Question

Halogenation is a substitution reaction, where halogen replaces one or more hydrogens of hydrocarbons.
Chlorine free radicals make \[{1^o},{2^o},{3^o}\] 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 bromine is,
\[{3^o} > {2^o} > {1^o}\]
$\left( {1600} \right)$ $\left( {82} \right)$ $\left( 1 \right)$
\[{3^o} > {2^o} > {1^o}\]
$\left( 5 \right)$ $\left( {3.8} \right)$ $\left( 1 \right)$
A) $1$ -halo-$2,3$ -dimethyl butane will 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

Hint: We must have to know that halogenation reaction is reaction which produces hydrocarbon derivative from alkanes. And here, one more halogen is replaced by a hydrogen atom. In a halogenation reaction there is an introduction of halogen into a substance. The chlorine free radicals form primary, secondary and tertiary radicals but the bromine radical only forms the tertiary free radicals.

Complete answer:
$1$ -halo-$2,3$ -dimethyl butane will be obtained in better yields, not in the presence of bromine atoms. Hence, option (A) is incorrect.
Formation of $1$ -halo-$2,3$ -dimethyl butane yields better yields, if the halogen is chlorine. Because, among these three halogens, chlorine is more reactive than bromine and iodine. Chlorine free radicals make \[{1^o},{2^o},{3^o}\] radicals. But in the case of bromine, which gives only tertiary radicals. The chlorine atom is less selective and the product is more stable and it is isomers. Therefore, there is a formation of 1-chloro-2,3-dimethyl butane. And the reaction can be written as,

Hence, option (B) is correct.
$1$ -halo-$2,3$ -dimethyl butane will not be obtained in the presence of iodine. Because iodine is least reactive. Hence, the option (C) is incorrect.
$1$ -halo-$2,3$ -dimethyl butane in better yields, in the presence of chlorine atoms which can be predicted. Hence, the option (D) is incorrect.

Note:
We have to know that among the halogens, chlorine is more reactive and it is less selective. $1$ -halo-$2,3$ -dimethyl butane is formed in better yields in the presence of chlorine atoms. And the chlorine atom is less selective and more effective. Hence, by the reaction of $2,3$ -dimethyl butane with chlorine atom, there is a formation of two products, that is, $1$ -chloro-$2,3$ -dimethyl butane and $2$ -chloro-$2,3$ -dimethyl butane.