Alcohols and Esters: Question 5

Worked solution

Classifying butan-2-ol

The carbon bearing the $-\text{OH}$ group in butan-2-ol, $\text{CH}_3\text{CH(OH)CH}_2\text{CH}_3$, is attached to two other carbon groups (a methyl group and an ethyl group), so this is a secondary alcohol.

Oxidation of a secondary alcohol

Acidified potassium dichromate(VI) oxidises a secondary alcohol by removing a hydrogen atom from the carbon bearing the $-\text{OH}$ group (together with the $-\text{OH}$ hydrogen), forming a carbon–oxygen double bond. Because that carbon has only one remaining hydrogen to lose, oxidation stops at the ketone stage. There is no further hydrogen available to allow oxidation to continue to a carboxylic acid.

$\text{CH}_3\text{CH(OH)CH}_2\text{CH}_3 \rightarrow \text{CH}_3\text{COCH}_2\text{CH}_3$

As the orange dichromate(VI) ion, $\text{Cr}_2\text{O}_7^{2-}$, is reduced to the green chromium(III) ion, $\text{Cr}^{3+}$, the solution changes colour from orange to green, and the product is butan-2-one.

Why the other options are wrong

• B describes the outcome for a primary alcohol oxidised under reflux with excess oxidant, which continues past the aldehyde all the way to a carboxylic acid; a secondary alcohol cannot reach this stage.
• C would only be correct for a tertiary alcohol, which has no hydrogen on the $-\text{OH}$-bearing carbon at all.
• D gives an aldehyde, which is the partial oxidation product of a primary alcohol (e.g. butan-1-ol), not of a secondary alcohol.

Final answer

A. The solution turns orange to green, and butan-2-one, $\text{CH}_3\text{COCH}_2\text{CH}_3$, is formed.