Answer:
Step-by-step explanation:
Radical chlorination of butane in the presence of light gives rise to the formation of two isomeric monochlorides B (1-chlorobutane) and C (2-chlorobutane). Both upon reaction with a bulky base (potassium tert-butoxide) give rise to D(but-1-ene) and E(but-2-ene) respectively, this is because the bulky base abstracts the less hindered proton. Compound B and C were later treated with aqueous C2H5OH which results in the hydrolysis of alkyl halides to produce Compound F(butan-1-ol) and compound G (butan-2-ol) respectively.
The diagrammatic expression of the whole reaction is shown in the attached image below.