Question

A $10g$ mixture of $iso$-butane and $iso$-butene requires $20g$ of $B{r}_2$ (in $CC{l}_4$) for complete addition. If $10g$ 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. $24.21g$
• B. $20.0g$
• C. $30.0g$
• D. $12g$

Answer 1

Answer: (A)

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

Now $160g$ of $B{r}_2$ reacts with $56g$ of $iso$-butene
$\therefore 20g$ of $B{r}_2$ reacts with $\frac{56\times 20}{160}=7g$ of $iso$-butene
$(b)$
Thus, amount of iso-butane $(a)$ is $10-7=3g$

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

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