Question

Halogenation is a substitution reaction, where halogen replaces one or more hydrogens of hydrocarbon.
Chlorine free radical make $1^{\circ },2^{\circ },3^{\circ }$ radicals with almost equal ease, whereas bromine free radicals have a clear preference for the formation of tertiary free radicals. So , bromine is less reactive, and more selective whereas chlorine is less selective and more effective.
The relative rate of abstraction of hydrogen by Br is
$\underset{\left(1600\right)}{3^{\circ }}>\underset{\left(82\right)}{2^{\circ }}>\underset{\left(1\right)}{1^{\circ }}$
The relative rate of abstraction of hydrogen by Cl is
$\underset{\left(5\right)}{3^{\circ }}>\underset{\left(3.8\right)}{2^{\circ }}>\underset{\left(1\right)}{1^{\circ }}$
$C{H}_3-\stackrel{C{H}_3}{\stackrel{|}{\underset{C{H}_3}{\underset{|}{C}}}}-C{H}_2-\stackrel{C{H}_3}{\stackrel{|}{C}H}-C{H}_3\stackrel{B{r}_2/hv}{\to }C{H}_3-\stackrel{C{H}_3}{\stackrel{|}{\underset{C{H}_3}{\underset{|}{C}}}}-C{H}_2-C{H}_2-\stackrel{C{H}_3}{\stackrel{|}{\underset{Br}{\underset{|}{C}}}}-C{H}_3$
2,Bromo-2,5,5 trimethyl hexane(x%)
What is the value of x(% yield of product) ?

• A. 18 %
• B. 82 %
• C. 90 %
• D. 60 %

Answer 1

The correct option is C 90 %

$C{H}_3-\stackrel{C{H}_3}{\stackrel{|}{\underset{C{H}_3}{\underset{|}{C}}}}-C{H}_2-\stackrel{C{H}_3}{\stackrel{|}{C}H}-C{H}_3\stackrel{B{r}_2/hv}{\to }C{H}_3-\stackrel{C{H}_3}{\stackrel{|}{\underset{C{H}_3}{\underset{|}{C}}}}-C{H}_2-\stackrel{C{H}_3}{\stackrel{|}{\underset{Br}{\underset{|}{C}}}}-C{H}_3$

Here, tertiary bromination occurs-

To find the percentage $\left(3^o\%\right)=\frac{3^o}{All}\times 100$

As, $3^o\to 1\times 1600=1600$

$2^o\to 2\times 82=164$

$1^o\to 15\times 1=15$

So, All is $1600+164+15=1779$

$\therefore \%=\frac{1600}{1779}\times 100=90\%$