The reaction of n-butane with bromine in the | vedclass

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(B) The reaction of $n$-butane with bromine in the presence of light is a free radical substitution reaction. At high temperatures like $725\,^oC$,the

Vedclass · Accepted Answer

$CH_3CH_2CH(Br)CH_3$

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(B) The reaction of $n$-butane with bromine in the presence of light is a free radical substitution reaction. At high temperatures like $725\,^oC$,the reaction is highly selective. Bromination is more selective than chlorination,and the reactivity order of hydrogen atoms is $3^\circ > 2^\circ > 1^\circ$. In $n$-butane $(CH_3-CH_2-CH_2-CH_3)$,the $2^\circ$ hydrogen atoms are more reactive than the $1^\circ$ hydrogen atoms. Therefore,the major product formed is $2$-bromobutane,which is $CH_3CH_2CH(Br)CH_3$.

The reaction of $n$-butane with bromine in the presence of light at $725\,^oC$ yields ...................

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