Identify the alcohol reactant needed to produce each of the following compounds as the major product of an alcohol dehydration reaction.
a. Alcohol \overset{H_{2}SO_{4}}{\underset{180°C}{\longrightarrow}} CH_{3}-CH=CH-CH_{3}
b. Alcohol \begin{matrix} \overset{H_{2}SO_{4}}{\underset{180°C}{\longrightarrow}} CH_{2}=CH-\underset{|}{C}H-CH_{3} \\ \quad\quad\quad\quad\quad\quad CH_{3} \end{matrix}
c.Alcohol \begin{matrix} \overset{H_{2}SO_{4}}{\underset{140°C}{\longrightarrow}} CH_{3}-\underset{|}{C}H-CH_{2}-O-CH_{2}-\underset{|}{C}H-CH_{3} \\ \quad\quad \ \ \ \ CH_{3} \ \quad\quad\quad\quad\quad\quad\quad\quad\quad CH_{3} \end{matrix}
The following is a summary of products obtained from alcohol dehydration reactions using H_{2}SO_{4} as a catalyst.
a. Both carbon atoms of the double bond are equivalent to each other. Add an H atom to one carbon atom of the double bond and an OH group to the other carbon atom of the double bond. It does not matter which goes where; you get the same molecule either way.
b. There are two possible parent alcohols: one with an —OH group on carbon 1 and the other with an —OH group on carbon 2.
\begin{matrix}\underset{|}{C}H_{2}-CH_{2}-\underset{|}{C}H-CH_{3} \\ OH \quad\quad\quad\quad\quad CH_{3} \quad\quad\quad \end{matrix} or \begin{matrix}CH_{3}-\underset{|}{C}H-\underset{|}{C}H-CH_{3} \\ OH \quad \ CH_{3}\end{matrix}
Using the reverse of Zaitsev’s rule, we fi nd that the hydrogen atom will go back on the double-bonded carbon that bears the most alkyl groups.
c. This is an ether. The primary alcohol from which the ether was formed will have the same alkyl group present as is in the ether. Thus the alcohol is
\begin{matrix}CH_{3}-\underset{|}{C}H-CH_{2}-OH \\ CH_{3} \quad\quad\quad \end{matrix}