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

There are two `1^(@)` alkyl halides with the molecular formula `C_(4)H_(9)Br`. These are:
So compound (A) is either n-butyl bromide or isobutyl bromide. N-butyl bromide reacts with metalic sodium (Wurtz reaction) to give n-octane.
`2CH_(3)CH_(2)CH_(2)CH_(2)Br+2Na rarr underset("n-octane")(CH_(3)CH_(2)CH_(2)CH_(2)CH_(2)CH_(2)CH_(2)CH_(3)+2NaBr)`
So, the structure of compound (A) will be such that it will not form n-octane of reaction with Na-metal. This shows that compound (A) is isobutyl bromide. Its reaction with Na-metal can be formulated as follows, thereby showing that compound (D) is 2,5-dimethylhexene.

The reaction of compound (A) with alcoholic KOH can be formulated as follow:
`CH_(3)-underset(H)underset(|)overset(CH_3)overset(|)(C)-underset((A))(CH_2)-Brunderset((-HBr))overset(KOH(alc.),Delta)(rarr)underset("Isobutene"^(B))(CH_3-overset(CH_3)overset(|)C = CH_(2))`
THis shows that compound (B) is isobutene. finally the reaction of compound (B) with HBr can be formulated as follows, thereby indicating that the compound (C) is tert-butyl bromide [which is an isomer of compound (A)].
`CH_(3)-underset((B))underset(|)overset(CH_3)overset(|)(C)-CH_(2)-HBr rarrCH_3-underset(Br(C))underset(|)overset(CH_3)overset(|)(C) -CH_(3)`.