^{14}CH_2=CH-CH_3 xrightarrow[text{or high te | vedclass

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(D) Under conditions of low concentration of $Br_2$ and high temperature,the reaction proceeds via a free radical mechanism.This mechanism favors ally

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both $(A)$ and $(B)$

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(D) Under conditions of low concentration of $Br_2$ and high temperature,the reaction proceeds via a free radical mechanism.This mechanism favors allylic substitution over electrophilic addition.The allylic radical formed is $CH_2=CH-CH_2^{\bullet}$ (where the label $^{14}C$ is at the terminal position).Due to resonance,the radical can be represented as $CH_2=CH-CH_2^{\bullet} \leftrightarrow ^{\bullet}CH_2-CH=CH_2$.Since the radical is delocalized,the bromine atom can attach to either terminal carbon,resulting in a mixture of $^{14}CH_2=CH-CH_2-Br$ and $CH_2=CH-^{14}CH_2-Br$.Therefore,both $(A)$ and $(B)$ are formed.

$^{14}CH_2=CH-CH_3 \xrightarrow[\text{or high temp}]{\text{low conc of } Br_2} (?)$
Product of the above reaction is

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$^{14}CH_2=CH-CH_3 \xrightarrow[\text{or high temp}]{\text{low conc of } Br_2} (?)$Product of the above reaction is