Exam Style Question - 1

QuestionPropan-1-ol, CH₃CH₂CH₂OH, can be oxidised to two different organic products.(a) Give the reagents and conditions needed to make propanal, and explain how the method stops further oxidation.(b) Give the conditions needed to make propanoic acid, and write an equation using [O].(c) State the colour change seen in both reactions and explain it.(d) Name the product formed when propan-2-ol is oxidised under the same conditions.Answer(a) Acidified potassium dichromate(VI) (potassium dichromate(VI) and dilute sulfuric acid), with the alcohol in excess, warmed. The propanal is distilled off as soon as it forms. Because it leaves the reaction mixture, it cannot be oxidised further to propanoic acid.CH₃CH₂CH₂OH + [O] → CH₃CH₂CHO + H₂O(b) Excess acidified potassium dichromate(VI), heated under reflux. Any propanal that forms condenses and returns to the flask, where it is oxidised to the acid.CH₃CH₂CH₂OH + 2[O] → CH₃CH₂COOH + H₂OIn words: propan-1-ol + 2[O] → propanoic acid + water.(c) Orange to green. The orange dichromate(VI) ions are reduced to green chromium(III) ions, Cr³⁺.(d) Propan-2-ol is a secondary alcohol, so it is oxidised to the ketone propanone, CH₃COCH₃.Examiner tip: for the aldehyde, the marks are for distilling off the product as it forms; for the acid, the marks are for excess oxidising agent and reflux.