Classifying & Testing for Alcohols (CIE AS Chemistry)

• Primary alcohols are alcohols in which the carbon atom bonded to the -OH group is attached to one other carbon atom (or alkyl group)
• Secondary alcohols are alcohols in which the carbon atom bonded to the -OH group is attached to two other carbon atoms (or alkyl groups)
• Tertiary alcohols are alcohols in which the carbon atom bonded to the -OH group is attached to three other carbon atoms (or alkyl groups)

Primary, secondary and tertiary alcohols

Classifying primary, secondary and tertiary alcohols and alcohols with more than one alcohol group

• Only primary and secondary alcohols can get oxidised when mildly oxidised with acidified K₂Cr₂O₇
  • Primary alcohols get mildly oxidised to aldehydes
  • Secondary alcohols get mildly oxidized to ketones

• Tertiary alcohols do not undergo oxidation with acidified K₂Cr₂O₇
• Therefore, only the oxidation of primary and secondary alcohols will change the colour of K₂Cr₂O₇ solution as the orange Cr₂O₇^2- ions are reduced to green Cr³⁺ ions

 Testing for alcohols

Only propan-1-ol and propan-2-ol, which are primary and secondary alcohols respectively, can get oxidised, turning the orange solution green; no colour change is observed with 2-methyl-propan-2-ol, which is a tertiary alcohol

• Tri-iodomethane (also called iodoform) forms a yellow precipitate with methyl ketones
  • Methyl ketones are compounds that have a CH₃CO-group
  • Ethanal also contains a CH₃CO- group and therefore also forms a yellow precipitate with iodoform

• The reagent is heated with an alkaline solution of iodine
• This reaction involves a halogenation and hydrolysis step
  • In the halogenation step, all three H-atoms in the -CH₃ (methyl) group are replaced for iodine atoms, forming -CI₃
  • The intermediate compound is hydrolysed by alkaline solution to form a sodium salt (RCO₂^- Na⁺) and a yellow precipitate of CHI₃

The iodoform reaction

The reaction of methyl ketones with iodoform results in the formation of a yellow CHI₃ precipitate

Iodoform & alcohols

• The position of a secondary alcohol can be deduced by reacting the compound with alkaline I₂
• If the -OH group is on the carbon atom next to a methyl group, it will firstly get oxidised to CH₃CH(OH)- by the alkaline solution
• This will result in the formation of a methyl ketone RCOCH₃
• The methyl ketone will then first get halogenated and then hydrolysed to form the sodium salt and the yellow precipitate
• If no yellow precipitate is formed, then this means that the secondary alcohol is not on a carbon next to a methyl group

Using the iodoform test

The secondary alcohol butan-2-ol will firstly get oxidised to the methyl ketone butanone which will form a yellow precipitate when reacted with alkaline I₂