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
$\underset{\left(5\right)}{3^{\circ }}>\underset{\left(3.8\right)}{2^{\circ }}>\underset{\left(1\right)}{1^{\circ }}$
A.1 -halo -2, 3 -dimethyl butane will be obtained in better yields, if halogen is :

• A. $B{r}_2$
• B. $C{l}_2$
• C. $I_2$
• D. Can't be predicted

Answer 1

The correct option is B $C{l}_2$

In case of Bromination, tertiary is the most stable whereas, terminal bromination is relatively very low.

In case of chlorine ratio is $5:3.8:1$, respectively obtainable. So, halogen is $C{l}_2$ (Chlorine).