2019 H2 Chem P1 Q22 — Hydroxyl Compounds Solution

The answer is C.

Hot acidified \(\text{K}_2\text{Cr}_2\text{O}_7\) in excess under reflux oxidises each OH group according to its type (see figure):
• the ring -OH on a carbon that carries one H is a secondary alcohol, oxidised to a ketone (C=O);
• the \(-\text{CH}_2\text{OH}\) group is a primary alcohol; under reflux with excess oxidant it is oxidised all the way to a carboxylic acid, \(-\text{CO}_2\text{H}\);
• the -OH on the ring carbon that also carries \(\text{CH}_3\) is a tertiary alcohol (no H on that carbon), so it is not oxidised.
The product therefore has C=O, \(-\text{CO}_2\text{H}\) and the unchanged tertiary C(OH)\(\text{CH}_3\): structure C.

• A is wrong: it stops the primary alcohol at the aldehyde, -CHO. An aldehyde is obtained only if it is distilled off as it forms; under reflux with excess dichromate it is oxidised further to \(-\text{CO}_2\text{H}\).
• B is wrong: it leaves the secondary alcohol unoxidised; secondary alcohols are oxidised to ketones by acidified dichromate.
• D is wrong: it shows the tertiary alcohol carbon converted into \(-\text{CO}_2\text{H}\) with loss of the methyl group. A tertiary alcohol has no H on the C-OH carbon and is not oxidised by acidified dichromate.