}$ with the given reagent is drawn in Figure 6.