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(D) The reaction of an alkane with $Br_2$ in the presence of $hv$ (ultraviolet light) proceeds via a free radical substitution mechanism.In the first

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Secondary free radical

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(D) The reaction of an alkane with $Br_2$ in the presence of $hv$ (ultraviolet light) proceeds via a free radical substitution mechanism.In the first step,the bromine molecule undergoes homolytic cleavage to form bromine radicals $(Br^{\bullet})$.These bromine radicals then abstract a hydrogen atom from the alkane to form an alkyl radical intermediate.For $2-	ext{methylbutane}$ $(CH_3-CH(CH_3)-CH_2-CH_3)$,the abstraction of a hydrogen atom from the $C-2$ position leads to a tertiary free radical,while abstraction from the $C-3$ position leads to a secondary free radical.The formation of $2-	ext{bromo}-3-	ext{methylbutane}$ involves the secondary free radical intermediate formed at the $C-3$ position.

Which reactive intermediate is involved in the following reaction?
$2-\text{methylbutane} \xrightarrow{Br_2, hv} 2-\text{bromo}-3-\text{methylbutane}$ (not the major product)

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Which reactive intermediate is involved in the following reaction?$2-\text{methylbutane} \xrightarrow{Br_2, hv} 2-\text{bromo}-3-\text{methylbutane}$ (not the major product)