Question

Primary alkyl halide $C_4{H}_{9}Br$ $\left(a\right)$ reacted with alcoholic $KOH$ to give compound $\left(b\right)$. Compound $\left(b\right)$ is reacted with $HBr$ to give $\left(c\right)$ which is an isomer of $\left(a\right)$. When $\left(a\right)$ is reacted with sodium metal it gives compound $\left(d\right)$, $C_8{H}_{18}$ which is different from the compound formed when $n$-butyl bromide is reacted with sodium. Give the structural formula of $\left(a\right)$ and write the equations for all the reactions.

Answer 1

$\text{Primary alkyl halide}$

There are two primary alkyl halides having the formula, $C_4{H}_{9}Br$. They are $n$-butyl bromide and isobutyl bromide.

Therefore, compound $\left(a\right)$ is either $n$-butyl bromide or isobutyl bromide.


Now, compound $\left(a\right)$ reacts with $Na$ metal to give compound $\left(d\right)$ of molecular formula, $C_8{H}_{18}$, which is different from the compound formed when $n$-butylbromide reacts with $Na$ metal. Hence, compound$\left(a\right)$ must be isobutyl bromide.

$\text{Reaction of n-butylbromide and isobutyl bromide}$




Thus, compound $\left(d\right)$ is $2$,$5$-dimethylhexane. Which is different from the product ($n$-octane) obtained by $n$-butylbromide.
Compound $\left(a\right)$ reacts with $Na$ metal to give compound $\left(b\right)$ of molecular formula, $C_8{H}_{18}$, which is different from the compound formed when $n$-butyl bromide reacts with $Na$ metal.
Hence, compound $\left(a\right)$ must be isobutyl bromide.

$\text{Reaction of isobutyl bromide with alcoholic KOH}$

It is given that compound $\left(a\right)$ reacts with alcoholic $KOH$ to give compound $\left(b\right)$.


Hence, compound $\left(b\right)$ is $2$-Methylpropene.

$\text{Reaction of 2-Methylpropene with HBr}$

Now, compound $\left(b\right)$ reacts with $HBr$ to give compound $\left(c\right)$ as follows:


The product is $2$-bromo-$2$-methylpropane, which is an isomer of $\left(a\right)$.