• 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)

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

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 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

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