2-Methylbutane on reacting with bromine in th | vedclass

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Question

(B) The reaction of alkanes with bromine in the presence of sunlight (or $UV$ light) proceeds via a free radical mechanism. Bromination is highly sele

Vedclass · Accepted Answer

$2-$bromo$-2-$methylbutane

Vedclass · Answer

(B) The reaction of alkanes with bromine in the presence of sunlight (or $UV$ light) proceeds via a free radical mechanism. Bromination is highly selective compared to chlorination. The reactivity order of hydrogen atoms towards free radical substitution is $3^\circ > 2^\circ > 1^\circ$. In $2-$methylbutane $(CH_3-CH(CH_3)-CH_2-CH_3)$,the hydrogen atom on the $C-2$ carbon is a tertiary $(3^\circ)$ hydrogen. Therefore,the abstraction of the $3^\circ$ hydrogen is the fastest,leading to the formation of the most stable tertiary free radical. Consequently,the major product is $2-$bromo$-2-$methylbutane: $CH_3-C(Br)(CH_3)-CH_2-CH_3$.

$2-$Methylbutane on reacting with bromine in the presence of sunlight gives mainly:

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