Chapter 9, Problem 9.51P

Interpretation Introduction

(a)

Interpretation: The major alkene formed from the dehydration of A with ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is to be stated.

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 secondary and tertiary alcohols occurs in three steps and it is $\text{E1}$ reaction. The first step is protonation of $\text{OH}$ to form good leaving group. The second step is heterolytic cleavage of $\text{C}-\text{O}$ bond to form a carbocation (carbocation rearrangement is possible). The third step is abstraction of proton by a base $\left({\text{HSO}}_{\text{4}}{}^{-}\text{or}{\text{H}}_{\text{2}}\text{O}\right)$ from a carbon adjacent to the carbocation to result the new $\text{π}$ bond.

Interpretation Introduction

(b)

Interpretation: The major alkene formed from the dehydration of A with ${\text{POCl}}_3,\text{and}\text{pyridine}$, and an explanation that corresponds to the different major products in these reactions are to be stated.

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. This dehydration reaction involves two steps. The first step is protonation of $\text{OH}$ to form good leaving group. The second step is loss of the leaving group and removal of the $\text{β}$ proton.