Classifying & Testing for Alcohols (Cambridge International AS Chemistry)

Classifying Alcohols

• 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

Test for Alcohols

• Tri-iodomethane (also called iodoform) is a yellow precipitate

• It is formed from the reaction of iodine in an alkaline solution 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 alkaline iodine

• 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 with iodine atoms, forming a -CI₃ group
  • The intermediate compound is hydrolysed by an alkaline solution to form a sodium salt (RCO₂^- Na⁺) and a yellow precipitate of CHI₃

The iodoform reaction

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

Iodoform & alcohols

• The position of a secondary alcohol group in a molecule can be identified using the iodoform test with iodine in alkaline solution

  • If the –OH group is attached to a carbon atom that is next to a methyl group, the alcohol can be oxidised by the alkaline iodine to form a methyl ketone (RCOCH₃)

  • The methyl ketone then undergoes:

    • Halogenation, where the three hydrogen atoms of the methyl group are replaced by iodine atoms (–CI₃), followed by
    • Hydrolysis, forming a sodium carboxylate salt and a yellow precipitate of iodoform (CHI₃)

• A positive result (yellow precipitate) indicates that the alcohol has the structure CH₃CH(OH)R

• No precipitate means the secondary alcohol is not adjacent to a methyl group and therefore cannot form a methyl ketone intermediate

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 iodine