$A$ primary alkyl halide $(a)$ with molecular formula $C_{4}H_{9}Br$ reacts with alcoholic $KOH$ to give compound $(b)$. Compound $(b)$ reacts with $HBr$ to give compound $(c)$,which is an isomer of $(a)$. When $(a)$ reacts with sodium metal,it gives compound $(d)$ with molecular formula $C_{8}H_{18}$,which is different from the compound formed when $n$-butyl bromide reacts with sodium. Identify the structural formula of $(a)$ and write the equations for all the reactions.

(A) There are two primary alkyl halides with the formula $C_{4}H_{9}Br$: $n$-butyl bromide $(CH_{3}CH_{2}CH_{2}CH_{2}Br)$ and isobutyl bromide $(CH_{3}CH(CH_{3})CH_{2}Br)$.
Since compound $(a)$ reacts with $Na$ to form a product $(d)$ different from the product of $n$-butyl bromide (which is $n$-octane),$(a)$ must be isobutyl bromide.
$1$. Reaction of $(a)$ with $Na$ (Wurtz reaction):
$2CH_{3}CH(CH_{3})CH_{2}Br + 2Na$ $\xrightarrow{\text{dry ether}} CH_{3}CH(CH_{3})CH_{2}CH_{2}CH(CH_{3})CH_{3} (2,5-\text{dimethylhexane}) + 2NaBr$
$2$. Reaction of $(a)$ with alcoholic $KOH$ (Dehydrohalogenation):
$CH_{3}CH(CH_{3})CH_{2}Br + KOH(\text{alc})$ $\xrightarrow{\Delta} CH_{3}C(CH_{3})=CH_{2} (2-\text{methylpropene}) + KBr + H_{2}O$
$3$. Reaction of $(b)$ with $HBr$ (Markovnikov addition):
$CH_{3}C(CH_{3})=CH_{2} + HBr \rightarrow CH_{3}C(Br)(CH_{3})CH_{3} (2-\text{bromo}-2-\text{methylpropane})$
Compound $(c)$ is $2-\text{bromo}-2-\text{methylpropane}$,which is an isomer of $(a)$.