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Question 14.4: Draw the structural formula(s) for the product(s) formed by ......

Draw the structural formula(s) for the product(s) formed by oxidation of the following alcohols with a mild oxidizing agent. If no reaction occurs, write “no reaction.”

a. \begin{matrix} CH_{3}-CH_{2}-CH_{2}-\underset{|}{C}H-CH_{3} \\ \quad\quad\quad\quad\quad\quad \ \ OH \end{matrix}             b. \begin{matrix} CH_{3}-\underset{|}{C}H-CH_{2}-OH \\ CH_{3} \quad\quad\quad \end{matrix}

c. \begin{matrix} CH_{3}-CH_{2}-\underset{|}{C}H-OH \\ \quad\quad\quad\quad CH_{3} \end{matrix}             d. 

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a. The oxidation product will be a ketone, as this is a 2° alcohol.

\begin{matrix} \quad\quad\quad\quad\quad\quad \ \ \ OH \ \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad O \\ CH_{3}-CH_{2}-CH_{2}-\overset{|}{C}H-CH_{3} \longrightarrow CH_{3}-CH_{2}-CH_{2}-\overset{||}{C}-CH_{3} \end{matrix}

b. A 1° alcohol undergoes oxidation first to an aldehyde and then to a carboxylic acid.

\begin{matrix} \ \ \ \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad O \ \ \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad O \\ CH_{3}-\underset{|}{C} H-CH_{2}-OH \longrightarrow CH_{3}-\underset{|}{C}H-\overset{||}{C}-H \longrightarrow CH_{3}-\underset{|}{C} H-\overset{||}{C}-OH \\ CH_{3} \ \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad CH_{3} \ \quad\quad\quad\quad\quad\quad\quad\quad\quad CH_{3} \quad \end{matrix}

c. A ketone is the product from the oxidation of a 2° alcohol.

\begin{matrix} \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \ \ \ O \\ CH_{3}-CH_{2}-\underset{|}{C}H-OH \longrightarrow CH_{3}-CH_{2}-\overset{||}{C}-CH_{3} \\ CH_{3} \quad\quad\quad\quad\quad\quad\quad\quad\quad \end{matrix}

d. This cyclic alcohol is a tertiary alcohol. The hydroxyl-bearing carbon atom is attached to two ring carbon atoms and a methyl group. Tertiary alcohols do not undergo oxidation with mild oxidizing agents. Therefore, “no reaction.”

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