Answer:
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.
KSO4 + HSO3 is your products. Brainliest would be much appreciated.
Since the structures are not given, I'll just provide the structure for these functional groups.
Alkane: You know the compound is an alkane if it conforms to the general formula CₓH₂ₓ₊₂, where x is any positive integer.
Alcohol: A compound is an alcohol if the compound contains an -OH functional group.
Carboxylic acid: The general formula for carboxylic acid is RC=OOH. The R is any hydrocarbon chain. The C atom is attached to one oxygen atom by a double bond, and one O atom by a single bond.