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Answer:
CH3-CH2-Br + KOH → CH3-CH2-OH + KBr
Hence, A = CH3-CH2-OH and B = KBr
This is a nucleophilic substitution reaction.
Substitution of halogen by hydroxyl group that is formation of alcohols from haloakane is called hydrolysis of haloalaknes.
Bromoethane is a primary haloalkane, hence it undergoes SN2 mechanism.
The OH- ion attack on the bromoethane occurs from the back side and the bromide ion leaves from the front side.
As a result, the product ethyl alcohol is formed with inversion of configuration.
This inversion of configuration is known as walden inversion.
The SN2 mechanisms are accompanied by inversion of configuration.
It implies that both reactant ethyl bromide and product ethyl alcohol are optically active.
Explanation: