Alcohol reactions, oxidation and purification

What you’ll learn

• Classify alcohols and predict oxidation products.
• Distinguish substitution, elimination and combustion.
• Explain reflux, distillation, extraction and drying conceptually.

Understand the idea

Classify an alcohol by the number of carbon groups bonded to the carbon carrying OH. A primary alcohol can oxidise to an aldehyde and then a carboxylic acid; a secondary alcohol gives a ketone. Tertiary alcohols resist oxidation by the usual acidified dichromate test because the OH-bearing carbon lacks the required hydrogen. Dichromate(VI) changes from orange to green when reduced during these oxidations.

For a primary alcohol, removing a volatile aldehyde as it forms limits further oxidation. Reflux keeps volatile material in the reacting mixture by condensing vapour back, favouring complete oxidation when oxidant is available. These descriptions explain the role of apparatus; chromium(VI) reagents and flammable organic liquids require a supervised laboratory and controlled handling.

Alcohols combust to carbon dioxide and water when oxygen is sufficient. Replacing OH with halogen gives a halogenoalkane; removing water gives an alkene. Phosphorus(V) chloride is one qualitative reagent for OH, producing hydrogen chloride along with a chloroalkane and phosphorus oxychloride, but it is not a uniquely selective test for an alcohol. Aldehydes can reduce suitable alkaline copper(II) tests, while carboxylic acids release carbon dioxide from carbonate.

Purification depends on physical properties. Extraction distributes a compound between immiscible layers; a drying agent removes traces of water from an organic liquid. Distillation separates components using volatility, whereas reflux returns them to the flask. A narrow boiling range near an expected value supports purity but is not conclusive structural identification.

Key formulae

$$\mathrm{CH_3CH_2OH+[O]\rightarrow CH_3CHO+H_2O}$$

$$\mathrm{CH_3CHO+[O]\rightarrow CH_3COOH}$$

$$\mathrm{CH_3CH(OH)CH_3+[O]\rightarrow CH_3COCH_3+H_2O}$$

$$\mathrm{CH_3CH_2OH\rightarrow CH_2{=}CH_2+H_2O}$$

Worked example

Compare oxidation of propan-1-ol and propan-2-ol using the usual acidified dichromate model.

1. Propan-1-ol is primary because its OH-bearing carbon is attached to one other carbon.
2. It can form propanal, then propanoic acid with further oxidation.
3. Propan-2-ol is secondary and forms propanone; it does not continue to a same-carbon-count carboxylic acid under the usual conditions.

Answer: Propan-1-ol gives propanal or propanoic acid; propan-2-ol gives propanone.