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
$\underset{\left(5\right)}{3^{\circ }}>\underset{\left(3.8\right)}{2^{\circ }}>\underset{\left(1\right)}{1^{\circ }}$
F. What would be the product ratio x/y in the chlorination of propane if all the hydrogen were abstracted at equal rate ?
$C{H}_3-C{H}_2-C{H}_3\underset{hv}{\overset{C{l}_2}{\to }}C{H}_3-C{H}_2-\underset{\left(x\right)}{C}{H}_2-Cl+C{H}_3-\underset{\left(y\right)}{\underset{Cl}{\underset{|}{C}}}H-C{H}_3$

• A. $\frac{1}{3}$
• B. $\frac{3}{1}$
• C. $\frac{9}{1}$
• D. $\frac{1}{9}$

Answer 1

The correct option is B $\frac{3}{1}$

$C{H}_3-C{H}_2-C{H}_3\underset{hv}{\overset{C{l}_2}{\to }}C{H}_3-C{H}_2-\underset{\left(x\right)}{C}{H}_2-Cl+C{H}_3-\underset{\left(y\right)}{\stackrel{Cl}{\stackrel{|}{C}}}H-C{H}_3$

We know, rate of reactivity is-

$3^o>2^o>1^o$

$\left(5\right)$ $\left(3.8\right)$ $\left(1\right)$

$x\to 1^o\Rightarrow 6\times 1=6$

$y\to 2^o\Rightarrow 2\times 3.8=6.4$

So, $\frac{x}{y}=\frac{6.4}{6}=\frac{3.2}{3.0}$

But, given rate is same.

So, $x\to 1^o\Rightarrow 6\times 1=6$

$y\to 2^o\Rightarrow 2\times 1=2$

$\frac{x}{y}=\frac{6}{2}=\frac{3}{1}$