Question

A $10\,g$ mixture of $iso$-butane and $iso$-butene requires $20\,g$ of $Br_2$ (in $CCl_4$) for complete addition. If $10\,g$ of the mixture is catalytically hydrogenated and the entire alkane is monobrominated in the presence of light at $127^°C$, how much of it would be formed?
(Atomic weight of bromine $= 80$)

• A. $24.21\,g$
• B. $20.0\,g$
• C. $30.0\,g$
• D. $12\,g$

Answer 1

Answer: (A)

Let $iso$-butane be $a$ $g$ and $iso$-butene $b$ $g$. Thus $a + b = 10 \,g$

Now $160\, g$ of $Br_2$ reacts with $56\, g$ of $iso$-butene
$\therefore 20\, g$ of $Br_{2}$ reacts with $\frac{56\times20}{160}=7\,g$ of $iso$-butene
$(b)$
Thus, amount of iso-butane $(a)$ is $10 - 7 = 3\, g$

Now $56\,g$ of $iso$-butene gives $58\,g$ of $iso$-butane
$\therefore 7\,g$ of $iso$-butene gives $=\frac{58\times7}{56}
=7.25\,g$ of $iso$-butane
Total amount of $iso$-butane available for $10\,g$ mixture
$= (7.25 + 3) = 10.25\, g$

Now $58 \,g$ of $iso$-butane gives $137\,g$ of $tert$-butyl bromide
$\therefore 10.25\,g$ of $iso$-butane gives $\frac{137\times10.25}{58}=24.21\,g$ $tert$-butyl bromide.