Chromic acid oxidation of an alcohol template

Propan-2-ol and acidified dichromate (H₂CrO₄): the alcohol displaces water from protonated chromic acid, the chromate ester loses H⁺, and water removes the C–H as Cr leaves with the O–Cr electrons to give propanone.

Mechanism step by step

A chromate ester forms; an E2-like elimination makes C=O and reduces Cr(VI) to Cr(IV)

1. Protonation – Chromic acid (H₂CrO₄) takes H⁺
2. Chromate ester – The alcohol O attacks Cr; water leaves (drawn as one step)
3. Deprotonation – The neutral chromate ester
4. Elimination – Water takes the C–H; C=O forms; Cr(VI) → Cr(IV)

Key points

• Orange Cr₂O₇²⁻ turns green (Cr³⁺) as it oxidises 1° and 2° alcohols; 3° alcohols have no C–H on the carbinol carbon and do not react.
• 1° alcohols give aldehydes (distil out at once) or, under reflux, carboxylic acids via the aldehyde hydrate.
• PCC in dry CH₂Cl₂ works the same way but, with no water, stops at the aldehyde.
• Cr(IV) goes on through Cr(V) to Cr(III); chromium(VI) compounds are carcinogenic.

Drawn with 77 atoms, 14 arrows, 8 shapes – every one editable in MolDraw.

Sources

• Clayden, Greeves & Warren, Organic Chemistry, 2nd ed., OUP 2012, ch. 23 Textbook
• Smith, March's Advanced Organic Chemistry, 8th ed., Wiley 2020, ch. 19 Textbook
• Westheimer, Chem. Rev. 1949, 45, 419 Review