Primary alkyl halide `C_4 H_9 Br` (A) is 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_8H_(18)` which is different from the compound when A-butyl bromide is reacted with sodium. Give the structural formula of (A) and write the equations for all the reactions.

Explanation

Two primary alkyl lialides having the molecular formula `C_4 H_9 Br` are possible.
` underset(n -Butyl bramide")(CH_3 CH_2 CH_2 CH_2 Br ) and underset("Isobutyl bromide ")(CH_3 - overset(CH_3) overset(|)(CH ) - CH_2 Br)`
Thus compound (A) is either n-butyl bromide or isobutyl bromide Since compound (A) when reacted with Na metal gave a compound obtained when n-butyl bromide is reacted with Na, metal, therefore, (A) must be isobutyl bromide and compound (D) must be 2,5-dimethylhexane.
` underset("n -Butylbromide")(2CH_3 CH_2 CH_2 cH_2 Br + 2Na ) overset("Wurtz reaction")to underset("n-Octane")(CH_3 CH_2 CH_2 CH_2 CH_2CH_2 CH_2 CH_3) +2NaBr`
` underset("Isobutyl bromide (A)")(CH_3 - overset(CH_3) overset(|)(CH) - CH_2 Br +2Na) overset("Wurtz reaction")to underset("2,5,-dimethylhexane (D) ")(CH_3 - overset(CH_3) overset(|) (CH) - CH_2 - CH_2 - overset(CH_3) overset(|)CH - CH_3 + 2NaBr)`
Reaction of (A) with alc. KOH (elimination) to give (B) followed by reaction with HBO (addition) to give (C)-an isomer of (A) is given ahead..