Primary alkyl halide `C_(4)H_(9)Br` (A) reacted with alcoholic KOH to give compound B. Compound B is reacted with HBr to give C which is an isomer of A. When A is reacted with sodium metal, it gives compound D, `C_(8)H_(18)` which is different from the compound formed when n-butyl bromide is reacted with sodium. Give the structural formula of A and write the equations for all the reactions.

Explanation

Two possible isomers of given primary alkyl halide, `C_(4)H_(9)Br` are:
`CH_(3)-CH_(2)-undersetunderset((1^(@)))("[I]")(CH_(2))-CH_(2)-Br,CH_(3)-undersetundersetundersetunderset((1^(@)))("[II]")(CH_(3))(|)(CH)-CH_(2)-Br`
According to the question, compound A on reaction with sodium does not give the same product produced by n-butyl bromide. So A cannot be [I].
`2CH_(3)-CH_(2)underset("n-butyl bromide")(-CH_(2)-CH_(2)-)Br+2Naoverset("Ether")rarrCH_(3)-underset("n-octane (not formed in the reaction) "C_(8)H_(18))(CH_(2)-CH_(2)-CH_(2)-CH_(2)-CH_(2)-CH_(2)-CH_(3))`
Now, [II] must be the correct isomer.
Hence, the overall equations for all the reactions are :