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(B) The reaction of $n$-butane with bromine at high temperatures (like $130^\circ C$) or under $UV$ light proceeds via a free-radical substitution mec

Vedclass · Accepted Answer

$CH_3CH_2CH(Br)CH_3$

Vedclass · Answer

(B) The reaction of $n$-butane with bromine at high temperatures (like $130^\circ C$) or under $UV$ light proceeds via a free-radical substitution mechanism.In $n$-butane $(CH_3-CH_2-CH_2-CH_3)$,there are two types of hydrogen atoms: primary $(1^\circ)$ and secondary $(2^\circ)$.The secondary hydrogen atoms are more reactive towards free-radical substitution because the resulting secondary free radical is more stable than the primary free radical.Therefore,the major product is $2$-bromobutane,which is $CH_3CH_2CH(Br)CH_3$.

What is the product of the reaction between $n$-butane and bromine at $130^\circ C$?

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